Solutions Manual for Fundamentals of Momentum, Heat, and Mass Transfer 7th Edition By James Welty, G by ACADEMIAMILL

SOLUTIONS MANUAL Fundamentals of Momentum, Heat, and Mass Transfer 7th Edition By James Welty, Gregory Rorrer, David Foster | All Chaters 1-31 Part 1: End of Chapter Problem Solutions Part 2: Instructor Only Problems Solutions Part 3: Show/Hide Problems Solutions

Part 1: End of Chapter Problem Solutions

Chapter 1 End of Chapter Problem Solutions 1.1 n = 4 x 1020 molecules/in3 =

= 1.32x104 in./s

A= (10-3 in)2 NA = n A=1.04x108 m/s

1.2 Flow Properties: Velocity, Pressure Gradient, Stress

Fluid Properties: Pressure, Temperature, Density, Speed of Sound, Specific Heat

1.3 mass of solid= ρsvs mass of fluid= ρfvf

mix=

1.4 Given

For P1= 1 atm

= 1.01

P= 3001(1.01)7 – 3000 = 217 atm

1.5 At Constant Temperature = constant

= constant

For 10% increase in P must also increase by 10 %

1.6 Since density varies as

=nM

(M=Molecular wt.)

n250,000= nS.L.

= 4 x 1020

= 2.5 x 1016

1.7 =

+ =

Q.E.D.

1.8

= = Q.E.D.

= =

1.9 Transformation from (x,y) to (r, )

r2 = x2+y2 = so:

= =

= +

1.10

)

=( Thus:

+ =

+( )

+ (

) +

+ )

1.11

P(a,b)= =

{[

+ + 2)

+ (

)

(sin1cos1)

1.12

T(x,y)= Toe-1/4 [ (cos cosh )

(sin sinh )

T(a,b)= Toe-1/4 [ (cos cosh1)

(sin sinh1)

= Toe-1/4[ =

+ [

(1+e-2)

] +

1.13 In problem 1.12 T(x,y) is dimensionally homogeneous (D.H.) P(x,y) in Prob 1.11 will be D.H. if ~

LBf s2/ ft4

or using the conversion factor gc

1.14 = 3x2y+4y2

A scalar field is given by the function: (a) Find at the point (3,5)

For the value of

at the point (3,5)

(b) Find the component of

that makes a -60 angle with the axis at the point (3,5)

Let the unit vector be represented by

At the point (3,5) this becomes:

1.15 For an ideal gas

P= From Prob 1.3:

1.16 = Arsin (1- ) a)

= Asin (1- ) 1/2

=A max is given by d

Requiring For

= 0 or

= = 0: -

And for

d =0

) = 0

(1)

=0 (1+ ) +

(1-

=0:

(2)

4a2/r2=0

From Eq. 2: If a 0, r

dr+

then

Subst. into Eq. 1

for which

= 0,

=0, 1-a2/r2=0

Giving a = r For =

1+ a2/r2 =0

Thus conditions for

impossible

max are

r = a

(3)

1.17

P=Po+

1.18 Vertical cylinder d=10m, h=6m V= (10m)2(6m)= 471.2 m3 @ 20

w=998.2 kg/m

m= wV= (998.2)(471.2)= 470350 kg 3 @ 80 w=971.8 kg/m m=(971.8)(471.2)= 457910 kg 12440 kg

1.19 V= 1200 cm3 @ 1.25 MPa V=1188 cm3 @ 2.5 MPa

Liquid

= -V

-V

V= 1194 cm3= 1.194 x 10-3 m3 ∆V= -12 cm3= -1.2 x 10-7 m3 = -1.194 x 10-3 = +12440 MPa

1.20 = -V

-V

V=0.25 m3 V= -0.005 m3 P= 10 mPa = -0.25

= 500 MPa

1.21 For H20:

=2.205 GPa

= -0.0075 -V

or ∆P =

∆P= (2.205 GPa)(0.0075) = 0.0165 GPa= 16.5 MPa

1.22 For H20:

P1=100kPa

= -V =

or =

P2=120 MPa

2.205

= = 0.999 x 10-3 = 0.0999 percent

1.23 H20@ 68

(341 K)

0.123 [1-0.00139(341)] = 0.0647 N/m In a clean tube- =0

= 9.37 x 10-3m = 9.37 mm

1.24 Parallel Glass Plates Gap=1.625 mm = 0.0735 N/m For a unit depth: Surface Tension Force = 2(1) cos Weight of H20 =

(1)(1.625x10-3)

For clean glass cos =1 Equating Forces: 2(1) = h=

(1)(1.625x10-3) = 0.00922m= 9.22 mm

1.25 1.26 H20-Air-Glass Interface @40 Tube Radius= 1 mm h=

cos

= 0.123[1-0.0139(313)]=0.0695 N/m h=

= 0.0143m (1.43 cm)

1.26 1.27 Soap Bubble- T=20 d=4mm = 0.025 N/m (Table 1.2) Force Balance for Bubble: r2

= 25 N/m2 = 25 Pa

1.27 1.29 @60 C

= 0.0662 N/m = 0.44 N/m

Tube Diameter= 0.55 mm h= For H20: For Hg:

= 0.0499m (4.99 cm Rise) = -0.0157 m (1.57 cm Depression)

1.28 1.30 H20/ Glass Interface T=30 C = 0.123[1-0.0139(303)] = 0.0712 N/m = 996 kg/m3 h

1 mm

h= r=

d = 2r = 2.915 cm

= 0.0146 m (1.457 cm)

1.29 1.31 Bubble Diameter = 0.25 cm = 0.0025 m, and so Radius = 0.00125 m Capillary Tube: Diameter = 0.2cm = 0.002 m, and so Radius = 0.001 m Beginning with:

Rearrange and remember the unit conversion Pa=kg/ms2,

Next, we can calculate the height of the fluid in the tube:

1.30 1.32 First, calculate the surface tension of water using the temperature of the water:

Next, using the equation for the height of a fluid in a capillary,

Rearranging and solving for the radius:

Diameter is 2r so D=1.50 mm

Chapter 2 End of Chapter Problem Solutions

2.1 Assume Ideal Gas Behavior =

For T = a + by = 530 - 24 y/n =

= ln

With P = 10.6 PSIA, Po = 30.1 in Hg h = 9192 ft.

2.2

(a)

=0 on tank (1)

At H20 Level in Tank: P=Patm+

wg(h-y)

From (1) & (2): h-y=1.275 ft.

(2) (3)

For Isothermal Compression of Air PatmVtank=P(Vair) P=

Patm

(4)

Combing (1) & (4): y=0.12 ft. and h=1.395 ft. (b) For Top of Tank Flush with H20 Level =0 P=Patm+ At H20 Level in Tank: P=Patm+

wg(3-y)

Combining Equations: F= 196(3-y)- 250 For Isothermal Compression of Air: (As in Part (a)) 3-y = 2.8 ft. F = 196(2.8) - 250 = 293.6 LBf

2.3 When New Force on Tank = 0 Wt. = Buoyant Force = 250 Lbf Vw Displaced = 250/ wg = 4.01 ft3 Assuming Isothermal Compression PatmA(3ft.)= P(4.01 ft3) = (Patm+ gy)(4.01) y=45.88 ft. Top is at Level: yor at 44.6 ft. Below Surface

2.4 =

= = 1ln(1-

) = 300,000 ln(1-0.0462) = 14190 Psi

Density Ratio: =

= 1.0484

so P= 1.0484

2.5 Buoyant Force: FB= V= For constant volume: F varies inversely with T

2.6 Sea H20: S.G.=1.025 At Depth y=185m Pg= 1.025 wgy = 1.025(1000)(9.81)(185) = 1.86x106 Pa = 1.86 MPa

2.7 r = Measured from Earth’s Surface R = Radius of Earth = g= go P-Patm= At Center of Earth: r = R PCtr-Patm= Since PCtr PCtr

Patm =

= 176x109 Pa = 176 MPa

2.8 =

g =

P-Patm= g(+h) = (1050)(9.81)(11034) = 113.7 MPa

1122 Atmospheres

2.9 As in Previous Problem P-Patm= gh For P-Patm= 101.33 kPa h=101.33/ g for H20: h = Sea H20: h = Hg: h =

= 10.33m = 10.08m = 0.80m

2.10

5 4

2 1

P1=Patm+ Hg g(12’’)

P1=P2

P2=P3+ K g(5’’)

P3=P4

P4=PA+ w g(2’’)

P4=P5

Patm+ Hg g(12’’)= PA

’’

PA= Patm+

K g(5’’)

g[(13.6)(12)-2-0.75(5)]=Patm+5.81

PA = 5.81 PSIG

2.11 Force Balance on Liquid Column: A=Area of Tube -3A + 14.7A – ghA = 0

= 26.6 in.

2.12 A

B D C

PA=PB- o g(10 ft.) PC=PB+ w g(5 ft.) PD=PC- Hg g(1 ft.)

PA-PD= PA-PAtm=

Hg g(1)-

w g(5)-

o g(10)

w g(13.6 x1-5-0.8 x 10 x 1) = 37.4 LBf/ft

2.13 P3=PA-d1g

w = PB+(

Hg g)x(d3+d4sin45)

PA-PB = = 245 LBf/ft2= 1.70 Psi 3

2.14

P3 = PA -

wgd1

P3= PB -

wg(d1+d2+d3) +

Hg gd2

Equating: PA-PB= Hg gd2- wg(d2+d3) =

wg[(13.6)(1/12)-7.3/12] = 32.8 LBf/ft

= 0.227 Psi

2.15

PI=PA+ wg(10”) PII=PB+

wg(4’’)+

Hg g(10”)

PI=PII PA-PB =

wg[-6+13.6(10)] = 56.3 psi

2.16

Pressure Gradient is in direction of - & isobars are perpendicular to ( - ) ( - ) String will assume the ( - ) direction & Balloon will move forward.

2.17

At Rest: P=

Accelerating: P= Equating: ya= Level goes down.

= (g+a)ya which

2.18 F = P6.6A - PatmA =

h( r2)

h=2m r=0.3m F=5546N yC.P.= + Ibb/A For a circle: Ibb= r4/4 yC.P.= 2m +

= 2.011m

2.19

Height of H20 column above differential element= h-4+y

For (a) - Rectangular gate- dA = 4dy dFw=[ wg(h-4+y)+Patm]dA dFA=[Patm+(6Psig)(144)]dA =

[ g(h-4+y)-864](4dy)=0 h=15.18 ft. For (b): dA=(4-y)dy [ g(h-4+y)-864](4-y)dy=0 h=15.85 ft.

2.20

Per unit depth: Fy|up =

r2/2 {buoyancy}

Fy|down = g r2+

2 wg(r -

r2/4)

Equating:

g r2+

= w

/ =

2 wgr (1-

/4)

= 0.432 w = 432 kg/m3

2.21 a) To lift block from bottom F = {wt. of concrete}+{wt. of H20} =

cgV+[

wg(22.75’)+Patm]A

= (150)g(3x3x0.5)+[62.4g(22.75)+14.7(144)]x(3x3) = 675+31828 = 32503 lbf b) To maintain block in free position F = {wt. of concrete}-{Buoyant force of H20)} = 675-

wgV = 675-[62.4g(3x3x0.5)] = 675 - 281 = 394 lbf

2.22

Distance z measured along gate surface from bottom

=500(15)-

g(h-zsin60)dz=0

=7500

(62.4)g

=7500

h3 = h=8.15 ft.

= 541

2.23

Using spherical coordinates for a ring At y=constant dA=2 r2sin d P= g[h-rcos +rcos ] dFy=dFcos Fy = (h-rcos +rcos )(2 r2sin =2 gr2 (sin Let: c=2 = c[

d +

d ]

+r )(1-

d )

gr2 sin

= c[(h-

d )

]

)- (0-

)]

Now for Fy=0 sin =

=[1-

0=(h-

)(

]1/2

& r=d/2

)+ (

) = h-

Giving h= =

1/2

For d=0.6m h=

1/2

2.24

12 ft

H2O, A

10 ft

Mud, B

PA-Patm= wg(12)=24g PB-Patm=24g+40g=64g Between 0 & A: P-Patm= wgy Between A & B: P= wg(12)+

mg(y-12)

Per unit depth: F= =

dy (192)+

(50)

=18790 lbf Force Location: Fxy= = =

dy+ (576+2040)+

=288,400 ft. LBf =

= 15.35 ft.

(2973-2040)

)dy

2.25

Force on gate=

A = (1000)(9.81)(12) (2)2 = 369.8 kN

yC.P.=

= 0.0208m (Below axis B)

= =0

P(1) = (369.8x103)(0.0208) = 7.70 kN

2.26

Fc Fw

10 m

Fw =

A= (1000)(9.81)(4)(10)(1) = 392 kN

ycp is 2/3 distance form water line to A ~ 6.66m down form H20 line ~ 3.33m up from A = Fc(9)=392(3.33) Fc = 145.2 kN

2.27

Width = 100m H20@27 =997 kg/m3 F=

A = (997)(9.81)(64)x(160)(100) = 10.016 x 109 N = 10.02x103 MN

For a free H20 surface ycp= (128m) = 85.3m {below H20 surface} =106.7 m {measured along dam surface}

2.28 Spherical Float Upward forces

F + FBuoyant

Downward forces

W= gV=

R3 )

Fb=

gVz=

R3)z

z= fraction submerged F= z=

R3)-

R3 )

2.29

= G is center of mass of solid = 2[1/2(L/2)(0.1L)(L)

(

)-(0.9L)(L)(L)

(0.05L

)]+M

{Part of original submerged volume is now out of H20} (Part that was originally out is now submerged} M=

[

+ 0.045] =

(0.02833) = 0.00556

Chapter 4 End of Chapter Problem Solutions 4.1 = 10 +7x At (2,2) = 10 +14

At -30 from x axis:

Unit vector: = Along this direction the component is : * * =(

)*( 10 +14

-7 = 1.66m/s

4.2 = 10

+ 7x

10y = 7

At (2,1) c = 10 – 14 = - 4 Eqn is: 7 Or x2

10y + C = 0 y-

= 0 (a)

Across the surface connecting points (1,0) and (2,2): (2,2)

(1,0)

= 10

=20+ (3)= 20+10.5 =30.5 m3/s (b)

4.3 CV

dA+

dA = 2A2=18

=18 2Avg =

dV=0

2AvgA2 -

)dr = 72 = 128 ft/s

(1-

) 2 rdr=0

4.4

dA = 0 = in =

cos30 dA2

out

Ain = 4Aout out =

= 46.2 ft/s

A = 10( x ) = 1.11 ft.3/s

dA1=0

4.5

Steady Flow:

A1 +

1A1=

A2 +

2A2+

6 (0.2)2=

dA=0

dx = [0.2(0.2+0.5)] = 1.71 m/s

2A2 +

(D+L)

4.6 For Steady, Incompressible Flow: =A Avg=

From given data set: Dist. from center in. 0 3.16 4.45 5.48 6.33 7.07 7.75 8.37 8.94 9.49 10.-

Ft./s 7.5 7.10 6.75 6.42 6.15 5.81 5.47 5.10 4.50 3.82 2.4

Ai in.2 7.844 37.64 31.96 32.10 31.48 29.85 33.22 33.22 31.44 31.57 15.82

Ai Ft. /s 0.4084 1.856 1.498 1.431 1.344 1.204 1.262 1.176 0.982 0.838 0.264

316.14

18.263

= 316.14 in.2 {exact area= 314.16 in.2} = 2.195 ft.2 Ai = 18.26 ft3/s

Avg =

/A =

= 8.32 ft/s

4.7 (a)

Let M = total mass in tank and S = salt in the tank.

Thus,

Integrating,

t = 1 hour and 40 minutes which equals 100 minutes. M = 68 lb/ ft3 x 1ft3/7.48 gal x 100 gals = 909.9 lb

(b)

Rearrange and solve for dt

Integrate,

4.8

For Piston & Cylinder Shown: At 1: = 1= 2 ft./s A1 1 = A2 2 2=

a2 = a1

a = a1 = 5ft/s2 )2(2) = 128 ft./s

= 5( )2 = 320 ft./s2

4.9 For steady flow: dA = 0 or

=constant =

Q.E.D.

4.10 dA= ~ ( ~ =

)dA=d dV= M +

Q.E.D.

4.11

For the C.V. shown: dA+ Steady flow:

dV=0 dV=0

C.V. moves to right with = Thus: 2=

w(1-

2A( w-

2) = 0

4.12 Avg =

dA =

For z = r/R Avg = 2

dz = dr/R

Max

For

d = -dz

Avg= -2

Max

Avg=

max= 0.817

max

4.13

dA+ Steady flow=

dV=0

dA= Horizontal=

dV=0

o(6d)-

o(6d)+

Horizontal + 2

o(3d) =

o(3d)

ydy

4.14

m= (2L)(b)(1)

=2 L

Where = -v =2

side=2

Giving: -2

or LV= (a) For = Avg (A constant) LV= Avgb avg = V (b) =ay+by2 with =4

(b/2)= max

LV= 4 max=

max

4.15

dA = w

(2 -

2000 cm3/m = (10)

(2)

= 150 cm/m= 2.5 cm/s

) dy = w h

4.16

v = wA = wh2cot20 = w cot(20) d h2/dt = = t=( -

t )[ cot(20)]

4.17

v = wA = wL2tan /2 =

= wL2

~ cos2 = ~ =

(tan ) = wL2sec2

4.18

Steady Flow: dA=0 Let the exit from the nozzle be position 3. 1=

A2 + A3

1.3x10-3 = (0.02)2(2.1) + (100) (10-3)2 = 8.15 m/s

4.19

Steady Flow:

A3 - A1 - A2 =0 = = 5.15 m/s

4.20 Volume displaced by plunger

=Ap =

Volume of H20 moving past P: = (A-Ap) = (D2-

)

In steady state operation these must be equal: V = (D2=V

)

(a)

Relative to plungerR=

+V = V

+1] (b)

4.21

Cons. of Mass - constant 3 out= 6 cm /s - constant

For no leakage = ApV = (2)2V = 6 v = 1.91 cm/s

For Leakage: = 6+0.6 = = 2.1 cm/s

4.22

Parallel Plates - Incompressible, Steady Flow = constant zo=

(zo-z)dz = a

is max at z=zo/2 = 6 [zo- ]

= 6 Vo/4= 12 cm/s

4.23

For steady incompressible flow: out- in=0

Q+b

(

)dy=

~ dy= ~ Q=

b(1-5/8)=

= 5/8

4.24 (See sketch for Prob. 4.14) Plates are circular +

m= b L2 =

L2 = -

L2v

out=

Lb = = Lv/2b (a) exit

L2v

As in Prob. 4.14- Parabolic Exit Profile 2 exit=ay+by

max[

–(

]

~ exit = max=

2 Ldy = v

Max

4.25

Since there are no leaks in the system, And the equation becomes

4.26 (a) Total mass in the tank after 2 hours 1 liter/min = 0.01667 liters/sec Total Mass = M

At the 2 hour point, which is 120 minutes,

(b) The amount of sodium sulfate in the tank after 2 hours

Rearrange,

Integrating,

The initial condition given in the problem was that at time = 0, there is 6000 grams of a 10% solution of sodium sulfate solution.

At the 2 hour point (2 hours = 120 minutes):

Chapter 5 End of Chapter Problem Solutions 5.1

Cons. of Mass:

dA=0

1A1=[2

4 =2 =

= = 26.7

5.2 System shown in Prob. 5.1 =

-Assuming unit depthFx + (P1-P2)4=2 [ =2 [ =2

+ [

]

+1)]-4

From 5.1

Fx+(P1-P2)4= Fx= -800N/m= 52.8 lbf/ft. P1-P2 =

= 157 lbf/ft2

7500 Pa = 7.5 kPa

5.3 Same general configuration except exit velocity distribution is = 2(1-cos ) As in 5.1 the expression to be used is: A1 = 2 4 =2

= 55 ft/s

5.4

Steady Flow Fx= =

dA 2

A 2-

(1.02 - ) =

1A1

= (0.0805 LBm/ft.3)(10.8 ft.2)(300 ft./s) =260.8 LBm/s

Fx = (260.8 lbm/s)[1.02(900 ft./s)-300 ft./s] = 5005 lbf

5.6

C.V. around boat- Steady, Incompressible flow =

Fx=

dA =

( - )

Tension in Rope= Fx/cos30 =215 lbf

= 186 lbf

5.7

Flow is Steady, Incompressible = dA = ( - ) =

= 0.8(62.4)(3 ft.3/s) =149.8 lbm/s

= Fx+P1A1-P2A2 {Atmospheric Pressure cancels} Equating: Fx + P 1

- P2

Fx= (P2

- P1

(

P1= 50 Psig P2= 5 Psig Fx= -5630+392 = -5238 lbf

) -

)

5.8 Fluid is H20 P1= 60 Psig P2= 14.7 Psia (units?) D1= 0.25 ft. D2 = ft. = 400 gal/m= 0.892 ft.3/s Steady, Incompressible Flow dA =0 =

= 18.17

Fx-P1A1-P2A2=

= 18.17 ft./s

= 72.7 ft./s ( - )

Fx= ( - )-P1A1+P2A2 =[62.4(0.892)(72.7-18.17)]/32.2-(60+14.7)(144) (0.25)2+(14.7)(144) =94.3-528.0+25.4 =-408 lbf

5.9

H20- Flow is Steady, Incompressible = dA = P 1A 1 - P 2A 2 dA = A2

- (As + Aj )

Equating: P1-P2= [

]

By Conservation of Mass: dA =0 A2 - As - Aj =0 = P1-P2 =

(10) +

(90) = 18 ft./s (a)

[(18)2 -

(10)2 -

(90)2] = -1116 lbf/ft.2= -7.75 psi

P2-P1= 7.75 psi

5.10

Flow is Steady, Incompressible, Frictionless For Frictionless Flow - No Drag on plate = Fn=

dA =0 ( cos )Aj = -

Fx = (-160)(4/5)= -128 lbf Fy = (-160)(3/5)= -96 lbf

= -2(100)(4/5)= -160

5.11

Steady, Incompressible, Frictionless Flow In x-direction: = ~ Cons. of mass:

dA = h=

(a+b)

h=a+b a= (1- cos

b= (1+ cos

~ =

dA =

Part (b): =Fy = Fy = 2

h sin

(

a)-

(

h(sin

(a)

h cos =0

5.12

Flow is Steady, Incompressible, Frictionless - Atmospheric pressure cancels C.V. moves to left with velocity, vp = dA Fx= Aj( j+vp)2 =

(5+30)2 = 237.4 lbf (for moving plate)

For vp = 0 Fx =

(30)2 = 174.4 lbf

5.13

Cons. of Mass: For unit cross section +

M= 2

2x+

Since

2( w -

2) -

2 2

2 2= 0

m= 0

(1)

x-momentum: = P2-P1 = From (1):

dA + 2

2 ( m-

2)=

Giving: P2 - P1 =

2x=

Q.E.D.

2 [ w-

5.14 For situation considered in Prob. 5.13 (a) Air

m= 1130 ft./s = 0.00238 Slug/ft.3 P2-P1= 1 2 m = (0.00238)(10)(1130) = 26.9 PSF = 0.187 psi

(b) H20

m= 4700 ft./s

= 1.938 Slugs/ft.3

P=(1.938)(10)(4700) = 91,080 PSF= 633 psi

5.15

Cons. of Mass: Technique is to let P2 = P1 + 2=

S S

etc.

By Conservation of Mass- { S= y->0} ( A )=0 By Momentum Theorem, using Cons. of mass result: dP+ d +gdy=0 -messy-

5.16

D1=0.3m 1=12 m/s P1=128 kPag

D2=0.38m P2=145 kPag 2=7.48 m/s

A1 = (0.3m)2 = 0.0707 m2 = A1

A2 = 0.1134 m2

3 1 = (0.0707)(12) = 0.8484 m /s

In x direction:

Fx + P1A1 - P2A2cos =

dA ( 2 cos -

Fx=(1000)(0.8484)[7.48(cos30 )-12]-(1000)[(128)(0.0707)+(145)(0.1134) cos30 ] = -505.5 N In y direction: Fy – P2A2sin =

( 2 sin )

Fy=(1000)(0.8484)(7.48sin30 )+(1000)(145)(0.1134)(sin30 ) = 11395 N=11.395 kN

5.17

Steady incompressible flow: = dA A1= (1.5)2=1.767 ft.2 A2= (1)2=0.785 ft.2 A3= (2)2=3.142 ft.2 For C.V. between (1)&(2) of the figure: = dA Fx+F2+P1A1-P2A2= Fx

( 2-

+(530-14.7)(144)(0.785)-(990-14.7)(144)(1.767)-F2 = -130,000 lbf –F2

For C.V. between (2)&(3) of the figure: Fx+P2A2-P3A3= ( 3- 2) Fx= (6700-3400)+(26-14.7)(144)(3.14)-(530-14.7)(144)(0.785) = 25777 lbf Stress at position 2 in the figure: = =

= 1823 psi Compression

At 1: F1= -130,000+25777 = -104220 lbf =

= 4915 psi Tension

5.18

Flow is steady & incompressible No net pressure force = Fx=

dA dA-

Ain = 2

( )2dy+

(3d)-

Momentum out top & bottom Fx=2

Force on cylinder=

(3d-6d) = d

(6d)

5.19 Fluid is H20 Vsonic=1433 m/s This is just like Prob. 5.13 for an observer moving with H20 stream: = 3 m/s Then P=

=(1000)(1433-3)(3) = 4287 kPa

5.20

Static pressure of H20: On left- P= H On right- P=

H2 = H=[

=( [ h

+ h2]1/2

+ h2

5.21

Conservation of Mass: =0 - hv1+ Lv2 = 0 v2=

(a)

x-momentum: =

P1A1-P2A2+Fx= ( 2Fx= ( 2-

1)+P2A2- P1A1 =

1h( 2-

1)+

h(h/L -1)+

(L2-h2)

(b)

5.22

Conservation of Mass: 1h 1= 2h 2 Momentum Thm: = dA P1h1-P2h2= ( 2P 1=

1h 1( 2-

From Cons. of Mass:

( -

1 h1/h2

Factoring & cancelling h1-h2 (h1+h2)=

+ h1h2 -

h 2= [

- 1]

& from continuity [1+

]

5.23

D1= 8 cm 1= 5 m/s h= 58 cm

D2= 5 cm P2= 1 atm

A1= (8 cm)2= 50.3 cm2 A2= (5 cm)2= 19.6 cm2 2= 5 m/s(

)= 12.83 m/s

x-momentum:

Fx+P1A1-P2A2=

( 2-

P1-P2=

dA 1)

wgh[13.6-1] = (1000)(9.81)(0.58)(12.6) = 71.69 kPa

Since P2= 1 atm P1= 71.69 kPa Fx+P1A1= (1000)(50.3x10-4)(5)(12.83-5) Fx= 197-71.69(50.3x10-4)(1000) = -163.7 N

5.24

x momentum: = + dV Fx= - wAj ( cos )+ wAj = wAj ( - cos ) =1000( )(0.1)2(20)[4.5-20cos45 ] = -1515 N Force on car by jet: Fx=1515 N y momentum: = Fy= - sin

( Aj)+ 0 = -(20sin45)(1000)(-20)* (0.1)2 = +2220 N

Force exerted by H20: Fy= -2220 N Total Force = = 1515 - 2220

5.25

Coordinates fixed to cart ~moving to right at Momentum thm. in x-direction = Aj (- )- As (- ) Fx= As - Aj 2 In y-direction Fy= Aj (0)- As (- s) = As

Force of fluid on car is negative of these

5.26

for C.V. shown as a red dotted line in the figure. + M=0 M= Ah = A =

- gAh= - A 2+ A (h -gh= - 2+ (h = -g

5.27

Steady in compressible flow Rotation is about z-axis= dA

out = (rx

)

out

At position on x-axis: rx=r2 and ry=0 = 800( = tan =

)= 1.783 ft.3/s

)(

= 8.51 ft./s =

= 8.51 ft./s

Abs. velocity @ r2 = -rw+ tan = -(8/12)( )+ t = -82.38+ 8.51= -73.87 ft./s Now- in momentum expression: Mz=(r2 )

Power= Mzw = 174(

= 174 ft. lbf )(

) = 39.1 Hp

5.28 Same configuration of Prob. 5.28 = 1.783 ft.3/s at inlet- =

Vr1

= 25.54 ft./s rw

r =-r1 =-(2/12) (

) = -20.6 ft./s

= 38.9

Part (B): = Fx=-

dA (

)A1 =

= 70.6 lbf

5.29

y x

For C.V. shown as a red dotted line. =

Mz=

=2 (

= 1.057 ft. lbf

5.30

x )z

( )

Mz=2() ry= Rsin = sin - R Mz=2 [-Rsin ( sin - r )] =

5.31

x )z

tdx

t xdx= =

t( )

= 64 ft./s

(

) = -5950 ft. lbf

5.32

x )z

MB=

+ r2P2A2 =(

=(30 gal/m)(

= 0.0668 ft.3/s

= = MB =

)

+ r 2P 2A 2

= 21.79 ft./s + (3)(24)( )(0.75)2 = 8.46+ 31.81= 40.3 ft. lbf

5.33

Linear Momentum: Coordinate System moves with cart = Fx= A[ - c)cos ]( - c)- A( - c)2 P=

cFx=

A ( - c)2[cos -1]

For m=

P= A[cos -1]

For P=Pmax

m(1-m)2

= A[~] 3

(m-2m2+m3)

={ } (1-4m+3m2)=0 m= 1, 1/3 m=1 ~ minimum m= 1/3~ maximum m= = 1/3 Part (b) Rotation about z-axis=

x )z

dA =

=r [ - c)cos + c- ] = r (cos -1)( c) P= Mzw= Mz =[ (cos -1)] c( for m= P=[ ]m 2(1-m) = [ ] 2(1-2m)=0 or Pmax occurs when m= = ½

(out)-

(in)

Chapter 6 End of Chapter Problem Solutions

6.1 For V=A+Br V(ro)=0=A+Bro V(ri)= = A+Bri A=-Bro= B(ro-ri)= B= A= V=

- Bri

6.2 Steady Flow: -

=0 dA

[(u2-u1)+

= 1025(21)= 215.25 kg/s

= 4.278 m/s

= 11.573 m/s

Since T=0

+ g(z2-z1)]

u2-u1=0

z2-z1= 1.8 m P2= 175 kPa P1= -0.15 m Hg = -19.9 kPa = =

= 190 m2/s2 = -57.7 m2/s2

g(z2-z1)=9.81 (1.8)= 17.7 m2/s2 - =(190-57.7+17.7)(215.3) = 32,295 w= 32.3 kw

6.3 =

dA +

=0 -

[h1+

+ gz1]+ [mu]sys=0

gz1=negligible

vcv = A ( ) =

= 21.8 F/s

6.4 Energy equation reduces to dA =0 u2-u1+

+ g(y2-y1)=0

= g(y2-y1)=0 u= c T= T= =

= 0.0297 F

6.5

Between A & B: =

= = u=

= 402,000 [

+ g(yB-yA)+ u]

=0 ={

+ g(yA-yB)+

}

PB=PA+ {

+ g(yA-yB)+

} = 46.5 Psia

6.6

= 4 ft.3/s P1= -6 Psig =

P2= 40 Psig

= 5.10 ft./s

= 7.33 ft./s

y2-y1= 5 ft. Energy Eqn. reduces to: =

[ +

g(y2-y1)] =

g[ +

+y2-y1] = 27850 ft. LBf/s = 50.6 Hp

6.7

CV is red dotted line. Entrance is 1 and Exit is 2.

For C.V. (1)- Energy Eqn. reduces to 0=

+ g(z2-z1)

(z2-z1)=

= [2 ]1/2= 20 m/s =A = (0.3)2(20)= 1.417 m3/s For C.V. (2) Energy Eqn. is: =

[ u+

=(1.22)(1.417)(20)2/2 =346 W

6.8

Steady Flow Energy Eqn: 1(u1+

+ )+

3(u3+

+ )=

2(u2+

Energy Eqn. can be written: A1

1[cvT1+

+ )

Cons. of Mass: 1A1+ 3A3= 2A2 (1)

Momentum: (P1-P2)A1=

A1-

+ ]+ A3

3[cvT3+

A3 cos (3)

For u= cvT, T1=T3, P1=P3 & Lots of Algebra cv(T2-T1)=

[1+

-1] [

]+ [

-2 cos ]

+ ] = A2

2[cvT2+

(2)

6.9

=3 ft.3/s Between A & B- Energy Eqn. is: ) dA=0 + uB-uA+ A=

= 3.82 ft./s

= 15.28 ft./s =

+ 0.45 = 2.15 ft. H20

PA= 2.15+2= 4.15 ft. of H20

6.10

C.V.is red dashed line in the figure. PA= 10 Psig = dA Flow rate must be determined Energy Eqn. for C.V. shown: u+ = A= B=

+ g z=0

u=0

= 743 ft.2/s2 = 2.865 = 3.82

= (2.8652-3.822) = -3.19 g y= 32.2(-5)= -61 ft.2/s2 743-3.19 -61=0 = = 14.6 ft.3/s A= 41.9 ft./s B= 55.9 ft./s Fy+PAAA- gV= ( A) PAAA=10( )(8)2= 502 lbf gV (wt.)= = 109 lbf ( A)= = -1185 lbf Fy= -502+ 109- 1185 = -1578 lbf (Force on lid is 1578 lbf )

6.11 = 6 m3/s P= 0.10 m Alcohol (S.G.= 0.8) = 0.08 m H20= 785 Pa A1= (0.6)2= 0.283 m2 1=

= 21.2 m/s

Energy Eqn. reduces to: +

+ g y=0

= [

= 640 m2/s2

640= A2= 0.144 m2 D2=0.428m

g y=0 ]= [

6.12

Energy Eqn. reduces to: u+

+ g y=0

& for u=

1=0

+ g(y2-y1)=0

2 = [2

)+g y]1/2

By Conservation of Mass: ATank( =Ajet 2 =[2

)+g y]1/2 =

k1=

k2=2g

t= k1= 344 ft.2/s2 [ [ t= 105 s

( ]1/2= 23.2 ft./s (yo)]1/2= 25.8 ft./s

6.13 Energy Eqn. reduces to: +

= +

P1=Patm= 29 “Hg(

)=14.25 Psi

P2=? =[(85 n /H)(

)]2=(124.7)2

= 1202 P2=14.25+ [(124.7)2-(120)2] ~ = = = 0.0770 lbm/ft.2 ~ P2= 14.25+1.37= 15.6 Psi

6.14

Energy Eqn:

+g(y2-y1)=0

In x-direction:

ocos

In y-direction:

osin

x=(

ocos

y=(

osin

t-gt2/2

Combining:

y=xtan -

y=0.6 m x=3.6 m

tan = 0.577 cos =0.866

0.6= 3.6(0.577)= 7.57 ft./s Total head= 0.6+

= 3.52 m

6.15

= 550 g/m= 1.225 ft.3/s = =

= 6.35 ft./s

Energy Eqn: 2 is at H20 level outside pipe + = P1=-

+ g(y1-y2)=0 - gy1 =

= =0

- 32.2(6) = -(20.16+193.2) = -213.4 ft.2/s2

= -2.87 Psig

6.16 With reference to Prob. 6.15 between H20 surface & pump inlet energy Eqn. is +

+ g(y1-y2)= hL -

-(y2-y1)- hL =

P2=Pv

+ y1-y2= hL

2=[2(32.2)(25.35)]

P1=PAtm

1/2

= 40.4 ft./s

(40.4)= 7.8 ft.3/s

-4-4 = 25.35 ft.

6.17 From Prob. 5.27

r= 10.22 ft./s

r2= 82.2 ft./s

t= 10.22 ft./s

At r2:

x= 82.2-10.22,

y= 10.22

=( + )1/2= 82.9 ft./s Head=

= 106.7 ft.

= 62.4(82.9)2/32.2(2) = 6660 LBf/ft.2 = 46.2 Psi

6.18

For the situation shown: Thrust= F= Power=

~ ~ ~ Favorable Choice:

High Volume, low pressure

6.19 From Prob. 5.7: P1= 50 Psig D1= 12 in. = 3 ft.3/s hL=

P2= 5 Psig D2= 2.5 in. S.G.= 0.8

~ 2 1=3/ (1) = 3.82 ft./s

2=3/

hL=

= 88 ft./s +

= 9.79 ft.

6.20 For a C.V. enclosing the falls: u+

+ g y=0

u= g y= 9.81(165)= 1620 m2/s2 =1620 m2/s2(1 For H20: Cp= 4184 J/kg*K T= =

0.39

)= 1620 J/kg

6.21

Assume Vertical Forces do not include momentum of incoming air: (P-PAtm)A=mg

{Pressure Force}= Wt.

Energy Eqn. becomes Bernoulli Eqn. between inside & exit=

=2 = 4885 m2/s2

= 69.9 m/s

=69.9(24)(0.03) =50.3 m3/s = 60.6 kg/s Energy Eqn: = With u=

( u+ g

)

Energy Equation becomes =

= 148 kW {per person}

6.22 From Prob. 5.22 h2= [

- 1]

For Bernoulli Eqn. to be vsolid-hL=0 Energy Eqn. for this case is hL=

+ y1-y2

& since P=Patm+ hL=

(h-y)

+ h1-h2

Writing solution to Prob. 5.22 as h2= [

- 1]

{B=

}

& Note that for h2>h1 B>8 Bernoulli Eqn. applies for B=8 & Obviously hL>0 for B>8

6.23 Energy Eqn. applies in form: =

(Power)

= 1.238x10-4 ft.2/s

= P=

= 1.584 Ft. lbf/s = 2.148 W

Per monthP= 2148 W (30)(24) = 1547 Wh= 1.547 kWh

6.24 Bernoulli Eqn. between free stream & a reference point(1) on car +

= =

6.25 Energy Eqn. is =

)

= [(u2-u1)+

dA +

+ g(y2-y1)]

200 kJ/kg =

= 340 kJ/kg

=0 g y= 9.81(15)= 0.147 kJ/kg = 200+340+0.15= 540 kJ/kg

6.26

B(2

)(1)(

)

2 2 A = 8.22 2 2 B =14.6

A=2.865 B=3.82

For negligible friction- Bernoulli Eqn. applies +

+ g(y2-y1)=0

= 2 =

=3.32 2 = 743 ft.2/s2

g(yA-yB)= 32.2(-5)= -161 3.32 2= 743-161= 582 = 13.2 ft.2/s

6.27

Bernoulli Eqn. between 1&2 +

1=1/

= 5.09 ft./s

2=1/

= 11.5 ft./s

= -1.65 ft. H20

Manometer reading = -1.457 “ Hg h[1-1/13.6]= 1.457 h= 1.63 inches

6.28

For a Control volume between 1&2(exit) in mixture region +

+ g(y2-y1)=0

P1-PAtm=

g y1

=0 (1)

Mass Balance around mixing chamber Air+ w= m As given: = /2 m=2

w+2

(2)

Control volume between H20 surface & 1 (H20 only) + P1-PAtm=

+ g y2=0 g y2-

(3)

Equating (1) & (3): g y1=

g y2-

)

= g( y2- y1/2) w=[2g(0.45m)]

1/2

Substituting Expression for m=2[g(0.9m)]

Since

>>1

1/2

w into (2)

2nd term is small

and 1/2 m=2[9.81(0.9)] = 5.94 m/s

6.29 For conditions of Prob. 6.28 Control volume around mixing chbr. (word?) = dA PA= A m[ m m] ~ This neglects momentum of air ~ From Prob. 6.28 Above mixerP=PAtm+

g y1 2

Below mixerP= g(y2-y1)-

P=PAtm+

g y2-

Equating with Momentum Expression 2 P= (1- ) = [g(y2-y1)- 2] 2

= g(y2-y1)-

For P=

=2 2

= 4.6 m/s

(1-1/2)= ½ (4.6)2(1000/2)

5.3 kPa

6.30

In both gases- Bernoulli Eqn. is: 2 2

2 1

+ 2

2 2

152

(a) P= 2g=2 9 1 = 11.47 m Air = 1.39 cm H20 2

242 15

(b) P= 2 9 1 = 17.9 m Air = 2.52 cm Oil

6.31

Between liquid surface & exit: Bernoulli Eqn: 2

=g y =(2 g y)1/2 = 2(32.2)(10)= 25.35 ft./s

=4 12 2(25.35)= 1.66 ft.3/s Between point B & exit: + g(yB-yExit)=0 PB=PAtm- g y = 14.7 PsiBy continuity: Tank= ATank(

1 2

2[10

) =Apipe 2g

dy=

2y1/2 17 = 1/2

2 2 144

2g y1/2

= 7 1

2 4 2 2 14

2g t 1/2

– 7 ]=

1 2 12 2

t= 1854 s= 0.515 h

2 22 t

= 8.63 Psi

6.32 Energy Eqn. for this case: 2 2

2 1

+ g(y2-y1)+ u2-u1=0

1=0

y2-y1= -10 ft. 2

u2-u1= 3.2 g 2 2

+ (-10)+ 3.2 g2 =0 2g 2 2 g

= 10

= 10.03 ft./s 1

=4 12 2(10.03)= 0.0547 ft.3/s

6.33

Between 1 (top of tank) & 2 (at the exit of the nozzle): 2 2

+ g(y2-y1)=0 1/2 2=[2(32.2)(20)] = 35.9 ft./s 2

= A =4 12 2(35.9)= 0.783 ft.3/s I 4” l

= 42 = 8.975 ft./s 2 = 80.55 ft.2/s2

Between 1 & A: 1

2 1

+ g(yA-y1)=0 24

PA=PAtm+ (- 2 /2)+ g(y1-yA)= PAtm+ 2 2 ( 2 )+ 62.4(23) = PAtm+ 1356 LB/ft.2= 3475 Psf (24.12 Psi) A= 8.975 ft./s Between A & B: 2

+ g(yA-yB)=0

PB=PA+ g(-3)= 3290 Psf (22.83 Psi) B= 8.975 ft./s Conditions at D & B are equal PD= 3290 Psf D= 8.975 ft./s Between B & C: 2

+ g(yB-yC)=0

PC=PB+ g(yB-yC) = PB+ 62.4(-20) = 2042 Psf (14.18 Psi) C= 8.975 ft./s

6.34 Between water level (1) & exit (2) 2 2

2 1

2 2=

2 1

+ g(y2-y1)=0 2g 2

[2g ]1/2

H20 in tank: =4 2 )2 2g y1/2

= 1 2

dy=

2y1/2 24 =

)2 2g t 1

)2 2g

2 2 2 42 2 2 12 2 15

= 6644 s= 1.846 hours 1 2

6.35

1 2

From 1 to 2 1

+ 2 2 1 + g(y1-y2)=0

+ 22 =0

1 1

(1)

From 3 to 4: 2

4 2 4

2 4

+ g(y3-y4)=0 4

= 2 + 2

– gL

(2)

Note that P4+ 1 gL=P1, giving 2 4

= 2 + 2

(3)

From 2 to 3: 2 1

+ 22

For 2 & P2=P3

=0 negligible

& from (2) 2

= -gL+ 1 gL= gL( 1 - 1) 2

6.36 From Prob. 6.28: 1

+ 22 =0

= 2 + gL 2

1 2

- 2

Cons. of Mass: = 2 3= 2 /R 1 2 ~ H =R ~ 2 2= 2

2 2

)(

= 2

Giving P2-P1=

2 2

)

2 2

P3-P1= 2 2 + gL( 2 - 1 - 2 2 2

From momentum theorem= ( )dA (P2-P3)A= 1 2A( 32

P2-P3= 2 2 1

)

Bernoulli Eqn: 2

) + 21

2 2

)

Doing the Algebra: 1 2

2g 2=

2 1

2 2

+ 2 + gL(1

1 2

2 2

6.37 Frictionless flow: From Bernoulli 2

+ 22g 1 + (y2-y1)=0 g

2 2 = 2g(y1-y2)

=[2(9.81)(10)]1/2= 14 m/s =(1000)(4 (0.04)2(14)= 17.6 kg/s with nozzle- = 14 m/s {still} =(1000)(4 (0.01)2(14)= 1.10 kg/s with u2-u1=3

Energy Eqn. reduces to 2 2

+ g = 2(y1-y2) 2g 1/2 2=[ 4/7(9.81)10] = 7.49 m/s

Pipe: = 9.42 kg/s Nozzle: = 0.589 kg/s

6.38 Same tank as in Prob. 6.37 but 2 exit pipesPipe 1: D= 0.04 m y= 10 m Pipe 2: D= 0.04 m y= 20 m Frictionless Flow: Pipe 1As in Prob. 6.37 = 2g

= 14 m/s

= 17.6 kg/s Pipe 2: Also = 2g =[2(9.81)(20)]1/2 = 19.81 kg/s =(1000)(4 (0.04)2(19.81) = 24.9 kg/s

6.39

,l ’

First, since the shaft work and heat must be 2 l 5 1

2 2 1

77 17

l 2 174 l

1 1 4 5 4

Assuming no viscous work and steady state,

Writing in terms of energy efflux,

Expanding the energy term, and using 2

,l ’ 2

l ul

2 2

2 1

2 2

u ll

55 7 9

21 74

l l

5 2 4

2 2

2 4 12 2

2 174

77 17

2 1

1 1

2 2

229 1

1 1 2

44 u

Plug these values back into 2 l 1 1 4 5 4

5 1 2

Note that the same value of left out since it cancels. 4 11 4 5 2

7 9

2 2

2 174 215

l 2

l l

l 2

55 7 9

21 74

was used in both places in the above equation, it could have been l

1, l

7255 9

6.40

2 2

12 4 2 2 12 4

2 1

77 17

1 19

91 7

Plug all known values unto equation 1

l 2 174 l

29 1

2 2

71 1

29 1 5

2 2

l 2 2 174 l

1 19 2

91 7

l 2

Chapter 7 End of Chapter Problem Solutions 7.1

(a)

(b)

7.2 For 2-D Flow-in x,y =0

zx= xz=0

zy= yz=0

7.3 I II

y x

I= (x) t II= (x+ x) t Axial Strain Rate: = lim ->0 ->0

Rate of Volume Change: =lim ->0 ->0 =lim ->0 ->0

In 3 Dimensions Both Axial Strain Rate and Volume Change are given by +

7.4

In r-z plane: = - lim ->0 =- lim [ ->0 = lim [ ->0 ->0

rz= zr=

[

]

In the -z plane Same procedure lim [ ->0 ->0 =

= [

]

& in r- plane lim [ ->0 ->0 =

= [

]

( )]

7.5 Nitrogen 175 K = 2.6693x106 T= 175 K M=28 =91.5 =1.91 11.55x10-6 Pa.s.

=3.681 =1.1942

7.6 Oxygen @ 350 K Eqn. 7.10 =

Yielding M=32

= 2.6693x106 = 3.097

=1.03 =3.433

= 2.327x10-5 Pa.s. Table value: = 2.318 Pa.s

7.7 For H20 = 0.76x10-3 LBm/s*ft. = 0.375x10-3 LBm/s*ft.

Percent change =

= 0.51 or 51%

7.8

Properties of Helium, Glycerin from Appendix T,F 60 0.00125 80 0.00132 100 0.00141

Plot the data and obtain an intersection very close to 100

,Glycerin 0.0177 0.00762 0.00128

7.9 For H20 ~1/ @ 120 F

w= 0.391x10

@ 32 F

w= 1.2x10

-3

lbm/s*ft.

= 3.07

Percent change =

= 3.07-1 = 2.07 or 207 %

7.10 For Air: @ 140 F @ 32 F

= 1.34x10-5 LBm/s*ft. = 1.15x10-5 LBm/s*ft.

For ~1/ =

= 0.852

Percent change =

= 0.852-1= -0.148 = -14.8 %

7.11 oil

Di= 3.175 cm Do= 3.183 cm = 0.1 Pa.s 1st Law:

{no flow in or out}

=0 =

Viscous =

-at moving boundary {t= gap width}

( DL)(r ) =

=1700( )= 178 Rad/s =

= 5.58 W

7.12 Refer to Prob. 7.13 For =

2=2 1

Percent increase =

= 4-1=3 = 300 %

7.13

Ship 1: Ship 2:

1= 4 m/s 2= 3.1 m/s

Choose control volume attached to Ship 1 = dA = Relative to moving ship =0 Fx= +

= -0.9 m/s

= 100 kg/s(0.9 m/s) = 90 N

This is force applied to maintain stated conditions Force exerted by fluid transfer= -90 N

7.14 t= Gap= F= =

= 0.01 cm

= =

{Assumes linear profile}

= F=

= 1676 N

7.15 Refer to conditions of Prob. 7.14 Load on Ram= 680 kg, L= 244 m F= mg= v=

= 0.768 m/s

7.16

M= dF= dA is on the conical surface = 2 rdL {dL is along slanted surface} dL= dr/sin so: dF=

= = rdF =

2 r

7.17 =

[1-(

]=2

[1-(

]

= =2

[

]

At r=R

= -4

= ~ w= 0.76x10

-3

lbm/s*ft. @ 60

~ =

= -0.0453 lbf/ft.2

7.18 For conditions of Prob. 7.17 = F= A= DL = P=

( DL) = (-0.0453) )(0.1/2)(1) = 0.00119 LBf

= 0.00119 lbf ( (0.1/12)2/4) = 21.75 Psf

7.19 Shear work rate= = For parabolic profile= max[1-(

2 max [1-(

( )= For

2 max [

( )=0 =

] ][

]= ]

2 max [

]

7.20

MOVING AT 3 ft/sec 0.03 inches

FIXED

(a) What is shear stress exerted on the fluid under these conditions?

Solving for shear stress,

(b) What is the force of the upper plate on the fluid?

Chapter 8 End of Chapter Problem Solutions

8.1 Hagen- Poiseulle Eqn. =

For D=Do For D1=2Do

o=[(

)

] Do4

4 1=[ ](2Do)

1=16 o

Percent Change =

-1= 15 = 1500 %

8.2 For single pipe: Po=[

]

o(40)

For single- parallel combination P1= [ ] (22){ single branch} P2= [ ] (18){parallel branch} P1+ P2= Po= 3.45x106 Pa =2 Case 2: P1+ P2= [ ](22+9) = (4000) o= = 5161 BBL/Day

8.3 For 1- Reservoir 2- Pipe Entrance 3- Pipe Exit Between 1 & 2: =

+ g(y1-y2)=0

=(y1-y2)=0

Between 2 & 3: +

+ g(y2-y3)-

=(y2-y3)= 0

For Inviscid Flow =0 For Laminar Viscous Flow: = Inviscid Case: =

~ Assuming fluid is hydraulic fluid @ 60 F – 15.9 K =849 kg/m3 =0.0165 Pa.s ~ =[ ]1/2= 22.08 m/s =A = (0.006352)(22.08)

7x10-4 m3/s

Viscous Case: =

-2

=0 =30.85 m2/s2

= 2

= 487.6 m2/s2 2

+30.85 -487.6=0

= 11.51 m/s

= (0.00635)2(11.51) = 3.645x10-4 m/s =

= 1.92

8.4

From 1 to 2 (Bernoulli) +

+ g(y2-y1)=0

with

+ g(y1-y2)

From 2 to 3 +

+ g(y2-y3) +

~ =

~ Combining expressions: =g(y1-y3)~ = ~ =g y

= 1.0605x10-5 = 0.000987x10-5

=(1.0605-0.001)x10-5 = 1.0595x10-5 m2/s

8.5

Using the same development as in Section 8.1: (r )=r For an element of length, L (r ) =r which becomes (r )=r Integrating: r=

+ c1

+ c1/r

For laminar flow, Newtonian = so

= =

+ c1/r

rdr+

Integrating: = r2+ lnr+c2 Boundary conditions: (r=R)=0 (r=kR)=0 k<1 Considerable algebra yields c1=

c2=

R2[1-(1-k2)

]

& with substitution & simplification: =

[1-

ln ]

8.6 This is same configuration as shown in Prob. 8.5 (r )-

r=0

Integrating:

& for laminar flow, Newtonian fluid: = -

Integrating: x-

r2= lnr+c2

Boundary conditions: (r=D/2)=0 (r=d/2)=v More algebra: c1= [v+ c2=

(D2-d2)]

- ln

Drag force per unit length: F= A= d)(1) = ( d) Giving: F= d [ + For the case with F=

= d [ =0

]

8.7 In -direction: = -r z

z component of force on(+ )face

Divide by r (r r

z & take limit as =0

+ 2 =0 + 2 =0

ln +2lnr=ln(constant) r2 = constant = r ( ) r2 = r3 ( )= constant(c) d( )= = c1( =

)+c2 + rc2

Boundary conditions: (R)=0 0= + Rc2 (kR)=V Algebra = (

V= - )

If profile is linear: (word?) = ar+b = ( – 1) Percent error= = Actual- Linear

+ kRc2

(

- )-

(

(r– R) + )-

Vmax occurs at = =1-

=0.01

Resulting in =0.99 k= 0.96

[

(

+ )- ] =0

8.8 For flow between 2 horizontal platesGoverning D.E.( yx)- =0 Laminar, Steady Newtonian yx= B.C.- At interface (y=0) (1) = (2) yxA= yxB (3) (-hB)= hA)=0

8.9 Fully developed, steady, laminar flow; Newtonian fluidyx -

yx=

B.C.

=0 for y= h =

= =

y+c1 y2+c1y+c2

Applying Boundary conditions c1=0 c2= Giving: =

(y2-h2)

8.10 Governing D.E. is yx -

Integrating: yx -

y=c1

Laminar flow, Newtonian fluid: yx= -

y=c1

For yx(0)=0 c1=0 -

y=0

= c2

@y=h=

c2=

Also

@y=0=0

Giving

c2=0

8.11 For horizontal pipe flow: Development in Section 8.1 results in =( For =0

+ c2 must=0 for all r

=constant=v

8.12 For an element in liquid film: y

- xy y - g x y=0 – =0

In limit as x->0 =0 xy xy=

- x= c1 x2=c1x+c2

Boundary conditions: (0)= o (h)=0 c1= - h c2= o Giving: = o - (2hx-x2) =

[2 -(

]

= =

[2 -( ohoh-

(h3-h3/3)

]dy

8.13

Treat fluid layer as a thin linear layer: In the usual way: Treat

=0 constant~

& = Giving =

+ c1y+c2

Boundary Conditions: (0)=R (h)=0 Giving R =c2 0=

+ c1h+ R

c1= = R (1-y/h) Flow rate=

[( )-(

= Giving:

)dy = =

[

Efficiency= = ~ evaluated at R(y=0) ~ = + After doing the algebra: =

]

8.14 Fluid enters at x=0 & flows equally in +x & -x directions exiting at x=L/2 where P=Patm Working with the R.H. flow(in +x direction) The applicable D.E. is = & as usual Giving

Integrating: =

y+c1

Boundary conditions: (0)=0 y2+ c2

Again =

Boundary conditions: ( =0 So c2=

)

Velocity expression is: =

)(

-y2)

= =

)

So the expression for

is:

= &

Po-PAtm= & for the plate of total length,L, Fy=( Po-PAtm)2L=

c1=0

8.15 Liquid flowing Down the outside of a cylinder: Governing D.E. is (r )+ =0 &, as usual (r )+ r +

= c1

B.C. =0 @ r=R+h r = [(R+h)2-r2] And again: = [(R+h)2 lnr- ]+c2 B.C. (R)=0 Giving = [(R+h)2 ln +

(1- )

8.16 For result of Prob. 8.15 max=

max occurs where

[2(1+ ) ln(1+ ) -

– ]

=0 which is at r=R+h

8.17

8.18

8.19

Calculate

Calculate viscosity,

Chapter 9 End of Chapter Problem Solutions 9.1

~ Substituting into C.V. relationship & evaluating in limit as r (r r)+

z go to 0

9.2 = x + =

* =

Note

* =

+ (

(

=1 For j=i =0 For j i x

(

)

9.3 4

4’

3’

1’

2’

For 2-Dimensional flow: Volume change= ( )(

)-(

= x = x+[

)(

)

= y x(x+

x,y)-

3 2 = y[ y(x+ x,y+ ( )( )= x y

x(x,y)]

y(x+

x,y)]

( )( )= x y+[ y(x+ x,y+ [ y(x+ x,y+ )- y(x+ x,y)] 2

x,y)]

Dividing by x y

& evaluating in limit as x y

->0

Volume change =

= *

But, from continuity * =0

y(x+

x(x+

x,y)-

x(x,y)]

9.4

d =

dr+

d + dt

= cos + sin =- sin =

= - sin

In similar fashion =0

Giving: =

- )

For

to be

then

= +( r

+ r) =

= =

- )

)

9.5 Navier-Stokes Eqn.- Incompressible form: = -

(a) For small- all terms involving ( & ) are small relative to other terms. (b) For small & large the product cannot be considered small relative to other terms

9.6

y x

Incompressible N.S. Eqn. in x-direction + With

=gx-

y=0

y+ c1

+ c1y+c2

B.C. c1=0

=0 @ y= c2=

(y2-L2)

9.7 = =

(r )+

~ * =

(

( =0

And continuity is satisfied

9.8 =

At y=100,000 ft. =

=20,000 ft./s = 0.0096

s-1

9.9 P= ( - ) =400[(y/L)2 +(x/L)2 ] =

= 400(y/L)2800

Evaluated at (L,2L) we get P=

- [2g+

]

+400(x/L)2800

[x2y +xy2 ]

9.10 In x-direction: = gx= gx-

(

[

( (

(

Similarly in y & zA total of 45 terms!

)+ ( + )+

+ (

)+ (

) [

(

)] )+

(

)

]

9.11 For = o+ r o- of coordinate origin r- relative to coordinate origin =

N.S. Eqn. reduces to =

- P and P= ( - )

9.12 Given that (r )+ a) For

(r )=0

and r ( )=F( or = F( /r b) For =0 =f(r)

9.13 N.S. for Incompressible, laminar flow = -

For negligible a) Vector properties~ and are independent by themselves but in same relationship must lie in same plane. b) If viscous force are negligible = is determined by

& is positive in direction of decreasing pressure.

c) In similar fashion, any fluid- either moving or static-will move or tend to move in direction of decreasing P.

9.14 For 1-D Steady flow: = (x) = =0

(

+ [

)=0

)

]

9.15

Continuity:

+ (

Momentum: (

)=0 )=-

9.16 Taking z as positive down with

=0 &

=f(r)

Eqn. E.6. yields z direction 2

(

+ gz+ [

(r )+ 2

]

Realizing that: 2

= = =

2 2

and since gz=-g = (r ) Proceed as was done in solutions to problems 8.17 & 8.18 9.17 For incompressible, steady flow, with r direction

= =0 Eqn. (E-4) has the form 2

(

+ gr+ [

Realizing that the following terms are zero: 2

= = =2

2 2

The initial equation becomes 2

(P+

)= gr

)+ 2

- 2

]

9.18 Governing Eqns. Are r direction 2

(

+ gr+ [

)+ 2

2 - 2

]

direction 2

(

+ [

)+ 2

2 - 2

z direction 2

(

When

=f(r) and with

= = Q.E.D.

(r )+ 2

2 2 +

]

= =0 the only non-zero term on the left-hand side of all 2

component eqns. is

+ gz+ [

]

9.19 Eqn. (E-5) is simplified for this case as

direction:

(

Realizing that the following terms are zero: 2

= 2=

& in the absence of gravity we have =

[

(r )]

+ [

)+ 2

2 - 2

]

9.20 From Prob. 9.19 & steady flow =0 Giving

=c1

Integrating again: r =c1lnr+c2 B.C.

(R1)=R1 (R2)=R2

= [R12

2 2 2 2

2 2

]

9.21

Fluid flow around center section

V0 Need to begin with z-direction Navier-Stokes equation in cylindrical coordinates: 2 2

Applying the assumptions given in the problem statement:

Boundary Conditions: 1. 2. Apply Boundary Condition #1:

2 2

Apply Boundary Condition #2: 2 2

Solve for

Solve for 2 by plugging back into equation for Boundary Condition #1 (if you plug into the Boundary Condition #2 equation your result will be slightly different and no less correct), 2

2 2

Thus, 2

9.22 VB

Fluid 1

Fluid 2

(I will show two different solutions, the second solution may be slightly easier. The only difference is the identification of the boundary conditions that are in red in the figures below.) Need y-direction Navier-Stokes Equation 2

2 2

Using the conditions in the problem statement this equation becomes, 2 2

Integrate twice solving for velocity, 2 2

Next, we develop equations for the velocity of Fluid 1 and Fluid 2 using the above equation.

Fluid 1 VB

Fluid 1

Fluid 2

x=0

x=L1

x=L1+L2 2 2

2 Boundary Conditions: BC#1: BC#2: 2 Apply BC#1: 2

Apply BC#2: 2

2 Solve for

, 2

2 Thus, plugging in the expression

Fluid 2 2 2

Boundary Conditions: BC#2: 2 BC#3: 2 Apply BC#2:

(this boundary condition applies to both fluids) 2

2 2

2 Apply BC#3: 2

2 Set

2 ,

Solve for

2 2

Now solve for

2 2

2 Insert the equation for

2 2

Thus, 2 2

And, plugging 2 2

and

2 back into the original equation for

2 2

2, 2

2 2

2 2 2 This equation could be simplified, but that is not necessary.

Solution 2 (this may be a slightly easier solution, but both are correct). Need y-direction Navier-Stokes Equation 2

2 2

Using the conditions in the problem statement this equation becomes, 2 2

Integrate twice solving for velocity, 2 2

2 2

Next, we develop equations for the velocity of Fluid 1 and Fluid 2 using the above equation. Fluid 1 VB

Fluid 1

Fluid 2

VA x= -L1

x=0

x= +L2

2 2

2 Boundary Conditions: BC#1: BC#2: 2 2

2 Apply BC#2: 2

2 2

Apply BC#1 (and insert the result of BC#2): 2 2

2 Thus, plugging in the expression for

2 ,

2 Fluid 2

BC#2: BC#3:

(this boundary condition applies to both fluids)

2 2

2 2 2

Apply BC#2: 2 2

Apply BC#3, inserting the value for 2 2

2: 2

2 2

Plugging the values of

and 2

2 2 back into the original equation: 2 2 2

9.23

Thin viscous fluid with thickness = h

Belt moving up with a velocity=Vw

Air

Begin with the y-direction equation in rectangular coordinates: 2

2 2

Thus, 2 2

Boundary Conditions: BC#1: at the moving wall: BC#2: at the free surface: Rearrange, 2 2

Integrate,

at x = 0 at x = h

2 2

Integrate again,

From BC#2:

From BC#1: 2

Thus, 2

9.24

z-direction Navier-Stokes Equation in Cylindrical Coordinates 2 2

To obtain the desired equation we must make the following assumptions:

2 2 2 2

Rearrange,

Integrate, 2

2 Rearrange, 2 Substitute in Newton's Law of Viscosity

Boundary condition:

2 , so 2

2 2

Chapter 10 End of Chapter Problem Solutions

10.1

Since

and

For reference, see problem 9.4

All remaining (non-zero) terms give:

10.2

In the limit: Note that tan(z)→z

10.3

10.4

10.5

since

, continuity can be expressed as

using

Check

10.6 In the core: Euler’s Eqn.

Since velocity variation is linear

Outside the central core – Bernoulli Eqn. applies:

v varies inversely with r

Adding (1) and (2)

For P = -10 psf

Pressure will fall from -10 to -38 psf in a distance of 138 ft At 60 mph = 88 ft/s, Time = 138/88 = 1.57 seconds Pressure at tornado center = -38 psf At edge of core Far from center Total ΔP = 38 psf

(c)

(b)

10.7

Along the stagnation streamline, θ=π

10.8 From continuity:

10.9

10.10 (a)

Flow configuration is:

When

(b)

(c)

When

10.11

In figure – for ψ=0, or any number – pick 3 –

Choose θ – solve for r – Plot looks like:

10.12

10.13 For source at Origin, Adding:

, where m=Source Strength. Free stream:

10.14

For Steady, Irrotational Flow, @ Stagnation point, where v=0,

10.15

Lift Force:

From Bernoulli Equation:

On Hut: Substitute into the expression for Fy,

10.16

10.17

10.18 For this case -

10.19 when origin is at vortex

10.20

a) Stagnation Point

& at Stagnation Point,

b) Body Height Stagnation Streamline

So when

c) For X Large – All flow is at

d) Maximum Surface Velocity

10.21 In this case,

In upper half plane they appear as:

Streamlines can be plotted for

10.22

10.23 To be irrotational, the flow must satisfy the equation, must be equal to zero.

For the equation of the stream function we solve for function,

And then,

So the condition of irrotational flow is satisfied.

and as a result, Equation 10.1

and

using the definition of the stream

10.24

We defined the stream function as

and

Thus,

We will choose to begin by integrating equation (5) partially with respect to x,

Where

is an arbitrary function of y.

Next, we take the other part of the definition of the stream function, equation (2), and differentiate equation (6) with respect to y,

Here,

since f is a function of the variable y.

The result is that we now have two equations for these and solve for .

Solving for

, equations (4) and (7). We can now equate

The integration constant C is added to the above equation since f is a function of y only. The final equation for the stream function is,

The constant C is generally dropped from the equation because the value of a constant in this equation is of no significance. The final equation for the stream function is,

Which agrees with the result of Example 3.

Chapter 11 End of Chapter Problem Solutions 11.1 Variable D ρ H g ω Q η P

Dimensions L M/L3 L L/t2 1/t L3/t ML2/t3

11.2

Variable v ρ D μ e

Dimensions L/t M/L3 L M/Lt L

11.3

Variable ΔP ρ ω D Q μ

Dimensions M/Lt2 M/L3 1/t L L3/t M/Lt

11.4 11.5

Variable k D d ω ρ μ

Dimensions L/t L2/t L 1/t M/L3 M/Lt

Plot π1 vs. π3 over a range in values of π2

11.5 11.6

Variable Q d σ ω ρ μ

Dimensions L3/t L M/t2 1/t M/L3 M/Lt

11.6 11.7

Variable M d σ g ρ

Dimensions M L M/t2 L/t2 M/L3

11.7 11.8

Variable n d L T ρ

Dimensions 1/t L L ML/t2 M/L3

11.8 11.9

Variable Q d P ω ρ μ

Dimensions L3/t L ML2/t3 1/t M/L3 M/Lt

11.9 11.10 Variable r t E ρ

Dimensions L t ML2/t2 M/L3

11.10 11.11 Variable D d V σ ρ μ

Dimensions L L L/t M/t3 M/L3 M/Lt

11.11 11.13

11.14 11.12

We could do all other terms in a like manner, but problem statement only asks for a ratio of gravity forces to inertial forces. The Gravitational (Buoyancy) Force is asked for is

Q.E.D.

so the ratio

11.13 11.15

Variable Dimensions Prototype D D 6D v v 20 kT ρ ρ ρ μ μ μ F 10 LBf Fp 2 A D (6D)2 Dynamic symmetry requires:

11.14 11.16

Similarity Requires:

Variable Model v vm L L/10

Prototype vp L

11.15 11.17 Model L=3m vm ρA νA Fm Am

Prototype 4L 16 m/s ρw νw Fp 16Am

11.16 11.18

11.17 11.19

Variable L a v τ g

Dimensions L L L/t t L/t2

11.18 11.20

For Equal Reynolds Numbers:

For Ideal Gas Behavior

11.19 11.21

Variable Model Prototype L 0.41 2.45 v 2.58 v Equating Froude numbers:

Equating

Thrust Force involved Euler No

Torque = FL – From Euler Equation

11.20 11.22

Variable Flow rate Density Viscosity Velocity Surface Tension Channel Length Q 3 0 -1

M L t

1 -3 0

Symbol Dimensions Q L3/t M/L3 M/Lt v L/t M/t2 L L

1 -1 -1

v 0 1 -1

1 0 -2

The rank is found to be 3 (rank of a matrix is the number of rows (columns) in the largest nonzero determinant that can be formed from it). Using Equation 11-4, the number of dimensionless parameters is

Thus there are 3 dimensionless parameters to be found. Find

Evaluate for M, L and t M: L: t:

Thus,

and

L 0 1 0

Find

Evaluate for M, L and t M: L: t: Thus,

Find

and

Evaluate for M, L and t M: L: t: Thus,

and

Chapter 12 End of Chapter Problem Solutions 12.1

12.2

12.3

12.4

From (1),

12.5

Comparison Approximate Exact

12.6

12.7

12.8

v, ft/s

50 100 150 200 250 300 350 400

0.47 0.47 0.46 0.45 0.41 0.35 0.20 0.08

Re (x104) 4.07 8.14 12.2 16.3 20.3 24.4 28.5 32.5

FD, LBf 0.0199 0.0797 0.175 0.305 0.434 0.534 0.415 0.217

Plot of Drag vs. Velocity 0.6

Drag (lbf)

0.5 0.4

0.3 0.2 0.1 0 50

100

150

200

250

300

Velocity (feet/sec)

350

400

12.9

12.10

12.11

12.12

12.13

12.14

12.15

12.16

12.17

12.18 Re (x104) CD

7.5 10 0.48 0.38

15 0.22

Re (x104) 7.5 10 15 20 25

92.2 122.9 184.4 245.8 307.3

0.48 0.48 0.47 0.44 0.10

FD, LBf 0.069 0.100 0.129 0.125 0.164

20 0.12

25 0.10

0.5 0.45 0.4 Drag (lbf)

0.35 0.3 0.25

Drag for Sphere

0.2

Drag for Golf Ball

0.15

0.1 0.05 0

100

200

300

Velocity (ft/s)

400

12.19

12.20

12.21

12.22

12.23

x,m 0 0.1 0.5 1 2

Rex δL,cm 0 0 5 2x10 0.111 0.249 0.352 0.498

δt,cm 0 0.327 1.186 2.063 3.591

12.24

12.25

12.26

12.27

0 0.1 0.3 0.5 0.7 0.9 1

0 0.156 0.455 0.707 0.89 0.99 1.0

0 0.0244 0.207 0.5 0.795 0.98 1.0

0 0.0038 0.094 0.355 0.708 0.97 1.0

0 0.1 0.3 0.5 0.7 0.9 1

0 0.518 0.709 0.820 0.903 0.970 1.0

0 0.373 0.600 0.743 0.858 0.956 1.0

12.28

12.29

12.30

12.31

12.32

12.33

Chapter 13 End of Chapter Problem Solutions

13.1

13.2

13.3

13.4

13.5

13.6

13.7

13.8

(5.9 in)

13.9

13.10

13.11

Figure 13.1 for a smooth Tube:

13.12 Rectangular duct: 8”x8”x25 ft Standard Air

Energy equation reduces to:

From Figure 13.1:

13.13

14 inch pipe is the cheapest.

13.14

With

13.15

13.16

13.17

13.18

13.19 13.20

(1) = surface of tank (2) =pipe exit

13.20 13.21

13.21 13.22

13.22 13.23

13.23 13.24

13.24 13.25 Pipe 1 2 3

Length, m 125 150 100

Diameter, cm 8 6 4

Roughness, mm 0.240 0.120 0.200

13.25 13.26

13.26 13.27

13.27 13.28 Pipe

Length, m

Diameter, cm

Roughness, mm

100

0.240

150

0.120

0.200

Chapter 14 End of Chapter Problem Solutions 14.1

14.2

14.3

Equating

14.4

14.5

14.6

14.7

14.8

14.9

14.10

14.11

14.12

14.13

14.14

14.15

14.16

14.17

14.18

14.19

0.1 0.15 0.2 0.25

h 93.48 95.93 100.54 106.5

14.20

(1)

v 0.20 0.25 0.30 0.35

hsyst 7.27 7.71 8.24 8.87

14.21

14.22 Pump Performance: Capacity, m3/s x 104 0 10 20 30 40 50

Developed Head, m 36.6 35.9 34.1 31.2 27.5 23.3

Efficiency, % 0 19.1 32.9 41.6 42.2 39.7

vx 104 20 30 40 50

h,m 20.61 22.63 25.45 29.07

14.23

14.24

Chapter 15 End of Chapter Problem Solutions 15.1

(

)

∫

[

]

[

]

(

)

15.2

15.3 ∫

∫ (

(

)

[

]

[

]

[

(

)]

](

) [

](

)

15.4

(

)( )(

)

15.5

( (

)(

) )

15.6

( (

)(

)

)(

(

)(

( )

)

)(

)

15.7

| ( )

(

)(

)

15.8 (

)

( [

(

) ]

)

15.9

(

)(

)

15.10

( ) (

)

(

)

(

)

(

)

( )

15.11

(a)

(b) (c)

(d)

15.12

(a)

(b)

15.13

(

) [ (

)

(

)

) ]

( )

( (

) (

)(

)

) ( )(

)

15.14

[ (

)

] ( ) [(

)

]

15.15

( (

) )(

)(

)

15.16 ( [

(

) )

[ (

] )]

15.17

[ (

(

)

(

)]

[(

)

]

15.18

(

)

[(

( ( )

) (

(

)

] ) ( )

)

( )

(

) ( )

15.19 [(

)(

)

(

)(

(

)(

)

(

)(

)]

15.20

(

) ( (

) )

( (

)

(

)

15.21

(

)(

(

)(

( (

)(

) )(

)

)(

)

15.22

15.23

( )

15.24

) ) )

15.25

(

)( )

( )

(

)( )

15.26

15.27

∫

( )

(

)

(

)

Chapter 16 End of Chapter Problem Solutions

16.1 (

)

( )

(

( )

)

( )

(

)

( )

(

)

16.2 (

)

( )

( (

)

(b) (

) (c)

16.3

( )

(

)

(

∫

)

( )

( )(

)

16.4 ( )

⃑

( )

[

( )

⃑

( )

⃑⃑⃑

( )

( )]

( )

16.5 ̇

̇ ̇ ̇ ̇

( ) ̇

( ) (

)

[

(

)]

16.6

( )

[

]

16.7

( )

[

]

16.8

⃑ ⃑

̇ ⃑

16.9 ( )

[ [

( ) ]

]

16.10 (

)

16.11

[

(

) ]

(

)

) (

∫

)

∫ [

(√

)

)](

( ∫ (

∫ [

(√

)][

)(

) (

)] (

)

16.12

(

) ̇

∫

(

∫ (

(

) )

)

(

)

16.13 ̇

(

)

( ) ) ̇

∫ (

(

∫[

]

)

∫ ̇

)

( ) )

∫ (

)

16.14

(

)

(

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∫ ̇

( ) )

∫ (

)

Chapter 17 End of Chapter Problem Solutions 17.1 Steady State X-directional conduction through a plane wall qx= -kA

= kA/L (T1-T2)

For T1-T2= 75 K = 7500 w/m2

q= dT/dx =

-250 K/m

For T1=300 K dT/dx= T=

q= -2000 W/m

= =

= 66.7 K/m = -20 K

T2= 320 K For T2=350 K

dT/dx= -300 K/m

q= -(30)(-300)= 9000 W/m2 T= -300 K/m (0.3m)= 90 K T1= 440 K For T1=250 K

dT/dx= 200 K/m

q = -(30)(200 K/m)= -6000 W/m2 T= -200(0.3m)= -60 K T2= 310 K

17.2 T=

% Error=

x 100

For

=1.5

% error= 1.3 %

=3 =5

% error= 10.0% % error= 20.7%

T (a)

17.3 q= Am=

(a) = =2

% Error=

= 1-(1/2)(

)

(b) = 1.5

% error=8.3 %

=3 =5

% error= 66.6% % error= 160 %

17.4

q”= -k

= -ko(1+bT)

From 0 to 12: q”

= -ko

q”=

[T+ T2

= 23(T-300)

Solving: TRH wall= 307.1 K q”= 163.3 W/m2 From 0 to L: q” L= & Solving: L=0.646 m

17.5

Input Parameters: Ro= 1.5 cm

(0.015m)

Rm= 1.8 cm

(0.018 m)

h= 80 W/m2 K To= 70 = 20 L= 6.0 cm

(0.06 m)

q= 15 W

(a) Source: heating rod

Sink: surrounding air

(b) B.C. system= biomaterial r=Ro, T=To; r=Rm, T=Ts (c) km Assumptions 

Constant Source & Sink

steady state



Sealed ends

1-D heat flux along “r”



Constant km, Ro, Rm, L, q

q= 2 RmLh(TsTs=

Ts= 20

)

Ts= 47.6

For a hollow cylinder

km =

km= 0.324 W/m*K

17.6

Input Parameters: h= 50 W/m2 K ks= 16 W/m-K (stainless steel) L= 1.6 cm Assumptions: 

Steady state



1-D heat flux along x by conduction



convection of surface (x=L)



ks & h not a function of T

(a) q=? Topic 17.1 =

(T2-T1)

= h(T1=

= T2-T1 at surface

= T1= 10,952

(b) T1= ?

T1=

+ 20

= 239

17.7

RGL= Rair= RConv=

= 0.06633 K/W q= 37/0.06633= 585 W Ti=27 -

22.94 C

17.8 Brick Size= 9”x 4.5”x 3” Brick #1

k=0.44

Tmax=1500 F

Brick # 2

k=0.94

Tmax=2200 F

Most economical arrangement is to use as much of #1 as possible (low k). Use #2 next to high temp such that its cooler surface has T . q”=

= 28.5 in

= 2.35 Ft. = 28.2 in. (9x2+4.5+2x3)

= 31.6 in.

Most economical:

17.9

Ri=1/115= 8.696x10-5 K/W R1= 0.25/1.13= 0.221 K/W R2= 0.20/1.45= 0.138 K/W R3= 0.10/0.66= 0.152 K/W Ro= 1/23 = 0.0435 K/W

q= 23(To-300)

= 1900 W/m2 = 176.5 W/ft2 To= 383 K

17.10

To= 325 K Ri=1/115= 8.696x10-5 K/W R1= 0.25/1.13= 0.221 K/W R2= L/1.45 R3= 0.10/0.166= 0.152 K/W = 0.416 + L/1.45= 0.416+ L/1.45= L= 2.03 m

q=23(325-300) = 575 W/m2

17.11

LSt= 1/8”, k= 10 BTU/Hr ft. F LASB= 1/8”, k= 0.15 BTU/Hr ft. F q=

= 1351 W/ft2 (a)

2305= h T= 768 F

=3 TSurf = 838 F (b)

17.12

R1 =

= 0.0694

R2 =

0.00104

R3 =

0.154

RConduction, Equiv =

= 0.0483

per side = 0.0483+ 1/3= 0.3817 New Ht. Flux = Increase =

2437 BTU/Hr-ft2 = 0.057= 5.7 %

17.13

Ti= 325 K

h1= 12 W/m2 K h2= “ k1= 2.42 W/ m K k2= 229 W/m K a) Applied to Plastic: q = 12(T1- 295) +

(T1-325)

(325-T2) + 12(T2-295) T2= 324.9 T1= 328.7 q = 359+404= 763 W

b) Applied to Al: q = 12(T2- 295) + (T2-325) +

(T2-325) (325-T1)

T2= 325 K T1= 322 K q = 320+361= 681 W

17.14

Bolts in a square array with 4 equiv. Bolts/ ft2

R1(bolts) =

= = 5.7 Arf/ BTU – steel = 0.475 Arf/ BTU - alum.

R2 SS=

= 0.0021

R3 CB=

= 10

R4 PL=

= 0.0278

= 10.03

Per ft2 of x-section REquiv= =

= 3.63 Hr F/BTU (a) = 0.454 Hr F/BTU (b)

17.15

For a section 36 cm wide & 1 m deep: R1 =

K/W

R2 =

= 0.0683 K/W

R3 =

= 16.67 K/W

R2 =

= 14.28 K/W

R5 =

K/W

R Stud Wall Equiv = q=

= 4.28 W

q3=

q4= =

q3=

= 7.691 K/W

= 32.92 K

= 1.975 W

q4= 2.305 W

17.16

q= = 292.7- 70= 227.7 F Room Temp (Assumed) For 2-in sched 40 RST =

ID= 2.067 in OD=2.375 in = 1.475x10-5

RINS =

= 0.0529

ROuts = 1/hAo= 0.00537 w/o Insul = 0.00237 w/o Insul = 0.0553 = 0.00537 = 37470 BTU/hr Cost = 37470(

= $0.255/hr

Time =

= 177 Hours

q= 4030 BTU/hr q= 41,500 BTU/hr

17.17

R1 = R2 = R3 =

= 0.0015 = 6.19x10-5

= 0.0725

= 0.072 Hr F/BTU q=

= 2530 BTU/Hr (a)

R1+R2= 0.00156 Hr F/BTU R3 =

= 0.461 Hr F/BTU

R4 =

= 0.0221

= 0.485 q=

= 386 BTU/Hr (b)

For Bare Pipe: =

= 2.71 LBm/Hr (c)

17.18

= 30 = 303 K h= 100 W/m2-K L1= 0.5 cm

k1= 1.0 W/m-K

L2= 0.3 cm

k2= 230 W/m-K

W= 4.0 cm V= 10.0 W Aside: not required for your calculations. Verify T2=T3 For System II x=L2 x=L1

T2=c2

T(x) =T2=T3

c1=0 =0

T(x) =c1x+c2 (a) x=0

q= AL1=Ah(T1-

1.25 W/cm3 * 0.5 cm = 100 T1= 365.5 K

)

System I

= 1.25 W/cm3 (T1-303 K)

(b) T2 For system I Assumptions 

Steady state



Const



1-D flux along x



X=L1,

(no heat flux)

From Topic 17.2 & assumptions given T(x) = T1+

(L1x – )

x=L1 T=T2= T1 +

B.C. x=0, T=T1

T2= 365.5 K +

= 381.1 K

x=L1,

17.19

Input Parameters kIC= 1.0 W/m-K = 10.0 W h= 100 W/m2 K L= 0.5 cm A= (4 cm) 2= 16cm2 Vchip= A*L1 Assumptions: 

Steady state (constant source & sink)



Constant power



1-D heat flux aligned along Source and Sink

Vchip

= = 1.25 x 106 W/m3

(1.25 W/cm3)

(a) T1 surface temp (x=0) From Topic 17.2

A*L1= h A (T1 -

)

T1=

T1= 365.5 K

(92.5

)

(b) T2 at base of IC chip

From Topic 17.2 heat balance is T(x) =

+ xc1 + c2

kIC =

+ c1

Boundary conditions x=0, T=T1

x=L,

T1=0+0+c2

=0 + c1

c2= T1 T(x)=

+ T1

at x=L T=T2 T2=

+ T1=

+ T1

T2=

+ 365.5 K

T2= 381.1 K Tav=

chip failure= 22%

(108 = 100

)

c1=

17.20

(Ti-To) = I2R= 3.43 W R=

= 2.95 x10-4

I2 = I = 107.9 Amp

(120-70) = 11.72 BTU/Hr-Ft= 3.43 W

17.21 =

I = 106.1 Amp 11.34 = 4[ T = 71.3 F

17.22 Same as Problem 17.21 except wire is aluminumR= 4.85x10-4 I= 83.9 Amp

17.23 Assume thin-walled inner vessel is 77 K throughout

17.24 For q = ½ of value in 17.23

Insulation thickness= 0.0555m Added thickness: 0.0055 m or 5.5 mm

17.25

L=0.06 m = qo[1-x/L] Poisson Eqn. applies: (Energy Balance) =0

+ c1

17.26

In waste material:

For Stainless steel:

Equating with equation (1):

Putting in values @ center of waste material:

17.27

17.28

Input Variables Ro= 0.5 m

R1=0.6 m

q= 2 x 105 W/m3 h= 1000 W/m2-K krw= 20 W/m-K ksteel= 15 W/m-K =25

(298 K)

(a) Model for T(r) System I Assumptions: System I 

Steady state



1-D heat flux along r (sealed ends)



Const. q



k is not a function of T

Energy Balance (System I) differential volume element (fix in Lect notes) IN-OUT+GEN=ACC ACC=0 2

2 *

+ *

=volume of vol. element

Divide by

, rearrange, r -> 0

+ (r )=

Ordinary diff. equation (O.D.E.) is perfect differential

B.C.

r=Ro

r=0

Qr=0

T=To =0

based on O.D.E., integrate twice r =

+ c1

T(r)=

+ c1ln(r) + c2

Apply B.C. At r=0

0*0=0+c1

At r=Ro

T=To

To=

+ 0*ln(Ro)+c2

c2= To+ T(r) =

(

-r2) +To

c1=0

(b) To & T1 Assumptions: System II 

Steady state



1-D flux along r



B.C.

r=Ro

T=To

r=R1

T=T1

From Topic 17.1

qr=

At surface by convection

qr= 2 LR1h(T1-

If =constant

= *

(To-T1)

=qr

= energy generation/volume

= vol. of radwaste =2

R1h(T1-

)

T1=

298 K= 339.7 K

(To-T1)

To=T1+ To=339.7 K +

= 643.5 K

(r=0)

)

Tmax(r=0)=

+ 643.5 K

= 1268.5 K (995.5

-ouch!)

17.29

Input Parameters: k=ktissue= 0.6 W/m-K L=0.5 cm Ro2= 0.5 mmol O2/cm3 tissue-hr R, O2= 468 J/mol O2

Ts= 30 Assumptions: 

steady state



1-D flux along x



Insulated boundary



k(T)= constant

From Topic 17.2 Heat balance is (on diff vol. element) -kA

Divide by k, A, rearrange, =

=0 ->0

B.C.

x=0 T=Ts, x=L

=RO2*

T(x) = Ts+ (L*x – x2/2)

At x=L TL=Ts+

=6.5 x 10-2 W/cm2

TL= 30

TL= 31.4

=1 K)

(tissue is reasonably at constant temp…)

* (1 W= 1 J/sec)

17.30

Assumptions 

steady state



1-D flux along x



kA & kB constant with T

L1= 0.5 cm L2= 0.7 cm kA=50 W/m-K kB= 20 W/m-K (a)

At steady state

= *L1= (b) T1=?

= T1=

* 0.5 cm = 10 W/cm2

At x=L1

(T1-Ts)

T1= +

+ Ts

= h(Ts-

)

h= 0.20 W/cm2-K= 200 W/m2-K (d) To=?

at x=0 (see aside for Model Development)

T(x) = To + (x2/2 –L1*x)

T1=To-

To=T1+

at x=L1, T=T1

= 90

17.31

Input Parameters: A= 2.0 m2

L1= 0.02 m (inert layer) Lc= 10 cm (compost layer) System I= inert layer k1= 0.4 W/m-K System II= compost layer kc= 0.1 W/m-K

(a) q= A*Lc= 600 W/m3* 2.0 m2* 0.1 m= 120 W (b) Tc=?

Tc=

System I

1 source-> sink

= 0.25 m2K/W

= 288 K + 120 W *

at base (x=Lc) dT/dx=0

TL= Tc +

System II

303 K (30 ) insulated boundary

TL= 303 K +

= 333 K (60 )

(d) As Lc increases, q increases

Tc increases and TL increases and Ts increases

17.32

Input Parameters: km= 0.419 W/m-K

Ro= 1.5 cm

= 15 k W/ m3 (15,000 W/m3)

T1= 37

Assumptions: 

Steady state



1-D flux along “r” only (neglect ends)



constant q



km not a function of “T”



Ro constant along z

Fourier’s Law:

(a) Model for T(r) Heat Balance on Differential Volume Element IN-OUT+GEN=ACC 2 Lr (

2 Lr= flux area = vol. of volume

element ,

, rearrange (watch signs!) +

ordinary differential equation (ODE)

Boundary conditions (make the O.D.E. unit) r= Ro, T=To r=0, Qr=0=

by symmetry

Based on “type” of O.D.E. (perfect differential) integrate twice 1) r

2) T

Apply B.C. at r=0,

for 1) 0*0=0+c1

At r=Ro

T(r) =

+ To

c1=0

At r=0.75 cm (0.0075 m) T=

+ 37

T= 38.5

(b) Where and what is Tmax? Find the maxima of curve =

Tmax=

maximum T at r=0

+ To =

(1.5 cm *

)2 + 37

Tmax= 39

(c)

= 112 W/m2

(d) Revised Physical System Figure

R1= 1.8 cm

For System I tissue T(r)=

+ To

For System II sheath cylindrical geometry, =0 Topic 17.1 qr=

(To-T1)

Eliminate To Realize qr= * To=

+T1

Ro=1.5 cm

17.33 Per foot of bare tube:

For longitudinal fins:

For circular fins: Per fin: Per foot:

17.34

Longitudinal case:

Circular case

Increase: Long: Circle:

17.35 Solution for Is in form

Long fin approximation:

17.36

Airside:

Waterside:

For

Without fins: With fins: Waterside: Airside: Fins added to airside only:

To waterside:

Both sides:

17.37 One dimensional conduction with internal heat generation

Or:

At L/2:

Mid-point temperature:

17.38

17.39 Energy balance: (d^2 )/(dx^2 )-m^2 =-W/kA {W in Btu/ft} =C_1 e^mx+C_2 e^(-mx)+W/hP (0)=0

C_1=-(W/hP (1-e^(-mx) ))/(e^mx+e^(-mx) )

(L)=0

C_2=-(W/hP (e^mx-1))/(e^mx-e^(-mx) )

Putting in values: m=2.28 C_1=-0.0017 W/hP C_2=-0.999 = _max @ 1.5 ft W_max=I_max^2 R R= L/A=(1.72x〖10〗^(-6) )(3)/( /4 (1/48)^2 (30.48) )=4.97x〖10〗^(-4) θ_max=90[-0.00107e^2.28(1.5) -0.999e^(-2.28(1.5) )+1] W/hP W/hP=96.3 W=96.3(6)( )(0.25/12)=37.8Btu/(hr ft)=11.08Btu/ft I_max^2=11.08/(4.97x〖10〗^(-4) )=2.23x〖10〗^4 A^2 I_max=150 A

/ft

17.40

Substituting:

17.41

Substituting:

17.42 For aluminum I-beam: Same procedure as previous problem

17.43 17.45

Without fins:

For longitudinal fins:

Increase = 2068 Btu/hr % increase = 290% For varying values of h:

2 0.731 0.84 412 5 1.156 0.71 346 8 1.462 0.60 290 15 2.00 0.48 229 50 3.65 0.27 122 100 5.17 0.19 81 i.e. fins are most effective when h is small

17.44 17.46 Same as previous problem except material is aluminum Without fins:

h % incr.

2 488

5 477

15 448

50 370

100 303

17.45 17.47

For known end temps:

Substituting:

17.46 17.48

Figure

For IC Chip: *Vchip= 15 W= qx

(a) T1 & chip failure rate without “extended surface” Assumptions: 

Steady state



1-D flux along x



is constant

Energy Balance At x=0 T1= T1= 413 K= 140

qx= *Vchip = Achip*h*(T1=

293 K chip failure rate > 80 % (off the chart)

(b) T1 with extended heat transfer surfaces

Af= 2WL= 2(5.0 cm)(1.0 cm)= 20 cm2

Nf= 5

For rectangular fin with ho=h, heat transfer rate is q= (Ao+ NfAf )h(T1with m=

“f in efficiency”

= if t<<w

for Al from Appendix H W3Rk=229 W/m-K (293-373K) for Aluminum = 24.12 m-1

m= mL= 24.12 m-1 * 1.0 cm (

0.241

= = 0.981 Ao= WW- Nf(t*w)= 5.0 cm*5.0 cm -5(0.03 cm)(5.0 cm)= 24.25 cm2 Back to heat transfer rate 15 W= (24.25 cm2+ 5* 10 cm2*0.981) T1= 395 K= 122

New chip failure rate

(T1-293 K) 50%

Chapter 18 End of Chapter Problem Solutions

18.1

 One can use lumped parameters Combination of Newton’s law of cooling and Fourier’s 1st Law:

Where If:

The solution is:

18.2

Clearly a lumped parameter case Energy Balance:

Letting:

18.3 Aluminum wire:

This shows the lumped parameter solution can be used Steady state case per mole:

Transient case:

18.4

A lumped parameter problem

18.5 Lumped parameter solution applies if or if

Therefore ut

must be se istributed arameter solution

18.6

Use lumped parameter solution

Case a b c d e

L(in) 3 6 12 24 60

V/A 0.5 0.6 0.67 0.706 0.732

T(min) 19.7 23.6 26.3 27.9 28.9

18.7

a) Cu – Bi<0.1 Lumped

b) Al – Bi<0.1Lumped

c) Zn - Bi<0.1Lumped

d) Steel - Bi<0.1Lumped

18.8 Water is well-stirred, therefore lumped solution applicable, thus T=T(t) only

t (hr.) 0 T ( ) 40

0.1 81

0.2 0.4 0.6 0.8 1.0 115 169 210 237 253

18.9

Lumped parameter case. Temperature may be considered uniform at any time.

18.10

Lumped parameter

18.11

A distributed parameter problem

By trial and error

18.12

Must use distributed parameter solution fig. f.8

18.13

18.14

Using chart for Cyl.

Interpolating @ 0.33

18.15

Distributed parameter solution

From chart

18.16 Using chart solution

By trial and error

18.17

Distributed parameter problem

By trial and error

18.18

Distributed Parameter problem

From chart

18.19

For a finite cylinder:

For both r & x directions:

Trial and error using charts:

18.20 Same cycle as in Prob. 18.15 but H varies: As With ends considered

For plane H 1.3 0.65 0.26 0.13

10 5 2 1

t,s 3114 3114 ~3000 ~1600 ~600

18.21

Assumptions: 

Unsteady state



1- flux along “x”



Constant



Small penetration depth ( Error function)

For “meat” (steak) = 1500 kg/m3 k= 0.6 W/m-K Cp= 4000 J/kg-K

What is time required for T(x, t) = 100

Based on Physical System and Assumptions, Consider = Appendix L

= erf (

) =erf ( )

= 0.60= erf ( ) at erf ( ) =0.6 -> =0.59

at x= 2 mm?

=> t=

= 1.0 x10-7 m2/sec

28.7 sec

Now consider Ts= 150 = From Appendix L, t=

= 0.400= erf ( ) ,

=0.37 73 sec

(by linear interpolation)

18.22

L= 10 cm

a=L/2= 5 cm

= 120

To=20

h=60 W/m2K

2cm=0.02 m Thermo physical Parameters = 1500 kg/m3

k= 0.6 W/m-K

Cp= 4000 J/kg-K

= 1.0 x10-7 m2/sec

Assumptions 

USS conduction



2-D flux r & z (cylindrical coordinates)



Constant h,



Neglect contribution of metal container to heat transfer resistance

What is T(r,z,t) at t=2500 sec, r= 0 cm, z= 0 cm? z-direction:

x a=

0.1

na=

ma=

0.2

Fig F.7

Ya=0.95 (does accounting for z direction matter in this calculation?)

r-direction:

Xcyl=

0.625

ncyl=

mcyl=

Fig F.8

Ycyl= 0.27

Y=YaYcyl= 0.95*0.27= 0.257= T(r,z,t)= 94

0.5

18.23

Input Parameters ksi=22 W/m-K si=2300 kg/m

Cp,si=1000 J/kg-K 3

h=147 W/m2-K =303 K (

R= )

0.075 m

To=1600 K

What is T(r,t) at t= 583 sec, r=R=0.075 m

Assumptions



Unsteady state conduction



Finite volume system



1-D flux along r (neglect end effects)



Cylindrical geometry

= = 9.57 x10-6 m2/sec

For cylinder, use Temp Time charts (Fig F.8 or Fig F.2 n= m=

= 2.0

1.0

Fig F.8

(Fig F.2 also works)

Y= 0.375= Ts= 789 K (still too hot to handle!)

)

18.24

Input Parameters: To= 25 =0 kliq= 0.3 W/m-K =1.0 g/cm3= 1000 kg/m3 Cp,liq= 2000 J/kg-K h= 0.86 W/m2-K Tc= 2.0

precipitation temperature of drug

D=0.70 cm

R=0.35 cm

Assumptions: 

Unsteady state



1-D flux along r



23>10 (neglect conduction)



Thermo hysical ro erties constant with “T”



Neglect conduction through syringe wall

Heat Transfer Model: = r=0

r=R

T(0,t)=Ts or = h(Ts -

)

T=0, T(r,0)=To se the analytical solution of the heat transfer model as contained in the “charts” to hel solve the problem. Two Approaches: 

Get “t” at center (r=0) where T(0,t) =2.0



Get T(0,t) at center for t= 2.0 hr

Time-Temperature Charts (core calculation) n= m=

0 = 99.7 convection resistance is significant!

Call “m”= 100 to facilitate use of gra h

0.080 Fig F.2 for cylinder is “off the chart”. Try Fig. F.5 with r=0 X=130=

t= = 1.5x10-7 m2/sec

= 10,616 sec (2.9 hr)

Problem 18.25 Cannot use “Charts”, as no thickness of water is s ecified

Assumptions: 

Unsteady state



1-D flux



Semi infinite medium

Heat transfer model: @ x=0, t= 330 sec = exp [

Input Parameters: = 1.716x10-5 m2/sec

erf(

= 0.05

)= 0.0564

Appendix L

Ts= (1020 -20 ) exp Ts= 966 q=hA(Ts q= 334 W

(1-0.0564)

(1239 K) = 20 W/m2K * 176.7 cm2 *

(966-20)

18.26 Use Semi-infinite wall solution.

@ surface: x=0

Approximation: Use 1st term in series expansion.

At this time: Solving for T:

18.27 Use semi-infinite solution:

Given:

For y=0

18.28 Use semi-infinite wall solution

At x=0 this reduces to

Where

t Substituting:

By trial and error

18.29

where So

is max when

is max

18.30

This is a semi infinite case

18.31

Solution is amenable to either numerical or analytical approach

Trial and error at each t T T( )

1 hr 580

6 hr 396

24 hr 275

18.32 18.36

18.33 18.37

Input Parameters: =2.5 x10-7 m2/sec Ts= 250 To=37 t=2.0 sec Assumptions:    

Constant source, transient sink 1-D flux along x =0 semi infinite medium

unsteady state

Model: = erf(

(a)

)

for living skin

= 9.165x10-8 m2/sec

Compare to = 2.8 x10-4 cm2/sec *

(b) Get x at t=2.0 sec for

= 2.8 x10-8 m2/sec

=2.5x10-7 m2/sec and T(x,t)= 62.5

for dry cadaverous epidermis

=0.8803=erf( ) 5th =1.1

From Appendix L =

=> x=2 = 1.56 x10-3m= 1.56 mm

x= 2* 1.1 Deep dermal burn

= 0.354

By Linear interpolation Appendix L erf ( ) = 0.3814= T(x, t) = 169

(ouch)

18.34 Problem 18.38

Input Parameters: For tissue k= 0.35 W/m-K Cp=3.8 kJ/kg K = 1005 kg/m3 Ts=55.5 To=37 x= 3 mm (0.003 m)

Assumptions: 

Constant source, transient sink



1-D flux along x



Semi-infinite medium



Skin tissue and tumor tissue has some thermo physical properties

(a) Thermal diffusivity

unsteady state

= 9.165 x10-8 m2/sec

(b) Active treatment time tact

ttotal=tact+ t

final tmin for T(x, t) = 45 , x= 3.0 mm Model:

=erf ( ) = erf ( =

= 0.568= erf ( )

= 0.556

Appendix L

= 79.5 sec

tactual= ttotal-t= 120 sec- 79.5 sec= 40.5 sec

5th by linear interpolation

Chapter 19 End of Chapter Problem Solutions

19.1 For a plane wall: Variables T

x L Α k t h

Dimension T T T L L

If temps are grouped as

19.2 Air 1.9e-5

k Re Pr Nu St

1.008e3 0.0293 2.3e5 0.699 348 2.16e-3

H2O 0.474e5 1.0

Benz. 0.473e-5

0.383 1.02e7 2.72 15.4 5.55e-7

0.0762 1.02e7 5.21 77.3 1.45e-6

0.45

Hg 1.06e6 0.033

Glyc. 0.128e2 0.598

5.03 4.57e7 0.021 1.17 1.22e6

0.165 37800 13.1 35.7 7.21e-5

19.3 ~Plots~

19.4

As per development in text:

In limit as

0 1 2 3 4

250 1310 1750 1310 250

0 27 80.3 134 161

19.5 For a single 4 ft long plate:

For stack of plates:

19.6

Energy balance Standard procedure, resulting expression is:

For

x 0 0.4 0.8 1.2 1.22

900 3040 3110 1030 900

T 100 171 285 360 361

116 226 340 378 377

19.7

19.8

Laminar flow for all values of

T,F 30 50 80

79.5 79 18.2

6.61e-2 1.52e-2 0.13e-2

TCl

TFl

30 50 80

80 100 130

55 75 105

T 30 50 80

hL 8.74 11.4 16

437 570 800

Re 303 1320 15400

fd 37.6 17.9 5.2

0.0419 21.9 7.25 0.0101 44.6 4.64 0.0011 134.8 2.16

19.9

x 0 0.5 1 1.5 2

10.92 15.46 18.92 21.85

8.74 6.19 5.05 4.37

22.2 31.45 38.5 44.5

11.37 8.06 6.57 5.70

67.5 95.2 117 135

16.06 11.32 9.29 8.03

x 0 0.5 1 1.5 2

19.10 N2 @

k Pr

a) b) c) d) e) f) g) h)

100 0.069 1.77e-3 0.0154 0.71

200 0.0583 0.236e-3 0.0174 0.71

150 0.209e-3 0.0164 0.71

19.11 For air @

1.008

b) c)

19.12

19.13

For conditions specified:

Probably turbulent boundary layer Use CouBurn analogy: From Chapter 13 for turbulent boundary layer

{Assuming all surface exposed to turbulent B.L.}

19.14

For O2 gas at 300 K = 1.3 kg/m3 Cp=920 J/kg-K = 2.06 x10-5 kg/m-sec k= 0.027 W/m-K Assumptions:  

Steady state Constant

(a) Pr=

= 0.702

(b) h required for Ts= 310 K Energy balance (Topic 17.2) *l*A= hA (Ts-

)

h= = qO2* h=

= = 32.5 W/m2-K

= 60.2

(d) Consider laminar flow with h= Nux= 0.332 Rex1/2 Pr1/3 = 0.332 Or h= = h=

2L1/2

Or NuL= = 0.664 ReL1/2Pr1/3

ReL=

(e) What is ? Nu= 0.664 ReL1/2Pr1/3 ReL= Re=

= 10,401

=> = = 3.3 m/sec

High flow rate! How could this system be redesigned? How is T s affected by O2 flow rate?

19.15

Input Parameters: Kpoly= 0.12 W/m-K l= 1.5 mm L= 2.0 m W= 1.5 m (a) Tav=

= 67 = 340 K ( film average temperature) 1.9553 x10-5 m2/sec

From Appendix I Prair= 0.699

kair= 0.629282 W/m-K

(b) q= A*h*( ReL=

L*W= 2.0 m* 1.5 m= 3.0 m2

) = 51,143 (<2x105

laminar)

Nu=0.664 ReL1/2Pr1/3= 0.664 (51,143)1/2(0.699)1/3 Nu= 133.3=

h= 1.95 W/m2-K q= 2*W*L*h (

)

= 2* (3.0 m2) (1.95 W/m2-K) (107 -27 ) (1 = 1K) = 936 W

19.16

Input Parameters: Ts= 60 =20 Tav=

=40

For fluid (water at Tav= 40 ) = 992.2 kg/m3 = 658x10-6 kg/m-sec Cp= 4175 J/kg-K k= 0.663 W/m-K =

= 0.10 m/sec

(a) Re & Pr ReL=

ReL= 120,632 < 2x105 Pr= Pr= 4.15

laminar flow

(b) hx at x= 0.5 m from leading edge Nux= 0.332 Rex0.5Pr1/3= 0.332 (75,395)1/2(4.15)1/3= 146.5 Rex=

= 75,395

Nux=

hx=

q/A=hx (

hx= 194 W/m2-K

= = 7770 W/m2

)

with A=W*L

= h= 0.664

ReL= (120,632)1/2(4.15)1/3

h=0.664 = 307 W/m2-K

q= 0.2 m *0.8 m * 307 W/m2-K (60

) = 1966 W

(d) Boundary Layer thickness at x=L Hydrodynamic =>

= 1.15 cm < 12 cm liq depth H=

Pr-1/3 = 1.15 cm (4.15)-1/3= 0.717 cm << 12 cm

19.17 Momentum theorem in x direction:

At low velocity: & Steady state

LHS: Buoyant force (B.F.):

{per unit volume}

Viscous Force (V.F.): RHS:

Equating: (LHS)=(RHS) & dividing by

In limit as

Energy equation: Same for both natural & forced convection

19.18

Into energy equation:

Equating: Into momentum equation:

Equating:

Letting: Previous two equations become:

Exponents on x must agree

Giving: So equations for A&B becomes

So we have

& finally upon substitution

19.19 Air@ 310 K

15 cm

30 cm

1.5 m

Gr=

4.31e7

3.45e8

4.31e10

0.985 cm

1.17 cm

1.75 cm

Nux=

30.5

51.2

171.3

Hx=

5.48

4.61

3.08

19.20 19.22

B.C.

B.C

Into momentum equation:

Equating & solving:

Energy equation:

Substituting & solving:

Giving:

If X=0

19.21 19.24

Assuming B.C.

Temperature profile becomes:

Similarly for velocity:

Into the integral expression:

Left hand side (LHS):

(RHS)

Substitute equations (1)&(2):

Giving Equating (LHS)=(RHS)

& letting

Substitution & some algebra give:

Separating variables:

Nusselt #:

19.22 19.25

B.C.

Into energy equation

{Presume

for integration}

19.23 19.27

Reynolds analogy: Assume

Close enough, doing over with Colburn Analogy: Assume

Summary Reynolds Colburn

174 141

43500 35300

will yield

as a result.

19.24 19.28

For air @ 180 :

Reynolds:

Colburn:

19.25 19.29

From Colburn analogy: Re=

{Laminar}

19.26 19.30

19.27 19.31

Reynolds analogy:

Colburn analogy:

Chapter 20 End of Chapter Problem Solutions 20.1

Ends are neglected

For vertical orientation By trial and error HTR surface temp Horizontal orientation

Trial and error: HTR surface temp

20.2 Bismuth As in problem 20.1 Vertical – same formula as above By trial and error HTR surface temp Horizontal – same formula as above By trial and error HTR surface temp Continued hydraulic fluid Same formulas and procedures Vertical

{Properties used at 200 Horizontal

- highest temp in tables}

20.3

16 cm (0.525 ft) – vertical height (20.3 continued)

By trial and error:

For 10 cm (0.328 ft) – height Trial and error:

English units used – tables easier to use

20.4 For

Use lumped parameter

Since lumped parameter solution is valid answers to parts (a) & (b) are the same when:

20.5

Air @ 355 K:

Horizontal Cylinder:

For

Remainder of input goes to electrical & conduction losses & to illumination

20.6 For a horizontal cylinder

Trial and error:

20.7 For Cu cylinder with: Height = 20.3 cm, Diameter= 2.54 cm

For

For an h value < 6340 Bi<0.1 ∴Lumped parameter

(20.7 continued)

20.8 For a sphere:

D, cm

7.5

615

0.0104

710

0.0135

1.5

1180

0.0271

L.E. surface resistance is very small ∴ ~Time 0

& Falls to this value almost instantaneously

20.9 For

to reach 320 K – use values calculated in problem 20.8

D, cm

7.5

615

0.16

1.19 hr

710

0.16

31.7 min

1.5

1180

0.16

2.86 min

at all times

20.10

a) Horizontal

b) Vertical

20.11 Same conditions as prob. 20.10 except fluid is H2O @ 60

Horizontal:

Vertical:

20.12 Spherical tank

For sphere:

Properties @ 260 K

20.13 Assuming each plate is independent

20.14

Fraction of total = With 1ftx1ft ridges

Same as in part (a) fraction = 0.54

20.15 For forced convection

Flow is Plate is mostly in transition regime Assume laminar boundary layer

Fraction due to F.C.=0.34 If B.L. is turbulent

Fraction = 1.49 For this case there is more capacity to transfer heat than there is solar energy supplied. Surface temperature will be <150

20.16

By trial and error

Resistance of insulation

20.17

Trial and error

{see problem 20.43}

(20.19 continued)

20.18

Assume

T in R Trial and error T=1147 R = 687

20.19 Forced convection outside

{B,n functions of Re} Assume T 650 =1110R Tf=360

Table 20.3 B=0.027 n=0.805 @ This temp {Pretty good}

20.20 Insulation on outside w/natural convection on surface

With

Trial and error

20.21

Equations to be solved are:

Trial and error: Insulation thickness =

20.22 For natural convection case plane upward – facing hot surface if 2x107<Re<1010 Assume top surface is square ~A=L2

Now for same Heat loss and L=0.982 m {forced convection} Assume

Transition regime Assume laminar B.L.

20.23

Process Input Parameters: W= 20 cm (0.2 m) L= 60 cm (0.6 m) ks= 16 W/m-K km= 2.0 W/m-K Air flow:

= 3.0 /sec

Tav=

=400 K

air= 2.5909x10

-5

m2/sec

Prair= 0.689 kair=0.033651 W/m-K

(a) & (b) From heat balance * qT=

Vm+ qloss

qloss= h*A (

Vm= 600 W )

A= W*L

Get h, for flow over a flat plate Re=

= 69,474 (<2x105) ∴ laminar

Use Nu= 0.664 Re1/2Pr1/3= 0.664 (69,474)1/2(0.689)1/3=154.6

Nu= h= 8.67 W/m2-K qloss= 8.67 W/m2-K * 0.2 m x 0.6 m (480 K-320 K) = 166.5 W qT= 600 W+ 166.5 W= 766.5 W (d) T2=? q=

(T1-T2)

T2=

T2= T2= 22.5

+ 217

(490 K)

(498 K)

*Energy Balance on stainless steel slab + load material Assumptions: Steady state energy balance IN-OUT+GEN=ACC qT-qloss- m m=0 qT= qloss* m m (c) T1=? Show T1=Ts+ Q=q/A m= -1.0 x10

T1= 217

W/m2

-qmVm (since measured)

20.24

20.25

a) Flow parallel to tube

B.L is laminar

B.L. is in transition

If laminar over total length

b) Crossflow case

20.26 Water: a) Flow parallel to tube

{Turbulent}

b) Crossflow case

20.27 a) Flow parallel to tube

{turbulent}

b) Crossflow

20.28

{From problem 20.18}

Insulation resistance = 0.842 {Prob. 20.43}

20.29 Sphere

20.30 (a)

Input Parameters: Gas=Air

540 K (247 )

kair= 0.0417 W/m-K = 5.94 x10-5 m2/sec 4.04 x10-5 m2/sec Solid = popcorn kernel D= 0.5 cm (0.005 m) k= 0.2 W/m-K Cp= 2000 J/kg-K = 1300 kg/m3

(b) h For fluid flow around a sphere in a Fluidized Bed NuD= ReD=

= 2.0 + 0.6 ReD1/2Pr1/3 = 371.3

Pr=

= 0.6801

NuD= 2.0 + 0.6 (371.3)1/2 (0.6801)1/3= 12.16 = 101.5 W/m2-K

(d) Time “t” required for popcorn kernel to pop T (0, t) = 175 = 448 K USS conduction in a sphere with finite “m” m=

= 0.78

call it m=0.8 n=

0 (center) =

= 0.32

Fig F.9 X= 0.5= =

= 7.7 x10-8 m2/sec

= 40.6 sec

20.31

Input Parameters: For liq H20 @ 30 = k= 0.633 W/m-K Pr=4.33 = 6.63 x10-7 m2/sec @ Ts= 100 =278 x10-6 kg/m-sec for H20 kpoly= 0.08 W/m-K for bead

Heat Transfer Model: q=

)

(A=

for sphere)

for external fluid flow around a sphere falling in a fluid Nu=

2.0+0.6 Re1/2Pr1/3 for the fluid

Re= Nu= 2.0 + (452.5)1/2(4.33)1/3= 22.8

= 452.5

= 1443 W/m2-K

∴ q= (0.01 m) 2 1443 W/m2-K (100 If Whitaker correlation is used

) = 31.7 W Nu= 2.0 +[0.4Re1/2+0.06Re2/3]Pr0.4

For ( ) = 1, h=1497 W/m2-K, q=32.9 W For ( ) = 2.97, h=1925 W/m2-K, q=42.3 W

20.32

Input Parameters: Ts=200 (383 K) d= 4.0

(4.0x10-6 m)

L=1.2 mm (1.2x10-3 m) P=I2R=q= 13 mW For N2 @ Tav= 110 = 0.8948 kg/m3 = 2.13 x10-5 kg/m-sec Cp= 1047.3 J/kg-K k= 0.032 W/m-K

(a) Pr=

(b) Nu=?

= 0.697

q= dLh(Ts-

)

h= h = 4789.4 W/m2-K Nu=

= 0.599

Small “d”  large h, small Nu

(c)

Nu= 0.30 + [1+

for flow around a cylinder ] 4/5

[1+ ] 4/5 1 for small Re

Back out Re 0.599= 0.30 +

Re= 0.601 =

= 3.6 m/sec Hot wire anemometer in tube with flowing N2 gas

20.33

Input Parameters: D= 30 cm (0.3m), L=3.0 m ksi= 30 W/m-K Ts= 860 K = 300 K = 0.50 m/sec

(a), (b) Heat transfer rate, q

For flow normal to a cylinder NuD= For air at Tav= (Ts+

[1+ )/2= (860 K+300 K)/2= 580 K

= 4.8512 x10-5 m2/sec, Prair= 0.68, kair= 0.45407 W/m-K ReD=

NuD= 0.30+

= 3092

[1+

)

NuD= 28.1 h=

= 4.25 W/m2-K

q= (0.3 m) (3.0 m) 4.25 W/m2-K (860 K-300 K) = 6734 W

=0.011

∴ lumped parameter analysis can be used (Bi<0.1) Asides: Glycerin at 38 = 1253 kg/m3

= 78.2 Cp=0.598 = k=

J/kg-K = 0.15 kg/m-sec = 0.285 W/m-K

20.34

From Figure 20.13 {extrapolation}

20.35

20.36 Figures 20.12&20.13 apply strictly for liquids flowing through tube banks but they will be used for lack of other resources. Using Fig 20.13

At this Re: j 0.01

For a bank of 10 tubes, 10 rows deep

20.37 For same conditions as Prob. 20.51 except staggered tube arrangement

& both arrangements give same value for j ∴

20.38 20.44

{Laminar}

Assume

20.39 20.45

Assume

{Laminar}

20.40 20.46

Putting everything together

By trial and error:

20.47 20.41

(20.44 continued)

Assume T0 is outside tube temperature

20.42 20.48 Outside of tube insulated ∴ all heat generated goes into H2O

At constant flux

Chapter 21 End of Chapter Problem Solutions

21.1 Plate is assumed to be copper For H2O @ 323 K

Natural convection

Nucleate boiling:

L=0.565 ft

Equating (1)=(2)

Part b): Plot q/A from (1), q/A from (2), and their sum

21.2

In English units:

For Ni & Brass For Cu & Pt Ts(K)

390

0.033 5.04

0.364

420

0.09

4.67

0.36

450

139

0.148 3.79

0.355

Ni, Brass

Cu, Pt

390

168

2e5

165

.533e5

420

3516

8.5e5

346

4.07e5

450

16340

24e5

1610

1610e5

21.3

21.4 Boiling H2O @ 1 atm: Burnout point is

Film boiling part:

(21.4 continued)

Substituting into formula:

Nucleate boiling part:

As it cools the cylinder is in:

With

21.5 Using English units:

Assume nucleate boiling:

Surface temp = 221

21.6

Per plate:

It appears that nucleate boiling on one plate can achieve this:

{Ok one plate will do it}

21.7

21.8 {Film boiling}

21.9

{at a given depth y}

t, hours

ft x102

inches

0.2

0.586

0.0681

0.4

0.804

0.0964

0.6

1.0

0.12

0.8

1.14

0.137

1.0

1.27

0.152

21.10 {for thickness y}

if pan is horizontal For pan inclined: Per unit detla:

Length of surface exposed:

Assume accumulation of condensate due principally to condensation on exposed surface

21.11

21.12 Vertical tube:

Horizontal:

21.13 Horizontal cylinder:

(21.16 continued)

21.14

Horizontal tube case {see prob. 21.16}

21.15 Single horizontal tube:

Chapter 22 End of Chapter Problem Solutions 22.1

Using dittus-boelter correlation

As diameter increases the required area increases as

22.2 Oil:

22.3

(22.4 continued)

Assuming turbulent flow:

Solving for

22.4 Water

Oil:

For

max

will be associated with minimum

If T0 out max=160 For H2O as minimum fluid:

{Fig 22.12 a

}

Problem 22.5

Biodiesel (30-60 ) = 880 kg/m3 Cp,B= 2400 J/kg-K

(a) qshell= = 2.5 m3/hr * 880 kg/m3 * 2400 J/kg-K (35 = -44,000 W qtube= =

= - qshell

at Tav=

= 293 K

Cp, H20= 4182 J/kg-K H20=

(b) A=?

= 27.4

= 0.526 kg/sec

* 1 hr/3600 sec

= 0.462 m/sec Re=

= 17,698 is turbulent (>10,000)

NU= 0.023 Re0.8 Pr0.4 = 0.023 (17,698)0.8(6.96)0.4 Nu= 125.06 =

hi= 1959.6 W/m2-K U=

= 289.7 W/m2-K

5.53 m2

recalculate

= 19.5 K ( ) A=

7.77 m2 (larger!)

Problem 22.6

Input Parameters: Tav=

= 1253 kg/m3 Cp= 2501 J/kg-K = 0.15 kg/m-sec k= 0.295 W/m-K Di= 0.0135 m (0.532 in) Do= 0.01905 m (0.75 in)

(a) Required steam flow,

qtube= qshell= -qtube=

550,220 W =

= 0.244 kg/sec

(b) (c)

= 59.3 tw=

= 2.969 x 10-3 m

Tube-side convective heat transfer Re=

= 203 (<2100)

Pr=

laminar

= 1316

For laminar flow inside a tube Nu= 1.86 (Re Pr)1/3( )1/3 = 1.86 (203*1316)1/3 (

)1/3= 16.68

= 352 W/m2-K

Nu=

U= 322 W/m2-K

(d) A

q= UA

Heat Exchanger Design Equation = 28.8 m2

A= n=

= 96 tubes

Note: the next level of analysis could be to adjust the desired so that A by the Heat Exchanger Design Equation matches the actual tube velocity in “n” tubes. Aside to part (c) Rigorously, for a “thick walled” tube + + Uo= 233 W/m2-K A= 39.8 m2

22.7

𝑤 ,𝑖𝑛 = 20

𝑤 ,𝑜𝑢𝑡

𝑠 = 120

(a) Change

𝑠𝑎𝑡 = 120

in problem statement

(b) Change desired Re in the problem to 40,000.

(c)

Since water side resistance is >> condensate side,

The heat transfer surface area is,

22.8

b) Water in shell; oil ~ 2 passes

can’t be done c)

Fig (22.12) ~

22.9

To find

Fig 22.9 a:

F 1

22.10 22.11

(22.11 Continued)

To find F:

Fig. 22.9a F 1

22.11 22.14

Counter flow:

Cross flow – air mixed:

Cross flow – both mixed:

Cross flow configuration with both mixed is most compact

22.12 22.15

Tubes:

{fig 22.12c}

2 Tubes passes will work 37 tubes per pass L=1.64 m per pass

22.13 22.16

Steam condensation rate:

22.14 Problem 22.17

Hot fluid: ethanol, shell side (B) Cold fluid: water, tube side (A) = 250 W/m2-K based on outer tube area

(a) What is

a (cooling water mass flowrate)

Energy balance shell=

B,1 –

out

B,2) = =

B,1 (sat liq)

B,2 (sat vap)

qshell= 2.0 kg EtOH/sec * (-838 kJ/kg)= -1676 kJ/sec qtube= -qshell=

ACp,A,liq (TC,2- TC,1)

A= 13.3 kg/sec

(b)

= 41.2

(d) ntubes=? For 16 ft, 1.5 in schedule 40 pipes Appendix , W3R 5th; Do= 1.90 inches= 0.048 m n=

= 220 tubes

What would be the diameter of the tube bundle of the

pitch was 3 inches?

22.15 Problem 22.18

(a) TC,2 qshell=

H20= 0.5 kg/sec,

CO2 Cp,CO2(TH,1-TH,2)

= 325 K Appendix I, W3R

Cp,CO2= 875.8 J/kg-K at 325 K

CO2= 𝑛CO2 * MW,CO2= 360 kgmol/hr * 44 kg/kgmol * 1 hr/3600 sec = 4.4 kg/sec

qshell= 4.4 kg/sec * 875.8 J/kg-K (300 K-350 K)= -1.93 x 105 J/sec (193 kW) qtube=

H20CP,H20(TC,2-TC,1)= -qshell

TC,2= TC,1+

+ 293 K= 385 K

TC,2 > TH,2 – ouch 2ND Law Thermo violated Recalc with TH,2-TC,2=17

(or 17 K)

= 350-17= 333 K =

(a) U=?

1.15 kg/sec

ho= 300 W/m2-K tw= 0.109 in= 4.291 x 10-4 m

0.5 inch sched 40 tube

Di= 0.546 in= 0.00215 m Do=0.622 in= 0.00245 m Tav=

313 K tube side 0.663 x 10-6 m2/sec

Pr= 4.333

Appendix I W3R k= 0.633 W/m-K

= 992.2 kg/m3

Re= 10,000= = 3.08 m/sec For Cu

kw= 386 W/m-K at 293 K ( Appendix H W3R)

For tube

Nu= 0.023 Re0.8Pr0.4 (fluid is heated) 0.023 (10,000)0.8(4.33)0.4= 40.86

Nu= h=

= 1.203 x 10-4 W/m2-K

U= 292.6 W/m2-K

shell side is obviously limiting

(d) A=? Design Equation

q= UA

(e) n=?

= 11.3 K (11.3

= 58.4 m2

by continuity

)

𝑛

= or n= also A=

= 104 tubes 𝑛

73 m

wow

Go multi-pass and limit Lp= 5 m # passes=

14 passes

Will this configuration work? Calculate heat-exchanger correction factor “F” Y= = Off chart

won’t work

22.16 Problem 22.19

Appendix I, M W3R kw= 386 W/m-K Schedule 40 1” pipe Di= 1.049 in= 2.67 x 10-2 m Do= 1.315 in= 3.34 x 10-2 m tw= 0.133 in= 3.378 x 10-3 m

Input Parameters: Tube side CO2

Tav=

Cp,CO2= 875.8 J/kg-K kCO2= 0.01851 W/m-K

325 K 1.67 kg/m3 PrCO2= 0.763

9.76 x 10-6 m2/sec (Appendix H, W3R 5th)

(a) qtube=

and q= qtube= - qshell CO2 on tube side

=𝑛

𝑠

= 0.44 kg/sec qtube= 0.44 kg/sec * 875.4 J/kg-K ( 350 K-300 K)= 1.927 x 104 J/sec (W) For water on shell side @ Tav= 40 , Cp,H20= 4184 J/kg-K qshell=

= -qtube 𝑠

= (b) as

(414.5 kg/hr)

, TC,1

(c)

Overall Heat Transfer Coeff. “U”

For tube side, Re= 20,000 is desired Nu= 0.023 Re0.8Pr0.3= 0.023 (20,000)0.8(0.763)0.3 Nu= 58.52= = 40.57 W/m2-K

U= 35.73 W/m2-K (why lower than hi?)

(d) A

q=UA =

= 11.3 K = 47.85 m2

(e) ntube and Ltube n= Re=

CO2 flows in one tube

= 7.31 m/sec n=

(call it 65)

A= (

𝑛

L= 7.0 m for each tube If 10 ft tubes were available, what are your options? 1) multipass 2) work backwards: change Re, n etc. until L=10 ft.

(one tube)

22.17 22.20 Problem

Thermo physical properties of CO2: (Tav 425 K) = 1.268 kg/m3 Cp= 960.9 J/kg-K = 2.0325 x 10-5 kg/m-sec k= 0.02678 W/m-K Tube Properties Do= 1.05 in= 0.0267 m Di= 0.824 in= 0.0209 m tw= 0.113 in= 0.0029 m kw= 42.9 W/m-K

(a) Heat Exchanger Area “A” q=U*A*

note TH>Ts!

=> U= 49.6 W/m2-K =

= 47.7

q=qshell=

= 1,128,500 W m2

(b)

for CO2 flow in tube

-qshell=qtube=

*Cp,CO2(TH,2-TH,1)

(c)

= 23.5 kg CO2/sec

for single tube, fluid= CO2 @ Tav= 425 K

For turbulent flow, Nu= Nu=

Pr=

= 0.729

Re=

12,279

Re= =

(d) For A->

= 9.4 m/sec

TH,2 ->Ts=100

22.18 22.21

22.19 22.22

22.20 22.23

To find F:

Fig. 22.10 a F 0.96

22.21 22.24 Input data – see problem 22.3

Crossflow:

Shell and tube: Same data

22.22 22.27

For cool fluid in tubes:

Hot fluid in tubes:

Both are off the charts. Neither is possible

can’t use this configuration.

Chapter 23 End of Chapter Problem Solutions 23.1 Radiant emission from sun All passes through a spherical surface of radius, L At the earth:

Flux at earth

23.2

Area subtended by earth

Incident solar energy = {from problem 23.1}

Energy balance:

23.3

23.4 Energy balance for collector:

) Equating: 800=

By trial and error T=322 K

23.5 a)

b) With window

23.6 From Wien’s displacement law:

Fraction in visible band:

{table 23.1}

23.7

T (K)

( m)

Sun

5790

1.998

L. bulb

2910

1.004

surface

1550

0.535

Skin

308

0.1063

23.8 Filament at 2910 K q=100 W

Visible range:

Fraction in V.R. = 0.1008

23.9

For

23.10

For solar irradiation: Plain glass:

Tinted glass:

In the visible range:

{For tinted glass

Plain glass: Tinted glass:

}

23.11

T (K) 500

-0

0.1613

2000

0.0197

0.9142

0.8945

3000

0.1402

0.9689

0.8287

4500

0.4036

0.9896

0.586

23.12

For:

Fraction in visible range = 0.3696 In U.V. range

Fraction in U.V. range = 0.12 In I.R. range

Fraction in I.R. range = 0.88 Wein’s law

23.13 For surroundings at 0 K:

In visible range 0.4< <0.7 m

Fraction = 0.0078 In U.V. range 0< <0.4

Fraction = 0 In I.R. range 0.4< <100

Fraction = 0.992

23.14 T=1500 K Peephole D=10 cm for 0< <3.2 m for 3.2< <

For:

Total heat loss

23.15

23.16 through hole with D=0.0025 m2

23.17 With no intervening plate:

With intervening plate present: Per unit area:

(

{Emissivity of intervening plate has no effect}

22.18

23.19 Entire hole interior is surf. 2 Opening (surroundings) is surf. 1

23.20 1 is inner cylinder 2 is outer cylinder

With radiation shield

23.21 23.22

With no reflector:

23.22 23.23

per ft {radiant loss} Convection:

For Tf=137

23.23 23.24

23.24 23.25 Diameter of hole = 5 cm

Gray case:

23.25 23.28 Opening diameter= 5 mm Equivalent surface (1) sees interior as a single surface

Analogous circuit:

All interior may be considered a single surface

23.26 23.39 Problem statement asks for radiant energy reaching tank bottom, i.e. the irradiation

23.27 23.30 Surroundings are considered and equivalent surface (3) at 0 K

(23.30 continued)

To surroundings

23.28 23.32

Solving:

23.29 23.33

From Figs 23.14 & 23.15

For black disks:

For gray bodies:

For loops shown:

Solving simultaneously:

23.30 23.36

(23.36 continued)

23.31 23.37 {surface 3 is surroundings}

{Fig 23.14 F12 0.37} → F13=0.63

Loops equation:

Solving:

23.37 (Alternate solution) Using equations 23.37 & 23.38 Applying them to each node:

Solving these equations simultaneously gives same result as above

23.32 23.38 Test specimen is 1 Tube is 2 Viewing port W

becomes:

From specimen

Loss through window

23.33 23.41

In limit as

By graphical integration:

1900

3.2

2.93

1700

4.0

3.0

1500

5.4

3.15

1300

8.0

3.33

1100

14.6

3.65

Radiant fraction For v doubled:

Integrate graphically until x=35.2 At this location T=1265

24.1 Assume: 1) Ideal gas mixture, 2) constant 𝜌𝜌 (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑), 3) constant c (total molar concentration). Given 𝐍𝐍𝐴𝐴 = −𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 ∇𝑦𝑦𝐴𝐴 + 𝑦𝑦𝐴𝐴 (𝐍𝐍𝐴𝐴 + 𝐍𝐍𝐵𝐵 )

With

𝑤𝑤𝐴𝐴 = And

𝑦𝑦𝐴𝐴 𝑀𝑀𝐴𝐴

𝑦𝑦𝐴𝐴 𝑀𝑀𝐴𝐴 +𝑦𝑦𝐵𝐵 𝑀𝑀𝐵𝐵

𝐧𝐧𝐴𝐴 = 𝜌𝜌𝐴𝐴 𝐯𝐯𝐴𝐴 = 𝜌𝜌𝐴𝐴

with M = 𝑦𝑦𝐴𝐴 𝑀𝑀𝐴𝐴 + 𝑦𝑦𝐵𝐵 𝑀𝑀𝐵𝐵 𝐍𝐍𝐴𝐴 𝜌𝜌𝐴𝐴 = 𝐍𝐍 = 𝑀𝑀𝐴𝐴 𝐍𝐍𝐴𝐴 𝑐𝑐𝐴𝐴 𝑐𝑐𝐴𝐴 𝐴𝐴

𝐍𝐍𝐴𝐴 𝑀𝑀𝐴𝐴 = −𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 𝑀𝑀𝐴𝐴 ∇𝑦𝑦𝐴𝐴 + 𝑦𝑦𝐴𝐴 𝑀𝑀𝐴𝐴 (𝐍𝐍𝐴𝐴 + 𝐍𝐍𝐵𝐵 ) 𝐧𝐧𝐴𝐴 = −𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 𝑀𝑀𝐴𝐴

∇𝑐𝑐𝐴𝐴 𝐧𝐧𝐴𝐴 𝐧𝐧𝐵𝐵 + 𝑤𝑤𝐴𝐴 𝑀𝑀( + ) 𝑐𝑐 𝑀𝑀 𝑀𝑀

𝐧𝐧𝐀𝐀 = −𝐷𝐷𝐴𝐴𝐴𝐴 𝑀𝑀𝐴𝐴 ∇𝑐𝑐𝐴𝐴 + 𝑤𝑤𝐴𝐴 (𝐧𝐧𝐴𝐴 + 𝐧𝐧𝐵𝐵 ) 𝐧𝐧𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴 ∇𝜌𝜌𝐴𝐴 + 𝑤𝑤𝐴𝐴 (𝐧𝐧𝐴𝐴 + 𝐧𝐧𝐵𝐵 )

24-1

24.2 a. Prove 𝐍𝐍𝐴𝐴 + 𝐍𝐍𝐵𝐵 = 𝑐𝑐𝐕𝐕 𝐍𝐍𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴 ∇𝑐𝑐𝐴𝐴 + 𝐍𝐍𝐴𝐴 = 𝐉𝐉𝐴𝐴 +

𝑐𝑐𝐴𝐴 (𝐍𝐍 + 𝐍𝐍𝐵𝐵 ) 𝑐𝑐 𝐴𝐴

𝐉𝐉𝐴𝐴 + 𝑐𝑐𝐴𝐴 𝐕𝐕 = 𝐉𝐉𝐴𝐴 + Thus,

𝑐𝑐𝐴𝐴 (𝐍𝐍 + 𝐍𝐍𝐵𝐵 ) 𝑐𝑐 𝐴𝐴

𝑐𝑐𝐴𝐴 (𝐍𝐍 + 𝐍𝐍𝐵𝐵 ) = 𝑐𝑐𝐴𝐴 𝐕𝐕 𝑐𝑐 𝐴𝐴 𝐍𝐍𝐴𝐴 + 𝐍𝐍𝐵𝐵 = 𝑐𝑐𝐕𝐕

b. Prove 𝐧𝐧𝑨𝑨 + 𝐧𝐧𝑩𝑩 = 𝜌𝜌𝐕𝐕 𝐍𝐍𝐴𝐴 + 𝐍𝐍𝐵𝐵 = 𝑐𝑐𝐕𝐕

𝐧𝐧𝐴𝐴 = 𝑀𝑀𝐴𝐴 𝐍𝐍𝐴𝐴 Then

𝐧𝐧𝐴𝐴 𝐧𝐧𝐵𝐵 + = 𝑐𝑐𝐕𝐕 𝑀𝑀𝐴𝐴 𝑀𝑀𝐵𝐵

1 𝐕𝐕 = (𝑐𝑐𝐴𝐴 𝐯𝐯𝐴𝐴 + 𝑐𝑐𝐵𝐵 𝐯𝐯𝐵𝐵 ) 𝑐𝑐

𝐧𝐧𝐴𝐴 𝐧𝐧𝐵𝐵 1 + = 𝑐𝑐 ∙ (𝑐𝑐𝐴𝐴 𝐯𝐯𝐴𝐴 + 𝑐𝑐𝐵𝐵 𝐯𝐯𝐵𝐵 ) 𝑀𝑀𝐴𝐴 𝑀𝑀𝐵𝐵 𝑐𝑐 𝜌𝜌𝐴𝐴 = 𝑐𝑐𝐴𝐴 𝑀𝑀𝐴𝐴

𝐧𝐧𝐴𝐴 𝐧𝐧𝐵𝐵 𝜌𝜌𝐴𝐴 𝐯𝐯𝐴𝐴 𝜌𝜌𝐵𝐵 𝐯𝐯𝐵𝐵 + =( + ) 𝑀𝑀𝐴𝐴 𝑀𝑀𝐵𝐵 𝑀𝑀𝐴𝐴 𝑀𝑀𝐵𝐵 Thus

𝐧𝐧𝐴𝐴 + 𝐧𝐧𝐵𝐵 = (𝜌𝜌𝐴𝐴 𝐯𝐯𝐴𝐴 + 𝜌𝜌𝐵𝐵 𝐯𝐯𝐵𝐵 ) 24-2

𝐯𝐯 =

1 (𝜌𝜌 𝐯𝐯 + 𝜌𝜌𝐵𝐵 𝐯𝐯𝐵𝐵 ) 𝜌𝜌 𝐴𝐴 𝐴𝐴

𝐧𝐧𝐴𝐴 + 𝐧𝐧𝐵𝐵 = 𝜌𝜌𝐯𝐯

c. Prove 𝐣𝐣𝐴𝐴 + 𝐣𝐣𝐵𝐵 = 0 𝐣𝐣𝐴𝐴 = −𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴 ∇𝑤𝑤𝐴𝐴 So

𝐣𝐣𝐴𝐴 + 𝐣𝐣𝐵𝐵 = −𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴 ∇𝑤𝑤𝐴𝐴 − 𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴 ∇𝑤𝑤𝐵𝐵 = −𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴 (∇𝑤𝑤𝐴𝐴 + ∇𝑤𝑤𝐵𝐵 )

∇𝑤𝑤𝐴𝐴 + ∇𝑤𝑤𝐵𝐵 = ∇𝑤𝑤𝐴𝐴 + ∇(1 − 𝑤𝑤𝐴𝐴 ) = 0 Thus

𝐣𝐣𝐴𝐴 + 𝐣𝐣𝐵𝐵 = (−𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴 ∙ 0) + (−𝜌𝜌𝐷𝐷𝐴𝐴𝐴𝐴 ∙ 0) = 0

24-3

24.3 Given 𝑛𝑛

𝑦𝑦𝑖𝑖 𝑦𝑦𝑗𝑗 𝑑𝑑𝑦𝑦𝑖𝑖 = � (𝑣𝑣 − 𝑣𝑣𝑖𝑖 ) 𝑑𝑑𝑑𝑑 𝐷𝐷𝑖𝑖𝑖𝑖 𝑗𝑗 𝑗𝑗=1,𝑗𝑗≠𝑖𝑖

i =A, j = B, and n = 2 (species A and B)

𝑑𝑑𝑦𝑦𝐴𝐴 𝑦𝑦𝐴𝐴 𝑦𝑦𝐵𝐵 = �𝑣𝑣 − 𝑣𝑣𝐴𝐴,𝑧𝑧 � 𝑑𝑑𝑑𝑑 𝐷𝐷𝐴𝐴𝐴𝐴 𝐵𝐵,𝑧𝑧 = = =

𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 Thus

𝑦𝑦𝐴𝐴 𝑦𝑦𝐵𝐵 𝑣𝑣𝐵𝐵,𝑧𝑧 − 𝑦𝑦𝐴𝐴 𝑦𝑦𝐵𝐵 𝑣𝑣𝐴𝐴,𝑧𝑧 𝐷𝐷𝐴𝐴𝐴𝐴

𝑦𝑦𝐴𝐴 (𝑐𝑐𝐵𝐵 ⁄𝑐𝑐 )𝑣𝑣𝐵𝐵,𝑧𝑧 − (𝑐𝑐𝐴𝐴 ⁄𝑐𝑐 )𝑦𝑦𝐵𝐵 𝑣𝑣𝐴𝐴,𝑧𝑧 𝐷𝐷𝐴𝐴𝐴𝐴 𝑦𝑦𝐴𝐴 (𝑐𝑐𝐵𝐵 𝑣𝑣𝐵𝐵,𝑧𝑧 ) − 𝑦𝑦𝐵𝐵 (𝑐𝑐𝐴𝐴 𝑣𝑣𝐴𝐴,𝑧𝑧 ) 𝑐𝑐𝑐𝑐𝐴𝐴𝐴𝐴 𝑦𝑦𝐴𝐴 𝑁𝑁𝐵𝐵,𝑧𝑧 − (1 − 𝑦𝑦𝐴𝐴 )𝑁𝑁𝐴𝐴,𝑧𝑧 𝑐𝑐𝑐𝑐𝐴𝐴𝐴𝐴

𝑑𝑑𝑦𝑦𝐴𝐴 = 𝑦𝑦𝐴𝐴 𝑁𝑁𝐵𝐵,𝑧𝑧 − 𝑁𝑁𝐴𝐴,𝑧𝑧 +𝑦𝑦𝐴𝐴 𝑁𝑁𝐴𝐴,𝑧𝑧 𝑑𝑑𝑑𝑑

= 𝑦𝑦𝐴𝐴 �𝑁𝑁𝐴𝐴,𝑧𝑧 + 𝑁𝑁𝐵𝐵,𝑧𝑧 � − 𝑁𝑁𝐴𝐴,𝑧𝑧

𝑁𝑁𝐴𝐴,𝑧𝑧 = −𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴

𝑑𝑑𝑦𝑦𝐴𝐴 + 𝑦𝑦𝐴𝐴 �𝑁𝑁𝐴𝐴,𝑧𝑧 + 𝑁𝑁𝐵𝐵,𝑧𝑧 � 𝑑𝑑𝑑𝑑

24-4

24.4 NA = 5.0 × 10-5 kgmole A/m2∙s NB = 0 kgmole B/m2∙s

𝑐𝑐𝐴𝐴 = 0.005 kgmole/m3

𝑐𝑐𝐵𝐵 = 0.036 kgmole/m3 Find 𝑣𝑣𝐴𝐴 , 𝑣𝑣𝐵𝐵 , and V NA = 𝑐𝑐𝐴𝐴 𝑣𝑣𝐴𝐴

𝑁𝑁𝐴𝐴 5.0 × 10−5 kgmole A/m2 ∙ s 𝑣𝑣𝐴𝐴 = = = 0.010 m/s 𝑐𝑐𝐴𝐴 0.005 kgmole/m3 NB = 𝑐𝑐𝐵𝐵 𝑣𝑣𝐵𝐵 = 0

Thus vB = 0 m/s

𝑁𝑁𝐴𝐴 1 1 𝑁𝑁𝐴𝐴 = 𝑉𝑉 = (𝑐𝑐𝐴𝐴 𝑣𝑣𝐴𝐴 + 𝑐𝑐𝐵𝐵 𝑣𝑣𝐵𝐵 ) = (𝑁𝑁𝐴𝐴 + 𝑁𝑁𝐵𝐵 ) = 𝑐𝑐 𝑐𝑐 𝑐𝑐 𝑐𝑐𝐴𝐴 + 𝑐𝑐𝐵𝐵 =

5.0 × 10−5 kgmole A/(m2 ∙ s) = 0.0012 m/s 0.005 kgmole/m3 + 0.036 kgmole/m3

24-5

24.5 Given: P = 6.1 × 10−3 bar, T = 210 K

yCO2 = 0.9532, yN2 = 0.027, yAr = 0.016, yO2 = 0.0013, yCO = 0.0008

a. Partial pressure 𝑝𝑝𝐴𝐴 for CO2 A = CO2

 105 Pa  −3 × 0.9532 6.1 10 bar ( )   = 581 Pa  1 bar  b. Molar concentration 𝑐𝑐𝐴𝐴 for CO2

(

p A y= = AP

cA =

)

pA 581 Pa gmole = = 0.333 3 RT m3 8.314 m ⋅ Pa/gmole ⋅ K ( 210 K )

(

)

c. Total molar concentration c for the Martian atmosphere

= c

P = RT

(

610 Pa gmole = 0.349 m3 8.314 m ⋅ Pa/gmole ⋅ K ( 210 K ) 3

)

d. Total mass concentration ρ for the Martian atmosphere

M avg= yCO2 M CO2 + yN2 M N2 + yAr M Ar + yO2 M O2 + yCO M CO    g  g  g  M avg = ( 0.9532 )  44.01  + ( 0.027 )  28.00  + ( 0.0163)  9.95 + gmole  gmole  gmole       g  g  ( 0.0013)  32.00  + ( 0.0008 )  28.01  gmole  gmole    M avg = 43.41

g gmole

 g  gmole  g c  43.41 = ρ M avg ⋅=   0.349  =15.2 3 3 gmole   m  m 

24-6

24.6 A = benzene solute (C6H6) in B = liquid ethanol solvent (C2H5OH) 𝑤𝑤𝐴𝐴 = 0.0050 T = 293 K

𝜌𝜌𝐵𝐵 = 789 kg/m3 𝑀𝑀𝐴𝐴 = 78.11 Estimate 𝑐𝑐𝐴𝐴

, 𝑀𝑀𝐵𝐵 = 46.07

gmole

= 46.07

kgmole

The solution is dilute with respect to A 𝜌𝜌𝐵𝐵 𝑐𝑐 ≈ = 𝑀𝑀𝐵𝐵

kg kmole m3 = 17.1 kg m3 46.07 kgmole 789

𝑤𝑤𝐴𝐴 + 𝑤𝑤𝐵𝐵 = 1

𝑤𝑤𝐵𝐵 = 1 − 𝑤𝑤𝐴𝐴 = 1‒ 0.0050 = 0.9950

0.0050 g 78.11 gmole 𝑥𝑥𝐴𝐴 = 𝑤𝑤 = 0.0030 𝑤𝑤 = 𝐴𝐴 0.0050 0.9950 + 𝐵𝐵 𝑀𝑀𝐴𝐴 𝑀𝑀𝐵𝐵 g + g 78.11 46.07 gmole gmole 𝑤𝑤𝐴𝐴 𝑀𝑀𝐴𝐴

𝑐𝑐𝐴𝐴 = 𝑥𝑥𝐴𝐴 𝑐𝑐 = (0.0030)(17.1 kgmole⁄m3 ) = 0.051 kgmole⁄m3 = 51 gmole⁄m3

24-7

24.7 a. Estimate 𝐷𝐷𝐴𝐴𝐴𝐴 for 𝐶𝐶𝐶𝐶2 in air

A = 𝐶𝐶𝐶𝐶2 and B = air 𝑀𝑀𝐴𝐴 = 44.01 T = 310 K

, 𝑀𝑀𝐵𝐵 = 28.97

gmole

𝑃𝑃 = (1.5 × 105 Pa) �

gmole

1atm � = 1.49 atm 1.01 × 105 Pa

Hirschfelder Equation

𝐷𝐷𝐴𝐴𝐴𝐴 =

1 1 1⁄2 0.001858 𝑇𝑇 3⁄2 �𝑀𝑀 + 𝑀𝑀 � 𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴

2Ω

𝐴𝐴

𝐷𝐷

𝐵𝐵

From Appendix K, Table K.2: 𝜎𝜎𝐴𝐴 = 3.996 Å, 𝜎𝜎𝐵𝐵 = 3.617 Å, 𝜎𝜎𝐴𝐴𝐴𝐴 =

𝜀𝜀𝐴𝐴 κ

𝜀𝜀

= 190 K, 𝐵𝐵 = 97 K κ

𝜎𝜎𝐴𝐴 + 𝜎𝜎𝐵𝐵 3.996 Å + 3.617 Å = = 3.807 Å 2 2

𝜅𝜅𝜅𝜅 𝜅𝜅 𝜅𝜅 1/2 1 1 1⁄2 = � ∙ � 𝑇𝑇 = � ∙ � ∙ 310 K = 2.28 𝜀𝜀𝐴𝐴𝐴𝐴 𝜀𝜀𝐴𝐴 𝜀𝜀𝐵𝐵 190 K 97 K From Table K.1, interpolate to find ΩD = 1.029

1/2

𝐷𝐷𝐴𝐴𝐴𝐴 =

1 1 0.001858 ∙ 310 K 3/2 � g + g � 44.01 28.97 gmole gmole 2

(1.49 atm)�3.807 Å� (1.029)

= 0.109 cm2 ⁄s

b. Estimate DAB for ethanol in air A = ethanol and B = air 𝑀𝑀𝐴𝐴 = 46.07

, 𝑀𝑀𝐵𝐵 = 28.97

gmole

24-8

T = 325 K 𝑃𝑃 = (2.0 × 105 Pa) �

1atm � = 1.98 atm 1.01 × 105 Pa

𝜀𝜀

From Appendix K, Table K.2: 𝜎𝜎𝐴𝐴 = 4.455 Å, 𝜎𝜎𝐵𝐵 = 3.617 Å, 𝐴𝐴 = 391 𝐾𝐾, 𝐵𝐵 = 97 𝐾𝐾 𝜎𝜎𝐴𝐴𝐴𝐴 =

𝜎𝜎𝐴𝐴 + 𝜎𝜎𝐵𝐵 4.455 Å + 3.617 Å = = 4.036 Å 2 2

𝜅𝜅𝜅𝜅 𝜅𝜅 𝜅𝜅 1/2 1 1 1⁄2 = � ∙ � 𝑇𝑇 = � ∙ � ∙ 325 K = 1.67 𝜀𝜀𝐴𝐴𝐴𝐴 𝜀𝜀𝐴𝐴 𝜀𝜀𝐵𝐵 391K 97K From Table K.1, interpolate to find ΩD = 1.148

𝐷𝐷𝐴𝐴𝐴𝐴 =

0.001858 ∙ 325 𝐾𝐾 3/2 �

1/2

1 g + g � 46.07 28.97 gmole gmole 2

(1.98 atm)�4.036 Å� (1.148)

= 0.0697 cm2 ⁄s

c. Estimate DAB for CCl4 in air A = CCl4 and B = air 𝑀𝑀𝐴𝐴 = 153.82 T = 298 K

, 𝑀𝑀𝐵𝐵 = 28.97

gmole

𝑃𝑃 = (1.913 × 105 Pa) �

gmole

1atm � = 1.894 atm 1.01 × 105 Pa

𝜀𝜀

From Appendix K, Table K.2, 𝜎𝜎𝐴𝐴 = 5.881 Å, 𝜎𝜎𝐵𝐵 = 3.617 Å, 𝐴𝐴 = 327 K, 𝐵𝐵 = 97 K 𝜎𝜎𝐴𝐴𝐴𝐴 =

𝜎𝜎𝐴𝐴 + 𝜎𝜎𝐵𝐵 5.881 Å + 3.617 Å = = 4.749 Å 2 2

𝜅𝜅𝜅𝜅 𝜅𝜅 𝜅𝜅 1/2 1 1 1⁄2 = � ∙ � 𝑇𝑇 = � ∙ � ∙ 298 K = 1.67 𝜀𝜀𝐴𝐴𝐴𝐴 𝜀𝜀𝐴𝐴 𝜀𝜀𝐵𝐵 327K 97K From Table K.1, ΩD = 1.148

24-9

𝐷𝐷𝐴𝐴𝐴𝐴 =

1 0.001858 ∙ 298 K 3/2 � 153.82

1/2

1 g � g + 28.97 gmole gmole 2

(1.894 atm)�4.749 Å� (1.148)

= 0.0395 cm2 ⁄s

24-10

24.8 Estimate 𝐷𝐷𝐴𝐴𝐴𝐴 for NH3 in air A = NH3, B = air

𝑀𝑀𝐴𝐴 = 17.03

, 𝑀𝑀𝐵𝐵 = 28.97

gmole

P = 1.0 atm, T = 373 K, (𝜇𝜇𝑝𝑝 )𝐴𝐴 = 1.46 debye, (𝑉𝑉𝑏𝑏 )𝐴𝐴 = 25.8 Brokaw equation

𝑐𝑐𝑐𝑐3

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

, (𝑇𝑇𝑏𝑏 )𝐴𝐴 = 239.7 K

1.94 × 103 (𝜇𝜇𝑝𝑝 )𝐴𝐴 2 1.94 × 103 (1.46 debye)2 𝛿𝛿𝐴𝐴 = = = 0.669 𝑐𝑐𝑐𝑐3 (𝑉𝑉𝑏𝑏 )𝐴𝐴 (𝑇𝑇𝑏𝑏 )𝐴𝐴 (25.8 )(239.7 K) 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

𝜀𝜀𝐴𝐴 = 1.18�1 + 1.3𝛿𝛿𝐴𝐴 2 �(𝑇𝑇𝑏𝑏 )𝐴𝐴 = 1.18(1 + 1.3(0.669)2 )(239.7 K) = 447 K 𝜅𝜅 𝜎𝜎𝐴𝐴 = �

1.585(𝑉𝑉𝑏𝑏 )𝐴𝐴 1 + 1.3𝛿𝛿𝐴𝐴 2

1/3

�

1/3

𝑐𝑐𝑚𝑚3 ) 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 =� � 1 + 1.3(0.669)2 1.585(25.8

= 2.957 Å

Since the air has no dipole, δB = 0, δAB = 0 From Appendix K, Table K.2,

𝜀𝜀𝐵𝐵 κ

= 97 K

From Table 24.4 of Ch.24, (𝑉𝑉𝑏𝑏 )𝐵𝐵 = 29.9 𝜎𝜎𝐵𝐵 = �

1.585(𝑉𝑉𝑏𝑏 )𝐵𝐵 1 + 1.3𝛿𝛿𝐵𝐵 2

1/3

�

𝑐𝑐𝑐𝑐3

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

1/3

𝑐𝑐𝑐𝑐3 1.585(29.9 ) 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 =� � 1 + 1.3(0)2

= 3.619 Å

𝜎𝜎𝐴𝐴𝐴𝐴 = (𝜎𝜎𝐴𝐴 𝜎𝜎𝐵𝐵 )1/2 = (2.957 Å ∙ 3.619 Å)1/2 = 3.271 Å 𝑇𝑇 ∗ =

𝜅𝜅𝜅𝜅 𝜅𝜅 𝜅𝜅 1/2 1 1 1⁄2 = � ∙ � 𝑇𝑇 = � ∙ � ∙ 373 K = 1.79 𝜀𝜀𝐴𝐴𝐴𝐴 𝜀𝜀𝐴𝐴 𝜀𝜀𝐵𝐵 447K 97K

Ω𝐷𝐷 = Ω𝐷𝐷,0 =

𝐴𝐴 𝐶𝐶 𝐸𝐸 𝐺𝐺 + + + ∗ 𝐵𝐵 ∗ ∗ (𝑇𝑇 ) exp(𝐷𝐷𝑇𝑇 ) exp(𝐹𝐹𝑇𝑇 ) exp(𝐻𝐻𝑇𝑇 ∗ )

24-11

1.06036 0.19300 1.03587 1.76474 + + + = 1.12 0.15610 (1.79) exp(0.47635 ∙ 1.79) exp(1.52996 ∙ 1.79) exp(3.89411 ∙ 1.79)

Hirschfelder Equation

𝐷𝐷𝐴𝐴𝐴𝐴 =

1 1 1⁄2 0.001858 𝑇𝑇 3⁄2 �𝑀𝑀 + 𝑀𝑀 � 𝐴𝐴

𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴 2 Ω𝐷𝐷

𝐵𝐵

0.001858 ∙ 373 K 3/2 � 17.03

1/2

1 g � g + 28.97 gmole gmole

= 0.341 cm2 ⁄s

(1.0 atm)�3.271 Å� (1.12)

For comparison, from Appendix J, Table J.1, DABP = 0.198

𝑐𝑐𝑐𝑐2 𝑎𝑎𝑎𝑎𝑎𝑎 s

𝑐𝑐𝑐𝑐2 𝑎𝑎𝑎𝑎𝑎𝑎 𝐷𝐷𝐴𝐴𝐴𝐴 𝑃𝑃 0.198 𝑐𝑐𝑐𝑐2 s 𝐷𝐷𝐴𝐴𝐴𝐴,1 = = = 0.198 1.0 atm s 𝑃𝑃

at T1 = 273 K

𝜅𝜅𝜅𝜅 𝜅𝜅 𝜅𝜅 1/2 1 1 1⁄2 𝑇𝑇 = = � ∙ � 𝑇𝑇 = � ∙ � ∙ 273 K = 1.31 𝜀𝜀𝐴𝐴𝐴𝐴 𝜀𝜀𝐴𝐴 𝜀𝜀𝐵𝐵 447K 97K ∗

𝐴𝐴 𝐶𝐶 𝐸𝐸 𝐺𝐺 + + + (𝑇𝑇 ∗ )𝐵𝐵 exp(𝐷𝐷𝑇𝑇 ∗ ) exp(𝐹𝐹𝑇𝑇 ∗ ) exp(𝐻𝐻𝑇𝑇 ∗ ) 1.06036 0.19300 1.03587 1.76474 = + + + = 1.27 0.15610 (1.31) exp(0.47635 ∙ 1.31) exp(1.52996 ∙ 1.31) exp(3.89411 ∙ 1.31) Ω𝐷𝐷,2 =

Table J.1: DAB = 0.198

𝑐𝑐𝑐𝑐2 s

at T = 273 K and P = 1.0 atm

𝑃𝑃1 𝑇𝑇2 3/2 Ω𝐷𝐷,1 𝑐𝑐𝑐𝑐2 1 atm 373 K 3⁄2 1.12 𝑐𝑐𝑐𝑐2 𝐷𝐷𝐴𝐴𝐴𝐴,2 = 𝐷𝐷𝐴𝐴𝐴𝐴,1 � � � � = �0.198 �� = 0.279 �� 𝑃𝑃2 𝑇𝑇1 Ω𝐷𝐷,2 s 1 atm 273 K 1.27 s Brokaw equation, scaled to 373K and 1.0 atm, DAB = 0.279

𝑐𝑐𝑐𝑐2 s

24-12

24.9 T = 1073 K 𝑃𝑃 = (1.5 × 105 Pa) �

1atm � = 1.49 atm 1.01 × 105 Pa

a. Estimate DAB for SiCl4 in H2 A = SiCl4 and B = H2 𝑀𝑀𝐴𝐴 = 169.9

𝑔𝑔

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

, 𝑀𝑀𝐵𝐵 = 2.02

𝑔𝑔

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

𝜀𝜀

From Appendix K, Table K.2, 𝜎𝜎𝐴𝐴 = 5.08 Å, 𝜎𝜎𝐵𝐵 = 2.968 Å, 𝐴𝐴 = 358 K, 𝐵𝐵 = 33.3 K 𝜎𝜎𝐴𝐴𝐴𝐴 =

𝜎𝜎𝐴𝐴 + 𝜎𝜎𝐵𝐵 5.08 Å + 2.968 Å = = 4.02 Å 2 2

𝜅𝜅𝜅𝜅 𝜅𝜅 𝜅𝜅 1/2 1 1 1⁄2 = � ∙ � 𝑇𝑇 = � ∙ � ∙ 1073 K = 9.83 𝜀𝜀𝐴𝐴𝐴𝐴 𝜀𝜀𝐴𝐴 𝜀𝜀𝐵𝐵 358K 33.3K From Table K.1, interpolate to find ΩD = 0.7448

𝐷𝐷𝐴𝐴𝐴𝐴 =

1 1 1⁄2 0.001858 𝑇𝑇 3⁄2 �𝑀𝑀 + 𝑀𝑀 � 𝐴𝐴

𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴 2 Ω𝐷𝐷

𝐵𝐵

0.001858 ∙ 1073 K 3/2 � 169.9

1/2

1 𝑔𝑔 + 𝑔𝑔 � 2.02 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 2

(1.49 atm)�4.02 Å� (0.7448)

= 2.58

𝑐𝑐𝑐𝑐2 s

b. Estimate DA-m for SiCl4 in mixture gas Find DAC for SiCl4 in HCl A = SiCl4 , B = H2, C = HCl 𝑀𝑀𝐴𝐴 = 169.9

𝑔𝑔

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

, 𝑀𝑀𝐶𝐶 = 36.46

𝑔𝑔

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

𝜀𝜀

From Appendix K, Table K.2, 𝜎𝜎𝐴𝐴 = 5.08 Å, 𝜎𝜎𝐶𝐶 = 3.305 Å, 𝐴𝐴 = 358 K, 𝐶𝐶 = 360 K κ

24-13

𝜎𝜎𝐴𝐴𝐴𝐴 =

𝜎𝜎𝐴𝐴 + 𝜎𝜎𝐶𝐶 5.08 Å + 3.305 Å = = 4.19 Å 2 2

𝜅𝜅 𝜅𝜅 1/2 1 1 1⁄2 𝜅𝜅𝜅𝜅 = � ∙ � 𝑇𝑇 = � ∙ � ∙ 1073 K = 2.99 𝜀𝜀𝐴𝐴 𝜀𝜀𝐶𝐶 358K 360K 𝜀𝜀𝐴𝐴𝐴𝐴 From Table K.1, interpolate to find ΩD = 0.9499

𝐷𝐷𝐴𝐴𝐴𝐴 = =

1 1 1⁄2 + � 𝑀𝑀𝐴𝐴 𝑀𝑀𝐶𝐶 𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴 2 Ω𝐷𝐷

0.001858 𝑇𝑇 3⁄2 �

0.001858(1073K)3/2 �

1 1 𝑔𝑔 + 𝑔𝑔 � 169.90 36.46 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 2

(1.49 atm)�4.19 Å� (0.9499)

𝑦𝑦𝐴𝐴 = 0.4, 𝑦𝑦𝐵𝐵 = 0.4, 𝑦𝑦𝐶𝐶 = 0.2 𝐷𝐷𝐴𝐴−𝑀𝑀 = 𝑦𝑦 ′ 𝐵𝐵 =

1/2

𝑦𝑦′𝐵𝐵 𝑦𝑦′𝐶𝐶 + 𝐷𝐷𝐴𝐴𝐴𝐴 𝐷𝐷𝐴𝐴𝐴𝐴

0.4

1−0.4

𝐷𝐷𝐴𝐴−𝑀𝑀 =

𝑦𝑦′𝐵𝐵 =

= 0.667 , 𝑦𝑦 ′ 𝐶𝐶 = 1

𝑦𝑦𝐵𝐵 , 1 − 𝑦𝑦𝐴𝐴

0.2

1−0.4

0.667 0.333 + 𝑐𝑐𝑐𝑐2 𝑐𝑐𝑐𝑐2 2.58 0.480 s s

𝑦𝑦′𝐶𝐶 =

= 0.480

𝑐𝑐𝑐𝑐2 s

𝑦𝑦𝐶𝐶 1 − 𝑦𝑦𝐴𝐴 𝑦𝑦𝐵𝐵 𝑦𝑦𝐶𝐶 , = + 1 − 𝑦𝑦𝐴𝐴 𝐷𝐷𝐴𝐴−𝑚𝑚 𝐷𝐷𝐴𝐴𝐴𝐴 𝐷𝐷𝐴𝐴𝐴𝐴

= 0.333

= 1.05

𝑐𝑐𝑐𝑐2 s

24-14

24.10 A = CO, B = H2 yA = 0.01, yB = 0.99

dpore = 15 × 10−7 cm, ε = 0.30, T = 673 K, P = 5.0 atm MA = 28.01

a. Estimate cA cA = yA c =

, MB = 2.02

gmole

yA P = RT

gmole

0.01 ∙ 5.0 atm mol = 9.054 × 10−7 3 cm ∙ atm cm3 ∙ 673 K 82.06 mol ∙ K

b. Estimate molecular diffusion coefficient, DAB Hirschfelder Equation

1 3 2 1 1 0.001858 ∙ T 2 ∙ �M + M � A B DAB = 2 P ∙ σAB ΩD

εA = 110 K, κ

εB = 33.3 K, κ

εAB εA εB = � ∙ = 60.52 K, κ κ κ

Use Appendix K. 1 and interpolate for ΩD = 0.7336 σAB =

σA + σB 3.590 Å + 2.968 Å = = 3.279 Å 2 2 3

DAB =

1 2

1 g + g � 2.02 cm2 gmole gmole = 0.599 s 5.0 atm ∙ 3.279 Å2 ∙ 0.7336

0.001858 ∙ 673 K 2 ∙ � 28.01

κT = 11.12 εAB

c. Estimate Knudsen diffusion and effective diffusion coefficient T 673 cm2 −7 DKA = 4850 ∙ dpore ∙ � = 4850 ∙ 15 × 10 ∙ � = 0.0357 MA 28.01 s 24-15

1 1 1 1 1 = � + + �=� � 2 cm cm2 DAe DAB DKA . 0357 0.599 s s DAe = 0.0337

cm2 s

′ 𝐷𝐷𝐴𝐴𝐴𝐴 = 𝜀𝜀 2 DAe = (0.3)2 �0.0337

cm2 s

� = 3.03 x 10−3

cm2 s

Knudsen diffusion is important at current conditions. DAe ≈ DKA 1

When DKA = DAB , P =? 2

1 3 2 1 1 2 1 0.001858 ∙ T ∙ �MA + MB � DKA = ∙ 2 P ∙ σ2AB ΩD

1 2

3 1 1 0.001858 ∙ 673 K 2 ∙ � g + g � 28.01 2.02 cm2 1 mol mol = 0.0357 = ∙ 2 s P ∙ 3.279 Å2∙ 0.7336

∴ P = 42.0 atm

24-16

24.11 Given: A = H2S, B = CH4 MA = 34.08 g/gmole, MB = 16.05 g/gmole yA = 0.010,

yB = 0.99

T = 303 K, P = 15.0 atm

ε = 0.50, dpore = 20 nm = 20 ×10−7 nm

a. Molar concentration of H2S, cA Ideal Gas Law

= c A y= Ac

( 0.010 )(15.0 atm ) = 6.033 × 10−6 gmole yA P = RT  cm3 cm3 ⋅ atm   82.06  ( 303 K ) gmole ⋅ K  

b. Molecular diffusion coefficient of H2S−CH4, DAB Hirschfelder Equation

σA = 3.623 Å, σB = 3.822 Å, Appendix K.2

σ AB =

σ A +σ B 2

3.623 Å+3.822 Å =3.723 Å 2

εA ε ε =301.1 K, B =136.5 K, AB = 301.1 K ⋅136.5 K = 202.7 K κ κ κ κT 303 K = =1.495 , Appendix K.1, ΩD = 1.198 ε AB 202.7 K 1/2

1/2  1 1  1 1  3/2  + 0.001858T  0.001858 ( 303)  +   MA MB  34.08 16.05    = 2 ΩD Pσ AB (15.0 ) ( 3.723) 2 (1.198) 3/2

DAB

DAB = 0.0119

cm 2 s

24-17

Fuller Schettler Giddings Equation Table 24.3

2 ⋅1.98 + 17.0 = 20.96 ( Σv ) A = ( Σv )B= 16.5 + 4 ⋅1.98= 24.42 1/2

 1 1  + 0.001T   MA MB   = 1/3 1/3 2 P ( Σv ) A + ( Σv ) B 1.75

DAB

(

)

0.001 ⋅ ( 303 )

1.75

1/2

1   1 +    34.08 16.05 

(15.0 ) ( ( 20.96 ) + ( 24.42 ) ) 1/3

1/3 2

cm 2 DAB = 0.0138 s b. Effective diffusion coefficient of H2S in porous material, DAe Knudsen diffusion coefficient

T 303 cm 2 = 4850 20 ×10−7 = 0.0289 34.08 s MA Effective diffusion coefficient       1 1 1 1 1 =  + + = 2 2  cm cm  DAe  DAB DKA   0.0289  0.0138  s s   cm 2 DAe = 9.37 ×10−3 s Effective diffusion coefficient corrected for porosity 2s cm 2  2 ' −5 cm = × DAe ε 2 DAe = = 2.34 10 ( 0.050 )  9.37 ×10−3  s  s  DKA

4850 d pore

(

)

24-18

24.12 A = SiH4, B = He yA = 0.01, yB = 0.99

dpore = 10 × 10−4 cm, T = 900 K

MA = 32.12

, MB = 4.00

gmole

κ = 1.38 × 10−16 ergs/K

gmole

a. Determine wA

g 0.01 ∙ 32.12 yA ∙ MA gmole wA = = = 0.750 yA ∙ MA + yB ∙ MB 0.01 ∙ 32.12 g + 0.99 ∙ 4.00 g gmole gmole

b. Estimate DAB at P = 1.0 atm and P = 100 Pa, T = 900 K P = 1.0 atm

εA κ

= 207.6 K,

εB κ

= 10.22 K ,

εAB κ

=� A∙ κ

εB κ

= 46.06 K,

Using Appendix K. 1, interpolate to get ΩD = 0.6676 σA = 4.08 Å,

σB = 2.576 Å, σAB =

σA +σB 2

= 3.328 Å

κT

εAB

= 19.54

1 2

3 1 1 1 0.001858 ∙ 900 K 2 ∙ � 3 2 g + g � 1 1 32.12 4.00 0.001858 ∙ T 2 ∙ �M + M � gmole gmole A B DAB = = 2 P ∙ σAB ΩD 1.0 atm ∙ 3.328 Å2 ∙ 0.6676

= 3.60

cm2 s

𝑃𝑃 = 100 Pa ∙

1 atm = 9.87 × 10−4 atm 101325 Pa

24-19

1 2

3 1 1 1 0.001858 ∙ 900 K 2 ∙ � 3 2 g + g � 1 1 32.12 4.00 0.001858 ∙ T 2 ∙ � + � gmole gmole MA MB DAB = = 2 P ∙ σAB ΩD 9.87 × 10−4 atm ∙ 3.328 Å2 ∙ 0.6676

= 3645

cm2 s

c. Assess importance of Knudsen Diffusion Pressure = 1.0 atm

J ∙ 900 K K λ= = = 1.66 × 10−7 cm −10 2 2 √2πσA P √2π ∙ (4.08 × 10 m) 101325 Pa 1.38 × 10−23

κT

Kn =

dpore

6.76×10−7 cm 10×10−4 cm

Pressure = 100 Pa

= 1.66 × 10−4 < 1, Knudsen diffusion is not important.

J ∙ 900 K K λ= = = 1.68 × 10−4 cm √2πσA 2 P √2π ∙ (4.08 × 10−10 m)2 100 Pa κT

Kn =

dpore

1.38 × 10−23

1.68 × 10−4 cm = 0.168, 10 × 10−4 cm

Knudsen diffusion plays moderate role

d. Gas velocity for Pe = 5.0 x 10-4 T 900 K cm2 DKA = 4850 ∙ dpore ∙ � = 4850 ∙ 10 × 10−4 cm ∙ � = 25.67 g MA s 32.13 gmole P = 1.0 atm

1 1 1 1 1 = � + + �=� � 2 cm2 cm DAe DAB DKA 3.60 25.67 s s DAe = 3.157

cm2 s

24-20

Pe =

v∞∙dpore DAe

v∞ = 1.58

V̇ = v∞ 1

= �

Pe =

cm s

cm πd2 cm3 = 1.58 ∙ π ∙ (5 × 10−4 cm)2 = 1.24 × 10−6 s 4 s

P = 100 Pa DAe

= 5 × 10−4

DAB

v∞∙dpore DAe

DKA

= 5 × 10−4

∴ v∞ = 12.75 V̇ = 12.75

� , ∴ DAe = 25.49 cm2 /s

cm s

cm π cm3 −4 2 −5 ∙ ∙ (10 × 10 cm) = 1.00 × 10 s 4 s

24-21

24.13 A = CH4, B = H2O T = 673 K, dpore = 5 × 10−3 cm, v∞ = 4.0 cm/s MA = 16.04 Pe =

, MB = 18.02

gmole

v∞ dpore =2 DAe

gmole

a. Knudsen diffusion coefficient for CH4 T 673 K 2 DKA = 4850 ∙ dpore ∙ � = 4850 ∙ 5 × 10−3 cm ∙ � g = 157 cm /s MA 16.04 gmole DAB will dominate DAe since DKA is large

b. Total system pressure for Pe = 2.0 𝑃𝑃𝑃𝑃 =

𝑣𝑣∞ 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝐷𝐷𝐴𝐴𝐴𝐴

𝐷𝐷𝐴𝐴𝐴𝐴 =

DAe = 1 = DAe

𝑣𝑣∞ 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑃𝑃𝑃𝑃

cm 4.0 s ∙ 5 × 10−3 cm 2

cm2 = 0.01 s

1 1 1 1 1 = � + + �=� � 2 cm DAB DKA DAB 157 cm2 /s 0.01 s

DAB = 0.01

cm2 s

Fuller, Schettler, and Giddings Equation: DAB =

1 1 0.001 ∙ T1.75 ∙ (M + M )1/2 A B 1 1 P ∙ ((Σv)3A + (Σv)3B )2

24-22

(Σvi )A = 16.5 + 4 ∙ 1.98 = 24.42 (Σvi )B = 12.7 DAB =

1 1/2 g + g ) 16.05 18.02 gmole gmole

0.001 ∙ 673 K1.75 ∙ (

1 3

P((24.42)

0.001 ∙ 673 K1.75 ∙ (

1 + (12.7)3 )2

1 1/2 g + g ) 16.04 18.02 gmole gmole 1

cm2 0.01 ∙ ((24.42)3 + (12.7)3 )2 s

= 0.01

cm2 s

= 111.6 atm

24-23

24.14 A = albumin, B = H2O DAB = 5.94 × 10−7 cm2 /s, T = 293K, κ = 1.38 × 10−16 ergs/K 1 Pa ∙ s = 1

kg m∙s

1 erg = 1 g·cm2/s2 μB (293 K) = 993 × 10−6 Pa ∙ s ∙ � Stokes Einstein Equation: DAB =

κT 6πrA μB

1000 g 1m g �∙� � = 993 × 10−5 1kg 100 cm cm ∙ s

κT diameterA = rA ∙ 2 = 2 ∙ =2∙ 6πDAB μB = 7.27 nm

cm2 1.38 × 10−16 g ∙ 2 ∙ 293K s ∙K g cm2 6 ∙ π ∙ 5.94 × 10−7 ∙ 993 × 10−5 s cm ∙ s

This answer is close to the known value of 7.22 nm

24-24

24.15 Given: A = O2 (dissolved solute), B = H2O (solvent) T = 310 K Solute: VA = 25.6 cm3/gmole (Table 24.4)  1000 cP  Solvent: ΦB = 2.6, MB = 18.02 g/gmole, µ B =7.083 ×10−4 Pa ⋅ s   =0.7083 cP  1 Pa ⋅ s 

Wilke Chang Equation DAB

7.4 ×10−8 (Φ B M B )1/2 T = VA0.6 µB

7.4 ×10−8 (2.6 ⋅18.02 )1/2

( 25.6 )

0.6

310 0.7083

D= 3.17 ×10−5 cm 2 / s AB Hayduk and Laudie Equation (for nonelectrolytes in liquid H2O) DAB = 13.26 ⋅10−5 µ B−1.14VA−0.589 = 13.26 ×10−5 ( 0.7083)

−1.14

( 25.6 )

−0.589

DAB = 2.91×10−5 cm 2 /s

24-25

24.16 A = Benzene (C6H6, liquid), B = Ethanol (C2H6O, liquid) MA = 78.11 g/gmole, MB = 46.07 g/gmole

Interpolate and convert Pa∙s to cP from Appendix I to find viscosity of liquid benzene at 288K (59 F) lbm 0.45359 kg 1 ft μA (59℉) = 44.5 × 10−5 ∙� �� = 0.662 × 10−3 Pa ∙ s = 0.662 cP ft ∙ s lbm 0.3048 m From Perry’s Chemical Engineering Handbook, interpolate and convert units to cP: μB (288 K) = 1.34 cP

Using Table 24.4 and 24.5 VA = 6 ∙ 14.8 + 6 ∙ 3.7 − 15 (ring) = 96.0 VB = 2 ∙ 14.8 + 6 ∙ 3.7 + 7.4 = 59.2

ΦA = 1.0, ΦB = 1.5

cm3 gmole

Wilke Chang correlation (liquid benzene in ethanol) 7.4 × 10−8 (ΦB MB )1/2 T DAB = ∙ μB VA 0.6 DAB =

7.4 × 10−8 (1.5 ∙ 46.07 96

3 0.6

cm gmole

g 1/2 ) gmole

∙

288 K = 8.55 × 10−6 cm2 /s 1.34 cP

Schiebel correlation (liquid benzene-solute in ethanol-solvent) K

T 3VB 3 DAB = 1/3 ∙ , K = 8.2 × 10−8 ∙ (1 + � � ) μB VA V A

cm3 3 3 ∙ 59.2 K 288 K ⎛ gmole ⎞ −8 DAB = ∙ ⎜1 + � � ⎟ 1 ∙ 1.34 cP , K = 8.2 × 10 3 cm 3 96.0 cm 3 gmole �96.0 � gmole ⎝ ⎠ DAB = 9.65 × 10−6 cm2 /s

24-26

Wilke Chang correlation (liquid ethanol in benzene) DBA =

7.4 × 10−8 (ΦA MA )1/2 T ∙ μA VB 0.6

7.4 × 10−8 (1.0 ∙ 78.11 g/mole)1/2 288 K cm2 −5 DBA = ∙ = 2.46 × 10 59.2 cm3 /gmole0.6 0.662 cP s Schiebel correlation (liquid ethanol-solute in benzene-solvent) Note VB < 2VA, benzene is the solvent DBA = DBA =

1/3 VB

∙

T , K = 18.9 × 10−8 μA

18.9 × 10−8 59.2

288 K

1 ∙ 0.662 cP, 3 cm 3

gmole

DBA = 2.11 × 10−5 cm2 /s DAB ≠ DBA

24-27

24.17 A = CuCl2 (an ionic solute), B = H2O T = 298 K, ℱ = 96,500 C/gmole R = 8.316

J gmole ∙ K

λ°+ (Cu2+ ) = 108 λ°− (Cl− ) = 76.3

A ∙ cm2 V ∙ gmole

n+ (valence of cation) = 2

n− (valence of anion) = 1 Nernst-Haskell Equation

1 1 ( + + − )RT n n DAB = 1 1 ( ° + ° )(ℱ)2 λ+ λ−

1 1 J ( + )(8.316 ) ∙ 298K cm2 2 1 gmole ∙ K = 1.78 × 10−5 DAB = 1 1 s ( + )(96,500 C/gmole)2 A ∙ cm2 A ∙ cm2 108 76.3 V ∙ gmole V ∙ gmole

24-28

24.18 For liquids

DAB µ B =constant T

 T  µ (T )  DAB (T2 ) = DAB (T1 )  2  B 1   T1  µ B (T2 )  2 cm 2  303K   1.14 cP  −5 cm = × DAB (T2 ) = 1.28 × 10−5 1.93 10    s  288 K   0.797 cP  s

24-29

24.19 a. DAB for glucose in water A = C6H12O6, B = H2O Stokes-Einstein equation DAB =

𝑟𝑟𝐴𝐴 =

κT 6π rA µ B

1 1 𝑑𝑑𝐴𝐴 = (0.86 𝑛𝑛𝑛𝑛) = 4.3 𝑥𝑥10−8 𝑐𝑐𝑐𝑐 2 2

Appendix I, µΒ = 0.00826 g cm ⋅ s at 303 K

) (1.38 ×10 g ⋅ cm s ⋅ K ) ( 303 K= D= 6.25 ×10 6π ( 4.3 ×10 cm ) ( 0.00826 g cm ⋅ s ) −16

−8

−6

cm 2 s

b. DAe of glucose through the membrane

DAe = D o AB F1 (ϕ ) F2 (ϕ ) = ϕ

ds 0.86 nm = = 0.287 d pore 3.0 nm

F1 (ϕ ) = (1 − ϕ ) 2 = (1 − 0.287 ) =0.508 2

F2 (ϕ ) = 1 − 2.104ϕ + 2.09ϕ 3 − 0.95ϕ 5 = 1 − 2.104 ( 0.287 ) + 2.09 ( 0.287 ) − 0.95 ( 0.287 ) = 0.444 3

DAe = 1.41×10−6 cm 2 s ( 6.25 ×10−6 cm2 s ) ( 0.508)( 0.444 ) =

24-30

24.20 Given: o = 3.46 × 10−7 cm2/s DAB

ds = 12.38 nm, dpore = 100 nm Effective diffusion coefficient Solute diffusion through solvent-filled pore o DAe = DAB F1 (ϕ ) F2 (ϕ )

Reduced pore diameter = ϕ

ds 12.38 nm = = 0.124 100 nm d pore

Correction factors F1(φ), F2(φ)

F1 (ϕ ) = (1 − ϕ ) 2 = (1 − 0.124 ) =0.767 2

F2 (ϕ ) = 1 − 2.104ϕ + 2.09ϕ 3 − 0.95ϕ 5 = 1 − 2.104 ( 0.124 ) + 2.09 ( 0.124 ) − 0.95 ( 0.124 ) = 0.743 3

DAe = ( 3.46 × 10−7 cm 2 s ) ( 0.767 )( 0.743) = 1.97 × 10−7 cm 2 s

24-31

24.21 Find dpore when DAe = 5.0 x 10-7 cm2/s

DAe = D o AB F1 (ϕ ) F2 (ϕ ) DoAB =DAB = 1.19 x 10-6 cm2/s (dilute solution) F1 (ϕ ) F2 = (ϕ )

DAe 5.0 ×10−7 cm 2 s = = 0.420 DAB 1.19 ×10−6 cm 2 s

0.420 = (1 − ϕ ) (1 − 2.104ϕ + 2.09ϕ 3 − 0.95ϕ 5 ) 2

Solve for ϕ ϕ = 0.183

ϕ=

ds d pore

= d pore

3.6 nm = 19.7 nm 0.183

24-32

24.22 a. Determine molecular diffusion coefficients of solute A and solute B in water at 293 K Wilke-Chang Equation DA− H 2O µ H 2O 7.4 ×10−8 (φ M H 2O )1/2 = T VA0.6 From Appendix I, µH2O = 9.93 x 10-4 Pa ⋅ s = 0.993 cP at 293 K MH2O = 18.02 g/gmole For solute A DA − H 2 O =

7.4 x 10-8 (2.6×18.02 g/gmole)1/2

( 233.7 cm gmole ) 3

0.6

( 293 K ) = 5.67 x 10-6 cm 2 s ( 0.993 cP )

DA− H2O = 5.67 x 10-6 cm2/s For solute B DB − H 2O =

7.4 x 10-8 (2.6×18.02 g/gmole)1/2

( 347.4 cm gmole ) 3

0.6

( 293 K ) = 4.47 x 10-6 cm 2 s ( 0.993 cP )

DB − H2O = 4.47 x 10-6 cm2/s b. Determine dpore to achieve a separation factor of α = 3

ϕ=

ds d pore

DAe = D o A− H 2O F1 (ϕ ) F2 (ϕ ) F1 (ϕ )= (1 − ϕ ) 2

= α

F2 (ϕ ) = 1 − 2.104ϕ + 2.09ϕ 3 − 0.95ϕ 5

DAe DA− H2O F1 (ϕ A ) F2 (ϕ A ) = DBe DB − H2O F1 (ϕ B ) F2 (ϕ B )

24-33

ϕA =

2.0 nm d pore

ϕB =

3.0 nm d pore

DA− H2O (1 − ϕ A ) (1 − 2.104ϕ A + 2.09ϕ A3 − 0.95ϕ A5 ) 2

α=

DB − H2O (1 − ϕ B ) (1 − 2.104ϕ B + 2.09ϕ B 3 − 0.95ϕ B 5 ) 2

2 3 5  2.0 nm    2.0 nm   2.0 nm   2.0 nm   DA − H 2 O  1 −  1 − 2.104   + 2.09   − 0.95     d d d d  pore    pore   pore   pore    2 3 5  3.0 nm    3.0 nm   3.0 nm   3.0 nm   DB − H2O 1 −  1 − 2.104   + 2.09   − 0.95     d d d d  pore pore pore pore          

Trial-and-error: guess dpore and solve for α Guess dpore = 7.5 nm

( 5.67 ×10 α=

−6

2 3 5  2.0 nm    2.0 nm   2.0 nm   2.0 nm   cm s ) 1 −  1 − 2.104   + 2.09   − 0.95     7.5 nm    7.5 nm   7.5 nm   7.5 nm   2

2 3 5  3.0 nm    3.0 nm   3.0 nm   3.0 nm   2 −6 × − − + 4.47 10 cm s 1 1 2.104 2.09 − 0.95 ( )  7.5 nm           7.5 nm   7.5 nm   7.5 nm   

α = 3.20 Guess dpore = 7.88 nm

( 5.67 ×10 α=

−6

2 3 5 2.0 nm     2.0 nm   2.0 nm   2.0 nm   cm s ) 1 −  1 − 2.104   + 2.09   − 0.95     7.88 nm    7.88 nm   7.88 nm   7.88 nm   2

2 3 5 3.0 nm     3.0 nm   3.0 nm   3.0 nm   −6 2 × − − + 4.47 10 cm s 1 1 2.104 2.09 − 0.95 ( )  7.88 nm           7.88 nm   7.88 nm   7.88 nm   

α = 3.00 To achieve a separation factor (α) of 3, dpore = 7.88 nm

24-34

24.23

a. CA

= C A y= yA AC C A = ( 0.05 )

P RT

(1.5 atm )  -5 m ⋅ atm   8.206×10  ( 353K ) gmole ⋅ K   3

=2.58

gmole A m3

' b. DAe at T = 80 oC (353 K) and 1.5 atm

 Pref  cm 2  1.0 atm  cm 2 = DAB (T , P) D= =  0.10 AB (T , Pref )    0.0.067 s  1.5 atm  s  P  DKA = 4850 d pore

T 353 cm 2 −5 4850(1.5 × 10 ) 0.202 = = MA 46 s

1 1 1 = + = DAe DKA DAB

1 0.202

cm s

1 0.067

cm 2 s

cm 2 DAe = 0.050 s  cm 2  cm 2 ' 2 DAe DAe (0.5) 2  0.05 = = ε= 0.0125  s  s 

24-35

24.24 A = solute, B = solvent Find solvent viscosity to achieve a D'Ae of 1.0 x 10-7 cm2/s DAB = 2.04 x 10-7 cm2/s D ' Ae = ε 2 DAe D ' Ae 1.0 × 10 −7 cm 2 s = = = 2.04 × 10 −7 cm 2 s D Ae 2 2 ε ( 0.70 )

For hindered diffusion of a solute in a solvent-filled pore DAe = D o AB F1 (ϕ ) F2 (ϕ ) Hindered diffusion of the solute within the pore can be neglected so DAe ≅ DAB Stokes-Einstein Equation DAB =

κT 6π rA µ B

rA = 10 nm = 1 x 10-6 cm Rearrange to solve for µΒ K) (1.38 ×10 g ⋅ cm s ⋅ K ) ( 303 = = = 0.0109 g= cm ⋅ s 0.00109 Pa ⋅ s µ 6π D r 6π (1×10 cm )( 2.04 ×10 cm s )

κT

−16

−6

−7

AB A

24-36

24.25 a. CB inside pipe

= CB y= yB BC CA

P RT

( 3.0 atm ) gmole B 0.20 ) 25.2 (= 3 m3  −5 m ⋅ atm  K × 8.206 10 290 )  ( gmole ⋅ K 



b. DAB inside pipe

Fuller Schettler Giddings Equation Table 24.3

+ 24.92 ( Σv= ) A 16.5 + 4 ⋅ 1.98= ( Σv )B = 2 ⋅ 16.5 + 6 ⋅ 1.98 = 44.88 1/ 2

 1 1  + 0.001T   MA MB   = 1/ 3 1/ 3 2 P ( Σv ) A + ( Σv ) B 1.75

DAB

(

DAB = 0.0505

)

1/2

0.001 ⋅ ( 290 )

1.75

1 1   +   16 30 

( 3.0 ) ( ( 24.42 ) + ( 44.88) ) 1/3

1/3 2

cm 2 s

 Pref  cm 2  1.0 atm  cm 2 DAB (T , P) D= = =  0.10 AB (T , Pref )    0.0.067 s  1.5 atm  s  P  ' c. DAe in porous layer

 Pref  cm 2  3.0 atm  cm 2 = DAB (T , P ) D= T P = ( , ) 0.505 0.151  AB ref    s  1.0 atm  s  P  T 290 cm 2 = DKA 4850 = d pore 4850(0.0010) = 20.6 MA 16 s

24-37

1 1 1 = + = DAe DKA DAB DAe = 0.150

1 1 + 2 cm cm 2 20.6 0.151 s s

cm 2 s

 cm 2  cm 2 D DAe (0.3)  0.150 = = ε=  0.0135 s  s  ' Ae

24-38

24.26 A = carbon, B = solid iron DAB for carbon diffusing into fcc iron and bcc iron at 1000 K Arrhenius equation  Q  = DAB Do exp  −   RT 

DAB for carbon diffusing into fcc iron at 1000 K From Table 24.8, Do = 2.5 mm2/s and Q = 144.2 kJ/gmole = 144200 J/gmole 

  mm 2  144200 J/gmole 7.34 ×10−8 mm 2 / s = 7.34 ×10−10 cm 2 / s  =  exp  − s   ( 8.314 J/gmole ⋅ K )(1000 K ) 

DAB =  2.5 

DAB for carbon diffusing into bcc iron at 1000 K From Table 24.8, Do = 2.0 mm2/s and Q = 84.1 kJ/gmole = 84100 J/gmole    mm 2  84100 J/gmole 8.09 ×10−5 mm 2 / s = 8.09 ×10−7 cm 2 / s DAB =  =  2.0  exp  − s    ( 8.314 J/gmole ⋅ K )(1000 K ) 

24-39

24.27 a. Comparison of diffusion coefficients For As-Si, Do = 0.658 cm2/sec, Q = 348.1 kJ/gmole (348,100 J/gmole) For P-Si, Do = 11.1 cm2/sec, Q = 356.2 kJ/gmole (356,200 J/gmole) DAB Do exp(−Q / RT ) = At T = 600 oC = 873 K For As(A)-Si(B), DAB = ( 0.658cm 2 /sec ) (exp(-348,100 J/gmole)/((8.314J/gmole-K)(873 K))) =9.76 x10-22cm 2 /sec For P(A)-Si(B), DAB = (11.1 cm 2 /sec ) (exp(-356,200 J/gmole)/((8.314J/gmole-K)(873 K))) =5.39 x 10-21cm 2 /sec At T = 1000 oC = 1273 K For As(A)-Si(B), DAB = ( 0.658cm 2 /sec ) (exp(-348,100 J/gmole)/((8.314J/gmole-K)(1273 K))) =3.42 x 10-15cm 2 /sec For P(A)-Si(B), DAB = (11.1 cm 2 /sec ) (exp(-356,200 J/gmole)/((8.314J/gmole-K)(1273 K))) =2.69 x 10-14cm 2 /sec At 600 oC, the diffusion coefficients are ~106 lower relative to 1000 oC

b. At what temperature might DAs-Si = 0.12 DP-Si? Let Do , As − Si exp(−QAs − Si / RT= ) 0.12 Do ,P − Si exp(−QP − Si / RT )

 0.12 Do ,P − Si  QP − Si − QAs − Si = ln    D RT o , As − Si   T

( 356,200-348,100 ) J/gmole = 1382 K QP − Si − QAs − Si =  0.12 Do ,P − Si  8.314 J  0.12×11.1 cm 2 /sec  ln  R ln    2  D o , As − Si   gmole-K  0.658 cm /sec 

24-40

25.1

 ∂ρ ∇ ⋅ n A + A − rA = 0 ∂t If ρ and DAB are assumed constant,

  nA = − DAB ∇ρ A + ρ A v   ∇ n A = − DAB ∇ 2 ρ A + ∇ ⋅ ρ A v Substitution yields

 ∂ρ − DAB ∇ 2 ρ A + ∇ ⋅ ρ A v + A = rA ∂t

25-1

25.2 impermeable barrier for A y=H y x

Control Volume still mixture of A and B (initially uniform conc. at cA = cAo = 0)

cAs

y=0 x=0

CAs

x=L

a. Assumptions 1. 2. 3. 4. 5.

Unsteady state process - constant source for A, but control volume is sink for A No reaction of A in control volume (RA = 0) 2-D flux in x, y directions Dilute with respect to A No bulk flow in control volume

b. Differential Equation for Mass Transfer Rectangular Coordinate System Flux Equation By Assumption 4 ∂c ∂c c − DAB A + A ( N A, x + N B , x )  − DAB A N A, x = ∂x ∂x c ∂c ∂c c − DAB A + A ( N A, y + N B , y )  − DAB A N A, y = ∂y ∂y c ∂c ∇N A + RA =A ∂t Assumptions 1-3

∂N A, x ∂x

∂N A, y

∂c A = ∂y ∂t

Assumption 4 CA(x,y)

25-2

∂  ∂c A  ∂  ∂c A  ∂c A =  − DAB  +  − DAB ∂x  ∂x  ∂y  ∂y  ∂t  ∂ 2c ∂ 2c  ∂c A DAB  2A + 2A  = ∂y  ∂t  ∂x c. Boundary Conditions

y= 0, 0 ≤ x ≤ L, c A ( x,0)= c As y H , 0 ≤ x < L, =

∂c A ( x, H ) = 0 ∂y

∂c A ( 0, H ) = 0 ∂x x= L, 0 ≤ y < H , c A ( L, y= ) cAs x= 0, 0 ≤ y ≤ H ,

25-3

25.3 a. Diagram liquid MEK

vx(y)

x=0

thin polymer film

x=L

exiting MEK + dissolved polymer

b. Assumptions 1. 2. 3. 4. 5.

Steady-state process No homogenous reaction, RA = 0 Dilute with respect to solute A No fluid motion in y-direction (vy = 0) Bulk flow dominates in the x-direction

c. Flux Equation

− DAB N A, y =

∂C A ∂C + v yC A = − DAB A (vy = 0) ∂y ∂y

  y 2  ∂C A N A, x = − DAB + v x (y) CA with = v x ( y ) v max 1 −     H  ∂x   d. Differential Equation for Mass Transfer Rectangular Coordinate System

∂N A, y ∂N A, y   ∂N ∂C A −  A, x + +  + RA = ∂y ∂y  ∂t  ∂x By assumptions 1 and 2 ∂N A, x ∂x

∂N A, y ∂y

= 0

25-4

∂  ∂C A ∂C A   ∂  0 + v x (y)  +  − DAB =  − DAB ∂x  ∂x ∂y   ∂y 

 ∂ 2C A ∂ 2C A  ∂ CA DAB  + - v x (y) = 0 2 2  ∂y  ∂x  ∂x

  y 2  ∂ CA  ∂ 2C A ∂ 2C A  DAB  0 + =  - v max 1 −    2 ∂y 2   ∂x   H   ∂x If bulk flow dominates in the x-direction, the final model is

  y 2  ∂ CA ∂ 2C A DAB - v max 1 −    0 =   H   ∂x ∂y 2   e. Boundary Conditions x = 0, 0 < y < H, CA(0,y) = 0 y = 0, 0 < x < L, CA(x,0) = CA* y = H, 0 < x < L,

∂C A (x,0) =0 ∂y

25-5

25.4 a. General differential equation for O2 transfer in liquid film in terms of the fluxes Assumptions: 1) Steady-state 2) The process is dilute 3) Falling liquid film has a flat velocity profile with velocity vmax 4) Gas space always contains 100% oxygen, 5) W >> L 6) No homogenous reaction of O2 (RA = 0) 7) δ is constant Mass transfer only in x-direction and z-direction (rectangular coordinates) ∂c A  ∂N A, x ∂N A, y ∂N A, z  = + + RA  = ∂y ∂t ∂z   ∂x

∂N A, x

∂N A, z

0 = ∂x ∂z General flux equation for O2 in liquid film

∂c ∂c N A, x = − DAB A + vx∗c A = − DAB A (x-direction) ∂x ∂x ∂c ∂c N A, z = − DAB A + vz∗c A = − DAB A + vmax c A (z-direction) ∂z ∂z 𝜕𝜕𝑐𝑐𝐴𝐴 If 𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚 ≫ −𝐷𝐷𝐴𝐴𝐴𝐴 , then N A, z = vmax c A 𝜕𝜕𝜕𝜕

b. Differential equation in terms of the oxygen concentration cA

Insert flux equations into differential equation for mass transfer

∂c  ∂c    ∂  − DAB A  ∂  − DAB A + vmax c A  ∂x  ∂z  = +  0 ∂x ∂z

 ∂ 2cA ∂ 2cA  ∂c 0 DAB  2 + 2  − vmax A = ∂z  ∂z  ∂x c. Boundary conditions for the oxygen mass-transfer process 2 boundary conditions on x-direction, 1 boundary condition on z-direction 25-6

At z= 0, 0 ≤ x ≤ δ , c A ( x, 0)= 0 At x = 0, 0 ≤ z ≤ L, c A (0, z ) = c As = c∗A = p A / H At x = δ , 0 ≤ z ≤ L, N A (δ , z ) = 0 ∴

∂c A (δ , z ) = 0 ∂x

25-7

25.5

bulk gas phase z = L = 2.0 cm 0.01 cm diameter channels catalytic surface A (g) → B (g) z=0 inert support a. Assumptions and simplification of General Differential Equation for Mass Transfer Physical system: gas space inside pore Source for A: bulk gas phase Sink for A: reaction of A to B at catalyst surface on cylindrical walls of pore Assumptions: 1. Constant source and sink – steady-state process 2. No homogeneous reaction of A within control volume 3. Rapid heterogeneous surface reaction so that flux of A to catalyst surface is diffusion limited with cAs = 0 at r = R for all z 4. Two-dimensional flux in r and z direction, given orientation of source and sink for A 5. Binary mixture of A and B, not dilute yA∞ = 0.60 yB∞ = 0.40 z = L = 2.0 cm

cAs = 0

CA(r,z)

impermeable barrier

z=0 r = R = 0.005 cm r=0

25-8

General Differential Equation for Mass Transfer in terms of NA ∂c ∇N A + RA =A ∂t

∂c A By assumptions 1 and 2, = RA 0 0,= ∂t ∂N A, z 1 ∂ 0 rN A,r ) + = ( r ∂r ∂z b. General Flux Equations and Differential Equation for Mass Transfer in terms of cA ∂c c − DAB A + A ( N A,r + N B ,r ) N A, r = ∂r c Note N A,r = − N B ,r ∂c A ∂r ∂c N A,z = − DAB A ∂z

N A,r = − DAB

Insert flux equations into differential equation for mass transfer

1 ∂  ∂c A  ∂  ∂c A   0 −  −rDAB  +  − DAB  +0= ∂r  ∂z  ∂z    r ∂r   ∂ 2 c A 1 ∂c A ∂ 2 c A  0 DAB  2 + + 2 = r ∂r ∂z   ∂r c. Boundary conditions for gas phase concentration cA

z= L, 0 ≤ r ≤ R, c A ( r , L= ) c A, ∞

bulk gas

∂c A ( r , 0 ) = 0 impermeable barrier ∂z ∂c A ( 0, z ) r= 0, 0 ≤ z ≤ L, = 0 symmetry condition ∂r r = R, 0 ≤ z ≤ L, c A ( R, z ) = c As = 0 rapid surface reaction z= 0, 0 ≤ r ≤ R

25-9

25.6 a. General Differential Equation for Mass Transfer Coordinate System: x, y, z Assumptions: 1) Rapid reaction, 𝑐𝑐𝐴𝐴𝐴𝐴 = 0 2) Gas velocity profile is flat, 𝑣𝑣𝑥𝑥 (𝑦𝑦) = 𝐯𝐯 3) Gas velocity (𝑣𝑣) is slow (doesn’t dominate over diffusion flux in “x”) 𝜕𝜕𝑐𝑐 4) Steady state process, constant source and sink for CO, 𝐴𝐴 = 0 𝜕𝜕𝜕𝜕 5) RA = 0 (no homogeneous reaction 6) 2-D flux in x and y 7) Dilute with respect to A Source: Inlet gas Sink: Catalyst surface −∇𝑁𝑁𝐴𝐴 + 𝑅𝑅𝐴𝐴 = 𝑁𝑁𝐴𝐴,𝑥𝑥 = −𝐷𝐷𝐴𝐴𝐴𝐴 𝑁𝑁𝐴𝐴,𝑦𝑦 = −𝐷𝐷𝐴𝐴𝐴𝐴 ∴

𝜕𝜕𝑐𝑐𝐴𝐴 , 𝜕𝜕𝜕𝜕

∴ −�

𝜕𝜕𝑐𝑐𝐴𝐴 + 𝑐𝑐𝐴𝐴 ∙ 𝑣𝑣 𝜕𝜕𝜕𝜕

𝜕𝜕𝑁𝑁𝐴𝐴,𝑥𝑥 𝜕𝜕𝑁𝑁𝐴𝐴,𝑦𝑦 + �=0 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

𝜕𝜕𝑐𝑐𝐴𝐴 𝜕𝜕𝜕𝜕

𝜕𝜕 𝜕𝜕𝑐𝑐𝐴𝐴 𝜕𝜕 𝜕𝜕𝑐𝑐𝐴𝐴 �−𝐷𝐷𝐴𝐴𝐴𝐴 + 𝑐𝑐𝐴𝐴 ∙ 𝑣𝑣� + �−𝐷𝐷𝐴𝐴𝐴𝐴 �=0 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

𝐷𝐷𝐴𝐴𝐴𝐴 �

𝜕𝜕 2 𝑐𝑐𝐴𝐴 𝜕𝜕 2 𝑐𝑐𝐴𝐴 𝜕𝜕𝑐𝑐𝐴𝐴 + ∙ 𝑣𝑣 = 0 �− 2 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

b. Boundary Conditions x = 0, 𝑐𝑐𝐴𝐴 (0, 𝑦𝑦) = 𝑐𝑐𝐴𝐴0

x = L, 𝑐𝑐𝐴𝐴 (L,y) = 𝑐𝑐𝐴𝐴𝐴𝐴 y=0, y=

𝐻𝐻 2

𝑐𝑐𝐴𝐴 (𝑥𝑥, 0) ≅ 0 𝐻𝐻

𝜕𝜕𝜕𝜕𝐴𝐴 (𝑥𝑥, 2 ) =0 𝜕𝜕𝜕𝜕

25-10

25.7 𝑘𝑘1

𝐴𝐴 → 𝐵𝐵, 𝑅𝑅𝐴𝐴 = −𝑘𝑘1 𝑐𝑐𝐴𝐴 , 𝑘𝑘𝑠𝑠

𝐴𝐴 → 2𝐶𝐶, 𝑟𝑟𝐴𝐴,𝑠𝑠 = −𝑘𝑘𝑠𝑠 𝑐𝑐𝐴𝐴𝐴𝐴 ,

1 𝑘𝑘1 [ ] 𝑠𝑠 𝑘𝑘𝑠𝑠 [

𝑐𝑐𝑐𝑐 ] 𝑠𝑠

4 species: A=1, B=2, C=3, D=4 (inert) ( 𝑅𝑅1 = −𝑘𝑘1 𝑐𝑐1 , 𝑟𝑟1,𝑠𝑠 = −𝑘𝑘𝑠𝑠 𝑐𝑐1𝑠𝑠 )

a. Assumptions, Source and Sink for species 1 (A) Assumptions: 1) Dilute process with respect to species 1 2) Constant source and sink for species 1 (steady state) 3) Right side and top side are impermeable barriers 4) 2-D flux of species 1 for catalyst II (source and sink antiparallel)

Source: Flowing fluid containing reactant 1, of constant concentration (𝑐𝑐1,∞ ) Sinks: First-order homogeneous reaction of 1 in porous layer (catalyst I) First-order heterogeneous surface reaction of 1 at nonporous boundary surface (catalyst II) b. Differential model for C1(x,y) (Shell balance on differential volume element for species 1) IN – OUT + GEN = ACC = 0

𝑛𝑛1,𝑥𝑥 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑖𝑖𝑖𝑖 𝑥𝑥 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑤𝑤 = 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 1 𝑖𝑖𝑖𝑖 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

[𝑛𝑛1,𝑥𝑥 ∆𝑦𝑦𝑦𝑦|𝑥𝑥,𝑦𝑦� + 𝑛𝑛1,𝑦𝑦 ∆𝑥𝑥𝑥𝑥|𝑥𝑥̅ ,𝑦𝑦 ] − [𝑛𝑛1,𝑥𝑥 ∆𝑦𝑦𝑦𝑦|𝑥𝑥+∆𝑥𝑥,𝑦𝑦� + 𝑛𝑛1,𝑦𝑦 ∆𝑥𝑥𝑥𝑥|𝑥𝑥̅ ,𝑦𝑦+∆𝑦𝑦 ] + 𝑟𝑟1 ∆𝑥𝑥∆𝑦𝑦𝑦𝑦 = 0 ÷ ∆𝑥𝑥, ∆𝑦𝑦; 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 lim ∆𝑥𝑥 → 0, 𝑡𝑡𝑎𝑎𝑘𝑘𝑘𝑘 lim ∆𝑦𝑦 → 0 ; ÷ 𝑤𝑤

25-11

−

𝜕𝜕𝑛𝑛1𝑦𝑦 𝜕𝜕𝑛𝑛1,𝑥𝑥 +− + 𝑟𝑟1 = 0 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

General Flux Equation

x-direction: 𝑛𝑛1,𝑥𝑥 = −𝐷𝐷1

y-direction: 𝑛𝑛1,𝑦𝑦 = −𝐷𝐷1

𝜕𝜕𝐶𝐶1 𝜕𝜕𝜕𝜕

(𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑒𝑒 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 ≈ 0, 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠)

Homogeneous reaction: 𝑟𝑟1 = −𝑘𝑘1 𝑐𝑐1 Combine: 𝐷𝐷1 �

𝜕𝜕2 𝐶𝐶1 𝜕𝜕𝑥𝑥

2 +

𝜕𝜕2 𝐶𝐶1 𝜕𝜕𝑦𝑦 2

� − 𝑘𝑘1 𝑐𝑐1 = 0, 𝑐𝑐1 (𝑥𝑥, 𝑦𝑦)

c. Boundary Conditions, 4 required: (2 for x, 2 for y) 𝑦𝑦 = 0, 0 ≤ 𝑥𝑥 ≤ 𝐿𝐿,

𝑐𝑐1 (𝑥𝑥, 0) = 𝑐𝑐1,∞

𝑦𝑦 = 𝐻𝐻, 0 ≤ 𝑥𝑥 ≤ 𝐿𝐿,

𝑛𝑛1 (𝑥𝑥, 𝐻𝐻) = 0,

𝑥𝑥 = 0, 0 < 𝑦𝑦 < 𝐻𝐻,

𝑟𝑟1,𝑠𝑠 = 𝑛𝑛1,𝑠𝑠 ,

𝑥𝑥 = 𝐿𝐿, 0 < 𝑦𝑦 < 𝐻𝐻,

𝑛𝑛1 (𝐿𝐿, 𝑦𝑦) = 0

∴

𝜕𝜕𝑐𝑐1 (𝑥𝑥, 𝐻𝐻) =0 𝜕𝜕𝜕𝜕

−𝑘𝑘𝑠𝑠 𝑐𝑐1,𝑠𝑠 = −𝐷𝐷1 ∴

𝜕𝜕𝑐𝑐1 (𝐿𝐿,𝑦𝑦) 𝜕𝜕𝜕𝜕

𝜕𝜕𝑐𝑐1 (0, 𝑦𝑦) , 𝜕𝜕𝜕𝜕

∴

𝜕𝜕𝑐𝑐1 (0, 𝑦𝑦) 𝑘𝑘𝑠𝑠 = 𝑐𝑐 (0, 𝑦𝑦) 𝜕𝜕𝜕𝜕 𝐷𝐷1 1

25-12

25.8

well-mixed flowing bulk fluid (cAo)

A dead tissue (B)

clump of unhealthy cells inert surface

A→D

r = R1

r = R2

a. Define physical system and assumptions Physical system: dead tissue between bulk fluid and unhealthy cells Source for A: bulk liquid phase Sink for A: reaction of A inside dead cells, flux of A into unhealthy cells 1. 2. 3. 4. 5. 6.

Constant source and sink – steady-state process First-order homogeneous reaction of species A within dead tissue (control volume) Dilute process with respect to species A and D One-dimensional flux in r direction, given orientation of source and sink for species A R1 and R2 constant No reaction of A at boundary surface (r = R1)

b. Differential Forms of General Flux Equation for species A

Fick’s Flux Equation

N A , r = DA − m

dc A c A + ( N A, r + N B , r + N D , r ) dr c

By assumption 3, dilute process with c A / c → 0 , also NB,r = 0 dc N A,r = DAB A dr c. General Differential Equation for Mass Transfer in terms of NA and cA

25-13

−∇N A + RA =

dc A dt

By assumptions 1 and 2, −

(

∂c A = 0, RA = −k c A ∂t

)

d 2 r N A, r − k c A = 0 dr

Insert simplified form of General Flux Equation dc A  1 d  2 − 2 0  −r DAB  − k cA = r dr  dr   d 2 c A 2 dc A  DAB  2 + 0  − k cA = r dr   dr d. Boundary conditions for species A and D = r R= c Ao 2 , cA

bulk fluid

= r R= c As 1 , cA

active concentration for drug A at interface to unheathy tissue

= r R= 0 2 , cD

bulk fluid

dcD 0 = r R= 1, dr

impermeable barrier to D

25-14

25.9 a. System for Analysis and Assumptions System: Tissue in quadrant IV (A = dissolved O2, B = tissue) Assumptions: 1) Steady state O2 transport, constant source and sink 2) 2-D flux along x and y, via orientation of source and sink 3) Homogeneous first-order reaction 𝑅𝑅𝐴𝐴 = −𝑘𝑘1 𝑐𝑐𝐴𝐴 4) Dilute solution, tissue acts as a homogeneous medium approximating water 5) Constant dimensions (𝐿𝐿1 , 𝐿𝐿2 , 𝐻𝐻1 , 𝐻𝐻2 , 𝑑𝑑), constant 𝐷𝐷1 Sources for O2: O2 gas in duct, liquid surrounding monolith

Sink for O2: homogeneous O2 consumption within tissue Differential Model (System IV, rectangular geometry, mass concentration equation) −� −�

𝜕𝜕𝑁𝑁𝐴𝐴,𝑥𝑥 𝜕𝜕𝑁𝑁𝐴𝐴,𝑦𝑦 𝜕𝜕𝑁𝑁𝐴𝐴,𝑧𝑧 𝜕𝜕𝑐𝑐𝐴𝐴 + + � + 𝑅𝑅𝐴𝐴 = =0 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝑁𝑁𝐴𝐴,𝑥𝑥 𝜕𝜕𝑁𝑁𝐴𝐴,𝑦𝑦 + � − 𝑘𝑘1 𝑐𝑐𝐴𝐴 = 0 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

General Flux Equation, Species A 𝑁𝑁𝐴𝐴,𝑥𝑥 = −𝐷𝐷𝐴𝐴𝐴𝐴 𝑁𝑁𝐴𝐴,𝑦𝑦 = −𝐷𝐷𝐴𝐴𝐴𝐴

𝜕𝜕𝑐𝑐𝐴𝐴 , 𝜕𝜕𝜕𝜕

𝑑𝑑𝑑𝑑𝐴𝐴 , 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠: 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠:

𝑐𝑐𝐴𝐴 ≅0 𝑐𝑐

Combine mass conservation equation and general flux equation 𝜕𝜕 2 𝑐𝑐𝐴𝐴 𝜕𝜕 2 𝑐𝑐𝐴𝐴 𝐷𝐷𝐴𝐴𝐴𝐴 � 2 + � − 𝑘𝑘 𝑐𝑐𝐴𝐴 = 0, 𝜕𝜕𝑥𝑥 𝜕𝜕𝜕𝜕 2 b. Boundary Conditions

(𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑡𝑡𝑡𝑡 𝑎𝑎𝑎𝑎𝑎𝑎 𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞 𝐼𝐼 − 𝐼𝐼𝐼𝐼)

Symmetry between quadrants I and II & III and IV System IV 𝑥𝑥 = 0,

0 ≤ 𝑦𝑦 ≤ 𝐻𝐻1 ,

𝑁𝑁𝐴𝐴 (0, 𝑦𝑦) = 0,

∴

𝜕𝜕𝜕𝜕𝐴𝐴 (0, 𝑦𝑦) =0 𝜕𝜕𝜕𝜕

25-15

𝑥𝑥 = 𝐿𝐿,

0 ≤ 𝑦𝑦 ≤ 𝐻𝐻1 ,

𝑦𝑦 = 𝐻𝐻1 ,

0 ≤ 𝑥𝑥 ≤ 𝐿𝐿1 ,

𝑦𝑦 = 0,

0 ≤ 𝑥𝑥 ≤ 𝐿𝐿1 ,

𝑐𝑐𝐴𝐴 (𝐿𝐿, 𝑦𝑦) =

𝑃𝑃𝐴𝐴 = 𝑐𝑐𝐴𝐴∗ 𝐻𝐻

𝑐𝑐𝐴𝐴 (𝑥𝑥, 0) = 𝑐𝑐𝐴𝐴,∞ 𝑃𝑃𝐴𝐴 𝑐𝑐𝐴𝐴 (𝑥𝑥, 𝐻𝐻1 ) = = 𝑐𝑐𝐴𝐴∗ 𝐻𝐻

System III 𝑥𝑥 = 𝐿𝐿1 + 𝑑𝑑, 𝑥𝑥 = 𝐿𝐿2 , 𝑦𝑦 = 0,

𝑦𝑦 = 𝐻𝐻1 ,

0 ≤ 𝑦𝑦 ≤ 𝐻𝐻1 , 0 ≤ 𝑦𝑦 ≤ 𝐻𝐻1 ,

𝐿𝐿1 + 𝑑𝑑 ≤ 𝑥𝑥 ≤ 𝐿𝐿2 ,

𝑐𝑐𝐴𝐴 (𝐿𝐿1 + 𝑑𝑑, 𝑦𝑦) =

𝜕𝜕𝜕𝜕𝐴𝐴 (𝐿𝐿2 , 𝑦𝑦) =0 𝜕𝜕𝜕𝜕

𝐿𝐿1 + 𝑑𝑑 ≤ 𝑥𝑥 ≤ 𝐿𝐿2 ,

𝑃𝑃𝐴𝐴 = 𝑐𝑐𝐴𝐴∗ 𝐻𝐻

𝑐𝑐𝐴𝐴 (𝑥𝑥, 0) = 𝑐𝑐𝐴𝐴,∞ 𝑐𝑐𝐴𝐴 (𝑥𝑥, 𝐻𝐻1 ) =

𝑃𝑃𝐴𝐴 = 𝑐𝑐𝐴𝐴∗ 𝐻𝐻

25-16

25.10 A = C6H6, B = H2O (liq) T = 298 K, 𝑐𝑐𝐴𝐴,∞ 𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

a. Differential model for cA(r,t) Assumptions: 1) Unsteady state (control volume is also the sink) 2) No homogeneous reaction (𝑅𝑅𝐴𝐴 = 0) 3) Dilute with respect to A 4) 1-D flux along r (unimolecular diffusion) General Differential Equation ′ 𝑁𝑁𝐴𝐴,𝑟𝑟 ≅ −𝐷𝐷𝐴𝐴𝐴𝐴

𝜕𝜕𝑐𝑐𝐴𝐴 , 𝜕𝜕𝜕𝜕

(𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)

Mass Conservation Equation −∇𝑁𝑁𝐴𝐴 =

𝜕𝜕𝑐𝑐𝐴𝐴 , (𝑅𝑅𝐴𝐴 = 0), 𝜕𝜕𝜕𝜕

Combine Equations

𝜕𝜕𝑐𝑐𝐴𝐴 1 ′ 𝜕𝜕 2 𝜕𝜕𝑐𝑐𝐴𝐴 𝐷𝐷𝐴𝐴𝐴𝐴 �𝑟𝑟 , �= 2 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

∴ ∇𝑁𝑁𝐴𝐴 =

1 𝜕𝜕 2 �𝑟𝑟 𝑁𝑁𝐴𝐴,𝑟𝑟 �, 𝑟𝑟 2 𝜕𝜕𝜕𝜕

′ 𝑜𝑜𝑜𝑜 𝐷𝐷𝐴𝐴𝐴𝐴 �

𝑐𝑐𝐴𝐴 (𝑟𝑟, 𝑡𝑡)

𝜕𝜕 2 𝑐𝑐𝐴𝐴 2 𝜕𝜕𝜕𝜕𝐴𝐴 𝜕𝜕𝑐𝑐𝐴𝐴 + �= 2 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

b. Boundary and Initial Conditions IC: 𝑡𝑡 = 0,

BC:

𝑟𝑟 = 0,

𝑐𝑐𝐴𝐴 (𝑟𝑟, 0) = 𝑐𝑐𝐴𝐴0 = 0

𝜕𝜕𝑐𝑐𝐴𝐴 (0, 𝑡𝑡) =0 𝜕𝜕𝜕𝜕

𝑟𝑟 = 𝑅𝑅, 𝑐𝑐𝐴𝐴 (𝑅𝑅, 𝑡𝑡) = 𝑐𝑐𝐴𝐴𝐴𝐴

25-17

25.11 a. System A = phenol, B = H2O (A = species 1 in analysis below) Adsorbent Surface = carbon Ong, S. (1984). Simplified Approach for Designing Carbon Adsorption Column. J. Environ. Eng., 110(6), 1184–1188.

b. Differential model for cA(z,t) Assumptions: 1) Unsteady state 2) 1-D flux along z 3) Dilute with respect to species 1 (A) 4) No homogeneous reaction (𝑅𝑅𝐴𝐴 = 0) 5) Linear adsorption isotherm is valid 6) Fast adsorption General Flux Equation 𝑁𝑁𝐴𝐴,𝑥𝑥 = −𝐷𝐷𝐴𝐴𝐴𝐴

𝜕𝜕𝑐𝑐𝐴𝐴 𝜕𝜕𝜕𝜕

Differential volume element:

25-18

Mass Conservation Equation Shell balance on Species 1 (A) IN – OUT + GEN = ACC 𝑁𝑁𝐴𝐴

÷ 𝜋𝜋,

𝑑𝑑2 , ∆𝑧𝑧 4

−

−

𝜕𝜕𝑁𝑁𝐴𝐴 (𝑧𝑧, 𝑡𝑡) 4 𝜕𝜕𝑞𝑞𝐴𝐴 𝜕𝜕𝑐𝑐𝐴𝐴 = + 𝜕𝜕𝜕𝜕 𝑑𝑑 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

lim∆𝑡𝑡 → 0

𝑞𝑞𝐴𝐴 =

𝑞𝑞𝐴𝐴,𝑚𝑚𝑚𝑚𝑚𝑚 𝑐𝑐𝐴𝐴 , 𝐾𝐾

∴

𝜕𝜕𝑞𝑞𝐴𝐴 𝑞𝑞𝐴𝐴,𝑚𝑚𝑚𝑚𝑚𝑚 𝜕𝜕𝑐𝑐𝐴𝐴 = 𝜕𝜕𝜕𝜕 𝐾𝐾 𝜕𝜕𝜕𝜕

𝜕𝜕 2 𝑐𝑐𝐴𝐴 4 𝑞𝑞𝐴𝐴,𝑚𝑚𝑚𝑚𝑚𝑚 𝜕𝜕𝜕𝜕𝐴𝐴 (𝑧𝑧, 𝑡𝑡) = � + 1� 𝐷𝐷𝐴𝐴𝐴𝐴 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 2 𝑑𝑑 𝐾𝐾

𝐷𝐷𝐴𝐴𝐴𝐴 𝜕𝜕 2 𝑐𝑐𝐴𝐴 (𝑧𝑧, 𝑡𝑡) 𝜕𝜕𝜕𝜕𝐴𝐴 (𝑧𝑧, 𝑡𝑡) � 𝑞𝑞 � = 2 4 𝐴𝐴,𝑚𝑚𝑚𝑚𝑚𝑚 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 +1 𝑑𝑑 𝐾𝐾

Boundary and Initial Conditions 𝑧𝑧 = 0,

𝑐𝑐1 (0, 𝑡𝑡) = 𝑐𝑐1,𝑠𝑠 25-19

𝑧𝑧 = 𝐿𝐿,

𝑡𝑡 = 0,

𝜕𝜕𝜕𝜕1 (𝐿𝐿, 𝑡𝑡) =0 𝜕𝜕𝜕𝜕 𝑐𝑐1 (𝑧𝑧, 0) = 𝑐𝑐1,0 = 0

25-20

25.12 Sealed Ends

Catalyst Surface

Porous Layer

Bulk Gas (cA = cA,∞)

A B

L = 5.0 cm

r=0

r = Ro (0.5 cm)

1.5 atm, 150 oC 4 mole% C2H4 (A) 94 mole% H2 (B) 2 mole% C2H6 (C)

r = R1

a. Species A, B, and C are undergoing mass transfer, as each has a source and sink b. Assumptions 1. 2. 3. 4. 5.

Steady-state, constant source and sink No homogeneous reaction within the control volume, RA = 0 1-D flux along coordinate r Dilute process with respect to species A Rapid reaction of A at catalyst surface

c. General Differential Equation for Mass Transfer (cylindrical system) Cylindrical Coordinate System By assumptions 1 and 2 ∂N A, z  1 ∂ ∂c A rN A, r ) + − (  + RA = ∂z  ∂t  r ∂r 1 d ( r ⋅ N A, r ) = 0 r dr d ( r ⋅ N A, r ) = 0 dr

By continuity N A,r r = R ⋅ R = N A,r ⋅ r NA,r r constant along r Flux Equation, 1-D flux along r

25-21

dc c dc c N A, r = − DAB A + A ( N A, r + N B , r + N C , r ) = − DAB A + A ( N A, r − N A, r + 0 + N A, r ) dr C dr C − DAB dc A N A, r = 1 − y A dr N A, r = − DAB

dc A (if dilute with respect to A) dr

d. Boundary Conditions r = Ro, cA = 0 (rapid reaction at surface) r = R1, cA = cA,∞ (bulk fluid)

25-22

25.13

flow A+B cAo cAo

z=0

A →B

z=L porous disk with sides and bottom sealed

impermeable barrier

r=0 r=R

a. Assumptions 1. Steady state process – constant source and sink for A and B 2. Homogeneous, first-order reaction of A within control volume 3. 1-D flux along z direction (sealed circumference and base)

b. General Flux Equation dc c dc c − DAB A + A ( N A, z + N B , z ) = − DAB A + A ( N A, z − N A, z ) N A, z = dz C dz C dc N A, z = − DAB A (equilimolar counter diffusion with respect to A and B) dz

c. Differential Equation for Mass Transfer ∂N A, z  1 ∂ ∂c A rN A, r ) + − (  + RA = ∂z  ∂t  r ∂r ∂N − A , z + RA = 0 ∂z

Homogeneous, first-order reaction, RA = −kc A DAB

d 2 cA − k cA = 0 dz 2

d. Boundary Conditions z = 0, cA = cAo z = L,

dc A =0 dz 25-23

26.1 A = SiH4 vapor, B = H2 gas 𝑇𝑇 = 900 𝐾𝐾, 𝑃𝑃 = 70 𝑃𝑃𝑃𝑃 (6.91 𝑥𝑥 10−4 𝑎𝑎𝑎𝑎𝑎𝑎), 𝛿𝛿 = 5.0 𝑐𝑐𝑐𝑐, 𝑦𝑦𝐴𝐴𝐴𝐴 = 0.20, 𝜎𝜎𝐴𝐴 = 4.08 Å, 𝜀𝜀𝐴𝐴 = 207.6 𝐾𝐾, 𝜎𝜎𝐵𝐵 = 2.968 Å, 𝜀𝜀𝐵𝐵 /𝜅𝜅 = 33.3 𝐾𝐾 𝜅𝜅 𝑔𝑔 𝜌𝜌𝑆𝑆𝑆𝑆 = 2.32 3 𝑐𝑐𝑐𝑐 a. Estimate the rate of Si film formation 𝑦𝑦𝐴𝐴𝐴𝐴 ≈ 0 𝑃𝑃 = 𝑐𝑐 = 𝑅𝑅𝑅𝑅

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 70 𝑃𝑃𝑃𝑃 −3 = 9.355 × 10 𝑚𝑚3 ∙ 𝑃𝑃𝑃𝑃 𝑚𝑚3 8.314 ∙ 900 𝐾𝐾 𝐾𝐾 ∙ 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 Determine DAB 1 1 2 0.001858 ∙ 𝑇𝑇 3/2 �𝑀𝑀 + 𝑀𝑀 � 𝐴𝐴 𝐴𝐴 𝐷𝐷𝐴𝐴𝐴𝐴 = 2 𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴 ΩD σAB = (2.968 + 4.08)/2 = 3.524 Å 𝜀𝜀𝐴𝐴𝐴𝐴 = √207.6 𝐾𝐾 ∙ 33.3 𝐾𝐾 = 83.145

900 𝐾𝐾 𝜅𝜅𝜅𝜅 = = 10.82, ∴ Ω𝐷𝐷 = 0.73597 𝜀𝜀𝐴𝐴𝐴𝐴 83.145 𝐾𝐾 𝐷𝐷𝐴𝐴𝐴𝐴 =

1 1 0.001858 ∙ 900 𝐾𝐾 3/2 [2.02 + 32.12]1/2 6.91 × 10−4 ∙ 3.524 2 ∙ 0.73597

Material balance on Si

= 5672

𝑐𝑐𝑐𝑐2 𝑠𝑠

IN – OUT + GEN = ACC dm 0 − N A ⋅ S + 0 = Si dt 𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 𝑁𝑁𝐴𝐴 = ln(1 + 𝑦𝑦𝐴𝐴𝐴𝐴 ) 𝛿𝛿 𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 𝜌𝜌𝑆𝑆𝑆𝑆 𝑑𝑑ℓ ln(1 + 𝑦𝑦𝐴𝐴𝐴𝐴 ) = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝛿𝛿 ≫ ℓ 𝛿𝛿 𝑀𝑀𝑆𝑆𝑆𝑆 𝑑𝑑𝑑𝑑 𝑔𝑔 𝑐𝑐𝑐𝑐2 −9 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 9.355 × 10 ∙ 5672 ∙ 28.09 ∙ ln(1 + 0.2) 3 𝑑𝑑ℓ 𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 𝑀𝑀𝑆𝑆𝑆𝑆 𝑠𝑠 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑐𝑐𝑐𝑐 = ln(1 + 𝑦𝑦𝐴𝐴𝐴𝐴 ) = 𝑔𝑔 𝑑𝑑𝑑𝑑 𝛿𝛿𝜌𝜌𝑆𝑆𝑆𝑆 5.0 𝑐𝑐𝑐𝑐 ∙ 2.32 3 𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐 𝜇𝜇𝜇𝜇 −5 = 2.34 𝑥𝑥 10 = 1406 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚

26-1

b. Estimate the rate of Si film formation with ks 𝑘𝑘𝑠𝑠 = 1.25

𝑐𝑐𝑐𝑐 𝑠𝑠

ln(1 + 𝑦𝑦) ≅ 𝑦𝑦

𝑁𝑁𝐴𝐴 =

𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 (𝑦𝑦𝐴𝐴𝐴𝐴 − 𝑦𝑦𝐴𝐴𝐴𝐴 ) 𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 1 + 𝑦𝑦𝐴𝐴𝐴𝐴 ln � �= 𝛿𝛿 1 + 𝑦𝑦𝐴𝐴𝐴𝐴 𝛿𝛿

𝑦𝑦𝐴𝐴𝐴𝐴 =

𝑁𝑁𝐴𝐴 𝑘𝑘𝑠𝑠 𝑐𝑐

𝑁𝑁𝐴𝐴 = 𝑘𝑘𝑠𝑠 𝑐𝑐𝐴𝐴𝐴𝐴 = 𝑘𝑘𝑠𝑠 𝑐𝑐𝑐𝑐𝐴𝐴𝐴𝐴

𝑁𝑁𝐴𝐴 =

𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 𝑦𝑦𝐴𝐴𝐴𝐴 𝐷𝐷 𝛿𝛿 �1 + 𝐴𝐴𝐴𝐴 � 𝛿𝛿𝑘𝑘𝑠𝑠

Material balance on Si IN – OUT + GEN = ACC 𝑁𝑁𝐴𝐴 𝑆𝑆 − 0 + 0 =

𝑑𝑑𝑑𝑑𝑆𝑆𝑆𝑆 𝑑𝑑𝑑𝑑

𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 𝑦𝑦𝐴𝐴𝐴𝐴 𝜌𝜌𝑆𝑆𝑆𝑆 𝑑𝑑ℓ 𝑆𝑆 = 𝑆𝑆 𝐷𝐷𝐴𝐴𝐴𝐴 𝑀𝑀 𝑑𝑑𝑑𝑑 𝑆𝑆𝑆𝑆 𝛿𝛿 �1 + � 𝛿𝛿𝑘𝑘𝑠𝑠 𝑑𝑑ℓ 𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 𝑦𝑦𝐴𝐴𝐴𝐴 𝑀𝑀𝑆𝑆𝑖𝑖 = 𝑑𝑑𝑑𝑑 𝛿𝛿 �1 + 𝐷𝐷𝐴𝐴𝐴𝐴 � 𝜌𝜌 𝛿𝛿𝑘𝑘𝑠𝑠 𝑆𝑆𝑆𝑆

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔 𝑐𝑐𝑐𝑐2 ∙ 5672 ∙ 0.2 ∙ 28.09 3 𝑑𝑑ℓ 𝑐𝑐𝑐𝑐 𝜇𝜇𝜇𝜇 𝑠𝑠 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑐𝑐𝑐𝑐 −8 = = 2.83 𝑥𝑥 10 = 1.7 𝑐𝑐𝑐𝑐2 𝑑𝑑𝑑𝑑 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 5672 𝑠𝑠 𝑔𝑔 5.0 𝑐𝑐𝑐𝑐 ∙ �1 + 𝑐𝑐𝑐𝑐� 2.32 𝑐𝑐𝑐𝑐3 5.0 𝑐𝑐𝑐𝑐 ∙ 1.25 𝑠𝑠 In this case, surface reaction controls the overall rate of Si thin film formation 9.355 × 10−9

26-2

26.2 A = drug, B = gel diffusion barrier 𝑐𝑐𝑐𝑐2 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐷𝐷𝐴𝐴𝐴𝐴 = 1.5 × 10−5 , 𝑐𝑐𝐴𝐴𝐴𝐴 = 0.01 𝑠𝑠 𝑐𝑐𝑐𝑐3 a. Develop model equation for WA

Assume: 1) Steady-state (PSS), 2) 1-D flux transfer along r, 3) No reaction, 4) Dilute system, 5) UMD, 6) Saturation concentration A ( cAs = cA*) Flux equation 𝑑𝑑𝑑𝑑𝐴𝐴 𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴 𝑑𝑑𝑑𝑑

General Differential Equation for Mass Transfer 1 d 2 d 0 ⇒ ( r 2 N A, r ) = 0 r N A, r ) = ( 2 r dr dr

∴WA = N A,R i ⋅ 4π Ri 2 = N A,R o ⋅ 4π Ro 2 = N A,r ⋅ 4π r 2 = const 𝑊𝑊𝐴𝐴 = 𝑁𝑁𝐴𝐴,𝑟𝑟 ∙ 4𝜋𝜋𝑟𝑟 2 �−𝐷𝐷𝐴𝐴𝐴𝐴

𝑑𝑑𝑑𝑑𝐴𝐴 � 𝑑𝑑𝑑𝑑

𝑅𝑅𝑜𝑜

𝑐𝑐𝐴𝐴𝐴𝐴

𝑅𝑅𝑖𝑖

𝑐𝑐𝐴𝐴𝐴𝐴

𝑑𝑑𝑑𝑑 𝑊𝑊𝐴𝐴 � 2 = −4𝜋𝜋𝐷𝐷𝐴𝐴𝐴𝐴 � 𝑑𝑑𝑑𝑑𝐴𝐴 𝑟𝑟 𝑊𝑊𝐴𝐴 =

4𝜋𝜋𝐷𝐷𝐴𝐴𝐴𝐴 (𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 ) 1 1 − 𝑅𝑅𝑖𝑖 𝑅𝑅𝑜𝑜

b. Determine WA

At maximum transfer rate,𝑐𝑐𝐴𝐴𝐴𝐴 ≈ 0 𝑊𝑊𝐴𝐴 =

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑐𝑐𝑐𝑐2 ∙ 0.01 𝑠𝑠 𝑐𝑐𝑐𝑐3 = 3.17 × 10−3 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 1 1 ℎ𝑟𝑟 − 0.2 𝑐𝑐𝑐𝑐 0.35 𝑐𝑐𝑐𝑐

4𝜋𝜋 ∙ 1.5 × 10−5

26-3

26.3

Sealed SealedEnds Ends

DAe = 1.2 x 10-5 cm2/s /sec CAA** == 0.05 c 0.05 mmole/cm mmole/cm33 gel layer DAe

solid solid solute solute AA (0.2 mmol)

L =L 1.5 = 1.5 cmc

CAA** c

CAA∞∞ == 0.01 c 0.01 mmole/cm mmole/cm33

CAA∞∞ c

r = Ro (0.75 cm)

r = 0 r = Ri (0.25 cm)

a. Boundary Conditions The system is the gel layer surrounding the source reservoir for solute A r = Ri, CA = CA* r = Ro, CA = CA∞ b. Assumptions 1. 2. 3. 4. 5.

Steady state - constant source and sink for A as long as solid A is not completely dissolved No homogeneous reaction of A within gel layer 1-D flux along position r - ealed top and bottom ends Constant CA* - source reservoir is well mixed, solid A is not yet completely dissolved Dilute with respect to solute A

c. Total transfer rate, WA Flux Equation based on assumption 3

N A, r = − DAe

dC A dr

General Differential Equation for Mass Transfer based on assumptions 1-3 26-4

d ( r N A, r ) = 0 dr By continuity N A,r r = R ⋅ R = N A,r ⋅ r NA,r r constant along r = WA 2= π rL N A, r 2π R= Ro WA is constant along r o L N A, r r C A∞ dr = − DAe ∫ dC A Ri r CA * 2π LDAe ( C A* − C A∞ ) WA = R  ln  o   Ri 

WA ∫

 cm 2  mmol A 2π(1.5 cm) 1.2×10-5  ( 0.05-0 ) s  cm3  WA =  0.75 cm  ln    0.25 cm  WA = 4.12×10-6 mmol A/s

d. Time t Mass balance on solute A in reservoir IN − OUT = ACCUMULATION dm 0 − WA =A dt mAo − mA = WAt

When mA = 0 0.2 mmol A mAo  3600 s  t = =   =13.5 h WA 4.12×10-6 mmol A  1 h  s

26-5

26.4 A = drug, B = gel diffusion barrier ℓ = 2.0 mm, 𝑆𝑆 = 9.0 𝑐𝑐𝑐𝑐2 , 𝑇𝑇 = 293 𝐾𝐾, 𝑐𝑐𝐴𝐴∗ = 0.50

𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 𝑐𝑐𝑐𝑐3

Assume: 1) Steady state, 2) 1-D flux along z, 3) dilute system, 4) UMD process a. Estimate diffusion coefficient of drug (A) in gel diffusion barrier (B)

d ( N A,z ) = 0 , constant flux along z dz 𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴 𝐿𝐿

𝑑𝑑𝑑𝑑𝐴𝐴 𝑑𝑑𝑑𝑑

𝑁𝑁𝐴𝐴 � 𝑑𝑑𝑑𝑑 = −𝐷𝐷𝐴𝐴𝐴𝐴 � 𝑑𝑑𝑐𝑐𝐴𝐴 , ∗ 𝑐𝑐𝐴𝐴

𝑊𝑊𝐴𝐴 = 𝑁𝑁𝐴𝐴 𝑆𝑆 = 𝐷𝐷𝐴𝐴𝐴𝐴 𝐷𝐷𝐴𝐴𝐴𝐴 =

𝑊𝑊𝐴𝐴 ∙ 𝐿𝐿 𝑆𝑆 ∙ 𝑐𝑐𝐴𝐴∗

𝑐𝑐𝐴𝐴∗ 𝑆𝑆 𝐿𝐿

𝑐𝑐𝐴𝐴∗ 𝑁𝑁𝐴𝐴 = 𝐷𝐷𝐴𝐴𝐴𝐴 𝐿𝐿

Linear drug release profile, from data 𝑊𝑊𝐴𝐴 =

∆𝑚𝑚𝐴𝐴 0.16 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 = = 0.008 Δ𝑡𝑡 20 ℎ𝑟𝑟 ℎ𝑟𝑟

𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 ∙ 0.2 𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐2 ℎ𝑟𝑟 −8 𝐷𝐷𝐴𝐴𝐴𝐴 = = 9.88 × 10 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 𝑠𝑠 9.0 𝑐𝑐𝑐𝑐2 ∙ 0.50 𝑐𝑐𝑐𝑐3 0.008

b. Determine new WA at 35oC

DAB (T2 ) T µ (T ) WA (T2 ) W= WA (T1 ) 2 B 2 = A (T1 ) DAB (T1 ) T1 µ B (T1 ) 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 308 𝐾𝐾 993 × 10−6 𝑃𝑃𝑃𝑃 ∙ 𝑠𝑠 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 𝑊𝑊𝐴𝐴 = 0.008 ∙� = 0.0113 = 0.270 � −6 ℎ𝑟𝑟 293 𝐾𝐾 742 × 10 𝑃𝑃𝑃𝑃 ∙ 𝑠𝑠 ℎ𝑟𝑟 𝑑𝑑𝑑𝑑𝑑𝑑 26-6

26.5 A = H2O vapor, B = air (inside pore) 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 10 𝜇𝜇𝜇𝜇, 𝐿𝐿 = 0.05 𝑐𝑐𝑐𝑐, 𝑛𝑛𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 20,000

𝑇𝑇 = 303 𝐾𝐾, 𝑃𝑃𝐴𝐴 = 0.044 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑝𝑝𝐴𝐴∞ = 0.022 𝑎𝑎𝑎𝑎𝑎𝑎

a. Determine the effective diffusion coefficient, DAe

A = H2O, B = air 𝑐𝑐𝑐𝑐2 303 𝐾𝐾 1.5 𝑐𝑐𝑐𝑐2 𝐷𝐷𝐴𝐴𝐴𝐴 = 0.260 ∙� � = 0.267 298 𝐾𝐾 𝑠𝑠 𝑠𝑠

𝑇𝑇 303 𝑐𝑐𝑐𝑐2 = 4850 ∙ 10 × 10−4 � = 19.89 𝑠𝑠 𝑀𝑀𝐴𝐴 18.02

𝐷𝐷𝐾𝐾𝐾𝐾 = 4850𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 �

1 1 1 = + = 𝐷𝐷𝐴𝐴𝐴𝐴 𝐷𝐷𝐴𝐴𝐴𝐴 𝐷𝐷𝐾𝐾𝐾𝐾

0.267

𝑐𝑐𝑐𝑐2 𝑠𝑠

19.89

𝑐𝑐𝑐𝑐2 𝑠𝑠

𝑐𝑐𝑐𝑐2 ∴ 𝐷𝐷𝐴𝐴𝐴𝐴 = 0.263 𝑠𝑠

Knudsen Diffusion is not important in this case. b. Determine WA for single egg

Assume: 1) Steady state, 2) 1-D flux in pore, 3) Dilute system, 4) Constant P and T 𝑑𝑑𝑑𝑑𝐴𝐴 𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴 𝑑𝑑𝑑𝑑 d ( N A,z ) = 0 , constant flux along z dz 𝐿𝐿

𝐶𝐶𝐴𝐴∞

∗ 𝑐𝑐𝐴𝐴

𝑁𝑁𝐴𝐴 � 𝑑𝑑𝑑𝑑 = −𝐷𝐷𝐴𝐴𝐴𝐴 � 𝑑𝑑𝑐𝑐𝐴𝐴 ,

𝑁𝑁𝐴𝐴 = 𝐷𝐷𝐴𝐴𝐴𝐴

(𝑐𝑐𝐴𝐴∗ − 𝑐𝑐𝐴𝐴∞ ) (𝑃𝑃𝐴𝐴 − 𝑝𝑝𝐴𝐴∞ ) = 𝐷𝐷𝐴𝐴𝐴𝐴 𝐿𝐿 𝐿𝐿 ∙ 𝑅𝑅𝑅𝑅

𝑐𝑐𝑐𝑐2 (0.044 − 0.022) 𝑎𝑎𝑎𝑎𝑎𝑎 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠 ∙ 𝑁𝑁𝐴𝐴 = = 4.654 𝑥𝑥 10−6 3 𝑐𝑐𝑐𝑐 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐2 ∙ 𝑠𝑠 0.05 𝑐𝑐𝑐𝑐 ∙ 82.06 ∙ 303 𝐾𝐾 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾 0.263

2 𝜋𝜋𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑛𝑛𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 4 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝜋𝜋(10−3 𝑐𝑐𝑐𝑐)2 18.02 𝑔𝑔 86400 𝑠𝑠 = 4.654 × 10−6 ∙ 20,000 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ∙ ∙ 2 𝑐𝑐𝑐𝑐 ∙ 𝑠𝑠 4 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑑𝑑𝑑𝑑𝑑𝑑

𝑊𝑊𝐴𝐴 = 𝑁𝑁𝐴𝐴 𝑆𝑆 = 𝑁𝑁𝐴𝐴 = 0.114 𝑔𝑔/𝑑𝑑𝑑𝑑𝑑𝑑

c. To reduce water loss by 50%, Increase 𝑝𝑝𝐴𝐴∞ to 0.033 atm or double temperature.

26-7

26.6 A = drug, B = water-swollen polymer 𝑆𝑆 = 16.0 𝑐𝑐𝑐𝑐2 , 𝑊𝑊𝐴𝐴 = 𝑐𝑐𝐴𝐴∗ = 0.50

0.02 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 = 5.56 × 10−6 ℎ𝑟𝑟 𝑠𝑠

𝑐𝑐𝑐𝑐2 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 −7 , 𝐷𝐷𝐴𝐴𝐴𝐴 = 2.08 × 10 𝑐𝑐𝑐𝑐3 𝑠𝑠

Estimate drug patch thickness, L

Assumptions: 1) Steady state, 2) 1-D flux along z, 3) dilute system, 4) UMD process d ( N A,z ) = 0 , constant flux along z dz 𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴 𝐿𝐿

𝑑𝑑𝑑𝑑𝐴𝐴 𝑑𝑑𝑑𝑑

𝑁𝑁𝐴𝐴 � 𝑑𝑑𝑑𝑑 = −𝐷𝐷𝐴𝐴𝐴𝐴 � 𝑑𝑑𝑐𝑐𝐴𝐴 0

𝑁𝑁𝐴𝐴 = 𝐷𝐷𝐴𝐴𝐴𝐴

𝑐𝑐𝐴𝐴∗ 𝐿𝐿

∗ 𝑐𝑐𝐴𝐴

𝑊𝑊𝐴𝐴 = 𝑁𝑁𝐴𝐴 ∙ 𝑆𝑆 = 𝐷𝐷𝐴𝐴𝐴𝐴 𝐿𝐿 = 𝐷𝐷𝐴𝐴𝐴𝐴

𝑐𝑐𝐴𝐴∗ ∙ 𝑆𝑆 𝑊𝑊𝐴𝐴

𝑐𝑐𝐴𝐴∗ ∙ 𝑆𝑆 𝐿𝐿

𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 2 0.50 𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐3 𝐿𝐿 = 2.08 × 10−7 ∙ 16.0 𝑐𝑐𝑐𝑐2 = 0.30 𝑐𝑐𝑐𝑐 𝑠𝑠 5.56 × 10−6 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 𝑠𝑠

26-8

26.7 A = solute, D = reaction product, B = gel diffusion barrier 𝑐𝑐𝑐𝑐2 𝑆𝑆 = 2.0 𝑐𝑐𝑐𝑐2 , 𝐿𝐿 = 0.30 𝑐𝑐𝑐𝑐, 𝐷𝐷𝐴𝐴𝐴𝐴 = 4.0 × 10−7 , 𝑇𝑇 = 293 𝐾𝐾 𝑠𝑠 𝑐𝑐𝑐𝑐3 𝑔𝑔𝑔𝑔𝑔𝑔 𝑐𝑐𝐴𝐴′ = 𝐾𝐾𝑐𝑐𝐴𝐴 , 𝐾𝐾 = 0.8 3 surrounding liquid 𝑐𝑐𝑐𝑐 𝑙𝑙𝑙𝑙𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐷𝐷 𝑐𝑐𝐷𝐷𝐷𝐷 ≈ 0, 𝑊𝑊𝐷𝐷 = 1.0 × 10−8 𝑠𝑠 Reaction at boundary surface: 𝐴𝐴 → 2𝐷𝐷 Mass Transfer Model

Assume: 1) Steady state, 2) 1-D flux transfer along z, 3) No homogenous reaction, 4) Diffusion limited reaction at the boundary surface z = L, 5) Dilute process, gel d ( N A,z ) = 0 , constant flux along z dz 𝑑𝑑𝑑𝑑𝐴𝐴 𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴 𝑑𝑑𝑑𝑑 𝐿𝐿

′ 𝑐𝑐𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴 = 𝐾𝐾

𝑁𝑁𝐴𝐴 � 𝑑𝑑𝑑𝑑 = −𝐷𝐷𝐴𝐴𝐴𝐴

�

𝑑𝑑𝑐𝑐𝐴𝐴 ,

𝑁𝑁𝐴𝐴 = 𝐷𝐷𝐴𝐴𝐴𝐴

Relate 𝑊𝑊𝐴𝐴 to 𝑊𝑊𝐷𝐷 by moles (stoichiometry).

′ 𝑐𝑐𝐴𝐴𝐴𝐴 𝐿𝐿𝐿𝐿

′ 1 𝑐𝑐𝐴𝐴𝐴𝐴 𝑊𝑊𝐴𝐴 = 𝑊𝑊𝐷𝐷 = 𝑁𝑁𝐴𝐴 ∙ 𝑆𝑆 = 𝐷𝐷𝐴𝐴𝐴𝐴 𝑆𝑆 𝐿𝐿𝐿𝐿 2 ′ 𝑐𝑐𝐴𝐴𝐴𝐴 =

𝑊𝑊𝐷𝐷 𝐿𝐿𝐿𝐿 = 2 ∙ 𝐷𝐷𝐴𝐴𝐴𝐴 ∙ 𝑆𝑆

= 1.5 × 10−3

1.0 × 10−8

𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑐𝑐𝑐𝑐3

𝑐𝑐𝑐𝑐3 𝑔𝑔𝑔𝑔𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐷𝐷 ∙ 1.0𝑐𝑐𝑐𝑐2 ∙ 0.30 𝑐𝑐𝑐𝑐 ∙ 0.8 𝑠𝑠 𝑐𝑐𝑐𝑐3 𝑙𝑙𝑙𝑙𝑙𝑙 2 ∙ 4.0 × 10−7 𝑐𝑐𝑐𝑐2 ∙ 2.0 𝑐𝑐𝑐𝑐2

26-9

26.8

well-mixed bulk fluid (+ drug A) CAO NA

healthy tissue (B)

CAs clump of cancer cells inert surface

r=0

R1 = 0.05 cm R2= 0.1 cm

Given: A = drug, B = tissue R1 = 0.05 cm, R2 = 0.10 cm, DAB = 2 × 10−7 cm2/s NA = −6.914 × 10 –4

mmole A day mmole = − 8.00 × 10−9 2 cm day 86,400 s cm 2s

a. System, source and sink for drug, assumptions Physical System: healthy tissue between bulk fluid and unhealthy cells Source for A: well mixed bulk fluid containing “A” Sink for A: clump of cancer cells Assumptions 1. Steady state 2. 1-D flux along r 3. No homogenous reaction of drug within tissue 4. Dilute system with respect to drug (A) b. Boundary conditions = r R1 , c= c As ≈ 0 A = r R= c Ao 2 , cA

surface of unhealthy tissue bulk fluid

26-10

c. Integral form of flux equation NA General Differential Equation for Mass Transfer ∂c A ∂c A 1 d 2 − 2 R= r N A,r ) + R= 0, = 0 ( A A ∂t ∂t r dr 1 d 2 ∴ 2 0 ( r N A, r ) = r dr d 2 ( r N A, r ) = 0 dr N A,r ⋅ r 2 is constant along r General Flux Equation dc N A = − DAB A dr Note N A,R1 ⋅ R12 = N A,R 2 ⋅ R2 2 = N A,r ⋅ r 2 (constant along r) dc   2 N A, R1 R= r 2  − DAB A  1 dr   Separate variables cA and r, pull N A, R1 R12 outside of integral, then integrate with respect to limits shown below c

Ao dr = − D AB ∫ dc A 2 r R1 0

N A, R1 R12 ∫ N A, R1 =

DAB c Ao  1 1  R12  −   R2 R1 

 1 1  R12  −  N A, R1  R2 R1  c Ao = DAB

d. cAo at r = R2

( 0.05 cm )  2

c Ao =

1 1  −9 mmole  −   −8.00 × 10  mmole cm 2s   0.10 cm 0.05 cm   =1.0 × 10−3 2 cm cm3 2.0 × 10−7 s

26-11

26.9 A = H2O vapor, B = O2 gas 𝑇𝑇 = 310 𝐾𝐾, 𝑃𝑃 = 1.2 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 50 × 10−7 𝑐𝑐𝑐𝑐

𝜀𝜀 = 0.40, 𝑃𝑃𝐴𝐴 = 0.0618 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑦𝑦𝐴𝐴∗ = 0.0515, 𝐻𝐻 = 800

a. Determine effective diffusion coefficient, DAe

𝐿𝐿 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

Fuller-Schettler-Giddings Correlation for DAB 1 1 1/2 1 1 1/2 1.75 0.001𝑇𝑇 1.75 � + � 0.001 ∙ 310 � + � 𝑐𝑐𝑐𝑐2 𝑀𝑀𝐴𝐴 𝑀𝑀𝐴𝐴 18.02 32.0 = = 0.236 𝐷𝐷𝐴𝐴𝐴𝐴 = 1 1 1/3 2 1/3 𝑠𝑠 𝑃𝑃�𝑉𝑉𝐴𝐴 + 𝑉𝑉𝐵𝐵 � 1.2 ∙ [12.73 + 16.63 ]2 Knudsen Diffusion Coefficient

𝑇𝑇 310 𝑐𝑐𝑐𝑐2 = 4850 ∙ 50 × 10−7 � = 0.1006 𝑀𝑀𝐴𝐴 18.02 𝑠𝑠

𝐷𝐷𝐾𝐾𝐾𝐾 = 4850𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 �

Effective Diffusion Coefficient 1 1 1 1 𝑐𝑐𝑐𝑐2 1 = + = + ∴ 𝐷𝐷 = 0.0705 𝐴𝐴𝐴𝐴 𝑐𝑐𝑐𝑐2 𝑐𝑐𝑐𝑐2 𝐷𝐷𝐴𝐴𝐴𝐴 𝐷𝐷𝐴𝐴𝐴𝐴 𝐷𝐷𝐾𝐾𝐾𝐾 𝑠𝑠 0.236 0.1006 𝑠𝑠 𝑠𝑠 𝑐𝑐𝑐𝑐2 𝑐𝑐𝑐𝑐2 ′ 𝐷𝐷𝐴𝐴𝐴𝐴 = 𝜀𝜀 2 𝐷𝐷𝐴𝐴𝐴𝐴 = 0.42 ∙ 0.0705 = 0.0113 𝑠𝑠 𝑠𝑠 b. Determine L

Assume: 1) Steady state, 2) no homogenous reaction, 3) 1-D flux along z, 4) constant T and P 𝑐𝑐 =

𝑃𝑃 = 𝑅𝑅𝑅𝑅

1.2 𝑎𝑎𝑎𝑎𝑎𝑎 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 4.72 × 10−5 3 𝑐𝑐𝑐𝑐 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐3 82.06 ∙ 310𝐾𝐾 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾

d ( N A,z ) = 0 , constant flux along z dz 𝑑𝑑𝑑𝑑𝐴𝐴 𝑐𝑐𝐴𝐴 𝑑𝑑𝑑𝑑𝐴𝐴 𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴 + �𝑁𝑁𝐴𝐴,𝑧𝑧 � = −𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐 + 𝑦𝑦𝐴𝐴 �𝑁𝑁𝐴𝐴,𝑧𝑧 � 𝑑𝑑𝑑𝑑 𝑐𝑐 𝑑𝑑𝑑𝑑 𝑁𝑁𝐴𝐴 = −

𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐 𝑑𝑑𝑑𝑑𝐴𝐴 1 − 𝑦𝑦𝐴𝐴 𝑑𝑑𝑑𝑑

Boundary Conditions: 𝑧𝑧 = 0, 𝑦𝑦𝐴𝐴𝐴𝐴 = 𝑦𝑦𝐴𝐴∗

26-12

𝑧𝑧 = 𝐿𝐿, 𝑦𝑦𝐴𝐴∞ = 0.2𝑦𝑦𝐴𝐴∗ 𝐿𝐿

𝑁𝑁𝐴𝐴 � 𝑑𝑑𝑑𝑑 = −𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 � 0

𝑁𝑁𝐴𝐴 = 𝐿𝐿 = 𝐿𝐿 =

𝑦𝑦𝐴𝐴∞

𝑦𝑦𝐴𝐴𝐴𝐴

𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 1 − 𝑦𝑦𝐴𝐴∞ 𝑙𝑙𝑙𝑙 � � 𝐿𝐿 1 − 𝑦𝑦𝐴𝐴𝐴𝐴

𝑑𝑑𝑦𝑦𝐴𝐴 1 − 𝑦𝑦𝐴𝐴

𝑐𝑐𝐷𝐷𝐴𝐴𝐴𝐴 1 − 𝑦𝑦𝐴𝐴∞ 𝑙𝑙𝑙𝑙 � � 𝑁𝑁𝐴𝐴 1 − 𝑦𝑦𝐴𝐴𝐴𝐴

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑐𝑐𝑐𝑐2 ∙ 4.72 × 10−5 𝑠𝑠 𝑐𝑐𝑐𝑐3 ln �1 − 0.2 ∙ 0.0515� = 0.20 𝑐𝑐𝑐𝑐 𝑚𝑚𝑚𝑚𝑚𝑚 1 − 0.0515 1.15 × 10−7 𝑐𝑐𝑐𝑐2 ∙ 𝑠𝑠

0.0113

∗ c. Determine 𝑐𝑐𝐵𝐵𝐵𝐵 ∗ 𝑐𝑐𝐵𝐵𝐵𝐵 =

𝑝𝑝𝐵𝐵 𝑃𝑃 − 𝑝𝑝𝐴𝐴 (1.2 − 0.062) 𝑎𝑎𝑎𝑎𝑎𝑎 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = = = 1.42 × 10−3 𝐿𝐿 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝐻𝐻 𝐻𝐻 𝐿𝐿 800 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

26-13

26.10 A = O2, B = silicone polymer layer 𝑐𝑐𝐴𝐴′ 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 , 𝑇𝑇 = 298 𝐾𝐾 𝑙𝑙 = 0.10 𝑐𝑐𝑐𝑐, 𝐴𝐴 = 50 𝑐𝑐𝑐𝑐2 , 𝑝𝑝𝐴𝐴 = , 𝑆𝑆 = 3.15 × 10−3 𝑆𝑆 𝑐𝑐𝑐𝑐3 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐2 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐷𝐷𝐴𝐴𝐴𝐴 = 1 × 10−7 , 𝑊𝑊𝐴𝐴 = 1.42 × 10−5 = 2.37 × 10−7 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠 Find 𝑝𝑝𝐴𝐴 given 𝑊𝑊𝐴𝐴

Assume: 1) Steady state, 2) 1-D flux along z, 3) Dilute UMD process, 4) No homogenous reaction, 5) Rapid consumption of O2

d ( N A,z ) = 0 , constant flux along z dz 𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴 𝐿𝐿

𝑑𝑑𝑑𝑑𝐴𝐴 𝑑𝑑𝑑𝑑

𝑁𝑁𝐴𝐴 � 𝑑𝑑𝑑𝑑 = −𝐷𝐷𝐴𝐴𝐴𝐴 � 𝑑𝑑𝑐𝑐𝐴𝐴 0

𝑁𝑁𝐴𝐴 =

′ 𝑐𝑐𝐴𝐴

𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴′ 𝐷𝐷𝐴𝐴𝐴𝐴 𝑝𝑝𝐴𝐴 𝑆𝑆 = 𝐿𝐿 𝐿𝐿

𝑊𝑊𝐴𝐴 = 𝑁𝑁𝐴𝐴 ∙ 𝐴𝐴 =

𝐷𝐷𝐴𝐴𝐴𝐴 𝑝𝑝𝐴𝐴 𝑆𝑆 ∙ 𝐴𝐴 𝐿𝐿

𝑊𝑊𝐴𝐴 ∙ 𝐿𝐿 𝑝𝑝𝐴𝐴 = = 𝐴𝐴 ∙ 𝐷𝐷𝐴𝐴𝐴𝐴 ∙ 𝑆𝑆

2.37 × 10−7

𝑐𝑐𝑐𝑐 50 𝑐𝑐𝑐𝑐2 ∙ 1 × 10−7 𝑠𝑠

𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ∙ 0.1 𝑐𝑐𝑐𝑐 𝑠𝑠

∙ 3.15 × 10−3

𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑐𝑐𝑐𝑐3 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎

= 1.5 𝑎𝑎𝑎𝑎𝑎𝑎

26-14

26.11 A = benzene, B = water, C = air 𝑇𝑇 = 298 𝐾𝐾, 𝐷𝐷𝐴𝐴𝐴𝐴 = 1.1 × 10−5

𝑐𝑐𝑐𝑐2 𝑚𝑚3 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐2 , 𝐷𝐷𝐴𝐴𝐴𝐴 = 0.093 , 𝐻𝐻𝐴𝐴 = 4.84 × 10−3 𝑠𝑠 𝑠𝑠 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

𝑃𝑃𝐴𝐴 = 0.13 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑃𝑃𝐵𝐵 = 0.031 𝑎𝑎𝑎𝑎𝑎𝑎, 𝜌𝜌𝐵𝐵 = 1000 a. System Information

𝑘𝑘𝑘𝑘 𝑔𝑔 𝑔𝑔 , 𝑀𝑀𝐴𝐴 = 78 , 𝑀𝑀𝐵𝐵 = 18 3 𝑚𝑚 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

Assume: 1) Steady state, 2) 1-D flux along z, 3) Dilute UMD system, 4) No homogenous reaction Benzene in air Source: Liquid benzene Sink: Air Boundary: Top of liquid to top of well Benzene in water Source: Liquid benzene Sink: Water vapor Boundary: Bottom of liquid to top of liquid Water in air Source: Liquid water Sink: Air Boundary: Top of liquid to top of well b. Boundary Conditions Benzene in air 𝑧𝑧 = 𝐿𝐿, 𝑐𝑐𝐴𝐴 = 𝑐𝑐𝐴𝐴∞ = 0 𝑧𝑧 = 0, 𝑐𝑐𝐴𝐴 = 𝑐𝑐𝐴𝐴∗

𝑃𝑃𝐴𝐴∗ = 𝐻𝐻𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴 = 𝐻𝐻𝐴𝐴 𝑐𝑐𝐴𝐴∗ =

𝑃𝑃𝐴𝐴∗ = 𝑅𝑅𝑅𝑅

𝜌𝜌𝐴𝐴 𝑚𝑚3 𝑎𝑎𝑎𝑎𝑎𝑎 156 𝑔𝑔 𝐴𝐴 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 4.83 𝑥𝑥 10−3 = 9.68 𝑥𝑥 10−3 𝑎𝑎𝑎𝑎𝑎𝑎 𝑀𝑀𝐴𝐴 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑚𝑚3 78 𝑔𝑔

9.68 𝑥𝑥 10−3 𝑎𝑎𝑎𝑎𝑎𝑎 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 −7 = 3.95 × 10 𝑐𝑐𝑐𝑐3 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐3 82.06 ∙ 298 𝐾𝐾 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾

26-15

Water in air 𝑧𝑧 = 𝐿𝐿, 𝑐𝑐𝐵𝐵 = 𝑐𝑐𝐵𝐵∞ 𝑧𝑧 = 0, 𝑐𝑐𝐵𝐵 = 𝑐𝑐𝐵𝐵∗

𝑝𝑝𝐵𝐵,∞ = 0.4 ∙ 0.031 𝑎𝑎𝑎𝑎𝑎𝑎 = 0.0124 𝑎𝑎𝑎𝑎𝑎𝑎

𝑐𝑐𝐵𝐵∞ = 𝑐𝑐𝐵𝐵∗ =

𝑝𝑝𝐵𝐵∞ = 𝑅𝑅𝑅𝑅

𝑃𝑃𝐵𝐵∗ = 𝑅𝑅𝑅𝑅

0.0124 𝑎𝑎𝑎𝑎𝑎𝑎 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 5.07 × 10−7 3 𝑐𝑐𝑐𝑐 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐3 82.06 ∙ 298 𝐾𝐾 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 0.031 𝑎𝑎𝑎𝑎𝑎𝑎 = 1.27 × 10−6 3 𝑐𝑐𝑐𝑐 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐3 82.06 ∙ 298 𝐾𝐾 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾

c. Determine WA, WB, and mA (t = 30 days) Benzene emitted to air

𝑐𝑐𝑐𝑐2 −7 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴∗ 𝜋𝜋𝑑𝑑 2 0.093 𝑠𝑠 ∙ 3.95 × 10 𝜋𝜋(10 𝑐𝑐𝑐𝑐)2 3 𝑐𝑐𝑐𝑐 𝑊𝑊𝐴𝐴 = 𝑁𝑁𝐴𝐴𝐴𝐴 ∙ 𝐴𝐴 = ∙ = ∙ 𝐿𝐿 4 300 𝑐𝑐𝑐𝑐 4 = 9.62 × 10−9

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 8.31 × 10−4 𝑑𝑑𝑑𝑑𝑑𝑑 𝑠𝑠

Total Benzene in 30 days

𝑚𝑚𝐴𝐴 = 𝑊𝑊𝐴𝐴 𝑀𝑀𝐴𝐴 𝑡𝑡 = �8.31 × 10−4 Water evaporation to air

78 𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 �� (30 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑) = 1.95 𝑔𝑔 𝐴𝐴 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑑𝑑𝑑𝑑𝑑𝑑

𝐷𝐷𝐵𝐵𝐵𝐵 (𝑐𝑐𝐵𝐵∗ − 𝑐𝑐𝐵𝐵∞ ) 𝜋𝜋𝑑𝑑 2 𝑊𝑊𝐵𝐵 = 𝑁𝑁𝐵𝐵 ∙ 𝐴𝐴 = ∙ 𝐿𝐿 4 =

0.260

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑒𝑒 𝑐𝑐𝑐𝑐 2 ∙ (1.27 × 10−6 − 5.07 × 10−7 ) 𝑠𝑠 𝑐𝑐𝑐𝑐3 ∙ 𝜋𝜋(10 𝑐𝑐𝑐𝑐) 300 𝑐𝑐𝑐𝑐 4

= 5.19 × 10−8

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 4.49 × 10−3 𝑠𝑠 𝑑𝑑𝑑𝑑𝑑𝑑

26-16

26.12 A = O2, B = silicone rubber 𝑅𝑅𝑖𝑖 = 0.635 𝑐𝑐𝑐𝑐, 𝑅𝑅𝑜𝑜 = 0.955 𝑐𝑐𝑐𝑐 𝑎𝑎𝑎𝑎𝑎𝑎 ∙ 𝑚𝑚3 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 , 𝑆𝑆 = 3.15 PA = 2.0 atm, 𝐻𝐻 = 0.78 𝑎𝑎𝑎𝑎𝑎𝑎 ∙ 𝑚𝑚3 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ′,∗ 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑐𝑐𝐴𝐴∞ = 0.005 , 𝑐𝑐 = 6.30 𝑎𝑎𝑎𝑎 𝑃𝑃𝐴𝐴 = 2.0 𝑎𝑎𝑎𝑎𝑎𝑎 𝐴𝐴 𝑚𝑚3 𝑚𝑚3 a. Develop model equation for NA from r = Ri to r = Ro

Assume: 1) PSS, 2) 1-D mass transfer along r, 3) Dilute UMD process, 4) No homogenous reaction, 5) Convective resistances neglected. Flux Equation 𝑑𝑑𝑑𝑑𝐴𝐴 𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴 𝑑𝑑𝑑𝑑 General Differential Equation for Mass Transfer (cylindrical system) d 1 d rN A,r ) = 0 ⇒ ( rN A,r ) = 0 ( r dr dr N A,R o ⋅ R= N A,r ⋅ r= const ∴ N A,R i ⋅ R= i o 𝑁𝑁𝐴𝐴,𝑅𝑅𝑜𝑜 ∙ 𝑅𝑅𝑜𝑜 = 𝑟𝑟 �−𝐷𝐷𝐴𝐴𝐴𝐴 𝑅𝑅𝑜𝑜

𝑑𝑑𝑑𝑑′𝐴𝐴 � 𝑑𝑑𝑑𝑑

𝑐𝑐′𝐴𝐴𝐴𝐴

𝑑𝑑𝑑𝑑 𝑁𝑁𝐴𝐴,𝑅𝑅𝑜𝑜 ∙ 𝑅𝑅𝑜𝑜 � = −𝐷𝐷𝐴𝐴𝐴𝐴 � 𝑑𝑑𝑐𝑐′𝐴𝐴 𝑟𝑟 𝑁𝑁𝐴𝐴,𝑅𝑅𝑜𝑜 =

𝑅𝑅𝑖𝑖

𝐷𝐷𝐴𝐴𝐴𝐴 (𝑐𝑐′𝐴𝐴,∗ − 𝑐𝑐′𝐴𝐴𝐴𝐴 )

′,∗ 𝑐𝑐𝐴𝐴

𝑅𝑅 𝑅𝑅𝑜𝑜 ln � 𝑜𝑜 � 𝑅𝑅𝑖𝑖 b. Determine NA at r = Ro

′ 𝑐𝑐𝐴𝐴𝐴𝐴 = 𝐻𝐻 ∙ 𝑆𝑆 ∙ 𝑐𝑐𝐴𝐴∞ = �0.78

𝑎𝑎𝑎𝑎𝑎𝑎 ∙ 𝑚𝑚3 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 � �3.15 � �0.0050 � = 0.012 3 3 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑎𝑎𝑎𝑎𝑎𝑎 ∙ 𝑚𝑚 𝑚𝑚 𝑚𝑚3

DAB = 1.0 x 10-7 cm2/s (Problem 26.11)

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 1.0 𝑚𝑚3 𝑐𝑐𝑐𝑐2 − 0.012 �1.0 𝑥𝑥 10−7 𝑠𝑠 � �6.30 � � � 𝑚𝑚3 𝑚𝑚3 1.0 𝑥𝑥 106 𝑐𝑐𝑐𝑐3 𝑁𝑁𝐴𝐴,𝑅𝑅𝑜𝑜 = 0.955 𝑐𝑐𝑐𝑐 0.955 𝑐𝑐𝑐𝑐 ∙ ln � � 0.635 𝑐𝑐𝑐𝑐 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 1.613 𝑥𝑥 10−12 𝑐𝑐𝑐𝑐𝑠𝑠 ∙ 𝑠𝑠 26-17

26.13 A = H2O vapor, B = air 𝑇𝑇 = 303 𝐾𝐾, 𝑃𝑃 = 1.0 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑃𝑃𝐴𝐴 = 0.042 𝑎𝑎𝑎𝑎𝑎𝑎, 𝐷𝐷𝐴𝐴𝐴𝐴 = 0.266

𝑐𝑐𝑐𝑐2 𝑠𝑠

𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 = 0.9952 𝑔𝑔/𝑐𝑐𝑐𝑐3 (0.042 𝑎𝑎𝑎𝑎𝑎𝑎) 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑃𝑃𝐴𝐴 = = 1.689 × 10−6 𝑐𝑐𝐴𝐴𝐴𝐴 = 3 𝑐𝑐𝑐𝑐 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑅𝑅𝑅𝑅 𝑐𝑐𝑐𝑐3 �82.06 � ∙ (303 𝐾𝐾) 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾 𝑐𝑐𝐴𝐴∞ = 0 𝑅𝑅𝑜𝑜 = 0.5 𝑐𝑐𝑐𝑐

a. Determine WA Develop model for WA

General Differential Equation for Mass Transfer (spherical system): 1 𝑑𝑑 2 �𝑟𝑟 𝑁𝑁𝐴𝐴,𝑟𝑟 � = 0 𝑟𝑟 2 𝑑𝑑𝑑𝑑 𝑁𝑁𝐴𝐴,𝑟𝑟 ∙ 𝑟𝑟 2 = 𝑁𝑁𝐴𝐴,𝑅𝑅 ∙ 𝑅𝑅2 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 Flux Equation: 𝑑𝑑𝑑𝑑𝐴𝐴 𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴 𝑑𝑑𝑑𝑑 Combine

𝑅𝑅

𝑑𝑑𝑑𝑑𝐴𝐴 � 𝑑𝑑𝑑𝑑

∞

𝑁𝑁𝐴𝐴 ∙ 𝑅𝑅2 = 𝑟𝑟 2 �−𝐷𝐷𝐴𝐴𝐴𝐴

𝑐𝑐𝐴𝐴𝐴𝐴

𝑑𝑑𝑑𝑑 𝑁𝑁𝐴𝐴 ∙ 𝑅𝑅2 � 2 = −𝐷𝐷𝐴𝐴𝐴𝐴 � 𝑑𝑑𝑑𝑑𝐴𝐴 𝑟𝑟 𝑁𝑁𝐴𝐴 =

𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴 𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴 = 1 1 𝑅𝑅 𝑅𝑅2 � − � 𝑅𝑅 ∞

𝑊𝑊𝐴𝐴 = 2𝜋𝜋𝜋𝜋 ∙ 𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴

𝑐𝑐𝑐𝑐2 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 3600 𝑠𝑠 1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 � �1.689 × 10−6 �� 𝑠𝑠 𝑐𝑐𝑐𝑐3 ℎ𝑟𝑟 1 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 5.08 ℎ𝑟𝑟

𝑊𝑊𝐴𝐴 = 2𝜋𝜋(0.50 𝑐𝑐𝑐𝑐) ∙ �0.266

26-18

b. Determine the time it takes for the water droplet to completely evaporate Material balance on water (PSS mass transfer) IN – OUT + GEN = ACC 0 − 𝑁𝑁𝐴𝐴,𝑟𝑟 𝑆𝑆 + 0 = −

𝑑𝑑𝑚𝑚𝐴𝐴 𝑑𝑑𝑑𝑑

𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑𝑑𝑑 𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑 2 3 𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴 𝑑𝑑𝑑𝑑 ∙ 2𝜋𝜋𝑅𝑅2 = = ∙ 2𝜋𝜋𝑅𝑅2 � 𝜋𝜋𝑅𝑅 � = 𝑅𝑅 𝑀𝑀𝐴𝐴 𝑑𝑑𝑑𝑑 𝑀𝑀𝐴𝐴 𝑑𝑑𝑑𝑑 3 𝑀𝑀𝐴𝐴 𝑑𝑑𝑑𝑑

−𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴 𝑅𝑅 =

𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 2 𝑑𝑑𝑑𝑑 ∙ 𝑅𝑅 𝑀𝑀𝐴𝐴 𝑑𝑑𝑑𝑑 𝑡𝑡

𝑅𝑅

𝑀𝑀𝐴𝐴 � 𝑑𝑑𝑑𝑑 = � 𝑅𝑅𝑅𝑅𝑅𝑅 −𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴 𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 𝑅𝑅2 𝑡𝑡 = = 2𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴 𝑀𝑀𝐴𝐴

𝑔𝑔 𝑐𝑐𝑐𝑐3 2 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔 𝑐𝑐𝑐𝑐 2 ∙ 0.266 ∙ 1.689 × 10−6 ∙ 18.02 3 𝑠𝑠 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑐𝑐𝑐𝑐 (0.5 𝑐𝑐𝑐𝑐)2 ∙ 0.995

t = 15,380 s = 4.3 hr

26-19

26.14 A = pheromone drug vapor, B = polymer diffusion layer 𝐿𝐿 = 0.15 𝑐𝑐𝑐𝑐, 𝑝𝑝𝐴𝐴 = 𝑐𝑐′𝐴𝐴 ∙ 𝑆𝑆, 𝑁𝑁𝐴𝐴 = 𝑘𝑘𝐺𝐺 (𝑝𝑝𝐴𝐴𝐴𝐴 − 𝑝𝑝𝐴𝐴∞ ), 𝑝𝑝𝐴𝐴∞ ≈ 0, 𝑝𝑝𝐴𝐴𝐴𝐴 = 𝑝𝑝𝐴𝐴∗ 𝑘𝑘𝐺𝐺 = 1 × 10−5

10−6 𝑐𝑐𝑐𝑐2 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 , 𝐷𝐷 = 1.0 𝑥𝑥 , 𝑆𝑆 = 0.80 𝑐𝑐𝑐𝑐3 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎/𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑐𝑐𝑐𝑐2 ∙ 𝑠𝑠 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝐴𝐴𝐴𝐴 𝑠𝑠

a. Develop model for NA

Assume: 1) Steady state, 2) 1-D flux along z, 3) Dilute system, 4) No homogenous reaction Differential Equation for Mass Transfer 𝑑𝑑𝑁𝑁𝐴𝐴,𝑧𝑧 = 0 (𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑎𝑎𝑎𝑎𝑜𝑜𝑛𝑛𝑛𝑛 𝑧𝑧) 𝑑𝑑𝑑𝑑 Flux Equation 𝑑𝑑𝑑𝑑′𝐴𝐴 𝑁𝑁𝐴𝐴 = −𝐷𝐷𝐴𝐴𝐴𝐴 𝑑𝑑𝑑𝑑

𝑝𝑝 𝑐𝑐 ′ 𝐴𝐴𝐴𝐴 = 𝑆𝑆𝐴𝐴𝐴𝐴

𝐿𝐿

𝑁𝑁𝐴𝐴 � 𝑑𝑑𝑑𝑑 = −𝐷𝐷𝐴𝐴𝐴𝐴 0

�

𝑐𝑐 ′ 𝐴𝐴𝐴𝐴

𝑝𝑝∗ = 𝐴𝐴 𝑆𝑆

𝑑𝑑𝑑𝑑𝐴𝐴

𝐷𝐷𝐴𝐴𝐴𝐴 ∗ (𝑝𝑝 − 𝑝𝑝𝐴𝐴𝐴𝐴 ) 𝑁𝑁𝐴𝐴 = 𝐿𝐿𝐿𝐿 𝐴𝐴 𝑁𝑁𝐴𝐴 = 𝑘𝑘𝐺𝐺 (𝑝𝑝𝐴𝐴𝐴𝐴 − 𝑝𝑝𝐴𝐴∞ ) 𝑁𝑁𝐴𝐴 𝑝𝑝𝐴𝐴𝐴𝐴 = + 𝑝𝑝𝐴𝐴∞ 𝑘𝑘𝐺𝐺 𝐷𝐷𝐴𝐴𝐴𝐴 ∗ 𝑁𝑁𝐴𝐴 𝑁𝑁𝐴𝐴 = − 𝑝𝑝𝐴𝐴∞ � �𝑝𝑝 − 𝐿𝐿𝐿𝐿 𝐴𝐴 𝑘𝑘𝐺𝐺 (𝑝𝑝𝐴𝐴∗ − 𝑝𝑝𝐴𝐴∞ ) 𝑁𝑁𝐴𝐴 = 𝐿𝐿𝐿𝐿 1 + 𝐷𝐷𝐴𝐴𝐴𝐴 𝑘𝑘𝐺𝐺 b. Determine NA 𝑁𝑁𝐴𝐴 =

(1.1 𝑎𝑎𝑎𝑎𝑎𝑎 − 0) 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 = 9.82 𝑐𝑐𝑐𝑐3 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐2 ∙ 𝑠𝑠 0.15 𝑐𝑐𝑐𝑐 ∙ 0.80 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 1 + 2 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑐𝑐𝑐𝑐 1 × 10−5 1 × 10−6 𝑠𝑠 𝑐𝑐𝑐𝑐2 ∙ 𝑠𝑠 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 26-20

26.15

bulk stomach fluid (essentially water) cA∞ ≈ 0

detail for one pore cA∞ ≈ 0 0.4 mm

initially

cross-section view of pill z

Lt = 2.0 mm

0.36 cm

Lo = 1.2 mm

at a later time solid A r Given: A = drug, B = water inside pore T = 310 K cA* = 2.0 ×10−4 gmole/cm3, ρA,solid = 1.10 g/cm3, MA = 120 g/gmole DAB = 2.0 × 10−5 cm2/s

Pore dimensions: d = 0.04 cm, Lt = 0.20 cm, Lo = 0.12 cm

a. Transfer rate WA at L = Lt = 0.20 cm Physical System: liquid inside pore (System I) Source for A: solid A at base of pore (dissolves into liquid) Sink for A: surrounding fluid at pore exit Assumptions 1. Pseudo-steady state (PSS) process for A 2. No homogenous reaction of A 3. 1-D flux along z 26-21

4. Dilute UMD process with respect to A 5. Surrounding fluid is semi-infinite sink for A with cA∞ ≈ 0 General Differential Equation for Mass Transfer

∂c ∂c −∇N A + RA =A RA = 0, A = 0,1-D flux along z ∂t ∂t dN ∴ A, z = 0 ( flux is constant along z ) dz General Flux Equation - for a dilute process with respect to A dc N A, z = − DAB A dz Since NA,z is constant along z, separate variables cA and z, pull NA,z outside of the integral, then integrate with respect to limits shown below z = 0, cA = cA*; z = L, cA = cA∞ = 0 L

c*A

N A ∫ dz = − DAB ∫ dc A NA =

DAB c*A L − Lo

2  −5 cm   −4 gmole  2.0 10 × 2    2.0 × 10  * 2 s  cm3  π ( 0.04 cm ) DAB c A π d  WA= N A, z ⋅ S= np = (16 pores ) 4 L − Lo 4 ( 0.2 cm - 0.12 cm )

WA = 1.01 × 10−9

gmole s

b. Time required for complete release of drug from pore Material balance on solid drug A loaded into base of pore (System II) IN – OUT + GEN = ACC

dm 0 − N AS + 0 = A dt 0−

ρ DAB c*A dL S + 0 = A,solid S Lt − L MA dt

Let δ = Lt − L,

dδ dL = − dt dt 26-22

At complete release, L = Lt DAB c*A M A

ρ A,solid

DAB c*A M A

ρ A,solid

Lt − Lo

∫dt = t=

∫ δ dδ

Lt 2 − ( Lt − Lo )

ρ A,solid DAB c*A M A

1.0 h  (( 0.2 cm ) − ( 0.2 cm − 0.12 cm ) ) 1.10 cmg   3600  s 2

 cm 2   g  −4 gmole   2  2.0 ×10−5 120   2.0 × 10 3  s  cm   gmole  

= 10.7 h

c. Increase required time by a factor of 2.0 Increase Lt  Lt 2 − ( Lt − Lo )2   Lt 2 − ( Lt − 0.12 cm )2  tnew   new   new = 2= = 2 2 2 told  Lt − ( Lt − Lo )   0.2 cm ) − ( 0.2 cm − 0.12 cm )2    old (  old Lt ,new = 0.34 cm

26-23

26.16 bulk fluid cA ∞ ≈ 0

detail for one pore cA ∞ ≈ 0 0.1 cm

cross-section view of pore array

NA Lt = 0.5 cm z

at the final time

solid A 4.0 cm (not to scale)

Lo =0.20 cm

Assumptions 1. 2. 3. 4. 5.

Pseudo-steady state (PSS) process for A No homogenous reaction of A 1-D flux along z Dilute UMD process with respect to A Surrounding fluid is semi-infinite sink for A with cA∞ ≈ 0

Total transfer rate, WA General Differential Equation for Mass Transfer

∂c −∇N A + RA =A ∂t Simplification based on assumptions 1-3 ∴

dN A, z 0 ( flux is constant along z ) = dz

General Flux Equation - for a dilute process with respect to A

N A, z = − DAB

dc A dz

c*A

N A ∫ dz = − DAB ∫ dc A

26-24

NA =

DAB c*A L − Lo

2  µ mole  −5 cm   1.6 10 × 2   90  * 2 s  cm3  π ( 0.19 cm ) DAB c A π d  WA= N A, z ⋅ S= n p= (16 pores ) 4 L − Lo 4 ( 0.25cm - 0.2 cm )

WA = 6.03 × 10−4

µ mole s

b. Time required to completely dissolve solid in pore Material balance on solid A loaded into base of pore (System II) IN – OUT + GEN = ACC

dm 0 − N AS + 0 = A dt 0−

ρ DAB c*A dL S + 0 = A,solid S Lt − L MA dt

Let δ = Lt − L,

dδ dL = − dt dt

At complete release, L = Lt DAB c*A M A

ρ A,solid

DAB c*A M A

ρ A,solid

Lt − Lo

∫dt = ∫ δ dδ t=

Lt 2 − ( Lt − Lo )

ρ A,solid DAB c*A M A

1.0 h  (( 0.5 cm ) − ( 0.5 cm − 0.2 cm ) )  2.10 cmg  3600  s 2

 cm 2   µ mole   g  2 1.6 × 10−5 74  90  3  s  cm   gmole  

= 438 h

26-25

26.17 Given: A = O2, B = air, C = carbon (s), D = CO2 T = 1000 K, P = 2.0 atm Electrode dimensions: L = 25 cm, d = 2.0 cm, δ = 0.5 cm boundary layer Electrode materials: ρC,solid = 2.25 g/cm3, MC = 12 g/gmole a. Determine WA Model for WA Physical System: gas space in boundary layer between carbon electrode surface and bulk gas (System I) Source for A: bulk gas Sink for A: consumption by carbon electrode to CO2 gas Assumptions 1. 2. 3. 4.

PSS process for A No homogenous reaction of A in gas space 1-D flux along r Instantaneous consumption of O2 gas (A) at electrode surface so that cAs = 0 at r = R (instantaneous surface reaction) 5. NA,r = − ND,r by reaction C ( s ) + O 2 ( g ) → CO 2 ( g )

CO2

solid carbon electrode bulk gas cA∞

O2 (A)

25 cm

NA,r

r = R+δ = 1.5 cm r = R = 1.0 cm cAs = 0

26-26

General Differential Equation for Mass Transfer (cylindrical system) ∂c A = RA 0,= 0 by assumptions 1 and 2 ∂t ∂c A 1 d rN A,r ) + RA = ( ∂t r dr 1 d ∴ 0 ( r ⋅ N A, r ) = r dr d ( r ⋅ N A, r ) = 0 dr −

By continuity N A,r r = R ⋅ R = N A,r ⋅ r NA,r r constant along r Note WA = N A,r r = R ⋅ 2π RL = N A,r ⋅ 2π rL Flux Equation dc A c A dc c + ( N A,r + N B ,r + N C ,r + N D,r ) =− DAB A + A ( N A,r + 0 + 0 + − N A,r ) dr C dr C dc N A,r = − DAB A dr dc   N A,r ⋅ 2π rL = 2π rL  − DAB A  = Note: WA N= A,r r = R ⋅ 2π RL dr   WA is not a function of r; separate variables cA, r, pull WA outside of the integral, then integrate with respect to limits shown below N A,r =− DAB

r = R, cA = cAs = 0 (rapid consumption of O2 at surface) r = R + δ , cA = cA∞ (bulk gas composition beyond boundary layer) R

dr = −2π LDAB ∫ dc A r R +δ c A∞

WA ∫

WA =

2π LDAB c A∞  R +δ  ln    R 

Calculate WA

26-27

c A∞ =

( 0.21)( 2.0 atm ) y A∞ P gmole = =4.65 × 10−6 3 RT  cm3 cm atm  82.06 1100 K ( )   gmole ⋅ K  

 P  T  DAB (T , P) = DAB (Tref , Pref )  ref     P   Tref 

3/2

= 0.136

cm 2  1.0 atm   1100 K  cm 2 = 0.55    s  2.0 atm   273 K  s

 cm 2   −6 gmole  2π ( 25 cm )  0.55   4.65 × 10  s  cm3  gmole  = 1.80 × 10−3 WA = s  2.5 cm  ln    2.0 cm  b. Time it takes for the rod to completely disappear (t at R = 0) Material balance on carbon (PSS mass transfer) to model “shrinkage” of carbon electrode (System II) IN – OUT + GEN = ACC

0 − N A, r ⋅ mC =

dm 1.0 mol C ⋅S +0 = C 1.0 mol O 2 dt

ρC π R 2 L

MC ρC DAB c A∞ dR − 2π RL = 2π RL MC dt  R +δ  R ⋅ ln    R  ρ A dR DAB c A∞ R − =  R + δ  M A dt ln    R  Separate variables R, t and integrate, then solve for time t at R = 0 t

ρC 0  R + δ 

− DAB c A∞ ∫dt = ln   RdR M C ∫R  R  0 R

 R +δ 

1 

 R +δ 

 1

 R +δ   

= R  R ⋅ ln   + δ  − δ ln  ∫ln  R  RdR 2   R   δ  2 0

26-28

  2  R +δ   R +δ  R  R ⋅ ln   + δ  − δ ln    R   δ  ρC  t=  2 DAB c A∞ MC    2  2.5 cm   2.5 cm    g   2.0 cm  ( 2.0 cm ) ln   +0.5 cm  − ( 0.5 cm ) ln     2.25 3   2.0 cm   0.5 cm    cm    t= = 5455 s 2   g  cm   −6 gmole  2  0.55   4.65 × 10 12.01 gmole   s  cm3    

26-29

26.18 Physical System: SiO2 solid layer (species B) containing “dissolved” O2 (species A) SOURCE for O2: 100% O2 gas over SiO2 solid film SINK for O2: unreacted Si at boundary surface SYSTEM I: dissolved O2 in SiO2 film SYSTEM II: unreacted Si Heterogeneous Surface Reaction: Si ( s ) + O2 ( g ) → SiO2 ( s ) 𝑇𝑇 = 1273 𝐾𝐾, 𝑃𝑃𝐴𝐴 = 1 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑐𝑐𝐴𝐴𝐴𝐴 = 9.6 × 10−8 Assumptions:

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑂𝑂2 𝑔𝑔 𝑔𝑔 , 𝑀𝑀𝐵𝐵 = 60.08 , 𝜌𝜌𝐵𝐵 = 2.27 3 3 𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑐𝑐𝑐𝑐

1. The oxidation of Si to SiO2 occurs only at the Si/SiO2 interface. The unreacted Si at the interface serves as the sink for molecular mass transfer of O2 through the film. 2. The O2 in the gas phase above the wafer represents an infinite source for O2 transfer. The O2 molecules “dissolve” into the nonporous SiO2 solid at the gas solid interface. 3. The rate of SiO2 formation is controlled by the rate of molecular diffusion of O2 (species A) through the solid SiO2 layer (species B) to the unreacted Si layer. The reaction is very rapid, so that the concentration of molecular O2 at the interface is equal to zero, i.e CA = 0. Furthermore, there are no mass transfer resistances in the gas film above the wafer surface, since O2 is a pure component in the gas phase. 4. The flux of O2 through the SiO2 layer is one-dimensional along coordinate z. 5. The rate of SiO2 film formation is slow enough so that at a given film thickness δ, there is no accumulation of reactants or products within the SiO2 film. However, the thickness of the film will still increase with time (the PSS Assumption). 6. The overall thickness of the wafer does not change as the result of the formation of the SiO2 layer. 7. The process is isothermal. General Differential Equation for Mass Transfer dN Az = 0 (NA,z is constant along z, therefore can integrate flux equation directly) dz

Flux Equation dc c dc c dc N Az = − DAB A + A ( N Az + N Bz ) = − DAB A + A N Az ≅ − DAB A dz c dz c dz

∫ N Az dz = − DAB ∫ dcA 0

N Az =

c As

DAB c As

26-30

Mass balance on SiO2 IN – OUT + GENERATION = ACCUMULATION

1.0 mole SiO 2 (B) dmB = 1.0 mole O 2 (A) dt δ M B D AB c As t d δ δ = ∫ ∫ dt or δ = ρ B 0 0

ρ Sδ  d B  dmB MB   = dt dt

0 − 0 + N A, z ⋅ S ⋅

2 M B D AB C As

ρB

A plot of δ vs. t 0.5 data will be linear with least-squares slope =

note: δ ∝ t

2 M B D AB C As

ρB

and intercept = 0

SiO2 Film Thickness, δ (µm) Measured Predicted

Time t (h)

t0.5 (h0.5)

(100)Si

(111)Si

(100)Si

(111)Si

1.0 2.0 4.0 7.0 16.0

1.000 1.414 2.000 2.646 4.000

0.049 0.078 0.124 0.180 0.298

0.070 0.105 0.154 0.212 0.339

0.069 0.098 0.138 0.183 0.277

0.081 0.115 0.163 0.215 0.326

(100)Si

(111)Si

slope =

0.0692

0.081

µm/h0.5

ρB =

2.270

g/cm3

MB =

60.0

g/mole

DAB cAs = 2.513E-16 3.486E-16 mol/cm-s cAs = 9.550E-08 9.550E-08 mol/cm3 DAB = 2.631E-09 3.650E-09 cm2/s Sample Calculation (with units) 2  ρB 2 0.5 2  1 cm   1 hr  DAB = ( slope )   = ( 0.0692 μm/h )  4     10 μm   3600 s   2M B CAs    3   ( 2.27 g/cm )   = 2.63 x 10-9 cm 2 /s   g  -8 gmol    2*60  9.550×10  3  gmol   cm   

26-31

0.4 Data (100)Si

SiO2 film thickness,δ (µm)

Best Fit (100)Si Data (111)Si 0.3

Best Fit (111)Si

0.2

0.1

0.0 0

2 Time, t

3 0.5

0.5

(hr)

26-32

26.19 A = acetone, B = air 𝑑𝑑 = 0.3 𝑐𝑐𝑐𝑐, 𝑇𝑇 = 293.9 𝐾𝐾 𝑃𝑃𝐴𝐴∗ = 0.254 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑀𝑀𝐴𝐴 = 58

𝑔𝑔 𝑔𝑔 , 𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 = 0.79 3 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑐𝑐𝑐𝑐

a. Determine DAB from Arnold Diffusion Cell data

Assume: 1) 1-D mass transfer along z, 2) No homogenous reaction, 3) PSS. Material balance on acetone (PSS mass transfer) IN – OUT + GEN = ACC 𝑑𝑑𝑚𝑚𝐴𝐴 0 − 𝑁𝑁𝐴𝐴,𝑧𝑧 𝑆𝑆 + 0 = 𝑑𝑑𝑑𝑑 𝑁𝑁𝐴𝐴 =

𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐 1 − 𝑦𝑦𝐴𝐴2 ln � � 1 − 𝑦𝑦𝐴𝐴1 𝑍𝑍

𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑𝑑𝑑 𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐 1 − 𝑦𝑦𝐴𝐴2 ln � �= ∙ 𝑍𝑍 1 − 𝑦𝑦𝐴𝐴1 𝑀𝑀𝐴𝐴 𝑑𝑑𝑑𝑑 𝑡𝑡

𝑍𝑍

𝑍𝑍𝑜𝑜

𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴𝐴𝐴 𝑀𝑀𝐴𝐴 � 𝑑𝑑𝑑𝑑 = � 𝑍𝑍𝑍𝑍𝑍𝑍 𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙

Z2-Zo2 [cm2]

𝑍𝑍 2 − 𝑍𝑍0 2 =

2𝐷𝐷𝐴𝐴𝐴𝐴 𝑐𝑐𝑀𝑀𝐴𝐴 1 − 𝑦𝑦𝐴𝐴∞ ln � � (𝑡𝑡 − 𝑡𝑡𝑜𝑜 ) 𝜌𝜌𝐴𝐴,𝑙𝑙𝑙𝑙𝑙𝑙 1 − 𝑦𝑦𝐴𝐴𝐴𝐴

160 140 120 100 80 60 40 20 0

y = 0.6472x + 0.5537

100

150

200

250

time [hr] 26-33

Slope = 0.6472 𝑐𝑐 =

𝑃𝑃 = 𝑅𝑅𝑅𝑅

𝑦𝑦𝐴𝐴𝐴𝐴 = 𝐷𝐷𝐴𝐴𝐴𝐴 =

𝑐𝑐𝑐𝑐2 𝑐𝑐𝑐𝑐2 = 1.80 × 10−4 ℎ𝑟𝑟 𝑠𝑠

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 1.0 𝑎𝑎𝑎𝑎𝑎𝑎 = 4.15 × 10−5 3 𝑐𝑐𝑐𝑐 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐3 82.06 ∙ 293.9 𝐾𝐾 𝐾𝐾 ∙ 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

𝑃𝑃𝐴𝐴∗ 0.254 𝑎𝑎𝑎𝑎𝑎𝑎 = = 0.254 𝑃𝑃 1.0 𝑎𝑎𝑎𝑎𝑎𝑎 𝜌𝜌𝐴𝐴 ∙ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 1−0 2 ∙ 𝑐𝑐 ∙ 𝑀𝑀𝐴𝐴 ln � � 1 − 𝑦𝑦𝐴𝐴𝐴𝐴

𝑔𝑔 𝑐𝑐𝑐𝑐2 ∙ 1.80 × 10−4 𝑐𝑐𝑐𝑐3 3 𝑠𝑠 𝑐𝑐𝑐𝑐 = 0.101 𝐷𝐷𝐴𝐴𝐴𝐴 = 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔 1−0 𝑠𝑠 2 ∙ 4.15 × 10−5 ∙ 58 ∙ ln � � 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 1 − 0.254 𝑐𝑐𝑐𝑐3 0.79

b. Determine DAB by correlation Hirschfelder Correlation

𝐷𝐷𝐴𝐴𝐴𝐴 =

1 1 2 0.001858 ∙ 𝑇𝑇 3/2 �𝑀𝑀 + 𝑀𝑀 �

𝜎𝜎𝐴𝐴 = 2.44 � 𝜀𝜀𝐴𝐴 𝜅𝜅

2 𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴 ΩD

508 𝐾𝐾

47.4 𝑎𝑎𝑎𝑎𝑎𝑎

𝐴𝐴

1 3

� = 5.380 Å, 𝜎𝜎𝐵𝐵 = 3.617 Å,, 𝜎𝜎𝐴𝐴𝐴𝐴 =

5.380+3.617

𝜀𝜀

= 4.50 Å

= 0.77(508 𝐾𝐾) = 391.16 𝐾𝐾, 𝐵𝐵 = 97 𝐾𝐾,𝜀𝜀𝐴𝐴𝐴𝐴 = √391.16 ∙ 97 𝐾𝐾 = 194.79 𝐾𝐾

𝜅𝜅𝜅𝜅 293.9 𝐾𝐾 = = 1.509, 𝜀𝜀𝐴𝐴𝐴𝐴 194.79 𝐾𝐾 𝐷𝐷𝐴𝐴𝐴𝐴 =

𝜅𝜅

∴ Ω𝐷𝐷 = 1.195

1 1 1/2 + ] 𝑐𝑐𝑐𝑐2 58.0 29.0 = 0.088 1.0 ∙ 4.50 2 ∙ 1.195 𝑠𝑠

0.001858 ∙ 293.93/2 [

26-34

26.20 A = O2 and B = CO2, 𝑦𝑦𝐴𝐴∞ = 1.0, 𝑐𝑐𝐴𝐴𝐴𝐴 = 0, 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 0.1 𝑐𝑐𝑐𝑐, 𝐿𝐿𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 0.5 𝑐𝑐𝑐𝑐 𝑃𝑃 = 2.0 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑇𝑇 = 873 𝐾𝐾, 𝜌𝜌𝐶𝐶 = 2.25

𝑔𝑔 𝑔𝑔 , 𝑀𝑀𝐶𝐶 = 12.01 3 𝑐𝑐𝑐𝑐 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

Assume: 1) PSS system, 2) No homogenous reaction, 3) 1-D flux along z, EMCD a. Determine WB at Z = 0.30 cm Flux model for NA d ( N A,z ) = 0 , constant flux along z dz 𝑑𝑑𝑑𝑑𝐵𝐵 𝑁𝑁𝐴𝐴,𝑧𝑧 = −𝐷𝐷𝐵𝐵𝐵𝐵 𝑐𝑐 𝑑𝑑𝑑𝑑 𝑧𝑧 = 0, 𝑦𝑦𝐴𝐴𝐴𝐴 = 0, 𝑦𝑦𝐵𝐵𝐵𝐵 = 1.0

𝑍𝑍 = 0.3 𝑐𝑐𝑐𝑐, 𝑦𝑦𝐴𝐴𝐴𝐴 = 1.0, 𝑦𝑦𝐵𝐵𝐵𝐵 = 0 𝑍𝑍

𝑦𝑦𝐵𝐵∞ =0

𝑦𝑦𝐵𝐵𝐵𝐵 =1

𝑁𝑁𝐵𝐵 � 𝑑𝑑𝑑𝑑 = −𝐷𝐷𝐵𝐵𝐵𝐵 𝑐𝑐

𝐷𝐷𝐵𝐵𝐵𝐵 𝑦𝑦𝐵𝐵𝐵𝐵 𝑐𝑐 𝑁𝑁𝐵𝐵 = 𝑍𝑍

�

𝑑𝑑𝑦𝑦𝐵𝐵

Hirschfelder Correlation

𝐷𝐷𝐴𝐴𝐴𝐴 =

1 1 2 0.001858 ∙ 𝑇𝑇 3/2 �𝑀𝑀 + 𝑀𝑀 � 2 𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴 ΩD

𝐴𝐴

𝜎𝜎𝐵𝐵 = 3.433 Å, 𝜎𝜎𝐴𝐴 = 3.996 Å,, 𝜎𝜎𝐵𝐵𝐵𝐵 = 𝜀𝜀𝐵𝐵 𝜅𝜅

𝜀𝜀

3.996+3.433 2

= 3.7145 Å

= 113 𝐾𝐾, 𝐴𝐴 = 190 𝐾𝐾,𝜀𝜀𝐵𝐵𝐵𝐵 = √113𝐾𝐾 ∙ 190 𝐾𝐾 = 146.53 𝐾𝐾 𝜅𝜅

𝜅𝜅𝜅𝜅 873 𝐾𝐾 = = 5.958, 𝜀𝜀𝐵𝐵𝐵𝐵 146.53 𝐾𝐾 𝐷𝐷𝐴𝐴𝐴𝐴 = 𝐷𝐷𝐵𝐵𝐵𝐵 =

∴ Ω𝐷𝐷 = 0.8137

1 1 1/2 2 44.01 + 32.0] = 0.496 𝑐𝑐𝑐𝑐 2.0 ∙ 3.7145 2 ∙ 0.8137 𝑠𝑠

0.001858 ∙ 873 3/2 [

26-35

𝑐𝑐 =

𝑃𝑃 = 𝑅𝑅𝑅𝑅

2.0 𝑎𝑎𝑎𝑎𝑎𝑎 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 2.79 × 10−5 3 𝑐𝑐𝑐𝑐 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐3 ∙ 873 𝐾𝐾 82.06 𝐾𝐾 ∙ 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

𝑊𝑊𝐵𝐵 = 𝑁𝑁𝐵𝐵 𝑆𝑆 =

𝐷𝐷𝐵𝐵𝐵𝐵 𝑐𝑐 𝜋𝜋𝑑𝑑 = 𝑍𝑍 4

= 3.62 × 10−7

0.496

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑐𝑐𝑐𝑐2 2 ∙ 2.79 × 10−5 𝑠𝑠 𝑐𝑐𝑐𝑐3 𝜋𝜋(0.1 𝑐𝑐𝑐𝑐) 0.3 𝑐𝑐𝑐𝑐 4

b. Determine time t for carbon in pore to be completely oxidized Material balance on solid carbon in pore IN – OUT + GEN = ACC 0 − 𝑁𝑁𝐵𝐵,𝑧𝑧 𝑆𝑆 ∙

1 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐶𝐶 𝑑𝑑𝑚𝑚𝐶𝐶 +0= 1 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐶𝐶𝐶𝐶2 𝑑𝑑𝑑𝑑

𝐷𝐷𝐵𝐵𝐵𝐵 𝑦𝑦𝐵𝐵𝐵𝐵 𝑐𝑐 𝜌𝜌𝐶𝐶 𝑑𝑑𝑑𝑑 = 𝑀𝑀𝐶𝐶 𝑑𝑑𝑑𝑑 𝑍𝑍 𝑡𝑡

𝑍𝑍

𝐷𝐷𝐵𝐵𝐵𝐵 𝑦𝑦𝐵𝐵𝐵𝐵 𝑐𝑐 𝜌𝜌𝐶𝐶 � 𝑑𝑑𝑑𝑑 = � 𝑑𝑑𝑑𝑑 𝑍𝑍 𝑀𝑀𝐶𝐶 𝑡𝑡 =

𝑍𝑍 2 𝜌𝜌𝐶𝐶 2𝐷𝐷𝐵𝐵𝐵𝐵 𝑦𝑦𝐵𝐵𝐵𝐵 𝑐𝑐 𝑀𝑀𝐶𝐶

𝑔𝑔 3 𝑐𝑐𝑐𝑐 𝑡𝑡 = = 609 𝑠𝑠𝑠𝑠𝑠𝑠 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 12.01 𝑔𝑔 𝑐𝑐𝑐𝑐2 2 ∙ 0.496 ∙ 1.0 ∙ 2.79 × 10−5 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠 𝑐𝑐𝑐𝑐3 (0.3 𝑐𝑐𝑐𝑐)2

2.25

26-36

26.21 drug patch (side view) drug reservoir impermeable barrier skin surface

diffusion barrier (water-filled micropores, dpore = 100 Å)

L infected tissue (sink for drug)

SIDE VIEW micropore array (not to scale) Given: A = drug, B = water T = 298 K cAs = 1.0 × 10−6 gmole/cm3, DAB = 1.0 × 10−6 cm2/s

Pore dimensions: ds = 25 Å, dpore = 100 Å, S = 0.25 (4.0 cm2) = 1.0 cm2

 μmol   1day 1 gmole  −13 gmole = WA N=  0.05   = 5.79 × 10 AS 6 day   86,400 s 10 μmole  s  a. Effective diffusion coefficient, DAe Solute diffusion through solvent-filled pore ° DAe = DAB F1 (ϕ ) F2 (ϕ )

Reduced pore diameter ds 25 A = ϕ = = 0.25 d pore 100 A F1 (ϕ ) = (1 − ϕ ) 2 = (1 − 0.25) 2 = 0.5625

F2 (ϕ ) = 1 − 2.104(ϕ ) + 2.09(ϕ )3 − 0.95(ϕ )5 F2 (ϕ ) = 1 − 2.104(0.25) + 2.09(0.25)3 − 0.95(0.25)5 = 0.5057

° DAe = DAB F1 (ϕ ) F2 (ϕ ) = 1.0 × 10−6

cm 2 cm 2 ( 0.5625)( 0.5057 ) = 2.84 × 10−7 s s 26-37

b. Thickness of the diffusion barrier, L Physical System: liquid inside pore Source for A: solid A at base of pore (dissolves into liquid) Sink for A: surrounding fluid at pore exit Assumptions 1. 2. 3. 4. 5. 6.

Steady-state (PSS) process for A No homogenous reaction of A 1-D flux along z Dilute UMD process with respect to A Surrounding fluid is semi-infinite sink for A with cA∞ ≈ 0 Hindered diffusion of solute in solvent-filled pores

General Differential Equation for Mass Transfer −∇N A + R= A ∴

∂c A ∂t

R= 0, A

∂c A = 0,1-D flux along z ∂t

d 0 ( flux is constant along z ) ( N A,z ) = dz

General Flux Equation—for a dilute process with respect to A dc N A = − DAe A dz Since NA,z is constant along z, separate variables cA and z, pull NA,z outside of the integral, then integrate with respect to limits shown below z = 0, cA = cAs; z = L, cA = cA∞ = 0 L

c As

N A ∫dz = − DAe ∫ dc A NA =

DAe c As L

WA N= = AS L=

DAe c As S L

DAe c As S WA

2  −7 cm   −6 gmole  2.84 10 ×   1.0 × 10  s  cm3   L= 1.0 cm 2 = 0.568 cm  −13 gmole   5.79 × 10  s  

(

)

26-38

26.22 2.0 mol% TCE in air

porous catalyst layer degrades TCE (k = 1.5 s-1)

well-mixed gas phase

nonporous barrier

CAo

z = 0 z = L = 0.5 cm 5.0 mol% TCE (A) in air

a. Boundary Conditions z = 0, CA = CAo z = L,

dC A =0 dz

b. yA at z = L k 1.5s -1 L= =0.5 cm 2.165 cm 2 DAe 0.080 s at z = L

 k  cosh  ( L − z )  DAe  c Ao  c A c= Ao cosh(φ ) cosh(φ ) 0.02 y Ao = = 0.0045 yA = cosh(φ ) cosh(2.165)

c. Surface area S at WA = 0.20 gmole A/s WA = N A S

26-39

= NA

DAec Ao DAe y AoC = φ tanh φ φ tanh φ

 cm  gmole  1m    0.080  ( 0.02 )  21.3   s  m3  100 cm    (2.165)tanh(2.165) NA = ( 0.5 cm ) 2

N A =1.44 × 10-3

= S

gmole A m2 ⋅ s

WA 0.20 gmole A/s = =139 m 2 -3 2 N A 1.44×10 gmoleA/m s

26-40

26.23

nonporous support well-mixed liquid phase

enzyme layer

cA∞= 1.0 mmol/m3

z=0

z = L = 0.2 cm cAL = 0.25 mmol/m3

a. Boundary Conditions z = 0, CA = CA∞ z = L,

dC A =0 dz

b. Rate constant k at z = L

c AL

 k  cosh  ( L − z )  DAe  c A∞  c= Ao cosh(φ ) cosh(φ ) c   c AL 

φ = cosh −1  A∞  φ=L

k DAe

 cm 2  2.0×10-5   3   s   DAe  -1  1.0 mmole/m −1  c A∞  = cosh k = 2 cosh     2 3  L  ( 0.2 cm )   0.25 mmole/m    c AL   k 2.13 × 10−3 s −1 =

c. Total transfer rate of product B, WB

26-41

k = φ L= 0.2 cm DAe

= NA

2.13×10-3s -1 2.063 = 2 -5 cm 2.0×10 s

DAec Ao DAe y AoC = φ tanh φ φ tanh φ

 mmole  1m  −5 cm    2.0 × 10 1.0   s  m3  100 cm  (2.063)tanh(2.063) NA =  ( 0.2 cm ) 2

mmole A m2 ⋅ s mmole A 1 mol B  1 mol B  2 -6 mmole A  4.0 × 10-6 = WB = NA S  )  2.0 × 10  ( 2.0 m= 2 s 1 mol A m ⋅s   1 mol A  N A =2.0 × 10-6

d. At φ = 2.063, process is between diffusion control and reaction control

26-42

26.24 a. Develop differential model for cA(r) using shell balance A = drug, B = gel material System: gel bead Source: the surrounding liquid Sink: gel bead Assume: 1) Steady State, 2) Uniform Distribution within bead, 3) 1-D Mass Transfer, 4) Dilute with respect to solute A Flux equation for first order reactions dc N A,r = − DAB A dr Material Shell Balance IN – OUT + GEN = ACC = 0 4π r 2 N A,r |r −4π r 2 N A,r |r +∆r +4π r 2 ∆r =0

÷ 4𝜋𝜋 △ 𝑟𝑟, rearrange with lim ∆r → 0

1 d 2 − 2 0 ( r N A , r ) + RA = r dr

dc A  1 d  2 0  r DAB  + RA = 2 r dr  dr  𝐷𝐷𝐴𝐴𝐴𝐴 �

2 𝑑𝑑𝑑𝑑𝐴𝐴 𝑑𝑑 2 𝑐𝑐𝐴𝐴 + � − 𝑘𝑘1 𝑐𝑐𝐴𝐴 = 0 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 2

Boundary Conditions 𝑟𝑟 = 0,

𝑟𝑟 = 𝑅𝑅,

𝑑𝑑𝑑𝑑𝐴𝐴 =0 𝑑𝑑𝑑𝑑 𝑐𝑐𝐴𝐴 = 𝑐𝑐𝐴𝐴∞

b. Determine WA c A (r ) = c A∞

(

R sinh r k1 / DAB r sinh R k / D AB 1

( (

) )) 26-43

( (

 d  R sinh r k1 / DAB c A∞ dr  r sinh R k1 / DAB 

cosh ( r k / D ) )  = c R sinh ( r k / D ) c R k /D − + sinh ( R k / D ) )  r sinh ( R k / D ) r 1

A∞ 2

A∞

  R k1 / DAB dc A 2 1 −  | 4 | 4 = = − = π π W= SN R D D c R A A, r r R AB r R AB A∞ =  tanh R k / D  dr 1 AB     0.019 s -1 ) (   ( 0.3 cm ) -6 2  ( 2x10 cm /s )  -6 2 3  WA = 4π ( 2 x 10 cm /s )( 0.02 μmol/cm ) ( 0.3 cm ) 1  ( 0.019 s-1 )    tanh  ( 0.3 cm )   ( 2x10-6cm2 /s )     μmol 3600 s μmol = 0.0153 WA = 4.26 x 10-6 s hr hr

(

)

Note

s ) ( 0.019 = φ R= k /D = 29.24 ( 0.3 cm ) ( 2x10 cm /s ) -1

-6

26-44

26.25 Given: A = TCE, B = biofilm T = 20 C (293 K) cAi = 0.25 mg TCE/L = 0.25 g TCE/m3 Biofilm: S = 800 m2, δ = 100 µm, DAB = 9.03 × 10−9 m2/s, k = 4.3 s−1 a. Volumetric inlet flow rate, vo Develop process model: Source for A: liquid inflow Sink for A: consumption by biofilm Assumptions 1. 2. 3. 4.

Constant source and sink for A—steady-state process Well-mixed liquid phase Constant liquid volume No consumption of A in bulk liquid phase

Material balance in TCE in biofilm reactor, bulk liquid phase

IN – OUT + GEN = ACC v 0 c Ai − v 0 c Ao − S ⋅ N A = 0 v0 =

S ⋅ NA c Ai − c A

Recall NA for 1-D flux along z, homogeneous first-order reaction with RA =−k ⋅ c A D c N A = AB Ao φ tanh φ

Therefore, v0 =

SDAB c Ao φ tanh φ SN A = c Ai − c Ao δ ( cAi − cAo )

Calculate φ and NA

26-45

 1.0 m  k 4.3 s −1 = φ δ= 100 μm  6  = 6.909 2 DAB  10 μm  9.03 × 10−10 m s 2  g TCE  −10 m     9.03 × 10   0.05 s  m3  g TCE NA  ( 6.909 ) tanh ( 6.909 ) = 3.12 × 10−6 2 −6 100 ×10 m ms Finally, back out vo  (800 m )  3.12 × 10 g mTCE  s   3600 s  2

−6

v0 =

g TCE g TCE    − 0.05  0.25  m3 m3  

m3  = 44.9 h 

b. TCE concentration at point biofilm is attached to surface, cA at z = δ Concentration profile of A in biofilm

 k1  g TCE c Ao cosh  (δ − z )  DAB  c A0 cosh ( 0 ) 0.25 m3 (1.0) g TCE  = 4.99 × 10−4 c A ( z ) |z =δ = = = cosh (φ ) cosh ( 6.909 ) m3  k1  cosh  δ  DAB  

26-46

26.26 A = atrazine, B = water in water-saturated soil a. Determine DAe Wilkes-Chang correlation 1

T 7.4 x10 [ ΦBM B ] 2 DAB = µB VA0.6 −8

-8 2 293 ) 7.4 x 10 ( 2.6 )(18.02 )  2 ( -6 cm = 6.859 x 10 DAB = 0.6 s ( 0.993 ) (170 ) 2 = DAe ε= DAB

( 0.6 ) 6.859 x 10-6 2

cm 2 cm 2 = 2.47 x 10-6 s s

b. Determine c A at z = δ cA ( z ) =

c A* cosh (δ − z ) cosh (φ )

k1 DAB

hr  ( 5.0 x 10 hr )  3600  s -4

k1 φ L= = 10.0 cm DAe

-1

cm 2 2.47 x 10 s mmol 0.139 * cA L = 0.026 mmol = c A (δ ) = cosh (φ ) cosh ( 2.371) L

= 2.371

-6

26-47

26.27 A = CO, B = CO2 a. Determine total molar flow rate out n2 Material balance on reactor for CO IN – OUT + GEN = ACC

y A,1n1 − y A,2 n2 + WA = 0

π d 2 DAe c A0 φ tan φ WA =S ⋅ N A = 4 δ k = φ δ= 1.0 cm DAe

6.0 s -1 = 3.87 cm 2 0.4 s

yCO ,2 + yO2 ,2 + yCO2 ,2 = 1

∴ yCO2 ,2 = 1 − yCO ,2 − yO2 ,2 = 1.0 − 0.05 − 0.025 = 0.925 = y A0 P 1.0 atm gmole = c A0 y= y A0 = 0.925 = 1.051 x 10-5 A 0C 3 RT cm3  cm ⋅ atm  82.06 1073 K ( )   gmole ⋅ K    π (10.0 cm )2   0.40 cm 2 /s   gmole  1.051 x 10-5 3.87 ) tanh ( 3.87 ) WA =    3 (    1.0 cm   4 cm    gmole WA = 1.277 x 10-3 s gmole CO   (0.0)n1 − 0.05n2 + 1.277 x 10-3 0 = s   gmole CO  60 s  gmole CO n2 = 0.0255   = 1.53 s min  1.0 min  b. Mole fraction y A on backside of catalyst layer at z = δ

= yA ( z)

(

)

y A0 cosh (δ − z ) k / DAe y A0 0.925 = = = 0.0386 cosh (φ ) cosh ( 3.87 ) cosh δ k / DAe

(

)

26-48

26.28 A = glucose, B = gel a. Develop model for cA(z) Assumptions: 1) Steady state; 2) 1-D flux along z; 3) Sero order reaction with RA = -k; 4) Dilute system Flux Equation dc N A, z = − DAB A dz General Differential Equation for Mass Transfer dN − A, z + k = 0 dz Plug Flux into Mass Transfer Equation and Integrate dc d k (− DAe A ) = dz dz Integrate twice k z2 = cA + c1 z + c2 DAe 2 z L= , c A c A0 = dc A z 0,= 0 = dz At z = 0, c1 = 0

At z = L, c2 = cA0 cA = c A0 +

k ( z 2 − L2 ) 2 DAe

b. Estimate DAe

mmole   -0.05   mmole   mmole   L-min   (0.25 - 0)  4.5  =  50 + L   L  2D Ae  𝑐𝑐𝑐𝑐2 −6 𝐷𝐷𝐴𝐴𝐴𝐴 = 2.29 × 10 𝑠𝑠

26-49

26.29 A = O2, B = tissue (approximates liquid water) a. Model to predict cA(r) From Chapter 25, Example 3, Differential Equation for Mass Transfer DAB

d 2 c A 1 dc A 0 + −m = dr 2 r dr

d  dc A  m r r = dr  dr  DAB Boundary Conditions: = r R1 , c= c= c A= * A As

pA H

dc A = r R= 0 (net flux NA = 0 at r = R2) 2, dr Analytical Solution: For this particular form of O.D.E., the general solution for cA( r) is obtained by “integrating twice” with respect to r First integration: dc A α m = r+ 1 dr 2 DAB r Second integration: m 2 cA (r ) = r + α1 ln(r ) + α 2 4 DAB Apply boundary conditions to solve for integration constants α1 and α2 at r = R2, α1 = −

mR12 mR2 2 mR 2 2 , at r = R1, α 2 = c As − + ln( R1 ) 2 DAB 4 DAB 2 DAB

Plug integration constants α1 and α2 back into general solution

c A (r ) =c As +

mR2 2  r  m (r 2 − R12 ) − ln   4 DAB 2 DAB  R1  26-50

valid cA(r ) > 0 and R1 < r < R2

For a constant O2 consumption rate m, at some point cA( r) = 0. The “critical radius” where cA( r) = 0 is found at R2 = Rc and r = Rc

0 c As + c A ( Rc ) ==

mRc 2  Rc  m ( Rc 2 − R12 ) − ln   4 DAB 2 DAB  R1 

Rc is found from the root of this nonlinear equation (m, R1, DAB known) b. Estimate Rc and plot of cA(r) 0.35

Variable Units m = 6.94E-05 µmol O2/cm3-sec 0.25 µmol O2/cm3-sec

0.30

0.25

PA =

1.0 atm

H= c As =

0.78 atm·m3/mole

D AB = R1 = Rc = f(R c ) =

C A( r ) 0.20 µmol O2 / cm3 0.15

0.32 µmol O2/cm3 2

2.10E-05 cm /sec 0.25 cm 0.62 cm 0.0000

0.10

0.05

0.00 0.25

0.35

0.45

0.55

0.65

0.75

Radial position from center of tube, r (cm)

c. Calculate WA dc A R12  m  = r −   dr 2 DAB  r 

26-51

dc   W SN 2π RL  − DAB A |R2  = = A |R2 A, r dr   2  m R  WA |R2 = 2π RL  −  R2 − 1   R2    2    -11 mol  2   6.94 x 10 cm3s     0.25 cm ( ) mol  0.75 cm  = -1.63 x 10-9 = 2π ( 0.75 cm )(15 cm ) -  (  ) 2 s ( 0.75 cm )      

26-52

26.30 A = drug, B = tissue a. Define system, source, and sink for mass transfer of species A, and five reasonable assumptions System: healthy tissue layer Source: bulk liquid Sink: tumor tissue Assume:1) Steady State, 2) Dilute with respect to A, 3) 1-D Mass Transfer along z, 4) Drug A is not very soluble in tissue, 5) Homogeneous first order reaction of drug in tissue. b. Develop differential model for cA(z) Differential Equation for Mass Transfer dN − A + RA = 0 dz Combine Flux Equation dc N A, z = − DAe A dz dc  d  DAe A  + RA = 0  dz  dz  d 2cA DAe − k1c A = 0 dz 2 c. Boundary Conditions z 0,= c A Kc A∞ z = 0;= z = L, N A, z = N A,0 = − DAe

dc A dz

d. Analytical Model for cA(z) General Solution = c A ( z ) α1 sinh( β z ) + α 2 cosh( β z )

β = k1 / DAe Apply Boundary Conditions

26-53

At z = 0, c A Kc = = α1 sinh(0) + α 2 cosh(0) A, ∞

α 2 = Kc A,∞ dc At z = L, A β [α1 cosh( β L) + α 2 sinh( β L) ] = dz N A, s = − DAe β α1 cosh( β L) + Kc A,∞ sinh( β L)  N  −  A, s + Kc A,∞ sinh( β L)  D β  α1 =  Ae cosh( β L) 



A, s − + Kc A,∞ tanh( β L)  α1 =  DAe β cosh( β L) 

N A, s   − + Kc A,∞ tanh( β L)  sinh( β z ) + Kc A,∞ cosh( β z ) cA ( z ) =  DAe β cosh( β L) 

e. Determine N A,0 dc A |z =0= β [α1 cosh(0) + α 2 sinh(0) ]= βα1 dz N A, s   dc A − + β Kc A,∞ tanh( β L)  |z 0 = dz  DAe cosh( β L)  N A, s = + β DAe KC A,∞ tanh( β L) N A,0 cosh( β L)

= β

= k1 / DAe

( 4 x 10 s ) / (1.0 x 10 cm /s ) = 2.0 cm -5

-1

-5

-1

 -6 mg   2.0 x 10  cm 2s   N A,0 = + cosh ( 2.0 cm -1 ) ( 0.5 cm )

(

)

2 cm3   mg  -1  -5 cm  2.0 cm 1.0 x 10 0.10 3.0 3  tanh ( 2.0 cm -1 ) ( 0.5 cm ) ( )  3  s  cm   cm   mg  86,400 s  mg N A,0 = 5.87 x 10-6  = 0.507 2 2  cm s  day  cm day

(

)

26-54

26.31 Develop Model for Fick’s flux equation General Flux Equation dy −CDAB A + y A ( N A, z + N B , z ) N A, z = dz Energy balance on adiabatic process N A, z ( ∆H v , A ) + N B , z ( ∆H v , B ) = 0

( ∆H ) = N ( ∆H ) ( ∆H ) = 30 kJ/mol =0.909 ( ∆H ) 33 kJ/mol

− N A, z

v, A

B,z

v,B

v, A v,B

Combine

 ( ∆H v, A ) N  dy dy −CDAB A + y A ( N A, z + N B , z ) = −CDAB A + y A  N A, z − N A, z =  dz dz ( ∆H v,B ) A, z    ( ∆H v , A )  dy −CDAB A + y A N A, z 1 − N A, z =   ( ∆H v , B )  dz  

Simplify for N A, z dy N A, z − 0.091 y A N A, z = −CDAB A dz δ

y As

dy A 1 − 0.091 y A ) y A∞ (

N A, z ∫ dz = −CDAB ∫ 0

N A, zδ = N A, z =

CDAB  1 − 0.091 y As  ln   0.091 1 − 0.091 y A∞ 

 1 − 0.091 y As  CDAB ln   0.091δ 1 − 0.091 y A∞ 

26-55

26.32 L = 4.0 cm (along x), H = 6.0 cm (along y), W = 4.0 cm (along z) PA 0.10 atm = c*A = = 4.062 gmole/m3 3 RT   -5 atm ⋅ m  8.206 x 10  (300 K) gmole ⋅ K   DAB =.020 cm2/sec a. Flux out of y-z and x-z planes Flux N A (0,y) on y-z plane, x = 0, 0 ≤ y ≤ H

∞ (−1)  nπ x   nπ y  −4 c*A ∑ c A ( x, y ) = n= sin  sinh  1,3,5  L  L   nπ H    n =1 nπ sinh    L  ∂c A ( x, y ) 4 c* ∞ (−1) n  nπ x   nπ y  = − A∑ n= cos  sinh  1,3,5  ∂x L n =1 L  L   nπ H    sinh    L  n

∂c A (0, y ) 4 DAB c*A ∞ (−1) n  nπ y  N A, x (0, y ) = − DAB = sinh  1,3,5 n= ∑ ∂x L L   nπ H   n =1 sinh    L  Flux N A (x,0) on x-z plane, y = 0, 0 < x < L

4 c*A ∞ ∂c A ( x, y ) (−1) n  nπ x   nπ y  sin  cosh  1,3,5 n= = − ∑  L n =1 L  L  ∂y  nπ H    sinh    L 

4 c* ∞ ∂c A ( x, y ) (−1) n  nπ x   nπ y  sin  cosh  1,3,5 n= = − A∑  L n =1 L  L  ∂y  nπ H    sinh    L  ∂c ( x, 0) 4 DAB c*A ∞ (−1) n  nπ x  sin  1,3,5 N A, y ( x, 0) = n= − DAB A = ∑ L L  ∂y  nπ H   n =1 sinh    L  b. Average flux

N A, x x = 0 = N A, x x = 0

1 y=H N A, x (0, y ) dy H ∫y = 0 1 y = H 4 DAB c*A ∞ (−1) n  nπ y  sinh  = ∑  dy n 1,3,5 ∫ H y =0 L L   nπ H   n =1 sinh    L 

26-56

4 DAB c*A ∞ ∑ H n =1

N A, x x = 0

 (−1) n  nπ H   = cosh   − 1 n 1,3,5  L    nπ H    nπ sinh    L 

N A, x x =0 = 6 x 10-4 gmole/m 2sec (slow convergence)

N A, y

y =0

N A, y

y =0

N A, y

y =0

N A, y

y =0

1 x=L N A, y ( x, 0) dx L ∫x =0 1 x = L 4 DAB c*A ∞ (−1) n  nπ x  sin  = ∑  dx n 1,3,5 ∫ 0 x = L L  nπ H   L  n =1 sinh    L  4 DAB c*A ∞ ∑ L n =1

(−1) n = ( cos ( n π ) − 1) n 1,3,5  nπ H  n π sinh    L 

= 9.29 x 10-6 gmole/m 2 sec

3.5E-03

1.6E-05

3.0E-03

1.4E-05

- NA(x,0), gmol/m2-sec

- NA(0,y), gmole/m2-sec

Average flux values were calculated by implementation of convergent infinite series solution on computer spreadsheet.

2.5E-03 2.0E-03 1.5E-03 1.0E-03 5.0E-04

1.2E-05 1.0E-05 8.0E-06 6.0E-06 4.0E-06 2.0E-06 0.0E+00

0.0E+00 0.0

1.0

2.0

3.0

4.0

5.0

0.0

6.0

2.0

3.0

4.0

x (cm)

y (cm)

Calculated local flux NA,x (0,y) through y-z plane along y from y = 0 to y = H (source) c. Total Release Rate, WA WA =2 ⋅ W ⋅ H ⋅ N A, x x =0 +W ⋅ L ⋅ N A, y

1.0

Calculated local flux NA,y (x,0) through x-z plane along x

y =0

(

) (4.0 cm)(4.0 cm) ( 9.29 x 10 gmole/m sec ) (1m/100 cm)

=2(4.0 cm)(6.0 cm) 6 x 10-4 gmole/m 2 sec (1m/100 cm) 2 + -6

= 2.9 x 10-5 gmole/sec Note that flux is dominated by y-z plane, particularly near the source at y = H 26-57

26.33 Additional process information provided in Problem 25.13 Assumptions 1. 2. 3. 4. 5.

Steady state - constant source and sink for A No homogeneous reaction of A within porous layer 1-D flux along position r - sealed top and bottom ends Rapid reaction at catalyst surface (CA ≈ 0 at r = Ro) Dilute with respect to solute A - H2 (B) is dominant species

Model for total transfer rate,WA Flux Equation based on assumption 3

N A, r = − DAe

dC A dr

General Differential Equation for Mass Transfer based on assumptions 1-3 d ( r N A, r ) = 0 dr By continuity N A,r r = R ⋅ R = N A,r ⋅ r NA,r r constant along r Boundary Conditions: r = Ro, CA =CAs ≈ 0, r = R1, CA = CA∞

= WA 2= π rL N A, r 2π R= Ro WA is constant along r o L N A, r r C A∞ dr = − DAe ∫ dC A Ro r C As 2π LDAe ( C As − C A∞ ) WA = R  ln  1   Ro 

WA ∫

Aside: estimation of effective diffusion coefficient DAe FSG Correlation

26-58

1/ 2

 1 1  + 0.001T   MA MB   = 1/ 3 2 1/ 3 P ( Σv ) A + ( Σv ) B 1.75

DAB

(

)

1/2

0.001 ⋅ ( 473 )

1.75

 1 1  +   28 2 

(1.5) ( ( 40.92 ) + ( 7.07 ) ) 1/3

1/3 2

cm 2 DAB = 0.669 s

T 423 cm 2 4850(0.00010) = DKA 4850 = d pore = 1.885 MA 28 s       1 1 1 1 2 + + DAe = ε 2   = ( 0.40 )  2 2  cm   DKA DAB  1.885 cm 0.669  s s  cm 2 DAe = 0.079 s

−1

Aside: = C A y= yA AC C A = ( 0.04 )

P RT

(1.5 atm )  -5 m ⋅ atm   8.206×10  ( 423 K ) gmole ⋅ K   3

=1.73

gmole A m3

Back out R1 for WA = 9.238 ×10-6 gmole A/s  cm 2  gmole A  1m  -2π(5.0 cm)  0.079  ( 0-1.73)   s  m3  100 cm   −6 gmole A = −9.238 × 10 s  R1  ln    0.50 cm  R1 = 0.8 cm

26-59

27.1 Concentration profile cA(z,t) ∞

𝑐𝑐𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 4 1 𝑛𝑛𝑛𝑛𝑛𝑛 −�𝑛𝑛𝑛𝑛�2𝑋𝑋𝐷𝐷 = � sin � � 𝑒𝑒 2 𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 𝜋𝜋 𝑛𝑛 𝐿𝐿 𝑛𝑛=1

∞

4 1 𝑛𝑛𝑛𝑛𝑛𝑛 −�𝑛𝑛𝑛𝑛�2 𝑋𝑋𝐷𝐷 � 𝑒𝑒 2 + 𝑐𝑐𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴 = (𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 ) � sin � 𝜋𝜋 𝑛𝑛 𝐿𝐿 𝑛𝑛=1

Average concentration:

∞

1 𝐿𝐿 4 1 𝑛𝑛𝑛𝑛𝑛𝑛 −�𝑛𝑛𝑛𝑛�2𝑋𝑋𝐷𝐷 𝑐𝑐�𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 = � �(𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 ) � � sin � � 𝑒𝑒 2 �� 𝑑𝑑𝑑𝑑 𝐿𝐿 0 𝜋𝜋 𝑛𝑛 𝐿𝐿 ∞

𝑛𝑛=1

𝑛𝑛𝑛𝑛 2 𝑛𝑛𝑛𝑛 2 4 1 𝑐𝑐�𝐴𝐴 = 𝑐𝑐𝐴𝐴𝐴𝐴 + (𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 ) 2 � � 2 �𝑒𝑒 −� 2 � 𝑋𝑋𝐷𝐷 − cos(𝑛𝑛𝑛𝑛) 𝑒𝑒 −� 2 � 𝑋𝑋𝐷𝐷 �� 𝜋𝜋 𝑛𝑛 𝑛𝑛=1

Since cos(nπ) = -1

𝑛𝑛𝑛𝑛 2

∞

𝑒𝑒 −� 2 � 𝑋𝑋𝐷𝐷 8 𝑐𝑐�𝐴𝐴 = 𝑐𝑐𝐴𝐴𝐴𝐴 + (𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 ) 2 � � � 𝜋𝜋 𝑛𝑛2 𝑛𝑛=1

∞

𝑌𝑌� =

𝑐𝑐�𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 8 𝑒𝑒 = 2�� 𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴𝐴𝐴 𝜋𝜋 𝑛𝑛=1

𝑛𝑛𝑛𝑛 2 −� 2 � 𝑋𝑋𝐷𝐷

𝑛𝑛2

�

𝑛𝑛 = 1,3,5 …

See attached spreadsheet and plot

27-1

27.1 continued XD Y

1.00E-06 9.86E-01

1.00E-04 9.84E-01

1.00E-02 8.87E-01

1.00E-01 6.43E-01

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29

8.11E-01 9.01E-02 3.24E-02 1.65E-02 1.00E-02 6.70E-03 4.79E-03 3.60E-03 2.80E-03 2.24E-03 1.84E-03 1.53E-03 1.29E-03 1.11E-03 9.62E-04

8.10E-01 8.99E-02 3.22E-02 1.63E-02 9.81E-03 6.50E-03 4.60E-03 3.41E-03 2.61E-03 2.05E-03 1.65E-03 1.34E-03 1.11E-03 9.29E-04 7.83E-04

7.91E-01 7.21E-02 1.75E-02 4.94E-03 1.36E-03 3.38E-04 7.41E-05 1.40E-05 2.24E-06 3.04E-07 3.46E-08 3.29E-09 2.60E-10 1.71E-11 9.38E-13

6.33E-01 4.95E-01 3.02E-01 2.36E-01 6.87E-02 5.83E-03 3.56E-06 9.78E-03 1.06E-03 1.25E-05 1.36E-06 2.04E-11 4.64E-21 5.41E-50 6.79E-05 1.42E-07 6.24E-13 1.31E-15 5.26E-29 8.55E-56 3.66E-136 9.29E-08 5.21E-13 1.64E-23 9.22E-29 5.14E-55 1.60E-107 4.81E-265 2.09E-11 4.37E-20 1.91E-37 3.99E-46 1.59E-89 2.54E-176 0.00E+00 7.24E-16 7.83E-29 9.15E-55 9.90E-68 1.46E-132 3.19E-262 0.00E+00 3.73E-21 2.89E-39 1.75E-75 1.36E-93 3.84E-184 0.00E+00 0.00E+00 2.79E-27 2.17E-51 1.30E-99 1.01E-123 2.83E-244 0.00E+00 0.00E+00 3.01E-34 3.24E-65 3.74E-127 4.03E-158 0.00E+00 0.00E+00 0.00E+00 4.65E-42 9.62E-81 4.12E-158 8.54E-197 0.00E+00 0.00E+00 0.00E+00 1.02E-50 5.64E-98 1.73E-192 9.58E-240 0.00E+00 0.00E+00 0.00E+00 3.15E-60 6.49E-117 2.75E-230 5.66E-287 0.00E+00 0.00E+00 0.00E+00 1.38E-70 1.46E-137 1.65E-271 0.00E+00 0.00E+00 0.00E+00 0.00E+00 8.47E-82 6.45E-160 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 7.32E-94 5.55E-184 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

Series Term n =

2.00E-01 4.96E-01

4.00E-01 3.02E-01

5.00E-01 1.00E+00 2.00E+00 5.00E+00 2.36E-01 6.87E-02 5.83E-03 3.56E-06

1.00 0.80

0.60 0.40 0.20 0.00 1.0E-06 1.0E-05 1.0E-04 1.0E-03 1.0E-02 1.0E-01 1.0E+00 1.0E+01

27-2

27.2 A = aluminum (Al), B = solid silicon (Si) T = 1250 K L = 0.50 µm = 5.0 x 10-5 cm wAS = 0.010 wA0 = 0 Assume back side is an impermeable barrier, so that x1 = L 𝐷𝐷𝐴𝐴𝐴𝐴 = Do e−Qo/RT Do = 2.61 cm2/sec (Table 24.7) Qo = 319,100 J/gmole (Table 24.11) R = 8.314 J/gmole-K 𝐷𝐷𝐴𝐴𝐴𝐴 = (2.61 cm2 / sec ) exp[−(319,100 J/gmole )/(8.314 J/gmole ∙ K) (1250 K)] DAB = 1.207 x 10-13 cm2/sec Determine wA(z,t) for z = 0.25 µm at t = 10.0 hr (36,000 sec) USS Diffusion in a slab m = 0 (Al metal in direct contact silicon surface) 𝑛𝑛 =

𝑥𝑥 0.25 𝜇𝜇𝜇𝜇 = = 0.50 𝑥𝑥1 0.50 𝜇𝜇𝜇𝜇

𝑋𝑋𝐷𝐷 =

𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 (1.207 x 10−13 cm2 /sec)(36,000 sec) = = 1.738 (5.0 𝑥𝑥 10−5 𝑐𝑐𝑐𝑐)2 𝑥𝑥12

Figure F.1, Y ≈ 0.010 𝑌𝑌 =

𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) 𝑤𝑤𝐴𝐴𝐴𝐴 − 𝑤𝑤𝐴𝐴 (𝑥𝑥, 𝑡𝑡) = = 0.01 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0 𝑤𝑤𝐴𝐴𝐴𝐴 − 𝑤𝑤𝐴𝐴0

𝑤𝑤𝐴𝐴 (𝑥𝑥, 𝑡𝑡) = 𝑤𝑤𝐴𝐴𝐴𝐴 − 𝑌𝑌 ∙ (𝑤𝑤𝐴𝐴𝐴𝐴 − 𝑤𝑤𝐴𝐴0 )

= 0.010 − 0.01 (0.010) = 0.0099

27-3

27.3 a. Determine temperature (T) A = Boron, B = solid silicon (Si) ρSi = 5.0 x 1022 atoms Silicon/cm3 CAs = 5.0 x 1020 atoms Boron/cm3 at z = 0 CA0 = 0.0 atoms Boron/cm3 CA(z,t) = 1.7 x 1019 atoms Boron/cm3 at z = 0.20 μm after t = 30 min 𝐷𝐷𝐴𝐴𝐴𝐴 = Do e−Qo/RT Do = 0.019 cm2/sec Qo = 2.74x105 J/gmole R = 8.314 J/gmole-K USS diffusion in semi-infinite medium 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 𝑧𝑧 = erf � � = erf(∅) 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0 2�𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡

5.0 × 1020 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵/𝑐𝑐𝑚𝑚3 − 1.7 × 1019 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵/𝑐𝑐𝑚𝑚3 𝑧𝑧 = 0.966 = erf � � 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 2�𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 5.0 × 1020 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 − 0 𝑐𝑐𝑚𝑚3

Appendix L, erf(φ) = 1.50 1.50 =

0.00002 𝑐𝑐𝑐𝑐

2�(0.0019 𝑐𝑐𝑐𝑐2 /𝑠𝑠𝑠𝑠𝑠𝑠)(1800 𝑠𝑠𝑠𝑠𝑠𝑠)𝑒𝑒𝑒𝑒𝑒𝑒 � �−2.74 𝑥𝑥 105

T = 1204 K

b. Determine NA(0,t) at t = 10 min, 30 min −𝑄𝑄 𝑐𝑐𝑚𝑚2 𝐷𝐷𝐴𝐴𝐴𝐴 = 𝐷𝐷𝑜𝑜 𝑒𝑒 𝑅𝑅𝑅𝑅 = 0.019 𝑒𝑒𝑒𝑒𝑒𝑒 �

𝑠𝑠𝑠𝑠𝑠𝑠

𝐷𝐷𝐴𝐴𝐴𝐴 = 8.455 × 10−13 𝑐𝑐𝑚𝑚2 /𝑠𝑠

�−2.74 𝑥𝑥 105

�8.314

𝐷𝐷𝐴𝐴𝐴𝐴 (C − CA0 ) 𝜋𝜋𝜋𝜋 AS

𝐽𝐽 � /(8.314 𝐽𝐽/𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐾𝐾)𝑇𝑇� 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

𝐽𝐽 � 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

� 𝐽𝐽 � (1383 𝐾𝐾) 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾

𝑁𝑁𝐴𝐴 (0, 𝑡𝑡) = � 𝑁𝑁𝐴𝐴 (0, 𝑡𝑡) = � 𝑁𝑁𝐴𝐴 (0, 𝑡𝑡) = �

8.455 × 10−13 𝑐𝑐𝑐𝑐2 /𝑠𝑠𝑠𝑠𝑠𝑠 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 (5.0 × 1020 − 0)𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎/𝑐𝑐𝑐𝑐3 = 1.059 × 1013 𝜋𝜋(600 𝑠𝑠𝑠𝑠𝑠𝑠) 𝑐𝑐𝑚𝑚2 𝑠𝑠𝑠𝑠𝑠𝑠 8.455 × 10−13 𝑐𝑐𝑐𝑐2 /𝑠𝑠𝑠𝑠𝑠𝑠 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 (5.0 × 1020 − 0)𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎/𝑐𝑐𝑐𝑐3 = 6.114 × 1012 𝜋𝜋(1800 𝑠𝑠𝑠𝑠𝑠𝑠) 𝑐𝑐𝑚𝑚2 𝑠𝑠𝑠𝑠𝑠𝑠

27-4

27.4 A = Phosphorous (P), B = solid silicon (Si) T = 1000 °C (1273 K) S = 100 cm2 square Si wafer CAS = 1.0 x 1021 P atoms/cm3 Si at z = 0 CA0 = 0.0 Determine CA(z,t) (atoms P/ cm3) doped into Si at z = 0.468 μm and t = 40 min USS diffusion in semi-infinite medium 𝑧𝑧 𝜙𝜙 = 2�𝐷𝐷𝐴𝐴𝐴𝐴 𝑇𝑇 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑧𝑧, 𝑡𝑡) = erf(𝜙𝜙) 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0 Estimate DAB from data

4𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 (𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0 ) 𝜋𝜋

𝑚𝑚𝐴𝐴 (𝑡𝑡) − 𝑚𝑚𝐴𝐴0 = 𝑆𝑆�

6.0 𝑥𝑥 1018 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑃𝑃

4𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 (𝐶𝐶𝐴𝐴𝐴𝐴 ) 1 𝜋𝜋 100 sec 2 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑃𝑃 6.0 ∗ 1016 1 2 (𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠) 𝜋𝜋 2 sec = 𝐷𝐷𝐴𝐴𝐴𝐴 = 4(𝑆𝑆𝐶𝐶𝐴𝐴𝐴𝐴 )2 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑃𝑃 2 4(100 𝑐𝑐𝑐𝑐2 ) �1.0 𝑥𝑥 1021 � 𝑐𝑐𝑚𝑚3 𝐷𝐷𝐴𝐴𝐴𝐴 = 2.827 𝑥𝑥 10−13 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠 1 𝑐𝑐𝑐𝑐 0.467 𝜇𝜇𝜇𝜇 � 4 � 10 𝜇𝜇𝜇𝜇 𝜙𝜙 = = 0.90 2 𝑐𝑐𝑚𝑚 60 𝑠𝑠𝑠𝑠𝑠𝑠 2��2.827 𝑥𝑥 10−13 �40 𝑚𝑚𝑚𝑚𝑚𝑚 � 𝑠𝑠𝑠𝑠𝑠𝑠 � 1 𝑚𝑚𝑚𝑚𝑚𝑚 Appendix L, erf(φ) = 0.797 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 − 𝐶𝐶𝐴𝐴 (𝑧𝑧, 𝑡𝑡) 1.0 x 1021 𝑐𝑐𝑐𝑐3 erf(𝜙𝜙) = 0.797 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 1.0 x 1021 − 0 𝑐𝑐𝑐𝑐3 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶𝐴𝐴 (𝑧𝑧, 𝑡𝑡) = 2.03 𝑥𝑥 1020 𝑐𝑐𝑚𝑚3

𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 =

= 𝑆𝑆�

27-5

27.5 Given: A = Arsenic (As), B = solid silicon (Si) T = 1050 C Wafer dimensions: L = 1.0 mm, d = 10 cm Solute A: cAs = cA* = 2.3 × 1021 atoms As/cm3 at z = 0

cAo = 2.3 × 1017 atoms As/cm3 initial As concentration in solid Si DAB = 5.0 × 10−13 cm2/s

a. Time required and mA(t) to achieve CA(z,t) = 2.065 × 1020 atoms As/cm3 at z = 0.5 µm

USS diffusion in semi-infinite medium—concentration profile

atoms atoms − 2.065 10 × 20 3 cm cm3 atoms atoms 2.3 10 − 2.3 10 × 21 × 17 3 cm cm3

2.3 10 × 21

c As − c A ( z , t ) = erf = (φ ) c As − c Ao erf (φ ) = 0.9103

φ=

z 2 DAB t

From Appendix L, φ = 1.2 z2 t = = 4φ 2 DAB

−5

4 (1.2 )

2  −13 cm  5.0 10 ×   s  

= 868 s

( 5.0 × 10 cm /s ) (868 s ) = 2.08 × 10 cm < < L −13

DAB= ⋅t

Note S=

( 5.0 ×10 cm )

−5

πd2 4

USS diffusion in semi-infinite medium—total mass loaded in medium = m A ( t ) − m Ao

πd2 4

4 DAB t

( cAs − cAo )

27-6

 cm 2  4  5.0 10 × −13  ( 868 s ) π (10 cm ) s  atoms  2.3 10 × 21 − 2.3 10 × 17 mA ( t ) − mAo = 4 cm3 π 4.25 10 × 18 atoms As mA ( t ) − mAo = 2

πd2

(

)

π (10cm )

)

mAo = c Ao L = 2.3 × 10 atoms As/cm ( 0.1cm ) = 1.8 ×1016 atoms As 4 4 × 18 atoms As = 4.25 10 × 18 atoms As mA ( t ) = 1.8 × 1016 atoms As + 4.25 10 17

b. Plot of z and vs. t at φ = 1.2 1.20

1.00

0.80

z (µm)

z1/2 (µm)1/2

1.00

0.60 0.40

0.80

0.60

0.20

0.40

0.00 1000

2000 3000 Time, t (s)

4000

t1/2 (s)1/2

Note plot of z vs. t1/2 is linear c. Transfer rate WA at t = 5.0 min, 10.0 min π d 2 DAB WA = S ⋅ N A z =0 = ( c − c ) for t > 0 4 π t As Ao At t = 5.0 min = 300 s WA =

π (10 cm )

cm /s ) × ( 5.0 10 × ( 2.3 10 π ( 300 s ) −13

× 17 − 2.3 10

) atoms cm

− 2.3 10 × 17

) atoms cm

WA = 4.16 × 1015 atoms As / s

At t = 10.0 min = 600 s WA =

π (10 cm ) 4

cm /s ) × ( 5.0 10 × ( 2.3 10 π ( 600 s ) −13

WA = 2.94 ×1015 atoms As / s

Note as t increases, WA decreases. At t = 0, the flux is not defined because no concentration gradient for As in the solid Si initially exists. 27-7

27.6 a. yA(0,t) at t = 30 s Use Concentration-Time Charts m = 0 (no convective mass transfer resistance) = n

r 0 = =0 R 1.5 cm

DAet = XD = R2

s) ( 0.0143 cm /s ) ( 30= 0.19 2

(1.5 cm )

From Figure F.9 (sphere, n = 0, m = 0), Y ≅ 0.28 = Y

c A,1 − c A ( r, t ) c A∞ − c A ( 0, t ) y A∞ − y A ( 0, t ) = = c A,1 − c Ao c A∞ − c Ao y A∞ − y Ao

= y A ( 0, t ) y A∞ − Y ( y A∞ − y Ao ) y A ( 0, t )= 0.05 − (0.28)(0.05 − 0)= 0.036 b. Time t at yA(0,t) = 0.048

y A∞ − y A ( 0, t ) y A∞ − y Ao

( 0.050 − 0.048) =0.04 ( 0.05 − 0.0 )

From Figure F.3 (sphere, n = 0, m = 0), XD ≅ 0.40 X R 2 ( 0.4 )(1.5 cm ) = 63 s t= D = DAe  cm 2   0.0143  s   2

c. New time t if R = 3.0 cm X R 2 ( 0.4 )( 3.0 cm ) t= D = = 252 s (4 times longer) DAe  cm 2   0.0143  s   2

27-8

27.7 Comments: Without any insight to the patch shape or dimensions, we will assume the patch is planar and it is capable of supplying a constant supply of drug. We will also assume that this is the first patch applied, and that all tissue has no drug initially, with the tissue acting as semiinfinite medium for drug diffusion. A = drug, B = tissue CAS = CA* = 2.0 mole/m3 CA0 = 0.0 mole/m3 DAB = 1.0 x 10-6 cm2/sec Determine time required (t) for cA(z,t) = 0.20 mole/m3 at z = 0.5 cm USS diffusion in semi-infinite medium 𝑧𝑧 𝐶𝐶𝐴𝐴 − 𝐶𝐶𝐴𝐴𝐴𝐴 = erf � � = 𝑒𝑒𝑒𝑒𝑒𝑒(∅) 𝐶𝐶𝐴𝐴0 − 𝐶𝐶𝐴𝐴𝐴𝐴 2�𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡

0.5 𝑐𝑐𝑐𝑐 0.20 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 − 2.0 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 = 0.90 = erf � � 0.0 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 − 2.0 𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 2�(1.0 𝑥𝑥 10−6 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠 ) 𝑡𝑡 From Appendix L, ∅ = 1.1631 1.1631 =

0.5 𝑐𝑐𝑐𝑐

2�(1.0 𝑥𝑥 10−6 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠 ) 𝑡𝑡

𝑡𝑡 = 46,201 𝑠𝑠𝑠𝑠𝑠𝑠 = 12.83 ℎ𝑟𝑟

27-9

27.8 dye solution (solute A) z=0 sealed edges

z = 0.05 cm Solid (B) CAo = 0

z = L = 1.0 cm

a. Time t required for CA(z,t) = 0.0062 gmole/m3 with z = 0.050 cm Semi-infinite diffusion medium Strategy: Back out DAB from observed flux NA(0,t) at t = 120 s, then use DAB to get CA(z,t) N A (0, t) =

DAB ( c − c ) for t > 0 π t As Ao

-9 2 N A (0, t ) ] π t ( 5.15×10 gmole/m ⋅ s ) π (120 s ) [ DAB = = 2 2 ( cAs − cAo ) ( 0.01gmole/m3 - 0 ) 2

-10 2 m /s 1.0×10-6 cm 2 /s DAB 1.0×10 = =

c As − c A ( z , t ) c As − c Ao c As − c A ( z , t ) c As − c Ao

= erf (φ ) 0.010-0.0062 ) gmole/m3 ( = = 0.38 ( 0.010-0.0 ) gmole/m3

Appendix L, at erf(φ) = 0.38, φ = 0.35

φ=

z 2 DAB t

z2 t = = 4φ 2 DAB

( 0.050 cm ) 4 ( 0.35 )

2  −6 cm  × 1.0 10   s  

= 5102 s

b. The system is best represented by a semi-infinite medium

27-10

27.9 food (contains liquid water) z = L = 0.2 cm porous film (air + water vapor) z=0 nonporous metal plate

Use Concentration-Time Charts m = 0 (no convective mass transfer resistance) n=

z 0 =0 = L 0.20 cm

c A,1 − c A ( z, t ) PA* − PA ( 0, t ) = Y = = c A,1 − c Ao PA* − PAo

( 0.030-0.015) atm = 0.50 ( 0.030-0.0 ) atm

Figure F.7 (slab, n = 0, m = 0), XD = 0.38 X L2 ( 0.38 )( 0.2 cm ) t= D = = 117 s DAe  cm 2   0.00013  s   2

27-11

27.10 A = H2O vapor, B = air R = 1.0 mm = 0.10 cm 𝐶𝐶𝐴𝐴𝐴𝐴 = 0.90 𝐶𝐶𝐴𝐴∗ CA0 = 0 DAB = 0.260 cm2/sec at T = 298 K and P = 1.0 atm (Table J.1) Determine time required for center of air bubble to reach water vapor concentration of 90% of saturation, 𝐶𝐶𝐴𝐴 (0, 𝑡𝑡) = 0.90 𝐶𝐶𝐴𝐴∗

USS Diffusion in a sphere, use Concentration -Time charts 𝑚𝑚 = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟) ∴ 𝐶𝐶𝐴𝐴𝐴𝐴 = 𝐶𝐶𝐴𝐴∞ 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑟𝑟, 𝑡𝑡) 𝐶𝐶𝐴𝐴∗ −0.90𝐶𝐶𝐴𝐴∗ = = 0.10 𝑌𝑌 = 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0 𝐶𝐶𝐴𝐴∗ − 0 𝑛𝑛 =

𝑟𝑟 0 𝑐𝑐𝑐𝑐 = =0 𝑅𝑅 0.10 𝑐𝑐𝑐𝑐

Figure F.3

𝑋𝑋𝐷𝐷 ≅ 0.40 =

𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 𝑅𝑅2

𝑋𝑋𝐷𝐷 𝑅𝑅2 (0.40)(0.1 𝑐𝑐𝑐𝑐)2 𝑡𝑡 = = = 0.015 𝑠𝑠𝑠𝑠𝑠𝑠 𝐷𝐷𝐴𝐴𝐴𝐴 0.260 𝑐𝑐𝑐𝑐2 /𝑠𝑠𝑠𝑠𝑠𝑠

27-12

27.11 A = CO2, B = air T = 25 °C, P = 1.0 atm x1=L = 2.0 cm CA∞ = 0 DAB = 0.161 cm2/sec DAe = 0.010 cm2/sec kc = 0.0025 cm/sec Determine time required for yA(x,t) = 0.02 at 1.6 cm from the surface of the slab USS Diffusion in a slab, use Concentration -Time charts 𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) 𝑦𝑦𝐴𝐴∞ − 𝑦𝑦𝐴𝐴 (𝑥𝑥, 𝑡𝑡) = 𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴0 𝑦𝑦𝐴𝐴∞ − 𝑦𝑦𝐴𝐴0

𝑌𝑌 =

0 − 0.02 = 0.2 0 − 0.10

𝑌𝑌 = 𝑛𝑛 =

𝑚𝑚 =

2.0 𝑐𝑐𝑐𝑐 − 1.6 𝑐𝑐𝑐𝑐 𝑥𝑥 = = 0.20 𝑥𝑥1 2.0 𝑐𝑐𝑐𝑐

𝐷𝐷𝐴𝐴𝐴𝐴 𝐷𝐷𝐴𝐴𝐴𝐴 0.010 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠 = = = 2.0 𝑘𝑘𝑐𝑐 𝑥𝑥1 𝑘𝑘𝑐𝑐 𝐿𝐿 �0.0025 𝑐𝑐𝑐𝑐 � (2.0 𝑐𝑐𝑐𝑐) 𝑠𝑠𝑠𝑠𝑠𝑠

Figure F.1

𝑋𝑋𝐷𝐷 = 4.0 = 𝑡𝑡 =

𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 𝑥𝑥12

𝑋𝑋𝐷𝐷 𝑥𝑥12 4.0(2.0 𝑐𝑐𝑐𝑐)2 = = 1600 𝑠𝑠𝑒𝑒𝑐𝑐 𝐷𝐷𝐴𝐴𝐴𝐴 0.01 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠

27-13

27.12 A = solute, B = adsorbent material D = 1.0 cm, R = 0.5 cm L = 5.0 cm K = 0.15 cm3 fluid, cm3 adsorbent DAB = 4.0 x 10-7 cm2/sec 𝐶𝐶𝐴𝐴∞ ′ = 2.0 gmole A/m3 fluid

a. Determine time (t) required for CA (r,t) = 2.94 gmol/m3, r = 0.10 cm (0.40 cm from surface), assuming no end effects USS Diffusion in a cylinder, use Concentration -Time charts ′ 𝐶𝐶𝐴𝐴𝐴𝐴 = 𝐶𝐶𝐴𝐴 (𝑅𝑅, 𝑡𝑡) = 𝐾𝐾 ∗ 𝐶𝐶𝐴𝐴∞ = 2.0

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐴𝐴 1.5 𝑚𝑚3 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐴𝐴 � � = 3.0 𝑚𝑚3 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑚𝑚3 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚3

𝑚𝑚 = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟), ∴ 𝐶𝐶𝐴𝐴𝐴𝐴 = 𝐶𝐶𝐴𝐴∞ 𝑌𝑌 =

𝑛𝑛 =

𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑟𝑟, 𝑡𝑡) (3.0 − 2.94)𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔/𝑚𝑚3 = = 0.020 (3.0 − 0)𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔/𝑚𝑚3 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴𝐴𝐴 𝑟𝑟 0.10 𝑐𝑐𝑐𝑐 = = 0.20 𝑅𝑅 0.50 𝑐𝑐𝑐𝑐

Figure F.2

𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 𝑅𝑅2 𝑋𝑋𝐷𝐷 𝑅𝑅2 0.75(0.5 𝑐𝑐𝑐𝑐)2 𝑡𝑡 = = = 468,750 𝑠𝑠𝑠𝑠𝑠𝑠 = 130.2 ℎ𝑟𝑟 (4.0 𝑥𝑥10−7 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠) 𝐷𝐷𝐴𝐴𝐴𝐴

𝑋𝑋𝐷𝐷 = 0.75 =

b. Determine time (t) required for CA (r,z,t) = 2.94 gmol/m3, r = 0.10 cm (0.40 cm from surface), z = 1.5 cm (1.0 cm from end) USS Diffusion in a cylinder, use Concentration -Time charts 𝑌𝑌 = 𝑌𝑌𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑌𝑌𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑌𝑌𝑎𝑎

Since XD will be different for Ycylinder and Ya, an iterative solution is required were time (t) is guessed, Ycylinder , Ya and Y are calculated, and CA (r,z,t) = 2.94 gmol/m3 is checked Guess t = 100 hr (360,000 sec) 𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 4.0 x 10−7 cm2 /sec(360,000 sec) 𝑋𝑋𝐷𝐷 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = 2 = 2 = = 0.576 (0.5𝑐𝑐𝑐𝑐)2 𝑅𝑅 𝑥𝑥1 27-14

𝑋𝑋𝐷𝐷 (𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠) =

𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 4.0 x 10−7 cm2 /sec(360,000 sec) = 2 = = 0.023 2 𝑎𝑎 𝑥𝑥12 0.5 � 𝑐𝑐𝑐𝑐� 2

𝑛𝑛 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) =

𝑟𝑟 0.10 𝑐𝑐𝑐𝑐 = = 0.20 𝑅𝑅 0.50 𝑐𝑐𝑐𝑐

𝑥𝑥 𝑥𝑥 2.5 𝑐𝑐𝑐𝑐 − 1.0 𝑐𝑐𝑐𝑐 𝑛𝑛(𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠) = � � = = = 0.60 𝑥𝑥1 2.5 𝑐𝑐𝑐𝑐 𝐿𝐿/2

𝑚𝑚 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 ) 𝑚𝑚 (𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠) = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟) Figures F.2 and F.1

𝑌𝑌𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 0.0541, 𝑌𝑌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 0.7073

𝑌𝑌 = 𝑌𝑌𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑒𝑒𝑒𝑒 𝑌𝑌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 0.03826 𝑌𝑌 =

𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑟𝑟, 𝑧𝑧, 𝑡𝑡) 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴𝐴𝐴

𝐶𝐶𝐴𝐴 (𝑟𝑟, 𝑧𝑧, 𝑡𝑡) = 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝑌𝑌(𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴𝐴𝐴 ) = 3.00 − (0.0386)(3.00 − 0) = 2.88 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔/𝑚𝑚3 Guess t = 110 hr = 396,000 sec, CA (r,z,t) = 2.91 gmole/m3

Guess t = 120 hr = 432,000 sec, CA (r,z,t) = 2.94 gmole/m3 (done)

27-15

27.13 groundwater flow

water flow (v∞)

DAB = 1. 5x10-6 cm2/s kc = 1.0 x 10-6 cm2/s

CAo= 2.0 mg/cm3 DAe = 5x10 DA-7 cm2/s

CAs

CA∞ ≈ 0

pesticide-filled polymer capsules (1.0 cm diameter) buried in loose soil

a. cA(0,t) at time t = 173.6 h Use Concentration-Time Charts DAB = m = kc R = n

(1.5×10 cm /s ) = 3.0 (1.0×10 cm/s ) ( 0.5 cm ) -6

-6

r 0 = =0 R 0.5 cm s ( 5.0×10 cm /s ) 173.6 h 3600  h  -7

= XD

DAet = R2

( 0.5 cm )

= 1.25

Figure F.9 (sphere, n = 0, m = 3.0, XD = 3.0), Y = 0.33 Y =

c A,1 − c A ( r, t ) c A∞ − c A ( 0, t ) = c A,1 − c Ao c A∞ − c Ao

c A ( 0, t ) c A∞ − Y ( c A∞ − c Ao ) = 0 - 0.33 ( 0 - 2.0 ) mg/cm3 = c A (0, t ) = 0.66 mg/cm3 b. Surface cA(R,t) and NA(R,t) at t = 173.6 h At sphere surface, cA(R,t) = cAs

= n

r R = =1 R R

Figure F.9 (sphere, n = 0, m = 3.0, XD = 3.0), Y = 0.30

27-16

= Y

c A,1 − c A ( r, t ) c A∞ − c A ( R, t ) = c A,1 − c Ao c A∞ − c Ao

= c A ( R, t ) c A∞ − 0.30 ( c A∞ − c Ao ) = 0 - 0.33 ( 0 - 2.0 ) mg/cm3 c A ( R, t ) = 0.60 mg/cm3

= N A ( R, t ) kc [ c A ( R, t ) − c A∞ ] N A ( R, t ) = 1.0×10-6

cm mg mg ( 0.60 - 0 ) 3 =6.0×10-7 2 s cm cm s

27-17

27.14 A = cadmium, B = biopolymer D = 0.50 cm, R = 0.25 cm DAB = 2.0 x 10–8 cm2/sec K = 150 m3 aqueous phase/m3 biopolymer phase CA∞ = 0.10 mol/m3 / 𝐶𝐶𝐴𝐴0 = 0 mol/m3 /

Determine time (t) required for 𝐶𝐶𝐴𝐴 (0, 𝑡𝑡) = 12 mol/m3 at center of spherical absorbent bead (r = 0) USS Diffusion in a sphere, use Concentration -Time charts 𝑚𝑚 = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟) /

𝐶𝐶𝐴𝐴𝐴𝐴 = 𝐾𝐾 ∙ 𝐶𝐶𝐴𝐴∞ = (150 𝑚𝑚3 /𝑚𝑚3 )(0.10 𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 ) = 15 𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 /

𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴𝐴𝐴 (𝑟𝑟, 𝑡𝑡)

𝑌𝑌 =

𝑛𝑛 =

𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0

𝑟𝑟 0 𝑐𝑐𝑐𝑐 = =0 𝑅𝑅 0.25 𝑐𝑐𝑐𝑐

𝐾𝐾𝐾𝐾𝐴𝐴∞ − 𝐶𝐶𝐴𝐴𝐴𝐴 (𝑟𝑟, 𝑡𝑡) /

𝐾𝐾𝐾𝐾𝐴𝐴∞ − 𝐶𝐶𝐴𝐴0

(15 − 12)𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 = = 0.20 15 𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 − 0

Figure F.3

𝑋𝑋𝐷𝐷 ≅ 0.25 = 𝑡𝑡 =

𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 𝑅𝑅2

𝑋𝑋𝐷𝐷 𝑅𝑅2 0.25(0.25 𝑐𝑐𝑐𝑐)2 = = 7.813 𝑥𝑥 105 𝑠𝑠𝑠𝑠𝑠𝑠 (2.0 𝑥𝑥10−8 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠) 𝐷𝐷𝐴𝐴𝐴𝐴

t = 217 hr

27-18

27.15 Given: A = acetic acid, B = H2O (pickle) T = 80 C(353 K) Pickle Dimensions: D = 2.5 cm, R = 1.25 cm, L = 12 cm cAo = 0

initial concentration of A inside pickle

cA∞ = 0.70 kgmole/m3 bulk fluid concentration of A DAB = 1.21 × 10−5 cm2/s at T = 293 K

a. Time (t) required for cA(0,t) = 0.864 kgmole/m3 USS diffusion in a cylinder, use concentration–time charts m = 0, no convection resistance assumed ∴cA,1 = cA∞ = 0900 kgmole/m3 = n

r 0 = = 0 relative position R 1.25 cm

Relative concentration at r = 0 (n = 0) and m = 0 c A,1 − c A ( r, t ) c A∞ − c A ( r, t ) c A∞ − c A ( 0, t ) = = = Y = c A∞ − c Ao c A,1 − c Ao c A∞ − c Ao

kgmole m3 = 0.040 kgmole ( 0.900 − 0 ) 3 m

( 0.900 − 0.864 )

From Figure F.2 cylinder, at Y = 0.040, n = 0, m = 0, read off XD = X D 0.625 =

DAet DAB t = R2 R2

Scale diffusion coefficient from reference temperature (Tref) to process temperature (T) assuming pickle approximates properties of liquid water DAB (T )

( (

) )

2 × −6 Pa ⋅ s   T   µ (Tref )   353 K  993 10 −5 cm  1.21 10 × D= T ( )       AB ref   s  × −6 Pa ⋅ s   Tref   µ (T )   293 K  352 10

× −5 DAB = 4.112 10

cm 2 s 27-19

X R 2 ( 0.625 )(1.25 cm ) =23,749 s = 6.6 h t= D = 2 DAB  −5 cm   4.112 ×10  s   2

b. Time (t) required for cA(0,t) = 0.864 kgmole/m3 for kc = 1.94 × 10−5 cm/s USS diffusion in a cylinder, use concentration–time charts with convection resistance Y = 0.040, n = 0 unchanged from part (a) cm 2 4.112 10 × −5 DAB s =1.7 m = = kC R  −5 cm  × 1.94 10  (1.25 cm ) s   Use Figure F.5—cylinder, read off XD from Y = 0.040, n = 0, m = 1.7 3.0 (1.25 cm ) = 113,996 s = 31.7 h 2 −5 cm 4.112 10 × s Note that if m > 0 that convective resistance increases the time required. X D R2 Read off XD = 3.0, and= so t = DAB

27-20

27.16 A = solvent, B = polymer x1 = L = 6.0 mm = 0.60 cm 𝐶𝐶𝐴𝐴∞ = 0 wA0 = 0.010 DAB = 2.0 x 10-6 cm2/sec Determine time (t) required for wA(x,t) = 0.00035 at 1.2 mm (0.12 cm) from the exposed surface USS Diffusion in a slab, use Concentration -Time charts 𝑚𝑚 = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟)

𝑌𝑌 = 𝑛𝑛 =

𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴 (0, 𝑡𝑡) 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) 𝑤𝑤𝐴𝐴𝐴𝐴 − 𝑤𝑤𝐴𝐴 (𝑥𝑥, 𝑡𝑡) 0 − 0.00035 = = = = 0.035 𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴0 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴𝑜𝑜 0 − 0.010 𝑤𝑤𝐴𝐴𝑆𝑆 − 𝑤𝑤𝐴𝐴0

𝑥𝑥 0.60 𝑐𝑐𝑐𝑐 − 0.12 𝑐𝑐𝑐𝑐 = = 0.80 𝑥𝑥1 0.60 𝑐𝑐𝑐𝑐

Figure F.1

𝑋𝑋𝐷𝐷 ≅ 1.0 = 𝑡𝑡 = 1.0

𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 𝑥𝑥12

(0.60 𝑐𝑐𝑐𝑐)2 𝑥𝑥12 = 1.0 = 180,000 sec (50 hr) 𝐷𝐷𝐴𝐴𝐴𝐴 2.0 𝑥𝑥 10−6 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠

27-21

27.17 A = glucose, B = gel (liquid water) L = 1.0 cm x1 = L/2 = 0.50 cm (symmetry) CA0 = 0 𝐶𝐶𝐴𝐴∞ = 50 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3

Determine DAB for CA(x,t) = 48.5 mole/m3 at x = 0 (center, symmetry) and t = 42 hr (151,200 sec)

USS Diffusion in a slab, use Concentration -Time charts 𝑚𝑚 = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟) 𝑌𝑌 = 𝑛𝑛 =

𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) (50.0 − 48.5)𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 = = = 0.030 (50.0 − 0.0)𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴0 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴𝑜𝑜

𝑥𝑥 0 = =0 𝑥𝑥1 0.50 𝑐𝑐𝑐𝑐

Figure F.1

𝑋𝑋𝐷𝐷 ≅ 1.55 = 1.55 =

𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 𝑥𝑥12

𝐷𝐷𝐴𝐴𝐴𝐴 (151200 𝑠𝑠𝑠𝑠𝑠𝑠) (0.5 𝑐𝑐𝑚𝑚)2

DAB = 2.56 x 10-6 cm2/sec

27-22

27.18 A = griseofulvin (drug), B = gel D = 0.10 cm, R = 0.050 cm CA0 = 0.200 mmol/L 𝐶𝐶𝐴𝐴∞ = 0 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚3 DAB = 1.5 x 10–7 cm2/sec a. Determine time required (t) for CA(0,t) = 0.10CA0 at r = 0 (center of bead) USS Diffusion in a sphere, use Concentration -Time charts 𝑚𝑚 = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟) 𝑌𝑌 = 𝑛𝑛 =

𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴 (𝑟𝑟, 𝑡𝑡) 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑟𝑟, 𝑡𝑡) 0 − 0.10𝐶𝐶𝐴𝐴0 = = = 0.10 𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴0 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0 0 − 𝐶𝐶𝐴𝐴0

0 𝑐𝑐𝑐𝑐 𝑟𝑟 = =0 𝑅𝑅 0.050 𝑐𝑐𝑐𝑐

Figure F.3 or F.9 𝑋𝑋𝐷𝐷 ≅ 0.30 = 𝑡𝑡 = 0.30

𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 𝑅𝑅2

(0.050 𝑐𝑐𝑐𝑐)2 𝑅𝑅2 = 0.30 = 5000 𝑠𝑠𝑠𝑠𝑠𝑠 𝐷𝐷𝐴𝐴𝐴𝐴 1.5 𝑥𝑥 10−7 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠

b. Determine time required (t) for CA(0,t) = 0.10CA0 at r = 0 (center of bead) using infinite series solution, and plot out concentration profile CA(r,t) At r = 0 ∞

𝐶𝐶𝐴𝐴 (0, 𝑡𝑡) − 𝐶𝐶𝐴𝐴0 2 2 2 𝑌𝑌 = = 0.90 = 1 + 2 �(−1)𝑛𝑛 𝑒𝑒 −𝐷𝐷𝐴𝐴𝐴𝐴𝑛𝑛 𝜋𝜋 𝑡𝑡/𝑅𝑅 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴0 𝑛𝑛=1

𝑛𝑛 = 1,2,3 …

See spreadsheet and plot. The required time (t) was guessed iteratively until CA(0,t) = 0.020 mmol/L using the Solver function in Microsoft Excel, t = 5058 sec.

27-23

27.18 continued 2

D AB = R= C Ao =

1.50E-07 cm /s 0.050 cm 0.20 mmol/L

C As =

0 mmol/L

t= t= ∆r =

1.405 hr 5058.635 sec 0.0083 cm (plot interval)

C A(0,t)/CAo = Solver =

0.10 -7.44E-08

r (cm) Y(r,t) C A(r,t) (mmol/L)

0.00E+00 9.00E-01 2.00E-02

1 8.33E-03 9.05E-01 1.91E-02

2 1.67E-02 9.17E-01 1.65E-02

3 2.50E-02 9.36E-01 1.27E-02

4 3.33E-02 9.59E-01 8.27E-03

5 4.17E-02 9.81E-01 3.82E-03

6 5.00E-02 1.00E+00 0.00E+00

1 2 3 4 5 6 7 8 9 10

-5.00E-02 6.25E-06 -1.96E-12 1.53E-21 -2.99E-33 1.46E-47 -1.79E-64 5.47E-84 -4.18E-106 7.99E-131

-2.50E-02 2.71E-06 -6.52E-13 3.31E-22 -2.99E-34 2.99E-64 1.28E-65 -5.92E-85 4.64E-107 -6.92E-132

-4.33E-02 2.71E-06 -7.99E-29 -3.31E-22 5.18E-34 -5.97E-64 -2.21E-65 5.92E-85 -1.71E-122 -6.92E-132

-5.00E-02 3.83E-22 6.52E-13 -9.37E-38 -5.98E-34 8.96E-64 2.55E-65 -3.35E-100 -4.64E-107 4.89E-147

-4.33E-02 -2.71E-06 1.60E-28 3.31E-22 5.18E-34 -1.19E-63 -2.21E-65 -5.92E-85 3.41E-122 6.92E-132

-2.50E-02 -2.71E-06 -6.52E-13 -3.31E-22 -2.99E-34 1.49E-63 1.28E-65 5.92E-85 4.64E-107 6.92E-132

-6.13E-18 -7.66E-22 -2.40E-28 -1.87E-37 -3.66E-49 -1.79E-63 -2.19E-80 -6.70E-100 -5.12E-122 -9.79E-147

Series Term n =

CA(r,t) (mmol/L)

0.03

0.02

0.01

0.00 0.00

0.01

0.02 0.03 0.04 Radial postion, r (cm)

0.05

0.06

27-24

27.19 A = drug (Dramamine), B = gel L = 0.652 cm x1 = L/2 = 0.326 cm (symmetry) CA0 = 64 mg A/cm3 CA∞ = 0 DAB = 3.0 x 10-7 cm2/sec a. Determine CA(x, t) at the center of the slab (x = 0, symmetry) at t = 96 hr, using Concentration -Time charts USS Diffusion in a slab

𝑋𝑋𝐷𝐷 =

𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 = 𝑥𝑥12

3.0 𝑥𝑥 10−7

𝑐𝑐𝑚𝑚2 3600 𝑠𝑠𝑠𝑠𝑠𝑠 ∗ 96 ℎ𝑟𝑟 � � 𝑠𝑠𝑠𝑠𝑠𝑠 ℎ𝑟𝑟 = 0.976 (0.326 𝑐𝑐𝑐𝑐)2

𝑚𝑚 = 0 (𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟) 𝑛𝑛 =

𝑥𝑥 0 = =0 𝑥𝑥1 0.326 𝑐𝑐𝑐𝑐

𝑌𝑌 =

𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) 𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) = = ≅ 0.12 𝐶𝐶𝐴𝐴∞ − 𝐶𝐶𝐴𝐴0 𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0 𝐶𝐶𝐴𝐴0

Figure F.1

𝐶𝐶𝐴𝐴 (𝑥𝑥, 𝑡𝑡) = 𝑌𝑌 ∙ 𝐶𝐶𝐴𝐴0 = 0.12(64 𝑚𝑚𝑚𝑚 𝐴𝐴/𝑐𝑐𝑚𝑚3 ) = 7.7 𝑚𝑚𝑚𝑚 𝐴𝐴/𝑐𝑐𝑚𝑚3

b. Determine CA(x, t) at the center of the slab (x = L/2) at t = 96 hr, using the infinite series solution approach (x = 0, surface; x = L/2 center) C A − C AS 4 ∞ 1  nπ x  −( nπ /2)2 X D = ∑ sin  , e C A0 − c AS π n =1 n  L 

n = 1,3,5,

See Microsoft Excel spreadsheet, CA(x,t) = 7.34 mg A/cm3

27-25

27.19 continued 2

D AB =

3.00E-07 cm /sec

C AS =

0.0 mg/cm 3 64.00 mg/cm 345600 sec

96.000 hr

x 1 = L/2 = x= C A (x,t) = XD = n

0.326 cm 0.326 cm (center of slab) 3 7.34 mg/cm 0.975573 Term

Summation

% Change

1.15E-01

1.147E-01

-1.66E-10

1.147E-01

0.000

1.87E-27

1.147E-01

0.000

-1.08E-52

1.147E-01

0.000

2.97E-86

1.147E-01

0.000

-3.71E-128

1.147E-01

0.000

2.08E-178

1.147E-01

0.000

-5.17E-237

1.147E-01

0.000

5.66E-304

1.147E-01

0.000

0.00E+00

1.147E-01

0.000

0.00E+00

1.147E-01

0.000

0.00E+00

1.147E-01

0.000

0.00E+00

1.147E-01

0.000

0.00E+00

1.147E-01

0.000

0.00E+00

1.147E-01

0.000

0.00E+00

1.147E-01

0.000

0.00E+00

1.147E-01

0.000

0.00E+00

1.147E-01

0.000

0.00E+00

1.147E-01

0.000

1.40

1.20

1.00

0.80

0.80 Summation

0.60

Term

0.60

0.40

0.20

0.00

-0.20

-0.40

Term

Summation of terms

C A0 =

-0.40 0 1 2 3 4 5 6 7 8 9 10111213141516171819202122232425

27-26

27.20 L = 0.125 cm, x1 = L/2 (symmetry) W = 0.50 cm, S = W2 𝐶𝐶𝐴𝐴∞ = 𝐶𝐶𝐴𝐴𝐴𝐴 = 0 CA0 = 64 mg/cm3 DAB = 3.0 x 10-7 cm2/sec USS Diffusion in a slab, total amount transferred mA(t) by infinite series solution, assuming sealed edges At surface (x = 0) ∞ 2 4 DAB = N Az ( 0 ,t) (C As − C Ao )∑ e − (n π / 2 ) X D L n =1 𝐷𝐷𝐴𝐴𝐴𝐴 𝑡𝑡 𝑋𝑋𝐷𝐷 = 𝑥𝑥12

m Ao = C Ao ⋅ S ⋅ L initial amount (moles A) loaded within slab at t = 0 mA∞= C AS ⋅ S ⋅ L

final (equilibrium) amount loaded within slab at t →∞

Flux is into both sides of the slab of thickness L t

2S ∫ N 2S ∫ = m= A (t) − m Ao Az ( 0 ,t)dt

 n 2π 2 DAB t   L2 

− mA∞ − mA (t ) 8 ∞ 1 = 2 ∑ 2 exp  mA∞ − mAo π n =1 n



∞ − (n π / 2 ) 4 DAB (C AS − C A0 )∑ e  L n =1

2 DAB t   x12 

n = 1, 3, 5…∞

𝑚𝑚𝐴𝐴𝐴𝐴 = 𝐶𝐶𝐴𝐴𝐴𝐴 𝑊𝑊 2 𝐿𝐿 = (64 𝑚𝑚𝑚𝑚/𝑐𝑐𝑚𝑚3 )(0.5 𝑐𝑐𝑐𝑐)2 (0.125 𝑐𝑐𝑐𝑐) = 2.00 mg A 𝑚𝑚𝐴𝐴∞ = 0 See Microsoft Excel spreadsheet At t = 1.0 hr (3600 sec) 𝑚𝑚𝐴𝐴0 − 𝑚𝑚𝐴𝐴 (𝑡𝑡) = 0.593 𝑚𝑚𝑚𝑚 𝐴𝐴 At t = 2.0 hr (7200 sec) 𝑚𝑚𝐴𝐴0 − 𝑚𝑚𝐴𝐴 (𝑡𝑡) = 0.839 𝑚𝑚𝑚𝑚 𝐴𝐴

27-27

27.20 continued 2

C AS =

3.00E-07 cm /sec 3 0.0 mg/cm

C Ao =

64.00 mg/cm

m Ao =

2.00 mg

m A,inf =

0.00 mg

m Ao - m A =

0.593 mg delivered Term Summation % Change 6.83E-01 6.835E-01 2.8 1.94E-02 7.029E-01 4.56E-04 7.033E-01 0.1 3.88E-06 7.033E-01 0.0 1.00E-08 7.033E-01 0.0 7.31E-12 7.033E-01 0.0 1.46E-15 7.033E-01 0.0 7.79E-20 7.033E-01 0.0 1.10E-24 7.033E-01 0.0 4.10E-30 7.033E-01 0.0 3.99E-36 7.033E-01 0.0 1.01E-42 7.033E-01 0.0 6.62E-50 7.033E-01 0.0 1.12E-57 7.033E-01 0.0 4.93E-66 7.033E-01 0.0 5.59E-75 7.033E-01 0.0 1.63E-84 7.033E-01 0.0 1.22E-94 7.033E-01 0.0 2.37E-105 7.033E-01 0.0

c As =

3.00E-07 cm /sec 3 0.0 mg/cm

c Ao =

64.00 mg/cm

mAo =

2.00 mg

mA,inf =

0.00 mg 7200 sec 2.00 hr 1.161 mg remaining

mAo - mA =

0.839 mg delivered

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37

1.00 Summation Term

Summation of terms

0.80

0.60

0.40

0.20

0.00

0.00 0

0.125 cm

0.50 cm

t= t= mA =

1.00

Term Summation % Change 5.76E-01 5.763E-01 4.18E-03 5.805E-01 0.7 6.42E-06 5.805E-01 0.0 9.12E-10 5.805E-01 0.0 1.00E-14 5.805E-01 0.0 7.97E-21 5.805E-01 0.0 4.43E-28 5.805E-01 0.0 1.68E-36 5.805E-01 0.0 4.33E-46 5.805E-01 0.0 7.49E-57 5.805E-01 0.0 8.65E-69 5.805E-01 0.0 6.64E-82 5.805E-01 0.0 3.38E-96 5.805E-01 0.0 1.14E-111 5.805E-01 0.0 2.53E-128 5.805E-01 0.0 3.70E-146 5.805E-01 0.0 3.57E-165 5.805E-01 0.0 2.27E-185 5.805E-01 0.0 9.45E-207 5.805E-01 0.0

1.00

1.00 Summation Term

0.80

0.60

0.40

0.20

0.00

Term

DAB =

0.50 cm

Term

3600 sec 1.00 hr 1.407 mg remaining

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37

0.125 cm

t= t= mA =

L= W=

Summation of terms

D AB =

0.00 0

27-28

27.21 Given: A = drug, B = gel, C = surrounding fluid Sphere dimensions: D = 0.5 cm, R = D/2 = 0.5 cm / 2 = 0.25 cm Assume cA,1 = cA∞ = 0 bulk fluid, cAs > 0 DAe = 3.0 × 10−6 cm2/s effective diffusion coefficient of drug inside gel

DAC = 1.5 × 10−5 cm2/s molecular diffusion coefficient of drug in surrounding fluid kc = 5.0 × 10−5 cm/s mass transfer coefficient for A around sphere

a. Biot number (Bi) × −5cm 2 /s ( 0.25 cm ) 5.0 10 kc R = Bi = = 4.1 DAe × −6 cm/s 3.0 10

(

)

(

)

b. CA(r,t) at center of sphere (r = 0) at t = 2.3 h USS diffusion in a sphere Get XD, n, m, then use to get Y from concentration–time charts  3600 s  × −6 cm 2 /s ( 2.3 h )  3.0 10  D t  1h  = 0.40 X D = Ae2 = 2 R ( 0.25 cm )

(

)

r 0 = =0 R 0.25 cm

DAe 3.0 10 × −6 cm 2 /s = = 0.24 kc R 5.0 10 × −5cm/s ( 0.25 cm )

(

)

Figure F.6 - sphere, for n = 0, m = 0.25, X= 0.40, read off Y = 0.15 c A,1 − c A ( r, t ) c A∞ − c A ( 0, t ) 0 − c A ( 0, t ) c A ( 0, t ) = = = Y ≅= 0.15 c Ao 0 − c Ao c A,1 − c Ao c A∞ − c Ao

= c Ao

mAo mAo 0.005 mmol mmole = =0.0764 = 4 4 3 V cm3 π R3 π ( 0.25 cm ) 3 3

mmole  mmole  0.15  0.0764 = 0.0115 c A ( 0, t ) = Y ⋅ c A,o = 3  cm  cm3  27-29

27.22 A = CO2, B = N2 T = 200 oC (473 K), P = 15.0 atm K/ = 1776.2 cm3 gas/cm3 bulk adsorbent ε = 0.60 cm3 gas/cm3 bulk adsorbent DAe = 0.018 cm2/sec S = 100 cm2 yA∞ = yAS = 0.10 bulk gas CA0 = 0 inside porous adsorbent Determine total CO2 loaded into adsorbent after t = 60 min (3600 sec) USS diffusion in semi-infinite medium 4𝐷𝐷′𝐴𝐴𝐴𝐴 𝑡𝑡 (𝐶𝐶𝐴𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴0 ) 𝜋𝜋

𝑚𝑚𝐴𝐴 (𝑡𝑡) − 𝑚𝑚𝐴𝐴0 = 𝑆𝑆� ′ = 𝐷𝐷𝐴𝐴𝐴𝐴

(0.60)2 (0.018 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠) 𝜀𝜀 2 𝐷𝐷𝐴𝐴𝐴𝐴 = = 3.65 𝑥𝑥 10−6 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠 (0.60 + 1776.2)𝑐𝑐𝑐𝑐3 /𝑐𝑐𝑐𝑐3 𝜀𝜀 + 𝐾𝐾′

𝐶𝐶𝐴𝐴𝐴𝐴 = 𝑦𝑦𝐴𝐴𝐴𝐴 ∙ 𝐶𝐶 = 𝑦𝑦𝐴𝐴𝐴𝐴 𝑚𝑚𝐴𝐴 (𝑡𝑡) = 𝑆𝑆�

𝑃𝑃 15 𝑎𝑎𝑎𝑎𝑎𝑎 = (0.10) � � = 3.86 𝑥𝑥 10−5 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔/𝑐𝑐𝑐𝑐3 (82.06 𝑐𝑐𝑐𝑐3 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎/𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾)(473 𝐾𝐾) 𝑅𝑅𝑅𝑅

4𝐷𝐷′𝐴𝐴𝐴𝐴 𝑡𝑡 4 ∙ (3.65 𝑥𝑥 10−6 𝑐𝑐𝑚𝑚2 /𝑠𝑠𝑠𝑠𝑠𝑠)(3600 sec) (3.86 𝑥𝑥 10−5 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔/𝑐𝑐𝑐𝑐3 ) 𝐶𝐶𝐴𝐴𝐴𝐴 = 100 𝑐𝑐𝑐𝑐2 � 𝜋𝜋 𝜋𝜋

mA(t) = 5.0 x 10-4 gmole CO2

27-30

28.1 Given: T = 300 K, P = 1.0 atm µ ν Sc = = ρ DAB DAB  P  T  DAB ,T2 , P 2 = DAB ,T1 , P 1  1   2   P2   T1 

3/2

⋅

Ω D |T1 Ω D |T2

Neglect temperature dependency on Ω D  P  T  DAB ,T2 , P 2 = DAB ,T1 , P 1  1   2   P2   T1 

3/2

Appendix I: ν H2O,liquid (300K ) = 8.788 × 10−7

m2 m2 , ν air (300 K) = 1.5689 × 10−5 s s

O2 gas (A) in Air (B) Appendix J: DAB (273 K,1.0 atm) = 0.175

cm 2 s

3/2

cm 2  1 atm   300 K  cm 2 = 0.202 DAB ,T2 , P 1 = 0.175    s  1 atm   273 K  s 2 m 1.5689 × 10−5 vB s = 0.786 Sc = = 2 m DAB 0.202 × 10−4 s O2 (A) dissolved in liquid H2O (B) cm3 µ L ,H O (300 K) = 8.76 ×10 Pa s = 0.876 cP , VA = 25.6 gmole Determine DAB using Hayduk and Laudie correlation −4

DAB = 13.26 × 10−5 µ L −1.14VA−0.589

(

)

DAB = 13.26×10 −5 ( 0.876 )

v Sc = B = DAB

8.788 × 10−3 2.28 × 10 −5

−1.14

( 25.6 )

−0.589

= 2.28 × 10 −5

cm 2 s

cm s = 385 2

cm s

CO2 gas (A) in air (B) 28-1

Appendix J: DAB (273 K,1 atm) = 0.136 DAB ,T2 , P 1 = 0.136

cm 2  1 atm   300 K     s  1 atm   273 K 

cm 2 s

3/2

= 0.157

cm 2 s

CO2 (A) dissolved in liquid H2O (B) Determine DAB using Hayduk and Laudie correlation cm3 VCO2 = 34.0 gmole DAB = 13.26 ⋅10−5 µ L −1.14VA−0.589 = DAB

(13.26 × 10 )( 0.876

v Sc = B = DAB

−5

8.788 × 10−7 1.93 × 10 −9

−1.14

) ( 34.0 )

−0.589

=1.93 × 10 −5

cm 2 s

m s =455 2

Schmidt numbers are higher in liquids relative to gases

28-2

28.2

St =

kc  kc L   ν   DAB  Sh =    =  v ∞  DAB   Lv ∞   ν  Re⋅ Sc

Pe =

v∞ L  v∞ L   ν  =  =  Re⋅ Sc  DAB  ν   DAB 

28-3

28.3 Property Char. Length = Diameter Velocity Fluid density Fluid viscosity Fluid Diffusivity Mass Transfer Coefficient

Symbol D 𝑣𝑣 𝜌𝜌 𝜇𝜇 DAB kc

Units L L/t M/L3 M/Lt L2/t L/t

2 L M  L  a b c = v ρ DAB ( L) a    3    π 1 D= t L   t  = M : 0 c= c 0 t : 0 =−b − 1 b =−1 L : 0 =a + b − 3c + 1 a =−1 D π 1 = AB Dv b

L M   M  = v ρ µ ( L)    3   π 2 D=   t   L   L ⋅t  t : 0 =−e − 1 e =−1 −1 M :0 = f +1 f = L : 0 =d + e − 3 f − 1 d =−1 µ 1 = π2 = D ρ v Re d

L M  L g h l = v ρ kc ( L ) g    3    π 1 D= t L  t = M : 0 l= l 0 t : 0 =−h − 1 h =−1 L : 0 = g + h − 3l + 1 g=0 k π3 = c v k π2 µ = = Sc → c = f [Re, Sc] v∞ π 1 ρ DAB

28-4

28.4

gas distributor

Drying Chamber cA,∞ ≈ 0 60 m3/min air H = 1.0 m 27°C, 1.0 atm (W = 1.5 m) L = 1.5 m

painted steel plate

heated surface (27°C) Given: A = solvent, B = air T = 300 K, P = 1.0 atm g , cm3 Plate dimensions: L = 150 cm, H = 100 cm, W = 150 cm m3 1.0 3 3 vo m m s= 0.667 m v∞ = = vo 60 = 1.0 = W ⋅ L 1.0 m ⋅1.5 m s min s Paint coating:  = 0.10 cm , wA = 0.30 , ρ A,liq = 1.5

Physical parameters: g m2 cm 2 M A = 78 , p*A = 0.138 atm , DAB = 0.097 , ν air (300K) = 1.5689 ×10−5 gmole s s Physical System: boundary layer between surface and bulk gas Source for A: solvent coating surface Sink for A: bulk flowing gas a. ShL over entire plate Re and Sc: m  0.667  (1.5 m )  v∞ L s   = = Re = 63, 771 laminar flow 2 ν  −5 m  1.5689 × 10  s  

28-5

m2 1.5689 × 10 vair s = 1.617 Sc = = 2 m DAB 0.097 × 10−4 s −5

Laminar flow over a flat plate: 1/3 = ShL 0.664 = Re1/2 L Sc

= ) (1.617 ) 197 ( 0.664 )( 63, 771 1/2

1/3

b. Emissions rate WA Transfer rate across boundary layer = WA N= kc (c As − c A,∞ )S AS

c A,∞ ≈ 0 in bulk air flow

S =L ⋅ W =(150 cm)(150 cm) = (150 cm) 2 cm 2 197 ⋅ 0.097 Sh L DAB cm s kc = = = 0.127 L 150 cm s c= As

( 0.138 atm ) p*A mol =5.61 × 10−6 = 3 RT  cm3 cm ⋅ atm   82.06  ( 300 K ) mol ⋅ K  

cm   g   60 s  g 2  −6 mol  150 cm )  78 = WA  0.127   5.61 × 10   =75.0 3 ( s  cm  min   mol   min  c. Time required for complete drying of paint Material balance on solvent in paint layer IN – OUT + GEN = ACC

dm 0 − WA + 0 = A dt 0

m Ao

∫ dmA = ∫ WAdt

mAo WA 28-6

g   mA,0 = V ⋅ ρ ⋅ x A =  ⋅ L2 ⋅ ρ ⋅ wA = ( 0.1 cm ) (150 cm) 2 1.5 3  (0.30) = 1012.5 g  cm  mA,0 1012.5g = t = = 13.5 min WA 75 g/min d. Boundary layer thickness at x = L = 1.5 m

( 5)(1.5 m ) = 2.97 cm 5⋅ x = Re x 63771 δ 2.97 cm δc = =2.53 cm = 1/3 1/3 Sc (1.617 )

δ =

These are both much smaller than the height of the drying chamber. e. Volumetric flowrate of air needed for WA = 150 g A/min, all other variables unchanged

WA,new kc ,new = = WA,old kc ,old

1/2

 v ∞ ,new  =    v ∞ ,old  2

1/2

 vo ,new     vo ,old 

 WA,new   m3  150 g/min  m3 vo ,new v= = = 60 240     o ,old    min  WA,old   min   75 g/min  2

28-7

28.5 W = 0.50 m, L = 2.5 m, 𝑣𝑣∞ = 1.5

0.080

𝑐𝑐𝑐𝑐2 𝑠𝑠

, 𝑀𝑀𝐴𝐴 = 86

υ air = 1.505 × 10 −5

𝑔𝑔

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

m2 s

𝑚𝑚 𝑠𝑠

, 𝑇𝑇 = 293 𝐾𝐾, 𝑃𝑃 = 1 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑝𝑝𝐴𝐴∗ = 0.159 𝑎𝑎𝑎𝑎𝑎𝑎, 𝐷𝐷𝐴𝐴𝐴𝐴 =

, 𝑝𝑝𝐴𝐴,∞ ≈ 0, ∴ 𝑐𝑐𝐴𝐴,∞ ≈ 0

a. Sc and Sh 2  −5 m  × 1.505 10   s  υair  Sc = = = 1.88 2 DAB  −4 m   0.080 ×10  s   m 1.5 ⋅ 0.5m v∞ L s = = = Re 4.98 x 104 laminar L 2 m ν 1.505 ×10−5 s 1/3 ShL = = 0.664 Re1/2 0.664 ( 4.98 ×104 ) L Sc

1/2

(1.88)

1/3

= 183

b. Solvent evaporation rate, WA  cm 2  (183)  0.08  s  Sh ⋅ DAB cm  = kc = = 0.293 W s ( 50 cm )

( 0.159 atm ) p*A gmole = = 6.61 x 10-6 3 RT  cm3 cm ⋅ atm  82.06 293 K ) (   gmole ⋅ K   c A, ∞ = 0 c= As

WA = N A S = kc ( c As − c A,∞ ) ⋅ 2WL =

cm   gmole    100 cm  -6 gmole -0  ( 2 ⋅ 0.5 m ⋅ 2.5 m )   0.293   6.61 x 10  = 0.048 3 s   cm s    1m 

c. Determine 𝑋𝑋𝐴𝐴𝐴𝐴 𝑔𝑔 𝐴𝐴 𝑔𝑔 𝑑𝑑𝑑𝑑𝑑𝑑 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑋𝑋𝐴𝐴,𝑂𝑂 = 0.10 , 𝑚𝑚̇𝑠𝑠 = 50.0 𝑔𝑔 𝑑𝑑𝑑𝑑𝑑𝑑 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑠𝑠 Mass balance on solvent in moving sheet IN – OUT – GEN = ACC

28-8

•

X A,0 m s − X A, f m s − WA M A + 0 = 0 gmole   gA    0.048   86 s   gmole  WA M A gA gA  X A, f = X A,0 − • = − = 0.10 0.017 g poly g poly  g poly  ms  50  s  

28-9

28.6 a. Mass transfer coefficients for AsH3 and Ga(CH3)3 𝐴𝐴 = 𝐴𝐴𝐴𝐴𝐻𝐻3 , 𝐵𝐵 = 𝐺𝐺𝐺𝐺(𝐶𝐶𝐶𝐶3 )3 , 𝐶𝐶 = 𝐻𝐻2, , 𝑇𝑇 = 800 𝐾𝐾, 𝑃𝑃 = 1 𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚2 𝑐𝑐𝑐𝑐 𝐿𝐿 = 15 𝑐𝑐𝑐𝑐, 𝑆𝑆 = 225 𝑐𝑐𝑐𝑐2 , 𝑣𝑣∞ = 100 , 𝜈𝜈𝐻𝐻2 = 5.6863 × 10−4 𝑠𝑠 𝑠𝑠 𝑐𝑐𝑐𝑐2 𝑦𝑦𝐴𝐴∞ = 0.001, 𝑦𝑦𝐵𝐵∞ = 0.001, 𝑐𝑐𝐴𝐴𝐴𝐴 = 𝑐𝑐𝐵𝐵𝐵𝐵 = 0, 𝐷𝐷𝐵𝐵𝐵𝐵 = 1.55 𝑠𝑠 𝑃𝑃 1 𝑎𝑎𝑎𝑎𝑎𝑎 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑐𝑐 = = = 1.52 × 10−5 𝑐𝑐𝑐𝑐3 𝑎𝑎𝑎𝑎𝑎𝑎 𝑅𝑅𝑅𝑅 𝑐𝑐𝑐𝑐3 82.06 ∙ 800 𝐾𝐾 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾 cm   100   (15 cm ) v∞ L  s  Re L = = = 263.8 Re , 2.0 x 105 laminar flow 2 νC   cm  5.6863  s   Ga(CH3)3  cm 2  5.6863   s  vC  = = Sc = 3.67 DBC  cm 2  1.55  s   DBC 1.55 cm 2 /s cm 1/2 1/3 1/3 0.664 Re1/2 = ( 0.664 )( 263.8) ( 3.67 ) = 1.72 L Sc L 15 cm s cm gmole gmole N B= kc ( cB∞ − cBs = ) kc yB∞ c= 1.72  ( 0.001) 1.52 x 10-5 3  = 2.61 x 10-8 2 cm  cm s s    AsH3 1 1/2 1.5 1 𝑇𝑇 � + � 𝑀𝑀𝐴𝐴 𝑀𝑀𝐶𝐶 𝐷𝐷𝐴𝐴𝐴𝐴 = 0.001858 2 𝑃𝑃𝜎𝜎𝐴𝐴𝐴𝐴 Ω𝐷𝐷 1 1 1/2 0.001858 ∙ 8001.5 � + � 𝑐𝑐𝑐𝑐2 77.95 2.02 = = 4.54 1 𝑎𝑎𝑎𝑎𝑎𝑎 ∙ (3.514) 2 ∙ 0.535 𝑠𝑠 2  cm   5.6863  s  vC  = = Sc = 1.252 DAC  cm 2   4.54  s   D 4.54 cm 2 /s cm 1/2 1/3 1/3 = kc = AC 0.664 Re1/2 Sc ( 0.664 )( 263.8) (1.25) = 3.52 L 15 cm s L cm gmole gmole N A= kc ( c A∞ − c As = ) kc y A∞ c=  3.52  ( 0.001) 1.52 x 10-5 3  = 5.35 x 10-8 2 s  cm  cm s   kc =

28-10

b. Ratio of AsH3 and Ga(CH3)3

gmole 5.35 x 10-8 NA cm 2s 2.05 = = N B 2.61 x 10-8 gmole cm 2s y c⋅k NA 1.0 Let = = A∞ c , A NB y B∞ c ⋅ k c , B kc , A yB∞ y= = A∞ kc , B

( 0.001)

3.52 cm/s = 0.00205 1.72 cm/s

28-11

28.7 A = polymer, B = MEK (liquid solvent) L = 20 cm, W = 10 cm, 𝑙𝑙𝑜𝑜 = 0.02 𝑐𝑐𝑐𝑐, 𝑣𝑣𝑜𝑜 = 30 𝐷𝐷𝐴𝐴𝐴𝐴 = 3 × 10−6

𝑐𝑐𝑐𝑐2 2

, 𝜈𝜈𝐵𝐵 = 6.0 × 10−3

a. Sc and Sh 2  −3 cm  ⋅ 6 10   s  υB  = = Sc = 2000 2 DAb   cm −6  3 ⋅10  s  

𝑐𝑐𝑐𝑐2 𝑠𝑠

𝑐𝑐𝑐𝑐3 𝑠𝑠

, 𝑐𝑐𝐴𝐴∞ ≈ 0

, 𝑐𝑐𝐴𝐴∗ = 0.04

𝑔𝑔

𝑐𝑐𝑐𝑐3

1/2

  cm ⋅ 20cm  1.5  1/3 s = = = Sh 0.664 Re1/2 Sc1/3 ( 0.664 )  ( 2000 ) 591.6 2   6 ⋅10−3 cm    s   b. Flux NA

= N A kc ( c As − c A,∞ ) 2  -6 cm  ( 591.6 )  3.0 x 10  s  Sh ⋅ DAB  kc = = = 8.87 x 10-5 cm/s 20 cm L ( )

gA gA   N A = ( 8.87 x 10-5 cm/s )  0.04 3 - 0  = 3.55 x 10-6 cm cm 2s   c. Local kc,x vs. position x

kc,x [cm/s] *104

2.1 1.8 1.5 1.2 0.9 0.6 0.3 0

x [cm]

28-12

d. Polymer coating thickness vs. position and time Material balance on solute IN – OUT + GEN = ACC mA t dm 0 − WA + 0 = A , ∫ dmA = ∫ WA dt dt m A ,0 0

mA (t ) = mA,0 − WA ⋅ t mA (t )= (t ) ⋅ S ⋅ ρ A, solid kc ( c As − c A,∞ ) S ⋅ t kc ( c As − c A,∞ ) ⋅ t (t ) = (to ) − (to ) − = S ⋅ ρ A, solid ρ A, solid

e. Boundary layer thickness at x = 10 cm = δ

5⋅ x = Re x

( 5)(10 cm ) = 1.0 cm 2500

= / Sc1/3 1.0 cm / ( 2000 ) = 0.079 cm δ c δ= 1/3

28-13

28.8 Shx =

kc x = 0.0292 Re 4x / 5 Sc1 / 3 , where Re L ≥ 3.0 ⋅106 D AB

kc =

0.0292 Re 4x / 5 Sc1/ 3 DAB dx ∫0 x L

∫ dx

0.0365 Re 4L/ 5 Sc1/ 3 DAB L

= ShL

kc L 1/3 = 0.0365 Re 4/5 L Sc DAB

28-14

28.9 a. Determine α, β, η, ξ

v= α + β y1/7 BC for velocity profile: v( y = 0) = 0, α = 0

v( y = δ ) = v∞ , β =  y v x = v∞   δ 

δ 1/ 7

1/ 7

Given that:

c A − c A,∞ = η + ξy1/ 7 BC for concentration profile:

y = 0; c A − c A,∞ = c A, s − c A,∞ y = δ c ; c A − c A,∞ = 0

η = c A, s − c A,∞ c

−c

ξ = − A, s 1 / 7 A,∞ δc c A − c A,∞

 y = 1 −   c A, s − c A,∞ δc 

1/ 7

b. Local mass transfer coefficient, kc δ

d c (c A − c A,∞ )v x dy = k c (c A, s − c A,∞ ) dx ∫0 δ

k d c (c A − c A , ∞ ) v x dy = c ∫ dx 0 (c A, s − c A,∞ ) v ∞ v∞

28-15

d  c   y   1−   = υ ∞ dx  ∫0   δ c    kc

1/ 7

δc 1/ 7 2/7     y   d    y   y dy   ∫  1 / 7  −  dy = 1/ 7 1/ 7     δ dx δ     0     δ c δ    δc

d  7 y8/ 7 7 y9/ 7  d  7 δ c8 / 7  = − =     dx  8 δ 1/ 7 9 δ c1/ 7δ 1/ 7  0 dx  92 δ 1/ 7 

If δ c =δ →

7 d = [δ ] υ∞ 72 dx

For turbulent flow: υ  δ= = 0.371  1/ 5  v∞   xv ∞     υ  0.371x

υ  dδ = 0.371  dx  v∞ 

1/ 5

x4/5

k 4 −1 / 5 0.291 7  0.291  0.0289 = 1 / 5 → c =  1 / 5  = x 5 v ∞ f 72  Re x  Re 1x/ 5 Re x

k c = 0.0289v ∞ Re −x 1 / 5

28-16

28.10 𝑐𝑐𝐴𝐴 − 𝑐𝑐𝐴𝐴,𝑠𝑠 = 𝑎𝑎 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐𝑦𝑦 2 + 𝑑𝑑𝑑𝑑 3 BC for concentration profile:

= y 0, = c A c A, s

= a 0

y =δ , c A =c A,∞

y= δc ,

c A,∞ − c A, s =bδ + cδ 2 + d δ 3

d 0 ( c A − c A, s ) = dy

d2 y= 0 ( cA − cA,s ) =0 dy 2 d =−

c A, ∞ − c A, s 2δ

c A − c A, s

3 c

b + 2cδ c + 3d δ c 3 0=

0= 2c + 6dy , y = 0, c = 0

3  c A, ∞ − c A, s    δc 2  

3 y  1 y  =   −   c A, ∞ − c A, s 2  δ c  2  δ c 

28-17

28.11 A = methylene chloride, B = air 𝑇𝑇 = 300 𝐾𝐾, 𝑃𝑃 = 1 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑣𝑣∞ = 7.5

𝑚𝑚 𝑠𝑠

𝑐𝑐𝑐𝑐2 𝑐𝑐𝑐𝑐2 , 𝜐𝜐𝑎𝑎𝑎𝑎𝑎𝑎 = 0.15 𝐷𝐷𝐴𝐴𝐴𝐴 = 0.085 𝑠𝑠 𝑠𝑠

= 750

𝑐𝑐𝑐𝑐 𝑠𝑠

, L = 100 m = 100 x 102 cm

A = methylene chloride, B = water (liquid) 𝐷𝐷𝐴𝐴𝐴𝐴 = 1.07 × 10−5

𝑐𝑐𝑐𝑐2 𝑐𝑐𝑐𝑐2 , 𝜐𝜐𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 0.010 𝑠𝑠 𝑠𝑠

a. Position for laminar flow transition, Lt

= Ret 2.0 = x 105

v∞ Lt

𝑐𝑐𝑐𝑐2 5 𝑅𝑅𝑅𝑅𝑡𝑡 𝜈𝜈 2.0 × 10 ∙ 0.15 𝑠𝑠 𝐿𝐿𝑡𝑡 = = = 40 𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐 𝑣𝑣∞ 750 𝑠𝑠

Since the laminar portion is so small compared to the length of the pond, it is assumed that turbulent flow mass transfer will dominate.

b. Average gas phase film mass transfer coefficient, kc 𝑐𝑐𝑐𝑐 2 𝑣𝑣∞ 𝐿𝐿 750 𝑠𝑠 ∙ 100 × 10 𝑐𝑐𝑐𝑐 𝑅𝑅𝑅𝑅 = = = 5.0 𝑥𝑥106 𝑐𝑐𝑐𝑐2 𝜈𝜈𝑎𝑎𝑎𝑎𝑎𝑎 0.15 𝑠𝑠 𝑐𝑐𝑐𝑐2 0.15 𝑠𝑠 𝜈𝜈𝑎𝑎𝑎𝑎𝑎𝑎 𝑆𝑆𝑆𝑆 = = = 1.765 𝑐𝑐𝑐𝑐2 𝐷𝐷𝐴𝐴𝐴𝐴 0.085 𝑠𝑠 6 Re > 3.0 x 10 , fully turbulent 1 𝐷𝐷𝐴𝐴𝐴𝐴 ∙ 0.0365𝑅𝑅𝑅𝑅𝐿𝐿0.80 𝑆𝑆𝑆𝑆 3 = 𝐿𝐿 𝑐𝑐𝑐𝑐2 0.085 𝑠𝑠 1 𝑐𝑐𝑐𝑐 0.80 (1.765)3 ∙ 0.0365(50,000,000) = 0.541 2 100 × 10 𝑐𝑐𝑐𝑐 𝑠𝑠

𝑘𝑘𝑐𝑐 =

28-18

𝑘𝑘𝑐𝑐 𝑘𝑘𝐺𝐺 = = 𝑅𝑅𝑅𝑅

82.06

𝑐𝑐𝑐𝑐 0.541 𝑠𝑠

𝑐𝑐𝑐𝑐3 𝑎𝑎𝑎𝑎𝑎𝑎

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐾𝐾

∙ 300 𝐾𝐾

= 2.2 × 10−5

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑐𝑐𝑐𝑐2 ∙ 𝑠𝑠 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎

c. Schmidt number for gas and liquid phase 𝑐𝑐𝑐𝑐2 0.15 𝑠𝑠 = 1.76 𝑆𝑆𝑆𝑆𝑔𝑔𝑔𝑔𝑔𝑔 = 𝑐𝑐𝑐𝑐2 0.085 𝑠𝑠 𝑐𝑐𝑐𝑐2 0.010 𝑠𝑠 𝑆𝑆𝑆𝑆𝑙𝑙𝑙𝑙𝑙𝑙 = = 935 𝑐𝑐𝑐𝑐2 1.07 × 10−5 𝑠𝑠

28-19

28.12 A = octane, B = air 𝑊𝑊 = 10.0 𝑚𝑚 , 𝐿𝐿 = 1.0 𝑚𝑚, 𝑇𝑇 = 288 𝐾𝐾, 𝑃𝑃 = 1.0 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑝𝑝𝐴𝐴 = 0.01025 𝑎𝑎𝑎𝑎𝑎𝑎, 𝜀𝜀 = 0.40 𝑚𝑚2 𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐2 , 𝑣𝑣∞ = 2.0 , 𝐷𝐷𝐴𝐴𝐴𝐴 (298 𝐾𝐾, 1 𝑎𝑎𝑎𝑎𝑎𝑎) = 0.0602 𝜐𝜐𝑎𝑎𝑎𝑎𝑎𝑎 (288 𝐾𝐾) = 1.46 × 10−5 𝑠𝑠 𝑠𝑠 𝑠𝑠 𝑐𝑐𝑐𝑐2 288 𝐾𝐾 3/2 𝑐𝑐𝑐𝑐2 𝐷𝐷𝐴𝐴𝐴𝐴 (𝑇𝑇, 𝑃𝑃) = 0.0602 � � = 0.0572 𝑠𝑠 298 𝐾𝐾 𝑠𝑠 a. Determine yAs v∞W ( 2.0 cm/s )(1000 cm ) = Re L = =1.37 x 104 2 υ 0.146 cm /s

(

)

0.146 cm 2 /s = = 2.55 Sc = DAB 0.0572 cm 2 /s Re < 2.0 x 105, therefore laminar flow 1/2 D 0.0572 cm 2 /s cm 1/3 1/3 = kc = AB ( 0.664 ) Re1/2 Sc ( 0.664 ) (1.37 x 104 ) ( 2.55) = 0.00607 L s L 1000 cm (1039 Pa ) pA mol = 4.34 x 10-7 c*A = = 3 3 cm3 RT  m Pa   100 cm  8.314 288 K )  (  mol ⋅ K   1m   101,325 Pa mol P = 4.23 x 10-5 = c = 3 3 cm3 RT  m Pa   100 cm   8.314  ( 288 K )   mol ⋅ K   1m   Convection through gas phase boundary layer = N A kc ( c As − c A∞ ) Diffusion flux through porous rock layer DAe * = NA ( cA − cAs ) L Equate flux ε 2 DAB * kc ( c As − c= ) ( cA − cAs ) A∞ L c A∞ ≈ 0

(0.40) 2 ( 0.0572 cm 2 /s )  -7 mol  c  4.34 x 10  100 cm cm3   L c As = = 2 2 ε 2 DAB cm (0.40) ( 0.0572 cm /s ) kc + 0.00607 + L s 100 cm mol c As = 6.45 x 10-9 cm3

ε 2 DAB

* A

28-20

mol 3 c As cm= y= = 1.52 x 10-4 As mol c 4.23 x 10-5 cm3 6.45 x 10-9

cm  mol   -9 mol − 0  = 3.97 x 10-11 N A = kc ( c As − c A∞ )=  0.00607  6.45 x 10 3 s  cm cm 2s  

b. If 𝑣𝑣∞ = 50

𝑐𝑐𝑐𝑐 𝑠𝑠

, Determine yAs

 cm   50  (1000 cm ) s   Re L = =342,466  cm 2   0.146  s   Re > Ret = 2.0 x 105 transition region D DAB 1/3 4/5  0.664 AB Sc1/3 Re1/2 kc = Sc  Re 4/5 t + 0.0365 L − Ret  L L   cm 2  cm 2  0.0572  0.0572    4/5 s  s  1/3 1/3 4/5  5 1/2 kc = ( 0.664 )  2.55 2.0 x 10 + 0.0365 ( ) ( ( ) ( 2.55) ( 342,466 ) - ( 2.0 x 105 )  ) (1000 cm ) (1000 cm )

= 0.050

cm s

(0.40) 2 ( 0.0572 cm 2 /s )  -7 mol  c  4.34 x 10  100 cm cm3   L = c As = 2 2 ε 2 DAB cm (0.40) ( 0.0572 cm /s ) kc + 0.050 + L s 100 cm mol c As = 7.93 x 10-10 cm3 mol 7.93 x 10-10 c As cm3 = 1.88 x 10-5 = y= As mol c 4.23 x 10-5 cm3

ε 2 DAB

* A

cm  mol   -10 mol − 0  = 3.97 x 10-11 N A = kc ( c As − c A∞ )=  0.050  7.93 x 10 3 2 s cm cm s   

c. Biot number At v∞ = 2.0 cm/s,= Bi

kc L kL = 2 c= DAe ε DAB

( 0.00607 cm/s ) (100 cm) = 66.3 (0.40) 2 (0.0572 cm 2 /s)

( 0.050 cm/s ) (100 cm) kc L kL = 2 c= = 546 DAe ε DAB (0.40) 2 (0.0572 cm 2 /s) Even at low gas flowrate, the Bi number is high and so the process is limited by the diffusion of gasoline vapor through the porous rock layer At v∞ = 50 cm/s,= Bi

28-21

28.13 A = lysozyme, B = water (liquid) 𝑐𝑐𝑐𝑐 , 𝐿𝐿 = 10 𝑐𝑐𝑐𝑐, 𝑙𝑙 = 0.10 𝑐𝑐𝑐𝑐, 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 30 × 10−7 𝑐𝑐𝑐𝑐 𝑇𝑇 = 298 𝐾𝐾, 𝑣𝑣∞ = 5.0 𝑠𝑠 𝜈𝜈𝐵𝐵 = 9.12 𝑥𝑥 10−3 𝑐𝑐𝑐𝑐2 /𝑠𝑠 𝑐𝑐𝑐𝑐2 𝑔𝑔 −6 , 𝑀𝑀𝐴𝐴 = 14,100 , 𝑑𝑑 = 4.12 × 10−7 𝑐𝑐𝑐𝑐 𝐷𝐷𝐴𝐴𝐴𝐴 = 1.04 × 10 𝑠𝑠 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝐴𝐴 𝑐𝑐𝑐𝑐2 −7 𝐷𝐷𝐴𝐴𝐴𝐴 = 5.54 × 10 𝑠𝑠 a. Flux model

Flux through membrane N A kc ( c Ao − c As ) = Flux through boundary layer over membrane DAe = NA ( cAs − cA∞ )  Eliminate cAs NA = c Ao − c As kc N A = c As − c A∞ DAe 1   NA  + c Ao − c A∞ = k D Ae   c (c − c ) N A = Ao A∞  1 + kc DAe b. Estimate NA cm   5.0   (10.0 cm ) v∞ L  s  Re L = = = 5482 Re < 2.0 x 105, laminar flow 2 νB  -3 cm   9.12 x 10  s   cm 2 νB s = 8769 = Sc = cm 2 DAB 1.04 x 10-6 s Estimate kc -6 2 D 1/2 1/3 1/3  1.04 x 10 cm /s  -4 kc = AB 0.664 Re1/2 Sc =   0.664 ( 5482 ) ( 8769 ) =1.05 x 10 cm/s L L 10 cm   9.12 x 10-3

28-22

N= N= A A

( cAo − cA∞ ) 1  + kc DAe

mmol   1 m   gmol    (1.0 − 0.40 )1.0    gmol m3   100 cm   1000 mmol  = 3.16 x 10-11 2 NA=  1 0.10 cm ms + 2 cm cm 1.05 x 10-4 5.54 x 10-7 s s Limiting case: all mass transfer resistance in boundary layer

N A ≈= kc ( c Ao − c A∞ ) 1.05 x 10-4

cm  mmol   1 m   gmol   (1.0 − 0.4 )1.0    m3   100 cm   1000 mmol  s 

gmol m 2s Limiting case: all mass transfer resistance in membrane cm 2 5.54 x 10-7 DAe s  (1.0 − 0.4 )1.0 mmol   1 m   gmol  NA ≈= ( cAo − cA∞ )      0.10 cm m3   100 cm   1000 mmol   gmol = 3.32 x 10-10 2 ms = 6.326 x 10-10

c. Biot number cm   1.05 x 10-4   (0.10 cm) kc   s  Bi = = = 19.03 2 DAe  -7 cm   5.54 x 10  s   The analysis suggests most of the resistance to mass transfer is through membrane, not through the boundary layer d. Concentration boundary layer thickness 5(10 cm) ( 8769 ) 5x = 3.3 x 10-2 cm δc = δ ⋅ Sc = 1/2 Sc −1/3 = 1/2  v∞ x    cm  10.0 cm     5.0  ( )  s   νB    2     cm -3   9.12 x 10   s     ∴δ c <  = 0.10 cm -1/3

−1/3

28-23

28.14 𝐴𝐴 = 𝐶𝐶𝐶𝐶, 𝐵𝐵 = 𝑂𝑂2 , 𝐶𝐶 = 𝐶𝐶𝐶𝐶2 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑔𝑔𝑔𝑔𝑔𝑔)

𝑐𝑐𝑐𝑐2 𝑠𝑠 600 𝐾𝐾 1.5 Ω𝐷𝐷,300 𝐾𝐾 𝐷𝐷𝐴𝐴𝐴𝐴 (600 𝐾𝐾) = 𝐷𝐷𝐴𝐴𝐴𝐴 (300 𝐾𝐾) � � ∙ 300 𝐾𝐾 Ω𝐷𝐷,600𝐾𝐾 Appendix K 𝑐𝑐𝑐𝑐2 𝑐𝑐𝑐𝑐2 0.978 𝑐𝑐𝑐𝑐2 𝐷𝐷𝐴𝐴𝐴𝐴 (300 𝐾𝐾) = 0.213 , 𝐷𝐷𝐴𝐴𝐴𝐴 (600 𝐾𝐾) = 0.213 ∙ 2.828 ∙ = 0.709 𝑠𝑠 𝑠𝑠 𝑠𝑠 0.8308 1.0615 𝑐𝑐𝑐𝑐2 𝑐𝑐𝑐𝑐2 𝑐𝑐𝑐𝑐2 , 𝐷𝐷𝐴𝐴𝐴𝐴 (600 𝐾𝐾) = 0.155 ∙ 2.828 ∙ = 0.531 𝐷𝐷𝐴𝐴𝐴𝐴 (300 𝐾𝐾) = 0.155 0.8764 𝑠𝑠 𝑠𝑠 𝑠𝑠 𝑐𝑐𝑐𝑐2 𝑐𝑐𝑐𝑐2 1.0665 𝑐𝑐𝑐𝑐2 𝐷𝐷𝐵𝐵𝐵𝐵 (300 𝐾𝐾) = 0.166 , 𝐷𝐷𝐴𝐴𝐴𝐴 (600 𝐾𝐾) = 0.166 ∙ 2.828 ∙ = 0.570 𝑠𝑠 𝑠𝑠 0.879 𝑠𝑠 𝑇𝑇 = 600 𝐾𝐾, 𝑃𝑃 = 1.0 𝑎𝑎𝑎𝑎𝑎𝑎, 𝜈𝜈𝐶𝐶𝐶𝐶2 = 0.30004

a. Sc for CO and O2 mass transfer CO

= Sc

= DAC

cm 2 s 0.566 = cm 2 0.531 s

0.30004

O2 cm 2 ν s = = 0.527 Sc = cm 2 DBC 0.570 s b. Average kc for CO mass transfer, local kc at x=L = 50 cm cm   4000   ( 50 cm ) v∞ L  s  = = Re = 6.67 ⋅105 transition region L 2 υ  cm   0.30004  s   Average kc D DAC 1/3 4/5  0.664 AC Sc1/3 Re1/2 kc = Sc  Re 4/5 tr + 0.0365 L − Retr  L L cm 2 cm 2 0.531 0.531 s ( 0.566 )1/3 2.0 x 105 1/2 + 0.0365 s ( 0.566 )1/3 kc = 0.664 ( ) 50 cm 50 cm 4/5 4/5 cm ⋅ ( 6.67 x 105 ) - ( 2.0 x 105 )  = 11.7   s 0.30004

28-24

Local kc DAC 0.0292 Re0.80 Sc1/3 x x  cm 2  0.534   s  ( 0.0292 ) 6.67 x 105 0.8 ( 0.562 )1/3 = 11.7 cm kc , x =  ( ) s  50 cm      kc , x =

c. Scale kc for CO mass transfer to kc for O2 mass transfer 2/3

2/3

 DBC  cm   0.570 cm 2 /s  cm  = 11.7 kc , B k=   = 12.3 c, A    2 s   0.531 cm /s  s   DAC  d. Flux NA For a first-order reaction on a convective surface kk N A = c A∞ c s kc + k s c= A, ∞

( 0.001)(1.0 atm ) yA P mol = = 2.03 x 10-8 3 RT  cm3 cm atm  82.06 600 K )  ( mol ⋅ K  

  cm   cm   11.7   1.5   mol s  s     -8 mol    = 2.7 x 10-8 N A =  2.03 x 10 3  cm   cm   cm    cm 2s  11.7 + 1.5     s   s    

28-25

28.15 𝑇𝑇 = 600 𝐾𝐾, 𝑃𝑃 = 1.0 𝑎𝑎𝑎𝑎𝑎𝑎, ℎ = 50 Appendix I, for CO2

𝑊𝑊

𝑚𝑚2 ∙ 𝐾𝐾

𝑘𝑘𝑘𝑘 𝐽𝐽 𝑚𝑚2 𝜌𝜌𝐶𝐶𝐶𝐶2 = 0.8941 3 , 𝑐𝑐𝑝𝑝,𝐶𝐶𝐶𝐶2 = 1076.1 , 𝜈𝜈 = 3.004 , 𝑃𝑃𝑃𝑃𝐶𝐶𝐶𝐶2 = 0.668 𝑚𝑚 𝑘𝑘𝑘𝑘 ∙ 𝐾𝐾 𝐶𝐶𝐶𝐶2 𝑠𝑠 a. Estimate kc for CO in CO2 by Reynolds and Chilton-Colburn analogies Reynolds Analogy J   50 2   v∞ C f h m m sK  = kc = = = 0.0520 ρcp  2 s kg   J   0.8941 3 1076.1  m  kg K   Chilton Colburn Analogy kc =

h  Pr  ρ ⋅ CP  Sc 

2/3

cm 2 ν s = = 0.566 Sc = cm 2 DAC 0.531 s J   2/3  50 2  m  0.668   m sK kc =   = 0.0581 s kg   J   0.566    0.8941 3 1076.1  m  kg K   0.30004

b. Scale kc for CO mass transfer to kc for O2 mass transfer 2/3

2/3

 DBC  m   0.570 cm 2 /s  m  = kc , B k= 0.0581    =0.0691 c, A   2 s   0.531 cm /s  s   DAC 

28-26

28.16 A = n-octane, B = air, T = 27 C = 300 K, P = 1.0 atm νB = 0.155 cm2/s = 1.55 x 10-5 m2/s, Appendix I FSG correlation for DAB 1/2

 1 1  0.001T  +  MA MB   = 1/3 1/3 2 P ( Σv ) A + ( Σv ) B 1.75

DAB

(

DAB = 0.066

)

1/2

0.001 ⋅ ( 300 )

1.75

1   1  +   29 114 

( 1.0 ) ( ( 20.1) + (167.64 ) ) 1/3

1/3 2

cm 2 s

Antoine Equation for PA*

B 1351.99 1351.99 = 6.91868 − = 6.91868 − = 1.194 C +T 209.155 + T 209.155 + 27  1.0 atm  PA* = 15.6 mm Hg   = 0.021 atm  760mm Hg 

log10 PA*= A −

( 0.021 atm ) PA* gmole = 0.853 = 3 m3 RT  -5 m ⋅ atm  8.206 10 300 K × )  ( gmole ⋅ K   a. Point of transition, hydrodynamic thickness, (δ) and boundary layer thickness (δc) = C A*

Ret = 2.0 x 105 for end of laminar flow Ret =

v ∞ Lt

νB

2  −5 m  5 × 1.55 10   ( 2 ×10 ) s  ν Re = L t = B t = 0.39 m m v∞   8.0  s   δ and δc at x = L m  8.0  (10 m )  v∞ L s  = =  Re = 5.16 ×106 x 2 νB   m −5 1.55 × 10  s  

28-27

cm 2 0.155 vB s = 2.35 = Sc = cm 2 DAB 0.066 s ( 5)(10 m ) = 0.022 m = 2.20 cm 5⋅ x δ = = Re x 5.16 ×106

δc =

= Sc1/3

2.2 cm

( 2.348)

1/3

= 1.66 cm

b. Re and flow regime Flow over a flat surface m  8.0  (10 m )  v∞ L s  = =  Re = 5.16 ×106 fully-developed turbulent flow (ReL > 3 x 106) L 2 νB   −5 m 1.55 × 10  s   c. n-octane vapor emissions rate WA = WA N= kc (C A* − C A,∞ )S AS

S =L ⋅ W =(10 m)(8 m) = 80 m 2 kc for fully-developed turbulent flow over a flat surface 0.8 kc L 1/3 1/3 ShL = = = 0.0365 Re0.8 1.138 ×104 ( 0.0365) ( 5.16 ×106 ) ( 2.35) = L Sc DAB

cm 2  4  1.138×10 0.066 ( )  s  Sh L DAB m  kc = = = 7.51×10−3 s L  100 cm  10 m    1m 

m  gmole   - 0  ( 80 m 2 ) WA =  7.51×10-3   0.853 3 s  m   WA = 0.51 gmole/s d. This is a combined heat and mass transfer process subject to both a material and energy balance. The liquid must be vaporized. If the heat transfer rate is not sufficient to sustain the surface temperature at the bulk gas stream temperature, then the liquid temperature will be lowered to a provide a vapor pressure where the simultaneous material and energy balance is satisfied. 28-28

28.17 A = H2O vapor, B = air

𝑘𝑘𝑘𝑘 , 𝑃𝑃 = 2.34 × 103 𝑃𝑃𝑃𝑃, 𝑐𝑐𝐴𝐴∞ ≈ 0 𝑔𝑔 𝐴𝐴 𝑐𝑐𝑐𝑐2 𝑔𝑔 𝐽𝐽 𝑃𝑃𝑃𝑃𝐵𝐵 = 0.708, 𝜈𝜈𝐵𝐵 = 0.1505 , 𝜌𝜌𝐵𝐵 = 0.001207 3 , 𝑐𝑐𝑝𝑝,𝐵𝐵 = 1.006 𝑠𝑠 𝑐𝑐𝑐𝑐 𝑔𝑔 ∙ 𝐾𝐾 Appendix J: 𝑐𝑐𝑐𝑐2 𝐷𝐷𝐴𝐴𝐴𝐴 = 0.260 𝑠𝑠 𝑐𝑐𝑐𝑐2 0.1505 𝑠𝑠 𝜈𝜈 𝑆𝑆𝑆𝑆 = = = 0.579 𝑐𝑐𝑐𝑐2 𝐷𝐷𝐴𝐴𝐴𝐴 0.260 𝑠𝑠 𝑃𝑃𝐴𝐴 2.34 × 103 𝑃𝑃𝑃𝑃 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 −7 = = 9.60 × 10 𝑐𝑐𝐴𝐴𝐴𝐴 = 𝑐𝑐𝑐𝑐3 𝑃𝑃𝑃𝑃 𝑅𝑅𝑅𝑅 𝑐𝑐𝑐𝑐3 8.314 × 106 ∙ 293 𝐾𝐾 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 ∙ 𝐾𝐾 𝑞𝑞 = ℎ(𝑇𝑇∞ − 𝑇𝑇𝑠𝑠 ) = ∆𝐻𝐻𝑣𝑣,𝐴𝐴 𝑀𝑀𝐻𝐻2𝑂𝑂 𝑁𝑁𝐻𝐻2𝑂𝑂 𝐴𝐴 ∆𝐻𝐻𝑣𝑣,𝐴𝐴 𝑀𝑀𝐻𝐻2𝑂𝑂 𝑁𝑁𝐻𝐻2𝑂𝑂 + 𝑇𝑇𝑠𝑠 𝑇𝑇∞ = ℎ 𝑘𝑘𝑐𝑐 𝑇𝑇∞ = ∆𝐻𝐻𝑣𝑣,𝐴𝐴 𝑀𝑀𝐻𝐻2𝑂𝑂 ∙ (𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴∞ ) + 𝑇𝑇𝑠𝑠 , ℎ 2/3 𝑘𝑘𝑐𝑐 1 𝑃𝑃𝑃𝑃 = � � ℎ 𝜌𝜌𝑐𝑐𝑝𝑝 𝑆𝑆𝑆𝑆 1 𝑃𝑃𝑃𝑃 2/3 𝑇𝑇∞ = ∆𝐻𝐻𝑣𝑣,𝐴𝐴 𝑀𝑀𝐻𝐻2𝑂𝑂 ∙ � � (𝑐𝑐𝐴𝐴𝐴𝐴 − 𝑐𝑐𝐴𝐴∞ ) + 𝑇𝑇𝑠𝑠 𝜌𝜌𝑐𝑐𝑝𝑝 𝑆𝑆𝑆𝑆 𝑇𝑇𝑆𝑆 = 293 𝐾𝐾, ∆𝐻𝐻𝑣𝑣,𝐴𝐴 = 2.45

𝐽𝐽 𝑔𝑔 1 0.708 3 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 2450 ∙ 18.02 ∙ − 0� � � �9.60 × 10−7 𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 0.001207 𝑔𝑔 ∙ 1.006 𝐽𝐽 0.579 𝑐𝑐𝑐𝑐3 𝑔𝑔 ∙ 𝐾𝐾 𝑐𝑐𝑐𝑐3 + 293 𝐾𝐾 𝑇𝑇∞ = 333 𝐾𝐾

28-29

28.18 Set L = Lt = 40 cm Ret =

v ∞ Lt

νB

 cm 2  5 0.00 20   ( 2×10 ) s  ν Re v∞ = B t =  = 10 cm/s Lt ( 40 cm ) cm 2 vB s = 133.3 Sc = = 2 cm DAB 1.5 ×10−5 s 0.0020

At x = 0, the local kc(x) is not defined. As x increases, kc(x) decreases with kc ( x) ∝ x −1/2 At x = L, the local kc(x) for laminar flow is 1/3 Shx = 0.332 Re1/2 = ( 0.332 ) ( 2 ×105 ) x Sc

1/2



kc= (x)

Sh x DAB = x

( 758.5) 1.5 ×10−5 

40 cm

(133.3)

1/3

= 758.5

cm 2   s  cm = 2.84 ×10−4 s

28-30

28.19 Laminar flow over a flat plate 𝑆𝑆ℎ = 0.664𝑅𝑅𝑅𝑅1/2 𝑆𝑆𝑆𝑆1/3 𝑐𝑐𝑐𝑐2 0.15869 𝑠𝑠 𝜈𝜈 𝑆𝑆𝑆𝑆 = = = 1.743 𝑐𝑐𝑐𝑐2 𝐷𝐷𝐴𝐴𝐴𝐴 0.090 𝑠𝑠 𝑅𝑅𝑅𝑅 =

𝑣𝑣∞ 𝐿𝐿 =� 𝜈𝜈

𝑘𝑘𝑐𝑐 𝐿𝐿 1 =� 1 0.664 ∙ 𝑆𝑆𝑆𝑆 3 𝐷𝐷𝐴𝐴𝐴𝐴 ∙ 0.664 ∙ 𝑆𝑆𝑆𝑆 3

𝑣𝑣∞ = 𝜈𝜈� = 0.046

𝑆𝑆ℎ

𝑘𝑘𝑐𝑐

𝐷𝐷𝐴𝐴𝐴𝐴 ∙ 0.664 ∙ 𝑆𝑆𝑆𝑆 3 ∙ 𝐿𝐿

𝑐𝑐𝑐𝑐 𝑠𝑠

𝑐𝑐𝑐𝑐2 = 0.15869 ∙� 𝑠𝑠

0.090

𝑐𝑐𝑐𝑐 0.090 𝑠𝑠

1 𝑐𝑐𝑐𝑐2 ∙ 0.664 ∙ 1.7433 ∙ 15 𝑐𝑐𝑐𝑐 𝑠𝑠

28-31

28.20 𝐿𝐿 = 150 𝑐𝑐𝑐𝑐, 𝑃𝑃 = 1.0 𝑎𝑎𝑎𝑎𝑎𝑎, 𝑇𝑇 = 298 𝐾𝐾, 𝑆𝑆 = 0.50 𝑐𝑐𝑐𝑐2 , 𝛿𝛿 = 0.2 𝑐𝑐𝑐𝑐, 𝑦𝑦𝐴𝐴∞ = 1.0, 𝑐𝑐𝐴𝐴𝐴𝐴 = 0 𝑚𝑚3 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐2 𝑐𝑐𝑐𝑐3 𝑎𝑎𝑎𝑎𝑎𝑎 −5 𝐻𝐻 = 29.5 = 29500 , 𝐷𝐷𝐴𝐴𝐴𝐴 = 2.0 × 10 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠 a. Average molar flux of A into falling liquid film At 25 oC, for liquid water, ρ = kg/m3, µ =

kg/m∙s

= N A kc ( c As − c A0 ) 4 DAB vmax πL kg   m 2  996.7 3   9.81 2  ( 0.002 m )  2 ρ gδ  m m  s  vmax = = = 21.5 -6 2µ s ( 2 ) ( 909.25 x 10 kg/m ⋅ s ) kc =

2  cm  -5 cm   4 2.0 x 10 ( )   2150  s  s  cm  kc = = 0.0191 s ( π )(150 cm )

c= As

p A y A P (1.0)(1.0 atm) kmol mol = = = 0.03389 3 = 3.39 x 10-5 3 H H m cm3  m atm  29.5   kmol  

cm   mol mol   N A =  0.0191   3.39 ⋅10−5 3 − 0  = 6.48 ×10−7 s  cm cm 2 s  

b. Average concentration in liquid film N A,ave = vmax c A,ave mol   6.48 ⋅10−7   N A,ave  mol cm 2 s  c A,ave = = = 3.0 ⋅10−10 cm  vmax cm3  2150   s  

28-32

29.1 a.

Equilibrium distribution curves

Cl2 − Water, T = 293 K,

𝑘𝑘𝑘𝑘 , 𝑀𝑀 = 0.01802 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚3 𝐻𝐻2 𝑂𝑂

yA - xA

0.25

0.20

0.15

pA [atm]

pA - cAL

𝜌𝜌𝐻𝐻2 𝑂𝑂(𝑙𝑙) = 998.2

0.10

0.05

0.00

0.00 0

0.01

0.02

0.03

0.0003

cAL [gmol/L] ClO2 − Water, T = 293 K

pA - cAL

yA - xA

0.200

0.150

pA [atm]

0.100 0.050

0.000

0.000 0

0.1

0.2

0.001 0.002 0.003 0.004

cAL [gmol/L] NH3 − Water, T = 303 K, 𝜌𝜌𝐻𝐻2 𝑂𝑂(𝑙𝑙) = 995.2

pA - cAL

𝑘𝑘𝑘𝑘

𝑚𝑚3

, 𝑀𝑀𝐻𝐻2 𝑂𝑂 = 0.01802 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚

yA - xA

1.0

0.8

0.6

pA [atm]

0.0006

0.4

0.2

0.0

0.0 0

cAL [gmol/L]

0.2

0.4

xA 29-1

SO2 − Water, T = 303 K

yA - xA

1.0

0.8

0.6

pA [atm]

pA - cAL

0.4

0.2

0.0

0.0 0

0.5

1.5

0.01

0.02

cAL [gmol/L]

Ammonia, 𝑁𝑁𝑁𝑁3 , is the most soluble in water. It has a lower range of gas mole fractions than the other species. Chlorine, 𝐶𝐶𝐶𝐶2 , is the easiest to operate liquid stripping. As seen in the 𝑦𝑦𝐴𝐴 𝑣𝑣𝑣𝑣. 𝑥𝑥𝐴𝐴 graph, the gas mole fraction is 20% when the liquid mole fraction is less than a tenth of a percent, 𝑥𝑥𝐴𝐴 ~0.0006, so there is a huge driving force to the gas phase. b. Henry’s law constants

Chlorine, 𝐶𝐶𝐶𝐶2

Chlorine dioxide, 𝐶𝐶𝐶𝐶𝐶𝐶2 Ammonia, 𝑁𝑁𝑁𝑁3

Sulfur dioxide, 𝑆𝑆𝑆𝑆2

H, based on 𝑝𝑝𝐴𝐴 = 𝐻𝐻 ∙ 𝐶𝐶 ∗𝐴𝐴𝐴𝐴 𝐿𝐿 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 � � 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔

m, based on 𝑦𝑦𝐴𝐴 = 𝑚𝑚 ∙ 𝑥𝑥𝐴𝐴∗

6.45

358

0.764

42.3

0.0378

2.09

0.779

43.0

29-2

29.2 a. Effect of temperature on the solubility of H2S gas in water At 40 oC, to maintain a given mole fraction of H2S dissolved in the liquid, the equilibrium mole fraction of H2S must be higher relative to 20 oC. Therefore, as temperature increases, H2S is less soluble in water. 0.25

40 C 20 C 0.20

0.15 yA

0.10

0.05

0.00 0.0000

0.0001 0.0002 0.0003 mole fraction A in solution, xA

0.0004

b. yA-xA equilibrium line at 40 oC for H2S in water vs. H2S in 15.9 wt% MEA solution Convert liquid phase units to mole fraction using the following conversion factor 15.9 𝑤𝑤𝑤𝑤% 𝑀𝑀𝑀𝑀𝑀𝑀 = =

15.9 𝑔𝑔 𝑀𝑀𝑀𝑀𝑀𝑀 100 𝑔𝑔 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆

1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑀𝑀𝑀𝑀𝑀𝑀 15.9 𝑔𝑔 𝑀𝑀𝑀𝑀𝑀𝑀 ∙ 61.08 𝑔𝑔 𝑀𝑀𝑀𝑀𝑀𝑀

1 𝑚𝑚𝑜𝑜𝑜𝑜 𝑀𝑀𝑀𝑀𝑀𝑀 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐻𝐻 𝑂𝑂 �15.9 𝑔𝑔 𝑀𝑀𝑀𝑀𝑀𝑀 ∙ 61.08 𝑔𝑔 𝑀𝑀𝑀𝑀𝑀𝑀 + 84.1 𝑔𝑔 𝐻𝐻2 𝑂𝑂 ∙ 18.02 𝑔𝑔 𝐻𝐻2 𝑂𝑂 � 2

= 0.053

𝑚𝑚𝑚𝑚𝑚𝑚 𝑀𝑀𝑀𝑀𝑀𝑀 𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 29-3

mol H2S/

(mm Hg) 0.00 0.96 3.00 9.10 43.10 59.70 106.00

0.0000 0.0013 0.0039 0.0120 0.0567 0.0786 0.1395

mol MEA 0.000 0.125 0.208 0.362 0.643 0.729 0.814

mol H2S/ mol H2S + MEA 0.0000 0.1111 0.1722 0.2658 0.3914 0.4216 0.4487

0.0000 0.0063 0.0094 0.0183 0.0325 0.0369 0.0412

0.25

water 0.20

15.9 wt% MEA

0.15 yA

0.10

0.05

0.00 0.000

0.010 0.020 0.030 0.040 mole fraction A in solution, xA

When compared to water, the solubility of H2S in 15.9 wt% MEA solution drastically increases at a given mole fraction of H2S in the gas phase.

29-4

29.3 Given: A = O2 (solute), B = H2O (solvent) T = 293 K, P = 2.0 atm Gas: y A = 0.21, yN2 = 0.78, yAr = 0.010 Solvent: ρB,liq = 1000 kg/m3, MB = 18 kg/kgmole Solute: H = 40,000 atm with p A,i = Hx A,i a. Determine xA* x*A =

pA yA P = = H H

( 0.21)( 2.0 atm ) = 1.05 × 10−5 ( 40,100 atm )

b. Determine cAL* kg   1000 3   ρ H O,liq ρ B ,liq kgmole m  c* AL = x*AcL  x*A 2 x*A = = 1.05 × 10−5  =5.83 × 10−4 M H2O MB m3  kg  18 kgmol    c. Determine cAL* if P = 4.0 atm x= A*

pA yA P = = H H

( 0.21)( 4.0atm ) = 2.10 × 10−5 ( 40,100 atm )

kg   1000 3   ρ H O,liq kgmole m  = 2.10 × 10−5  = 1.17 × 10−3 c* AL = x*AcL = x*A 2 m3 M H2O  kg  18.0  kgmole  

(

)

29-5

29.4 Given: A = ClO2 (solute), B = H2O (solvent) T = 293 K, P 1.5 atm Operating point: yA = 0.040, xA = 0.00040 Solute: MA = 67.5 kg/gmole Solvent: ρB,liq = 992.3 kg/m3, MB = 18 kg/kgmole Film Mass Transfer Coefficients: kgmole kgmole k x =1.0 (liquid film), kG = 0.010 (gas film) 2 m ⋅s m 2 ⋅ s ⋅ atm a.

pA vs. CAL equilibrium line and operating point

Operating point: c AL = x AcL  x A

ρ H O ,liq 2

M H2O

= xA

ρ B ,liq MB

( 992.3 kg/m ) = 0.022 kgmole = ( 0.00040 ) 3

(18 kg/kgmole )

= p A y= (0.06)(1.5 atm) = 0.090 atm AP 0.12

cAL , pA

0.10

pA (atm)

0.08

0.06

0.04

cAL,i , pA,i 0.02

pA* cAL*

0.00 0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

cAL (kgmole/m 3)

Gas Absorption process since the operating point is above the equilibrium line 29-6

b. Determine m Get H from slope of PA vs. CAL data, H = 0.772 atm∙m3/kgmole

(

)(

)

0.772 atm ⋅ m3 /kgmole 55.1 kgmole/m3 HcL = = 28.4 P (1.5 atm ) c. Determine kL  kg  18 kgmole  kx MB  m gmole    = 1.81 × 10−5 = k L =≅ kx 1.0  2 ρ B ,liq  s m s  CL kg   1000 gmole   992.3 3    m   kgmole   ∗ d. Determine 𝑐𝑐𝐴𝐴𝐴𝐴 m=

* c= AL

pA = H

( 0.06 atm )

( 0.772 atm ⋅ m /kgmole ) 3

= 0.078

kgmole m3

e. Compositions at the gas-liquid interface, pA,i and cAL,i Recall −

k L ( p A − p A ,i ) = kG ( c AL − c AL ,i )

For linear equilibrium line, p A,i = Hc AL ,i   atm ⋅ m3 m  − c AL ,i  0.090 atm 0.772 − 1.81 × 10−5   kgmole ( pA − HcAL,i ) ⇒  k s    = − L = kgmole kgmole kG    ( cAL − cAL,i ) 1.0 × 10−5 2 − c AL ,i    0.022 3 m ⋅ s ⋅ atm  m    kgmole m3  atm ⋅ m3   kgmole  0.772 = p A,i Hc =    0.0503 AL ,i  = 0.039 atm kgmole   m3   c AL ,i = 0.0503

f. Determine Ky by two approaches Path 1: Convert kG to ky, get Ky from kx, ky, m kgmole   −5 −5 kgmole ky = kG P = 1.0 × 10  (1.5 atm ) =1.5 × 10 2 m ⋅ s ⋅ atm  m 2s 

29-7

-1

   1 m   1 28.4 kgmole + = 1.05 × 10−5 K y = +  =  kgmole kgmole m 2s  k y k x  1.5 × 10−5 2  1.0 × 10−3 2 m s ⋅ atm ms   −1

Path 2: KG from kL kG, H, then convert KG to Ky -1

 atm ⋅ m3  −1 0.772   1 H kgmole 1 kgmole  K G = +  = + = 7.10 × 10−6 2 m ⋅ s ⋅ atm 1.0 × 10−5 kgmole 1.81 × 10−5 m   kG k L  2   ms s   kgmole   −5 kgmole K y K G P =  7.10 × 10−6 2 =  (1.5 atm) = 1.05 × 10 m ⋅ s ⋅ atm  m 2s  g.

Flux NA

Base on overall gas phase mole fraction driving force as an example  atm ⋅ m3   kgmole  0.772 = p A * Hc =    0.022 AL  = 0.017 atm kgmole   m3   p A * 0.017 atm = yA* = = 0.0113 P 1.5 atm kgmole   N A = K y ( y A − y* A ) = 1.05 × 10−5  ( 0.040 − 0.0113) m 2s   kgmole N A = 3.0 × 10−5 m 2s

29-8

29.5 P = 2.20 atm, T = 293 K, 𝜌𝜌𝐻𝐻2 𝑂𝑂 = 992.3 𝑥𝑥𝐴𝐴 = 0.0040, 𝑦𝑦𝐴𝐴 = 0.10 𝑘𝑘𝑥𝑥 = 0.125

𝑘𝑘𝑘𝑘 𝑚𝑚3

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 , 𝑘𝑘 = 0.010 𝑦𝑦 𝑚𝑚2 ∙ 𝑠𝑠 𝑚𝑚2 ∙ 𝑠𝑠

Equilibrium: 𝑦𝑦𝐴𝐴,𝑖𝑖 = 𝑚𝑚𝑥𝑥𝐴𝐴,𝑖𝑖 , 𝑚𝑚 = 50.0

a. xA vs. yA equilibrium line and operating point 0.5 0.4

0.3 0.2 (Operating Point)

0.1 0 0

0.002

0.004

0.006

0.008

0.01

xA The operation is liquid stripping, because at the specified liquid mole fraction, the gas mole fraction is below the equilibrium line.

b. Estimate H for linear equilibrium line 𝐻𝐻 =

𝑚𝑚𝑚𝑚 (50)(2.2 𝑎𝑎𝑎𝑎𝑎𝑎) 𝑎𝑎𝑎𝑎𝑎𝑎 ∙ 𝑚𝑚3 = = 1.995 55.1 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘/𝑚𝑚3 𝐶𝐶𝐿𝐿 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 29-9

c. Determine 𝐾𝐾𝐿𝐿

𝑘𝑘𝑘𝑘 0.018 𝑘𝑘𝑥𝑥 𝑀𝑀𝐵𝐵 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑘𝑘𝐿𝐿 = ≅ 𝑘𝑘𝑥𝑥 = ∙ 0.125 2 = 2.27 ∙ 10−6 𝑘𝑘𝑘𝑘 𝐶𝐶𝐿𝐿 𝜌𝜌𝐵𝐵 𝑚𝑚 ∙ 𝑠𝑠 𝑠𝑠 992.3 3 𝑚𝑚 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑘𝑘𝑦𝑦 0.010 𝑚𝑚2 ∙ 𝑠𝑠 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑘𝑘𝐺𝐺 = = = 4.55 ∙ 10−3 2 𝑃𝑃 2.2 𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚 ∙ 𝑠𝑠 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝐾𝐾𝐿𝐿 = �

𝑘𝑘𝐿𝐿

𝐾𝐾𝑥𝑥 = �

𝑘𝑘𝑥𝑥

𝐻𝐻𝑘𝑘𝐺𝐺

�

−1

𝑚𝑚𝑚𝑚𝑦𝑦

�

=�

−1

𝑚𝑚 2.27∙10−6 𝑠𝑠

=�

d. Determine 𝑁𝑁𝐴𝐴

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 0.125 2 𝑚𝑚 ∙𝑠𝑠

𝑁𝑁𝐴𝐴 = 𝐾𝐾𝑥𝑥 (𝑥𝑥𝐴𝐴 − 𝑥𝑥 ∗𝐴𝐴 ) = 0.1 e. Determine xA,i and yA,i

�

𝑎𝑎𝑎𝑎𝑎𝑎∙𝑚𝑚3 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 2.0×10−3 ∙ 4.55∙10−3 2 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑚𝑚 ∙𝑠𝑠∙𝑎𝑎𝑎𝑎𝑎𝑎

−1

�

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 50.0 ∙0.010 2 𝑚𝑚 ∙𝑠𝑠

= 0.1

−1

= 1.82 × 10−6

𝑚𝑚 𝑠𝑠

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑚𝑚2 ∙𝑠𝑠

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 (0.004 − 0.002) = 0.0002 𝑚𝑚2 ∙ 𝑠𝑠 𝑚𝑚2 ∙ 𝑠𝑠 𝑘𝑘

Equate the equilibrium line 𝑦𝑦 𝐴𝐴 = 𝑚𝑚𝑥𝑥 𝐴𝐴 to a line with slope − 𝑥𝑥 through the operating point to find the intersection (𝑥𝑥𝐴𝐴𝐴𝐴 , 𝑦𝑦𝐴𝐴𝐴𝐴 ). 𝑘𝑘

− 𝑥𝑥 = −12.5, 𝑦𝑦 𝐴𝐴 = −12.5 𝑥𝑥𝐴𝐴 + 𝑏𝑏, 𝑘𝑘𝑦𝑦

𝑘𝑘

𝑘𝑘𝑦𝑦

0.1 = −12.5 ∗ 0.004 + 𝑏𝑏, 𝑏𝑏 = 0.15

𝑚𝑚𝑥𝑥𝐴𝐴,𝑖𝑖 = − 𝑥𝑥 𝑥𝑥𝐴𝐴,𝑖𝑖 + 0.15, 50𝑥𝑥𝐴𝐴,𝑖𝑖 = −12.5𝑥𝑥𝐴𝐴,𝑖𝑖 + 0.15, (𝑥𝑥𝐴𝐴𝐴𝐴 , 𝑦𝑦𝐴𝐴𝐴𝐴 ) = (0.0024 , 0.12) 𝑘𝑘𝑦𝑦

29-10

29.6 a. Determine 𝐾𝐾𝑦𝑦 and % resistance in the gas phase 𝑘𝑘𝑥𝑥 = 0.01

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 , 𝑚𝑚2 ∙ 𝑠𝑠

𝑘𝑘𝑦𝑦 = 0.02

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑚𝑚2 ∙ 𝑠𝑠

𝑘𝑘

From the operating point, use the slope − 𝑥𝑥 = −0.5 towards the equilibrium line to find 𝑘𝑘𝑦𝑦

(𝑥𝑥𝐴𝐴𝐴𝐴 , 𝑦𝑦𝐴𝐴𝐴𝐴 ) on the intersection. (𝑥𝑥𝐴𝐴𝐴𝐴 , 𝑦𝑦𝐴𝐴𝐴𝐴 ) = (0.027 , 0.016)

The operating point is below the equilibrium line (liquid stripping). 𝑦𝑦𝐴𝐴∗ − 𝑦𝑦𝐴𝐴,𝑖𝑖 0.0425 − 0.016 𝑚𝑚 = = = 2.04 𝑥𝑥𝐴𝐴 − 𝑥𝑥𝐴𝐴𝐴𝐴 0.04 − 0.027 ′′

1 𝑚𝑚′′ 𝐾𝐾𝑦𝑦 = � + � 𝑘𝑘𝑥𝑥 𝑘𝑘𝑦𝑦

−1

1 2.04 −1 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 =� + � = 0.0039 2 0.02 0.01 𝑚𝑚 ∙ 𝑠𝑠

Finally, the gas phase percent resistance is 1 1 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 � � 0.02 2 𝑘𝑘𝑦𝑦 𝑔𝑔𝑔𝑔𝑔𝑔 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑚𝑚 ∙ 𝑠𝑠 ∗ 100% = 19.5% = = 1 1 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 �𝐾𝐾 � . 0039 𝑚𝑚2 ∙ 𝑠𝑠 𝑦𝑦 b. Determine ky for 10% gas film controlling resistance 1/𝑘𝑘𝑦𝑦 = 0.10 = 1/𝐾𝐾𝑦𝑦

1 � 𝑘𝑘𝑦𝑦 , 1 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 �. 0039 2 � 𝑚𝑚 ∙ 𝑠𝑠 �

∴ 𝑘𝑘𝑦𝑦 = 0.039

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑚𝑚2 ∙ 𝑠𝑠

c. Determine xAi and yAi −

𝑘𝑘𝑥𝑥

𝑘𝑘𝑦𝑦,𝑛𝑛𝑛𝑛𝑛𝑛

= −0.256, (𝑥𝑥𝐴𝐴𝐴𝐴 , 𝑦𝑦𝐴𝐴𝐴𝐴 ) = (0.0253 , 0.0138)

The interface point moves down along the equilibrium line indicating that the liquid resistance begins to control more of the system.

29-11

29.7 𝑙𝑙𝑏𝑏

𝑤𝑤𝐴𝐴 = 0.003, 𝜌𝜌𝐿𝐿 = 61.8 𝑚𝑚3 , 𝑇𝑇 = 303 𝐾𝐾, 𝑃𝑃 = 1 𝑎𝑎𝑎𝑎𝑎𝑎 𝑓𝑓𝑓𝑓 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑘𝑘𝐿𝐿 = 2.5 , 𝑘𝑘𝐺𝐺 = 0.125 2 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟 ∙ ( ) 𝑓𝑓𝑓𝑓 3 a. pA vs. CAL equilibrium line and operating point 𝐶𝐶𝐴𝐴𝐴𝐴 ≅ 𝑥𝑥𝐴𝐴 𝐶𝐶𝐿𝐿 = (8.46 𝑥𝑥 10−4 ) �3.43 𝑦𝑦𝐴𝐴 =

𝑝𝑝𝐴𝐴 𝑃𝑃

, 𝑃𝑃𝐴𝐴 = 0.1 𝑎𝑎𝑎𝑎𝑎𝑎

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓 3

� = 2.9 𝑥𝑥 10−3

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓 3

pA - cAL

0.14 0.12

pA [atm]

0.10 0.08 0.06 0.04

Equilibrium

0.02

Operating Point

0.00 0

0.005

0.01

0.015

cAL [lbmole/ft3] ∗ 𝑐𝑐𝐴𝐴𝐴𝐴 = 0.00944

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓3

𝑝𝑝𝐴𝐴∗ = 0.0261 𝑎𝑎𝑎𝑎𝑎𝑎

b. Determine pA,i and cAL,i 𝑘𝑘

− 𝐿𝐿 is the slope of the line going from the operating point to �𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖 , 𝑝𝑝𝐴𝐴,𝑖𝑖 �. 𝑘𝑘𝐺𝐺

Use this slope backwards from the operating point to find the y-intercept (b): 𝑝𝑝 𝐴𝐴 = −

𝑘𝑘𝐿𝐿 𝑐𝑐 + 𝑏𝑏, 𝑘𝑘𝐺𝐺 𝐴𝐴𝐴𝐴

𝑏𝑏 = 0.1 𝑎𝑎𝑎𝑎𝑎𝑎 +

2.5 0.00289 = 0.158 𝑎𝑎𝑎𝑎𝑎𝑎 0.125

Polynomial fit of equilibrium line, equate to operating point line to find 𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖 :

3 4 2 −20𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖 + 0.158 = 12422𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖 − 2033.5𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖 + 113.97𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖 + 10.121𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖 − 0.0041

29-12

𝑐𝑐𝐴𝐴𝐴𝐴,𝑖𝑖 = 0.00529

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 , 𝑓𝑓𝑓𝑓 3

𝑝𝑝𝐴𝐴,𝑖𝑖 = 0.0523 𝑎𝑎𝑎𝑎𝑎𝑎

c. Determine 𝐾𝐾𝐺𝐺 , 𝐾𝐾𝐿𝐿 , 𝐾𝐾𝑦𝑦 , 𝐾𝐾𝑥𝑥, 𝑎𝑎𝑎𝑎𝑎𝑎 𝑁𝑁𝐴𝐴 Assume linear equilibrium, 𝐻𝐻 =

𝑝𝑝𝐴𝐴𝐴𝐴

𝐶𝐶𝐴𝐴𝐴𝐴𝐴𝐴

= 9.887

𝑎𝑎𝑎𝑎𝑎𝑎∙𝑓𝑓𝑓𝑓 3 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙

𝑎𝑎𝑎𝑎𝑎𝑎 ∙ 𝑓𝑓𝑓𝑓 3 ⎞ 9.887 1 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 + ⎟ 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 ⎟ 2.5 2 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟 ∙ ( ) 𝑓𝑓𝑓𝑓 3 ⎝ ⎠ 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 0.0836 2 𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎

1 𝐻𝐻 −1 ⎛ 𝐾𝐾𝐺𝐺 = � + � = ⎜ ⎜ 𝑘𝑘𝐺𝐺 𝑘𝑘𝐿𝐿 0.125

1 1 −1 𝐾𝐾𝐿𝐿 = � + � 𝑘𝑘𝐿𝐿 𝐻𝐻𝑘𝑘𝐺𝐺 ⎛ =⎜ ⎜

−1

⎞ 1 1 + ⎟ 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑎𝑎𝑎𝑎𝑎𝑎 ∙ 𝑓𝑓𝑓𝑓 3 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 ⎟ 2.5 9.887 � � ∙ 0.125 2 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟 ∙ ( ) 3 𝑓𝑓𝑓𝑓 ⎝ ⎠ = 0.827 𝑓𝑓𝑓𝑓/ℎ𝑟𝑟

𝐾𝐾𝑦𝑦 = 𝐾𝐾𝐺𝐺 𝑃𝑃 = 0.0836

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙

𝑓𝑓𝑓𝑓2 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎

∙ 1 𝑎𝑎𝑎𝑎𝑎𝑎 = 0.0836

−1

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙

𝑓𝑓𝑡𝑡2 ∙ ℎ𝑟𝑟

𝑙𝑙𝑙𝑙 0.827 𝑓𝑓𝑓𝑓/ℎ𝑟𝑟 ∙ 61.8 � 𝑚𝑚3 � 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓 𝐾𝐾𝑥𝑥 = 𝐾𝐾𝐿𝐿 𝐶𝐶𝐿𝐿 = = 0.798 2 𝑙𝑙𝑙𝑙𝑚𝑚 𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 64.07 � � 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑁𝑁𝐴𝐴 = 𝐾𝐾𝐺𝐺 (𝑝𝑝𝐴𝐴 − 𝑝𝑝𝐴𝐴∗ ) = 0.0836 = 0.00618

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙

𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎

∙ (0.1 − 0.0261) 𝑎𝑎𝑎𝑎𝑎𝑎

29-13

d. Gas phase resistance 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 1 𝐾𝐾𝐺𝐺 0.0836 𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑘𝑘𝐺𝐺 = = ∙ 100% 1 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑘𝑘𝐺𝐺 0.125 2 𝐾𝐾𝐺𝐺 𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 = 66.9% 𝑔𝑔𝑔𝑔𝑔𝑔 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

29-14

29.8 𝑦𝑦𝐴𝐴 = 0.030, 𝑥𝑥𝐴𝐴 = 0.010 𝑘𝑘𝑦𝑦 = 1.0

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟

a. xA vs. yA equilibrium line and operating point 0.05 Equilibrium 0.04

Operating Point

0.03 0.02 0.01 yA *

xA*

0 0 From graph 𝑥𝑥 ∗𝐴𝐴 ≈

0.01 0.036,

b. Determine xAi, yA,i

0.02 ∗

0.03

0.04

𝑦𝑦 𝐴𝐴 ≈ 0.003

1 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑖𝑖𝑖𝑖 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝐾𝐾𝑥𝑥 �𝑥𝑥𝐴𝐴,𝑖𝑖 − 𝑥𝑥𝐴𝐴 � 𝑘𝑘𝑥𝑥 = 0.80 = = = 1 (𝑥𝑥𝐴𝐴∗ − 𝑥𝑥𝐴𝐴 ) 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑖𝑖𝑖𝑖 𝑏𝑏𝑏𝑏𝑏𝑏ℎ 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑘𝑘𝑥𝑥 𝐾𝐾𝑥𝑥 0.80 =

(𝑥𝑥𝐴𝐴𝐴𝐴 − 0.010) �𝑥𝑥𝐴𝐴,𝑖𝑖 − 𝑥𝑥𝐴𝐴 � = (𝑥𝑥𝐴𝐴∗ − 𝑥𝑥𝐴𝐴 ) (0.036 − .0.010)

xAi = 0.031, from graph, yAi = 0.021 c. Determine Ky and NA −

�𝑦𝑦𝐴𝐴 − 𝑦𝑦𝐴𝐴,𝑖𝑖 � (0.030 − 0.021) 𝑘𝑘𝑥𝑥 = = = −0.43 𝑘𝑘𝑦𝑦 �𝑥𝑥𝐴𝐴 − 𝑥𝑥𝐴𝐴,𝑖𝑖 � (0.010 − 0.031) 29-15

𝑘𝑘𝑥𝑥 = 0.43𝑘𝑘𝑦𝑦 = 0.43 𝑚𝑚′ =

(𝑦𝑦𝐴𝐴 − 𝑦𝑦𝐴𝐴∗ )

�𝑥𝑥𝐴𝐴,𝑖𝑖 − 𝑥𝑥𝐴𝐴, �

1 𝑚𝑚′ 𝐾𝐾𝑦𝑦 = � + � 𝑘𝑘𝑦𝑦 𝑘𝑘𝑥𝑥

−1

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟

0.021 − 0.0030 = 0.86 0.031 − 0.010

1 0.86 −1 𝑙𝑙𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 =� + = 0.33 2 � 2 1 0.43 𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟

𝑁𝑁𝐴𝐴 = 𝐾𝐾𝑦𝑦 �𝑦𝑦𝐴𝐴 − 𝑦𝑦 ∗ 𝐴𝐴 � = 0.33

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 ∙ (0.03 − 0.003) = 9.0 ∙ 10−3 2 2 𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟

29-16

29.9 𝑇𝑇 = 293 𝐾𝐾,

𝑃𝑃 = 1.0 𝑎𝑎𝑎𝑎𝑎𝑎,

a. Diagram of tower

𝐶𝐶𝐶𝐶2 = 𝐴𝐴,

𝑘𝑘𝑦𝑦 = 5

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 , 𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟

𝑘𝑘𝑥𝑥 = 5

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟

b. Plot of equilibrium line and terminal stream operating points

Gas Absorption Tower 0.25

Equilibrium 0.20

2 - Top

0.15

1 - Bottom

0.10 0.05 0.00 0.0000

0.0002

0.0004

0.0006

c. Determine overall mass transfer coefficients 𝐾𝐾𝑦𝑦 for top and bottom of the column 3

Polynomial fit of equilibrium line, 𝑦𝑦𝐴𝐴 = −6 ∙ 108 𝑥𝑥𝐴𝐴 + 106 𝑥𝑥𝐴𝐴 2 − 34.998𝑥𝑥𝐴𝐴 − 0.0013

29-17

𝑘𝑘

Using a slope of − 𝑥𝑥 = −0.25 , move from the operating points at 1 and 2 to find 𝑘𝑘𝑦𝑦

�𝑥𝑥𝐴𝐴1,𝑖𝑖 , 𝑦𝑦𝐴𝐴1,𝑖𝑖 � 𝑎𝑎𝑎𝑎𝑎𝑎 �𝑥𝑥𝐴𝐴2,𝑖𝑖 , 𝑦𝑦𝐴𝐴2,𝑖𝑖 � located on the equilibrium line. Find (𝑥𝑥𝐴𝐴2𝑖𝑖 , 𝑦𝑦𝐴𝐴2𝑖𝑖 ) for Top of Tower:

−0.25𝑥𝑥𝐴𝐴2,𝑖𝑖 + 0.05 = −6 ∙ 108 𝑥𝑥𝐴𝐴2,𝑖𝑖 + 106 𝑥𝑥𝐴𝐴2,𝑖𝑖 2 − 34.998𝑥𝑥𝐴𝐴2,𝑖𝑖 − 0.0013, ∴ 𝑥𝑥𝐴𝐴2,𝑖𝑖 = 0.000269, 𝑦𝑦𝐴𝐴2,𝑖𝑖 = 0.0499

Find (𝑥𝑥𝐴𝐴1𝑖𝑖 , 𝑦𝑦𝐴𝐴1𝑖𝑖 ) for Bottom of Tower: 3

−0.25𝑥𝑥𝐴𝐴1,𝑖𝑖 + 0.20 = −6 ∙ 108 𝑥𝑥𝐴𝐴1,𝑖𝑖 + 106 𝑥𝑥𝐴𝐴1,𝑖𝑖 2 − 34.998𝑥𝑥𝐴𝐴1,𝑖𝑖 − 0.0013, ∴ 𝑥𝑥𝐴𝐴1,𝑖𝑖 = 0.000584, 𝑦𝑦𝐴𝐴1,𝑖𝑖 = 0.1998

Find m/ for Top and Bottom using 𝑚𝑚′ = ∗ (𝑥𝑥𝐴𝐴2 = 0) = −0.0013 𝑦𝑦𝐴𝐴2 ∗ (𝑥𝑥 𝑦𝑦𝐴𝐴1 𝐴𝐴1 = 0.0005) = 0.156

𝑦𝑦𝐴𝐴,𝑖𝑖 −𝑦𝑦𝐴𝐴 ∗ 𝑥𝑥𝐴𝐴,𝑖𝑖 −𝑥𝑥𝐴𝐴

𝑦𝑦𝐴𝐴2,𝑖𝑖 − 𝑦𝑦𝐴𝐴2 ∗ 0.0499 + 0.0013 𝑚𝑚 𝑡𝑡𝑡𝑡𝑡𝑡 = = = 190.3 𝑥𝑥𝐴𝐴2,𝑖𝑖 − 𝑥𝑥𝐴𝐴2 0.000269 − 0 ′

𝑚𝑚′ 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 =

𝑦𝑦𝐴𝐴1,𝑖𝑖 − 𝑦𝑦𝐴𝐴1 ∗ 0.1998 − 0.156 = = 521.4 𝑥𝑥𝐴𝐴1,𝑖𝑖 − 𝑥𝑥𝐴𝐴1 0.000584 − 0.0005

Find 𝐾𝐾𝑦𝑦 for Top and Bottom 𝐾𝐾𝑦𝑦,𝑡𝑡𝑡𝑡𝑡𝑡 = �

1 𝑚𝑚′ 𝑡𝑡𝑡𝑡𝑡𝑡 + � 𝑘𝑘𝑦𝑦 𝑘𝑘𝑥𝑥

𝐾𝐾𝑦𝑦,𝑏𝑏𝑏𝑏𝑏𝑏 = �

′

−1

= �

1 𝑚𝑚 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 + � 𝑘𝑘𝑦𝑦 𝑘𝑘𝑥𝑥

−1

1 190.3 + � 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 5 2 20 2 𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟

= �

−1

= 0.103

1 521.4 + � 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 5 2 20 2 𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟 𝑓𝑓𝑓𝑓 ∙ ℎ𝑟𝑟

𝐾𝐾𝑦𝑦 varies due to nonlinearity of the equilibrium line

−1

𝑙𝑙𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚 𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟

= 0.0381

𝑙𝑙𝑙𝑙 𝑚𝑚𝑚𝑚𝑚𝑚 𝑓𝑓𝑓𝑓 2 ∙ ℎ𝑟𝑟

29-18

29.10 a. xA* and yA* From equilibrium distribution plot at P = 2.0 atm At xA = 0.020, yA* = 0.008; at yA = 0.060, xA* = 0.040 0.07

xA, yA 0.06 0.05 0.04

yA 0.03

xAi, yAi

0.02

yA* 0.01

xA* 0.00 0.000

0.010 0.020 0.030 mole fraction H2S in solution, xA

0.040

b. xA,i and yA,i

(

)

N A= K y y A − y*A = k y ( y A − y A,i ) Ky ky

(y − y ) A

A,i

* A

(y − y )

gmole m 2s = ( 0.06 − y A,i ) gmole ( 0.06 − 0.008 ) 0.040 2 ms

0.030

yAi = 0.021, xAi = 0.030 (from equilibrium distribution plot) 29-19

c. KG K= G

K y 0.030 gmole/m 2s gmole = = 0.015 2 P 2.0 atm m s ⋅ atm

d. NA

(

)

N= K y y A − y*A= 0.030 A

gmole gmole ( 0.060 − 0.008=) 1.56×10-3 2 2 ms ms

29-20

29.11 a. cAL* and pA* Plot not provided H = 4.0 m3atm/kgmole = 0.0040 m3atm/kgmole * c= AL

pA = H

( 0.0 atm )

( 0.0040 m atm/gmole ) 3

= 0.0

gmole m3

 m3atm   gmole  p*A =H ⋅ c AL = 0.0040   3.0  = 0.012 atm gmole   m3   b. cAL* < cAL, therefore stripping mass transfer process with operating point (pA, cAL) below the equilibrium line c. KG 1 1 H = + K G kG k L cL 

ρ B,liq MB

(1000 kg/m ) = 55.6 kgmole = 3

(18 kg/kgmole )

gmole   2.0   kx m m 2s   = kL = = 3.6 ×10-5 CL  s kgmole   1000 gmole   55.6   3 m   kgmole  

-1

 m3atm  −1 0.0040   1 H 1 kgmole gmole  K G = +  = + = 2.25 × 10−3 2 m ⋅ s ⋅ atm  3.0 × 10−3 kgmole 1.81 × 10−5 m   kG k L  2   m s ⋅ atm s   d. NA

29-21

gmole   N A = K G p A − p* A =  2.25 × 10−3 2  ( 0.012 atm - 0.0 atm ) m s ⋅ atm   gmole N A = 2.7 × 10−5 m 2s

(

)

e. pAi

m 3.6 ×10-5 kL m3atm s = = 0.012 kG 3.0 × 10−3 kgmole gmole m 2s ⋅ atm − p A ,i ) ( pA = k = − L and p A,i Hc AL ,i kG ( c AL − c AL ,i ) kL c AL kG = k 1+ L kG H

pA + = ∴ p A ,i



m3atm   3.0 gmole    gmole   m3   = 0.0090 atm 3  m atm   0.012  gmole   1+  m3atm  0.0040   gmole  

( 0.0 atm ) +  0.012

29-22

29.12 A = NH3 p A = 0.20 atm, x A = 0.04 atm, P = 2.0 atm, T = 303 K, CL = 55.6

kG = 1.0

kgmole m3

kgmole m , k L = 0.045 2 s m ⋅ s ⋅ atm

a. pA vs. xA equilibrium line and operating point 0.5 0.4

0.3

(Operating Point)

0.2 0.1 0 0.00

0.05

0.10

0.15

0.20

0.25

xA b. Determine k x (55.6 = k x C= L kL

kgmole m kgmole )(0.045= ) 2.50 2 3 m s m ⋅s

c. Determine x A,i , p A,i −

kx = −2.50 atm kG

Find intercept through the equilibrium plot to find interface compositions. x A,i = 0.0761 , p A,i = 0.110 atm

29-23

Determine ( x*A , p*A ) Use plot to find x*A = 0.1216 , p*A = 0.053 atm d. Determine K G H = mP H' =

p A,i − p*A 0.110 atm − 0.053 atm = = 1.58 atm 0.0761 − 0.0400 x A ,i − x A

1 1 H' = + K G kG k x −1

   1 1.58 atm  kgmole + = 0.613 2 KG =   kgmole kgmole m ⋅ s ⋅ atm 1.0 2  2.50 2 m ⋅ s ⋅ atm m ⋅s  

e. Determine N A N= K G ( p A − p*A= ) 0.613 A

kgmole kgmole = 0.0901 (0.20 atm − 0.053 atm) 2 m ⋅ s ⋅ atm m2 ⋅ s

29-24

29.13

waste water A = H2S vo = 20 m3/h cAL,o = 2.5 gmole/m3

Open Tank yA ≈ 0.005, P = 1.0 atm D = 5.0 m cAL

well-mixed

effluent waste water cAL = ?

Given; A = H2S (solute), B = H2O (solvent) T = 293 K, P = 1.0 atm Solute: H = 9.34

m3 ⋅ atm kgmole

Solvent: ρB,liq = 1000 kg/m3, MB = 18 kg/kgmole

m3 gmole Stirred Tank: c AL ,o = 2.50 , vo = 20 , D = 5.0 m (diameter), depth = 1.0 m hr m3 Operating point: yA = 0.0050; xA and cAL unknown Film Mass Transfer Coefficients: k L = 2.0 × 10−4

m kgmole (liquid film), kG = 5.0 × 10−4 2 s m ⋅ s ⋅ atm

a. Estimate m kg   1000 3   ρ kgmole m  cL = B ,liq =  = 55.56 MB  m3 kg  18  kgmole  

kgmole   m3 ⋅ atm   55.56 9.34    m3   kgmole  C H  m= L = = 518.9 P (1.0 atm )

Operating point in mole fraction coordinates (yA, xA) 29-25

0.035 0.030 0.025

yA 0.020 yA* 0.015 0.010

xA, yA 0.005

xA* 0.000 0.0E+00

2.0E-05 4.0E-05 Mole fraction of H2S in water, xA

6.0E-05

Need cAL from part (c) to estimate xA

 gmole  1.7  c AL m3   = = 3.06 × 10−5 xA = cL  kgmole   1000 gmole   55.5  m3   kgmole   b. Determine K y and KL

kgmole   −4 kgmole ky = kG P =  5.0 × 10−4 2  (1.0 atm) = 5.0 × 10 m ⋅ s ⋅ atm  m2 ⋅ s  kgmole   kgmole  −4 m  = k x C= L kL  55.56   2.0 × 10  = 0.011 3 m s  m 2 〈s   −1

 1 m K y = +   k y k x 

−1

    1 518.9 kgmole = + = 2.03 × 10−5  kgmole kgmole m2 ⋅ s  5.0 × 10−4  0.011 m2 ⋅ s m2 ⋅ s   -1

  −1   1 1  1 1   =1.92 × 10−4 m + KL =  +  = 3  s  −4 m m ⋅ atm   −4 kgmole   k L HkG   2.0 × 10 s  9.34    5.0 × 10 2 kgmole   m ⋅ s ⋅ atm     29-26

c. Determine cAL Physical System: liquid in tank with liquid inflow and outflow Source for A: liquid containing H2S dissolved in liquid Sink for A: gas space over liquid Assumptions 1. Constant source and sink for A—steady-state process 2. Well-mixed liquid phase, inflow or outflow of liquid 3. Constant liquid volume with constant inflow and outflow of liquid 4. No homogeneous reaction of A in liquid 5. Dilute UMD process with respect to dissolved A Mass balance on dissolved H2S, liquid phase IN – OUT + GENERATION = ACCUMULATION (moles A/time) S = surface area between the two phases c AL ,o v o − c AL v o − N A S = 0

c AL ,o vo − c AL vo − SK L (c AL −c AL *) = 0 c AL =

S= * c= AL

c AL ,o v o + SK L c AL * v o + SK L

π D 2 π (5.0 m) 2 = =19.6 m 2 4 4 pA yA P = = H H

(0.005)(1.0 atm) gmole = 0.53 3 −3 m3 ( 9.34 × 10 atm ⋅ m /gmole )

kgmole   m3   h  gmole   −4 m   2   0.00250   20  +(19.6 m ) 1.92 × 10   0.53   3 m  h   3600 s  s  m3    c AL =  m3   h  −4 m  2  20  + (19.6 m ) 1.92 × 10    h   3600 s  s    gmole c AL = 1.7 m3

29-27

29.14

gmole gmole m3 c AL ,o = 0.5 , vo = 2.0 , c AL = 0.35 , L = 10.0 m, p A ≈ 0 , c*AL ≈ 0 3 3 m m s m3 ⋅ atm m3 ⋅ atm = = 500 T= 293 K, H 0.50 , P = 1.0 atm gmole kgmole

kc 2.67 ×10−4 =

m m 5.5 ×10−3 , k= L s s

a. Determine K L m kc gmole s kG = = = 0.0111 2 3 m ⋅ atm RT m ⋅ s ⋅ atm (8.206 x 10−5 )(293 K) gmole ⋅ K 2.67 x 10−4

1 1 1 = + K L k L H kG −1

    1 1 m   = KL = 0.00276 + 3 m ⋅ atm gmole  s  5.5 x 10−3 m (0.50 )(0.0111 2 )  s gmole m ⋅ s ⋅ atm  

b. Develop material balance model and determine surface area S Assume: 1) steady state, 2) no reaction, 3) constant T and P, 4) liquid stripping process. A = Species A, B = Air, C = Water IN – OUT – FLUX OUT = 0

c AL ,o vo − c AL vo − N A S = 0 c AL ,o vo − c AL vo − SK L (c AL − c*AL ) = 0 Determine S , the surface area between the two phases

29-28

m3 gmole gmole 2.0 (0.5 ) − 0.35 3 vo (c AL ,o − c AL ) s m m3 311 m 2 S = = = * m gmole gmole K L (c AL − c AL ) 0.00276 (0.35 ) −0 s m3 m3 c. Determine v∞ Laminar flow over a flat plate:

ν= B

µB = ρB

kg 2 m ⋅= s 1.503 x 10−5 m kg s 1.206 3 m

1.813 x 10−5

νB 

kc L  DAB   v∞ =   L  0.664 DAB  ν B  

  

13 2    m2  cm 2 −4 m −4 m 1.503 x 10 (2.67 x 10 )(10 m) )(1 x 10 )    (0.08 2  m s s s cm  v∞ =    0.249 2 2 2 cm cm s 10 m  −4 m    1.503 x 10−5    0.664(0.08 s )(1 x 10 cm 2 )  s    d. Determine p A,i −5

p A − p A ,i kL m3 ⋅ atm − = = −0.495 kG c AL − c AL ,i gmole 0 atm − p A,i m3 ⋅ atm = gmole 0.35 gmole − c AL ,i m3 m3 ⋅ atm 0.50 p A,i Hc c AL ,i = = AL ,i gmole −0.495

atm, c AL ,i 0.174 = = p A,i 0.087 Two equations, two unknowns:

gmole m3

29-29

29.15 A = ozone, B = water, P = 1.5 atm, S = 4.0 m2, T = 293 K, ρ = 992.3

kg m3

gmole mg gmole m3 1.0 c = c AL 3.0 = = 238 , , = v 0.050 AL , o o m3 m3 L hr m3 ⋅ atm m −3 m k= 5.0 ×10 3.0 ×10 , k= , H = 68.2 , p A= H ⋅ c AL ,i L c ,i kgmole s s −6

Assume: 1) dilute solution, 2) constant P and T, 3) linear equilibrium. a. Determine m

kg m3 ⋅ atm (68.2 ) CL H ρH m3 kgmole m = = = = 2.51×106 P M B P (0.018 kg ) ⋅1.5 atm kgmole 992.3

m is large, so ozone is sparingly soluble in water. b. Determine K G

1 1 H RT H = + = + K G kG k L k c k L −1

 m3 ⋅ atm m3 ⋅ atm  −8 (8.206 x 10 )(293 K) 68.2  kgmole K ⋅ kgmole kgmole  KG  4.40 x 10−8 2 = += m m m ⋅ s ⋅ atm  5.0 x 10−3 3.0 x 10−6   s s   c. Determine K L 1 1 1 1 RT = + = + K L k L HkG k L Hkc −1

  m3 ⋅ atm −8 (8.206 x 10 )(293 K)   1 m K ⋅ kgmole   = 3.00 x 10−6 KL = + 3 m ⋅ atm m  s  3.0 x 10−6 m (68.2 )(5.0 x 10−3 )   s kgmole s  

29-30

d. Develop material balance model Assume: 1) steady state, 2) no reaction, 3) gas absorption process. IN – OUT + FLUX IN = 0

c AL ,o vo − c AL vo + N A S = 0 vo (c AL ,o − c AL ) + SK G ( p A − p*A= ) vo (c AL ,o − c AL ) + SK G ( p A − Hc AL= ) 0

e. Determine WA Determine p A and y A

= pA

vo (c AL ,o − c AL ) SK G

+ Hc AL

m3 gmole gmole − 1.0 (0.050 )(3.0 ) 3 m3 ⋅ atm kgmole hr m m3 = + (68.2 pA )(0.001= ) 0.226 atm 3 gmole s kgmole m −5 2 (4.0 m )(4.40 x 10 )(3600 ) m 2 ⋅ s ⋅ atm hr p A 0.226 atm = = 0.151 yA = P 1.5 atm = WA SN = SK G ( p A − Hc AL ) A WA = (4.0 m 2 )(4.40 x 10−5

gmole  m3 ⋅ atm gmole  ) 0.226 atm-(0.0682 )(1.0 )  2 m ⋅ s ⋅ atm  gmole m3 

gmole s f. Resistance in liquid phase = 2.78 x 10−5

𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑖𝑖𝑖𝑖 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝐾𝐾𝐿𝐿 3.0 𝑥𝑥 10−6 𝑚𝑚/𝑠𝑠 = = = 1.0 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑖𝑖𝑖𝑖 𝑏𝑏𝑏𝑏𝑏𝑏ℎ 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑘𝑘𝐿𝐿 3.0 𝑥𝑥 10−6 𝑚𝑚/𝑠𝑠 Therefore the process is liquid phase controlling

29-31

29.16 H = 170 cm3 aqueous phase/cm3 organic phase / H ⋅ cA At equilibrium, c= A m m / 3.5 ×10−6 , k= 2.5 ×10−5 k= L L s s

a. Determine K L/ = Aqueous Film: N A k L ( c AL − c AL ,i )

/ / = Organic Film: N= k L / ( c AL ) kL / ( HcAL,i − cAL/ ) A ,i − c AL

(

)

/ / / Organic Overall:= N A K L / ( c AL K L / ( Hc AL − c AL ) ) − c= AL *

Combine: 1 1 H = + / K L k L/ k L −1

    1 170 m K L/ = + 1.41 x 10−7   = m m s  3.5 x 10−6 (2.5 x 10−5 )  s s   b. Determine K L 1 1 1 = + K L k L H ⋅ k L/ −1

    1 1 m + = KL = 2.40 x 10−5   m m s  2.5 x 10−5 170 ⋅ 3.5 x 10−6  s s  

c. Determine percent resistance to mass transfer encountered in the aqueous liquid film

1k K 2.40 x 10−5 m/s ⋅100% = Percent Resistance = L = L = 96.0% 1 K L kL 2.5 x 10−5 m/s

29-32

29.17

gmole , D = 20 m, P = 1 atm, T = 293 K m3 gmole gmole gmole , k x = 200 2 , k y = 0.1 2 , p A= y A = 0.04 , c AL = 10.0 H ⋅ x A ,i , ,i 3 m m ⋅s m ⋅s

c AL ,o = 50

H = 550 atm a. Determine K L A = TCE, B = water

C C P ρ ρB P 1 1 1 = + =L + L = B + K L k L HkG k x Hk y M B k x M B Hk y −1

  g g (998, 200 3 ) (998, 200 3 )(1 atm)   m m m KL = 7.78 x 10−4 +   = g gmole g gmole s  (18 )(200 2 ) (18 )(550 atm)(0.1 2 )   gmole  m ⋅s gmole m ⋅s  b. Determine N A  p c  yA PρB   N= K L (c AL − c*AL= ) K L  c AL − A L=  A  K L  c AL − H  MBH     g  (0.04)(1 atm)(998, 200 3 )   gmole m  = 4.64 x 10−3 gmole − N A = 7.78 x 10−4 m/s 10.0 3 g m m2 ⋅ s   (18 )(550 atm)   gmole   c. Develop material balance model and determine vo Assume: 1) steady state, 2) well-mixed, 3) no reaction, 4) liquid stripping. IN – OUT – FLUX OUT = 0 c AL ,o vo − c AL vo − N A ⋅ S = 0

(c AL ,o − c AL )vo − N A

vo =

π D2 N A

= 4(c AL ,o − c AL )

π D2 4

= 0

gmole ) m2 ⋅ s 0.0364 m3 /s = gmole gmole 4(50 ) − 10 m3 m3

π (20 m) 2 (4.64 x 10−3

29-33

29.18 D = 4 m, p A = 0.020 atm, P = 1.0 atm, T = 293 K

m3 ⋅ atm L m3 0.020 H = , c AL ,o = 0 , p A= , = vo 200 = 0.20 H ⋅ c ,i AL ,i kgmole hr hr kgmole m kG = 1.25 2 , k L = 0.05 m ⋅ hr ⋅ atm hr a. Develop material balance and determine cAL

c AL ,o vo − c AL vo + N A S = 0 c AL ,o vo − c AL vo + K L (c*AL − c AL )π

c AL ,o vo − c AL vo +

D2 = 0 4

π D 2 K L  pA  − c AL  = 0  4 H 

1 1 1 = + K L k L HkG −1

   1  1 m   = + KL = 0.0167 3 m m ⋅ atm kgmole  hr  0.05 (0.020 )(1.25 2 )  hr kgmole m ⋅ hr ⋅ atm   Aside: 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑖𝑖𝑖𝑖 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝐾𝐾𝐿𝐿 0.0167 𝑚𝑚/ℎ𝑟𝑟 = = = 0.334 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑖𝑖𝑖𝑖 𝑏𝑏𝑏𝑏𝑏𝑏ℎ 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑘𝑘𝐿𝐿 0.050 𝑚𝑚/ℎ𝑟𝑟 Therefore 33% resistance in liquid, 67% resistance in gas Determine c AL m )(0.020 atm) hr 0+ m3 ⋅ atm π D 2 K L pA 4(0.020 ) vo c AL ,o + kgmole kgmole H 4 c AL = = = 0.512 2 m π D KL m3 (4 m) 2 (0.0167 ) π 3 vo + m hr 4 (0.20 )+ hr 4

π (4 m) 2 (0.0167

29-34

b. Determine p A,i and c AL ,i

p A − p A ,i k = − L = kG c AL − c AL ,i

p A − p A ,i P c AL − A,i H 3 0.020 atm − p A,i k m ⋅ atm − L = −0.04 = p A ,i kgmole kG kgmole 0.512 − 3 m3 ⋅ atm m 0.020 kgmole

p A,i = 0.0135 atm c AL ,i =

p AL ,i = H

0.0135 atm kgmole = 0.675 3 m ⋅ atm m3 0.020 kgmole

c. Determine WA

WA = SN A =

WA = 0.102

π D K L  PA 2

  − c AL  = H 

π (4 m) 2 (0.0167 4

 m  )  hr  (0.020 atm) − 0.512 kgmole  m3 ⋅ atm m3    (0.020 kgmole )   

kgmole hr

d. Would the flux NA increase by the following: Increasing volume level of tank with fixed surface area: does not affect flux. Increasing the agitation intensity of the bulk liquid: will increase the flux since kL will increase. Increasing the agitation intensity of the bulk gas: will increase the flux since kG will increase and the overall mass transfer coefficient is determined by contributing resistances in both the gas and liquid films. Increasing the system temperature: will likely increase the flux since the Henry’s constant will increase with increasing temperature, which will lower the solubility of the solute in the gas and lower the concentration driving force for mass transfer.

29-35

29.19 bulk gas y A∞ = y A ≈ 0 gas film

kc = 5.0 x 10-3 m/s

liquid film

kL = 4.0 x 10-5 m/s

bulk liquid CAL = 0.15 gmole A/m3

bulk air flow v∞ = ? m/s Waste water IN CAL,o = 0.25 gmole A/m3 Vo = ? m3/h

1.0 m

wwopen pond well-mixed

Waste water OUT CAL = 0.15 gmole A/m3 Vo = ? m3/h

x=0

10 m

a. H, cAL*, pA* , cAL,i, pA,i Plot not provided H

3  m3 Pa   1.0 atm  -4 m atm = 3.6 ×10  36.6  gmole   101,250 Pa  gmole 

Alternatively, calculate analytically using value for H At pA = 0.0 atm, cAL = 0.15 gmole/m3

( 0.0 atm ) pA gmole = = 0.0 3 H  m3 -4 m atm   3.6 ×10  gmole   3  gmole  * -4 m atm   -5 p A =H ⋅ c AL = 3.6 ×10   0.15  = 5.4 ×10 atm 3 gmole   m  

* c= AL

(

)

5.0×10-3 m/s kc gmole = kG = = 0.208 2 3 RT  m s ⋅ atm -5 m atm   8.206×10  ( 293K ) gmole K  

29-36

m 3 kL s= 1.92 ×10-4 m atm = gmole kG 0.208 kgmole 2 m s ⋅ atm − p A ,i ) ( pA = k and p A,i Hc AL ,i = − L kG ( c AL − c AL ,i ) 4.0 ×10-5

kL c AL kG = k 1+ L kG H

pA + = ∴ p A ,i



m3atm   0.15 gmole    gmole   m3   = 1.9 ×10-5 atm 3  -4 m atm  1.92 ×10  gmole  1+  3  -4 m atm   3.6 ×10  gmole  

( 0.0 atm ) + 1.92 ×10-4

b. KL

  −1   1 1  1 1   = + K  +  = L 3 m  k Hk   -5 m atm gmole   -4 G   L  4.0 ×10 s  3.6 ×10    0.208 2 gmole   m s ⋅ atm     m K L = 2.61×10−5 s

-1

c. v∞, Sc, Re for air flow over pond cm 2 vB s = 2.32 = Sc = 2 cm DAB 0.065 s m  5.0×10-3  (10 m )  kc L s   = = Sh = 7692 2 DAB  cm 2   1m   0.065   s   100 cm   Assume that flow is turbulent, check ReL 0.151

ShL =

kc L 1/3 = 0.0365 Re0.8 L Sc DAB 5/4

  ShL Re L = = 1/3   0.0365Sc 

 ( 7692 )     0.0365 ( 2.32 ) 

5/4

=3.16×106 (> 3.0×106 , ∴ turbulent)

29-37

Re L =

v∞ L

νB

 cm 2   1m  3.16×10 )  0.151 (   s   100 cm  Re L ν B m  = v∞ = = 4.8 s L (10 m ) 2

d. vo based on mass balance S = surface area between the two phases 0 c AL ,o v o − c AL v o − N A S = 0 c AL ,o v o − c AL v o −  K L ( c AL − c*AL )  S = vo =

K L ( c AL − c*AL ) S c AL ,o − c AL

gmole   -5 m    2.61×10   ( 0.15 - 0 )  (10 m x10 m ) m3 m3 s  m3  = =3.9×10-3 14.0 gmole s h ( 0.25 - 0.15) 3 m

29-38

30.1 Given: A = solvent, B = air T = 298 K, P = 1.0 atm Sphere: D = 1.0 cm, 0.12 g A/cm2 liquid solvent on sphere at t = 0 Physical parameters: MA = 78 g/gmole, pA* = 1.17 × 104 Pa (298 K), DAB = 0.0962 cm2/s, νB = 0.156 cm2/s Physical System: boundary layer between surface of sphere and bulk gas Source for A: solvent coating surface of sphere Sink for A: bulk flowing gas a. Time to dry for still air Mass balance on solvent evaporating from paint-coated sphere Initial mass of solvent on sphere

π D 2 0.12 g

= mAo

= M A cm 2

π(1.0 cm) 2  0.12 g  = 4.83 x 10-3 gmole  2  ( 78 g/gmole )  cm 

IN – OUT + GEN = ACCUMULATION (moles A/time) dm 0 − N AS + 0 = A dt t

m Ao

−WA ∫ dt = ∫ dmA WA ⋅= t mAo − mA

= WA kc ( c*A − c A∞ ) π D 2 Determine kc, WA, then t For still air, Sh = 2.0 DAB 0.0962 cm 2 /s = kc Sh = 2.0 =0.192 cm/s D 1.0 cm 30-1

(1.17 x 10 Pa ) = 4.72 gmole/m = 4.72x10 gmole/cm (8.314 m ⋅ Pa/gmole ⋅ K ) (298 K) 4

pA = c = RT * A

-6

WA =π D 2 kc (c*A − c A∞ ) = π (1.0 cm ) ( 0.192 cm/s ) ( 4.72 x 10-6 - 0 ) gmole/cm3 = 2.85 x 10-6 gmole/s 2

Final mass of solvent on sphere − mA = 0 for dried paint coating on sphere t =

mAo 4.83 x 10-3 gmole =1697 s = WA 2.85 x 10-6 gmole/s

b. Time to dry for flowing air with v∞ = 1.0 m/s Determine kc, WA, then t For air at 298 K and 1.0 atm, νair = 1.56 × 10−5 m2/s (Appendix I) Re = Sc =

v∞ D

ν air ν air DAB

(1.0 m/s )( 0.01m ) = 641 1.56 x 10-5 m 2 /s

0.156 cm 2 /s =1.62 0.0962 cm 2 /s

Froessling equation for gas flow around a single sphere

Sh =

kc D = 2 + 0.552 Re1/2 Sc1/3 =2.0+0.552(641)1/2 (1.62)1/3 =18.4 DAB

DAB 0.0962 cm/s 18.4 = 1.77 cm/s = = kc Sh 1.0 cm D

WA =π D 2 kc (c*A − c A∞ ) = π (1.0 cm ) (1.77 cm/s ) ( 4.72 x 10-6 - 0 ) gmole/cm3 = 2.63 x 10-5 gmole/s 2

= t

mAo 4.83 x 10-3 gmole = 184 s = WA 2.63 x 10-5 gmole/s

30-2

30.2 A = naphthalene, B = air a. Initial evaporation rate (as flux NA) , ν air 1.3876 ×10−5 m 2 /s T = 280 K= From Appendix J, T 298 K , DAB P 0.619 m 2 -Pa/s = = 3

DAB

0.619 m 2 -Pa/s  280  2 5.57×10-6 m 2 /s =   1.013×105 Pa  298 

ν air 1.3876 x 10-5 m 2 /s Dv ∞ ( 0.0175 m )(1.4 m/s ) Sc = = 1766, = = 2.49 = ν air DAB 1.3876 × 10-5 m 2 /s 5.57×10-6 m 2 /s

Sh =

kc D = 2.0 + 0.552 Re1/2 Sc1/3 = 2.0+0.552(1766)1/2 (2.49)1/3 =33.4 DAB

33.4 DAB 33.4(5.57×10-6 m 2 /s) = = 0.011m/s kC = D ( 0.0175 m ) PA 2.8 Pa = = 1.203 x 10-3 mol/m3 3 RT  Pa-m   8.314  (280 K) mol-K   N A = kc (C As − C A∞ ) = (0.011m/s)(1.203 x 10-3 mol/m3 ) = 1.32 x 10-5 mol/m 2 -s

C= As

=1.32 x 10-8 kgmol/m 2 -s b. Time to reach half of the original diameter -2 -3 -3 m, R1 8.75×10 m, D2 8.75×10 m, R2 4.38×10-3 m = D1 1.75×10 = = = 4 V = π R3 3 dV dR = 4π R 2 dt dt let "A" be Naphthalene, M A = 128 lb/lbmol

ρ A dV = W= kc C As 4π R 2 A M A dt ρA dR 4π R 2 = kc C As 4π R 2 MA dt ρ A 71.136 lb/ft 3 = 0.556 lbmol/ft 3 = 8.90 kgmol/m3 = M A 128 lb/lbmol R2 1 t ρA = dR ∫= dt t ∫ 0 M A C As R1 kc

30-3

ρA 8.90 × 103 mol/m3 = = 7.4 × 106 -3 3 M A C As 1.203 x 10 mol/m R2 1 dR = t (7.4 × 106 ) ∫ R1 k c Evaluate by graphical integration, plot 1/k c vs. R (R = D/2) = Re Sh =

( D )(1.4 m/s ) Dv∞ = = ( D)(1.009 × 105 ) m −1 -5 2 ν air 1.3876 x 10 m /s

kc D = 2.0 + 0.552 Re1/2 Sc1/3 DAB

k c as a function of D = 2R (5.57×10-6 m 2 /s)  2.0 + 0.552 (( D)(1.009 × 105 ) m −1 )1/2 (2.49)1/3  D 6 (7.4 × 10 )(0.3505) = 2.59 × 106 sec = 720 hr t=

kc =

30-4

30.3

z = L = 25 m water CAL,o = 0 gmole A/m3 v∞ = 0.50 m/s 293 K

D = 2.0 cm

CAL

solid contaminant layer lining inside bottom half of tube

CAL* tube cross section

a. Solute A undergoes mass transfer. The SOURCE for A is the solid A in the contaminant layer, and dissolved solute A carried in with the inlet water (zero). The SINK for A is the flowing fluid. A = solute A, B = water b. Mass flowrate of inlet water at 20 C, ρB = 998.2 kg/m3

π D2  m  kg   π(0.02 m) 2  kg wB v= ρ 0.50 998.2 =  = 0.157 ∞ B ,liq   3  4 s  m  4 s   c. kL 2 µ B 9.93×10-4 kg/m ⋅ s -7 m = 9.95×10 = ν= B 998 kg/m3 s ρB m2 ν B 9.95×10 s 199 = Sc = = 2 DAB -9 m 5.0×10 s -7

= Re

v ∞ D (0.5m/s)(0.02m) = = 10, 050 laminar regime (1.47 cm 2 /s) νB

0.83 = = Sh 0.023Re Sc1/3 0.023 (10,= 050 ) (199 ) 281.7 0.83

Sh =

1/3

kL D DAB

2  -9 m  281.7 5.0×10 ( )  s  Sh ⋅ DAB m  = kL = = 7.04 ×10-5 D s ( 0.02 m )

30-5

d. Differential model for cAL(z) Differential material balance on dissolved solute A in liquid phase IN – OUT + GEN = ACCUMULATION (moles A/time) v∞

π D2 4

c AL z − v∞

π D2 4

(

= N A k L c*AL − c AL Note

c AL z +∆z − N A

π D∆z 2

+0= 0

)

Rearrange, lim Δz→0 c AL z +∆z − c AL z  2k − − L c*AL − c AL = 0 ∆z v∞ D

(

)

Differential model −

(

)

dc AL 2k L * c AL − c AL = 0 − dz v∞ D

z = 0, cAL = cAL,o (entrance); z = L, cAL = cAL Integrated model c AL

dc A 2k = − L ∫ dz ∫ * v∞ D 0 c AL ,o c AL − c AL

(

)

 c* − c AL ,o  2 L k L ln  AL*  = − c c  AL AL  D v ∞ e. cAL  2L kL  c AL =c*AL − ( c*AL − c AL ,o ) exp  −   D v∞   2(25 m) ( 7.04×10-5 m/s )  mmole  mmole   c AL = 10 - 10 - 0  exp m3  m3 (0.02 m)(0.5 m/s)     mmole c AL = 2.97 m3

30-6

30.4 stent vo = 10 cm3 /sec 1.0 cm

blood vessel

stent detail support ring drug (Taxol) loaded on surface of stent

1.0 cm

D = 0.2 cm

Given: A = taxol, B = H2O (liquid bodily fluid)

taxol: cA* = 2.5 × 10–4 mg/cm3, DAB = 1.0 × 10–6 cm2/s Body fluid: µB = 0.040 g/cm∙s, ρB =1.05 g/cm3, MB = 18 g/gmole

Stent: Dcyl = 0.2 cm, L = 1.0 cm, mAo = 5.0 mg Physical System: boundary layer between surface of cylindrical stent and body fluid Source for A: drug coating external surface of cylindrical stent Sink for A: body fluid a. Convective mass transfer coefficient kL External flow of liquid around a cylinder k x Sc 0.56 = 0.281( Re ') −0.4 GM 4vo 4 ⋅10.0 cm3 / s = = 12.73 cm/s v∞ = π Dtube 2 π (1.0 cm) 2 = GM

v ∞ ρ B (12.73 cm/s)(1.05g/cm3 ) = =0.743 gmol/cm 2 ⋅ s MB 18g/gmol

30-7

v ∞ Dcyl v ∞ Dcyl ρ L (12.73cm/s)(0.2 cm)(1.05 g/cm3 ) =66.83 Re ' = = = µL vL ( 0.040 g/cm ⋅ s )

µL 0.040 g/(cm-s) = = 38, 095 ρ L DAB (1.05 g/cm3 )(1.0×10-6 cm 2 /s)

= Sc

0.281( = Re ') −0.4 ( Sc) −0.56 GM 0.281(66.83) −0.4 (38, 095) −0.56 (0.743gmole/cm 2s)

k x =1.06 x 10-4 gmole/cm 2s

 18 g/gmol  kx M = = 1.06 ×10−4 gmol/cm 2s )  1.81 x 10-3 cm/s kL = kx L = ( 3  ρL CL  1.05 g/cm 

b. Time required for complete Taxol release Mass balance on Taxol coating stent IN – OUT + GEN = ACCUMULATION (moles A/time) dm 0 − N AS + 0 = A dt t

m Ao

−WA ∫ dt = ∫ dmA WA ⋅= t mAo − mA Determine WA, then t WA = k L (c*AL − c AL )π Dcyl L = k L (c*AL )π Dcyl L WA = (1.81×10-3 cm/s )( 2.5×10-4 mg/cm3 ) ( π )( 0.2 cm )(1.0 cm ) WA = 2.84 x 10-7 mg/s t =

mAo − mA mAo 5.0 mg = = WA WA 2.84 x 10-7 mg/s

= 1.76 x 107 sec = 4890 h

30-8

30.5 Let A = O2, B = H2O (liquid) a. Material balance In − Out + Generation = Accumulation [ S ⋅ N A + v∞ LWC Alo ] − [v∞ LWC Al ] + 0 = 0 * [( N tπ DL)(k L (C AL 0, combineterms to solve for C AL − C AL )) + v∞ LWC Alo ] − [v∞ LWC Al ] = * (v∞ LW )C Alo + ( N tπ DLk L )C AL C AL = (v∞ LW ) + ( N tπ DLk L )

b. Mass transfer coefficient, kL = ShD

kL D = 0.281( Re ') 0.6 ( Sc ) 0.44 DAB

v∞ =

 w nM = LW ρ

0.44

 ν   DAB       D   DAB  ( 50 mol/s )( 0.018 kg/mol ) = 1.80×10-3 m/s (1 m )( 0.50 m ) ( 998.2 kg/m3 )

v D k L = 0.281 ∞   ν 

0.6

 (1.80×10-3 m/s)(0.02 m)  k L = 0.281  (0.995×10-6 m 2 /s)  

0.6

 (0.995×10-6 m 2 /s)    -9 2  (2×10 m /s) 

0.44

 (2×10-9 m 2 /s)  −6 =   3.72 × 10 m/s (0.02 m)  

c. Outlet concentration, CAL For CAL,o = 0 C AL

* ( N tπ DLk L )C AL (10)π(0.02 m)(1 m)(3.72×10-6 m/s)(5 mol O 2 /m3 ) = (v∞ LW ) + ( N tπ DLk L ) (1.80×10-3 m/s)(0.02 m)(1 m)+(10)π(0.02 m)(1 m)(3.72×10-6 m/s)

C AL = 0.0129 mol/m3

Increase CAL* by increasing pressure on the oxygen side to increase CAL.

30-9

30.6 Let A = H2O vapor, B = air Attempt to use Concentration-Time Charts with m, n, XD, and Y For m, Estimate kc first Using Froessling Equation for liquid flow around a sphere with 2 ≤ Re ≤ 800 and 0.6 ≤ Sc ≤ 2.7

Re =

v∞ D (50 cm/ s)(0.20 cm) = = 63.74 (0.15689 cm 2 / s ) ν

Sc =

= DAB

= Sh

kc D = 2.0 + 0.552 Re1/2 Sc1/3 DA− air

0.15689cm 2 / s = 0.603 0.260cm 2 / s

= kc  2.0 + 0.552 Re1/2 Sc1/3 

DA− air D

kc =  2.0+0.552(63.74)1/2 (0.603)1/3  = m

0.260 cm 2 /sec = 7.049 cm/sec 0.20 cm

DAB 1.0 x 10-6 cm 2 /sec = =1.42 x 10-6 ≈ 0 kc R (7.049 cm/sec)(0.10 cm)

n = 0 (center of the sphere)

sec  (1.0 x 10 cm /sec )  2.78 hr 3600  1 hr  -6

DAB t = XD = R2

(0.10 cm) 2

= 1.00

Off the range of the Charts for a sphere. Therefore, use Infinite Series Solution for center of sphere

C As= C A* =

S ·PA=

S ⋅ yA P

=2.0 gmole O 2 /(cm3 silica)(atm H 2 O vapor)(0.01)(1.0 atm) = 0.02 gmole/cm3 ∞ 2 2 c −c n Y = A Ao = 1 + 2∑ ( −1) e − n π X D , r = 0, n = 1, 2, 3, c As − c Ao n =1 c −0 Y= A  1.0 , CA = 0.02 gmole/cm3 0.02 − 0

30-10

30.7 Let A = Dicyclomine, B = intestinal fluid (H2O liquid) a. Concentration of A in center of sphere Use Concentration-Time charts for a sphere, Fig. F.3 DAet (4.0×10-6 cm 2 /sec)(15,625 sec) =1.00 = (0.25 cm) 2 R2 0 cm r n = = = 0 R 0.25 cm 1 1 DAe m= = = = 0.20 kc R Bi 5

XD =

0.002= Y=

C A (0, t ) − C A∞ C Ao − C A∞

2.0 mg mAo mAo = = = 30.56 mg/cm3 4 3 4 V π(0.25 cm)3 πR 3 3 C A (0, t )= Y (C Ao − C A∞ ) + C A∞ = 0.002(30.56 - 0.20) mg/cm 2 +0.20 mg/cm3 = 0.26 mg/cm3 C = Ao

b. Estimate of Bi  ρ v D   µB  = Pe (= Re)( Sc)  B ∞   =   µ B   ρ B DAB   (1.0 g/cm3 )(0.50 cm/sec)(0.50 cm)    (0.007 g/cm-sec)  =  3 -5 2 (0.007 g/cm-sec)    (1.0 g/cm )(1.0×10 cm /sec)  = = 25, 000 Pe (35.7)(700) kR kD = c , c = Sh = (4.0 + 1.21Pe 2/ 3 )1/2 Bi DAe DAB  R   DAB  2/3 1/2 = Bi    (4.0 + 1.21Pe ) D D   Ae      (1.0×10-5 cm 2 /sec)  (0.25 cm) 2/3 1/2 Bi  40.3 =  (4.0 + 1.21(25, 000) )  -6 2 (0.50 cm)  (4.0×10 cm /sec)   

30-11

30.8 Let A = solute, B = solvent a. Estimate the film convective mass transfer coefficient, kL v ∞ D (10 cm/sec)(1.0 cm) Re = = 1005 = ν (9.95 x 10-3 cm 2 /sec)

9.95 x 10-3 cm 2 /sec = = 829.17 Sc = DAB 1.2 x 10-5 cm 2 /sec = Pe (Re)( = Sc) 8.33 x105 > 10, 000 Using Levich Equation for liquid flow around a sphere with Pe > 10,000 1/3

1/3

v D υ   v∞ D  k D 1/3 Sh = L =1.01Pe1/3 =1.01 ∞  1.01  AB =1.01(Re ⋅ Sc) DAB  υ DAB   DAB 

1.01( Pe ) DAB 1.01(8.33 x 105 )1/3 (1.2 x 10-5 cm 2 /sec) = 1.14 x 10-3 cm/sec = kL = 1.0 cm D b. Dissolution rate of the pellet at initial diameter of D = 1.0 cm (D = 2R) 1/3

* = W A 4π R 2 ( C AL − C AL ,∞ )

= 4π ( 0.50 cm ) (1.14 x 10-3 cm/sec )( 7.0 x 10-4 - 1.0 x 10-4 ) gmol/cm3 = 2.15 x 10-6 /gmol/sec 2

c. Time t at D = 0.50 cm In – Out +Generation = Accumulation (moles A/time) ρ  ρA d  A V 4π R 2 dR M dmA dmA A =  MA 0 − N= , = A ⋅S +0 dt dt dt dt ρ dR A 4π R 2 −k L (C AL* − C AL∞ )4π R 2 = MA dt M dR * = −k L (C AL − C AL ,∞ ) A ρA dt 1/3

kL −

 v ∞ 2 R  DAB (v ∞ )1/3 ( DAB ) 2/3 1.01 = 1.01   (2 R) 2/3  DAB  2 R

M 1 * dR =(C AL − C AL ,∞ ) A dt kL ρA

30-12

1.01(v ∞ )1/3 ( DAB ) 2/3 * M = − ∫ R dR (C AL − C AL ,∞ ) A ∫ dt 2/3 2 ρA 0 Ro 2/3

Let D = 2R, integration and simplification yields ρA 5 t Do5/3 − D 5/3 1/3 3 v  * − C AL ,∞ ) 2.02 DAB M A  ∞ Sc  ( C AL ν 

(

5 3

5 t= 3

ρA 1/3

v  2.02 DAB M A  ∞ Sc  ν 

(C − C ) * AL

5/3 o

)

− D 5/3

)

AL , ∞

( 2.0 g/cm ) 3

1/3

10 cm/sec   2.02 (1.2 x 10 cm /sec ) (110 g/gmol )  ⋅ 829  -3 2  9.95 x 10 cm /sec  1 (1.0 cm)5/3 -(0.5 cm)5/3 ⋅ -4 -4 3 ( 7.0 x 10 - 1.0 x 10 ) gmol/cm 5

(

)

t = 7583 sec = 2.1 hr

30-13

30.9 sunlight wastewater CAL,o = 10 g A/m3 Re = 5000 313 K

z = L = 25 m CAL

D = 1.5 cm

transparent TiO2 catalyst layer lining inside of glass tube

CAL,s ≈ 0

catalyst surface

a. Solute A (benzene) undergoes mass transfer. The SOURCE for benzene is the benzene dissolved in the inlet liquid. The SINK for benzene is the consumption by photocatalytic reaction at the catalyst surface lining the inner wall of the tube. A = benzene, B = water b. The photocatalytic decomposition of benzene is a heterogeneous reaction at a boundary surface. c. kL at Re = 5000 DAB 13.26 ×10−5 µ B−1.14VA−0.589 = VA = 6 ⋅ VC + 6 ⋅ VH − 15 = 6 ⋅14.8 + 6 ⋅ 3.7 − 15 = 96cm3 /gmole DAB = 13.26×10-5 ( 0.658 ) Sc =

B = ρ B DAB

-1.14

( 96 )

-0.589

=1.45×10-5

cm 2 s

( 0.00658g/cm ⋅ s ) = 457

( 0.993 g/cm )(1.45×10 cm /s ) 3

-5

Re > 2000 therefore turbulent flow inside tube ρ v D Re = B ∞

µB

v∞ =

µ B Re ( 0.00658 g/cm ⋅ s )( 5000 ) cm = = 22.1 3 ρB D s ( 0.993 g/cm ) (1.5 cm )

0.83 Sh 0.023Re Sc1/3 0.023 ( 5000 = = = ) ( 457 ) 208.2 0.83

Sh =

kL D DAB 

kL =

1/3

Sh ⋅ DAB = D

( 208.2 ) 1.45×10-5 

(1.5 cm )

cm 2   s 

= 2.013 ×10-3

cm s

30-14

d. Differential model for cAL(z) rapid surface reaction Differential material balance on dissolved solute A in liquid phase IN – OUT + GEN = ACCUMULATION (moles A/time) v∞

π D2 4

c AL z − v∞

π D2 4

(

c AL z +∆z − N Aπ D∆z + 0 =0

)

Note N A = k L c AL − c AL , s ≈ k L c AL assuming rapid reaction of A at catalyst surface Rearrange, lim Δz→0 c AL z +∆z − c AL z  4k − − L c AL = 0 ∆z v∞ D Differential model dc 4k 0 − AL − L c AL = v∞ D dz

( )

z = 0, cAL = cAL,o (entrance); z = L, cAL = cAL Integration (needed for part e) c AL

dc A 4k = − L ∫ dz ∫ c v∞ D 0 c AL ,o AL Integrated model  c  4L kL ln  AL ,o  =  c AL  D v ∞ e. cAL for rapid surface reaction  4L kL  c AL c AL ,o exp  − =   D v∞   4(2500 cm) ( 2.013×10-3 cm/s )  gA  c AL = 10 3 exp m (1.5 cm)(22.1 cm/s)   gA c AL = 5.45 3 m

30-15

f. cAL for surface reaction with ks = 0.010 cm/s (cAL,s > 0) At surface

(

)

N A = k L c AL − c AL , s = ks c AL , s c AL , s =

k L c AL ks + kL

∴N = A

ks k L c AL = k ' c AL ks + kL

ks kL = k' = ks + kL

( 0.10 cm/s ) ( 2.013×10-3 cm/s ) = 6.68×10-4 cm/s -3 ( 0.10 cm/s ) + ( 2.013×10 cm/s )

Modified model  4L k '  c AL c AL ,o exp  − =   D v∞   4(2500 cm) ( 6.68×10-4 cm/s )  gA  c AL = 10 3 exp m (1.5 cm)(22.1 cm/s)   gA c AL = 8.17 3 m

30-16

30.10 Let A = O2, B = H2O (liquid) a. Material balance on O2 (component A) in the liquid phase Steady state material balance for species A on well mixed liquid phase of constant volume: C AL ,o vo + N A Ai − C AL vo + RAV = 0 * = − C AL ) N A K L (C AL

Ai V

v * − C AL ) + RA = (C AL ,o − C AL ) o + k L a (C AL 0 V

b. kL for db = 2 mm

µL

825×10-6 kg/m-sec = = 412.5 Sc = ρ L DAB (1000 kg/m3 )(2.0×10-9 m 2 /sec) Gr =

d b 3 ρ L g ( ρ L − ρG )

µL2 3

1m   3 2 3  2 mm ⋅  (1000 kg/m )(9.81 m/s )( (1000-1.2 ) kg/m ) 1000 mm  =  = 1.153 × 105 -6 2 (825 × 10 kg/m-sec) = = 112.3 Sh 0.31( Sc)1/3 (Gr )1/3 0.31(412.5)1/3 (1.153 × 105 )1/3 = Sh ⋅ DAB (112.3)(2.0×10-9 m 2 /sec) = = kL =1.123×10-4 m/sec db  1m  (2 mm)    1000 mm  c. Outlet concentrations CAL PA 0.21 atm * =262.5 mmol/m3 C= = AL 3 H m -atm 8×10-4 mmol

30-17

vo * + k L aC AL + RA V C AL = v kL + o V 3 mmol O 2 3  0.05 m /sec  -4 2 3 3 (10.0 mmol/m )   + (1.123×10 m/sec)(10 m /m )(262.5 mmol/m ) - 0.2 3 3 m -sec  1000 m  = 3  0.05 m /sec  (1.123×10-4 m/sec)(10 m 2 /m3 ) +   3  1000 m  C AL ,o

= 81.23 mmol/m3

30-18

30.11

flue gas 10 mole% CO2 (A) 90 mole% N2

Open Pond (no inflow or outflow of liquid) well-mixed pond water + algae + dissolved CO2 RA

CAL

Given: A = CO2, B = H2O (liquid)

Cells: k/1 = 0.06435 m3/g cells∙h, X = 50 g cells/m3 Liquid: νB = 0.995 x 10-6 m2/s, ρB =1000 kg/m3 Gas: db = 7.0 mm, a = 5.0 m2/m3 , ρG = 1.2 kg/m3 a. DAB for dissolved CO2 (solute) in water (solvent) Wilke-Chang Correlation (assume infinite dilute solution) DAB µ B 7.4 x10 [φB M B ] = T VA0.6 −8

1/2

7.4x10-8 ( 2.6 )(18 )  (293) 7.4 x10−8 [φB M B ] T = =1.8 x 10-5cm 2 /s VA0.6 µB (34.0)0.6 (993x10-3 ) 1/2

1/2

DAB

b. Determine kL for CO2 transfer through liquid film surrounding gas bubble = Gr

db3 ρ L g ∆ρ (7x10-3 m)3 (998.2 kg/m3 )(9.81m/s 2 )(998.2 kg/m3 - 1.7967 kg/m3 ) = 3.39 x 106 = 2 -6 2 (993 x 10 kg/m ⋅ s) µL

( 0.995 x 10 m /s ) = 553 = Sc = D (1.8 x 10 m /s ) ν

-6

-9

30-19

= Sh

k L db = 0.42Gr1/3 Sc1/2 =0.42(3.39 x 106 )1/3 (553)1/2 = 1483 DAB

D  k L = 0.42Gr1/3 Sc1/2  AB   db   1.8 x 10-9 m/s  -4 k L = (0.42)(3.39x106 )(553)1/2   = 3.82 x 10 m/s -3  7.0 x 10 m  c. Material balance model for predicting dissolved CO2 concentration cAL Physical System: liquid in pond—no liquid inflow or outflow Source for A: gas containing CO2 bubbled into liquid Sink for A: consumption of CO2 by cells suspended in pond Assumptions 1. Constant source and sink for CO2—steady-state process 2. First-order homogeneous reaction of CO2 in pond by cells suspended in liquid 3. Well-mixed liquid phase, but no inflow or outflow of liquid 4. Constant liquid volume 5. Dilute UMD process with respect to dissolved CO2 6. Liquid-phase controlling mass transfer process with kL ≈ KL Mass balance on dissolved CO2, liquid phase IN – OUT + GEN = ACCUMULATION (moles A/time) N A ⋅ Ai − 0 + RA ⋅ V = 0

K L a (c*AL − c AL ) ⋅ V − k1c AL ⋅ V = 0 K L a(

pA − c AL ) − k1c AL = 0 H

Solve for outlet concentration cAL = c AL

K L a ⋅ PA k L a ⋅ PA = H ( K L a + k1 ) H (k L a + k1 )

d. Determine cAL

 m3   g cells  -1 -4 -1 = k1 k= ' X  0.06435  50  = 3.22 hr = 8.938 x 010 s 3 g cells hr   m   30-20

c AL =

kL a ⋅ pA H (k L a + k1 )

( 3.82 x 10 m/s )( 5.0 m /m ) (0.10 atm) = 2.64 gmole/m c = ( 0.025 atm·m /gmole ) (3.82 x 10 m/sec×5.0 m /m + 8.938 x 10 s ) -4

-4

-1

e. Determine the total CO2 removal rate in kg CO2/h if the total pond volume is 1000 m3 WA = N A Ai = RA ⋅ V = −k1C AL ⋅ V = −k1' X ⋅ C AL ⋅ V

gmole   3  0.044 g CO 2  WA = − ( 3.22 hr -1 )  2.64   (1000 m )  3 m    1 gmole  WA = −274 g CO 2 /hr

30-21

30.12 Let A = TCE, and B = H2O (liq) a. Determine kL DAB = 8.9x10-10 m2/sec ρL = 998.2 kg/m3, µL = 9.93x10-4 kg/m-sec, νL = 0.995x10-6 m2/sec ρG = 1.19 kg/m3 Gr =

Sc =

-3 3 3 2 3 db3 ρ L g ∆ρ (5.0 x 10 m) (998.2 kg/m )(9.81m/sec )( ( 998.2-1.19 ) kg/m ) = 1.24 x 106 = 2 -6 2 (993 x 10 kg/m ⋅ sec) µL

DAB db = 5 mm

Sh =

0.995 x 10-6 m 2 /sec =1118 8.9 x 10-10 m 2 /sec

k L db = 0.42Gr1/3 Sc1/2 DAB

D  k L = 0.42Gr1/3 Sc1/2  AB   db 

 8.9 x 10-10 m 2 /sec  -4 k L = (0.42)(1.24x106 )1/3 (1118)1/2   =2.69 x 10 m/sec -3 5 x 10 m   3 b. Time required for CAL = 0.005 g/m , no inflow or outflow Material balance model (A = TCE) Assumptions: 1) PA ≈ 0, 2) no reaction occurs, 3) unsteady state, 4) dilute system, 5) kL≈KL In – Out +Generation = Accumulation (liquid phase – mole A/time) d ( C ⋅V ) 0 − N A ⋅ Ai + 0 = AL dt A dC * ) ⋅ i = AL −k L (C AL − C AL V dt P * A C= ≈0 AL H C AL A t 1 − ∫ dC AL = k L i ∫ dt C AL V 0 C AL ,0

 C AL ,o  Ai ln   = kL t V  C AL 

30-22

Ai Va 6 6 = ⋅ = 0.015 m 2 /m 2 = 18 m 2 /m3 V V db 0.005 m

 C AL ,0  1  50g TCE/m3  1 1hr = × × =0.528 hr t ln   ⋅ = ln  3  -4 2 3  0.005g TCE/m  (2.69 x 10 m/sec)(18m /m ) 3600 sec  C AL  k L a c. Determine CAL at steady state Material balance model In – Out +Generation = Accumulation (liquid phase – mole A/time) C AL ,0V0 − C ALV0 − k L a ⋅ V ⋅ C AL + 0 = 0 C AL ,o ⋅ V0 (50gTCE/m3 )(1.0m3 /sec) C AL = = V0 + k L a ⋅ V (1.0 m3 /sec)+(2.69 x 10-4 m/sec)(18m 2 /m3 )(2 m ⋅ 10 m ⋅ 10 m)

= 25.4 g TCE/m3

30-23

30.13 Let A = CO2, B = H2O (liquid) a. Determine kL = Gr

-3 3 3 2 3 db3 ρ L g ∆ρ (2.0 x 10 m) (998.2 kg/m )(9.81m/sec )( ( 998.2-1.7967 ) kg/m ) = 79161 = (993 x 10-6 kg/m ⋅ sec) 2 µ L2

ν 0.995 x10−6 m 2 / sec = = 562 DAB 1.77 x10−9 m 2 / sec db = 2.0 mm

= Sc

= Sh

k L db = 0.31Gr1/3 Sc1/3 DAB

D  k L = 0.31Gr1/3 Sc1/3  AB   db 

 1.77 x10−9 m 2 / sec  −5 = (0.31)(79161) (562)   9.721x10 m / sec −3 2 x10 m   1/3

1/3

b. Determine if the inlet flow rate of CO2 gas is sufficient PA = 2.0 atm PA 2.0 atm =0.06757 kmol/m3 = 3 H 29.6 atm ⋅ m /kmol 6φg A * − C AL ,o )= WA= N A i V= k L V ( C AL V db * C= AL

(9.721 x 10-5 m/sec)

6(0.05) 2.0 m3 ) (0.06757 kmol /m3 ) = 1.97 x 10-3 kmol CO 2 /sec ( -3 2.0 x 10 m

Convert WA in kmol/sec to volumetric flowrate of STD m3 CO2/ min VCO 2 = WA M A / ρ A = (1.97 x 10-3 kmol CO 2 /sec)(44kg/kmol)(1m3 /1.7967 kg)(60 s/min) = 2.89 m3 CO 2 /min

The inlet flow rate of CO2 from the problem statement (4.0m3/min) is larger than 2.89 m3/min, thus the inlet flow rate of CO2 gas is sufficient to ensure that the CO2 dissolution is mass transfer limited. c. Determine CAL,out Assumptions : 1 ) no chemical reaction, 2) steady state, 3) dilute system 30-24

In – Out +Generation = Accumulation (liquid phase – mole A/time) N A ⋅ Ai − C ALV0 + 0 = 0

−C ALV0 + k L a ⋅ V ⋅ (C*AL − C AL ) + 0 = 0 6(0.05) (0.06757 kmolCO 2 /m3 ) -3 k L a ⋅ V ⋅ (C ) 2.0 x10 m C AL ,out = = (0.45m3 /min)(1min/60 sec) 6(0.05) V0 +(9.721 x 10-5 m/sec) + kL a 3 2.0 m 2.0 x 10-3 m V C AL ,out = 0.0537 kmol/m3 * AL

(9.721 x 10-5 m/sec)

30-25

30.14 Let A = TCE, B = water Determine kL DAB = 8.9x10-10 m2/sec ρL = 998.2 kg/m3, µL = 9.93x10-4 kg/m∙sec ρG = 1.19 kg/m3 = Gr

db3 ρ L g ∆ρ (0.01m)3 (998.2kg / m3 )(9.81m / s 2 )(998.2 − 1.19kg / m3 ) = = 9.89 x106 −6 2 2 (993 x10 kg / m − sec) µL

µL 993 x10−6 kg / m 2 sec = = 1118 ρ L DAB (998.2 kg/ m3 )(8.9 x10−10 m 2 / sec) db = 10 mm

Sc =

= Sh

k L db = 0.42Gr1/3 Sc1/2 DAB

D  k L = 0.42Gr1/3 Sc1/2  AB   db 

 8.9 x10−10 m 2 / sec  −4 (0.42)(9.89 x106 )1/3 (1118)1/2  =  2.683 x10 m / sec 0.01m   Material balance model (A = TCE) kL

Assumptions: 1) steady state, 2) dilute system, 3) concentration along z-direction only, 4) No reaction, 5) PA≈0 In – Out +Generation = Accumulation (liquid phase – mole A/time) A v∞ C AL − N A ⋅ i ⋅ ∆z − v∞ C AL + 0 =0 z z +∆z V ÷ Δz, rearrangement, ∆z → 0 A dC AL v∞ + NA ⋅ i = 0 dz V A PA dC AL * * ) i = 0 , also C= ≈0 + k L (C AL − C AL AL H dz V A dC AL + k L C AL i = 0 dz V C AL ,out dC k L Ai L AL −∫ = dz C AL ,o C AL v∞ V ∫0

 C AL ,o  k L Ai L ln   = C v V ∞  AL ,out  30-26

Determine L Ai Va 6 0.02 m3 TCE 6 = ⋅ = a= × =12 m 2 /m3 3 V V db 0.01 m m water  C AL ,o  v∞  50 mg/L  0.05 m/sec L ln= ln = 107 m =    -4 2 3 C k a 0.05 mg/L (2.683x10 m/sec)(12 m /m )   , AL out L  

30-27

30.15 Let A = POCl3, B = He a. Develop material balance model Assumptions: 1) Steady state; 2) Concentration profile along z-direction only; 3) No reaction; 4) Dilute solution and constant fluid velocity. Let A = POCl3, B = He (g) Differential material balance π D2 π D2 v∞ c A + N A π D ∆z − v∞ cA + 0 =0 4 4 z z +∆z ÷ ∆z, rearrangement, ∆z → 0 4k dc − A + c c*A − c A = 0 dz v∞ Integration: c A ,out − dc 4k L ∫cA,o c*A − cAA = − v∞ Dc ∫0 dz

(

)

 c*A − c A,o  4 kc L ln  *  = −  c −c v∞ D  A A,out  b. Determine mass transfer coefficient, kc 4Q g 4 (110 cm3 /s) v∞ = = 35 cm/s = π (2.0 cm) 2 π D2 v∞ D ( 35 cm/s )( 2.0 cm ) = = 50 (laminar) Re = νB 1.4 cm 2 /s

(

)

νB

1.4 cm 2 /s = = 3.78 DAB 0.37 cm 2 /s Sieder-Tate Correlation for laminar flow in tube 1/3 1/3 D   2.0 cm  = Sh 1.86 = 1.86  ( 50 )( 3.78)  = 3.44  Re Sc   L   60 cm  Sh 3.44 = kc = DAB 0.37 cm 2 /s = 0.636 cm/s D 2.0 cm = Sc

c. Determine yA,out * = y A,out c= PA / P =0.15 atm/1.0 atm = 0.15 A / c, y A

30-28

   4k L    -4(0.636 cm/s)(60 cm)   y A,out = y*A 1 − exp  − c   = 0.15 1 − exp    = 0.133 v (35 cm/s)(2.0 cm) D   ∞      d. kc and yA,out at D = 4.0 cm v∞ = 8.75 cm/s, Re = 25; k c ∝ D −1 , D1/3 , Re1/3 −1

+1/3

 Dnew   Dnew   Re new  cm  4.0 cm   4.0 cm   25  kc ,new k= 0.636      c , old        s  2.0 cm   2.0 cm   50   Dold   Dold   Reold  cm = 0.318 s    4k L    -4(0.318 cm/s)(60 cm)   y A,out = y*A 1 − exp  − c   = 0.15 1 − exp    = 0.133  (8.75 cm/s)(4.0 cm)    v∞ D     -1

+1/3

No net change from part c. e. yA,out →yA* as z→∞

30-29

30.16 Liquid TEOS flow In

TEOS vapor +He gas Out

liquid film L = 2.0 m

D = 5 cm

z+Δz NA

cAs z

cA v∞

Liquid TEOS Out ( = 0)

100% He flow in 333 K, 1.0 atm 2000 cm3/s

Given: A = TEOS, B = He T = 333 K, P = 1.0 atm TEOS: PA = 2133 Pa, DAB = 1.315 cm2/s He gas: νB = 1.47 cm2/s, Vo = 2000 cm3/s Wetted wall column: D = 5.0 cm, L = 2.0 m Physical System: gas phase inside tube Source for A: volatile liquid TEOS flowing down inner surface of tube which vaporizes into the gas phase Sink for A: He carrier gas flowing up tube a. Mass transfer coefficient for TEOS vapor in He gas, kc Estimate Reynolds number to establish flow regime inside tube Vo 2000 cm3 /s = = 101.86 cm/s π 2 π 2 D (5.0 cm) 4 4 v ∞ D (5.0 cm)(101.86 cm/s) = = 346.4, laminar regime Re = (1.47 cm 2 /s) νB

= v∞

30-30

For laminar flow inside tube 1/3

kc D D  = 1.86  ReSc  DAB L 

νB

= Sc

1.47 cm 2 /s = = 1.118 DAB 1.315 cm 2 /s

 1.315 cm 2 /s   DAB  D   5.0 cm  (346.4)(1.118)  1.042 cm/s  1.86  = = 1.86  ReSc  L   200 cm   D   5.0 cm  1/3

1/3

Convert kc to kG kG =

kc RT

( 0.01042 m/s ) (8.314 J/mol ⋅ K)(333 K)

= 3.76 x 10-6 mol/m 2s ⋅ Pa

b. Mole fraction TEOS in outlet gas Develop process model for prediction of cA(z), then input kc and other parameters posed by model Assumptions: 1. 2. 3. 4. 5. 6.

Constant source and sink for TEOS vapor—steady-state process No radial concentration gradient of TEOS vapor in gas phase Constant temperature and pressure, therefore cAs constant Dilute process with respect to TEOS vapor in He TEOS transfer from liquid to gas limited by boundary layer mass transfer through gas Falling liquid film of pure TEOS is thin (<< D) and uniformly distributed

Differential material balance on TEOS in gas phase IN – OUT + GEN = ACCUMULATION (moles A/time) v∞

π D2 4

cA z − v∞

π D2 4

c A z +∆z − N Aπ D∆z + 0 =0

N A kc ( c As − c A ) Note= Rearrange, lim Δz→0 cA − c A z  4kc −  z +∆z − c As − c A = 0 ∆z v∞ D

(

)

30-31

−

(

)

dc A 4kc − c As − c A = 0 dz v∞ D

Separate variables cA, z and integrate with limits shown below z = 0, cA = cAo (entrance); z = L, cA = cA,out c A ,out

∫ (

c Ao

4k dc A = − c ∫ dz v∞ D 0 c As − c A

)

 c − c  4 L kc ln  As Ao  = c −c   As A,out  D v ∞ Determine cA,out, then yA,out PA 2133 Pa = = 0.770 mol/m3 RT (8.314 J/mol-K)(333 K) c Ao = 0 c= As

  4 L kc   c A,out = c As 1 − exp  −    D v∞    4 L kc 4(200 cm) (1.042 cm/s) = 1.6368 D v∞ ( 5.0 cm ) (101.86 cm/s) = c A,out c=

P RT

y= A, out

( 0.770 mol/m ) ⋅ (1 − exp ( −1.6368) ) = 0.620 mol/m 3

(101,250 Pa ) = (8.314 J/mol-K)(333 K)

36.6 mol/m3

c A,out 0.620 mol/m3 = = 0.017 c 36.6 mol/m3

c. Transfer rate of TEOS (g/s), WA Overall gas phase material balance for TEOS—the rate of input of TEOS liquid flowing down the inner wall of the tube must balance its mass-transfer limited evaporation to the He carrier gas W = Vo (c A,out −c Ao = ) Vo c A,out A

WA = ( 2000 cm3 /s )( 0.620 mol/m3 )(1.0m3 /106 cm3 ) = 1.24 x 10-3 mol/sec WA M A = (1.24 x 10-3 mol/s)(208.33 g/mol) = 0.258 g TEOS/s

30-32

30.17 Let A = O3, B = H2O a. Determine kL Vo V 100cm3 / sec v= = = = 50 cm/sec ∞ A hw (1.0cm)(2.0cm) 4hw 4(1.0cm)(2.0cm) d eq = 1.33 cm = = 2h + 2 w 2(1.0cm) + 2(2.0cm) v ∞ d eq (50 cm / sec)(1.33 cm) = Re = = 6650 ν 0.01 cm 2 / sec

νL

0.01 cm 2 / sec = = 575 DAB 1.74 x10−5 cm 2 / sec Gilliland-Sherwood correlation for turbulent flow of liquid flowing inside of a tube kL D = Sh = 0.023Re0.83 Sc1/3 DAB Sc =

D  k L = 0.023Re0.83 Sc1/3  AB   D   1.74 x10−5 cm 2 / sec  −3 0.83 = k L 0.023(6650) (575)1/3   3.73 x10 cm / sec 1.33cm   b. Determine L Assumptions: 1) Steady state, 2) Dilute system, 3) Concentration along z-direction only, 4) No homogeneous chemical reaction Material balance In – Out +Generation = Accumulation (liquid phase – mole A/time)  C  C V + N A (2 w ⋅ ∆z ) − V 0 AL 0 AL z

z +∆z

+ 0 =0

÷ Δz, rearrangement, ∆z → 0 dC 2w - AL + k L (C*AL − C AL ) 0 =  V dz 0

∫

C AL ,out

C AL ,o

 C*AL − C AL ,o  2 wk L dC AL 2 wk L L = dz → ln  *  =  L  ∫0 C −C C*AL − C AL V V0 0 AL ,out   AL

PA 15 atm ⋅ CL = (0.056 mol/cm3 ) = 223.40 gmole/m3 H 3760 atm 100 cm3 /sec   223.4 gmole/m3 -0 L = ln = 60.3cm 3 3  -3 223.4 gmole/m -2.0 gmole/m 2 2.0cm 3.73x10 cm/sec ( )  

* C AL =

(

)

30-33

c. Determine new L when PA = 30atm DAB will not change with respect to P in this case because this is diffusion in liquid phase. * will change because it is dependent on P However, C AL * = C AL

PA 30 atm ⋅ CL = (0.056 mol/cm3 ) = 446.81 gmole/m3 H 3760 atm

(

)

100 cm3 /sec   446.81 gmole/m3 -0 L = ln  =30 m 3 3  -3  446.81 gmole/m -2.0 gmole/m  2 ( 2.0 cm ) 3.73x10 cm/sec

(

)

30-34

30.18 Let A = CO2, and B = H2O (liquid) DAB = 1.77x10-9 m2/sec w 36 g / sec  V = = = 3.61x10−5 m3 / sec 0 ρ L 998.2 x103 g / m3 a. Determine CAL* PA 2.54 atm * =0.10 kgmol/m3 C= = AL 3 H atm ⋅ m 25.4 kgmol b. Determine kL 4 w 4(36 g / sec) Re = = = 769 π Dµ L π (6.0cm)(9.93x10−3 g/ cm − sec) Sc =

µL 993 x10−6 kg / m 2 sec = = 562 ρ L DAB (998.2 kg/ m3 )(1.77 x10−9 m 2 / sec) 1/6

 ρ 2 g ⋅ z3  kL z = Sh = 0.433( Sc)1/2  L 2  (Re L )0.4 DAB  µL  1/6

 ρ 2 g ⋅ z3  D  k L = 0.433( Sc)  L 2  (Re L )0.4  AB   z   µL  1/2

1/6

-9 2  (998.2 kg/m3 ) 2 (9.81 m/sec 2 )(2m)3  0.4  1.77 x 10 m /sec  (769) k L = 0.433(562)     2.0 m (993 x 10-6 kg/m-sec) 2     -5 =2.69 x 10 m/sec 1/2

c. Determine CAL,out Assumption: 1) no reaction, 2) Steady state, 3) dilute system Material balance on differential volume element (A = CO2) In – Out +Generation = Accumulation (liquid phase – mole A/time)

V0C AL + N A ⋅ π D ⋅ ∆z − V0C AL z

÷ Δz, rearrangement, ∆z → 0  dC AL + N ⋅ π D = V 0 A 0 dz

z +∆z

+ 0 =0

30-35

dC AL π D * k L (C AL + − C AL ) = 0  dz V0 C AL ,out L dC AL kL −∫ = π D ∫0 dz C AL ,o C * − C V0 AL AL *  C AL − C AL ,o  k L ln  *  =  π DL C −C AL AL out ,   V0    −kL   -π(0.06 m)(2.69x10-5 m/sec)(2 m)   * 3 π − 1 exp = 0.10 kgmol/m 1-exp C= C DL )     (  AL , out AL  3.61x10-5 m3 /sec     V0    3 C AL ,out = 0.0245 kgmol/m

30-36

30.19 Let A = O2, B = blood a. Determine kL 4 w 4V 4(300 cm3 /min)(1min/60 sec) Re = = = 79.6 = π D µ L π Dν L π(2.0 cm)(0.040 cm 2 /sec)

νL

0.040 cm 2 /sec = 2000 = Sc = DAB 2.0 x 10-5 cm 2 /sec 1/6

 g ⋅ z3  kL z Sh = = 0.433( Sc)1/2  2  (Re L )0.4 DAB  νL  1/6

 g ⋅ z3  D  k L = 0.433( Sc)  2  (Re L )0.4  AB   z   νL  1/2

1/6

-9  (9.81 m/s 2 )×(0.5m)3  m 2 /sec  0.4  2.0 x 10 -5 (79.6) k L = 0.433(2000)    = 2.91 x 10 m/sec -6 2 2  0.5 m    (4 x 10 m /sec)  1/2

b. Determine CAL,out Assumption: 1) No reaction, 2) Steady state, 3) Dilute system, 4) kL≈KL Material balance on differential volume element (A = CO2) In – Out +Generation = Accumulation (liquid phase – mole A/time)

V0C AL + N A ⋅ π D ⋅ ∆z − V0C AL z

z +∆z

+ 0 =0

÷ Δz, rearrangement, ∆z → 0  dC AL + N ⋅ π D = V 0 A 0 dz dC AL π D * + − C AL ) = 0 k L (C AL  dz V0 C AL ,out L dC AL kL −∫ = π D ∫ dz  C AL ,o C * − C 0 V0 AL AL  C * − C AL ,0  k L ln  *AL  =  π DL  C −C AL AL out ,   V0

 −k  * * − (C AL − C AL ,o ) exp  L π DL  C AL ,out =C AL   V0 

30-37

Also,

PA k ⋅ PA * C AL C AL ,max = = + H 1 + k ⋅ PA 1atm (28 atm -1 )(1 atm) + 9.3 mol O 2 /m3 ) =10.23 mol O 2 /m3 ( 3 -1 0.80 atm ⋅ m /mol 1 + (28 atm )(1 atm)

C AL ,out = 10.23 mol O 2 /m3 - (10.23 mol O 2 /m3 -1.0 mol O 2 /m3 )  -π(0.02 m)(2.91 x 10-5 m/sec)(0.5m)  exp   -4 3  (3 x 10 m /min)(1min/60 sec)  C AL ,out = 2.54 mol O 2 /m3

30-38

30.20 Model 1 kc 1.17 ( Sc) 2/3 = = jD 0.415 v∞ Re 3 At 311 K, ,υ 1.673 × 10−5 m 2 /s = ρ 1.136 kg/m= 3/2

DAB

2.634 m 2 -Pa/ssec  311 K  2.772×10-5 m 2 /s =   -5 1.013 x 10 Pa  298 K 

v= ∞

G 0.816 kg/m 2 -s = 0.72 m/s = ρ 1.136 kg/m3

= Re

Dv∞ (5.71×10-3 m)(0.72 m/s) = = 246 υ (1.673×10-5 m 2 /s)

Re0.415 = 9.82 = Sc

1.673 x 10-5 m 2 /sec υ = = 0.60 DAB 2.772 x 10-5 m 2 /sec

Sc 2/3 = 0.711 1.17v∞ 1.17(0.72 m/sec) = = = 0.12 m/sec kc 0.415 2/3 (9.82)(0.711) Re Sc kc 0.12 m/sec = = kG =4.64×10-8 kgmol/m 2 -sec-Pa = 4.70×10-3 kgmol/m 2 -sec-atm RT (8.314 J/mol-K)(311 K) Model 2 k 2.06 = ε jD = ε c ( Sc) 2/3 0.575 v∞ Re 0.575 = = 23.7 Re0.575 (246) 2.06v∞ 2.06(0.72 m/sec) = =0.117 m/sec kc = 0.575 2/3 (0.75)(23.7)(0.711) ε Re Sc k 0.117 m/sec = 4.52×10-8 kgmol/m 2 -sec-Pa = 4.58 x 10-3 kgmol/m 2 -sec-atm kG = c = RT (8.314 J/mol-K)(311 K)

30-39

30.21 Let A = TCE, B = air a. Determine kc, gas-film mass transfer coefficient for TCE vapor in air d p G (3 x 10-3 m)(0.10 kg/m 2 -sec) = = 16.55 Re = µ (1.813 x 10-5 kg/m-sec) Re0.31 = 2.39 Sc =

ν DAB

1.505 x 10-5 m 2 /sec = 1.863 8.08 x 10-6 m 2 /sec

Sc 2/3 = 1.514 G 0.1 kg/m 2 -sec = = 0.083 m/sec ρ 1.206 kg/m3 k 0.25 = ε c ( Sc) 2/3 0.31 v∞ Re

v= ∞

0.25  0.083 m/sec   1  =    0.011m/sec 2.39  1.514   0.5 

b. Length required, L Material balance on a differential volume element (moles A/time)  A v∞= C AWD z kc   ( C A* − C A ) WD∆z + v∞C AWD z +∆z V  dC A  A kc   ( C A* − C A ) = v∞ dz V  A π D2 π where= = V D3 D π  L 0.999 C *A dC A kc   ∫ dz = v∞ ∫ * 0 d  0 (CA − CA )  p L

v∞ d p   (0.083 m/sec)(3 x 10-3m)  1  C A* = ln  * ln    = 0.050 m π kc  C A − 0.999C A  π(0.011m/sec)  1-0.999 

30-40

30.22 Film theory, k L α DAB1.0 1.0

 DCO2 − H 2O   1.46 x 10-5 cm 2 /sec  -1 (k L a )CO2 (= (300 hr -1 )  = k L a )O2    =158 hr -5 2  DO − H O  2.77 x 10 cm /sec    2 2  1.0

Penetration theory, k L α DAB1/2 = (k L a )CO2 (300 = hr -1 ) ( 0.527 ) 217.8 hr -1 1/2

Boundary theory, k L α DAB 2/3

= (k L a )CO2 (300 = hr -1 ) ( 0.527 ) 195.8 hr -1 2/3

30-41

30.23 Let A = O2, B = H2O (liquid) a. Volumetric mass transfer coefficient for O2, ( k L a )O

= Rei

di N =

Po= 6=

(0.3 m) (4 rev/s) = 4.83 x 105 -6 2 (0.746 x 10 m /s)

Pg c ρ L N 3 di5

6(993.7 kg/m3 )(4 rev/sec)3 (0.3 m)5 = 927.2 N ⋅ m/sec 1kg ⋅ m/sec 2 ⋅ N Aerated power input, Pg P

 Pg  d  log10   = −192  i   dT   P  0.3 m  = −192    1m 

4.38

 di2 N     υL 

0.115

d  1.96 i   dT 

 di N 2     g 

 (0.3 m) 2 (4.0 rev/sec)    -6 2  (0.746 x 10 m /sec) 

0.115

 Qg   3   di N   0.3 m  1.96   1m 

 (0.3 m)(4 rev/sec) 2    9.8 m/sec 2  

  0.02 m3 /sec   3  (0.3 m) (4 rev/sec) 

=-0.539 Pg =10(-0.539) (P)=(0.289)(927.2 N ⋅ m/sec)=268 N ⋅ m/sec Volumetric mass transfer coefficient, kLa 0.4 P −2  g  ( k L a= )O2 (2.6 × 10 )   (u gs )0.5 V      P Q   g g  = (2.6 × 10−2 )    2  V   π dt     4  b. O2 transfer rate, WA 0.4

0.5

  0.4  3 0.02 m /sec   268 N ⋅ m/sec  =(2.6 x 10-2 )     3 2 2.0 m    π(1.0 m)      4

0.5

= 0.0294 sec-1

 PO  WA = k L a (C A* − C A∞ )V = k L a  2 CL − 0  V  H   0.21 atm  3  3 =(0.0294 sec-1 )   (55.6 mol/L)(1000 L/m )  (2.0 m ) 4 5.05 x 10 atm    = 0.01359 mol/sec = 0.82 mol/min

30-42

30.24 Let A = CO2, B = H2O (liquid) a. Determine kLa = Sc = S L

993×10-6 kg/m-s µ = = 562 ρ DAB (998.2 kg/m3 )(1.77×10-9 m 2 /s)

π D2

π (0.25) 2 = = 0.0491m 2 = 0.534 ft 2 4 4 1  5.0 kgmol   18 kg   2.204 lb   60 min    =22,288 lb/ft 2 ⋅ hr      2   min   kgmol   kg   hr   0.534 ft 

µ = 993 x 10-6 kg/m-sec = 2.40 lb/hr ⋅ ft -9 = = m 2 /s 6.86×10-5 ft 2 /hr DAB 1.77×10 1− n

1− 0.22

 22,288 lb/hr ⋅ ft 2  L 1/2 -5 2 k L a D=  AB   ( Sc) (α ) (6.86×10 ft /hr)  µ  2.40 b/hr ⋅ ft  = 202 = hr -1 0.056 sec-1

(562)1/2 (100)

b. Determine length required, L

 m3 1  5.0 kgmol     1min  v∞ =     =0.0306 m/s 2   min   55.5 kgmol   0.0491 m  60 s  dC A k L a ( C A* − C A ) = v∞ dz v C AL dC A L= ∞ ∫ k L a 0 ( C A* − C A ) L

 ( 0.0306 m/sec)  1  v∞  C A* ln  * ln  =  = 1.64 m *  k L a  C A − 0.95C A  (0.056 sec-1 )  1-0.95 

30-43

30.25 Let A = O2, B = blood a. Determine kLa Liquid mass flow rate per cross-sectional area of the empty tower (L), lbm/ft2-hr νL 0.040cm 2 / sec = = 2000 Sc = DAB 2.0 x10−5 cm 2 / sec w ρ LV0 (1.025 g/cm3 )(300 cm 2 /min)(1.0 lb m /454g)(60 min/1 hr) = = = L = 12017 lb m /ft ⋅ hr A π D2  π(2.0 cm) 2  2   (1ft/30.48 cm) 4 4   1lb m 30.48 cm 3600 sec (1.025 g/cm3 )(0.040 cm 2 /sec) = 9.9 lb m /ft ⋅ hr = µ L ρ= Lν L 454 g 1ft 1hr DAB = (2.0 x 10-5 cm 2 /sec)(1ft/30.48 cm) 2 (3600 sec/1 hr) = 7.75 x 10-5 ft 2 /hr 1− n

 L  kL a =α  DAB  µL 

Sc 0.5

1− n

 L  kL a α  =   µL 

Sc 0.5 ⋅ DAB 1-0.46

 12017 lb m /ft 2 hr  k L a = 550    9.91 lb m /ft-hr 

20000.5 ( 7.75 x 10-5 ft 2 /hr ) = 88.2 hr -1

b. Determine CAL,out Assumption: 1) No reaction, 2) Steady state, 3) Dilute system, 4) kL ≈ KL Material balance for species A on a differential volume element In – Out +Generation = Accumulation (liquid phase – mole A/time) π D2 V0 C AL + N A ⋅ a ⋅ ⋅ ∆z − V0 C AL + 0 =0 z z +∆z 4 ÷ Δz, rearrangement, ∆z → 0 2  dC AL + N ⋅ a ⋅ π D = V 0 0 A dz 4 dC AL π D 2 * k L a (C AL + − C AL ) = 0  dz 4V0 −∫

C AL ,out

C AL ,o

L dC AL π D2 k L a ∫ dz = * 0 C AL − C AL 4V0

30-44

 C * − C AL ,0  π D 2 ln  *AL = k a⋅L   C −C  L 4 V AL AL out , 0  

 −π D 2  * * C AL ,out =C AL kL a ⋅ L  − (C AL − C AL ,o ) exp    4V0  PA k ⋅ PA * C AL = + C AL ,max = H 1 + k ⋅ PA Also, 1atm (28 atm -1 )(1 atm) + ( 9.3 molO2 /m3 ) =10.23 molO2 /m3 0.8 atm-m3 /mol 1+(28 atm -1 )(1 atm)  -π(0.02 m) 2 (88.2 hr -1 )(0.5 m)  C AL ,out =10.23 mol O 2 /m3 - (10.23 mol O 2 /m3 -1.0 mol O 2 /m3 ) exp   -4 3  4(3 x 10 m /min)(60 min/1hr)  C AL ,out = 5.96 mol O 2 /m3 c. Performance comparison The packed bed tower performs better than the wetted wall column in terms of overall mass transfer because of the increase in surface area for interphase mass transfer when using packed bed.

30-45

30.26 Let A = Cu+2, B = H2O a. Determine kc 2 d 2ω ( 8 cm ) ( 2.0 rev/sec ) 2π Re = = 80425 = v 0.01 cm 2 /sec ν 0.01 cm 2 /sec = 833.33 Sc = L = DAB 1.20 x 10-5 cm 2 /sec kc d 0.62 Re1/2 Sc1/3 Sh = = DAB DAB d D 1.2 x 10-5 cm 2 /sec 0.62 Re1/2 Sc1/3 AB 0.62(80425)1/2 (833.33)1/3 =2.48 x 10-3 cm/sec = 8 cm d

kc = 0.62 Re1/2 Sc1/3

b. Determine NA = N A kc (C A,∞ − C As ) (1) NA = ks C As − RA =

(2)

Combine (1) and (2) kc (C A,∞ − C As ) = ks C As kc C A, ∞ kc + k s Combine (2), (3) kk N A = c s C A, ∞ kc + k s C As =

(3)

( 2.48 x 10 cm/sec ) ( 3.2 cm/sec ) (0.005 gmole/L)(1L/1000 cm ) = 1.24 x 10 mol N = cm sec ( 2.48 x 10 cm/sec ) + ( 3.2 cm/sec ) -3

-8

-3

Since ks >> kc, boundary layer diffusion controls c. Determine time required, t Mass Balance on Cu2+: VC A,0 + mA,0 = VC A,t + mA,t , also mA,0 = 0 and mA,t =

ρA π d 2 MA 4

30-46

VC A,0 − C A, f =

ρA π d 2 MA V

 8.96 g/cm3   π(8 cm) 2  ( 500 cm )( 5 x 10 mole/cm ) -  64 g/gmole   4  ( 2 x 10-4 cm )    = 500 cm3 C A, f = 2.185 x 10-3 mol/L 3

-6

Assumptions: 1) Unsteady state, 2) Dilute system, 3) Well-mixed liquid phase Unsteady-state material balance for species A on well-mixed liquid phase In – Out + Generation = Accumulation (liquid phase – mole A/time) dC A πd2 V 0 − NA ⋅ = dt 4 CA, f t 2 π d kc k s 1 − dt = dC A 4V kc + ks ∫0 C A,∞ C∫A ,0 ln

C A, f C A,0

−π d 2 kc k s t 4V kc + k s

 C A, f  4V kc + k s t = − ln   C  π d 2 k k c s  A,0 

 2.185 x 10-3 mol/L  4(500 cm3 ) ( 2.48 x 10 cm/sec ) + ( 3.2 cm/sec ) t = -ln   2 -3  0.005 mol/L  π ( 8.0 cm ) ( 2.48 x 10 cm/sec ) ( 3.2 cm/sec ) -3

= 3323 sec = 55 min

30-47

30.27 Let A = O2, B = H2O Liquid film mass transfer coefficient, kL Re =

v∞ D (5.0 cm/sec)(1.0 cm) = 548 = νL 9.12 x 10-3 cm 2 /sec

νL

9.12x10−3cm 2 / sec = = 434 2.1x10−5 cm 2 / sec DAB Sieder-Tate correlation for laminar flow of liquid flowing inside of a tube (function of tubing length L)

Sc =

1/3

kL D D  D  D  = → k L 1.86  Re⋅ Sc   AB  Sh = 1.86  Re⋅ Sc  = DAB L  L   D  -5 2  1.0cm   2.1 x 10 cm /sec  -3 -1/3 (548)(434)   k L = 1.86   =2.42 x 10 L cm/sec L 1.0 cm     1/3

Determine the overall mass transfer coefficient KL 1 1 1 = + K L k L km  atm ⋅ m3  -6 2 0.78   ( 5.0 x 10 cm /sec ) gmole H ⋅ DAe  =  = 6.72 x 10-4 cm/sec km = 3 S m ⋅ lm (0.029 atm ⋅ m silicone/gmole)(0.20 cm) −1

 1   1  1 1 + KL =  +  =   −3 −1/3 −4  2.42 x10 L (cm / sec) 6.72 x10 cm / sec   k L km  1 (cm / sec) KL = 1/3 413.223L + 1488.1

−1

Mass transfer model Assumptions: 1) Steady state, 2) Dilute system, 3) Concentration along z-direction only, 4) No reaction Material balance on differential volume element In – Out +Generation = Accumulation (liquid phase – mole A/time)

π D2

v∞ C AL + N A ⋅ π D ⋅ ∆z −

π D2

4 4 z ÷ Δz, rearrangement, ∆z → 0

+ 0 =0

v∞ C AL z +∆z

30-48

dC AL 4 + NA ⋅ = 0 dz v∞ D dC 4 * K L (C AL − AL + − C AL ) = 0 dz v∞ D −

−∫

C AL , out

C AL , o

L dC AL 4 L 4 1 (cm / sec)dz K dz = = L * 1/3 ∫ ∫ v∞ D 0 413.223 z + 1488.1 C AL − C AL v∞ D 0

*   C AL  3 L + 3.6012   − C AL ,0   v∞ D  2/3 3 f L L L − = = − + ln  * ( ) 0.00363 0.0261 0.0942 ln       C −C 4   AL , out    3.6012    AL

Determine the length of tubing (L) * C AL =

PA = H

2.0 atm = 2.56 gmol/m3 3 atm-m 0.78 gmol

*  C AL − C AL ,0   v∞ D   (2.56 - 0) gmole/m3   5.0 cm × 1.0 cm  2 ln  * ln − =   = 0.446 cm  3  C −C   4  4 (2.560 .768) gmole/m    AL , out   AL   3 L +3.6012   0.446cm 2 = 0.00363L2/3 -0.02613 L +0.0942ln      3.6012   Using Solver, L = 23.9 m

30-49

30.28 NA

yA(z)

v∞

yAs = 1.0

100% O2 yAo = 0

shell side

100% H2

H2 permeable tube wall tube side

yAL

v∞

Di = 1.5 cm Do = 2.5 cm

Re = 10,000 z = 0 cm

z = 10 cm

a. v∞ on shell side (annular space) at Re =10,000 D= Do − Di = 2.5 cm - 1.5 cm = 1.0 cm e Re = v∞

v ∞ De

νB

( 0.159 cm /s ) (10,000 ) = 1590 cm 15.9 m = =

ν B Re De

(1.0 cm )

b. Sc for shell side A = H2 (g), B = O2 (g) DAB = 0.697 cm2/s at 273 K, 1.0 atm (Appendix J) 1.5

1.5  T   cm 2   300 K  cm 2 = = ( , ) 0.697 0.803 DAB (T , P) D= T P     AB ref   Tref  s   273 K  s    cm 2 0.159 νB s = 0.198 = Sc = cm 2 DAB 0.803 s

c. kc for shell side Re > 2000 turbulent flow through tube

30-50

0.83 = = = Sh 0.023Re Sc1/3 0.023 (10, 000 ) ( 0.198) 28 0.83

Sh =

1/3

kc De DAB

 cm 2  28 0.803 ( )  s  Sh ⋅ DAB cm  = 22.5 = kc = s De (1.0 cm ) d. Model for yA(z) shell side Differential material balance on dissolved solute A in liquid phase IN – OUT + GEN = ACCUMULATION (moles A/time)

π ( Do2 − Di2 )

cA z − v∞ c A z +∆z + N Aπ Di ∆z + 0 =0 4 4 = N A k y y As − y A assume that yAs ≈1.0 at outer surface of inner tube (constant) Note v∞

(

)

cA = yAC if constant T and P, then C is constant ky = kcC

v∞

π ( Do2 − Di2 ) C 4

y A z − v∞

π ( Do2 − Di2 ) C 4

(

)

y A z +∆z + k y y As − y A π Di ∆z + 0 =0

Rearrange, lim Δz→0

 y A z +∆z − y A z  4kc CDi − − 0 y As − y A = ∆z v∞ Do2 − Di2 C

(

)

(

)

Differential model −

4kc Di dy A − dz v∞ Do2 − Di2

β=

(

4kc Di

(

v∞ Do2 − Di2

)

0 (y − y )= As

)

z = 0, yA = yAo (entrance); z = L, yA = yAL Integration y AL L dy A = − β ∫ dz ∫ y Ao ( y As − y A ) 0 30-51

Integrated model  y − y Ao  ln  As  = −β L  y As − y AL  e. yAL   4kc Di  y AL =y As − ( y As − y Ao ) exp  − L 2 2  v ∞ ( Do − Di )    cm   4  22.5  (1.5 cm )(10 cm )   s    y AL =1.0 − (1.0 − 0) exp  − 2 2  ( 2.5 cm ) - (1.5 cm ) (1590 cm/s )    y AL = 0.191

(

)

30-52

31.1 a. Determine KLa for O3 transfer 3

m3  3.28 ft   1.0 hr  ft 3 Qg = 17.8     =10.5 hr  1.0 m   60 min  min At depth = 10.5 ft, from Eckenfelder plot KL

A ft 3 V ≅ 400 V hr

K L ( A / V )V K La = n = V

( 400 ft /hr ) (8 spargers ) =1.132 hr ft  (80 m )  3.28  1.0 m  3

-1

1/ 2

1/ 2 K L a O  DO − H O   1.7 x 10-5 cm 2 /sec  3 3 2 = 0.90 =  =  -5 2 K La O 2.1 x 10 cm /sec D     − O H O  2 2  2

K L a O = (1.132 hr -1 ) ( 0.9 ) =1.02 hr -1 3

b. Estimate time required for CA = 0.25 CA* (A = O3) Ptop + Pbottom 1 P= = Ptop + ρ L gh ave 2 2   1  1000 kg  9.81 m  1.0 atm Pave = 1.0atm +  3.2 m )  =1.155 atm  3 2 ( 5 2  2 m  sec   1.0123 x 10 kg/m ⋅ sec  = PA y= AP

( 0.04 )(1.155atm ) = 0.0462 atm

gmol mg PA 0.0462 atm = = 0.683 3 = 327 3 m L H 0.0667atm ⋅ m /gmol gmol mg C A,t = 0.150 3 =7.2 m L  C A* − C Ao   1   ( 0.683 - 0.0 ) gmol/m3   1  t ln= ln = 0.246 hr = 886 sec  *    3  -1   ( 0.683 - 0.150 ) gmol/m   1.02 hr   C A − C A,t   K L a  * C= A

31-1

31.2 a. Required KLa values for H2S transfer and O2 transfer at t = 2.5 hr and CA = 0.050 gmole/m3 (A = H2S)

 C * − C A0  ln  A*  CA − CA   K La = t PA * = = C A* x= C 0 (= PA 0) AC H  0.0 - 0.30 gmol/m3  ln  0.0 - 0.05 gmol/m3   = 0.717 hr -1 KLa H S = 2 2.5 hr KLa H S 0.717 hr -1 2 KLa O = = = 1.075 hr -1 1/ 2 1/2 2 -5 2  DH 2 S − H 2 O   1.4 x 10 cm /sec     2.1 x 10-5cm 2 /sec     DO2 − H 2 O  b. Aeration rate to each sparger (based on O2 trasnfer)  A  K L ( A / V )V KL   = n V V  K L ( A / V ) V (15 spargers) 1.075 hr -1 = (425 m3 ) (3.28 ft/1.0 m)3 K L ( A / V ) V = 1075 ft 3 /hr Using the Eckenfelder plot, at 3.2 m (10.5 ft) depth, Qg = 20 SCFM

31-2

31.3 Given: Counter-Current Flow

Co-Current Flow XA2 = 0 xA2 = 0 L2 Ls = 1.4 Ls,min

YA2 yA2 G2 Gs R = 0.80

XA1 xA1 L1 Ls = 1.4 Ls,min

YA1 = 0.05 yA1 G1 = 10 lbmole/ft2-hr Gs

YA2 yA2 G2 Gs R = 0.80

YA1 = 0.05 yA1 G1 = 10 lbmole/ft2-hr Gs

XA2 xA2 L2 Ls = 1.4 Ls,min

XA1 = 0 xA1 = 0 L1 Ls = 1.4 Ls,min

a. See YA-XA plots after parts (b) and (c) are completed b. Determine Ls,min, Ls, XA1 for counter-current flow YA 2= YA1 (1 − R )= 0.05 (1 − 0.8 )= 0.010 y A1 =

YA1 0.05 = = 0.0476 1 + YA1 1.05

Gs = G1 (1 − y A1 ) = 10 (1 − 0.0476 ) = 9.52

lbmol ft 2 h

X A1,min 0.23 from YA − X A plot = YA1Gs + X A 2 Ls ,min = YA 2Gs + X A1,min Ls ,min  YA1 − YA 2   0.050 − 0.010  Ls ,min = Gs  =   ( 9.52 ) X   0.23 − 0.0   A1,min − X A 2  lbmol Ls ,min = 1.66 2 ft h Ls =

= (1.4 )(1.66 ) 2.32

lbmol ft 2 h

31-3

YA1Gs + X A 2 Ls =YA 2 Gs + X A1 Ls

9.52 ) + 0 ( 0.01)( 9.52 ) + X A1 ( 2.32 ) ( 0.05)(= X A1 = 0.164 0.06

YA1 0.05 0.04 0.03

YA 0.02 0.01

YA2

XA1

0.00 0.00

0.05

0.10

0.15

0.20

XA1,min 0.25

0.30

XA (mole A / mole solvent)

c. Determine Ls,min, Ls, XA1 for cocurrent flow = At YA1 0.01, = X A1,min 0.03 YA 2Gs + X A 2 Ls ,min = YA1Gs + X A1,min Ls ,min

9.52 ) + 0 ( 0.01)( 9.52 ) + ( 0.03) ( Ls ,min ) ( 0.05)(= Ls ,min = 12.7

= Ls

lbmol ft 2 h

= (1.4 )(12.7 ) 17.78

lbmol ft 2 h

YA 2Gs + X A 2 Ls =YA1Gs + X A1 Ls

9.52 ) + 0 ( 0.01)( 9.52 ) + X A1 (17.78 ) ( 0.05)(= X A1 = 0.0214

31-4

31.3 cont. 0.06 0.05

XA2,YA2

0.04 0.03

YA 0.02 0.01

YA1

XA1,min

0.00 0.00 XA1 0.05

0.10

0.15

0.20

0.25

0.30

XA (mole A / mole solvent)

31-5

31.4 a. Determine AGs,min x A1 AL1 (1 − R ) xA2 AL2 = lbmol hr

0.2 x A1 AL1 = (1 − 0.8)( 0.01)(100 ) = ALs = AL2 (1 − x A 2 ) =

(100 )(1 − 0.01) = 99

lbmol hr

 x  x A1 AL1 = ALs  A1   1 − x A1   x  0.20 = ( 99 )  A1   1 − x A1  x A1 2.02 ⋅ 10−4 = AL1 =

AL s 99 lbmol 99.02 = = −4 1 − x A1 1 − 2.02 ⋅ 10 hr

y A1 AG1,min + x A 2 AL2 = x A1 AL1 + y A 2,min AG2,min * y= A 2,min

H  515 atm  xA2  =  ( 0.01) = 0.412 PT  12.5 atm 

0 + ( 0.01)(100 ) = ( 0.20 ) + ( 0.412 ) ( AG2,min ) AG2,min = 1.94

lbmol hr

lbmol 1.14 AGs ,min = (1 − 0.412 )(1.94 ) = (1 − y A2,min ) AG2,min = hr

b. Determine yA2,min From part (a) above, y*A 2,min = 0.412

31-6

31.5 a. Determine yA2,min, see YA-XA plot xA2 0.107 = = 0.12 X A2 = 1 − x A 2 1 − 0.107 kgmol ALS = 1.786 (1 − xA2 ) AL2 = (1 − 0.107 )( 2.0 ) = sec X A1 = 0.020 (1 − R ) X A2 = (1 − 0.833)( 0.12 ) = YA 2,min = 0.195 (see YA -X A plot) = YA1 AGs ,min + X A 2 AL YA 2,min AGS ,min + X A1 ALs s 0 + ( 0.12 )(1.786= ) AGs ,min = 0.916 = y A 2,min

( 0.195) ( AGS ,min ) + ( 0.02 )(1.786 )

kgmol sec

YA 2,min 0.195 = = 0.163 1 + YA 2,min 1 + 0.195

b. Determine AG1 and yA2 YA1 = 0 = = = AG AG 1.5 ( AGs ,min ) s 1

= ) 1.374 (1.5)( 0.916

kgmol sec

YA1 AGs + X A 2 ALs =YA 2 AGs + X A1 ALs

YA 2 (1.374 ) + ( 0.02 )(1.786 ) 0 + ( 0.12 )(1.786 ) =

YA 2 = 0.13 = y A2

YA 2 0.130 = = 0.115 1 + YA 2 1.0 + 0.130

AG2 =

AGs 1.374 = = 1.55 kgmol/sec 1 − y A 2 1.0-0.115

31-7

31.5 continued 0.25

0.20

YA2,min

0.15

YA2

YA 0.10

0.05

0.00 0.00

XA2 XA1, YA1

0.05

0.10

0.15

XA (mole dissolved NH3 / mole water)

31-8

31.6 a.

Determine L2 and mole fraction compositions of terminal streams

(1 − y A1 ) G1 ( 5)(1 − 0.12 ) kgmol = = 4.44 2 1 − y A2 m hr (1 − 0.01)

= G2

y A1G1 + x A 2 L2 = y A 2 G2 + x A1 L1

( 0.12= )( 5) + 0 ( 0.01)( 4.44 ) + ( 0.03) L1 L1 = 18.519

kgmol m 2 hr

Ls = L1 (1 − x A1 ) = (18.519 )(1 − 0.03) = 17.96

kgmol m 2 hr

Ls = L2 = y A 2 0.01,= x A 2 0,= y A1 0.12,= x A1 0.03 b. See YA-XA plot, all mole fraction values scaled to mole ratios c. Determine packing height z z = H OG N OG G1 + G2 kgmol kgmol = 4.72 2 , K y' a = K G' a ⋅ PT = 2.0 3 2 m hr m hr kgmol   4.72 2   G m hr   = = H = 2.36 m OG ' kgmol  Kya   2.0 3  m hr   y −y N OG = A1 * A 2 ( yA − yA )

* A2

= 0.0 at x A2 = 0.0

y*A1 = 0.09 at x A1 = 0.090

( y − y ) − ( y − y ) ( 0.12 − 0.090 ) − ( 0.01 − 0 ) = = 0.018 (y − y ) = (y − y )  ( 0.12 − 0.090 )  ln    ln   ( 0.010 − 0 )   ( y − y )  A

* A lm

* A1

* A2

0.12 − 0.01 = 6.042 0.018 z = ( 6.042 )( 2.36 m ) = 14.26 m

= N OG

31-9

31.6 continued 0.15

XA1, YA1

0.10

YA 0.05

XA2, YA2 0.00 0.00

0.01

0.02

0.03

0.04

0.05

XA (mole A / mole solvent)

31-10

31.7 Given: XA2 = 0 xA2 = 0 L2 Ls = 2.0 Ls,min

YA2 yA2 = 0.005 G2 Gs R

XA1 xA1 L1 Ls = 2.0 Ls,min

YA1 yA1 = 0.06 G1 = 10 lbmole/ft2-hr Gs

a. Molar flowrate and mole fraction composition of all terminal streams

x A 2 0,= y A 2 0.005, y A1 0.06 = = x A1,min 0.00085 from y A − x A plot = lbmol ft 2 h Gs 9.4 lbmol G2 = = = 9.45 2 ft h (1 − y A2 ) (1 − 0.005) Gs = G1 (1 − y A1 ) = 10 (1 − 0.06 ) = 9.4

y A1G1 + x A 2 L2 = y A 2G2 + x A1,min L1,min L1,min = L1,min

y A1G1 + x A 2 L2 − y A 2G2 x A1,min 0.06 )(10 ) + 0 − ( 0.005 )( 9.45 ) (= lbmol 650.3 0.00085

ft 2 h

Ls ,min = L1,min (1 − x A1,min ) = ( 650.3)(1 − 0.00085 ) = 649.74

lbmol ft 2 h

31-11

Equilibrium Line xA

kg SO2 /

100 kg H2O

(mm Hg, SO2)

0.00

0.20 0.30

T= PT =

o 30 C

1 atm

0.00000

0.000

Mw, SO2 =

64 g/gmole

0.00056

0.038

Mw, H2O =

18 g/gmole

0.00084

0.061

0.50

0.00140

0.109

0.70

119

0.00196

0.157

0.16 0.14 0.12 0.10

yA0.08

xA1, yA1

0.06 0.04 0.02

xA1,min

xA2,yA2 0.00 0.00E+00

5.00E-04

1.00E-03

1.50E-03

2.00E-03

xA (mole fraction SO2 in water)

lbmol ft 2 h y A1G1 + x A 2 L2 = y A 2G2 + x A1 L1

= Ls 2= Ls ,min 1299.48

= ( 0.06 )(10 ) + 0 ( 0.005)( 9.45) + ( xA1 )( L1 ) x A1 L1 = 0.5528 G1 + L2 = G2 + L1 L1 = 10 + 1299.48 − 9.45 = 1300

lbmol ft 2 h

x A1 L1 0.5528 = = 4.25 ⋅10−4 L1 1300 Material Balance Summary: G2 (lbmol/ft2 h) 9.45 G1 (lbmol/ft2 h) 10 2 L2 (lbmol/ft h) 1299.48 L1 (lbmol/ft2 h) 1300 x1 0.000425 x2 0 y1 0.06 y2 0.005 = x A1

31-12

b. Packing height z z = H OG N OG

( y A1 − y A2 ) = ( y A1 − y*A1 ) − ( y A2 − y*A2 )

N OG

 ( y A1 − y*A1 )   ln  *  ( y A 2 − y A 2 ) 

H OG = = G

( 0.06 − 0.005) = 3.94 ( 0.06 − 0.03) − ( 0.005 − 0 )  ( 0.06 − 0.03)  ln    ( 0.005 − 0 ) 

G K y' a

G1 + G2 lbmol = 9.73 2 2 ft h

m ' 1 ( 70.5 ) 1 1 = + = + K y a k y a k x a 15 250 K y a = 2.87

lbmol ft 3 h

lbmol lbmol (1 − 0.06 )= 2.70 3 3 ft h ft h lbmol lbmol Top: K y' a = K y a (1 − y A 2 ) = 2.87 3 (1 − 0.005 ) = 2.85 3 ft hr ft h lbmol Average: K y' a = 2.78 3 ft h lbmol 9.73 2 ft h =3.51 ft H OG = lbmol 2.78 3 ft h z = ( 3.51 ft )( 3.94 ) =13.8 ft Bottom: K y' a= K y a (1 − y A1 )= 2.87

31-13

31.8 a. Determine xA1  std m3   1.0 kmol  kmol AG1 =  880 = 39.3  3  hr   22.4 std m  hr  AGs = AG1 (1 − y A1 )= AG2 =

( 39.3)(1-0.01) =38.9

( 38.9 ) AGs kmol = =38.9 1 − y A 2 (1-0.0005 ) hr

kmol hr

AG1 + AL2 = AG2 + AL1 39.3 + 50 = 38.9 + AL1 kmol hr y A1 AG1 + x A 2 AL2 = y A 2 AG2 + x A1 AL1

AL1 = 50.4

= ( 39.3 )( 0.01) + 0 ( 38.9 )( 0.0005) + ( 50.4 )( xA1 ) x A1 = 0.0074 Aside: (1 − R) y A1G1 = y A 2 G2

( 0.0005)( 38.9 ) (1 − R )( 0.01)( 39.3) = R = 0.95 (95%)

b. See yA-xA plot c. Determine Ls,min = ALs ,min AL1,min (1 − x A1,min ) y A1 AG1 + x A 2 AL2 = y A 2 AG2 + x A1,min AL1,min x A1,min = 0.028

( 39.3)( 0.01) + 0 = ( 38.9 )( 0.0005) + ( AL1,min ) ( 0.028)

AL1,min = 13.3

kmol hr

ALs ,min = (13.3)(1-0.028 ) =12.9 A=

kmol hr

π D2 4

Ls ,min =

ALs ,min A

kmol   12.9  kmol hr   = = 4.1 2 2 m hr  π ( 2.0 m )      4  

31-14

d. Determine KGa a z = 10 m z = H OG N OG N OG

( y A1 − y A2 ) = ( y A1 − y*A1 ) − ( y A2 − y*A2 )

 ( y A1 − y*A1 )   ln   ( y A 2 − y*A 2 )  z 10 m = H= = 3.22 m OG N OG 3.06 H OG =

( 0.01 − 0.005) = 3.06 ( 0.01 − 0.0005) − ( 0.005 − 0 )  ( 0.01 − 0.0005 )  ln    ( 0.005 − 0 ) 

G K y' a

G1 + G2 39.3+38.9 kgmol = =39.1 2 2 2 m hr kgmol 39.1 2 G ' m hr 12.15 kgmol = K= ya = 3.22 m H OG m3 hr

K y' a ' K= = Ga P

kgmol kgmol m3 hr = 3.04 3 1.0 atm  m ⋅ hr ⋅ atm ( 405 kPa )    101.25 kPa  12.15

31-15

31.8 continued Equilibrium Line T= P= MEA Conc. = MW MEA =

PA mm Hg

40 oC 4.0 atm 15.3 61

wt% MEA = g/gmole

mol H2S/ mol MEA

mol H2S/ mol H2S+MEA

5.061

mole% MEA

yA*

YA*

0.000

0.0000

0.960 3.000

0.125 0.208

0.111 0.172

0.0063 0.0094

0.0003 0.0010

0.0064 0.0095

0.0003 0.0010

9.100

0.362

0.266

0.0183

0.0030

0.0187

0.0030

43.100

0.643

0.391

0.0325

0.0142

0.0336

0.0144

59.700

0.729

0.422

0.0369

0.0196

0.0383

0.0200

106.000

0.814

0.449

0.0412

0.0349

0.0430

0.0361

0.04

0.03

0.02

xA1, yA1

0.01

xA2, yA2 0.00 0.00

XA1,min 0.01

0.02

0.03

0.04

0.05

31-16

31.9 a. Characterize molar flowate and mole fraction composition of all terminal streams kgmol kgmol AGs = AG2 (1 − y A1 )= 2.0 (1 − 0.10)= 1.8 sec sec AGS 1.8 kgmol/sec kgmol = =1.836 AG2 = 1 − y A2 1-0.02 sec AG1 + AL2 = AG2 + AL1 kgmol AL1 = AG1 + AL2 − AG2 = 2.0 - 1.837 + 3.0 = 3.163 sec AG1 y A1 + AL2 x A 2 = AG2 y A 2 + AL1 x A1 (2.0)(0.10) + (3.0)(0.01) = (1.837)(0.020) + (3.163) x A1 x A1 = 0.061 Material Balance Summary: AG1 (kgmol/sec) 2.0 AG2 (kgmol/sec) 1.84 AL1 (kgmol/sec) 3.163 AL2 (kgmol/sec) 3.0 yA1 0.1 yA2 0.02 xA1 0.061 xA2 0.01 b. Tower diameter (D) at 50% of flooding condition Cf = 98 for 1.0 inch ceramic Intalox saddles (Table 31.2) = AL '1 AL1 ( x A1 M A + (1 − x= A1 ) M B )

( 0.061(17)+(1-0.061)(18) )( 3.163) =56.75 kg/sec

1/2 kg   kg   1/2 2.8 3  56.75    AL1'  ρG  sec   m 0.056 x − axis = =   =  AG1  ρ L − ρG  kgmol   kg  1000 kg − 2.8 kg     2.0   26.9 m3 m3  kgmol   sec   

0.25 y − axis = Y= 1/2

   ρ ( ρ − ρ ) g  ( 2.8)(1000 − 2.8)(1.0 )  kg 0.25 =  3.77 2 G Y  G L 0.1 G c   = 0.1   C f µL J m sec    ( 98)( 0.001) (1)    kg kg 53.8 53.8 AG1' sec =14.27 m 2 at flooding, sec = 28.54 m 2 at 50% of flooding = = A = A ' kg kg Gf 3.77 2 1.885 2 m sec m sec 1/2

' f

31-17

1/2

 4A  D=   π 

1/2

 4 ⋅ 14.27 m 2  =  = 4.25 m at flooding π   1/2

 4 ⋅ 28.54 m 2  D=  = 6.03 m at 50% of flooding π   c. Log-mean mass transfer driving force (yA – yA*)lm y A1 − y*A1 ) − ( y A 2 − y*A 2 ) ( * ( y A − y A )lm =  y A1 − y*A1 )  (  ln   ( y A 2 − y*A 2 )  x A1 0.061, y*A1 0.035 = = x A 2 0.01, y*A 2 0.005 = =

( y − y ) = ( 0.10 − 0.035) = 0.065 ( y − y ) = ( 0.02 − 0.005) = 0.015 A1

* A1

* A2

−y ) (y = A

* A lm

( 0.065 − 0.015)

= 0.034  0.065  ln    0.015 

0.12

XA1, YA1 0.10 0.08

YA0.06 0.04

XA2, YA2

0.02 0.00 0.00

0.02

0.04

0.06

0.08

XA (mole dissolved NH3 / mole water) 31-18

31.10 a. Determine L2 kgmol  kgmol  AGs = AG1 (1 − y A1 )= 1.25  (1 − 0.08 )= 1.15 hr  hr  AGs kgmol 1.1512 AG2 = = hr (1 − y A2 ) y A1 AG1 + x A 2 AL2 = x A1 AL1 + y A 2 AG2

1.25 ) + 0 ( 0.02 )( AL1 ) + ( 0.001)(1.1512 ) ( 0.08)(=

AL1 = 4.942

kgmol hr

ALs = AL1 (1 − x A1 ) =

( 4.942 )(1 − 0.02 ) = 4.845

kgmol hr

4.843 ALs kgmol = = 4.845 (1 − x A 2 ) (1 − 0) hr b. See yA-xA plot

= AL2

x A1,min = 0.35 from y A -x A plot x A1 AL1 = x A1,min

= AL1,min

( 0.02 )( 4.942 ) 0.35

= 0.282

kgmol hr

ALs ,min = AL1,min (1 − x A1,min )= 0.282(1 − 0.35)= 0.184

kgmol hr

c. Log-mean mass transfer driving force (yA – yA*)lm

(y − y )−(y − y ) (y − y ) =  ( y − y )  ln   ( y − y )  A

* A lm

* A1

* A2

x A1 0.02, y*A1 0.0 = = x A 2 0.00, y*A 2 0.0 = =

( y − y ) = ( 0.08 − 0.0 ) = 0.080 ( y − y =) ( 0.0010 − 0.0=) 0.0010 A1

* A1

* A2

−y ) ( y= A

* A lm

( 0.080 − 0.0010 )

= 0.018  0.080  ln   0.0010 

31-19

Aside: y A1

dy A * yA 2 y A − y A

N OG = ∫

m ≅ 0, y*A =mx A =0 N OG =

 y A1  dy A  0.08  ln 4.38 = = ∫y y A − 0  y A2  ln = 0.001  A2 y A1

d. Determine ∆P/z for tower diameter D = 0.50 m Cf = 52 for 1.5 inch ceramic Intalox saddles (Table 31.2)

(

) (1.25) ( 0.08)( 36.5) + (1 − 0.08)( 2 )

= AG1' AG1 y A1 M HCl + (1 − = y A1 ) M H 2 AG1' = 5.95

(

kg hr

) ( 4.94 ) ( 0.02 )( 36.5) + (1 − 0.02 )(18)

= AL1' AL1 x A1 M HCl + (1 − x= A1 ) M H 2O AL1' = 90.75

kg hr 

(1.0 atm )  4.76

kg   kgmol 

kg  PT   = 0.195 3 M w,ave =  3 m  m atm   RT   0.08206  ( 298 K ) kgmol-K   Flooding Correlation

ρG = 

1./2

0.195 AL'  ρG   90.75    x − axis :  =    = 0.21 '  AG  ρ L − ρG   5.95   1033 − 0.195  1/2

kg   5.95   AG kg hr   1hr   8.418 ⋅10−3 2 y − axis : G ' = = = 2  A m sec  π ( 0.5m )   3600sec    4   ' 1

( G ') C f µ L0.1 J (8.418 ⋅10−3 ) ( 52 ) (1.2 ⋅10−3 ) = 9.33 ⋅10−6 = Y = ρG ( ρ L − ρG ) g c ( 0.195)(1033 − 0.195)(1) 2

0.1

∴∆P / Z << 50 Pa / m tower will not flood

31-20

31.10 contined 0.15

0.10

xA1, yA1

yA1

yA 0.05

0.00 0.00

xA1,min

xA2, yA2 0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

xA (mole fraction HCl in water)

31-21

31.11 Given: XA2 xA2 = 0.0020 L2 = 100 kgmole/hr Ls

YA2 yA2 = 0.015 G2 = 12 kgmole/hr Gs

1.5 inch plastic Pall rings

XA1 xA1 = 0.00020 L1 Ls

YA1 yA1 G1 Gs

a. AGs,min Get yA2,min from equilibrium curve

H  41.25 atm  −3 y A 2,min =m ⋅ x A 2 = ⋅ x A 2 =  ( 2.0 × 10 ) = 0.050 P 1.65 atm   Terminal stream material balance for AG2,min

x A 2 AL2 + y A1 AG1,min = x A1 AL1 + y A 2,min AG 2,min 3 100 (1 − 2 ⋅10−= AL = AL2 (1 − x A= ) 99.8 s 2)

= AL1

ALs = (1 − xA1 )

kgmol h

99.8 kgmol 99.82 = −4 h (1 − 2 ⋅10 )

x A 2 AL2 + y A1 AG 1,min = x A1 AL1 + y A 2,min AG 2,min

( 2 ⋅10 ) (100 ) + 0 = ( 2 ⋅10 ) ( 99.82 ) + ( 0.05) ( AG ) −3

−4

2,min

AG2,min = 3.60

kgmol hr

kgmol  AGs ,min = AG2,min (1 − y A 2,min= ) 3.42 )  3.60 kgmol  (1 − 0.05= h  h  31-22

b. Gf’ and tower diameter (D) at flooding for AG2 = 12 kgmol/h and yA2 = 0.015 Use Flooding Correlation for packed towers in counter-current gas–liquid flow Cf = 39 for 1.5 inch plastic Pall rings (Table 31.2) AL'2

kgmol   kg  kg  100  1800 =  18 h   kgmol  h 

kg  kg  ' AG2= AG2 ⋅ M w,ave= AG2 ( y A 2 M W , A + (1 − y A 2 ) M w,air = ) 12 kgmol = 366.8   30.56 hr   kgmol  hr      kg kg P kg  1.65atm  2.1 = = , ρ L 1000 3 = = 30.56 ρG M w ,ave 3 3   m m RT kgmol  m atm    0.0821  ( 293K )  kgmol K    Flooding Correlation 1/2

kg   kg   2.1 3  1800 h    AL'2  ρG  m x − axis 0.225 = = =     '  kg kg  AG2  ρ L − ρG   366.8   (1000 − 2.1) 3  h  m   y − axis, Y ≅ 0.13 1/2

Gas flooding velocity Gf’

 Y ⋅ ρG ( ρ L − ρG ) g c  G =  C f ⋅ µ L0.1 ⋅ J  

1/2

' f

J = g c =1 for SI units 1/2

' f

 0.13 ⋅ 2.1(1000 − 2.1)1  kg 3.7 2 =  0.1 ms  39 ⋅ ( 0.001) ⋅1 

Tower diameter D at the flooding condition kg   h   366.8     AG h   3600 s  =0.027 m 2 A= ' = kg  Gf   3.7 2  m sec   ' 2

1/2  4 ( 0.027 m 2 )   4A   =0.2 m D=  =   π  π    1/2

31-23

31.12 a. Determine xA1 AG1 + AL2 = AG2 + AL1 AL1 = AG1 + AL2 − AG2 = 32 + 18 − 30.7 = 19.3

lbmol hr

y A1 AG1 + x A 2 AL2 = y A 2 AG2 + x A1 AL1

( 0.05)( 32= ) + ( 0 ) ( 0.01)( 30.7 ) + ( xA1 )(19.3)

x A1 = 0.067 b. Determine packing height, z z = H OG N OG = A

π D2

π(2.0 ft) 2 =3.1415 ft 2 4

= 4

( AG1 + AG2 ) / 2 A=

G H = = OG K ya

 y + y A2  K G a ⋅ P 1 − A1 2  

( 32+30.7 ) lbmol/hr/2 ( 3.1415 ft 2 )  lbmol  0.05 + 0.01      2.15 ft 3hr ⋅ atm  (1.4 matm ) 1 2    

H OG = 3.41 ft = N OG

y A1 − y A 2 y A1 − y A 2 = = * ( y A − y A ) ( y A1 − y*A1 ) − ( y A2 − y*A2 ) lm

 ( y A1 − y*A1 )   ln  *  ( y A 2 − y A 2 )  z = (1.76 )( 3.41 ft ) = 6.01 ft

( 0.05 − 0.01) = 1.76 ( 0.05 − 0.0067 ) − ( 0.01 − 0 )  0.05 − 0.0067  ln   0.01 − 0 

c. Pressure drop (∆P/z) for 0.25-inch ceramic Raschig rings Flooding Correlation ∆P / z = 300 N/m 5/8-inch ceramic Raschig rings For 0.25" Raschig Rings, C f = 1600 G' =

AG '1 1006 lb m /hr =320.2 lb m /ft 2 hr = 2 A π (2.0 ft) /4

( G ) C µ J ( 320.2 ) ( 580 )( 2 ) (1.503) = 0.038 Y = = ρ (ρ − ρ ) g ( 0.11)( 55 − 0.11) ( 4.18 ⋅ 10 ) ' 2

0.1 L

0.1

C f ,new = scaled y-axis, Ynew Y= old C f ,old

 ( 0.038= )   0.104 1600  580 

∆P / z = 1200 N / m 31-24

31.13 a. Determine ALs,min y A1 AG1 + x A 2 AL2 = y A 2 AG2 + x A1,min AL1,min

H 0.46 atm =0.023 = 2.0 atm P y A1 0.03 = = = 0.130 x A1,min m 0.23 AG1 (1 − y A1 ) 2.0(1-0.03) lbmol = 1.95 = AG2 = 1-0.005 hr (1 − y A2 ) = m

= ( 0.03 )( 2 ) + 0 ( 0.005)(1.95) + ( 0.130 ) AL1,min AL1,min = 0.387

lbmol hr

ALs ,min = AL1,min (1 − x A1,min ) = ( 0.387 )(1-0.130 ) =0.336

lbmol hr

b. Estimate required K’ya for z = 6.0 ft z = H OG N OG H OG =

Z N OG

AL2 + AG1 = AL1 + AG2 1.0 + 2.0 = AL1 + 1.95 lbmol hr x A 2 AL2 + y A1 AG1 = x A1 AL1 + y A 2 AG2 AL1 = 1.05

0 + 0.03(2.0) = x A1 (1.05) + 0.005(1.95) x A1 = 0.048

( AG1 + AG2 )

lbmol = 1.975 2 hr ( AL1 + AL2 ) lbmol = AL = 1.025 2 hr * 0.23(0.048) = 0.011 = = y A1 mx A1

AG =

0.23(0) = y*A 2 mx = = 0.0 A2

31-25

( y A1 − y A2 ) = ( y A1 − y*A1 ) − ( y A2 − y*A2 )

N OG

 ( y A1 − y*A1 )   ln   ( y A 2 − y*A 2 ) 

H OG =

( 0.03 − 0.005) = 2.39 ( 0.03 − 0.011) − ( 0.005 − 0 )  ( 0.030 − 0.011)  ln    ( 0.005 − 0 ) 

G 6.0 ft =2.51ft = ' 2.39 Kya

( AG1 + AG2 )

lbmol = 10.07 2 2A ft hr lbmol 10.07 2 G ' ft hr = 4.0 lbmol = Kya = H OG 2.51 ft ft 3 hr

= G

c. New size of packing at 45% of flooding  G1'   '  = 0.26  G f old G 'f ,old =

G1' 0.26

 G1'   '  = 0.45  G f new G1' 0.45  G1'    G 'f ,new  0.45  = = G 'f ,old  G1'     0.26  G 'f ,new =

1/2

 380     C f ,new 

C f ,new = 1138 Therefore 3/8 inch packing size is suitable

31-26

31.14 a. See yA-xA plot b. Determine the overall mass transfer driving force (xA* – xA) and NOL x A,2 − x A,1 N OL = ( xA − x*A ) lm

(x − x ) −(x − x ) (x − x ) =  ( x − x )  ln   ( x − x )  x − x = (1.6 ⋅ 10 − 8 ⋅ 10 )= 8.0 ⋅ 10 x − x = ( 2 ⋅ 10 − 0 ) = 2.0 ⋅ 10 (8.0 ⋅10 ) − ( 2.0 ⋅10 =) 6.0 ⋅10 = 4.33 ⋅10 ( x − x )= 1.39  ( 8.0 ⋅ 10 )   ln   ( 2.0 ⋅ 10 )  1.6 ⋅ 10 − 2 ⋅ 10 ) (= = 3.235 N A

* A2

* A1

* A lm

* A2

* A lm

* A1

* A2

* A1

−4

−5

−4

−5

.433 ⋅ 10−5

c. Determine packing height, z z = H OL N OL H OL =

L K x' a

K x' a ≅ K x a L=

( L1 + L2 )

2 L2 (1 − x A 2 ) Ls lbmol L1 = = = 99.99 2 ft hr (1 − xA1 ) (1 − xA1 ) lbmol ft 2 hr 1 1 1 = + K x a k x a mk y a

L = 100

31-27

lb   62.4 3   = k x a k= (17.4 hr -1 )  lbft  =60.32 lbmol L aC L ft 3 hr  18   lbmol  lbmol lbmol k y a= kG a ⋅ P= 7.4 3 1.2 atm = 8.88 3 ft hr atm ft hr H 150 atm = = 125 m = P 1.2 atm 1 1 1 = + K x a 60.32 (125 )( 8.88 ) lbmol ft 3 hr lbmol 100 2 ft hr =1.75ft H OL = lbmol 57.21 3 ft hr z = (1.75ft )( 3.235 ) =5.65 ft K x a = 57.21

2.00E-02

1.50E-02

xA2, yA2 1.00E-02

yA 5.00E-03

0.00E+00 0.00E+00

xA1, yA1 5.00E-05

1.00E-04

1.50E-04

2.00E-04

xA (mole fraction benzene in water)

31-28

31.15 a. See yA-xA plot b. Determine xA2 for z = 4.0 ft ( 5.0 atm ) PA * = =1.142 x 10-4 x= A 4 H ( 4.38 x 10 atm ) lb   lbmol   min  lbmol  AL1 ≈ AL2 ≈ AL ≈  200 m     60  =666.7 min   18 lb m   hr  hr  lbmol ' K= k= k= 194 3 (no gas phase mixture) xa xa xa ft hr lbmol 666.7 AL hr = 3.346 ft 3 = AH= OL ' lbmol kx a 194 3 ft hr 3 3.346 ft =1.094 ft H OL = 3.1415 ft 2 xA 2  x*A − x A1  dx A 4.0 ft Z ln  * = =3.657 N =  = OL ∫x x*A −= xA  x A − x A 2  H OL 1.094 ft A1  x*A − x A1   1.142 ⋅ 10−4 − 0  e = =    −4 *  x A − x A 2  1.142 ⋅ 10 − x A 2  x A 2 1.113 ⋅ 10−4 = 3.657

1.20

yA1 = yA2 = 1.0 1.00 0.80 0.60

yA 0.40 0.20

xA1 = 1.113 x 10-4 0.00 0.0E+00

2.0E-05

4.0E-05

6.0E-05

8.0E-05

1.0E-04

1.2E-04

xA (mole fraction O2 in water)

31-29

31.16 a. Tower diameter (D) at ∆P/z = 200 Pa/m Cf = 98 for 1.0 inch ceramic Intalox saddles (Table 31.2)

kg  kg  kgmol   AG1' = 1  =44   44 min  min   kgmol  kgmol   kg  kg  AL'1 =  4.0  = 72  18 min   kgmol  min      P  kg   2.0 atm  = 3.66 kg ρG M = =  44  A 3  RT  kgmol   m3 m atm  293 K   0.08206  ( )   kgmol K    Flooding Correlation 1/2

AL'  ρG  3.66  72    : 0.10 x − axis=  =   '  AG  ρ L − ρG   44   998.2 − 3.66  y − axis= 0.04 at ∆P/z = 200 Pa/m

(G ) C µ J ' 2

= = Y 0.04

1/2

0.1 L

ρG ( ρ L − ρG ) g c

 ρG ( ρ L − ρG ) g c  G' = Y   C f µ L0.1 J  

1/2

 0.04 3.66 998.2-3.66  [ ][ ][ ]  =1.722 kg  = 0.1  [98] 993×10-6  [ 0.01]  m 2 sec    

kg   1min   44  AG '  min   60 sec  = A = = 0.426 m 2 kg  G'  1.722 2  m sec   1/2

 4 ⋅ 0.426 m 2   4A  = D =   = 0.74 m  π  π    1/2

31-30

b. Estimate kLa by Sherwood and Holloway correlation 1− n

0.5

 µL     ρ L DAB  cm 2 m2 ft 2 kg lb DAB = 1.92 x 10-5 = 2.4 = 1.92 x 10-9 =7.43 x 10-5 , μ L = 993 x 10-6 sec sec hr m×sec ft ×hr α = 170, n = 0.28  L  kL a =α  DAB  µL 

lb   min  kg     72   2.2 kg   60 hr  lb min     =2073 2 × L=  2 2 ft hr ft  ( 0.426 m )   3.28  m  1-0.28

 2073  k L a = 170    2.4 

  993 x 10-6 kg/m ⋅ sec ) (    ( 998.2 kg/m3 )(1.92 x 10-9 m 2 /sec )   

1/2

( 7.43 x 10 ) = 37.4 hr -5

-1

c. Estimate volume of packing required for separation V = A ⋅ H OL ⋅ N OL kgmol   min   4.0    60  AL2 min   hr   A ⋅ H OL = = =0.1156 m3 kgmol k L a ⋅ CL , r ( 37.4 hr -1 )  55.5 m3   x*A − x A 2   1.419 ⋅ 10−3  dx A = = = N OL ∫= ln ln 1.22    −3 −3  * * 1.419 ⋅ 10 − 10   x A − x A1  xA 2 x A − x A x A1

V = (1.22 ) ( 0.1156 m3 ) = 0.141 m3

31-31

30.17 a. G2

G2 =

G1 (1 − y A1 ) 10.0(1-0.05) kgmol = 9.6 = m 2 hr (1-0.01) (1 − y A2 )

b. Ls,min at 40 oC and 100 oC Develop equilibrium distribution plot in yA vs. xA coordinates Sample calculation for xA (A = H2S) For 15.3 wt% MEA, wMEA = 0.153 wMEA / M w, MEA

(1 − wMEA ) / M w,H O 2

xH 2 S

kg MEA kg MEA+kg H 2 O

0.153/61 mol MEA = 0.053 (1-0.153)/18 mol H 2 O

mol H 2S = mol H 2S+mol MEA+mol H 2 O

mol H 2S mol MEA mol H 2S mol H 2 O +1+ mol MEA mol MEA

For 0.365 mol H2S/mol MEA

mol H 2S 0.365 mol MEA = 0.018 mol H 2S mol H 2 O 0.365 +1+ mol MEA 0053 mol MEA

(see plot next page at 40 oC and 100 oC) At 40 oC, xA1,min = 0.0315 = Ls ,min L1,min (1 − x A1,min ) y A1G1 + x A 2 L2 = y A 2G2 + x A1,min L1,min

( 0.05)(10.0 ) + 0 = ( 0.01)( 9.6 ) + ( 0.0315) ( L1,min ) L1,min = 12.8

kgmol m 2 hr

Ls ,min = (12.8 )(1 - 0.0315 ) =12.4

kgmol m 2 hr

At 100 oC, xA1,min = 0.0090

31-32

= Ls ,min L1,min (1 − x A1,min ) y A1G1 + x A 2 L2 = y A 2G2 + x A1,min L1,min

( 0.05)(10.0 ) + 0 = ( 0.01)( 9.6 ) + ( 0.009 ) ( L1,min ) L1,min = 44.9

kgmol m 2 hr

Ls ,min = ( 44.9 )(1 - 0.009 ) =44.5

kgmol m 2 hr

As temperature is lowered, minimum solvent rate goes down due to higher solubility of solute in solvent. T=

40 C

mol H2S/

CAL

mm Hg 0.00 0.96

0.0000 0.0013

mol MEA 0.000 0.125

0.0000 0.0063

(kgmol/m3) 0.00 0.33

3.00 9.10 43.10 59.70 106.00

0.0039 0.0120 0.0567 0.0786 0.1395

0.208 0.362 0.643 0.729 0.814

0.0104 0.0180 0.0315 0.0356 0.0396

0.54 0.94 1.65 1.86 2.07

100 C

1.00 3.00 5.00 10.00

0.0013 0.0039 0.0066 0.0132

0.029 0.050 0.065 0.091

0.0015 0.0025 0.0033 0.0046

0.08 0.13 0.17 0.24

30.00 50.00 70.00 100.00

0.0395 0.0658 0.0921 0.1316

0.160 0.203 0.238 0.279

0.0080 0.0102 0.0119 0.0139

0.42 0.53 0.62 0.73

0.060 40 C

yA1

100 C

Mole fraction H2S in gas, yA

0.050 0.040 0.030 0.020

xA1,min

xA1,min 0.010 0.000 0.000

0.010

0.020

0.030

0.040

Mole fraction H2S in solvent (water+15 wt% MEA), xA

30.18 31-33

a. Tower diameter D A = H2S, B = N2 Cf = 52 for 1.5 inch ceramic Intalox saddles (Table 31.2) kgmol   kg   = AG '1 AG1= M w,ave AG1 ( y A1 M A + (1 − y A=  1)MB )  360   ( 0.10(64)+(1-0.10)(28) ) hr   kgmol   kg hr  (1.5 atm ) kg  kg  P  ρG =  T  M w,ave =  31.6  = 1.9 3 3 kgmol  m  m atm   RT    0.08206  ( 305 K ) kgmol-K   Flooding Correlation 1/2 kg   kg  1/2 18, 000   1.9  '  AL1  ρG  hr    m3 = = x − axis 0.066 =   kg kg  kg   AG1  ρ L − ρG    − 1090 1.9 11,376   m3 m3  hr   0.25 y − axis = Y= = 11,376

1/2

   ρ ( ρ − ρ ) g  (1.9 )(1090 − 1.9 )(1.0 )  4.4 kg G 'f = Y  G L 0.1 G c   0.25 =  0.1  C f µL J m 2 sec 52 0.0012 1   ( )( ) ( )     kg   1hr   11,376   '   AG1 hr   3600sec   A = =1.19 m 2 at 60% flooding = 0.6G 'f  kg  0.6  4.4 2  m sec   1/2

 4 ⋅1.19 m 2   4A  D= =   = 1.23 m  π  π    1/2

b. Will tower “flood” if AL1/ = 54,000 kg/hr (3x old AL1/) Get new xaxis; gas flow unchanged but now need yaxis at 0.6Gf/ xaxis ,new = 3 ⋅ xaxis ,old = (0.066(3)) = 0.20 2

 G'  2 yaxis yaxis,f= 0.090 = =   0.25(0.6) G '  f  Locus of intersection is below flooding line with ∆P/z between 50 and 100 Pa/m, therefore tower will not flood.

31-34

Part 2: Instructor Only Problems Chapter 1 Instructor Only Problems 1.23 A clean glass capillary tube contains water at 40oC. Please calculate how high (in millimeters) the water will rise in the capillary tube if the diameter of the tube is 0.25 cm. Solution H2O at 60oC = 40 + 273 = 313 K Equation 1-17: σ = 0.123(1 − 0.00139 T) = 0.123 [1-0.00139 (313)] = 0.0695 N/m For a clean capillary: θ = 0o Height, using Equation 1-20: h =

2 cos θ ρgr

σ = 0.0695 N/m ρ = 992.2 kg/m3 (from data in Appendix I) ℎ=

2𝜎𝑐𝑜𝑠𝜃 2(0.0695 𝑁/𝑚)cos (0) = = 0.01144 𝑚 = 𝟏𝟏. 𝟒𝟒 𝒎𝒎 0.25 𝑐𝑚 𝑚 𝜌𝑔𝑟 992.2 𝑘𝑔/𝑚3 (9.8𝑚/𝑠 2 ) ( 2 𝑥 100 𝑐𝑚)

1.33 A beaker of water with a density of 987 kg/m3 has a capillary tube inserted into it. The water is rising in the capillary tube to a height of 1.88 cm. The capillary tube is very clean and has a diameter of 1.5 mm. What is the temperature of the water? Solution First, calculate the surface tension of water using the equation for the height: 2𝜎𝑐𝑜𝑠𝜃 ℎ= 𝜌𝑔𝑟 Rearrange, solving for 𝜎, 𝑚 1.5 𝑚𝑚 𝑚 (1.88 𝑐𝑚) ( ) (987 𝑘𝑔/𝑚3 )(9.81 𝑚/𝑠 2 ) ( 2 ) (1000 𝑚𝑚) ℎ𝜌𝑔𝑟 100 𝑐𝑚 𝜎= = 2𝑐𝑜𝑠𝜃 2 cos 0 𝑘𝑔 = 0.0683 2 = 0.0683 𝑁/𝑚 𝑠 Now use the surface tension to calculate the temperature: 𝜎 = 0.123(1 − 0.00139𝑇) 0.0683 𝑁/𝑚 = 0.123(1 − 0.00139𝑇) 𝑇 = 320.2 𝐾 = 47.02 ℃

1.34 The water in a lake has an average temperature of 60oF. If the barometric pressure of the atmosphere is 760 mm Hg (which is equal to 2.36 feet). Determine the gage pressure and the absolute pressure at a water depth of 46 feet. Solution From Appendix I, 𝜌 = 62.3 𝑙𝑏𝑚 /𝑓𝑡 3 at 60oF. (62.3 𝑃𝑔 = 𝜌𝑤 𝑔ℎ =

(847 𝑃𝑎𝑡𝑚 = 𝜌𝐻𝑔 𝑔ℎ =

𝑓𝑡 𝑙𝑏𝑚 ) (32.2 2 ) (46 𝑓𝑡) 3 𝑙𝑏𝑓 𝑓𝑡 𝑠 = 2868 2 𝑙𝑏 𝑓𝑡 𝑓𝑡 32.174 𝑚 2 𝑙𝑏𝑓 𝑠

𝑓𝑡 3.28 𝑓𝑡 𝑙𝑏𝑚 760 𝑚𝑚 ) (32.2 2 ) ( )( ) 𝑙𝑏𝑓 𝑚 1000 𝑚𝑚/𝑚 𝑓𝑡 3 𝑠 = 2113 𝑙𝑏 𝑓𝑡 𝑓𝑡 2 32.174 𝑚 2 𝑙𝑏𝑓 𝑠

𝑃𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 = 𝑃𝑎𝑡𝑚 + 𝑃𝑔 = 644.24

𝑙𝑏𝑓 𝑙𝑏𝑓 𝑙𝑏𝑓 + 2868 2 = 3512 2 2 𝑓𝑡 𝑓𝑡 𝑓𝑡

1.35 Air initially fills a very long vertical capillary tube of inside diameter D. The tube is suddenly immersed in a large body of water, still in the vertical position. Water wets the tube surface and as soon as the ends of the tube are submerged, water enters the tube. When equilibrium is reached, what is the length, if any, of the air column that remains in the tube? For this problem we know that the tube diameter is 0.1 cm and the surface tension of water is 0.072 N/m and the density of water is 1000 kg/m3. We can assume that the figure is not drawn to scale and that the capillary tube is drawn much bigger than reality, such that D<<H1, and that the maximum bubble pressure occurs when the radius of curvature equals the tube radius so that R1=R2=D/2. Assume the system is at 273 K. After immersion, the system looks like this: Air/water interface Air Pocket

H1 H2

Solution Begin with the Young-Laplace Equation: 1 1 𝑃𝐻2 − 𝑃𝐻2 = 𝜎 ( + ) 𝑅1 𝑅 2 Since R1 = R2 = D/2 𝑃𝐻2 − 𝑃𝐻2 =

2𝜎 4𝜎 = 𝐷/2 𝐷

Realizing that ∆𝑃 = 𝜌𝑔ℎ 𝜌𝑔𝐻2 − 𝜌𝑔𝐻1 =

4𝜎 𝐷

Realizing that 𝐻2 − 𝐻1 = 𝐿 and 0.072

𝑁 𝑘𝑔 𝑔 = 0.072 2 = 72 2 𝑚 𝑠 𝑠

So, 𝐻2 − 𝐻1 = 𝐿 =

4𝜎 4(72 𝑔/𝑠 2 ) = = 2.94 𝑐𝑚 𝜌𝑔𝐷 (1 𝑔/𝑐𝑚3 )(980 𝑐𝑚/𝑠 2 )(0.1 𝑐𝑚)

1.36 Hydrometers are inexpensive and easy to use to measure specific gravity and then density. You want to measure the density of an unknown liquid but all you have is a test tube that is 1 cm in diameter. You fill the test tube with water and drop it into a known sample of water with a temperature of 20℃, and measure the distance from the bottom of the test tube to the surface of the water to be 8 cm. Then you take this same test tube (after cleaning on the outside) and drop into an unknown liquid which is also at 20℃ and measure the distance submerged to be 7.4 cm. Hydrometers floating in liquids are in static equilibrium. What is the density of the unknown liquid?

8 cm submerged

7.4 cm submerged

Water

Unknown

Solution A hydrometer floating in water is in static equilibrium and the buoyant force FB exerted by the liquid must always be equal to the weight W of the hydrometer, so that F B = W. F = PA = ρghA = ρgW FB = ρgVsubmerged = ρghATT Where h is the height of the submerged portion of the test tube and A TT is the cross-sectional area of the test tube which is constant. For the pure water: 𝑊 = 𝜌𝑤 𝑔ℎ𝑤 𝐴𝑇𝑇 For unknown: 𝑊 = 𝜌𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑔ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝐴𝑇𝑇 Setting the equations equal since the weight of the test tube does not change, 𝜌𝑤 𝑔ℎ𝑤 𝐴𝑇𝑇 = 𝜌𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑔ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝐴𝑇𝑇 Solve for the density of the unknown liquid, ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛 =

ℎ𝑤 ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛

𝜌𝑤 =

8 𝑐𝑚 (998.2 𝑘𝑔/𝑚3 ) = 1079 𝑘𝑔/𝑚3 7.4 𝑐𝑚

Chapter 2 Instructor Only 2.30 A large industrial waste collection tank contains butyl alcohol, benzene and water at 80℉ that have separated into three distinct phases as shown in the figure. The diameter of the circular tank is 10 feet and it has total depth is 95 feet. The gauge at the top of the tank reads 2116 lbf /ft2. Please calculate (a) the pressure at the butyl alcohol/benzene interface, (b) the pressure at the benzene/water interface and (c) the pressure at the bottom of the tank. Solution (a) the pressure at the butyl alcohol/benzene interface ∆𝑃 = 𝜌𝑔ℎ 𝑃𝑡𝑜𝑝 − 𝑃𝐴𝑖𝑟−𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = −𝜌𝑎𝑖𝑟 𝑔(17𝑓𝑒𝑒𝑡) 𝑃𝐴𝑖𝑟−𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 − 𝑃𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙−𝐵𝑒𝑛𝑧𝑒𝑛𝑒 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = −𝜌𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙 𝑔(19𝑓𝑒𝑒𝑡) 𝑃𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙−𝐵𝑒𝑛𝑧𝑒𝑛𝑒 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = 𝑃𝑡𝑜𝑝 + 𝜌𝑎𝑖𝑟 𝑔(17𝑓𝑒𝑒𝑡) + 𝜌𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙 𝑔(19𝑓𝑒𝑒𝑡) 𝑙𝑏𝑚 2 𝑙𝑏𝑓 (0.0735 𝑓𝑡 3 ) (32.2 𝑓𝑡/𝑠 )(17 𝑓𝑡) = 2116 2 + 𝑙𝑏 𝑓𝑡 𝑓𝑡 32.174 𝑚 2 𝑙𝑏𝑓 𝑠 𝑙𝑏𝑚 (50.0 3 ) (32.2 𝑓𝑡/𝑠 2 )(19 𝑓𝑡) 𝑙𝑏𝑓 𝑓𝑡 + = 3068.0 2 = 21.3 𝑝𝑠𝑖 𝑙𝑏 𝑓𝑡 𝑓𝑡 32.174 𝑚 2 𝑙𝑏𝑓 𝑠 (b) the pressure at the benzene/water interface 𝑃𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙−𝐵𝑒𝑛𝑧𝑒𝑛𝑒 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 − 𝑃𝐵𝑒𝑛𝑧𝑒𝑛𝑒−𝑊𝑎𝑡𝑒𝑟 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = −𝜌𝐵𝑒𝑛𝑧𝑒𝑛𝑒 𝑔(34𝑓𝑒𝑒𝑡) 𝑃𝐵𝑒𝑛𝑧𝑒𝑛𝑒−𝑊𝑎𝑡𝑒𝑟 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = 𝑃𝐵𝑢𝑡𝑦𝑙𝐴𝑙𝑐𝑜ℎ𝑜𝑙−𝐵𝑒𝑛𝑧𝑒𝑛𝑒 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 + 𝜌𝐵𝑒𝑛𝑧𝑒𝑛𝑒 𝑔(34𝑓𝑒𝑒𝑡) 𝑙𝑏𝑚 2 𝑙𝑏𝑓 (54.6 𝑓𝑡 3 ) (32.2 𝑓𝑡/𝑠 )(34 𝑓𝑡) 𝑙𝑏𝑓 = 3068.0 2 + = 4925.9 2 = 34.2 𝑝𝑠𝑖 𝑙𝑏 𝑓𝑡 𝑓𝑡 𝑓𝑡 32.174 𝑚 2 𝑙𝑏𝑓 𝑠 (c) the pressure at the bottom of the tank 𝑃𝐵𝑒𝑛𝑧𝑒𝑛𝑒−𝑊𝑎𝑡𝑒𝑟 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 − 𝑃𝐵𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑇𝑎𝑛𝑘 = −𝜌𝑊𝑎𝑡𝑒𝑟 𝑔(25𝑓𝑒𝑒𝑡) 𝑃𝐵𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑇𝑎𝑛𝑘 = 𝑃𝐵𝑒𝑛𝑧𝑒𝑛𝑒−𝑊𝑎𝑡𝑒𝑟 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 + 𝜌𝐵𝑒𝑛𝑧𝑒𝑛𝑒 𝑔(25𝑓𝑒𝑒𝑡) 𝑙𝑏𝑚 2 𝑙𝑏𝑓 (62.2 𝑓𝑡 3 ) (32.2 𝑓𝑡/𝑠 )(25 𝑓𝑡) 𝑙𝑏𝑓 = 4925.9 2 + = 6482.16 2 = 45 𝑝𝑠𝑖 𝑙𝑏 𝑓𝑡 𝑓𝑡 𝑓𝑡 32.174 𝑚 2 𝑙𝑏𝑓 𝑠

2.31 The maximum blood pressure in the upper arm of a healthy person is about 120 mm Hg (this is a gauge pressure). If a vertical tube open to the atmosphere is connected to the vein in the arm of a person, determine how high the blood will rise in the tube. Take the density of the blood to be constant and equal to 1050 kg/m3. (The fact that blood can rise in a tube explains why IV tubes must be placed high to force fluid into the vein of a patient.) Assume the system is at 80oF. Solution ∆𝑃 = 𝜌𝑔ℎ For Blood:

𝑃 = 𝜌𝐵 𝑔ℎ𝐵

For Mercury: 𝑃 = 𝜌𝑀 𝑔ℎ𝑀 So, ∆𝑃 = 𝜌𝐵 𝑔ℎ𝐵 = 𝜌𝑀 𝑔ℎ𝑀 Solve for the height of the blood, 𝜌𝐵 𝑔ℎ𝐵 = 𝜌𝑀 𝑔ℎ𝑀 Eliminate the gravity terms, 𝜌𝐵 ℎ𝐵 = 𝜌𝑀 ℎ𝑀 ℎ𝐵 =

𝜌𝑀 ℎ 𝜌𝐵 𝑀

We take 120 mm Hg as the height here, and 120 mm = 0.12 m, and the density of mercury is 845 𝑙𝑏𝑚 /ft 3 . (13535.6 𝑘𝑔/𝑚3 ) (0.12 𝑚) = 1.55 𝑚 ℎ𝐵 = (1050 𝑘𝑔/𝑚3 )

2.32 If the nitrogen tank in the figure below is at a pressure of 4500 lbf/ft2 and the entire system is at 100℉, please calculate the pressure at the bottom of the tank of glycerin.

2 ft

Nitrogen

12 ft

4 ft

7 ft

2.5 ft

Benzene

4 ft

7.5 ft

10 ft 5 ft 8 ft 5 ft

11 ft 10 ft 7 ft

7.5 ft

Water

5.8 ft

Mercury

Glycerin

Solution PN − P1 = −ρNitrogen g(5 ft) so P1 = PN + ρNitrogen g(5 ft) P2 − P1 = −ρH2O g(8 − 5 ft) so P2 = P1 − ρH2O g(8 − 5 ft) P2 − P3 = −ρFreon g(7 − 4 ft) so P3 = P2 + ρFreon g(7 − 4 ft) P3 − P4 = −ρMercury g(10 − 7 ft) so P4 = P3 + ρMercury g(10 − 7 ft) P5 − P4 = −ρBenzene g(10 − 7.5 ft) so P5 = P4 − ρBenzene g(10 − 7.5 ft) P5 − PA = −ρA g(11 ft) so PA = P6 + ρA g(11 ft) PA = ρA g(11 ft) − ρBenzene g(10 − 7.5 ft) + ρMercury g(10 − 7 ft) + +ρFreon g(7 − 4 ft) − ρH2O g(8 − 5 ft) + ρNitrogen g(5 ft) +PN (78.2 lb /ft3 )

(53.6 lb /ft3 )

PA = (32.174 lb mft/lb s2) (32.2 ft/s2 )(11 ft) − (32.174 lb mft/lb s2) (32.2 ft/s2 )(10 − 7.5 ft) + m

(843 lbm

/ft3 )

(32.174 lbm ft/lbf s2 ) lb (62.1 m 3)

ft lb ft (32.174 m 2 ) lbf s

(32.2 ft/s 2 )(10 − 7 ft) + ft

(32.2 s2) (8 − 5 ft) +

lb (78.7 m 3)

ft lb ft (32.174 m 2 ) lbf s lbm (0.0685 3 ) ft ft lbm ft s2 (32.174 ) lbf s2

(32.2 s2) (7 − 4 ft) − lb

(32.2 ) (5 ft) + 4500 ft2f = 7808.0 lbf /ft 2

2.33 The water in a lake has an average temperature of 60oF. If the barometric pressure of the atmosphere is 760 mm Hg (which is equal to 2.36 feet). Determine the gage pressure and the absolute pressure at a water depth of 46 feet. Solution From Appendix I, 𝜌 = 62.3 𝑙𝑏𝑚 /𝑓𝑡 3 at 60oF. (62.3 𝑃𝑔 = 𝜌𝑤 𝑔ℎ =

(847 𝑃𝑎𝑡𝑚 = 𝜌𝐻𝑔 𝑔ℎ =

𝑓𝑡 𝑙𝑏𝑚 ) (32.2 2 ) (46 𝑓𝑡) 3 𝑙𝑏𝑓 𝑓𝑡 𝑠 = 2868 2 𝑙𝑏 𝑓𝑡 𝑓𝑡 32.174 𝑚 2 𝑙𝑏𝑓 𝑠

𝑓𝑡 3.28 𝑓𝑡 𝑙𝑏𝑚 760 𝑚𝑚 ) (32.2 2 ) ( )( ) 𝑙𝑏𝑓 𝑚 1000 𝑚𝑚/𝑚 𝑓𝑡 3 𝑠 = 2113 𝑙𝑏 𝑓𝑡 𝑓𝑡 2 32.174 𝑚 2 𝑙𝑏𝑓 𝑠

𝑃𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 = 𝑃𝑎𝑡𝑚 + 𝑃𝑔 = 644.24

𝑙𝑏𝑓 𝑙𝑏𝑓 𝑙𝑏𝑓 + 2868 2 = 3512 2 2 𝑓𝑡 𝑓𝑡 𝑓𝑡

2.34 Air initially fills a very long vertical capillary tube of inside diameter D. The tube is suddenly immersed in a large body of water, still in the vertical position. Water wets the tube surface and as soon as the ends of the tube are submerged, water enters the tube. When equilibrium is reached, what is the length, if any, of the air column that remains in the tube? For this problem we know that the tube diameter is 0.1 cm and the surface tension of water is 0.072 N/m and the density of water is 1000 kg/m3. We can assume that the figure is not drawn to scale and that the capillary tube is drawn much bigger than reality, such that D<<H1, and that the maximum bubble pressure occurs when the radius of curvature equals the tube radius so that R1=R2=D/2. Assume the system is at 273 K. After immersion, the system looks like this: Air/water interface Air Pocket

H1 H2

Solution Begin with the Young-Laplace Equation: 1 1 𝑃𝐻2 − 𝑃𝐻2 = 𝜎 ( + ) 𝑅1 𝑅 2 Since R1 = R2 = D/2 𝑃𝐻2 − 𝑃𝐻2 =

2𝜎 4𝜎 = 𝐷/2 𝐷

Realizing that ∆𝑃 = 𝜌𝑔ℎ 𝜌𝑔𝐻2 − 𝜌𝑔𝐻1 =

4𝜎 𝐷

Realizing that 𝐻2 − 𝐻1 = 𝐿 and 0.072

𝑁 𝑘𝑔 𝑔 = 0.072 2 = 72 2 𝑚 𝑠 𝑠

So, 𝐻2 − 𝐻1 = 𝐿 =

4𝜎 4(72 𝑔/𝑠 2 ) = = 2.94 𝑐𝑚 𝜌𝑔𝐷 (1 𝑔/𝑐𝑚3 )(980 𝑐𝑚/𝑠 2 )(0.1 𝑐𝑚)

2.35 Hydrometers are inexpensive and easy to use to measure specific gravity and then density. You want to measure the density of an unknown liquid but all you have is a test tube that is 1 cm in diameter. You fill the test tube with water and drop it into a known sample of water with a temperature of 20℃, and measure the distance from the bottom of the test tube to the surface of the water to be 8 cm. Then you take this same test tube (after cleaning on the outside) and drop into an unknown liquid which is also at 20℃ and measure the distance submerged to be 7.4 cm. Hydrometers floating in liquids are in static equilibrium. What is the density of the unknown liquid?

8 cm submerged

7.4 cm submerged

Water

Unknown

Solution A hydrometer floating in water is in static equilibrium and the buoyant force F B exerted by the liquid must always be equal to the weight W of the hydrometer, so that F B = W. F = PA = ρghA = ρgW FB = ρgVsubmerged = ρghATT Where h is the height of the submerged portion of the test tube and A TT is the cross-sectional area of the test tube which is constant. For the pure water: 𝑊 = 𝜌𝑤 𝑔ℎ𝑤 𝐴𝑇𝑇 For unknown: 𝑊 = 𝜌𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑔ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝐴𝑇𝑇 Setting the equations equal since the weight of the test tube does not change, 𝜌𝑤 𝑔ℎ𝑤 𝐴𝑇𝑇 = 𝜌𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑔ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝐴𝑇𝑇 Solve for the density of the unknown liquid, ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛 =

ℎ𝑤 ℎ𝑢𝑛𝑘𝑛𝑜𝑤𝑛

𝜌𝑤 =

8 𝑐𝑚 (998.2 𝑘𝑔/𝑚3 ) = 1079 𝑘𝑔/𝑚3 7.4 𝑐𝑚

Chapter 4 Instructor Only Problems 4.27 A well-mixed tank initially contains pure water. The tank has inlet and outlet ports both with a diameter of 1 centimeter, and the inlet port is 20 centimeters above the outlet port. Into the tank through the inlet port flows a 10% solution of sodium sulfate dissolved in water with a velocity of 0.1 meters/second that has a density of 1000 kg/m3. (a) After 60 seconds of flow, the solution at the outlet port is measured to contain 14 grams of sodium sulfate per liter of solution at a constant flow rate of 0.01 liters per second. Please calculate the number of grams of sodium sulfate in the tank at this time. (b) What is the mass flow rate at the exit port once the system has reached steady state? Solution (a) Grams of sodium sulfate in the tank after 60 seconds 𝑑𝑆 𝑚̇𝑜𝑢𝑡 − 𝑚̇𝑖𝑛 + =0 𝑑𝑡 𝜋(0.01𝑚)2 𝑑𝑆 3 (14 𝑔/𝐿)(0.01 𝐿/𝑠) − (1000 𝑘𝑔/𝑚 )(1000 𝑔/𝑘𝑔)(0.1 𝑚/𝑠) ( ) (0.1) + =0 4 𝑑𝑡 𝑑𝑆 (0.14 𝑔/𝑠) − (0.7854 𝑔/𝑠 ) + =0 𝑑𝑡 𝑑𝑆 = 0.6454 𝑔/𝑠 𝑑𝑡 𝑆

∫ 𝑑𝑆 = 0.6454 𝑔/𝑠 ∫ 𝑑𝑡 0

𝑆 = 38.7 𝑔𝑟𝑎𝑚𝑠 (b) What is the mass flow rate at the exit port once the system has reached steady state? At Steady state, 𝑚̇𝑜𝑢𝑡 = 𝑚̇𝑖𝑛 = 0.7854 grams/sec

4.28

A cylindrical water tank is 4 feet high, with a 3-foot diameter, open to the atmosphere at the top is initially filled with water with a temperature of 60oF. The outlet at the bottom has a diameter of 0.5 in. is opened and the tank empties. The average velocity of the water exiting the tank is given by the equation, v = √2gh, where h is the height of the water in the tank measured from the center of the outlet port and g is the gravitational acceleration. Determine: (a) how long will it take for the water level in the tank to drop to 2 feet from the bottom, and (b) how long will it take to drain the entire tank. Solution 𝑑𝑆 𝑚̇𝑜𝑢𝑡 − 𝑚̇𝑖𝑛 + =0 𝑑𝑡 Since there is no fluid entering the tank: 𝑚̇𝑖𝑛 = 0 𝑑𝑆 = −𝑚̇𝑜𝑢𝑡 𝑑𝑡 𝑑ℎ 𝜋𝑑𝑡2 𝜋𝑑𝑜2 𝜌 ( ) = −𝜌√2gh ( ) 𝑑𝑡 4 4 𝑑ℎ 𝑑𝑜2 = √2gh 𝑑𝑡 𝑑𝑡2 𝑑𝑡2 𝑑ℎ 1 𝑑𝑡 = − 2 𝑑𝑜 √ℎ √2𝑔 𝑡

∫ 𝑑𝑡 = − 0

ℎ2 𝑑𝑡2 1 𝑑ℎ ∫ 2 𝑑𝑜 √2𝑔 ℎ0 √ℎ 1/2

1/2

𝑑𝑡2 1 ℎ2 − ℎ0 𝑡=− 2 𝑑𝑜 √2𝑔 (1/2)

(a) How long will it take for the water level in the tank to drop to 2 feet from the bottom?

𝑡=−

(3 𝑓𝑡)2 1 21/2 − 41/2 = 745 𝑠𝑒𝑐 = 12.4 𝑚𝑖𝑛 (0.042𝑓𝑡)2 √2(32.2𝑓𝑡/𝑠 2 ) (1/2)

(b) How long will it take to drain the entire tank? (3 𝑓𝑡)2 1 01/2 − 41/2 𝑡=− = 2543 𝑠𝑒𝑐 = 42.4 𝑚𝑖𝑛 (0.042𝑓𝑡)2 √2(32.2𝑓𝑡/𝑠 2 ) (1/2) 4.29 A tank with a total volume of 75 liters is initially empty. Into the tank flows pure water with a temperature of 273K, density of 999.3 kg/m3 and surface tension of 0.0763 N/m. This pure water

is being delivered from an inlet port with a diameter of 0.01m at a velocity of 2.0 m/s. The exit port is initially closed but it too has a diameter of 0.01m. (a) How long does it take to fill the tank with water (assuming the exit port remains closed)? Solution ∬ ρ(v ∙ n) dA +

∂ ∭ ρdV = 0 ∂t

Since exit port is closed −ṁin +

∂ V ∫ ρdV = 0 ∂t 0

The tank has a total volume of 75 liters, (75 liters)

0.028317 m3 = 0.075m3 28.32 liters 3

kg π(0.01 m)2 kg ∂ 0.075 m − (999.3 3 ) (2 m/s) ( ) + (999.3 3 ) ∫ dV = 0 m 4 m ∂t 0 1.57x10−4 m3 t ∫ dt s 0 t = 447.5 sec

0.075 m3 =

(b) Once the tank is completely filled with pure water, into the inlet port (still with a diameter of 0.01m) flows at 2.0 m/s a 60% solution of sodium chloride with a specific gravity of 1.07. What is the total mass (water plus sodium chloride) in the tank after 600 seconds if a sample at the exit port at that time contains 50 g/liter of sodium chloride and the flow rate is 0.5 liters/s? Solution The tank is now filled with pure water which has a mass of: (75 liters)

0.028317 m3 999.3 𝑘𝑔 ( ) = 74.93 kgs 28.32 liters 𝑚3 𝑀

𝑑𝑀 =0 𝑀0 𝑑𝑡

𝑚̇𝑜𝑢𝑡 − 𝑚̇𝑖𝑛 + ∫

𝑀 𝑘𝑔 𝑚 π(0.01 m)2 𝑔 𝑑𝑀 ) (1000 ) (2 ) ( ) + ∫ =0 3 𝑚 𝑠 4 𝑘𝑔 𝑀0 𝑑𝑡 𝑀 𝑔 𝑑𝑀 25 − 168 𝑔/𝑠 + ∫ 𝑠 74.93 𝑘𝑔 𝑑𝑡

(0.5 𝐿/𝑠)(50 𝑔/𝐿) − (1.07) (999.3

𝑚 = 160.7 𝑘𝑔 4.30 A large tank of unknown total volume is initially filled with 6000 grams of a 10% by mass sodium sulfate solution. Into this tank a 50% sodium sulfate solution is added at a rate of 40 g/min. At the

single outlet to the tank flows a 20 g/Liter solution at a rate of 0.01667 liters/sec. Please calculate (a) the total mass in the tank after 2 hours and (b) the amount of sodium sulfate in the tank after 2 hours. Solution (a) Total mass in the tank after 2 hours Total Mass = M 𝜕 ∭ 𝜌𝑑𝑉 = 0 𝜕𝑡 𝑑𝑀 𝑚̇𝑜𝑢𝑡 − 𝑚̇𝑖𝑛 + =0 𝑑𝑡 𝑀 𝑔 𝑑𝑀 (20 𝑔/𝐿)(0.01667 𝐿/𝑠)(60 𝑠/𝑚) − 40 +∫ =0 𝑚𝑖𝑛 6000 𝑑𝑡 𝑀 𝑑𝑀 20 𝑔/𝑚𝑖𝑛 − 40 𝑔/𝑚𝑖𝑛 + ∫ =0 6000 𝑑𝑡 𝑔 𝑑 −20 + (𝑀 − 6000) = 0 𝑚𝑖𝑛 𝑑𝑡 𝑀 = 6000 + 20𝑡 At the 2 hour point, which is 120 minutes, 𝑀 = 6000 + 20𝑡 = 6000 𝑔𝑟𝑎𝑚𝑠 + 20 𝑔𝑟𝑎𝑚𝑠/𝑚𝑖𝑛(120𝑚𝑖𝑛) = 8400.5 𝑔𝑟𝑎𝑚𝑠 ∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 +

(b) The amount of sodium sulfate in the tank after 2 hours 𝑑𝑆 𝑚̇𝑜𝑢𝑡 − 𝑚̇𝑖𝑛 + =0 𝑑𝑡 𝑆 𝑆 = 𝑀 6000 + 20𝑡 𝑆 𝑑𝑆 (20 𝑔/𝐿)(0.01667 𝐿/𝑠)(60 𝑠/𝑚) ( ) − (0.50)(40 𝑔/𝑚𝑖𝑛) + =0 6000 + 20𝑡 𝑑𝑡 𝑆 𝑑𝑆 (20 𝑔/𝑚𝑖𝑛) ( ) − (0.50)(40 𝑔/𝑚𝑖𝑛) + =0 6000 + 20𝑡 𝑑𝑡 𝑆 𝑑𝑆 − 20 𝑔/𝑚𝑖𝑛 + =0 300 + 𝑡 𝑑𝑡 Rearrange, 𝑑𝑆 𝑆 + = 20 𝑑𝑡 300 + 𝑡 Integrating, 𝑑𝑆 (300 + 𝑡) = (300 + 𝑡)20 𝑑𝑡 20𝑡 2 𝑆(300 + 𝑡) = 6000𝑡 + +𝐶 2 20𝑡 2 6000𝑡 + 2 𝐶 𝑆= + (300 + 𝑡) (300 + 𝑡)

6000𝑡 + 10𝑡 2 𝐶 𝑆= + (300 + 𝑡) (300 + 𝑡) The initial condition given in the problem was that at time = 0, there is 6000 grams of a 10% solution of sodium sulfate solution. 6000(0) + 10(0)2 𝐶 (6000 𝑔𝑟𝑎𝑚𝑠)(0.1) = + (300 + 0) (300 + 0) 𝐶 = 180,000 6000𝑡 + 10𝑡 2 180,000 𝑆= + (300 + 𝑡) (300 + 𝑡) At the 2 hour point (2 hours = 120 minutes): 6000(120) + 10(120)2 180,000 𝑆= + = 2485.7 𝑔𝑟𝑎𝑚𝑠 (300 + 120) (300 + 120)

4.31 A spherical metal tank has a constant volume of 450 in3 and contains air at an absolute pressure of 85 psi and a temperature of 20oC. You are told that the air is leaking through a 0.1 inches in

diameter whole in the side. The air exits at 900 ft/s with a density of 0.075 lbm/ft3. Calculate the change in density (loss) with respect to time.

Solution ∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 +

𝜕 ∭ 𝜌𝑑𝑉 = 0 𝜕𝑡

𝑚̇𝑜𝑢𝑡 − 𝑚̇𝑖𝑛 +

𝜕𝑀 =0 𝜕𝑡

Since nothing is going into the tank, 𝑚̇𝑖𝑛 = 0 𝑚̇𝑜𝑢𝑡 +

𝜕𝑀 =0 𝜕𝑡

𝑚̇𝑜𝑢𝑡 +

𝜕𝜌𝑉 =0 𝜕𝑡

𝑓𝑡 0.1 𝑙𝑏𝑚 𝜕𝜌 𝑚̇𝑜𝑢𝑡 𝜌𝑣𝐴 0.075 𝑓𝑡 3 (900 𝑠 ) ( 12 𝑓𝑡) 0.5625 =− =− = = = 2.16 𝑙𝑏𝑚 𝑓𝑡 2 /𝑠 𝑓𝑡 3 𝜕𝑡 𝑉 𝑉 0.260 3 450 𝑖𝑛 ( 3 3 ) 12 𝑖𝑛

4.32 You are working on a project for lab when you realize that you forgot to measure the volume of your mixing tank. Luckily, you took excellent notes in lab and have a lot of information about the

system. You were running an experiment to determine the concentration of salt in a mixing tank at unsteady state. You measured the flow rate in and out of the mixing tank to be 1 L/min. The 25 𝑔 𝑠𝑎𝑙𝑡 tank was supplied with salt water at a concentration of 𝐶 = 100 𝑚𝑙 𝑠𝑎𝑡𝑒𝑟. You also observed that the solution in the tank was uniform and well mixed. Use a computational model (equation 1) and your data in the table below from the experiment to determine the volume of the mixing tank. Time 0 2 4 6 8 10 12 14 16 18 20 22 24 Concentration 0 0.059 0.106 0121 0.187 0.170 0.205 0.202 0.237 0.219 0.242 0.238 0.243

The equation describing the concentration is: 𝑡

𝐶(𝑡) = 𝐶𝑠𝑜𝑢𝑟𝑐𝑒 (1 − 𝑒 −𝜏 )

(equation 1)

In the above equation, t is time and 𝜏 is the volume of the tank. (Hint: A residual is the difference between a measured value, as in the data in the table, and the predicted value of a regression model, which is what you will calculate with the equation. It is important to understand residuals because they show how accurate a mathematical function, such as a line, is in representing a set of data. You can find the volume of the tank by minimizing the sum of the squares of all the residuals of all data points. One way to do this is to put in guesses of the tank volume to find when the sum of the squares of the residuals is the smallest. There are other more elegant ways to do this, but a guess a tank value and calculate the sum of the squares of the residuals is a fine way. This is easy to do in MSExcel)

Solution For this problem the data in the table was put into MSExcel. The sum of the residuals squared was calculated and then minimized using a solver by varying 𝜏 to determine the volume of the tank. 𝜏 = 7.45 𝐿, based on the graph and data.

Time

Concentration 0 0.059 0.106 0.121 0.187 0.17 0.205 0.202 0.237 0.219 0.242 0.238 0.243

0 2 4 6 8 10 12 14 16 18 20 22 24 Co Tau

Fitted Line (using given equation) 0.000 0.059 0.104 0.138 0.165 0.185 0.200 0.212 0.221 0.228 0.233 0.237 0.240

0.25 7.45

Residual (difference between actual concentration and that of the fitted line). 0.000 0.000 0.002 -0.017 0.022 -0.015 0.005 -0.010 0.016 -0.009 0.009 0.001 0.003 Total:

Residual Squared 0.00000 0.00000 0.00000 0.00030 0.00050 0.00022 0.00002 0.00010 0.00026 0.00008 0.00008 0.00000 0.00001 0.0015718

Problem 3 Concentration [g/ml]

0.3 0.25

0.2 0.15

Data

0.1

Fitted Line

0.05 0 0

15 Time [min]

Chapter 5 Instructor Only Problems

5.5 (a) Determine the magnitude of the x and y components of the force exerted on the fixed blade shown by a 3 ft3/s jet of water flowing at 25 ft/s. Assume the water is at 100oF. (b) If the blade is moving to the right at 15 ft/s, find the magnitude and velocity of the water jet leaving the blade. Assume steady state. Solution (a) Q = vA = 3 ft3/s, v1 = v2 = 25 ft/s ∑ 𝐹 = ∬ 𝑣𝑥 𝜌(𝑣 ∙ 𝑛)𝑑𝐴 +

𝜕 ∭ 𝑣𝑥 𝜌𝑑𝑉 𝜕𝑡

𝐹𝑥 = 𝜌𝑣2 𝐴2 𝑣2 𝑐𝑜𝑠𝜃 − 𝜌𝑣1 𝐴1 (−𝑣1 ) Where the negative sign on the v1 is because the blade is moving in the negative x-direction. 𝐹𝑥 = 𝜌𝑄𝑣2 𝑐𝑜𝑠𝜃 − 𝜌𝑄(−𝑣1 ) (62.1 𝑙𝑏𝑚 /𝑓𝑡 3 ) (3 𝑓𝑡 3 /𝑠)(25 𝑓𝑡/𝑠)𝑐𝑜𝑠30 = 32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2 (62.1 𝑙𝑏𝑚 /𝑓𝑡 3 ) (3 𝑓𝑡 3 /𝑠)(−25 𝑓𝑡/𝑠) = 270.12 𝑙𝑏𝑓 − 32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2 y-direction, remembering that the water velocity is flowing out in the minus y-direction, (62.1 𝑙𝑏𝑚 /𝑓𝑡 3 ) (3 𝑓𝑡 3 /𝑠)(−25 𝑓𝑡/𝑠)𝑠𝑖𝑛30 = −72.4𝑙𝑏𝑓 𝐹𝑦 = 𝜌𝑣2 𝐴2 𝑣2 𝑠𝑖𝑛𝜃 = 32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2 (b) If the blade moves to the right at 15 ft/s then, 𝑣1 = 25 + 15 = 40 𝑓𝑡/𝑠 𝑣2 = −40 𝑓𝑡/𝑠 So the magnitude of the velocity leaving the blade is: 𝑣 = 𝑣2 (𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑏𝑙𝑎𝑑𝑒) + 𝑣𝑏𝑙𝑎𝑑𝑒 𝑣 = [𝑣𝑥 𝑐𝑜𝑠𝜃𝑒𝑥 + 𝑣𝑦 𝑠𝑖𝑛𝜃𝑒𝑦 ] + 𝑣𝑥 𝑒𝑥 = [40𝑐𝑜𝑠(30)𝑒𝑥 + 40 sin(30) 𝑒𝑦 ] + 15𝑒𝑥 = [34.64𝑒𝑥 + 15𝑒𝑦𝑥 ] − 20𝑒𝑦 = 49.64𝑒𝑥 − 20𝑒𝑦 𝑣 ∙ 𝑒𝑠 = (49.64𝑒𝑥 − 20𝑒𝑦 ) ∙ (𝑐𝑜𝑠𝜃𝑒𝑥 + 𝑠𝑖𝑛𝜃𝑒𝑦 ) = 49.64 cos(−30) − 20 sin(−30) = 52.98 𝑓𝑡/𝑠

5.34 You have been given a project where you must attach a particular reducing pipe fitting into a Freon-12 delivery system. Freon-12 is a potentially hazardous material so you must be certain that the reducing pipefitting is attached properly and will withstand the force due to the reduction in pipe diameter. The direction of the flow, the inlet and outlet positions are labeled in the figure below. The inlet has a diameter of 1 foot, and a pressure of 1000 lb f /ft2. The outlet port has a diameter of 0.2 feet and a pressure of 200 lbf /ft2. The system must maintain a constant flow rate of 4.5 ft3/s and constant temperature of 80℉. The weight of the Freon in the fitting is 6 lbf. In your analysis you may assume that the system is at steady state and that the fluid is incompressible. Please calculate the forces in all directions necessary to hold the fitting stationary.

Outlet from Reducing Fitting

y x

60 degrees

Inlet to Reducing Fitting

Solution 𝐹𝑥 + 𝑃2 𝐴2 cos 𝜃 = 𝜌𝑣22 𝐴2 cos 𝜃 𝐹𝑥 = 𝜌𝑣22 𝐴2 cos 𝜃 − 𝑃2 𝐴2 cos 𝜃 𝐹𝑥 =

81.3 𝑙𝑏𝑚 /𝑓𝑡 3 𝜋(0.2𝑓𝑡)2 (143.3 𝑓𝑡/𝑠)2 ( ) cos(60) 𝑙𝑏𝑚 𝑓𝑡 4 32.174 𝑙𝑏𝑓 𝑠 2 𝜋(0.2𝑓𝑡)2 2 − (200 𝑙𝑏𝑓 /𝑓𝑡 ) ( ) cos(60) = 811.93 𝑙𝑏𝑓 4 𝐹𝑦 + 𝑃1 𝐴1 − 𝑃2 𝐴2 sin 𝜃 − 𝑊 = 𝜌𝑣22 𝐴2 sin 𝜃 − 𝜌𝑣12 𝐴1 𝐹𝑦 = 𝜌𝑣22 𝐴2 sin 𝜃 − 𝜌𝑣12 𝐴1 − 𝑃1 𝐴1 + 𝑃2 𝐴2 sin 𝜃 + 𝑊

81.3 𝑙𝑏𝑚 /𝑓𝑡 3 𝜋(0.2𝑓𝑡)2 2 (143.3 𝑓𝑡/𝑠) ( 𝐹𝑦 = ) sin 60 𝑙𝑏 𝑓𝑡 4 32.174 𝑚 2 𝑙𝑏𝑓 𝑠 81.3 𝑙𝑏𝑚 /𝑓𝑡 3 𝜋(1𝑓𝑡)2 𝜋(1𝑓𝑡)2 (5.73𝑓𝑡/𝑠)2 ( − ) − 1000 ( ) 𝑙𝑏𝑚 𝑓𝑡 4 4 32.174 𝑙𝑏𝑓 𝑠 2 𝜋(0.2𝑓𝑡)2 + 200 ( ) sin 𝜃 + 6 𝑙𝑏𝑓 = 572.6 𝑙𝑏𝑓 𝑑𝑜𝑤𝑛 4

5.35 The horizontal Y-fitting splits 80℉ water into two equal parts. The first part goes through Exit 1 and the second part goes through Exit 2. The volumetric flow rate at the entrance is 8 ft3/sec, and the gage pressures at the three positions are 25 lbf/in2 at the entrance, 1713 lbf/ft2 at exit 1, and 3433 lbf/ft2 at exit 2. The diameters at the entrance, exit 1, and exit 2 are 6.5, 4 and 3.5 inches respectively. Please determine the forces in all directions required to keep the fitting in place. Solution ventry =

v1 =

Q 8 ft 3 /sec = = 34.72 ft/sec A1 π 6.5 in 2 ( ) 4 12in/ft

Q 4 ft 3 /sec = = 45.83 ft/sec A2 π 4in 2 ( ) 4 12in/ft

Q 4 ft 3 /sec v2 = = = 59.87 ft/sec A2 π 3.5 in 2 ( ) 4 12in/ft

First, calculate the force in the x-direction: Fx + Pentry Aentry 2 − P1 A1 cos 50 − P2 A2 cos 45 = ρv12 A1 cos 50 + ρv22 A2 cos 45 − ρ ventry Aentry

Fx = −Pentry Aentry 2 + P1 A1 cos 40 + P2 A2 cos 45 + ρv12 A1 cos 40 + ρv22 A2 cos 45 − ρ ventry Aentry 2 2

= − (25 + 14.7

lbf 144 in π 6.5in )( ) ) )( ( 2 2 in ft 4 12in ft 2

lbf π 4in ) cos 40 + (3433 + (1713 + 2116.2 2 ) ( ft 4 12in ft 2

lbf π 3.5in ) ( ) cos 45 ft 2 4 12in ft 2 lbm ft 2 62.2 ft 3 π 4in ) ) ] cos 40 + (45.83 [ ( sec 32.174 lbm ft 4 12in ft lbf s2 2 lb m 2 62.2 ft ft 3 [π ( 3.5in ) ] cos 45 ) + (59.87 sec 32.174 lbm ft 4 12in ft lbf s2 2 lbm ft 2 62.2 ft 3 π 6.5in ) ) ] − (34.72 [ ( sec 32.174 lbm ft 4 12in ft lbf s2 Fx = −1317.37 + 255.98 + 262.17 + 271.45 + 327.38 − 537.03 = −737.42 lbf (to left) + 2116.2

Next, calculate the force in the y-direction: Fy − P1 A1 sin 40 + P2 A2 sin 45 = v1 ρv1 A1 sin 40 + v2 ρ(−v2 )A2 sin 45 Fy = v1 ρv1 A1 sin 40 + v2 ρ(−v2 )A2 sin 45 + P1 A1 sin 40 − P2 A2 sin 45 Fy = (45.83

ft 2 62.2 lbm /ft 3 π 4in 2 ) ( )[ ( ) ] sin 40 lbm ft 4 12in/ft sec 32.174 lbf s2 ft 2 62.2 lbm /ft 3 π 3.5in 2 ) ( )[ ( ) ] sin 45 − (59.87 lb ft sec 4 12in/ft 32.174 m 2 lbf s lbf π 4in 2 ) sin 40 − (3433 + (1713 + 2116.2 2 ) ( ft 4 12in/ft lbf π 3.5in 2 ) sin 45 = 227.77 − 327.38 + 214.8 − 262.17 + 2116.2 2 ) ( ft 4 12in/ft = −146.98 lbf

5.36 Water at 100℃ flows through a 10 cm diameter pipe that has a 180 degree vertical turn as shown in the figure below. The volumetric flow rate is constant at 0.2 m3/s. The absolute pressure at position 1 is 64,000 Pa and the absolute pressure at position 2 is 33,000 Pa. The weight of the fluid and pipe together is 10 kg. Please calculate the total force that vertical turn must withstand to remain in place assuming the system is at steady state. Solution 𝑄 0.2 𝑚3 /𝑠 𝑣1 = 𝑣2 = = 𝜋 = 25.46 𝑚/𝑠 𝐴 (0.10 𝑚)2 4 𝜕 ∑ 𝐹 = ∬ 𝑣𝑥 𝜌(𝑣 ∙ 𝑛)𝑑𝐴 + ∭ 𝑣𝑥 𝜌𝑑𝑉 𝜕𝑡 Since we are working at steady state, ∑ 𝐹 = ∬ 𝑣𝑥 𝜌(𝑣 ∙ 𝑛)𝑑𝐴 𝐹𝑥 + 𝑃1 𝐴1 − 𝑃2 𝐴2 cos 180 = 𝜌2 𝑣22 𝐴2 cos 180 − 𝜌1 𝑣12 𝐴1 𝐹𝑥 = −𝑃1 𝐴1 + 𝑃2 𝐴2 cos 180 + 𝜌2 𝑣22 𝐴2 cos 180 − 𝜌1 𝑣12 𝐴1 𝜋 𝜋 𝐹𝑥 = −(64,000 𝑃𝑎) ( (0.1 𝑚)2 ) + (33,000 𝑃𝑎) ( (0.1 𝑚)2 ) cos 180 4 4 𝑘𝑔 𝑚 2 𝜋 + (958.4 3 ) (25.46 ) ( (0.1 𝑚)2 ) cos (180) 𝑚 𝑠 4 𝑘𝑔 𝑚 2 𝜋 − (958.4 3 ) (25.46 ) ( (0.1 𝑚)2 ) = −10,520.3 𝑁 𝑚 𝑠 4 𝑘𝑔∙𝑚 𝐹𝑦 = 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑃𝑖𝑝𝑒 + 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐹𝑙𝑢𝑖𝑑 = (10 𝑘𝑔)(9.8 𝑚/𝑠 2 ) = 98 𝑠2 = 98 𝑁

5.37 Benzene is flowing at steady state through the nozzle shown below at 80oF and at a constant flow rate of 3 ft3/sec. Note that the nozzle exit is at an angle of 50 degrees as shown. At the inlet the diameter is 5 inches and the liquid is at a pressure of 500 lbf /ft2. At the outlet, the diameter is 2 inches and the pressure is 300 lbf/ft2. For this problem the weight of the benzene in the fitting is 30 lbf. Calculate the forces in all directions necessary to hold the pipe bend stationary. Solution 𝐹𝑥 + 𝑃1 𝐴1 − 𝑃2 𝐴2 cos 𝜃 = 𝜌2 𝑣22 𝐴2 cos 𝜃 − 𝜌1 𝑣12 𝐴1 𝐹𝑥 = 𝑃2 𝐴2 cos 𝜃 − 𝑃1 𝐴1 + 𝜌2 𝑣22 𝐴2 cos 𝜃 − 𝜌1 𝑣12 𝐴1 𝜋 2 2 𝜋 5 2 2 2 𝐹𝑥 = (300 𝑙𝑏𝑓 /𝑓𝑡 ) ( ( ) ) cos 𝜃 − (500 𝑙𝑏𝑓 /𝑓𝑡 ) ( ( ) ) 4 12 4 12 2

54.6 𝑙𝑏𝑚 /𝑓𝑡 3 𝑓𝑡 /𝑠 𝜋 2 2 ( ) ( ) ) cos 𝜃 ( 𝑙𝑏𝑚 𝑓𝑡 𝜋 2 2 4 12 32.174 𝑙𝑏𝑓 𝑠 2 4 (12) 3

54.6 𝑙𝑏𝑚 /𝑓𝑡 3 3 𝑓𝑡 3 /𝑠 𝜋 5 2 ( ) ( ( ) ) = 501.6 𝑙𝑏𝑓 𝑡𝑜 𝑙𝑒𝑓𝑡 − 𝑙𝑏𝑚 𝑓𝑡 𝜋 5 2 4 12 32.174 𝑙𝑏𝑓 𝑠 2 4 (12) 𝐹𝑦 + 𝑃2 𝐴2 sin 𝜃 − 𝑊 = −𝜌2 𝑣22 𝐴2 𝑠𝑖𝑛 𝜃 𝐹𝑦 = −𝑃2 𝐴2 sin 𝜃−𝜌2 𝑣22 𝐴2 𝑠𝑖𝑛 𝜃 + 𝑊 2

𝜋 2 54.6 𝑙𝑏𝑚 /𝑓𝑡 3 𝑓𝑡 /𝑠 𝜋 2 2 ( 𝐹𝑦 = −(300 𝑙𝑏𝑓 /𝑓𝑡 ) ( ( ) ) sin 𝜃 − 2 ) ( 4 (12) ) sin 𝜃 + 30 𝑙𝑏𝑓 𝑙𝑏 𝑓𝑡 4 12 32.174 𝑚 2 𝜋 ( 2 ) 𝑙𝑏𝑓 𝑠 4 12 = −215.24 𝑙𝑏𝑓 𝑢𝑝 2

5.39 A rectangular plate 5 feet long and 3 feet deep, hangs in the air from a hinge at the top as shown in the figure. The plate is struck in the center by a horizontal jet of water at 60oF that is 4 inches in diameter and moving at 20 ft/s, causing the plate to swing outward at an angle 𝜃. If the plate weighs 12 lbs., calculate the angle 𝜃 that the plate will hang from the vertical. hinge 𝜃

Fjet Solution From Appendix I: Water at 60oF, ρ = 62.3 lbm /ft 3 ∑ F = ∬ vx ρ(v ∙ n)dA +

∂ ∭ vx ρdV ∂t

Steady state reduces to, 2 Fx = ∬ vx ρ(v ∙ n)dA = ρin vin Ain =

2 (62.3 lbm /ft 3 ) π 4 2 (20 ft/s) ( ft) cosθ 32.174 lbm ft/lbf s2 4 12

= 67.6(cos θ) lbf 5ft 5ft ∑ M = 0 = ∑ F X r = −67.6(cos θ) ( ) + (12 lbs)(32.2 ft/s) ( ) sin θ 2 2

0 = −67.6(cos θ) (

5ft 5ft ) + (12 lbs)(32.2 ft/s) ( ) sin θ 2 2

5ft 5ft 67.6(cos θ) ( ) = (12 lbs)(32.2 ft/s) ( ) sin θ 2 2 5ft 67.6 ( 2 ) sin θ = = 0.175 cos θ (12 lbs)(32.2 ft/s) (5ft) 2 tan θ = 0.175 θ = 9.92o

5.40 A circular water jet 2 inches in diameter strikes a concrete (SG = 2.6) slab as shown which rests freely on a level floor. The slab is 40 inches tall, 10 inches deep and 5 inches wide into the paper. Please calculate the jet velocity which will just begin to tip the slab over if the water is at 62.4 lbm/ft3 and the system is at 273K. 10 inches Water jet Concrete Slab

2 inches

40 inches 30 inches

B Need to find the water force and then the moments about the lower left corner of the slab at B. The slab will tip when the center of mass is ½ the width at the tipping point B. ∂ ∭ vx ρdV ∂t ∂ ∑ M = ∬(rxv)z ρ(v ∙ n)dA + ∭(rxv)z ρdV ∂t ∑ F = ∬ vx ρ(v ∙ n)dA +

∑ Fx = Fjet = ∑ ṁout vout − ṁin vin = 0 − ṁin vin = −ρAv(−v) = ρv 2 A Wslab = ρVg =

2.6(62.4 lbm /ft 3 ) 10 40 5 ( ft) ( ft) ( ft) (32.2 ft/s2 ) = 187.9 lbf 2 32.174 lbm ft/lbf s 12 12 12 ∑ MB = ρv 2 A (

30 5 ft) − Wslab ( ft) 12 12

2 62.4 lbm /ft 3 π 2 30 + 2/2 10 1 ( ft) v 2 ( ft) = (187.9 lbf ) ( ft) 2 32.174 lbm ft/lbf s 4 12 12 12 2

Notes 2 2

30+( )

1. The center of the water jet is at a height of 30 inches + 2 inch/2 or ( 10

2. The center of the slab is ½ it’s width or (12 ft) 2 0.1093 v 2 = 78.3 lbf ft v = 26.7 ft/s

ft)

Chapter 6 Instructor Only Problems

6.41 You have been asked to do an analysis of a steady state pump that will transfer Aniline, an organic material from one area to another in a chemical plant. Please determine the temperature change that the fluid undergoes during this process. The flow rate is constant in the system at 1.0 ft3/sec. The inlet to the pump has a diameter of 8 inches and the pressure to the pump is 250 lb f /ft2. The outlet from the pump has a diameter of 5 inches and a pressure of 600 lbf /ft2. The fluid that will be pumped has a density of 63.0 lbm/ft3, a heat capacity of 0.490 BTU/lbmoF, viscosity of 1.8x103 lbm/ft sec, and kinematic viscosity of 2.86x10-5 ft2/s. The outlet from the pump is located 75 feet higher than the inlet to the pump. The pump provides work to the fluid at a rate of 8.3x106 lbmft2/s3 and the heat transferred to the CV from the pump is 1.40x106 BTU/hr. In this analysis you may assume steady state and ignore viscous work. Solution δWμ δQ δWs P ∂ − = ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV + ∂t ∂t ρ ∂t ∂t δQ BTU lbm ft hr lbm ft 2 6 ) (778.17 ft lbf /BTU) (32.174 )( ) = 9736549.5 = (1.4x10 ∂t hr lbf s2 3600 s s2 2 δWs lbm ft = 8.3x106 ∂t s2 δWμ = 0 since there air is a very low viscosity fluid ∂t ∂ ∭ eρdV = 0 since the system is a steady state ∂t δQ δWs P − = ∬ (e + ) ρ(v ∙ n) dA ∂t ∂t ρ δQ δWs P − = ∬ (e + ) ρ(v ∙ n) dA ∂t ∂t ρ δQ δWs p p − = (e + ) ṁ − (e + ) ṁ ∂t ∂t ρ 2 ρ 1 2 2 δQ δWs v2 − v1 P2 P1 − = ρQ [U2 − U1 + + ( − ) + g(z2 − z1 )] ∂t ∂t 2 ρ2 ρ1 Q 1 ft 3 /s v1 = = 2 = 2.865 ft/s A1 π 8 4 (12 ft) v2 =

Q 1 ft 3 /s = 2 = 7.33 ft/s A2 π 5 4 (12 ft)

δQ δWs v22 − v12 P2 P1 − = ρQ [U2 − U1 + + ( − ) + g(z2 − z1 )] ∂t ∂t 2 ρ2 ρ1 9736549.5

lbm ft 2 lb ft 2 6 m + 8.3x10 s2 s2 = (63 lbm /ft 3 )(1 ft 3 /s) [U2 − U1 + +(

(7.33ft/s)2 − (2.865ft/s)2 2

600 − 250 lbm /ft 2 ) (32.174 lbm ft/lbf s 2 ) + 32.2 ft/s2 (75ft)] 63 lbm /ft 3 ft 2 U2 − U1 = 283666.0 2 s ∆U = Cv dT

Solve for dT 283666.0 dT =

ft 2 s2

BTU ft lb 0.49 (778.17 BTUf ) (32.174 lbm ft/lbf s2 ) lbm ℉

= 23.12℉

6.42 Air flows steadily through a turbine that produces 3.5x105 ft-lbf/s of work. Using the data below at the inlet and outlet, where the inlet is 10 feet below the outlet, please calculate the heat transferred in units of BTU/hr. You may assume steady flow and ignore viscous work. Inlet diameter = 0.962 ft pressure = 150 lbf/in2 Turbine temperature = 300oF density = 0.0534 lbm/ft3 heat capacity = 0.243 BTU/lbmoF viscosity = 1.6x10-5 lbm/ft-s kinematic vicsocity = 3.06x10-4 ft2/s velocity = 100 ft/s

Outlet diameter = 0.5 ft pressure = 400 lbf/in2 temperature = 35oF density = 0.0810 lbm/ft3 heat capacity = 0.240 BTU/lbmoF viscosity = 1.5x10-5 lbm/ft-s kinematic vicsocity = 1.42x10-4 ft2/s velocity = 244 ft/s

Solution δWμ δQ δWs P ∂ − = ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV + ∂t ∂t ρ ∂t ∂t δWμ = 0 since there air is a very low viscosity fluid ∂t ∂ ∭ eρdV = 0 since the system is a steady state ∂t δQ δWs P − = ∬ (e + ) ρ(v ∙ n) dA ∂t ∂t ρ δQ δWs p p − = (e + ) ṁ − (e + ) ṁ ∂t ∂t ρ 2 ρ 1 2 2 δQ δWs v2 − v1 P2 P1 − = ṁ [U2 − U1 + + ( − ) + g(z2 − z1 )] ∂t ∂t 2 ρ2 ρ1 U2 − U1 = Cv2 T2 − Cv1 T1 = [(0.240 BTU/lbm ℉)(35℉) − (0.243 BTU/lbm ℉)(300℉)](778.17 ft lbf /BTU)(32.174 lbm ft/lbf s 2 ) = −1.61x106 ft 2 /s2 v22 − v12 (244 ft/s)2 − (100 ft/s)2 = = 24768 ft 2 /s 2 2 2 lb lb 2 400 2f 150 2f P2 P1 in − in ) (32.174 lbm ft) (144 in ) = 9.865x106 ft 2 /s 2 ( − )=( lb lbm ρ2 ρ1 lbf s2 ft 2 0.0810 m 0.0534 ft 3 ft 3

ft g(z2 − z1 ) = 32.2 2 = 322 ft 2 /s2 s (10 ft) Plug back into original equation, δQ δWs v22 − v12 P2 P1 − = ṁ [U2 − U1 + + ( − ) + g(z2 − z1 )] ∂t ∂t 2 ρ2 ρ1 δQ − 3.5x105 ft ∙ lbf /s ∂t π(0.962 ft)2 (0.0534 lbm /ft 3 )(100 ft/s) ( ) 4 [−1.61x106 ft 2 /s2 = lbf s2 ( ) 32.174 lbm ft + 24768 ft 2 /s 2 + 9.865x106 ft 2 /s2 + 322 ft 2 /s2 ] δQ lbf − 3.5x105 ft ∙ = 9.9887x105 ft ∙ lbf /s ∂t s δQ ft ∙ lbf = 1.348x106 ∂t s Convert to BTU/hr δQ ft ∙ lbf 3600 s 1 )( ) = (1.348x106 = 6.24x106 BTU/hr ∂t s hr 778.17 ft lbf /BTU

6.43 You have been asked to do an analysis of a pump that will transfer a fluid. The inlet to the pump has a diameter of 0.35 m and the pressure to the pump is 2500 kg/m-sec2. The outlet from the pump has a diameter of 0.15 m and a pressure of 6000 kg/m-sec2. The fluid that will be pumped has a viscosity of 1.09 x 10-3 kg / meter-sec and a density of 600 kg/m3. The outlet from the pump is located 5 m higher than the inlet to the pump. In your analysis assume that there is no temperature change between the inlet and outlet and that the system is running at steady state. Please determine the amount of power that the pump must add to the fluid to maintain a constant flow rate of 10 m3/sec. Solution δWμ δQ δWs P ∂ − = ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV + ∂t ∂t ρ ∂t ∂t δWμ = 0 since there is no viscous work ∂t ∂ ∭ eρdV = 0 since the system is a steady state ∂t δQ = 0 from problem statement ∂t δWs P − = ∬ (e + ) ρ(v ∙ n) dA ∂t ρ δWs p p v22 − v12 P2 − P1 ) + g(y2 − y1 )] − = ρQ (e + ) − ρQ (e + ) = ρQ [ +( ∂t ρ 2 ρ 1 2 ρ Q 10 m3 /s v1 = = = 104 m/s A1 π(0.35m)2 4 Q 10 m3 /s v2 = = = 565.85 m/s A2 π(0.15m)2 4 δWs v22 − v12 P2 − P1 ) + g(y2 − y1 )] − = ρQ [ +( ∂t 2 ρ (565.85 m/s)2 − (104 m/s)2 3 )(10 3 (600 = kg/m m /s) [ 2 6000 − 2500 kg/ms 2 + + (9.8 m/s2 )(5 m)] = 9.28x108 J/s 600 kg/m3

6.44 You have just been hired into a new manufacturing facility where you are the process engineer. You are touring the facility with your new boss and she mentions to you that there is some concern in the plant about the change in pressure (in units of lbf /ft2) in a new system you will be working on, and asks you calculate this pressure change as soon as possible before the process is started. The system will be pumping a fluid at steady state from Tank#1, which is at 30℉ with a viscosity of 7.2 lbm/ft-s, heat capacity of 0.54 BTU/lbm℉ and kinematic viscosity of 9.03 ft2/s, up to Tank#2 that is at 100℉ with a viscosity of 7.10 lbm/ft-s, heat capacity of 0.598 BTU/lbm℉ and kinematic viscosity of 0.128 ft2/s. The bottom tank is connected to the pump with a pipe that is 10 inches in diameter. The outlet to the pump is connected to a pipe that is 4 inches in diameter. The flow rate through the system is constant at 0.3 ft3/sec. At steady state the pump provides work to the fluid of 4.3x107 lbmft2/s3 and the heat transfer in the system is 4.6 x106 BTU/hr. You may assume that viscous work is negligible, and that Tank #1 is 40 feet above Tank #2, and that the system is at a constant density of density of 79.7 lbm/ft3 (the assumption of constant density is a simplification to allow the problem to be solved more easily). Solution From the problem statement, this requires the energy equation to solve. 𝛿𝑊𝜇 𝛿𝑄 𝛿𝑊𝑠 𝑃 𝜕 − = ∬ (𝑒 + ) 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 + ∭ 𝑒𝜌𝑑𝑉 + 𝜕𝑡 𝜕𝑡 𝜌 𝜕𝑡 𝜕𝑡 𝛿𝑄 𝐵𝑇𝑈 𝑙𝑏𝑚 𝑓𝑡 ℎ𝑟 𝑙𝑏𝑚 𝑓𝑡 2 6 7 ) (778.17 𝑓𝑡 𝑙𝑏𝑓 /𝐵𝑇𝑈) (32.174 ) = 3.2𝑥10 = (4.6𝑥10 )( 𝜕𝑡 ℎ𝑟 𝑙𝑏𝑓 𝑠 2 3600 𝑠 𝑠3 𝜕 ∭ 𝑒𝜌𝑑𝑉 = 0 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑠 𝑎 𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒 𝜕𝑡 𝛿𝑄 𝛿𝑊𝑠 𝑃 − = ∬ (𝑒 + ) 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 𝜕𝑡 𝜕𝑡 𝜌 2 𝛿𝑄 𝛿𝑊𝑠 𝑣22 𝑃2 𝑣21 𝑃1 )] − = 𝜌𝑄 [𝑈2 + + ( ) + 𝑔(𝑦2 − 𝜌𝑄 [𝑈1 + + ( ) + 𝑔(𝑦1 )] 𝜕𝑡 𝜕𝑡 2 𝜌 2 𝜌

𝒗𝟐𝟐 𝑷𝟐 𝝆𝟐 𝑸𝟐 [𝑼𝟐 + + ( ) + 𝒈(𝒚𝟐 )] 𝟐 𝜌

𝑙𝑏𝑓 𝑙𝑏𝑚 𝑓𝑡 3 𝐵𝑇𝑈 𝑙𝑏𝑚 𝑓𝑡 ) (100℉) (778.17 𝑓𝑡 = 79.7 3 (0.3 ) (0.598 ) (32.174 ) 𝑓𝑡 𝑠𝑒𝑐 𝑙𝑏𝑚 ℉ 𝐵𝑇𝑈 𝑙𝑏𝑓 𝑠 2 [ 2 𝑓𝑡 3 (0.3 ) 𝑠𝑒𝑐 ( ) 𝜋 4 2 2 𝑃2 𝑓𝑡 4 (12) 𝑓𝑡 + + + 32.2 2 (40 𝑓𝑡) 2 𝜌 𝑠

] 𝑙𝑏𝑚 𝑓𝑡 2 𝑓𝑡 2 𝑓𝑡 4 𝑓𝑡 = 23.91 [1497203 2 + 5.91 2 + 𝑷𝟐 (0.4114) + 1288 2 ] 2 𝑠 𝑠 𝑠 𝑙𝑏𝑓 𝑠 𝑠 2 𝑙𝑏𝑚 𝑓𝑡 𝑃2 = 23.91 [1.5𝑥106 2 + ] 𝑠 𝑠 𝜌2

𝒗𝟐𝟏 𝑷𝟏 𝝆𝟏 𝑸𝟏 [𝑼𝟏 + + ( ) + 𝒈(𝒚𝟏 )] 𝟐 𝜌

𝑙𝑏𝑓 𝑙𝑏𝑚 𝑓𝑡 3 𝐵𝑇𝑈 𝑙𝑏𝑚 𝑓𝑡 ) (30℉) (778.17 𝑓𝑡 = 79.7 3 (0.3 ) (0.54 ) (32.174 ) 𝑓𝑡 𝑠𝑒𝑐 𝑙𝑏𝑚 ℉ 𝐵𝑇𝑈 𝑙𝑏𝑓 𝑠 2 [ 2 𝑓𝑡 3 (0.3 ) 𝑠𝑒𝑐 ( ) 𝜋 10 2 2 𝑷𝟏 𝑓𝑡 4 (12) 𝑓𝑡 + + + 32.2 2 (0) 2 𝜌 𝑠

] 𝑙𝑏𝑚 𝑓𝑡 2 𝑓𝑡 2 𝑓𝑡 4 = 23.91 [405596.8 2 + 0.1513 2 + 𝑷𝟏 (0.4036) +0] 𝑠 𝑠 𝑠 𝑙𝑏𝑓 𝑠 2 𝑙𝑏𝑚 𝑓𝑡 2 𝑷𝟏 = 23.91 [4.05𝑥105 2 + ] 𝑠 𝑠 𝝆𝟏 Combine everything, 2 𝛿𝑄 𝛿𝑊𝑠 𝑣22 𝑃2 𝑣21 𝑃1 − = 𝜌2 𝑄 [𝑈2 + + ( ) + 𝑔(𝑧2 )] − 𝜌1 𝑄 [𝑈1 + + ( ) + 𝑔(𝑧1 )] 𝜕𝑡 𝜕𝑡 2 𝜌2 2 𝜌1

3.2𝑥10

7.5𝑥107

𝑙𝑏𝑚 𝑓𝑡 2 𝑙𝑏 𝑓𝑡 2 7 𝑚 + 4.3𝑥10 𝑠3 𝑠3 𝑙𝑏𝑚 𝑓𝑡 2 𝑃2 𝑙𝑏𝑚 𝑓𝑡 2 𝑃1 = 23.91 [1.5𝑥106 2 + ( )] − 23.91 [4.05𝑥105 2 + ( )] 𝑠 𝑠 𝜌 𝑠 𝑠 𝜌

𝑙𝑏𝑚 𝑓𝑡 2 𝑙𝑏𝑚 𝑓𝑡 2 𝑃2 𝑙𝑏𝑚 𝑓𝑡 2 𝑃1 6 5 ( )] ( )] = 23.91 [1.5𝑥10 + − 23.91 [4.05𝑥10 + 𝑠3 𝑠 𝑠2 𝜌 𝑠 𝑠2 𝜌

3.14𝑥106

𝑓𝑡 2 𝑓𝑡 2 𝑃2 𝑓𝑡 2 𝑃1 6 5 ( ) ( ) = 1.5𝑥10 + − 4.05𝑥10 + 𝑠2 𝑠2 𝜌 𝑠2 𝜌 𝑓𝑡 2 𝑃2 − 𝑃1 2.04𝑥10 = 𝑠2 𝜌 6

𝑃2 − 𝑃1 = 2.04𝑥106

𝑙𝑏𝑓 𝑓𝑡 2 𝑙𝑏𝑚 lbf s2 (79.7 ) ( ) = 5.05𝑥106 2 2 3 𝑠 𝑓𝑡 32.174 lbm ft 𝑓𝑡

This is a high pressure, so it is a good thing that your boss asked you to calculate it so that proper safety measures could be taken, the system probably should not be operating at this pressure and changes will need to be made to reduce it!

6.45 As part of your responsibilities as a new hire in a chemical plant, you have been asked to buy a pump. So that you know which pump to purchase, you need to calculate the work being done on the system and they you will contact the supplier to discuss the size needed. The important data is included in the table below: Position

Density (lbm/ft3)

Inlet Outlet 1 Outlet 2

55.2 53.6 51.8

Heat Capacity (Btu/lbmoF) 0.395 0.420 0.450

Area of pipe (ft2) 0.34 1.00 0.25

Volumetric Flow Rate (ft3/s) 85.7 40.0 50.0

Temperature (oF)

Pressure (lbf/ft2)

60 100 150

Distance pipe is from floor (ft)

20.0 30.0 21.5

Please calculate the work done by the system assuming steady flow of a Newtonian fluid, that viscous work is negligible and that the heat transferred to the CV from the pump is 1.40x106 Btu/hr. Solution δWμ δQ δWs P ∂ − = ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV + ∂t ∂t ρ ∂t ∂t δQ BTU lbm ft hr lbm ft 2 ) (778.17 ft lbf /BTU) (32.174 ) ( ) = (1.4x106 = 9736549.5 ∂t hr lbf s2 3600 s s3 δQ δWs P − = ∬ (e + ) ρ(v ∙ n) dA ∂t ∂t ρ δQ δWs p p p − = (e + ) ṁ2 + (e + ) ṁ3 − (e + ) ṁ 1 ∂t ∂t ρ 2 ρ 3 ρ 1 δQ δWs v22 P2 v32 P3 − = (U2 + + gz2 + ) ṁ2 + (U3 + + gz3 + ) ṁ3 ∂t ∂t 2 ρ2 2 2 ρ3 3 2 v1 P1 − (U1 + + gz1 + ) ṁ 1 2 ρ1 1

v1 =

Q1 85.7 ft 3 /s = = 252 ft/s A1 0.34 ft 2

v2 =

Q2 40 ft 3 /s = = 40 ft/s A2 1.0 ft 2

Q3 50 ft 3 /s v3 = = = 200 ft/s A3 0.25 ft 2

0.0 4.0 1.5

Position

Density (lbm/ft3)

Inlet Outlet 1 Outlet 2

55.2 53.6 51.8

Heat Capacity (Btu/lbmoF) 0.395 0.420 0.450

Area of pipe (ft2) 0.34 1.00 0.25

Volumetric Flow Rate (ft3/s) 132.5 40.0 50.0

Temperature (oF) 60 100 150

Pressure (lbf/ft2)

Distance pipe is from floor (ft)

20.0 30.0 21.5

𝑣12 𝑃1 𝑣12 𝑃1 (𝑈1 + + 𝑔𝑧1 + ) 𝑚̇1 = (𝐶𝑣1 𝑇1 + + 𝑔𝑧1 + ) 𝜌1 𝑣1 𝐴1 2 𝜌1 1 2 𝜌1 1 (252 𝑓𝑡/𝑠)2 𝑓𝑡 𝑙𝑏𝑓 𝐵𝑇𝑈 𝑙𝑏𝑚 𝑓𝑡 ) (60℉) (778.17 = [(0.395 ) (32.174 ) + 𝑙𝑏𝑚 ℉ 𝐵𝑇𝑈 𝑙𝑏𝑓 𝑠 2 2 2 2 20 𝑙𝑏𝑓 /𝑓𝑡 (32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 ) + (32.2 𝑓𝑡/𝑠 2 )(0 𝑓𝑡) + ] (55.2 𝑙𝑏𝑚 55.2 𝑙𝑏𝑚 /𝑓𝑡 3 1 /𝑓𝑡 3 )(252 𝑓𝑡/𝑠)(0.34 𝑓𝑡 2 ) 𝑓𝑡 2 𝑓𝑡 2 𝑓𝑡 2 𝑙𝑏𝑚 ) = [593373.15 2 + 31752 2 + 0 + 11.66 2 ] (4729.6 𝑠 𝑠 𝑠 𝑠 𝑙𝑏𝑚 𝑓𝑡 2 = 2.95655𝑥109 𝑠3 𝑣22 𝑃2 𝑣22 𝑃2 (𝑈2 + + 𝑔𝑧2 + ) 𝑚̇2 = (𝐶𝑣2 𝑇2 + + 𝑔𝑧2 + ) 𝜌2 𝑣2 𝐴2 2 𝜌2 2 2 𝜌2 2 (40 𝑓𝑡/𝑠)2 𝑓𝑡 𝑙𝑏𝑓 𝐵𝑇𝑈 𝑙𝑏𝑚 𝑓𝑡 ) (100℉) (778.17 = [(0.420 ) (32.174 ) + 𝑙𝑏𝑚 ℉ 𝐵𝑇𝑈 𝑙𝑏𝑓 𝑠 2 2 2 2 30 𝑙𝑏𝑓 /𝑓𝑡 (32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 ) + (32.2 𝑓𝑡/𝑠 2 )(4 𝑓𝑡) + ] (53.6 𝑙𝑏𝑚 53.6 𝑙𝑏𝑚 /𝑓𝑡 3 1 /𝑓𝑡 3 )(40 𝑓𝑡/𝑠)(1.0 𝑓𝑡 2 ) 𝑓𝑡 2 𝑓𝑡 2 𝑓𝑡 2 𝑓𝑡 2 𝑙𝑏𝑚 ) = [1051547.3 2 + 800 2 + 128.8 2 + 18.0 2 ] (2144 𝑠 𝑠 𝑠 𝑠 𝑠 𝑙𝑏𝑚 𝑓𝑡 2 = 2.2565𝑥109 𝑠3

0.0 4.0 1.5

𝑣32 𝑃3 𝑣32 𝑃3 (𝑈3 + + 𝑔𝑧3 + ) 𝑚̇3 = (𝐶𝑣3 𝑇3 + + 𝑔𝑧3 + ) 𝜌3 𝑣3 𝐴3 2 𝜌3 3 2 𝜌3 3 (200 𝑓𝑡/𝑠)2 𝑓𝑡 𝑙𝑏𝑓 𝐵𝑇𝑈 𝑙𝑏𝑚 𝑓𝑡 ) (150℉) (778.17 = [(0.450 ) (32.174 ) + 𝑙𝑏𝑚 ℉ 𝐵𝑇𝑈 𝑙𝑏𝑓 𝑠 2 2 2 2 21.5 𝑙𝑏𝑓 /𝑓𝑡 (32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 ) + (32.2 𝑓𝑡/𝑠 2 )(1.5 𝑓𝑡) + ] (51.8 𝑙𝑏𝑚 51.8 𝑙𝑏𝑚 /𝑓𝑡 3 1 /𝑓𝑡 3 )(200 𝑓𝑡/𝑠)(0.25 𝑓𝑡 2 ) 𝑓𝑡 2 𝑓𝑡 2 𝑓𝑡 2 𝑓𝑡 2 𝑙𝑏𝑚 ) = [1689986.8 2 + 20,000 2 + 48.3 2 + 13.35 2 ] (2590 𝑠 𝑠 𝑠 𝑠 𝑠 𝑙𝑏𝑚 𝑓𝑡 2 = 4.43𝑥109 𝑠3

9736549.5

𝑙𝑏𝑚 𝑓𝑡 2 𝛿𝑊𝑠 𝑙𝑏 𝑓𝑡 2 𝑙𝑏 𝑓𝑡 2 𝑙𝑏 𝑓𝑡 2 9 𝑚 9 𝑚 9 𝑚 − = 4.43𝑥10 + 2.2565𝑥10 − 2.95655𝑥10 𝑠3 𝜕𝑡 𝑠3 𝑠3 𝑠3 𝛿𝑊𝑠 𝑙𝑏 𝑓𝑡 2 9 𝑚 − = 3.73𝑥10 𝜕𝑡 𝑠3 𝛿𝑊𝑠 𝑙𝑏𝑚 𝑓𝑡 2 = −3.73𝑥109 𝜕𝑡 𝑠3

Could have also expressed as, 𝑙𝑏𝑓 𝑓𝑡 𝛿𝑊𝑠 𝐵𝑡𝑢 = −1.16𝑥108 = −1.5𝑥105 = −5.38𝑥108 𝐵𝑡𝑢/ℎ𝑟 𝜕𝑡 𝑠 𝑠

6.46 A Newtonian fluid is being pumped through a pumping station that has fluid in inlet and outlet pipes. The inlet fluid is at a temperature of 80oF, density of 50 lbm/ft3, viscosity of 1.80 x 10-3 lbm/ft sec, heat capacity of 0.58 Btu/lbmoF and kinematic viscosity of 3.60 x 10-5 ft2/sec. The pressure at the inlet is 300 psi. The outlet fluid is at a temperature of 100oF, density of 49.6 lbm/ft3, viscosity of 1.30 x 10-3 lbm/ft sec, heat capacity of 0.61Btu/lbmoF and kinematic viscosity of 2.62 x 10-5 ft2/sec. The pressure at the outlet is 750 psi. The outlet is 167 ft above the inlet, and the pump provides work to the fluid at 1 x 108 lbm ft2/s3. The inlet has a diameter of 0.5 feet, but the outlet diameter is unknown. The inlet velocity is 86 ft/s and the outlet velocity is 184 ft/s. For this analysis you may assume that the fluid undergoes no viscous work, and that the system is at steady state. Find the number of Btu’s added to the system after 4 hours. Solution δWμ δQ δWs P ∂ − = ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV + ∂t ∂t ρ ∂t ∂t ∂ ∭ eρdV = 0 since the system is at steady state ∂t δWs − = 1x108 lbm ft 2 /s3 ∂t δQ δWs p p − = (e + ) ṁ − (e + ) ṁ ∂t ∂t ρ 2 ρ 1 δQ δWs v22 P2 v12 P1 − = ρ2 v2 A2 [U2 + + ( ) + g(z2 )] − ρ1 v1 A1 [U1 + + ( ) + g(z1 )] ∂t ∂t 2 ρ2 2 ρ1 We are not given the exit diameter, we could calculate it, but we don’t need to since we are told that the system is at steady state so we could just use, ρ2 v2 A2 = ρ1 v1 A1 Or we could calculate it, πD22 π(0.5ft)2 ) = (50 lbm /ft 3 )(86 ft/s) ( ) 4 4 πD22 = 0.0925 ft 2 4

(49.6 lbm /ft 3 )(184 ft/s) (

D2 = 0.343 ft

Or you could simply say that ṁ1 = ṁ2 = ṁ, and not bother to calculate D2. δQ δWs v22 − v12 P2 P1 − = ṁ [U2 − U1 + + ( − ) + g(z2 − z1 )] ∂t ∂t 2 ρ2 ρ1 Plugging in the values, δQ + 1x108 lbm ft 2 /s3 ∂t = (50 lbm /ft 3 )(86 ft π(0.5ft)2 𝐵𝑡𝑢 ) (100℉) /s) ( ) [{(0.61 4 𝑙𝑏𝑚 ℉ 𝑓𝑡 𝑙𝑏𝑓 𝐵𝑡𝑢 lbm ft ) (80℉)} (778.17 ) − (0.58 ) (32.174 𝑙𝑏𝑚 ℉ 𝐵𝑡𝑢 lbf s 2 (184 ft/s)2 − (86 ft/s)2 + 2 750 lbf /in2 300 lbf /in2 144 in2 lbm ft (32.174 ) +( − ) ( ) 49.6 lbm /ft 3 50 lbm /ft 3 ft 2 lbf s 2 + 32.2 ft/s 2 (167 ft)] δQ lbm ft 2 lbm ft 2 ft 2 ft 2 ft 2 + 1x108 = 844.3 [365537.88 + 13230 + 42257.95 + 5377.4 ] ∂t s3 s s2 s2 s2 s2 δQ lb ft 2 lbm ft 2 8 m + 1x10 = 844.3 [426403.23 2 ] ∂t s3 s s δQ lbm ft 2 lb ft 2 8 m + 1x108 = 3.6 x 10 ∂t s3 s3 2 2 δQ lbm ft lbf s BTU Btu ) = 10380 = 2.6x108 ( )( 3 ∂t s 32.174 lbm ft 778.17 ft lbf s 10380

Btu 60 sec 60 min x x x 4 hr = 1.49x108 Btu s min hr

6.47 Venturi meters are commonly used to measure the flow rate of gases and liquids where the gas or liquid is considered frictionless and incompressible under the conditions of the flow. Air at a constant temperature of 75℉, is flowing steadily from left to right (as illustrated in the figure by the arrows). At position labeled 1 the diameter is 3 inches and the absolute pressure is 14 psi. At the position labeled 2, the diameter is 2 inches and the absolute pressure is 12 psi. Assuming that the flow is Newtonian, isothermal and inviscid with no change in internal energy, please calculate the volumetric flow rate of the air. (Note: at these pressures, air can be considered an incompressible fluid.) 1

Solution P1 v12 P2 v22 + + gy1 = + + gy2 ρ 2 ρ 2 Q = v1 A1 = v2 A2

A1 =

A2 =

v1 =

Q A1

v2 =

Q A2

1 ft 2 π (3in x 12 in) 4

1 ft 2 π (2 in x 12 in) 4

= 0.05 ft 2

= 0.022 ft 2

𝑙𝑏 144 𝑖𝑛2 𝑙𝑏 𝑃1 = 14 2 𝑥 = 2016 2 2 𝑖𝑛 𝑓𝑡 𝑓𝑡 𝑃2 = 12

𝑙𝑏 144 𝑖𝑛2 𝑙𝑏 𝑥 = 1728 2 2 2 𝑖𝑛 𝑓𝑡 𝑓𝑡

P1 v12 P2 v22 + + gy1 = + + gy2 ρ 2 ρ 2 y1 = y2 P1 v12 P2 v22 + = + ρ 2 ρ 2 v22 − v12 P1 − P2 = ρ ( ) 2 Plug in values for v1 and v2 in terms of Q which is what we are solving for, Q 2 Q 2 (A ) − (A ) 1 ) P1 − P2 = ρ ( 2 2 Solve for Q: P1 − P2 Q 2 Q 2 )=( ) −( ) 2( ρ A2 A1

𝑙𝑏 𝑙𝑏 − 1728 2 2 1 1 𝑓𝑡 𝑓𝑡 2 = 𝑄2 [ 2 − 2 ] 2 𝑙𝑏𝑓 𝑠 𝐴2 𝐴1 𝑙𝑏 ) 0.075 𝑚3 ( 32.174 𝑙𝑏 𝑓𝑡 𝑓𝑡 ( ) 𝑚 2016

𝑙𝑏 𝑙𝑏 − 1728 2 𝑓𝑡 2 𝑓𝑡 ) 2( 𝑙𝑏𝑓 𝑠 2 𝑙𝑏𝑚 𝑓𝑡 2 ) 0.075 3 ( 247096 2 𝑓𝑡 32.174 𝑙𝑏𝑚 𝑓𝑡 𝑠 = 12.2 𝑓𝑡 3⁄ 𝑄= =√ 𝑠 1 1 1 1666 [ − ] 2 2 (0.05 𝑓𝑡 2 )2 𝑓𝑡 4 √ (0.022 𝑓𝑡 ) 2016

6.48 Water at 293 K is flowing steadily from a hose with a diameter of 1.6 cm. You cover the hose outlet with your thumb causing the area of the opening to be 25% of what it was originally, resulting a thin jet of high-speed water to emerge at a velocity of 3 m/s. If the hose is held upward (perfectly vertical), and the pressure of the water in the hose just before it exits is 40 kPa gauge, what is the height above the nozzle opening that the water jet will achieve? You may assume inviscid, isothermal flow with no change in internal energy and no shaft work occurs in the control volume. Solution Assumptions: incompressible, Newtonian fluid, inviscid, isothermal flow with no change in internal energy and no shaft work P1 v12 P2 v22 + + gy1 = + + gy2 ρ 2 ρ 2 v2 = 0 at the top of the jet, y1 = 0 a the top of hose and bottome of jet 𝑃𝑔𝑎𝑢𝑔𝑒 = 𝑃𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 − 𝑃𝐴𝑇𝑀 = 40 𝑘𝑃𝑎 (𝑔𝑎𝑢𝑔𝑒) 𝑦2 =

(3 𝑚/𝑠)2 𝑃1 − 𝑃2 𝑣12 40𝑥103 𝑘𝑔/𝑚𝑠 2 + = + = 4.5 𝑚 𝜌𝑔 2𝑔 (998.2 𝑘𝑔/𝑚3 )(9.8 𝑚/𝑠 2 ) 2(9.8 𝑚/𝑠 2 )

6.49 A system is pumping a fluid at steady state from tank #1 to tank #2. The data for the two tanks is included in the table below. Tank #1 is connected to the pump with a pipe that is 10 inches in diameter and is at a pressure of 300 lbf/ft2 as it enters the pump. The outlet to the pump is connected to a pipe that is 4 inches in diameter and at a pressure of 850 lbf/ft2. The flow rate through the system is constant at 0.3 ft3/min. Please calculate the work that the pump provides to the fluid if the heat transfer in the system is 4.65x103 BTU/hr, and tank #2 is 100 ft above tank #1. Viscous work is assumed negligible. (You may assume that the pump and tank#1 are at the same height.) Tank

Density (lbm/ft3)

Viscosity (lbm/ft s)

Heat Capacity (BTU/lbmoF) 0.563

Temperature (oF)

1.4

Kinematic Viscosity (ft2/s) 1.77x10-2

79.1

78.7

0.6

0.762x10-2

0.580

Solution δWμ δQ δWs P ∂ − = ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV + ∂t ∂t ρ ∂t ∂t δQ BTU lb ft hr lbm ft 2 m = (4.65x103 ) (778.17 ft lbf /BTU) (32.174 )( ) = 3.23x104 2 ∂t hr lbf s 3600 s s3 ∂ ∭ eρdV = 0 since the system is a steady state ∂t δQ δWs P − = ∬ (e + ) ρ(v ∙ n) dA ∂t ∂t ρ δQ δWs v22 P2 v22 P1 − = ρ2 Q [U2 + + ( ) + g(z2 )] − ρ1 Q [U1 + + ( ) + g(z1 )] ∂t ∂t 2 ρ2 2 ρ1 𝛒𝟐 𝐯𝟐 𝐀𝟐 [𝐔𝟐 +

𝐯𝟐𝟐 𝐏𝟐 + ( ) + 𝐠(𝐳𝟐 )] 𝟐 𝛒𝟐

= 78.7

lbm ft 3 min BTU lbf lbm ft (0.3 )( ) (0.598 ) (80℉) (778.17 ft ) (32.174 ) ft 3 min 60 sec lbm ℉ BTU lbf s2

[ 2 ft 3 min ) (0.3 )( min 60 sec ( ) 2 π 4 850lbf ( ) ft 2 lbm ft ft 4 12 ft 2 + + (32.174 ) + 32.2 2 (y2 ) 2 lb 2 lbf s s 78.7 m ft 3 ] lbm ft 2 ft 2 ft 2 ft lbm ft 2 6 −3 (100 = 0.3935 [1.2x10 2 + 1.64x10 + 347.5 2 + 32.2 2 ft)] = 471319.5 2 s s s s s s s2

𝛒𝟏 𝐯𝟏 𝐀𝟏 [𝐔𝟏 +

𝐯𝟏𝟐 𝐏𝟏 + ( ) + 𝐠(𝐳𝟏 )] 𝟐 𝛒𝟏

= 79.1

lbm ft 3 min BTU lbf lbm ft (0.3 )( ) (0.563 ) (60℉) (778.17 ft ) (32.174 ) ft 3 min 60 sec lbm ℉ BTU lbf s2

[ 2 ft 3 min ) (0.3 )( min 60 sec ( ) 2 π 10 300lbf ( ) ft 2 lbm ft ft 4 12 ft 2 (0 ft) + + (32.174 2 ) + 32.2 s 2 lbm 2 lb s f 79.1 3 ft ] lbm ft 2 ft 2 ft 2 lbm ft 2 −5 = 0.3955 [845744.5 2 + 4.2x10 + 122 + 0] = 334491.95 s s s2 s2 s s2 Combine everything, 3.23x104

lbm ft 2 δWs lbm ft 2 lbm ft 2 lbm ft 2 + = (471319.5 − 334491.95 ) = 136827.55 3 2 2 s ∂t s s s s s s2

δWs lb ft 2 = 1.045x105 m3 ∂t s

Chapter 7 Instructor Only Problems 7.21 A thin coating is to be applied to both sides of a piece of thin plastic that is being mechanically transported. The plastic is 4.5 μm thick and 0.0254 meters wide and is very long. We want to coat a specific length of the plastic that is 1 meter. This thin plastic will break if the force applied exceeds 20 lbf. The thin coating is made of transporting it mechanically through a narrow gap that determines the thickness of the coating. The plastic is centered in the gap with a clearance of 1. 0 μm on the top and bottom (as a result the coating is 1. 0 μm thick on both the top and bottom). The incompressible Newtonian fluid coating fluid a temperature of 80oF, density of 52.5 lbm/ft3, heat capacity of 0.453 BTU/lbmoF, viscosity of 6.95x10-3 lbm/ft-s, and kinematic viscosity of 1.33x10-4 ft2/s and completely fills the space between the plastic and gap for a length of 0.75 inches along the plastic. Calculate the velocity with which the tape can be transported through the gap using the maximum force (so that the tape does not break).

Solution Area = length x width = (1 meter)(0.0254 meter) = 0.0254 m2 = 0.2734 ft2 Since we have a Newtonian fluid we can use Newton’s law: F dv =τ=μ A dy Rearrange, solving for dv and since the tape has two sides we must multiply the area by 2, and integrating, v

∫ dv = 0

y F ∫ dy 2μA 0

1.0 μm = 3.28x10-6 ft on each side

(20 lbf )(3.28x10−6 ft)(32.174 lbm ft/lbf s2 ) F ft (y) = = 0.55 = 33.08 ft/min −3 2 2μA 2(6.95x10 lbm /ft s)(0.2754 ft ) s

7.22 A viscometer is used to measure viscosity when the fluid in common laboratory settings. A rheometer is more sophisticated instrement in that it measures the way a liquid flows in response to an applied force. A rheometer is used to study fluids which cannot be defined by a single viscosity value and thus measures the rheology of the fluid. In a concentric cylinder (couette geometry) rheometer, the rotating of the cup produces a strain on the sample, which in turn applies a torque on the center cylinder. A transducer measures the magnitude of the torque, and (based on the geometry of the system) determines the resultant shear stress from the applied strain. Figure 1 shows the geometry of the experiment. The fluid lies between R 0 and R i where R i = 16.00 mm and R 0 = 17.00 mm. The gap between the two cylinders can be modeled as two parallel plates separated by a fluid. The torque is the force times the moment arm, which is the radius 𝑅𝑖 of the inner cylinder. The tangential velocity is equal to the angular velocity times the radius. The inner cylinder is rotated at 6000 rpm and the torque is measured to be 0.03 Nm. The length of the inner cylinder is 33.4 mm. Determine the viscosity of the fluid. We can assume that the inner cylinder is completely submerged in the fluid, that the fluid is Newtonian and that the viscous end effects of the two ends of the inner cylinder are negligible. Solution We are given that “The torque is the force times the moment arm, which is the radius R i of the inner cylinder. The tangential velocity is equal to the angular velocity times the radius.” In equation format these statements become: Torque = T = FR i Velocity = v = ωR We know that the wetted surface area of the inner cylinder is: A = 2πRL. As a result, we can express the torque as: μ(2πRL)(ωR) T = FR = ℓ We can rewrite this equation noting that, if the cylinder length is L and the number of revolutions per unit time is ṙ , which is expressed here as revolutions per minute (rpm). One note, the angular distance traveled during one rotation is (2π rad) and the rpm is ω = 2πṙ . Thus, T = FR = τAR =

μv(2πRL)R μωR(2πRL)R μ(2πṙ )R(2πRL)R μ4π2 R3 = = = ṙ L ℓ ℓ ℓ ℓ

Rearranging and solving for viscosity, Tℓ μ= 2 3 4π R ṙ L m (0.03 Nm)(1 mm) ( 1000 mm) = 3 1m rotations min m (33.4 mm) ( ) (6000 4π2 (16 mm)3 ( ) ( ) 3 3 minute 60 sec 1000 mm) 1000 mm Ns = 0.0555 2 = 55.54 centipoise m

7.23 A Non-Newtonian fluid is being sheared between two parallel plates. The top plate is moving at a velocity of 0.09 ft/sec and the bottom plate is stationary. The two plates are separated by a distance of 0.008 ft. The fluid is at 100℉, with a density of 84.3 lbm/ft3, heat capacity of 0.0331 BTU/lbm℉, and surface tension of 2.5x10-3 lbf /inch. If the force on the fluid by the moving plate is 2.3 𝑙𝑏𝑓 and the area of contact is 0.05 ft2, and the yield stress of 38.45 lbf/ft2, please calculate the viscosity of this fluid in this situation. Solution Begin with the appropriate equation for a Non-Newtonian fluid, incorporating the yield stress. τ=μ

dv + τ0 dy

F dv = μ + τ0 A dy 2.3 𝑙𝑏𝑓 𝑙𝑏𝑓 𝑠 2 𝑙𝑏𝑓 0.09 ft/s = μ ( ) + 38.45 2 2 0.05 ft 0.008 ft 32.174 𝑙𝑏𝑚 𝑓𝑡 ft 46

𝑙𝑏𝑓 𝑙𝑏𝑓 𝑠 𝑙𝑏𝑓 = μ(0.35) + 38.45 2 2 ft 𝑙𝑏𝑚 𝑓𝑡 ft μ = 21.57

𝑙𝑏𝑚 ft − s

7.24 A Non-Newtonian fluid is being sheared between two parallel plates. The top plate is moving at a velocity of 0.09 ft/sec and the bottom plate is stationary. The two plates are separated by a distance of 0.008 ft. The fluid is at 100℉, with a density of 84.3 lbm/ft3, heat capacity of 0.0331 BTU/lbm℉, surface tension of 2.5x10-3 lbf /inch, and kinematic viscosity of 0.256 ft2/sec. If the force on the fluid by the moving plate is 6 𝑙𝑏𝑓 and the area of contact is 0.05 ft2, please calculate the yield stress in this situation. Solution Begin with the appropriate equation for a Non-Newtonian fluid, dv τ = μ + τ0 dy F dv = μ + τ0 A dy 6 𝑙𝑏𝑓 𝑙𝑏𝑓 𝑠 2 0.09 ft/s 3 )(3.56 2 (84.3 = lbm /ft ft /s) ( ) + τ0 0.05 ft 2 0.008 ft 32.174 𝑙𝑏𝑚 𝑓𝑡 120

𝑙𝑏𝑓 𝑙𝑏𝑓 = 104.93 2 + τ0 2 ft ft τ0 = 15.07

𝑙𝑏𝑓 ft 2

7.25 A Bingham plastic is being sheared between two parallel plates. This material has a yield stress of 90 lbf/ft2, a density of 71 lbf/ft3. If the velocity of the upper plate is 0.041 ft/s and weight 0.5 lbf. The plates are 0.5 inches apart and have an area of 1.2 ft2 what is the viscosity of this fluid? (Hint: remember that the units of viscosity are lbm/ft s). Solution 𝜏𝑦𝑥 = 𝜇

𝑑𝑣𝑥 + 𝜏0 𝑑𝑦

𝐹 𝑑𝑣𝑥 =𝜇 + 𝜏0 𝐴 𝑑𝑦 7 𝑙𝑏𝑓 𝑙𝑏𝑓 𝑠 2 𝑙𝑏𝑓 0.041 𝑓𝑡/𝑠 = 𝜇 + 3.3 2 2 0.5 0.75 𝑓𝑡 32.174 𝑙𝑏𝑚 𝑓𝑡 𝑓𝑡 12 𝑓𝑡 0.5 7 𝑙𝑏𝑓 𝑙𝑏𝑓 12 𝑓𝑡 32.174 𝑙𝑏𝑚 𝑓𝑡 = 33.7 𝑙𝑏𝑚 𝜇=( − 3.3 ) 0.75 𝑓𝑡 2 𝑓𝑡 2 0.24 𝑓𝑡/𝑠 𝑙𝑏𝑓 𝑠 2 𝑓𝑡 𝑠

Chapter 8 Instructor Only Problems

8.20 You have been asked to calculate the density of an incompressible Newtonian fluid in steady flow that is flowing continuously at 250℉ down a 2500 foot pipe with a constant diameter of 4 inches and a volumetric flow rate of 2.5 ft3/s. The only fluid properties known is that the kinematic viscosity is 7.14x10-6 ft2/sec and a surface tension is 0.0435 N/m. Assuming that the flow is laminar and fully developed with a pressure drop of 256 lbf/ft2, please calculate the density of this fluid if the no-slip boundary condition applies. (Hint: the units of density are lbm/ft3.) Solution dP 32μvavg = dx D2 ∆P 32μvavg = L D2 3 Q 2.5 ft /s vavg = = 2 = 28.65 ft/s A π 4 4 (12 ft) Rearrange the Hagen-Poiseuille Equation and solve for viscosity: 2 4 (256 lbf /ft 2 ) ( ft) ∆PD2 lbf s 12 μ= = = 1.24x10−5 2 32Lvavg 32(2500 ft)(28.65 ft/s) ft Realizing the relationship between density, viscosity and kinematic viscosity: lb s 1.24x10−5 f2 μ ft (32.174 lbm ft) = 55.93 lb /ft 3 ρ= = m ν 7.14x10−6 ft 2 /s lbf s2 −

8.21 Determine the pressure drop across the ends of a pipe that is 10 meters long, 0.1 meters in diameter through which is flowing a Newtonian oil with a viscosity of 6 Pa-s and a flow rate of 8.5 cubic meters per hour. Solution −

𝑑𝑃 8𝜇𝑣𝑎𝑣𝑔 = 𝑑𝑥 𝑅2

∆𝑃 8𝜇𝑣𝑎𝑣𝑔 = 𝐿 𝑅2 𝜋𝐷2 𝑄 = 𝑣𝐴 = 𝑣 ( ) = 𝑣𝜋𝑅 2 4 ∆𝑃 =

8𝐿𝜇𝑄 8(10 𝑚)(6 𝑃𝑎 ∙ 𝑠)(8.5 𝑚3 ⁄ℎ𝑟)(1 ℎ𝑟 ⁄3600𝑠𝑒𝑐 ) = = 5.77𝑥104 𝑃𝑎 𝜋𝑅 4 𝜋(0.05𝑚)4

8.22 Please calculate the density of an incompressible, Newtonian fluid in steady flow that is flowing continuously at 250℉ down a 2500 foot pipe with a constant diameter of 4 inches and a volumetric flow rate of 2.5 ft3/s. This fluid has a heat capacity of 0.527 BTU/lbm℉, a kinematic viscosity of 7.14x10-6 ft2/sec and a surface tension of 0.0435 N/m. Assuming that the flow is laminar and fully developed with a pressure drop of 256 lbf/ft2, please calculate the density of this fluid if the noslip boundary condition applies. Solution Assumptions: incompressible, Newtonian, Steady, Continuous, Fully Developed, Laminar and the No-Slip Boundary Condition suggests the use of the Hagen-Poiseuille Equation. −

dP 32μvavg = dx D2

∆P 32μvavg = L D2 Q 2.5 ft 3 /s vavg = = 2 = 28.65 ft/s A π 4 4 (12 ft) Rearrange the Hagen-Poiseuille Equation and solve for viscosity: 2 2) 4 2 (256 lb /ft ( ft) f ∆PD lbf s 12 μ= = = 1.24x10−5 2 32Lvavg 32(2500 ft)(28.65 ft/s) ft Realizing the relationship between density, viscosity and kinematic viscosity: lb s 1.24x10−5 f2 μ ft (32.174 lbm ft) = 55.93 lb /ft 3 ρ= = m −6 ν 7.14x10 ft 2 /s lbf s2

8.23. Water at 293 K, heat capacity of 4182 J/kg-K, density of 998.2 kg/m3, kinematic viscosity of 9.95 x10-7 m2/s and viscosity of 9.93 x10-4 Pa-s in fully developed, laminar flow is flowing continuously and steadily through a tube with a diameter of 10 cm and velocity of 45 m/s where the pressure drop is measured to be 15,000 Pa, please estimate the length of the pipe over which the pressure drop was determined.

Solution

−

dP 32μvavg = dx D2

∆P 32μvavg = L D2 2 m ) ∆PD 100cm = 104.9 m L= = −4 32μvavg 32(9.93 x10 Pa ∙ s)(45 m/s) 2

(15,000 Pa) (10 cm x

8.24 Benzene flows steadily and continuously at 100oF through a 4500 ft horizontal pipe with a constant cross-sectional area of 12.56 in2. The pressure drop across the pipe under these conditions is 300 lbf/ft2. Assuming fully developed, parabolic, laminar, incompressible flow, calculate the maximum velocity and volumetric flow rate of this fluid.

Solution 𝜋𝐷2 𝐴= 4 So, 𝐷=√

4𝐴 4(12.56 𝑖𝑛2 ) √ = = 4 𝑖𝑛 𝜋 𝜋 −

dP 32μvavg = dx D2

∆P 32μvavg = L D2 2

vavg =

2 4 (300 𝑙𝑏𝑓 /𝑓𝑡 2 ) (12 𝑓𝑡)

∆PD lbm ft = 32.174 = 22.56 𝑓𝑡/𝑠 32𝜇𝐿 32 (3.3𝑥10−4 𝑙𝑏𝑚 ) (4500 𝑓𝑡) lbf s 2 𝑓𝑡 𝑠 12.56 𝑖𝑛2 (22.56 Q = vavg 𝐴 = 𝑓𝑡/𝑠) ( ) = 1.97 𝑓𝑡 3 /𝑠 2 2 144 𝑖𝑛 /𝑓𝑡 v𝑚𝑎𝑥 = 2vavg = 45 𝑓𝑡/𝑠

Chapter 9 Instructor Only Problems 9.25 A Newtonian fluid is flowing through a cylindrical conduit. Beginning with the appropriate form of the Navier-Stokes Equations (see Appendix E), derive the following equation, 0 = −r

∂P ∂ + (r τrz ) ∂z ∂r

State the reason why you eliminated any terms in your derivation. 9.26 A film in free stream flow is flowing down an inclined plan as shown in the figure. Beginning with the appropriate form of the Navier-Stokes equations, please derive the following equation: ∂ τ + ρgsin(θ) = 0 ∂y yx As always, please state all assumptions you make in your derivation. 9.27 An incompressible Newtonian fluid is confined between two concentric circular cylinders of infinite length, a solid inner cylinder of radius R, and a hollow stationary outer cylinder of radius Ro. See the figure below where the z-axis is coming out of the page. The inner cylinder rotates with an angular velocity ωi . The flow is steady, laminar and two-dimensional in the rθ-plane. The flow is also rotationally symmetric, meaning that nothing is a function of coordinate θ (so vθ and P are functions of the radius, r only) and that gravity can be ignored. We can also assume that the flow is circular, meaning that the velocity component vr = 0 everywhere. Please derive an exact expression for the velocity component vθ as a function of the radius r and the other parameters in the problem.

i R0

Liquid: ,

Rotating inner cylinder Stationary outer cylinder

9.28 Beginning with the appropriate form of the Navier-Stokes Equation, please derive an expression for the velocity of a fluid flowing through a vertical pipe with diameter equal to D. For this derivation, please assume steady, incompressible, fully developed flow and that the fluid is not open to the atmosphere. Please be sure to indicate the reason why you eliminate any terms in the original equation as you proceed with your derivation. (Hint: assume that at the center of the pipe, r = 0) 9.29 Fluid Direction

𝜃

A Newtonian fluid is flowing down an incline between two vertical, infinitely long parallel plates separated by a distance h as shown in the figure. The plate on the right is moving up with a velocity of VT and the plate on the left is moving up with a velocity of VB . The velocities of the plates are different such that VT ≠ VB . The fluid is not open to the atmosphere and is flowing in one direction only as shown in the figure at an angle 𝜃 with the y-axis. Assuming steady, incompressible, fully developed, laminar flow, please derive an equation for the velocity of the fluid. You must use the coordinate system shown in the figure in your solution to the problem (which means that you cannot draw or make another coordinate system).

(1) Using the appropriate form of the Navier-Stokes Equations, please derive equations for the velocity of Fluid 1 and the velocity of Fluid 2 assuming that both fluids are in fully developed, continuous, incompressible, adiabatic, isothermal flow. You may also assume that the no-slip boundary condition applies. (2) Please draw the velocity profiles for both Fluid 1 and Fluid 2 in the figure provided below. Be careful in your drawing so it is clear. 9.32. Consider steady, incompressible, laminar, fully developed flow of a Newtonian fluid in an infinitely long, stationary pipe of diameter D inclined at an angle of 𝛽 to the horizontal plane as shown in the figure for problem 9.24. You may assume that the No-Slip boundary condition applies. (a) Show that the continuity equation holds in this system, (b) Using the appropriate form of the Navier-Stokes Equation, derive an expression to describe the velocity of this fluid. Be sure to indicate the reason why you eliminate any terms in the original equation as your proceed with your derivation.

9.33 V0 Fluid 2

h2 Interface

Fluid 1

Two immiscible viscous fluids are sandwiched between two infinitely long and wide parallel flat plates as shown in the figure above. Because the fluids are immiscible they do not mix at any point. At time t = 0, the fluids and the plates on either side are stationary. The upper plate is set in motion at a velocity V0, and the bottom plate remains stationary at all times. The flow is incompressible, parallel, fully developed and laminar. The flow is caused solely by the viscous effects created by the moving upper plate. You may ignore gravity, and assume that the fluids are not open to the atmosphere and that the interface between the fluids is horizontal at all times. (a) In the figure above please draw the velocity profiles for both Fluid 1 and Fluid 2 in this system. (b) Please derive equations for the velocity profiles of both fluids when the flow is at steady state.

9.25 A Newtonian fluid is flowing through a cylindrical conduit. Beginning with the appropriate form of the Navier-Stokes Equations (see Appendix E), derive the following equation, 0 = −r

∂P ∂ + (r τrz ) ∂z ∂r

State the reason why you eliminated any terms in your derivation.

Solution Begin with the z-direction equation in cylindrical coordinates, ∂vz ∂vz vθ ∂vz ∂vz ∂P 1 ∂ ∂vz 1 ∂2 vz ∂2 vz )=− (r )+ 2 2 + 2) ρ( + vr + + vz + ρg z + μ ( ∂t ∂r r ∂θ ∂z ∂z r ∂r ∂r r ∂θ ∂z To obtain the desired equation we must make the following assumptions: ∂vz Steady State: =0 ∂t ∂vz vθ ∂vz No velocity in the r or θ directions: vr = =0 ∂r r ∂θ ∂vz ∂2 vz Fully developed flow: vz = =0 ∂z ∂z 2 Gravity is negligable: ρg z = 0 1 ∂2 vz There is no θ direction contribution to the flow: 2 2 = 0 r ∂θ 0=−

∂P 1 ∂ ∂vz (r )) + μ( ∂z r ∂r ∂r ∂v

If we assume that the fluid is Newtonian, so τrz = μ ∂rz, Then after multiplying through by r, 0 = −r

∂P ∂ + (rτ ) ∂z ∂r rz

9.26 A film in free stream flow is flowing down an inclined plan as shown in the figure. Beginning with the appropriate form of the Navier-Stokes equations, please derive the following equation: ∂ τ + ρgsin(θ) = 0 ∂y yx As always, please state all assumptions you make in your derivation.

Solution ρ(

∂vx ∂vx ∂vx ∂vx ∂P ∂2 vx ∂2 vx ∂2 vx )=− + vx + vy + vz + ρg x + μ ( 2 + + 2) ∂t ∂x ∂y ∂z ∂x ∂x ∂y 2 ∂z

Evaluating the equation term by term: ∂vx = 0 since the flow is steady ∂t ∂vx ∂2 vx vx = = 0 since the flow is fully developed ∂x ∂x 2 ∂vx vy = 0 since there is no y direction velocity ∂y ∂P = 0 since this is free stream flow ∂x ∂2 vx = 0 since we have two dimensional flow ∂z 2 Taking the angle of the flow into consideration, ρg x = ρg x sin(θ) we now have, ∂2 vx 0 = ρg x sin(θ) + μ 2 ∂y Next, we must assume that we have a Newtonian Fluid, ∂τyx 0 = ρg x sin(θ) + ∂y Rearranging,

∂τyx + ρg x sin(θ) = 0 ∂y 9.27 An incompressible Newtonian fluid is confined between two concentric circular cylinders of infinite length, a solid inner cylinder of radius R, and a hollow stationary outer cylinder of radius Ro. See the figure below where the z-axis is coming out of the page. The inner cylinder rotates with an angular velocity ωi . The flow is steady, laminar and two-dimensional in the rθ-plane. The flow is also rotationally symmetric, meaning that nothing is a function of coordinate θ (so vθ and P are functions of the radius, r only) and that gravity can be ignored. We can also assume that the flow is circular, meaning that the velocity component vr = 0 everywhere. Please derive an exact expression for the velocity component vθ as a function of the radius r and the other parameters in the problem.

i R0

Liquid: ,

Rotating inner cylinder Solution

Stationary outer cylinder

We begin with the Navier-Stokes Equation for cylindrical coordinates in the θ-direction. ∂vθ ∂vθ vθ ∂vθ vr vθ ∂vθ ) ρ( + vr + + + vz ∂t ∂r r ∂θ r ∂z 1 ∂P ∂ 1∂ 1 ∂2 vθ 2 ∂vr ∂2 vθ (rvθ )) + 2 =− + ρg θ + μ ( ( + + ) r ∂θ ∂r r ∂r r ∂θ2 r 2 ∂θ ∂z 2 Evaluation of the terms, ∂vθ = 0 since we have steady state ∂t ∂vθ vr vθ ∂vθ vr = = vz = 0 since there is no z or r direction velocity ∂r r ∂z vθ ∂vθ = 0 since we have fully developed flow r ∂θ 1 ∂P 1 ∂2 vθ = = 0 since we have rotational symmey r ∂θ r 2 ∂θ2

2 ∂vr = 0 since we have no r direction velocity change in the θ direction r 2 ∂θ ρg θ = 0 since we are told we can neglect gravity Thus, ∂ 1∂ (rv ))) 0 = μ( ( ∂r r ∂r θ We can convert from partial differentials to ordinary differentials since vθ = f(r) only, and integrate, μd (rv ) = C1 r dr θ Rearrange and integrate again, vθ = C1

r C2 + 2μ r

Determine the boundary conditions to find values for C1 and C2, BC#1: at the inner wall the No-Slip boundary condition applies: r = R i at vθ = ωi R i BC#2: at the outer wall the No-Slip boundary condition applies: r = R o at vθ = 0 Apply BC#2: 0 = C1

R 0 C2 + 2μ R 0

C1 R20 C2 = − 2μ Apply BC#1: R i C1 R20 Ri R20 ωi R i = C1 − = C1 ( − ) 2μ 2μR i 2μ 2μR i So, 2μωi R2i C1 = − 2 R 0 − R2i And, C1 R20 2μωi R2i R20 ωi R2i R20 = = 2μ 2μ(R20 − R2i ) (R20 − R2i ) Plug back into the original equation, 2μωi R2i r ωi R2i R20 rωi R2i ωi R2i R20 ( )+ vθ = − 2 =− 2 + (R 0 − R2i ) 2μ (R 0 − R2i ) r(R20 − R2i ) r(R20 − R2i ) C2 = −

ωi R2i R20 vθ = 2 ( − r) (R 0 − R2i ) r

∂vz =0 ∂t

∂vz vθ ∂vz = =0 ∂r r ∂θ ∂vz ∂2 vz Fully developed flow: vz = =0 ∂z ∂z 2

No velocity in the r or θ directions: vr

1 ∂2 vz There is no θ direction contribution to the flow: 2 2 = 0 r ∂θ Since the pipe is vertical, the gravity acts in the negative z-direction: ρg z = −ρg z ∂P 1 ∂ ∂vz (r )) 0=− − ρg z + μ ( ∂z r ∂r ∂r Rearrange, ∂ ∂vz r ∂P (r ) = ( + ρg z ) ∂r ∂r μ ∂z Integrate, ∂vz r 2 ∂P ( + ρg z ) + C1 r = ∂r 2μ ∂z Rearrange (divide through by r and by μ), μ

∂vz r ∂P C1 = ( + ρg z ) + μ ∂r 2 ∂z r

𝜕𝑣

Boundary Condition #1: r = 0 in center of tube would result in 𝜇 𝜕𝑟𝑧 = 𝜏𝑟𝑧 = 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒, which is impossible, so 𝐶1 = 0 (we have seen this boundary condition before). Could have also said that 𝜕𝑣𝑧 cannot be infinite. 𝜕𝑟

Integrate again, vz =

r 2 ∂P ( + ρg z ) + C2 4μ ∂z

BC#2: At vz = 0, r = D/2 = R (No Slip Boundary Condition) Apply BC#2: (−R)2 ∂P ( + ρg z ) + +C2 0= 4μ ∂z Thus, C2 = −

R2 ∂P ( + ρg z ) 4μ ∂z

Resulting in, r 2 ∂P R2 ∂P 1 ∂P ( + ρg z ) − ( + ρg z ) = ( + ρg z ) (r 2 − R2 ) vz = 4μ ∂z 4μ ∂z 4μ ∂z

9.29. Fluid Direction

𝜃

Solution 1 Here the key points in the problem statement were that the flow is in one direction only and that it was vertical flow. Need x-direction Navier-Stokes Equation 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕𝑣𝑦 𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦 𝜕 2 𝑣𝑦 𝜕𝑃 𝜌( + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 )=− + 𝜌𝑔𝑦 + 𝜇 ( 2 + + ) 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑦 𝜕𝑥 𝜕𝑦 2 𝜕𝑧 2 Using the conditions in the problem statement this equation becomes, 𝜕 2 𝑣𝑦 𝜕𝑃 0=− − 𝜌𝑔𝑦 cos 𝜃 + 𝜇 ( 2 ) 𝜕𝑦 𝜕𝑥 Integrate twice solving for velocity, 𝜕 2 𝑣𝑦 1 𝜕𝑃 = ( + 𝜌𝑔𝑦 cos 𝜃) 𝜕𝑥 2 𝜇 𝜕𝑦 𝜕𝑣𝑦 𝑥 𝜕𝑃 = ( + 𝜌𝑔𝑦 cos 𝜃) + 𝐶1 𝜕𝑥 𝜇 𝜕𝑦 𝑣𝑦 = Boundary Conditions:

𝑥 2 𝜕𝑃 ( + 𝜌𝑔𝑦 cos 𝜃) + 𝐶1 𝑥 + 𝐶2 2𝜇 𝜕𝑦

BC#1: 𝑣𝑦 = −VB 𝑎𝑡 𝑥 = 0 BC#2: 𝑣𝑦 = −VT 𝑎𝑡 𝑥 = ℎ Apply Boundary Condition #1 (0)2 𝜕𝑃 ( + 𝜌𝑔𝑦 cos 𝜃) + 𝐶1 (0) + 𝐶2 −VB = 2𝜇 𝜕𝑦 𝐶2 = −VB Apply Boundary Condition #2 ℎ2 𝜕𝑃 ( + 𝜌𝑔𝑦 cos 𝜃) + 𝐶1 ℎ + VB −VT = 2𝜇 𝜕𝑦 VB − VT ℎ 𝜕𝑃 ( + 𝜌𝑔𝑦 cos 𝜃) − h 2𝜇 𝜕𝑦

𝐶1 = And Finally

𝑥 2 𝜕𝑃 VB − VT ℎ 𝜕𝑃 ( + 𝜌𝑔𝑦 cos 𝜃) + 𝑥 [ ( + 𝜌𝑔𝑦 cos 𝜃)] − VB 𝑣𝑦 = − 2𝜇 𝜕𝑦 h 2𝜇 𝜕𝑦

Alternative Solution (and perhaps a better solution): 𝑣𝑥 =

𝑥 2 𝜕𝑃 ( + 𝜌𝑔𝑦 cos 𝜃) + 𝐶1 𝑥 + 𝐶2 2𝜇 𝜕𝑦

Boundary Conditions: BC#1: 𝑣𝑥 = −VB cos 𝜃 𝑎𝑡 𝑥 = 0 BC#2: 𝑣𝑥 = −VT cos 𝜃 𝑎𝑡 𝑥 = ℎ Apply B.C. #1: −VB sin 𝜃 = 𝐶2 Apply B.C. #2: −V𝑇 sin 𝜃 =

𝐶1 =

ℎ2 𝜕𝑃 ( + 𝜌𝑔𝑦 cos 𝜃) + 𝐶1 ℎ − VB cos 𝜃 2𝜇 𝜕𝑦

VB cos 𝜃 − VT cos 𝜃 ℎ 𝜕𝑃 ( + 𝜌𝑔𝑦 cos 𝜃) − h 2𝜇 𝜕𝑦

𝑥 2 𝜕𝑃 VB cos 𝜃 − VT cos 𝜃 ℎ 𝜕𝑃 ( + 𝜌𝑔𝑦 cos 𝜃) + [ ( + 𝜌𝑔𝑥 cos 𝜃)] 𝑥 − VB cos 𝜃 𝑣𝑦 = − 2𝜇 𝜕𝑦 h 2𝜇 𝜕𝑦

9.30 Beginning with the appropriate Navier-Stokes equation, please derive the appropriate equation to describe the flux of 𝐳-momentum in the y-direction assuming assume steady, fully developed flow between two infinitely long vertical parallel plates separated by a distance L, where the fluid is not open to the atmosphere and parabolic flow can be assumed. Please indicate the reason why you eliminate any terms in the equations as you proceed with your derivation. [Hint: we want an equation to describe 𝐳-momentum in the y-direction and NOT 𝐲-momentum in the 𝐳-direction.] Solution dv The flux of z-momentum in the y-direction is defined as τyz = μ dyz , so we need to begin with the Navier-Stokes equation in the z-direction of rectangular coordinates. As a result, the coordinate system is chosen for us in the problem statement. Since the problem states that the flow is vertical, we keep the gravity term, and since it is not open to the atmosphere, we keep the pressure term. ∂vz ∂vz ∂vz ∂vz ∂P ∂2 vz ∂2 vz ∂2 vz )=− ρ( + vx + vy + vz + ρg z + μ ( 2 + 2 + 2 ) ∂t ∂x ∂y ∂z ∂z ∂x ∂y ∂z Applying all the simplifying assumptions of the problem statement this equation simplifies to: ∂P ∂2 vz 0=− − ρg z + μ ( 2 ) ∂z ∂y 2 ∂ vz ∂P μ( 2 ) = + ρg z ∂y ∂z dτyz ∂P = + ρg z dy ∂z ∂P τyz = ( + ρg z ) y + C1 ∂z Next, we need to determine the boundary condition. We know that the shear is negligible in the center, away from the boundary, and are given that a distance, L, separates the plates. So at the center, L Boundary condition: τyz = 0 at y = 2 Substitute and solve for C1 , ∂P L 0 = ( + ρg z ) + C1 ∂z 2 C1 = − ( Plug this into the original equation,

∂P L + ρg z ) ∂z 2

∂P ∂P L + ρg z ) y − ( + ρg z ) ∂z ∂z 2 ∂P L τyz = ( + ρg z ) (y − ) ∂z 2

τyz = (

9.31 Two different Newtonian fluids, so that they have different densities and viscosities (Fluid 1 and Fluid 2 in the figure), are flowing steadily down between infinitely long vertical plates as shown in the figure below. The plate at the left is moving down at a velocity of VA, the center plate is moving up at a velocity of VB and the plate at the right is stationary. In the figure, the distances between the plates are not equal so that L1≠L2. (1) Using the appropriate form of the Navier-Stokes Equations, please derive equations for the velocity of Fluid 1 and the velocity of Fluid 2 assuming that both fluids are in fully developed, continuous, incompressible, adiabatic, isothermal flow. You may also assume that the no-slip boundary condition applies. (2) Please draw the velocity profiles for both Fluid 1 and Fluid 2 in the figure provided below. Be careful in your drawing so it is clear.

Solution We first assign an appropriate coordinate system:

Fluid 1

Fluid 2

y x

Need y-direction Navier-Stokes Equation ∂vy ∂vy ∂vy ∂vy ∂2 vy ∂2 vy ∂2 vy ∂P ρ( + vx + vy + vz )=− + ρg y + μ ( 2 + + 2) ∂t ∂x ∂y ∂z ∂y ∂x ∂y 2 ∂z Using the conditions in the problem statement this equation becomes, ∂2 vy ∂P 0=− − ρg y + μ ( 2 ) ∂y ∂x Integrate twice solving for velocity, ∂2 vy 1 ∂P = ( + ρg y ) ∂x 2 μ ∂y ∂vy x ∂P = ( + ρg y ) + C1 ∂x μ ∂y 2 x ∂P ( + ρg y ) + C1 x + C2 vy = 2μ ∂y This equation can be used for both sides using appropriate boundary conditions. There are two ways to do this problem depending on how you choose the boundary conditions. METHOD #1 For Fluid #1: v1 =

x 2 ∂P ( + ρg y ) + C1 x + C2 2μ ∂y

Boundary Conditions: BC#1: v1 = −V𝐴 at x = −L1 BC#2: v1 = V𝐵 at x = 0 Apply BC#1: −V𝐴 =

(−L1 )2 ∂P ( + ρg y ) + C1 (−L1 ) + C2 2μ ∂y

Apply BC#2: V𝐵 = 0 + 0 + C2 → C2 = V𝐵 Plug the result of BC2 into BC1 equation and solve for C1

(−L1 )2 ∂P ( + ρg y ) + C1 (−L1 ) + V𝐵 −V𝐴 = 2μ ∂y C1 =

V𝐴 + V𝐵 𝐿1 2 ∂P ( + ρg y ) + 𝐿1 2μ ∂y

x 2 ∂P V𝐴 + V𝐵 𝐿1 2 ∂P ( + ρg y ) + [ ( + ρg y )] x + V𝐵 v1 = + 2μ ∂y 𝐿1 2μ ∂y

For Fluid #2: x 2 ∂P ( + ρg y ) + C1 x + C2 v2 = 2μ ∂y

Boundary Conditions: BC#1: v2 = V𝐵 at x = 0 BC#2: v2 = 0 at x = L2 Apply BC#1: V𝐵 = 0 + 0 + C2

→

C2 = V𝐵

Apply BC#2 (and incorporating the result of BC#1): 0=

L22 ∂P ( + ρg y ) + C1 L2 + V𝐵 2μ ∂y

C1 = −

L2 ∂P V𝐵 ( + ρg y ) − 2μ ∂y L2

Substituting this back to obtain the final equation for v2 : v2 =

x 2 ∂P L2 ∂P V𝐵 ( + ρg y ) − [ ( + ρg y ) + ] x + V𝐵 2μ ∂y 2μ ∂y L2

METHOD #2

For Fluid #1: v1 =

x 2 ∂P ( + ρg y ) + C1 x + C2 2μ ∂y

Boundary Conditions: BC#1: v1 = −V𝐴 at x = 0 BC#2: v1 = V𝐵 at x = L1 Apply BC#1: −V𝐴 = 0 + 0 + C2

→

𝐶2 = −𝑉𝐴

Apply BC#2: V𝐵 =

L21 ∂P 𝑉𝐴 + V𝐵 L1 ∂P ( + ρg y ) + C1 L1 − 𝑉𝐴 → 𝐶1 = − ( + ρg y ) 2μ ∂y 𝐿1 2μ ∂y

So, the equation for the velocity of Fluid 1 is, x 2 ∂P 𝑉𝐴 + V𝐵 L1 ∂P ( + ρg y ) + x [ v1 = − ( + ρg y )] − 𝑉𝐴 2μ ∂y 𝐿1 2μ ∂y For Fluid #2: x 2 ∂P ( + ρg y ) + C1 x + C2 v2 = 2μ ∂y Boundary Conditions: BC#1: v2 = V𝐵 at x = L1 BC#2: v2 = 0 at x = L1 + L2 Apply BC#1: V2 =

𝐿21 ∂P ( + ρg y ) + C1 L1 + C2 2μ ∂y

Apply BC#2: 0= Set C2 = C2

(L1 + L2 )2 ∂P ( + ρg y ) + 𝐶1 (L1 + L2 ) + C2 2μ ∂y

(L1 + L2 )2 ∂P 𝐿21 ∂P ( + ρg y ) − C1 L1 = − ( + ρg y ) − 𝐶1 (L1 + L2 ) 2μ ∂y 2μ ∂y 2 (L1 + L2 )2 ∂P 𝐿1 ∂P ( + ρg y ) 𝐶1 (L1 + L2 ) − C1 L1 = −VB + ( + ρg y ) − 2μ ∂y 2μ ∂y

VB −

C1 L2 = −VB +

(L1 + L2 )2 ∂P 𝐿21 ∂P ( + ρg y ) − ( + ρg y ) 2μ ∂y 2μ ∂y

(L1 + L2 )2 ∂P −VB 𝐿21 ∂P ( + ρg y ) − ( + ρg y ) C1 = + L2 2μL2 ∂y 2μL2 ∂y So, (L1 + L2 )2 ∂P x 2 ∂P −VB 𝐿21 ∂P ( + ρg y ) + [ ( + ρg y ) − ( + ρg y )] x + C2 v2 = + 2μ ∂y L2 2μL2 ∂y 2μL2 ∂y Next, solve for C2 , could pick either result form application of BC#1 or BC#2, we’ll use the result of BC#2 here since velocity is zero in that boundary condition, (L1 + L2 )2 ∂P (L1 + L2 )2 ∂P −VB 𝐿21 ∂P ( + ρg y ) + [ ( + ρg y ) − ( + ρg y )] (L1 + L2 ) 0= + 2μ ∂y L2 2μL2 ∂y 2μL2 ∂y + C2 (L1 + L2 )2 ∂P ( + ρg y ) C2 = − 2μ ∂y (L1 + L2 )2 ∂P −VB 𝐿21 ∂P ( + ρg y ) + ( + ρg y )] (L1 + L2 ) −[ + L2 2μL2 ∂y 2μL2 ∂y And the final equation for the velocity is, v2 =

(L1 + L2 )2 ∂P x 2 ∂P −VB 𝐿21 ∂P ( + ρg y ) + [ ( + ρg y ) − ( + ρg y )] x+ + 2μ ∂y L2 2μL2 ∂y 2μL2 ∂y 2 (L1 + L2 ) ∂P ( + ρg y ) =− 2μ ∂y 2 (L1 + L2 )2 ∂P −VB 𝐿1 ∂P ( + ρg y ) + ( + ρg y )] (L1 + L2 ) −[ + L2 2μL2 ∂y 2μL2 ∂y

This could be further simplified if desired.

9.32 Consider steady, incompressible, laminar, fully developed flow of a Newtonian fluid in an infinitely long, stationary pipe of diameter D inclined at an angle of 𝛽 to the horizontal plane as shown in the figure for problem 9.24. You may assume that the No-Slip boundary condition applies. (a) Show that the continuity equation holds in this system, (b) Using the appropriate form of the Navier-Stokes Equation, derive an expression to describe the velocity of this fluid. Be sure to indicate the reason why you eliminate any terms in the original equation as your proceed with your derivation. Solution (a) We begin with the general form of continuity given in the equation sheets (or something similar), ∇ ∙ ρv +

∂ρ =0 ∂t

which expands, in cylindrical coordinates to, ∂ρ 1 ∂ 1 ∂ ∂ (ρrvr ) + (ρvθ ) + (ρvz ) = 0 + ∂t r ∂r r ∂θ ∂z ∂ρ = 0 since we have steady flow ∂t 1∂ (ρrvr ) = 0 since we do not have any r − direction flow r ∂r 1 ∂ (ρvθ ) = 0 since we don′ thave any θ − direction flow r ∂θ ∂ (ρvz ) = 0 since we have fully developed flow ∂z Thus, 0=0 And we have proved that continuity holds. (b) For flow down a pipe we need the z-direction of the Navier-Stokes equations in cylindrical coordinates: ∂vz ∂vz vθ ∂vz ∂vz ∂P 1 ∂ ∂vz 1 ∂2 vz ∂2 vz )=− (r )+ 2 2 + 2) ρ( + vr + + vz + ρg z + μ ( ∂t ∂r r ∂θ ∂z ∂z r ∂r ∂r r ∂θ ∂z

Applying the assumptions given in the problem statement including that it is on an angle, this equation simplifies to, 0=−

∂P 1 ∂ ∂vz (r )) + ρg z sinβ + μ ( ∂z r ∂r ∂r

Rearrange and integrate, 1 ∂ ∂vz 1 ∂P (r ) = ( + ρg z sinβ) r ∂r ∂r μ ∂z ∂ ∂vz r ∂P (r ) = ( + ρg z sinβ) ∂r ∂r μ ∂z

∂vz r 2 ∂P ( + ρg z sinβ) + C1 = ∂r 2μ ∂z

∂vz r ∂P C1 ( + ρg z sinβ) + = ∂r 2μ ∂z r The first boundary condition is that, because of fully developed flow, there is no change in the z-direction velocity with respect to r in the center of the pipe where r = 0, so, B.C. #1:

∂vz ∂r

= 0 at r = 0

Application of this boundary condition gives, C1 0= 0+ 0

→ C1 = 0

So the equation is now, ∂vz r ∂P ( + ρg z sinβ) = ∂r 2μ ∂z Integrate, vz =

r 2 ∂P ( + ρg z sinβ) + C2 2μ ∂z

For the second boundary condition, the No-Slip boundary conditions applies, B.C. #2: vz = 0 at r = R Apply this boundary condition,

R2 ∂P ( + ρg z sinβ) + C2 0= 2μ ∂z Giving, R2 ∂P C2 = − ( + ρg z sinβ) 2μ ∂z So the final equation for the velocity is r 2 ∂P R2 ∂P ( + ρg z sinβ) − ( + ρg z sinβ) vz = 2μ ∂z 2μ ∂z This can be simplified if desired.

9.33 V0 Fluid 2

h2 Interface

Fluid 1

Two immiscible viscous fluids are sandwiched between two infinitely long and wide parallel flat plates as shown in the figure above. Because the fluids are immiscible they do not mix at any point. At time t = 0, the fluids and the plates on either side are stationary. The upper plate is set in motion at a velocity V0, and the bottom plate remains stationary at all times. The flow is incompressible, parallel, fully developed and laminar. The flow is caused solely by the viscous effects created by the moving upper plate. You may ignore gravity, and assume that the fluids are not open to the atmosphere and that the interface between the fluids is horizontal at all times. (c) In the figure above please draw the velocity profiles for both Fluid 1 and Fluid 2 in this system. (d) Please derive equations for the velocity profiles of both fluids when the flow is at steady state.

Chapter 10 Instructor Only Problems

10.25 The stream function for steady, incompressible flow is given by, Ψ = y 2 − xy − x 2 . Determine the velocity components for this flow. Solution We define the stream function by, ∂Ψ = vx ∂y and − Thus,

∂Ψ = vy ∂x

∂Ψ = 2y − x ∂y ∂Ψ vy = − = y + 2x ∂x vx =

10.26 The stream function for steady, incompressible flow is given by Ψ(x, y) = 2x 2 − 2y 2 − xy. Determine the velocity potential for this flow. Solution Since a condition for the velocity potential to exist is irrotational flow, we must first determine whether the flow in this problem is irrotational. We first find the components of the velocity using the definition of the stream function. We define the stream function by, ∂Ψ = vx ∂y and ∂Ψ − = vy ∂x Thus, ∂Ψ vx = = −4y − x ∂y ∂Ψ vy = − = −4x + y ∂x Next, we want to determine whether the flow is rotational or irrotational. To do this we must satisfy Equation 10.1 1 ∂vy ∂vx 1 wz = ( − ) = (−4 − (−4)) = 0 2 ∂x ∂y 2 So the flow is irrotational as desired. Next we determine the equation for the velocity potential. The velocity potential is defined by Equation 10-15. v = ∇∅ or ∂ϕ = vx = −4y − x ∂x x2 ϕ = −4xy − + f(y) 2 Differentiating with respect to y and equating to vy ∂ϕ d = −4x + y = −4x + f(y) ∂y dy Thus, d f(y) = y dy y2 f(y) = 2 So that the final equation for the velocity potential is x2 y2 ϕ = −4xy − + 2 2

10.27 Determine the velocity components for the flow in Problem 10.23. Solution We defined the stream function as ∂Ψ = −vy ∂x And ∂Ψ = vx ∂y Thus, ∂Ψ = −12y ∂y ∂Ψ vy = − = −12x ∂x vx =

The equations for the velocity components are vx = −12y and vy = −12x.

10.28 The steady, incompressible flow field for two-dimensional flow is given by the following velocity components, 𝑣𝑥 = 16𝑦 − 𝑥 and 𝑣𝑦 = 16𝑥 + 𝑦. Determine the equation for the stream function and the velocity potential. Solution First, let’s check to make sure continuity is satisfied. 𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕 𝜕 (16𝑦 − 𝑥) + (16𝑥 + 𝑦) = −1 + 1 = 0 + = 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 So, continuity is satisfied which is a necessary condition for us to proceed.

(1)

We defined the stream function as ∂Ψ = vx ∂y

(2)

and −

∂Ψ = vy ∂x

(3)

Thus, ∂Ψ = 16𝑦 − 𝑥 (4) ∂y ∂Ψ vy = − = 16x + y (5) ∂x We can begin by integrating equation (4) or equation (5). Either will result in the same answer. (Problem 2 of the Homework will let you verify this.) We will choose to begin by integrating equation (4) partially with respect to y, Ψ = 8y 2 − xy + f1 (x) (6) Where f1 (x) is an arbitrary function of x. Next, we take the other part of the definition of the stream function, equation (3), and differentiate equation (6) with respect to x, ∂Ψ vy = − = y − f2 (x) (7) ∂x df Here, f2 (x) is dx since f is a function of the variable x. The result is that we now have two equations for vy , equations (5) and (7). We can now equate these and solve for f2 (x). vy = y − f2 (x) = 16x + y Solving for f2 (x), f2 (x) = −16x So x2 f1 (x) = −16 = −8x 2 + C 2 The integration constant C is added to the above equation since f is a function of x only. The final equation for the stream function is, Ψ = 8y 2 − xy − 8x 2 + C (6) The constant C is generally dropped from the equation because the value of a constant in this equation is of no significance. The final equation for the stream function is, vx =

Ψ = 8y 2 − xy − 8x 2 One interesting point is that the difference in the value of one stream line in the flow to another is the volume flow rate per unit width between the two streamlines. Next we want to find the equation for the velocity potential. Since a condition for the velocity potential to exist is irrotational flow, we must first determine whether the flow in this example is irrotational. To do this we must satisfy Equation 10.1 1 ∂vy ∂vx 1 ( − ) = (16 − 16) = 0 2 ∂x ∂y 2 So the flow is irrotational as required. We now want to determine the equation for the velocity potential. The velocity potential is defined by Equation 10-15. v = ∇∅ or ∂ϕ = vx = 16y − x ∂x x2 ϕ = 16xy − + f(y) 2 Differentiating with respect to y and equating to vx ∂ϕ d = 16x + f(y) = 16x + y ∂y dy Thus, d f(y) = y dy And, y2 f(y) = 2 So that the final equation for the velocity potential is x2 y2 ϕ = 16xy − + 2 2 wz =

10.29 Repeat Problem 2 but instead begin by integrating the following equation, ∂Ψ vx = . ∂y and show that it does not matter which equation you begin with; the results are identical. Solution We defined the stream function as ∂Ψ = vx ∂y

(2)

and −

∂Ψ = vy ∂x

(3)

Thus, ∂Ψ = 16𝑦 − 𝑥 (4) ∂y ∂Ψ vy = − = 16x + y (5) ∂x We will choose to begin by integrating equation (5) partially with respect to x, Ψ = −8x 2 − xy + f1 (x) (6) Where f1 (y) is an arbitrary function of y. Next, we take the other part of the definition of the stream function, equation (2), and differentiate equation (6) with respect to y, ∂Ψ vx = = −x + f2 (y) (7) ∂y df Here, f2 (y) is dy since f is a function of the variable y. vx =

The result is that we now have two equations for vx , equations (4) and (7). We can now equate these and solve for f2 (y). vy = −x + f2 (y) = 16𝑦 − 𝑥 Solving for f2 (y), f2 (y) = 16y So y2 f1 (x) = 16 = 8y 2 + C 2 The integration constant C is added to the above equation since f is a function of y only. The final equation for the stream function is, Ψ = −8x 2 − xy + 8y 2 + C (6) The constant C is generally dropped from the equation because the value of a constant in this equation is of no significance. The final equation for the stream function is, Ψ = 8y 2 − xy − 8x 2

10.30 In 2-dimensional potential flow the stream function Ψ(x,y) and the velocity potential Φ(x.y) are geometrically related. Sketch this relationship on an x-y plane. What is the physical significance of Ψ(x,y)=C1 and Φ(x.y)=C2 where C1 and C2 are arbitrary constants? Solution vy =

 ( x, y ) −  ( x, y ) = y x

vx =

 ( x, y )  = x y

dy 1 = dx  =cons tan t − dy / dx  =cons tan t These equations suggest that the stream lines intersect the velocity potential lines at 90o at all times and that they are perpendicular to each other at the intersection. 



     = C2 y

90o angle

 = C1

Chapter 11 Instructor Only Problems

11.4 The maximum pitching moment that is developed by the water on a flying boat as it lands is noted as Cmax. The following are the variables involved in this action: = angle made by flight path of plane with horizontal =angle defining attitude of plane M=mass of plane L=length of hull =density of water g=acceleration due to gravity R=radius of gyration of plane about axis of pitching (a) According to the Buckingham  theorem, how many independent dimensionless groups should there be which characterize this problem? Cmax = ML2t-2 = angle =angle M = Mass L = Length  = density g = gravity R = radius

dimensionless dimensionless M L ML-3 Lt-2 L

Dimension  0 and the rank =3 i = n – r = 8 – 3 = 5 independent dimensionless groups (b) What is the dimensional matrix of this problem? What is its rank?

M L t

Cmax 1 2 -2

 0 0 0

 0 0 0

M 1 0 0

L 0 1 0

 1 -3 0

g 0 1 -2

R 0 1 0

 1= 

2 = 

3=MaLbgcCmax

2  L  ML MoLoto = 1 = (M )a (L )b  2  t  t2 M: 0 = a+1 L: 0 = b+c+2 T: 0 = -2c-2 Thus: a= -1, b= -1, c= -1

1=-1D-2-2 →

 C max   1=   M Lg  c

4=MaLbgc

L M MoLoto = 1 = (M )a (L )b  2  2 t  L M: 0 = a+1 L: 0 = b+c-3 T: 0 = -2c Thus: a= -1, b= 3, c= 0

1=-1D-2-2 →

 L3    1=    M 

5=MaLbgcR

L MoLoto = 1 = (M )a (L )b  2  L t  M: 0 = a L: 0 = b+c+1 T: 0 = -2c Thus: a= 0, b= 3, c= 0

1=-1D-2-2 →

R 1=   L 

11.12 Identify the variables associated with Problems 8.13 and find the dimensionless parameters. Solution Variable ΔP Q h Ω L μ R

Dimensions M/Lt2 L3/t L 1/t L M/Lt L i = 7 − 3 = 4 , Choose core as h, Ω, μ L π1 = ha Ωb μc L ∴ = h R π2 = hd Ωe μf R ∴ = h ΔP π3 = hg Ωh μi ΔP ∴ = Ωμ Q π4 = hj Ωk μl Q ∴ = 3 h Ω

11.23 A pump in a manufacturing plant is transferring viscous fluids to a series of delivery tanks. This critical transfer requires careful monitoring of the solution mass flow rate, the power (work) that the pump adds to the fluid, the internal energy of the system, the viscosity and density of the solution. Please determine the dimensionless groups formed from the variables involved using the Buckingham method. Carefully choose your core group based on the description of the system.

Solution

Variable Mass flow rate Work of pump Internal energy viscosity Density

M L t

ṁ

δWs / ∂t

1 0 -1

1 2 -3

Symbol

ṁ δWs / ∂t U

μ ρ

Dimensions M/t ML2/t3 L2/s2 M/Lt M/L3

U 0 2 -2

1 -1 -1

1 -3 0

𝑖 =5−3=2 Thus there are 2 dimensionless parameters to be found. Choose the core group to be δWs / ∂t and viscosity. Find π1 : b

π1 = ρa ṁ U c

δWs ∂t

M a M b L2 ML2 M L T = 1 = ( 3) ( ) ( 2 ) 3 L L t t 0 0 0

Evaluate for M, L and t M: 0=a+b+1 L: 0 = −3a − b + 2c + 2 t: 0 = −2c − 3

Thus, a = − 2 , b = − 2 and c = −3/2 and −1/2

π1 = π1 = ρ−1/2 ṁ

U −3/2

δWs ∂t = 1/2 3/2 ρ1/2 ṁ U

Find π2 :

δWs ∂t

π2 = ρd ṁ U f μ f M d M e L2 ML 0 0 0 M L T = 1 = ( 3) ( ) ( 2 ) L L t t

Evaluate for M, L and t M: 0=d+e+1 L: 0 = −3d − e + 2f + 1 t: 0 = −2f − 1 1

Thus, d = − 3 , e = − 3 , f = − 2 and

−

π1 = ρ−3 ṁ 3 U −2 μ =

μ 2/3 1/2 ρ1/3 ṁ U

11.24 The performance of a fluid flow system is affected by variables of pressure, density, angular velocity, impeller diameter, volumetric flow rate and viscosity. Determine the dimensionless groups formed from the variables involved using the Buckingham Method. Choose the groups so that , Q and P each appear in one group only, so that the core group is , D and . Solution Core Group = ρ, D, ω

M L t

P 1 -1 -2

ρ 1 -3 0

ω 0 0 1

D 0 1 0

𝜋1 = 𝜌𝑎 𝐷𝑏 ω𝑐 𝑃 𝑀 𝑎 1 𝑐 𝑀 𝑏 𝑀 𝐿 𝑡 = 1 = ( 3 ) (𝐿) ( ) 𝐿 𝑡 𝐿𝑡 2 0 0 0

𝑀: 0 = 𝑎 + 1 𝐿: 0 = −3𝑎 + 𝑏 − 1 𝑡: 0 = −𝑐 − 2 𝑐 = −2,

𝑎 = −1,

𝜋1 = 𝜌−1 𝐷−2 ω−2 𝑃 =

𝑏 = −2 𝑃 𝜌𝐷2 𝜔 2

𝜋2 = 𝜌𝑎 𝐷𝑏 ω𝑐 𝑄 𝑀 𝑎 1 𝑐 𝐿3 𝑀0 𝐿0 𝑡 0 = 1 = ( 3 ) (𝐿)𝑏 ( ) 𝐿 𝑡 𝑡 𝑀: 0 = 𝑎 𝐿: 0 = −3𝑎 + 𝑏 + 3 𝑡: 0 = −𝑐 − 1 𝑐 = −1,

𝑎 = 0,

𝜋2 = 𝜌0 𝐷 −3 ω−1 𝑄 =

𝑏 = −3 𝑄 𝐷3 𝜔

Q 0 3 -1

 1 -1 -1

𝜋3 = 𝜌𝑎 𝐷𝑏 ω𝑐 𝜇 𝑀 𝑎 1 𝑐𝑀 𝑏 𝑀 𝐿 𝑡 = 1 = ( 3 ) (𝐿) ( ) 𝐿 𝑡 𝐿𝑡 0 0 0

𝑀: 0 = 𝑎 + 1 𝐿: 0 = −3𝑎 + 𝑏 − 1 𝑡: 0 = −𝑐 − 1 𝑐 = −1,

𝑎 = −1,

𝜋3 = 𝜌0 𝐷−3 ω−1 𝑄 =

𝑏 = −2 μ 𝜌𝐷2 𝜔

Units check: 𝑀 3 μ 𝐿𝑡 = 𝑀𝐿 𝑡 = 1 = 𝜌𝐷2 𝜔 𝑀 𝐿2 1 𝐿𝑡𝑀𝐿2 𝐿3 𝑡

11.25 The performance of a fluid flow system is affected by the following variables: force, F (units are MLt-2), fluid density,  (units are ML-3), velocity, v (units are L/t), coating diameter, D (units are L) fluid viscosity,  (units are ML-1t-1). Determine the dimensionless groups formed from the variables involved using the Buckingham Method. Solution Core Group = ρ, D, 

M L t

F 1 1 -2

ρ 1 -3 0

v 0 1 1

D 0 1 0

5 variables, rank is 3 so i=2. π1 = ρa Db c F M a M c ML b (L) ( ) 2 M L t = 1 = ( 3) L Lt t 0 0 0

𝑀: 0 = 𝑎 + 𝑐 + 1 𝐿: 0 = −3𝑎 + 𝑏 − 𝑐 + 1 𝑡: 0 = −𝑐 − 2 𝑐 = −2, 𝑎 = 1, 𝑏 = 0

π1 = ρ1 D0 −2 P =

𝑃ρ 2

Units Check: ML M 𝑃ρ 2 3 𝐿𝑡𝐿𝑡 = t L = 2 2 =1 2 MM  t 𝐿 Lt Lt π2 = ρa Db c v M a M cL M 0 L0 t 0 = 1 = ( 3 ) (L)b ( ) L Lt t

 1 -1 -1

M: 0 = a + c L: 0 = −3a + b − c + 1 t: 0 = −c − 1 c = −1,

a = 1,

π1 = ρ1 D1 −1 P =

b=1 vρD 

Unit Check LM vρD t 3 𝐿 𝐿𝐿𝐿𝑡 = L = =1 M  𝑡𝐿𝐿𝐿 Lt

Chapter 12 Instructor Only Problems

12.34 Air, water and glycerin are flowing through separate tubes with diameters of 0.5 inches and a velocity of 40 ft/sec and are all at 80℉. Determine in each case whether the flow is laminar or turbulent. Solution We must determine the Reynolds Number for each situation. Appendix I contains all the values for density and viscosity that we need. Air lbm ft ρvD (0.0735 ft 3 ) (40 sec) (0.5 in) ft Re = = = 9879 lbm μ 12 in −5 (1.24x10 s) ft Turbulent Flow (Re > 2300) Water lbm ft ρvD (62.2 ft 3 ) (40 sec) (0.5 in) ft Re = = = 1.8 x 105 lb μ 12 in (0.578x10−3 m s) ft Turbulent Flow (Re > 2300) Glycerin lbm ft ρvD (78.7 ft 3 ) (40 sec) (0.5 in) ft Re = = = 218 lbm μ 12 in (0.6 s) ft Laminar Flow (Re < 2300)

12.35 The determination of the boundary layer thickness is important in many applications. Aniline at 100℉ is flowing down a flat plate at 2.0 ft/s. When the fluid is 0.1 ft beyond the leading edge, what is the thickness of the boundary layer? Solution Re =

ρvx (63.0 lbm /ft 3 )(2 ft/s)(0.1 ft) = = 7000 (180x10−5 lbm /ft s) μ

Since Re is in the laminar region, we choose the laminar equation for the boundary layer thickness. 𝛿 5 5 = = 0.1 𝑓𝑡 √𝑅𝑒𝑥 √7000 δ = 0.006 ft = 0.072 inches

12.36 Water at 20oC is flowing along a 5-m-long flat plate with a velocity of 50 m/s. What will be the boundary layer thickness at a position 5 m from the leading edge? Is the boundary-layer flow at this location laminar or turbulent? If each of the plate surfaces measures 500 m2, determine the drag force if both sides are exposed to the flow.

Solution (a) Is this laminar or turbulent flow? ρvx (998.2 kg/m3 )(50 m/s)(5 m) Re = = = 2.5 x 108 kg μ (9.93 x10−4 Pa − s) ( ) Pa m s2 The flow is turbulent because Re > 3 x 106 (b) Calculate the boundary layer thickness. δ 0.376 = x Re1/5 0.376(5 m) δ= = 0.039 m (2.5x108 )1/5 (c) Calculate the drag force on the surface. 0.0576 Cfx = = 0.0012 Re1/5 FD = ACfx ρ

v∞2 (500 m2 )(0.0012)(998.2 kg/m3 )(50 m/s)2 = x 2 sides = 1.5x106 kg m/s 2 2 2

12.37 In the case of carbon monoxide at 100oF flowing along a plane surface at a velocity of 4 ft/s, will the boundary-layer flow at a position 0.5 ft from the leading edge be laminar or turbulent? What will be the boundary-layer thickness at this location? For a plane surface 0.5 ft long with an effective area of 200 ft2 on each side, determine the total drag force exerted if both sides are exposed to the flow of CO. Solution (a) Is this laminar or turbulent flow? Re =

ρvx (0.0684 lbm /ft 3 )(4 ft/s)(0.5 ft) = = 44487.8 lbm μ −5 (1.23x10 ) ft s

The flow is laminar because Re < 2 x 105 (b) Calculate the boundary layer thickness. δ 5 = 1/2 x Re δ=

5(0.5 ft) = 0.012 ft (44487.8)1/2

1.328 = 0.0063 Re1/2

FD = ACfL ρ

v∞2 (200 ft 2 )(0.0063)(0.0684 lbm /ft 3 )(4ft/s)2 = x 2 sides = 1.38 lbm ft/s 2 2 2

12.38 Graph the results of Problem 12.8 and 12.18 and in one sentence describe what the data describes. Solution The sphere without the dimples has significantly more drag than that of the golf ball with dimples, so the dimples on the golf ball significantly reduce drag. 0.5 0.45 0.4

Drag (lbf)

0.35 0.3

0.25

Drag for Sphere

0.2

Drag for Golf Ball

0.15 0.1 0.05 0 0

100

200 300 Velocity (ft/s)

400

12.39 Using the data in Appendix E, calculate the Reynolds number for water flowing through a 0.25inch pipe at 0.5 and 1 ft/s in the temperature range from 32 to 600℉. Graph the Re vs. temperature and state approximately where the flow goes from laminar or turbulent. Solution We must determine the Reynolds Number for each situation. Appendix I contains all the values for density and viscosity that we need. Re =

ρvD μ

Re at Temperature Density Viscosity 0.5 ft/s Re at 1 ft/s 32 62.4 1.20E-03 541.7 1083.3 60 62.3 7.60E-04 853.9 1707.8 80 62.2 5.78E-04 1121.0 2241.9 100 62.1 4.58E-04 1412.4 2824.8 150 61.3 2.90E-04 2201.9 4403.7 200 60.1 2.06E-04 3039.0 6078.1 250 58.9 1.60E-04 3834.6 7669.3 300 57.3 1.30E-04 4591.3 9182.7 400 53.6 9.30E-05 6003.6 12007.2 500 49 7.00E-05 7291.7 14583.3 600 42.4 5.79E-05 7628.1 15256.2 For 0.5 ft/s, the flow reaches the transition point just above 150oF, and just about 80oF for the 1 ft/s velocity.

Temperature effect on Re 18000.0

Reyolds Number

16000.0 14000.0 12000.0 10000.0 8000.0

Re at 0.5 ft/s

6000.0

Re at 1 ft/s

4000.0 2000.0 0.0 0

100

200

300

400

500

TemperatureAxis Title

600

700

12.40 Water, air and benzene all at 60oF are flowing over a flat plate at 2.5 ft/s. For each fluid, determine the boundary layer thickness at a point 0.25 ft beyond the leading edge. Solution Water Re =

ρvx (62.3 lbm /ft 3 )(1.5 ft/s)(0.25 ft) = = 30740 (0.76x10−3 lbm /ft s) μ

Since Re is in the laminar region, we choose the laminar equation for the boundary layer thickness. 𝛿 5 5 = = 0.25 𝑓𝑡 √𝑅𝑒𝑥 √30740 δ = 0.007 ft = 0.085 inches Air ρvx (0.0764 lbm /ft 3 )(1.5 ft/s)(0.25 ft) e= = = 2368 (1.21x10−5 lbm /ft s) μ Since Re is in the laminar region, we choose the laminar equation for the boundary layer thickness. 𝛿 5 5 = = 0.25 𝑓𝑡 √𝑅𝑒𝑥 √2368 δ = 0.026 ft = 0.31 inches Benzene Re =

ρvx (55.2 lbm /ft 3 )(1.5 ft/s)(0.25 ft) = = 46517 (44.5x10−5 lbm /ft s) μ

Since Re is in the laminar region, we choose the laminar equation for the boundary layer thickness. 𝛿 5 5 = = 0.25 𝑓𝑡 √𝑅𝑒𝑥 √46517 δ = 0.0058 ft = 0.07 inches

Chapter 13 Instructor Only Problems 13.19 A 2.20-in diameter pipe carries water at 15oC. The head loss due to friction is 0.500 m per 300 m of pipe. Determine the volumetric flow rate of the water leaving the pipe. Solution H2 O@15℃:

ΔP = 0.5m, ρ

L = 300m,

D = 2.2m,

hL = 2ff Re =

ff v 2 = 0.01799

ν = 1.195x10−6

L 2 v D

(2.2)(v) Dv = = 1.841x106 v, −6 ν 1.195x10 300 2 hL = 9.81(0.5) = 2ff v , 2.2

Trial & Error − Assume turbulent flow smooth pipe, ff = 0.003 v = 2.448 v = 2.86

m , Re = 4.508x106 , ff = 0.0022 s

m m , Re = 5.26x106 , ff = 0.0021, v = 2.93 − CLOSE ENOUGH s s

π m3 2 v = 2.93 ( ) (2.2) = 11.13 4 s

m2 , s

13.29 You have been hired to consult on a pilot plant project that requires the installation of a rough horizontal pipe made from commercial steel into a flow system. The pipe has a diameter of 4 inches and a length of 30 feet. A Newtonian fluid (heat capacity = 0.149 BTU / lbmºF, density=0.161 lbm/ft3, viscosity=8.88x10-4 lbm/ft-sec and kinematic viscosity = 5.52x10-3 ft2/sec) will flow through the pipe at 300 ft3/min. Assume that the flow is without edge effects and that the no-slip boundary condition applies. Determine the pressure drop of the fluid as it travels through the pipe. Solution ft 3 min ft 3 Q = 300 x =5 min 60 sec s ft 3 5 s Q v= =π = 58.5 ft/s A (0.33 ft)3 4 ft Dv (0.33 ft) (58.5 s ) Re = = = 3497.3 ft 2 ν −3 5.52x10 s This Reynolds number indicates that the flow is turbulent. Figure 13.2 for commercial steel give a value for e of 0.00015. Calculating the relative roughness, e 0.00015 = = 4.54x10−4 D 0.33 Now we use Equation 13-14 for turbulent flow in a rough pipe, 1 D 0.33 ) + 2.28 = 15.65 = 4 log ( ) + 2.28 = 4 log ( e 0.00015 √ff ff = 0.00408 Leq v 2 ΔP hL = = 2ff ρg D g Rearranging, ΔP = 2ρff

Leq 2 (0.166 lbm /ft 3 )(0.00408) 30 ft ( ) (58.5 ft/s)2 = 13.0 lbf /ft 2 v =2 D 32.174 lbm ft/lbf s2 0.33 ft

13.30 Your boss comes to you with an important project where you are to determine the pressure drop in a rough horizontal high pressure pipe made of cast iron. The pipe has a diameter of 0.25 feet and a length of 10 feet. Benzene at 150℉ flows through the pipe at 250 ft3/min. Solution This problem will illustrate the use of figures 13.1 and 13.2. We begin by making some basic calculations for flow rate, velocity, and most importantly the Reynolds Number. ft 3 min ft 3 Q = 250 x = 4.17 min 60 sec sec ft 3 4.17 Q sec = 85 ft/s v= =π A (0.25 ft)2 4 ρvD (51.8 lbm /ft 3 )(85 ft/s)(0.25 ft) Re = = = 4.5x106 lbm μ −5 24.5 x 10 ft s Figure 13.2 contains roughness parameters for various materials including cast iron which the figure gives as e=0.00085. We will use Figure 13.1 for that. We next calculate the relative roughness, 𝑒 0.00085 = = 0.0034 D 0.25 ft Next we turn to Figure 13.1. Along the right hand side y-axis are values for the relative roughness. We find 0.0034 on the axis. This is obviously an approximation and with some practice is straightforward. The value for 0.0034 is between 0.003 and 0.004 so we pick a point approximately half way between and we follow an imaginary line in between the line for 0.003 and 0.004 until we cross the line for the Reynolds Number we calculated above to be 4.5x106 . We find that point and then look to the y-axis on the left to find a value for the friction factor to be 0.0068. Since the flow is Turbulent, we use Equation 13-3 and 13-16 in combination to calculate the pressure drop. Leq v 2 ∆P hL = = 2ff ρg D g Rearrange and solve for the pressure change, Leq 2 (51.8 lbm /ft 3 ) 10 ft (0.0068) (85 ft/s)2 = 6328 lbf /ft 2 ∆P = 2ρff v =2 2 D 32.174 lbm ft/lbf s 0.25 ft

Problem 13.31 Freon-12 is flowing at 10 ft/sec through a pipe made of galvanized iron at 100℉. If the pipe is 100 feet long at 4 inches in diameter, what is the headloss in this system? Solution From Figure 13.2, e=0.0005 for galvanized iron. ρvD (78.7 lbm /ft 3 )(10 ft/s)(4/12 ft) Re = = = 3.0x105 lbm μ −5 88.4 x 10 ft s 𝑒 0.0005 = = 0.0015 D 4/12 ft From Figure 13.1, 𝑓𝑓 = 0.0056. Leq v 2 100 ft (10 ft/s)2 ) hL = 2ff = 2(0.0056) ( = 10.4 ft D g 4/12 ft 32.2 ft s2

13.32 Water at a rate of 0.1 ft3/min (1.6x10-3 ft3/sec) and 60oF flows through a smooth horizontal tube 200 ft long. The pressure drop is 300 lbf/ft2. Determine the pipe diameter.

Solution Δ𝑃 𝐿 𝑣2 = ℎ𝐿 = 2𝑓𝑓 𝜌𝑔 𝐷 𝑔 Δ𝑃 𝐿 = 2𝑓𝑓 𝑣 2 𝜌 𝐷 300 𝑙𝑏𝑓 /𝑓𝑡 3 (32.174 62.3 𝑙𝑏𝑚 /𝑓𝑡 3

𝑙𝑏𝑚 𝑓𝑡 ) 𝑙𝑏𝑓 𝑠 2

200𝑓𝑡 (1.6𝑥10−3 𝑓𝑡 3 /𝑠) = 2𝑓𝑓 [ ] 𝜋 2 𝐷 𝐷 4

𝑓𝑓 = 99329 𝐷5

(1.6𝑥10−3 𝑓𝑡 3 /𝑠) 167 𝐷𝑣 𝐷 𝑅𝑒 = = = 𝜋 2 𝜈 1.22𝑥10−5 𝑓𝑡 2 /𝑠 𝐷 𝐷 4 Use trial and error to solve for D: 167 1 𝑓𝑓 Guess 𝑅𝑒 = = 4 log10 (𝑅𝑒√𝑓𝑓 ) − 0.4 𝐷5 = 𝐷 99329 √𝑓𝑓 0.025 6680 0.00862 𝑓𝑓 = 0.001 0.0387 4315 0.00976 𝑓𝑓 = 0.00862 0.0397 4209 0.00984 𝑓𝑓 = 0.00976 0.0397 4202 0.00984 𝑓𝑓 = 0.00984 Note: A solver was used to do the calculation in the far-right column (MatLab, Python or a calculator with a solver will all work for this). D=0.04 feet.

13.33 You want to understand how the piping material will affect the pressure drop for benzene flowing at 80oF with a flow rate of 1𝑥10−3 𝑓𝑡 3 /𝑠. All the pipes are 100 ft long with a diameter of 0.25 inches and considered rough. Calculate the pressure drop for Commercial Steel, Galvanized Iron, and Cast Iron. Make a table with the material, relative roughness and pressure drop to compare the results. Solution Q = 1𝑥10−3 𝑓𝑡 3 /𝑠 Q 1𝑥10−3 𝑓𝑡 3 /𝑠 v= = 2 = 2.93 ft/s A π 0.25 4 ( 12 ft) 0.25 Dv ( 12 ft) (2.93 ft/s) Re = = = 8783 ft 2 ν 0.695x10−5 s This Reynolds number indicates that the flow is turbulent. Commercial Steel Figure 13.2 for commercial steel give a value for e of 0.00015. Calculating the relative roughness, e 0.00015 = = 7.2x10−3 D 0.0208 Now we use Equation 13-14 for turbulent flow in a rough pipe, 1 D 0.0208 ) + 2.28 = 10.84 = 4 log ( ) + 2.28 = 4 log ( e 0.00015 √ff ff = 0.0085 Leq v 2 ΔP hL = = 2ff ρg D g Rearranging, Leq 2 (54.6 lbm /ft 3 )(0.0085) 100 ft ( ) (2.93 ft/s )2 = 1190.7 lbf /ft 2 v =2 D 32.174 lbm ft/lbf s2 0.0208 ft Galvanized Iron Figure 13.2 for commercial steel give a value for e of 0.0005. Calculating the relative roughness, e 0.0005 = = 2.4x10−2 D 0.0208 Now we use Equation 13-14 for turbulent flow in a rough pipe, 1 D 0.0208 ) + 2.28 = 8.75 = 4 log ( ) + 2.28 = 4 log ( e 0.0005 √ff ff = 0.01306 Leq v 2 ΔP hL = = 2ff ρg D g ΔP = 2ρff

Rearranging, ΔP = 2ρff

Leq 2 (54.6 lbm /ft 3 )(0.01306) 100 ft ( ) (2.93 ft/s )2 = 1829.5 lbf /ft 2 v =2 D 32.174 lbm ft/lbf s2 0.0208 ft

Cast Iron Figure 13.2 for commercial steel give a value for e of 0.0005. Calculating the relative roughness, e 0.00085 = = 4.0x10−2 D 0.0208 Now we use Equation 13-14 for turbulent flow in a rough pipe, 1 D 0.0208 ) + 2.28 = 7.83 = 4 log ( ) + 2.28 = 4 log ( e 0.00085 √ff ff = 0.0163 Leq v 2 ΔP hL = = 2ff ρg D g Rearranging, Leq 2 (54.6 lbm /ft 3 )(0.0163) 100 ft ( ) (2.93 ft/s )2 = 2283.4 lbf /ft 2 ΔP = 2ρff v =2 D 32.174 lbm ft/lbf s2 0.0208 ft Material Relative Roughness Pressure drop (lbf /ft 2 ) −3 Commercial Steel 1190.7 7.2x10 −2 Galvanized Iron 1829.5 2.4x10 −2 Cast Iron 2283.4 4.0x10

13.34 A fluid moves through a 50 meter horizontal tube with a 3 cm diameter at a temperature of 20oC, density of 870 kg/m3 and viscosity of 0.10 kg/m-s. What is the pressure drop if the velocity is 3 m/s? Solution 𝐿 v2 𝐷𝑔 ∆𝑃 ℎ𝐿 = 𝜌𝑔

ℎ𝐿 = 2𝑓𝑓

Equating these gives, ∆𝑃 𝐿 v2 = 2𝑓𝑓 𝜌𝑔 𝐷𝑔 Solving for ∆𝑃, ∆𝑃 = 2𝜌𝑓𝑓

𝐿 2 50 𝑚 (3 𝑚/𝑠)2 = 2.6𝑥107 𝑓𝑓 v = 2(870 𝑘𝑔/𝑚3 )𝑓𝑓 𝐷 3/100 𝑚

Next, need to find 𝑓𝑓 , and to do this we must determine whether the flow is laminar or turbulent. 3 3 𝜌v𝐷 (870 𝑘𝑔/𝑚 )(3 𝑚/𝑠) (100 𝑚) 𝑅𝑒 = = = 783 𝑘𝑔 𝜇 (0.1 ) 𝑚∙𝑠 16 𝑓𝑓 = = 0.02043 𝑅𝑒 Thus, ∆𝑃 = 2.6𝑥107 𝑓𝑓 = 2.6𝑥107 (0.02043) = 5.5𝑥105

𝑘𝑔 = 5.5𝑥105 𝑃𝑎 𝑚𝑠 2

Chapter 14 Instructor Only Problems

14.25 A centrifugal pump with an impeller diameter of 0.18 m is to be used to pump water (density = 1000 kg/m3) with the pump inlet located 3.8 m above the surface of the supply reservoir. At a flow rate of 0.760 m3/s, the head loss between the reservoir surface and the pump inlet is 1.8 m of water. The performance curves are shown below. Would you expect cavitation to occur?

Solution Pump: D = 0.18 m, inlet at y = 3.8m above supply reservoir, v̇ = 0.760

m3 s

Between reservoir survace & 𝑝𝑢𝑚𝑝 𝑖𝑛𝑙𝑒𝑡: hL = 1.8 m H2 O Patm − Pv Energy Equation: NPSH = − y2 − hL at 20℃ PV = 2.34 kPa ρg Patm − Pv (101.3 − 2.34)(103 ) = = 10.09m ρg 1000(9.81) NPSH = 10.09 − 3.8 − 1.8 = 4.49 mH2 O m3 From Performance curve @ v̇ = 0.760 , NPSH ≅ 3.9m, Cavitation should NOT occur s

14.26 Pumps used in an aqueduct operate at 400 rpm and deliver a flow rate of 220 m3/s against a head of 420 m. What types of pumps are they? Solution m3 = 3.487x106 gpm, h = 420 m = 1378 ft s 1 (400)(3.487x106 )2 NS = = 3302 3 (1378)4 According to Fig 14.11 This is probably a high capacity Centrifugal Pump v̇ = 220

14.27 A pump is required to deliver 60,000 gpm against a head of 300 m when operating at 2000 rpm. What type of pump should be specified? Solution Pump to deliver 60,000 gpm with h = 300m @ 2000 rpm 1

NS =

(2000)(6x104 )2 3

≅ 2790

300 4 (0.3048) Accoding to Fig 14.11 This is probably a high capacity Centrifugal Pump

14.28 An axial-flow pump has a specified specific speed of 6.0. The pump must deliver 2400 gpm against a head of 18 m. Determine the required operating rpm of the pump. Solution 1

Axial Flow Pump − NS = 6.0 = NS =

CQ2

v2w

3 = 3 3 4 h4 g 4 CH

This ratio is (obviously)dimensionless − by converting to units on abscissa of Figure 14.11 − the ratio of NS is given by this equation to the value on Figure 14.11 is 2733 − So A value of 6 for the equation is equivalent to 6(2733) = 1.64x104 on abscissa of Figure 14.11 1

∴ 1.64x104 =

n(2400)2 3 (18)4

∴ n = 2925 rpm

14.29 A pump operating at 520 rpm has the capability of producing 3.3m3/s of water flow against a head of 16 m. What type of pump is this?

Solutions Pump @ 520 rpm,

m3 v̇ = 3.3 , s

h = 16m

3 1 ) (7.48)(60) = 52302 gpm v̇ = (3.3) ( 0.3048

16 = 42.65 ft 0.3048 1

Ns =

(520)(5.23x105 )2 3

(42.65)4

= 22532 ∴ from Fig. 14.11 − Axial Flow

14.30 A pump operating at 2400 rpm delivers 3.2 m3/s of water against a head of 21 m. Is this pump an axial-flow, mixed-flow, or radial-flow machine? Solution Pump @ 2400 rpm, v̇ = 3.2

m3 , h = 21m s

3 1 ) (7.48)(60) = 50720 gpm v̇ = (3.2) ( 0.3048

21 = 68.9 ft 0.3048 1

Ns =

(2400)(5.072x104 )2

= 22601 3 (68.9)4 ∴ from Fig. 14.11 − Axial Flow

Chapter 15 Instructor Only Problems 15.28 Water at 40°F is to flow through a 11=2-in: schedule 40 steel pipe. The outside surface of the pipe is to be insulated with a 1-in.-thick layer of 85% magnesia and a 1-in.-thick layer of packed glass wool, k = 0.022 Btu/h ft °F. The surrounding air is at 100°F. a. Which material should be placed next to the pipe surface to produce the maximum insulating effect? b. What will be the heat flux on the basis of the outside pipe surface area? The convective heattransfer coefficients for the inner and outer surfaces are 100 and 5Btu/hft°F, respectively. Solution Per unit length: 1 R ins = = 0.0238 2πri hi

R1 =

R2 =

R3 = R4 =

r ln (r2 ) 1

2πk 2 r ln (r3 ) 2

2πk 2 r ln (r4 ) 3

2πk 3

= 0.00105

0.115 , k2

0.0659 , k3

1 = 0.1296 2πr4 ho

ΣR = 0.1545 +

0.115 0.0659 + k2 k3

Case 1: k 2 for Magnesia: ΣR = 6.134 Case 2: k 2 for Glass Wool: ΣR = 7.096 (BEST) q=

60 BTU = 8.47 7.096 hrft

q 8.47 BTU = = 5.55 A 2π (2.95) hrft 2 12

15.29 A 1-in.-nominal-diameter steel pipe with its outside surface at 400°F is located in air at 90°F with the convective heat-transfer coefficient between the surface of the pipe and the air equal to 1.5 Btu/h-ft-°F. It is proposed to add insulation having a thermal conductivity of 0.06 Btu/h ft °F to the pipe to reduce the heat loss to one-half that for the bare pipe. What thickness of insulation is necessary if the surface temperature of the steel pipe and ho remain constant? Solution For Base Pipe: q = πD1 hΔT = π ( For Insulated Pipe: 80 =

Ts − T∞ D ln (D2 ) 1 1 + 2πk πD2 h

𝜋ℎ 𝐷2 1 𝜋ℎ ln + = ∆𝑇 2𝜋𝑘 𝐷1 𝐷2 80 D2 1 )+ 12.5 ln ( = 18.25 0.1905 D2 By Trial & Error: D2 = 0.382ft, 2t = D2 − D1 = 3.265in, t = 1.63in

1.315 BTU ) (1.5)(310) = 160 per ft 12 hr

15.30 1/4 If, for the conditions of Problem 15.29, ho in Btu/h ft °F varies according to ℎ𝑜 = 0.575/𝐷𝑜 , where Do is the outside diameter of the insulation in feet, determine the thickness of insulation that will reduce the heat flux to one-half that of the value for the bare pipe. Solution For Base Pipe, Per Foot: q = hAΔT =

ΔT D ln ( Do ) i

2πk 53.3 =

1 πDo ho 310

D ln ( o ) 1.315 + 2π(0.06)

D 4 (12o ) D 0.575π 12o

By trial and error: Do = 9.22in Insulation thickness =

1.315 BTU ) (310) = 106.7 12 hr

1π( 1.315 4

( 12 )

With Insulation: q = 53.3 q=

0.575

9.22 − 1.315 = 3.95in 2

15.31 Liquid nitrogen at 77 K is stored in a cylindrical container having an inside diameter of 25 cm. The cylinder is made of stainless steel and has a wall thickness of 1.2 cm. Insulation is to be added to the outside surface of the cylinder to reduce the nitrogen boil-off rate to 25% of its value without insulation. The insulation to be used has a thermal conductivity of 0.13 W/m K. Energy loss through the top and bottom ends of the cylinder may be presumed negligible. Neglecting radiation effects, determine the thickness of insulation when the inner surface of the cylinder is at 77 K, the convective heat-transfer coefficient at the insulation surface has a value of 12 W/m2 K, and the surrounding air is at 25°C. Solution qo =

ΔT ΣR

13.7 1 ln (12.5) 1 0.6136 K without insulation: ΣR = [ + ]= (12)(0.137) 2πL 17.3 2πL W 13.7 ro ln (13.7 ) 1 ln (12.5) 1 With insulation: ΣR = [ + + ] (12)(ro ) 2πL 17.3 0.13 q wo If q w = 4 2.454 ΣR w = 4ΣR wo = 2πL 13.7 ro ln ( ) ln (13.7 ) 1 12.5 + + = 2.454 (12)(ro ) 17.3 0.13 By trial and error, ro = 0.177m, insulation thickness = ro − ri = 0.177 − 0.137 = 0.04m = 4 cm

15.32 Air at 100oC is moving over a stainless-steel plate. Thermocouples are positioned at 15 and 25 mm below the surface and measure temperatures at 60 and 50oC respectively. Calculate the convective heat transfer coefficient (assume negligible radiation). Solution Stainless Steel: k=17.3 𝑊/𝑚𝐾 (Appendix H). 𝑞𝑖𝑛 = 𝑞𝑜𝑢𝑡 ℎ𝐴∆𝑇 = −𝑘𝐴𝛻⃑𝑇 ℎ(𝑇∞ − 𝑇𝑠𝑠 ) = −𝑘 At steady state, 𝑞 is constant in the plate.

𝑑𝑇 𝑑𝑥

𝑑𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑑𝑥 𝑑𝑇 𝑞 =− 𝑑𝑥 𝑘 𝑑𝑇 𝑇1 − 𝑇2 𝑇𝑆𝑆 − 𝑇1 = = 𝑑𝑥 𝑥1 − 𝑥2 0 − 𝑥1 𝑇𝑆𝑆 = 75℃ 𝑞 = −𝑘

Back to equation to solve for h: ℎ(𝑇∞ − 𝑇𝑠𝑠 ) = −𝑘

ℎ=−

𝑊 (17.3 𝑚 ∙ 𝐾)

𝑑𝑇 𝑑𝑥

60 − 50 = 692 𝑊/𝑚 ∙ 𝐾 (100 − 75) 0.015 − 0.025𝑚

15.33 A steel rod at a temperature of 450oF with an outside diameter of 2 inches is suspended in air. The air temperature is 100oF and the heat transfer coefficient between the rod and the air is 2.0 BTU/h ft2 oF. The rod is coated with insulation with a thermal conductivity of 0.05 BTU/h ft oF in an attempt to minimize heat loss from the rod. Determine the thickness of the insulation necessary to reduce the heat loss by 40%. Solution

q ins = 0.4 (= 40%) q bare r0 = ri + ∆r (where the unknown is ∆r and represents the insulation thickness) Tp − T∞ (Tp − T∞ )L ∆T q ins = = = r r 0 IR ln( 0⁄ri ) 1 ln( ⁄ri ) 1 + + hA 2πkL 2πk ins h(2πr0 ) q bare = hA∆T = h(2πri L)(Tp − T∞ ) q ins = 0.4 = q bare

1 r ln( 0⁄ri ) 1 ( ) (h(2πri L)) + 2πk ins h(2πr0 ) 2 0.167 ft D = 2 in = ft = 0.167 ft, R= = 0.083 ft 12 2 1 0.4 = r ln( 0⁄0.083 ft) 1 BTU ( + BTU BTU ) (2 h ft 2 ℉ (2π(0.083 ft))) (2 ) (2πr0 ) 2π (0.05 ) h ft ℉ h ft 2 ℉ r ln( 0⁄0.083 ft) 1 BTU ( ) (2 + (2π(0.083 ft))) (0.4) = 1 BTU BTU h ft 2 ℉ (2πr ) (2 ) 2π (0.05 ) 0 h ft ℉ h ft 2 ℉ ro = 0.15 ft r0 = ri + ∆r ∆r = 0.15 ft − 0.083 ft = 0.067ft = 0.8 in

15.34 Derive an expression for the thermal resistance, as described in Equation 15-16, for onedimensional heat conduction through a spherical shell assuming steady state. Solution ⃑T q = −kA∇ q r = −k(4πr 2 )

dT dr

T2 dr = −4πk ∫ dT 2 r1 r T1

qr ∫

q r [(−

1 1 ) − (− )] = −4πk(T2 − T1 ) r2 r1 qr =

4πk(T2 − T1 ) 1 1 (r ) − (r ) 1 2

Using the form of Equation 15-16, R=

1 1 (r ) − (r ) 1

4πk

Chapter 16 Problem Assignments Show/Hide problems: 16.1, 16.5 (solutions provided) Instructor Only Problems: (16.15, 16.16, 16.17) (3 new problems)

Chapter 17 Instructor Only Problems

17.43 Circular fins are employed around the cylinder of a lawn mower engine to dissipate heat. The fins are made of aluminum are 0.3-m thick, and extend 2 cm from base to tip. The outside diameter of the engine cylinder is 0.3 m. Design operating conditions are T∞ = 30°C and h = 12 W/m2 K. The maximum allowable cylinder temperature is 300°C. Estimate the amount of heat transfer from a single fin. How many fins are required to cool a 3-kW engine, operating at 30% thermal efficiency, if 50% of the total heat given off is transferred by the fins?

Solution rL = t=

0.3 = 0.15 m r0 = 0.15 + 0.02 = 0.17 m 2

0.003 = 0.0015 m 2

A = 2π(r02 − ri2 ) + Aend = 2π(0.172 − 0.152 ) + 2π(0.17)(0.003) = 0.0434 m2 1

2 h 2 12 (r0 − ri ) ( ) = 0.02 [ ] = 0.263 (46.4)(0.0015) kt

r0 = 1.13 ri From fig 17.11 ηf ≅ 0.96 q = Af ηf h(T0 − T∞ ) = 0.0434(0.96)(12)(270) = 135 W per fin For a 30% 3 kW engine (3 kW) Qin = = 10 kW 0.3 Qout = Qin − W = 7 kW Amount Tx from fins = 0.5(7) = 3.5 kW Number of fins required: n=

3500 W = 25.92 26 fins required W 135 fin

17.44 Heat from a flat wall is to be enhanced by adding straight fins, of constant thickness, made of stainless steel. The following specifications apply:

h = 60 W⁄m2 K Tb (base) = 120℃ T∞ (air) = 20℃ Fin base thickness, t = 6mm Fin length, L = 20 m m Determine the fin efficiency and heat loss per unit width for the finned surface. Solution

a) L = 20 mm t = 6 mm For both cases: Tb = 120℃ T∞ = 20℃ h = 60

W , m2 K

k s.s. = 15.3

W mK

For case a) straight fin q = ηf hAf θb Using text – Fig 17.11 3 2

1 2

3 2 t h 60 2[ (0.02 (L + ) [ = + 0.003) = 0.588 ] ] t (15.3)(0.006)(0.023) 2 kt (L + 2)

ηf ≅ 0.80

Per meter of width {neglecting ends}: q f = 0.8(60)(100)(2)(0.020) = 192 For case b) triangular

W m

3 (L)2 [

3 2 h 2 60 ] = 0.022 [ ] = 0.723 (15.3)(0.003)(0.02) kt(L)

ηf ≅ 0.81 q = ηf Af hθb = (0.81)(2)(0.02)(60)(100) = 194.4

W m

17.49 Find the rate of heat transfer from a 3-in.-OD pipe placed eccentrically inside a 6-in.-ID cylinder with the axis of the smaller pipe displaced 1 in. from the axis of the large cylinder. The space between the cylindrical surfaces is filled with rock wool (k = 0.023 Btu/h ft °F). The surface temperatures at the inside and outside surfaces are 400°F and 100°F, respectively.

Solution Using table 17.1 S=

2π 1 + ρ2 − ϵ 2 ) cosh−1 ( 2ρ

ρ = 0.5 ϵ = S=

1 6 2π

1 1 1 + 4 − 36 −1 ) cosh ( 1

2π 11 cosh−1 ( )

= 9.6

q Btu = kS∆T = (0.023)(9.6)(300) = 66.3 L hr ft

17.50 A cylindrical tunnel with a diameter of 2 m is dug in permafrost (k = -0.341 W/m2 K) with its axis parallel to the permafrost surface at the depth of 2.5 m. Determine the rate of heat loss from the cylinder walls, at 280 K, to the permafrost surface at 220 K.

Solution Table 17.1 S=

2π

2π = = 1.311 −1 ρ cosh−1 ( r ) cosh (2.5)

q W = kS∆T = (0.341)(1.311)(60) = 26.83 L m

17.51 Determine the heat flow per foot for the configuration shown, using the numerical procedure for a grid size of 1.5 ft. The material has a thermal conductivity of 0.15 Btu/h ft °F. The inside and outside temperatures are at the uniform values of 200°F and 0°F, respectively.

Solution

200 + 0 + 2T2 − 4T1 = 0 200 + 0 + T1 − T3 − 4T2 = 0 0 + 0 + 2T2 − 4T3 = 0 T1 = 91.7 ℉ T2 = 83.3 ℉ T4 = 41.7℉ q 200 − 91.7 Btu = 8k [ + 200 − 83.3] = 205 L 2 hr ft

17.52 Repeat the previous problem, using a grid size of 1 ft. Solution

By iteration: Node 1 2 3 4 5 6 7

T,℉ 123 112 74 58 51.5 36 18

q Btu = 8k[(200 − 123) + (200 − 112)] = 198 L hr ft

17.53 A 5-in. standard-steel angle is attached to a wall with a surface temperature of 600°F. The angle supports a 4.375-in. by 4.375-in. section of building bring whose mean thermal conductivity may be taken as 0.38 Btu/h ft °F. The convective heat- transfer coefficient between all surfaces and the surrounding air is 8 Btu/h ft2 °F. The air temperature is 80°F. Using numerical methods, determine

(a) the total heat loss to the surrounding air (b) the location and value of the minimum temperature in the brick

Solution Numerical solution using a 12x12 mesh yields the following: q Btu = 1400 L hr ft 2 Tmin = 91.8 ℉ @ i, j = 12,12

17.54 Saturated steam at 400°F is transported through the 1-ft pipe shown in the figure, which may be assumed to be at the steam temperature. The pipe is centered in the 2-ft-square duct, whose surface is at 100°F. If the space between the pipe and duct is filled with powdered 85% magnesia insulation, how much steam will condense in a 50-ft length of pipe?

Solution S=

2π = 8.17 r0 ln ( ri ) − 0.271

q = kSL∆T = (0.037)(8.17)(300)(50) = 2180 q

2180

Steam condensed = h = 826 = 65 ( hrm ) fg

Btu hr

17.55 A 32.4-cm-OD pipe, 145-cm long, is buried with its centerline 1.2 m below the surface of the ground. The ground surface is at 280 K and the mean thermal conductivity of the soil is 0.66 W/m ? K. If the pipe surface is at 370 K, what is the heat loss per day from the pipe?

Solution S=

2π

ρ = cosh−1 ( r )

q = kS∆T = (0.66

2π 2π

= 3.17

1.2 cosh−1 (0.324) W ) (3.17)(90 K)(145 m) = 273 W mK

Chapter 18 Instructor Only Problems

18.32 A thick wall of oak, initially at a uniform temperature of 25°C, is suddenly exposed to combustion exhaust at 800°C. Determine the time of exposure required for the surface to reach its ignition temperature of 400°C, when the surface coefficient between the wall and combustion gas is 20 W/m2 K. Solution

T − T0 x hy h2 αt x h √αt ) − [exp ( + 2 )] [erfc ( = erfc ( + )] T∞ − T0 k k k 2√αt 2√αt T − T0 400 − 25 = − 0.484 T∞ − T0 800 − 25 @ surface (x=0) x 2√αt

= 0 erf(0) = 0 erfc(0) = 1

h √αt 20 = (1.07x10−7 )2 = 0.0311t 2 = z k 0.21 2

0.484 = 1 − ez (1 − erf(z)) 1

Trial and error z ≅ 0.73 = 0.0311t ⁄2 t = 551 s = 9.18 min

18.33 Air at 65°F is blown against a pane of glass 1/8 in. thick. If the glass is initially at 30°F, and has frost on the outside, estimate the length of time required for the frost to begin to melt.

Solution For glass: α=

k ρĈp

0.45 = 0.0132 ft 2 ⁄hr (170)(0.2)

Using semi-infinite wall expression T − T0 32 − 30 x ) = = 0.0572 = erfc ( Ts − T0 65 − 30 2√αt x 2√αt

= 1.38

t = 3.9 s

18.34 18.34 How long will a 1-ft-thick concrete wall subject to a surface temperature of 1500°F on one side maintain the other side below 130°F? The wall is initially at 70°F.

Solution Charts apply but are difficult to read – check validity of infinite wall solution L 2√αt

1 ft >2 2(0.0231t)2

Works for t > 27 ℎ𝑟𝑠 x 2√αt

1 3.29 = 1 2 2(0.0231t) t ⁄2

Ts − T 3.29 = erf ( 1 ) Ts − T0 t ⁄2 t ≅ 5.2 hrs

18.35 A stainless-steel bar is initially at a temperature of 25°C. Its upper surface is suddenly exposed to an air stream at 200°C, with a corresponding convective coefficient of 22 W/m2 K. If the bar is considered semi-infinite, how long will it take for the temperature at a distance of 50 mm from the surface to reach 100°C?

Solution Applicable expression is T∞ − T hx = erfc(A) + exp ( + B2 ) [1 − erf(A + B)] T∞ − T0 k A=

x 2√αt

0.05 1 2(0.444x10−5 t) ⁄2

= 11.92t

−1⁄ 2

hx 22(0.05) = = 0.0636 k 17.3 B=

h √αt 22 1 1 = (0.444x10−5 t) ⁄2 = 0.00267t ⁄2 k 17.3

B2 = 7.11x10−6 t Trial and error t ≅ 49000 s ≅ 13.6 hrs

18.39 IF the heat flux into a solid is given as F(t), show that the penetration depth 𝛿 for a semi-infinite solid is of the form 𝑡

1/2

∫ 𝐹(𝑡)𝑑𝑡 𝛿 = (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)√𝛼 [ 0 ] 𝐹(𝑡) Solution Eqn. (18-33) qx d δ dδ = ∫ ρĈp T dx − ρĈp T0 A dt 0 dt T − T0 x = Φ( ) Ts − T0 δ ∂T 1 ∂Φ | = (Ts − T0 ) | ∂x x=0 δ ∂x x=0 ∂Φ ∂x

(0) = K (a constant)

qx ∂T = −K (0) = F(t) A ∂x  Ts − T0 = F(t) ρĈp

F(t)δ kK

d δ dδ ∫ T dx − T0 dt 0 dt

δ d δ dδ d dδ d ∫ T dx = T0 + ( Ts − T0 ) ∫ Φ dx = T0 + [(T − T0 )Bδ] dt 0 dt dt dt dt s 0

(B is a constant) d d Bδ2 F(t) [(T )Bδ] = = [ ] s − T0 dt kK ρĈp dt

F(t)

F(t)kK ρĈp B

d 2 [δ F(t)] dt

t K δ F(t) = α ∫ F(t) dt B 0 2

∫ F(t) dt 1⁄ δ = (constant)√α( 0 ) 2 F(t)

18.40 If the temperature profile through the ground is linear, increasing from 35°F at the surface by 0.5°F per foot of depth, how long will it take for a pipe buried 10 ft below the surface to reach 32°F if the outside air temperature is suddenly dropped to 0°F. The thermal diffusivity of soil may be taken as 0.02 ft2/h, its thermal conductivity is 0.8 Btu/h ft °F, and the convective heat- transfer coefficient between the soil and the surrounding air is 1.5 Btu/h ft2 °F.

Solution Numerical solution required Initial temperature profile – T = 35 + 0.5x T in ℉, x in ft Algorithms For all nodes except surface: Tit+1 =

Ti+1 + Ti−1 2

For surface node: T0t+1 = T1t − ~

α∆t 1 = ; ∆x 2 2

h∆x t T k 0 2h∆t ρĈp ∆x

h∆x ; k

T∞ = 0

Result – Using a spreadsheet or program t ≅ 1800 hours

18.41 A brick wall (α=0.016ft2/h) with a thickness of 1.51 ft is initially at a uniform temperature of 80°F. How long, after the wall surfaces are raised to 300°F and 600°F, respectively, will it take for the temperature at the center of the wall to reach 300°F?

Solution Numerical solution required Algorithms: Node 1: T1t+1 =

T0 + T2n 2

t+1 Tn−1 =

t Tn + Tn−2 2

Tit+1 =

Ti+1 + Ti−1 2

α∆t

For ∆x2 = 2 If ∆x = 0.25 ft; ∆t = 1.95 hr No of increments ≅7.4 t = 7.4(1.95) ≅ 144 hrs

18.42 A masonry brick wall 0.45 m thick has a temperature distribution at time, t=0 which may be approximated by the expression T(K)=520+330sinπ(x/L) where L is the wall width and x is the distance from either surface. How long after both surfaces of this wall are exposed to air at 280 K will the center temperature of the wall be 360 K? The convective coefficient at both surfaces of the wall may be taken as 14 W/m2 K. What will the surface temperature be at this time?

Solution T = 520 + 330 sin ( α

k ρĈp

πx ) L

0.66 = 4.72x10−5 m2 ⁄s (1670)(8.38)

α∆t 1 = ∆x 2 2 Same as problem (18.34) α∆t

For ∆x2 = 2 If ∆x = 0.225 m; ∆t = 536 s No of increments ≅2.4 Time ≅ 2.4(536) =1286 s =21.4 min At this time Tsurface ≅ 360 K

Chapter 19 Instructor Only Problems

19.20 Determine the total heat transfer from the vertical wall described in Problem 19.19 to the surrounding air per meter of width if the wall is 2.5 m high. Solution L L k βg 4 3 5 1 −1 q = ∫ hx ∆T dx = ∫ Nux ∆T dx = k∆T ⁄4 (0.508)Pr ⁄2 (Pr + 0.954) ⁄4 ( 2 ) ( L ⁄4 ) x ν 3 0 0 = 995 W

19.21 Simplified relations for natural convection in air are of the form ℎ = 𝛼(∆𝑇/𝐿)𝛽 where α, β are constants; L is a significant length, in ft; ΔT is Ts - T∞, in °F; and h is the convective heat-transfer coefficient, Btu/h ft2 °F. Determine the values for α and β for the plane vertical wall, using the equation from Problem 19.14. Solution hx x 1 −1 1 Nux = = 0.508Pr ⁄2 (Pr + 0.954) ⁄4 Gr ⁄4 k For air: Pr = 0.72, k = 0.015 Btu⁄hr ft ℉ 1

∆T 4 ∆T 4 h = k(0.38) (2x106 ) = 0.445 ( ) x x L

q = hL A∆T = ∫ hx ∆T dx 0 1

1 L ∆T 4 ∆T β hL = ∫ hx dx = 0.286 ( ) = α ( ) L 0 L L α = 0.286 β = 1⁄4

19.23 Repeat Problem 19.22 for velocity and temperature profiles of the from, 𝑣 = 𝑎 + 𝑏𝑦 + 𝑐𝑦 2 T − Ts = α + βy + γy 2 Solution For v = a + by + cy 2 B.C. v(0) = 0 v(δ) = v∞ dv (δ) = 0 dy v y y = 2 − ( )2 v∞ δ δ For T − Ts = α + βy + γy 2 B.C. (T − Ts )|0 = 0 (T − Ts )|δt = 0 ∂ (T − Ts )| = 0 ∂y δ t

T − Ts y y = 2 − ( )2 T∞ − Ts δt δt Into momentum equation to get: νx (1) δ2 = 30 v∞ Into energy equation to get: 1 α δξd (δξ2 − δξ3 ) = 12 dx 5 v∞ δ

Where ξ = δt Solution gives ξ ≅ Pr ∴ δt = Pr

−1⁄ 3δ

−1⁄ 3

(2)

Since A = −k dy (0) = h(Ts − T∞ ) −1

h 2 2Pr ⁄3 v∞ ⁄2 −1 = = = 0.365Pr ⁄3 ( ) k δt δ νx hx

Or: Nux = k = 0.365Pr

−1⁄ 1 3 (Rex ) ⁄2

19.26 For the case of a turbulent boundary layer on a flat plate, the velocity profile has been shown to follow closely the form 𝑣 𝑦 1/7 =( ) 𝑣∞ 𝛿 Assuming a temperature profile of the same form – that is, 1

T − Ts y 7 =( ) T∞ − Ts δt And assuming that 𝛿 = 𝛿1 , use the integral relation for the boundary layer to solve for hx and Nux. The temperature gradient at the surface may be considered similar to the velodity gradient at y=0 given by equation (13-26). Solution 1

v y 7 =( ) v∞ δ

T − Ts y 7 =( ) T∞ − Ts δt

Energy equation: ∂T d δt vx T − Ts (1 − ) dy α (0) = (T∞ − Ts )v∞ ∫ ∂y dx 0 v∞ T∞ − Ts 1

0.0225(T∞ − Ts )v∞ ν 4 ( ) LHS = ν⁄ δv∞ α RHS = (T∞ − Ts )v∞

7 dδ 72 dx

{Assumes δ = δt for integration} Equating & some algebra: δ −1⁄ 4 = 0.371Rex 5 Pr − ⁄5 x 1

q ∂T k(0.0225)(∆T)v∞ ν 4 ( ) = h∆T = −k (0) = − A ∂y ν δv∞ Nux =

19⁄ hx 1 = 0.0288Rex 20 Pr ⁄5 k

19.32 Work Problem 19.29 for the case in which the flowing fluid is sodium entering the tube at 200oF. Solution q For contant ⁄A q ⁄ 500 T = T0 + A = 200 + ≅ 200℉ (58.1)(12.25 ∗ 3600)(0.332) ρvCp

Chapter 20 Instructor Only Problems

20.38 A tube bank employs tubes that are 1.30 cm in outside diameter at ST = SL = 1.625 cm. There are eight rows of tubes, which are held at a surface temperature of 90°C. Air, at atmospheric pressure and a bulk temperature 27°C, flows normal to the tubes with a free-stream velocity of 1.25 m/s. The tube bank is eight rows deep, and the tubes are 1.8 m long. Estimate the heat-transfer coefficient.

Solution Using figure 20.12 Same caveats as for problem 20.51 Dequi v =

Re =

4 π [(0.032)(0.032) − (0.013)2 ] = 0.0873 m π(0.013) 4

(0.0873)(1.25) = 6.95x103 −5 1.569x10

Out of laminar range, must use Fig. 20.13 Re =

(0.013)(1.25) = 1.04x103 1.569x10−5

For in-line configuration j ≅ 0.017 h = 0.017(1006.3)(1.177)(1.25)(0.708)

−

2 2.143 −0.14 3( ) = 30.95

A = 64π(0.013)(1.8) = 4.70 m2 q = hA∆T = 30.95(4.70)(63) = 9.164 kW

1.813

W m2 K

20.39 Rework Problem 20.38 for a staggered arrangement. All other conditions remain the same.

Solution Same conditions as Prob. 20.53 except tubes are in staggered configuration All calculations the same as in prob. 20.54 except j = 0.035 Giving h = 63.7 W/m2 K & q = (63.7)(4.70)(63) = 18.87 kW

20.40 Air at 60°F and atmospheric pressure flows inside a 1-in., 16-BWG copper tube whose surface is maintained at 240°F by condensing steam. Find the temperature of the air after passing through 20 ft of tubing if its entering velocity is 40 fps.

Solution Assume TL = 235℉ Tavg = 148℉ 0.87 Dv ( 12 ) (40) Re = = = 1.39x104 ν 0.209x10−3 {turbulent} −0.2 St = 0.023Re−0.8 = 4.33x10−3 D Pr L TL − Ts = e−4(D)St T0 − Ts

TL = 240 − 180(0.0083) = 239 ℉ Close enough

20.41 A valve on a hot-water line is opened just enough to allow a flow of 0.06 fps. The water is maintained at 180°F, and the inside wall of the 1/2-in. schedule-40 water line is at 80°F. What is the total heat loss through 5 ft of water line under these conditions? What is the exit water temperature?

Solution G = ρν = 3.64 GD

Re = μ =

Lbm s ft 2

0.622 ) 12 0.29x10−3

3.64(

= 650 {Laminar}

Use Sieder-Tate eqn. assume Tb avg = 150℉ 1

0.14

D 3 μ Nu = 1.86 (RePr ) ( b ) L μs

k 0.383(1.86) 0.622 3 0.29 0.14 Btu )] ( ) h = ( ) Nu = [(650)(2.71) ( = 32.9 0.622 D 12(5) 0.578 hrft 2 ℉ 12 St =

Nu = 0.00252 RePr

L TL − Ts = e−4(D)St = e−0.972 = 0.378 T0 − Ts

T = 80 + 0.378(100) = 117.8 ℉ Tb avg =

117.8 + 180 = 149 ℉ 2

{using steam tables} q = ṁCp ∆T = 3.64(0.0021)(149.7 − 85.7) = 1710 Btu/hr

20.42 When the valve on the water line in Problem 20.41 is opened wide, the water velocity is 35 fps. What is the heat loss per 5 ft of water line in this case if the water and pipe temperatures are the same as specified in Problem 20.41?

Solution G = 60.6(35) = 2120

Lbm s ft 2

Re = 307000 {turbulent} Tf ≅ 130℉ use Colburn eq: 2

St = 0.023(307000)−0.2 (3.44)−3 = 0.000807 St(4)(5)

T = 80 + 100e

− 0.622 12

= 153.3℉ {First guess}

80 + 154 + 180] 2 Tf = ≅ 149℉ 2 [

At this temp: Re=443000 Pr=4.51 St=0.000625 T ≅ 155℉

20.43 Steam at 400 psi, 800°F flows through a 1-in. schedule- 40 steel pipe at a rate of 10,000 lbm/h. Estimate the value of h that applies at the inside pipe surface.

Solution GA = 10000 G=

Lbm hr

10000 Lbm = 37400 0.276 hrft 2

Re =

7.001 37400 ( 12 ) 1.63x10−5 (3600)

= 3.72x105 {Turbulent}

Use Dittus-Boelter Eqn.: k 0.0321 Btu (0.023)(3.72x105 )0.8 (0.912)0.3 = 35.2 h = ( ) (0.023)Re0.8 Pr 0.3 = 7.001 D hrft 2 ℉ 12

20.49 Air at 25 psia is to be heated from 60°F to 100°F in a smooth, 3/4-in.-ID tube whose surface is held at a constant temperature of 120°F. What is the length of the tube required for an air velocity of 25 fps? At 15 fps?

Solution L T − Ts = e−4(D)St T0 − Ts

St =

D 60 − 120 0.017171 )= ln ( 4L 100 − 120 L

a) v = 25 ft/s

Re =

0.75 ( 12 ) (25) 0.181x10−3

= 8640 C

Use Colburn analogy: St = 2f Pr −3 St = L=

0.0078 (1.257) = 0.0049 2

0.01717 = 3.5 ft 0.0049

b) v = 15 s

Re = 5190

St = 0.00566 L = 3.03 ft

20.50 Air is transported through a rectangular duct measuring 2 ft by 4 ft. The air enters at 120°F and flows with a mass velocity of 6 lbm/s ? ft2. If the duct walls are at a temperature of 80°F, how much heat is lost by the air per foot of duct length? What is the corresponding temperature decrease of the air per foot? Solution

Dequiv =

Re =

4(2)(4) 16 = ft (2)(6) 6

16 ( 6 ) (6) 1.28x10−5

= 1.25x106 2

St = 0.023Re−0.2 Pr −3 = 0.023(1.25x106 )−0.2 (0.703)−3 = 0.00176 For the short distance involved: h = St(ρvCp ) = (0.00176)(6)(924) = 2.53x10−3 q = hA∆T = (2.53x10−3 )(12)(40) = 1.215 ∆T =

q 1.215 = = 0.105 ℉ per ft ṁCp (0.24)(6)(8)

Btu s ft 2 ℉

Btu Btu per ft = 4370 per ft s hr

20.51 Cooling water flows through thin-walled tubes in a condenser with a velocity of 1.5 m/s. The tubes are 25.4 mm in diameter. The tube-wall temperature is maintained constant at 370 K by condensing steam on the outer surface. The tubes are 5 m long and the water enters at 290 K. Estimate the exiting water temperature and the heat- transfer rate per tube.

Solution Assume Tin = 290 K

Tout = 350 K

Tsurf = 370 K Tb avg = 320 K ν = 0.596x10

−6

m2 Pr = 3.87 s

L T − Ts = e−4(D)St T0 − Ts

Re =

(0.0254)(1.5) = 63900 0.596x10−6

Use Dittus-Boelter equation: St = 0.023 = 0.023(63900)−0.2 (3.87)−0.6 = 0.00112 L

e−4(D)St = 0.414Re−0.2 Pr −0.6 Tout = 370 − (0.414)(80) ≅ 337 Second try Tout = 337 Tb avg = 313.5 ν=0.663x10−6 Pr = 4.33 Re = 57460 St = 0.00107 L

e−4(D)St = 0.432 Tout = 370 − (0.432)(80) = 335 K

π q = ṁCp ∆T = (992) ( ) (0.0254)2 (1.5)(4175)(45) = 141.6 kW 4

20.52 Air, at 322 K, enters a rectangular duct with a mass velocity of 29.4 kg/s m2. The duct measures 0.61 m by 1.22 m and its walls are at 300 K. Determine the rate of heat loss by the air per meter of duct length and the corresponding decrease in air temperature per meter.

Solution Rectangular duct: 0.61m x 1.22m Dequi v =

4(0.61)(1.22) = 0.813 m 2(0.61 + 1.22)

q = hA∆T Use Dittus-Boelter Equation Re =

DG (0.813)(29.4) = = 1.227x106 −5 μ 1.948x10

Pr = 0.703 h=

k 0.0279 W (0.023)Re0.8 Pr 0.3 = (0.023)(1.227x106 )0.8 (0.703)0.3 = 52.8 2 D 0.813 m K

q = hA∆T = 52.8(2)(0.61 + 1.22)(22) = 4250 q = ṁCp ∆T = 6ACp ∆T ∆T =

4250 = 0.193 K per m 29.4(0.61)(1.22)(1007)

W m

Chapter 21 Instructor Only Problems

21.16 If eight tubes of the size designated in Problem 21.13 are arranged in a vertical bank and the flow is assumed laminar, determine a. the average heat-transfer coefficient for the bank b. the heat-transfer coefficient for the first, third, and eighth tubes

Solution 1

1 4 h havg = h̅ ( ) = 8 1.681 W

hhoriz = 2250 m2K {from prob. 21.16} 2250

For bank: havg = 1.681 = 1341 m2K For n tubes q = havg,n nAtube ∆T = havg,(n−1) (n − 1)Atube ∆T ̅̅̅n − (n − 1)h ̅̅̅̅̅̅ nth tube: hn = nh n−1 W

top tube: hi = 2250 m2 K W ̅̅̅3 = 1710 W 3rd tube: ̅̅̅ h2 = 1890 m2K h m2 K

h3 = 3(1710) − 2(1890) = 1350

W m2 K

W ̅̅̅7 = 1383 W 8th tube: ̅̅̅ h8 = 1341 m2K h m2 K

h8 = 8(1341) − 7(1383) = 1047

W m2 K

21.17 Given the conditions of Problem 21.16, what height of vertical wall will cause the film at the bottom of the tube to be turbulent?

Solution 4AΓc = Rec = 2000 Pμf 4AΓc A ṁ 1 4 hPL∆T = 4( )( )( ) = ( )( ) Pμf P A μ Pμ hfg 1

μhfg (0.0206x10−3 )(970)(2000)L4 (3600) L= = 1 4h∆T 4(100) [2250(1.3)(0.02)4 ] 3

L4 = 3.27 L = 4.85 ft

21.18 A vertical flat surface 2 ft high is maintained at 60°F. If saturated ammonia at 85°F is adjacent to the surface, what heat-transfer coefficient will apply to the condensation process? What total heat transfer will occur?

Solution 1

1 4 3 ρL (∆ρ)g (hfg + 8 Cp ∆T) 37.2(32.2)(37.2)(0.294)3 (505) 4 L h = 0.943 [ ] = 0.943 [ ] Lμ∆T 2(14x10−5 )(25)

= 694

Btu hr ft 2 ℉

q = hA∆T = 694(2)(25) = 34700

Btu per foot of witdh hr

Chapter 22 Instructor Only 22.10 Consider the exchanger in Problem 22.9. After 4 years of operation, the outlet of the oil reaches 28°C instead of 30°C with all other conditions remaining the same. Determine the fouling resistance on the oil side of the exchanger.

Solution Same exchanger & entrance conditions as problem 22.8 {Problem statement should say T0 out = 28℃} q = ṁ0 Cp ∆T0 = (12)(2200)(8) = ṁw Cp ∆Tw = 3.26(4180)∆Tw 0

∆Tw ≅ 16℃ T0 out ≅ 59℃ ∆TLM =

47 − 39 = 42.9℃ 47 ln (39)

To find F: Y=

8 = 0.145 55

16 =2 8

Fig 22.9 a F≅1 U=

(12)(2200)(8) q W = = 802 2 UF∆TLM 6.14(42.9) m K

UA|0 =

Fouling Resistance

1 ∑ R0

UA|1 =

1 ∑ R1

∑ R0 =

1 = 1.508x10−4 (1080)(6.14)

∑ R1 =

1 = 2.031x10−4 (802)(6.14)

= ∑ R1 − ∑ R 0 = 0.5230x10−4

K W

22.12 A shell-and-tube exchanger having one shell pass and eight tube passes is to heat kerosene from 80°F to 130°F. The kerosene enters at a rate of 2500 lbm/h. Water entering at 200°F and at a rate of 900 lbm/h is to flow on the shell side. The overall heat-transfer coefficient is 260 Btu/h ft2 °F. Determine the required heat-transfer area.

Solution ṁw = 900 U = 260 ∆Tw =

Lbm Lbm ṁk = 2500 hr hr

Btu hr ft 2 ℉

2500(130 − 80)(0.51) = 70.8 900(1)

Tw,out = 129℉ ∆TLM =

AF =

70 − 49 = 58.9℉ 70 ln (49)

q 2500(50)(0.51) = = 4.16 U∆TLM 58.9(260)

Figure 22.9a F≅ Y=

130 − 80 = 0.416 200 − 80

200 − 129 = 1.4 130 − 80

4.16 = 5.01 ft 2 0.83

22.13 A condenser unit is of a shell-and-tube configuration with steam condensing at 85°C in the shell. The coefficient on the condensate side is 10,600 W/m2 ? K. Water at 20°C enters the tubes, which make two passes through the single-shell unit. The water leaves the unit at a temperature of 38°C. An overall heat- transfer coefficient of 4600 W/m2 ? K may be assumed to apply. The heat-transfer rate is 0.2 106 kW. What must be the required length of tubes for this case? Solution

2x108 q = ṁw Cp ∆Tw = q = Cw ∆Tw Cw = ρAvCp = = 1.11x107 w 18 UA 4600A = = 4.14x10−4 A Cmin 1.11x107 A = nπDL = nπ (

1.37 ) L = 0.359nL 12

UA = 1.485x10−4 nL Cmin Neglect tube resistance: U=

1 1 1 = − hi 4600 10600

1 1 +( ) hi h0

hi = 8130

Nui =

1.37 8130 ( 12 ) 0.615

= 1509

Using Dittus Boelter equation: 1.37 0.8 ρv ( 12 ) Nu = 1509 = 0.023 [ ] (5.65)0.4 −6 825x10 q 2x108 ρv = = = 3192 nπD2 ACp ∆T ( ) (4179)(18) 4 n = 81 tubes q = εCmin (65)

ε = 0.277

Fig. 22.2c

UA Cmin

≅ 0.38 L=

0.38 = 31.7 m (1.48x10−4 )(81)

22.25 Determine the required heat-transfer surface area fro a heat exchanger constructed from 10-cm OD tubes. A 95% ethanol solution (cp = 3.810 kJ/kg K), flowing at 6.93 kg/s is cooled from 340 to 312 K by 6.30 kg/s of water that is available at 283 K. The overall heat-transfer coefficient based on outside tube area is 568 W/m2 K. Three different exchanger configurations are of interest: a. Counterflow, single pass b. Parallel flow, single pass c. Crossflow with one tube pass and one shell pass, shell-side fluid mixed

Solution ṁEth = 6.93

kg s

q = ṁCp ∆T|Eth = (6.93)(3810)(28) = ṁCp ∆T|w = 6.30(4182)∆T ∆Tw = 28.1 (a) Counter flow: ∆TLM ≅ 29℉ A=

q 6.93(3810)(28) = = 44.9 m2 U∆TLM 568(29)

(b) Parallel flow: ∆TLM =

57 − 0.9 = 13.52 57 ln (0.9)

q = 96.3 m2 568(13.52)

Cunmixed = ṁCp = 26403 E

312 − 340 = 0.491 283 − 340

28.1 ≅1 28

F ≅ 0.85

44.9 = 52.8 m2 0.85

22.26 Water flowing at a rate of 10 kg/s through 50 tubes in a double-pass shell-and-tube heat exchanger heats air that flows on the shell side. The tubes are made of brass with outside diameters of 2.6 cm and are 6.7 m long. Surface coefficients on the inside and outside tube surfaces are 470 and 210 W/m2 ? K, respectively. Air enters the unit at 15°C with a flow rate of 16 kg/s. The entering water temperature is 350K. Determine the following: (a) Heat-exchanger effectiveness (b) Heat-exchanger rate to the air (c) Exiting temperatures of the water and air systems If, after a long period of operation, a scale has been built up inside the tubes resulting in an added fouling resistance of 0.0021 m2 K/W, determine the new results for parts (a), (b), and (c), above.

Solution hw = 470

W m2 K

ṁw = 10

hA = 210

W m2 K

ṁA = 16

q = (10

kg s

kg J kg J ) (4181 ) (350 − Tw out ) = (16 ) (1007 ) (TA out − 288) s kgK s kgK

Cw = 4180

CA = 16112 = Cmin

Cmin = 0.385 Cmax A = πDL(50) = π(0.026)(6.7)(50) = 27.36 m2 {Assumes total length of each tube is 6.7 m} U=

1 1 1 ( ) + R cond + ( ) hi h0

1 1 1 (470) + (210)

UA 145(27.36) = = 0.246 Cmin 16112 ε ≅ 0.20

= 145

q = εCmin (Tw i − TA i ) = 0.2(16112)(62) = 199800 W Tw out = 347.6

TA out = 294.2

22.28 For the heat exchanger described in Problem 22.27, it is observed, after a long period of operation, that the cold stream leaves at 184°F instead of at the design value of 220°F. This is for the same flow rates and entering temperatures of both streams. Evaluate the fouling factor that exists at the new conditions.

Solution If counter flow: AUold =

1 1 1 )] [( ) + RT + ( Ai hi A0 h0

= 36.7(300)

For new operating conditions: q = 5000(1)(184 − 75) = 2400(1)∆TH = 545000 ∆TH = 227 ∆TLM =

216 − 98 = 149.3 216 ln ( 98 )

AUnew =

[

TH out = 173

q 545000 1 = = 3650 = 1 1 ∆TLM 149.3 )] + R F [( ) + RT + ( Ai hi A0 h0 1 1 [36.7 (300)] + R F

1 1 (300)] + R F = 36.7 3650

R F = 1.83x10−4

K W

Or −3

AR F = 36.7R = 6.27x10

m2 K W

Chapter 23 Instructor Only Problems

23.21 A circular duct 2 ft long with a diameter of 3 in. has a thermocouple in its center with a surface area of 0.3in.2. The duct walls are at 200°F, and the thermocouple indicates 310°F. Assuming the convective heat-transfer coefficient between the thermocouple and gas in the duct to be 30 Btu/h ft2 °F, estimate the actual temperature of the gas. The emissivity of the duct walls may be taken as 0.8 and that of the thermocouple as 0.6.

Solution Assuming thermocouple at geometric center of duct Solid angle of duct opening 1.5 2 ( 12 ) πR2 duct area Ω≅ 2= = = 0.0156 12 πR 0 hemisphere surface {Thermocouple sees duct primarily} For thermocouple: q rad = q conv Aℱcw (Ebc − Ebw ) = hA(TG − Tc ) Ac ℱcw =

ℱcw =

1 ρ ρ 1 (A cϵ ) + (A F ) + (A wϵ ) c c c cw w w 1

ρ Ac ρ 1 (ϵc ) + (F ) + (A ϵw ) c

ℱcw ≅ 1

w w

Ac ≅0 Aw

∴ ℱcw =

1 ≅ ϵc = 0.6 ϵ 1 − (ϵc ) + 1 c

30(TG − Tc ) = ϵc (0.1714) [( Tc = 316℉

Tc 4 Tw 4 ) −( ) ] 100 100

23.26 A room measuring 12 ft by 20 ft by 8 ft high has its floor and ceiling temperatures maintained at 85°F and 65°F, respectively. Assuming the walls to be reradiating and all surfaces to have an emissivity of 0.8, determine the net-energy exchange between the floor and ceiling.

Solutions {Walls assumed to be at a uniform temperature} R1 =

0.2 = 0.00104 12(20)(0.8)

R 2 = 0.00104 R3 =

1 1 = = 0.00903 AF FF−c 12(20)(0.45)

R4 =

1 1 = = 0.0076 AF FF−w 12(20)(0.55)

R 5 = R 4 = 0.0076 1

R equiv = (

1 1 )+( ) R3 R4 + R5

= 0.0058

∑ R = R1 + R 2 + R equiv = 0.00785 q=

σ(TF4 − Tc4 ) 0.1714(5.454 − 5.254 ) Btu = = 2680 ∑R 0.00785 hr

23.27 A dewar flask, used to contain liquid nitrogen, is made of two concentric spheres separated by an evacuated space. The inner sphere has an outside diameter of 1 m and the outer sphere has an inside diameter of 1.3 m. These surfaces are both diffuse-gray with ε = 0.2. Nitrogen, at 1 atmosphere, has a saturation temperature of 78 K and a latent heat of vaporization of 200 kJ/kg. Under conditions when the inner sphere is full of liquid nitrogen and the outer sphere is at a temperature of 300 K, estimate the boil-off rate of nitrogen.

Solution {Equivalent circuit} R1 =

ρ1 ρ 1 1 R2 = = R3 = 2 A1 ϵ1 A1 F12 A2 F21 A 2 ϵ2

T1 = 300 K T2 = 78 K A1 = πD12 = π(1.3)2 = 1.69π m2 A2 = πD22 = π(1)2 = π m2 R1 =

0.8 2.37 −1 = m (1.69π)(0.2) π

R2 =

1 1 = m−1 π(1) π

(23.7 continued) R3 =

0.8 4 = m−1 π(0.2) π

∑R =

7.37 = 2.35 m−1 π

Eb1 − Eb2 σ(T14 − T24 ) 5.676(34 − 0.784 ) = = = 194.8 W ∑R ∑R 2.35

Boil off rate: ṁ = h

ṁ =

1948 kg kg = 9.74x10−4 = 3.51 5 2x10 s hr

23.31 Two parallel rectangles have emissivities of 0.6 and 0.9, respectively. These rectangles are 1.2 m wide and 2.4 m high and are 0.6 m apart. The plate having ε = 0.6 is maintained at 1000 K and the other is at 420 K. The surroundings may be considered to absorb all energy that escapes the twoplate system. Determine a. The total energy lost from the hot plate b. The radiant-energy interchange between the two plates Solution

ϵ1 = 0.6 T1 = 1000 K A1 = 2.88 m2 ϵ2 = 0.9 T2 = 400 K

A2 = 2.88 m2

R1 =

ρ1 = 0.231 A1 ϵ1

R3 =

ρ2 1 = 0.039 R 4 = R 5 = = 0.694 A 2 ϵ2 A1 F13

R2 =

1 = 0.694 A1 F12

Writing equations for loops as shown: Eb1 − 0 = (I1 + I2 )R1 + I1 R 4 Eb2 − 0 = (I2 + I3 )R 3 + I2 R 5 Eb1 − Eb2 = (I1 + I3 )R1 + I3 R 2 + (I3 −I2 )R 3 Substituting values & solving simultaneous equations I1 = 59550 I2 = 4695 I3 = 42970 q1 net = I1 + I3 = 102.5 kW q12 = I3 = 42.97 kW {These results presume no HT TX from other sides of plates}

23.34 Evaluate the net heat transfer between the disks described in Problem 23.33 if they are bases of a cylinder with the side wall considered a nonconducting, reradiating surface. How much energy will be lost through the hole?

Solution R1 = 470 R 2 = 11.94 R 3 = 30.6 R 4 = 735 R 5 = 19.6 Eb1 = 1460 Eb2 = 345 Eb3 = 0 4 is adiabatic For black surfaces q12 =

Eb1 − Eb2 R equiv,12

R equiv,12 =

q12 =

1 1 1 (R ) + R + R 1 2 4

1460 − 345 Btu = 3.87 288 hr

q lost through hole = q13 =

R equiv =

Eb1 − 0 R equiv

1 1 1 (R ) + (R + R ) 5

q lost =

1460 Btu = 77.5 18.83 hr

For grey surfaces: ϵ1 = 0.6 ϵ2 = 0.3 Additional resistance RA, RB

= 288

= 18.83

RA =

RB =

q12 =

ρ1 0.4 = = 7.64 A1 ϵ1 π 4 2 4 (12) 0.6 0.3 π 4 2.5 2 [( ) − ( 4 12 12 ) ] 0.7 2

= 8.06

1460 − 345 Btu = 3.67 288 + 7.64 + 8.06 hr

q lost =

1460 Btu = 55.2 18.83 + 7.64 hr

23.35 Evaluate the heat transfer leaving disk 1 for the geometry shown in Problem 23.33. In this case the two disks comprise the bases of a cylinder with side wall at constant temperature of 350°F. Evaluate for the case where a. The side wall is black b. The side wall is gray with ε=0.2 Determine the rate of heat loss through the hole in each case.

Solution R1 = 470 R 2 = 11.94 R 3 = 30.6 R 4 = 735 R 5 = 19.6 Eb1 = 1460 Eb2 = 345 Eb3 = 0 Eb4 = 738 Writing loop equations: Eb1 − Eb4 = R 2 (I1 + I4 + I5 ) Eb4 − 0 = R 3 (I3 − I5 ) Eb1 − Eb2 = R1 (I2 − I4 ) 0 = R1 (I4 − I2 ) + R 2 (I4 + I1 − I5 ) + I4 R 4 0 = R 3 (I5 −I3 ) + R 2 (I5 − I1 − I4 ) + I5 R 5 Solving: I1 = 134.8

I2 = 2.84

I4 = 0.47

I3 = 98.7

I5 = 74.6

q12 = I2 = 2.84

Btu hr

q lost = I3 + I5 = 173.3

Btu hr

23.39 A duct with square cross section measuring 20 cm by 20 cm has water vapor at 1 atmosphere and 600 K flowing through it. One wall of the duct is held at 420 K and has an emissivity of 0.8. The other three walls may be considered refractory surfaces. Determine the rate of radiant-energy transfer to the cold wall from the water vapor.

Solution q gas−wall direct = A1 F1G αG σ(TG4 − T14 ) q gas to reradiating walls = A2 F2G αG σ(TG4 − T24 ) q gas reradiating walls to (1) = A1 F1G τG σ(TG4 − T14 ) q G2 = q 21 = q R =

σ(TG4 − T14 ) 1 1 (A F τ ) + (A F α ) 1 1G G

2 2G G

q total to 1 = q G1 + q R L = 3.4(0.2)(0.2)(1)4(0.2)(1) = 0.17 m p = 1 atm

αG = 0.22

PL = 0.558 atm ft

τG = 0.78

R1 =

1 = 22.7 0.2(1)(0.22)

R2 =

0.2 = 1.25 0.2(1)(0.8)

R3 =

1 = 7.58 3(0.2)(1)(1)(0.22)

R4 =

1 = 6.41 0.2(1)(1)(0.78)

R equiv =

1 1 1 (R ) + (R + R ) 1 3 4

= 8.66

∑ R = 8.66 + 1.25 = 9.91

5.676(64 − 4. 24 ) W = 564 9.91 m

23.40 A gas of mixture at 1000 K and a pressure of 5 atm is introduced into an evacuated spherical cavity with a diameter of 3 m. The cavity walls are black and initially at a temperature of 600 K. What initial rate of heat transfer will occur between the gas and spherical walls if the gas contains 15% CO2 with the remainder of the gas being nonradiating?

Solution q net = σA(ϵG TG4 − αG Tw4 ) A = 4πr 2 = π(3 m)2 = 28.27 TG = 1000 K Tw = 600 K L=

2 D=2m 3

pL = 0.15(5)(6.56) = 4.92 atm ft αG = 0.18 ϵG = 0.22 q net = 5.676(9π)[0.22(104 ) − 0.18(64 )] = 316 kW

24.1 A = H2S, B = N2, C = SO2 "

34.08

0.03,

g‰Š‹Œ

0.92, Q

28.01 g‰Š‹Œ, "Q 0.05

T = 350 K, P = 1.0 atm.

64.07

a. Determine c, cA, and ρA Z 1.0 atm mM · atm ab u8.206 > 10AB v 350 K gmole · K gmole v mM

0.03 u34.8

u1.04

€ •

373.2K, N+

0.841 NQ

¥jn

cmM */M 0.841 98.5 gmole

*/M

0.77 bQ

0.77 373.2K

3.884 Å

•b €

• • */K u · v b € €

287.4K

3.681 Å, ~

3.681 Å

}

3.78 Å

1 " Zy K Ω{

0.001858 b M⁄K w

1 *⁄K x "

gmole mM

g m3

3.884 Å 91.5 K .

1 1 *⁄K u · v · 350 K 287.4K 91.5K

From Table K.1, interpolate to get ΩD = 1.047.

g‰Š‹Œ

34.8

35.4

98.5 ijklm.

From Appendix K, Table K.2, y y

gmole mM

gmole g v u34.08 v m3 gmole

b. Estimate DAB for H2S in N2 b+

1.04

2.16

0.001858 · 350 K M/K • 34.08

g gmole

28.01

1.0 atm 93.78 Å; 1.047 K

*/K

g ‚ gmole

0.207 cmK ⁄s

c. Estimate DA-m for H2S in mixture gas Find DAC for H2S in SO2

From Appendix K, Table K.2, yQ y

•b €Q

3.884 Å

• • */K · v b € €Q

4.290 Å

4.09 Å

1 *⁄K x "Q

1 " Zy Q K Ω{

0.001858 b M⁄K w

1 0.001858 · 350 K M/K u 34.08 g/gmole

0.03,

0.03

252 K

}

1 1 *⁄K u · v · 350 K 287.4K 252K

From Table K.1, ΩD = 1.273. Q

4.290 Å, ¡

0.92, ¢Q

1.30

*/K 1 v 64.07 g/gmole

0.12 cmK ⁄s

1.0 atm 94.09 Å; 1.273

0.92 0.21 cmK ⁄s

0.05 ¢

0.05 , 0.12 cmK ⁄s

¢Q §

ŽK 0.20 s

•¦

A‰

This diffusion coefficient is slightly different than that of H2S-N2 because H2S diffuses with N2 and SO2 in the mixture gas. But, since N2 gas dominates, the diffusion coefficient is very similar.

0.181

M¨

32.00

D¨©

dªk«m

0.181

¥jS Ÿ

at T = 273 K, P = 1.0 atm

10 > 10A• cm

*.B

0.201

28.01

cmK s

g gmole

a. Effective diffusion coefficient for straight 10 nm pores in parallel array D¹¨

T 4850 · dªk«m · º M¨

D¨m

0.085 cmK /s

1 D¨m

1 u D¨©

1 v D¹¨

293 K 4850 · 10 > 10A• cm · â g 32.00 gmole

1 u 0.201 cmK /s

1 v 0.147 cmK /s

0.147 cmK /s

b. Effective diffusion coefficient for random pores of 10 nm diameter with void fraction of ε 0.40 See part (a) D¢¨m

εK D¨m

0.40K · 0.085

cmK s

0.0136

cmK s

c. Effective diffusion coefficient for random pores of 1.0 µm diameter with void fraction of ε 0.40 dªk«m

1 > 10A¤ cm

D¹¨

T 4850 · dªk«m · º M¨

D¨m

0.177 cmK /s

1 D¨m

D¢¨m

1 u D¨©

εK D¨m

1 v D¹¨

4850 · 1 > 10A¤ cm · â

1 u 0.201 cmK /s

0.40K · 0.177

cmK s

293 K g 32.00 gmole

1 v 1.47 cmK /s

0.0283

cmK s

1.468

cmK s

24.3 T = 288 K Physical property information Interpolate and convert Pa·s to cP from Appendix I to find viscosity of liquid water at 288 K: µìKí 288 K

1.193 cP (water)

For viscosity of liquid ethanol and n-butanol, From Perry’s Chemical Engineering Handbook, interpolate and convert to cP:

µîì¤í 288 K 0.641 cP (ethanol) µî¤ì*•í 288 K 3.41 cP (n-butanol)

From Table 24.4 and Table 24.5 VìKí 18.9 cmM /gmole (water) Vîì¤í 14.8 4 · 3.7 7.4 37.0 cmM /gmole (ethanol) Vî¤ì*•í 4 · 14.8 10 · 3.7 7.4 103.6 cmM /gmole (n-butanol) ΦìKí = 2.6 (water) Φîì¤í = 1.9 (ethanol) Φî¤ì*•í = 1.0 (n-butanol)

MìKí 18 g/gmole (water) Mîì¤í 32.04 g/gmole (ethanol) Mî¤ì*•í 74.12 g/gmole (n-butanol)

Hayduk and Laudie Equation (used for nonelectrolyte solutes in H2O water solvent)

D¨©

Wilke Chang Equation

a. Molecular diffusion coefficient for liquid methanol solute (A) in solvent H2O (B)

Appendix J.2 Value: DAB = 1.28 > 10AB cmK /s

D¨©

13.26 > 10AB 1.193 A*.*¤ · 37.0 A•.BÌ¿

1.29 > 10AB

cmK

D¨©

7.4 > 10AÌ 1.9 · 18 */K 288 · 18.9•.• 0.641

4.45 > 10

cmK s

c. Molecular diffusion coefficient for liquid n-butanol solute (A) in solvent H2O (B)

D¨©

7.7 > 10A• cmK /s

13.26 > 10AB 1.193 A*.*¤ · 103.6 A•.BÌ¿

7.05 > 10A•

cmK s

D¨©

7.4 > 10AÌ 1.0 · 74.12 */K 288 · 18.9•.• 3.41

9.23 > 10A•

cmK s

25.1 Control-volume expression for the conservation of mass

{net rate of efflux of A } + {rate of accumulation} − {rate of production} = 0 Net rate of A efflux

N A , x ∆y ∆z | x + ∆x − N A , x ∆y ∆z | x

(x-direction)

N A, y ∆x∆z | y +∆y − N A, y ∆x∆z | y

(y-direction)

N A, z ∆x∆y |z +∆z − N A, z ∆x∆y |z

(z-direction)

Rate of accumulation in control volume ∂c A ∆x∆y∆z ∂t

Rate of production in control volume

∂cA ∆x∆y∆z − RA ∆x∆y∆z = 0 ∂t

Divide by volume, ∆x∆y∆z, to obtain

N A, x ∆y∆z |x +∆x − N A, x ∆y∆z |x

∆x ∆y ∆z +

N A, y ∆x∆z | y +∆y − N A, y ∆x∆z | y ∆x∆y∆z

N A, z ∆x∆y |z +∆z − N A, z ∆x∆y |z

∂cA − RA = 0 ∂t

This equation reduces to

N A, x |x +∆x − N A, x |x ∆x +

N A, y | y +∆y − N A, y | y ∆y

N A, z |z +∆z − N A, z |z ∆z

∂c A − RA = 0 ∂t

Evaluated in the limit as ∆x, ∆y, and ∆z approach zero

∂N A, x ∂x

∂N A , y

∂y

∂N A, z ∂z

∂c A − RA = 0 ∂t

∇ ⋅ NA +

∂c A − RA = 0 ∂t

∆x ∆y ∆z

25.2

Spherical coordinates Assumptions: 1) 1D mass transfer (in r-direction only), 2) Steady-state, 3) No homogeneous reaction in gas phase, RA = 0, 4) Temperature and pressure are constant, 5) Constant concentration at r = R General differential equation (r-direction only) for spherical coordinates

∂c A 1 ∂ 2 + 2 ( r N A , r ) = RA ∂t r ∂r ∂c Since A =0 and RA = 0 ∂t 1 ∂ 2 ( r N A, r ) = 0 r 2 ∂r r 2 N A,r is constant Fick’s flux equation for A

N A , r = − cD AB

(

N A , r = − cD AB

dy A + y A ( N A ,r + 0 ) dr

dy A + y A N A ,r + N B ,r dr Since N B , r = 0

N A, r =

−cDAB dy A 1 − y A dr

Boundary conditions

PAo At r = R, y A = RT At r = ∞, y A = y A,∞

)

25.3 System: Air Bubble

P|}||~•

0.2 cm, T = 298 K

M@B € 298 ‚!

0.03 6 1, M|}||~• 298 ‚!

16 1

A = H2O, B = Air Assume: 1) 1-D (radial) flux, 2) Unsteady State, 3) Constant bubble diameter, 4) Slow bubble velocity (diffusion dominates), 5) Dilute water in air, 6) No homogeneous reaction, 0. a. Fick’s Flux Equation P , P5

P7O3 2 4/O3 7/.:

Differential Model, Spherical Geometry 8

g o l dB

^5 %

^ dB B

% · d

b. Boundary and Initial Conditions BC: 5 5

, !

0, ! 5

IC: t = 0, cA(r,0) = 0

M@B € „ 0

26.1 A = MeOH vapor, B = O2 gas

T = 293 K, P = 1.0 atm, G€ [w

a. Determine effective diffusion coefficient, DAe 1 3 1 0.001858 · )/3 45 6 5 7 2" 3 " Ω9 ).LXLp). )) 3.585 , " 3.433 ,, " qb r

507 , r 293 239.36

113 ,

1.224,

√113

> Ω?

· 507

1.308

1 1 B/3 0.001858 · 293 )/3 @ 6 A 32.04 32.00 1.0 · 3.5093 · 1.308

Knudsen Diffusion Coefficient

3.509

239.36 0.145

3 293 7.33 32.04 C Effective Diffusion Coefficient 1 1 1 1 1 6 6 3 3 >2 w 2 w 2 " 2t 0.145 7.33 C C

4850 · 5 1 10

0.142

b. Boundary conditions for system I and II: •

For system I:

„ h

nB ,

For system II: At z = L1, yA = yA1, at z = L2, yA = yA2 c. Model for WA Assumptions: 1) Steady state, 2) Dilute UMD process, 3) No reaction, 4) 1-D transfer along z, 5) Constant T and P, 6) No sink for O2 D

T2 "

G Gh

d ( N A,z ) = 0 , constant flux along z dz System I: i…

D ] Gh

T2 "

T nB

System II:

2 "

D ] Gh

T2 w

i†

i…

kb…

] G

kb_

kb†

T nB

] G

\G3 4

kb…

n3 T nB

2 w

n3 T nB

„€

Eliminate yA1 and solve for for WA T 3 4nB n Tn 6 „3 2 B 3 \G 2 " € w

1.0 )· 82.06 ( ./0 ·

d. Estimate WA

„€ /O

0.5

0.005

‡ „T ,„ ˆ6 0.129 1.0 T

4nB n T nB 6 3 3 „€ 2 w \G 2 "

· 293

, „B

4.16 1 10 L

\G3 4

11.9673, ‡

0.129

\ 0.1

\ 10

)

3626.55, ˆ

4.16 1 10 L 4 · 0.3

3 · 0.145

WA = 9.55 x 10-9 gmole/sec = 35 µmol/hr

( ./0 3

T34.29

( ./0 )

7.854 1 10 u

0.129 T 0.020

0.45 T 0.30

0.005

3 · 0.142

26.2 A = naphthalene, B = air 347

% ,* |'Ÿ *

1.145

1. 0 (

, )

666

8.314 1 10o

1.0

666

0.25

)·

( ./0 ·

· 347

128

( ,2 " ( ./0

2.31 1 10

0.0819

( ./0 u )

General Differential Equation for Mass Transfer (spherical system): 1 G 3 ”Y D ,[ – 0 Y 3 GY

D ,[ · Y 3

D ,^ · , 3

Combine

Y 3 RT2 "

Flux Equation: G D T2 " GY

D · ,3

GY D · ,3 ] 3 Y e

G S GY abc

T2 " ] G

2" 1 1 , 3 • , T ∞•

.OC O

2" ,

Material balance on naphthalene (PSS mass transfer) IN – OUT + GEN = ACC 0 T D ,[ U 6 0

2" ,

T2 "

· 4\, 3

G G

% ,* |'Ÿ G’ 5 G

% ,* |'Ÿ G, ·, 5 G

% ,* |'Ÿ G 4 ) R \, S 5 G 3

% ,* |'Ÿ G, · 4\, 3 5 G

T2 "

% ,* |'Ÿ

]G j

% ,* |'Ÿ ”,'3 T , 3 – 22 " * 5 t = 221,636 s = 61.6 hr

^£

] ,G,

1.145

( 3

)

1.0

T 0.25

2 · 0.0819 C · 2.31 1 10 u

( ./0 )

· 128

( ( ./0

26.3 A = TCE, B = biofilm a. Determine volumetric flow rate vo Material balance in TCE in biofilm reactor

IN – OUT + GEN = ACC = 0 v0cAi − v0 cA0 − SN A = 0 NA = v0 =

DAB cA0

φ tanh φ

SN A SDAB cA0 φ tanh φ = c Ai − c A0 δ ( cAi − cA0 )

φ =δ

 1m  k1 = 100 µm  6  DAB  10 µm 

4.3 s -1 9.03 x 10

-10

m2 s

= 6.909

2  g TCE  -10 m    9.03 x 10   0.05  s  m3  g TCE  NA = ( 6.909 ) tanh ( 6.909 ) = 3.12 x 10-6 2 -6 100 x 10 m ms  (800 m2 )  3.12 x 10-6 g mTCE  2 m3 s   3600 s  v0 = = 44.9 g TCE g TCE  hr  hr 0.25 0.05 3 3 m m

b. Determine TCE concentration at the postion biofilm is attached to surface (z = δ)  k1  g TCE c A0 cosh  (δ − z )  DAB  c A0 cosh ( 0 ) 0.25 m3 (1.0) g TCE  c A ( z ) |z =δ = = = = 4.99 x 10-4 cosh (φ ) cosh ( 6.909 ) m3  k1  cosh  δ  DAB  

27.1 A = H atoms, B = solid iron (Fe) 56

‹6 W X

124 N 10 f P /c

CA0 = 0

‹6 W |

Determine ‹6 W

2.2 N 10 €

Pbk Œ j•

Ž •• 1.76N 10 € ‘ ’“ aS

0.1 P

USS diffusion in semi-infinite medium

W X W ,S ^ erf _ erf \ WX W) 2]5 6 S Or ‹6 W , S ^ 1 e” \ ‹6 W | 2]5 6 S

Pbk Œ 1.76 N 10 € j • 2.2 N10 €

erf _

Pbk Œ j•

0.200

From Appendix L, _

S 1 $ % $ % _ 45 6

0.8

0.17925

e” _

0.1 P 1 1 •e $ % q r$ % P 0.17925 3600 cec f 4 $124 N 10 % sec

174.3 •e

27.2 A = solvent, B = polymer D = 0.3 cm, R = 0.15 cm W• 0 wA0 = 0.2 wt% solvent in polymer DAe = 4.0 x 10-7 cm2/sec a. Determine time (t) for wA(r,t) = 0.002 wt% for center of bead (r = 0) USS Diffusion in a sphere, use Concentration -Time charts P 0 b b › S•b e c•cSa

W ∞ W e, S W∞ W)

e ž

0 P 1.25 P

0.55

ž 5“

QR Ÿ 0.55 Figure F.3

5 “S ž

0.55

W e, S W

0.15 P

4.0 N 10 €

P sec

Y X Y e, S YX Y)

0 0

0.002 0.200

30,937.5 sec

8.6 •e

0.010

b. Determine time (t) for wA(r,t) = 0.18 wt% for center of bead 0.01 cm from the surface /

YX Y e, S Y) YX

e ž

0.18 0.200

1.25 P 0.01 P 1.25 P

0 0

0.90

0.99 Ÿ 1.0

Figure F.9, off the chart. The charts will not work, but penetration depth is small, therefore the process can be approximated as USS diffusion in semi-infinite medium Y

erf _

From Appendix L, erf _

2]5 6 S

1 ^ 2]5 6 _

0.90

0.90; _

0.01 P

1.17

2]4.0 N 10 € P /c

1 ^ 1.17

46 c

28.1 Local mass transfer coefficient at position x W ,&X YZX[

W ,^_[`_&'Z^

0.332;GH] =K/> @* 8

0.0292;GH] =E/J @* 8

Average mass transfer coefficient

kc =

4/5  0.664 DAB Ret1/2 Sc1/3 + 0.0365DAB Sc1/3 Re4/5 L − Ret 

 0.664 DAB Re1/ 2 Sc1/3    kc ,lam L   = 1/2 1/3 1/3 4/5 kc  0.664 DAB Ret Sc + 0.0365 DAB Sc  Re 4/5   L − Ret    L  

kc ,lam kc

0.664 Re1/2 t 1/2 4/5   0.664 Ret + 0.0365  Re 4/5 L − Ret 

ReL = 3.0 x 106, Ret = 2.0 x 105

kc ,lam kc

0.664(2.0 x 105 )1/2 =0.0606 (6.06%) 0.664(2.0 x 105 )1/2 +0.0365 (3.0 x 106 )4/5 -(2.0 x 105 ) 4/5 

28.2 ?

200 * , j

1.97 C 10DE 0.065

200 * , k ,

0.01 * ,

900

, cXY[ ;400 =

," 8

400 ,

300

2.5909 C 10

1.0

. , ∆7m, . 34H

a. Position for transition from laminar flow, Lt 50 · ?^ 6 GH 2 C 10J 10.4 * > ) ?^ 2.5909 C 10DJ 6 50 6 · 2.0 3.86 C 10V GH > 2.5909 C 10DJ 6 turbulent flow

200

L = 2.0 m >> Lt = 0.104 m, therefore laminar portion can be neglected b. Average mass transfer coefficient, kc W

0.25909

0.0365GHe J @* 8 *

0.065 >

3.986

0.065 6 E * 0.0365;3.86 C 10V =J ;3.986=K/8 3.50 200 * 6 c. Time required for local evaporation at x = 1.2 cm W

GH] W ,]

50 6 · 1.2

2.5909 C 10DJ \

0.0292GH] J @* 8

2.32 C 10V

* > 0.065 6 E 0.0292;2.32 C 10V =J ;3.986=K/8 120 *

3.1

* 6

n .

5,000

* 6

* p

W ,] ;*

1.86 C 10Dg -

900

1.97 C 10DE * 8· 82.06 · 400 . 34H · a*

. 34H * >·6 @ℓ" p@

W ,] · *

ℓ" p

6.0 C 10Do

3.1

* . 34H q6.0 C 10Do r 6 * 8

W. 34H 1000 . 34H · 300 W. 1 W. 34H . 34H 1.86 C 10DE > ·6

8 · 100 C 10

* 8 . 34H

1613 6

d. Heat transfer coefficient (h) and surface temperature Ts Chilton-Colburn analogy @* >/8 k *v W q r 1 *v,XY[ ;400 = k

900

1.0142 C 10D8

· 0.5 C 10 8

n W. ·

3.986 8 0.035 q r 6 0.689

0.0506

n >·6·

Energy Balance

Q = N A ∆H v , A M A = h(Ts − T∞ ) A gmole   200 J  300 g   1.86 x 10-4 2    N A ∆H v , A M A m ⋅ s   g  gmole   Ts = T∞ + = 400 K + J h    0.0506 2  m ⋅s⋅K   Ts = 621 K

29.1 𝑔 1003 𝑐𝑚3 𝑔 5 𝜌𝐵 = 0.8 3 ∙ = 8 × 10 𝑐𝑚 𝑚3 𝑚3 𝑀𝐵 = 180

𝑔 , 𝑚𝑜𝑙𝑒

𝑇 = 293 𝐾,

𝑝𝐴 = 0.015 𝑎𝑡𝑚,

𝑦𝐴 =

𝑘𝑔𝑚𝑜𝑙𝑒

𝑃 = 1.5 𝑎𝑡𝑚

0.015 𝑎𝑡𝑚 = 0.01, 1.5 𝑎𝑡𝑚 𝑘𝑔𝑚𝑜𝑙𝑒

𝑘𝑥 = 0.01 𝑚2 ∙𝑠 , 𝑘𝑦 = 0.02 𝑚2 ∙𝑠 𝑝𝐴,𝑖 = 𝐻𝑥𝐴,𝑖 ,

𝐻 = 0.15 𝑎𝑡𝑚

a. Determine 𝐾𝐿 𝑘

𝑀

𝑘𝐿 = 𝐶𝑥 ≅ 𝜌 𝐵 𝑘𝑥 = 𝐿

𝑥𝐴 = 0.05

𝐵

𝑔 𝑔𝑚𝑜𝑙 𝑔 8×105 3 𝑚

180

𝑔𝑚𝑜𝑙

𝑚

∙ 10 𝑚2 ∙𝑠 = 2.25 ∙ 10−3 𝑠

𝑘𝑔𝑚𝑜𝑙 𝑘𝑦 0.02 � 𝑚2 ∙ 𝑠 � 𝑘𝑔𝑚𝑜𝑙 𝑘𝐺 = = = 1.33 ∙ 10−2 2 1.5(𝑎𝑡𝑚) 𝑃 𝑚 ∙ 𝑠 ∙ 𝑎𝑡𝑚 −1

1 1 𝐾𝐿 = � + � 𝑘𝐿 𝐻𝑘𝐺 = 1.06 ∙ 10−3

𝑚 𝑠

1 1 =� + 𝑚 2.25 ∙ 10−3 𝑠 0.15 𝑎𝑡𝑚 ∙ 1.33 ∙ 10−2

𝑃𝐴 𝑁𝐴 = 𝐾𝐿 (𝑐 ∗𝐴𝐿 − 𝑐𝐴𝐿 ) = 𝐾𝐿 � ∙ 𝑐𝐿 − 𝑥𝐴 𝑐𝐿 � 𝐻 𝑚

𝑔

𝑔𝑚𝑜𝑙

𝑘𝑔𝑚𝑜𝑙 𝑚2 ∙ 𝑠 ∙ 𝑎𝑡𝑚

−1

�

5 8×105 3 0.015 𝑎𝑡𝑚 8×10 𝑚3 𝑔𝑚𝑜𝑙 𝑚 − 0.05 ∙ 𝑔 𝑔 � = 0.236 𝑚2 ∙𝑠 180 180

= 1.06 ∙ 10−3 𝑠 ∙ � 0.15 𝑎𝑡𝑚 ∙

b. Determine 𝑥𝐴𝑖 , and 𝑝𝐴𝑖 .

𝑘

Equate the equilibrium line 𝑝𝐴,𝑖 = 𝐻𝑥𝐴,𝑖 to a line with slope − 𝑘𝑥 through the operating point to find the intersection (𝑥𝐴𝑖 , 𝑦𝐴𝑖 ). 𝐻

Equilibrium line: 𝑦 𝐴𝑖 = 𝑃 ∙ 𝑥 𝐴𝑖

𝑦

𝑘

Operating point line: 𝑦 𝐴 = − 𝑘𝑥 𝑥𝐴 + 𝑏, 0.01 = −0.5 ∗ 0.05 + 𝑏, ∴ 𝑏 = 0.035 𝐻 𝑃

𝑦

∙ 𝑥 𝐴𝑖 = −0.5𝑥𝐴,𝑖 + 0.035,

(𝑥𝐴𝑖 , 𝑦𝐴𝑖 ) = (0.0583 , 0.00585)

and 𝑃𝐴𝑖 = 𝑦𝐴𝑖 ∙ 𝑃 = 0.00585 ∙ 1.5 𝑎𝑡𝑚 = 0.00878 𝑎𝑡𝑚

29.2

gmole m3 c A= 1.0 ×10 , vo = 0.20 , p A ≈ 0 , P = 1 atm, D = 4 m, depth = 1.0 m ,o m3 min −3

k L = 5 ×10−4

m3 ⋅ atm m kgmole H = 10.0 k = 0.01 H ⋅ c AL ,i , G , , p A= ,i s m 2 ⋅ s ⋅ atm kgmole

a. Determine % resistance to mass transfer in the liquid film

1 1 1 = + K L k L HkG −1

    1 1   = 4.98 x 10−4 m/s + KL = 3 m m ⋅ atm kgmole   5 ×10−4 (10.0 )(0.01 2 )  s kgmole m ⋅ s ⋅ atm  

% resistance (liquid) =

1 k L K L 4.98 x 10−4 m/s = = ⋅100% = 99.5% 1 K L kL 5 x 10−4 m/s

b. Develop material balance model and determine c AL and c AL,i Assume: 1) steady state, 2) no reaction, 3) liquid stripping. IN – OUT – FLUX OUT = 0

c AL ,o vo − c AL vo − N A S = 0 vo (c AL ,o − c AL ) − SK L (c AL − c*AL ) = 0 vo (c AL ,o − c AL ) − π

p D2 0 K L (c AL − A ) = 4 H

π D2 K L pA vo c AL ,o + 4H = c AL = π D2 K L vo + 4

m )(0 atm) s m3 ⋅ atm 4(10.0 ) kgmole m s π (4 m) 2 (4.98 x 10−4 )(60 ) 3 m s min (0.20 )+ min 4

m3 gmole (0.20 )(0.001 )+ min m3

π (4 m) 2 (4.98 x 10−4

c AL = 3.48 x 10−4

gmole m3

Determine c AL ,i m 5 x 10−4 p A − p A ,i p − Hc AL ,i kL s = − = − =A kgmole kG c AL − c AL ,i c AL − c AL ,i 0.01 2 m ⋅ s ⋅ atm m3 ⋅ atm − 0 (10.0 ) ⋅ c AL ,i atm ⋅ m3 kgmole −0.05 = kgmole 3.48 x 10−7 kgmole − c AL ,i m3

c AL ,i 1.73 x 10−9 kgmole/m3 1.73 x 10−6 gmole/m3 = = c. Would the flux N A increase by the following: Increasing volume level of tank with fixed surface area: does not affect flux. Increasing the agitation intensity of the bulk liquid: will increase the flux since k L will increase. Increasing the agitation intensity of the bulk gas: will not increase the flux since the mass transfer is liquid film controlling. Increasing the system temperature: will likely increase the flux since the Henry’s constant will increase with increasing temperature, which will lower the solubility of the solute in the gas and lower the concentration driving force for mass transfer.

30.1 Diffusion through stagnant air ( N B = 0 ) cDAB dy A N Ar = − 1 − y A dr ∞

A∞ −dy A dr ( N Ar ⋅ R ) ∫ 2 = cDAB ∫ 1 − yA R r y As

1 − y A∞  1 1 ( N Ar ⋅ R 2 )  −  = cDAB ln   , at the surface R ∞  1 − y As  cD N Ar R = AB ( y As − y A∞ ) yB ,lm N Ar

D DAB = ( C As − C A∞ ) 2 yB ,lm

N Ar =

2 DAB ( C As − C A∞ ) yB ,lm D

Recognize that N Ar = kc ( C As − C A∞ ) such that kc =

kc D 2 = DAB yB ,lm

Sh ≈ 2 , when yB ,lm ≈ 1.0

2 DAB yB ,lm D

30.2 Let A = O3, B = H2O (liquid) a. Determine kLa P 4.0 atm C A* = A = =14.29 gmole/m3 3 H atm ⋅ m 0.07 gmole DAB = 1.74 x10 −9 m 2 / s at T = 20oC and P = 1.0 atm ρL = 998 kg/m3, ρA = 2.14 kg/m3, µ L = 993x10-6 kg/m-sec, νL = 0.995x10-6 m2/sec

Material balance model on dissolved O2 Assumptions: 1) unsteady state, 2) dilute system, 3) kL≈KL, 4) No reaction occurs. In – Out +Generation = Accumulation (moles A/time) dC A N A ⋅ Ai − 0 + 0 = V dt dC A k L a (C A* − C A ) = dt CA t dC ∫0 CA* − ACA = kL a ∫0 dt

 C*  ln  * A  = kL a ⋅ t  CA − CA   C*   14.29 gmole/m 3  ln  * A  ln   CA − CA  14.29-4.00 gmole/m 3    kL a = = = 0.033 min -1 = 5.473 x 10-4 sec -1 t 10.0 min b. Bubble diameter, db 3 3 2 3 db3 ρ L g ∆ρ d b (998 kg/m )(9.81 m/sec )( ( 998-1.2 ) kg/m ) Gr = = =9.8x1012 d 3b 2 -6 2 µL (993 x 10 kg/m ⋅ sec)

Sc =

ν 0.995 x 10-6 m 2 /sec = = 572 DAB 1.74 x 10-9 m 2 /sec

Assume db < 2.5 mm k d Sh = L b = 0.31Gr1/3 Sc1/3 DAB

D  kL = 0.31Gr1/3 Sc1/3  AB   db 

 1.71x10−9 m2 / sec  −5 k L = (0.31)(9.8 x1012 db3 )1/3 (572)1/3   = 9.6 x10 m / sec db   −4 −1 k a 5.473 x10 s a= L = = 5.7 m 2 /m3 −5 kL 9.6 x10 m / sec 6φg a= db db =

6φg

3 3 6(0.005mgas / m liquid ) −1

= 5.3 x10−3 m = 5.3 mm

a 5.7 m 5.3 mm > 2.5 mm, therefore assume db ≥ 2.5 mm D  k d Sh = L b = 0.42Gr1/3 Sc1/2 → kL = 0.42Gr1/3 Sc1/2  AB  DAB  db 

 1.74 x10−9 m 2 / sec  −4 k L = (0.42)(9.8 x1012 db3 )1/3 (572)1/2   = 3.75 x10 m / sec db   −4 −1 k a 5.473 x10 s = 1.46m −1 a= L = −4 kL 3.75 x10 m / sec 6φg a= db db =

6φg a

6(0.005 m 3gas /m 3liquid ) 1.46 m -1

= 0.0205 m = 20.5 mm

20.5 mm > 2.5 mm, therefore correct correlation used c. Determine how bubble diameter, liquid viscosity, and aeration change in order to increase the kLa: 6φ A k L a = k L i = k L g , k L a ↑⇒ d b ↓ V db 1/3

 1  k L ∝  2  µ L1/3 = µ L −1/3 for db < 2.5mm  µL  1/3

 1  k L ∝  2  µ L1/2 = µ L −1/6 for db ≥ 2.5mm  µL  k L a ↑⇒ µ L ↓

WA = k L aV ∆C AL , k L a ↑⇒ WA ↑

30.3 Let A = CO2, B = N2 P  1atm  2 DAB = DAB , ref ref = ( 0.166cm 2 /sec )   = 0.0415 cm /sec P  4atm  wɺ (0.54 kgmol/hr)(1000 gmol/1kgmol)(1hr/3600 sec) Vɺ0 = = = 923.175 cm 3 /sec 4atm C  82.06atm-cm 3    (300 K)  gmol-K 

4 ( 923.175 cm /sec ) 4Vɺ0 = v∞ = = 1.88 cm/sec π D2 π(25 cm 2 ) 3

a. Determine kc Re = Sc =

v∞ D ρ L

µL µL

ρ L DAB

(1.88 cm/sec)(25 cm)(4.68x10-3 g/cm3 ) = 1236 1.78x10-4 g/cm-sec 1.78x10-4 g/cm-sec

(4.68x10-3 g/cm3 ) ( 0.0415 cm 2 /sec )

= 0.916

Sieder-Tate correlation for laminar flow of liquid flowing inside of a tube 1/3 kc D D  = 1.86  Re⋅ Sc  Sh = DAB L  1/3

D  D  kc = 1.86  Re⋅ Sc   AB  L   D  1/3

2  25 cm   0.0415 cm /sec  kc = 1.86  (1236)(0.916)    25 cm  3000 cm    = 0.0941(3000 cm)-1/3 = 6.52 x 10-3 cm/sec

b. Determine yA,out Assumptions: 1) Steady state, 2) Dilute system, 3) Concentration along z-direction only, 4) No reaction, 5) PA ≈ 0, 6) Kc ≈ kc Material balance on differential volume element (A = CO2) In – Out +Generation = Accumulation (liquid phase – mole A/time)

π D2 4

v∞C A − N A ⋅ π D ⋅ ∆z − z

π D2 4

+0=0

v∞C A z +∆z

÷ ∆z, rearrangement, ∆z → 0 dC 4 − A − NA ⋅ =0 dz vɺ∞ D dC 4 − A− kc C A = 0 dz v∞ D

∫

C A ,out

C A ,o

dC A 4k L = − c ∫ dz CA v∞ D 0

C  4k ln  A, out  = − c L  C  v∞ D  A, o   4k  y A, out = y A, o ⋅ exp  − c L   v∞ D 

 4 ( 6.52 x 10-3 cm/sec )  y A, out = ( 0.05 ) ⋅ exp  ( 3000 cm )  = 9.5 x 10-3  (1.88 cm/sec)(25 cm)  c. Determine DAe T 300 K DKA = 4850d pore =4850(0.8x10-4 cm) =1.013 cm 2 /sec MA 44 g/gmole -1

1 1  −1 −1  2 DAe = (D −AB1 + DKA ) = +  = 0.0399 cm /sec 2 2  0.0415 cm /sec 1.013 cm /sec  ' 2 2 DAe = ε DAe = (0.6) (0.0399 cm 2 /sec) = 0.0144 cm 2 /sec

d. Determine Kc D 0.0144 cm 2 /sec km = Ae = = 0.012 cm/sec lm 1.2 cm −1

1   1  1 1 Kc =  + +  =  -1/3  0.0941 L (cm/sec) 0.012 cm/sec   kc k m  1 Kc = = 4.23x10-3 cm/sec 1/3 10.63(3000cm) +83.33

-1

e. Determine new yA,out  4 ( 4.23 x 10-3 cm/sec )  y A, out = ( 0.05 ) ⋅ exp ( 3000 cm ) = 0.017  (1.88 cm/sec)(25 cm) 

 0.141(3000 cm) 2/3     -4 3 y A,out = 0.05 ⋅ exp    3000 cm +7.839   3     (1.88 cm/sec)(25 cm)  2.212 3000 cm +17.343ln  7.839      y A,out = 0.013

31.1 a. Determine CA at t = 5.0 min (A = O2) 3

m3  3.28 ft   60 sec  ft 3 =16.5 sec  1.0 m   1min  min At depth = 4.55 m (15 ft), from Eckenfelder plot

Qg = 0.0078

A ft 3 V ≅ 1200 V hr K L ( A / V )V  ft 3  ( 6 spargers ) K La = n =  1200 =0.72 hr -1  3 V hr 10,000 ft ( )   KL

Ptop + Pbottom

1 ρ L gh 2   1  1000 kg  9.81 m  1.0 atm Pave = 1.0atm +  =1.22 atm   ( 4.55 m )  5 2  2  m3  sec 2  1.0123 x 10 kg/m sec ⋅  

Pave =

= Ptop +

PA y A P 0.21(1.22 atm ) = 7.83 x 10-6 = = H H 3.27 x 10 4atm ρ 1000 g/L gmol C= B = = 55.56 M B 18 g/gmole L x*A =

gmol  gmol  C *A = x*AC = ( 7.83 ⋅ 10−6 )  55.56 = 4.35 ⋅ 10−4  L  L  t = 5.0 min = 0.083 hr C A = C A* − ( C A* − C A0 ) e(

− K L a ⋅t )

gmol  gmol gmol  -1 -  4.35 x 10-4 - 0.05 x 10-3  exp ( -0.72 hr ) ( 0.083hr )  L L L   gmol mmol C A = 7.23 x 10-5 = 0.0723 L L C A = 4.35 x 10-4

b. Determine time (t) required for CA = 0.20 mmole/L

 ( C A* − C A )   ln  *  ( C A − C A0 )  t= ( −K La )  ( 0.435 mmol/L - 0.20 mmol/L )  ln   ( 0.435 mmol/L - 0.05 mmol/L )   t= = 0.685 hr ( -0.72 hr -1 )

31.2 a. Determine AG1 and YA2

From YA -X A plot, YA2,min = 0.152 x A2 =

X A2 0.20 = = 0.167 1 + X A2 1.0 + 0.20

x A1 =

X A1 0.05 = = 0.0476 1 + X A1 1.0 + 0.05

ALs = AL2 (1 − x A2 ) = 100 (1.0 − 0.167 ) = 83.3

lbmol hr

YA1 AGs ,min + X A 2 ALs = YA2 AGs ,min + X A1 ALs

0 + ( 0.20 )( 83.3) = ( 0.152 ) ( AGs ,min ) + ( 0.05 )( 83.3)

AGs ,min = 82.2

lbmol hr

lbmol hr YA1 AGs + X A2 ALs = YA2 AGs + X A1 ALs AGs = (1.4 )( 82.2 ) = 115.1

0 + ( 0.20 )( 83.3) = (YA 2 )(115.1) + ( 0.05 )( 83.3)

YA 2 = 0.109 b. See YA-XA plot 0.16

YA2,min 0.14 0.12

YA2 0.10 0.08

YA 0.06 0.04 0.02 0.00 0.00

XA2

XA1, YA1 0.05

0.10

0.15

XA (mole A / mole solvent)

0.20

0.25

31.3 a. Determine AL1  std m3   kgmol  kgmol AG1 =  224 = 10.0  3  hr   22.4 std m  hr 

AGs = AG1 (1 − y A1 ) = 10(1-0.09)=9.1

(1 − R ) yA1 AG1 = y A2 AG2 =

kgmol hr

y A1 AGs 1 − y A1

 y A1   ( 9.10 )  1 − y A1 

(1 − 0.898)( 0.09 )(10 ) =  y A2 = 0.01 AG2 =

AGs 9.1 kgmol = =9.19 1 − y A 2 1-0.01 hr

y A1 AG1 + x A2 AL2 = y A 2 AG2 + xA1 AL1

( 0.09 )(10 ) + 0 = ( 0.01)( 9.19 ) + ( 0.02 ) AL1 AL1 = 40.4

kgmol hr

b. Determine ALs,min

y A2 AG2 + 0 = y A 2 AG2 + xA1,min AL1,min x A1,min = 0.092 at y A1 = 0.090

( 0.09 )(10 ) = ( 0.01)( 9.19 ) + ( 0.092 ) AL1,min AL1,min = 8.78

kgmol hr

ALs ,min = AL1,min (1 − x A1,min ) = 8.78 (1 − 0.092 ) = 7.97 c. Determine packing height, z kgmol 10 AG1 hr = 0.50 m 2 A= = kgmol G1 20 2 m hr H OG = G=

G K 'y a

AG1 + AG2 (10+9.19 ) kgmol/hr kgmol = =19.19 2 2 2A 2 ⋅ 0.5 m m hr

kgmol hr

1 1 H = ' + ' ' K G a kG a k L a

mP (1.2 )(1.0 atm ) m3atm = =0.0216 kgmol  CL  kgmol  55.6  3 m   1 1 0.0216 = + ' K G a 20 50

kgmol m ⋅ hr ⋅ atm kgmol  kgmol  K y' a = K G' a ⋅ P =  19.58 3  (1.0 atm ) = 19.58 3 m ⋅ hr ⋅ atm  m ⋅ hr ⋅ atm  kgmol 19.19 2 m hr H OG = = 0.98 m kgmol 19.58 3 m ⋅ hr ⋅ atm y −y N OG = A1 *A2 ( yA − yA ) K G' a = 19.58

* A1

y = 0.023 at x A1 = 0.020 y*A 2 = 0.0 at x A2 = 0.0

( yA − y*A ) = lm

( y − y ) − ( y − y ) = ( 0.09 − 0.023) − ( 0.01 − 0.0 ) = 0.030 (y − y )  ( 0.090 − 0.023)  ln  ln     ( y − y )   ( 0.010 − 0.0 )  A1

* A1

* A2

0.09 − 0.01 = 2.67 0.030 z = H OG N OG

N OG =

z = ( 0.98 m )( 2.67 ) = 2.6 m

* A2

d. Gas flooding velocity, Gf

  kg  kg    kmol   AG1' = AG1 (1 − y A1 ) M W ,air + y A1 M W , A  = 10  + ( 0.09 )  58   (1-0.09 )  29 hr     kgmol   kgmol  

AG1' = 316

kg hr

  kg  kg    kmol   AL1' = AL1 (1 − x A1 ) M W , H 2O + xA1 M W , A  =  40  + ( 0.02 )  58   (1-0.02 ) 18 hr     kgmol   kgmol  

AL1' = 759.5

kg hr

    P  kg   1.0 atm  = 1.34 kg =  31.6 ρG = M W ,ave  3  RT  kgmol   m3 m atm    0.08206  ( 288 K )  kgmol ⋅ K    Flooding Correlation 1/2

AL1'  ρG    AG1'  ρ L − ρG 

x − axis:

1/2

759.5  1.34    316  1000 − 1.34 

= 0.09

( G ) C µ J = ( G ) (100)(1.0) ⋅ 1 y − axis: Y = 0.22 = ' f

0.1 L

ρ G ( ρ L − ρG ) g c

G 'f = 1.7

' f

0.1

1.34 (1000 − 1.34 ) ⋅ 1

kg m 2 sec 1/2

1/2  C f ,old  kg   100  kg  G =G = 1.7      = 1.2 2 2 m sec   200  m sec   C f ,new  Intalox Saddles have lower G'f , and so the packing is not better ' f

' f , old

e. The operating mole fraction inlet gas is above the saturation mole fraction, so some acetone will condense to lower the outlet mole fraction.

31.3 continued 0.10

0.08

xA1, yA1

yA1

0.06

yA 0.04

0.02

xA2, yA2 0.00 0.00

xA1,min 0.02

0.04

0.06

0.08

xA (mole fraction acetone in water)

0.10

Part 3: Show/Hide Problems

Chapter 1 Show/Hide Problems 1.2 Which of the quantities listed below are flow properties and which are fluid properties? Solution Flow Properties: velocity, pressure gradient, stress Fluid Properties: pressure, temperature, density, speed of sound, specific heat.

1.18 A vertical cylindrical tank having a base diameter of 10 meters and a height of 5 meters is filled to the top with water at 20℃. How much water will overflow if the water is heated to 80℃? Solution Diameter = 10 m Height = 5 m V=

π 2 π d h = (10m)2 (5m) = 392.7 m3 4 4

At 20℃, ρWater = 998.2 kg/m3 So the mass of this is: m = ρW V = (998.2 kg/m3 )(392.7 m3 ) = 391992.2 kg At 80℃, ρWater = 971.8 kg/m3 So the mass of this is: m = ρW V = (971.8 kg/m3 )(392.7 m3 ) = 381625.86 kg So the mass of water at 80℃ that can be in the tank is lower than at 20℃. The difference is: ∆m = 10366.34 kg

1.21 The bulk modulus of elasticity for water is 2.205 GPa. Determine the change in pressure required to reduce a given volume of 0.75%. Solution 𝛽𝐻2𝑂 = 2.205 𝐺𝑃𝑎 ∆𝑉 = −0.75% = −0.0075 𝑉 dP ∆P 𝛽𝐻2𝑂 = V ( ) ≅ −V dV ∆V Rearrange, ∆𝑃 = 𝛽𝐻2𝑂

∆𝑉 = −2.205 𝐺𝑃𝑎(−0.0075) = 0.0165𝐺𝑃𝑎 = 16.5𝑀𝑃𝑎 𝑉

1.30 A colleague is trying to measure the diameter of a capillary tube, something that is very difficult to physically accomplish. Since you are a Fluid Dynamics student, you know that the diameter can be easily calculated after doing a simple experiment. You take a clean glass capillary tube and place it in a container of pure water and observe that the water rises in the tube to a height of 17.5 millimeters. You take a sample of the water and measure the mass of 100 mls to be 97.18 grams and you measure the temperature of the water to be 80℃. Please calculate the diameter of your colleague’s capillary tube. Solution First, calculate the surface tension of water using the temperature of the water: 𝑇 = 80℃ = 353𝐾 𝜎 = 0.123(1 − 0.00139𝑇) 𝜎 = 0.123(1 − 0.00139(353)) = 0.06265 𝑁/𝑚 Next, using the equation for the height of a fluid in a capillary, 2𝜎𝑐𝑜𝑠𝜃 ℎ= 𝜌𝑔𝑟 Rearranging and solving for the radius: 2𝜎𝑐𝑜𝑠𝜃 𝑟= 𝜌𝑔ℎ 2(0.06265 𝑁/𝑚) cos(0) = 97.18 𝑔 𝑘𝑔 1000 𝑚𝑙 28.32 𝑙 𝑚 )( (( )( )( )) (9.81 𝑚/𝑠 2 )(17.5𝑚𝑚) (1000𝑚𝑚 ) 100 𝑚𝑙𝑠 1000𝑔 𝑙𝑖𝑡𝑒𝑟 0.028317 𝑚3 = 7.509𝑥10−4 𝑚𝑒𝑡𝑒𝑟𝑠 = 0.7509 𝑚𝑚 Diameter is 2r so D = 1.50 mm

Chapter 2 Show/Hide Problems 2.6 The practical depth limit for a suited diver is about 185 meters. What is the gage pressure of sea water at that depth? Assume that the specific gravity of sea water is constant at 1.025. Solution 𝑑𝑃 = −𝜌𝑔 𝑑𝑦 𝑃

∫

𝑑𝑃 = −𝜌𝑔 ∫ 𝑑𝑦

𝑃𝑎𝑡𝑚

ℎ

𝑃 − 𝑃𝑎𝑡𝑚 = − 𝜌𝑔(0 − ℎ) = 𝜌𝑔ℎ 𝑃 − 𝑃𝑎𝑡𝑚 = 1.025(1000 𝑘𝑔/𝑚3 )(9.8 𝑚/𝑠 2 )(185 𝑚𝑒𝑡𝑒𝑟𝑠) = 1.86𝑥106

𝑘𝑔 = 1.86𝑥106 𝑃𝑎 𝑚𝑠 2

2.9 Determine the depth change to cause a pressure increase of 1 atm for (a) water, (b) sea water (SG=1.0250) and (c) mercury (SG=13.6). (Assume the temperature is 30℃.) Solution 𝑃 − 𝑃𝑎𝑡𝑚 = 𝜌𝑔ℎ 𝜌𝑤𝑎𝑡𝑒𝑟 = 995.2 𝑘𝑔/𝑚3 𝑁 1 𝑎𝑡𝑚 = 101325 2 = 101325 𝑘𝑔/𝑠 2 𝑚 𝑚 ℎ=

𝑃 − 𝑃𝑎𝑡𝑚 𝜌𝑔

(a) water ℎ=

𝑃 − 𝑃𝑎𝑡𝑚 101325 𝑘𝑔/𝑠 2 𝑚 = = 10.4 𝑚 (995.2 𝑘𝑔/𝑚3 )(9.8 𝑚/𝑠 2 ) 𝜌𝑔

(b) sea water (SG=1.0250)

𝑃 − 𝑃𝑎𝑡𝑚 101325 𝑘𝑔/𝑠 2 𝑚 ℎ= = = 10.13 𝑚 (995.2 𝑘𝑔/𝑚3 )(1.0250)(9.8 𝑚/𝑠 2 ) 𝜌𝑔

𝑃 − 𝑃𝑎𝑡𝑚 101325 𝑘𝑔/𝑠 2 𝑚 = = 0.764 𝑚 (995.2 𝑘𝑔/𝑚3 )(13.6)(9.8 𝑚/𝑠 2 ) 𝜌𝑔

2.12 What is the pressure pA in the figure? The oil in the middle tank has a specific gravity of 0.8. Assume that the entire system is at 80oF.

D Solution 𝑃𝐸 = 𝑃𝐵 − 𝜌𝑜𝑖𝑙 𝑔(10 𝑓𝑒𝑒𝑡) (1) 𝑃𝐶 = 𝑃𝐵 + 𝜌𝐻2𝑂 𝑔(15 − 10 𝑓𝑒𝑒𝑡) (2) 𝑃𝐷 = 𝑃𝐶 − 𝜌𝐻𝑔 𝑔(1 𝑓𝑜𝑜𝑡) (3) Insert (2) into (3) then into (1): 𝑃𝐷 = 𝑃𝐵 + 𝜌𝐻2𝑂 𝑔(15 − 10 𝑓𝑒𝑒𝑡) − 𝜌𝐻𝑔 𝑔(1 𝑓𝑜𝑜𝑡) 𝑃𝐸 = 𝑃𝐷 − 𝜌𝐻2𝑂 𝑔(15 − 10 𝑓𝑒𝑒𝑡) + 𝜌𝐻𝑔 𝑔(1 𝑓𝑜𝑜𝑡) − 𝜌𝑜𝑖𝑙 𝑔(10 𝑓𝑒𝑒𝑡) From Appendix I, at 80oF: ρH2O=62.2 lbm/ft3, ρHg=845 lbm/ft3 Thus, (62.2 𝑙𝑏𝑚 /ft 3 )(32.2 𝑓𝑡/𝑠 2 )(15 − 10 𝑓𝑡) 𝑃𝐴 = 𝑃𝐸 −𝑃𝐷 = − 32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2 (845 𝑙𝑏𝑚 /ft 3 )(32.2 𝑓𝑡/𝑠 2 )(1 𝑓𝑡) 0.8(62.2 𝑙𝑏𝑚 /ft 3 )(32.2 𝑓𝑡/𝑠 2 )(10 𝑓𝑡) + − 32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2 32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2 𝑙𝑏𝑓 = 36.45 2 (𝑔𝑎𝑢𝑔𝑒) 𝑓𝑡 PA is a gauge pressure because PD is open to the atmosphere.

2.13 Assume that the entire system is at 80oF.

Referring to the figure at left, please find the pressure difference between tanks A and B if d1=2 feet, d2=6 inches, d3=2.4 inches and d4=4 inches.

Solution 𝑃𝐴 − 𝑃3 = −𝜌𝐻2 𝑂 𝑔𝑑1

(1)

𝑃𝐵 − 𝑃3 = −𝜌𝐻𝑔 𝑔𝑑3 − 𝜌𝐻𝑔 𝑔𝑑4 𝑠𝑖𝑛𝜃

(2)

Subtract (1) -(2): 𝑃𝐴 − 𝑃𝐵 = −𝜌𝐻2 𝑂 𝑔𝑑1 + 𝜌𝐻𝑔 𝑔𝑑3 + 𝜌𝐻𝑔 𝑔𝑑4 sin (45) From Appendix I, at 80oF: ρH2O=62.2 lbm/ft3, ρHg=845 lbm/ft3 𝑃𝐴 −𝑃𝐵 =

−(62.2 𝑙𝑏𝑚 /ft 3 )(32.2 𝑓𝑡/𝑠 2 )(2 𝑓𝑡) (845 𝑙𝑏𝑚 /ft 3 )(32.2 𝑓𝑡/𝑠 2 )(0.2 𝑓𝑡) + 32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2 32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2 (845 𝑙𝑏𝑚 /ft 3 )(32.2 𝑓𝑡/𝑠 2 )(0.333 𝑓𝑡)sin (45) 𝑙𝑏𝑓 𝑙𝑏𝑓 + = 244 2 = 1.70 2 2 32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 𝑓𝑡 𝑖𝑛 = 1.7 𝑝𝑠𝑖

2.15 Assume the system is at 80oF. Points A and 3 are not necessarily at the same height.

4 3

A differential manometer is used to measure the pressure change caused by a flow constriction in a piping system as shown. Determine the pressure difference between points A and B in psi. Which section has the higher pressure? Solution 𝑃1 − 𝑃𝐴 = 𝜌𝐻2 𝑂 𝑔(𝑑𝑥 + 𝑑1 ) → 𝑃1 = 𝑃𝐴 + 𝜌𝐻2 𝑂 𝑔(𝑑1 + 𝑑1 ) = 𝑃𝐴 + 𝜌𝐻2 𝑂 𝑔(10 + 𝑥 𝑖𝑛) 𝑃2 − 𝑃𝐵 = 𝜌𝐻2 𝑂 𝑔𝑑1 + 𝜌𝐻2 𝑂 𝑔𝑑𝑥 + 𝜌𝐻𝑔 𝑔𝑑2 → 𝑃2 = 𝑃𝐵 + 𝜌𝐻2 𝑂 𝑔𝑑1 + 𝜌𝐻2 𝑂 𝑔𝑑𝑥 + 𝜌𝐻𝑔 𝑔𝑑2 = 𝑃𝐵 + 𝜌𝐻2 𝑂 𝑔(4 𝑖𝑛) + 𝜌𝐻2 𝑂 𝑔(𝑥 𝑖𝑛) + 𝜌𝐻𝑔 𝑔(10 𝑖𝑛) Equate the two equations by realizing that 𝑃1 = 𝑃2 𝑃𝐴 + 𝜌𝐻2 𝑂 𝑔(10 𝑖𝑛) + 𝜌𝐻2 𝑂 𝑔(𝑥 𝑖𝑛) = 𝑃𝐵 + 𝜌𝐻2 𝑂 𝑔(4 𝑖𝑛) + 𝜌𝐻2 𝑂 𝑔(𝑥 𝑖𝑛) + 𝜌𝐻𝑔 𝑔(10 𝑖𝑛) Rearrange and cancel 𝜌𝐻2 𝑂 𝑔(𝑥 𝑖𝑛), 𝑃𝐴 − 𝑃𝐵 = −𝜌𝐻2 𝑂 𝑔(10 𝑖𝑛) + 𝜌𝐻2 𝑂 𝑔(4 𝑖𝑛) + 𝜌𝐻𝑔 𝑔(10 𝑖𝑛) = −𝜌𝐻2 𝑂 𝑔(6 𝑖𝑛) + 𝜌𝐻𝑔 𝑔(10 𝑖𝑛) From Appendix I: 𝜌𝐻2 𝑂 𝑎𝑡 80℉ = 62.2 𝑙𝑏𝑚 /𝑓𝑡 3 𝜌𝐻𝑔 𝑎𝑡 80℉ = 845 𝑙𝑏𝑚 /𝑓𝑡 3 In addition, we know that 10 inches = 0.83 ft and 6 inches = 0.5 ft.

−(62.2 𝑙𝑏𝑚 /𝑓𝑡 3 )(32.2 𝑓𝑡/𝑠 2 )(0.5 𝑓𝑡) + (845 𝑙𝑏𝑚 /𝑓𝑡 3 )(32.2 𝑓𝑡/𝑠 2 )(0.83 𝑓𝑡) 𝑃𝐴 − 𝑃𝐵 = 32.174 𝑙𝑏𝑚 𝑓𝑡/𝑙𝑏𝑓 𝑠 2 2 = 670.82 𝑙𝑏𝑓 /𝑓𝑡 We were asked to state the problem in terms of psi (pounds per square inch): 𝑃𝐴 − 𝑃𝐵 = 670.82

𝑙𝑏𝑓 𝑙𝑏𝑓 𝑓𝑡 2 𝑥 = 4.66 2 = 4.66 𝑝𝑠𝑖 2 2 𝑓𝑡 144 𝑖𝑛 𝑖𝑛

Since 𝑃𝐴 − 𝑃𝐵 = 4.66 𝑝𝑠𝑖 The pressure at A must be greater than the pressure at B.

Chapter 4 Show/Hide Problems

4.3 Water is flowing through a large circular conduit with a velocity profile given by the 𝑟2

equation 𝑣 = 9 (1 − 16) fps. What is the average velocity in the 1.5 ft. pipe?

Solution The control volume is the dashed red line in the figure above. For the above control volume, ∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 +

𝜕 ∭ 𝜌𝑑𝑉 = 0 𝜕𝑡

Assume steady flow, ∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 = 0 ∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴2 − ∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴1 = 0 𝜌2 𝑣2,𝑎𝑣𝑔 𝐴2 − 𝜌1 𝑣1,𝑎𝑣𝑔 𝐴1 = 0 𝑟2

Assuming incompressible flow and substituting 𝑣 = 9 (1 − 16) and integrating from 0 → 𝑅, 𝑣2,𝑎𝑣𝑔 𝐴2 − 𝑣1,𝑎𝑣𝑔 𝐴1 = 0 𝑅

𝑟2 𝑣2,𝑎𝑣𝑔 𝐴2 − ∫ 9 (1 − ) 2𝜋𝑟𝑑𝑟 = 0 16 0 𝐴2 = 𝑣2 (

𝜋𝐷2 𝜋(1.5)2 = 4 4

𝜋(1.5)2 9 ) − [9𝜋𝑅 2 − 𝜋𝑅 4 ] = 0 4 32

Solve for 𝑣2 , 𝑣2 =

9 [9𝜋42 − 32 𝜋44 ] 𝜋(1.5)2 ( ) 4

= 128 𝑓𝑡/𝑠

4.10 Beginning with the integral mass balance equation, 𝜕 ∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 + ∭ 𝜌𝑑𝑉 = 0 𝜕𝑡 and using the symbol M for the mass in the control volume, show that equation (1) above can be written as: 𝜕𝑀 + ∬ 𝑑𝑚̇ = 0 𝜕𝑡

Solution We can solve this problem in pieces. First, let's examine: ∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴. If we let 𝑚̇ = 𝜌𝑣𝐴 = 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒. We can now write: ∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 = ∬ 𝑑(𝜌𝑣𝐴) = ∬ 𝑑𝑚̇ 𝜕

Next, let's examine the accumulation term: 𝜕𝑡 ∭ 𝜌𝑑𝑉. If we rearrange this term, 𝜕 𝜕 𝜕 ∭ 𝜌𝑑𝑉 = ∭ 𝑑(𝜌𝑉) = 𝑀 𝜕𝑡 𝜕𝑡 𝜕𝑡 As a result, ∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 +

𝜕 ∭ 𝜌𝑑𝑉 = 0 𝜕𝑡

Becomes, 𝜕𝑀 + ∬ 𝑑𝑚̇ = 0 𝜕𝑡

4.18 Water flows steadily through a piping junction shown in the figure below. It enters section 1 at 0.0013 m3/s. The average velocity at section 2 is 2.1 m/s. A portion of the flow is diverted through the showerhead that contains 100 holes of 1-mm diameter. Assuming uniform shower flow, estimate the exit velocity from the showerhead jets.

Solution The Control Volume for this problem is shown as a dotted red line in the figure above. Since we are told that we have steady flow, we can do the following mass balance: Mass Flow In = Mass Flow Out Mass flow in = 𝜌1 𝑣1 𝐴1 Mass flow out = 𝜌2 𝑣2 𝐴2 + 𝜌𝑗 𝑣𝑗 𝐴𝑗 (where j stands for the flow from the showerhead jets). So, 𝜌1 𝑣1 𝐴1 = 𝜌2 𝑣2 𝐴2 + 𝜌𝑗 𝑣𝑗 𝐴𝑗 𝜌1 𝑄 = 𝜌2 𝑣2 𝐴2 + 𝜌𝑗 𝑣𝑗 𝐴𝑗 Since in this system the fluid is water, we can assume incompressible flow and that density is constant. Thus, 𝑄 = 𝑣2 𝐴2 + 𝑣𝑗 𝐴𝑗

𝜋 1.3𝑥10−3 𝑚3 /𝑠 = (2.1 𝑚/𝑠)(𝜋(10−2 𝑚)2 ) + (𝑣𝑗 )(100) ( (10−3 )2 ) 4 Rearrange and solve for 𝑣𝑗 , 1.3𝑥10−3 𝑚3 /𝑠 − (2.1 𝑚/𝑠)(𝜋(10−2 𝑚)2 ) 𝑣𝑗 = 𝜋 (100) ( (10−3 )2 ) 4 𝑣𝑗 = 8.2 𝑚/𝑠

4.19 The jet pump injects water at V1=40 m/s through a 7.6 cm pipe and entrains a secondary flow of water V2=3 m/s in the annular region around the small pipe. The two flows become fully mixed downstream, where V3 is approximately constant. For steady incompressible flow, compute V3.

Solution The control volume for the problem is in the red dashed line in the figure. Mass flow in = Mass flow out 𝜌1 𝑣1 𝐴1 + 𝜌2 𝑣2 𝐴2 = 𝜌3 𝑣3 𝐴3 Since all the fluid is water, 𝜌1 = 𝜌2 = 𝜌3 𝑣1 𝐴1 + 𝑣2 𝐴2 = 𝑣3 𝐴3 2 𝜋 1𝑚 (40 𝑚/𝑠) [(7.6 𝑐𝑚) ( )] 4 100 𝑐𝑚 2 2 𝜋 1𝑚 𝜋 1𝑚 )] − (3 𝑚/𝑠) [(7.6 𝑐𝑚) ( )] } + {(3 𝑚/𝑠) [(25 𝑐𝑚) ( 4 100 𝑐𝑚 4 100 𝑐𝑚 2 𝜋 1𝑚 )] = (𝑣3 ) [(25 𝑐𝑚) ( 4 100 𝑐𝑚

0.1814

𝑚 𝑚 + {0.1473 − 0.0136 } = 0.0490(𝑣3 ) 𝑠 𝑠 𝑣3 = 6.48 𝑚/𝑠

4.21 The hypodermic needle shown in the figure below contains an incompressible liquid serum with a density of 1 g/cm3. If the serum is to be injected steadily at 6 cm3/s, please calculate how fast the plunger must be advanced: (a) if leakage in the plunger clearance is neglected and (b) if leakage is 10% of the needle flow. Solution (a) if leakage in the plunger clearance is neglected 𝜕 ∭ 𝜌𝑑𝑉 = 0 𝜕𝑡 Rearrange and insert mass and 𝑚̇ into the equation, ∬ 𝜌(𝑣 ∙ 𝑛) 𝑑𝐴 +

𝜕𝑀 + ∬ 𝑑𝑚̇ = 0 𝜕𝑡

−𝜚𝑣𝐴 + 𝜚𝑄 = 0 𝜚𝑣𝐴 = 𝜚𝑄 Since the serum is incompressible, 𝑄 6 cm3 /s 𝑣= = = 1.91 𝑐𝑚/𝑠 𝐴 𝜋(1 𝑐𝑚)2

(b) if leakage is 10% of the needle flow. 𝜕𝑀 + ∬ 𝑑𝑚̇ = 0 𝜕𝑡

−𝜚𝑣𝐴 + 𝜚𝑄 + 𝜚𝑄𝑙𝑒𝑎𝑘 = 0 Since the serum is incompressible, and 10% = 0.10, 𝑄 + 𝑄𝑙𝑒𝑎𝑘 = 𝑣𝐴 cm cm3 (6 ) + (0.10) (6 ) = v(π(1 cm2 )) s s 3

𝑣 = 2.10 𝑐𝑚/𝑠 (So leakage from the plunger end necessitates a higher velocity.)

Chapter 5 Show/Hide Problems 5.1 A two-dimensional object is placed in a 4-ft-wide water tunnel as shown. The upstream velocity, v1, is uniform across the cross section. For the downstream velocity profile as shown, find the value of v2. (Assume steady state.)

Solution ! 𝜌(𝑣 ∙ 𝑛)𝑑𝐴 +

𝜕 - 𝜌𝑑𝑉 = 0 𝜕𝑡

If have steady state (a.k.a. steady flow), ! 𝜌(𝑣 ∙ 𝑛)𝑑𝐴 = 0 4

4 1 6 1 9 1 𝜌𝑣2 𝑑𝑥 = 1 𝜌𝑣6 𝑑𝑥 + 1 𝜌𝑣6 𝑑𝑥 + 1 𝜌𝑣6 𝑑𝑥 + 1 𝜌𝑣6 𝑑𝑥 2 2 2 6 5 5 9

1 1 4𝑣2 𝜌 = 1𝜌𝑣6 + (2 − 1)𝜌𝑣6 + (3 − 2)𝜌𝑣6 + (4 − 3)𝜌𝑣6 2 2 Divide through by 𝜌 and simplify, 4𝑣2 = 3𝑣6 Solve for 𝑣6 4 4 𝑓𝑡 𝑓𝑡 𝑣6 = 𝑣2 = =20 @ = 26.67 3 3 𝑠 𝑠

5.7 Fluid is H20 P1= 60 Psig P2= 14.7 Psia (units?) D1= 0.25 ft. D2 = ft. = 400 gal/m= 0.892 ft.3/s Steady, Incompressible Flow dA =0 =

= 18.17

Fx-P1A1-P2A2=

= 18.17 ft./s

= 72.7 ft./s ( - )

Fx= ( - )-P1A1+P2A2 =[62.4(0.892)(72.7-18.17)]/32.2-(60+14.7)(144) (0.25)2+(14.7)(144) =94.3-528.0+25.4 =-408 lbf

5.16

Steady incompressible flow: = dA A1= (1.5)2=1.767 ft.2 A2= (1)2=0.785 ft.2 A3= (2)2=3.142 ft.2 For C.V. between (1)&(2) of the figure: = dA Fx+F2+P1A1-P2A2= Fx

( 2-

+(530-14.7)(144)(0.785)-(990-14.7)(144)(1.767)-F2 = -130,000 lbf –F2

For C.V. between (2)&(3) of the figure: Fx+P2A2-P3A3= ( 3- 2) Fx= (6700-3400)+(26-14.7)(144)(3.14)-(530-14.7)(144)(0.785) = 25777 lbf Stress at position 2 in the figure: = =

= 1823 psi Compression

At 1: F1= -130,000+25777 = -104220 lbf =

= 4915 psi Tension

5.23

x momentum: = + dV Fx= - wAj ( cos )+ wAj = wAj ( - cos ) =1000( )(0.1)2(20)[4.5-20cos45 ] = -1515 N Force on car by jet: Fx=1515 N y momentum: = Fy= - sin

( Aj)+ 0 = -(20sin45)(1000)(-20)* (0.1)2 = +2220 N

Force exerted by H20: Fy= -2220 N Total Force = = 1515 - 2220

5.29

x )z

( )

Mz=2() ry= Rsin = sin - R Mz=2 [-Rsin ( sin - r )] =

Chapter 6 Show/Hide Problems 6.2 Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3/s. The pump inlet is 0.25 m in diameter. At the inlet the pressure is 81326 N/m2. The pump outlet is 0.152 m in diameter and is 1.8 meters above the inlet. The outlet pressure is 175 kPa. If the inlet and exit temperatures are equal, how much power does the pump add to the fluid? (Note, inlet pressure changed for this problem relative to book problem.) Solution Begin with the overall energy balance equation, δWμ δQ δWs P ∂ − = ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV + ∂t ∂t ρ ∂t ∂t ∂ ∭ eρdV = 0 because we have steady state ∂t

δQ = 0 since we have no heat transfer out of the system ∂t δWμ = 0 since there is no viscous work being done ∂t δWs p p − = ∬ eρ(v ∙ n) dA = (e + ) ṁ − (e + ) ṁ ∂t ρ 2 ρ 1 2 2 v2 − v1 P2 − P1 = ṁ [U2 − U1 + + + g(z2 − z1 )] 2 ρ Since the inlet and exit temperatures are equal, U2 = U1. Next, we calculate some quantities, ṁ = ρQ = (1025 kg/m3 )(0.21 m3 /s) = 215.25 kg/s Q 0.21 m3 /s v1 = = = 4.28 ft/s A1 π(0.25 m)2 4 Q 0.21 m3 /s v2 = = = 11.57 ft/s A2 π(0.15 m)2 4 δW Substituting into the above equation for − ∂t , and remembering that Pa=N/m2 and N=kg m/s2, −

δWs kg (11.57 ft/s)2 − (4.28 ft/s)2 175000 − 81326 N/m2 = 215.25 [ + ∂t s 2 1025 kg/m3 + (9.81 m/s2 )(1.8m)] = 35,908 Nm/s δW

The negative sign on − ∂t s indicates work done to the fluid.

6.3 Air at 70oF flows into a 10 ft3 resevoir at a velocity of 110 fps. If the reservoir pressure is 14 psig and the reservoir temperature is 70oF, find the rate of temperature increase in the reservoir. Assume the incoming air is at reservoir pressure and flows through an 8 in diameter pipe. Solution δQ dt δQ

δW

δWμ

δW

δWμ

− dts − dt = ∬ (e + ρ ) ρ(v ⃑ ∙n ⃑ )dA + dt ∭ eρdV

= dts = dt = 0 dt v2

- ṁ[h1+ 21 + gz1]+ dt[mu]sys=0

gz1=negligible

ρvcv dt = ρA v( 2 ) dT

Av v2

= vc dt

= 2

π 8 2 2 ft. ( ) ft. (110 )3 4 12 s 778 ft.LBf 32.2 LBm ft. BTU 3 )(0.17 ) 2(10ft. )( )( LBm F BTU s2 LBf

= 21.8 F/s

6.4 Water flows through a 2 inch diameter horizontal pipe at a flow rate of 35 gal/min. Assume steady flow, that the heat transfer to the pipe is negligible, and that frictional forces cause a pressure drop of 10 psi. What is the temperature change of the water? Solution δWμ δQ δWs P ∂ − = ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV + ∂t ∂t ρ ∂t ∂t δWs = 0 since there is no shaft work ∂t ∂ ∭ eρdV = 0 since we assume steady state ∂t δWμ = 0 since there is no viscous work ∂t δQ = 0 since no heat transfer ∂t P ∬ (e + ) ρ(v ∙ n) dA = 0 ρ P1 v12 P2 v22 (U1 + + + gy1 ) − (U2 + + + gy2 ) = 0 ρ 2 ρ 2 v2

Since we have a horizontal pipe, gy1 = gy2 , and we are given, 21 = 22 (U1 +

P1 P2 ) − (U2 + ) = 0 ρ ρ

Recalling that dU = Cv dT, (10 lb/in2 )(144 in2 /ft 2 ) U2 − U1 P1 − P2 ∆T = = = = 0.030℉ (62.4 lbm /ft 3 )(1 BTU/lbm ℉)(778.18 ft ∙ lb/BTU) Cv ρCv

6.7 A fan draws air from the atmosphere at an inlet velocity of 5.0 m/s through a 0.30 meter diameter round duct that has a smoothly rounded entrance. A differential manometer connected to an opening in the wall of the duct shows a vacuum pressure of 2.50 cm of water. The density of air is 1.22 kg/m3. Assume steady flow and that there is no change in internal energy and no shaft or viscous work. (a) Calculate the volume rate of air flow in the duct in cubic feet per second. (b) Calculate the horsepower output of the fan. Solution Begin with Bernoulli's Equation: P1 v12 P2 v22 + + gy1 = hL + + + gy2 ρ 2 ρ 2 Since the body is horizontal, y1 = y2, and assume no head loss P1 v12 P2 v22 + = + ρ 2 ρ 2 P1 − P2 v22 − v12 = ρ 2 Given, P1 − P2 = 2.5 cm H2 O = 0.025 m H2 O atm 101325 Pa x = 245.22 Pa = 245.22 kg m/s 2 10.33 m H2 O atm m 1/2 1/2 2(245.22 kg m/s 2 ) + (5.0 s )2 2(P1 − P2 ) + v12 v2 = [ ] =[ ] = 20.55 m/s ρ 1.22 kg/m3 0.025 m H2 O x

Flow Rate is: Q = Av =

πD2 π(0.3m)2 m3 (v2 ) = (20.55 m/s) = 1.452 = 51.3 ft 3 /s 4 4 s

(b) Calculate the horsepower output of the fan. The energy equation simplifies to: δWs P = ∬ (e + ) dṁ ∂t ρ m With vout = 20.05 s , vin = 0 and Pout = Patm −

δWs v22 − v12 P2 − P1 ṁ 2 ρQv22 − = ṁ [U2 − U1 + + + g(y2 − y1 )] = v2 = ∂t 2 ρ 2 2 2 (20.55 m/s) kgm2 3 3 = (1.22 kg/m )(1.452 m /s) ( ) = 374.0 3 = 374.0 J/s 2 s J sec hp − hr 347.5 x 3600 x 3.725 x10−7 = 0.501 hp s hr J

6.9 A liquid flows from point A to point B in a horizontal pipe line shown in the figure. The liquid flows at a rate of 3 ft3/s with a friction loss of 0.45 ft of flowing fluid. If the pressure head at point B is 24 inches, what will be the pressure head at point A?

Solution U −U We are given that friction loss = 0.45 ft = B g A We want to find the pressure at A. P1 v12 P2 v22 (U1 + + + gy1 ) = (U2 + + + gy2 ) ρ 2 ρ 2 UB − UA vB2 − vA2 PB − PA + + =0 g 2g ρg Q 3 ft 3 /s vA = = = 3.83 ft/s π(1)2 AA 4 Q 3 ft 3 /s v2 = = = 15.28 ft/s A2 π(0.5 m)2 4 (15.28 ft/s)2 − (3.83 ft/s)2 PB − PA UB − UA vB2 − vA2 = + = 0.45 ft + = 3.85 ft of fluid ρg g 2g 2(32.2 ft/s2 ) PA = 3.85 ft + PB = 3.85 + (24 in x ∆P

Remember, ∆P = ρgh so that h = ρg

1 ft ) = 5.85 feet of fluid 12 in

6.11 A Venturi meter with an inlet diameter of 0.6 m is designed to handle 6 m3/s of standard air. What is the required throat diameter if this flow is to give a reading of 0.10 m of alcohol in a differential manometer connected to the inlet and the throat? The specific gravity of alcohol may be taken as 0.8. Solution V̇= 6 m3/s ∆P= 0.10 m Alcohol (S.G.= 0.8) = 0.08 m H20= 785 Pa π

A1= 4(0.6)2= 0.283 m2 V̇

v1=A = 0.283= 21.2 m/s 1

Energy Eqn. reduces to: P2 −P1 ρ

v2 −v2

+ 2 2 1 + g∆y=0

P1 −P2 v22 −v21 ρ

g∆y=0 v2

V̇2 1

= 2 [A2 - A2]= 21[(A1)2 − 1]

P1 −P2

785

= 1.226= 640 m2/s2

(21.2)2

640=

[(A1)2 − 1] 2

A2= 0.144 m2 D2=0.428m

6.39 A client has asked you to find the pressure change in a pumping station. The outlet from the pump is 20 feet above the inlet. A Newtonian fluid is being pumped at steady state. At the inlet to the pump where the diameter of 6 inches, the temperature of the fluid is 80℉, the viscosity is 1.80x103 lbm/ft sec, the density is 50 lbm/ft3, the heat capacity is 0.580 BTU/lbm F, and the kinematic viscosity is 3.60 x 10-5 ft2/sec. At the outlet to the pump where the diameter of 4 inches, the temperature of the fluid is 100℉, the viscosity is 1.30x10-3 lbm/ft sec, the density is 49.6 lbm/ft3, the heat capacity is 0.610 BTU/lbm F, and the kinematic viscosity is 2.62 x 10-5 ft2/sec. The flow rate through the system is constant at 20 ft3/sec. The pump provides work to the fluid at 3.85 x 108 lbm ft2/s3 and the heat transferred is 2.32x106 BTU/hr. You may neglect viscous work in your analysis. Under these circumstances, please calculate the pressure change between the inlet and the outlet of the pumping station. Solution δWμ δQ δWs P ∂ − = ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV + ∂t ∂t ρ ∂t ∂t First, since the shaft work and heat must be added, let’s make them have the same units: δWs lbm ft 2 8 − = 3.85x10 ∂t s3 δQ BTU ft − lb lb ft hr lbm ft 2 m 6 (778.17 ) (32.174 )( ) = 16134853.46 3 = 2.32x10 ∂t hr BTU lbf s2 3600 sec s δWμ δQ δWs P ∂ − = ∬ (e + ) ρ(v ∙ n) dA + ∭ eρdV + ∂t ∂t ρ ∂t ∂t Assuming no viscous work and steady state, δQ δWs P − = ∬ (e + ) ρ(v ∙ n) dA ∂t ∂t ρ Writing in terms of energy efflux, δQ δWs P P − = (e + ) ρvA − (e + ) ρvA ∂t ∂t ρ 2 ρ 1 Expanding the energy term, and using ρvA = ρQ, δQ δWs v22 − v12 P2 − P1 − = ρQ (U2 − U1 + + + g(y2 − y1 )) (equation 1) ∂t ∂t 2 ρ Next, let’s calculate each of the parts individually and then combine them. U2 − U1 = Cv2 T2 − Cv1 T1 BTU ) (100℉) = [(0.61 lbm ℉ BTU ft − lb lbm ft ft 2 ) (80℉)] (778.17 ) (32.174 ) − (0.58 = 365537.89 lbm ℉ BTU lbf s2 s2

2 3

20ft /s 20ft /s Q 2 Q 2 ( (4/12)2 ) − ( (0.5)2 ) π π (229.18 ft/s)2 − (101.86 ft/s)2 v22 − v12 (A2 ) − (A1 ) 4 4 = = = 2 2 2 2 ft 2 = 21074.0 2 s ft 2 2 g(y2 − y1 ) = (32.2 ft/s )(20 ft) = 644 2 s Plug these values back into equation 1, lbm ft 2 lbm ft 2 16134853.46 3 + 3.85x108 s s3 lbm ft 2 ft 2 P2 − P1 ft 2 3 (20ft ) /s) [365537.89 + 21074.0 + + 644 ] ft 3 s2 s 2 (50 lbm ) s2 ft 3 Note that the same value of ρ was used in both places in the above equation, it could have been left out since it cancels. = (50

lbm ft 2 lbm ft 2 P2 − P1 = 1000 [387255.89 + ] 3 2 lbm s s s (50 3 ) ft 2 P2 − P1 ft = 13878.96 2 lb s (50 m ) ft 3 ft 2 lb 13878.96 2 (50 m ) s ft 3 = 21568.6 lbf P2 − P1 = lb ft ft 2 32.174 m 2 lbf s

40113485305

Chapter 7 Show/Hide Problems 7.6 Calculate the viscosity of oxygen at 350 K and compare with the value given in Appendix I. Solution Begin with the equation (7-10) for viscosity of a pure gas, 𝜇 = 2.6693𝑥10−6

√𝑀𝑇 𝜎 2 Ω𝜇

Appendix K, Table K.2 has for O2 at 350 𝐾: 𝜖𝐴⁄ 𝜅 = 113 𝐾 𝑎𝑛𝑑 𝜎 = 3.433 Å 𝜅𝑇 𝑇 350 =𝜖 = = 3.097 ⁄𝜅 113 𝜖 Go to Table K.1 on and find the value for Ω𝜇 . Ω𝜇 = 1.030 The molecular weight of O2 is 16+16 = 32. Plug in the data, √(32)(350) √𝑀𝑇 𝜇 = 2.6693𝑥10−6 2 = 2.6693𝑥10−6 = 2.327𝑥10−5 Pa − s (3.433)2 (1.030) 𝜎 Ω𝜇 For comparison, Appendix I on page 736 gives viscosity results for Oxygen at 350K as 2.3176 x 10-5 Pa-s, which is in reasonable agreement.

7.10 Repeat the preceding problem for air. Solution For Air: @ 140oF @ 32oF

μ= 1.34x10-5 LBm/s*ft. μ= 1.15x10-5 LBm/s*ft.

For V̇~1/μ V̇140 1.15 = = 0.852 V̇32 1.34

Percent change =

1.15−1.34 1.34

= 0.852-1= -0.148 = -14.8 %

7.20 A Newtonian oil with a density of 60 lbm/ft3, viscosity of 0.206 x 10-3 lbm/ft-sec and kinematic viscosity of 0.342 x 10-5 ft2/sec undergoes steady shear between a horizontal fixed lower plate and a moving horizontal upper plate. The upper plate is moving with a velocity of 3 feet per second. The distance between the plates is 0.03 inches, and the area of the upper plate in contact with the fluid is 0.1 ft2. Assume incompressible, isothermal, inviscid, frictionless flow.

MOVING AT 3 ft/sec 0.03 inches

(a) What is shear stress exerted on the fluid under these conditions? (b) What is the force of the upper plate on the fluid? Solution (a) What is shear stress exerted on the fluid under these conditions?

𝜏𝑦𝑥 = 𝜇

𝑑𝑣𝑥 𝑑𝑦

0.03 𝑖𝑛𝑐ℎ𝑒𝑠

3 𝑓𝑡/𝑠

𝜏𝑦𝑥 ∫

𝑑𝑦 = 𝜇 ∫

𝑑𝑣𝑥

𝜏𝑦𝑥 (0.03 𝑖𝑛𝑐ℎ𝑒𝑠) = 𝜇(3 𝑓𝑡/𝑠) Solving for shear stress, 𝜏𝑦𝑥 =

(0.206 𝑥10−3 𝑙𝑏𝑚 /𝑓𝑡 ∙ 𝑠)(3 𝑓𝑡/𝑠) 𝑙𝑏𝑓 = 7.68𝑥10−3 2 𝑓𝑡 𝑙𝑏 𝑓𝑡 𝑓𝑡 (0.03 𝑖𝑛) ( ) (32.174 𝑚 2 ) 12 𝑖𝑛 𝑙𝑏𝑓 𝑠

(b) What is the force of the upper plate on the fluid?

𝜏𝑦𝑥 = 𝐹 = 𝜏𝑦𝑥 𝐴 = (7.68𝑥10−3

𝐹 𝐴

𝑙𝑏𝑓 ) (0.1 𝑓𝑡 2 ) = 7.68𝑥10−4 𝑙𝑏𝑓 𝑓𝑡 2

FIXED

Chapter 8 Show/Hide Problems

8.1 Express equation 8-9 in terms of the flow rate and the pipe diameter. If the pipe diameter is doubled at constant pressure drop, what percentage change will occur in the flow rate? Solution

Equation 8-9: − Substitute: 𝑄 = 𝑣𝐴 𝑎𝑛𝑑 𝐴 =

𝜋𝐷 2

dP 32μvavg = dx D2

−

dP 128𝜇𝑄 = dx 𝜋𝐷4

Which is an equation in terms of flow rate and diameter. Rearrange and solve for Q: dP 𝜋𝐷4 𝑄=− dx 128𝜇 We are given that 𝐷2 = 2𝐷1 , and it follows that 𝑄2 = 16𝑄1 The percentage change will be: 𝑄2 − 𝑄1 16𝑄1 − 𝑄1 𝑥100 = 𝑥100 = 15𝑥100 = 1500% 𝑄1 𝑄1 So the flow rate increases 1500% by doubling the pipe diameter.

8.9 Determine the velocity profile for fluid flowing between two parallel plates separate by a distance 2h. The pressure drop is constant.

Solution dp

(Hint: Begin with the following equation: dx = dy τyx ) This analysis is similar to what is done in the text in section 8.1 except in rectangular coordinates. dp d = τ dx dy yx Substitute Newton's Law of viscosity, dp d2 vx =μ 2 dx dy Rearrange, d2 vx 1 dp = dy 2 μ dx Integrate, dvx 1 dp = y + C1 dy μ dx integrate again, vx =

1 dp 2 y + C1 y + C2 2μ dx

Boundary conditions for this problem, 1. y = +h at vx = 0 (No Slip Boundary Condition) 2. y = −h at vx = 0 (No Slip Boundary Condition) Substitute in the first boundary condition and determine that C1 = 0. 1 dp Next, substitute in the second boundary condition to determine C2 = − 2μ dx ℎ2 Thus, vx =

1 dp 2 1 dp 2 y − h 2μ dx 2μ dx

8.18 A Newtonian fluid in continuous, incompressible, laminar flow is moving steadily through a very long 700 meter, horizontal pipe. The inside radius is 0.25 meter for the entire length and the pressure drop across the pipe is 1000 Pa. The average velocity of the fluid is 0.5 m/s. What is the viscosity of this fluid?

Solution 𝑑𝑃 8𝜇𝑣𝑎𝑣𝑔 = 𝑑𝑥 𝑅2 ∆𝑃 8𝜇𝑣𝑎𝑣𝑔 = 𝐿 𝑅2 (1000 𝑃𝑎)(0.25 𝑚)2 ∆𝑃𝑅 2 𝜇= = = 0.022 𝑃𝑎 ∙ 𝑠 8𝑣𝑎𝑣𝑔 𝐿 (8)(0.5 𝑚/𝑠)(700 𝑚) −

8.19 Benzene, which is an incompressible Newtonian Fluid, flows steadily and continuously at 150oF through a 3000 foot pipe with a constant diameter of 4 inches with a volumetric flow rate of 3.5 ft3/s. Assuming fully developed, laminar flow and that the no-slip boundary condition applies, calculate the change in pressure across this pipe system. Solution dP 32μvavg = dx D2 ∆P 32μvavg = L D2

−

Calculate vavg , vavg =

Q 3.5 ft 3 /s = = 40.11 ft/s A 4 2 π (12) 4

Calculate viscosity, 32μvavg L 32(2.45x10−4 lbm /ft s)(40.11 ft/s)(3000 ft) 1 ∆P = = 2 2 lb ft D 4 32.174 m 2 (12) ft 2 lbf s 2 = 263.89 lbf /ft

Chapter 9 Show/Hide Problems

9.2 ∂

∂

In Cartesian coordinates, show that vx ∂x + vy ∂y + vz ∂z may be written as (v ∙ ∇). What is the physical meaning of the term (v ∙ ∇)? Solution Start with the equations for ∇ and v, v = vx ex + vy ey + vz ez ∇=

v ∙ ∇= vx

∂ ∂ ∂ ex + e y + ez ∂x ∂y ∂z

∂ ∂ ∂ (ex ∙ ex ) + vy (ey ∙ ey ) + vz (ez ∙ ez ) ∂x ∂y ∂z

Thus, v ∙ ∇= vx

∂ ∂ ∂ + vy + vz ∂x ∂y ∂z

Physical Meaning: v ∙ ∇ is the rate of change due to motion.

9.4 𝐷𝑣

Find an equation for 𝐷𝑡 in polar coordinates by taking the derivative of the velocity. This can be done by using 𝑣 = 𝑣𝑟 (𝑟, 𝜃, 𝑡)𝑒𝑟 + 𝑣𝜃 (𝑟, 𝜃, 𝑡)𝑒𝜃 And 𝑣 = 𝑣𝑟 𝑒𝑟 + 𝑣𝜃 𝑒𝜃 Remember that unit vectors have derivatives. It might be helpful to consult Appendix A of the text. Solution 𝜕𝑣 𝜕𝑣 𝜕𝑣 𝑑𝑟 + 𝑑𝜃 + 𝑑𝑡 𝜕𝑟 𝜕𝜃 𝜕𝑡

(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2)

𝑑𝑣 𝜕𝑣 𝑑𝑟 𝜕𝑣 𝑑𝜃 𝜕𝑣 𝑑𝑡 = + + 𝑑𝑡 𝜕𝑟 𝑑𝑡 𝜕𝜃 𝑑𝑡 𝜕𝑡 𝑑𝑡 Partial differentiating 𝑣 by 𝜕𝑟:

(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3)

𝑑𝑣 = Divide by dt

𝜕𝑒𝜃 𝜕𝑒𝑟 𝜕𝑣 𝜕𝑣𝑟 𝜕𝑣𝜃 = 𝑒𝑟 + 𝑒𝜃 + 𝑣𝑟 + 𝑣𝜃 𝜕𝑟 𝜕𝑟 𝜕𝑟 𝜕𝑟 𝜕𝑟 Partial differentiating 𝑣 by 𝜕𝜃: 𝜕𝑒𝜃 𝜕𝑒𝑟 𝜕𝑣 𝜕𝑣𝑟 𝜕𝑣𝜃 = 𝑒𝑟 + 𝑒𝜃 + 𝑣𝑟 + 𝑣𝜃 𝜕𝜃 𝜕𝜃 𝜕𝜃 𝜕𝜃 𝜕𝜃

(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 4)

(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 5)

Realizing that 𝑒𝑟 and 𝑒𝜃 are related to 𝑒𝑥 and 𝑒𝑦 by: 𝑒𝑟 = 𝑒𝑥 cos 𝜃 + 𝑒𝑦 sin 𝜃

(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 6)

𝑒𝜃 = −𝑒𝑥 sin 𝜃 + 𝑒𝑦 cos 𝜃

(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 7)

Differentiate (𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 6) with respect to 𝑟 𝑎𝑛𝑑 𝜃 (see Appendix A of text): 𝜕𝑒𝑟 𝜕𝑟

𝜕𝑒𝑟 𝜕𝜃 𝜕𝜃 𝜕𝑟 𝜕𝑒𝑟 𝜕𝜃

= 𝑒𝜃

𝜕𝜃 =0 𝜕𝑟

(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 8)

(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 9)

Differentiate (𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 7) with respect to 𝑟 𝑎𝑛𝑑 𝜃 (see Appendix A of text):

𝜕𝑒𝜃 𝜕𝑟

𝜕𝑒𝜃 𝜕𝜃

𝜕𝜃 𝜕𝑟 = −𝑒𝑟

(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 10)

(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 11)

Now, substitute the values from equations 8, 9, 10, and 11 into 4 and 5: 𝜕𝑒𝜃 𝜕𝑒𝑟 𝜕𝑣 𝜕𝑣𝑟 𝜕𝑣𝜃 = 𝑒𝑟 + 𝑒𝜃 + 𝑣𝑟 + 𝑣𝜃 𝜕𝑟 𝜕𝑟 𝜕𝑟 𝜕𝑟 𝜕𝑟 So, 𝜕𝑣 𝜕𝑣𝑟 𝜕𝑣𝜃 = 𝑒𝑟 + 𝑒 𝜕𝑟 𝜕𝑟 𝜕𝑟 𝜃

(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 12)

𝜕𝑒𝜃 𝜕𝑒𝑟 𝜕𝑣 𝜕𝑣𝑟 𝜕𝑣𝜃 = 𝑒𝑟 + 𝑒𝜃 + 𝑣𝑟 + 𝑣𝜃 𝜕𝜃 𝜕𝜃 𝜕𝜃 𝜕𝜃 𝜕𝜃 Becomes, 𝜕𝑣 𝜕𝑣𝑟 𝜕𝑣𝜃 = 𝑒𝑟 + 𝑒 + 𝑣𝑟 𝑒𝜃 − 𝑣𝜃 𝑒𝑟 𝜕𝜃 𝜕𝜃 𝜕𝜃 𝜃 Which simplifies to 𝜕𝑣 𝜕𝑣𝑟 𝜕𝑣𝜃 =( − 𝑣𝜃 ) 𝑒𝑟 + ( + 𝑣𝑟 ) 𝑒𝜃 𝜕𝜃 𝜕𝜃 𝜕𝜃

(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 13)

From the hint given in the problem statement: 𝑑𝑟 = 𝑣𝑟 (𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 14) 𝑑𝑡 𝑑𝜃 𝑣𝜃 = (𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 15) 𝑑𝑡 𝑟 From equation 3, 𝑑𝑣 𝜕𝑣 𝑑𝑟 𝜕𝑣 𝑑𝜃 𝜕𝑣 = + + 𝑑𝑡 𝜕𝑟 𝑑𝑡 𝜕𝜃 𝑑𝑡 𝜕𝑡

(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3)

Substitute in equations 14, 15, 12 and 13 and rearrange: 𝐷𝑣 𝑑𝑣 𝜕𝑣𝑟 𝑣𝜃 𝜕𝑣𝑟 𝑣𝜃2 𝜕𝑣𝜃 𝑣𝜃 𝜕𝑣𝜃 𝑣𝜃 𝑣𝑟 ) 𝑒𝜃 = + (𝑣𝑟 + − ) 𝑒𝑟 + (𝑣𝑟 + + 𝐷𝑡 𝑑𝑡 𝜕𝑟 𝑟 𝜕𝜃 𝑟 𝜕𝑟 𝑟 𝜕𝜃 𝑟

9.5 For flow at very low speeds and with large viscosity (the so-called creeping flows) such as occur in lubrication, it is possible to delete the inertia terms, Dv/Dt from the Navier-Stokes equation. For flows at high velocity and small viscosity, it is not proper to delete the viscous term 𝜈∇2 𝐯. Explain this. Solution Navier-Stokes Eqn.- Incompressible form: Dυ⃑ Dt

∇P

= ⃑g - ρ +ν∇2 υ⃑ dυ⃑

(a) For υ⃑ small- all terms involving υ⃑ (~ dt & ν∇2 ⃑υ) are small relative to other terms. (b) For ν small & υ⃑ large the product νυ ⃑ cannot be considered small relative to other terms

9.6 Using the Navier-Stokes equations and the continuity equation, obtain an expression for the velocity profile between two flat, parallel plates. Solution First we sketch a picture of the problem.

flow direction

We begin with the Navier-Stokes equation in rectangular coordinates for the x-direction. ∂vx ∂vx ∂vx ∂vx ∂P ∂2 vx ∂2 vx ∂2 vx )=− ρ( + vx + vy + vz + ρg x + μ ( 2 + + 2) ∂t ∂x ∂y ∂z ∂x ∂x ∂y 2 ∂z Evaluating the equation term by term: ∂vx = 0 since the flow is steady ∂t ∂vx ∂2 vx vx = = 0 since the flow is fully developed ∂x ∂x 2 ∂vx vy = 0 since there is no y direction velocity ∂y ∂vx vz = 0 since there is no z direction velocity ∂z ρg x = 0 since there is no gravity in the x direction ∂2 vx = 0 since we have two dimensional flow ∂z 2 Thus, ∂P ∂2 vx 0=− +μ( 2 ) ∂x ∂y Rearrange, ∂2 vx 1 ∂P = ∂y 2 μ ∂x Integrate, ∂vx 1 ∂P = y + C1 ∂y μ ∂x Integrate again, 1 ∂P 2 vx = y + C1 y + C2 2μ ∂x Boundary conditions:

BC#1: when v=0, y=0 (No-Slip boundary condition) BC#2: when v=0, y=L (No-Slip boundary condition) From BC#1: C2=0. L ∂P

From BC#2: C1 = − 2μ ∂x Thus,

vx =

1 ∂P 2 1 ∂P 1 ∂P 2 (y − Ly) y − Ly = 2μ ∂x 2μ ∂x 2μ ∂x

9.21 Beginning with the appropriate form of the Navier-Stokes equations, develop an equation in the appropriate coordinate system to describe the velocity of a fluid that is flowing in the annular space as shown in the figure. The fluid is Newtonian, and is flowing in steady, incompressible, fully developed, laminar flow through an infinitely long vertical round pipe annulus of inner radius RI and outer radius RO. The inner cylinder (shown in the figure as a dotted line) is solid and the fluid flows between the inner and outer walls as shown in the figure. The center cylinder moves downward in the same direction as the fluid with a velocity V0. The outside wall of the annulus is stationary. In developing your equation, please specifically state the reason for eliminating any terms in the original equation. Fluid flow around center section

Solution Need to begin with z-direction Navier-Stokes equation in cylindrical coordinates:

𝜌(

𝜕𝑣𝑧 𝜕𝑣𝑧 𝑣𝜃 𝜕𝑣𝑧 𝜕𝑣𝑧 + 𝑣𝑟 + + 𝑣𝑧 ) 𝜕𝑡 𝜕𝑟 𝑟 𝜕𝜃 𝜕𝑧 𝜕𝑃 1𝜕 𝜕𝑣𝑧 1 𝜕 2 𝑣𝑧 𝜕 2 𝑣𝑧 =− + 𝜌𝑔𝑧 + 𝜇 ( + (𝑟 )+ 2 ) 𝜕𝑧 𝑟 𝜕𝑟 𝜕𝑟 𝑟 𝜕𝜃 2 𝜕𝑧 2

Applying the assumptions given in the problem statement:

0=−

𝜕𝑃 1𝜕 𝜕𝑣𝑧 − 𝜌𝑔𝑧 + 𝜇 ( (𝑟 )) 𝜕𝑧 𝑟 𝜕𝑟 𝜕𝑟

1𝜕 𝜕𝑣𝑧 𝜕𝑃 𝜇( + 𝜌𝑔𝑧 (𝑟 )) = 𝑟 𝜕𝑟 𝜕𝑟 𝜕𝑧 𝜕 𝜕𝑣𝑧 𝑟 𝜕𝑃 (𝑟 ) = ( + 𝜌𝑔𝑧 ) 𝜕𝑟 𝜕𝑟 𝜇 𝜕𝑧 2 𝜕𝑣𝑧 𝑟 𝜕𝑃 𝑟 = ( + 𝜌𝑔𝑧 ) + 𝐶1 𝜕𝑟 𝜇 𝜕𝑧 𝜕𝑣𝑧 𝑟 𝜕𝑃 𝐶1 = ( + 𝜌𝑔𝑧 ) + 𝜕𝑟 2𝜇 𝜕𝑧 𝑟 2 𝑟 𝜕𝑃 𝑣𝑧 = ( + 𝜌𝑔𝑧 ) + 𝐶1 ln 𝑟 + 𝐶2 4𝜇 𝜕𝑧 Boundary Conditions: 1. 𝑣𝑧 = 0 𝑎𝑡 𝑟 = 𝑅0 2. 𝑣𝑧 = −𝑣0 𝑎𝑡 𝑟 = 𝑅𝐼 Apply Boundary Condition #1:

𝑅0 2 𝜕𝑃

( + 𝜌𝑔𝑧 ) + 𝐶1 ln 𝑅0 + 𝐶2 4𝜇 𝜕𝑧

Apply Boundary Condition #2: 𝑅𝐼 2 𝜕𝑃 −𝑣0 = ( + 𝜌𝑔𝑧 ) + 𝐶1 ln 𝑅𝐼 + 𝐶2

4𝜇 𝜕𝑧

Solve for 𝐶1 , 𝑅0 2 𝜕𝑃

𝑅𝐼 2 𝜕𝑃

( + 𝜌𝑔𝑧 ) + 𝐶1 ln 𝑅0 = ( + 𝜌𝑔𝑧 ) + 𝐶1 ln 𝑅𝐼 + 𝑣0 4𝜇 𝜕𝑧 4𝜇 𝜕𝑧

𝐶1 =

𝜕𝑃 + 𝜌𝑔𝑧 ) + 𝑣0 𝜕𝑧 4𝜇(ln 𝑅0 − ln 𝑅𝐼 )

𝑅𝐼 2 − 𝑅0 2 (

Solve for 𝐶2 by plugging back into equation for Boundary Condition #1 (if you plug

into the Boundary Condition #2 equation your result will be slightly different and no less correct), 𝜕𝑃 𝑅𝐼 2 − 𝑅0 2 ( + 𝜌𝑔𝑧 ) + 𝑣0 𝑅0 2 𝜕𝑃 𝜕𝑧 0= ln 𝑅0 + 𝐶2 ( + 𝜌𝑔𝑧 ) + 4𝜇 𝜕𝑧 4𝜇(ln 𝑅0 − ln 𝑅𝐼 ) 𝜕𝑃 𝑅𝐼 2 − 𝑅0 2 ( + 𝜌𝑔𝑧 ) + 𝑣0 𝑅0 2 𝜕𝑃 𝜕𝑧 𝐶2 = − ln 𝑅0 ( + 𝜌𝑔𝑧 ) − 4𝜇 𝜕𝑧 4𝜇(ln 𝑅0 − ln 𝑅𝐼 ) Thus,

𝜕𝑃 𝑅𝐼 2 − 𝑅0 2 ( + 𝜌𝑔𝑧 ) + 𝑣0 𝑟 2 𝜕𝑃 𝑅0 2 𝜕𝑃 𝜕𝑧 𝑣𝑧 = ln 𝑟 − ( + 𝜌𝑔𝑧 ) + ( + 𝜌𝑔𝑧 ) 4𝜇 𝜕𝑧 4𝜇(ln 𝑅0 − ln 𝑅𝐼 ) 4𝜇 𝜕𝑧 𝜕𝑃 𝑅𝐼 2 − 𝑅0 2 ( + 𝜌𝑔𝑧 ) + 𝑣0 𝜕𝑧 − ln 𝑅0 4𝜇(ln 𝑅0 − ln 𝑅𝐼 )

Chapter 10 Show/Hide Problems

10.3 Find the stream function for a flow with a uniform free-stream velocity v∞ . The free-stream velocity intersects the x-axis at an angle α. Solution dψ = −vy dx + vx dy = (−v∞ sinα)dx + (v∞ cosα)dy ψ = (−v∞ sinα)x + (v∞ cosα)y + ψ0

10.5 5 The velocity potential for a given two-dimensional flow field is ϕ = 3 x 3 − 5xy 2 . Show that the continuity equation is satisfied and determine the corresponding stream function. Solution 5 3 x − 5xy 2 3 since v = ∇ψ, continuity can be expressed as ∇ ∙ v ⃑ = 0 or ∇2 ϕ = 0 ϕ=

using ∇2 ϕ = 0 ∂2 ϕ ∂2 ϕ + =0 ∂x 2 ∂y 2 10x − 10x = 0 vx =

∂ϕ ∂ψ = = 5x 2 ∂x ∂y

or ψ = 5x 2 y ∂ϕ

∂ψ

~ ∂y = vy = − ∂x ~ Check

10.17 Sketch the streamlines and potential lines of the flow due to a line source at (a,0) plus an equivalent sink at (-a,0) Solution

10.23 The stream function is given by

ψ = 6x 2 − 6y 2 Determine whether this flow is rotational or irrotational. Solution To be irrotational, the flow must satisfy the equation, ∇ x v = 0 and as a result, Equation 10.1 must be equal to zero. 1 𝜕𝑣𝑦 𝜕𝑣𝑥 𝑤𝑧 = ( − )=0 2 𝜕𝑥 𝜕𝑦 For the equation of the stream function we solve for 𝑣𝑥 and 𝑣𝑦 using the definition of the stream function, 𝜕𝜓 𝑣𝑥 = = −12𝑦 𝜕𝑦 𝑣𝑦 = − And then,

𝜕𝜓 = −12𝑥 𝜕𝑥

𝜕𝑣𝑥 = −12 𝜕𝑦 𝜕𝑣𝑦 = −12 𝜕𝑥 1 𝑤𝑧 = (−12 − (−12)) = 0 2 So the condition of irrotational flow is satisfied.

Chapter 11 Show/Hide Problems 11.1 The power output of a hydraulic turbine depends on the diameter D of the turbine, the density  of water, the height H of water surface above the turbine, the gravitational acceleration g, the angular velocity  of the turbine wheel, the discharge Q of water through the turbine, and the efficiency  of the turbine. By dimensional analysis, generate a set of appropriate dimensionless groups. Solution D = diameter  = density H = height g = gravity  = angular velocity Q = flow rate  = efficiency P = power

M L t

D 0 1 0

 1 -3 0

L ML-3 L Lt-2 t-1 L3t-1 unitless ML2t-3

H 0 1 0

g 0 1 -2

 0 0 -1

 0 0 0

Q 0 3 -1

P 1 2 -3

i = n-r = 8 - 3 = 5 core group = , D,  a

1=aDbcH

H  1=   D

1=0D-10H → a

2=aDbcQ

M  1 MoLoto = 1 =  3  (L )b   L L  t  M: 0 = a+0 L: 0 = -3a+b+1 T: 0 = -c+0 Thus: a= 0, b= -1, c= 0

3 M  1 L MoLoto = 1 =  3  (L )b   L  t  t

M: 0 = a L: 0 = b+3 T: 0 = -c+1 Thus: a= 0, b= -3, c= -1  Q   1=  3  D 

2=0D-10H →

3=aDbcg

M  MoLoto = 1 =  3 

(L )   L2

L  t  t M: 0 = a L: 0 = -3a+b+1 T: 0 = c-2 Thus: a= 0, b= -1, c= -2  g   1=  2   D 

3=0D-10g → a

4=aDbcP

b1

2 M   1  ML MoLoto = 1 =  3  (L )b   3 L  t  t M: 0 = a+1 L: 0 = -3a+b+2 T: 0 = -c-3 Thus: a= -1, b= -5, c= -3

4=0D-10P →

5=  (by inspection)

 P   1=  5 3  D  

11.3 The pressure rise across a pump P (this term is proportional to the head developed by the pump) may be considered to be affected by the fluid density , the angular velocity , the impeller diameter D, the volumetric rate of flow Q, and the fluid viscosity . Find the pertinent dimensionless groups, choosing them so that P, Q and  each appear in one group only.

Solution Variable ΔP ρ ω D Q μ

Dimensions M/Lt2 M/L3 1/t L L3/t M/Lt

i = 6 − 3 = 3 , Choose core as ρ, D, ω ΔP π1 = ρa Db ωc ΔP ∴ = 2 2 ρD ω Q π2 = ρd De ωf Q ∴ = 3 D ω μ g h i π3 = ρ D ω μ ∴ = 2 ρD ω

11.7 The functional frequency n of a stretched string is a function of the string length L, it’s diameter D, the mass density ρ, and the applied tensile force T. Suggest a set of dimensionless parameters relating these variables. Solution Variable n d L T ρ

Dimensions 1/t L L ML/t2 M/L3

i = 5 − 3 = 2 , Choose core as d, n, ρ L π1 = da nb ρc L ∴ = d T π2 = dd ne ρf T ∴ = 2 4 n Lρ

11.8 The power P required to run a compressor varies with compressor diameter D, angular velocity ω, volume flow rate Q, fluid density ρ, and fluid viscosity μ. Develop a relation between these variable by dimensional analysis, where fluid viscosity and angular velocity appear in only one dimensionless parameter. Solution Variable Q d P ω ρ μ

Dimensions L3/t L ML2/t3 1/t M/L3 M/Lt i = 6 − 3 = 3 , Choose core as d, Q, ρ d3 ω π1 = da Qb ρc ω ∴ = Q dμ π2 = dd Qe ρf μ ∴ = ρQ d4 P g h i π3 = d Q ρ P ∴ = 3 ρQ π3 = f(π1 , π2 )

Chapter 12 Show/Hide Problems

12.1 If Reynolds’s experiment was performed with a 38-mm ID pipe, what flow velocity would occur at transition? Solution At transiton Red =

Dv = 2300 ν

H2 O@20℃: ν = 0.995x10−6 v=

m2 s

2300(0.995x10−6 ) m cm = 0.060 = 6 0.038m s s

12.2 Modern subsonic aircraft have been refined to such an extent that 75% of the parasite drag (portion of total aircraft drag not directly associated with producing lift) can be attribute to friction along the external surfaces. For a typical subsonic jet, the parasite drag coefficient based on wing area is 0.011. Determine the friction drag on such an aircraft. (a) at 500 mph at 35,000 ft (b) at 200 mph at sea level Solution v2 fD = CD Aρ 2 For 35000 ft: ρ = 0.0237

lbm lbm , for S. L. , ρ = 0.0766 3 3 ft ft

a) @35000 ft, 500mph = 733ft/s 0.011(2400)(0.0237)(733)2 fDp = = 5220 lb𝑓 2(32.2) b)@ Sea Level, 200mph = 293ft/s 0.011(2400)(0.0766)(293)2 fD = = 2700 lb𝑓 2(32.2)

12.3 Consider the flow of air at 60oF and 30 m/s along a flat plate. At what distance from the leading edge will transition flow occur?

Solution Transition over a flat plate begins at Re=2x105 𝑅𝑒 =

𝐷𝜌v 𝜇

From Appendix I, Air at 60oF: kinematic viscosity =  = 0.159x10-3 ft2/s Density =  = 0.0764 lbm/ft3 viscosity =  = 1.21x10-5 lbm/ft-s Transition occurs, 𝑅𝑒𝜇 (2𝑥105 )(1.21𝑥10−5 𝑙𝑏𝑚 /𝑓𝑡 ∙ 𝑠) 𝑚2 𝑥= = ( ) = 0.1 𝑚 (0.0765 𝑙𝑏𝑚 /𝑓𝑡 3 )(30 𝑚/𝑠) 𝜌v 10.76 𝑓𝑡 2

12.9 For what wind velocities will a 12.7-mm-diameter cable be in the unsteady wake region of Figure 12.2? Solution In the unsteady wake region, 1 < 𝑅𝑒 < 103 , Re = @ 20℃, ν = 1.505x10−5 @Re = 1 =

m2 s

0.0127v m ∴ v = 0.00119 1.505x10−5 s

@Re = 103 =

0.0127v m ∴ v = 1.185 −5 1.505x10 s

These are the lower & 𝑢𝑝𝑝𝑒𝑟 𝑏𝑜𝑢𝑛𝑑𝑠 𝑓𝑜𝑟 𝑣

Dv ν

12.10 Estimate the drag force on a 3-ft radio antenna with an average diameter of 0.2 in at a speed of 40 mph. Solution ft 2 lbm For Air @ 80℉, ν = 0.169x10−3 , ρ = 0.0735 3 s ft 0.2 ( 12 ) (88) Re = = 8680 0.169x10−3 From 12.2, CD ≅ 1.2 FD = CD Aρ

v2 0.2 0.0735 882 )( = 1.2 ( ) (3) ( ) = 0.530 lbf 2 12 32.2 2

12.11 A 2007 Toyota Prius has a drag coefficient of 0.26 at road speeds, using a reference area of 2.33 m2. Determine the horsepower required to overcome drag at a velocity of 30 m/s. Compare this figure with the case of head and tail winds of 6 m/s. Solution 100mph = 44.7

m s

v 2 0.21(2.33)(1.2048)(44.7)2 FL = CL Aρ = = 589 N 2 2

Chapter 13 Show/Hide Problems

13.1 An oil with a kinematic viscosity of 0.08 x 10-3 ft2/s, viscosity of 0.00456 lbm/ft-s and a density of 57 lbm/ft3 flows through a horizontal tube 0.24 inches in diameter at the rate of 10 gallons per hour. Determine the pressure drop in 50 feet of tube.

Solution Q = 10

gal ft 3 min hr x x x = 3.714x10−4 ft 3 /s hr 7.480 gal 60 sec 60 min v=

Q 3.714x10−4 ft 3 /s = = 1.18 ft/s A ft 2 π (0.24in x 12 in) 4

Next, we need to determine the change in pressure in a tube with a length of 50 ft. We can do this by equating the head loss equations, L v2 hL = 2ff Dg ∆P hL = ρg Equating these gives, ∆P L v2 = 2ff ρg Dg Solving for ∆P, (57 lbm /ft 3 ) L 50 ft (1.18 ft/s)2 = 12334ff ∆P = 2ρff v2 = 2 ff 2 (32.174 lbm ft/lbf s ) 0.24/12 ft D Next, need to find ff , and to do this we must determine whether the flow is laminar or turbulent. Re =

ρvD (57 lbm /ft 3 )(1.18 ft/s)(0.02 ft) = = 295 (0.00456 lbm /ft ∙ s) μ 16 ff = = 0.054 Re

Thus, ∆P = 12334ff = 12334(0.054) = 666.0 lbf /ft 2

13.2 A lubricating line carrying oil has an inside diameter of 0.1 inches and is 30 inches long. If the pressure drop is 15 psi, determine the flow rate of the oil if the oil has a kinematic viscosity of 0.08 x10-3 ft2/s and a density of 57 lbm/ft3. (Hint, you many assume laminar flow to begin the problem but you must confirm this once you have the velocity calculated.) Solution hL =

∆P 32μvavg L = ρg ρgD2

Solve for vavg , vavg =

∆PρgD2 ∆P D2 ∆P D2 = = 32μLρg 32μL 32νρL

Where ν is the kinematic viscosity. 2

vavg =

∆P D2 = 32νρL

lb 144 in2 ft ) (0.1 in ( (15 2f ) ( 12 in)) ft in 32 (57

lbm ft 2 ft ) (0.08x10−3 s ) (30 in (12 in)) 3 ft

(32.174

lbm ft ) = 13.23 ft/s lbf s2

Before we proceed further, we need to confirm that the flow is laminar. ft vD (13.23 ft/s) [0.1 in (12 in)] Re = = = 1378 (so we confirm laminar flow) ft 2 ν 0.08x10−3 s 2

ft π (0.1 in (12 in))

πD2 Q = Av = ( ) (13.23 ft/s) = 4

(13.23 ft/s) = 7.2x10−4

4 (

)

ft 3 s

13.3 The pressure drop in a section of a pipe is determined from tests with water. A pressure drop of 13 psi is obtained at a flow rate of 28.3 lbm/s. If the flow is fully turbulent, what will be the pressure drop when liquid oxygen (density = 70 lbm/ft3) flows through the pipe at the rate of 35 lbm/s. Solution L 2 v D e For a specified pipe: ΔP~fF v 2 , and if fully turbulent, fF ~ only ∴ ΔP = v 2 D lbm For H2 O: ΔP = 13 PSI for ṁ = 28.3 s ΔP = 2fF

For Lox: ρ = 70

lbm lbm and m ̇ = 35 ft 3 s

m 2 (ρA)

ΔPlox 35 2 62.4 2 Lox ( ) ( ) = 1.21 = = m 2 ΔPH2O 70 28.3 (ρA) H O 2

For Lox: ΔP = 13(1.21) = 158 PSI

13.8 Water at a rate of 118ft3/min flows through a smooth horizontal tube 250 ft long. The pressure drop is 4.55 psi. Determine the tube diameter. Solution P2 − P1 v22 − v12 0= + + gΔy + hL ρ 2 ΔP 4.55(144)(32.2) ft 2 − =− = −338.1 , ρ 62.4 s Δv 2 = 0, gΔy = 0, L hL = 2ff v 2 D 118 2.50 v= = , π D2 (60) ( ) (D2 ) 4 250 2.5 2 3125 ( ) = 5 ff hL = 2ff D D2 D 3125 Governing Eqn is − 338.1 + 5 ff = 0 ∴ ff = 0.1082D5 , other constraint is ff (Re) D Dv 118 2.052x105 Re = = = πD ν D 60 ( 4 ) (1.22x10−5 ) Trial & Error − Assume ff = 0.004 0.004 ] = 0.517ft, 0.1082 2.052x10−5 Re = = 3.97x10−5 , 0.517 ∴ ff = 0.00325 using this value, D = 0.496 ft, Re = 4.137x10−5 , ff = 0.0031 → D = 0.491 ft (5.9 in) D=[

13.12 A galvanized rectangular duct 8 in. square is 25 ft long and carries 600 ft 3/min of standard air. Determine the pressure drop in inches of water.

Solution Rectangular duct: 8”x8”x25 ft v̇ = 600ft 3 /m Standard Air 4(8)(8) Deq = = 8in 4 8 600/60 v= = 22.5 ft/s (8)8/144 Energy equation reduces to: ∆P L = 2ff v 2 ρ D 8 ( ) (25) Re = 32 = 9.59x104 1.56x10−5 e 0.005 = = 0.00075 D 8/12 From Figure 13.1: ff = 0.0054 ∆P 25 ) (22.5)2 = 205 ft 2 /s2 = 2(0.0054) ( ρ 8/12 ∆P =

205(0.0766) 0.0766 = 0.4876 psf = 6.366 ft air = 76.4 in air = 76.4 = 0.0938 in H2 O 32.2 62.4

Chapter 14 Show/Hide Problems

14.1 A centrifugal pump delivers 02 m3/s of water when operating at 850 rpm. Relevant impeller dimensions are as follows: Outside diameter = 0.45 m, blade length = 50 cm, and blade exit angle = 24o. Determine (a) the torque and power required to drive the pump and (b) the maximum pressure increase across the pump. Solution Centrifugal Pump: m3 v̇ = 0.2 , s ω = 850rpm (89.0 r2 = 0.225m, kg ρ = 1000 3 , m L = 0.05m, β2 = 24°

rad ), s

Torque − Eqn. 14.9: mz = ρv̇ r2 [r2 ω −

v̇ cotβ2 ] 2πr2 L

2π = 89.0 rad/s 60 v̇ 0.2cot24 mz = ρv̇ r2 [r2 ω − cotβ2 ] = (1000)(0.2)(0.225)x [(0.225)(89) − ] 2πr2 L 2π(0.225)(0.05) = 615Nm w = 850

ẇ = mz ω = 615 ∗ 89 = 54.75kW Nm 54.75x103 s ΔP ẇ ẇ | = − = − ∴ ΔPmax = − = −274kPa m2 ρ max ṁ ρv̇ 0.2 s

14.5 A centrifugal pump is being used to pump water at a flow rate of 0.018 m3/s and the required power is measured to be 4.5 kW. If the pump efficiency is 63%, determine the head generated by the pump. Solution kg Centrifugal Pump: ρ = 1000 3 , m η=

ṁΔP ρẇ

∴ ΔP =

ẇ = 4.5kW,

m3 v̇ = 0.018 , s

η = 63%

ηρẇ ηẇ 0.63(4500) = = = 157.5 kPa ṁ v̇ 0.018

ΔP 157500 = = 16.05 m H2 O ρg 1000(9.81)

14.17 The pump having the characteristics shown in Problem 14.14 is to be operated at 800 rpm. What discharge rate is to be expected if the head developed is 410 m? Solution Same Pump family as Prob 14.14 New Pump: η = 800 rpm = 83.8

CH =

rad , h = 410m s

gh 9.81(410) = = 4.161 2 2 (83.8)2 (0.371)2 n D

At this value of CH , CQ = 0.16 =

CQ ≅ 0.16

v nD3

∴ v̇ = 0.16n1 D13 = 0.16(83.8)(0.371)2 = 0.685

m3 s

14.18 If the pump having the characteristics shown in Problem 14.14 is tripled in size but halved in rotational speed, what will be the discharge rate and head when operating at maximum efficiency? Solution Same Pump family as Prob 14.14 New Pump: D2 = 3D1 , n2 = 0.5n1 @ηmax , CQ ≅ 0.12, CH ≅ 5.3 h2 ω2 2 D2 2 = ( ) ( ) = 0.52 (3)2 = 2.25 h1 ω1 D1 v2̇ n2 D2 3 1 = ( ) ( ) = (3)3 = 13.5 v1̇ n1 D1 2 2 (0.371)2 5.3(37.70) h1 = = 105.7 m 9.81 ∴ h2 = 105.7(2.25) = 238m m3 2 v1̇ = 0.12(37.70)(0.371) = 0.231 s m3 ∴ v2 = 0.231(13.5) = 3.12 s

Chapter 15 Show/Hide Problems 15.5 A sheet of insulating material, with thermal conductivity of 0.22W/m K, is 2 cm thick and has a surface area of 2.97m2. If 4 kW of heat are conducted through this sheet and the outer (cooler) surface temperature is measured at 55°C (328K), what will be the temperature on the inner (hot) surface?

Solution kA ΔT L 4000W(0.02m) ΔT = = 122.4K 0.22W ( m ∙ K ) (297m2 ) q=

Ti = 55 + 122.4 = 177.4 C

15.7 Plate glass, k = 1.35 W/m K, initially at 850 K, is cooled by blowing air past both surfaces with an effective surface coefficient of 5 W/m2 ? K. It is necessary, in order that the glass not crack, to limit the maximum temperature gradient in the glass to 15 K/cm during the cooling process. At the start of the cooling process, what is the lowest temperature of the cooling air that can be used?

Solution

dt 1.35W 15K 100cm W ΔT )( )( ) = 2025 2 = | =( dx max m∙K cm m m R 1 ΔT = 2025 ( ) = 405K 5 Tmin = 850 − 405 = 445K q max = −k

15.8 Solve Problem 15.7 if all specified conditions remain the same but radiant-energy exchange from glass to the surroundings at the air temperature is also considered. Solution

W ΔT 4 = + σ(Tsurf − TA4 ) 2 m R 4 850 − T T 4 ) ] = + 5.676 [8.5 − ( 1 100 5 by Trial and Error, T = 836K q max (From previous problem) = 2025

15.9 The heat loss from a boiler is to be held at a maximum of 900 Btu/h ft2 of wall area. What thickness of asbestos (k = 0.10 Btu/h ft °F) is required if the inner and outer surfaces of the insulation are to be 1600 and 500°F, respectively?

Solution q kΔT = A L

(0.10

kΔT L= q A

Btu ) (1000F) hrftF = 0.122ft = 1.47in Btu 900 hrft 2

15.14 If, in Problem 15.13, the plate is made of asbestos, k = 0.10 Btu/h ft °F, what will be the temperature of the top of the asbestos if the hot plate is rated at 800 W? Solution

BTU hr If all heat leaves top surface 800W = 2730

q = hAΔT + σAϵ[T 4 − Ts4 ] 2700 T 4 ) − 5.44 ] = 5(T − 40) + 0.1714(1) [( A 100 By trial & error: T = 1086R = 626F

Chapter 16 Show’Hide Problems

16.1 The Fourier Field Equatio in cylindrical coordinate is 𝜕𝑇 𝜕 2 𝑇 1 𝜕𝑇 1 𝜕 2 𝑇 𝜕 2 𝑇 = 𝛼( 2 + + + ) 𝜕𝑡 𝜕𝑟 𝑟 𝜕𝑟 𝑟 2 𝜕𝜃 2 𝜕𝑧 2 (a) What form does this equation reduce to for the case of steady-state, radial heat transfer? (b) Given the boundary conditions 𝑇 = 𝑇𝑖 𝑎𝑡 𝑟 = 𝑟𝑖 𝑇 = 𝑇𝑜 𝑎𝑡 𝑟 = 𝑟𝑜 (c) Generate an expression for the heat flow rate, 𝑞𝑟 , using the result from part (b). Solution d2 T 1 dT 1 d dT (r ) = 0 In Cylindrical Coordinates: 2 + = 0 or dr r dr r dr dr r

dT = c1 dr

T = c1 lnr + c2 BC: Ti = c1 lnri + c2 ,

To = c1 lnro + c2

Ti − To c2 = TL − c1 lnri r ln ( ro ) i r ln (r ) i T = Ti − (Ti − To ) ro ln ( r ) c1 = −

q = −kA

dT dT 2πkL = −k(2πrL) = −2πkLc1 = r (Ti − To ) dr dr ln ( ro ) i

16.5 Solve equation (16-19) for the temperature distribution in a plane wall if the internal heat generation per unit volume varies according to 𝑞̇ = 𝑞̇ 𝑜 𝑒 −𝛽/𝐿 . The boundary conditions that apply are 𝑇 = 𝑇𝑜 𝑎𝑡 𝑥 = 0 𝑇 = 𝑇𝐿 𝑎𝑡 𝑥 = 𝐿 Solution @ Steady State ∇2 T +

q̇ =0 k

d2 T q̇ −βx + e L =0 dx 2 k dT q ȯ L −βx = e L + c1 dx k β q ȯ L2 −βx T=− e L + c1 x + c2 k β2 q ȯ L2 −βx BC: T(O) = To , To = − e L + c2 , k β2 q ȯ L2 −β T(L) = TL , TL = − e + c1 L + c2 k β2 βx x q ȯ L2 x − L − (1 − e−β )] (T ) T = To + L − T0 + [1 − e L k β2 L

Chapter 17 Instructor Only Problems 17.1 Solution dT

qx= -kAdx = kA/L (T1-T2) For T1-T2= 75 K q=

(30 W/m∗K)(1 m2 )(75 K) 0.30 m ∆T

= 7500 w/m2

75 K

dT/dx = L = 0.30 m = -250 K/m For T1=300 K q

q= -2000 W/m

dT/dx= − kA = qL

∆T = kA =

2000 W/m2 W ) m∗K

(30

−2000 (0.3) 30

= 66.7 K/m

= -20 K

T2= 320 K For T2=350 K

dT/dx= -300 K/m

q= -(30)(-300)= 9000 W/m2 ∆T= -300 K/m (0.3m)= 90 K T1= 440 K For T1=250 K

dT/dx= 200 K/m

q = -(30)(200 K/m)= -6000 W/m2 ∆T= -200(0.3m)= -60 K T2= 310 K

17.2 Solution ̅ kA

r −r

2πk

ln o

q = r −r ∆T = r −r 2π o roi ∆T = o

% Error=

ALM −AAM ALM

x100

2π(ro −ri ) − π (ro +ri ) r ln o ri 2π(ro − ri ) r ln o ri r (ro +ri ) ln o ri

=|1 −

2(ro − ri ) r ri

r ri

x 100

| x 100

( o +1) ln o

=|1 − r

r ri

2( o −1)

| x 100

For ro =1.5

% error= 1.3 %

=3 =5

% error= 10.0% % error= 20.7%

∆T (a)

17.3 Solution 4πkr r

̅=4πro ri (a) A

q = r −or i ∆T o

Am =

4π(r2o +r2i ) 2

% Error =

= 2π(ro2 + ri2 )

4πro ri −2π (r2o +r2i ) 4πro ri

= 1-(1/2)( ro − r i ) i

(b) ro ri

= 1.5

% error=8.3 %

=3 =5

% error= 66.6% % error= 160 %

17.4 Solution

q”= -k dx = -ko(1+bT) dx From 0 to 12: 1.2

q” ∫0 dx= -ko∫925(1 + bT)dT k

q”= 1.2o [T+2T2]T925 = 23(T-300) Solving: TRH wall= 307.1 K q”= 163.3 W/m2 From 0 to L: L

650

q”∫0 = −k o ∫925 (1 + bT)dT k

L= q"o [T + 2 T 2 ]925 650 & Solving: L=0.646 m

17.8 Solution Brick Size= 9”x 4.5”x 3” Brick #1 Brick # 2

BTU

k=0.44 Hr Ft F

Tmax=1500 F

k=0.94 Hr Ft F

Tmax=2200 F

BTU

Most economical arrangement is to use as much of #1 as possible (low k). Use #2 next to high temp such that its cooler surface has T≤ 1500 F. q”=

2000−T

Lz =

L/k

LActual = 28.5 in

k(2000−Tm ) 200

= 2.35 Ft. = 28.2 in. (9x2+4.5+2x3) q′′

TInterface = TH − k2 = 1495 F L2

Lmin =

0.44(1495−300) 200

= 2.63 Ft = 31.6 in.

LAct = 33 in Most economical:

L1 = 33 in L2 = 28.5 "

17.9 Solution

Ri=1/115= 8.696x10-5 K/W R1= 0.25/1.13= 0.221 K/W R2= 0.20/1.45= 0.138 K/W R3= 0.10/0.66= 0.152 K/W Ro= 1/23 = 0.0435 K/W ∑ R = 0.563 1070

q= ∆T/ ∑ R= 0.563 = 1900 W/m2 = 176.5 W/ft2 q= 23(To-300)

To= 383 K

17.10 Solution

To= 325 K

q=23(325-300) = 575 W/m2

Ri=1/115= 8.696x10-5 K/W R1= 0.25/1.13= 0.221 K/W R2= L/1.45 R3= 0.10/0.166= 0.152 K/W ∑ R= 0.416 + L/1.45= ∆T/q 1045

0.416+ L/1.45= 573 L= 2.03 m

17.12 Solution

0.125 12

R1 = 0.15 = 0.0694 R2 = R3 =

0.125/12

=0.00104

10 0.25/12 π 4

(22)(2)( )(

0.75 2 ) 12

RConduction, Equiv =

= 0.154 1

1 1 + R1 +R2 R3

1 1 1 + 0.07044 0.154

= 0.0483

∑ R per side = 0.0483+ 1/3= 0.3817 New Ht. Flux = Increase =

930 0.3817

2437−2305 2305

= 2437 BTU/Hr-ft2

= 0.057= 5.7 %

Chapter 18 Show/Hide Problems

18.4 Solution Bi =

hV 16 6/12 = ( ) = 0.058 kS 23 6

A lumped parameter problem Fo =

αt 23 ft 2 t = ∗ = 72t 2 (V/S) 460(0.1) hr (6/12)2 6

T − T∞ 600 −(0.058)(72t) = e To − T∞ 2000 t = 0.228 hr = 17.3 min

18.5 Solution Lumped parameter solution applies if Bi =

hV < 0.1 or if h < 0.1 kS

kS kπD2 = 3 = 0.47(6) = 2.82 V πD /6 W

Therefore h must be < 0.1(2.82) = 0.282 m2K But h = 15 ∴ Use Distributed parameter solution (0.47)t αt = = 5.26x10−5 t ro 2 (940)(3800)(0.05)2 T − T∞ = 0.5 To − T∞

k 0.47 = = 0.627 hro 15(0.05)

X ≅ 0.17 = 5.26x10−5 t t = 3230 s = 53.9 min

18.7 Solution T − T∞ 500 − 1000 = = 0.538 To − T∞ 70 − 1000 V D 1 = = A 4 + 2 D 17 L v hs

4 17 Bi = = k k a) Cu – Bi<0.1 Lumped ρĈp V 1 t= ln = 27.9 min hA 0.538 b) Al – Bi<0.1Lumped t = 0.345 hr = 20.7 min c) Zn - Bi<0.1Lumped t = 0.381 hr = 22.9 min d) Steel - Bi<0.1Lumped t = 0.502 hr = 30.2 min

18.9 Solution

Bi =

v hs k

πD2 L 4⁄ 15 [ πDL] =

12.4

hD D 4k (L + 2 )

85(0.6)(0.6) = 0.0371 (229)(4)(0.9)

Lumped parameter case. Temperature may be considered uniform at any time. Fo =

(9.16x10−5 )(3600) αt = = 32.98 (V/S)2 (0.1)2

T − T∞ = e−Bi Fo = e−(0.0371)(32.98) = 0.294 To − T∞ T = 345 + 0.294(130) = 383.2 K

18.10 Solution 3 hV 40 πD ⁄6 Bi = = = 0.00575 kS 19.3πD2

Lumped parameter 0.8(15⁄3600) αt Fo = = = 432 (V/S)2 [0.2⁄2(6)]2 T − T∞ = e−Bi Fo = 0.0834 To − T∞ T = 115.9 ℉

Chapter 19 Show/Hide Problems

19.1 Solution For a plane wall: Variables T T0 T∞ x L Α k t h

Dimension T T T L L L2 ⁄t Q⁄LtT t Q⁄LtT

i=n−r=5 If temps are grouped as T − T∞ , T0 − T∞ ,

i=n−r=4

π1 = ∆T a Lb k c αd (T − T∞ ) π2 = ∆T a Lb k c αd (x) π3 = ∆T a Lb k c αd (t) π4 = ∆T a Lb k c αd (h) π1 =

T − T∞ , T0 − T∞

x π2 = , L

π3 =

αt , L2

π4 =

hL k

19.5 Solution For a single 4 ft long plate: L

q = W ∫ (α + β sin ( 0

πx )) dx L

βL πx q = W (αx + (cos ( ))) π L

β 2(1500) q = WL (α + 2 ) = 16 ft 2 (250 + ) = 19300 Btu⁄hr π π For stack of plates: q = 19300

(640) = 772,000 Btu⁄hr 16

19.7 Solution L q πx πx βL 15 3000 = α + β sin ( ) = πD ∫ [α + β sin ( )] dx = πD [αL + 2 ] = π ( ) [250 + ] A L L π 12 π 0 = 4730 Btu⁄Hr

TExit = Ti +

q 4730 = 60 + = 60.3 ℉ ρvCp A 60(1)(0.067)(3600)

Tw = 60.3 +

250 ≅ 60.6 ℉ 976

19.11 Solution For air @ Tf = 325 K: ρ = 1.087 kg⁄m3 Cp =1.008 kJ⁄kg K k = 2.816 W⁄m K ν = 1.807x10−5 m2 ⁄s Pr = 0.702 −1⁄ 2

a) CfL = 1.328ReL

m (1 m) (2.8 ) Lv s Re = = = 1.55x105 −5 2 ⁄ ν 1.807x10 m s −1 CfL = 1.328(1.55x105 ) ⁄2 = 0.00337 δv2

b) fd = CfL A 2 =

(0.00337)(0.25)(1)(1.087)(2.8)2 2

= 3.59x10−3 N

c) q = hA∆T Using Colburn analogy

St L =

(0.00337) CfL h −2 −2 (Pr) ⁄3 = (0.702) ⁄3 = 2.133x10−3 = 2 2 ρvCp

h = (2.133x10−3 )(1.087)(1008)(2.8) = 6.54 W⁄m2 K

q = 6.54(1)(0.25)(55) = 90.0 W

Chapter 20 Show/Hide Problems 20.1 Solution q 750(3.413) = = 26100 Btu⁄hr ft 2 A π(3⁄ )(1⁄ ) 48 2 Ends are neglected 2 1

k k h = NuL = 0.825 + L L [

0.387ReL 6 8 9 27 0.492 16

{1 + ( Pr ) } ]

(a) For vertical orientation By trial and error ∆𝑇 ≅ 103 ℉ HTR surface temp = 198 ℉ (b) Horizontal orientation 2

k h = 0.6 + D [

0.387ReD 6 8 9 27 16

0.559 {1 + ( Pr ) } ]

Trial and error: ∆T = 99℉ HTR surface temp = 194 ℉

20.5 Solution Ts = 140℃ T∞ = 25℃

Tf = 82.5℃

Air @ 355 K: ν2 = 0.625x108 (m3 K)−1 Re = (0.625x108 )(0.035)3 (115) = 3.08x105 Horizontal Cylinder: 2 1

Nu = 0.6 + [

0.387ReD 6 8 9 27 0.559 16

{1 + ( Pr ) } ]

For Pr = 0.696 NuD = 10.47 q = hA∆T =

k (πDL)(∆T) = (0.0304)(π)(0.8)(115)(10.47) = 92 W D

Remainder of input goes to electrical & conduction losses & to illumination

20.9 Solution k

For Tc to reach 320 K – use values calculated in problem 20.8 α = ρC = 2.1x10−7 m2 ⁄s p

D, cm

αt⁄ x1 2

7.5

615

0.16

1.19 hr

710

0.16

31.7 min

1.5

1180

0.16

2.86 min

Tsurf ≅ 295 K at all times

20.13 Solution Assuming each plate is independent 2

k k h = Nu = 0.825 + D D [

0.387ReD 6 8 9 27 0.492 16

{1 + ( Pr ) } ]

Tplate = 200℉ T∞ = 80℉ Tf = 140℉ Pr = 3.08 Re = (540x106 )(3)3 (120)(3.08) = 5.39x1012 h = 287 Btu⁄hr ft 2 ℉ q = hA∆T = (278)(30 ∗ 1 ∗ 3 ∗ 2)(120) = 6.19x106 Btu⁄hr = 1.81 kW

Chapter 21 Show/Hide Problems 21.3 Solution CL ∆T = 0.0709 hfg q 0.0709 3 Btu Btu ) = 190 2 = 680000 =( A 0.01235 s ft hr ft 2 1 2 Btu q = (680000)(π) ( ) = 178000 24 hr h=

680000 Btu = 10000 68 hr ft 2 ℉

21.5 Solution Using English units: A = πDL +

2πD2 π = π(0.02)(0.15) + (2)(0.02)2 = 0.1082 ft 2 4 4

q 500(3.413) Btu = = 15800 A 0.1082 hr ft 2 Assume nucleate boiling: 1 1 3 3.79x10−3 2

1∆T 15800 = 0.006 [ ( 1.7 (1.8) (970) (0.195x10−3 )(970)(3600) ∆T = 9.0℉ Surface temp = 221℉ h=

15800 Btu = 1760 9 hr ft 2 ℉

) ]

21.7 Solution

h = 0.62 [

1 4 3 (∆ρ)g k v ρv (hfg + 0.4Cp ∆T) v

Dμv (Ts − Tsat )

= 0.62 [

]

(0.0153)3 (0.0341 ∗

40 ) (58.4) 14.7

1 (12) (0.914x10−5 )(933) 1 4

∗ 32.3(3600)(934 + 0.4 ∗ 0.481 ∗ 933)] = 26.9

q 0.2 Btu = h∆T = 43.3π ( ) (1)(1960) = 25000 A 12 hr ft 2

Btu hr ft 2 ℉

Chapter 22 Show/Hide Problems

22.1 Solution q = ṁCp ∆T|H2O = UA∆TLM UA = constant =

q ∆TLM 1

1 ∆i r ∆i ) ( ) + (2πK ln ( r0 )) + ( hi A i 0 h0

≅ hi

hi Ai = constant Using dittus-boelter correlation 0.8

k k k DQ hi = (constant)Re0.8 Pr 0.4 = (Dv)0.8 = ( 2 ) D D D πD 4 hA = constant = A(constant)D−1.8 As diameter increases the required area increases as D−1.8

= (constant)D−1.8

22.2 Solution Oil: Tin = 400K Tout = 350K ṁ = 2

kg J q = 1880 s kgK

q = ṁCp ∆T = 2(1880)(50) = 188000 W ∆Tw =

q 188000 = = 22.5 K ṁCp 2(4187)

Tw in = 280K Tw out = 302.5 K TLM =

97.5 − 70 = 83K 97.5 ln ( 70 )

q 188000 = = 9.85 m2 U∆TLM 230(83)

22.3 Solution Dequiv =

4(0.1)(0.2) = 0.0667 m 2(0.1 + 0.2)

(22.4 continued) Tb avg = 295 K RE = 0.667

Tf = 345 K

ρv ṁ = 0.0667 μ Aμ

q = hA∆TLM TLM =

105 − 95 = 99.9 K 105 ln ( ) 95

Assuming turbulent flow: 2

ṁCp ∆T = ρvCp [0.023Re−0.2 Pr −3 ] As ∆TLM 2 ṁ 0.0667 ṁ −0.2 − 3 ] (2)(0.3)(2.5)(99.9) (0.698) ) ṁ(10) = [0.023 ( A 0.205x10−5 A kg

Solving for ṁ: ṁ = 105 s

q = ṁCp ∆T = 105(1009)(10) = 1060 kW

22.16 Solution NTU = 1.25

Cmin = 0 ε ≅ 0.72 Cmax

q = εCmin (Tw in − TA in ) = 0.72(0.07)(4.18)(93) = 19.59 kW = Cw ∆Tw = 4.18(0.07)∆Tw ∆Tw =

0.72(0.07)(4.18)(93) = 67 K 4.18(0.07)

Tw out = 280 + 67 = 347 K Steam condensation rate: ṁcond =

q 19.59 kg = = 8.68x10−3 hfg 2256 s

Chapter 23 Show/Hide Problems 23.1 Solutions Radiant emission from sun = As Ebs All passes through a spherical surface of radius, L At the earth: q πD2sEbs D 2 = = ( ) Ebs A 4πL2 2L Btu

Flux at earth= 360 + 90 = 450 hr ft2 2

8.6x105 450 = [ ] σTs4 2(93x106 ) Ts = 10530 R

23.3 Solution q q q Ts 4 T∞ 4 ) ) −( ) ] ( )| = ( )| − ( )| = 1000 − h(Ts − T∞ − ϵσ [( A net A in A out 100 100 = 1000 − 12(30 − 20) − 5.676(0.3)(106) = 862

W m2

23.8 Solution Filament at 2910 K q=100 W λmax =

2897.6 = 0.999 μm 2910

Visible range: 0.38 < λ < 0.76 λ1 T1 = 0.38(2910) = 1102 F0−λT = 0.0009 λ2 T2 = 0.74(2910) = 2204 F0−λT = 0.1017 Fraction in V.R. = 0.1008

23.14 Solution T=1500 K Peephole D=10 cm τ = 0.78 for 0<λ<3.2μm τ = 0.08 for 3.2<λ<∞ π q max = 5.676(15)4 ( ) (0.01)2 = 22.57 W 4 For: λ1 T1 = 0 F0−λT1 = 0 λ2 T2 = 4800 F0−λT2 = 0.6075 λ3 T3 = ∞ F0−λT3 = 1 Total heat loss = 22.57[0.78(0.6075) + 0.8(0.3925)] = 11.40 W

23.15 Solution q 1200 W W T 4 ) − (2.8)4 ] = = 490 2 = ϵσ [( 2 ) A 5(0.49 m m 100 T 4 ) − (2.8)4 ] 490 = 0.7(5.676) [( 100 T = 369 K

23.16 Solution q = 8 W through hole with D=0.0025 m2 Eb =

8 W = 3200 2 = σT 4 0.0025 m

T = 4.87 K

24.5 Given: P = 6.1 × 10−3 bar, T = 210 K yCO2 = 0.9532, yN2 = 0.027, yAr = 0.016, yO2 = 0.0013, yCO = 0.0008

a. Partial pressure 𝑝𝐴 for CO2 A = CO2

 105 Pa  p A = y A P = ( 0.9532 ) 6.110−3 bar   = 581 Pa  1 bar  b. Molar concentration 𝑐𝐴 for CO2

(

cA =

)

pA 581 Pa gmole = = 0.333 3 RT m3 8.314 m  Pa/gmole  K ( 210 K )

(

)

c. Total molar concentration c for the Martian atmosphere

P 610 Pa gmole = = 0.349 3 RT m3 8.314 m  Pa/gmole  K ( 210 K )

(

)

d. Total mass concentration ρ for the Martian atmosphere

M avg = yCO2 M CO2 + yN2 M N2 + yAr M Ar + yO2 M O2 + yCO M CO    g  g  g  M avg = ( 0.9532 )  44.01  + ( 0.027 )  28.00  + ( 0.0163)  9.95 + gmole  gmole  gmole       g  g  ( 0.0013)  32.00  + ( 0.0008 )  28.01  gmole  gmole    M avg = 43.41

g gmole



 = M avg  c =  43.41 

g  gmole  g  0.349  =15.2 3 3 gmole   m  m

24-1

24.12 A = SiH4, B = He yA = 0.01, yB = 0.99 dpore = 10 × 10−4 cm, T = 900 K g

MA = 32.12 gmole, MB = 4.00 gmole κ = 1.38 × 10−16 ergs/K a. Determine wA

g 0.01 ∙ 32.12 yA ∙ MA gmole wA = = = 0.750 yA ∙ MA + yB ∙ MB 0.01 ∙ 32.12 g + 0.99 ∙ 4.00 g gmole gmole

b. Estimate DAB at P = 1.0 atm and P = 100 Pa, T = 900 K P = 1.0 atm εA κ

= 207.6 K,

εB κ

= 10.22 K ,

εAB κ

εB

=√ A∙

= 46.06 K,

κT εAB

= 19.54

Using Appendix K. 1, interpolate to get ΩD = 0.6676 σA = 4.08 Å,

σB = 2.576 Å, σAB =

σA +σB 2

= 3.328 Å 1 2

3 1 1 1 0.001858 ∙ 900 K 2 ∙ ( + 3 2 g g ) 1 1 32.12 4.00 0.001858 ∙ T 2 ∙ (M + M ) gmole gmole A B DAB = = 2 P ∙ σAB ΩD 1.0 atm ∙ 3.328 Å2 ∙ 0.6676

= 3.60

cm2 s

𝑃 = 100 Pa ∙

1 atm = 9.87 × 10−4 atm 101325 Pa

24-2

1 2

3 1 1 1 0.001858 ∙ 900 K 2 ∙ ( + 3 2 g g ) 1 1 32.12 4.00 0.001858 ∙ T 2 ∙ (M + M ) gmole gmole A B DAB = = 2 P ∙ σAB ΩD 9.87 × 10−4 atm ∙ 3.328 Å2 ∙ 0.6676

= 3645

cm2 s

c. Assess importance of Knudsen Diffusion Pressure = 1.0 atm λ=

κT √2πσA

Kn = d

pore

J 1.38 × 10−23 K ∙ 900 K √2π ∙ (4.08 × 10−10 m)2 101325 Pa

= 1.66 × 10−7 cm

6.76×10−7 cm

= 10×10−4 cm = 1.66 × 10−4 < 1, Knudsen diffusion is not important.

Pressure = 100 Pa λ=

κT √2πσA 2 P

Kn =

λ dpore

J 1.38 × 10−23 K ∙ 900 K √2π ∙ (4.08 × 10−10 m)2 100 Pa

1.68 × 10−4 cm = 0.168, 10 × 10−4 cm

= 1.68 × 10−4 cm

Knudsen diffusion plays moderate role

d. Gas velocity for Pe = 5.0 x 10-4

DKA = 4850 ∙ dpore ∙ √

T 900 K cm2 = 4850 ∙ 10 × 10−4 cm ∙ √ = 25.67 g MA s 32.13 gmole

P = 1.0 atm 1 1 1 1 1 )=( ) = ( + + 2 cm cm2 DAe DAB DKA 3.60 s 25.67 s DAe = 3.157

cm2 s 24-3

v∞∙dpore

= 5 × 10−4 DAe cm v∞ = 1.58 s πd2 cm cm3 V̇ = v∞ = 1.58 ∙ π ∙ (5 × 10−4 cm)2 = 1.24 × 10−6 4 s s Pe =

P = 100 Pa 1 DAe

= (D

Pe =

+ D ) , ∴ DAe = 25.49 cm2 /s

v∞∙dpore DAe

= 5 × 10−4

∴ v∞ = 12.75 V̇ = 12.75

cm s

cm π cm3 −4 2 −5 ∙ ∙ (10 × 10 cm) = 1.00 × 10 s 4 s

24-4

24.17 A = CuCl2 (an ionic solute), B = H2O T = 298 K, ℱ = 96,500 C/gmole R = 8.316

J gmole ∙ K

λ°+ (Cu2+ ) = 108 λ°− (Cl− ) = 76.3

A ∙ cm2 V ∙ gmole

n+ (valence of cation) = 2 n− (valence of anion) = 1 Nernst-Haskell Equation 1 1 ( + + n− )RT n DAB = 1 1 ( ° + ° )(ℱ)2 λ+ λ− 1 1 (2 + 1)(8.316

J ) ∙ 298K cm2 gmole ∙ K −5 DAB = = 1.78 × 10 1 1 s 2 ( + )(96,500 C/gmole) A ∙ cm2 A ∙ cm2 108 76.3 V ∙ gmole V ∙ gmole

24-5

24.23

a. CA

C A = y AC = y A C A = ( 0.05 )

P RT

(1.5 atm )  -5 m  atm   8.206×10  ( 353K ) gmole  K   3

=2.58

gmole A m3

' b. DAe at T = 80 oC (353 K) and 1.5 atm

P  cm2  1.0 atm  cm 2 DAB (T , P) = DAB (T , Pref )  ref  = 0.10 = 0.0.067   s  1.5 atm  s  P  DKA = 4850 d pore

T 353 cm 2 −5 = 4850(1.5  10 ) = 0.202 MA 46 s

1 1 1 = + = DAe DKA DAB

1 0.202

cm s

1 0.067

cm 2 s

cm 2 DAe = 0.050 s  cm 2  cm 2 ' DAe =  2 DAe = (0.5) 2  0.05 = 0.0125  s  s 

24-6

24-7

25.7 𝑘1

𝐴 → 𝐵, 𝑅𝐴 = −𝑘1 𝑐𝐴 , 𝑘𝑠

𝐴 → 2𝐶, 𝑟𝐴,𝑠 = −𝑘𝑠 𝑐𝐴𝑠 ,

𝑘1 [

1 ] 𝑠

𝑘𝑠 [

𝑐𝑚 ] 𝑠

4 species: A=1, B=2, C=3, D=4 (inert) ( 𝑅1 = −𝑘1 𝑐1, 𝑟1,𝑠 = −𝑘𝑠 𝑐1𝑠 ) a. Assumptions, Source and Sink for species 1 (A) Assumptions: 1) Dilute process with respect to species 1 2) Constant source and sink for species 1 (steady state) 3) Right side and top side are impermeable barriers 4) 2-D flux of species 1 for catalyst II (source and sink antiparallel) Source: Flowing fluid containing reactant 1, of constant concentration (𝑐1,∞) Sinks: First-order homogeneous reaction of 1 in porous layer (catalyst I) First-order heterogeneous surface reaction of 1 at nonporous boundary surface (catalyst II) b. Differential model for C1(x,y) (Shell balance on differential volume element for species 1) IN – OUT + GEN = ACC = 0

𝑛1,𝑥 = 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑥 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑤 = 𝑚𝑎𝑠𝑠 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑝𝑒𝑐𝑖𝑒𝑠 1 𝑖𝑛 𝑚𝑖𝑥𝑡𝑢𝑟𝑒 [𝑛1,𝑥 ∆𝑦𝑤|𝑥,𝑦̅ + 𝑛1,𝑦 ∆𝑥𝑤|𝑥̅ ,𝑦 ] − [𝑛1,𝑥 ∆𝑦𝑤|𝑥+∆𝑥,𝑦̅ + 𝑛1,𝑦 ∆𝑥𝑤|𝑥̅ ,𝑦+∆𝑦 ] + 𝑟1 ∆𝑥∆𝑦𝑤 = 0 ÷ ∆𝑥, ∆𝑦; 𝑡𝑎𝑘𝑒 lim ∆𝑥 → 0, 𝑡𝑎𝑘𝑒 lim ∆𝑦 → 0 ; ÷ 𝑤 25-1

−

𝜕𝑛1𝑦 𝜕𝑛1,𝑥 +− + 𝑟1 = 0 𝜕𝑥 𝜕𝑥

General Flux Equation 𝜕𝐶

x-direction: 𝑛1,𝑥 = −𝐷1 𝜕𝑥1 (𝑜𝑡ℎ𝑒𝑟 𝑡𝑒𝑟𝑚𝑠 ≈ 0, 𝑑𝑖𝑙𝑢𝑡𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛) 𝜕𝐶

y-direction: 𝑛1,𝑦 = −𝐷1 𝜕𝑦1 (𝑜𝑡ℎ𝑒𝑟 𝑡𝑒𝑟𝑚𝑠 ≈ 0, 𝑑𝑖𝑙𝑢𝑡𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛) Homogeneous reaction: 𝑟1 = −𝑘1 𝑐1 𝜕2𝐶

𝜕2𝐶

Combine: 𝐷1 [ 𝜕𝑥 21 + 𝜕𝑦 21] − 𝑘1 𝑐1 = 0, 𝑐1 (𝑥, 𝑦) c. Boundary Conditions, 4 required: (2 for x, 2 for y) 𝑦 = 0, 0 ≤ 𝑥 ≤ 𝐿,

𝑐1 (𝑥, 0) = 𝑐1,∞

𝑦 = 𝐻, 0 ≤ 𝑥 ≤ 𝐿,

𝑛1 (𝑥, 𝐻) = 0,

𝑥 = 0, 0 < 𝑦 < 𝐻,

𝑟1,𝑠 = 𝑛1,𝑠 ,

𝑥 = 𝐿, 0 < 𝑦 < 𝐻,

𝑛1 (𝐿, 𝑦) = 0

∴

𝜕𝑐1 (𝑥, 𝐻) =0 𝜕𝑥

−𝑘𝑠 𝑐1,𝑠 = −𝐷1 ∴

𝜕𝑐1 (𝐿,𝑦) 𝜕𝑦

𝜕𝑐1 (0, 𝑦) , 𝜕𝑥

∴

𝜕𝑐1 (0, 𝑦) 𝑘𝑠 = 𝑐 (0, 𝑦) 𝜕𝑥 𝐷1 1

25-2

25.10 A = C6H6, B = H2O (liq) T = 298 K, 𝑐𝐴,∞ 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 a. Differential model for cA(r,t) Assumptions: 1) Unsteady state (control volume is also the sink) 2) No homogeneous reaction (𝑅𝐴 = 0) 3) Dilute with respect to A 4) 1-D flux along r (unimolecular diffusion) General Differential Equation ′ 𝑁𝐴,𝑟 ≅ −𝐷𝐴𝑒

𝜕𝑐𝐴 , 𝜕𝑟

(𝑑𝑖𝑙𝑢𝑡𝑒 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛)

Mass Conservation Equation −∇𝑁𝐴 =

𝜕𝑐𝐴 , (𝑅𝐴 = 0), 𝜕𝑡

∴ ∇𝑁𝐴 =

1 𝜕 2 (𝑟 𝑁𝐴,𝑟 ), 𝑟 2 𝜕𝑟

𝑐𝐴 (𝑟, 𝑡)

Combine Equations 1 ′ 𝜕 2 𝜕𝑐𝐴 𝜕𝑐𝐴 )= 𝐷𝐴𝑒 (𝑟 , 2 𝑟 𝜕𝑟 𝜕𝑟 𝜕𝑡

𝜕 2 𝑐𝐴 2 𝜕𝑐𝐴 𝜕𝑐𝐴 ′ 𝑜𝑟 𝐷𝐴𝑒 [ 2 + ]= 𝜕𝑟 𝑟 𝜕𝑟 𝜕𝑡

b. Boundary and Initial Conditions IC: 𝑡 = 0,

𝑐𝐴 (𝑟, 0) = 𝑐𝐴0 = 0

BC: 𝑟 = 0,

𝜕𝑐𝐴 (0, 𝑡) =0 𝜕𝑟

𝑟 = 𝑅, 𝑐𝐴 (𝑅, 𝑡) = 𝑐𝐴𝑠

25-3

26.9 A = H2O vapor, B = O2 gas 𝑇 = 310 𝐾, 𝑃 = 1.2 𝑎𝑡𝑚, 𝑑𝑝𝑜𝑟𝑒 = 50 × 10−7 𝑐𝑚 𝜀 = 0.40, 𝑃𝐴 = 0.0618 𝑎𝑡𝑚, 𝑦𝐴∗ = 0.0515, 𝐻 = 800

𝐿 ∙ 𝑎𝑡𝑚 𝑔𝑚𝑜𝑙𝑒

a. Determine effective diffusion coefficient, DAe Fuller-Schettler-Giddings Correlation for DAB 1 1 1/2 1 1 1/2 0.001𝑇 1.75 [𝑀 + 𝑀 ] 0.001 ∙ 310 1.75 [18.02 + 32.0] 𝑐𝑚2 𝐴 𝐴 𝐷𝐴𝐵 = = = 0.236 1 1 1/3 1/3 2 𝑠 2 3 3 𝑃[𝑉𝐴 + 𝑉𝐵 ] 1.2 ∙ [12.7 + 16.6 ] Knudsen Diffusion Coefficient 𝐷𝐾𝐴 = 4850𝑑𝑝𝑜𝑟𝑒 √

𝑇 310 𝑐𝑚2 = 4850 ∙ 50 × 10−7 √ = 0.1006 𝑀𝐴 18.02 𝑠

Effective Diffusion Coefficient 1 1 1 1 1 𝑐𝑚2 = + = + ∴ 𝐷 = 0.0705 𝐴𝑒 𝑐𝑚2 𝑐𝑚2 𝐷𝐴𝑒 𝐷𝐴𝐵 𝐷𝐾𝐴 𝑠 0.236 0.1006 𝑠 𝑠 𝑐𝑚2 𝑐𝑚2 ′ 𝐷𝐴𝑒 = 𝜀 2 𝐷𝐴𝑒 = 0.42 ∙ 0.0705 = 0.0113 𝑠 𝑠 b. Determine L Assume: 1) Steady state, 2) no homogenous reaction, 3) 1-D flux along z, 4) constant T and P 𝑐=

𝑃 = 𝑅𝑇

1.2 𝑎𝑡𝑚 𝑔𝑚𝑜𝑙𝑒 = 4.72 × 10−5 3 𝑐𝑚 ∙ 𝑎𝑡𝑚 𝑐𝑚3 82.06 ∙ 310𝐾 𝑔𝑚𝑜𝑙𝑒 ∙ 𝐾

d ( N A,z ) = 0 , constant flux along z dz 𝑑𝑐𝐴 𝑐𝐴 𝑑𝑦𝐴 𝑁𝐴 = −𝐷𝐴𝐵 + (𝑁𝐴,𝑧 ) = −𝐷𝐴𝐵 𝑐 + 𝑦𝐴 (𝑁𝐴,𝑧 ) 𝑑𝑧 𝑐 𝑑𝑧 𝑁𝐴 = −

𝐷𝐴𝐵 𝑐 𝑑𝑦𝐴 1 − 𝑦𝐴 𝑑𝑧

Boundary Conditions: 𝑧 = 0, 𝑦𝐴𝑠 = 𝑦𝐴∗ 26-1

𝑧 = 𝐿, 𝑦𝐴∞ = 0.2𝑦𝐴∗ 𝐿

𝑦𝐴∞

𝑁𝐴 ∫ 𝑑𝑧 = −𝑐𝐷𝐴𝐵 ∫ 0

𝑁𝐴 =

𝐿=

𝑦𝐴𝑠

𝑑𝑦𝐴 1 − 𝑦𝐴

𝑐𝐷𝐴𝐵 1 − 𝑦𝐴∞ ) 𝑙𝑛 ( 𝐿 1 − 𝑦𝐴𝑠

𝑐𝐷𝐴𝐵 1 − 𝑦𝐴∞ ) 𝑙𝑛 ( 𝑁𝐴 1 − 𝑦𝐴𝑠

𝑔𝑚𝑜𝑙𝑒 𝑐𝑚2 0.0113 𝑠 ∙ 4.72 × 10−5 𝑐𝑚3 ln [1 − 0.2 ∙ 0.0515] = 0.20 𝑐𝑚 𝐿= 𝑚𝑜𝑙 1 − 0.0515 1.15 × 10−7 𝑐𝑚2 ∙ 𝑠 ∗ c. Determine 𝑐𝐵𝐿 ∗ 𝑐𝐵𝐿 =

𝑝𝐵 𝑃 − 𝑝𝐴 (1.2 − 0.062) 𝑎𝑡𝑚 𝑔𝑚𝑜𝑙𝑒 = = = 1.42 × 10−3 𝐿 ∙ 𝑎𝑡𝑚 𝐻 𝐻 𝐿 800 𝑔𝑚𝑜𝑙𝑒

26-2

26.17 Given: A = O2, B = air, C = carbon (s), D = CO2 T = 1000 K, P = 2.0 atm Electrode dimensions: L = 25 cm, d = 2.0 cm,  = 0.5 cm boundary layer Electrode materials: C,solid = 2.25 g/cm3, MC = 12 g/gmole a. Determine WA Model for WA Physical System: gas space in boundary layer between carbon electrode surface and bulk gas (System I) Source for A: bulk gas Sink for A: consumption by carbon electrode to CO2 gas Assumptions 1. 2. 3. 4.

CO2

solid carbon electrode bulk gas cA∞

O2 (A)

25 cm

NA,r

r = R+ = 1.5 cm r = R = 1.0 cm cAs = 0

26-3

General Differential Equation for Mass Transfer (cylindrical system)

RA = 0,

c A = 0 by assumptions 1 and 2 t

c 1 d rN A,r ) + RA = A ( r dr t 1 d  ( r  N A, r ) = 0 r dr d ( r  N A, r ) = 0 dr −

By continuity N A,r r = R  R = N A,r  r NA,r r constant along r Note WA = N A,r r = R  2 RL = N A,r  2 rL Flux Equation

dc A c A dc c + ( N A,r + N B ,r + NC ,r + N D,r ) = − DAB A + A ( N A,r + 0 + 0 + − N A,r ) dr C dr C dc N A,r = − DAB A dr dc   Note: WA = N A,r r = R  2 RL = N A,r  2 rL = 2 rL  − DAB A  dr   WA is not a function of r; separate variables cA, r, pull WA outside of the integral, then integrate with respect to limits shown below N A,r = − DAB

r = R, cA = cAs = 0 (rapid consumption of O2 at surface) r = R +  , cA = cA∞ (bulk gas composition beyond boundary layer) R

dr WA  = −2 LDAB  dc A r R + c A

WA =

2 LDAB c A  R +  ln    R 

Calculate WA

26-4

c A =

( 0.21)( 2.0 atm ) y A P gmole = =4.65  10−6 3 RT  cm3 cm atm  82.06 1100 K ( )   gmole  K  

 P  T  DAB (T , P) = DAB (Tref , Pref )  ref     P   Tref 

3/2

= 0.136

cm 2  1.0 atm   1100 K  cm 2 = 0.55    s  2.0 atm   273 K  s

 cm 2   −6 gmole  2π ( 25 cm )  0.55   4.65  10  s  cm3  gmole  WA = = 1.80  10−3 s  2.5 cm  ln    2.0 cm 

b. Time it takes for the rod to completely disappear (t at R = 0) Material balance on carbon (PSS mass transfer) to model “shrinkage” of carbon electrode (System II) IN – OUT + GEN = ACC

0 − N A, r 

mC =

dm 1.0 mol C S +0 = C 1.0 mol O 2 dt

C  R 2 L

MC  DAB c A dR − 2 RL = C 2 RL MC dt  R +  R  ln    R  DAB c A  dR − = A R  R +   M A dt ln    R  Separate variables R, t and integrate, then solve for time t at R = 0 t

− DAB c A dt = 0

 R + 

C 0  R +  

ln   RdR M C R  R  1 

 R + 

 1

 R +   

ln  R  RdR = 2 R  R  ln  R  +   − 2  ln   0

26-5

   R +   R +  2 R  R  ln   +   −  ln    R     C   t= 2 DAB c A MC

   2  2.5 cm   2.5 cm    g   2.0 cm  ( 2.0 cm ) ln   +0.5 cm  − ( 0.5 cm ) ln     2.25 3   2.0 cm   0.5 cm    cm    t= = 5455 s 2   g  cm   −6 gmole  2  0.55   4.65  10 12.01 gmole   s  cm3    

26-6

26.19 A = acetone, B = air 𝑑 = 0.3 𝑐𝑚, 𝑇 = 293.9 𝐾 𝑃𝐴∗ = 0.254 𝑎𝑡𝑚, 𝑀𝐴 = 58

𝑔 𝑔 , 𝜌𝐴,𝑙𝑖𝑞 = 0.79 3 𝑔𝑚𝑜𝑙𝑒 𝑐𝑚

a. Determine DAB from Arnold Diffusion Cell data Assume: 1) 1-D mass transfer along z, 2) No homogenous reaction, 3) PSS. Material balance on acetone (PSS mass transfer) IN – OUT + GEN = ACC 𝑑𝑚𝐴 0 − 𝑁𝐴,𝑧 𝑆 + 0 = 𝑑𝑡 𝑁𝐴 =

𝐷𝐴𝐵 𝑐 1 − 𝑦𝐴2 ln [ ] 𝑍 1 − 𝑦𝐴1

𝜌𝐴,𝑙𝑖𝑞 𝑑𝑍 𝐷𝐴𝐵 𝑐 1 − 𝑦𝐴2 ln [ ]= ∙ 𝑍 1 − 𝑦𝐴1 𝑀𝐴 𝑑𝑡 𝑡

𝑍

𝑍𝑜

𝐷𝐴𝐵 𝑐𝐴𝑠 𝑀𝐴 ∫ 𝑑𝑡 = ∫ 𝑍𝑑𝑍 𝜌𝐴,𝑙𝑖𝑞

Z2-Zo2 [cm2]

𝑍 2 − 𝑍0 2 =

2𝐷𝐴𝐵 𝑐𝑀𝐴 1 − 𝑦𝐴∞ ln [ ] (𝑡 − 𝑡𝑜 ) 𝜌𝐴,𝑙𝑖𝑞 1 − 𝑦𝐴𝑠

160 140 120 100 80 60 40 20 0

y = 0.6472x + 0.5537

100

150

200

250

time [hr] 26-7

Slope = 0.6472 𝑐=

𝑃 = 𝑅𝑇

𝑐𝑚2 𝑐𝑚2 = 1.80 × 10−4 ℎ𝑟 𝑠

1.0 𝑎𝑡𝑚 𝑔𝑚𝑜𝑙𝑒 = 4.15 × 10−5 3 𝑐𝑚 ∙ 𝑎𝑡𝑚 𝑐𝑚3 82.06 ∙ 293.9 𝐾 𝐾 ∙ 𝑔𝑚𝑜𝑙𝑒

𝑃𝐴∗ 0.254 𝑎𝑡𝑚 𝑦𝐴𝑠 = = = 0.254 𝑃 1.0 𝑎𝑡𝑚 𝐷𝐴𝐵 =

𝜌𝐴 ∙ 𝑠𝑙𝑜𝑝𝑒 1−0 2 ∙ 𝑐 ∙ 𝑀𝐴 ln [1 − 𝑦 ] 𝐴𝑠

𝑔 𝑐𝑚2 ∙ 1.80 × 10−4 𝑠 3 𝑐𝑚3 𝑐𝑚 𝐷𝐴𝐵 = = 0.101 𝑔𝑚𝑜𝑙𝑒 𝑔 1−0 𝑠 2 ∙ 4.15 × 10−5 ∙ 58 ∙ ln [ ] 𝑔𝑚𝑜𝑙𝑒 1 − 0.254 𝑐𝑚3 0.79

b. Determine DAB by correlation Hirschfelder Correlation

𝐷𝐴𝐵 =

1 1 2 0.001858 ∙ 𝑇 3/2 [𝑀 + 𝑀 ] 𝐴

𝐴

2 𝑃𝜎𝐴𝐵 ΩD 508 𝐾

1 3

𝜎𝐴 = 2.44 (47.4 𝑎𝑡𝑚) = 5.380 Å, 𝜎𝐵 = 3.617 Å,, 𝜎𝐴𝐵 = 𝜀𝐴 𝜅

5.380+3.617 2

= 4.50 Å

𝜀

= 0.77(508 𝐾) = 391.16 𝐾, 𝜅𝐵 = 97 𝐾,𝜀𝐴𝐵 = √391.16 ∙ 97 𝐾 = 194.79 𝐾

𝜅𝑇 293.9 𝐾 = = 1.509, 𝜀𝐴𝐵 194.79 𝐾

𝐷𝐴𝐵 =

∴ Ω𝐷 = 1.195

1 1 1/2 + ] 𝑐𝑚2 58.0 29.0 = 0.088 1.0 ∙ 4.50 2 ∙ 1.195 𝑠

0.001858 ∙ 293.93/2 [

26-8

26.23

nonporous support

well-mixed liquid phase

enzyme layer

cA∞= 1.0 mmol/m3

z=0

z = L = 0.2 cm cAL = 0.25 mmol/m3

a. Boundary Conditions z = 0, CA = CA z = L,

dC A =0 dz

b. Rate constant k at z = L

 k  cosh  ( L − z )  DAe  c A  c AL = c Ao = cosh( ) cosh( ) c   c AL 

 = cosh −1  A  =L

k DAe

2  -5 cm  2.0×10  3   s   D  −1  c A  -1  1.0 mmole/m k = Ae cosh = cosh     2 2  3  L  ( 0.2 cm )   0.25 mmole/m    c AL  

k = 2.13  10−3 s −1 c. Total transfer rate of product B, WB

26-9

=L

k = 0.2 cm DAe

2.13×10-3s -1 = 2.063 2 -5 cm 2.0×10 s

NA =

DAec Ao

DAe y AoC



 tanh  =



 tanh 

2  mmole  1m  −5 cm   2.0  10  1.0   s  m3  100 cm  NA =  (2.063)tanh(2.063) ( 0.2 cm )

mmole A m2  s 1 mol B  1 mol B  -6 mmole A  2 -6 mmole A WB = NA S =   2.0  10  ( 2.0 m ) = 4.0  10 2 1 mol A m s  s  1 mol A  N A =2.0  10-6

d. At  = 2.063, process is between diffusion control and reaction control

26-10

26.27 A = CO, B = CO2 a. Determine total molar flow rate out n2 Material balance on reactor for CO IN – OUT + GEN = ACC

y A,1n1 − y A,2 n2 + WA = 0

 d 2 DAe cA0 WA = S  N A =  tan  4  k  = = 1.0 cm DAe

6.0 s -1 = 3.87 cm 2 0.4 s

yCO ,2 + yO2 ,2 + yCO2 ,2 = 1  yCO2 ,2 = 1 − yCO,2 − yO2 ,2 = 1.0 − 0.05 − 0.025 = 0.925 = y A0 P 1.0 atm gmole = 0.925 = 1.051 x 10-5 3 RT cm3  cm  atm  82.06 1073 K ( )   gmole  K    π (10.0 cm )2   0.40 cm 2 /s   gmole  WA =  1.051 x 10-5 3.87 ) tanh ( 3.87 )   3 (    1.0 cm   4 cm    gmole WA = 1.277 x 10-3 s gmole CO   (0.0)n1 − 0.05n2 + 1.277 x 10-3 =0 s   gmole CO  60 s  gmole CO n2 = 0.0255   = 1.53 s min  1.0 min  c A0 = y A0C = y A0

b. Mole fraction y A on backside of catalyst layer at z = 

yA ( z) =

(

y A0 cosh ( − z ) k / DAe

(

cosh  k / DAe

)

y A0 0.925 = = 0.0386 cosh ( ) cosh ( 3.87 ) 26-11

27.5 Given: A = Arsenic (As), B = solid silicon (Si) T = 1050 C Wafer dimensions: L = 1.0 mm, d = 10 cm Solute A: cAs = cA* = 2.3 × 1021 atoms As/cm3 at z = 0 cAo = 2.3 × 1017 atoms As/cm3 initial As concentration in solid Si DAB = 5.0 × 10−13 cm2/s a. Time required and mA(t) to achieve CA(z,t) = 2.065 × 1020 atoms As/cm3 at z = 0.5 m USS diffusion in semi-infinite medium—concentration profile atoms atoms − 2.065  1020 3 cm cm3 = erf ( ) = atoms atoms c As − c Ao 2.3  1021 − 2.3  1017 3 cm cm3 erf ( ) = 0.9103 2.3  1021

c As − c A ( z , t )

=

z 2 DAB t

From Appendix L,  = 1.2 z2 t= 2 = 4 DAB

−5

4 (1.2 )

2  −13 cm  5.0  10   s  

= 868 s

( 5.0  10 cm /s ) (868 s ) = 2.08  10 cm < < L −13

DAB  t =

Note

( 5.0 10 cm )

−5

d2 4

USS diffusion in semi-infinite medium—total mass loaded in medium mA ( t ) − mAo =

d2 4

4 DAB t



( cAs − cAo )

27-1

2  −13 cm  4 5.0  10 2   (868 s ) π (10 cm ) s  atoms  mA ( t ) − mAo = 2.3  1021 − 2.3  1017 4  cm3 mA ( t ) − mAo = 4.25  1018 atoms As

(

d2

(

)

 (10cm )

)

L = 2.3  10 atoms As/cm ( 0.1cm ) = 1.8 1016 atoms As 4 4 mA ( t ) = 1.8  1016 atoms As + 4.25  1018 atoms As = 4.25  1018 atoms As

mAo = c Ao

b. Plot of z and vs. t at  = 1.2 1.20

1.00

0.80

z (m)

z1/2 (m)1/2

1.00

0.60 0.40

0.80

0.60

0.20

0.40

0.00 0

1000

2000 3000 Time, t (s)

4000

t1/2 (s)1/2

Note plot of z vs. t1/2 is linear c. Transfer rate WA at t = 5.0 min, 10.0 min  d 2 DAB WA = S  N A z =0 = ( c − c ) for t > 0 4  t As Ao At t = 5.0 min = 300 s WA =

 (10 cm )

( 5.0  10 cm /s ) 2.3  10 ( π ( 300 s ) −13

− 2.3  1017

) atoms cm

− 2.3  1017

) atoms cm

WA = 4.16  1015 atoms As / s

At t = 10.0 min = 600 s WA =

 (10 cm ) 4

( 5.0  10 cm /s ) 2.3  10 ( π ( 600 s ) −13

WA = 2.94 1015 atoms As / s

Note as t increases, WA decreases. At t = 0, the flux is not defined because no concentration gradient for As in the solid Si initially exists. 27-2

27.16 A = solvent, B = polymer x1 = L = 6.0 mm = 0.60 cm 𝐶𝐴∞ = 0 wA0 = 0.010 DAB = 2.0 x 10-6 cm2/sec Determine time (t) required for wA(x,t) = 0.00035 at 1.2 mm (0.12 cm) from the exposed surface USS Diffusion in a slab, use Concentration -Time charts 𝑚 = 0 (𝑛𝑜 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒) 𝑌=

𝐶𝐴∞ − 𝐶𝐴 (0, 𝑡) 𝐶𝐴𝑠 − 𝐶𝐴 (𝑥, 𝑡) 𝑤𝐴𝑆 − 𝑤𝐴 (𝑥, 𝑡) 0 − 0.00035 = = = = 0.035 𝐶𝐴∞ − 𝐶𝐴0 𝐶𝐴𝑠 − 𝐶𝐴𝑜 𝑤𝐴𝑆 − 𝑤𝐴0 0 − 0.010

𝑛=

𝑥 0.60 𝑐𝑚 − 0.12 𝑐𝑚 = = 0.80 𝑥1 0.60 𝑐𝑚

Figure F.1 𝑋𝐷 ≅ 1.0 = 𝑡 = 1.0

𝐷𝐴𝐵 𝑡 𝑥12

(0.60 𝑐𝑚)2 𝑥12 = 1.0 = 180,000 sec (50 hr) 𝐷𝐴𝐵 2.0 𝑥 10−6 𝑐𝑚2 /𝑠𝑒𝑐

27-3

28.1 Given: T = 300 K, P = 1.0 atm   Sc = =  DAB DAB  P  T  DAB ,T2 , P 2 = DAB ,T1 , P 1  1  2   P2  T1 

3/2



 D |T1  D |T2

Neglect temperature dependency on  D  P  T  DAB ,T2 , P 2 = DAB ,T1 , P 1  1   2   P2   T1 

3/2

Appendix I:  H2O,liquid (300K ) = 8.788  10−7

m2 m2 ,  air (300 K) = 1.5689  10−5 s s

O2 gas (A) in Air (B) Appendix J: DAB (273 K,1.0 atm) = 0.175

cm 2 s

3/2

cm2  1 atm  300 K  cm2 DAB ,T2 , P 1 = 0.175 = 0.202    s  1 atm  273 K  s 2 m 1.5689  10−5 vB s = 0.786 Sc = = 2 m DAB 0.202  10−4 s O2 (A) dissolved in liquid H2O (B)  L ,H O (300 K) = 8.76  10−4 Pa s = 0.876 cP , VA = 25.6 2

cm3 gmole

Determine DAB using Hayduk and Laudie correlation DAB = 13.26  10 −5  L −1.14VA−0.589

(

)

DAB = 13.26×10 −5 ( 0.876 )

v Sc = B = DAB

8.788  10−3 2.28  10−5

−1.14

( 25.6 )

−0.589

= 2.28  10 −5

cm 2 s

cm s = 385 2

cm s

CO2 gas (A) in air (B) 28-1

Appendix J: DAB (273 K,1 atm) = 0.136 DAB ,T2 , P 1 = 0.136

cm2  1 atm  300 K     s  1 atm  273 K 

cm 2 s

3/2

= 0.157

cm2 s

CO2 (A) dissolved in liquid H2O (B) Determine DAB using Hayduk and Laudie correlation cm3 VCO2 = 34.0 gmole DAB = 13.26 10−5  L −1.14VA−0.589

(

DAB = 13.26  10

v Sc = B = DAB

−5

)( 0.876

8.788  10−7 1.93  10−9

−1.14

) ( 34.0 )

−0.589

=1.93  10

−5

cm 2 s

m s =455 2

m s

Schmidt numbers are higher in liquids relative to gases

28-2

28.4

gas distributor

Drying Chamber cA, ≈ 0

60 m3/min air H = 1.0 m 27°C, 1.0 atm (W = 1.5 m) L = 1.5 m

painted steel plate

heated surface (27°C) Given: A = solvent, B = air T = 300 K, P = 1.0 atm

g , cm3 Plate dimensions: L = 150 cm, H = 100 cm, W = 150 cm m3 1.0 3 3 v m m m s v = o = = 0.667 vo = 60 = 1.0 W  L 1.0 m 1.5 m s min s Paint coating:

= 0.10 cm , wA = 0.30 ,  A,liq = 1.5

Physical parameters: 2 cm2 g * −5 m p = 0.138 atm , , ,  air (300K) = 1.5689 10 DAB = 0.097 M A = 78 A s s gmole Physical System: boundary layer between surface and bulk gas Source for A: solvent coating surface Sink for A: bulk flowing gas a. ShL over entire plate Re and Sc: m  0.667  (1.5 m )  v L s  Re =  =  = 63, 771 laminar flow 2   −5 m  1.5689  10  s  

28-3

m2 1.5689  10 v s = 1.617 Sc = air = 2 m DAB 0.097  10−4 s −5

Laminar flow over a flat plate: 1/3 ShL = 0.664 Re1/2 = ( 0.664 )( 63, 771) L Sc

1/2

(1.617 )

1/3

= 197

b. Emissions rate WA Transfer rate across boundary layer

WA = N A S = kc (cAs − cA, )S cA,  0 in bulk air flow

S = L W = (150 cm)(150 cm) = (150 cm)2 cm 2 197  0.097 Sh D s = 0.127 cm kc = L AB = L 150 cm s

( 0.138 atm ) p*A mol cAs = = =5.61  10−6 3 3 RT  cm cm  atm   82.06  ( 300 K ) mol  K   cm  g  60 s  g 2  −6 mol  WA =  0.127 150 cm )  78  5.61  10   =75.0 3 ( s  cm  min   mol  min  c. Time required for complete drying of paint Material balance on solvent in paint layer IN – OUT + GEN = ACC

0 − WA + 0 = 0

mAo

dmA dt

 dmA =  WAdt

mAo WA 28-4

g   mA,0 = V    x A =  L2    wA = ( 0.1 cm ) (150 cm) 2 1.5 3  (0.30) = 1012.5 g  cm  m 1012.5g t = A,0 = = 13.5 min WA 75 g/min d. Boundary layer thickness at x = L = 1.5 m

( 5)(1.5 m ) = 2.97 cm 5 x = Re x 63771  2.97 cm  c = 1/3 = =2.53 cm 1/3 Sc (1.617 )

=

These are both much smaller than the height of the drying chamber. e. Volumetric flowrate of air needed for WA = 150 g A/min, all other variables unchanged 1/2

v  = =   ,new  WA,old kc ,old  v,old 

WA,new

kc ,new

1/2

v  =  o ,new  v   o,old 

W   m3  150 g/min  m3 vo ,new = vo ,old  A,new  =  60 = 240   W  min  A,old   min   75 g/min  2

28-5

29.3 Given: A = O2 (solute), B = H2O (solvent) T = 293 K, P = 2.0 atm Gas: y A = 0.21, yN2 = 0.78, yAr = 0.010 Solvent: B,liq = 1000 kg/m3, MB = 18 kg/kgmole Solute: H = 40,000 atm with p A,i = HxA,i a. Determine xA*

x*A =

p A y A P ( 0.21)( 2.0 atm ) = = = 1.05  10−5 H H ( 40,100 atm )

b. Determine cAL* c* AL = x*AcL

kg   1000 3  kgmole m  xA 2 = xA = 1.05  10−5  =5.83  10 −4 M H2O MB m3  kg  18 kgmol   

 H O,liq *

 B ,liq *

c. Determine cAL* if P = 4.0 atm

xA * =

p A y A P ( 0.21)( 4.0atm ) = = = 2.10  10−5 H H ( 40,100 atm )

kg   1000 3    H O,liq kgmole m  c* AL = x*AcL = x*A 2 = 2.10  10−5  = 1.17  10 −3 M H2O m3  kg  18.0 kgmole   

(

)

29-1

29.4 Given: A = ClO2 (solute), B = H2O (solvent) T = 293 K, P 1.5 atm Operating point: yA = 0.040, xA = 0.00040 Solute: MA = 67.5 kg/gmole Solvent: B,liq = 992.3 kg/m3, MB = 18 kg/kgmole Film Mass Transfer Coefficients:

k x =1.0 a.

kgmole kgmole (liquid film), kG = 0.010 2 (gas film) 2 m s m  s  atm

pA vs. CAL equilibrium line and operating point

Operating point: c AL = x AcL

 H O ,liq 2

M H2O

= xA

 B ,liq MB

( 992.3 kg/m ) = 0.022 kgmole = ( 0.00040 ) 3

(18 kg/kgmole )

p A = y A P = (0.06)(1.5 atm) = 0.090 atm 0.12

cAL , pA

0.10

pA (atm)

0.08

0.06

0.04

cAL,i , pA,i 0.02

pA* cAL*

0.00 0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

cAL (kgmole/m 3)

Gas Absorption process since the operating point is above the equilibrium line 29-2

b. Determine m Get H from slope of PA vs. CAL data, H = 0.772 atm∙m3/kgmole

(

)(

)

0.772 atm  m3 /kgmole 55.1 kgmole/m3 HcL = = 28.4 P (1.5 atm ) c. Determine kL  kg  18 kgmole  k MB  gmole  m   kL = x  kx = 1.0 = 1.81  10−5  2 CL  B ,liq  m s  s kg   1000 gmole   992.3 3   m   kgmole   ∗ d. Determine 𝑐𝐴𝐿 m=

c*AL =

( 0.06 atm ) pA kgmole = = 0.078 3 H m3 0.772 atm  m /kgmole

(

)

e. Compositions at the gas-liquid interface, pA,i and cAL,i Recall −

k L ( p A − p A ,i ) = kG ( c AL − c AL ,i )

For linear equilibrium line, pA,i = HcAL ,i   atm  m3 m  0.090 atm − 0.772 c AL ,i  − 1.81  10−5   kgmole ( pA − HcAL,i )   k s   − L = = kgmole kG    ( cAL − cAL,i ) 1.0  10−5 kgmole − c AL ,i    0.022 2 3 m  s  atm  m    kgmole m3  atm  m3   kgmole  p A,i = HcAL ,i =  0.772  0.0503  = 0.039 atm kgmole   m3  

c AL ,i = 0.0503

f. Determine Ky by two approaches Path 1: Convert kG to ky, get Ky from kx, ky, m

kgmole   −5 kgmole k y = kG P = 1.0  10−5 2  (1.5 atm ) =1.5  10 m  s  atm  m 2s 

29-3

-1

   1 m   1 28.4 −5 kgmole Ky =  +  =  +  = 1.05  10 kgmole kgmole m 2s  k y k x  1.5  10−5 2  1.0  10−3 2 m s  atm ms   −1

Path 2: KG from kL kG, H, then convert KG to Ky -1

 atm  m3  −1 0.772   1 H 1 kgmole kgmole  KG =  +  =  + = 7.10  10−6 2 m  s  atm 1.0  10−5 kgmole 1.81  10−5 m   kG k L  2  ms s   kgmole   −5 kgmole K y = K G P =  7.10  10−6 2  (1.5 atm) = 1.05  10 m  s  atm  m 2s  g. Flux NA Base on overall gas phase mole fraction driving force as an example  atm  m3   kgmole  p A * = Hc AL =  0.772  0.022  = 0.017 atm kgmole   m3   p * 0.017 atm yA* = A = = 0.0113 P 1.5 atm kgmole   N A = K y ( y A − y* A ) = 1.05  10−5  ( 0.040 − 0.0113) m 2s   kgmole N A = 3.0  10−5 m 2s

29-4

29.14

gmole gmole m3 c AL ,o = 0.5 , vo = 2.0 , c AL = 0.35 , L = 10.0 m, p A  0 , c*AL  0 3 3 m m s T = 293 K, H = 0.50

kc = 2.67 10−4

m3  atm m3  atm = 500 , P = 1.0 atm gmole kgmole

m m , k L = 5.5 10−3 s s

a. Determine K L m k gmole s kG = c = = 0.0111 2 3 m  atm RT m  s  atm (8.206 x 10−5 )(293 K) gmole  K 2.67 x 10−4

1 1 1 = + K L k L H kG −1

    1 1 m  = 0.00276 KL =  + 3 m  atm gmole  s  5.5 x 10−3 m (0.50 )(0.0111 2 )  s gmole m  s  atm  

cAL ,o vo − cAL vo − N A S = 0 cAL ,o vo − cAL vo − SK L (c AL − c*AL ) = 0 Determine S , the surface area between the two phases

29-5

m3 gmole gmole 2.0 (0.5 − 0.35 ) 3 vo (cAL ,o − cAL ) s m m3 S= = = 311 m 2 * m gmole gmole K L (cAL − cAL ) 0.00276 (0.35 −0 ) s m3 m3 c. Determine v Laminar flow over a flat plate:

B =

B = B

kg 2 m  s = 1.503 x 10−5 m kg s 1.206 3 m

1.813 x 10−5

B 

kc L  DAB   v =   L  0.664 DAB   B  

  

2   m2  cm 2 −4 m −4 m (2.67 x 10 )(10 m) (0.08 )(1 x 10 )   2  s  s s cm   2 2 cm 2 m cm  −4   1.503 x 10 −5   0.664(0.08 s )(1 x 10 cm 2 )  s  

v =

1.503 x 10

−5

10 m

   = 0.249 m s   

d. Determine pA,i

p A − p A ,i kL m3  atm − = = −0.495 kG c AL − c AL ,i gmole 0 atm − p A,i m3  atm = gmole 0.35 gmole − c AL ,i m3 m3  atm p A,i = HcAL ,i = 0.50 cAL ,i gmole

−0.495

Two equations, two unknowns: p A,i = 0.087 atm, cAL ,i = 0.174

gmole m3

29-6

29-7

30.1 Given: A = solvent, B = air T = 298 K, P = 1.0 atm Sphere: D = 1.0 cm, 0.12 g A/cm2 liquid solvent on sphere at t = 0 Physical parameters: MA = 78 g/gmole, pA* = 1.17 × 104 Pa (298 K), DAB = 0.0962 cm2/s, B = 0.156 cm2/s Physical System: boundary layer between surface of sphere and bulk gas Source for A: solvent coating surface of sphere Sink for A: bulk flowing gas a. Time to dry for still air Mass balance on solvent evaporating from paint-coated sphere Initial mass of solvent on sphere mAo =

 D 2 0.12 g cm 2

π(1.0 cm) 2  0.12 g  = 4.83 x 10-3 gmole  2  ( 78 g/gmole )  cm 

IN – OUT + GEN = ACCUMULATION (moles A/time)

0 − N AS + 0 = t

m Ao

dmA dt

−WA  dt =  dmA WA  t = mAo − mA

WA = kc ( c*A − c A )  D 2 Determine kc, WA, then t For still air, Sh = 2.0 D 0.0962 cm 2 /s kc = Sh AB = 2.0 =0.192 cm/s D 1.0 cm 30-1

1.17 x 104 Pa ) ( pA c = = = 4.72 gmole/m3 = 4.72x10-6 gmole/cm3 3 R T ( 8.314 m  Pa/gmole  K ) (298 K) * A

WA = D 2 kc (c*A − c A ) = π (1.0 cm ) ( 0.192 cm/s ) ( 4.72 x 10-6 - 0 ) gmole/cm3 = 2.85 x 10-6 gmole/s 2

Final mass of solvent on sphere − mA = 0 for dried paint coating on sphere

mAo 4.83 x 10-3 gmole = =1697 s WA 2.85 x 10-6 gmole/s

b. Time to dry for flowing air with v∞ = 1.0 m/s Determine kc, WA, then t For air at 298 K and 1.0 atm, air = 1.56 × 10−5 m2/s (Appendix I)

Re = Sc =

v D

 air  air DAB

(1.0 m/s )( 0.01m ) = 641 1.56 x 10-5 m 2 /s

0.156 cm2 /s =1.62 0.0962 cm 2 /s

Froessling equation for gas flow around a single sphere

Sh =

kc D = 2 + 0.552 Re1/2 Sc1/3 =2.0+0.552(641)1/2 (1.62)1/3 =18.4 DAB

kc = Sh

DAB 0.0962 cm/s = 18.4 = 1.77 cm/s D 1.0 cm

WA = D 2 kc (c*A − c A ) = π (1.0 cm ) (1.77 cm/s ) ( 4.72 x 10-6 - 0 ) gmole/cm3 = 2.63 x 10-5 gmole/s 2

mAo 4.83 x 10-3 gmole = = 184 s WA 2.63 x 10-5 gmole/s

30-2

30.5 Let A = O2, B = H2O (liquid) a. Material balance In − Out + Generation = Accumulation [ S  N A + v LWC Alo ] − [v LWC Al ] + 0 = 0 * [( N t DL)(k L (C AL − C AL )) + v LWC Alo ] − [v LWC Al ] = 0, combineterms to solve for C AL

C AL =

* (v LW )C Alo + ( N t DLk L )C AL (v LW ) + ( N t DLk L )

b. Mass transfer coefficient, kL ShD =

kL D = 0.281( Re ') 0.6 ( Sc) 0.44 DAB

v =

0.44

    DAB     D     DAB  ( 50 mol/s )( 0.018 kg/mol )

v D k L = 0.281     

0.6

nM w = = 1.80×10-3 m/s LW  (1 m )( 0.50 m ) ( 998.2 kg/m3 )

 (1.80×10-3 m/s)(0.02 m)  k L = 0.281  (0.995×10-6 m 2 /s)  

0.6

 (0.995×10-6 m 2 /s)    -9 2  (2×10 m /s) 

0.44

 (2×10-9 m 2 /s)  −6   = 3.72  10 m/s (0.02 m)  

c. Outlet concentration, CAL For CAL,o = 0 C AL =

* ( N t DLk L )C AL (10)π(0.02 m)(1 m)(3.72×10-6 m/s)(5 mol O 2 /m3 ) = (v LW ) + ( N t DLk L ) (1.80×10-3 m/s)(0.02 m)(1 m)+(10)π(0.02 m)(1 m)(3.72×10-6 m/s)

C AL = 0.0129 mol/m3

Increase CAL* by increasing pressure on the oxygen side to increase CAL.

30-3

30.13 Let A = CO2, B = H2O (liquid) a. Determine kL Gr =

db3  L g 

 L2



(2.0 x 10-3 m)3 (998.2 kg/m 3 )(9.81m/sec 2 )( ( 998.2-1.7967 ) kg/m3 ) (993 x 10-6 kg/m  sec) 2

= 79161

0.995 x10−6 m2 / sec = 562 DAB 1.77 x10−9 m2 / sec db = 2.0 mm Sc =

Sh =

k L db = 0.31Gr1/3 Sc1/3 DAB

D  k L = 0.31Gr1/3 Sc1/3  AB   db 

 1.77 x10−9 m2 / sec  −5 k L = (0.31)(79161) (562)   = 9.721x10 m / sec −3 2 x10 m   1/3

1/3

b. Determine if the inlet flow rate of CO2 gas is sufficient PA = 2.0 atm PA 2.0 atm = =0.06757 kmol/m3 3 H 29.6 atm  m /kmol 6g A * WA = N A i V = k L V ( C AL − C AL ,o ) = V db * C AL =

(9.721 x 10-5 m/sec)

6(0.05) 2.0 m 3 ) (0.06757 kmol /m 3 ) = 1.97 x 10-3 kmol CO 2 /sec ( -3 2.0 x 10 m

Convert WA in kmol/sec to volumetric flowrate of STD m3 CO2/ min VCO 2 = WA M A /  A = (1.97 x 10-3 kmol CO 2 /sec)(44kg/kmol)(1m3 /1.7967 kg)(60 s/min) = 2.89 m3 CO 2 /min

In – Out +Generation = Accumulation (liquid phase – mole A/time)

N A  Ai − C ALV0 + 0 = 0 −CALV0 + kL a V  (C*AL − C AL ) + 0 = 0 6(0.05) (0.06757 kmolCO 2 /m 3 ) -3 k L a  V  (C ) 2.0 x10 m C AL ,out = = (0.45m3 /min)(1min/60 sec) 6(0.05) V0 +(9.721 x 10-5 m/sec) + kL a 3 2.0 m 2.0 x 10-3 m V C AL ,out = 0.0537 kmol/m3 * AL

(9.721 x 10-5 m/sec)

30-5

30.18 Let A = CO2, and B = H2O (liquid) DAB = 1.77x10-9 m2/sec w 36 g / sec V0 = = = 3.61x10−5 m3 / sec  L 998.2 x103 g / m3 a. Determine CAL* P 2.54 atm * C AL = A = =0.10 kgmol/m3 3 H atm  m 25.4 kgmol b. Determine kL 4w 4(36 g / sec) Re = = = 769  D L  (6.0cm)(9.93x10−3 g/ cm − sec)

Sc =

L

 L DAB

993x10−6 kg / m2 sec = 562 (998.2 kg/ m3 )(1.77 x10−9 m2 / sec) 1/6

  2 g  z3  k z Sh = L = 0.433( Sc)1/2  L 2  (Re L ) 0.4 DAB  L  1/6

  2 g  z3  D  k L = 0.433( Sc)  L 2  (Re L ) 0.4  AB   z   L  1/2

1/6

-9 2  (998.2 kg/m 3 ) 2 (9.81 m/sec 2 )(2m)3  0.4  1.77 x 10 m /sec  k L = 0.433(562)    (769)  2.0 m (993 x 10-6 kg/m-sec) 2     -5 =2.69 x 10 m/sec 1/2

V0C AL + N A   D  z − V0C AL z

z +z

+0=0

÷ Δz, rearrangement, z → 0 dC AL V0 + N A  D = 0 dz

30-6

dC AL  D * + k L (C AL − C AL ) = 0 dz V0 C AL ,out L dC AL k − = L  D  dz C AL ,o C * − C 0 V0 AL AL *  C AL − C AL ,o  k L ln  *  =  DL C −C V0 AL ,out   AL    −k L   -π(0.06 m)(2.69x10-5 m/sec)(2 m)   * 3 C AL ,out = C AL 1 − exp  DL = 0.10 kgmol/m 1-exp )     (  3.61x10-5 m3 /sec     V0    3 C AL ,out = 0.0245 kgmol/m

30-7

31.4 a. Determine AGs,min

(1 − R ) xA2 AL2 = xA1 AL1 (1 − 0.8)( 0.01)(100 ) = xA1 AL1 = 0.2

lbmol hr

ALs = AL2 (1 − x A 2 ) = (100 )(1 − 0.01) = 99

lbmol hr

 x  x A1 AL1 = ALs  A1   1 − x A1   x  0.20 = ( 99 )  A1   1 − x A1  x A1 = 2.02  10−4 AL1 =

AL s 99 lbmol = = 99.02 −4 1 − x A1 1 − 2.02  10 hr

y A1 AG1,min + x A 2 AL2 = x A1 AL1 + y A 2,min AG2,min y*A 2,min =

H  515 atm  xA2 =   ( 0.01) = 0.412 PT  12.5 atm 

0 + ( 0.01)(100 ) = ( 0.20 ) + ( 0.412 ) ( AG2,min ) AG2,min = 1.94

lbmol hr

AGs ,min = (1 − y A2,min ) AG2,min = (1 − 0.412 )(1.94 ) = 1.14

lbmol hr

b. Determine yA2,min From part (a) above, y*A 2,min = 0.412

31-1

31.9 a. Characterize molar flowate and mole fraction composition of all terminal streams kgmol kgmol AGs = AG2 (1 − y A1 ) = 2.0 (1 − 0.10) = 1.8 sec sec AGS 1.8 kgmol/sec kgmol AG2 = = =1.836 1 − y A2 1-0.02 sec AG1 + AL2 = AG2 + AL1 kgmol AL1 = AG1 + AL2 − AG2 = 2.0 - 1.837 + 3.0 = 3.163 sec AG1 y A1 + AL2 x A 2 = AG2 y A 2 + AL1 x A1

(2.0)(0.10) + (3.0)(0.01) = (1.837)(0.020) + (3.163) xA1 xA1 = 0.061 Material Balance Summary: AG1 (kgmol/sec) 2.0 AG2 (kgmol/sec) 1.84 AL1 (kgmol/sec) 3.163 AL2 (kgmol/sec) 3.0 yA1 0.1 yA2 0.02 xA1 0.061 xA2 0.01 b. Tower diameter (D) at 50% of flooding condition Cf = 98 for 1.0 inch ceramic Intalox saddles (Table 31.2) AL '1 = AL1 ( x A1 M A + (1 − x A1 ) M B ) = ( 0.061(17)+(1-0.061)(18) )( 3.163) =56.75 kg/sec 1/2

x − axis =

AL1'  G    AG1   L − G 

1/2 kg   kg   56.75 2.8     sec   m3 =   = 0.056 kgmol   kg  1000 kg − 2.8 kg     2.0   26.9 m3 m3  sec   kgmol   

y − axis = Y = 0.25 1/2

    (  −  ) g  ( 2.8)(1000 − 2.8 )(1.0 )  kg  = 3.77 2 G = Y  G L 0.1 G c   = 0.25 0.1  C f L J m sec    ( 98)( 0.001) (1)   kg kg 53.8 53.8 AG1' sec =14.27 m 2 at flooding, A = sec = 28.54 m 2 at 50% of flooding A= ' = kg kg Gf 3.77 2 1.885 2 m sec m sec 1/2

' f

31-2

1/2

 4A  D=    

1/2

 4  14.27 m 2  =  = 4.25 m at flooding π   1/2

 4  28.54 m 2  D=  = 6.03 m at 50% of flooding π  

c. Log-mean mass transfer driving force (yA – yA*)lm y A1 − y*A1 ) − ( y A 2 − y*A 2 ) ( * ( y A − y A )lm =  ( y A1 − y*A1 )   ln   ( y A 2 − y*A 2 )  x A1 = 0.061, y*A1 = 0.035 x A 2 = 0.01, y*A 2 = 0.005

( y − y ) = ( 0.10 − 0.035) = 0.065 ( y − y ) = ( 0.02 − 0.005) = 0.015 A1

* A1

* A2

(y − y ) = A

* A lm

( 0.065 − 0.015)  0.065  ln    0.015 

= 0.034

0.12

XA1, YA1 0.10 0.08

YA0.06 0.04

XA2, YA2

0.02

0.00 0.00

0.02

0.04

0.06

0.08

XA (mole dissolved NH 3 / mole water) 31-3

31.13 a. Determine ALs,min y A1 AG1 + x A 2 AL2 = y A 2 AG2 + x A1,min AL1,min

H 0.46 atm = =0.023 P 2.0 atm y 0.03 x A1,min = A1 = = 0.130 m 0.23 AG1 (1 − y A1 ) 2.0(1-0.03) lbmol AG2 = = = 1.95 1-0.005 hr (1 − y A2 )

( 0.03)( 2 ) + 0 = ( 0.005 )(1.95 ) + ( 0.130 ) AL1,min AL1,min = 0.387

lbmol hr

ALs ,min = AL1,min (1 − x A1,min ) = ( 0.387 )(1-0.130 ) =0.336

lbmol hr

b. Estimate required K’ya for z = 6.0 ft z = H OG N OG H OG =

Z N OG

AL2 + AG1 = AL1 + AG2 1.0 + 2.0 = AL1 + 1.95 lbmol hr x A 2 AL2 + y A1 AG1 = x A1 AL1 + y A 2 AG2 AL1 = 1.05

0 + 0.03(2.0) = x A1 (1.05) + 0.005(1.95) x A1 = 0.048

( AG1 + AG2 )

lbmol 2 hr ( AL1 + AL2 ) lbmol AL = = 1.025 2 hr * y A1 = mx A1 = 0.23(0.048) = 0.011

AG =

= 1.975

y*A 2 = mx A 2 = 0.23(0) = 0.0

31-4

N OG =

( y A1 − y A2 )

(y − y )−(y − y ) (y − y )  ln   ( y − y )  * A1

H OG = G=

* A1

* A2

( 0.03 − 0.005) = 2.39 ( 0.03 − 0.011) − ( 0.005 − 0 )  ( 0.030 − 0.011)  ln    ( 0.005 − 0 ) 

6.0 ft G =2.51ft = ' 2.39 Kya

( AG1 + AG2 )

K y' a =

* A2

2A G = H OG

= 10.07

lbmol ft 2 hr

lbmol ft 2 hr = 4.0 lbmol 2.51 ft ft 3 hr

10.07

c. New size of packing at 45% of flooding

 G1'   '  = 0.26  G f old G 'f ,old =

G1' 0.26

 G1'   '  = 0.45  G f new G 'f ,new =

G1' 0.45  G1'     0.45 

1/2

 380  = =   G 'f ,old  G1'   C f ,new     0.26  C f ,new = 1138 G 'f ,new

Therefore 3/8 inch packing size is suitable

31-5