Introduction to Chemical Engineering Thermodynamics 8th Edition Smith Solutions Manual

Page 1

Introduction to Chemical Engineering Thermodynamics 8th Edition Smith Solutions Manual

richard@qwconsultancy.com

1|Pa ge


SVNAS 8th Edition Annotated Solutions

power =

Chapter 1

energy J Nm kg m2 = = = time s s s3

charge time charge = current*time = A s

current =

energy = current*electric potential time energy kg m2 electrical potential = = current*time A s3 power =

electrical potential resistance electrical potential kg m 2 resistance = = 2 3 current A s

current =

Updated 16/01/2017

p. 1 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

charge electrical capacitance charge As A2 s4 electrical capacitance =   electrical potential kg m 2 kg m 2 A s3

electrical potential =

P

Sat

b    a   t / C  c   / torr  10  7.5  P Sat / kPa=7.5exp A 

 

B   T /K C 

 b B     exp  2.303  a   =7.5exp  A   T / K - 273.15  c   T / K  C    

b B   2.303 a   =ln  7.5 +A  T / K - 273.15  c  T /K C 

Updated 16/01/2017

p. 2 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

The point of this problem is just for you to practice evaluating and plotting a simple function. You will do a number of problems over the course of the semester (and many more over the course of your career) in which the results are best presented in graphical form. The only thing to be careful of in plotting the Antoine equation is to pay attention to the units of T and Psat and to whether the constants are given for use with the base 10 logarithm or the natural logarithm. The parameters in Table B.2 are for use with T in °C, P in kPa, and the natural logarithm. I used MS Excel to prepare the plots for diethyl ether. A portion of the spreadsheet (containing the plots) is shown below: 13.9891 B

2463.93 C

223.24

Vapor pressure of Diethyl Ether

Vapor pressure of Diethyl Ether

200 180

100

160

Updated 16/01/2017

Vapor Pressure (kPa)

T (deg C) Psat (kPa) -30 3.450361508 -29 3.684455127 -28 3.931786307 -27 4.192943086 -26 4.468531566 -25 4.759176154 -24 5.065519787 -23 5.388224158 -22 5.727969934 -21 6.085456959 -20 6.461404464 -19 6.856551255 -18 7.271655907 -17 7.707496937 -16 8.16487298 -15 8.644602953 -14 9.147526209 -13 9.674502686 -12 10.22641305 -11 10.80415881 -10 11.40866246 -9 12.0408676 -8 12.70173902 -7 13.39226279 -6 14.11344639 -5 14.86631874 -4 15.65193033 -3 16.4713532 -2 17.3256811 -1 18.21602942 0 19.14353533 1 20.10935774

Vapor Pressure (kPa)

A

140

120 100 80 60

10

40 20 0

1

-30

-20

-10

0

10

20

30

Temperature (°C)

40

50

60

-30

-10

10

30

50

Temperature (°C)

p. 3 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

SVNAS, problem 1.4 300

T (deg C or deg F)

200 100 0

deg C

-100 -200

deg F

-300 -400 -500 0

50

100

150

200

250

300

350

T (K)

Updated 16/01/2017

p. 4 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

  

Updated 16/01/2017

 

p. 5 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

Updated 16/01/2017

p. 6 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

Updated 16/01/2017

p. 7 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

dP MP  g dz RT

dP Mg  dz P RT

P Mg ln      z  zo  RT  Po   Mg P  Po exp    z  zo    RT 

 0.029*9.8 P  101325exp   1609    83430 Pa  8.314*283  Updated 16/01/2017

p. 8 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

Updated 16/01/2017

p. 9 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

Updated 16/01/2017

p. 10 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

Updated 16/01/2017

p. 11 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

 

 

Updated 16/01/2017

p. 12 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

Updated 16/01/2017

p. 13 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

m   q  1.25 kg m3 *55880 m3 s1 = 69850 kg s1

Updated 16/01/2017

p. 14 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

Updated 16/01/2017

p. 15 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

Updated 16/01/2017

p. 16 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

Updated 16/01/2017

p. 17 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

A 12.74

B 2017.87

C -80.87

T (°C) -18.5 -9.5 0.2 11.8 23.1 32.7 44.4 52.1 63.3 75.5

T(K) 254.65 263.65 273.35 284.95 296.25 305.85 317.55 325.25 336.45 348.65

Psat (kPa)

ln (Psat)

ln (Psat)fit

Error2

Psatfit (kPa)

3.18 5.48 9.45 16.9 28.2 41.9 66.6 89.5 129 187

1.157 1.701 2.246 2.827 3.339 3.735 4.199 4.494 4.860 5.231

1.125 1.697 2.253 2.849 3.368 3.767 4.211 4.479 4.841 5.201

1.02E-03 2.03E-05 4.77E-05 4.61E-04 7.96E-04 1.02E-03 1.43E-04 2.24E-04 3.50E-04 9.19E-04

3.08 5.46 9.52 17.27 29.01 43.26 67.40 88.17 126.61 181.42

Error Sum

5.01E-03

6.000

200 180

5.000

160

4.000

Pressure(kPa)

ln(Pressure(kPa))

Relative Error -3.2% -0.4% 0.7% 2.2% 2.9% 3.3% 1.2% -1.5% -1.9% -3.0%

3.000 2.000

140 120 100 80

60 40

1.000

20 0.000

0 200

250

300 Temperature (K)

ln P sat / kPa  A 

Updated 16/01/2017

350

400

-50

0 50 Temperature (°C)

100

B T / KC

p. 18 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 1

Updated 16/01/2017

p. 19 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

 1 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

 

2 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

3 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

4 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

  

5 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

6 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

 

7 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

  

8 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

9 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

10 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

11 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

(b) 

12 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

13 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

14 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

15 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

16 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

17 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

  

 

18 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

m'

19 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

 

20 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

21 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

22 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

23 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

24 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

 

25 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

26 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

 

 

27 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

28 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

29 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

30 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

31 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

32 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

33 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

34 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

∆𝑇

35 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

36 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

37 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

38 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 2

39 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


NAS 8th Edition Annotated Solutions

Chapter 3

Updated 5/23/2017

p. 1 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

Updated 5/23/2017

p. 2 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

k  44.18*106

Updated 5/23/2017

p. 3 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

1  V      V  P T

AP   V  V0 1    BP  A  B  P   AP  V0 AB V   V0  2   B  P   B  P 2 P   V0 AB 1 V    V P

 B  P

2

AP   V0 1    BP

AB  B  P  AP  B  P 

dV   dP V

dV c 0.125 cm3 g-1   dP  dP  dP V V  P  b V  P  2700 bar 

dV 

0.125 cm3 g-1 dP  P  2700 bar 

dW   PdV  P

Updated 5/23/2017

0.125 cm3 g-1 dP  P  2700 bar 

p. 4 of 182


NAS 8th Edition Annotated Solutions

500 bar

W

P

1 bar

Chapter 3

500 bar 0.125 cm3 g -1 dP  0.125 cm3 g -1  P   2700 bar  ln  P  2700 bar   1 bar  P  2700 bar 

  3200 bar   3 -1 W  0.125 cm3 g -1  499 bar   2700 bar  ln     5.1591 bar cm g 2701 bar    W  0.51591 Pa m3 g -1  515.91 Pa m3 kg -1  515.91 J kg -1

1  V      V  P T

1 dV   dP   3.9  106  0.1 109 P dP V V  ln    3.9  106  Po  P   5  1011 P 2  Po2  Vo 

V  Vo exp 3.9  106  Po  P   5  1011 P 2  Po2



  

 dP dW   3.9  10  0.1  10 P  PV exp  3.9  10  P  P   5  10  P  P   dP

dW   PVdP   PVo exp 3.9  106  Po  P   5  1011 P 2  Po2 6

9

6

o

Updated 5/23/2017

11

o

2

2 o

p. 5 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

   dW   3.9  10 P  0.1  10 P  exp  3.9  10 P  5  10

  P  dP

dW  3.9  106  0.1  109 P P exp 3.9  106 1  P   5  1011 P 2  1 dP 6

9

6

2

11

2

3000

W

 3.9  10 P  0.1 10 P  exp  3.9  10 P  5  10 P  dP 6

9

6

2

11

2

1

dV   dT  1.2  103 K -1 dT V

V  ln  2    T2  T1   1.2  103 K -1 T2  T1   V1  V  ln  2   1.2  103 K -1  20 K   0.024  V1 

V2  V1 exp  0.024  6.29 10-4 m3 kg-1 *exp  0.024   6.44 10-4 m3 kg-1

Updated 5/23/2017

p. 6 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 

 

dV   dT   dP V

V  ln    T  P  Vo 

V  ln    (104 K 1 ) *(5 K)  (4.8 105 bar -1 ) * 20 bar  5 104  9.6 104  0.0015  Vo 

Updated 5/23/2017

p. 7 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

𝜕 𝜕

Updated 5/23/2017

p. 8 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

°

Updated 5/23/2017

°

°

1  V      V  P T

1 dV   dP V

p. 9 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

V  ln      P  P1   V1 

V  A T  exp   P 

V  V1 exp    P  P1  

A T   V1 exp  P1 

 dW   PVdP   PV1 exp    P  P1   dP   A T  P exp   P  dP P

W   A T  P exp   P  dP 

A T 

  P1  1 exp   P1    P  1 exp   P  

P1

 W

V V

1  P1  1V1   P  1 V   PV  1 1  PV   

1

 

Updated 5/23/2017

p. 10 of 182


NAS 8th Edition Annotated Solutions

R

Chapter 3

2

T2  P2  C p  1  7       0.5520 T1  P1  8  

8 bar

(a) P (b) (c)

1 bar

0.00624 m3 mol-1

0.0275 m3 mol-1

0.0499 m3 mol-1

V

Updated 5/23/2017

p. 11 of 182


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 12 of 182


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 13 of 182


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 14 of 182


NAS 8th Edition Annotated Solutions

10 bar

Chapter 3

1

P

2 3 bar 2 bar

4 0.00499 m3 mol-1

3 0.0166 m3 mol-1

V

V1 

RT1 8.314 Pa m3mol-1K -1 *600 K   0.00499 m3 /mol P1 106 Pa

RT2 8.314 Pa m3mol-1K -1 *600 K V2    0.0166 m3 /mol 6 P2 0.3*10 Pa

Updated 5/23/2017

p. 15 of 182


NAS 8th Edition Annotated Solutions

V4 

Chapter 3

RT4 8.314 Pa m3mol-1K -1 *400 K  P4 0.2*106 Pa

J mol-1K-1

0.0166 m3 /mol 0.00499m3 /mol  

J mol-1K-1

J mol-1K-1

  

J mol-1K-1

J mol-1K-1 

  Updated 5/23/2017

p. 16 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

J mol-1K-1

J mol-1K-1 

V1 

RT1 8.314 Pa m3mol-1K -1 *300 K  P1 105 Pa

Updated 5/23/2017

p. 17 of 182


NAS 8th Edition Annotated Solutions

V1 

Chapter 3

RT1 8.314 Pa m3mol-1K -1 *300 K  P1 5*105 Pa

J mol-1K-1

 

J mol-1K-1

J mol-1K-1  

J mol-1K-1

 

J mol-1K-1

 

 Updated 5/23/2017

p. 18 of 182


NAS 8th Edition Annotated Solutions

Cv

Vi  P1    V1  P2 

W (adiabatic) 

Cp

Chapter 3

3

1 5     0.225  12 

P2Vi  PV (12 bar * 2.70 m3 )  (1 bar*12 m3 ) 1 1   30.6 bar m3  30.60 *105 J = 3060 kJ  1 (5 / 3)  1

Cp

Pi  V1    P1  V2 

Cv

 12  3  62.90 5

PV (62.9bar *1 m3 )  (1 bar *12 m3 ) i 2  PV 1 1 W (adiabatic)    76.3 bar m3  76.30 *105J = 7630 kJ  1 5 / 3 1

Updated 5/23/2017

p. 19 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

(c)

12 bar

(b)

P

(d)

(a) (e)

1 bar 1 m3

12 m3

V

∗𝑃 Updated 5/23/2017

−1

𝑑𝑃 + 𝑃 𝑑𝑧

𝑑𝑇 = 𝑑𝑧

𝑃 𝑃

𝑑𝑃 + 𝑃 𝑑𝑧

𝑑𝑇 =0 𝑑𝑧 p. 20 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

𝑑𝑃 𝑑𝑇 + 𝑃 𝑑𝑧 𝑑𝑧

=0

𝑑𝑃 𝑑𝑇 + =0 𝑑𝑧 𝑑𝑧

∗ −𝑀 𝑔 +

𝑀𝑔 ( 𝑅

dt

d  mU  dt

𝑑𝑇 𝑑𝑧

Inlet mass flow rate = m ' enthalpy = H’

dm  m' dt d  mU 

)=

𝑑𝑇 =0 𝑑𝑧

Inside tank, m = 0 at t = 0

 m'H '

dm H' dt

 mU t   mU t 0   mH 't   mH 't 0 Updated 5/23/2017

p. 21 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 mU t   mH 't

Tsupply =318 K Psupply =1500 kPa T= 298 K

dn n dt

n Updated 5/23/2017

p. 22 of 182


NAS 8th Edition Annotated Solutions

d  nU  dt

Chapter 3

 nHsupply  Q

Q

n

dU dn dn U  n Usupply  RTsupply   Q  Usupply  RTsupply   Q dt dt dt dn dt

n

Q Q U

dn dn dn  Usupply  RTsupply   U  Usupply  RTsupply   dt dt dt

Q   Qdt   U  Usupply  RTsupply 

dn dt  U  Usupply  RTsupply   n final  ninitial  dt

dn  n' dt

Updated 5/23/2017

p. 23 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

n' dm  m' dt

d  nU  dt

d  nU  dt

 n ' H ' Q

dn H ' Q dt

n2U 2  n1U1   n2  n1  H ' Q

n2 U 2  H '  n1 U1  H '  Q

U 2  H '  U 2  U ' P 'V '  U 2  U ' RT '  Cv T2  T '  RT '  CvT2  C pT '

n2 CvT2  C pT '  n1 CvT1  C pT '  Q

n2 CvT2  C pT '  Q Updated 5/23/2017

p. 24 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

n2 CvT2  C pT '  0

CvT2  C pT ' T2   T '

T2   T '

n

PV 300000 Pa  4 m3   346 mol RT 8.3145 Pa m3 mol-1 K  417.2 K

n2 CvT2  C pT '  mtank C p,tank T2  298 K  n2  2.5T2  3.5  298 K   8.3145 J mol-1 K  400 kg  460 J kg -1 K -1 T2  298 K 

n2 

P2V2 300000 Pa  4 m3  RT2 8.3145 Pa m3 mol-1 K -1  T2

1200000 T2

 2.5T2  3.5  298 K   400  460 T2  298 K 

3000000T2  1251600000  184000T2 2  54832000T2 T2 2  281.70T2  6802.2  0 T2  304.1 K or -44.7 K (of course only the first answer is physically meaningful) Updated 5/23/2017

p. 25 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

T = f(P) ?

dn  n dt

dU t  nH dt

Updated 5/23/2017

p. 26 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

dU d dU dn dn dn U   n U  RT   U  RT Utank  nU   tank  n dt dt dt dt dt dt or dUtank dU dn n  RT dt dt dt

n

dn  RT  C tank  nC  dT dt dt v,

v

1 dT R dn  T dt  Cv ,tank  nCv  dt  T  R  Cv ,tank  n2 Cv  ln  2   ln     T1  Cv  Cv ,tank  n1Cv 

PV   Cv ,tank  2 tank Cv   T  R RT2  ln  2   ln  PV T C   1 tank  1 v  Cv ,tank  RT Cv   1  R

P2Vtank   Cv C  C v , tank v  T2  RT2   PV T1   1 tank  Cv ,tank  RT Cv   1  Updated 5/23/2017

p. 27 of 182


NAS 8th Edition Annotated Solutions

V 

RT 8.314 Pa m3mol-1K -1 *303 K  P 100*103 Pa

V2 

RT2 8.314 Pa m3mol-1K -1 *480 K  P2 500*103 Pa

Updated 5/23/2017

Chapter 3

p. 28 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 

 

 

Updated 5/23/2017

p. 29 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 

  

 

Updated 5/23/2017

p. 30 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

    

 

 Updated 5/23/2017

p. 31 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 

 

 

T2  T1  

T2  T1 

T2  423.15K 

Updated 5/23/2017

u22  u12 2C p

u22  u12 2C p

(50 m/s) 2  (2.5 m/s) 2 *0.028 kg/mol  421.95 K or 148.8 C J 2*3.5*8.314 molK

p. 32 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

  

 

 

  

 

 Updated 5/23/2017

p. 33 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 

 

 

 

 

  

Updated 5/23/2017

p. 34 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 

Updated 5/23/2017

p. 35 of 182


NAS 8th Edition Annotated Solutions

R

Chapter 3

2

 P  CP  10 bar  5 Tintermidiate = T1  2  = 298.15K ×   = 335.947 K  1 bar   P1 

U = W = CV T = 1.5 × 8.314 J·mol1·K 1 × (335.947 K - 298.15 K) = 471.366 J·mol1

H = U + W = 471.366 J·mol1 + 471.366 J·mol1 = 942.732 J·mol1

U = CV T = 1.5 × 8.314 J·mol1·K 1 × (573.15 K - 335.947 K) = 2958.16 J·mol1

H = CP T = 2.5 × 8.314 J·mol1·K1 × (573.15 K - 335.947 K) = 4930.26 J·mol1 

W = -RT = -8.314 J·mol1·K1 × (573.15 K - 335.947 K) = -1972.11 J·mol1

P  Tintermidiate = T1  2   P1 

R CP

2

 10 bar  5 = 298.15K ×   = 335.947 K  1 bar 

U = W = CV T = 1.5 × 8.314 J·mol1·K 1 × (335.947 K - 298.15 K) = 471.366 J·mol1 Updated 5/23/2017

p. 36 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

H = U + W = 471.366 J·mol1 + 471.366 J·mol1 = 942.732 J·mol1

Q = U = CV T = 1.5 × 8.314 J·mol1·K 1 × (573.15 K - 335.947 K) = 2958.16 J·mol1 H = CP T = 2.5 × 8.314 J·mol1·K1 × (573.15 K - 335.947 K) = 4930.26 J·mol1

R

2

 P  CP  10 bar  5 Tintermidiate = T1  2  = 298.15K ×   = 335.947 K  1 bar   P1  U = W = CV T = 1.5 × 8.314 J·mol1·K 1 × (335.947 K - 298.15 K) = 471.366 J·mol1

H = U + W = 471.366 J·mol1 + 471.366 J·mol1 = 942.732 J·mol1 

 Updated 5/23/2017

p. 37 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

RTA RTB 2 RTo   p p po R

2

TB  p  C p  p  7     To  po   po 

Updated 5/23/2017

p. 38 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

   

 

2

TB  p  7   To  po 

TA TB 2 p   To To po TA 2.5 319.75    1.4342 300 1 300

2

TB  p  7   To  po 

TA TB 2 p   To To po

2

TA  p  7 2 p    To  po  po

Updated 5/23/2017

p. 39 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

2 p 1   p  7 TA      po 2   po  To   

2

TB  p  7   To  po 

TA TB 2 p   To To po TA 2 p TB 2*1.3233 325      1.563 To po To 1 300

TA TB 2 p   To To po 744.3  TB TB 2 p   300 300 po 7

p  TB  2   po  To 

Updated 5/23/2017

2

2  p 7 TB  To    300 1.2405 7  319.05 K  po 

p. 40 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

   PV  RT   b  P RT   V

RT     b   P  RT 

Updated 5/23/2017

p. 41 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 1  V      V  P T

 RT  V     2 P  P T RT



1 P2  RT     P     b  b  P 1    P  RT  RT  RT   

    P V   b    RT RT     RT P  V b RT

  1 d     d     V b R  RT    V b R       2 RT  RT dT   RT  dT T RT  P        2 2  T V       V  b   V  b   RT  RT       RT

V b

RT

P

RT d  2 R   P  P dT T   P    P     d           2  T V  T   RT   T dT   RT     P 

Updated 5/23/2017

p. 42 of 182


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 43 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

Z

PV B C D  1   2  3  ...  1  B   C  2  D  3  ... RT V V V

B

dZ d   0

C

1 d2Z 2 d2

 0

P

a T  RT  V  b V   b V   b 

Z

a T  V PV V   RT V  b RT V   b V   b  1

Z

1 b

Z

a T  1

   1    b

RT 1   b

a T   1  1  b  RT 1   b  1   b  

Updated 5/23/2017

p. 44 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

a T   1   b  1   b      b 1   b     b 1   b     dZ b     2 2 d  1  b  2 RT   1   b  1   b         a T   dZ b 1   b 2  2     d  1  b  2 RT  1   b  2 1   b  2    a T  dZ b d   0 RT

B

a T  dZ b d   0 RT



2 2 2 2 2 2 2  2 b 1   b  1   b    2 b 1   b   1   b    a T   2 b  1   b   1   b    1   b      4 4 RT  d  2 1  b  3 1   b   1   b     

d2Z

2b 2

d2Z

2b 2

d

1  b  

 2

d2Z d

2

 3

 2b 2 

a T  RT

 0

1 d 2Z C 2 d 2

2 2 2   a T   2 b  1   b  1   b    1   b   2 b 1   b    2 b 1   b      RT  1   b  3 1   b  3  

 2     b   2  b2  

 b2 

a T      b   RT 

a T     b

 0

RT

PV B C 1  2 RT V V

12 bar  V

1

140 cm3 mol-1 7200 cm 6 mol-2  V V2

83.145 bar cm3 mol-1 K -1  298 K  140 cm3 mol-1 7200 cm6 mol-2  V  2064.8 cm3 mol-1 1     V V2   Updated 5/23/2017

p. 45 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

 0.3040  0.00218

BPc  B 0   B1  0.3040  0.087  0.00218  0.3038 RTc  BP  P 0.2381 Z  1   c  r  1  0.3038  0.9315 1.056  RTc  Tr

P

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 1.05560.5  83.1452  282.32 a(T )  0.42748 r  0.42748  4548100 bar cm 6 mol-2 Pc 50.40 b  0.08664

Updated 5/23/2017

RTc 83.145  282.3  0.08664  40.35 cm3 mol-1 Pc 50.40

p. 46 of 182


NAS 8th Edition Annotated Solutions

V

Chapter 3

a T  V  b RT b P P V V  b 

 4548100 V  40.35   3 -1 V   2064.8  40.35   cm mol  12 V V  40.35      379008 V  40.35   3 -1 V   2105.2   cm mol  V V  40.35   

1   0.480  1.574  0.176 1  T  R T a(T )  0.42748 2

2

0.5 r

2 2 c

Pc

1  0.480  1.574  0.087  0.176 0.087   1  1.0556   83.145  282.3 a(T )  0.42748 2

2

2

2

50.40

a(T )  0.42748

0.9731  83.145  282.3 50.40 2

2

a(T )  4547150 bar cm6 mol-2 b  0.08664

V

RTc 83.145  282.3  0.08664  40.35 cm3 mol-1 Pc 50.40

a T  V  b RT b P P V V  b 

 4547150 V  40.35   3 -1 V   2064.8  40.35   cm mol  12 V V  40.35      378929 V  40.35   3 -1 V   2105.2   cm mol  V V  40.35    

Updated 5/23/2017

p. 47 of 182


NAS 8th Edition Annotated Solutions

P

Chapter 3

a T  RT  V  b V  2.4142b V  0.4142b 

1   0.37464  1.54226  0.2699 1  T  R T a(T )  0.45724 2

0.5 r

2 2 c

Pc

  a(T )  0.45724

1  0.37464  1.54226  0.087  0.2699  0.087 

a(T )  0.45724

2

2



1  1.0556

  83.1452  282.32 2

50.40

0.9724  83.145  282.3 50.40 2

2

a(T )  4860171 bar cm6 mol-2 RT 83.145  282.3 b  0.07779 c  0.07779  36.23 cm3 mol-1 Pc 50.40

V

a T  RT V b b P P V  0.4142b V  2.4142b 

  4860171V  36.23 3 -1 V   2064.8  36.23   cm mol  12 V  15.005 V  87.46      405014 V  36.23  3 -1 V   2101.0   cm mol  V  15.005V  87.46   

PV B C 1  2 RT V V

15 bar  V

1

156.7 cm3 mol-1 9650 cm 6 mol-2  V V2

83.145 bar cm3 mol-1 K -1  323.15 K  156.7 cm3 mol-1 9650 cm6 mol-2  V  1791.2 cm3 mol-1 1     V V2   Updated 5/23/2017

p. 48 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

 0.3026  0.003266

BPc  B 0   B1  0.3026  0.100  0.003266  0.3023 RTc  BP  P 0.3079 Z  1   c  r  1  0.3023  0.9120 1.058  RTc  Tr

P

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 1.0580.5  83.1452  305.32 a(T )  0.42748 r  0.42748  5496600 bar cm 6 mol-2 Pc 48.72 b  0.08664

V

RTc 83.145  305.3  0.08664  45.14 cm3 mol-1 Pc 48.72

a T  V  b RT b P P V V  b 

 5496600 V  45.14   3 -1 V  1791.2  45.14   cm mol  15V V  45.14     366440 V  45.14   3 -1 V  1836.3   cm mol  V V  45.14     Updated 5/23/2017

p. 49 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

1   0.480  1.574  0.176 1  T  R T a(T )  0.42748 2

0.5 r

2

2 2 c

Pc

1   0.480  1.574  0.100  0.176  0.100   1  1.058    83.145  305.3  a(T )  0.42748 2

2

2

2

48.72

a(T )  0.42748

0.96398  83.145  305.3 48.72 2

2

a(T )  5450090 bar cm6 mol-2

V

a T  V  b RT b P P V V  b 

 5450090 V  45.14   3 -1 V  1791.2  45.14   cm mol  15 V V  45.14      363399 V  45.14   3 -1 V  1836.3   cm mol  V V  45.14   

P

a T  RT  V  b V  2.4142b V  0.4142b 

Updated 5/23/2017

p. 50 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

1   0.37464  1.54226  0.2699 1  T  R T a(T )  0.45724 2

0.5 r

2

2

2 c

Pc

1  0.37464  1.54226  0.100  0.2699 0.100  1  1.058   83.145  305.3 a(T )  0.45724 2

2

2

48.72

a(T )  0.45724

0.97014  83.145  305.3 48.72 2

2

a(T )  5866378 bar cm6 mol-2 b  0.07779

V

RTc 83.145  305.3  0.07779  40.53 cm3 mol-1 Pc 48.72

a T  RT V b b P P V  0.4142b V  2.4142b 

 5866378 V  40.53  3 -1 V  1791.2  40.53   cm mol  15 V  16.78 V  97.85       391091 V  40.53  3 -1 V  1831.7   cm mol  V  16.78 V  97.85     

PV B C 1  2 RT V V

15 bar  V

1

194 cm3 mol-1 15300cm 6 mol-2  V V2

83.145 bar cm3 mol-1 K -1  348 K  194 cm3 mol-1 15300cm6 mol-2  V  1929.0 cm3 mol-1 1     V V2  

Updated 5/23/2017

p. 51 of 182

2


NAS 8th Edition Annotated Solutions

Chapter 3

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

 0.2836  0.02011

BPc  B 0   B1  0.2836  0.286  0.02011  0.2779 RTc  BP  P 0.3989 Z  1   c  r  1  0.2779  0.8985 1.0919  RTc  Tr

P

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 1.09190.5  83.1452  318.7 2 a(T )  0.42748 r  0.42748  7639640 bar cm 6 mol-2 Pc 37.6 b  0.08664

V

RTc 83.145  318.7  0.08664  61.06 cm3 mol-1 Pc 37.6

a T  V  b RT b P P V V  b 

 7639640 V  61.06   3 -1 V  1929.0  61.06   cm mol  15V V  61.06     509309 V  61.06   3 -1 V  1990.1   cm mol  V V  61.06    

Updated 5/23/2017

p. 52 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

1   0.480  1.574  0.176 1  T  R T a(T )  0.42748 2

2

0.5 r

2 2 c

Pc

1  0.480  1.574  0.286  0.176 0.286  1  1.0919   83.145  318.7 a(T )  0.42748 2

2

2

2

37.6

a(T )  0.42748

0.9193  83.145  318.7 36.7 2

2

a(T )  7519399 bar cm6 mol-2

V

a T  V  b RT b P P V V  b 

 7519399 V  61.06   3 -1 V  1929.0  61.06   cm mol  15 V V  61.06      501293 V  61.06   3 -1 V  1990.1   cm mol  V V  61.06    

P

a T  RT  V  b V  2.4142b V  0.4142b 

1   0.37464  1.54226  0.2699 1  T  R T a(T )  0.45724 2

0.5 r

2

2 2 c

Pc

1   0.37464  1.54226  0.286  0.2699  0.286   1  1.0919    83.145  318.7  a(T )  0.45724 2

2

2

37.6

a(T )  0.45724

0.92994  83.145  318.7 37.6 2

2

a(T )  7940485 bar cm6 mol-2 RT 83.145  318.7 b  0.07779 c  0.07779  54.82 cm3 mol-1 Pc 37.6

Updated 5/23/2017

p. 53 of 182

2


NAS 8th Edition Annotated Solutions

V

Chapter 3

a T  RT V b b P P V  0.4142b V  2.4142b 

 7940485 V  54.82   3 -1 V  1929.0  54.82   cm mol  15 V  22.71 V  132.35      529365 V  54.82   3 -1 V  1983.8   cm mol  V  22.71V  132.35   

Updated 5/23/2017

p. 54 of 182


NAS 8th Edition Annotated Solutions

P

Chapter 3

a T  RT  V  b V V  b 

Updated 5/23/2017

p. 55 of 182


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 56 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 cm3 cm6  724 93866  cm  mol  mol 2  V  16629 1  mol  V V2      3

Updated 5/23/2017

p. 57 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 BP  P 0.38750 Z  1   c  r 1  0.8010*  0.9532 0.6637  RTc  Tr

P

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 0.66370.5 *83.1452 *4522 a T   0.42748 r  0.42748*  19149726 bar cm6 mol 2 Pc 38.7

b  0.08664

RTc 83.145*452  0.08664*  84.14 cm3mol 1 Pc 38.7

V  16629  84.14 

Updated 5/23/2017

19149726 V  84.14  1.5 V V  84.14 

p. 58 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

1   0.48  1.574  0.176 1  T  R T a T   0.42748 2

0.5 r

2

2

2 c

Pc

1   0.48  1.574*0.086  0.176*0.086 1  0.6637  * 83.145 * 452 a T   0.42748* 2

0.5

2

2

2

38.7

a T   19350625.74 bar cm6 mol 2

b  0.08664

RTc 83.145*452  0.08664*  84.14 cm3mol 1 Pc 38.7

1   0.37464  1.54226*0.086  0.2699*0.086 1  0.6637  *83.145 * 452 a T   0.45724* 2

0.5

2

2

38.7

a T  19956222.093 bar cm6 mol 2 Updated 5/23/2017

p. 59 of 182

2


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 60 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 BP  P 2.13 Z  1   c  r 1  0.8010*  0.6833 1.282  RTc  Tr

P

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 1.2820.5 *83.1452 *2342 a T   0.42748 r  0.42748*  3204293 bar cm6 mol 2 Pc 44.6

b  0.08664

RTc 83.145*234  0.08664*  37.80 cm3mol 1 Pc 44.6

V  262.56  37.80 

Updated 5/23/2017

3204293 V  37.80  95 V V  37.80 

p. 61 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

1   0.48  1.574  0.176 1  T  R T a T   0.42748 2

0.5 r

2

2

2 c

Pc

1   0.48  1.574*0.126  0.176*0.126 1  1.282  * 83.145 * 234 a T   0.42748* 2

0.5

2

2

2

44.6

a T   3008454.99 bar cm6 mol 2

b  0.08664

RTc 83.145*234  0.08664*  37.80 cm3mol 1 Pc 44.6 V  262.56  37.80 

3008454.99 V  37.80  95 V V  37.80 

1   0.37464  1.54226*0.126  0.2699*0.126 1  1.282  *83.145 * 234 a T   0.45724* 2

0.5

2

2

2

44.6

a T   3322337.2 bar cm6 mol 2

b  0.07779

RTc  33.93 cm3mol 1 Pc

Updated 5/23/2017

p. 62 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

PV B C 1  2 RT V V

18 bar  V

1

152.5 cm3 mol-1 -5800 cm 6 mol-2  V V2

83.145 bar cm3 mol-1 K -1  523 K  152.5 cm3 mol-1 5800 cm6 mol-2  V  2415.8 cm3 mol-1 1     V V2  

Updated 5/23/2017

p. 63 of 182


NAS 8th Edition Annotated Solutions

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

Chapter 3

 0.5103  0.2816

BPc  B 0   B1  0.5103  0.345  0.2816  0.6075 RTc  BP  P 0.08161 Z  1   c  r  1  0.6075  0.9387 0.8082  RTc  Tr

Z  1  BP  CP2  DP3

 Z  ' ' 2    B  2C P  3D P  P T

Z  1  B   C  2  D 3

Updated 5/23/2017

p. 64 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 P (bar)

Zi

Updated 5/23/2017

Z1i

Z2i

p. 65 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

1 0.95 0.9

Zi

0.85 0.8

Z1i 0.75

Zi

0.7

Z1i

0.65

Z2i

Z2i

0.6 0.55 0.5 0

50

100

150

200

P (bar)

P

a T  RT  V  b V V  b 

T T 0.5 R 2Tc2 a(T )  0.42748 r  0.42748 Pc

0.5

R 2Tc2.5 Pc

 0.42748

T 0.5  83.1452  369.82.5 42.48

1.8295  108

bar cm6 mol-2

T

RT 83.145  369.8 b  0.08664 c  0.08664  62.71 cm3 mol-1 Pc 42.48

 Updated 5/23/2017

 p. 66 of 182


NAS 8th Edition Annotated Solutions

V

Chapter 3

a T  V  b RT b P P V V  b 

 10338000 V  62.71  3 -1 V  1899.1  62.71   cm mol  13.71  V V  62.71    754082 V  62.71  3 -1 V  1961.8   cm mol  V V  62.71    

 RT  bP  VP  V  b  V V  b     a T     83.14  313.15  62.17  13.71  13.71V  V  62.71  V V  62.71   10338000    1963.4  V  V  62.71  V V  62.71    754048 

0.2857 1T V sat  Vc Zc r 

10.8468 V sat  200  0.276

0.2857

Z  1  B0 Updated 5/23/2017

Pr P   B1 r Tr Tr p. 67 of 182


NAS 8th Edition Annotated Solutions

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

B 0  0.083  B1  0.139 

Chapter 3

0.422 0.84681.6 0.172 0.84684.2

 0.4676  0.2068

Z  1   0.4676  0.152  0.2068

0.3227  0.8098 0.8468

 

V

a T  V  b RT b P P V V  b 

 10177000 V  62.71  3 -1 V  1565.8  62.71   cm mol  17.16  V V  62.71      593080 V  62.71  3 -1 V  1628.5   cm mol  V V  62.71    

Updated 5/23/2017

p. 68 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V V  b     a T     83.14  323.15  62.17  17.16  17.16V  V  62.71  V V  62.71   10177000    1627.8  V  V  62.71  V V  62.71    593065 

0.2857 1T V sat  Vc Zc r 

10.8739 V sat  200  0.276

0.2857

Z  1  B0

Pr P   B1 r Tr Tr

B 0  0.083 

0.422

B1  0.139 

0.172

B 0  0.083  B1  0.139 

Updated 5/23/2017

Tr1.6 Tr4.2

0.422 0.87391.6 0.172 0.87394.2

 0.4406  0.1640

p. 69 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

Z  1   0.4406  0.152  0.1640

0.4040  0.8014 0.8739

V

a T  V  b RT b P P V V  b 

 10023300 V  62.71  3 -1 V  1305.4  62.71   cm mol  21.22  V V  62.71      472351 V  62.71  3 -1 V  1368.1   cm mol  V V  62.71  

 RT  bP  VP  V  b  V V  b     a T     83.14  333.15  62.17  21.22  21.22V  V  62.71  V V  62.71   100233000    1367.5  V  V  62.71  V V  62.71    472351 

0.2857 1T V sat  Vc Zc r 

V

sat

10.9009  200  0.276

Updated 5/23/2017

0.2857

p. 70 of 182


NAS 8th Edition Annotated Solutions

Z  1  B0

Chapter 3

Pr P   B1 r Tr Tr

B 0  0.083 

0.422

B1  0.139 

0.172

B 0  0.083  B1  0.139 

Tr1.6 Tr4.2

0.422 0.90091.6 0.172 0.90094.2

 0.4157  0.1276

Z  1   0.4157  0.152  0.1276

P

0.4995  0.7588 0.9009

a T  RT  V  b V V  b 

Updated 5/23/2017

p. 71 of 182


NAS 8th Edition Annotated Solutions

T T 0.5 R 2Tc2 a(T )  0.42748 r  0.42748 Pc b  0.08664

Chapter 3

0.5

R 2Tc2.5 Pc

 0.42748

T 0.5  83.1452  369.82.5 42.48

1.8295  108

bar cm6 mol-2

T

RTc 83.145  369.8  0.08664  62.71 cm3 mol-1 Pc 42.48

V

a T  V  b RT b P P V V  b 

 9876000 V  62.71  3 -1 V  1099.9  62.71   cm mol  25.94  V V  62.71      380725 V  62.71  3 -1 V  1162.6   cm mol  V V  62.71    

 RT  bP  VP  V  b  V V  b     a T     83.14  343.15  62.17  25.94  25.94V  V  62.71  V V  62.71   9876000    1162.0  V  V  62.71  V V  62.71    380725 

0.2857 1T V sat  Vc Zc r 

10.9279 V sat  200  0.276

0.2857

Updated 5/23/2017

p. 72 of 182


NAS 8th Edition Annotated Solutions

Z  1  B0

Chapter 3

Pr P   B1 r Tr Tr

B 0  0.083 

0.422

B1  0.139 

0.172

B 0  0.083  B1  0.139 

Tr1.6 Tr4.2

0.422 0.92791.6 0.172 0.92794.2

 0.3927  0.09652

Z  1   0.3927  0.152  0.0965

Updated 5/23/2017

0.6106  0.7319 0.9279

p. 73 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

P

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 T 0.5 R 2Tc2.5 T 0.5  83.1452  425.12.5 2.9006  108 a(T )  0.42748 r  0.42748  0.42748  bar cm6 mol-2 Pc Pc 37.96 T RT 83.145  425.1 b  0.08664 c  0.08664  80.67 cm3 mol-1 Pc 37.96

V

a T  V  b RT b P P V V  b 

 15016000 V  80.67   3 -1 V   2013.3  80.67   cm mol  15.41  V V  80.67     974430 V  80.67   3 -1 V   2094.0   cm mol  V V  80.67   

Updated 5/23/2017

p. 74 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V V  b     a T     83.14  373.15  80.67  15.41  15.41V  V  80.67  V V  80.67    15016000    2094.1  V  V  80.67  V V  80.67     974430 

0.2857 1T V sat  Vc Zc r 

10.8778 V sat  255  0.274

0.2857

Z  1  B0

Pr P   B1 r Tr Tr

B 0  0.083 

0.422

B1  0.139 

0.172

B 0  0.083  B1  0.139 

Updated 5/23/2017

Tr1.6 Tr4.2

0.422 0.87781.6 0.172 0.87784.2

 0.4369  0.1583

p. 75 of 182


NAS 8th Edition Annotated Solutions

Z  1   0.4369  0.200  0.1583

P

Chapter 3

0.4059  0.7833 0.8778

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 T 0.5 R 2Tc2.5 T 0.5  83.1452  425.12.5 2.9006  108 a(T )  0.42748 r  0.42748  0.42748  bar cm6 mol-2 Pc Pc 37.96 T RT 83.145  425.1 b  0.08664 c  0.08664  80.67 cm3 mol-1 Pc 37.96

Updated 5/23/2017

p. 76 of 182


NAS 8th Edition Annotated Solutions

V

Chapter 3

a T  V  b RT b P P V V  b 

 14818000 V  80.67   3 -1 V  1707.2  80.67   cm mol  18.66  V V  80.67     794105 V  80.67   3 -1 V  1787.9   cm mol  V V  80.67   

 RT  bP  VP  V  b  V V  b     a T     83.14  383.15  80.67  18.66  18.66V  V  80.67  V V  80.67    14818000    1787.9  V  V  80.67  V V  80.67     794105 

0.2857 1T V sat  Vc Zc r 

10.9013 V sat  255  0.274

0.2857

Z  1  B0

Updated 5/23/2017

Pr P   B1 r Tr Tr

p. 77 of 182


NAS 8th Edition Annotated Solutions

B 0  0.083 

0.422

B1  0.139 

0.172

B 0  0.083  B1  0.139 

Chapter 3

Tr1.6 Tr4.2

0.422 0.90131.6 0.172 0.90134.2

 0.4153  0.1271

Z  1   0.4153  0.200  0.1271

Updated 5/23/2017

0.4916  0.7596 0.9013

p. 78 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

P

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 T 0.5 R 2Tc2.5 T 0.5  83.1452  425.12.5 2.9006  108 a(T )  0.42748 r  0.42748  0.42748  bar cm6 mol-2 Pc Pc 37.96 T RT 83.145  425.1 b  0.08664 c  0.08664  80.67 cm3 mol-1 Pc 37.96

V

a T  V  b RT b P P V V  b 

 14628800 V  80.67   3 -1 V  1460.6  80.67   cm mol  22.38  V V  80.67     653655 V  80.67   3 -1 V  1541.3   cm mol  V V  80.67    

 RT  bP  VP  V  b  V V  b     a T     83.14  393.15  80.67  22.38  22.38V  V  80.67  V V  80.67    14628800    1541.3  V  V  80.67  V V  80.67     653655 

Updated 5/23/2017

p. 79 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

0.2857 1T V sat  Vc Zc r 

10.9248 V sat  255  0.274

0.2857

Updated 5/23/2017

p. 80 of 182


NAS 8th Edition Annotated Solutions

P

Chapter 3

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 T 0.5 R 2Tc2.5 T 0.5  83.1452  425.12.5 2.9006  108 a(T )  0.42748 r  0.42748  0.42748  bar cm6 mol-2 Pc Pc 37.96 T RTc 83.145  425.1 3 -1 b  0.08664  0.08664  80.67 cm mol Pc 37.96

V

a T  V  b RT b P P V V  b 

 14446000 V  80.67   3 -1 V  1260.6  80.67   cm mol  26.59  V V  80.67      543287 V  80.67   3 -1 V  1341.3   cm mol  V V  80.67    

 RT  bP  VP  V  b  V V  b     a T    83.14  403.15  80.67  26.59  26.59V   V  80.67  V V  80.67    14446000    1341.3  V  V  80.67  V V  80.67     543287 

Updated 5/23/2017

p. 81 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

0.2857 1T V sat  Vc Zc r 

10.9484 V sat  255  0.274

0.2857

Updated 5/23/2017

p. 82 of 182


NAS 8th Edition Annotated Solutions

T (K) 403.15

P (bar) 26.59

Chapter 3

Tc (K) 425.1

Pc (bar) 37.96

Tr 0.93 0.93 0.95 0.95

Pr 0.40 0.60 0.40 0.60

Z0 0.8059 0.6635 0.8206 0.6967

0.2

Tr 0.9484

Pr 0.7005

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm3/mol)

P

0.6310

Z1 -0.0763 -0.1662 -0.0589 -0.1110 -0.1432

0.6023 759.4

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 T 0.5 R 2Tc2.5 T 0.5  83.1452  408.12.5 2.7255  108 a(T )  0.42748 r  0.42748  0.42748  bar cm6 mol-2 Pc Pc 36.48 T RT 83.145  408.1 b  0.08664 c  0.08664  80.59 cm3 mol-1 Pc 36.48

Updated 5/23/2017

p. 83 of 182


NAS 8th Edition Annotated Solutions

V

Chapter 3

a T  V  b RT b P P V V  b 

 14302000 V  80.59   3 -1 V  1825.5  80.59   cm mol  16.54  V V  80.59     864692 V  80.59   3 -1 V  1906.1   cm mol  V V  80.59   

 RT  bP  VP  V  b  V V  b     a T     83.14  363.15  80.59  16.54  16.54V  V  80.59  V V  80.59    14302000    1906.0  V  V  80.59  V V  80.59     864692 

0.2857 1T V sat  Vc Zc r 

10.8899 V sat  262.7  0.282

0.2857

Updated 5/23/2017

p. 84 of 182


NAS 8th Edition Annotated Solutions

Z  1  B0

Chapter 3

Pr P   B1 r Tr Tr

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

B 0  0.083  B1  0.139 

0.422 0.88991.6 0.172 0.88994.2

 0.4256  0.1417

Z  1   0.4256  0.302  0.1417 

0.4534  0.7614 0.8899

V

a T  V  b RT b P P V V  b 

 14109000 V  80.59   3 -1 V  1549.0  80.59   cm mol  20.03  V V  80.59     704406 V  80.59   3 -1 V  1629.6   cm mol  V V  80.59   

Updated 5/23/2017

p. 85 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V V  b     a T     83.14  373.15  80.59  20.03  20.03V  V  80.59  V V  80.59    14109000    1629.6  V  V  80.59  V V  80.59     704406 

0.2857 1T V sat  Vc Zc r 

10.9144 V sat  262.7  0.282

0.2857

Z  1  B0

Pr P   B1 r Tr Tr

B 0  0.083 

0.422

B1  0.139 

0.172

Updated 5/23/2017

Tr1.6 Tr4.2

p. 86 of 182


NAS 8th Edition Annotated Solutions

B 0  0.083  B1  0.139 

0.422 0.91441.6 0.172 0.91444.2

Chapter 3

 0.4040  0.1115

Z  1   0.4040  0.302  0.1115

Updated 5/23/2017

0.5491  0.7372 0.9144

p. 87 of 182


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 88 of 182


NAS 8th Edition Annotated Solutions

P

Chapter 3

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 T 0.5 R 2Tc2.5 T 0.5  83.1452  408.12.5 2.7255  108 a(T )  0.42748 r  0.42748  0.42748  bar cm 6 mol-2 Pc Pc 36.48 T RT 83.145  408.1 b  0.08664 c  0.08664  80.59 cm3 mol-1 Pc 36.48

V

a T  V  b RT b P P V V  b 

 13920000 V  80.59   3 -1 V  1326.8  80.59   cm mol  24.01  V V  80.59      579922 V  80.59   3 -1 V  1407.4   cm mol  V V  80.59   

Updated 5/23/2017

p. 89 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V V  b     a T     83.14  383.15  80.59  24.01  24.01V  V  80.59  V V  80.59    13920000    1407.4  V  V  80.59  V V  80.59     579922 

0.2857 1T V sat  Vc Zc r 

10.9389 V sat  262.7  0.282

0.2857

Z  1  B0

Pr P   B1 r Tr Tr

B 0  0.083 

0.422

B1  0.139 

0.172

Updated 5/23/2017

Tr1.6 Tr4.2

p. 90 of 182


NAS 8th Edition Annotated Solutions

B 0  0.083  B1  0.139 

0.422 0.93891.6 0.172 0.93894.2

Chapter 3

 0.3838  0.0851

Z  1   0.3838  0.181  0.0851

Updated 5/23/2017

0.6582  0.7201 0.9389

p. 91 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

V

a T  V  b RT b P P V V  b 

 13746000 V  80.59   3 -1 V  1145.8  80.59   cm mol  28.53  V V  80.59      481800 V  80.59   3 -1 V  1226.3   cm mol  V V  80.59    

 RT  bP  VP  V  b  V V  b     a T    83.14  393.15  80.59  28.53  28.53V   V  80.59  V V  80.59    13746000    1226.3  V  V  80.59  V V  80.59     481800 

0.2857 1T V sat  Vc Zc r 

V

sat

10.9632  262.7  0.282

Updated 5/23/2017

0.2857

p. 92 of 182


NAS 8th Edition Annotated Solutions

Z  1  B0

Chapter 3

Pr P   B1 r Tr Tr

B 0  0.083 

0.422

B1  0.139 

0.172

B 0  0.083  B1  0.139 

Tr1.6 Tr4.2

0.422 0.96321.6 0.172 0.96324.2

 0.3651  0.0623

Z  1   0.3651  0.181  0.0623

Updated 5/23/2017

0.7821  0.6944 0.9632

p. 93 of 182


NAS 8th Edition Annotated Solutions

P

Chapter 3

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 T 0.5 R 2Tc2.5 T 0.5  83.1452  417.22.5 1.362  108 a(T )  0.42748 r  0.42748  0.42748  bar cm 6 mol-2 Pc Pc 77.10 T b  0.08664

RTc 83.145  417.2  0.08664  38.98 cm3 mol-1 Pc 77.10

V

a T  V  b RT b P P V V  b 

 7466000 V  38.98   3 -1 V  1521.1  38.98   cm mol  18.21  V V  38.98      409980 V  38.98   3 -1 V  1560.1   cm mol  V V  38.98    

Updated 5/23/2017

p. 94 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V V  b     a T     83.14  333.15  38.98  18.21  18.21V  V  38.98  V V  38.98    7466000    1560.1  V  V  38.98  V V  38.98     409980 

0.2857 1T V sat  Vc Zc r 

10.7985 V sat  124  0.265

0.2857

Z  1  B0

Pr P   B1 r Tr Tr

B 0  0.083 

0.422

B1  0.139 

0.172

Updated 5/23/2017

Tr1.6 Tr4.2

p. 95 of 182


NAS 8th Edition Annotated Solutions

B 0  0.083  B1  0.139 

Chapter 3

0.422 0.79851.6 0.172 0.79854.2

 0.5219  0.3036

Z  1   0.5210  0.069  0.3036 

T (K) 333.15

P (bar) 18.21

0.2362  0.8397 0.7985

Tc (K) 417.2

Pc (bar) 77.1

Tr 0.80 0.80 0.85 0.85

Pr 0.10 0.20 0.10 0.20

ω 0.069

Tr 0.7985

Pr 0.2362

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm3/mol)

0

Z 0.9319 0.8539 0.9436 0.8810 0.8247

1

Z -0.0487 -0.1160 -0.0319 -0.0715 -0.1419

0.8149 1239.7

Updated 5/23/2017

p. 96 of 182


NAS 8th Edition Annotated Solutions

V

Chapter 3

a T  V  b RT b P P V V  b 

 7353000 V  38.98   3 -1 V  1268.6  38.98   cm mol  22.49  V V  38.98     326900 V  38.98   3 -1 V  1307.6   cm mol  V V  38.98   

 RT  bP  VP  V  b  V V  b     a T     83.14  343.15  38.98  22.49  22.49V  V  38.98  V V  38.98    7353000    1307.6  V  V  38.98  V V  38.98     326900 

0.2857 1T V sat  Vc Zc r 

10.8225 V sat  124  0.265

0.2857

Z  1  B0

Updated 5/23/2017

Pr P   B1 r Tr Tr

p. 97 of 182


NAS 8th Edition Annotated Solutions

B 0  0.083 

0.422

B1  0.139 

0.172

B 0  0.083  B1  0.139 

Chapter 3

Tr1.6 Tr4.2

0.422 0.82251.6 0.172 0.82254.2

 0.4939  0.2518

Z  1   0.4939  0.069  0.2518

T (K) 343.15

P (bar) 22.49

0.2917  0.8187 0.8225

Tc (K) 417.2

Pc (bar) 77.1

Tr 0.80 0.80 0.85 0.85

Pr 0.10 0.20 0.10 0.20

ω 0.069

Tr 0.8225

Pr 0.2917

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm3/mol)

Updated 5/23/2017

0

Z 0.9319 0.8539 0.9436 0.8810 0.8009

1

Z -0.0487 -0.1160 -0.0319 -0.0715 -0.1462

0.7908 1003.4

p. 98 of 182


NAS 8th Edition Annotated Solutions

P

Chapter 3

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 T 0.5 R 2Tc2.5 T 0.5  83.1452  417.22.5 1.362  108 a(T )  0.42748 r  0.42748  0.42748  bar cm 6 mol-2 Pc Pc 77.10 T b  0.08664

RTc 83.145  417.2  0.08664  38.98 cm3 mol-1 Pc 77.10

V

a T  V  b RT b P P V V  b 

 7248000 V  38.98   3 -1 V  1070.5  38.98   cm mol  27.43  V V  38.98     264224 V  38.98   3 -1 V  1109.4   cm mol  V V  38.98    

 RT  bP  VP  V  b  V V  b     a T     83.14  353.15  38.98  27.43  27.43V  V  38.98  V V  38.98    7248000    1109.4  V  V  38.98  V V  38.98     264224 

Updated 5/23/2017

p. 99 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

0.2857 1T V sat  Vc Zc r 

10.8465 V sat  124  0.265

0.2857

Z  1  B0

Pr P   B1 r Tr Tr

B 0  0.083 

0.422

B1  0.139 

0.172

B 0  0.083  B1  0.139 

Tr1.6 Tr4.2

0.422 0.84651.6 0.172 0.84654.2

 0.4680  0.2073

Z  1   0.4680  0.069  0.2073

Updated 5/23/2017

0.3558  0.7973 0.8465

p. 100 of 182


NAS 8th Edition Annotated Solutions

T (K) 353.15

P (bar) 27.43

Chapter 3

Tc (K) 417.2

Pc (bar) 77.1

Tr 0.80 0.80 0.85 0.85

Pr 0.10 0.20 0.10 0.20

ω 0.069

Tr 0.8465

Pr 0.3558

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm3/mol)

Z0 0.9319 0.8539 0.9436 0.8810 0.7799

Z1 -0.0487 -0.1160 -0.0319 -0.0715 -0.1394

0.7703 824.6

V

a T  V  b RT b P P V V  b 

 7147000 V  38.98   3 -1 V   912.8  38.98   cm mol  33.08  V V  38.98      216057 V  38.98   3 -1 V   951.7   cm mol  V V  38.98    

Updated 5/23/2017

p. 101 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V V  b     a T     83.14  363.15  38.98  33.08  33.08V  V  38.98  V V  38.98    7147000    951.7  V  V  38.98  V V  38.98     216057 

0.2857 1T V sat  Vc Zc r 

10.8704 V sat  124  0.265

0.2857

Z  1  B0

Pr P   B1 r Tr Tr

B 0  0.083 

0.422

B1  0.139 

0.172

B 0  0.083  B1  0.139 

Updated 5/23/2017

Tr1.6 Tr4.2

0.422 0.87041.6 0.172 0.87044.2

 0.4439  0.1691

p. 102 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

Z  1   0.4439  0.069  0.1691

T (K) 363.15

P (bar) 33.08

0.4291  0.7754 0.8704

Tc (K) 417.2

Pc (bar) 77.1

Tr 0.90 0.90 0.93 0.93

Pr 0.20 0.40 0.20 0.40

ω 0.069

Tr 0.8704

Pr 0.4291

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm3/mol)

T (K) 363.15

P (bar) 33.08

Z0 0.9015 0.7800 0.9115 0.8059

Z1 -0.0442 -0.1118 -0.0326 -0.0763

0.7346

-0.1600

ω 0.069

Tr 0.8704

0.7235 660.5

Tc (K) 417.2

Pc (bar) 77.1

Tr 0.85 0.85 0.90 0.90

Pr 0.10 0.20 0.10 0.20

Pr 0.4291

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm3/mol) Updated 5/23/2017

Z0 0.9436 0.8810 0.9528 0.9015 0.7566

Z1 -0.0319 -0.0715 -0.0205 -0.0442 -0.1361

0.7472 682.1 p. 103 of 182


NAS 8th Edition Annotated Solutions

P

Chapter 3

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 T 0.5 R 2Tc2.5 T 0.5  83.1452  430.82.5 1.4439  108 a(T )  0.42748 r  0.42748  0.42748  bar cm6 mol-2 Pc Pc 78.84 T b  0.08664

RTc 83.145  430.8  0.08664  39.363 cm3 mol-1 Pc 78.84

V

a T  V  b RT b P P V V  b 

 7683000 V  39.363  3 -1 V  1573.6  39.363   cm mol  18.66  V V  39.363    411736 V  39.363  3 -1 V  1613.0   cm mol  V V  39.363    

 RT  bP  VP  V  b  V V  b     a T     83.14  353.15  39.363  18.66  18.66V  V  39.363  V V  39.363   7683000    1613.0  V  V  39.363  V V  39.363    411736  Updated 5/23/2017

p. 104 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

0.2857 1T V sat  Vc Zc r 

10.8198 V sat  122  0.269

0.2857

Z  1  B0

Pr P   B1 r Tr Tr

B 0  0.083 

0.422

B1  0.139 

0.172

B 0  0.083  B1  0.139 

Tr1.6 Tr4.2

0.422 0.81981.6 0.172 0.81984.2

 0.4969  0.2572

Z  1   0.4969  0.245  0.2572 

Updated 5/23/2017

0.2367  0.8282 0.8198

p. 105 of 182


NAS 8th Edition Annotated Solutions

T (K) 353.15

P (bar) 18.66

Chapter 3

Tc (K) 430.8

Pc (bar) 78.84

Tr 0.80 0.80 0.85 0.85

Pr 0.10 0.20 0.10 0.20

ω 0.245

Tr 0.8198

Pr 0.2367

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm 3/mol)

Z0 0.9319 0.8539 0.9436 0.8810

Z1 -0.0319 -0.0715 -0.0205 -0.0442

0.8382

-0.0729

0.8204 1291.0

V

a T  V  b RT b P P V V  b 

 7577000 V  39.363  3 -1 V  1295.3  39.363   cm mol  23.31  V V  39.363    325053 V  39.363  3 -1 V  1334.7   cm mol  V V  39.363    

Updated 5/23/2017

p. 106 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V V  b     a T     83.14  363.15  39.363  23.31  23.31V  V  39.363  V V  39.363   7577000    1334.7  V  V  39.363  V V  39.363    325053 

0.2857 1T V sat  Vc Zc r 

10.8430 V sat  122  0.269

0.2857

Z  1  B0

Pr P   B1 r Tr Tr

B 0  0.083 

0.422

B1  0.139 

0.172

Updated 5/23/2017

Tr1.6 Tr4.2

p. 107 of 182


NAS 8th Edition Annotated Solutions

B 0  0.083  B1  0.139 

0.422 0.84301.6 0.172 0.84304.2

Chapter 3

 0.4726  0.2134

Z  1   0.4726  0.245  0.2134 

T (K) 363.15

P (bar) 23.11

0.2957  0.8159 0.8430

Tc (K) 430.8

Pc (bar) 78.84

Tr 0.80 0.80 0.85 0.85

Pr 0.10 0.20 0.10 0.20

ω 0.245

Tr 0.8430

Pr 0.2931

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm 3/mol)

Z0 0.9319 0.8539 0.9436 0.8810 0.8169

Z1 -0.0319 -0.0715 -0.0205 -0.0442 -0.0722

0.7992 1044.3

Updated 5/23/2017

p. 108 of 182


NAS 8th Edition Annotated Solutions

P

a T  RT  V  b V V  b 

a(T )  0.42748 b  0.08664

Chapter 3

 Tr  R 2Tc2  Tr   83.1452  369.82  0.42748  9.5134  106 Tr  bar cm 6 mol-2 Pc 42.48

RTc 83.145  369.8  0.08664  62.71 cm3 mol-1 Pc 42.48

     T   1   0.480  1.574 * 0.152  0.176 * 0.152 1  0.9279    1.0532 2

 Tr   1  0.480  1.574  0.176 2 1  Tr0.5

2

0.5

2

r

V

a T  V  b RT b P P V V  b 

 10020000 V  62.71  3 -1 V  1099.9  62.71   cm mol  25.94  V V  62.71      386259 V  62.71  3 -1 V  1162.6   cm mol  V V  62.71    

 RT  bP  VP  V  b  V V  b     a T     83.14  343.15  62.17  25.94  25.94V  V  62.71  V V  62.71   10020000    1162.0  V  V  62.71  V V  62.71    386259 

Updated 5/23/2017

p. 109 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

P

a T  RT  V  b V V  b 

a(T )  0.42748 b  0.08664

 Tr  R 2Tc2  Tr   83.1452  425.12  0.42748  1.4068  107 Tr  bar cm6 mol-2 Pc 37.96

RTc 83.145  425.1  0.08664  80.67 cm3 mol-1 Pc 37.96

     T   1   0.480  1.574 * 0.200  0.176 * 0.200 1  0.8778    1.1019 2

 Tr   1  0.480  1.574  0.176 2 1  Tr0.5

2

0.5

2

r

V

a T  V  b RT b P P V V  b 

 15502000 V  80.67   3 -1 V   2013.3  80.67   cm mol  15.41  V V  80.67     1005940 V  80.67   3 -1 V   2094.0   cm mol  V V  80.67   

Updated 5/23/2017

p. 110 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V V  b     a T     83.14  373.15  80.67  15.41  15.41V  V  80.67  V V  80.67    15502000    2094.1  V  V  80.67  V V  80.67     1005940 

P

a T  RT  V  b V V  b 

a(T )  0.42748 b  0.08664

 Tr  R 2Tc2  Tr   83.1452  425.12  0.42748  1.4068  107 Tr  bar cm6 mol-2 Pc 37.96

RTc 83.145  425.1  0.08664  80.67 cm3 mol-1 Pc 37.96

     T   1   0.480  1.574 * 0.200  0.176 * 0.200 1  0.9013    1.0814 2

 Tr   1  0.480  1.574  0.176 2 1  Tr0.5

2

0.5

2

r

V

a T  V  b RT b P P V V  b 

 15212000 V  80.67   3 -1 V  1707.2  80.67   cm mol  18.66  V V  80.67      815252 V  80.67   3 -1 V  1787.9   cm mol  V V  80.67    

Updated 5/23/2017

p. 111 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V V  b     a T     83.14  383.15  80.67  18.66  18.66V  V  80.67  V V  80.67    15212000    1787.9  V  V  80.67  V V  80.67     815252 

P

a T  RT  V  b V V  b 

a(T )  0.42748 b  0.08664

 Tr  R 2Tc2  Tr   83.1452  425.12  0.42748  1.4068  107 Tr  bar cm6 mol-2 Pc 37.96

RTc 83.145  425.1  0.08664  80.67 cm3 mol-1 Pc 37.96

     T   1   0.480  1.574 * 0.200  0.176 * 0.200 1  0.9248    1.0613 2

 Tr   1  0.480  1.574  0.176 2 1  Tr0.5

2

0.5

2

r

V

a T  V  b RT b P P V V  b 

 14930000 V  80.67   3 -1 V  1460.6  80.67   cm mol  22.38  V V  80.67      667113 V  80.67   3 -1 V  1541.3   cm mol  V V  80.67    

Updated 5/23/2017

p. 112 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V V  b     a T     83.14  393.15  80.67  22.38  22.38V  V  80.67  V V  80.67    14930000    1541.3  V  V  80.67  V V  80.67     667113 

     T   1   0.480  1.574 * 0.200  0.176 * 0.200 1  0.9484    1.0416 2

 Tr   1  0.480  1.574  0.176 2 1  Tr0.5

2

0.5

2

r

V

a T  V  b RT b P P V V  b 

 14654000 V  80.67   3 -1 V  1260.6  80.67   cm mol  26.59  V V  80.67      551101V  80.67   3 -1 V  1341.3   cm mol  V V  80.67    

Updated 5/23/2017

p. 113 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V V  b     a T     83.14  403.15  80.67  26.59  26.59V  V  80.67  V V  80.67    14654000    1341.3  V  V  80.67  V V  80.67     551101 

P

a T  RT  V  b V V  b 

a(T )  0.42748 b  0.08664

 Tr  R 2Tc2  Tr   83.1452  430.82  0.42748  6.9565  106 Tr  bar cm6 mol-2 Pc 78.84

RTc 83.145  430.8  0.08664  39.363 cm3 mol-1 Pc 78.84

     T   1   0.480  1.574 * 0.245  0.176 * 0.245 1  0.8662    1.1220 2

 Tr   1  0.480  1.574  0.176 2 1  Tr0.5

2

0.5

2

r

V

a T  V  b RT b P P V V  b 

 7805500 V  39.36   3 -1 V  1079.5  39.36   cm mol  28.74  V V  39.36      271591 V  39.36   3 -1 V  1118.9   cm mol  V V  39.36     Updated 5/23/2017

p. 114 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V V  b     a T     83.14  373.15  39.36  28.74  28.74V  V  39.36  V V  39.36    7805500    1118.9  V  V  39.36  V V  39.36     271591 

     T   1   0.480  1.574 * 0.245  0.176 * 0.245 1  0.8894    1.0997 2

 Tr   1  0.480  1.574  0.176 2 1  Tr0.5

2

0.5

2

r

V

a T  V  b RT b P P V V  b 

 7650000 V  39.36   3 -1 V   909.9  39.36   cm mol  35.01  V V  39.36     218515 V  39.36   3 -1 V   949.3   cm mol  V V  39.36    

Updated 5/23/2017

p. 115 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V V  b     a T     83.14  383.15  39.36  35.01  35.01V  V  39.36  V V  39.36    7650000    949.3  V  V  39.36  V V  39.36     218515 

P

a T  RT  V  b V  2.4142b V  0.4142b 

a(T )  0.45724 a(T )  0.45724

 Tr  R 2Tc2 Pc

 Tr   83.1452  369.82 42.48

a(T )   Tr   1.0176  107 bar cm 6 mol-2 b  0.07779

RTc 83.145  369.8  0.07779  56.3 cm3 mol-1 Pc 42.48

     T   1   0.37464  1.54226 * 0.152  0.2699 * 0.152 1  0.9279    1.0448 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

Updated 5/23/2017

p. 116 of 182


NAS 8th Edition Annotated Solutions

V

Chapter 3

a T  RT V b b P P V  0.4142b V  2.4142b 

  10632000 V  56.3 3 -1 V  1099.9  56.3   cm mol  25.94 V  23.32 V  135.92       409851 V  56.3 3 -1 V  1156.2   cm mol  V  23.32 V  135.92   

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  343.15  56.3  25.94  25.94V  V  56.3  V  23.32 V  135.92    10632000    1156.2  V  V  56.3  V  23.32 V  135.92     409851 

 P

a T  RT  V  b V  2.4142b V  0.4142b 

a(T )  0.45724 a(T )  0.45724

 Tr  R 2Tc2 Pc

 Tr   83.1452  425.12 37.96

a(T )   Tr   1.5048  107 bar cm 6 mol-2 b  0.07779

RTc 83.145  425.1  0.07779  72.43 cm3 mol-1 Pc 37.96

Updated 5/23/2017

p. 117 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

     T   1   0.37464  1.54226 * 0.200  0.2699 * 0.200 1  0.8778    1.0866 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

V

a T  RT V b b P P V  0.4142b V  2.4142b 

  16352000 V  72.43 3 -1 V   2013.3  72.43   cm mol  15.41 V  30.00 V  174.86     1061129 V  72.43  3 -1 V   2085.7   cm mol  V  30.00 V  174.86   

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  373.15  72.43  15.41  15.41V  V  72.43  V  30.00 V  174.96    16352000    2085.7  V  V  72.43  V  30.00 V  174.96     1061129 

P

a T  RT  V  b V  2.4142b V  0.4142b 

Updated 5/23/2017

p. 118 of 182


NAS 8th Edition Annotated Solutions

a(T )  0.45724 a(T )  0.45724

Chapter 3

 Tr  R 2Tc2 Pc

 Tr   83.1452  425.12 37.96

a(T )   Tr   1.5048  107 bar cm 6 mol-2 b  0.07779

RTc 83.145  425.1  0.07779  72.43 cm3 mol-1 Pc 37.96

     T   1   0.37464  1.54226 * 0.200  0.2699 * 0.200 1  0.9013    1.0692 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

V

a T  RT V b b P P V  0.4142b V  2.4142b 

  16090000 V  72.43 3 -1 V  1707.2  72.43   cm mol  18.66 V  30.00 V  174.86       862226 V  72.43  3 -1 V  1779.6   cm mol  V  30.00 V  174.86    

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  383.15  72.43  18.66  18.66V  V  72.43  V  30.00 V  174.96    16090000    1779.6  V  V  72.43  V  30.00 V  174.96     862226 

Updated 5/23/2017

p. 119 of 182


NAS 8th Edition Annotated Solutions

P

Chapter 3

a T  RT  V  b V  2.4142b V  0.4142b 

a(T )  0.45724 a(T )  0.45724

 Tr  R 2Tc2 Pc

 Tr   83.1452  425.12 37.96

a(T )   Tr   1.5048  107 bar cm 6 mol-2 b  0.07779

RTc 83.145  425.1  0.07779  72.43 cm3 mol-1 Pc 37.96

     T   1   0.37464  1.54226 * 0.200  0.2699 * 0.200 1  0.9248    1.0522 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

V

a T  RT V b b P P V  0.4142b V  2.4142b 

  15833000 V  72.43 3 -1 V  1460.6  72.43   cm mol  22.38 V  30.00 V  174.86       707471V  72.43  3 -1 V  1533.0   cm mol  V  30.00 V  174.86    

Updated 5/23/2017

p. 120 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  393.15  72.43  22.38  22.38V  V  72.43  V  30.00 V  174.96    15833000    1533.0  V  V  72.43  V  30.00 V  174.96     707041 

     T   1   0.37464  1.54226 * 0.200  0.2699 * 0.200 1  0.9484    1.0355 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

V

a T  RT V b b P P V  0.4142b V  2.4142b 

  15582000 V  72.43 3 -1 V  1260.6  72.43   cm mol  26.59 V  30.00 V  174.86     586008 V  72.43  3 -1 V  1333.0   cm mol  V  30.00 V  174.86     

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  393.15  72.43  26.59  26.59V  V  72.43  V  30.00 V  174.96    15582000    1333.0  V  V  72.43  V  30.00 V  174.96     586008  Updated 5/23/2017

p. 121 of 182


NAS 8th Edition Annotated Solutions

P

Chapter 3

a T  RT  V  b V  2.4142b V  0.4142b 

a(T )  0.45724 a(T )  0.45724

 Tr  R 2Tc2 Pc

 Tr   83.1452  408.12 36.48

a(T )   Tr   1.4431  107 bar cm 6 mol-2 b  0.07779

RTc 83.145  408.1  0.07779  72.36 cm3 mol-1 Pc 36.48

     T   1   0.37464  1.54226 * 0.302  0.2699 * 0.302 1  0.8899    1.0946 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

V

a T  RT V b b P P V  0.4142b V  2.4142b 

  15796000 V  72.36  3 -1 V  1825.5  72.36   cm mol  16.54 V  29.97 V  174.69      955018 V  72.36   3 -1 V  1897.9   cm mol  V  29.97 V  174.69   

Updated 5/23/2017

p. 122 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  363.15  72.36  16.54  16.54V  V  72.36  V  29.97 V  174.69    15796000    1897.9  V  V  72.36  V  29.97 V  174.69     955018 

     T   1   0.37464  1.54226 * 0.302  0.2699 * 0.302 1  0.9144    1.0727 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

V

a T  RT V b b P P V  0.4142b V  2.4142b 

  15480000 V  72.36  3 -1 V  1549.0  72.36   cm mol  20.03 V  29.97 V  174.69       772841 V  72.36   3 -1 V  1621.3   cm mol  V  29.97 V  174.69    

Updated 5/23/2017

p. 123 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  373.15  72.36  20.03  20.03V  V  72.36  V  29.97 V  174.69    15480000    1621.3  V  V  72.36  V  29.97 V  174.69     772841 

P

a T  RT  V  b V  2.4142b V  0.4142b 

a(T )  0.45724 a(T )  0.45724

 Tr  R 2Tc2 Pc

 Tr   83.1452  408.12 36.48

a(T )   Tr   1.4431  107 bar cm 6 mol-2 b  0.07779

RTc 83.145  408.1  0.07779  72.36 cm3 mol-1 Pc 36.48

     T   1   0.37464  1.54226 * 0.181  0.2699 * 0.181 1  0.9389    1.0405 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

Updated 5/23/2017

p. 124 of 182


NAS 8th Edition Annotated Solutions

V

Chapter 3

a T  RT V b b P P V  0.4142b V  2.4142b 

  15015000 V  72.36  3 -1 V  1326.8  72.36   cm mol  24.01 V  29.97 V  174.69      625364 V  72.36   3 -1 V  1399.2   cm mol  V  29.97 V  174.69   

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  383.15  72.36  24.01  24.01V  V  72.36  V  29.97 V  174.69    15015000    1399.2  V  V  72.36  V  29.97 V  174.69     625364 

     T   1   0.37464  1.54226 * 0.181  0.2699 * 0.181 1  0.9634    1.0240 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

Updated 5/23/2017

p. 125 of 182


NAS 8th Edition Annotated Solutions

V

Chapter 3

a T  RT V b b P P V  0.4142b V  2.4142b 

  14777000 V  72.36  3 -1 V  1145.8  72.36   cm mol  28.53 V  29.97 V  174.69      517946 V  72.36   3 -1 V  1218.1   cm mol  V  29.97 V  174.69   

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  393.15  72.36  28.53  28.53V  V  72.36  V  29.97 V  174.69    14777000    1218.1  V  V  72.36  V  29.97 V  174.69     517946 

P

a T  RT  V  b V  2.4142b V  0.4142b 

Updated 5/23/2017

p. 126 of 182


NAS 8th Edition Annotated Solutions

a(T )  0.45724 a(T )  0.45724

Chapter 3

 Tr  R 2Tc2 Pc

 Tr   83.1452  417.22 77.10

a(T )   Tr   7.136  10 bar cm 6 mol-2 6

b  0.07779

RTc 83.145  417.2  0.07779  35.00 cm3 mol-1 Pc 77.10

     T   1   0.37464  1.54226 * 0.069  0.2699 * 0.069 1  0.7985    1.1047 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

V

a T  RT V b b P P V  0.4142b V  2.4142b 

  7883000 V  35.0  3 -1 V  1521.1  35.00   cm mol  18.21 V  14.50 V  84.49       432903 V  35.00   3 -1 V  1556.1   cm mol  V  14.50 V  84.49    

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  333.15  35.00  18.21  18.21V  V  35.00  V  14.50 V  84.49    7883000    1556.1  V  V  35.00  V  14.50 V  84.49     432903  Updated 5/23/2017

p. 127 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

     T   1   0.37464  1.54226 * 0.069  0.2699 * 0.069 1  0.8225    1.091 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

V

a T  RT V b b P P V  0.4142b V  2.4142b 

  7788000 V  35.0  3 -1 V  1268.6  35.00   cm mol  22.49 V  14.50 V  84.49      346265 V  35.00   3 -1 V  1303.6   cm mol  V  14.50 V  84.49   

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  343.15  35.00  22.49  22.49V  V  35.00  V  14.50 V  84.49    7788000    1303.6  V  V  35.00  V  14.50 V  84.49     317293 

Updated 5/23/2017

p. 128 of 182


NAS 8th Edition Annotated Solutions

P

Chapter 3

a T  RT  V  b V  2.4142b V  0.4142b 

a(T )  0.45724 a(T )  0.45724

 Tr  R 2Tc2 Pc

 Tr   83.1452  417.22 77.10

a(T )   Tr   7.136  10 bar cm 6 mol-2 6

b  0.07779

RTc 83.145  417.2  0.07779  35.00 cm3 mol-1 Pc 77.10

     T   1   0.37464  1.54226 * 0.069  0.2699 * 0.069 1  0.8465    1.078 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

V

a T  RT V b b P P V  0.4142b V  2.4142b 

  7694000 V  35.0  3 -1 V  1070.5  35.00   cm mol  27.43 V  14.50 V  84.49       280490 V  35.00   3 -1 V  1105.5   cm mol  V  14.50 V  84.49   

Updated 5/23/2017

p. 129 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  353.15  35.00  27.43  27.43V  V  35.00  V  14.50 V  84.49    7694000    1105.5  V  V  35.00  V  14.50 V  84.49     280490 

     T   1   0.37464  1.54226 * 0.069  0.2699 * 0.069 1  0.8705    1.065 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

V

a T  RT V b b P P V  0.4142b V  2.4142b 

  7602000 V  35.0  3 -1 V   912.8  35.00   cm mol  33.08 V  14.50 V  84.49      229807 V  35.00   3 -1 V   947.8   cm mol  V  14.50 V  84.49   

Updated 5/23/2017

p. 130 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  363.15  35.00  33.08  33.08V  V  35.00  V  14.50 V  84.49    7602000    047.8  V  V  35.00  V  14.50 V  84.49     229807 

P

a T  RT  V  b V  2.4142b V  0.4142b 

a(T )  0.45724 a(T )  0.45724

 Tr  R 2Tc2 Pc

 Tr   83.1452  430.82 78.84

a(T )   Tr   7.441  10 bar cm 6 mol-2 6

b  0.07779

RTc 83.145  430.8  0.07779  35.34 cm3 mol-1 Pc 78.84

     T   1   0.37464  1.54226 * 0.245  0.2699 * 0.245 1  0.8198    1.1441 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

Updated 5/23/2017

p. 131 of 182


NAS 8th Edition Annotated Solutions

V

Chapter 3

a T  RT V b b P P V  0.4142b V  2.4142b 

  8513000 V  35.34  3 -1 V  1573.6  35.34   cm mol  18.66 V  14.64 V  85.32      456217 V  35.34   3 -1 V  1608.9   cm mol  V  14.64 V  85.32   

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  353.15  35.34  18.66  18.66V  V  35.34  V  14.64 V  85.32    8513000    1608.9  V  V  35.34  V  14.64 V  85.32     456217 

     T   1   0.37464  1.54226 * 0.245  0.2699 * 0.245 1  0.8430    1.1242 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

Updated 5/23/2017

p. 132 of 182


NAS 8th Edition Annotated Solutions

V

Chapter 3

a T  RT V b b P P V  0.4142b V  2.4142b 

  8365000 V  35.34  3 -1 V  1306.5  35.34   cm mol  23.11 V  14.64 V  85.32      361969 V  35.34   3 -1 V  1341.9   cm mol  V  14.64 V  85.32   

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  363.15  35.34  23.11  23.11V  V  35.34  V  14.64 V  85.32    8365000    1341.9  V  V  35.34  V  14.64 V  85.32     361969 

P

a T  RT  V  b V  2.4142b V  0.4142b 

Updated 5/23/2017

p. 133 of 182


NAS 8th Edition Annotated Solutions

a(T )  0.45724 a(T )  0.45724

Chapter 3

 Tr  R 2Tc2 Pc

 Tr   83.1452  430.82 78.84

a(T )   Tr   7.441  10 bar cm 6 mol-2 6

b  0.07779

RTc 83.145  430.8  0.07779  35.34 cm3 mol-1 Pc 78.84

     T   1   0.37464  1.54226 * 0.245  0.2699 * 0.245 1  0.8662    1.1047 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

V

a T  RT V b b P P V  0.4142b V  2.4142b 

  8220000 V  35.34  3 -1 V  1079.5  35.34   cm mol  28.74 V  14.64 V  85.32       285997 V  35.34   3 -1 V  1114.9   cm mol  V  14.64 V  85.32    

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  373.15  35.34  28.74  28.74V  V  35.34  V  14.64 V  85.32    8220000    1114.9  V  V  35.34  V  14.64 V  85.32     285997  Updated 5/23/2017

p. 134 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

     T   1   0.37464  1.54226 * 0.245  0.2699 * 0.245 1  0.8894    1.0856 2

 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5

2

0.5

2

r

V

a T  RT V b b P P V  0.4142b V  2.4142b 

  8078000 V  35.34  3 -1 V   909.9  35.34   cm mol  35.01 V  14.64 V  85.32      230772 V  35.34   3 -1 V   945.2   cm mol  V  14.64 V  85.32   

 RT  bP  VP  V  b  V  0.4142b V  2.4142b     a T     83.14  383.15  35.34  35.01  35.01V  V  35.34  V  14.64 V  85.32    8078000    945.2  V  V  35.34  V  14.64 V  85.32     230772 

Updated 5/23/2017

p. 135 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

Updated 5/23/2017

p. 136 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

B 0  0.083  B1  0.139 

Updated 5/23/2017

0.422 1.46561.6 0.172 1.46564.2

 0.1459  0.1045

p. 137 of 182


NAS 8th Edition Annotated Solutions

Z  1   0.1459  0.645  0.1045

Chapter 3

0.9759  0.9477 1.4656

0.2857 0.2857 1T V sat  Vc Zc r   200.0*0.276(1320 / 369.8)  96.8 cm3/mol

Updated 5/23/2017

p. 138 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 

Updated 5/23/2017

p. 139 of 182


NAS 8th Edition Annotated Solutions

T (K) 333.15

Chapter 3

P (bar) 140

Tc (K) 305.3

Pc (bar) 48.72

Tr 1.05 1.05 1.10 1.10

Pr 2.00 3.00 2.00 3.00

Z0 0.3452 0.4604 0.3953 0.4770

0.1

Tr 1.0912

Pr 2.8736

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm3/mol)

Updated 5/23/2017

0.4630

Z1 -0.0432 -0.0838 0.0698 -0.0373 -0.0334

0.4597 91.0

p. 140 of 182


NAS 8th Edition Annotated Solutions

T (K) 388.8

Chapter 3

P (bar) 200

Tc (K) 305.3

Pc (bar) 48.72

Tr 1.40 1.40 1.50 1.50

Pr 3.00 5.00 3.00 5.00

Z0 0.7202 0.7761 0.7887 0.8200

0.1

Tr 1.2735

Pr 4.1051

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm3/mol)

0.6816

Z1 0.2397 0.1737 0.2433 0.2309 0.1612

0.6978 112.8

454g 33.9975g/mol

P

a T  RT  V  b V V  b  .4278∗ Tr −0.5 ∗R2 ∗Tc2 Pc .4278∗ 0.9021 −0.5 ∗83.1452 ∗324.82 65.4 R∗Tc 83.145∗324.8 0.08664 ∗ Pc 65.4

P

a T  RT  V  b V V  b  5022832.543 179.7(179.7+35.776)

Updated 5/23/2017

p. 141 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

454g 67.825g/mol

P

a T  RT  V  b V V  b  .4278∗ Tr −0.5 ∗R2 ∗Tc2 Pc .4278∗ 1.1230 −0.5 ∗83.1452 ∗260.92 49.9 R∗Tc 83.145∗260.9 0.08664 ∗ Pc 49.9

P

a T  RT  V  b V V  b  358.5(358.5+37.6641)

454g 32.117 g/mol

P

a T  RT  V  b V V  b  .4278∗ Tr −0.5 ∗R2 ∗Tc2 Pc .4278∗ 1.0864 −0.5 ∗83.1452 ∗269.72 48.4

Updated 5/23/2017

p. 142 of 182


NAS 8th Edition Annotated Solutions

R∗Tc Pc

P

0.08664 ∗

Chapter 3

83.145∗269.7 48.4

a T  RT  V  b V V  b  169.78(169.78+40.14)

454g 76.62 g/mol

P

a T  RT  V  b V V  b  .4278∗ Tr −0.5 ∗R2 ∗Tc2 Pc .4278∗ 0.9385 −0.5 ∗83.1452 ∗312.22 49.5 R∗Tc 83.145∗312.2 0.08664 ∗ Pc 49.5

P

a T  RT  V  b V V  b  405.06(405.06+45.434)

454g 77.945 g/mol

Updated 5/23/2017

p. 143 of 182


NAS 8th Edition Annotated Solutions

P

Chapter 3

a T  RT  V  b V V  b  .4278∗ Tr −0.5 ∗R2 ∗Tc2 Pc .4278∗ 0.7855 −0.5 ∗83.1452 ∗3732 65.5 R∗Tc 83.145∗373 0.08664 ∗ Pc 65.5

P

a T  RT  V  b V V  b  412.044(412.044+41.022)

454g 71 g/mol

P

a T  RT  V  b V V  b  .4278∗ Tr −0.5 ∗R2 ∗Tc2 Pc .4278∗ 1.2521 −0.5 ∗83.1452 ∗2342 44.6 R∗Tc 83.145∗234 0.08664 ∗ Pc 44.6

P

a T  RT  V  b V V  b  375.35(375.35+37.795)

Updated 5/23/2017

p. 144 of 182


NAS 8th Edition Annotated Solutions

B 0 = 0.083 

0.422 =  0.415 Tr1.6

B1 = 0.139 

0.172 =  0.126 Tr4.2

Chapter 3

BPc = B 0 +  B1 =  0.415 + 0.045 ×  0.126  =  0.42067 RTc  BP  P 0.3822 Z =1 +  c  r =1  0.42067 × = 0.8218 0.9021  RTc  Tr MWphosphine = 33.997 g·mol1

V = 2.4 L = 2400 cm3

Using this,

PV 25 bar × 2400 cm3 1 mass = MWPhosphine × = 33.997 g·mol × = 101.89 g ZRT 0.8218 × 83.14 bar·cm3·mol1·K 1 × 293 K

Updated 5/23/2017

p. 145 of 182


NAS 8th Edition Annotated Solutions

B 0 = 0.083 

0.422 =  0.2675 Tr1.6

B1 = 0.139 

0.172 = 0.03333 Tr4.2

Chapter 3

BPc = B 0 +  B1 =  0.2675 + 0.434 ×  0.03333 =  0.2530 RTc  BP  P 0.5010 Z =1 +  c  r =1  0.2530 × = 0.887 1.1230  RTc  Tr MWBF3 = 67.82 g·mol1

V = 2.4 L = 2400 cm3

Using this,

PV 25 bar × 2400 cm3 1 mass = MWBF3 × = 67.82 g·mol × = 188.32 g ZRT 0.887 × 83.14 bar·cm3·mol1·K 1 × 293 K

Updated 5/23/2017

p. 146 of 182


NAS 8th Edition Annotated Solutions

B 0 = 0.083 

0.422 =  0.2866 Tr1.6

B1 = 0.139 

0.172 = 0.01756 Tr4.2

Chapter 3

BPc = B 0 +  B1 =  0.2866 + 0.094 ×  0.01756  =  0.2849 RTc  BP  P 0.51652 Z =1 +  c  r =1  0.2849 × = 0.8645 1.0864  RTc  Tr MWSiH4 = 32.12 g·mol1

V = 2.4 L = 2400 cm3

Using this,

PV 25 bar × 2400 cm3 1 mass = MWSiH4 × = 32.12 g·mol × = 91.499 g ZRT 0.8645 × 83.14 bar·cm3·mol1·K 1 × 293 K

Updated 5/23/2017

p. 147 of 182


NAS 8th Edition Annotated Solutions

B 0 = 0.083 

0.422 =  0.384 Tr1.6

B1 = 0.139 

0.172 =  0.0855 Tr4.2

Chapter 3

BPc = B 0 +  B1 =  0.384 + 0.151 ×  0.0855  =  0.3969 RTc  BP  P 0.5051 Z =1 +  c  r =1  0.3969 × = 0.7863 0.9385  RTc  Tr MWGeH4 = 76.62 g·mol1

V = 2.4 L = 2400 cm3

Using this,

PV 25 bar × 2400 cm3 1 mass = MWGeH4 × = 76.62 g·mol × = 239.99 g ZRT 0.7863 × 83.14 bar·cm3·mol1·K 1 × 293 K

Updated 5/23/2017

p. 148 of 182


NAS 8th Edition Annotated Solutions

B 0 = 0.083 

0.422 =  0.5379 Tr1.6

B1 = 0.139 

0.172 =  0.3351 Tr4.2

Chapter 3

BPc = B 0 +  B1 =  0.5379 + 0.011 ×  0.3351 =  0.5416 RTc  BP  P 0.38168 Z =1 +  c  r =1  0.5416 × = 0.7368 0.7855  RTc  Tr MWAsH3 = 77.945 g·mol1

V = 2.4 L = 2400 cm3

Using this,

mass = MWAsH3 ×

Updated 5/23/2017

PV 25 bar × 2400 cm3 = 77.945 g·mol1 × = 260.54 g ZRT 0.7368 × 83.14 bar·cm3·mol1·K 1 × 293 K

p. 149 of 182


NAS 8th Edition Annotated Solutions

B 0 = 0.083 

0.422 =  0.2115 Tr1.6

B1 = 0.139 

0.172 = 0.07209 Tr4.2

Chapter 3

BPc = B 0 +  B1 =  0.2115 + 0.120 × (0.07209) = -0.2028 RTc  BP  P 0.5605 Z =1 +  c  r =1  0.2028 × = 0.9092 1.2521  RTc  Tr MWNF3 = 71.00 g·mol1

V = 2.4 L = 2400 cm3

Using this,

mass = MWNF3 ×

Updated 5/23/2017

PV 25 bar × 2400 cm3 = 71 g·mol1 × = 192.33 g ZRT 0.9092 × 83.14 bar·cm3·mol1·K 1 × 293 K

p. 150 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

P  Patmosphere   gh  1.01325bar  1027kg / m3 *9.8m / s 2 *300m

 

P

a T  RT  V  b V V  b 

Updated 5/23/2017

p. 151 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

0.5

1.0561 T 0.5 R 2Tc2 a(T )  0.42748 r  0.42748 Pc b  0.08664

P

 83.1452  282.32 50.4

 4547019 bar cm 6 mol-2

RTc 83.145  282.3  0.08664  40.35 cm3 mol-1 Pc 50.4

83.145 * 298.15 4547019   85.3 bar 105.2  40.35 105.2 105.2  40.35 

Updated 5/23/2017

p. 152 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 

Updated 5/23/2017

p. 153 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

Updated 5/23/2017

p. 154 of 182


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 155 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

BP Bˆ  c  B 0   B1 RTc B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

 Updated 5/23/2017

p. 156 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

P

a T  RT  V  b V V  b 

T T 0.5 R 2Tc2 a(T )  0.42748 r  0.42748 Pc

0.5

R 2Tc2.5 Pc

 0.42748

T 0.5  83.1452  126.22.5 34.00

1.5551  107

bar cm6 mol-2

T

RT 83.145  126.2 b  0.08664 c  0.08664  26.74 cm3 mol-1 Pc 34.00

P

RT 900619 83.145 * 298.15 900619     450.0 bar V  26.74 V V  26.74  69.03  26.74 69.03  69.03  26.74 

Updated 5/23/2017

p. 157 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

Z  Z0   Z 1 = 0.322

Updated 5/23/2017

p. 158 of 182


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 159 of 182


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 160 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

For Z 0 9 and for approximately equal error in the five variables, a 1% maximum error in B requires errors in the variables of less than about 0.02%. This is because the divisor Z 1 0 1. In the limit as Z 1, the error in B approaches infinity.

Updated 5/23/2017

p. 161 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

Updated 5/23/2017

p. 162 of 182


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 163 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 (Tr ) 1/2 (Tr ) 3/2 0.42748 × (0.821) 1.5 q= = = = 6.6326 Tr  0.08664

P 0.294  =  r = 0.08664 × = 0.03102 Tr 0.821 Z =1+ -q × ×

Z - Z Z +  

Z = 1 + 0.03102 - 6.6326 × 0.03102 ×

Updated 5/23/2017

Z - 0.03102 = 0.7904 Z  Z + 0.03102 

p. 164 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

n  1.5 108 / 0.8366  1.79 108 mol/day = 7470 kmol/hr

Updated 5/23/2017

p. 165 of 182


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 166 of 182


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 167 of 182


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 168 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

180 160 140 120

Zc

100 80

y = -0.4578x + 92.712

60 40 20 0 -20 0

50

100

150

200

250

ω

Updated 5/23/2017

p. 169 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

 

 

   

 

 

 Updated 5/23/2017

 p. 170 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

  

 

Updated 5/23/2017

p. 171 of 182


NAS 8th Edition Annotated Solutions

Z-1/ρ

P Kpa

V (cm^3/g)

-118 -119 -120 -121 -122 -123 -124 -125 -126 -127

Chapter 3

/cm^3)

Z-1/

Z

2.63188E-05 0.996686

-125.9023437

3.1602E-05 0.996072

-124.2881683

3.6891E-05 0.995477

-122.6047371

4.21856E-05 0.994901

-120.8796292

4.74895E-05 0.994258

-120.9085917

5.28002E-05 0.993616

-120.9160244

5.81137E-05 0.993044

-119.697015

6.34358E-05 0.992433

-119.2916276

y = 166681x - 129.29

0

0.00002

0.00004

0.00006

0.00008

ρ (g/cm^3)

P Kpa

V (cm^3/g)

Updated 5/23/2017

/cm^3)

Z

Z-1/

2.41874E-05

0.997494

-103.6016566

2.90371E-05

0.997077

-100.6644948

3.38917E-05

0.996634

-99.32289962 p. 172 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

3.87526E-05

0.996138

-99.64691421

4.36138E-05

0.995747

-97.50696627

4.84847E-05

0.995235

-98.28634491

5.33532E-05

0.994861

-96.32176525

5.82313E-05

0.994386

-96.4013883

-95 -96

y = 189330x - 106.77

-97

Z-1/ρ

-98 -99 -100 -101 -102 -103 -104 0

0.00001

0.00002

0.00003

0.00004

0.00005

0.00006

0.00007

ρ (g/cm^3)

P Kpa

V (cm^3/g)

Updated 5/23/2017

/cm^3)

Z

Z-1/

2.23782E-05

0.998061

-86.63762145

2.68638E-05

0.997691

-85.94703438

3.13504E-05

0.997393

-83.14166567

3.58409E-05

0.997064

-81.92792085

4.03347E-05

0.996726

-81.17728694

4.48322E-05

0.996372

-80.92926009

4.93334E-05

0.99601

-80.88365086

5.38343E-05

0.995712

-79.65046047

p. 173 of 182


Z-1/ρ

NAS 8th Edition Annotated Solutions

-78 -79 -80 -81 -82 -83 -84 -85 -86 -87 -88

Chapter 3

y = 216165x - 90.772

0

0.00001

0.00002

0.00003

0.00004

0.00005

0.00006

ρ (g/cm^3)

1 + (1 - 0 -1) =3Zc

Updated 5/23/2017

p. 174 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

0 × 1 ×  2 -  0 +1 ( + 1) +  = 3(0.333) 2 or

1

( + 1) +  = 0.333 and

0 × 1 ×  2   + 1 +  = (0.333)3  = 0.03693

(2)

solving equations 1 and  2  , we get  = 0.086640  = 0.427480

1 + (1 - (1 - 2) - (1 + 2)) = 3Z c 1 + (1 - 1 + 2 - 1 - 2)) = 3Z c 1 -  = 3Z c 1

(1 - 2) 1 + 2  2 - 1 - 2 + 1 + 2 ( + 1) +  = 3Z c2 (1 - 2) 2 - 1 + 1  ( + 1) +  = 3Z c2 3 2 - 2 +  = 3Z c2 1 -   3 - 2 +  = 3    3  9 2 - 6 + 3 =1 +  2 - 2 2

2

10 2 - 4 + 3 =1

Updated 5/23/2017

p. 175 of 182


NAS 8th Edition Annotated Solutions

=

1 + 102 + 4 3

Chapter 3

(2)

(1 - 2) 1 + 2 2   + 1 +  = Z c3  1 + 102 + 4   1  3 -  +   =   3  3   

3

(3)

By solving equation (3), we get

 = 0.33202 Also, from equation (1), we get

1 - 0.33202 = 3Z c Z c =0.2227 and from equation (2), we get

=

1 + 10(0.33202) 2 + 4  0.33202  3

 = 1.1435

Updated 5/23/2017

p. 176 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

-0.2

(Z-1)*Z*Tr/ Pr

-0.22

y = 0.0364x - 0.333

-0.24 -0.26 -0.28 -0.3 -0.32 -0.34 0

0.5

1

1.5

2

2.5

3

3.5

Pr/Z*Tr

Updated 5/23/2017

p. 177 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

ˆ  0.339  0* 0.033  0.339 B

Updated 5/23/2017

p. 178 of 182


NAS 8th Edition Annotated Solutions

Updated 5/23/2017

Chapter 3

p. 179 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

Z

1 a b a   1  1  b RT 1  b RT

Z rep 

b a and Z atr  1  b RT

First, let’s put a and b equal to zero and check if it reduces to the ideal gas equation or not. If equation reduces to the ideal gas equation, then the modification is reasonable.

a)

P=

RT a V -b V

when a = 0 and b = 0, then P=

RT V

Updated 5/23/2017

p. 180 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

The equation reduces to the ideal gas equation. Therefore, the modification is reasonable.

b) P =

RT a (V - b)2 V

when a = 0 and b = 0, then P=

RT V2

The equation does not reduce to the ideal gas equation. Therefore, the modification is not reasonable. c) P =

RT a - 2 V V - b  V

when a = 0 and b = 0, then P=

RT V2

The equation does not reduce to the ideal gas equation. Therefore, the modification is not reasonable. d) P =

RT a - 2 V V

when a = 0 and b = 0, then

P=

RT V

The equation reduces to the ideal gas equation. Therefore, the modification is reasonable.

Updated 5/23/2017

p. 181 of 182


NAS 8th Edition Annotated Solutions

Chapter 3

dV   dT   dP V

V  ln    T  P  Vo 

 

 

 Updated 5/23/2017

p. 182 of 182


SVNAS 8th Edition Annotated Solutions

Chapter 4

4.1.(7th edition prob. 4.1) If we have set up a spreadsheet to evaluate the heat capacity integral like the example spreadsheet from the lecture notes, we can use it to evaluate the heat capacity integral in each case, and simply enter the heat capacity parameters and read out the value of the integral: T

 T  T0   0 

 R dT = A T  T  + 2 T  T  + 3 T  T  + D  TT CP

B

2

0

C

2 0

3

3 0

T0

T

CP

 R dT  ICPH T , T , A, B, C , D  0

T0

Q = H = n × R × ICPH

(a) If 10 mol of SO2 is heated from 200 to 1100C, the heat requirement will be 10*R*ICPH, where ICPH is the heat capacity integral. Using our spreadsheet to evaluate ICPH, we get T1 (K)

T2 (K)

A

B (1/K)

C (1/K ) 2

D (K ) 2

ICPH (K)

473.15 1373.15 5.699 8.01E-04 0 -1.02E+05 5.65E+03 -1 -1 So, the heat requirement is 10 mol*8.314 J mol K *5650 K = 469700 J = 470 kJ (b) Similarly, if we heat 12 moles of propane from 250 to 1200 C, the heat requirement is 12*R*ICPH. Putting the heat capacity parameters for propane and these temperatures into the ICPH spreadsheet gives T1 (K)

T2 (K)

A

B (1/K)

C (1/K2)

D (K2)

ICPH (K)

523.15 1473.15 1.213 2.88E-02 -8.82E-06 0.00E+00 1.95E+04 So, the total heat requirement is 12 mol*8.314 J mol-1 K-1*19500 K = 1.95106 J = 1950 kJ 4.2 (7th edition Prob. 4.2) If we have set up a spreadsheet to evaluate the heat capacity integral like the example spreadsheet from the lecture notes, we can use it to evaluate the heat capacity integral in each case, and simply vary the final temperature until we get the desired Q. (a) If 800 kJ is added to 10 mol of ethylene, then Q = H = 800000 J/10 mol = 80000 J mol . So, H/R = -1

80000 J mol-1 /8.314 J mol-1 K-1 = 9622 K, so the integral of Cp/R from 200C (473 K) to the final temperature should be 9622 K. Using our spreadsheet to evaluate this, we get Updated 18/01/2017 1 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

T1 (K)

T2 (K)

A

473

1374.4

1.424

B (1/K)

C (1/K2)

D (K2)

1.44E-02 -4.39E-06 0.00E+00

ICPH (K) 9.622E+03

So, the final temperature is 1374 K. (b) Similarly, if we add 2500 kJ to 15 mol of 1-butene, we have H/R = 2500000 J / (15 mol *8.314 J mol1

K-1 )= 20045 K. Putting the heat capacity coefficients for 1-butene into the spreadsheet and trying

final temperatures until we get this value, we get T1 (K)

T2 (K)

A

B (1/K)

C (1/K2)

D (K2)

ICPH (K)

533 1414 1.967 3.16E-02 -9.87E-06 0.00E+00 2.005E+04 So, the final temperature is 1414 K. (c) This is the same, but in English units, in which we can use R = 1.986 Btu lbmol-1 R-1, so H/R = 106 Btu / (40 lbmol *1.986 Btu lbmol-1 R-1) = 12588 R = 6993 K. The initial temperature of 500 F is 533 K. So, we get T1 (K)

T2 (K)

A

533

1202.7

1.424

B (1/K)

C (1/K2)

D (K2)

1.44E-02 -4.39E-06 0.00E+00

ICPH (K) 6.994E+03

And the final temperature is 1203 K = 1705 F. 4.3 (New) To compute the outlet temperature, use

But first the heat capacity at 373.15 K must be determine for each compound using the Table C.1 and Eqn 4.5.

This can easily be done in a spreadsheet.

a b c

species Methane Ethane Propane

d

n-Butane

e

n-Hexane

f

n-Octane

A 103 B 106 C 10−5 D 1.702 9.081 -2.164 0 1.131 19.225 -5.561 0 1.213 28.785 -8.824 0 1.935 36.915 11.402 0 3.025 53.722 16.791 0 4.108 70.567 22.208 0

T out Cp *R (K) 4.79 39.82 674.52 7.53 62.61 564.82 10.73 89.17 507.72 14.12 117.41

475.35

20.73 172.38

442.76

27.35 227.37

425.93

Updated 18/01/2017 2 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

g

Propylene

1.637 22.706

h

1-Pentene

2.691 39.753

i

1-Heptene

3.768 56.588

j k

1-Octene Acetylene

l m

Benzene Ethanol

4.324 64.96 6.132 1.952 0.206 39.064 3.518 20.001

n o p

Styrene Formaldehyde Ammonia Carbon monoxide Carbon dioxide Sulfur dioxide Water Nitrogen Hydrogen cyanide

q r s t u v

2.05 50.192 2.264 7.022 3.578 3.02

-6.915 12.447 17.847 20.521 0 13.301 -6.002 16.662 -1.877 0

0

9.15

76.05

530.95

0 15.79 131.29

464.55

0 22.40 186.22

437.59

0 25.71 213.72 -1.299 6.86 57.04

429.30 583.54

0 12.52 104.08 0 10.15 84.35

488.45 515.41

0 18.46 153.47 0 4.62 38.43 -0.186 4.70 39.12

451.34 685.37 679.92

3.376 5.457 5.699 3.47 3.28

0.557 1.045 0.801 1.45 0.593

0 0 0 0 0

-0.031 -1.157 -1.015 0.121 0.04

3.58 5.85 6.00 4.01 3.50

29.80 48.61 49.87 33.35 29.11

775.89 620.01 613.79 732.99 785.38

4.736

1.359

0

-0.725

5.24

43.59

648.43

4.4 (7th edition Prob. 4.3) Looking in table A.2, we are reminded that the gas constant is R = 0.7302 ft3 atm (lb mol)-1 R-1 in the sort of units used in this problem, so we can convert the volumetric flow rate to a molar flow rate using the ideal gas law (air at atmospheric conditions is very nearly an ideal gas). n = PV/RT = 1 atm * 250 ft s / (10.7302 ft atm lbmol R * (122R + 459.7R ) = 3

-1

3

-1

-1

n = 0.59 lbmol s-1 To find the heat needed to heat the air at constant pressure from 122F to 932 F, we need to integrate the heat capacity over that temperature range 932 F

Q

 nC dT p

122 F

The ideal gas heat capacity for air is given in table C.1 as

C p  R A  BT  DT 2

Updated 18/01/2017 3 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

T must be in Kelvins for use in this expression, so we convert 122F to 323.15 K and 932 F to 773.15 K. We then have 773 K

 A  BT  DT dT 2

Q  nR

323 K

773 K

B D  Q  nR  AT  T 2   2 T  323 K  -1 Putting in n = 0.59 lbmol s , R = 1.986 Btu lbmol-1 R-1, A = 3.355, B = 0.57510-3 K-1, D = -1600 K2 gives

 1   1 Q  0.59 lbmol s-1 *1.986 Btu lbmol-1 R -1 3.355  773  323  2.88  104 7732  3232  1600     K  1.8 R/K 773 323    Q  3478 Btu s-1

Note that because the heat capacity integral comes out with units of K, while we are using R with units of R, we have to multiply by 1.8 R/K. 4.5 (7th edition Prob. 4.4) How much heat is required when 10,000 kg of CaCO3 is heated at atmospheric pressure from 50°C to 880°C? The number of moles of CaCO3 is 10 g/ 100.09 g mol = 99910.08 mol CaCO3. Evaluating the heat 7

-1

capacity integral from 323 to 1153 K, we get H/R = 11350 K T1 (K)

T2 (K)

A

B (1/K)

323

1153

12.572

C (1/K2)

D (K2)

ICPH (K)

2.64E-03 0.00E+00 -3.12E+05 1.182E+04

So, 99910.08 mol * 8.314 J mol K * 11350 K = 9.43109 J = 9.43106 kJ. -1

-1

4.6. (7 edition prob. 4.5) th

For consistency with the problem statement, we rewrite Eq. (4.8) as:

C   A  B2 T   1  C3 T     1 p

where

1

2 1

2

T2/T1. DefineCPam as the value of CP evaluated at the arithmetic mean temperature Tam.

Then: Where

Whence, Updated 18/01/2017 4 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Define δ as the difference between the two heat capacities:

This readily reduces to: Making the substitution τ = T2/T1 yields the required answer. 𝑇2/𝑇1

=

𝑇2

𝑇1

4.7. (7th edition prob. 4.6) For consistency with the problem statement, we rewrite Eq. (4.8) as

C   A  B2 T   1   DT p

1

2 1

where τ ≡ T2/ T1. Define CPam as the value of CP evaluated at the arithmetic mean temperature Tam.

As in the preceding problem,

Whence,

Define δ as the difference between the two heat capacities:

This readily reduces to:

Updated 18/01/2017 5 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Making the substitution τ = T2/ T1 yields the required answer.

𝑇1 𝑇1 4.8. (7th edition prob. 4.7) Let step 12 represent the initial reversible adiabatic expansion, and step 23 the final constant-volume heating.

Given

P  T2  T3  2   P3 

R /Cp

Solve for Cp

4.9 (7th edition prob. 4.8) Except for the noble gases [Fig. (4.1)], CP increases with increasing T . Therefore, the estimate is likely to be low. 4.10 (7th edition Prob. 4.9) (a) In this part, we use equation 4.14 reduced temperature at another reduced temperature. The critical temperature for n-pentane (from appendix B) is 469.7 K. So, 25C corresponds to a reduced temperature of Tr1 = 298.15/469.7 = 0.6348. The normal boiling point for n-pentane (also from appendix B) is 309.2 K, which corresponds to a reduced temperature Tr2 = 309.2/469.7 = 0.6583. So, using equation 4.14, we have

Updated 18/01/2017 6 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

 1  Tr 2  H 2  H1    1  Tr1 

0.38

Chapter 4

 1  0.6583   366.3    1  0.6348 

0.38

 357.1 J g -1

This is essentially exact agreement with the handbook value (it differs by 0.03%). (b) In this part, we estimate the heat of vaporization at the normal boiling point without using the information on the heat of vaporization at 25C, using equation 4.13 also use the critical pressure, Pc = 33.70 bar for n-pentane. Then, using equation 4.13:

 1.092  ln Pc  1.013   1.092  ln 33.7  1.013  -1 H n  RTn    8.314*309.2    25876 J mol 0.930  T 0.930  0.6583 rn     Dividing by the molecular weight of n-pentane (72.15 g/mol) gives Hn = 358.6 J g . This is just 0.4% -1

higher than the handbook value, and it only required knowledge of the critical properties and the normal boiling point temperature. 4.11. (Like 7th edition 4.10, but in different units, solutions below are for 7th edition version and We want to evaluate the latent heat of vaporization using the Clapeyron equation: dP sat dT (a) At -16C, we see that the vapor volume is 0.12551 m3kg-1 and the liquid volume is 0.000743 m3kgH  T V

, so V = 0.124767 m3kg-1. Now, we can estimate the slope of the vapor pressure curve from the

1

values at -18, -16, and -14 C, which are 1.446,1.573, and 1.708 bar, respectively. So, we could estimate the slope as dP sat P 1.708bar  1.446bar    00.0655bar.K -1  0.0655*102 kPa.K 1 dT T (273.15  (14)) K  (273.15  (18)) K

So, we have H  (273.15  (16))K *0.124767m3.kg 1 *0.0655*102 kPa.K1  210.4kPa.m3 .kg1  210.4kJ.kg1

(b) At 0C, we see that the vapor volume is 0.069309 m3kg-1 and the liquid volume is 0.000722 m3kg, so V = 0.068587 m3kg-1. Now, we can estimate the slope of the vapor pressure curve from the

1

values at -2, 0, and 2 C, which are 2.722,2.928, and 3.146 bar, respectively. So, we could estimate the slope as dP sat P 3.146bar  2.722bar    0.106bar.K -1  0.106*102 kPa.K 1 dT T (273.15  (2)) K  (273.15  (2)) K

So, we have Updated 18/01/2017 7 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

H  (273.15  (0))K *0.068587m3.kg 1 *0.106*102 kPa.K 1  198.59kPa.m3 .kg1  198.59kJ.kg1

(c) At 12C, we see that the vapor volume is 0.046332 m3kg-1 and the liquid volume is 0.000797 m3kg1 , so V = 0.045535 m3kg-1. Now, we can estimate the slope of the vapor pressure curve from the values at 10, 12, and 14 C, which are 4.146,4.43, and 4.732 bar, respectively. So, we could estimate the slope as dP sat P 4.732bar  4.146bar    0.1465bar.K -1  0.1465*102 kPa.K 1 dT T (273.15  (14)) K  (273.15  (10)) K So, we have H  (273.15  (12))K *0.045535m3.kg 1 *0.1465*102 kPa.K 1  190.22kPa.m3 .kg1  190.22kJ.kg1

(d) At 26C, we see that the vapor volume is 0.029998 m3kg-1 and the liquid volume is 0.000831 m3kg, so V = 0.029167 m kg . Now, we can estimate the slope of the vapor pressure curve from the

1

3

-1

values at 24, 26, and 28 C, which are 6.458,6.854, and 7.269 bar, respectively. So, we could estimate the slope as dP sat P 7.269bar  6.458bar    0.20275bar.K -1  0.20275*102 kPa.K 1 dT T (273.15  (28)) K  (273.15  (24)) K

So, we have H  (273.15  (26)K *0.029167m3.kg 1 *0.20275*102 kPa.K 1  176.91kPa.m3 .kg1  176.91kJ.kg1

(e) At 40C, we see that the vapor volume is 0.019966 m3kg-1 and the liquid volume is 0.000872 m3kg1 , so V = 0.019094 m3kg-1. Now, we can estimate the slope of the vapor pressure curve from the values at 35, 40, and 45 C, which are 8.87, 10.166 and 11.599 bar, respectively. So, we could estimate the slope as dP sat P 11.599bar  8.87bar    0.2729bar.K -1  0.2729*102 kPa.K 1 dT T (273.15  (45)) K  (273.15  (35)) K

So, we have H  (273.15  (40)K *0.019094m3.kg 1 *0.2729*102 kPa.K 1  163.17kPa.m3 .kg1  163.17kJ.kg1

4.12 (7th edition Prob. 4.11) Updated 18/01/2017 8 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

(a) Equation 4.14 is used to estimate the latent heat at a temperature of interest from the (known) latent heat at some other temperature, and may be written as: 0.38

 1  Tr 2  H 2  H1    1  Tr1  For chloroform, the critical temperature is 536.4 K and the normal boiling point is 334.3 K. At 273 K, Tr1 = 0.5089 and at 334.3 K, Tr2 = 0.6232. So, 0.38

 1  0.6232  H 2  270.9 J g-1   245.0 J g -1   1  0.5089  This is 0.8% lower than the handbook value. Similarly for methanol, Tr1 = 0.5326 and Tr2 = 0.6592, so  1  0.6592  H 2  1189.5 J g-1    1  0.5326  This is 4.1% below the handbook value.

0.38

 1055.0 J g-1

Similarly for tetrachloromethane (also known as carbon tetrachloride!), Tr1 = 0.4909 and Tr2 = 0.6287, so -1  1  0.6287 

0.38

H 2  217.8 J g    1  0.4909  This is 0.5% below the handbook value.

 193.2 J g-1

(b) Equation 4.13 is used to estimate the heat of vaporization at the normal boiling point without knowing the heat of vaporization at any temperature. It can be written as  1.092  ln Pc  1.013  H n  RTn     0.930  Trn   where Pc, the critical pressure, must be given in bar. For chloroform, we get

 1.092  ln  54.72   1.013    29570 J mol-1 H n  8.314 J mol-1 K -1 334.3 K    0.930  0.6232   dividing by the molecular weight of 119.4 g mol-1 gives 247.7 J g-1. This is just 0.3% above the handbook

value. Likewise, for methanol:  1.092  ln 80.97   1.013    38302 J mol-1 H n  8.314 J mol-1 K -1* 337.9 K    0.930  0.6592   -1 and dividing by the molecular weight of 32.042 g mol gives 1194.2 J g-1. This is 8.6% above the handbook

value. Updated 18/01/2017 9 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Finally, for carbon tetrachloride,  1.092  ln  45.6   1.013    29587 J mol-1 H n  8.314 J mol-1 K -1 349.8 K    0.930  0.6287   -1 and dividing by the molecular weight of 153.822 g mol gives 192.3 J g-1. This is 1% below the handbook

value. 4.13 (7th edition Prob. 4.12) Again, we want to use the Clapeyron equation to find the latent heat of vaporization from the vapor pressure curve and molar volumes. First, we should find the normal boiling point by setting Psat = 101.3 kPa. Taking the derivative of the Antoine equation for the vapor pressure gives dP sat d  B  B B  BP sat    exp A   exp A        dT dT  T  C   T  C 2 T  C  T  C 2  

or, using the numbers for benzene dP sat 2772.78P sat  dT T  53.00 2

For T = 353.2 K and P = 101.325 kPa this gives sat

dP sat 2772.78  101.325   3.118 kPa K -1 2 dT  353.2  53.00 

Now, we can use generalized correlations to find the liquid and vapor volumes. The critical properties of 3 -1 benzene are Tc = 562.2 K, Pc = 48.98 bar,  = 0.210, Zc = 0.271, and Vc = 259 cm mol . For the volume of the saturated liquid, we can use the Rackett equation, which gives  353.2  1   259  0.271 562.2 

0.2857

1T V sat  Vc Zc r   96.8 cm3 mol-1 Since the reduced pressure is only 1.013/48.98 = 0.0207, we expect that the vapor will not be far from 0.2857

ideal, even though the reduced temperature is only 353.2/562.2 = 0.6282. We can get the compressibility from the Pitzer correlation as in the previous homework: 0.422 B 0  0.083   0.8049 0.62821.6 B1  0.139 

0.172 0.62824.2

 1.0730

and Updated 18/01/2017 10 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

0.02068  0.9661 0.6282 3 -1 So, V = ZRT/P = 0.9661*83.145*353.2/1 = 28370 cm mol . Z  1   0.8049  0.210  1.0730

So, V = 28370 --- 82.6 = 28288 cm3 mol-1. Finally, putting this all together, we have

H  353.2 K  28288cm3 mol-1  3.118 kPa K-1  3.115  107 kPa cm3 mol-1  3.115  104 J/mol So, our final answer is H = 31.15 kJ/mol. 7th edition Prob. 4.12, choose ethylbenzene We want to use the Clapeyron equation to find the latent heat of vaporization from the vapor pressure curve and molar volumes. First, we should find the normal boiling point by setting P = 101.3 kPa. That gives sat

ln 101.325   13.9726  T

3259.93 T  212.3

3259.93  212.3  136.2 C 13.9726  ln 101.325 

Reassuringly, this agrees with the value given in Table B.1 exactly (to the precision given). Taking the derivative of the Antoine equation for the vapor pressure gives dP sat d  B  B B  BP sat    exp A    exp  A      dT dT  T  C   T  C 2 T  C  T  C 2  

or, using the numbers for ethylbenzene

dP sat 3259.93P sat  dT T  212.32 For T = 136.2 °C and Psat = 101.325 kPa this gives

dP sat 3259.93  101.325   2.720 kPa K -1 2 dT 136.2  212.3 Now, we can use generalized correlations to find the liquid and vapor volumes. The critical properties of ethylbenzene are Tc = 617.2 K, Pc = 36.06 bar,  = 0.303, Zc = 0.263, and Vc = 374 cm3 mol-1. For the volume of the saturated liquid, we can use the Rackett equation, which gives Updated 18/01/2017 11 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

 409.3  1   374  0.263 617.2 

0.2857

1T V sat  Vc Zc r   140.5 cm3 mol-1 Because the reduced pressure is only 1.013/36.06 = 0.0281, we expect that the vapor will not be far from 0.2857

ideal, even though the reduced temperature is only 353.2/562.2 = 0.6632. We can get the compressibility from the Pitzer correlation:

B 0  0.083  B1  0.139 

0.422 0.66321.6 0.172 0.66324.2

 0.7311  0.8262

and

0.02809  0.9584 0.6632 So, V = ZRT/P = 0.9584*83.145*409.3/1.013 = 32198 cm3 mol-1. Z  1   0.7311  0.303  0.8262 

So, V = 32198 --- 140.5 = 32058 cm3 mol-1. Finally, putting this all together, we have

H  409.35 K  32058 cm3 mol-1  2.720 kPa K-1  3.569  107 kPa cm3 mol-1  3.569  104 J/mol So, our final answer is H = 35.7 kJ/mol. This agrees with the value in the table to within a few tenths of a percent --- essentially perfect agreement. 7th edition Prob. 4.12: Do this for toluene and phenol. Here, we want to use the Clapeyron equation to find the latent heat of vaporization from the vapor pressure curve and molar volumes. The Clapeyron equation is: dP sat dT We will use it, with the known vapor pressure curve and volume change to compute the enthalpy of H  T V

vaporization. (a) Toluene First, we should find the normal boiling point by setting Psat = 101.3 kPa. Using the Antoine parameters for toluene this gives

Updated 18/01/2017 12 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

3056.96 T  217.625 3056.96 T  217.625   110.6 C 13.9320  ln 101.325  ln 101.325   13.9320 

Reassuringly, this agrees with the value given in Table B.2 essentially exactly. Taking the derivative of the Antoine equation for the vapor pressure gives dP sat d  B  B B  BP sat    exp A   exp A        dT dT  T  C   T  C 2 T  C  T  C 2  

or, using the numbers for toluene

dP sat 3056.96 P sat  dT T  217.652 For T = 110.6 C and Psat = 101.325 kPa this gives

dP sat 3056.96  101.325   2.875 kPa K -1 2 dT 110.6  217.65 Now, we can use generalized correlations to find the liquid and vapor volumes. The critical properties of toluene are Tc = 591.8 K, Pc = 41.06 bar,  = 0.262, Zc = 0.264, and Vc = 316 cm3 mol-1. For the volume of the saturated liquid, we can use the Rackett equation, which gives  383.8  1   316  0.264 591.8 

0.2857

1T V sat  Vc Zc r   117.7 cm3 mol-1 Since the reduced pressure is only 1.013/41.06 = 0.0247, we expect that the vapor will not be far from 0.2857

ideal, even though the reduced temperature is only 383.8/591.8 = 0.6485. We can get the compressibility from the Pitzer correlation as in the previous homework:

B 0  0.083  B1  0.139 

0.422 0.64851.6 0.172 0.64854.2

 0.7608  0.9215

and

0.0247  0.9618 0.6485 3 -1 So, V = ZRT/P = 0.9618*83.145*383.8/1.013 = 30297 cm mol . Z  1   0.7608  0.262  0.9215

So, V = 30297 --- 117.7 = 30180 cm3 mol-1. Finally, putting this all together, we have

Updated 18/01/2017 13 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

H  383.8 K  30180 cm3 mol-1  2.875 kPa K-1  3.330 107 kPa cm3 mol-1  3.330 104 J/mol So, our final answer is H = 33.3 kJ/mol. This agrees with the value in table B.2 to within about 0.4%, which is about the precision of our calculations --- essentially exact agreement. (b) Now, we repeat the whole thing for phenol --- just for practice: First, we should find the normal boiling point by setting Psat = 101.3 kPa. Using the Antoine parameters for phenol, this gives

3507.8 T  175.4 3507.8 T  175.4   181.8 C 14.4387  ln 101.325  ln 101.325   14.4387 

Reassuringly, this agrees with the value given in Table B.2 essentially exactly. Taking the derivative of the Antoine equation for the vapor pressure gives dP sat d  B  B B  BP sat    exp A   exp A        dT dT  T  C   T  C 2 T  C  T  C 2  

or, using the numbers for phenol dP sat 3507.8P sat  dT T  175.4 2

For T = 181.8 C and Psat = 101.325 kPa this gives dP sat 3507.8  101.325   2.786 kPa K -1 2 dT 181.8  175.4 

Now, we can use generalized correlations to find the liquid and vapor volumes. The critical properties of phenol are Tc = 694.3 K, Pc = 61.3 bar,  = 0.444, Zc = 0.243, and Vc = 229 cm3 mol-1. For the volume of the saturated liquid, we can use the Rackett equation, which gives 455   1  694.3    229  0.243

0.2857

1T V sat  Vc Zc r   80.7 cm3 mol-1 Since the reduced pressure is only 1.013/61.3 = 0.01653, we expect that the vapor will not be far from 0.2857

ideal, even though the reduced temperature is only 455/694.3 = 0.6553. We can get the compressibility from the Pitzer correlation as in the previous homework:

Updated 18/01/2017 14 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

B 0  0.083  B1  0.139 

0.422 0.65531.6 0.172 0.65534.2

Chapter 4

 0.7469  0.8760

and

0.01653  0.9713 0.6553 So, V = ZRT/P = 0.9713*83.145*455/1.013 = 36275 cm3 mol-1. Z  1   0.7469  0.444  0.8760

So, V = 36275 --- 80.7 = 36195 cm3 mol-1. Finally, putting this all together, we have

H  455 K  36195 cm3 mol-1  2.786 kPa K-1  4.588 107 kPa cm3 mol-1  4.588 104 J/mol So, our final answer is H = 45.88 kJ/mol. This agrees with the value in table B.2 to within about 0.7%, which is about the precision of our calculations --- essentially exact agreement. 4.14 (7th edition Prob. 4.13) Using the equation above, solve for the pressure

Differentiate the above equation and determine dP/dT

Using the Clapeyron equation, solve for V

Then using Eq. 3.37, Solve for B

4.15 (7th edition Prob. 4.14)

Updated 18/01/2017 15 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

To find the total heat exchanger duty, we must consider three process steps: (1) heating of the subcooled liquid from 300 K to its saturation temperature at 3 bar, (2) vaporization at the saturation temperature, and (3) heating of the vapor from the saturation temperature to 500 K. For the first step, we do a heat capacity integral from 300 K to 368.0 K, using the liquid phase heat capacity: T2

ICPH T0,T;A,B,C,D  

1

B

2

0

2 0

C

3

3 0

0

T1

T0(K) 300

T (K) 368.00

1 

 R dT  AT  T   2 T  T   3 T  T   D  T  T  Cp

A 13.431

ig C (1/K2) D (K2) B (1/K) ICPH (K) H (J/mol) -5.13E-02 1.31E-04 0.00E+00 746.8 6209.3

For the second step, we need the heat of vaporization at 3 bar and 368.0 K. In table B.2, we find the heat of vaporization at the normal boiling point (337.9 K) is 35.21 kJ/mol. To find the heat of vaporization at 368 K, we can apply the Watson equation (p. 134 of SVNA). This gives:

 1.092(ln Pc - 1.013)  H 2 = RTn   0.930 - Tr      1.092(ln(80.97 bar) - 1.013)  H 2 = 8.314 J·mol1·K 1 × 337.9 K ×   368 K   0.930 512.6 K   1 H 2 = 48905.00 J·mol Finally, for the third step, we have: T2

ICPH T0,T;A,B,C,D  

Cp

B

0

2

2 0

C

3

3 0

T (K) 500.00

1  0

T1

T0(K) 368

1

 R dT  AT  T   2 T  T   3 T  T   D  T  T 

A 2.211

ig C (1/K2) D (K2) B (1/K) ICPH (K) H (J/mol) 1.22E-02 -3.45E-06 0.00E+00 905.2 7526.6

Comparing these, we see that the vaporization step has the largest contribution to the overall heat requirement. Adding them gives Q = H = 62.64 kJ/mol/. If the total methanol flow is 100 kmol/hr = 27.78 mol/s, then the total heat exchanger duty is 62.64*27.78 = 1740.13 kW.

4.16. (New) Need to gather several things for this proble, the boiling point, the latent heat of vaporization, and the heat capacities for each compound. Using the equation Updated 18/01/2017 16 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

And solving it for Tfinal. (a) Methanol --- Tboiling = 337.85 K , Cp,liq = 81.46 J/mol K , Cp,vap = 46.11 J/mol K , ΔH latent = 35210 J/mol So for the Liquid:

Now plugging into the first equation and solving for Tfinal gives:

(b) Ethanol --- Tboiling = 351.4 K , Cp,liq = 111.77 J/mol K , Cp,vap = 74.39 J/mol K , ΔH latent = 38560 J/mol So for the Liquid:

Now plugging into the first equation and solving for Tfinal gives:

(c) Benzene --- Tboiling = 353.15 K , Cp,liq = 134.33 J/mol K , Cp,vap = 85.29 J/mol K , ΔH latent = 30720 J/mol So for the Liquid:

Now plugging into the first equation and solving for Tfinal gives:

(d) Toluene --- Tboiling = 383.75 K , Cp,liq = 154.73 J/mol K , Cp,vap = 107.43 J/mol K , ΔH latent = 33180 J/mol So for the Liquid:

Now plugging into the first equation and solving for Tfinal gives:

(e) Water --- Tboiling = 373.15 K , Cp,liq = 75.40 J/mol K , Cp,vap = 33.57 J/mol K , ΔH latent = 40660 J/mol So for the Liquid:

Now plugging into the first equation and solving for Tfinal gives:

Updated 18/01/2017 17 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

4.17 (7th edition Prob. 4.15)

Estimate ΔHv using Riedel equation (4.13) and Watson correction (4.14)

Assume the throttling process is adiabatic and isenthalpic.

4.18 (7th edition Prob. 4.16) In each case, we can look up the heat of formation of the compound as a gas in table C.4, where we find Hf,(acetylene(g)) = 227480 J mol-1, Hf(1,3-butadiene(g)) = 109240 J mol-1, Hf(ethylbenzene(g)) = 29920 J mol , Hf,(n-hexane(g)) = -166920 J mol , Hf(styrene(g)) = 147360 J mol . The heat of formation of the liquid is equal to the heat of formation of the gas minus the heat of vaporization at 25C. If we only -1

-1

-1

know the critical properties and normal boiling point of each species, we could estimate this by first using equation 4.12 to estimate the heat of vaporization at the normal boiling point of each species, and then using equation 4.13 to estimate the heat of vaporization at 25C. We could, if we like, combine the equations to get 0.38

0.38

 1  Tr (25 C)   1.092  ln Pc  1.013  1  Tr (25 C)  H 25 C  H n   RTn      1  Tr    1  Tr  0.930  T r n n n      This is getting a bit cumbersome to type into the old calculator, so let's evaluate it in a spreadsheet. Doing so gives: Species

Tc

Pc

Tn

(K)

(bar)

(K)

Trn

Tr25C

n

25C

(J mol )

(J mol )

-1

-1

f

(g)

(J mol ) -1

f

(l)

(J mol-1)

acetylene

308.3

61.39

189.4

0.6143

0.9671

16911

6638

227480

220842

1,3-butadiene Ethylbenzene

425.2 617.2

42.77 36.06

268.7 409.4

0.6319 0.6633

0.7012 0.4831

22450 35852

20740 42196

109240 29920

88500 -12276

Updated 18/01/2017 18 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

n-hexane

507.6

30.25

341.9

0.6736

0.5874

29010

31712

-166920

-198632

Styrene

636

38.4

418.3

0.6577

0.4688

36753

43434

147360

103926

In the final column above, we have subtracted the latent heat of vaporization at 25 C from the heat of formation of each species in the gas phase to get the heat of formation of each species in the liquid phase. 4.19 (7 edition Prob. 4.17) th

1st law: dQ = dU

dW = CV dT + P dV

(A)

Ideal gas: P V = R T and P dV + V dP = R dT

Whence V dP = R dT

Since

P V = const

from which V dP = P

P dV

then

(B)

P V

1

dV =

V dP

dV

Combines with (B) to yield:

Combines with (A) to give:

Which reduces to: Or

Updated 18/01/2017 19 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Since CP is linear in T, the mean heat capacity is the value of CP at the arithmetic mean temperature. Thus

With

, and integrating (C):

4.20 (7th edition Prob. 4.18) The heat of combustion of 6 CH3OH(g) is the heat of reaction for 6 CH3OH(g) + 9 O2(g)  6 CO2(g) + 12 H2O(g) This is given by Hrxn = 6*Hf(CO2(g) ) + 12*Hf(H2O(g)) - 6*Hf(CH3OH(g) ) = 6*(-393.51) + 12*(241.82) -6*(-200.66) = -4059 kJ mol-1. The heat of combustion of C6H12(g) (assumed to be 1-hexene, though it doesn't really say and there are many other possible C6H12 isomers) is given by the heat of reaction for C6H12(g) + 9 O2  6 CO2(g) + 6 H2O(g) This is given by Hrxn = 6*Hf(CO2(g) ) +6*Hf(H2O(g)) - Hf(C6H12(g) ) = 6*(-393.51) + 6*(-241.82) (-41.95) = -3770 kJ mol-1. The difference between these two heats of combustion is the heat of reaction for the condensation reaction given in the problem statement. That is, subtracting the second combustion reaction above from the first gives 6 CH3OH(g)  C6H12(g) + 6 H2O(g) For which the heat of reaction is -4059 - (-3770) = -289 kJ/mol. The heat of combustion of ethylene at 25C (with water vapor product) is the heat of reaction for C2H4(g) +3 O2(g)  2 CO2(g) + 2 H2O(g) This is given by Hrxn = 2*Hf(CO2(g) ) + 2*Hf(H2O(g)) - Hf(C2H4(g) ) = 2*(-393.51) + 2*(-241.82) (52.51) = -1323 kJ mol-1 = -1.323106 J mol-1. The only difference in each case is how much excess O2 and N2 have to be heated up by the heat released by reaction. (a) The products (per mole of ethylene burned) will be 2 moles of CO2 + 2 moles of H2O + 3*(79/21) = 11.3 moles of N2 that was included with the 3 moles of O2. So, we integrate the total heat capacity of Updated 18/01/2017 20 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

this mixture from 298 K to some final temperature and vary the final temperature until we get H/R = 1.32310 J mol /8.314 J mol K = 159120 K. Doing so, using our handy-dandy heat capacity 6

-1

-1

-1

integrating spreadsheet gives 2533 K. (b) Similarly, here the products are be 2 moles of CO2 + 2 moles of H2O + 0.25*3 = 0.75 moles of O2 + 3*1.25*(79/21) = 14.1 moles of N2, and integrating the heat capacity of this mixture to get get H/R = 159120 K gives a final temperature of 2198 K. (c) and (d) are the same as (a) and (b), but with different numbers that lead to final temperatures of 1951 and 1609 K, respectively (e) Here, we can add the heat required to heat the air from 298K to 773K to the heat of reaction. We can imagine a path in which we cool the air from 773K to 298K (removing heat H1 = 309424 J per mol C2H4 burned), carry out the reaction at 298K (releasing the standard heat of reaction, Hrxn = -1.32310 ) 6

and then heat the products with both the heat removed to cool the air to 298K and the heat of reaction 6 (Htot = 1.66310 J per mol C2H4 burned), so the total Htot /R is 196330 K. Both the heat removed to cool the air from 773 to 298 K and the heat required to heat the reaction products to the final temperature are again computed using the heat capacity integrating spreadsheet. This gives a final temperature of 2282 K. To download the spreadsheet I used to solve this problem, click here. NOTE: A problem with this homework problem is that several of the final temperatures are above 2000 K, whereas the heat capacity expressions used (from table C.1 in the book) are only valid up to 2000 K. So, the results for parts (a), (b), and (e) are probably all incorrect (they are the correct answers to the homework problem, but probably are not good values for the actual adiabaic flame temperature of ethylene). 4.21 (7th edition Prob. 4.19) C2H4 + 3O2 = 2CO2 + 2H2O(g)

ΔH298= [2⋅(−241818) + 2⋅(−393509) − 52510] = 1323164 J/mol Parts (a) - (d) can be worked exactly as Example 4.7. However, with Mathcad capable of doing the iteration, it is simpler to proceed differently. Index the product species with the numbers: 1 = oxygen 2 = carbon dioxide 3 = water (g) Updated 18/01/2017 21 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

4 = nitrogen (a) For the product species, no excess air:

For the products

The integral is given by Eq. (4.8). Moreover, by an energy balance, 298

+

P

=0

Given Solving for

Parts (b), (c), and (d) are worked the same way, the only change being in the numbers of moles of products. (b) nO2 = 0.75 nN2 = 14.107 T = 2198.6⋅K Ans. (c) nO2 = 1.5

nN2 = 16.929

T = 1950.9⋅K Ans.

(d) nO2 = 3.0

nN2 = 22.571

T = 1609.2⋅K Ans.

(e) 50% xs air preheated to 500 degC. For this process,

ΔHair + ΔH298 + ΔHP = 0 ΔHair = MCPH⋅(298.15 − 773.15) For one mole of air: MCPH (773.15 ,298.15, 3.355, 0.575⋅10−3, 0.0, −0.016⋅105) = 3.65606 For 4.5/0.21 = 21.429 moles of air: Updated 18/01/2017 22 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

ΔHair = 21.429⋅8.314⋅3.65606⋅(298.15 − 773.15) = -309399 L/mol The energy balance here gives: ΔH298 + ΔHair + ΔHP = 0

Given Solving for

(f)theoretical amount of pure oxygen,

ΔH298 + ΔHP = 0 1323164 J/mol + ΔHP = 0

Starting with a guess and solving iteratively yields:

4.22 (similar to 7th edition Prob. 4.20)

4.22

(a) the heat of combustion of methane gas in the heat of reaction for CH4 + + 2 O2 (g)  CO2 (g) + 2 H2O (l) Updated 18/01/2017 23 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

This is given by Hrxn = *Hf(CO2(g) ) + 2*Hf(H2O(l)) - Hf(CH4(g)) = 1*( -393.51) + 2*(-285.83)-(74.520) = -890.65 kJ mol . -1

(b) The heat of combustion of ethane gas is the heat of reaction for 2 C2H6 + + 7 O2 (g) 4 CO2 (g) + 6 H2O (l) This is given by Hrxn = 4*Hf(CO2(g) ) + 6*Hf(H2O(l)) --- 2*Hf (C2H6 (g)) = 4*( -393.51) + 6*(-285.83)-2*(-83.820) = -3121.38 kJ mol-1. (c) The heat of combustion of ethylene gas is the heat of reaction for C2H4 + + 3 O2 (g) 2 CO2 (g) + 2 H2O (l) This is given by Hrxn = 2*Hf(CO2(g) ) + 2*Hf(H2O(l)) --- *Hf (C2H6 (g)) = 2*( -393.51) + 2*(-285.83)-(-52.510) = -1306.17 kJ mol-1. (d) The heat of combustion of propane gas is the heat of reaction for C3H8 + + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) This is given by Hrxn = 3*Hf(CO2(g) ) + 4*Hf(H2O(l)) --- *Hf (C3H8 (g)) = 3*( -393.51) + 4*(-285.83)-(-104.680) = -2219.47 kJ mol-1. (e) The heat of combustion of propylene, gas is the heat of reaction for 2 C3H6 + + 9 O2 (g) 6 CO2 (g) + 6 H2O (l) This is given by Hrxn = 6*Hf(CO2(g) ) + 6*Hf(H2O(l)) --- 2*Hf (C3H6 (g)) = 6*( -393.51) + 6*(-285.83)-2*(-19.710) = -4036.62 kJ mol-1.

(f) The heat of combustion of n-butane , gas is the heat of reaction for C4H10 + + 13/2 O2 (g) 4 CO2 (g) + 5 H2O (l) This is given by Hrxn = 4*Hf(CO2(g) ) + 5*Hf(H2O(l)) --- *Hf (C4H10 (g)) = 4*( -393.51) + 5*(-285.83)-(-125.790) = -2877.4 kJ mol-1. (g) The heat of combustion of 1-butene, gas is the heat of reaction for C4H8 + + 6 O2 (g) 4 CO2 (g) + 4 H2O (l) Updated 18/01/2017 24 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

This is given by Hrxn = 4*Hf(CO2(g) ) + 4*Hf(H2O(l)) --- *Hf (C4H8 (g)) = 4*( -393.51) + 4*(-285.83)-(-0.540) = -2716.82 kJ mol-1. (h) The heat of combustion of ethylene oxide, gas is the heat of reaction for C4H8 + + 6 O2 (g) 4 CO2 (g) + 4 H2O (l) This is given by Hrxn = 4*Hf(CO2(g) ) + 4*Hf(H2O(l)) --- *Hf (C4H8 (g)) = 4*( -393.51) + 4*(-285.83)-(-0.540) = -2716.82 kJ mol-1. (i) The heat of combustion of acetaldehyde gas is the heat of reaction for CH3CHO+ 2.5 O2 (g) 2 CO2 (g) + 2 H2O (l) This is given by Hrxn = 2*Hf(CO2(g) ) + 2*Hf(H2O(l)) --- *Hf (CH3CHO (g)) = 2*( -393.51) + 2*(-285.83)-(-166.190) = -1192.49 kJ mol-1. (j) The heat of combustion of methanol gas is the heat of reaction for 2 CH3OH+ 3 O2 (g) 2 CO2 (g) + 4 H2O (l) This is given by Hrxn = 2*Hf(CO2(g) ) + 4*Hf(H2O(l)) --- 2*Hf (CH3OH (g)) = 2*( -393.51) + 4*(-285.83)-2*(-200.660) = -1529.02 kJ mol . -1

(k) The heat of combustion of ethanol gas is the heat of reaction for C2H6 O+ 6 O2 (g) 2 CO2 (g) + 3 H2O (l) This is given by Hrxn = 2*Hf(CO2(g) ) + 3*Hf(H2O(l)) --- *Hf (C2H6O (g)) = 2*( -393.51) + 4*(-285.83)-(-235.100) = -1695.24 kJ mol-1.

4.23 (7th edition Prob. 4.21) (a) Hrxn = 2*Hf(NH3(g) ) - Hf(N2(g)) --- 3*Hf(H2(g)) = 2*(-46.110) ---0-0 = -92.22 kJ mol-1. (b) Hrxn = 4*Hf(NO(g)) + 6*Hf(H2O(g)) --- 4*Hf(NH3(g) --- 5*Hf(O2(g)) Hrxn = 4*(90.25) + 6*(-241.82) --- 4*(-46.110) --- 5*(0)= -905.5 kJ mol-1. (c) Hrxn = 2*Hf(HNO3(l)) + Hf(NO(g)) --- 3*Hf(NO2(g)) - Hf(H2O(l)) Hrxn = 2*(-174.1) + (90.25) --- 3*(33.18) - (-285.830) = -71.66 kJ mol-1. (d) Hrxn = Hf(CaO(s) ) + Hf(C2H2(g)) --- Hf(CaC2(s)) - Hf(H2O(l)) Hrxn = -635.090 + 227.480 + 59.800 + 285.830 = -61.98 kJ mol-1. (e) Hrxn = 2*Hf(NaOH(s)) + Hf(H2(g)) - 2*Hf(Na(s)) --- 2*Hf(H2O(g)) Updated 18/01/2017 25 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Hrxn = 2*(-425.609) + 0 --- 2*(0) --- 2*(-241.818)= -367.6 kJ mol-1. (f) Hrxn = 7*Hf(N2(g)) + 12*Hf(H2O(g)) --- 6*Hf(NO2(g)) --- 8*Hf(NH3(g)) Hrxn = 7*(0) + 12*(-241.818) --- 6*(33.18) --- 8*(-46.110) = -2732 kJ mol-1. (g) Hrxn = Hf

)

2 2

- Hf(C2H4(g)) --- ½ Hf(O2(g))

Hrxn = -52630 --- 52510 --- 0 = -105140 J mol (h) Hrxn = Hf

)

2 2

-1

- Hf(C2H2(g)) --- Hf(H2O(g))

Hrxn = -52630 --- 227480 --- -241818 = -38292 J mol-1 (i) Hrxn = Hf(CO2(g)) + 4*Hf(H2(g)) - Hf(CH4(g)) --- 2*Hf(H2O(g)) Hrxn = - 4*(92307)+ 2*(-241818) = -1.144 * 105 J mol-1 (j) Hrxn = Hf(CH3OH(g)) + Hf(H2O(g)) - Hf(CO2(g)) --- 3*Hf(H2(g)) Hrxn = -200660 - 241818 - -393509 --- 3*0 = -48969 J mol

-1

(k) Hrxn = Hf(HCHO(g)) + Hf(H2O(g)) - Hf(CH3OH(g)) --- ½ *Hf(O2(g)) -1 Hrxn = -108570 - 241818 - -200660 --- ½*0 = -149728 J mol (l) Hrxn = 2*Hf(H2O(g)) + 2*Hf(SO2(g)) - 2*Hf(H2S(g)) --- 3 *Hf(O2(g)) Hrxn = 2*-241818 -2*296830 +2*20630 --- 3*0 = -1036036 J mol-1 (m) Hrxn = 3*Hf(H2(g)) + Hf(SO2(g)) - Hf(H2S(g)) --- 2*Hf(H2O(g)) -1 Hrxn = 3*0 + -296830 + 20630 --- 2*-241818 = 207436 J mol (n) Hrxn = 2*Hf(NO(g)) - Hf(N2(g)) --- Hf(O2(g)) Hrxn = 2*90250 --- 0 --- 0 = 180500 J mol (o) Hrxn = Hf(CaO(s)) + Hf(CO2(g)) --- Hf(CaCO3(s)) Hrxn = -635090 --- 393509 + 1206920 =178321 J mol-1 -1

(p) Hrxn =Hf(H2SO4(l)) - Hf(SO3(g)) --- Hf(H2O(l)) Hrxn = -813989 + 395720 + 285830 = -132439 J mol-1 (q) Hrxn = Hf(C2H5OH(l)) - Hf(C2H4(g)) --- Hf(H2O(l)) Hrxn = -277690 --- 52510 + 285830 = -44370 J mol-1 (r) Hrxn = Hf(C2H5OH(g)) - Hf(CH3CHO(g)) --- Hf(H2(g)) Hrxn = -235100 +166190 --- 0 = -68910 J mol-1 (s) Hrxn = Hf(CH3COOH(l)) + Hf(H2O(l)) - Hf(C2H5OH(l)) --- Hf(O2(g)) Hrxn = -484500 -285830 + 277690 --- 0 = -492640 J mol-1 (t) Hrxn = Hf(CH2:CHCH:CH2(g)) + Hf(H2(g)) - Hf(C2H5CH:CH2(g)) Hrxn = 109240 + 0 + 540 = 109780 J mol-1 (u) Hrxn = Hf(CH2:CHCH:CH2(g)) + 2*Hf(H2(g)) - Hf(C4H10(g)) Hrxn = 109240 + 0 + 125790 = 235030 J mol-1 (v) Hrxn = Hf(CH2:CHCH:CH2(g)) + Hf(H2(g)) - Hf(C2H5CH:CH2(g))- Hf(O2(g))Hrxn = 109240 +(-241818)-(-540) = -132038 J mol-1 (w) Hrxn = 4*Hf(NH3(g)) -6* Hf(NO(g)) --- 6*Hf(H2O(g))- 5*Hf(N2(g))Updated 18/01/2017 26 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Hrxn =4(-46110) --- 6(90250) --- 6*(241818) - 0 = -1807968 J mol-1

(x) Hrxn = 2*Hf(HCN(g) ) - Hf(C2H2(g)) = 2*(135.1) - (227.48) = 42.72 kJ mol-1. (y) Hrxn = Hf(styrene) - Hf(ethylbenzene) = (147.36) - (29.92) = 117.44 kJ mol . -1

(z) Hrxn = Hf(CO(g)) - Hf(H2O(l)) = (-110.525) - (-285.830) = 175.305 kJ mol-1. 4.24 (7th edition Prob. 4.22) (a) A heat of reaction at a temperature other than standard temperature is given by the heat of reaction at standard temperature plus the integral of the difference in heat capacity between products and reactants: o o H rxn ,T  H rxn,298.15 K 

T

298.15 K

C

total total p ,products  C p ,reactants dT

A handy spreadsheet for evaluating the integral was provided in the lecture notes. Putting in the parameters for the reaction N2 + 3 H2 3 We have T

1 1  1 B 2 C 3 IDCPH  T0 ,T; A, B, C, D  C p dT  A T  T0   T  T02  T  T03  D    R 2 3  T T0 

To

T0 (K) 298.15

T (K) 873.15

A -5.871

B (1/K) 4.18E-03

C (1/K2) 0.00E+00

D (K2) IDCPH (K) -6.61E+04 -2113.9

Species Name N2 H2 NH3

Stoichiometric coefficient -1 -3 2

A 3.28 3.249 3.578

B (1/K) 5.93E-04 4.22E-04 3.02E-03

C (1/K2) 0.00E+00 0.00E+00 0.00E+00

D (K2) n i Ai 4.00E+03 -3.28E+00 8.30E+03 -9.75E+00 -1.86E+04 7.16E+00

Multiplying the result by R gives -

Cp. The heat of reaction at 298.15 K

is just twice the heat of formation of ammonia, because this is the formation reaction for ammonia, written with 2 NH3 H298 = -46110 J/mol *2 = -92.22 kJ/mol. Adding on the Cp gives H600°C = -109.80 kJ/mol for the reaction as written. (b) Repeating this for the reaction 4 NH3 + 5 O2

O gives

2

Updated 18/01/2017 27 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

T

1 1  1 B 2 C 3 IDCPH  T0 ,T; A, B, C, D  C p dT  A T  T0   T  T02  T  T03  D    R 2 3  T T0 

To

T0 (K) 298.15

T (K) 773.15

A 1.861

B (1/K) -3.39E-03

C (1/K2) 0.00E+00

D (K2) IDCPH (K) 2.66E+05 568.8

Species Name NH3 O2 NO H2O

Stoichiometric coefficient -4 -5 4 6

A 3.578 3.639 3.387 3.47

B (1/K) 3.02E-03 5.06E-04 6.29E-04 1.45E-03

C (1/K2) 0.00E+00 0.00E+00 0.00E+00 0.00E+00

D (K2) n i Ai -1.86E+04 -1.43E+01 -2.27E+04 -1.82E+01 1.40E+03 1.35E+01 1.21E+04 2.08E+01

Multiplying the result by R gives 4.729 kJ/mol for the Cp. The heat of reaction at 298.15 K is given by: H298 H298(H2 H298(NO) --H298(NH3) H298 = 6*(-241.818) + 4*90.250 --- 4*(-46.11) = -905.47 kJ/mol. H500°C = -900.74 kJ/mol. (f) A heat of reaction at a temperature other than standard temperature is given by the heat of reaction at standard temperature plus the integral of the difference in heat capacity between products and reactants: o o H rxn ,T  H rxn,298.15 K 

T

298.15 K

C

total total p ,products  C p ,reactants dT

A handy spreadsheet for evaluating this was provided in the lecture notes. Putting in the parameters for the reaction 6 NO2(g) + 8 NH3(g)  7 N2(g) + 12 H2O (g) gives: Reference Temperature T 0 (K) 298.15

Species Name NO2

Temperature of Interest T (K) 923.15

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T 0 kJ/mol (dimensionless) kJ/mol -2732.016 2.037 -2716.380 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -6 33.18

T

1 RT

1

1

1 

 C dT  T  A T  T   2 T  T   3 T  T   D  T  T   B

p

2

0

C

2 0

3

3 0

0

To

 1 B 2 C 3 1  o o  H rxn T  T02  T  T 03   D    ,T   H rxn ,T0  R   A  T  T 0   2 3  T T0   

2

2

A 4.982

B (1/K) 1.20E-03

0.00E+00

niHf,i niAi niBi -7.92E+04 -1.99E+02 -2.99E+01 -7.17E-03

niCi 0.00E+00

niDi 4.75E+05

C (1/K )

D (K )

NH3

-8

-46.11

3.578

3.02E-03

0.00E+00

-1.86E+04 3.69E+02 -2.86E+01 -2.42E-02

0.00E+00

1.49E+05

N2

7

0

3.28

5.93E-04

0.00E+00

4.00E+03 0.00E+00

2.30E+01

4.15E-03

0.00E+00

2.80E+04

H2O

12

-241.818

3.47

1.45E-03

0.00E+00

1.21E+04 -2.90E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

4.16E+01 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1.74E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1.45E+05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

B (1/K) -9.78E-03

C (1/K ) 0.00E+00

D (K ) 7.97E+05

A 6.084 Note: Light blue fields are inputs, pink fields are the final output.

2

2

Updated 18/01/2017 28 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

The heat of reaction at 650C is -2716 kJ/mol. This is only slightly different from the heat of reaction at standard conditions. It just happens that the average heat capacity of the reactants and products is about the same over this temperature range. 4.25 (7th edition Prob. 4.23)

This is a simple application of a combination of Eqs. (4.19) & (4.20) with evaluated parameters. In each case the value of ΔH0298 is calculated in Pb. 4.23. The values of ΔA, ΔB, ΔC and ΔD are given for all cases except for Parts (e), (g), (h), (k), and (z) in the preceding table. Those missing are as follows: Part No. e g h k z

ΔA 103 ΔB 106 ΔC -7.425 20.778 0 -3.629 8.816 -4.904 -9.987 20.061 -9.296 1.704 -3.997 1.573 -3.858 -1.042 0.18

10-5 ΔD 3.737 0.114 1.178 0.234 0.919

4.26 (New) Use table C.5 to find the standard enthalpies. (a)Reactants = -1262.2 kJ/mol + -3627.9 kJ/mol = -4890.1 kJ/mol Products = -2274.6 kJ/mol + -2638.5 kJ/mol = -4913.1 kJ/mol Products --- Reactants = -4913.1 J/mol --- (-4881.4) J/mol = -23 J/mol (b) Reactants = -2274.6 kJ/mol Products = -2265.9kJ/mol Products --- Reactants = 8.7 kJ/mol (c) Reactants = -5893.8 kJ/mol Products = -5956.8 kJ/mol Products --- Reactants = -63 kJ/mol (d) Reactants = -9144.4 J/mol Products = -9083.66J/mol Products --- Reactants = 60.74 kJ/mol (e) Reactants = -9144.4 J/mol Products = -9200.66 J/mol Products --- Reactants = -56.26 kJ/mol (f) Reactants = -9144.4 kJ/mol Updated 18/01/2017 29 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Products = -9231.66 kJ/mol Products --- Reactants = -87.26 J/mol (g) Reactants = -75.5 kJ/mol Products = -571.66 kJ/mol Products --- Reactants = -496.16 kJ/mol (h) Reactants = -3941.1 kJ/mol Products = -3913.73 kJ/mol Products --- Reactants = -27.37 kJ/mol (i) Reactants = -7957 kJ/mol Products = -8399.12 kJ/mol Products --- Reactants = --441.42 kJ/mol (j) Reactants = -9141.6 kJ/mol Products = -9231.66 kJ/mol Products --- Reactants = -90.06 kJ/mol (k) Reactants = -9137.4 kJ/mol Products = -9231.66 kJ/mol Products --- Reactants = -94.26 kJ/mol (l) Reactants = -4703.8 kJ/mol Products = -4707.2 kJ/mol Products --- Reactants = - 3.4 kJ/mol

4.27 (New) First determine how many moles of ethanol we have, MWetoh = 46.068 g/mol. Moles of EtOH = 10 g/ 46.068 g/mol = 0.217 mol. Now, Balance the equation C2H5OH + NAD+ → C2H4O + NADH + H

 

H f 298 = (-212.2 J·mol1 ) + (-31.9 J·mol1 ) + 0 J·mol1 - (-288.3 J·mol1 ) + 0 J·mol1 = 44.2 J·mol1 H = 0.217 mol × H f 298 = 0.217 mol × 44.2 J·mol1 = 9.59 J For the second part, write the equation Updated 18/01/2017 30 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

C2H5OH + NAD+ → (2CO2 + 2H2O) + NADH + H

 

H f 298 = 2(-393509 J·mol1 ) + 2(-285830 J·mol 1 ) + (-31.9 J·mol1 ) + 0 J·mol1 - -288.3 J·mol 1 + 0 J·mol1 H = -0.217 mol × H f 298 = 0.217 mol × (-1358421.6 J·mol1 ) = -294777.4872 J This reaction gives off heat as it proceeds in the reaction.

4.28 (7th edition Prob. 4.24) T = 288.71 K

P= 1 atm

The higher heating value is the negative of the heat of combustion with water as liquid product. Calculate methane standard heat of combustion with water as liquid product: CH4 + 2O2  CO2 +2H2O Standard Heats of Formation:

Assuming methane is an ideal gas at standard conditions:

4.29 (7th edition Prob. 4.25)

Calculate methane standard heat of combustion with water as liquid product Standard Heats of Formation:CH4 + 2O2 --> CO2 +2H2O Updated 18/01/2017 31 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Calculate ethane standard heat of combustion with water as liquid product: Standard Heats of Formation:C2H6 + 7/2O2 --> 2CO2 +3H2O

− ⋅

Calculate propane standard heat of combustion with water as liquid product Standard Heats of Formation:C3H8 + 5O2 --> 3CO2 +4H2O

Calculate the standard heat of combustion for the mixtures (a) ⋅

(b)

(c) Gas b) has the highest standard heat of combustion.

4.30 (7th edition Prob. 4.26) The heat of combustion is defined as the heat of reaction for CO2(g) + 2 H2O (l) + N2(g)  (NH2)2CO(s) + 3/2 O2(g) Because the number given for the heat of combustion is positive, we know that it is for formation of urea plus oxygen from the combustion products (the reverse of the combustion reaction). Updated 18/01/2017 32 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

So, the heat of combustion at 25C is Hcomb(298) = ½*Hf(O2(g)) + Hf((NH2)2CO(s)) - Hf(CO2(g)) - 2*Hf(H2O(l)) - Hf(N2(g)) We know all of the quantities in the above equation except the heat of formation of urea. So, we can put the numbers in and solve for Hf((NH2)2CO(s)) = Hf(CO2(g)) + 2*Hf(H2O(l)) + Hf(N2(g)) ---1.5*Hf(O2(g)) + Hcomb(298) Hf((NH2)2CO(s)) = -393.509 + 2*(-285.83) + 0 --- 1.5*0 + 631.66 = -333.5 kJ/mol.

4.31 (7th edition Prob. 4.27) (a) There is nothing profound here. If we compute the heat of combustion with water as a liquid product, then the value we get is higher (larger in absolute value) than if we compute it with water as a vapor product. They differ by the value of the heat of vaporization of water at 25C, which is 44012 J mol-1. (b) The heat of combustion for methane with liquid water as the product is the heat of reaction for CH4(g) + 3 O2(g)  CO2(g) + 2 H2O(l), which is Hrxn = Hf(CO2(g) ) + 2*Hf(H2O(l)) - Hf(CH4(g) ) = 393.51 + 2*(-285.83) -(-74.52) = -890.7 kJ mol-1. With water vapor as the product, it is -802.6 kJ/mol. Heating values for gaseous fuels are more commonly expressed in Btu per standard cubic foot. If we treat methane as an ideal gas and choose the commonly used 'standard conditions' for natural gas of 60F and 1 atm, then at there will be n/V = P/RT = 1 atm/(0.7302 SCF atm lbmol-1 R-1 * 519.7 R) = 0.00264 lbmol /SCF= 1.195 mol /SCF. Multiplying this by the higher heating value in kJ/mol gives 1.195*-890.7= -1064 kJ/SCF = -1009 Btu/SCF (Perry's handbook lists 1012 Btu ft-3, so we probably did it right). Multiplying it by the lower heating value gives -909 Btu/SCF. So, the HHV = 1009 Btu/SCF and the LHV = 909 Btu/SCF. (c) Similarly, the heat of combustion with liquid water product for n-decane is the heat of reaction for C10H22(l) + 15.5 O2(g) 10 CO2(g) + 11 H2O(l), which is Hrxn = 10*Hf(CO2(g) ) + 11*Hf(H2O(l)) Hf(C10H22(l) ) = 10*(-393.51) + 11*(-285.83) -(-249.7) = -6829.5 kJ mol-1. Similarly, the value with gas phase water product is -6345.4 kJ mol-1. For liquid fuels, the heating value is usually expressed as Btu per gallon of the liquid. The density of n-decane (from Perry's handbook) is 0.73 g cm , and the -3

molecular weight is 142.3 g mol-1, so we get HHV = 6829.5 kJ mol-1 / 143 g mol-1 * 0.73 g cm-3 = 34.86 kJ/cm3 * 3785 cm3 gal-1 * 0.9478 Btu kJ-1 = 125000 Btu gal-1. Likewise, the lower heating value is 116000 Btu gal-1.

4.32 (7th edition Prob. 4.28)

On the basis of 1 mole of C10H18 (molar mass = 162.27) Updated 18/01/2017 33 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

This value is for the constant-volume reaction: C10H18(l) + 14.5O2(g) = 10CO2(g) + 9H2O(l) Assuming ideal gases and with symbols representing total properties,

This value is for the constant-V reaction, whereas the STANDARD reaction is at const. P. However, for ideal gases H = f(T), and for liquids H is a very weak function of P. We therefore take the above value as the standard value, and for the specified reaction: C10H18(l) + 14.5O2(g) = 10CO2(g) + 9H2O(l)

ΔH

9H2O(l) = 9H2O(g) ___________________________________________________ C10H18(l) + 14.5O2(g) = 10CO2(g) + 9H2O(g) 298

:=

+

vap

= -6748436 J

4.33 (7th edition Prob. 4.29)

FURNACE: Basis is 1 mole of methane burned with 30% excess air. CH4 + 2O2 = CO2 + 2H2O(g) Entering: Total moles of dry gases entering At 30 degC the vapor pressure of water is 4.241 kPa. Moles of water vapor entering:

Leaving:

CO2 -- 1 mol H2O -- 2.585 mol O2 -- 2.6 - 2 = 0.6 mol N2 -- 9.781 mol

Updated 18/01/2017 34 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

By an energy balance on the furnace: Q = ΔH = ΔH303 + ΔHP From Example 4.7: ΔH298 = -802625 J/mol

ΔH303 = ΔH298 + MCPH * R * (T – T0) = -802279 J/mol For evaluation of ΔHP we number species as above.

The TOTAL value for MCPH of the product stream:

ΔHP = R⋅MCPH(303.15K,1773.15K,A,B,C,D)⋅(1773.15 − 303.15)K ΔHP = 731982 kJ/mol Q =ΔHP + ΔH303 = −70.30 kJ

HEAT EXCHANGER: Flue gases cool from 1500 C to 50 C. The partial pressure of the water in the flue gases leaving the furnace (in kPa) is

The vapor pressure of water at 50 degC (exit of heat exchanger) is 12.34 kPa, and water must condense to lower its partial pressure to this value. Moles of dry flue gases: Moles of water vapor leaving the heat exchanger:

Moles of water condensing: Latent heat of water at 50 C in J/mol: Updated 18/01/2017 35 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Sensible heat of cooling the flue gases to 50 degC with all the water as vapor (we assumed condensation at 50 degC): ⋅ ⋅ ⋅ ⋅ −

4.34 (7th edition Prob. 4.30)

4NH3(g) + 5O2(g) = 4NO(g) + 6H2O(g) BASIS: 4 moles ammonia entering reactor Moles O2 entering = (5)(1.3) = 6.5 Moles N2 entering = (6.5)(79/21) = 24.45 Moles NH3 reacting = moles NO formed = (4)(0.8) = 3.2 Moles O2 reacting = (5)(0.8) = 4.0 Moles water formed = (6)(0.8) = 4.8 ENERGY BALANCE: ΔH = ΔHR + ΔH298 + ΔHP = 0 REACTANTS: 1=NH3; 2=O2; 3=N2

TOTAL mean heat capacity of reactant stream: ⋅

The result of Pb. 4.23(b) is used to get )= -724374.4 J/mol

Products

Updated 18/01/2017 36 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Given

4.35 (7th edition Prob. 4.31) The enthalpy balance for this steady-state flow process in which no non-flow work is done simply gives Q = H, where H is the enthalpy of the ethanol stream leaving the reactor minus the enthalpy of the ethylene and water entering the reactor. To compute the enthalpy change, we can use an imaginary 2-step path in which we first cool the ethane and water from 320C to 25C, and then carry out the reaction. The total enthalpy change will be the sum of the entropy change for the two steps:

H 

298.15 K

o C pfeed dT  H rxn ,298.15 K

593.15 K

The choice of doing the heat capacity integral first (from 320 to 25C) is particularly convenient because the final temperature happens to be the standard temperature for heats of formation. In general, if we have reactants and products at different temperatures, we would have to do two heat capacity integrals --- one from the feed temperature to 298.15 K and one from 298.15 K back to the product temperature. The heat of reaction is Hrxn = Hf(C2H5OH(l)) - Hf(H2O(g)) --- Hf(C2H4(g)) Hrxn = -277690 --- (-241818) --- (52510) = -88382 J/mol We can use the handy spreadsheet from the lecture notes to evaluate the heat capacity integral: T1 (K)

T2 (K)

A

593.15

298.15

4.894

T2

Cp ICPH  dT A T2 T1  R T1

B (1/K)

D (K2)

ICPH (K)

1.58E-02 -4.39E-06 1.21E+04

-3.28E+03

C (1/K2)

B 2 T2 T12 2

 C3 T

3 2

T13

 DT1 T1 2

1

Updated 18/01/2017 37 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Where for A, B, C, and D, we use the sum of the values for water and ethylene. Multiplying this by R gives

298.15 K

C pfeed dT  8.314*(3820)  27270 J mol-1

593.15 K

Thus, the total heat load of the process is Q = H = -88380 --- 27270 = -115560 J/mol = -115.6 kJ/mol We must remove 115.6 kJ of heat from the reactor for each mole of ethanol produced. 4.36 (7th edition Prob. 4.32) Let x1 be the extent of the first reaction and x2 be the extent of the second reaction, defined so that they are extensive extents of reaction with units of moles, so that the number of moles of each species coming out of the reactor are nCH4 = nCH4,o - x1 nH2O = nH2O,o - x1 - x2

and

nCO = x1 - x2 nH2 = 3x1 + x2 nCO2 = x2

Given the reactor outlet composition in mole fractions, and doing the problem on a 'per mole of outlet gas' basis, the number of moles of each species coming out is just the given mole fractions, so we have nCH4 = nCH4,o - x1 = 0 nH2O = nH2O,o - x1 - x2 = 0.1725 nCO = x1 - x2 = 0.1725 and

nH2 = 3x1 + x2 = 0.6275 nCO2 = x2 = 0.0275

Note that because the four given mole fractions sum to 1, the mole fraction of CH4 in the outlet (not explicitly given) is zero. Above, we have 5 equations in 4 unknowns (x1, x2, nCH4,o, and nH2O,o), but fortunately they are consistent. From them, we get x2 = 0.0275, x1 = 0.2000, nCH4,o = 0.2000, and nH2O,o = 0.4000. That is, for each mole of the product stream, 0.4 moles of H2O and 0.2 moles of CH4 are fed to the reactor, and 0.2 moles of reaction 1 occur and 0.0275 moles of reaction 2 occur. H = Q. The heat input is equal to the enthalpy difference between the product stream and the reactant stream. This can be computed as the sum of three steps: (1) The change in enthalpy when 0.4 mol H2O and 0.2 mol CH4 are cooled from 500 ºC to 25 º C Updated 18/01/2017 38 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

(2) 0.2 times the standard heat of reaction at 25ºC for CH4 + H2 standard heat of reaction for CO + H2

2

2

plus 0.0275 times the

+ H2

(3) The change in enthalpy when a mixture of 0.0275 mol CO2, 0.1725 mol CO, 0.1725 mol H2O, and 0.6275 mol H2 is heated from 25 ºC to 850 ºC. For step (1), we can use one of our handy spreadsheets. A version adapted for gas mixtures is shown below: T

IDCPH T0,T; A,B,C,D  

1 1  1 B 2 C 3 C p dT  A T  T0   T  T02  T  T03  D    R 2 3  T T0 

To

T 0 (K) 773.15

T (K) 298.15

ΔA 1.7284

ΔB (1/K) 2.40E-03

ΔC (1/K2) -4.33E-07

ΔD (K2) IDCPH (K) 4.84E+03 -1377.787

ΔH (J) -11455

Species Name CH4 H2O CO H2 CO2

Amount (moles) 0.2 0.4 0 0 0

A 1.702 3.47 3.376 3.249 5.547

B (1/K) 9.08E-03 1.45E-03 5.57E-04 4.22E-04 1.05E-03

C (1/K2) -2.16E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00

D (K2) 0.00E+00 1.21E+04 -3.10E+03 8.30E+03 -1.16E+05

niBi 1.82E-03 5.80E-04 0.00E+00 0.00E+00 0.00E+00

niAi 3.40E-01 1.39E+00 0.00E+00 0.00E+00 0.00E+00

niCi -4.33E-07 0.00E+00 0.00E+00 0.00E+00 0.00E+00

niDi 0.00E+00 4.84E+03 0.00E+00 0.00E+00 0.00E+00

For step (3), we do the same thing, but with the composition of the product stream and the appropriate temperatures: T

IDCPH T0,T; A,B,C,D  

1 1  1 B 2 C 3 C p dT  A T  T0   T  T02  T  T03  D    R 2 3  T T0 

To

T 0 (K) 298.15

T (K) 1123.15

ΔA 3.372225

ΔB (1/K) 6.40E-04

ΔC (1/K2) 0.00E+00

ΔD (K2) IDCPH (K) 3.58E+03 3165.980

ΔH (J) 26322

Species Name CH4 H2O CO H2 CO2

Amount (moles) 0 0.1725 0.1725 0.6275 0.0275

A 1.702 3.47 3.376 3.249 5.547

B (1/K) 9.08E-03 1.45E-03 5.57E-04 4.22E-04 1.05E-03

C (1/K2) -2.16E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00

D (K2) niAi 0.00E+00 0.00E+00 1.21E+04 5.99E-01 -3.10E+03 5.82E-01 8.30E+03 2.04E+00 -1.16E+05 1.53E-01

niBi 0.00E+00 2.50E-04 9.61E-05 2.65E-04 2.87E-05

niCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

niDi 0.00E+00 2.09E+03 -5.35E+02 5.21E+03 -3.18E+03

For step (2), the relevant heats of formation are: Reaction 1: Species Name CH4 H2O CO H2

Reaction 2: Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -1 -74.52 -1 -241.818 1 -110.53 3 0

Ho298(rxn

Species Name CO H2O CO2 H2

Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -1 -110.53 -1 -241.818 1 -393.509 1 0

Ho298(rxn 2) = -41161 J/mol H = 0.2*205808 + 0.0257*-41161 = 40030 J.

Thus, the total for the three steps is -11455 + 40030 + 26322 = 54897 J. 4.37 (7th edition Prob. 4.33) Updated 18/01/2017 39 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

le of fuel (0.75 mol of methane and 0.25 mol of ethane). Because we use excess air, we can safely assume complete combustion of the fuel to give CO2 and water. So, we have two combustion reactions: CH4 + 2 O2  CO2 + 2 H2O C2H6 + 7/2 O2  2CO2 + 3 H2O Burning the 0.75 mol of CH4 requires 2*0.75 = 1.5 mol of O2, and burning the 0.25 mol of C2H6 requires 0.25*7/2 = 7/8 = 0.875 mol O2. So, the total O2 required for stoichiometric combustion of the fuel is 2.375 mol O2 per mole of fuel. If we use 80% excess air, that means we have 1.8 times this much O2, and the O2 feed rate is 1.8*2.375 = 4.275 mol O2 per mole of fuel. Because we are using air, we get 79/21 moles of N2 for every mole of O2, and therefore the N2 feed rate is 79/21*4.275 = 16.08 mol N2 per mole of fuel. So, the feed to the furnace is (per mole of fuel) 0.75 mol CH4 0.25 mol C2H6 4.275 mol O2 16.08 mol N2 The flue gas has the same amount of N2 as the feed, but all of the CH4 and C2H6 (and some of the O2) have been converted to CO2 and H2O. The 0.75 mol of CH4 produces 0.75 mol of CO2 and 1.5 mol of H2O. The 0.25 mol C2H6 produces 0.5 mol CO2 and 0.75 mol H2O. We already figured out that the stoichiometric O2 (the amount used up in burning the fuel) is 2.375 mol O2, so we have 4.275 --- 2.375 = 1.9 mol O2 remaining in the flue gas (this is the 80% excess O2). So the flue gas leaving the furnace is (per mole of fuel) 1.9 mol O2 1.25 mol CO2 2.25 mol H2O 16.08 mol N2 The energy balance is simply Q = H, where H is the enthalpy change between the feed stream and the flue gas stream and Q is specified as -8105 kJ per kgmol of fuel (or J per mol of fuel). We can compute the total H from feed to flue gas using a hypothetical three-step process in which we cool the feed from 30C to 25C, then react it at 25C, then heat it to the flue gas outlet temperature (which we are supposed to find). So, we have

H 

298.15 K

303.15 K

o C pfeed dT  H rxn ,298.15 K 

T flue

C pflue dT  Q

298.15 K

Updated 18/01/2017 40 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

We know Q, and we can compute the first heat capacity integral and the standard enthalpy of reaction, then we can find the flue gas temperature (by trial and error) that satisfies the above energy balance. For the integral of the feed heat capacity, we can use the same spreadsheet as always: T1 (K)

T2 (K)

303.15

298.15

A

B (1/K)

C (1/K )

D (K )

ICPH (K)

68.70413 0.023316

-3E-06

-32722.5

-3.75E+02

T2

Cp ICPH  dT A T2 T1  R

2

B 2 T2 T12 2

2

 DT1 T1

 C3 T

T13

0 0

3 2

2

T1

CH4 C2H6

1.702 1.131

9.08E-03 -2.16E-06 1.92E-02 -5.56E-06

O2 N2 CO2

3.369 3.28 5.457

5.06E-04 5.93E-04 1.05E-03

0 0 0

-2.27E+04 4.00E+03 -1.16E+05

H2O

3.47

1.45E-03

0

1.21E+04

1

For convenience, I have entered the heat capacity coefficients of the individual species, which I then add up (multiplied by the number of moles of each species) to get the total heat capacity coefficients of the mixture. Note that I can type in the formula once (for A) and then fill right to get B, C, and D. So, this gives us

298.15 K

C pfeed dT  375R  3118 J mol-1

303.15 K

The total heat of reaction is 0.75 times the enthalpy of reaction for CH4 + 2 O2  CO2 + 2 H2O, plus 0.25 times the enthalpy of reaction for CH4 + 2 O2  CO2 + 2 H2O, or Hrxn =0.75*(Hf(CO2) +2Hf(H2O) --- Hf(CH4)) + 0.25*(2Hf(CO2) +3Hf(H2O) --- Hf(C2H6)) Hrxn =1.25*Hf(CO2) +2.25Hf(H2O) ---0.75Hf(CH4)) --- 0.25Hf(C2H6)) Hrxn =1.25*(-393509)+2.25(-241818) ---0.75(-74520) --- 0.25(-83820) = -959132 J mol

-1

So, our energy balance becomes

H  3118 J mol-1  959132 J mol-1 

T flue

C pflue dT  800000 J mol-1

298.15 K

So,

T flue

C pflue dT = 162250 J mol-1. So, we can put the flue gas composition into the heat capacity

298.15 K

integral and try different final temperatures until we get this value: Updated 18/01/2017 41 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

ICPH*R T1 (K)

T2 (K)

A

298.15

543.5519 73.77225 0.015066

T2

Cp ICPH  dT A T2 T1  R

B (1/K)

C (1/K )

D (K )

ICPH (K)

(J/mol)

0

-96210

1.95E+04

162250.00

B 2 T2 T12 2

2

2

 DT1 T1

 C3 T

T13

0

3 2

T1

2

CH4

1.702

9.08E-03 -2.16E-06

C2H6 O2

1.131 3.369

1.92E-02 -5.56E-06 0 5.06E-04 0 -2.27E+04

N2 CO2 H2O

3.28 5.457 3.47

5.93E-04 1.05E-03 1.45E-03

0 0 0

1

4.00E+03 -1.16E+05 1.21E+04

A flue gas temperature of 543.6 K = 270C satisfies the energy balance. 4.38 (7th edition Prob. 4.34) On a basis of 1 mol of gas entering, we have 0.15 mol SO2, 0.2 mol O2, and 0.65 mol N2 in the inlet stream. If 86% of the SO2 is further oxidized to SO3, then 0.86*0.15 = 0.129 mol SO2 is oxidized, producing 0.129 mol SO3 by the reaction SO2 + ½ O2  SO3. This also consumes ½ * 0.129 = 0.0645 mol O2. So, in the exit stream we have 0.021 mol SO2, 0.129 mol SO3, 0.1355 mol O2, and 0.65 mol N2. Notice that for every 1 mol of gas entering, we only have a total of 0.9355 mol leaving. Since the entrance and exit temperatures are different, to compute the net heat removal requirement we could either compute the heat required to heat the reactants from 400C to 500C and then compute the heat of reaction at 500C, or we could compute the heat of reaction at 400 C and then compute the heat required to heat the products from 400C to 500C. The net heat requirement (the sum of these 2 steps) will be the same either way. Of course, we could also cool the whole inlet stream to 298K, add the heat of reaction at that temperature, and then heat the whole product stream to 500°C. That would also give the same answer.

Updated 18/01/2017 42 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Using the handy spreadsheet from the notes, we can compute the heat of reaction at 400C as follows: Reference Temperature T 0 (K) 298.15

Species Name SO2 O2 SO3

T Heat Capacity Heat of Heat of 1 1 1 B 2 C 3 1  C p dT    A  T  T0   T  T02  T  T03   D   Integral Reaction at T   Reaction at T 0 RT T  2 3  T T0   To kJ/mol (dimensionless) kJ/mol -98.890 0.053 -98.596   1 1  B 2 C 3 o o Heat Capacity Coefficients  H rxn T  T02  T  T03   D    ,T   H rxn ,T0  R   A  T  T0   2 3  T T0    Standard Heat Stoichiometric of Formation coefficient C (1/K2) D (K2) at T 0 (kJ/mol) A B (1/K) niHf,i niAi niBi niCi niDi -1 -296.83 5.699 8.01E-04 0.00E+00 -1.02E+05 2.97E+02 -5.70E+00 -8.01E-04 0.00E+00 1.02E+05 -0.5 0 3.639 5.06E-04 0.00E+00 -2.27E+04 0.00E+00 -1.82E+00 -2.53E-04 0.00E+00 1.14E+04 1 -395.72 8.06 1.06E-03 0.00E+00 -2.03E+05 -3.96E+02 8.06E+00 1.06E-03 0.00E+00 -2.03E+05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

Temperature of Interest T (K) 673.15

A 0.5415 Note: Light blue fields are inputs, pink fields are the final output.

B (1/K) 2.00E-06

C (1/K2) 0.00E+00

D (K2) -9.00E+04

So, the heat of reaction at 400C is -98.596 kJ per mol of SO2 reacted. If 0.129 mol reacts, then the heat released is -98.596*0.129 = -12.719 kJ. Now, we want to find the heat input required to increase the temperature of the products from 400C to 500C. We can make a modified heat capacity integral spreadsheet that adds up the total heat capacity of the mixture components to get the total heat capacity and then computes the integral. This is shown below, and can be downloaded by clicking here.

T1 (K) 673.15

T2 (K) 773.15

Total Values for Mixture T2 Cp  1 1 B C 3 C (1/K2) D (K2) ICPH (K*mol) A B (1/K) ICPH  dT  A T2  T1   T22  T12  T2  T13  D    3.78E+00 6.07E-04 0.00E+00 -2.88E+04 416.822 R 2 3  T2 T1 

T1

Species SO2 O2 SO3 N2

Moles 0.021 0.1355 0.129 0.65

Species Values A B (1/K) 5.699 0.000801 3.639 0.000506 8.06 0.001056 3.28 5.93E-04

Values * Moles 2 2 2 2 C (1/K ) D (K ) A mol B (mol/K) C (mol/K ) D (mol K ) 0.00E+00 -1.02E+05 1.20E-01 1.68E-05 0.00E+00 -2.13E+03 0.00E+00 -2.27E+04 4.93E-01 6.86E-05 0.00E+00 -3.08E+03 0.00E+00 -2.03E+05 1.04E+00 1.36E-04 0.00E+00 -2.62E+04 0.00E+00 4.00E+03 2.13E+00 3.85E-04 0.00E+00 2.60E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

So, the heat capacity integral is 416.8 K mol. Notice that since we multiplied the polynomial coefficients by the number of moles of each species, the integral is now in terms of the total heat capacity for the number of moles specified and has units of K mol, rather than just K. Multiplying by R = 0.008314 kJ mol K gives the heat requirement as 3.465 kJ. So, the reaction releases 12.719 kJ, and 3.465 kJ of this is required to heat the products to 500C. Therefore, -12.719 + 3.465 = -9.25 kJ has to be removed per mole of feed -1

gas fed to the catalytic converter. Updated 18/01/2017 43 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

4.39 (7th edition Prob. 4.35) This problem is very much like the previous one. We will work it on the basis of 1 mol of CO fed to the reactor. Thus, the feed is 1 mol CO and 1 mol H2O. If 65% of the H2O (and by stoichiometry therefore alse 65% of the CO) is converted, then the product is 0.35 mol CO, 0.35 mol H2O, 0.65 mol CO2, and 0.65 mol H2 reactants from 125C to the final temperature of 425C (698.15 K), and then compute the heat of reaction at 425C. We can use a spreadsheet just like the one used in the last problem: T1 (K) 398.15

T2 (K) 698.15

Total Values for Mixture C (1/K2) D (K2) A B (1/K) ICPH (K*mol) 6.85E+00 2.01E-03 0.00E+00 9.00E+03 2393.554

T2

ICPH 

Cp

B

2

1

2 2

2 1

C

3 2

3 1

T1

Species CO H2O

Moles

1

1

 R dT  AT  T   2 T  T   3 T  T   D  T  T 

Species Values Values * Moles 2 2 2 2 A B (1/K) C (1/K ) D (K ) A mol B (mol/K) C (mol/K ) D (mol K ) 1 3.376 5.57E-04 0.00E+00 -3.10E+03 3.38E+00 5.57E-04 0.00E+00 -3.10E+03 1 3.47 1.45E-03 0.00E+00 1.21E+04 3.47E+00 1.45E-03 0.00E+00 1.21E+04 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

Multiplying the heat capacity integral by R gives Q = 2393.5*0.008314 = 19.90 kJ required to heat the feed from 398.15 K to 698.15 K. Now, we compute the heat of reaction at 698.15 K in the same way as in the lecture notes and the previous problem:

Updated 18/01/2017 44 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

2

1


SVNAS 8th Edition Annotated Solutions

Reference Temperature T 0 (K) 298.15

Species Name CO H2O CO2 H2

Chapter 4

T Heat Capacity Heat of Heat of 1 1 1 B 2 C 3 1  C p dT    A  T  T0   T  T02  T  T03   D   Integral Reaction at T   Reaction at T 0 RT T  2 3  T T0   To kJ/mol (dimensionless) kJ/mol -41.166 0.591 -37.734   1 1  B 2 C 3 o o Heat Capacity Coefficients  H rxn T  T02  T  T03   D    ,T   H rxn ,T0  R   A  T  T0   2 3   T T0    Standard Heat Stoichiometric of Formation coefficient C (1/K2) D (K2) at T 0 (kJ/mol) A B (1/K) niHf,i niAi niBi niCi niDi -1 -110.525 3.376 5.57E-04 0.00E+00 -3.10E+03 1.11E+02 -3.38E+00 -5.57E-04 0.00E+00 3.10E+03 -1 -241.818 3.47 1.45E-03 0.00E+00 1.21E+04 2.42E+02 -3.47E+00 -1.45E-03 0.00E+00 -1.21E+04 1 -393.509 5.457 1.05E-03 0.00E+00 -1.16E+05 -3.94E+02 5.46E+00 1.05E-03 0.00E+00 -1.16E+05 1 0 3.249 4.22E-04 0.00E+00 8.30E+03 0.00E+00 3.25E+00 4.22E-04 0.00E+00 8.30E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

Temperature of Interest T (K) 698.15

A 1.86 Note: Light blue fields are inputs, pink fields are the final output.

B (1/K) -5.40E-04

C (1/K2) 0.00E+00

D (K2) -1.16E+05

So, the heat of reaction is ---37.734 kJ per mol of CO reacted. If 0.6 mol of CO reacts, then 37.734*0.6 = 22.64 kJ is released by the reaction. So, the net heat removal rate is 22.64 --- 19.90 = 2.74 kJ per mol of CO fed to the reactor (1.37 kJ per mole of feed). The exothermicity of the reaction is greater than the heat needed for the temperature increase, so the net result is that heat has to be removed. 4.40 (7 edition Prob. 4.36) th

BASIS: 100 lbmol DRY flue gases containing 3.00 lbmol CO2 and 11.80 lbmol CO x lbmol O2 and 100-(14.8-x)= 85.2-x lbmol N2. The oil therefore contains 14.80 lbmol carbon; a carbon balance gives the mass of oil burned:

The oil also contains H2O:

Also H2O is formed by combustion of H2 in the oil in the amount:

Find amount of air entering by N2 & O2 balances. N2 entering in oil:

lbmol N2 entering in the air=(85.2-x)-0.149 =85.051-x lbmol O2 in flue gas entering with dry air = 3.00 + 11.8/2 + x + 12.448/2 = 15.124 + x lbmol (CO2) (CO) (O2) (H2O from combustion) Updated 18/01/2017 45 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Total dry air = N2 in air + O2 in air = 85.051 - x + 15.124 + x = 100.175 lbmol Since air is 21 mol % O2,

O2 in air = 15.124 + x = 21.037 lbmol N2 in air = 85.051 - x = 79.138 lbmol N2 in flue gas = 79.138 + 0.149 = 79.287 lbmol [CHECK: Total dry flue gas = 3.00 + 11.80 + 5.913 + 79.287 = 100.00 lbmol] Humidity of entering air, sat. at 77 F in lbmol H2O/lbmol dry air, P(sat)=0.4594(psia)

lbmol H2O entering in air: 0.03227⋅100.175⋅lbmol = 3.233 lbmol If y = lbmol H2O evaporated in the drier, then lbmol H2O in flue gas: 0.116+12.448+3.233+y = 15.797 + y Entering the process are oil, moist air, and the wet material to be dried, all at 77 F. The "products" at 400 F consist of: 3.00 lbmol CO2 11.80 lbmol CO 5.913 lbmol O2 79.287 lbmol N2 (15.797 + y) lbmol H2O(g) Energy balance: Q = ΔH = ΔH298 + ΔHP where Q = 30% of net heating value of the oil:

Reaction upon which net heating value is based: OIL + (21.024)O2 = (14.8)CO2 + (12.448 + 0.116)H2O(g) + (0.149)N2 − ⋅ ⋅ − × To get the "reaction" in the drier, we add to this the following: Updated 18/01/2017 46 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

(11.8)CO2 = (11.8)CO + (5.9)O2 ⋅ −

(y)H2O(l) = (y)H2O(g) ⋅ ⋅ ⋅ [The factor 0.42993 converts from joules on the basis of moles to BTU on the basis of lbmol.] Addition of these three reactions gives the "reaction" in the drier, except for some O2, N2, and H2O that pass through unchanged. Addition of the corresponding delta H values gives the standard heat of reaction at 298 K: For the product stream we need MCPH:

Given Solve for y = 49.782 lbmol 49.782*18.015/209.13 = 4.288 lb H2O evap. per lb oil burned

4.41 (7th edition Prob. 4.37) If we work from a basis of 1 mole of N2 and 1 mole of C2H2 entering the reactor, then we have 2 moles of gas entering the reactor, and (since there is no mole change in the reaction) 2 moles of gas leaving the reactor. If this case is 24.2 mol-% HCN, then the total number of moles of HCN is 2*0.242 = 0.484. The number of moles of reaction that has occurred (as written) is 0.242 moles, and the number of moles of N2 and C2H2 remaining are (1-0.242) = 0.758 moles each. The total heat supplied to the reactor, on this basis, will therefore be 0.242 times the heat of reaction at 298 K (conveniently also the temperature at Updated 18/01/2017 47 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

which the reactants enter the reactor) plus the heat required to heat 0.748 mol N2, 0.748 mol C2H2 and 0.484 mol HCN from 298 K to 873 K. The heat of reaction is Hrxn = 2*Hf(HCN(g)) - Hf(N2(g)) - Hf(C2H2(g)) Hrxn = 2*135100 --- 0 --- 227480 = 42720 J mol-1 Since 0.242 mol of reaction occurs, 0.242*42720 = 10338 J of heat input is required to carry out the reaction at 298.15 K We can use a handy-dandy spreadsheet like the one provided in the notes to compute the heat capacity integral for the products: T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

T0(K) 298.15

T (K) 873.15

A 9.42652

B (1/K) 0.002587

C (1/K2) 0

C2H2 N2 HCN

0.758 0.758 0.484

6.132 3.28 4.736

1.95E-03 5.93E-04 1.36E-03

0 0 0

ig D (K2) ICPH (K) H (J/mol) -130522.2 6003.1 49912.6

-1.30E+05 4.00E+03 -7.25E+04

So, the total heat input is 10338 + 49913 = 60250 J per mole of N2 and C2H2 entering the reactor. This is 60250/2 = 30125 J per mole of total inlet or outlet stream, or 60250/0.484 = 124500 J per mole of HCN produced. 4.42 (7th edition Prob. 4.38) Chlorine is produced by the reaction: ( g ) + 2 C 1 2 ( g ) . The feed stream to the reactor consists of 60 mol-% HCl, 36 mol-% O2, and 4 mol-% N2, and it enters the reactor at 550°C. If the conversion of HCl is 75% and if the process is isothermal, how much heat must be transferred to or from the reactor per mole of the entering gas mixture?

BASIS: 1 mole gas entering reactor, containing 0.6 mol HCl, 0.36 mol O2, and 0.04 mol N2. HCl reacted = (0.6)(0.75) = 0.45 mol 4HCl(g) + O2(g) = 2H2O(g) + 2Cl2(g) Evaluate ΔH823 by Eq. (4.23) with Updated 18/01/2017 48 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Heat transferred per mol of entering gas mixture:

4.43 (7 edition Prob. 4.39) th

CO2 + C = 2CO 2C + O2 = 2CO Eq. (4.22) applies to each reaction: For (a)

H1148  H 298  MCPH * R * T  T0   1.696*105

J mol

For (b)

H1148  H 298  MCPH * R * T  T0   2.249*105

J mol

The combined heats of reaction must be 0

Updated 18/01/2017 49 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

r H1148b / H1148a  1.327 For 100 mol flue gas and x mol air, moles are:

CO2 CO O2 N2

Flue gas 12.8 3.7 5.4 78.1

Air 0 0 0.21x 0.79x

Feed mix 12.8 3.7 5.4 + 0.21x 78.1 + 0.79x

Whence in the feed mix: Flue gas to air ratio = 100/20.218 = 4.946 Product composition nCO := 3.7 + 2⋅(12.8 + 5.4 + 0.21⋅19.155 ) = 48.145 nN2= 78.1 + 0.79* 19.155 = 93.232 mol % CO = 34.07 % mol % N2 = 65.92 % 4.44 (7th edition Prob. 4.40)

CH4 + 2O2 = CO2 + 2H2O(g) CH4 + (3/2)O2 = CO + 2H2O(g) BASIS: 1 mole of fuel gas consisting of 0.94 mol CH4 and 0.06 mol N2 Air entering contains

1.35 2 0.94 = 2.538 mol O2 2.538 79/21= 9.548 mol N2

Moles CO2 formed by reaction = 0.94⋅0.7 = 0.658 Moles CO formed by reaction = 0.94⋅0.3 = 0.282 Updated 18/01/2017 50 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Moles H2O formed by reaction = 0.94⋅2.0 = 1.88 Moles O2 consumed by reaction = 2⋅0.658 + (3/2)⋅0.282 = 1.739 Product gases contain the following numbers of moles: (1) CO2: 0.658 (2) CO: 0.282 (3) H2O: 1.880 (4) O2: 2.538 - 1.739 = 0.799 (5) N2: 9.548 + 0.060 = 9.608

H P : R  MCPH  298.15K , 483.15K , A, B, C, D    483.15K  298.15K   0.7541*105 J/mol Energy balance: H rx  H 298 H P  599.2

KJ mol

⋅ From Table C.1:

=16.635 mol/s Volumetric flow rate of fuel, assuming ideal gas:

4.45 (7 edition Prob. 4.41) th

C4H8(g) = C4H6(g) + H2(g)

Using the basis of 1 mole C4H8 entering and only 33% reacts. The unreacted C4H8 and the diluent

H2O pass throught the reactor unchanged, and need not be included in the energy balance. MCPH A 3.2638

B C D T0 tau T 4.016 -0.004 0.000000991 8300 298.15 2.677008217 798.15

Then Updated 18/01/2017 51 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

4.46 (7th edition Prob. 4.42)

Assume Ideal Gas and P = 1 atm

(a) ⋅

(b)

4.47 (7th edition Prob. 4.43)

Assume Ideal Gas and P = 1 atm (a)

= -91.8 R

(b)

Updated 18/01/2017 52 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

4.48 (7th edition Prob. 4.44)

First calculate the standard heat of combustion of propane C3H8 + 5O2 = 3CO2(g) + 4H2O (g) Cost = 2.20 Dollars/gal Estimate the density of propane using the Rackett equation

4.49 (7 edition Prob. 4.45) th

The heat required to take an ideal gas from one temperature to another is given by T2

T2

 

H  C p dT  R A  BT  CT 2  DT 2 dT T1

T1

  1 1  B C 3 H  R  A T2  T1   T22  T12  T2  T13  D     2 3   T2 T1   where the temperature dependence of the heat capacity has been expressed in the polynomial form used in

SVNA. If we have set up a spreadsheet to evaluate the heat capacity integral like the example spreadsheet from the lecture notes, we can use it to evaluate the heat capacity integral in each case. (a) Using the coefficients for acetylene in the heat capacity polynomial, we have:

Updated 18/01/2017 53 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T 1 (K) 298.15

T 2 (K) 773.15 T2

ICPH 

R

Cp

Chapter 4

A 6.132

C (1/K2) 0

B (1/K) 1.95E-03

dT  A  T2  T1  

T1

D (K2) ICPH (K) H (kJ/mol) -1.30E+05 3.14E+03 26.12

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 

26.12 kJ/mol is required to heat acetylene from 25C to 500 C. (b) Using the coefficients for ammonia in the heat capacity polynomial, we have: T 1 (K) 298.15

T 2 (K) 773.15 T2

ICPH 

A 3.578

C (1/K2) 0

B (1/K) 3.02E-03

D (K2) ICPH (K) H (kJ/mol) -1.86E+04 2.43E+03 20.20

 1

1

 R dT  AT  T   2 T  T   3 T  T   D T  T  Cp

B

2

2 2

1

C

2 1

3 2

3 1

2

T1

1

20.20 kJ/mol is required to heat ammonia from 25C to 500 C. (c ) the heat required to take an ideal gas from one temperature to another is given by T2

T2

 

H  C p dT  R A  BT  CT 2  DT 2 dT T1

T1

  1 1  B C 3 H  R  A T2  T1   T22  T12  T2  T13  D     2 3   T2 T1   Where the temperature dependence of the heat capacity has been expressed in the polynomial form. If we

have set up a spreadsheet to evaluate the heat capacity integral, we can use it to evaluate the heat capacity integral in each cases. Using the coefficients for n-butane in the heat capacity polynomial, we have T0 = 298.15 K and T = 773.15 K H 71.96 kJ/mol is required to heat n-butane from 250 C to 5000 C. (d) c ) the heat required to take an ideal gas from one temperature to another is given by T2

T2

 

H  C p dT  R A  BT  CT 2  DT 2 dT T1

T1

  1 1  B C 3 H  R  A T2  T1   T22  T12  T2  T13  D     2 3   T2 T1  

Updated 18/01/2017 54 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

Where the temperature dependence of the heat capacity has been expressed in the polynomial form. If we have set up a spreadsheet to evaluate the heat capacity integral, we can use it to evaluate the heat capacity integral in each cases. Using the coefficients for n-butane in the heat capacity polynomial, we have T0 = 298.15 K and T = 773.15 K H 0 0 71.96 kJ/mol is required to heat n-butane from 25 C to 500 C.

(e) Using the coefficients for carbon monoxide in the heat capacity polynomial, we have: T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T 0(K) 298.15

T (K) 773.15

A 3.376

1 1  B 2 C T  T02  T 3  T03  D    2 3  T T0 

C (1/K ) D (K ) B (1/K) ICPH (K) H (J/mol) 5.57E-04 0.00E+00 -3.10E+03 1738.9 14457.5 2

2

14.5 kJ/mol is required to heat carbon monoxide from 25C to 500 C. (f) Using the coefficients for ethane in the heat capacity polynomial, we have: T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T 0(K) 298.15

T (K) 773.15

A 1.131

1 1  B 2 C T  T02  T 3  T03  D    2 3  T T0 

C (1/K ) D (K ) B (1/K) ICPH (K) H (J/mol) 1.92E-02 -2.16E-06 0.00E+00 5114.5 42521.6 2

2

42.5 kJ/mol is required to heat ethane from 25C to 500 C. 4.50 (7th edition Prob. 4.46) This is just like the previous problem, except that instead of being given the final temperature and computing Q, we are given Q and must compute the final temperature. To do so, we can just try different values of the final temperature in the spreadsheet for evaluating the heat capacity integral until we get the desired Q

el to automate this trial and error.

(a) For acetylene starting at 25C, the final temperature for a heat input of 30 kJ/mol is 835 K or 562C:

Updated 18/01/2017 55 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T 1 (K) 298.15

T 2 (K) 835.36 T2

ICPH 

R

Cp

A 6.132

dT  A  T2  T1  

T1

Chapter 4

B (1/K) 1.95E-03

C (1/K2) 0

D (K2) ICPH (K) H (kJ/mol) -1.30E+05 3.61E+03 30.00

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 

(b) For ammonia starting at 25C, the final temperature for a heat input of 30 kJ/mol is 964 K or 691C: T 1 (K) 298.15

T 2 (K) 964.00 T2

ICPH 

R

Cp

A 3.578

dT  A  T2  T1  

T1

B (1/K) 3.02E-03

C (1/K2) 0

D (K2) ICPH (K) H (kJ/mol) -1.86E+04 3.61E+03 30.00

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 

(c) For n-butane starting at 25C, the final temperature for a heat input of 30 kJ/mol is 534 K or 261 C: T 1 (K) 298.15

T 2 (K) 534.4073

A 1.935

T2

Cp ICPH  dT A T2 T1  R

C (1/K2) D (K2) B (1/K) ICPH (K) Q (J/mol) 3.69E-02 -1.14E-05 0.00E+00 3608 30000

B 2 T2 T12 2

 C3 T

3 2

T13

 DT1 T1 2

T1

1

(d) For CO2 starting at 25C, the final temperature for a heat input of 30 kJ/mol is 933 K or 660 C: T 1 (K) 298.15

T 2 (K) 932.9464 T2

A 5.457

Cp ICPH  dT A T2 T1  R T1

C (1/K2) D (K2) B (1/K) ICPH (K) Q (J/mol) 1.05E-03 0.00E+00 -1.16E+05 3608 30000

B 2 T2 T12 2

 C3 T

3 2

T13

 DT1 T1 2

1

(e) For carbon monoxide starting at 25C, the final temperature for a heat input of 30 kJ/mol is 1248 K or 975C:

Updated 18/01/2017 56 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T2

ICPH T0,T;A,B,C,D  

R

Cp

Chapter 4

dT  A T  T0  

T1

T 0(K) 298.15

T (K) 1248.143

A 3.376

1 1  B 2 C T  T02  T 3  T03  D    2 3  T T0 

C (1/K ) D (K ) B (1/K) ICPH (K) H (J/mol) 5.57E-04 0.00E+00 -3.10E+03 3608.4 30000.0 2

2

(f) For ethane starting at 25C, the final temperature for a heat input of 30 kJ/mol is 664 K or 391C: T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T 0(K) 298.15

T (K) 664.2338

A 1.131

1 1  B 2 C T  T02  T 3  T03  D    2 3  T T0 

C (1/K ) D (K ) B (1/K) ICPH (K) H (J/mol) 1.92E-02 -2.16E-06 0.00E+00 3608.4 30000.0 2

2

(g) For hydrogen starting at 25C, the final temperature for a heat input of 30 kJ/mol is 1298 K or 1025C: T2

ICPH T0,T;A,B,C,D  

1

B

2

0

C

2 0

3

3 0

0

T1

T0(K) 298.15

T (K) 1298.38

1 

 R dT  AT  T   2 T  T   3 T  T   D  T  T  Cp

A 3.249

B (1/K) 4.22E-04

C (1/K2) 0

ig D (K2) ICPH (K) H (J/mol) 8.30E+03 3608.2 30000.0

(h) For hydrogen chloride starting at 25C, the final temperature for a heat input of 30 kJ/mol is 1277 K or 1004C: T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T0(K) 298.15

T (K) 1276.95

A 3.156

B (1/K) 6.23E-04

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

C (1/K2) 0

ig D (K2) ICPH (K) H (J/mol) 1.51E+04 3608.2 30000.0

4.51 (7th edition Prob. 4.47) Given (a) Guess mole fraction of methane: y=0.5 Given

y = 0.637 Updated 18/01/2017 57 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

(b) Guess mole fraction of benzene: y=0.5 Given

y = 0.245 (c) Guess mole fraction of benzene: y=0.5 Given

y = 0.512

4.52 (7th edition Prob. 4.48)

To solve the problem, apply an energy balance equation TH 1

mc × H = nH  CP dT TH 2

mc × H = nH × R × ICPHair (a) TH 1  1273.15 K and TH 2  308.15 K Tin = 25°C = 298.15 K, Tboiling = 100°C = 373.15 K, and Tout = 100°C = 373.15 K ΔH = ΔHliquid + ΔHboiling + ΔHvapor

Updated 18/01/2017 58 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

H liquid = R × ICPH = 8.314 J·mol1·K 1 × 683.3399 K = 5681.288 J·mol1 H vapor = R × ICPH = 8.314 J·mol1·K 1 × 0 K = 0 J·mol1 H boiling = H latent × mH = 2257 J·g 1 × 18 g·mol1 = 40626 J·mol1 H = 46307.29 J·mol1

For air,

Assume as a basis

= 1 mol/s.

ICPHair = 3672.35 K mc =

nH × R × ICPH air = 0.6593 mol·s 1 H

·

mc ·

= 0.6593

nH

(b)

To solve the problem, apply an energy balance equation TH 1

mc × H = nH  CP dT TH 2

mc × H = nH × R × ICPHair (a) TH 1  773.15 K and TH 2  308.15 K Tin = 25°C = 298.15 K, Tboiling = 100°C = 373.15 K, and Tout = 100°C = 373.15 K Updated 18/01/2017 59 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

ΔH = ΔHliquid + ΔHboiling + ΔHvapor

H liquid = R × ICPH = 8.314 J·mol1·K 1 × 683.3399 K = 5681.288 J·mol1 H vapor = R × ICPH = 8.314 J·mol1·K 1 × 0 K = 0 J·mol1 H boiling = H latent × mH = 2257 J·g 1 × 18 g·mol1 = 40626 J·mol1 H = 46307.29 J·mol1

For air,

Assume as a basis

= 1 mol/s.

ICPHair = 1701.51 K mc =

nH × R × ICPH air = 0.30548 mol·s 1 H

·

mc ·

= 0.30548

nH

4.53 (7th edition Prob. 4.49) Saturated because the large ΔHlv overwhelms the sensible heat associated with superheat. Water because it is cheap, available, non-toxic, and has a large ΔHlv. The lower energy content is a result of the decrease in ΔHlv with increasing T, and hence P. However, higher pressures allow higher temperature levels. 4.54 (7th edition Prob. 4.50) Updated 18/01/2017 60 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

(a) The reaction for oxidation of glucose to CO2 and water is the same as the combustion reaction: C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l) The heat of reaction for this reaction is Hrxn = 6*Hf(CO2(g)) + 6*Hf(H2O(l)) --- Hf(C6H12O6(s)) --- 6*Hf(O2(g)) Hrxn = 6*(-393.509) + 6*(-285.83) --- (-1274.4) --- 6*(0) = -2801.6 kJ/mol (b) The total energy consumption for a 57 kg person (not a very big person) is 150 kJ/kg * 57 kg = 8550 kJ. Given that oxidation of glucose releases 2801.6 kJ/mol, this requires 8550/2802 = 3.05 moles of glucose per day. The molecular weight of glucose is 180.16 g/mol, so this is 550 g of glucose. (c) Oxidation of 3.05 moles of glucose produces 18.3 moles of CO2, or 806 g of CO2. Multiplying this by 275 million people gives 2.2210 g/day = 2.2210 kg/day = 222000 metric tons per day = 81 million metric tons per year. 11

8

4.55 (7 edition Prob. 4.51) th

Assume as a basis, 1 mole of fuel. 0.85 (CH4(g) + 2 O2(g) = CO2(g) + 2 H2O(g)) 0.10(C2H6 (g) + 3.5 O2(g) = 2 CO2(g) + 3 H2O(g)) -----------------------------------------------------------------0.85 CH4(g) + 0.10 C2H6(g) + 2.05 O2(g) = 1.05 CO2(g) + 2 H2O(g)

(a)

(b) For complete combustion of 1 mole of fuel and 50% excess air, the exit gas will contain

the following numbers of moles:

Air and fuel enter at 25 C and combustion products leave at 600 C.

Updated 18/01/2017 61 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 4

⋅ ⋅

⋅ ⋅

⋅ ⋅

Updated 18/01/2017 62 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

Shown at the bottom is a PV diagram with two adiabatic lines 1 → 2 and 2 → 3, assumed to intersect at point 2. A cycle is formed by an isothermal line from 3 → 1. An engine traversing this cycle would produce work. For the cycle U = 0, and therefore by the first law, Q + W = 0. Since W is negative, Q must be positive, indicating that heat is absorbed by the system. The net result is therefore a complete conversion of heat taken in by a cyclic process into work, in violation of Statement 1a of the second law (Pg. 160). The assumption of intersecting adiabatic lines is therefore false.

W   QH

 1

TC TH

 T   323 K  -1 -1 W  1  C  QH  1   250 kJ s  148.8 kJ s T 798 K   H  

QC  QH  W  250 kJ s-1  148.8 kJ s-1  101.2 kJ s-1

Updated 4/5/2017

p. 1 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

 1

Chapter 5

TC 300 K 1  0.6 TH 750 K

thermal  1 

QC QH

 1

TC 303  1  0.514 TH 623 

thermal  0.55 1  

 303  TC    0.55 1    0.35 TH  TH  

The energy balance for the over-all process is written: Q = Ut + EK + EP Assuming the egg is not scrambled in the process, its internal-energy change after it returns to its initial temperature is zero. So too is its change in kinetic energy. The potential-energy change, however, is negative, and by the preceding equation, so is Q. Thus heat is transferred to the surroundings. The total entropy change of the process is: Stotal = St + Stsurr Just as Ut for the egg is zero, so is St . Therefore,

Since Q is negative, Stotal is positive, and the process is irreversible.

By Eq. (5.9) the thermal efficiency of a Carnot engine is: Differentiate: Since TC/TH is less unity, the efficiency changes more rapidly with TC than with TH. So in theory it is Updated 4/5/2017

p. 2 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

more effective to decrease TC. In practice, however, TC is fixed by the environment, and is not subject to control. The practical way to increase η is to increase TH. Of course, there are limits to this too.

 1

TC 113.7 K 1  0.625 TH 303.15 K

 

  

 

  Updated 4/5/2017

p. 3 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

 

  

S  C p ln

P T  R ln   T0  P0 

T2

Q  U  n Cv dT t

T1

Q  U t  nCv T

T 

Q 15000 J   500 K nCv 1.443 mol  2.5  8.31451J mol-1 K -1

S  C p ln

P 7 T 5  R ln    R ln 2  R ln 2  R ln 2  1.733R  14.4 J mol-1 K -1 T0 2  P0  2 

Updated 4/5/2017

p. 4 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

 

 

 

  

 

  

For an ideal gas with constant heat capacities, and for the changes T1 → T2 and P1 → P2, Eq. (5.14) can be rewritten as:

V2 = V1, then Whence,

Since CP > CV , this demonstrates that SP > SV . V2 = V1, then Whence,

This demonstrates that the signs for ST and SV are opposite. Updated 4/5/2017

p. 5 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

Start with the equation just preceding Eq. (5.10):

For an ideal gas PV = RT, and lnP + ln V = ln R + ln T . Therefore,

********************** As an additional part of the problem, one could ask for the following proof, valid for constant heat capacities. Return to the original equation and substitute dT/T = dP/P + dV/V:

As indicated in the problem statement the basic differential equations are: − − (A) − (B) where QC and QH refer to the reservoirs.

Updated 4/5/2017

p. 6 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

(a) With

and

Whence, d ln TC = − d ln TH

Chapter 5

, Eq. (B) becomes:

where

Integration from TH0 and TC0 to TH and TC yields:

With

and

, Eq. (B) becomes

Eliminate TC by the boxed equation of Part (a) and rearrange slightly:

For infinite time, TH =TC ≡T, and the answer of Part (a) becomes:

Because /(

Updated 4/5/2017

+ 1) − 1 = −1/(

+ 1), then:

p. 7 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

Because TH =T , substitution of these quantities in the answer of Part (b) yields:

As indicated in the problem statement the basic differential equations are: − − (A) − (B) where QC and QH refer to the reservoirs. (a) With

, Eq. (B) becomes:

Substitute for dQH and dQC in Eq. (A):

Integrate from TC0 to TC:

For infinite time, TC = TH, and the boxed equation above becomes:

Updated 4/5/2017

p. 8 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

Write Eqs. (5.9) in rate form and combine to eliminate |

|

H|=

H|:

this becomes:

Differentiate, noting that the quantity in square brackets is constant:

Equating this equation to zero, leads immediately to: 4r = 3 or r = 0.75 Hence

Updated 4/5/2017

p. 9 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

thermal  1 

QC QH

 1

Chapter 5

TC 300  1  0.5 TH 600

 

 

Q T W 300  H 1  H 1   1  0.2 QC QC TC 250

P  T S  C p ln 2  R ln  2  T1  P1 

 7 T  P  S  R  ln 2  ln  2    8.314  3.5ln1.5  ln 5   1.58 J mol-1 K -1 2 T  1  P1   

 7 T  P  S  R  ln 2  ln  2    8.314  3.5ln1.667  ln 5   1.48 J mol-1 K -1 2 T  1  P1   

P  T S  C p ln 2  R ln  2  T1  P1 

Updated 4/5/2017

p. 10 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

5 T  P  S  R  ln 2  ln  2    8.314  2.5ln 0.667  ln 0.2   4.96 J mol-1 K -1 2 T  1  P1   

P  T S  C p ln 2  R ln  2  T1  P1 

 9 T  P  S  R  ln 2  ln  2    8.314  4.5ln 0.75  ln 0.2   2.62 J mol-1 K -1 2 T  1  P1   

P  T S  C p ln 2  R ln  2  T1  P1 

   P  T S  R  3ln 2  ln  2    8.314  4ln 0.6  ln 0.2   3.61 J mol-1 K -1   T1  P1   

This cycle is shown below. Assuming constant specific heats, the efficiency is given by:

Temperature T4 is not given and must be calculated. The following equations are used to derive and expression for T4.

Updated 4/5/2017

p. 11 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

For adiabatic steps 1 to 2 and 3 to 4:

Updated 4/5/2017

p. 12 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

Because W =0, Eq. (2.3) here becomes: Q=

Ut = m CV T

A necessary condition for T to be zero when Q is non-zero is that m =∞. This is the reason that natural bodies (air and water) that serve as heat reservoirs must be massive (oceans) or continually

renewed (rivers).

T  P  Sad ,rev  C p ln  rev   R ln  2   0  T1   P1  T  P  C p ln  rev   R ln  2   T1   P1  R

Trev  P2  C p   T1  P1 

R

2

 P  Cp 7 7 Trev  T1  2   298.15    426.46 K  2  P1 

 

T  P   471.4  7 S  C p ln  2   R ln  2   7 / 2* R ln   R ln    0.3506R   298.15   2  T1   P1   Updated 4/5/2017

p. 13 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

T  P  Sad ,rev  C p ln  rev   R ln  2   0  T1   P1    T   P   T   P  S  Sad ,rev  S  C p ln  2   R ln  2    C p ln  rev   R ln  2    T1   P1     T1   P1     T  T  T  S  C p ln  2   C p ln  rev   C p ln  2   T1   T1   Trev 

An appropriate energy balance here is: Q = Ht = 0 Applied to the process described, with T as the final temperature, this becomes:

If m1 = m2 = m,

.

Since

St is positive

Isentropic processes are not necessarily reversible and adiabatic. The term isentropic denotes a process for which the system does not change in entropy. There are two causes for entropy changes in a system: The process may be internally irreversible, causing the entropy to increase; heat may be transferred between system and surroundings, causing the entropy of the system to increase or decrease. For processes that are internally irreversible, it is possible for heat to be transferred out of the system in an amount such that the entropy changes from the two causes exactly compensate each other. One can imagine irreversible processes for which the state of the system is the same at the end as at the Updated 4/5/2017

p. 14 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

beginning of the process. The process is then necessarily isentropic, but neither reversible nor adiabatic. More generally, the system conditions may change in such a way that entropy changes resulting from temperature and pressure changes compensate each other. Such a process is isentropic, but not necessarily reversible. Expansion of gas in a piston/cylinder arrangement is a case in point. It may be reversible and adiabatic, and hence isentropic. But the same change of state may be irreversible with heat transfer to the surroundings. The process is still isentropic, but neither reversible nor adiabatic. An isentropic process must be either reversible and adiabatic or irreversible and non-adiabatic.

By definition, By inspection, one sees that for both T > T0 and T0 > T the numerators and denominators of the above fractions have the same sign. Thus, for both cases <CP>H is positive. Similarly,

By inspection, one sees that for both T > T0 and T0 > T the numerators and denominators of the above fractions have the same sign. Thus, for both cases <CP>S is positive. When T = T0, both the numerators and denominators of the above fractions become zero, and the fractions are indeterminate. Application of l’Hˆopital’s rule leads to the result: <CP>H = <CP>S = CP.

Step 1-2: Volume decreases at constant P. Heat flows out of the system. Work is done on the system.

Step 2-3: Isothermal compression. Work is done on the system. Heat flows out of the system.

Step 3-1: Expansion process that produces work. Heat flows into the system. Since the PT product is constant,

Combine with (A) this becomes Updated 4/5/2017

p. 15 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

Moreover

Updated 4/5/2017

p. 16 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T

S  C p T0

Chapter 5

P dT  R ln   T  P0 

T

P S D A    B  CT  3  dT  ln   R T  T  P0 

T0

T  P S C 2 D 2  A ln    B T  T0   T  T02  T  T02  ln   R 2 2  T0   P0 

T1 (K) 473.15

T2 (K) 1373.15 T2

ICPS 

 T2 

C (1/K2) D (K2) B (1/K) 8.01E-04 0.00E+00 -1.02E+00

ICPS 6.793

S (J/(mol K)) 56.48

 RT dT  A ln  T   B T  T   2 T  T   2 T  T  Cp

T1 (K) 523.15

T2 (K) 1473.15 T2

T1

C

2

2 2

1

D

2 1

2 2

2 1

1

T1

ICPS 

A 5.699

A 1.213

C (1/K2) D (K2) B (1/K) 2.88E-02 -8.82E-06 0.00E+00

ICPS 20.234

T  C D 2 dT  A ln  2   B T2  T1   T22  T12  T2  T1 2 RT T 2 2  1

Cp

Updated 4/5/2017

S (J/(mol K)) 168.23

p. 17 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

T

S  C p

dT T

T0

T2

Q  H  n C p dT t

T1

Q 800000 J   9622 K= nR 10 mol  8.314 J mol-1 K -1

T1 (K) 473.15

T2 (K) 1374.5 T2

ICPH 

A 1.424

T2

 R dT Cp

473 K

C (1/K2) D (K2) B (1/K) 1.44E-02 -4.39E-06 0.00E+00

ICPH (K) 9.622E+03

 1

1

 R dT  AT  T   2 T  T   3 T  T   D  T  T  Cp

B

2

1

2 2

C

2 1

3 2

3 1

2

T1

S  R

1374.5 K

1

C p dT R T

473.15 K

T1 (K) 473.15

T2 (K) 1374.5 T2

ICPS 

T1

A 1.424

C (1/K2) D (K2) B (1/K) 1.44E-02 -4.39E-06 0.00E+00

ICPS 10.835

T  C 2 D 2 dT  A ln  2   B T2  T1   T2  T12  T2  T12 RT T 2 2  1

Cp

Updated 4/5/2017

p. 18 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

1374.5 K

S  R

C p dT R T

Chapter 5

 10.835

473.15 K

Q 2500000 J   20045 K= nR 15 mol  8.314 J mol-1 K -1

T1 (K) 533.15

T2 (K) 1413.7 T2

ICPH 

R

Cp

A 1.967

dT  A T2  T1  

T1

T1 (K) 533.15

T2 (K) 1413.7 T2

ICPS 

A 1.967

 T2 

T2

Cp R

dT

533.15 K

C (1/K2) D (K2) B (1/K) 3.16E-02 -9.87E-06 0.00E+00

ICPH (K) 2.004E+04

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 

C (1/K2) D (K2) B (1/K) 3.16E-02 -9.87E-06 0.00E+00

ICPS 21.307

 RT dT  A ln  T   B T  T   2 T  T   2 T  T  Cp

1

2 2

2 1

D

2 2

2 1

1

T1

S  R

C

2

1413.7 K

C p dT R T

 21.307

533.15 K

Q 1000000 Btu   12588 R = 6993 K = nR 40 lbmol  1.986 Btu lbmol-1 R -1

T2

Cp R

dT

533.15 K

 Updated 4/5/2017

p. 19 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T1 (K) 533.15

T2 (K) 1202.7 T2

ICPH 

R

Cp

A 1.424

dT  A T2  T1  

T1

Chapter 5

C (1/K2) D (K2) B (1/K) 1.44E-02 -4.39E-06 0.00E+00

ICPH (K) 6.993E+03

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 

T1 (K) 533.15

T2 (K) 1202.7 T2

ICPS 

T1

A 1.424

C (1/K2) D (K2) B (1/K) 1.44E-02 -4.39E-06 0.00E+00

ICPS 8.244

T  C 2 D 2 dT  A ln  2   B T2  T1   T2  T12  T2  T12 RT T 2 2  1

Cp

S  R

1202.7 K

C p dT R T

 8.244

533.15 K

248.15 K 298.15 K

1 bar

5 bar

348.15 K 1 bar

 Updated 4/5/2017

p. 20 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

The process involves three heat reservoirs: the house, a heat source; the tank, a heat source; and the surroundings, a heat sink. Notation is as follows: |Q| Heat transfer to the house at temperature T |QF| Heat transfer from the furnace at TF |Qσ | Heat transfer from the surroundings at Tσ The first and second laws provide the two equations:

Updated 4/5/2017

p. 21 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

Combine these equations to eliminate |Qσ |, and solve for |QF|:

With T = 295 K , TF = 810 K, Tσ = 265 K, and |Q’| = 1000 kJ Shown to the right is a scheme designed to accomplish this result. A Carnot heat engine operates with the furnace as heat source and the house as heat sink. The work produced by the engine drives a Carnot refrigerator (reverse Carnot engine) which extracts heat from the surroundings and discharges heat to the house. Thus the heat rejected by the Carnot engine (|Q1|) and by the Carnot refrigerator (|Q2|) together provide the heat |Q| for the house. The energy balances for the engine and refrigerator are:

|W|engine = |Q| − |Qσ1| |W|refrig = |Qσ2| − |Q’| Equation (5.7) may be applied to both the engine and the refrigerator:

Updated 4/5/2017

p. 22 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

The process involves three heat reservoirs: the house, a heat source; the tank, a heat source; and the surroundings, a heat sink. Notation is as follows: |Q| Heat transfer from the tank at temperature T |Q‘| Heat transfer from the house at T ‘ |Qσ | Heat transfer to the surroundings at Tσ The first and second laws provide the two equations:

Combine these equations to eliminate |Qσ |, and solve for |Q|:

With T = 448.15 K , T’ = 297.15 K, Tσ = 306.15 K, and |Q’| = 1500 kJ Shown below is a scheme designed to accomplish this result. A Carnot heat engine operates with the tank as heat source and the surroundings as heat sink. The work produced by the engine drives a Carnot refrigerator (reverse Carnot engine) which extracts heat |Q_| from the house and discharges heat to the surroundings. The energy balances for the engine and refrigerator are:

Updated 4/5/2017

p. 23 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

|W|engine = |Q| − |Qσ1| |W|refrig = |Qσ2| − |Q’| Equation (5.7) may be applied to both the engine and the refrigerator:

Updated 4/5/2017

p. 24 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Q  Sg T

Sg  

Chapter 5

135 J s1  0.45 J K -1 s-1 300 K

Q  Sg T

Updated 4/5/2017

p. 25 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Sg  

Chapter 5

2500 J s1  8.33 J K -1 s-1 300 K

For a closed system the first term of Eq. (5.16) is zero, and it becomes:

Where j is here redefined to refer to the system rather than to the surroundings. Nevertheless, the second term accounts for the entropy changes of the surroundings, and can be written simply as dStsurr/dt:

Multiplication by dt and integration over finite time yields:

The general equation applicable here is Eq. (5.17):

(a)

: For a single stream flowing within the pipe and with a single heat source in the surroundings, this becomes:

(b)

(c)

Updated 4/5/2017

p. 26 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

(d) (e) (f) (g)

P  T S  C p ln 2  R ln  2  T1  P1 

 7 T  P  S  R  ln 2  ln  2    8.314  3.5ln 0.742  ln 0.2   4.698 J mol-1 K -1 2 T  1  P1   

Updated 4/5/2017

p. 27 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

S   j

Qj Tj



 Sg

Wlost  T S g

P  T S  C p ln 2  R ln  2  T1  P1 

   P  T S  R  4ln 2  ln  2    8.314  4ln 0.8356  ln 0.4   1.643 J mol-1 K -1   T1  P1   

 Updated 4/5/2017



p. 28 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

S   j

Qj Tj

Chapter 5

 Sg

Wlost  T S g

P  T S  C p ln 2  R ln  2  T1  P1 

  11 T  P   11  S  R  ln 2  ln  2    8.314  ln 0.8724  ln 0.3   3.767 J mol-1 K -1 2 T  2  1  P1    

S   j

Updated 4/5/2017

Qj Tj



 Sg p. 29 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

Wlost  T S g

P  T S  C p ln 2  R ln  2  T1  P1 

 9 T  P  S  R  ln 2  ln  2    8.314  4.5ln 0.7832  ln 0.2143  3.663 J mol-1 K -1 2 T  1  P1   

S   j

Qj Tj



 Sg

Wlost  T S g

Updated 4/5/2017

p. 30 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

P  T S  C p ln 2  R ln  2  T1  P1 

 5 T  P  S  R  ln 2  ln  2    8.314  2.5ln 0.7327  ln 0.3  3.546 J mol-1 K -1 2 T  1  P1   

S   j

Qj Tj



 Sg

Wlost  T S g

The figure below indicates the direct, irreversible transfer of heat |Q| from a reservoir at T1 to a reservoir at T2. The figure on the right depicts a completely reversible process to accomplish the same changes in the heat reservoirs at T1 and T2.

Updated 4/5/2017

p. 31 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

The entropy generation for the direct heat-transfer process is:

For the completely reversible process the net work produced is Wideal:

This is the work that is lost, Wlost, in the direct, irreversible transfer of heat |Q|. Therefore,

Note that a Carnot engine operating between T1 and T2 would not give the correct Wideal or Wlost, because the heat it transfers to the reservoir at T2 is not Q.

Updated 4/5/2017

p. 32 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Sg  



Chapter 5

QH QC 1 0.55     6.7  105 kJ/K TH TC 523.15 298.15

W QH

1

TC 293.15 1  0.5016 TH 588.15

Equation (5.14) can be written for both the reversible and irreversible processes:

Updated 4/5/2017

p. 33 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

Since Sirrev must be greater than zero, Tirrev must be greater than Trev.

T 1 (K) 293.15

T 2 (K) 300.15

A 3.355

T2

Cp ICPH  dT A T2 T1  R

B (1/K) 5.75E-04

C (1/K2) 0

D (K2) ICPH (K) -1.60E+03 24.55

B 2 T2 T12 2

 C3 T

T13

B (1/K) 5.75E-04

C (1/K2) 0

D (K2) ICPH (K) -1.60E+03 -146.57

 C3 T

T13

3 2

T 1 (K) 293.15

T 2 (K) 251.15 T2

A 3.355

Cp ICPH  dT A T2 T1  R T1

Updated 4/5/2017

B 2 T2 T12 2

3 2

 DT1 T1 2

T1

DH (J/mol) 204.12

1

DH (J/mol) -1218.58

 DT1 T1 2

1

p. 34 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

A 3.355

B (1/K) 5.75E-04

C (1/K2) 0

Cp T ICPS  dT A ln  2 RT T1

B T2 T1 

T 1 (K) 293.15

A 3.355

Cp T ICPS  dT A ln  2 RT T1

T 1 (K) 293.15

T 2 (K) 303.15

D (K2) -1.60E+03

ICPS 0.118

C 2 T2 T12 2

 D2 T

T1 2

B (1/K) 5.75E-04

C (1/K2) 0

D (K2) -1.60E+03

ICPS -0.540

B T2 T1 

C 2 T2 T12 2

 D2 T

T1 2

T2

 2 2

T1

T 2 (K) 251.15 T2

 2 2

T1

n

pV 1*100000   258.6 lbmol hr 1 RT 0.7302*529.67

T2

ICPH T0,T;A,B,C,D  

R

T1

Updated 4/5/2017

Cp

dT  A T  T0  

1 1  B 2 C T  T02  T 3  T03  D    2 3  T T0 

p. 35 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T

ICPS  T0,T;A,B,C,D 

Chapter 5

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

n

pV 101325*3000   122629 mol hr 1  34.06 mol s 1 RT 8.314*298.15

T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T0(K) 298.15

T (K) 265.15

A 3.355 T

B (1/K) 5.75E-04

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

ig C (1/K2) D (K2) ICPH (K) H (J/mol) 0 -1.60E+03 -115.4 -959.4

 T  D  T 2  T02  ICPS  T0,T;A,B,C,D  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

T0 (K) 298.15

T (K) 265.15

Updated 4/5/2017

A 3.355

B (1/K) 5.75E-04

C (1/K2) D (K2) 0 -1.60E+03

ICPS -0.410

Sig (J/(mol K)) -3.4099

p. 36 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5



Calculate the rate of entropy generation in the boiler. This is the sum of the entropy generation of the steam and the gas.

Calculate entropy generation per lbmol of gas:

(b)

Calculate lbs of steam generated per lbmol of gas cooled.

Use ratio to calculate ideal work of steam per lbmol of gas Updated 4/5/2017

p. 37 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

Calculate the rate of entropy generation in the boiler. This is the sum of the entropy generation of the steam and the gas.

Calculate entropy generation per lbmol of gas:

(b)

Calculate lbs of steam generated per lbmol of gas cooled. Updated 4/5/2017

p. 38 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 5

Use ratio to calculate ideal work of steam per lbmol of gas

Now place a heat engine between the ethylene and the surroundings. This would constitute a reversible process, therefore, the total entropy generated must be zero. calculate the heat released to the surroundings for ΔStotal = 0.

Now apply an energy balance around the heat engine to find the work produced. Note that the heat gained by the heat engine is the heat lost by the ethylene.

The lost work is exactly equal to the work that could be produced by the heat engine.

Updated 4/5/2017

p. 39 of 39

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


=

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


   G  V   RT  P  RT  T 

   G      RT  T  RT   P H

2

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


H RT

 2

 T  RT

H   T   T

S

1 d  T  RT dT d  T 

 2

dT

 d  T  1   T   RT ln P    T   T T dT 

S 

d  T  dT

 R ln P

U   T   T U   T   T

d  T  dT d  T  dT

 RT   P   P   RT

d  T   d  T  d  T  d  T  d  T    T   T   dT  dT  dT dT dT 2 2

Cp 

C p  T

Cv 

d 2  T  dT 2

 d  T  d  T  d 2  T  d  T  d   T  T  RT   T  R     dT  dT dT dT dT 2 

Cv  T

d 2  T  dT 2

 R  Cp  R

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


 dS      V  dP T  dH     1   T V  dP T

 S   V P H  1   T V P

S  2.095  103 K -1  1.551 103 m3 kg -1  819000 Pa = -2.66 Pa m3 kg-1 K -1 = -2.66 J kg-1 K -1

H  1  2.095  103 K -1  270 K  1.551 103 m3 kg -1  819000 Pa=552 J kg-1

dH  C p dT  1   T VdP

dT  

V 1   T  Cp

dP

T  

V 1   T  Cp

P

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


 2

 T  7 1  7 1  T T V sat  Vc Zc r   Vc Zc c  2

2    1 V   ln V    T  7    ln Vc  1   ln Z c   V T T T   Tc   

2ln Z c  T    1   7Tc  Tc 

5

7

2

 360  7 1  V  262.7 *0.282 408.1   132 cm3 mol-1  0.000132 m3 mol-1

 

2ln 0.282  360  1  7 * 408.1  408.1 

T  

5

7

 0.00408 K -1

0.000132 m3 mol-1 1  0.00408*360  2.78 J g -1 K -1 *58.123 g mol-1

 2000000 Pa   0.766 K

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


 dS      V  dP T  dH     1   T V  dP T

 1  V      V  P T

1 dV   dP V

V  ln      P  P1   V1 

V  V1 exp    P  P1  

V2  1003 cm3 kg-1 exp 45 106 bar -1 1499 bar  937.6 cm3 kg-1  9.376  104 m3 kg-1

 P2

W  Vave PdP 

Vave 2 P2  P12 2

P1

S  Vave P H  1   T Vave P

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


 W  45 10 bar *970.3 cm kg / 2* 1500  1  bar  49120 bar cm kg  4.91 kJ kg

V2  1003 cm3 kg-1 exp 45  106 bar -1  1499 bar  937.5741 cm3 kg-1  9.375741  104 m3 kg-1 6

1

3

-1

2

2

2

3

-1

1

S  250 106 K-1 *970.3 cm3 kg 1 *1499 bar  363.6 bar cm3 kg1 K1  36.4 J kg1 K1

H  1  250 106 K-1 *298.15 K *970.3 cm3 kg 1 *1499 bar  1.346 106 bar cm3 kg 1  134.6 kJ kg 1

 

 

 

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


P

RT a RT a  RT  2     2a 1 1 V b V 1  b b 2  

Z

P 1 a    RT 1  b  RT

GR  RT

d

  Z  1   Z  1  ln Z 0

 1 GR a  d      1  Z  1  ln Z RT  1  b  RT  

 0

GR  RT GR  RT

 1 a 1    d   Z  1  ln Z    1  b    RT   0

 

 1  1  b a     d   Z  1  ln Z 1  b    RT   0

 

 b GR a      d   Z  1  ln Z RT  1  b  RT 

 0

G a   ln 1  b     Z  1  ln Z RT RT R



p ZRT

GR bP  aP    ln 1    2 2  Z  1  ln Z RT  ZRT  ZR T

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.




bP a aP , and q   RT bRT  R 2T 2

GR    q   ln 1     Z  1  ln Z RT  Z Z

GR  Z    q   ln   Z  1  ln Z  RT  Z  Z

q GR  ln Z  ln  Z      Z  1  ln Z RT Z q GR  Z  1  ln  Z     RT Z

HR  Z  d   Z 1 T   RT  T   

 0

a  Z      T   RT 2

HR  Z 1 T RT

a d 

a

 RT   Z  1  RT  Z  1  ZR T 2

aP 2

0

q HR  Z 1 RT Z

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


S R H R GR   R RT RT q  q  SR  Z 1   Z  1  ln  Z     R Z  Z  SR  ln  Z    R

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


θ≡

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T 1 (K) 382.02

T 2 (K) 810.93

C (1/K ) D (K ) B (1/K) ICPH (K) H (J/mol) 1.45E-03 0.00E+00 1.21E+04 1876.0 15597.3 2

A 3.47

T2

2

1 T1

C (1/K ) D (K ) B (1/K) 1.45E-03 0.00E+00 1.21E+04

ICPS 3.266

S (J/(mol K)) 27.15

MW (g/mol) 18.02

B 2 T2 T12 2

 C3 T

3 2

T13

T1

T 1 (K) 382.02

T 2 (K) 810.93

A 3.47

2

T2

Cp T ICPS º dT A ln  2 RT T1

T1

MW (g/mol) 18.02

1 D  T2

Cp ICPH º dT A T2 T1  R

B T2 T1 

C 2 T2 T12 2

2

 D2 T

 2 2

T1 2

H (J/g) 865.6 H (Btu/lbm) 372.1

S (J/(g K)) 1.507 S (Btu/(lbm R)) 0.3599

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


 

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T2

ICPH T0,T;A,B,C,D  º

R

Cp

dT  A T  T0  

T1

T0(K) 723.15

T (K) 413.15

A 3.47

T

ICPS  T0,T;A,B,C,D º

B (1/K) 1.45E-03

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

C (1/K2) 0

ig D (K2) ICPH (K) H (J/mol) 1.21E+04 -1343.6 -11171.7

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

T0 (K) 723.15

T (K) 413.15

A 3.47

B (1/K) 1.45E-03

C (1/K2) 0

D (K2) 1.21E+04

ICPS -2.416

Sig (J/(mol K)) -20.0858

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 723.15

Tr 1.1175

P (bar) 30

Pr 0.1360

0.422

B 0.083 0

Tr1.6 0.172

B1 0.139

Tr4.2

Tc (K) 647.1

0

B -0.2703

Pc (bar) 220.55

1

B 0.0311

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

 0.345

dB0 dTr

dB1 dTr

0.5056

0.4051

 HR dB0  Pr B 0 Tr  RTc dTr 

HR (J mol -1) -717.78

HR /RTc -0.1334



  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

T (K) 723.15

Tr 1.1175

P (bar) 30

Pr 0.1360

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

Tc (K) 647.1

Pc (bar) 220.55

dB0 dTr

dB1 dTr

0.5056

0.4051

 0.345

SR /R -0.0878

dB 0 SR Pr  dT R  r

SR (J mol -1 K-1) -0.73

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 413.15

Tr 0.6385

P (bar) 2.35

Pr 0.0107

0.422

B 0.083 0

B1 0.139

Tr1.6 0.172 Tr4.2

Tc (K) 647.1

0

B -0.7822

Pc (bar) 220.55

1

B -0.9933

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

 0.345

dB0 dTr

dB1 dTr

2.1675

7.4444

 HR dB0  Pr B 0 Tr  RTc dTr 

HR (J mol -1) -237.82

HR /RTc -0.0442



  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

dB0 dTr

dB1 dTr

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr dB 0 SR Pr  dT R  r

dB1 dTr

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (°C) 127.41 127.54 127.67 127.80 127.94 128.07 128.19 128.32 128.45 128.58 128.71 128.84 128.96 129.09 129.22 129.34 129.47 129.59 129.72 129.84 129.97

P (MPa) H l (kJ/kg) S l (kJ/(kg K)) H v (kJ/kg) S v (kJ/(kg K)) H (kJ/kg) S (kJ/(kg K)) 0.250 0.251 0.252 0.253 0.254 0.255 0.256 0.257 0.258 0.259 0.260 0.261 0.262 0.263 0.264 0.265 0.266 0.267 0.268 0.269 0.270

535.34 535.91 536.46 537.02 537.58 538.13 538.68 539.23 539.78 540.33 540.87 541.42 541.96 542.50 543.04 543.57 544.11 544.64 545.18 545.71 546.24

1.6072 1.6086 1.6100 1.6114 1.6128 1.6142 1.6155 1.6169 1.6183 1.6196 1.6210 1.6223 1.6237 1.6250 1.6264 1.6277 1.6290 1.6303 1.6317 1.6330 1.6343

2716.5 2716.7 2716.9 2717.0 2717.2 2717.4 2717.6 2717.8 2717.9 2718.1 2718.3 2718.5 2718.6 2718.8 2719.0 2719.2 2719.3 2719.5 2719.7 2719.9 2720.0

7.0524 7.0511 7.0498 7.0485 7.0472 7.0458 7.0445 7.0432 7.0419 7.0407 7.0394 7.0381 7.0368 7.0355 7.0343 7.0330 7.0318 7.0305 7.0293 7.0280 7.0268

2607.4 2607.7 2607.9 2608.0 2608.2 2608.4 2608.7 2608.9 2609.0 2609.2 2609.4 2609.6 2609.8 2610.0 2610.2 2610.4 2610.5 2610.8 2611.0 2611.2 2611.3

6.7801 6.7790 6.7778 6.7766 6.7755 6.7742 6.7731 6.7719 6.7707 6.7696 6.7685 6.7673 6.7661 6.7650 6.7639 6.7627 6.7617 6.7605 6.7594 6.7583 6.7572

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


 

 

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


 

   

 

H  T V

dP sat -1  818.1* 2.996 *1.888  4628 ft 3psia lb-1 m  856.5 Btu lb m dT

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 468.15

P (bar) 135

Tc (K) 369.8

Pc (bar) 42.48

Tr 1.20 1.20 1.30 1.30

Pr 3.00 5.00 3.00 5.00

 0.152

Tr 1.2660

Pr 3.1780

Z1 0.1095 -0.0141 0.2079 0.0875

(HR )0/(RTc) -2.801 -3.166 -2.274 -2.825

0.1636

-2.4968

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Z0 0.5425 0.7069 0.6344 0.7358

Interpolated Values Final Values Z

0.6389

3

V (cm /mol)

184.2

T2

ICPH T0,T;A,B,C,D  º

R

Cp

T (K) 468.15

A 1.213 T

ICPS  T0,T;A,B,C,D º

HR /(RTc) R

H (J/mol)

dT  A T  T0  

T1

T0(K) 308.15

0.6140

(HR )1/(RTc) -0.934 -1.84 -0.3 -1.066

(SR )0/R -1.727 -1.827 -1.299 -1.554

(SR )1/R -0.991 -1.767 -0.29 -0.73

-1.4627

-0.5780

-0.5883

-2.58623

SR /R

-1.55056

-7951.4

SR (J/(mol K))

-12.8914

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

ig C (1/K2) D (K2) B (1/K) ICPH (K) H (J/mol) 2.88E-02 -8.82E-06 0.00E+00 1766.0 14683.6

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


H   H R  400 K,38 bar   

Tf

400 K

H   H R  400 K,38 bar   

C igp dT  H R T f ,1 bar 

Tf

400 K

Tf

400 K

C igp dT  0

C igp dT  H R  400 K,38 bar 

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 400.15

P (bar) 38

Tc (K) 365.6

Pc (bar) 46.65

Tr 1.05 1.05 1.10 1.10

Pr 0.80 1.00 0.80 1.00

Z0 0.7130 0.6026 0.7649 0.6880

Tr 1.0945

Pr 0.8146

Z1 -0.0032 0.0220 0.0236 0.0476

(H ) /(RT c) -0.995 -1.359 -0.827 -1.120

0.7533

0.0224

-0.8674

HR/(RT c) R H (J/mol)

-0.94242 -2864.57

0.14

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z 3 V (cm /mol)

T 1 (K) 400

T 2 (K) 362.9176 T2

0.7565 662.4

A 1.637

Cp ICPH º dT A T2 T1  R

(SR)1/R -0.642 -0.820 -0.470 -0.577

-0.5359

-0.5659

-0.4973

S (J/(mol K))

-0.63548 -5.28337

R 1

(H ) /(RT c) -0.691 -0.877 -0.507 -0.617

SR/R R

C (1/K2) D (K2) B (1/K) ICPH (K) H (J/mol) 2.27E-02 -6.92E-06 0.00E+00 -344.5 -2864.6

B 2 T2 T12 2

 C3 T

3 2

T1

Tf

C igp

400 K

T

S  S  400 K,38 bar    R

(SR)0/R -0.656 -0.965 -0.537 -0.742

R 0

dT  R ln

T13

 DT1 T1 2

1

1  S R  363 K,1 bar  38

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T 1 (K) 400

T 2 (K) 362.9176

A 1.637

C (1/K2) D (K2) B (1/K) 2.27E-02 -6.92E-06 0.00E+00

T2

Cp T ICPS º dT A ln  2 RT T1

C 2 T2 T12 2

B T2 T1 

 D2 T

 2 2

ICPS -0.903

T1 2

S (J/(mol K)) -7.51

T1

H   H R  423 K, 22 bar   

Tf

423 K

H   H R  423 K, 22 bar   

Tf

423 K

Tf

423 K

C igp dT  H R T f ,1 bar 

C igp dT  0

C igp dT  H R  423 K, 22 bar 

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 423

P (bar) 22

Tr 1.1439

Pr 0.5179

B 0 0.083

0.422 Tr1.6 0.172

B1 0.139

Tr4.2

 0.152

Tc (K) Pc (bar) 369.8 42.48

0

B -0.2573

dB0 dTr

1

B 0.0412

dB0 0.675  2.6 dTr Tr

dB1 dTr

0.4759

0.3589

 HR dB0  Pr B 0 Tr  RTc dTr 

dB1 0.722  5.2 dTr Tr

HR/RT c -0.4443



  B1 Tr  

HR (J mol-1) -1366.04

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

T 1 (K) 423

T 2 (K) 408.9078 T2

A 1.213

Cp ICPH º dT A T2 T1  R

C (1/K2) D (K2) B (1/K) ICPH (K) H (J/mol) 2.88E-02 -8.82E-06 0.00E+00 -164.3 -1366.0

B 2 T2 T12 2

 C3 T

3 2

T13

T1

408.9

C igp

423 K

T

S  S  423 K, 22 bar    R

dT  R ln

 DT1 T1 2

1

1  S R  408.9 K,1 bar  22

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 423

P (bar) 22

Tr 1.1439

Pr 0.5179

 0.152

Tc (K) Pc (bar) 369.8 42.48

dB0 dTr

dB1 dTr

0.4759

0.3589

dB0 0.675  2.6 dTr Tr

R

S /R -0.2747

dB 0 SR Pr  dT R  r

dB1 0.722  5.2 dTr Tr

S

R

-1

-1

(J mol K ) -2.28

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

T 1 (K) 423

T 2 (K) 408.9078

A 1.213

C (1/K2) D (K2) B (1/K) 2.88E-02 -8.82E-06 0.00E+00

T2

Cp T ICPS º dT A ln  2 RT T1

B T2 T1 

C 2 T2 T12 2

 D2 T

 2 2

ICPS -0.395

T1 2

S (J/(mol K)) -3.28

T1

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 373.15

Tr 1.0091

B 0.083 0

B1 0.139

P (bar) 1

Pr 0.0235

0.422 Tr1.6 0.172 Tr4.2

Tc (K) 369.8

0

B -0.3330

Pc (bar) 42.48

1

B -0.0266

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

 0.152

dB0 dTr

dB1 dTr

0.6594

0.6889

 HR dB0  Pr B 0 Tr  RTc dTr 

HR (J mol -1) -80.20

HR /RTc -0.0261



  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

T (K) 373.15

Tr 1.0091

B 0 0.083 B1 0.139

P (bar) 10

Pr 0.2354

0.422 Tr1.6 0.172 Tr4.2

Tc (K) 369.8

B0 -0.3330

Pc (bar) 42.48

B1 -0.0266

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

 0.152

dB0 dTr

dB1 dTr

0.6594

0.6889

 HR dB0  Pr B 0 Tr  RTc dTr 

HR (J mol -1) -801.97

HR /RTc -0.2608



  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 373.15

Tr 1.0091

P (bar) 1

Pr 0.0235

Tc (K) 369.8

Pc (bar) 42.48

dB0 dTr

dB1 dTr

0.6594

0.6889

dB0 0.675  2.6 dTr Tr

 0.152

SR /R -0.0180

dB 0 SR Pr  dT R  r

dB1 0.722  5.2 dTr Tr

SR (J mol -1 K-1) -0.1496

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output. T (K) 373.15

Tr 1.0091

P (bar) 10

Pr 0.2354

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

Tc (K) 369.8

Pc (bar) 42.48

dB0 dTr

dB1 dTr

0.6594

0.6889

 0.152

SR /R -0.1799

dB 0 SR Pr  dT R  r

SR (J mol -1 K-1) -1.4955

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 318.15

Tr 1.0459

B 0.083 0

B1 0.139

P (bar) 16

Pr 0.2167

Tc (K) 304.2

0

B -0.3098

Pc (bar) 73.83

1

B -0.0035

dB0 0.675  2.6 dTr Tr

0.422 Tr1.6

dB1 0.722  5.2 dTr Tr

0.172 Tr4.2

 0.224

dB0 dTr

dB1 dTr

0.6007

0.5718

 HR dB0  Pr B 0 Tr  RTc dTr 

HR (J mol -1) -588.04

HR /RTc -0.2325



  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

T2

ICPH T0,T;A,B,C,D  º

R

Cp

T1

T0(K) 318.15

T (K) 302.71

A 5.457

dT  A T  T0  

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

ig C (1/K2) D (K2) B (1/K) ICPH (K) H (J/mol) 1.05E-03 0.00E+00 -1.16E+05 -70.7 -588.0

dB0 dTr

dB1 dTr

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


B 0 0.083

0.422

B1 0.139

0.172

Tr1.6 Tr4.2

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

T2

ICPH T0,T;A,B,C,D  º

 HR dB0  Pr B 0 Tr  RTc dTr 



  B1 Tr  

dB1 dTr

1

1 

 R dT  AT  T   2 T  T   3 T  T   D  T  T  Cp

B

2

0

2 0

C

3

3 0

0

T1

T0(K) 318.15

T (K) 303.83

A 5.457

ig C (1/K2) D (K2) B (1/K) ICPH (K) H (J/mol) 1.05E-03 0.00E+00 -1.16E+05 -65.7 -546.0

T (K) 318.15

P (bar) 16

Tc (K) 304.2

Pc (bar) 73.83

dB0 dTr

dB1 dTr

0.6007

0.5718

Tr 1.0459

Pr 0.2167

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

 0.224

SR /R -0.1579

dB 0 SR Pr  dT R  r

SR (J mol -1 K-1) -1.31

dB1 dTr

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 303.8

Tr 0.9987

P (bar) 1.0133

Pr 0.0137

Tc (K) 304.2

Pc (bar) 73.83

dB0 dTr

dB1 dTr

0.6773

0.7270

dB0 0.675  2.6 dTr Tr

 0.224

SR /R -0.0115

dB 0 SR Pr  dT R  r

dB1 0.722  5.2 dTr Tr

T

ICPS  T0,T;A,B,C,D º

SR (J mol -1 K-1) -0.10

dB1 dTr

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

T0 (K) 318.15

T (K) 303.83

A 5.457

S  

Tf

C igp

523 K

T

Tf

C

ig p

523 K

T

C (1/K2) D (K2) B (1/K) 1.05E-03 0.00E+00 -1.16E+05

dT  R ln

ICPS -0.211

Sig (J/(mol K)) -1.7557

120 0 3800

dT  28.73 J/(mol K)

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T 1 (K) 523.15

T 2 (K) 308.167

C (1/K2) D (K2) B (1/K) 1.44E-02 -4.39E-06 0.00E+00

A 1.424

T2

ICPS -3.456

S (J/(mol K)) -28.73

Cp T ICPS º dT A ln  2 RT T1

B T2 T1 

T 1 (K) 523.15

C (1/K2) D (K2) B (1/K) ICPH (K) H (J/mol) 1.44E-02 -4.39E-06 0.00E+00 -1425.6 -11852.5

C 2 T2 T12 2

 D2 T

 2 2

T1 2

T1

T 2 (K) 308.167

A 1.424

T2

Cp ICPH º dT A T2 T1  R

B 2 T2 T12 2

 C3 T

T1

Tf

C igp

523 K

T

S   S R  523 K,38 bar   

Tf

Tf

523 K

523 K

C

ig p

T C igp T

dT  R ln

dT  S R  523 K,38 bar   R ln

3 2

T13

 DT1 T1 2

1

1.2  S R T f ,1.2 bar   0 38

1.2  S R T f ,1.2 bar  38

dT  S R  523 K,38 bar   28.73 J/(mol K)  S R T f ,1.2 bar 

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 523.15

P (bar) 38

Tr 1.8532

Pr 0.7540

 0.87

Tc (K) Pc (bar) 282.3 50.4

dB0 dTr

dB1 dTr

0.1357

0.0292

dB0 0.675  2.6 dTr Tr

SR/R -0.1215

dB 0 SR Pr  dT R  r

dB1 0.722  5.2 dTr Tr

SR (J mol-1 K-1) -1.01

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Tf

C igp

523 K

T

dT  1.01  28.73  29.74 J/(mol K)

T 1 (K) 523.15

T 2 (K) 301.3147

A 1.424

C (1/K2) D (K2) B (1/K) 1.44E-02 -4.39E-06 0.00E+00

T2

ICPS -3.577

S (J/(mol K)) -29.74

Cp T ICPS º dT A ln  2 RT T1

B T2 T1 

T 1 (K) 523.15

C (1/K2) D (K2) B (1/K) ICPH (K) H (J/mol) 1.44E-02 -4.39E-06 0.00E+00 -1462.6 -12160.3

C 2 T2 T12 2

 D2 T

 2 2

T1 2

T1

T 2 (K) 301.3147 T2

A 1.424

Cp ICPH º dT A T2 T1  R T1

B 2 T2 T12 2

 C3 T

3 2

T13

 DT1 T1 2

1

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 523.15

P (bar) 38

Tr 1.8532

Pr 0.7540

B 0 0.083

0.422

B1 0.139

Tr1.6 0.172 Tr4.2

 0.87

Tc (K) Pc (bar) 282.3 50.4

0

dB0 dTr

1

B -0.0743

B 0.1261

dB0 0.675  2.6 dTr Tr

0.1357

dB1 dTr

0.0292

 HR dB0  Pr B 0 Tr  RTc dTr 

dB1 0.722  5.2 dTr Tr

HR/RT c -0.1984



  B1 Tr  

R

H

-1

(J mol ) -465.77

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

H   H R  523 K,38 bar   

301 K

523 K

C igp dT  465.8  12160.3  11695 J/mol

Tf

C igp

493 K

T

S  S R  493 K,30 bar   

dT  R ln

2.6 R  S T f K, 2.6 bar   0 30

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Tf

C igp

493 K

T

 T 1 (K) 493

dT  20.33 J/(mol K)

T 2 (K) 367.472

A 1.131

C (1/K2) D (K2) B (1/K) 1.92E-02 -5.56E-06 0.00E+00

T2

ICPS -2.445

S (J/(mol K)) -20.33

Cp T ICPS º dT A ln  2 RT T1

B T2 T1 

T 1 (K) 493

C (1/K2) D (K2) B (1/K) ICPH (K) H (J/mol) 1.92E-02 -5.56E-06 0.00E+00 -1050.1 -8730.7

C 2 T2 T12 2

 D2 T

 2 2

T1 2

T1

T 2 (K) 367.472 T2

A 1.131

Cp ICPH º dT A T2 T1  R T1

B 2 T2 T12 2

 C3 T

3 2

T13

 DT1 T1 2

1

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Tf

C igp

493 K

T

S   S R  493 K,30 bar   

Tf

C igp

493 K

T

T (K) 493.15

Pr 0.6158

dB0 dTr

dB1 dTr

0.1940

0.0597

493 K

T

T 1 (K) 493

SR/R -0.1231

SR (J mol-1 K-1) -1.02

dB1 dTr

dT  21.35 J/(mol K)

T 2 (K) 361.4254

A 1.131

C (1/K2) D (K2) B (1/K) 1.92E-02 -5.56E-06 0.00E+00

T2

Cp T ICPS º dT A ln  2 RT T1

0.1

dB 0 SR Pr  dT R  r

dB1 0.722  5.2 dTr Tr

C igp

Tc (K) Pc (bar) 305.3 48.72

dB0 0.675  2.6 dTr Tr

Tf

2.6 0 30

dT  S R  493 K,30 bar   20.33 J/(mol K)

P (bar) 30

Tr 1.6153

dT  R ln

B T2 T1 

C 2 T2 T12 2

 D2 T

 2 2

ICPS -2.568

T1 2

S (J/(mol K)) -21.35

T1

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 361

P (bar) 2.6

Tr 1.1824

Pr 0.0534

dB0 dTr

dB1 dTr

0.4366

0.3021

dB0 0.675  2.6 dTr Tr

 

Tf

C igp

493 K

493 K

T 1 (K) 493

T C igp T

SR/R -0.0249

SR (J mol-1 K-1) -0.21

dB1 dTr

dT  S R  493 K,30 bar   20.33 J/(mol K)  S R T f , 2.6 bar  dT  1.02  20.33  0.21  21.14 J/(mol K)

T 2 (K) 362.6735

A 1.131

C (1/K2) D (K2) B (1/K) 1.92E-02 -5.56E-06 0.00E+00

T2

Cp T ICPS º dT A ln  2 RT T1

0.1

dB 0 SR Pr  dT R  r

dB1 0.722  5.2 dTr Tr

Tf

Tc (K) Pc (bar) 305.3 48.72

B T2 T1 

C 2 T2 T12 2

 D2 T

 2 2

ICPS -2.543

T1 2

S (J/(mol K)) -21.14

T1

H   H R  493 K,30 bar   

363 K

493 K

C igp dT  H R  363 K, 2.6 bar 

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 493

P (bar) 30

Tr 1.6148

Pr 0.6158

B 0 0.083

0.422 Tr1.6

Tc (K) Pc (bar) 305.3 48.72

0

B -0.1130

1

B1 0.139

0.172 Tr4.2

dB1 0.722  5.2 dTr Tr

T 1 (K) 493

T 2 (K) 363

A 1.131

T2

Cp ICPH º dT A T2 T1  R

0.1

dB0 dTr

B 0.1160

dB0 0.675  2.6 dTr Tr

0.1942

P (bar) 2.6

Tr 1.1890

Pr 0.0534

B 0 0.083

0.422

B1 0.139

Tr1.6 0.172 Tr4.2

B -0.2369



  B1 Tr  

R

H

-1

(J mol ) -663.72

dB1 dTr

C (1/K2) D (K2) B (1/K) ICPH (K) H (J/mol) 1.92E-02 -5.56E-06 0.00E+00 -1083.3 -9006.2

B 2 T2 T12 2

Tc (K) Pc (bar) 305.3 48.72

0

R

H /RT c -0.2615

0.0597

 HR dB0  Pr B 0 Tr  RTc dTr 

 C3 T

3 2

T1

T (K) 363

dB1 dTr

1

B 0.0559

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

T13

 DT1 T1 2

1

 0.1

dB0 dTr 0.4304

dB1 dTr

R

H /RT c -0.0415

0.2935

 HR dB0  Pr B 0 Tr  RTc dTr 



  B1 Tr  

R

H

-1

(J mol ) -105.38

dB1 dTr

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T2

T2

Cp  7.8  S   dT  R ln  dT  17.08 J mol1 K 1   T  1  323.15 T 323.15 ig

T (K) 323.15

Tr 0.7602

Cp

P (bar) 1

Pr 0.0263

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

Tc (K) 425.1

Pc (bar) 37.96

dB0 dTr

dB1 dTr

1.3770

3.0046

 0.2

SR /R -0.0521

dB 0 SR Pr  dT R  r

SR (J mol -1 K-1) -0.4332

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

T2

Cp

T 323.15

T2

Cp

T 323.15

dT  17.08  0.43  S R T2 , 7.8 bar   16.65  S R T2 , 7.8 bar  J mol1 K 1

dT  17.08 J mol1 K 1

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T

ICPS  T0,T;A,B,C,D º

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

T0 (K) 323.15

T (K) 376.56

A 1.935

C (1/K2) D (K2) B (1/K) 3.69E-02 -1.14E-05 0.00E+00

T (K) 376.56

P (bar) 7.8

Tc (K) 425.1

Pc (bar) 37.96

dB0 dTr

dB1 dTr

0.9251

1.3563

Tr 0.8858

Pr 0.2055

dB0 0.675  2.6 dTr Tr

Sig (J/(mol K)) 17.0803

 0.2

SR /R -0.2458

dB 0 SR Pr  dT R  r

dB1 0.722  5.2 dTr Tr

ICPS 2.054

SR (J mol -1 K-1) -2.0440

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

T2

Cp

 T dT  17.08  0.43  S T , 7.8 bar   16.65  2.044  18.69 J mol K 1

R

1

2

323.15

T

ICPS  T0,T;A,B,C,D º

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

T0 (K) 323.15

T (K) 381.69

A 1.935

C (1/K2) D (K2) B (1/K) 3.69E-02 -1.14E-05 0.00E+00

ICPS 2.248

Sig (J/(mol K)) 18.6893

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 381.69

Tr 0.8979

P (bar) 7.8

Pr 0.2055

Tc (K) 425.1

Pc (bar) 37.96

dB0 dTr

dB1 dTr

0.8932

1.2641

dB0 0.675  2.6 dTr Tr

 0.2

SR /R -0.2355

dB 0 SR Pr  dT R  r

dB1 0.722  5.2 dTr Tr

SR (J mol -1 K-1) -1.9579

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

T2

Cp

T 323.15

dT  17.08  0.43  S R T2 , 7.8 bar   16.65  1.958  18.604 J mol1 K 1

T

ICPS  T0,T;A,B,C,D º

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

T0 (K) 323.15

T (K) 381.42

A 1.935

C (1/K2) D (K2) B (1/K) 3.69E-02 -1.14E-05 0.00E+00

ICPS 2.238

Sig (J/(mol K)) 18.6041

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T2

ICPH T0,T;A,B,C,D  º

R

Cp

dT  A T  T0  

T1

T0(K) 323.15

T (K) 381.42

T (K) 323.15

P (bar)

Tr 0.7602

B 0.083 0

B1 0.139

1

Pr 0.0263

0.422 Tr1.6 0.172 Tr4.2

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

A 1.935

ig C (1/K2) D (K2) B (1/K) ICPH (K) H (J/mol) 3.69E-02 -1.14E-05 0.00E+00 787.9 6550.6

Tc (K) 425.1

Pc (bar) 37.96

0

B -0.5714

1

B -0.4051

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

 0.2

dB0 dTr

dB1 dTr

1.3770

3.0046

 HR dB0  Pr B 0 Tr  RTc dTr 

HR (J mol -1) -200.75

HR /RTc -0.0568



  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

T (K) 381.42

Tr 0.8972

B 0 0.083 B1 0.139

P (bar) 7.8

Pr 0.2055

0.422 Tr1.6 0.172 Tr4.2

Tc (K) 425.1

B0 -0.4189

Pc (bar) 37.96

B1 -0.1322

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

 0.2

dB0 dTr

dB1 dTr

0.8948

1.2688

 HR dB0  Pr B 0 Tr  RTc dTr 

HR (J mol -1) -1071.92

HR /RTc -0.3033



  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


56.0056 59.04121 61.1392 62.579174 63.39535 63.32491 62.15734

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


P2 (kPa)

T2 (K)

V2 (cm^3/g)

mass (kg)

1

384.09

303316

0.00576956

100

384.82

3032.17 0.577144421

200

385.57

1515.61 1.154650603

300

386.31

1010.08 1.732536037

400

387.08

757.34 2.310719096

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


2.5

387.5 387 386.5 386

1.5

385.5 1

T2 (K)

Mass (kg)

2

385 384.5

0.5

384 0

383.5 0

50

100

150

200

250

300

350

400

P2 (kPa)

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


T (K) 500

P (bar) 20

Tc (K) 425.2

Pc (bar) 42.77

Z 0.19

0

Tr 1.1759

Pr 0.4676

0.422

B 0.083 0

B1 0.139

Tr1.6 0.172 Tr4.2

0

B -0.2426

1

B 0.0519

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

0.9074

Z  1  B0   B1

 TP

1

dB dTr

dB dTr

0.4429

0.3109

 HR dB0  Pr B 0 Tr  RTc dTr 

HR (J mol -1) -1360.65

HR /RTc -0.3849



  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

T (K) 500

Tr 1.1759

P (bar) 20

Pr 0.4676

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

Tc (K) 425.2

Pc (bar) 42.77

dB0 dTr

dB1 dTr

0.4429

0.3109

 0.19

SR /R -0.2347

dB 0 SR Pr  dT R  r

SR (J mol -1 K-1) -1.9517

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

r

r


T (K) 400

P (bar) 200

Tc (K) 304.2

Pc (bar) 73.83

Tr 1.30 1.30 1.40 1.40

Pr 2.00 3.00 2.00 3.00

 0.224

Tr 1.3149

Pr 2.7089

Z1 0.1991 0.2079 0.1894 0.2397

(HR )0/(RTc) -1.56 -2.274 -1.253 -1.857

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Z0 0.6908 0.6344 0.7753 0.7202

Interpolated Values Final Values Z 3

V (cm /mol)

T

P

Tc1

0.7102 118.1

Tc2

0.6636

0.2083

(HR )1/(RTc) -0.178 -0.3 -0.07 -0.044

-2.0087

-0.2327

(SR )0/R -0.891 -1.299 -0.663 -0.99

(SR )1/R -0.3 -0.481 -0.22 -0.29

-1.1376

-0.4046

-2.06084

SR /R

-1.22828

R

-5212.12

SR (J/(mol K))

-10.2119

ω

ω

HR /(RTc) H (J/mol)

Pc1

Pc2

ω1

ω2

650

60

562.2

553.6

48.98

40.73

0.21

0.21

300

100

304.2

132.9

73.83

34.99

0.224

0.048

600

100

304.2

568.7

73.83

24.9

0.224

0.4

350

75

305.3

282.3

48.72

50.4

0.1

0.087

400

150

373.5

190.6

89.63

45.99

0.094

0.012

200

75

190.6

126.2

45.99

34

0.012

0.038

450

80

190.6

469.7

45.99

33.7

0.012

0.252

250

100

126.2

154.6

34

50.43

0.038

0.022

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Tpc

Ppc

ω

Tpr

Ppr

557.9

44.855

0.21 1.165083 1.337644

218.55

54.41

0.136 1.372684 1.837897

436.45

49.365

0.312 1.374728 2.025727

293.8

49.56

0.0935 1.191287 1.513317

282.05

67.81

0.053 1.418188 2.212063

158.4

39.995

0.025 1.262626 1.875234

330.15

39.845

0.132 1.363017

140.4

42.215

0.03 1.780627 2.368826

H0

H1

2.00778

Z0

Z1

S0

S1

0.6543

0.1219

-1.395

-0.461

-0.89

-0.466

0.7706

0.1749

-1.214

-0.116

-0.658

-0.235

0.7527

0.1929

-1.346

-0.097

-0.729

-0.242

0.6434

0.1501

-1.51

-0.4

-0.944

-0.43

0.7744

0.199

-1.34

-0.049

-0.704

-0.224

0.6631

0.1853

-1.623

-0.254

-0.965

-0.348

0.7436

0.1933

-1.372

-0.11

-0.75

-0.25

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


0.9168

0.1839

-0.82

0.172

-0.361

-0.095

Z

HR (J/mol)

SR (J/mol K)

0.680

-6919.58

-8.21307

0.794

-2234.53

-5.73633

0.813

-4993.97

-6.68865

0.657

-3779.76

-8.18268

0.785

-3148.34

-5.95176

0.668

-2145.75

-8.09534

0.769

-3805.81

-6.50986

0.922

-951.151

-3.02505

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


=

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Tc := 647.1K

Pc := 220.55bar

At Tr = 0.7:

T := 0.7 Tc

T = 452.97K

Find Psat in the Saturated Steam Tables at T = 452.97 K T1 := 451.15K

P1 := 957.36kPa

T2 := 453.15K

Psat = 998.619 kPa

Psatr:=

Psatr := 0.045

P2 := 1002.7kPa

Psat = 9.986 bar

ω := −1 − log(Psatr) ω = 0.344 Ans.

This is very close to the value reported in Table B.1 (ω = 0.345).

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


ω

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Estimated Value (kJ/mol)

Table B.2 (kJ/mol)

Benzene

30.8

30.72

iso-butane

21.39

21.3

Carbon tetrachloride

29.81

29.82

cyclohexane

30.03

29.97

n-Decane

39.97

38.75

n-Hexane

29.27

28.85

n-Ocatane

34.7

34.41

Toluene

33.72

33.18

o-Xylene

37.23

36.24

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 1 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Ϻ

Ϻ

Ϻ Updated 4/5/2017

p. 2 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 3 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

7.14 7.12

Area (cm^2)

7.10 7.08 7.06 7.04 7.02 7.00 400

420

440

460

480

500

Pressure (kPa)

Updated 4/5/2017

p. 4 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

=

and

or

Updated 4/5/2017

p. 5 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

˙ ( / /

=

)

/

=

Updated 4/5/2017

p. 6 of 73


SVNAS 8th Edition Annotated Solutions

P (kPa) 800 775 750 725 700

H2 (kJ/kg) V2 (cm3/g) u2 (m/sec) 2956 294.81 530.09 2948.5 302.12 544.06 2940.8 309.82 558.03 2932.8 317.97 572.19 2924.9 326.69 585.83

Chapter 7

A2 (cm2 s/kg) 5.5615 5.5531 5.5520 5.5571 5.5765

Updated 4/5/2017

p. 7 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 8 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 9 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 10 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 11 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 12 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 13 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 14 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 15 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 16 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 17 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 18 of 73


SVNAS 8th Edition Annotated Solutions

S  

C igp

Tf

673 K

Tf

C igp

673 K

T

Chapter 7

1 dT  R ln    0 T 8

dT  17.29 J/(mol K)

T

ICPS  T0,T;A,B,C,D 

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

Updated 4/5/2017

p. 19 of 73


SVNAS 8th Edition Annotated Solutions

H  

465.9 K

673.15 K

Chapter 7

C igp dT

T2

ICPH T0,T;A,B,C,D  

1

B

2

0

C

2 0

3

3 0

0

T1

T0(K) 673.15

T (K) 466.40

A 5.457

T2

ICPH T0,T;A,B,C,D  

R

Cp

T (K) 520.00

C (1/K2) D (K2) B (1/K) ICPH (K) 1.05E-03 0.00E+00 -1.16E+05 -1175.1

dT  A T  T0  

T1

T0(K) 673.15

1 

 R dT  AT  T   2 T  T   3 T  T   D  T  T  Cp

A 5.457

Hig (J/mol) -9770.7

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

C (1/K2) D (K2) B (1/K) ICPH (K) 1.05E-03 0.00E+00 -1.16E+05 -880.6

Hig (J/mol) -7321.8

Updated 4/5/2017

p. 20 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 21 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 22 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Δ

Updated 4/5/2017

p. 23 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Δ

Updated 4/5/2017

p. 24 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Part

T0 (K)

P0 (bar)

P (bar)

ndot (mol/sec)

a

753.15

6

1

200

b

673.15

5

1

150

c

773.15

7

1

175

d

723.15

8

2

100

e

755.37 95 psi

15 psi

226.795

η

T (K)

ΔH'

ΔH

T final (K)

Wdot (kW)

0.8 459.72 -9042.15 -7233.72 515.85 1446.74 0.75 430.41 -7394.54 -5545.91 489.74 831.886 0.78 452.45 -9911.01 -7730.59 520.02 1352.85 0.85 493.82 -7056.97 -5998.42 526.34 599.842 0.8 454.13 -9285.79 -7428.63 511.64 1684.78

Updated 4/5/2017

p. 25 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 26 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 27 of 73


SVNAS 8th Edition Annotated Solutions

=

Chapter 7

=

+

=

+

=

Updated 4/5/2017

p. 28 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

=

S   S R  298.15 K,1.0133 bar   

Tf

298.15 K

Tf

298.15 K

C igp

 3.75  R dT  R ln    S T f ,3.75 bar   0 T  1.0133 

C igp

 3.75  R R dT  R ln    S  298.15 K,1.0133 bar   S T f ,3.75 bar  T  1.0133 

C igp

 3.75  dT  R ln    10.880 J/(mol K) 298.15 K T  1.0133 

Tf

Updated 4/5/2017

p. 29 of 73


SVNAS 8th Edition Annotated Solutions

T

ICPS  T0,T;A,B,C,D 

Chapter 7

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

C (1/K2) D (K2) B (1/K) 5.75E-04 0.00E+00 -1.60E+03

T0 (K) 298.15

T (K) 431.1

A 3.355

T (K) 431.1

P (bar) 3.75

Tc (K) Pc (bar) 132.2 37.45

Tr 3.2610

Pr 0.1001

dB0 dTr

dB1 dTr

0.0312

0.0015

dB0 0.675  2.6 dTr Tr

S (J/(mol K)) 10.880

w 0.035

SR/R -0.0031

dB 0 SR Pr  dT R  r

dB1 0.722  5.2 dTr Tr

ICPS 1.309

w

SR (J mol-1 K-1) -0.026

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Tf

C igp

298.15 K

T

dT  10.880  0.026  0.018  10.872 J/(mol K)

Ws (isentropic)   H S   H R  298.15 K,1.013 bar  

431.0 K

298.15 K

T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T0(K) 298.15

T (K) 431.0

A 3.355

C pig dT  H R  431.0 K,3.75 bar

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

C (1/K2) D (K2) B (1/K) ICPH (K) H (J/mol) 5.75E-04 0.00E+00 -1.60E+03 471.7 3922.1

Updated 4/5/2017

p. 30 of 73


SVNAS 8th Edition Annotated Solutions

T (K) 431.1

P (bar) 3.75

Tr 3.2610

Pr 0.1001

B 0 0.083

0.422

B1 0.139

Chapter 7

w 0.035

Tc (K) Pc (bar) 132.2 37.45

0

1

B 0.0193

Tr1.6 0.172 Tr4.2

dB0 dTr

B 0.1378

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

0.0312

dB1 dTr

HR/RTc -0.0078

0.0015

 HR dB0  Pr B 0 Tr  RTc dTr 



w  B1 Tr  

HR (J mol-1) -8.57

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Ws (isentropic)   H S  6.30  3922.1 8.57  3919.8 J/mol

Ws  H   H R  298.15 K,1.013 bar  

Tf

298.15 K

Tf

298.15 K

C ig p dT  5266.4 J/mol

T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T0(K) 298.15

R C ig p dT  H  Tf , 3.75 bar  5266.4 J/mol

T (K) 475.8

A 3.355

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

C (1/K2) D (K2) B (1/K) ICPH (K) H (J/mol) 5.75E-04 0.00E+00 -1.60E+03 633.4 5266.4

Updated 4/5/2017

p. 31 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Ws  nWS  100 mol/s  5266 J/mol = 526.6 kW

S   S

R

 353.15 K,3.75 bar   353.15 K Tf

C igp

 10  R dT  R ln    S T f ,10 bar   0 T 3.75  

C igp

 10  R R dT  R ln    S  353.15 K,3.75 bar   S T f ,10 bar  353.15 K T 3.75  

Tf

C igp

Tf

353.15 K

 10  dT  R ln    8.155 J/(mol K) T  3.75  T

ICPS  T0,T;A,B,C,D 

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

T0 (K) 353.15

T (K) 464.51

A 3.355

B (1/K) 5.75E-04

C (1/K2) D (K2) 0 -1.60E+03

ICPS 0.981

Sig (J/(mol K)) 8.1551

Updated 4/5/2017

p. 32 of 73


SVNAS 8th Edition Annotated Solutions

T (K) 464.51

Tr 3.5137

P (bar) 10

Pr 0.2670

Chapter 7

Tc (K) 132.2

Pc (bar) 37.45

dB0 dTr

dB1 dTr

0.0257

0.0010

dB0 0.675  2.6 dTr Tr

w 0.035

SR /R -0.0069

dB 0 SR Pr  dT R  r

dB1 0.722  5.2 dTr Tr

SR (J mol -1 K-1) -0.0572

w

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Tf

C igp

353.15 K

T

dT  8.1551  0.0572  0.0438  8.1685 J/(mol K)

Ws (isentropic)   H S   H R  353.15 K,3.75 bar  

464.7 K

353.15 K

T2

ICPH T0,T;A,B,C,D  

T (K) 464.72

1

1 

 R dT  AT  T   2 T  T   3 T  T   D  T  T  Cp

B

0

A 3.355

2

2 0

C

3

3 0

0

T1

T0(K) 353.15

C pig dT  H R  464.7 K,10 bar

B (1/K) 5.75E-04

ig C (1/K2) D (K2) ICPH (K) H (J/mol) 0 -1.60E+03 399.5 3321.2

Updated 4/5/2017

p. 33 of 73


SVNAS 8th Edition Annotated Solutions

T (K) 464.72

Tr 3.5153

P (bar) 10

Pr 0.2670

B 0.083 0

B1 0.139

Tc (K) 132.2

0

B 0.0265

Chapter 7

1

B 0.1381

dB0 0.675  2.6 dTr Tr

0.422 Tr1.6

dB1 0.722  5.2 dTr Tr

0.172 Tr4.2

w 0.035

Pc (bar) 37.45

dB0 dTr

dB1 dTr

0.0257

0.0010

 HR dB0  Pr B 0 Tr  RTc dTr 

HR (J mol-1) -17.34

HR /RTc -0.0158



w  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Ws (isentropic)   H S  15.45  3321.2  17.34  3349.3 J/mol

Ws  H   H R  353.15 K, 3.75 bar  

Tf

353.15 K

Tf

353.15 K

C ig p dT  4784.7 J/mol

T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T0(K) 353.15

T (K) 513.21

Tf

353.15 K

R C ig p dT  H  Tf ,10 bar  4784.7 J/mol

A 3.355

B (1/K) 5.75E-04

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

ig C (1/K2) D (K2) ICPH (K) H (J/mol) 0 -1.60E+03 575.5 4784.7

C ig p dT  4789.2 J/mol

Ws  nWS  100 mol/s  4784 J/mol = 478.4 kW Updated 4/5/2017

p. 34 of 73


SVNAS 8th Edition Annotated Solutions

S   S

R

Tf

C igp

303.15 K

T

 303.15 K,1.00 bar   303.15 K Tf

C igp

 500  R dT  R ln    S T f ,5 bar   0 T  100 

dT  R ln  5   S R  303.15 K,1.00 bar   S R T f ,5 bar 

Tf

C igp

303.15 K

T

Chapter 7

dT  R ln  5  13.382 J/(mol K)

T

ICPS  T0,T;A,B,C,D 

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

T0 (K) 303.15

T (K) 476.21

T (K) 476.21

P (bar)

Tr 3.6022

5

Pr 0.1335

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

A 3.355

B (1/K) 5.75E-04

Tc (K) 132.2

Pc (bar) 37.45

dB0 dTr

dB1 dTr

0.0241

0.0009

C (1/K2) D (K2) 0 -1.60E+03

ICPS 1.610

Sig (J/(mol K)) 13.3820

w 0.035

SR /R -0.0032

dB 0 SR Pr  dT R  r

w

SR (J mol -1 K-1) -0.0268

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output. Updated 4/5/2017

p. 35 of 73


SVNAS 8th Edition Annotated Solutions

Tf

C igp

298.15 K

T

Chapter 7

dT  13.382  0.0268  0.0174  13.372 J/(mol K)

Ws (isentropic)   H S   H R  303.15 K,1.00 bar  

476.1 K

303.15 K

T2

ICPH T0,T;A,B,C,D  

C pig dT  H R  476.1 K,5.00 bar

1

B

2

0

2 0

C

3

3 0

0

T1

T0(K) 303.15

T (K) 476.06

T (K) 303.15

P (bar)

Tr 2.2931

B 0 0.083 B1 0.139

1

Pr 0.0267

0.422 Tr1.6 0.172 Tr4.2

1 

 R dT  AT  T   2 T  T   3 T  T   D  T  T  Cp

A 3.355

B (1/K) 5.75E-04

Tc (K) 132.2

Pc (bar) 37.45

B0 -0.0288

B1 0.1337

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

ig C (1/K2) D (K2) ICPH (K) H (J/mol) 0 -1.60E+03 616.9 5129.3

w 0.035

dB0 dTr

dB1 dTr

0.0780

0.0096

 HR dB0  Pr B 0 Tr  RTc dTr 

HR (J mol-1) -5.98

HR /RTc -0.0054



w  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Ws (isentropic)   H S  7.85  5129.3  5.98  5131 J/mol

Updated 4/5/2017

p. 36 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Ws  H   H R  303.15 K,1.00 bar  

Tf

303.15 K

Tf

303.15 K

C ig p dT  6413.96 J/mol

T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T0(K) 303.15

R C ig p dT  H  Tf , 5.00 bar  6413.96 J/mol

T (K) 518.57

A 3.355

B (1/K) 5.75E-04

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

ig C (1/K2) D (K2) ICPH (K) H (J/mol) 0 -1.60E+03 771.4 6414.0

Ws  nWS  150 mol/s  6414 J/mol = 962.1 kW

C igp

 13  dT  R ln    S R T f ,13 bar   0 373.15 K T 5

S   S R  373.15 K,5.00 bar    Tf

C igp

373.15 K

T

dT  R ln 13 / 5   S R  373.15 K,5.00 bar   S R T f ,13 bar 

Tf

C igp

373.15 K

T

Tf

dT  R ln 13 / 5  7.945 J/(mol K)

T

ICPS  T0,T;A,B,C,D 

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

T0 (K) 373.15

T (K) 486.89

A 3.355

B (1/K) 5.75E-04

C (1/K2) D (K2) 0 -1.60E+03

ICPS 0.956

Sig (J/(mol K)) 7.9453

Updated 4/5/2017

p. 37 of 73


SVNAS 8th Edition Annotated Solutions

T (K) 486.89

Tr 3.6830

P (bar) 13

Pr 0.3471

Chapter 7

Tc (K) 132.2

Pc (bar) 37.45

dB0 dTr

dB1 dTr

0.0228

0.0008

dB0 0.675  2.6 dTr Tr

w 0.035

SR /R -0.0079

dB 0 SR Pr  dT R  r

dB1 0.722  5.2 dTr Tr

w

SR (J mol -1 K-1) -0.0658

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Tf

C igp

298.15 K

T

dT  7.945  0.0506  0.0658  7.960 J/(mol K)

Ws (isentropic)   H S   H R  373.15 K,6.00 bar  

487.1 K

373.15 K

T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T0(K) 373.15

T (K) 487.13

A 3.355

B (1/K) 5.75E-04

C pig dT  H R  487.1 K,13.00 bar

1 1  B 2 C T  T02  T 3  T03  D    2 3  T T0 

ig C (1/K2) D (K2) ICPH (K) H (J/mol) 0 -1.60E+03 409.6 3405.5

Updated 4/5/2017

p. 38 of 73


SVNAS 8th Edition Annotated Solutions

T (K) 487.1

Tr 3.6846

P (bar) 13

Pr 0.3471

0

B 0.0306

B 0.083

Tr1.6

1

B 0.1383

dB1 0.722  5.2 dTr Tr

0.172 Tr4.2

w 0.035

Pc (bar) 37.45

dB0 0.675  2.6 dTr Tr

0.422

0

B1 0.139

Tc (K) 132.2

Chapter 7

dB0 dTr

dB1 dTr

0.0227

0.0008

HR (J mol-1) -18.47

HR /RTc -0.0168

 HR dB0  Pr B 0 Tr  RTc dTr 



w  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Ws (isentropic)   H S  18.47  3405.5  17.77  3406.2 J/mol

Ws  H   H R  373.15 K, 5.00 bar  

Tf

373.15 K

Tf

373.15 K

R C ig p dT  H  Tf ,13.00 bar  4541.6 J/mol

C ig p dT  4541.6 J/mol

T2

ICPH T0,T;A,B,C,D  

B

0

2

2 0

C

T (K) 524.67

Tf

373.15 K

A 3.355

3

3 0

1 0

T1

T0(K) 373.15

1

 R dT  AT  T   2 T  T   3 T  T   D  T  T  Cp

B (1/K) 5.75E-04

ig C (1/K2) D (K2) ICPH (K) H (J/mol) 0 -1.60E+03 546.2 4541.6

C ig p dT  4541.6  17.77  12.63 = 4536.5 J/mol

Ws  nWS  50 mol/s  4541.6 J/mol = 227.1 kW Updated 4/5/2017

p. 39 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

C igp

 3.791  R dT  R ln    S T f ,3.791 bar   0 299.82 K T  1.013 

S   S R  299.82 K,1.013 bar    Tf

C igp

299.82 K

T

dT  R ln  3.791/1.013  S R  298.82 K,1.013 bar   S R T f ,3.791 bar 

Tf

C igp

299.82 K

T

Tf

dT  R ln  3.791/1.013  10.973 J/(mol K)

T

ICPS  T0,T;A,B,C,D 

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

T0 (K) 299.82

T (K) 434.78

A 3.355

B (1/K) 5.75E-04

C (1/K2) D (K2) 0 -1.60E+03

ICPS 1.320

Sig (J/(mol K)) 10.9730

Updated 4/5/2017

p. 40 of 73


SVNAS 8th Edition Annotated Solutions

T (K) 434.78

Tr 3.2888

P (bar) 3.791

Pr 0.1012

Chapter 7

Tc (K) 132.2

Pc (bar) 37.45

dB0 dTr

dB1 dTr

0.0305

0.0015

dB0 0.675  2.6 dTr Tr

w 0.035

SR /R -0.0031

dB 0 SR Pr  dT R  r

dB1 0.722  5.2 dTr Tr

w

SR (J mol -1 K-1) -0.0258

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Tf

C igp

299.82 K

T

dT  10.973  0.0181 0.0258  10.980 J/(mol K)

Ws (isentropic)   H S   H R  299.82 K,1.013 bar  

434.88 K

299.82 K

T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T0(K) 299.82

T (K) 434.88

A 3.355

B (1/K) 5.75E-04

C pig dT  H R  434.88 K,3.791 bar

1 1  B 2 C T  T02  T 3  T03  D    2 3  T T0 

ig C (1/K2) D (K2) ICPH (K) H (J/mol) 0 -1.60E+03 480.0 3990.9

Updated 4/5/2017

p. 41 of 73


SVNAS 8th Edition Annotated Solutions

T (K) 299.82

Tr 2.2679

P (bar) 1.013

Pr 0.0270

B 0.083 0

B1 0.139

0.422 Tr1.6 0.172 Tr4.2

Tc (K) 132.2

0

B -0.0308

Chapter 7

Pc (bar) 37.45

1

B 0.1335

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.035

dB0 dTr

dB1 dTr

0.0803

0.0102

 HR dB0  Pr B 0 Tr  RTc dTr 

HR (J mol-1) -6.22

HR /RTc -0.0057



w  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

T (K) 434.88

Tr 3.2896

P (bar) 3.791

Pr 0.1012

B 0 0.083 B1 0.139

0.422 Tr1.6 0.172 Tr4.2

Tc (K) 132.2

B0 0.0202

Pc (bar) 37.45

B1 0.1378

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.035

dB0 dTr

dB1 dTr

0.0305

0.0015

 HR dB0  Pr B 0 Tr  RTc dTr 

HR (J mol-1) -8.41

HR /RTc -0.0076



w  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Ws (isentropic)   H S  6.22  3990.9  8.41  3988.7 J/mol

Ws  H   H R  299.82 K,1.013 bar  

Tf

299.82 K

Tf

299.82 K

R C ig p dT  H  Tf , 3.791 bar  5318.3 J/mol

C ig p dT  5318.3 J/mol

Updated 4/5/2017

p. 42 of 73


SVNAS 8th Edition Annotated Solutions

T2

ICPH T0,T;A,B,C,D  

R

Cp

Chapter 7

dT  A T  T0  

T1

T0(K) 299.82

T (K) 479.10

Tf

299.82 K

A 3.355

B (1/K) 5.75E-04

1 1  B 2 C T  T02  T 3  T03  D    2 3  T T0 

ig C (1/K2) D (K2) ICPH (K) H (J/mol) 0 -1.60E+03 639.6 5318.3

C ig p dT  5318.3  6.22  5.79 = 5317.9 J/mol

Ws  nWS  226.8 mol/s  5318.3 J/mol = 1.206 106 W  1.21MW

C igp

 9.308  R dT  R ln    S T f ,9.308 bar   0 338.71 K T  3.792 

S   S R  338.71 K,3.792 bar    Tf

C igp

338.71 K

T

dT  R ln  9.308 / 3.792   S R  338.71 K,3.792 bar   S R T f ,9.308 bar 

Tf

C igp

338.71 K

T

Tf

dT  R ln  9.308 / 3.792  7.466 J/(mol K)

T

ICPS  T0,T;A,B,C,D 

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2 T T0      T

Cp

0

T0 (K) 338.71

T (K) 435.72

A 3.355

B (1/K) 5.75E-04

C (1/K2) 0

D (K2) -1.60E+03

ICPS 0.898

Sig (J/(mol K)) 7.4660

Updated 4/5/2017

p. 43 of 73


SVNAS 8th Edition Annotated Solutions

T (K) P (bar) 338.71 3.792

Tr 2.5621

Pr 0.1013

Chapter 7

Tc (K) 132.2

Pc (bar) 37.45

dB0 dTr

dB1 dTr

0.0585

0.0054

dB0 0.675  2.6 dTr Tr

w 0.032

SR/R -0.0059

dB 0 SR Pr  dT R  r

dB1 0.722  5.2 dTr Tr

w

SR (J mol-1 K-1) -0.0494

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output. T (K) P (bar) 435.72 9.308

Tr 3.2959

Pr 0.2485

dB0 0.675  2.6 dTr Tr

Tc (K) 132.2

Pc (bar) 37.45

dB0 dTr

dB1 dTr

0.0304

0.0015

w 0.032

SR/R -0.0076

dB 0 SR Pr  dT R  r

dB1 0.722  5.2 dTr Tr

w

SR (J mol-1 K-1) -0.0629

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Tf

C igp

338.71 K

T

dT  7.466  0.0494  0.0629  7.480 J/(mol K)

Ws (isentropic)   H S   H R  338.71 K,3.792 bar  

435.93 K

338.71 K

C pig dT  H R  435.93 K,9.308 bar

Updated 4/5/2017

p. 44 of 73


SVNAS 8th Edition Annotated Solutions

T2

ICPH T0,T;A,B,C,D  

R

Cp

Chapter 7

dT  A T  T0  

T1

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

T (K) 435.93

A 3.355

B (1/K) 5.75E-04

C (1/K2) 0

T (K) P (bar) 338.71 3.792

Tc (K) 132.2

Pc (bar) 37.45

w 0.032

T0(K) 338.71

Tr 2.5621

B 0 0.083 B1 0.139

Pr 0.1013

0.422 Tr1.6 0.172 Tr4.2

0

B -0.0107

Hig (J/mol) 2883.1

D (K2) ICPH (K) -1.60E+03 346.8

dB1 dTr

0.0585

0.0054

HR/RTc -0.0159

 HR  Pr B 0  RTc 

Tr

dB0 dTr

w  B1 Tr 

B 0.1357

dB1 0.722  5.2 dTr Tr

dB0 dTr

1

dB0 0.675  2.6 dTr Tr

 

HR (J mol-1) -17.43

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

T (K) P (bar) 435.93 9.308

Tr 3.2975

B 0 0.083 B1 0.139

Pr 0.2485

0.422 Tr1.6 0.172 Tr4.2

Tc (K) 132.2

Pc (bar) 37.45

0

1

B 0.0205

B 0.1379

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.032

dB0 dTr

dB1 dTr

0.0303

0.0015

HR/RTc -0.0187

 HR  Pr B 0  RTc 

Tr

dB0 dTr

w  B1 Tr 

 

HR (J mol-1) -20.58

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Ws (isentropic)   H S  17.43  2883.1 20.58  2880.0 J/mol Updated 4/5/2017

p. 45 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Ws  H   H R  338.71 K, 3.791 bar  

Tf

338.71 K

Tf

338.71 K

C ig p dT  4114.2 J/mol

T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T0(K) 338.71

T (K) 476.94

Tf

338.71 K

R C ig p dT  H  Tf , 9.308 bar  4114.2 J/mol

A 3.355

B (1/K) 5.75E-04

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

C (1/K2) 0

D (K2) ICPH (K) -1.60E+03 494.8

Hig (J/mol) 4114.2

C ig p dT  17.43  4114.2  14.61 = 4117.0 J/mol

Ws  nWS  226.8 mol/s  4114.2 J/mol = 9.331 105 W  0.933 MW

Updated 4/5/2017

p. 46 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 47 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

C igp

 18  R dT  R ln    S T f ,isentropi ,18 bar   0 303.15 K T 11.5  

S   S  303.15 K,11.5 bar    R

T f ,ientropic ,ig

303.15 K

T f ,ientropic

C igp

 18  dT  R ln    3.725 J/(mol K) T  11.5 

T0 (K) 303.15

T (K) 320.52

A 1.637

C (1/K2) D (K2) B (1/K) 2.27E-02 -6.92E-06 0.00E+00

T (K) 303.15

P (bar) 11.5

Tc (K) 356.6

Pc (bar) 46.65

dB0 dTr

dB1 dTr

1.0296

1.6798

Tr 0.8501

Pr 0.2465

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

ICPS 0.448

Sig (J/(mol K)) 3.7250

w 0.14

SR/R -0.3118

dB 0 SR Pr  dT R  r

w

SR (J mol-1 K-1) -2.59

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Updated 4/5/2017

p. 48 of 73


SVNAS 8th Edition Annotated Solutions

T (K) P (bar) 320.51 18

Tr 0.8988

Pr 0.3859

Chapter 7

Tc (K) 356.6

Pc (bar) 46.65

dB0 dTr

dB1 dTr

0.8908

dB0 0.675  2.6 dTr Tr

1.2575

w 0.14

SR/R -0.4117

dB 0 SR Pr  dT R  r

1

dB 0.722  5.2 dTr Tr

SR (J mol-1 K-1) -3.42

w

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

T f ,ientropic

C igp

303.15 K

T

2.59   T f ,ientropic

C igp

303.15 K

T

dT  3.725  3.42  0

dT  4.555

T0 (K) 303.15

T (K) 324.42

A 1.637

C (1/K2) D (K2) B (1/K) 2.27E-02 -6.92E-06 0.00E+00

T (K) 324.41

P (bar) 18

Tc (K) 356.6

Pc (bar) 46.65

dB0 dTr

dB1 dTr

0.8632

1.1808

Tr 0.9097

Pr 0.3859

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

ICPS 0.548

Sig (J/(mol K)) 4.5547

w 0.14

SR/R -0.3969

dB 0 SR Pr  dT R  r

w

SR (J mol-1 K-1) -3.30

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Updated 4/5/2017

p. 49 of 73


SVNAS 8th Edition Annotated Solutions

T f ,ientropic

C igp

303.15 K

T

Chapter 7

dT  4.435

T

ICPS T0 ,T;A,B,C,D  

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T

Cp

0

T0 (K) 303.15

T (K) 323.86

A 1.637

C (1/K2) D (K2 ) B (1/K) 2.27E-02 -6.92E-06 0.00E+00

Ws ,isentropic  H isentropic   H R  303.15 K,11.5 bar  

323.9 K

303.15 K

T (K) 303.15

P (bar) 11.5

Tr 0.8501

Pr 0.2465

B 0 0.083

0.422

B1 0.139

Tc (K) 356.6

0

B -0.4642

Pc (bar) 46.65

1

B -0.2012

dB0 0.675  2.6 dTr Tr

Tr1.6

dB1 0.722  5.2 dTr Tr

0.172 Tr4.2

 Sig (J/(mol K)) 4.4351

ICPS 0.533

C igp dT  H R  323.9 K,18 bar

w 0.14

dB0 dTr

dB1 dTr

1.0296

1.6798

 HR dB0  Pr B 0 Tr  RTc dTr 

HR (J mol-1) -1145.75

HR/RTc -0.3864



w  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output. T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T0(K) 303.15

T (K) 323.95

A 1.637

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

C (1/K2) D (K2) B (1/K) ICPH (K) 2.27E-02 -6.92E-06 0.00E+00 168.0

Hig (J/mol) 1396.8

Updated 4/5/2017

p. 50 of 73


SVNAS 8th Edition Annotated Solutions

T (K) 323.9

P (bar) 18

Tr 0.9083

Pr 0.3859

B 0 0.083

0.422

Tc (K) 356.6

B0 -0.4092

Tr4.2

0.14

B1 -0.1186

dB1 0.722  5.2 dTr Tr

0.172

w

Pc (bar) 46.65

dB0 0.675  2.6 dTr Tr

Tr1.6

B1 0.139

Chapter 7

dB0 dTr

dB1 dTr

0.8668

1.1905

 HR dB0  Pr B 0 Tr  RTc dTr 

HR (J mol-1) -1561.03

HR/RTc -0.5265



w  B1 Tr  

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Ws,isentropic  H isentropic  1145.8  1396.8  1561.0  981.5 J/mol Ws ,actual  H actual   H R  303.15 K,11.5 bar   

T f ,actual

303.15 K

Ws ,actual  H actual  1145.8  

T f ,actual

303.15 K

T f ,actual

303.15 K

C igp dT  1561.0  1226.9 J/mol

C igp dT  1642.1 J/mol

T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T0(K) 303.15

C igp dT  H R T f ,actual ,18 bar   1226.9 J/mol

T (K) 327.51

A 1.637

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

C (1/K2) D (K2) B (1/K) ICPH (K) 2.27E-02 -6.92E-06 0.00E+00 197.5

Hig (J/mol) 1642.1

Updated 4/5/2017

p. 51 of 73


SVNAS 8th Edition Annotated Solutions

T (K) 327.51

P (bar) 18

Tr 0.9184

Pr 0.3859

B 0 0.083

0.422

B1 0.139

Chapter 7

Tc (K) 356.6

B0 -0.4006

Tr1.6 0.172 Tr4.2

w

Pc (bar) 46.65

0.14

B1 -0.1069

dB0 0.675  2.6 dTr Tr

dB0 dTr

dB1 dTr

0.8422

1.1239

 HR dB0  Pr B 0 Tr  RTc dTr 

dB1 0.722  5.2 dTr Tr

HR/RTc -0.5145



w  B1 Tr  

HR (J mol-1) -1525.54

dB1 dTr

Pale blue boxes are input fields, pink boxes are the final output.

Ws ,actual  H actual  1145.8  

T f ,actual

303.15 K

T f ,actual

303.15 K

C igp dT  1525.5  1226.9 J/mol

C igp dT  1606.6 J/mol

T2

ICPH T0,T;A,B,C,D  

R

Cp

dT  A T  T0  

T1

T0(K) 303.15

T (K) 326.99

A 1.637

B (1/K) 2.27E-02

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 

C (1/K2) -6.92E-06

D (K2) 0.00E+00

ICPH (K) 193.2

Hig (J/mol) 1606.6

Updated 4/5/2017

p. 52 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 53 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 54 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

= /

Updated 4/5/2017

p. 55 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 56 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 57 of 73


SVNAS 8th Edition Annotated Solutions

Part a b c d e

T1 (K) 300 290 295 300 305

P1 (bar) 2 1.5 1.2 1.1 1.5

Chapter 7

T2 (K) 464 547 455 505 496

P2 (bar) 6 5 6 8 7

Cp

H

Hs

η

29.099 20.785 37.413 45.727 33.256

(J/mol) 4772.236 5341.745 5986.08 9374.035 6351.896

(J/mol) 3218.973 3728.973 4745.499 5959.231 4764.994

0.675 0.698 0.793 0.636 0.750

Updated 4/5/2017

p. 58 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

T  S  C p ln  2   V P  0  T1 

 V P  T2  T1 exp   C  p  

 257.2 106 K 1 *0.001003 m3 kg 1 *1900 kPa  T2  298.15exp    298.185 4.184 kJ kg 1 K 1  

H  C p T  V 1   T  P H  4.184*0.035  0.001003* 1  257.2 106 *298.15 *1900  1.906 kJ kg 1

Updated 4/5/2017

p. 59 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

H  C p T  V 1   T  P  2.541 T 

2.541  0.001003* 1 257.2 106 * 298.15 *1900 4.184

 0.187 K

 T  S  C p ln  2   V P  0  T1 

 V P  T2  T1 exp   C  p     696.2 106 K 1 *0.001036 m3 kg 1 * 4800 kPa  T2  363.15exp    363.45 4.205 kJ kg 1 K 1  

H  C p T  V 1   T  P H  4.205*0.30  0.001036* 1  696.2 106 *363.15  *4800  4.976 kJ kg 1

H  C p T  V 1   T  P  7.108 T 

7.108  0.001036* 1  696.2 106 *363.15  *4800 4.205

 0.807 K

Updated 4/5/2017

p. 60 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

 T  S  C p ln  2   V P  0  T1 

 V P  T2  T1 exp   C  p  

  523.1  106 K 1 * 0.001017 m3 kg 1 * 4980 kPa  T2  333.15exp    333.36 4.204 kJ kg 1 K 1  

H  C p T  V 1   T  P H  4.204 * 0.21  0.001017 * 1  523.1  106 *333.15 * 4980  5.065 kJ kg 1

H  C p T  V 1   T  P  6.753 T 

6.753  0.001017 * 1  523.1  106 * 333.15  * 4980 4.204

 0.612 K

Updated 4/5/2017

p. 61 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

T  S  C p ln  2   V P  0  T1 

 V P  T2  T1 exp   C  p  

  217.3 106 K 1 *0.001002 m3 kg 1 *1925 kPa  T2  294.26 exp    294.29 K 4.182 kJ kg 1 K 1  

H  C p T  V 1   T  P H  4.182*0.029  0.001002* 1  217.3 106 *294.26  *1925  1.927 kJ kg 1

H  C p T  V 1   T  P  2.753 T 

2.753  0.001002* 1  217.3 106 *294.26  *1925 4.182

 0.226 K

 T  S  C p ln  2   V P  0  T1  Updated 4/5/2017

p. 62 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

 V P  T2  T1 exp   C  p  

  714.3  10 T2  366.5exp  

6

K * 0.001039 m kg 1 *10239 kPa    367.16 K 4.221 kJ kg 1 K 1  1

3

H  C p T  V 1   T  P H  4.221* 0.66  0.001039 * 1  714.3  106 *366.5 *10239  10.64 kJ kg 1

H  C p T  V 1   T  P  14.19 T 

14.19  0.001039 * 1  714.3  106 * 366.5  *10239 4.221

 1.50 K

Updated 4/5/2017

p. 63 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

if if

then

Where

then

also let

Updated 4/5/2017

p. 64 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 65 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

η= .

= =

η=

.

=

η=

.

=

Updated 4/5/2017

p. 66 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 67 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 68 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 69 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Δ

Updated 4/5/2017

p. 70 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Δ

Δ

Δ

Δ

Updated 4/5/2017

p. 71 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 72 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 7

Updated 4/5/2017

p. 73 of 73


SVNAS 8th Edition Annotated Solutions

Chapter 8

QH  3.322*103 kJ/kg

Updated 4/5/2017

p. 1 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8



Updated 4/5/2017

W23  W41 Q12

 0.368

p. 2 of 41


SVNAS 8th Edition Annotated Solutions

Ws  isentropic    H S 

Chapter 8

10000 kPa

VdP  V4  9990 kPa 

10 kPa

Updated 4/5/2017

p. 3 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 4 of 41


SVNAS 8th Edition Annotated Solutions

Ws  isentropic    H S 

Chapter 8

7000 kPa

 VdP  V  6980 kPa  4

10 kPa

Updated 4/5/2017

p. 5 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 6 of 41


SVNAS 8th Edition Annotated Solutions

Ws  isentropic    H S 

Chapter 8

10000 kPa

VdP  V4  8490 kPa 

10 kPa

Updated 4/5/2017

p. 7 of 41


SVNAS 8th Edition Annotated Solutions

Ws  isentropic    H S 

Chapter 8

6500 kPa

VdP  V4  6398.67 kPa 

101.33 kPa

Updated 4/5/2017

p. 8 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 9 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

H2 (kJ/kg) S2 (kJ/kg K) 3340.6 7.0373 3565.3 7.3282 3792.9 7.5891

Updated 4/5/2017

p. 10 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

x'3 0.914 0.959 0.999

η 0.297 0.314 0.332

P1 (kPa) 5000 7500 10000

Wpump (kJ/kg) 5.079 7.634 10.189

H1 (kJ/kg) 294.381 Updated 4/5/2017

p. 11 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

296.936 299.491

H2 (kJ/kg) S2 (kJ/kg K) 3664.5 7.2578 3643.7 7.0526 3622.7 6.9013

x'3 0.925 0.895 0.873

Updated 4/5/2017

η 0.359 0.375 0.386

p. 12 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

P2 (kPa) 725 750 775 800

H'2 (kJ/kg) 3023.9 3032.5 3040.9 3049

W12 (kJ/kg) H2 (kJ/kg) S2 (kJ/kg K) -579.15 3187.25 7.4939 -572.442 3193.958 7.4898 -565.89 3200.51 7.4851 -559.572 3206.828 7.4797

X'3 0.941 0.941 0.940 0.939

Updated 4/5/2017

H'3 (kJ/kg) 2471.438 2470.072 2468.506 2466.706

W23 (kJ/kg) -558.333 -564.631 -570.963 -577.295

W (kJ/kg) -20.817 -7.811 5.073 17.723

p. 13 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 14 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 15 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 16 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 17 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 18 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 19 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 20 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 21 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 22 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 23 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 24 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 25 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 26 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 27 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

VP = 450 kPa AVP  13.8254, BVP  2181.79, CVP  248.870         BVP 2181.79 T sat =  - CVP  °C =  - 248.870  °C = 34°C = 307.15 K  A - ln  VP    13.8254 - ln  450 kPa        VP     1 kPa   1 kPa      Vc = 262.7 cm3·mol1 Z c = 0.282

Trsat =

T sat 307.15 K = = 0.753 Tc 408.1 K  2  

Vliq = Vc Z

(1 - Trsat ) 7  c

Updated 4/5/2017

3

1

= 262.7 cm ·mol × 0.282

2  

(1 - 0.753) 7 

= 112.36 cm3·mol1

p. 28 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Tn  261.4 K

Trn =

Tn 261.4 K = = 0.641 Tc 408.1 K

 P   36.48 bar  RTn × 1.092 (ln  c  - 1.013) 8.314 J·mol1·K 1 × 261.4 K × 1.092 (ln   - 1.013) 1 bar  1 bar    H n = = 0.930 - Trn 0.930 - 0.641 H n = 2.118 × 104 J·mol1  1 - Trsat  H b = -H n    1 - Trn 

Updated 4/5/2017

0.38

 1 - 0.753  = -2.118 × 10 J·mol    1 - 0.641  4

1

0.38

= -18378 J·mol1 = -18.378 kJ·mol1

p. 29 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 30 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

 

Updated 4/5/2017

p. 31 of 41


SVNAS 8th Edition Annotated Solutions

 1

P    1  A   PB 

P    1  A   PB 

0.25926

Chapter 8

  1  0.333330.25926  0.248

Updated 4/5/2017

p. 32 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 33 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

WAB  WCD

Updated 4/5/2017

p. 34 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 35 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Δ

n

A

1 2 45.53402 10.10398

5.457 3.47 3.28 3.639

B D 0.001045 0.00145 0.000593 0.000506

Sum 58.638

Sum 198.52

Sum 0.04

Updated 4/5/2017

-115700 12100 4000 -22700 Sum -138724.27

p. 36 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 37 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 38 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 39 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 40 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 8

Updated 4/5/2017

p. 41 of 41


SVNAS 8th Edition Annotated Solutions

Chapter 9

| |

Updated 4/5/2017

|

|

p. 1 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 9

Updated 4/5/2017

p. 2 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 9

p. 3 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 9

ω

> ω < ωσ

Updated 4/5/2017

σ

< σ

ω

p. 4 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 9

Updated 4/5/2017

p. 5 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T4 (K) Part a b c d e

Chapter 9

299.15

H4 (kJ/kg) 235.97 T2 (K) η QdotC (kJ/s) H2 (kJ/kg) S2 (kJ/kg K) 273.15 0.79 600 398.6 1.727 267.15 0.78 500 395.06 1.731 261.15 0.77 400 391.46 1.735 255.15 0.76 300 387.79 1.74 248.15 0.75 200 383.45 1.746

H3' (kJ/kg) 414.5 415.9 417.3 419 Updated 4/5/2017

ΔH23 20.12658 26.71795 33.55844 41.06579

H3 (kJ/kg) 418.7266 421.7779 425.0184 428.8558 p. 6 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

and

Chapter 9

421.1 50.2 433.65 can be found through the equations below. This values have been

calculated above.

mdot (kg/s) 3.69 3.14 2.57 1.98 1.36

QdotH (kJ/s) -674.25 -583.97 -486.33 -381.15 -268.08

ω 8.080358 5.954424 4.633409 3.696995 2.937849

Updated 4/5/2017

Wdot (kJ/s) 74.25 83.97 86.33 81.15 68.08

ω carnot 10.50577 8.348438 6.872368 5.798864 4.865686

p. 7 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 9

Updated 4/5/2017

p. 8 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 9

Updated 4/5/2017

p. 9 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 9

Updated 4/5/2017

p. 10 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 9

Updated 4/5/2017

p. 11 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 9

.

Updated 4/5/2017

p. 12 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 9

Updated 4/5/2017

p. 13 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 9

Updated 4/5/2017

p. 14 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 9

Updated 4/5/2017

p. 15 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 9

Updated 4/5/2017

p. 16 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 9

Δ

Updated 4/5/2017

p. 17 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 9

so:

Δ

Updated 4/5/2017

p. 18 of 18

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

Siig T , P   Siig T , pi   Siig T , P   R ln yi N

N

N

i 1

i 1

i 1

ig Smix  S ig   yi Siig   R yi ln yi  R yi ln

1 yi

Updated 4/5/2017

p. 1 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

  

 

Siig T , P   Siig T , pi   Siig T , P   R ln yi N

N

N

i 1

i 1

i 1

ig Smix  S ig   yi Siig   R yi ln yi  R yi ln

Updated 4/5/2017

1 yi

p. 2 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

Wideal  H  T S

  

 N

ig Smix  R yi ln i 1

1 yi

  

Updated 4/5/2017

p. 3 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T1 (K) 448.15

T2 (K) 308.15 T2

ICPH 

R

Cp

A 1.702

dT  A  T2  T1  

T1

T1 (K) 448.15

T2 (K) 308.15 T2

ICPS 

A 1.702

Chapter 10

C (1/K2) D (K2) B (1/K) 9.08E-03 -2.16E-06 0.00E+00

H (J/mol) -5614

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 

C (1/K2) D (K2) B (1/K) 9.08E-03 -2.16E-06 0.00E+00

 T2 

S (J/(mol K)) -14.92

ICPS -1.794

 RT dT  A ln  T   B T  T   2 T  T   2 T  T  Cp

T1 (K) 448.15

T2 (K) 308.15 T2

ICPH 

C

2

2 2

1

D

2 1

2 2

2 1

1

T1

A 1.131

C (1/K2) D (K2) B (1/K) 1.92E-02 -5.56E-06 0.00E+00

ICPH (K) -1.06E+03

 1

H (J/mol) -8843

1

 R dT  AT  T   2 T  T   3 T  T   D T  T  Cp

B

2

1

2 2

C

2 1

3 2

3 1

2

T1

T1 (K) 448.15

T2 (K) 308.15 T2

ICPS 

ICPH (K) -6.75E+02

T1

A 1.131

C (1/K2) D (K2) B (1/K) 1.92E-02 -5.56E-06 0.00E+00

1

T  C D 2 dT  A ln  2   B T2  T1   T22  T12  T2  T1 2 RT T 2 2  1

Cp

Updated 4/5/2017

S (J/(mol K)) -23.45

ICPS -2.821

p. 4 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10



Wideal  H  T S

t  work required  

Wideal Ws

 0.05

Ws  20Wideal  20  H  T S 

N

N

N

i 1

i 1

i 1

ig Smix  S ig   yi Siig   R yi ln yi  R yi ln

1 yi

Updated 4/5/2017

p. 5 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

Ϻ Ϻ

Ϻ

Ϻ

Ϻ

Ϻ

Ϻ

Ϻ

Ϻ

Ϻ

Ϻ

Ϻ

Ϻ

Updated 4/5/2017

Ϻ

Ϻ Ϻ

p. 6 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

M1  M  x2

dM dM  M  1  x1  dx1 dx1

M 2  M  x1

dM dx1

V

1

1 a0  a1 x1  a2 x12

dV a1  2a2 x1  2 dx1  a  a x  a x2  0

1 1

V1  V  1  x1 

2 1

1  x1  a1  2a2 x1  dV 1   2 dx1 a0  a1 x1  a2 x1  a  a x  a x 2 2 0

V1 

1 1

2 2 2 1

a0  a1  2a1 x1  2a2 x1  3a2 x12

a  a x  a x  0

V2  V  x1

2 2 2 1

1 1

x1  a1  2a2 x1  dV 1   2 dx1 a0  a1 x1  a2 x1  a  a x  a x 2 2 0

V2 

2 1

2 2 1

a  a x  a x  0

V1 

1 1

a0  a1 x1  a x  a1  2a2 x1  a1 x1  2a x 2 2 1

1 1

2 1

a0  2a1 x2  3a x

2 2 1 2 2

a  a x  a x  0

Updated 4/5/2017

1 1

2 1

p. 7 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

   nM   Mi     ni  T , P ,n ji

   nM   M1     n1  T , P ,n2 ,n3

M2

M3 M1 M2

M3

x1M1  x2 M 2  x3M 3

lim x1 0 M1 lim x2 0 M 2

lim x3 0 M 3 Updated 4/5/2017

p. 8 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

Updated 4/5/2017

p. 9 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

and

   nV   Vi     ni  T , P ,n j  i Updated 4/5/2017

p. 10 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

V1  V1  V1  V1  V1 

V2  V2  V2  V2  V2 

x1V1  x2V2

Updated 4/5/2017

p. 11 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

N

 x dV  0 i 1

i

i

x1dV1  x2 dV2  0

V1  V1  V1  V2  V2  V2  dV1 

dV2 

x1dV1  x2 dV2  0

x1dV1  x2 dV2 x1dV1  x2 dV2

dV1 

dV2 

dV1  2  40 1  42 1  0 dx1 x 1 1

dV2  2  0   42  0   0 dx1 x 0 1

Updated 4/5/2017

p. 12 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

Total or Partial Molar Volume (cm 3/mol)

SVA problem 11.13(e) 140

V1,  128

130

V1  120

V1

120 110 100

V

V2,  85

90 80

V2  70 V2

70 60 0

0.2

0.4

0.6

0.8

1

x1

   nH   Hi     ni  T , P ,n ji

  n1  n2  nH  n1  a1  b1   n2  a2  b2  n1  n2  n1  n2   

   nH    n1  n1n2 n22 H1    a  b  b  b   1 1  1 2 2 2 n1  n2   n1  n2   n1  n2   n1  T , P ,n2 

H1   a1  b1 x1   b1 x1 x2  b2 x22

H 2   a2  b2 x2   b2 x1 x2  b1 x12 Updated 4/5/2017

p. 13 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

x1 H1  x2 H 2  x1  a1  b1 x1   b1 x12 x2  b2 x1 x22  x2  a2  b2 x2   b2 x1 x22  b1 x12 x2 x1 H1  x2 H 2  x1  a1  b1 x1   x2  a2  b2 x2 

H1  H  x2

dH dx1

H 2  H  x1

dH dx1

Ϻ

Ϻ

Ϻ

Updated 4/5/2017

p. 14 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

P

ln i 

Chapter 10

Zi  1 dP P

0

0.000000 -0.000200

0

100

200

300

400

500

600

-0.000400

(Z - 1)/P (1/bar)

P (Z-1)/P f (bar) Z (1/bar) ln(Phi) Phi (bar) 10 0.9850 -0.001500 -0.0150 0.9851 9.85 20 0.9700 -0.001500 -0.0300 0.9704 19.41 40 0.9420 -0.001450 -0.0595 0.9422 37.69 60 0.9130 -0.001450 -0.0885 0.9153 54.92 80 0.8850 -0.001438 -0.1174 0.8893 71.14 100 0.8690 -0.001310 -0.1449 0.8652 86.52 200 0.7650 -0.001175 -0.2691 0.7641 152.81 300 0.7620 -0.000793 -0.3675 0.6925 207.74 400 0.8240 -0.000440 -0.4292 0.6510 260.42 500 0.9100 -0.000180 -0.4602 0.6312 315.58

-0.000600 -0.000800 -0.001000 -0.001200 -0.001400 -0.001600 Pressure (bar)

Updated 4/5/2017

p. 15 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T (K)

P (bar)

423.15

Chapter 10

Tc (K) Pc (bar)

10

304.2

73.83

 0.224

Tr

Pr

B0

B1

1.3910

0.1354

-0.1659

0.0960

0.9860

 TP 

B0  0.083 

0.422

B1  0.139 

0.172

 

Tr1.6

  exp  B 0   B 1

r

r

Tr4.2

Pale blue boxes are input fields, pink box is the final output.

T (K)

P (bar)

423.15

200

Tc (K) Pc (bar)

Tr

Pr

304.2

73.83

0.224 1.391026 2.708926

Tr

Pr

Table Points Tr (1) Tr (1) Tr (2) Tr (2)

Pr(1) Pr(2) Pr(1) Pr(2)

1.30 1.30 1.40 1.40

2.00 3.00 2.00 3.00

0.7345 0.6383 0.7925 0.7145

1.1776 1.2853 1.1858 1.2942

Interpolated for Pr=2 Interpolated for Pr=3 Interpolated to Tr, Pr

0.7873 0.7077 0.7308 0.7699

1.1851 1.2934 1.2619

Final Value of 

Updated 4/5/2017

0

p. 16 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

Using appropriate Using 2nd virial method coefficient for all P f Percent f Percent (bar) Phi (bar) difference Phi (bar) difference 10 0.9860 9.86 0.09% 0.9860 9.86 0.09% 20 0.9723 19.45 0.19% 0.9723 19.45 0.19% 40 0.9453 37.81 0.33% 0.9453 37.81 0.33% 60 0.9191 55.15 0.41% 0.9191 55.15 0.41% 80 0.8936 71.49 0.49% 0.8936 71.49 0.49% 100 0.8689 86.89 0.43% 0.8689 86.89 0.43% 200 0.7699 153.98 0.76% 0.7549 150.98 -1.20% 300 0.7074 212.22 2.16% 0.6559 196.77 -5.28% 400 0.6626 265.04 1.78% 0.5699 227.96 -12.46% 500 0.6422 321.10 1.75% 0.4952 247.60 -21.54%

T (K) 600

P (bar) 300

Tc (K) 430.8

Pc (bar) 78.84

Tr 1.30 1.30 1.40 1.40

Pr 3.00 5.00 3.00 5.00

 0.245

Tr 1.3928

Pr 3.8052

0

Z 0.2079 0.0875 0.2397 0.1737

1

(H ) /(RT c) -2.274 -2.825 -1.857 -2.486

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Z 0.6344 0.7358 0.7202 0.7761

Interpolated Values Final Values Z 3 V (cm /mol)

0.7891 131.2

0.7378

0.2092

H /(RT c) R H (J/mol)

-2.20103 -7883.38

R

R 0

R 1

R 0

R 1

(S ) /R -1.299 -1.554 -0.99 -1.303

(S ) /R -0.481 -1.147 -0.29 -0.73

-0.2567

-1.1367

-0.4876

S (J/(mol K))

-1.25615 -10.4436

(H ) /(RT c) -0.3 -1.066 -0.044 -0.504

-2.1382 R

S /R R

 0.6383 0.5383 0.7145 0.6237

 1.2853 1.3868 1.2942 1.4488

0

1

0.6722  f (bar)

0.723998 217.1995

R

-0.32419 0.723111 216.9334

G /(RT)  f (bar)

 ln i 

1.3542

GiR H R  TS R  RT RT

 

Updated 4/5/2017

p. 17 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

 

T (K)

P (bar)

Tc (K)

Pc (bar)

20

417.9

40

553.15

 0.194

Tr

Pr

B

0

B

1

1.3236

0.5000

-0.1865

0.0860

0.9379

 TP 

B 0  0.083  B1  0.139 

0.422 Tr1.6 0.172

 

  exp  B 0   B 1

r

r

Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

Updated 4/5/2017

p. 18 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T (K)

P (bar)

553.15

100

Chapter 10

Tc (K) Pc (bar) 417.9

40

Tr

0.194 1.323642

Pr 2.5

Table Points Tr (1) Tr (1) Tr (2) Tr (2)

Pr(1) Pr(2) Pr(1) Pr(2)

2.00 3.00 2.00 3.00

0.7345 0.6383 0.7925 0.7145

1.1776 1.2853 1.1858 1.2942

Interpolated for Pr=2 Interpolated for Pr=3 Interpolated to Tr, Pr

0.7482 0.6563 0.7023 0.7314

1.1795 1.2874 1.2335

Pr

1.30 1.30 1.40 1.40

Final Value of 

Updated 4/5/2017

Tr

0

  

p. 19 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Vi fi  isat Pi sat exp  

sat

Chapter 10

 P  P   sat

i

 

 

RT

T (K)

P (bar)

Tc (K)

Pc (bar)

383.15

5.267

511.8

45.02

Tr

Pr

B

0

B

1

0.7486

0.1170

-0.5876

-0.4412

0.9000

 TP 

B 0  0.083  B1  0.139 

0.196

0.422 Tr1.6 0.172

  exp  B 0   B 1 

r

r

Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

 107.5 cm3 mol-1  275  5.267  bar   fi  0.9000 *5.267 bar*exp   83.145 cm3 bar mol-1 K -1 *383.15 K    fi  11.78 bar

Updated 4/5/2017

p. 20 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

 V sat P  P sat i i  sat sat  exp  sat  RT fi i Pi  fi

fi

   

 19.65 cm3 mol-1 150  4.76  bar     1.085  exp  83.145 cm3 bar mol-1 K -1 * 423.15 K  fi sat   fi

Gi  i T   RT ln fi G final  Ginitial  i T   RT ln f final  i T   RT ln f initial  f final  G final  Ginitial  RT ln    finitial 

 G final  Ginitial   G   H S   exp   exp     exp   finitial RT R   RT   RT   f final

 

Updated 4/5/2017

p. 21 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

 

2775.08 31.396   H S    exp    exp     finitial R   RT  8.3145* 673.15 8.3145  f final  exp  0.4958  3.7761  0.0376 finitial f final

Gi  i T   RT ln fi G final  Ginitial  i T   RT ln f final  i T   RT ln f initial  f final  G final  Ginitial  RT ln    finitial 

 G final  Ginitial   G   H S   exp   exp     exp   finitial RT R   RT   RT   f final

 

758.6 6.397   H S    exp      exp   finitial R   RT  1.986 *1259.7 1.986  f final  exp  0.30323  3.2210   0.0541 finitial f final

 V sat P  P sat i i fi  isat Pi sat exp   RT 

Updated 4/5/2017

   

p. 22 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

 T (K)

P (bar)

Tc (K)

Pc (bar)

309.2

1.013

469.7

33.7

Tr

Pr

B

0

B

1

0.6583

0.0301

-0.7408

-0.8568

0.9573

 TP 

B 0  0.083  B  0.139  1

0.252

0.422

 

Tr1.6

  exp  B 0   B 1

0.172

r

r

Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

 119.4 cm3 mol-1  200  1.01 bar   fi  0.9573*1.013 bar*exp   83.145 cm3 bar mol-1 K -1 *309.2 K    fi  2.4 bar

Vi fi  isat Pi sat exp    

sat

 P  P   sat

i

RT

 

Updated 4/5/2017

p. 23 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T (K)

Chapter 10

P (bar)

266.3

1

Tc (K) Pc (bar)

417.9

0.194

40

Tr

Pr

B0

B1

0.6372

0.0250

-0.7848

-1.0025

0.9623

B 0 0.083

0.422

 TP 

B1 0.139

Vi fi  isat Pi sat exp  

sat

 

 

Tr1.6

  exp  B0   B1

0.172

r

r

Tr4.2

 P  P   sat

i

 

RT

 82.0 cm3 mol-1  200  1.01 bar   fi  0.9623 *1.013 bar*exp   83.145 cm3 bar mol-1 K -1 * 266.3 K    fi  2.04 bar

Vi fi  isat Pi sat exp    

sat

 P  P   sat

i

RT

 

Updated 4/5/2017

p. 24 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

P (bar) Tc (K) Pc (bar)

266.9

1.01325

420

40.43

0.191

Tr

Pr

B0

B1

0.6355

0.0251

-0.7887

-1.0158

0.9620

 TP 

T (K)

0.422

B 0 0.083

 

Tr1.6

  exp  B0   B1

0.172

B1 0.139

r

r

Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

Vi fi  isat Pi sat exp    

sat

 P  P   sat

i

RT

 

 91.43 cm3 mol-1  200  1.01 bar   fi  0.9620 *1.013 bar*exp   83.145 cm3 bar mol-1 K -1 * 266.9 K    fi  2.21 bar

Updated 4/5/2017

p. 25 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Psat (bar)

T (K) 473.15

Chapter 10

Tc (K)

22.27

536.4

54.72

Vsat (cm3 /mol)

Tr

B0

B1

0.8821

-0.4328

-0.1523

Pc (bar)

Vc (cm3/mol)

0.222

 P    exp  B0   B1 r  T 0.172 r   B1 0.139 4.2

Tr

Above P sat

Vi fi  isat Pi sat exp    

sat

 P  P   sat

i

RT

 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5

0.422 Tr1.6

0.293

P (bar)

122.7

Below P sat

B 0 0.083

Zc

239.0

 

f (bar) 1.000 0.995 0.990 0.986 0.981 0.976 0.971 0.967 0.962 0.957 0.953 0.948 0.944 0.939 0.935 0.930 0.926 0.921 0.917 0.912

0.00 0.50 0.99 1.48 1.96 2.44 2.91 3.38 3.85 4.31 4.76 5.22 5.66 6.10 6.54 6.98 7.40 7.83 8.25 8.67

20

1.0

18

0.9

16

0.8

14

0.7

12

0.6

10

0.5

8

0.4

6

0.3

4

0.2

2

0.1

0

Fugacity Coefficient

Fugacity (bar)

Fugacity and Fugacity Coefficient of Chloroform at 200 °C

0.0 0

5

10

15

20

25

30

35

40

Pressure (bar)

Updated 4/5/2017

p. 26 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

Spreadsheet for computing fugacity coefficients in a binary mixture (Applied to Problem 11.25) T (K) P (bar) 423.15 30 Vc Mole 3 Fraction Name Tc (K) Pc (bar) (cm /mol) Zc Species 1 Ethylene 0.35 282.3 50.4 131 0.281 Species 2 Propylene 0.65 365.6 46.65 188.4 0.289

Cross-Parameters

kij

Vc Tc (K) Pc (bar) (cm3/mol) 0 321.26139 48.19 158.0

2nd Virial Coefficients (cm3/mol) B11 -60 B22 -159 B12 -99 12 (cm /mol) 3

ln ˆ

1

ln ˆ2

ij  Pcij 

20.83

 ˆ1

0.957 0.875

i   j 2 Zcij RTcij

Tr 1.4989 1.1574

Pr 0.5952 0.6431

B0 -0.1378 -0.2510

B1 0.1076 0.0459

0.1135

Tr 1.3172

Pr 0.6225

B0 -0.1886

B1 0.0849

Tcij  TciTcj 1 kij 

B 0  0.083 

0.422

Zci  Zcj

B1  0.139 

0.172

Zcij 

Vcij

2

Tr1.6 Tr4.2

3

 V 1/3  Vcj1/3  Vcij   ci   2  

-0.04351 -0.13371

ˆ 2

Zc 0.2850

0.087 0.14

Bij 

RTcij Pcij

B  B  0

P P  B11  y22 12   RT  B11  y22  2 B12  B11  B22   RT P P ln ˆ2   B22  y12 12   RT  B22  y12  2 B12  B11  B22   RT ln ˆ1 

fˆethylene  ˆethylene yethylene P  0.957*0.35*30 bar  10.05 bar fˆpropylene  ˆpropylene y propylene P  0.875*0.65*30 bar  17.06 bar

fˆiid  yi fi  yii P

Updated 4/5/2017

p. 27 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

1

ij


SVNAS 8th Edition Annotated Solutions

T (K)

Chapter 10

P (bar)

Tc (K)

Pc (bar)

30

282.3

50.4

423.15

 0.087

Tr

Pr

B

0

B

1

1.4989

0.5952

-0.1378

0.1076

0.9503

 TP 

B 0  0.083 

0.422

B  0.139 

0.172

1

Tr1.6

  exp  B 0   B 1 

r

r

Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

T (K)

P (bar)

Tc (K)

Pc (bar)

30

365.6

46.65

423.15

 0.14

Tr

Pr

B

0

B

1

1.1574

0.6431

-0.2510

0.0459

0.8729

 TP 

B 0  0.083 

0.422

B  0.139 

0.172

1

Tr1.6

  exp  B 0   B 1 

r

r

Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

id fˆethylene  yethyleneethylene P  0.35* 0.9503*30 bar  9.98 bar id fˆpropylene  y propylene propylene P  0.65* 0.8729 *30 bar  17.02 bar

φ

Updated 4/5/2017

p. 28 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

ln ˆk 

Chapter 10

 P  1 N N B  yi y j  2 ik   ij    kk  RT  2 i 1 j 1 

 ik   ki  2 Bik  Bii  Bkk  ij   ji  2 Bij  Bii  B jj  ii   jj   kk  0

P B11  y2212  y3213  y2 y3 12  13   23    RT P ln ˆ2  B22  y32 23  y1212  y3 y1  23  12  13    RT P ln ˆ3  B33  y1213  y22 23  y1 y2 13   23  12    RT ln ˆ1 

Updated 4/5/2017

p. 29 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

Spreadsheet for computing fugacity coefficients in a binary mixture (problem 11.27 in SVA) T (K) P (bar) 373.15 35

Species 1 Species 2 Species 3

Mole Fraction 0.21 0.43 0.36

Name Methane Ethane Propane Cross-Parameters 1,2 1,3 2,3

kij

Tc (K) Pc (bar) 0 241.22641 47.01 0 265.488 43.26 0 336.00586 45.26

3

2nd Virial Coefficients (cm /mol) B11 -21 B22 -113 B33 -244 B12 -52 B13 -78 B23 -167 12 (cm /mol) 3 13 (cm /mol) 3 23 (cm /mol) 3

 ij  Pcij 

ln ˆ3

0.01895 -0.12320 -0.25478

ˆ1 ˆ2 ˆ3

1.019 0.884 0.775

i   j

Tr 1.9578 1.2222 1.0091

Pr 0.7610 0.7184 0.8239

0

0.012 0.1 0.152

B -0.0610 -0.2231 -0.3330

B 0.1288 0.0650 -0.0266

Tr 1.5469 1.4055 1.1105

Pr 0.7446 0.8090 0.7734

0

0.0560 0.0820 0.1260

B -0.1270 -0.1618 -0.2738

B 0.1115 0.0978 0.0283

Zc 0.286 0.279 0.276

Zc 0.2825 0.2810 0.2775

Tcij  TciTcj 1  kij 

2 Z cij RTcij

Z cij 

Vcij

 V 1/ 3  Vcj1/ 3  Vcij   ci   2  

30.33 107.52 23.10

ln ˆ1 ln ˆ2

Vc 3 (cm /mol) 98.6 145.5 200 Vc (cm3/mol) 120.5 143.4 171.3

Tc (K) Pc (bar) 190.6 45.99 305.3 48.72 369.8 42.48

Z ci  Z cj

B 0  0.083 

0.422

B1  0.139 

0.172

2

1

1

Tr1.6 Tr4.2

3

Bij 

RTcij Pcij

B  B  0

1

ij

P  B11  y2212  y3213  y2 y3 12  13   23   RT P ln ˆ2   B22  y32 23  y1212  y3 y1  23  12  13   RT P ln ˆ3   B33  y1213  y22 23  y1 y2 13   23  12   RT ln ˆ1 

f1 (bar) f2 (bar) f3 (bar)

7.49 13.31 9.77

fˆiid  yi fi  yii P

T (K)

P (bar)

373.15

35

Tc (K) Pc (bar) 190.6

45.99

0.012

Tr

Pr

B

0

B

1

1.9578

0.7610

-0.0610

0.1288

0.9771

 TP 

B 0  0.083  B  0.139  1

0.422 Tr1.6 0.172

 

  exp  B 0   B1

r

r

Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

Updated 4/5/2017

p. 30 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T (K)

Chapter 10

P (bar)

373.15

Tc (K) Pc (bar)

35

305.3

48.72

0.1

Tr

Pr

B

0

B

1

1.2222

0.7184

-0.2231

0.0650

0.8805

 TP 

B 0  0.083  B  0.139  1

0.422

 

Tr1.6

  exp  B 0   B1

0.172

r

r

Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

T (K)

P (bar)

373.15

Tc (K) Pc (bar)

35

369.8

42.48

0.152

Tr

Pr

B

0

B

1

1.0091

0.8239

-0.3330

-0.0266

0.7594

 TP 

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6

  exp  B 0   B1 

r

r

Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

f1 (bar) f2 (bar) f3 (bar)

Updated 4/5/2017

7.18 13.25 9.57

p. 31 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

,

Part a b c d e f g

and

I II G1E H1E S1E CPE S2E H2E G2E S2E H2E G2E -622.0 -1920.0 -4.354 4.2 -3.951 -1794 -497.4 -4.354 -1920.0 -491.4 1095.0 1595.0 1.677 3.3 1.993 1694 1039.9 1.677 1595.0 1044.7 407.0 984.0 1.935 -2.7 1.677 903 352.8 1.935 984.0 348.9 632.0 -208.0 -2.817 23.0 -0.614 482 683.5 -2.817 -208.0 716.5 1445.0 605.0 -2.817 11.0 -1.764 935 1513.7 -2.817 605.0 1529.5 734.0 -416.0 -3.857 11.0 -2.803 -86 833.9 -3.857 -416.0 849.7 759.0 1465.0 2.368 -8.0 1.602 1225 699.5 2.368 1465.0 688.0

x1 0.02715 0.09329 0.1749 0.3276 0.40244 0.56689 0.63128 0.66233 0.69984 0.72792 0.77514

VE 87.5 265.6 417.4 534.5 531.7 421.1 347.1 321.7 276.4 252.9 190.7

Updated 4/5/2017

p. 32 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

0.79243 0.82954 0.86835 0.93287 0.98233

Chapter 10

178.1 138.4 98.4 37.6 10

600 500

VE

400

VE actual

300 200 100 0 0

0.2

0.4

0.6

0.8

1

x1

Updated 4/5/2017

p. 33 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

3500 3000 2500

VE1 VE2

VE

2000 1500 1000 500 0 -500 0

0.2

0.4

0.6

0.8

1

x1

B  y12 B11  2 y1 y2 B12  y22 B22

BP 504.25* 2 1  0.965 RT 83.14*348.15 H R BP P dB 504.25* 2 2     *3.55  0.1202 RT RT R dT 83.14*348.15 83.14 SR P dB 2   *3.55  0.08540 R R dT 83.14

Z 1

Updated 4/5/2017

p. 34 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

1 0.95 0.9

ɸhat1 ɸhat2

0.85 0.8 0.75

ɸhat1

0.7

ɸhat2

0.65 0.6 0.55 0.5 0

0.2

0.4

0.6

0.8

1

y1

Updated 4/5/2017

p. 35 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

isa function of T only: ,

; substitution yields:

x1 HE 0.0426 -23.3 0.0817 -45.7 0.1177 -66.5 0.151 -86.6 0.2107 -118.2 0.2624 -144.6 0.3472 -176.6 0.4158 -195.7 0.5163 -204.2 0.6156 -191.7 0.681 -174.1 0.7621 -141 0.8181 -116.8 0.865 -85.6 0.9276 -43.5 0.9624 -22.6

Updated 4/5/2017

p. 36 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

0 0

0.2

0.4

0.6

0.8

1

-50

HE

-100 -150 HE actual

-200

equation -250

Updated 4/5/2017

x1

p. 37 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

100 0 -100

0

0.2

0.4

0.6

1 HEbar1

-200

HEbar

0.8

HEbar2 -300 -400 -500 -600 -700

x1

N

N

B   yi y j Bij i 1 j 1

B

B  y12 B11  2 y1 y2 B12  y22 B22

RTc 0 0.422 0.172 B   B1 , with B0  0.083  1.6 and B1  0.139  4.2 Pc Tr Tr

Updated 4/5/2017

p. 38 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

i   j

ij 

2 Tcij  TciTcj 1  kij  Pcij 

Z cij RTcij

Z cij 

Vcij Z ci  Z cj 2

V Vcij    

1/ 3 ci

 Vcj1/ 3   2 

3

dB dB dB dB  y12 11  2 y1 y2 12  y22 22 dT dT dT dT dBij dT

V

0 dBij1  R  dBij0 dBij1  dBij0 0.675 RTc  dBij     , with    2.6 and Pc  dT dT  Pc  dTr dTr  dT Tr    

RT B, P

Updated 4/5/2017

G R  BP ,

H R  BP  TP

dB , dT

and S R   P

dBij1 dT

0.722 Tr5.2

dB dT

p. 39 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

Spreadsheet for computing residual properties of a binary mixture, for SVA problem 11.37(a) T (K) P (bar) 333.15 1.7 Vc Mole 3 Fraction Name Tc (K) Pc (bar) (cm /mol) Zc  Species 1 Acetone 0.28 508.2 47.01 209 0.233 0.307 Species 2 1,3-butadiene 0.72 425.2 42.77 220.4 0.267 0.190

Cross-Parameters

Vc Tc (K) Pc (bar) (cm3/mol) 464.8512 45.02 214.6

kij 0 3

2nd Virial Coefficients (cm /mol) B11 -912 B22 -500 B12 -665

i   j

ij  Pcij  3

Derivatives of 2nd Virial Coefficients (cm /((mol K)) dB11/dT 7.10 dB22/dT 3.42 dB12/dT 4.84

Zc 0.2500

 0.2485

Tr 0.6555 0.7835

Pr 0.0362 0.0397

B -0.7464 -0.5405

0

B -0.8744 -0.3402

Tr 0.7167

Pr 0.0378

B -0.6361

0

B -0.5579

Tcij  TciTcj 1  kij 

2 Zcij RTcij

Zcij 

Vcij

Zci  Zcj 2

1

1

B 0  0.083 

0.422

B  0.139 

0.172

1

0

dB dT 6.4892 2.5674

0

dB dT 4.0817

dB dT 2.0236 1.2729 dB dT 1.6049

1

1

Tr1 .6 Tr4 .2

3

 Vci1/ 3  Vcj1/ 3  Vcij     2  

Bij 

RTcij Pcij

B   B  0

1

ij

dBij RTc  dBij0 dBij1  dBij0 0.675 dBij1 0.722   , with    2.6 and  5.2 3 Mixture 2nd Virial Coefficient (cm /mol) dT Pc  dT dT  dT dT T Tr r   B -599 3 2 2 Derivative of Mixture 2nd Virial Coefficient (cm /((mol K)) B  y1 B11  2 y1 y2 B12  y2 B22 dB/dT 4.28

Mixture Properties: V (cm3/mol) Z R G (J/mol) R H (J/mol) R S (J/(mol K))

dB dB dB dB  y12 11  2 y1 y2 12  y22 22 dT dT dT dT

15695 0.9632 -101.8 -344.3 -0.728

N

V

N

B   yi y j Bij i 1 j 1

B

RT B, P

GR  BP ,

H R  BP  TP

dB , dT

and S R  P

dB dT

B  y12 B11  2 y1 y2 B12  y22 B22

RTc 0 0.422 0.172 B   B1 , with B0  0.083  1.6 and B1  0.139  4.2 Pc Tr Tr

Updated 4/5/2017

p. 40 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

i   j

ij 

2 Tcij  TciTcj 1  kij  Pcij 

Z cij RTcij

Z cij 

Vcij Z ci  Z cj 2

V Vcij    

1/ 3 ci

 Vcj1/ 3   2 

3

dB dB dB dB  y12 11  2 y1 y2 12  y22 22 dT dT dT dT dBij dT

V

0 dBij1  R  dBij0 dBij1  dBij0 0.675 RTc  dBij     , with    2.6 and Pc  dT dT  Pc  dTr dTr  dT Tr    

RT B, P

G R  BP ,

N

N

B   yi y j Bij i 1 j 1

B

H R  BP  TP

dB , dT

and S R   P

dBij1 dT

0.722 Tr5.2

dB dT

B  y12 B11  2 y1 y2 B12  y22 B22

RTc 0 0.422 0.172 B   B1 , with B0  0.083  1.6 and B1  0.139  4.2 Pc Tr Tr

Updated 4/5/2017

p. 41 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

i   j

ij 

2 Tcij  TciTcj 1  kij  Pcij 

Z cij RTcij

Z cij 

Vcij Z ci  Z cj 2

V Vcij    

1/ 3 ci

 Vcj1/ 3   2 

3

dB dB dB dB  y12 11  2 y1 y2 12  y22 22 dT dT dT dT dBij dT

V

0 dBij1  dBij0 0.675 RTc  dBij   , with   2.6 and Pc  dT dT  dT Tr  

RT B, P

Updated 4/5/2017

G R  BP ,

H R  BP  TP

dB , dT

dBij1 dT

0.722 Tr5.2

and S R   P

dB dT

p. 42 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

Spreadsheet for computing residual properties of a binary mixture, for SVA problem 11.37(c) T (K) P (bar) 298.15 1 Vc Mole 3 Fraction Name Tc (K) Pc (bar) (cm /mol) Zc  Species 1 Methyl chloride 0.45 416.3 66.8 143 0.276 0.153 Species 2 Ethyl chloride 0.55 460.4 52.7 200 0.275 0.190

Cross-Parameters kij

Tc (K) 437.7951

0

Vc Pc (bar) (cm3/mol) 59.02 169.9

3

2nd Virial Coefficients (cm /mol) B11 -374 B22 -682 B12 -507

i   j

ij  Pcij 

Derivatives of 2nd Virial Coefficients (cm3/((mol K)) dB11/dT 2.78 dB22/dT 5.37 dB12/dT 3.87

Pr 0.0150 0.0190

B0 -0.6369 -0.7627

B1 -0.5599 -0.9278

dB0 dT 1.6078 2.0889

dB1 dT 4.0963 6.9148

Tr 0.6810

Pr 0.0169

B0 -0.6973

B1 -0.7245

dB0 dT 1.8326

dB1 dT 5.3221

Tcij  TciTcj 1  kij 

2 Z cij RTcij

Z cij 

Vcij

 V 1/ 3  Vcj1/ 3  Vcij   ci   2  

 0.1715

Zc 0.2755

Tr 0.7162 0.6476

Z ci  Z cj 2

B1 0.139

0.172

Tr1.6 Tr4.2

Bij 

dBij1 dT

RTcij Pcij

B   B  0

1

ij

0.722 Tr5.2

dB dB dB dB  y12 11  2 y1 y2 12  y22 22 dT dT dT dT

24257 0.9785 -53.3 -175.6 -0.410

V

N

N

RT B, P

B   yi y j Bij i 1 j 1

B

0.422

3

dBij RTc  dBij0 dBij1  dBij0 0.675   , with    2.6 and Mixture 2nd Virial Coefficient (cm3/mol) dT Pc  dT dT  dT Tr   B -533 3 2 2 Derivative of Mixture 2nd Virial Coefficient (cm /((mol K)) B  y1 B11  2 y1 y2 B12  y2 B22 dB/dT 4.10

Mixture Properties: V (cm3/mol) Z GR (J/mol) HR (J/mol) SR (J/(mol K))

B 0 0.083

G R  BP ,

H R  BP  TP

dB , dT

and S R   P

dB dT

B  y12 B11  2 y1 y2 B12  y22 B22

RTc 0 0.422 0.172 B   B1 , with B0  0.083  1.6 and B1  0.139  4.2 Pc Tr Tr

ij 

i   j

2 Tcij  TciTcj 1  kij  Pcij  Z cij 

Z cij RTcij Vcij Z ci  Z cj 2

 V 1/ 3  Vcj1/ 3  Vcij   ci   2  

Updated 4/5/2017

3

p. 43 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

dB dB dB dB  y12 11  2 y1 y2 12  y22 22 dT dT dT dT 0 dBij1  dBij0 0.675 RTc  dBij     , with  2.6 and dT Pc  dT dT  dT Tr  

dBij

V

RT B, P

G R  BP ,

H R  BP  TP

0

Vc Tc (K) Pc (bar) (cm3/mol) 226.2727 62.00 80.6

i   j 2 Z RT Pcij  cij cij Vcij

2nd Virial Coefficients (cm 3/mol) B11 -7 B22 -228 B12 -56 Derivatives of 2nd Virial Coefficients (cm /((mol K)) dB11/dT 0.19 dB22/dT 1.89 dB12/dT

V Vcij    

1/ 3 ci

Z cij 

V

1/ 3 cj

2

dT

  

0.722 Tr5.2

and S R   P

dB dT

Tr 2.3229 0.7226

Pr 0.0882 0.0266

B0 -0.0266 -0.6267

B1 0.1340 -0.5343

dB0 dT 0.0754 1.5711

dB1 dT 0.0090 3.9114

Tr 1.2956

Pr 0.0484

B0 -0.1959

B1 0.0810

dB0 dT 0.3443

dB1 dT 0.1878

Tcij  TciTcj 1  kij 

ij 

3

 0.1455

Zc 0.2655

dB , dT

Spreadsheet for computing residual properties of a binary mixture, for SVA problem 11.37(d) T (K) P (bar) 293.15 3 Vc Mole 3 Fraction Name Tc (K) Pc (bar) (cm /mol) Zc  Species 1 Nitrogen 0.83 126.2 34 89.2 0.289 0.038 Species 2 Ammonia 0.17 405.7 112.8 72.5 0.242 0.253

Cross-Parameters kij

dBij1

Z ci  Z cj 2

B 0 0.083

0.422

B 0.139

0.172

1

Tr1.6 Tr4.2

3

Bij 

RTcij Pcij

B   B  0

1

ij

0.50

dBij RTc  dBij0 dBij1  dB0 0.675 dBij1 0.722   , with ij    and  5.2 2.6 Mixture 2nd Virial Coefficient (cm 3/mol) dT Pc  dT dT  dT dT T Tr r   B -27 3 Derivative of Mixture 2nd Virial Coefficient (cm /((mol K)) B  y12 B11  2 y1 y2 B12  y22 B22 dB/dT 0.32 Mixture Properties: V (cm3/mol) Z GR (J/mol) HR (J/mol) SR (J/(mol K))

Updated 4/5/2017

8098 0.9967 -8.1 -36.5 -0.097

dB dB dB 2 dB11  y1  2 y1 y2 12  y22 22 dT dT dT dT V

RT B, P

G R  BP ,

H R  BP  TP

dB , dT

and S R   P

dB dT

p. 44 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

N

Chapter 10

N

B   yi y j Bij

B  y12 B11  2 y1 y2 B12  y22 B22

i 1 j 1

B

RTc 0 0.422 0.172 B   B1 , with B0  0.083  1.6 and B1  0.139  4.2 Pc Tr Tr

ij 

i   j

2 Tcij  TciTcj 1  kij  Pcij  Z cij 

Z cij RTcij Vcij Z ci  Z cj 2

 V 1/ 3  Vcj1/ 3  Vcij   ci   2  

3

dB dB dB dB  y12 11  2 y1 y2 12  y22 22 dT dT dT dT 0 dBij1  dBij0 0.675 RTc  dBij     , with  2.6 and dT Pc  dT dT  dT Tr  

dBij

V

RT B, P

Updated 4/5/2017

G R  BP ,

H R  BP  TP

dB , dT

dBij1 dT

0.722 Tr5.2

and S R   P

dB dT

p. 45 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

Spreadsheet for computing residual properties of a binary mixture, for SVA problem 11.37(e) T (K) P (bar) 293.15 4.2 Vc Mole 3 Fraction Name Tc (K) Pc (bar) (cm /mol) Zc  Species 1 sulfur dioxide 0.32 430.8 78.84 122 0.269 0.245 Species 2 ethylene 0.68 282.3 50.4 131 0.281 0.087

Cross-Parameters kij 0

Vc Tc (K) Pc (bar) (cm3/mol) 348.7332 63.06 126.4

i   j 2 Z cij RTcij Pcij  Vcij

3

2nd Virial Coefficients (cm /mol) B11 -398 B22 -147 B12 -235

dB12/dT

 V 1/ 3  Vcj1/ 3  Vcij   ci   2  

1.79

dBij Mixture 2nd Virial Coefficient (cm 3/mol) dT B -211 3 Derivative of Mixture 2nd Virial Coefficient (cm /((mol K)) dB/dT 1.62 Mixture Properties: V (cm3/mol) Z GR (J/mol) HR (J/mol) SR (J/(mol K))

5593 0.9637 -88.5 -288.4 -0.682

B0 -0.6983 -0.3143

B1 -0.7274 -0.0078

dB0 dTr 1.8365 0.6120

dB1 dTr 5.3445 0.5934

Tr 0.8406

Pr 0.0666

B0 -0.4741

B1 -0.2176

dB0 dTr 1.0601

dB1 dTr 1.7809

Z cij 

Z ci  Z cj 2

B 0 0.083

0.422

B1 0.139

0.172

Tr1.6 Tr4.2

3

Bij 

RTcij Pcij

B   B  0

1

ij

0 dBij1  dB0 0.675 dBij1 0.722 R  dBij   , with ij   and  5.2 2.6 Pc  dTr dTr  dTr dTr Tr Tr  

B  y12 B11  2 y1 y2 B12  y22 B22 dB dB dB 2 dB11  y1  2 y1 y2 12  y22 22 dT dT dT dT

V

and Updated 4/5/2017

Pr 0.0533 0.0833

Tcij  TciTcj 1  kij 

ij 

Derivatives of 2nd Virial Coefficients (cm 3/((mol K)) dB11/dT 3.32 dB22/dT 1.09

 0.1660

Zc 0.2750

Tr 0.6805 1.0384

RT B, P

G R  BP ,

H R  BP  TP

then

dB , dT

and S R   P

dB dT

and p. 46 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

Updated 4/5/2017

p. 47 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

has the same sign over the whole composition range, both

Updated 4/5/2017

Chapter 10

and

p. 48 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

2

ME MEbar1

ME values

1.5

MEbar2

1

0.5

0

-0.5 0

0.2

0.4

0.6

0.8

1

xi

35 30

ME values

25 ME 20

MEbar1

15

MEbar2

10 5 0 0

0.2

0.4

0.6

0.8

1

xi

Updated 4/5/2017

p. 49 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

)

Updated 4/5/2017

p. 50 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

Updated 4/5/2017

p. 51 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

. Therefore the sign of

Chapter 10

is the same as the sign of

, and by the same argument the sign of

T (K) 283.15 303.15 323.15

Updated 4/5/2017

. Similarly, at

is of opposite sign as the

GE (J/mol) 544 513 494.2

HE (J/mol) 932.1 893.4 845.9

p. 52 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 10

Updated 4/5/2017

p. 53 of 53

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

M 1  M  x2

dM dx1

M 2  M  x1

dM dx1

V1  V  x2

dV dx1

V2  V  x1

dV dx1

Chapter 11

V  x1V1  x2V2  V E  x1V1  x2V2  V V  110 x1  90 x2  x1 x2  45 x1  25 x2 

V  110 x1  90 1  x1   x1 1  x1   45 x1  25 1  x1   V  20 x1  90   x1  x12   20 x1  25  V  90  45 x1  5 x12  20 x13 dV  45  10 x1  60 x12 dx1

dV dx1

V1  V  x2

dV  105.92  0.6*31.4  124.76 cm 3 mol -1 dx1

V2  V  x1

dV  105.92  0.4*31.4  93.36 cm3 mol-1 dx1

Updated 4/5/2017

p. 1 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

   nV   Vi     ni  T , P ,n ji V  x1V1  x2V2  V E  x1V1  x2V2  V V  110 x1  90 x2  x1 x2  45 x1  25 x2  nV  110n1  90n2 

n1n2

 n1  n2 

2

 45n1  25n2 

    nV   n1n2 n2 2n1n2  V1    110  45   45n1  25n2      2 2 3    n n  n n  n n  n       1   T , P ,n2 1 2 1 2  1 2  V  110  45 x1 x2   45 x1  25 x2  x2  2 x1 x2  V  110  90 x1 x2 1  x1   25 x22  50 x1 x22 V  110  40 x1 x22  25 x22

V1

V  x1V1  x2V2 V2 

V  x1V1 105.92  0.4*124.76   93.36 cm3 mol-1 x2 0.6

 

Updated 4/5/2017

p. 2 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

   

 

 

 

  

Updated 4/5/2017

p. 3 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

 

 

 

Updated 4/5/2017

p. 4 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

  

Updated 4/5/2017

p. 5 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

H

Updated 4/5/2017

p. 6 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

4 moles water

Chapter 11

Enthalpy Change = -25.5 kJ

1 mol LiCl Enthalpy Change = -20.8 kJ

3 moles water

Enthalpy Change = -25.5 kJ +20.8 kJ = -4.7 kJ 5 moles 20% LiCl in water

5 moles 20% LiCl in water

1 mol LiCl

5 moles 20% LiCl in water

1 mole water

 

   

Updated 4/5/2017

p. 7 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

Updated 4/5/2017

p. 8 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

 

 

    

Updated 4/5/2017

 p. 9 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

Updated 4/5/2017

p. 10 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

n 10 Updated 4/5/2017

Hf (kJ) -862.74

ΔHf HfCaCl2 -66.94 p. 11 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

15 20 25 50 100 300 500 1000

-867.85 -870.06 -871.07 -872.91 -873.82 -874.79 -875.13 -875.54

-72.05 -74.26 -75.27 -77.11 -78.02 -78.99 -79.33 -79.74

-66 -68

Hf -HfCaCl2

-70 -72 -74 -76 -78 -80 10

100

1000

ni

Updated 4/5/2017

p. 12 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

Updated 4/5/2017

p. 13 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

Updated 4/5/2017

p. 14 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

Updated 4/5/2017

p. 15 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

Updated 4/5/2017

p. 16 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

Updated 4/5/2017

p. 17 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

Updated 4/5/2017

p. 18 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

Updated 4/5/2017

p. 19 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

Updated 4/5/2017

p. 20 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

Updated 4/5/2017

p. 21 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

Updated 4/5/2017

p. 22 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

M: Differentiate:

Updated 4/5/2017

p. 23 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

0 0

0.2

0.4

0.6

0.8

1

H (kJ/kg)

-500

-1000

Hebar1 Hebar2

-1500

H(x1)

-2000

-2500

Updated 4/5/2017

X1

p. 24 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

θ θ

Updated 4/5/2017

p. 25 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

3600 3100

r(x1) (kg/h)

2600 2100 1600 1100 600 0

1

2

3

4

5

6

Θ (x1)

Updated 4/5/2017

p. 26 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

Updated 4/5/2017

p. 27 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 11

Updated 4/5/2017

p. 28 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

As the system is closed, extensive coordinates remain unchanged. The intensive variables will change value as temperature changes.

Updated 4/5/2017

p. 1 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 2 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 3 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 4 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 5 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 6 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 7 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 8 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 9 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 10 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 11 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 12 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 13 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 14 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 15 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 16 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 17 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 18 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 12

Updated 4/5/2017

p. 19 of 19

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

N

P   xi Pi sat  x1 P1sat  x2 P2sat i 1

P  0.33  180.45  0.67  74.26  109.3 kPa

xi Pi sat yi  P x P sat 0.33  180.45 y1  1 1   0.545 P 109.3 y2  1  y1  0.455

P N

1 yi

P i 1

P

x1 

sat

i

1 y1 y2  sat sat P1 P2

1 0.33 0.67  180.45 74.26

 92.16 kPa

y1 P 0.33  92.16   0.169 180.45 P1sat

x2  1  x1  0.831

Updated 4/5/2017

p. 1 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

N

P   xi Pi sat  x1 P1sat  x2 P2sat i 1

  B1  B2  P  x1 exp  A1    x2 exp  A2   T  C1  T  C2    2726.81  3056.96    120  0.33exp 13.7819    0.67 exp 13.9320   T  217.572  T  217.625    

x1 P1sat 0.33  197.15 y1    0.542 P 120 y2  1  y1  0.458

1

P N

yi

P i 1

i

sat

1 y1 y2  sat sat P1 P2 1

120 

0.33 2726.81   exp 13.7819   T  217.572  

0.67 3056.96   exp 13.9320   T  217.625   

x1 

y1 P 0.33  120.00   0.173 229.40 P1sat

x2  1  x1  0.827 Updated 4/5/2017

p. 2 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

 N

P   xi Pi sat  x1 P1sat  x2 P2sat i 1

P  x1  206.14  1  x1   86.20  120 kPa 120  86.20  0.282 206.14  86.20 x2  1  x1  0.718 x1 

P1sat 206.14  0.282  0.484 P 120 y2  1  y1  0.516 y1  x1

V

z1  x1 0.33  0.282   0.238 y1  x1 0.484  0.282

L  1  V  0.762

Updated 4/5/2017

p. 3 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Spreadsheet to compute a P-x-y diagram for SVA problem 10.2(a) Species 1 is benzene, species 2 is ethylbenzene Antoine Coefficients (for T in deg C and P in kPa) T (deg C) A1 B1 C1 A2 B2 C2 90 13.8594 2773.78 220.07 14.0045 3279.47 P (kPa)

x2 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

1.00 0.95 0.90 0.85 0.80 0.75 0.70 0.65 0.60 0.55 0.50 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00

Updated 4/5/2017

24.25 29.84 35.44 41.03 46.63 52.22 57.82 63.41 69.01 74.60 80.20 85.79 91.39 96.98 102.58 108.17 113.77 119.36 124.96 130.55 136.15

y1

P1sat (kPa) P2sat (kPa) 136.148 24.24732

P (kPa)

y2 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

1.00 0.95 0.90 0.85 0.80 0.75 0.70 0.65 0.60 0.55 0.50 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00

24.25 25.29 26.42 27.66 29.02 30.52 32.18 34.04 36.12 38.48 41.16 44.25 47.84 52.06 57.10 63.21 70.80 80.45 93.16 110.62 136.15

P-x-y Diagram for Benzene(1)/Ethylbenzene(2) at 90 deg C 140.00 120.00 100.00 P (kPa)

x1

213.2

80.00 60.00 40.00 20.00 0.00

0.20

0.40 0.60 x1 or y1

0.80

1.00

p. 4 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Spreadsheet to compute a T-x-y diagram for SVNA problem 10.2(a) Species 1 is 1-benzene, species 2 is ethylbenzene Antoine Coefficients (for T in deg C and P in kPa) P (kPa) A1 B1 C1 A2 B2 C2 90 13.7819 2726.81 217.572 13.9726 3259.93 x2 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29

2

P (kPa) (P-Pspec) y 1 T (deg C) P 1 sat P 2 sat 1.00 131.84 394.58 90.00 90.00 0.00 0.99 130.62 383.98 87.03 90.00 0.00 0.98 129.44 373.84 84.21 90.00 0.00 0.97 128.28 364.14 81.52 90.00 0.00 0.96 127.15 354.85 78.96 90.00 0.00 diagram for 0.95 126.04 345.95 T-x-y76.53 90.00 0.00 Benzene1)/Ethylbenzene(2) kPa 0.94 124.97 337.41 74.21 90.00 at 90 0.00 0.93 123.91 329.23 71.99 90.00 0.00 0.92140.00 122.88 321.37 69.88 90.00 0.00 0.91 121.88 313.83 67.86 90.00 0.00 0.90 120.89 306.58 65.94 90.00 0.00 130.00 0.89 119.93 299.62 64.09 90.00 0.00 0.88 118.99 292.92 62.33 90.00 0.00 120.00 0.87 118.07 286.48 60.64 90.00 0.00 0.86 117.16 280.27 59.03 90.00 0.00 110.00 0.85 116.28 274.30 57.48 90.00 0.00 0.84 115.42 268.54 55.99 90.00 0.00 100.00 0.83 114.57 262.99 54.57 90.00 0.00 0.82 113.74 257.64 53.20 90.00 0.00 90.00 0.81 112.93 252.48 51.89 90.00 0.00 0.80 112.13 247.50 50.63 90.00 0.00 80.00 0.79 111.35 242.68 49.41 90.00 0.00 0.78 110.58 238.03 48.25 90.00 0.00 70.00 0.77 109.83 233.54 47.12 90.00 0.00 0.00 229.190.20 46.04 0.4090.00 0.60 0.76 109.10 0.00 0.75 108.37 224.99 45.00 90.00 x1 or y1 0.00 0.74 107.66 220.92 44.00 90.00 0.00 0.73 106.97 216.98 43.03 90.00 0.00 0.72 106.28 213.17 42.10 90.00 0.00 0.71 105.61 209.47 41.20 90.00 0.00

Temperature (deg C)

x1

212.3

SUM((P-Pspec)2) 0.00

Updated 4/5/2017

y2 0.00 0.04 0.08 0.12 0.16 0.19 0.22 0.26 0.29 0.31 0.34 0.37 0.39 0.41 0.44 0.46 0.48 0.50 0.52 0.53 0.55 0.57 0.58 0.60 0.80 0.61 0.62 0.64 0.65 0.66 0.67

1.00 0.96 0.92 0.88 0.84 0.81 0.78 0.74 0.71 0.69 0.66 0.63 0.61 0.59 0.56 0.54 0.52 0.50 0.48 0.47 0.45 0.43 0.42 0.40 1.00 0.39 0.38 0.36 0.35 0.34 0.33

p. 5 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Spreadsheet to compute a P-x-y diagram for SVA problem 10.2(b) Species 1 is 1-chlorobutane, species 2 is chlorobenzene Antoine Coefficients (for T in deg C and P in kPa) T (deg C) A1 B1 C1 A2 B2 C2 P1sat (kPa) P2sat (kPa) 90 13.96 2826.26 224.1 13.9926 3295.12 217.55 142.8846 26.53606 P (kPa)

x2 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20 0.21 0.22

1.00 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.92 0.91 0.90 0.89 0.88 0.87 0.86 0.85 0.84 0.83 0.82 0.81 0.80 0.79 0.78

Updated 4/5/2017

26.54 27.70 28.86 30.03 31.19 32.35 33.52 34.68 35.84 37.01 38.17 39.33 40.50 41.66 42.82 43.99 45.15 46.32 47.48 48.64 49.81 50.97 52.13

y1

y2 0.00 0.05 0.10 0.14 0.18 0.22 0.26 0.29 0.32 0.35 0.37 0.40 0.42 0.45 0.47 0.49 0.51 0.52 0.54 0.56 0.57 0.59 0.60

1.00 0.95 0.90 0.86 0.82 0.78 0.74 0.71 0.68 0.65 0.63 0.60 0.58 0.55 0.53 0.51 0.49 0.48 0.46 0.44 0.43 0.41 0.40

P-x-y Diagram for 1-Chlorobutane(1)/Chlorobenzene(2) at 90 deg C 140.00 120.00

P (kPa)

x1

100.00 80.00 60.00 40.00 20.00 0.00

0.20

0.40 0.60 x1 or y1

0.80

1.00

p. 6 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Spreadsheet to compute a T-x-y diagram for SVNA problem 10.2(b) Species 1 is 1-chlorobutane, species 2 is chlorobenzene Antoine Coefficients (for T in deg C and P in kPa) P (kPa) A1 B1 C1 A2 B2 C2 90 13.96 2826.26 224.1 13.9926 3295.12 217.55 x2 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.30 0.31 0.32 0.33

Updated 4/5/2017

2

sat P (kPa) (P-Pspec) y 1 P 2 sat y2 T (deg C) P 1 1.00 129.57 391.01 90.00 90.00 0.00 0.00 0.99 128.36 380.45 87.07 90.00 0.00 0.04 0.98 127.18 370.35 84.28 90.00 for 0.00 0.08 T-x-y diagram 0.97 126.031-Chlorobutane(1)/Chlorobenzene(2) 360.71 81.63 90.00 0.00 at 90 0.12 kPa 0.96 124.91 351.48 79.11 90.00 0.00 0.16 0.95 123.82 342.64 76.70 90.00 0.00 0.19 140.00 0.94 122.75 334.17 74.41 90.00 0.00 0.22 0.93 121.71 326.06 72.23 90.00 0.00 0.25 130.00 0.92 120.69 318.28 70.15 90.00 0.00 0.28 0.91 119.69 310.82 68.16 90.00 0.00 0.31 120.00 0.90 118.72 303.65 66.26 90.00 0.00 0.34 0.89 117.77 296.76 64.44 90.00 0.00 0.36 110.00 0.88 116.84 290.15 62.71 90.00 0.00 0.39 0.87 115.93 283.78 61.04 90.00 0.00 0.41 100.00 0.86 115.04 277.66 59.45 90.00 0.00 0.43 0.85 114.17 271.76 57.92 90.00 0.00 0.45 0.84 90.00 113.32 266.09 56.46 90.00 0.00 0.47 0.83 112.48 260.61 55.06 90.00 0.00 0.49 0.82 80.00 111.66 255.34 53.71 90.00 0.00 0.51 0.81 110.86 250.25 52.41 90.00 0.00 0.53 0.80 70.00 110.08 245.34 51.17 90.00 0.00 0.55 0.00 0.20 49.97 0.40 90.00 0.600.00 0.80 0.79 109.31 240.59 0.56 0.78 108.55 236.01 48.82 90.00 0.00 0.58 x1 or y1 0.77 107.81 231.59 47.71 90.00 0.00 0.59 0.76 107.09 227.31 46.64 90.00 0.00 0.61 0.75 106.38 223.17 45.61 90.00 0.00 0.62 0.74 105.68 219.16 44.62 90.00 0.00 0.63 0.73 104.99 215.28 43.66 90.00 0.00 0.65 0.72 104.32 211.52 42.74 90.00 0.00 0.66 0.71 103.66 207.89 41.85 90.00 0.00 0.67 0.70 103.01 204.36 40.99 90.00 0.00 0.68 0.69 102.37 200.94 40.16 90.00 0.00 0.69 0.68 101.74 197.62 39.35 90.00 0.00 0.70 0.67 101.13 194.40 38.58 90.00 0.00 0.71

Temperature (deg C)

x1

SUM((P-Pspec)2) 0.00

1.00 0.96 0.92 0.88 0.84 0.81 0.78 0.75 0.72 0.69 0.66 0.64 0.61 0.59 0.57 0.55 0.53 0.51 0.49 0.47 0.45 1.00 0.44 0.42 0.41 0.39 0.38 0.37 0.35 0.34 0.33 0.32 0.31 0.30 0.29

p. 7 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 8 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 10.3(a), V vs. z1

1

Vapor phase fraction, V

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.5

0.6

0.7

0.8

0.9

Overall mole fraction z1

Updated 4/5/2017

p. 9 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

A1 13.8183

B1 2477.07

C1 233.21

P1sat (kPa)P2sat (kPa) 185.61 23.13

Chapter 13

A2 13.8587

B2 2911.32

C2 T (deg C) 216.64 55

Pbub (kPa) Pdew (kPa) 104.37 41.13

z1 0.5 P vs. V for Problem 10.3(b)

110.0000

x1 0.1108 0.1303 0.1497 0.1692 0.1886 0.2081 0.2276 0.2470 0.2665 0.2859 0.3054 0.3249 0.3443 0.3638 0.3832 0.4027 0.4222 0.4416 0.4611 0.4805 0.5000

y1 0.5000 0.5458 0.5856 0.6204 0.6511 0.6783 0.7027 0.7247 0.7446 0.7627 0.7792 0.7943 0.8082 0.8211 0.8330 0.8440 0.8543 0.8639 0.8729 0.8813 0.8892

V 1.0000 0.8897 0.8036 0.7332 0.6733 0.6207 0.5733 0.5296 0.4884 0.4490 0.4107 0.3731 0.3356 0.2979 0.2596 0.2205 0.1801 0.1383 0.0945 0.0486 0.0000

100.0000

Pressure (kPa)

P (kPa) 41.1340 44.2957 47.4575 50.6192 53.7809 56.9426 60.1043 63.2660 66.4278 69.5895 72.7512 75.9129 79.0746 82.2363 85.3980 88.5598 91.7215 94.8832 98.0449 101.2066 104.3683

90.0000 80.0000 70.0000 60.0000 50.0000 40.0000 0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

Vapor phase fraction (V)

x1 and y1 vs. V for Problem 10.3(b) 1.0000 0.9000 0.8000

x1 or y1

0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.2000

0.4000

0.6000

0.8000

Vapor phase fraction (V)

Updated 4/5/2017

p. 10 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

1.0000


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 11 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 10.4(a), V vs. z 1 1

Vapor phase fraction, V

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.5000

0.6000

0.7000

0.8000

0.9000

Overall mole fraction z 1

Updated 4/5/2017

p. 12 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 13 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

A1 13.7667

B1 2451.88

C1 232.014

P1sat (kPa)P2sat (kPa) 247.55 33.83

A2 13.8622

Chapter 13

B2 2910.26

C2 T (deg C) 216.43 65

Pbub (kPa) Pdew (kPa) 140.69 59.52

z1 0.5 P vs. V for Problem 10.4

160.0000

x1 0.1202 0.1392 0.1582 0.1772 0.1962 0.2152 0.2342 0.2531 0.2721 0.2911 0.3101 0.3291 0.3481 0.3671 0.3861 0.4051 0.4240 0.4430 0.4620 0.4810 0.5000

y1 0.5000 0.5420 0.5790 0.6118 0.6411 0.6674 0.6911 0.7127 0.7323 0.7503 0.7669 0.7821 0.7962 0.8093 0.8215 0.8328 0.8434 0.8534 0.8627 0.8715 0.8798

V 1.0000 0.8957 0.8123 0.7428 0.6829 0.6299 0.5818 0.5372 0.4951 0.4549 0.4157 0.3772 0.3390 0.3006 0.2617 0.2219 0.1811 0.1388 0.0948 0.0486 0.0000

140.0000

Pressure (kPa)

P (kPa) 59.5227 63.5810 67.6393 71.6975 75.7558 79.8140 83.8723 87.9305 91.9888 96.0470 100.1053 104.1635 108.2218 112.2800 116.3383 120.3965 124.4548 128.5130 132.5713 136.6295 140.6878

120.0000 100.0000 80.0000 60.0000 40.0000 0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

Vapor phase fraction (V)

x1 and y1 vs. V for Problem 10.4 1.0000 0.9000 0.8000

x1 or y1

0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

Vapor phase fraction (V)

Updated 4/5/2017

p. 14 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 15 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 16 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 17 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 18 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 19 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 20 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

z3 K3 z1 K1 z2 K 2   1 1  V  K1  1 1  V  K 2  1 1  V  K3  1 1  V  K 1 1  V  K 1 1  V  K 1  z K 1  V  K 1 1  V  K 1  1  V  K 1 z K 1  V  K 1  1 V  K  1 1 V  K  1 z K  0 1

2

3

1

1

2

3

1

2

2

3

1

2

3

3

V 3  4.288587V 2 10.43753V  5.148752  0

Updated 4/5/2017

p. 21 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

yi 

Chapter 13

zi K i L  VK i

0.3333  2.642  0.3701 0.1600  0.8400  2.642 0.3333 1.106 y2   0.3385 0.1600  0.8400 1.106 0.3333  0.5261 y3   0.2913 0.1600  0.8400  0.5261 y1 

z3 K3 z1 K1 z2 K 2   1 1  V  K1  1 1  V  K 2  1 1  V  K3  1 0.3333* 2.3789 0.3333* 0.9956 0.3333* 0.4735   1 1  1.3789V 1  0.0044V 1  0.5265V

Updated 4/5/2017

p. 22 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

yi 

Chapter 13

zi K i 1  V  K i  1

0.3333* 2.3789  0.4395 1  1.3789 * 0.5833 0.3333* 0.9956 y2   0.3327 1  0.0044 * 0.5833 0.3333* 0.4735 y3   0.2278 1  0.5265 * 0.5833 y1 

z3 K3 z1 K1 z2 K 2   1 1  V  K1  1 1  V  K 2  1 1  V  K3  1 0.3333* 2.1626 0.3333* 0.9051 0.3333* 0.4305   1 1  1.1626V 1  0.0949V 1  0.5695V

Updated 4/5/2017

p. 23 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

yi 

Chapter 13

zi K i 1  V  K i  1

0.3333* 2.1626  0.5064 1  1.1626 * 0.3641 0.3333* 0.9051 y2   0.3125 1  0.0949 * 0.3641 0.3333* 0.4305 y3   0.1810 1  0.5695 * 0.3641 y1 

z3 K3 z1 K1 z2 K 2   1 1  V  K1  1 1  V  K 2  1 1  V  K3  1 0.3333*1.9544 0.3333* 0.8307 0.3333* 0.3948   1 1  0.9544V 1  0.1693V 1  0.6052V

Updated 4/5/2017

p. 24 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

yi 

Chapter 13

zi K i 1  V  K i  1

0.3333*1.9544  0.572 1  0.9544 * 0.145 0.3333* 0.8307 y2   0.284 1  0.1693* 0.145 0.3333* 0.3948 y3   0.144 1  0.6052 * 0.145 y1 

z3 K3 z1 K1 z2 K 2   1 1  V  K1  1 1  V  K 2  1 1  V  K3  1

yi 

zi Ki 1  V  Ki  1

Updated 4/5/2017

p. 25 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

2756.22   P1sat (kPa)  exp 14.3145    143.8 kPa 66.85  228.06   3413.10   P2sat (kPa)  exp 14.8950    62.9 kPa 66.85  250.523  

N

zi Ki

i 1

i

 1  V  K  1  1 z1K1 z2 K 2  1 1  V  K1  1 1  V  K 2  1

z1 K1 1  V  K 2  1   z2 K 2 1  V  K1  1   1  V  K 2  1  1  V  K1  1  z1 K1  z2 K 2   z1 K1 K 2  z2 K 2 K1  z1K1  z2 K 2 V  1   K1  K 2  2 V   K 2  1 K1  1V 2   K 2  1 K1  1V 2   K 2 K1  z1 K1  z2 K 2  K1  K 2  2   z1K1  z2 K 2  1  0

0.11325V 2  0.1875V  0.07425  0 Updated 4/5/2017

p. 26 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

yi 

Chapter 13

zi Ki 1  V  Ki  1

2726.81   P1sat (kPa)  exp 13.7819    180.45 kPa 100  217.572   3259.93   P2sat (kPa)  exp 13.9726    34.27 kPa 100  212.3  

N

zi Ki

i 1

i

 1  V  K  1  1 z1K1 z2 K 2  1 1  V  K1  1 1  V  K 2  1

1.1874 0.2255  1 1  1.3747V 1  0.5491V 1.1874  0.6519V  0.2255  0.3100V  1  0.8256V  0.7873V 2 0.7548V 2  1.1675V  0.4129  0 Updated 4/5/2017

p. 27 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

zi Ki 1  V  Ki  1 1.1874 y1   0.6775 1  1.3747V

Chapter 13

yi 

y2 

0.2255  0.3224 1  0.5491V

3795.17   P1sat (kPa)  exp 16.8958    141.54 kPa 86.85  230.918   3483.67   P2sat (kPa)  exp 16.1154    67.48 kPa 86.85  205.807  

N

zi Ki

i 1

i

 1  V  K  1  1 z1K1 z2 K 2  1 1  V  K1  1 1  V  K 2  1

0.4365 0.6243  1 1  0.7461V 1  0.1676V 0.4365  0.0731V  0.6243  0.4658V  1  0.5786V  0.1250V 2 0.1250V 2  0.1859V  0.061  0 Updated 4/5/2017

p. 28 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

zi Ki 1  V  Ki  1 0.4365 y1   0.3202 1  0.7461V yi 

y2 

0.6243  0.6798 1  0.1676V

M

Updated 4/5/2017

p. 29 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

=

H H

Updated 4/5/2017

p. 30 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 31 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

P  x1 1 P1sat  x2 2 P2sat

P  x1 exp  0.95 x22  P1sat  x2 exp  0.95 x12  P2sat

P  0.05exp 0.95  0.95  79.80  0.95exp 0.95  0.05  40.50 2

2

P  47.97 kPa x1 1 P1sat x1 exp  0.95 x2  P1  P P 2

y1  y1 

0.05exp 0.95  0.95  79.80 2

47.97 y2  1  0.196  0.804

P

P

sat

 0.196

1 y1

P

sat 1 1

y2

 2 P2sat 1

y1

Updated 4/5/2017

exp  0.95 x22  P1sat

 y2

exp  0.95 x12  P2sat

p. 32 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

x1 

Chapter 13

y1 P  1 P1sat y1

exp  0.95 x22  P1sat x1  y1 y2 2 sat  exp  0.95 x2  P1 exp  0.95 x12  P2sat y1 x1 

y1

y1

sat 1

P

P1sat  y2

0.05

sat 1

y2

exp  0.95 x12  P2sat

0.05 0.05

sat 2

P

0.05

P

exp 0.95 1  x1  P

x1, guess 

x1 

2

2

y1

exp 0.95 1  x1  P1sat

79.80

79.80  0.95

 0.026 40.50

exp 0.95 1  x1  79.80 2

exp 0.95 1  x1  79.80 2

 0.95

exp  0.95 x12  40.50

1 0.05

exp  0.95  0.98962  79.80

 0.95

exp  0.95  0.0104  40.50

P  42.19 kPa

x1 P  x1 1 P1sat x2 P  x2 2 P2sat Updated 4/5/2017

p. 33 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

P   1 P1sat P   2 P2sat

 1 P1sat   2 P2sat

exp  0.95 x22   79.80  exp  0.95 x12   40.50

exp 0.95  x22  x12   40.50 / 79.80

 

exp 0.95 1  x1   x12

2

  0.50752

0.95 1  2 x1  x12   x12  ln  0.50752  1  2 x1  ln  0.50752  / 0.95 x1 

1  ln  0.50752  / 0.95  0.8570 2

P  exp  0.95  0.14302   79.80  81.37 kPa

x1 P  x1 1 P1sat x2 P  x2 2 P2sat

P   1 P1sat P   2 P2sat

Updated 4/5/2017

p. 34 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

 1 P1sat   2 P2sat

exp  Ax22  P1sat  exp  Ax12  P2sat

exp A  x22  x12   P2sat / P1sat

 

exp A 1  x1   x12

2

  P / P sat 2

sat 1

A 1  2 x1  x12   x12  ln  P2sat / P1sat  A 1  2 x1   ln  P2sat / P1sat  A

ln  P2sat / P1sat 

1  2 x1 

ln  31.66 / 75.20 

1  2  0.294  

 2.0998

P  x1 1 P1sat  x2 2 P2sat

P  0.6 exp 2.0998  0.4  75.20  0.4 exp 2.0998  0.6  31.66 2

2

P  38.18 kPa

y1  y1 

x1 1 P1sat P

0.6 exp 2.0998  0.4  75.20

38.18 y2  1  0.845  0.155

Updated 4/5/2017

2

 0.845

p. 35 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

P  x1 1 P1sat  x2 2 P2sat

Chapter 13

x1 1 P1sat P 0.65 *1.2467 *1.24 y1   0.601 1.67 y1 

x1 P  x1 1 P1sat x2 P  x2 2 P2sat

P   1 P1sat P   2 P2sat Updated 4/5/2017

p. 36 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

 1 P1sat   2 P2sat

exp 1.8 x22  P1sat  exp 1.8 x12  P2sat

exp 1.8  x22  x12   P2sat / P1sat

 

exp 1.8 1  x1   x12 2

  P / P sat 2

sat 1

1.8 1  2 x1  x12   x12  ln  P2sat / P1sat  1.8 1  2 x1   ln  P2sat / P1sat  sat sat 1  ln  P2 / P1   1  ln  0.89 /1.24      1  x1  1    0.592  2 2 1.8 1.8   

yi P  xi i Pi sat N

for all species  i  1, 2,..., N 

N

 yi P   xi i Pi sat i 1

i 1

N

P   xi i Pi sat i 1

 

Updated 4/5/2017

p. 37 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

 Acetone (1) A B 14.3145

C 2756.22

Methanol(2) A B C 228.06 16.5785 3638.27

T (deg C) P1sat (kPa) P2sat (kPa) 59.5 113.4 82.4

Updated 4/5/2017

x1 0.175

x2 0.825

239.5 1 1.55

2 P (kPa) 1.02 100

y1 0.307

y2 0.693

p. 38 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 39 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

ni 

i

1 ki M s

i

1  1632 bar 0.034 mol kg bar -1 0.01802 kg mol-1

ki xi

Updated 4/5/2017

M s  n  ni  n 1   ki ni ki M s  n  ni  ki M s 1  xi  n

i

i

Chapter 13

-1

p. 40 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Using the equations

Updated 4/5/2017

p. 41 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 42 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 43 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Component 1 Benzene 2-butanol Acentonitrile

Updated 4/5/2017

Chapter 13

𝑃𝑠𝑎𝑡1 (𝑇) = 39.591⁡𝑘𝑃𝐴⁡⁡⁡⁡⁡⁡⁡𝑇 = 52.321⁡𝐶 Component 2 T (deg C) P (kPA) Cyclohexane 52.3 39.6 water 87.7 64.2 Ethanol 65.8 60.6

p. 44 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.1(a) Data Fit 0.7000

y = -0.2075x + 0.6828 R² = 0.828 0.6500

GE/(x1x2RT)

0.6000

0.5500

0.5000

0.4500

0.4000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol Mole Fraction

ln  1  x22  A12  2  A21  A12  x1 

ln  2  x12  A21  2  A12  A21  x2 

GE  x1 x2  A21 x1  A12 x2  RT

Updated 4/5/2017

p. 45 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.1(a), Fit Results 0.7000

ln(1),ln(2), and GE/(RT)

0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol Mole Fraction

P  x1 1 P1sat  x2 2 P2sat y1 

x1 1 P1sat x1 1 P1sat  x2 2 P2sat P-x-y diagram with Margules Fit for Problem 12.1a

90.000

80.000

70.000

P(kPa)

60.000

50.000

40.000

30.000

20.000

10.000

0.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

Updated 4/5/2017

p. 46 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.1(b) Data Fit 2.2000

y = 0.6414x + 1.4185 R² = 0.8148 2.0000

(x1x2RT)/GE

1.8000

1.6000

1.4000

1.2000

1.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

Methanol Mole Fraction

  A' x  ln  1  A '21 1  12 1   A '21 x2 

2

 A' x  ln  2  A '12 1  21 2  A '12 x1  

Updated 4/5/2017

2

p. 47 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.1(b), Fit Results 0.8000

ln(1),ln(2), and GE/(RT)

0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol Mole Fraction

Updated 4/5/2017

p. 48 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

P-x-y diagram with Van Laar Fit for Problem 12.1(b) 90.000 80.000 70.000

P(kPa)

60.000 50.000 40.000 30.000 20.000 10.000 0.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

GE   x1 ln  x1  x2 12   x2 ln  x2  x1 21  RT

Updated 4/5/2017

p. 49 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Data Fit for Problem 12.1(c) 0.1600 0.1400 0.1200

GE/(RT)

0.1000 0.0800 0.0600 0.0400 0.0200 0.0000 0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

x1

 12  21  ln  1   ln  x1  x2 12   x2     x1  x2 12 x2  x1 21   12  21  ln  2   ln  x2  x1 21   x1     x1  x2 12 x2  x1 21 

Updated 4/5/2017

p. 50 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.1(c), Fit Results 0.7000

ln(1),ln(2), and GE/(RT)

0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

Methanol Mole Fraction

P-x-y diagram with Wilson Fit for Problem 12.1(c) 90.000 80.000 70.000

P(kPa)

60.000 50.000 40.000 30.000 20.000 10.000 0.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

Updated 4/5/2017

p. 51 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 52 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.1(d), Fit Results 0.7000

ln(1),ln(2), and GE/(RT)

0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol Mole Fraction

P-x-y diagram with Margules Fit for Problem 12.1d 90.000 80.000 70.000

P(kPa)

60.000 50.000 40.000 30.000 20.000 10.000 0.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

Updated 4/5/2017

p. 53 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.1 (d) - Pressure Residuals 0.400

0.300

d (P) kPa

0.200

0.100

0.000

-0.100

-0.200

-0.300

-0.400 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

0.9000

1.0000

x1

Problem 12.1 (d) - Vapor Composition Residuals 0.0080

0.0060

d (y1)

0.0040

0.0020

0.0000

-0.0020

-0.0040

-0.0060 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

x1 Updated 4/5/2017

p. 54 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

 

Updated 4/5/2017

p. 55 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.1(e), Fit Results 0.9000

ln(1),ln(2), and GE/(RT)

0.8000 0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol Mole Fraction

P-x-y diagram with Van Laar Fit for Problem 12.1(e) 90.000 80.000 70.000

P(kPa)

60.000 50.000 40.000 30.000 20.000 10.000 0.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

Updated 4/5/2017

p. 56 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.1 (e) - Pressure Residuals 0.400

0.300

d (P) kPa

0.200

0.100

0.000

-0.100

-0.200

-0.300

-0.400 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1

Updated 4/5/2017

p. 57 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.1 (e) - Vapor Composition Residuals 0.0100

0.0080

0.0060

d (y1)

0.0040

0.0020

0.0000

-0.0020

-0.0040

-0.0060 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1

Updated 4/5/2017

p. 58 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.1(f), Fit Results 0.7000

ln(1),ln(2), and GE/(RT)

0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol Mole Fraction

P-x-y diagram with Wilson Fit for Problem 12.1(f) 90.000 80.000 70.000

P(kPa)

60.000 50.000 40.000 30.000 20.000 10.000 0.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

Updated 4/5/2017

p. 59 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.1 (f) - Pressure Residuals 0.600

0.400

d (P) kPa

0.200

0.000

-0.200

-0.400 0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

1.2000

-0.600

x1

Problem 12.1 (f) - Vapor Composition Residuals 0.0120 0.0100 0.0080

d (y1)

0.0060 0.0040 0.0020 0.0000 -0.0020 -0.0040 -0.0060 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 Updated 4/5/2017

p. 60 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 61 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.3(a) Data Fit 1.2000

y = -0.0182x + 0.7077 R2 = 0.0034 1.0000

GE/(x1x2RT)

0.8000

0.6000

0.4000

0.2000

0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

ln  1  x22  A12  2  A21  A12  x1 

ln  2  x12  A21  2  A12  A21  x2 

GE  x1 x2  A21 x1  A12 x2  RT

Updated 4/5/2017

p. 62 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.3(a), Fit Results 0.8000

E

ln(1),ln(2), and G /(RT)

0.7000

0.6000

0.5000

0.4000

0.3000

0.2000

0.1000

0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

P  x1 1 P1sat  x2 2 P2sat y1 

x1 1 P1sat x1 1 P1sat  x2 2 P2sat

Updated 4/5/2017

p. 63 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

P-x-y diagram with Margules Fit for Problem 12.3a 105.000

100.000

95.000

P(kPa)

90.000

85.000

80.000

75.000

70.000

65.000

60.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

Updated 4/5/2017

p. 64 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.3(b) Data Fit 1.8000 1.6000 1.4000

(x1x2RT)/GE

1.2000

y = 0.015x + 1.4424 R2 = 0.0008

1.0000 0.8000 0.6000 0.4000 0.2000 0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

  A' x  ln  1  A '21 1  12 1   A '21 x2 

2

 A' x  ln  2  A '12 1  21 2  A '12 x1  

Updated 4/5/2017

2

p. 65 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.3(b), Fit Results 0.8000

ln(1),ln(2), and GE/(RT)

0.7000

0.6000

0.5000

0.4000

0.3000

0.2000

0.1000

0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

Updated 4/5/2017

p. 66 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

P-x-y diagram with Van Laar Fit for Problem 12.3(b) 105.000

100.000

95.000

P(kPa)

90.000

85.000

80.000

75.000

70.000

65.000

60.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

GE   x1 ln  x1  x2 12   x2 ln  x2  x1 21  RT

Updated 4/5/2017

p. 67 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Data Fit for Problem 12.3(c) 0.1800 0.1600 0.1400

GE/(RT)

0.1200 0.1000 0.0800 0.0600 0.0400 0.0200 0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1

 12  21  ln  1   ln  x1  x2 12   x2     x1  x2 12 x2  x1 21   12  21  ln  2   ln  x2  x1 21   x1     x1  x2 12 x2  x1 21 

Updated 4/5/2017

p. 68 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.3(c), Fit Results 0.7000

ln(1),ln(2), and GE/(RT)

0.6000

0.5000

0.4000

0.3000

0.2000

0.1000

0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

P-x-y diagram with Wilson Fit for Problem 12.3(c) 105.000

100.000

95.000

P(kPa)

90.000

85.000

80.000

75.000

70.000

65.000

60.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

Updated 4/5/2017

p. 69 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.3(d), Fit Results 0.8000

E

ln(1),ln(2), and G /(RT)

0.7000

0.6000

0.5000

0.4000

0.3000

0.2000

0.1000

0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

Updated 4/5/2017

p. 70 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

P-x-y diagram with Margules Fit for Problem 12.3d 105.000

100.000

95.000

P(kPa)

90.000

85.000

80.000

75.000

70.000

65.000

60.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

 

Updated 4/5/2017

p. 71 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.3(e), Fit Results 0.8000

ln(1),ln(2), and GE/(RT)

0.7000

0.6000

0.5000

0.4000

0.3000

0.2000

0.1000

0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

P-x-y diagram with Van Laar Fit for Problem 12.3(e) 105.000

100.000

95.000

P(kPa)

90.000

85.000

80.000

75.000

70.000

65.000

60.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

Updated 4/5/2017

p. 72 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Problem 12.3(f), Fit Results 0.7000

ln(1),ln(2), and GE/(RT)

0.6000

0.5000

0.4000

0.3000

0.2000

0.1000

0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

Updated 4/5/2017

p. 73 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

P-x-y diagram with Wilson Fit for Problem 12.3(f) 105.000

100.000

95.000

P(kPa)

90.000

85.000

80.000

75.000

70.000

65.000

60.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

Updated 4/5/2017

p. 74 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

γ γ

γ −

γ

( ‘)

/

Updated 4/5/2017

p. 75 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

=

Chapter 13

γ

=

γ

> =∞

=

/

> γ

=

=

/ =

Updated 4/5/2017

γ

=

γ < /

= =

γ

/

p. 76 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

<

Chapter 13

γ

γ

=

=

GE  x1 x2  A21 x1  A12 x2  Cx1 x2  RT ln  1  x22  A12  2  A21  A12  C  x1  3Cx12 

ln  2  x12  A21  2  A12  A21  C  x2  3Cx22 

GE  A21 x1  A12 x2  Cx1 x2 x1 x2 RT GE  A21 x1  A12 1  x1   Cx1 1  x1  x1 x2 RT GE  Cx12   A21  A12  C  x1  A12 x1 x2 RT

Updated 4/5/2017

p. 77 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

-0.35

GE/(x1x2RT)

-0.4

y = 0.0760x2 - 0.2393x - 0.3771 R2 = 0.9216 -0.45

-0.5

-0.55 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x1

Updated 4/5/2017

p. 78 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

0

GE/RT or ln(1) or ln(2)

-0.1

-0.2

-0.3

-0.4

-0.5

-0.6 0

0.2

0.4

0.6

0.8

1

x1

Updated 4/5/2017

p. 79 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

90 85 80

P (kPa)

75 70 65 60 55 50 45 40 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x1 or y1

Updated 4/5/2017

p. 80 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 81 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

x1 0.0523 0.1299 0.2233 0.2764 0.3482 0.4187 0.5001 0.5637 0.6469 0.7832 0.8576 0.9388 0.9813

ϒ1 1.202 1.307 1.295 1.228 1.234 1.18 1.129 1.12 1.076 1.032 1.016 1.001 1.003

Updated 4/5/2017

ϒ2 1.002 1.004 1.006 1.024 1.022 1.049 1.092 1.102 1.17 1.298 1.393 1.6 1.404

Chapter 13

GE/RT 0.0115 0.0383 0.0624 0.0739 0.0874 0.0971 0.1047 0.1063 0.1028 0.0812 0.0608 0.0297 0.0093

p. 82 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

1.8000 1.6000

ϒ's and GE/RT

1.4000 1.2000

GE/RT

1.0000

GE/RT(x1,x2)

0.8000

ϒ1(x1,x2)

0.6000

ϒ2(x1,x2)

0.4000

ϒ1

0.2000

ϒ2

0.0000 0

0.2

0.4

0.6

0.8

1

x1

del GE/RT del lnϒ1/ϒ2 0.0033 0.098001 Updated 4/5/2017

p. 83 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

-0.0023 -0.0032 -0.0030 -0.0029 -0.0022 -0.0022 -0.0015 -0.0008 0.0003 0.0001 0.0001 0.0004

-0.00017 -0.0206 0.026484 -0.01917 0.006071 0.028597 -0.00945 0.00924 -0.00057 -0.01079 0.027934 -0.16868

0.1

del lnϒ1/ϒ2

0.05 0 -0.05 -0.1 -0.15 -0.2 0

0.2

0.4

0.6

0.8

1

x1

Updated 4/5/2017

p. 84 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

⁡ ⁡ Updated 4/5/2017

p. 85 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 86 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 87 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

38 37 36

P and Pcalc

35 34

P/kPa

33

y1

32

Pcalc

31

Y1calc

30 29 28 0

0.2

0.4

0.6

0.8

1

x1,y1,X1,Y1calc 0.06

P and Y1 Residuals

0.04 0.02 0 -0.02

Y1 residual

-0.04

P residual

-0.06 -0.08 -0.1 0

0.2

0.4

0.6

0.8

1

x1

Updated 4/5/2017

p. 88 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 89 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

 B1  2795.82   P1sat  exp  A1    exp 14.3916    115.68 kPa T  C1  60  230.00     B2  3799.89   P2sat  exp  A2    exp 16.2620    19.924 kPa T  C 60  226.35    2   Updated 4/5/2017

p. 90 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

12 

 291.27  V2  a  18.07 exp  12   exp    0.1572 V1  RT  74.05  1.987  333.15 

 21 

 1448.01  V1  a  74.05 exp  21   exp    0.4598 V2 1.987 333.15    RT  18.07  

P-x-y Plot for Acetone (1) / Water (2) 120.00

Total Pressure (kPa)

100.00

80.00

60.00

40.00

20.00

0.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone mole fraction (x1 or y1)

 B1  3644.3   P1sat  exp  A1    exp 16.5938    121.07 kPa T  C1  60  239.76     B2  3799.89   P2sat  exp  A2    exp 16.2620    19.924 kPa T  C2  60  226.35     Updated 4/5/2017

p. 91 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

12 

 107.38  V2  a  18.07 exp  12   exp    0.3772 V1  RT  40.73  1.987  333.15 

 21 

 469.55  V1  a  40.73 exp  21   exp    1.1089 V2 1.987 333.15    RT  18.07  

P-x-y Plot for Methanol (1) / Water (2) 140.00

Total Pressure (kPa)

120.00

100.00

80.00

60.00

40.00

20.00

0.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol mole fraction (x1 or y1)

 B1  3483.67   P1sat  exp  A1    exp 16.1154    20.275 kPa T  C 60  205.807    1   B2  3885.70   P2sat  exp  A2    exp 16.3872    20.007 kPa T  C2  60  230.17   

Updated 4/5/2017

p. 92 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

 12 

 775.48  V2  a  18.07 exp  12   exp    0.07453 V1  RT  75.14  1.987  333.15 

 21 

 1351.90  V1  a  75.14 exp  21   exp    0.5395 V2 1.987 333.15    RT  18.07  

P-x-y Plot for 1-Propanol (1) / Water (2) 32.00

Total Pressure (kPa)

30.00

28.00

26.00

24.00

22.00

20.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

1-propanol mole fraction (x1 or y1)

 B1  3885.70   P1sat  exp  A1    exp 16.3872    20.007 kPa T  C1  60  230.17     B2  3579.78   P2sat  exp  A2    exp 15.0967    23.986 kPa T  C2  60  240.337    Updated 4/5/2017

p. 93 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

 12 

 1696.98  V2  a  85.71 exp  12   exp    0.3654 V1  RT  18.07  1.987  333.15 

 21 

  V1 219.39  a  18.07 exp  21   exp    0.2937 V2 1.987 333.15    RT  85.71  

P-x-y Plot for Water (1) / 1,4-Dioxane (2) 34.00

32.00

Total Pressure (kPa)

30.00

28.00

26.00

24.00

22.00

20.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Water mole fraction (x1 or y1)

Updated 4/5/2017

p. 94 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

 B1  3795.17   P1sat  exp  A1    exp 16.8598    45.345 kPa T  C1  60  230.918     B2  3056.96   P2sat  exp  A2    exp 13.932    18.558 kPa T  C2  60  217.625     12 

 1556.45  V2  a  106.85 exp  12   exp    0.1734 V1 1.987 333.15    RT  58.68  

 21 

 210.52  V1  a  58.68 exp  21   exp    0.3996 V2 1.987 333.15    RT  106.85  

P-x-y Plot for Ethanol (1) / Toluene (2) 55.00

50.00

Total Pressure (kPa)

45.00

40.00

35.00

30.00

25.00

20.00

15.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Ethanol mole fraction (x1 or y1)

Updated 4/5/2017

p. 95 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

 B1  3638.27   P1sat  exp  A1    exp 16.5785    88.989 kPa T  C1  60  239.50     B2  2726.81   P2sat  exp  A2    exp 13.7819    52.377 kPa T  C2  60  215.582     12 

 1734.42  V2  a  89.41 exp  12   exp    0.1598 V1 1.987 333.15    RT  40.73  

 21 

 183.04  V1  a  40.73 exp  21   exp    0.3455 V2 1.987 333.15    RT  89.41  

P-x-y Plot for methanol(1)/benzene2) 110.00

Total Pressure (kPa)

100.00

90.00

80.00

70.00

60.00

50.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone mole fraction (x1 or y1)

Updated 4/5/2017

p. 96 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

 B1  3638.27   P1sat  exp  A1    exp 16.5785    83.989 kPa T  C1  60  239.50     B2  3413.10   P2sat  exp  A2    exp 14.895    49.578 kPa T  C2  60  250.523     12 

 504.31  V2  a  66.30 exp  12   exp    0.7599 V1 1.987 333.15    RT  40.73  

 21 

 196.75  V1  a  40.73 exp  21   exp    0.4564 V2  RT  66.30  1.987  333.15 

P-x-y Plot for methanol(1)/acetonitrile(2) 90.00 85.00

Total Pressure (kPa)

80.00 75.00

70.00 65.00

60.00 55.00

50.00 45.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol mole fraction (x 1 or y1)

Updated 4/5/2017

p. 97 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Antoine Parameters A1 B1 16.3872

3885.7

Wilson Parameters V1

V2

(cm 3/mol) 18.07

Chapter 13

C1

(cm 3/mol) 85.71

230.17

A2 B2 C2 15.0967 3579.78 240.337

a 12 a 21 (cal/mol) (cal/mol) 1696.98 -219.39

T (°C) P1sat (kPa) P2sat (kPa) x1 x2 101.3 106.1 101.3 0.0000 1.0000 99.3 98.8 95.2 0.0200 0.9800 97.6 93.1 90.5 0.0400 0.9600 96.3 88.7 86.7 0.0600 0.9400 95.2 85.2 83.7 0.0800 0.9200 94.3 82.3 81.3 0.1000 0.9000 93.5 80.0 79.3 0.1200 0.8800 92.8 78.1 77.7 0.1400 0.8600 92.3 76.5 76.3 0.1600 0.8400

Updated 4/5/2017

ln( 1) ln( 2) P/kPa y1  12  21 Wilson Wilson Wilson Wilson 0.4848 0.2831 1.4410 0.0000 101.31 0.0000 0.4788 0.2836 1.3920 0.0006 101.33 0.0784 0.4740 0.2840 1.3417 0.0025 101.33 0.1406 0.4700 0.2843 1.2907 0.0055 101.33 0.1909 0.4668 0.2845 1.2398 0.0098 101.32 0.2323 0.4641 0.2847 1.1893 0.0152 101.33 0.2668 0.4618 0.2849 1.1395 0.0218 101.33 0.2960 0.4599 0.2851 1.0907 0.0296 101.33 0.3211 0.4583 0.2852 1.0429 0.0384 101.34 0.3427

Error2 2.66E-04 3.65E-06 2.17E-06 1.74E-05 4.45E-05 2.10E-05 3.86E-08 1.34E-05 2.89E-05

p. 98 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

T-x-y Plot for water(1)/1,4-dioxane(2) 103.0 101.0

Temperature (deg. C)

99.0 97.0 95.0 93.0 91.0 89.0 87.0 85.0 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol mole fraction (x1 or y1)

 B1  3638.27   P1sat  exp  A1    exp 16.5785    88.989 kPa T  C 60  239.50    1   B2  2726.81   P2sat  exp  A2    exp 13.7819    52.377 kPa T  C2  60  215.582    

Updated 4/5/2017

p. 99 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

 12 

b12 730.09   1.1029 RT 1.987  333.15

 21 

b21 1175.41   1.7756 RT 1.987  333.15

Chapter 13

G12  exp   12   exp  0.4743*1.1029   0.5927 G21  exp   21   exp  0.4743*1.7756   0.4308

G12⁡ 2   G   G12 21  ln  1  x  21     x1  x2G21   x2  x1G12 2    2   G   G21 12  ln  2  x12  12     x2  x1G12   x1  x2G21 2    2 2

Updated 4/5/2017

p. 100 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

P-x-y Plot for methanol(1)/benzene2) 110.00

Total Pressure (kPa)

100.00

90.00

80.00

70.00

60.00

50.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone mole fraction (x1 or y1)

 B1  2662.78   P1sat  exp  A1    exp 14.2456    112.745 kPa T  C1  60  219.69     B2  3638.27   P2sat  exp  A2    exp 16.5785    84.748 kPa T  C2  60  239.50    

Updated 4/5/2017

p. 101 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

 12 

b12 381.46   0.5763 RT 1.987  333.15

 21 

b21 346.54   0.5235 RT 1.987  333.15

Chapter 13

G12  exp   12   exp  0.2965 * 0.5763  0.8429 G21  exp   21   exp  0.2965 * 0.5235   0.8562

Ɣ1 2   G   G12 21  ln  1  x  21     x1  x2G21   x2  x1G12 2    2   G   G21 12  ln  2  x12  12     x2  x1G12   x1  x2G21 2    2 2

Updated 4/5/2017

p. 102 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

P-x-y Plot for methanol(1)/benzene2) 130.00 125.00

Total Pressure (kPa)

120.00 115.00 110.00 105.00

100.00 95.00 90.00 85.00

80.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

methyl acetate mole fraction (x1 or y1)

Updated 4/5/2017

p. 103 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

x 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 Updated 4/5/2017

y(eq) 0 0.32 0.377 0.394 0.402 0.408 0.415 0.424 0.434 0.447 0.462 0.48 0.5 0.524 0.552 0.586 0.629 0.682 0.754 0.853 1

Chapter 13

T(eq) 373.149 363.606 361.745 361.253 361.066 360.946 360.843 360.757 360.697 360.676 360.709 360.807 360.985 361.262 361.66 362.215 362.974 364.012 365.442 367.449 370.349 p. 104 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

T(eq)

374 372

x

370

y(eq)

368 366 364 362 360 0

0.2

0.4

0.6

0.8

1

x1, y(eq)

Updated 4/5/2017

p. 105 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 106 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Updated 4/5/2017

p. 107 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 108 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 109 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Ʌ

Ʌ Ʌ

Ʌ Ʌ

Ʌ Ʌ

Ʌ

Ʌ Ʌ

Ʌ Ʌ

Updated 4/5/2017

p. 110 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 111 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 112 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

V 74.05 40.73 18.07

A 14.3145 16.5785 16.3872

B 2756.22 3638.27 3885.7

Chapter 13

C 228.06 239.5 230.17

Psati 135.575914 102.5303094 25.10135396

Updated 4/5/2017

p. 113 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Wislon Parameters (cal/mol) 1 2 0 -161.88 583.11 0 1448.01 469.55

Chapter 13

3 291.27 107.38 0

Ʌ

Ʌ

Ʌ Ʌ

Updated 4/5/2017

p. 114 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

V 74.05 40.73 18.07

A 14.3145 16.5785 16.3872

B 2756.22 3638.27 3885.7

Chapter 13

C 228.06 239.5 230.17

Psati 135.575914 102.5303094 25.10135396

NRTL Parametes a 1 2 3 0 0.3084 0.5343 0.3084 0 0.2994 0.5343 0.2994 0 NRTL Parametes b (cal/mol) 0 184.7 631.05 222.64 0 −253.88 1197.41 845.21 0 ⁡

Updated 4/5/2017

p. 115 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 116 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

V 74.05 40.73 18.07

A 14.3145 16.5785 16.3872

B 2756.22 3638.27 3885.7

Chapter 13

C 228.06 239.5 230.17

Psati 135.575914 102.5303094 25.10135396

Wislon Parameters (cal/mol) 1 2 0 -161.88 583.11 0 1448.01 469.55

3 291.27 107.38 0

Ʌ

Ʌ

Ʌ Ʌ

Updated 4/5/2017

p. 117 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

V 74.05 40.73 18.07

A 14.3145 16.5785 16.3872

B 2756.22 3638.27 3885.7

Chapter 13

C 228.06 239.5 230.17

Psati 135.575914 102.5303094 25.10135396 Updated 4/5/2017

p. 118 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

NRTL Parametes a 1 2 3 0 0.3084 0.5343 0.3084 0 0.2994 0.5343 0.2994 0 NRTL Parametes b (cal/mol) 0 184.7 631.05 222.64 0 −253.88 1197.41 845.21 0 ⁡

Updated 4/5/2017

p. 119 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

GE  x1 ln  1  x2 ln  2 RT

Chapter 13

E

G  x1 x22  a  bx1   x2 x12  a  bx2  RT GE  x1 x2  ax2  bx1 x2  ax1  bx1 x2  RT GE  x1 x2  a  x1  x2    ax1 x2  0.273x1 x2 RT

Updated 4/5/2017

p. 120 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

 d  nG E   ln  1      dn1  RT   T , P ,n2

 d  nG E   ln  2      dn2  RT   T , P ,n1

nG E an n  nax1 x2  1 2 RT n1  n2  n2  n1  n2   n1n2   d  an1n2   ln  1    a      2   n  n    dn1  n1  n2   T , P ,n2 1 2   ln  1 

an22

 n1  n2 

2

 ax22  0.273x22

 n n  n   n n   d  an1n2   ln  2    a 1 1 2 2 1 2       n1  n2   dn2  n1  n2   T , P ,n2   ln  2 

an12

 n1  n2 

2

 ax12  0.273x12

x1d  ln  1   x2 d  ln  2   0 x1d  x22  a  bx1    x2 d  x12  a  bx2    0

x1d 1  x1   a  bx1   1  x1  d x12  a  b 1  x1    0

Updated 4/5/2017

2

p. 121 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

x1

Chapter 13

d d 2 2 x1  a  b 1  x1    0 1  x1   a  bx1   1  x1  dx1 dx1

x1 2 1  x1  a  bx1   b 1  x1   1  x1  2 x1  a  b 1  x1    x12b  0 2

x1  2 x2  a  bx1   bx22   x2  2 x1  a  bx2   x12b   0 2 x1 x2 a  2 x12 x2b  x1 x22b  2 x1 x2 a  2 x1 x22b  x12 x2b  0  x12 x2b  x1 x22b  0

 x1 x2  x1  x2  b  0  x1 x2b  0

γ =

Updated 4/5/2017

=

p. 122 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 123 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 124 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 125 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

=

Updated 4/5/2017

=

p. 126 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 127 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Ʌ

Chapter 13

Ʌ

Ʌ

Ʌ Ʌ

Updated 4/5/2017

Ʌ

Ʌ Ʌ

p. 128 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Ʌ Ʌ

Ʌ

Ʌ

Ʌ

Ʌ Ʌ

Ʌ Ʌ

Ʌ

Ʌ Ʌ

Ʌ Ʌ

Ʌ Ʌ

Ʌ

Ʌ Ʌ

Ʌ

Updated 4/5/2017

Ʌ

Ʌ

Ʌ

p. 129 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

⁡ ⁡

⁡ ⁡

Updated 4/5/2017

p. 130 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

γ

Chapter 13

γ

⁡ ⁡

Updated 4/5/2017

p. 131 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

 1  exp  x22  A12  2  A21  A12  x1  

 2  exp  x12  A21  2  A12  A21  x2  

SVNA Problem 14.1, part (a) P1

sat

(kPa)

P2

sat

(kPa)

82.37

37.31

A21 1.42

 1  exp  x22  A12  2  A21  A12  x1   x1

x2 0.25 0.5 0.75

A12 0.59

 2  exp  x12  A21  2  A12  A21  x2  

1 2 P (kPa) y1 y2 0.75 1.759998 1.010998 64.53299 0.561616 0.438384 0.5 1.426181 1.158933 80.35715 0.730952 0.269048 0.25 1.121523 1.759998 85.70126 0.808446 0.191554

yi i P   i xi Pi sat P N N   sat  Bii  P  Pi   2  y j yk  2d ji  d jk   ˆ  j 1 k 1   i  sati  exp    i RT      B11  P  P1sat   Py22d12   1  exp    RT   sat 2  B22  P  P2   Py1 d12    2  exp    RT    

Updated 4/5/2017

p. 132 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

P

y1 

 1 x1 P1sat 1

Chapter 13

 2 x2 P2sat 2

 1 x1 P1sat 1 P 

 

Updated 4/5/2017

p. 133 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 134 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

P(kPA) 15.51 18.61 21.63 24.01 25.92 27.96 30.12 31.75 34.15

x1 0.0895 0.1981 0.3193 0.4232 0.5119 0.6096 0.7135 0.7934 0.9102

Updated 4/5/2017

x2 0.9105 0.8019 0.6807 0.5768 0.4881 0.3904 0.2865 0.2066 0.0898

Chapter 13

y1 0.2716 0.4565 0.5934 0.6815 0.744 0.805 0.8639 0.9048 0.959

y2 0.7284 0.5435 0.4066 0.3185 0.256 0.195 0.1361 0.0952 0.041

ϒ1 1.304 1.188 1.114 1.071 1.044 1.023 1.01 1.003 0.997

ϒ2 1.009 1.026 1.05 1.078 1.105 1.135 1.163 1.189 1.268

p. 135 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

0.080

GERT and GERTcalc

0.070 0.060 0.050 0.040 0.030 0.020 0.010 0.000 0

0.2

0.4

0.6

0.8

1

x1

Updated 4/5/2017

p. 136 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

0.080

GERT and GERTcalc

0.070 0.060 0.050 0.040 0.030 0.020 0.010 0.000 0

0.2

0.4

0.6

0.8

1

x1

P

a T  RT  V  b V V  b 

Tr0.5 R 2Tc2 0.61430.5  83.1452  308.32 a(T )  0.42748  0.42748  5837599 bar cm 6 mol-2 Pc 61.39 b  0.08664



RTc 83.145  308.3  0.08664  36.18 cm3 mol-1 Pc 61.39

bP 36.18 * P(bar)   0.002297 P RT 83.145 *189.4

Updated 4/5/2017

q

a T  bRT

5837599  10.247 36.18 * 83.145 *189.4

p. 137 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Zv  1   

Chapter 13

q   Zv    Zv  Zv   

Zl    Zl  Zl   

1    Zl q

Z   ln   Z  1  ln  Z     q ln    Z 

 

Updated 4/5/2017

p. 138 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T

P (bar) 189.4 1.603982

a(T)

b

Tc (K)

Pc (K)

308.3 beta

Chapter 13

61.39

Tr

Pr 0.6143

q T 0.5 R 2Tc2 a (T )  0.42748 r Pc

5837599 36.17683 0.003685 10.24679 Zv

Zl Phi(v) 1 0.003685

0.0261

Phi(l)

Phi(v)-Phi(l) q

a T 

0.966205 0.004404 0.965993 1.012135 0.964904 0.004628 0.965992 0.970095

-0.04614 -0.0041

bRT

0.964852 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485

-0.00045 -5.3E-05 -6.3E-06 -7.7E-07 -9.6E-08 -1.6E-08 -5.9E-09 -4.7E-09 -4.5E-09 -4.5E-09 -4.5E-09 -4.5E-09 -4.5E-09 -4.5E-09 -4.5E-09 -4.5E-09 -4.5E-09 -4.5E-09

Zv  1   

0.004703 0.004728 0.004737 0.00474 0.004741 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742

0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992

0.966445 0.966045 0.965999 0.965993 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992

 q   Zv    Zv  Zv   

b  0.08664

RTc Pc

bP RT Zl    Zl  Zl   

1    Zl q

Z   ln   Z  1  ln  Z     q ln    Z 

Updated 4/5/2017

p. 139 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

P

Chapter 13

a T  RT  V  b V V  b 

0.6282 Tr0.5 R 2Tc2 a(T )  0.42748  0.42748 Pc b  0.08664



 83.1452  562.22 48.98

 24060400 bar cm 6 mol-2

RTc 83.145  562.2  0.08664  82.68 cm3 mol-1 Pc 48.98

bP 82.68 * P(bar)   0.002815P RT 83.145 * 353.2

Zv  1   

0.5

q

a T  bRT

24060400  9.909 82.68 * 83.145 * 353.2

q   Zv    Zv  Zv   

Zl    Zl  Zl   

1    Zl q

Z   ln   Z  1  ln  Z     q ln    Z 

 

Updated 4/5/2017

p. 140 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T

Chapter 13

P (bar) Tc (K) Pc (K) Tr Pr 353.2 1.600138 562.2 48.98 0.6282 0.0327

a(T)

b

beta

q

24059470 82.68498 0.004505 9.908378 Zv 1 0.960265 0.958452 0.958365 0.958361 0.958361 0.958361 0.958361 0.958361 0.958361 0.958361 0.958361

Zl Phi(v) Phi(l) Phi(v)-Phi(l) 0.004505 0.005415 0.959958 1.009294 -0.049336 0.005708 0.959957 0.964648 -0.004691 0.00581 0.959957 0.960514 -0.000558 0.005846 0.959957 0.960027 -7.04E-05 0.005859 0.959957 0.959966 -8.79E-06 0.005864 0.959957 0.959958 -8.29E-07 0.005866 0.959957 0.959957 2.09E-07 0.005866 0.959957 0.959956 3.45E-07 0.005866 0.959957 0.959956 3.63E-07 0.005866 0.959957 0.959956 3.65E-07 0.005866 0.959957 0.959956 3.65E-07

T 0.5 R 2 Tc2 a (T )  0.42748 r Pc

q

a T  bRT

Zv  1   

 q   Zv    Zv  Zv   

b  0.08664

RTc Pc

bP RT Zl    Zl  Zl   

1    Zl q

Z   ln   Z  1  ln  Z     q ln    Z 

P

a T  RT  V  b V V  b 

Updated 4/5/2017

p. 141 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

0.6415 Tr0.5 R 2Tc2 a(T )  0.42748  0.42748 Pc b  0.08664



 83.1452  425.12 37.96

 17565000 bar cm 6 mol-2

RTc 83.145  425.1  0.08664  80.67 cm3 mol-1 Pc 37.96

bP 80.67 * P(bar)   0.003558P RT 83.145 * 272.7

Zv  1   

0.5

q

a T  bRT

17565000  9.603 80.67 * 83.145 * 272.7

q   Zv    Zv  Zv   

Zl    Zl  Zl   

1    Zl q

Z   ln   Z  1  ln  Z     q ln    Z 

 

Updated 4/5/2017

p. 142 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T

Chapter 13

P (bar) Tc (K) Pc (K) Tr Pr 272.7 1.517526 425.1 37.96 0.6415 0.0400

a(T)

b

beta

q

17564956 80.67138 0.005399 9.602986 Zv 1 0.954107 0.951668 0.951532 0.951524 0.951524 0.951524 0.951524 0.951524 0.951524 0.951524 0.951524 0.951524 0.951524 0.951524 0.951524

Zl Phi(v) Phi(l) Phi(v)-Phi(l) 0.005399 0.006524 0.953679 1.006166 -0.05249 0.006898 0.953676 0.958988 -0.00531 0.007033 0.953676 0.954353 -0.00068 0.007083 0.953676 0.953768 -9.2E-05 0.007101 0.953676 0.953689 -1.3E-05 0.007109 0.953676 0.953677 -1.8E-06 0.007111 0.953676 0.953676 -2.8E-07 0.007112 0.953676 0.953676 -5.7E-08 0.007113 0.953676 0.953676 -2.6E-08 0.007113 0.953676 0.953676 -2.1E-08 0.007113 0.953676 0.953676 -2.1E-08 0.007113 0.953676 0.953676 -2E-08 0.007113 0.953676 0.953676 -2E-08 0.007113 0.953676 0.953676 -2E-08 0.007113 0.953676 0.953676 -2E-08

T 0.5 R 2 Tc2 a (T )  0.42748 r Pc

q

a T 



bRT

Zv  1   

q   Zv    Zv  Zv   

b  0.08664

RTc Pc

bP RT

Zl    Zl  Zl   

1    Zl q

Z   ln   Z  1  ln  Z     q ln    Z 

P

a T  RT  V  b V V  b 

Updated 4/5/2017

p. 143 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

T 0.5 R 2Tc2 0.6147 0.5  83.1452  132.92 a(T )  0.42748 r  0.42748  1902592 bar cm 6 mol-2 Pc 34.99 b  0.08664



RTc 83.145  132.9  0.08664  27.36 cm3 mol-1 Pc 34.99

bP 27.36* P(bar)   0.004028P RT 83.145*81.7

Zv  1   

q

a T  bRT

1902592  10.237 27.36 *83.145*81.7

q   Zv    Zv  Zv   

Zl    Zl  Zl   

1    Zl q

Z   ln   Z  1  ln  Z     q ln    Z 

 

Updated 4/5/2017

p. 144 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T

Chapter 13

P (bar) Tc (K) Pc (K) Tr Pr 81.7 0.920433 132.9 34.99 0.6147 0.0263

a(T)

b

beta

q

1902592 27.36123 0.003707 10.23651 0.004028 Zv 1 0.966037 0.964723 0.96467 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668

Zl Phi(v) Phi(l) Phi(v)-Phi(l) 0.003707 0.004432 0.965823 1.012058 -0.04623 0.004657 0.965822 0.969942 -0.00412 0.004733 0.965822 0.966278 -0.00046 0.004759 0.965822 0.965876 -5.4E-05 0.004768 0.965822 0.965829 -6.5E-06 0.004771 0.965822 0.965823 -8.8E-07 0.004772 0.965822 0.965822 -2E-07 0.004773 0.965822 0.965822 -1.2E-07 0.004773 0.965822 0.965822 -1.1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07

Updated 4/5/2017

T 0.5 R 2 Tc2 a (T )  0.42748 r Pc

q

a T  bRT

Zv  1   

 q   Zv    Zv  Zv   

b  0.08664

RTc Pc

bP RT Zl    Zl  Zl   

1    Zl q

Z   ln   Z  1  ln  Z     q ln    Z 

p. 145 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

P

Chapter 13

a T  RT  V  b V V  b 

T 0.5 R 2Tc2 0.68790.5  83.1452  540.22 a(T )  0.42748 r  0.42748  37947574 bar cm 6 mol-2 Pc 27.40 b  0.08664



RTc 83.145  540.2  0.08664  142.02 cm3 mol-1 Pc 27.40

bP 142.02 * P(bar)   0.004597 P RT 83.145 * 371.6

Zv  1   

q

a T  bRT

37947600  8.648 142.02 * 83.145 * 371.6

q   Zv    Zv  Zv   

Zl    Zl  Zl   

1    Zl q

Z   ln   Z  1  ln  Z     q ln    Z 

 

Updated 4/5/2017

p. 146 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T

Chapter 13

P (bar) Tc (K) Pc (K) Tr Pr 371.6 2.064241 540.2 27.4 0.6879 0.0753

a(T) 37947757 Zv 1 0.928973 0.922943 0.922389 0.922338 0.922333 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332

b

beta

q

142.023 0.009489

8.648

Zl Phi(v) Phi(l) Phi(v)-Phi(l) 0.009489 0.011683 0.927772 0.99209 -0.06432 0.012496 0.927752 0.935764 -0.00801 0.012827 0.927752 0.92904 -0.00129 0.012965 0.927752 0.927977 -0.00022 0.013024 0.927752 0.927792 -4E-05 0.013049 0.927752 0.927759 -7.4E-06 0.01306 0.927752 0.927753 -1.4E-06 0.013065 0.927752 0.927752 -2.6E-07 0.013067 0.927752 0.927752 -5.5E-08 0.013067 0.927752 0.927752 -1.8E-08 0.013068 0.927752 0.927752 -1.1E-08 0.013068 0.927752 0.927752 -9.5E-09 0.013068 0.927752 0.927752 -9.3E-09 0.013068 0.927752 0.927752 -9.2E-09 0.013068 0.927752 0.927752 -9.2E-09 0.013068 0.927752 0.927752 -9.2E-09 0.013068 0.927752 0.927752 -9.2E-09 0.013068 0.927752 0.927752 -9.2E-09 0.013068 0.927752 0.927752 -9.2E-09 0.013068 0.927752 0.927752 -9.2E-09

P

T 0.5 R 2Tc2 a (T )  0.42748 r Pc

q

a T  bRT

Zv  1   

 q   Zv    Zv  Zv   

b  0.08664

RTc Pc

bP RT Zl    Zl  Zl   

1    Zl q

Z   ln   Z  1  ln  Z     q ln    Z 

a T  RT  V  b V V  b 

Updated 4/5/2017

p. 147 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

T 0.5 R 2Tc2 0.61250.5  83.1452  126.22 a(T )  0.42748 r  0.42748  1768755 bar cm 6 mol-2 Pc 34.00 b  0.08664



RTc 83.145  126.2  0.08664  26.74 cm3 mol-1 Pc 34.00

bP 26.74* P(bar)   0.00416 P RT 83.145*77.3

Zv  1   

q

a T  bRT

1768755  10.292 26.74 *83.145* 77.3

q   Zv    Zv  Zv   

Zl    Zl  Zl   

1    Zl q

Z   ln   Z  1  ln  Z     q ln    Z 

 

Updated 4/5/2017

p. 148 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T

P (bar) 77.3 0.861996

a(T)

b

Tc (K)

Pc (K)

126.2 beta

Chapter 13

Tr 34

Pr 0.6125

q T 0.5 R 2Tc2 a (T )  0.42748 r Pc

1768755 26.73838 0.003586 10.29241 Zv

Zl Phi(v) 1 0.003586

0.0254

Phi(l)

Phi(v)-Phi(l) q

a T 

0.96694 0.004283 0.966739 1.012471 0.965696 0.004499 0.966738 0.970768

-0.04573 -0.00403

bRT

0.965648 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646

-0.00044 -5E-05 -5.1E-06 1.84E-07 8.17E-07 8.93E-07 9.02E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07

Zv  1   

0.004571 0.004595 0.004604 0.004607 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608

Updated 4/5/2017

0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738

0.967178 0.966788 0.966743 0.966738 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737

 q   Zv    Zv  Zv   

b  0.08664

RTc Pc

bP RT Zl    Zl  Zl   

1    Zl q

Z   ln   Z  1  ln  Z     q ln    Z 

p. 149 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 150 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Acetylene Argon Benzene n-Butane Carbon monoxide n-Decane Ethylene n-Heptane Methane Nitrogen

Updated 4/5/2017

Chapter 13

Tn(K) 1.894 87.3 353.2 272.7

Psat(bar) @ Tn 1.073 0.976 1.007 1.008

0.85Tc (K) 262.1 128.3 477.9 361.3

Psat (bar) @0.85 Tc 20.016 18.79 15.658 12.239

Psat (bar) Lit value 19.78 18.7 15.52 12.07

% Difference 1.2 0.5 0.9 1.4

81.7 447.3 169.4 371.6 111.4 77.3

1.019 1.014 1.004 1.011 0.959 0.992

113 525 240 459.2 162 107.3

12.871 5.324 17.918 7.779 17.46 12.617

12.91 5.21 17.69 7.59 17.33 12.57

-0.3 2.1 1.3 2.5 0.8 0.3

p. 151 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

=

<

Updated 4/5/2017

p. 152 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 153 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 154 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 155 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Ʌ

Ʌ

Ʌ

Updated 4/5/2017

Ʌ

Ʌ Ʌ

Ʌ p. 156 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 13

Ʌ

Updated 4/5/2017

Ʌ

Ʌ Ʌ

Ʌ

p. 157 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 13

p. 158 of 158

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 n  7 nNH3  2  4 nO2  5  5 nNO  4 nH 2O  6

2  4 7 5  5 yO2  7 4 y NO  7 6 y H 2O  7 y NH3 

 n  8 nH 2 S  3  2 nO2  5  3 nH 2O  2 nSO2  2

3  2 8 5  3 yO2  8 2 y H 2O  8 2 ySO2  8 yH 2 S 

Updated 4/5/2017

 p. 1 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 n  8  5 nNO2  3  6 nNH 3  4  8 nN 2  1  7 nH 2O  12

3  6 8  5 4  8 y NH3  8  5 1  7 yN2  8  5 12 y H 2O  8  5 y NO2 

 

 

 

  

Updated 4/5/2017

p. 2 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 

 

 

 

  

 

 

       

 

Updated 4/5/2017

   

p. 3 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

-208200 -208250 -208300 -208350

Equlibrium Point at 0.45308

Ge

-208400 -208450 -208500

-208550 -208600 -208650 -208700 0.3

0.35

0.4

0.45

0.5

0.55

0.6

εe

Updated 4/5/2017

p. 4 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

-209800 -209900 -210000 -210100

Ge

-210200 -210300 -210400

-210500 -210600 -210700 -210800 0

0.2

0.4

0.6

0.8

εe

Updated 4/5/2017

p. 5 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

-212100 -212200

Ge

-212300 -212400 -212500 -212600 -212700 0.4

0.45

0.5

0.55

0.6

0.65

0.7

0.75

εe

Updated 4/5/2017

p. 6 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

-214250 -214300 -214350

Ge

-214400 -214450

-214500 -214550 -214600 -214650 -214700 0.4

0.45

0.5

0.55

0.6

0.65

0.7

0.75

εe

Updated 4/5/2017

p. 7 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 8 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 9 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 GAo  3GHo 2O 

5 o o o GN  2GNH  3GNO 3 2 2

5  o o  K A  exp  GAo   exp  3GHo 2O  GNo 2  2GNH  3GNO  3 2   3

5

 fˆH O   fˆN  2  o2   o2   fH O   fN  2   2 KA   2 3  fˆNH   fˆ  NO 3  o   o   f NH   f NO  3    o o GBo  6GHo 2O  5GNo 2  4GNH  6GNO 3 o o K B  exp  GBo   exp  6GHo 2O  5GNo 2  4GNH  6GNO  3

6

5

 fˆH O   fˆN   o2   o2   fH O   fN  2   2 KB   4 6  fˆNH   fˆ  NO 3  o   o   f NH   f NO  3   Updated 4/5/2017

p. 10 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 o o GCo  2GNH  3GNO  3GHo 2O  3

5 o GN 2 2

5 o   o o o KC  exp  GCo   exp  2GNH  3 G  3 G  GN  NO H O 3 2 2 2 

KC 

2

3

3

5

 fˆNH   fˆ   o 3   NO o    f NH   f NO  3  

 fˆH O   fˆN  2  o2   o2   fH O   fN   2   2

o o K B  exp 6GHo 2O  5GNo 2  4GNH  6GNO 3

  5 o o  K B  exp  2  3GHo 2O  GNo 2  2GNH  3GNO  3 2    2

 5 2  o o  K B   exp  3GHo 2O  GNo 2  2GNH  3GNO   KA  3 2    5  o  o K C  exp  2GNH  3GNO  3GHo 2O  GNo 2  3 2     5 o o  K C  exp    3GHo 2O  GNo 2  2GNH  3GNO  3 2    1 1 KC   5 KA  o o  exp  3GHo 2O  GNo 2  2GNH  3GNO  3 2  

Updated 4/5/2017

p. 11 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

6

5

 fˆH O   fˆN   o2   o2   fH O   fN  2 2   2 KB   K A    4 6  fˆNH   fˆ   o 3   NO o    f NH   f NO  3   5

3

 fˆH O   fˆN  2  o2   o2   fH O   fN  2   2 KA   2 3  fˆNH   fˆ  NO 3  o   o   f NH   f NO  3  

2

 fˆNH   fˆ   o 3   NO o    f NH   f NO 1  3  KC    K A  ˆ 3  ˆ  5 2 fH O fN  o2   o2   fH O   fN   2   2 3

5

3

 fˆH O   fˆN  2  o2   o2   fH O   fN  2   2 KA   2 3  fˆNH   fˆ   o 3   NO o    f NH   f NO  3  

 

K  Ko K1K2 Updated 4/5/2017

p. 12 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 G o To   Ko  exp    RTo    H o To   T   K1  exp   1    RT  To   

T  1 T  C po o K 2  exp    C dT  dT  RT   RT T p T o o  

75948   13 Ko  exp    2.0205 10  8.3145  298.15   114408  773.15   13 K1  exp  1     4.8527 10  8.3145  773.15  298.15  

Updated 4/5/2017

p. 13 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

T

IDCPH  T0 ,T; A, B, C, D  

1 R

1

1 

 C dT  A T  T   2 T  T   3 T  T   D  T  T  B

p

0

2

C

2 0

3

3 0

0

To

T0 (K) 298.15

T (K) 773.15

A -0.439

B (1/K) 8.00E-05

C (1/K2) 0.00E+00

D (K2) IDCPH (K) -8.23E+04 -357.8

Species Name HCl O2 H2O Cl2

Stoichiometric coefficient -4 -1 2 2

A 3.156 3.639 3.47 4.442

B (1/K) 6.23E-04 5.06E-04 1.45E-03 8.90E-05

C (1/K2) 0.00E+00 0.00E+00 0.00E+00 0.00E+00

D (K2) 1.51E+04 -2.27E+04 1.21E+04 -3.44E+04

T

IDCPS  T0 ,T; A, B, C, D 

C p

iAi iBi -1.26E+01 -2.49E-03 -3.64E+00 -5.06E-04 6.94E+00 2.90E-03 8.88E+00 1.78E-04

0

2 2 D   T  T0   2 2  2 0 

T 

IDCPH/T -0.463

iCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iDi -6.04E+04 2.27E+04 2.42E+04 -6.88E+04

iCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iDi -6.04E+04 2.27E+04 2.42E+04 -6.88E+04

 RT dT  A ln  T   B T  T    C  T T   0

T0

T0 (K) 298.15

T (K) 773.15

A -0.439

B (1/K) 8.00E-05

C (1/K2) 0.00E+00

D (K2) -8.23E+04

Species Name HCl O2 H2O Cl2

Stoichiometric coefficient -4 -1 2 2

A 3.156 3.639 3.47 4.442

B (1/K) 6.23E-04 5.06E-04 1.45E-03 8.90E-05

C (1/K2) 0.00E+00 0.00E+00 0.00E+00 0.00E+00

D (K2) 1.51E+04 -2.27E+04 1.21E+04 -3.44E+04

IDCPS -0.774

iAi iBi -1.26E+01 -2.49E-03 -3.64E+00 -5.06E-04 6.94E+00 2.90E-03 8.88E+00 1.78E-04

K2  exp  0.463  0.774  0.7327 

i

 fˆ  K    io    i 1  f i  N

N

 y 

i

i

i 1

yH2 2O yCl2 2

 P  K o  P 



 P  K   1 bar 



1

 2 bar   7.184    14.37 4 yHCl yO2  1 bar 

Updated 4/5/2017

p. 14 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

n  6 nHCl  5  4 nO2  1   nH 2O  2 nCl2  2

5  4 6 1  yO2  6 2 y H 2O  6 2 yCl2  6 yHCl 

 2     6   14.368 4  5  4   1         6   6  4

 2   1      14.368  0  5  4   6  4

  2   1      14.368    5  4   6  4

 Problem 13.11

zero at equilibrium

20

15

10

5

0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-5

extent of reaction

Updated 4/5/2017

p. 15 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

5  4  0.35 6 1  yO2   0.040 6 2 y H 2O   0.305 6 2 yCl2   0.305 6 yHCl 

 

 

 

 

  

 H o To   T   42720  923   5 K1  exp   1     exp    1     1.1668  10   RT  To    8.314 *923  298    Updated 4/5/2017

p. 16 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

T  1 T C po  C po K 2  exp    dT   dT   TT R  RT To o  

T 1 (K) 298.15

T 2 (K) 923.15 T2

ICPH 

R

Cp

A 0.06

C (1/K2) D (K2) B (1/K) 1.73E-04 0.00E+00 1.91E-06

dT  A  T2  T1  

T1

T 1 (K) 298.15

T 2 (K) 923.15 T2

ICPS 

T1

ICPH (K) 1.04E+02

 1 B 2 C 3 1 T2  T12  T2  T13  D    2 3  T2 T1 

C (1/K2) D (K2) B (1/K) 1.73E-04 0.00E+00 1.91E-06

A 0.06

ICPS 0.176

T  C D 2 dT  A ln  2   B T2  T1   T22  T12  T2  T12 RT T 2 2  1

Cp

 104  K 2  exp    0.176   1.066  923  

 

 

Updated 4/5/2017

p. 17 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 

K  Ko K1K2  G o To   Ko  exp    RTo    H o To   T   K1  exp   1    RT  To   

T  1 T  C po o K 2  exp    C dT  dT  RT   RT T p T o o  

 

 

 

39630   6 Ko  exp    8.7669 10  8.3145  298.15 

Updated 4/5/2017

p. 18 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 68910  623.15   7 K1  exp  1     5.0546 10  8.3145  623.15  298.15  

Reference Temperature T 0 (K) 298.15

Temperature of Interest T (K) 623.15

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T 0 kJ/mol (dimensionless) kJ/mol -0.363 Heat Capacity Coefficients

Standard Heat Species Stoichiometric of Formation Name coefficient at T 0 (kJ/mol) acetaldehyde -1 H2 -1 ethanol 1

A 1.693 3.249 3.518

A -1.424 Note: Light blue fields are inputs, pink fields are the final output.

T

B

p

2

C

2 0

0

3

0

 1 B 2 C 3 1  o o  H rxn T  T02  T  T 03   D    ,T   H rxn ,T0  R   A  T  T 0   2 3  T T0   

2

2

B (1/K) 1.80E-02 4.22E-04 2.00E-02

C (1/K ) -6.16E-06 0.00E+00 -6.00E-06

D (K ) iHf,i iAi 0.00E+00 0.00E+00 -1.69E+00 8.30E+03 0.00E+00 -3.25E+00 0.00E+00 0.00E+00 3.52E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

B (1/K) 1.60E-03

C (1/K ) 1.56E-07

D (K ) -8.30E+03

2

Temperature of Interest T (K) 1073.15

A -1.424 Note: Light blue fields are inputs, pink fields are the final output.

1 

1

3 0

To

Entropy Entropy Change of Heat Capacity Change of Reaction at To Integral Reaction at T J/(mol K) (dimensionless) J/(mol K) -0.543 Heat Capacity Coefficients Standard Species Stoichiometric Entropy at T 0 Name coefficient (J/(mol K)) A B (1/K) acetaldehyde -1 1.693 1.80E-02 H2 -1 3.249 4.22E-04 ethanol 1 3.518 2.00E-02

Reference Temperature T 0 (K) 298.15

1

 C dT  T  A T  T   2 T  T   3 T  T   D  T  T  

1 RT

B (1/K) 1.60E-03

T

1 R

iBi -1.80E-02 -4.22E-04 2.00E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iCi 6.16E-06 0.00E+00 -6.00E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iDi 0.00E+00 -8.30E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

2

T 

1 

D  1

 T dT   A ln  T   B T  T   2 T  T   2  T  T   Cp

C

2

o

2 o

2

o

To

2 o

 T  C 2 D  1 1  o o  S rxn T  To2     B T  To   ,T  S rxn ,T0  R  A ln   T 2 T 2    T 2 2 o   o   

2

2

C (1/K ) -6.16E-06 0.00E+00 -6.00E-06

D (K ) iHf,i iAi 0.00E+00 0.00E+00 -1.69E+00 8.30E+03 0.00E+00 -3.25E+00 0.00E+00 0.00E+00 3.52E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

C (1/K ) 1.56E-07

D (K ) -8.30E+03

2

iBi -1.80E-02 -4.22E-04 2.00E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iCi 6.16E-06 0.00E+00 -6.00E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iDi 0.00E+00 -8.30E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

2

K2  exp  0.363  0.543  0.8353 

Updated 4/5/2017

p. 19 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

i

 fˆ  K    io    i 1  f i  N

N

 y 

i

i

i 1

 P  K o  P 



 P  K   1 bar 



 3 bar   3.701   11.104 yH 2 yCH3CHO  1 bar  yC2 H5OH

n  no   nCH3CHO  nCH3CHO ,o   nH 2  nH 2 ,o   nC2 H5OH  nC2 H5OH ,o   yCH3CHO  yH 2 

nCH3CHO ,o   no  

nH 2 ,o   no  

yC2 H 5OH 

nC2 H5OH ,o   no  

1  2.5   1.5   yH 2  2.5   yCH 3CHO 

yC2 H 5OH 

2.5  

  2.5     11.104 1.5   1   

2.5   2  11.104 1.5  2.5   2  12.104 2  30.26  16.656  0 Updated 4/5/2017

p. 20 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 1  0.8182  0.1081 2.5  0.8182 1.5  0.8182 yH 2   0.4054 2.5  0.8182 0.8182 yC2 H5OH   0.4865 2.5  0.8182 yCH3CHO 

 1 bar   3.701   3.701 yH 2 yCH3CHO  1 bar  yC2 H5OH

  2.5     3.701 1.5   1   

2.5   2  3.7011.5  2.5   2  4.701 2  11.7525  5.5515  0

1  0.6323  0.1969 2.5  0.6323 1.5  0.6323 yH 2   0.4646 2.5  0.6323 0.6323 yC2 H5OH   0.3385 2.5  0.6323 yCH3CHO 

Updated 4/5/2017

p. 21 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

     

K

 

yC8 H10

yC8 H8 yH 2

  2.5    1   1.5   

  2.5     K 1   1.5     0 ( K  1) 2  2.5  K  1   1.5K  0 2.5  K  1  6.25  K  1  6 K  K  1 2

 

2  K  1

2.5  K  1  0.25 K 2  6.5 K  6.25 2  K  1

  1.25 

0.25K 2  6.5K  6.25 2  K  1

 



K  Ko K1K2  G o To   Ko  exp    RTo   Updated 4/5/2017

p. 22 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 H o To   T   K1  exp   1    RT  To   

T  1 T  C po o K 2  exp    C dT  dT  RT   RT T p T o o  

 83010  14 Ko  exp    3.496 10  8.314*298   117440  923   14 K1  exp  1     1.174 10  8.314*923  298  

Reference Temperature T 0 (K) 298.15

Temperature of Interest T (K) 923.15

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T 0 kJ/mol (dimensionless) kJ/mol -117.440 -1.375 -127.990 Heat Capacity Coefficients

Standard Heat Species Stoichiometric of Formation Name coefficient at T 0 (kJ/mol) styrene -1 147.36 H2 -1 0 ethylbenzene 1 29.92

A 2.05 3.249 1.124

A -4.175 Note: Light blue fields are inputs, pink fields are the final output.

Updated 4/5/2017

T

1 RT

1

1

1 

 C dT  T  AT  T   2 T  T   3 T  T   D  T  T   B

p

2

0

C

2 0

3

3 0

0

To

o o  H rxn H rxn ,T ,T0

 R  A T  

T0 

 B 2 T 2

T02

B (1/K) 5.02E-02 4.22E-04 5.54E-02

C (1/K2) -1.67E-05 0.00E+00 -1.85E-05

D (K2) iHf,i iAi 0.00E+00 -1.47E+02 -2.05E+00 8.30E+03 0.00E+00 -3.25E+00 0.00E+00 2.99E+01 1.12E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

B (1/K) 4.77E-03

C (1/K2) -1.81E-06

D (K2) -8.30E+03

C 3 T 3

iBi -5.02E-02 -4.22E-04 5.54E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

T03

1 D  T

iCi 1.67E-05 0.00E+00 -1.85E-05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1 T0

iDi 0.00E+00 -8.30E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

p. 23 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Entropy Entropy Change of Heat Capacity Change of Reaction at To Integral Reaction at T J/(mol K) (dimensionless) J/(mol K) -115.48 -2.474 -136.048 Heat Capacity Coefficients Standard Species Stoichiometric Entropy at T 0 Name coefficient (J/(mol K)) A B (1/K) styrene -1 -223.2 2.05 5.02E-02 H2 -1 0.0 3.249 4.22E-04 ethylbenzene 1 -338.7 1.124 5.54E-02

Reference Temperature T 0 (K) 298.15

Temperature of Interest T (K) 923.15

A -4.175

B (1/K) 4.77E-03

T

1 R

T 

1 

D  1

 T dT   A ln  T   B T  T   2 T  T   2  T  T   Cp

C

2

o

2 o

2

o

To

2 o

 T  C 2 D  1 1  o o Srxn T  To2   2  2     B T  To   ,T  S rxn ,T0  R  A ln    T 2 2 T T  o o   

2

2

C (1/K ) -1.67E-05 0.00E+00 -1.85E-05

D (K ) iHf,i iAi 0.00E+00 2.23E+02 -2.05E+00 8.30E+03 0.00E+00 -3.25E+00 0.00E+00 -3.39E+02 1.12E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

C (1/K2) -1.81E-06

D (K2) -8.30E+03

iBi -5.02E-02 -4.22E-04 5.54E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iCi 1.67E-05 0.00E+00 -1.85E-05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iDi 0.00E+00 -8.30E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

T  1 T  C po o K 2  exp    C dT  dT p  RT   RT T To o   K 2  exp 1.375  2.474   0.3332

        

  

 Updated 4/5/2017

p. 24 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

    

  

 

 P  K o  P 

0.5

K

ySO3

ySO2  yO2 

0.5

 1  0.5  0.15    0.20  0.5

K  Ko K1K2  G o To   Ko  exp    RTo  

 H o To   T   K1  exp   1    RT  To   

T  1 T  C po o K 2  exp    C dT  dT  p    RT T  RT To o  

Updated 4/5/2017

p. 25 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

70866   12 Ko  exp    2.605 10 8.314*298.15    98890  753.15   11 K1  exp  1     3.412 10  8.314*753.15  298.15  

Reference Temperature T 0 (K) 298.15

Species Name SO2 O2 SO3

Temperature of Interest T (K) 753.15

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T 0 kJ/mol (dimensionless) kJ/mol -98.890 0.086 -98.353 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -1 -296.83 -0.5 0 1 -395.72

A 5.699 3.639 8.06

A 0.5415 Note: Light blue fields are inputs, pink fields are the final output. Reference Temperature T 0 (K) 298.15

Species Name SO2 O2 SO3

B

p

2

0

3

1 

1

3 0

0

o o  H rxn H rxn ,T ,T0

 R  A T  

T0 

D (K2) iHf,i -1.02E+05 2.97E+02 -2.27E+04 0.00E+00 -2.03E+05 -3.96E+02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

B (1/K) 2.00E-06

C (1/K ) 0.00E+00

D (K ) -9.00E+04

B (1/K) 2.00E-06

C

2 0

To

C (1/K2) 0.00E+00 0.00E+00 0.00E+00

Entropy Entropy Change of Heat Capacity Change of Reaction at To Integral Reaction at T J/(mol K) (dimensionless) J/(mol K) -93.99 0.076 -93.361 Heat Capacity Coefficients Standard Stoichiometric Entropy at T 0 coefficient (J/(mol K)) A B (1/K) -1 11.28 5.699 8.01E-04 -0.5 0 3.639 5.06E-04 1 -82.71 8.06 1.06E-03

Updated 4/5/2017

1

 C dT  T  AT  T   2 T  T   3 T  T   D  T  T  

B (1/K) 8.01E-04 5.06E-04 1.06E-03

Temperature of Interest T (K) 753.15

A 0.5415 Note: Light blue fields are inputs, pink fields are the final output.

T

1 RT

2

T

1 R

 B 2 T 2

T02

iAi -5.70E+00 -1.82E+00 8.06E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

C 3 T 3

T03

iBi -8.01E-04 -2.53E-04 1.06E-03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1 D  T

iCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1 T0

iDi 1.02E+05 1.14E+04 -2.03E+05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

2

T 

1 

D  1

 T dT   A ln  T   B T  T   2 T  T   2  T  T   Cp

C

2

o

To

2 o

2

o

2 o

 T  C 2 D  1 1  o o  Srxn T  To2     B T  To   ,T  S rxn ,T0  R  A ln   T 2 T 2    T 2 2 o   o   

C (1/K2) 0.00E+00 0.00E+00 0.00E+00

D (K2) iHf,i -1.02E+05 -1.13E+01 -2.27E+04 0.00E+00 -2.03E+05 -8.27E+01 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

C (1/K2) 0.00E+00

D (K2) -9.00E+04

iAi -5.70E+00 -1.82E+00 8.06E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iBi -8.01E-04 -2.53E-04 1.06E-03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iDi 1.02E+05 1.14E+04 -2.03E+05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

p. 26 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 1  0.5 K  0.15    0.20  0.5



K  0.15    0.20  0.5 1  0.5

 1  0.5 K 0.20  0.5

  0.15 

88.0  0.15    0.20  0.5 1  0.5

  0.15   

 1  0.5 K 0.20  0.5

  

K  Ko K1K2  G o To   Ko  exp    RTo   Updated 4/5/2017

p. 27 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 H o To   T   K1  exp   1    RT  To   

T  1 T  C po o K 2  exp    C dT  dT p  RT   RT T To o  

42290   8 Ko  exp    3.901  10  8.3145  298.15   82670  T   82670  625   7 K1  exp  1     exp  1     3.750  10  8.3145  T  298.15    8.3145  625  298.15  

Updated 4/5/2017

p. 28 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Reference Temperature T 0 (K) 298.15

Species Name Propane Ethylene Methane

Temperature of Interest T (K) 625

Chapter 14

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T 0 kJ/mol (dimensionless) kJ/mol 0.000 -0.018 -0.094 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -1 1 1

A 1.213 1.424 1.702

A 1.913 Note: Light blue fields are inputs, pink fields are the final output.

Reference Temperature T 0 (K) 298.15

Species Name Propane Ethylene Methane

1

B

p

Updated 4/5/2017

3

3 0

 B 2 C 3 o o  H rxn T  T02  T  ,T   H rxn ,T0  R   A  T  T 0   2 3 

2

B (1/K) -5.31E-03

C (1/K ) 2.27E-06

D (K ) 0.00E+00

2

T

1 R

2

2

T 

D  1

 T dT    A ln  T    B T  T   2 T  T   2  T Cp

C

o

2

2 o

o

To

2

 T  C 2 o o S rxn T  To2   B T  To   ,T  S rxn ,T0  R  A ln   T 2 o   

2

2

C (1/K ) -8.82E-06 -4.39E-06 -2.16E-06

D (K ) 0.00E+00 0.00E+00 0.00E+00

C (1/K ) 2.27E-06

D (K ) 0.00E+00

2

2

K2  exp  0.018  0.023  1.042 

C

2 0

To

D (K ) 0.00E+00 0.00E+00 0.00E+00

B (1/K) -5.31E-03

2

0

C (1/K ) -8.82E-06 -4.39E-06 -2.16E-06

Entropy Entropy Change of Heat Capacity Change of Reaction at To Integral Reaction at T J/(mol K) (dimensionless) J/(mol K) 0.00 0.023 0.187 Heat Capacity Coefficients Standard Stoichiometric Entropy at T 0 coefficient (J/(mol K)) A B (1/K) -1 1.213 2.88E-02 1 1.424 1.44E-02 1 1.702 9.08E-03

1

 C dT  T  A T  T   2 T  T   3 T  T   D  T

B (1/K) 2.88E-02 1.44E-02 9.08E-03

Temperature of Interest T (K) 625

A 1.913 Note: Light blue fields are inputs, pink fields are the final output.

T

1 RT

p. 29 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

i

 fˆ  K    io    i 1  f i  N



 y 

 P  K o  P 

yC2 H 4 yCH 4

 1 bar   K   1.524  1 bar 

N

i

i

i 1

 P  K   1 bar 



1

yC3H8

n  1  nC3 H8  1   nC2 H 4   nCH 4   1  1 

yC3 H8  yC2 H 4  yCH 4 

1 

1 

     2 2 1     K  1    1   1    1   2    1    2 1  K   K 2



K 1.524   0.777 1 K 2.524

Updated 4/5/2017

p. 30 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

1  0.736  0.152 1  0.736 0.736 yC2 H 4   0.424 1  0.736 0.736 yCH 4   0.424 1  0.736 yC3 H8 

      2 2 1     K  1    1   1    1   2    1    2

42920   8 Ko  exp    3.901  10 8.3145  298.15    82670  T  K1  exp  1    8.3145  T  298.15  

 82670  T  K  3.901  108 exp  1    *1.04  2.604  8.3145  T  298.15    82670  1 1  7 exp       6.418  10  8.3145  T 298.15   82670  1 1      17.977 8.3145  T 298.15  1 1   0.001808  0.001546 T 298.15

K2  exp  0.036  0.002  1.035 Updated 4/5/2017

p. 31 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 32 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 33 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 34 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

n  4 nN2  1  0.5 nH 2  3  1.5 nNH3   1  0.5 4 3  1.5 yH 2  4 yN2 

y NH3 

Updated 4/5/2017

4

p. 35 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14



P P    yi  i  K  o   K    P   1 bar  i 1 yNH3  P  K  0.5 1.5  1 bar   y N2   yH 2  N

 4   

1  0.5   3  1.5  0.5

1.5



 KP

   4   

1  0.5   3 2

1.5

 KP

 

16450   K  Ko  exp    761.9  8.3145  298.15   46110  300   K1  exp  1     0.8833  8.3145  300  298.15  

 4   

1  0.5   3 2

1.5

 673

Updated 4/5/2017

p. 36 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

  4    673  3

1.5

  2 1   

 1  0.5 

2

 4    

  4         2 1  1.5   3497 673  3   

1  0.5 1.932   0.016 4  1.932 3  1.5 1.932  yH 2   0.049 4  1.932 1.932 yNH3   0.934 4  1.932  yNH   0.5 4

yN2 

3

 4   

 KP 1.5

1  0.5   3 2

1.333  4  1.333

1  0.5 1.333  3 2

1.5

 6.158  KP

 46110  T  K  K 0 K1  761.9 exp  1    8.3145  T  298.15   46110    46110  K  761.9 exp   exp    8.3145  298.15   8.3145  T   46110  K  6.365 106 exp    8.3145  T 

46110  6.158  ln   6   6.365 10  8.3145  T 46110 T  402 K  6.158  8.3145  ln  6   6.365 10 

Updated 4/5/2017

p. 37 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

1.333  4  1.333

1  0.5 1.333 3 2

1.5

Chapter 14

 6.158  KP  100 K

 46110  K  6.365 106 exp    8.3145  T  46110  0.06158  ln   6   6.365 10  8.3145  T 46110 T  604 K  0.06158  8.3145  ln  6   6.365 10 

Reference Temperature T 0 (K) 298.15

Species Name N2 H2 NH3

Reference Temperature T 0 (K) 298.15

Species Name N2 H2 NH3

Temperature of Interest T (K) 604

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T 0 kJ/mol (dimensionless) kJ/mol -0.997 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -0.5 -1.5 1

1

B

p

2

0

C

2 0

1 

 1 B 2 C 3 1  o o  H rxn T  T02  T  T 03   D    ,T   H rxn ,T0  R   A  T  T 0   2 3  T T0   

2

2

D (K ) iHf,i iAi 4.00E+03 0.00E+00 -1.64E+00 8.30E+03 0.00E+00 -4.87E+00 1.86E+04 0.00E+00 3.58E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

A -2.9355

B (1/K) 2.09E-03

C (1/K ) 0.00E+00

D (K ) 4.15E+03

B (1/K) 2.09E-03

3 0

0

C (1/K ) 0.00E+00 0.00E+00 0.00E+00

A -2.9355

3

To

B (1/K) 5.93E-04 4.22E-04 3.02E-03

Entropy Entropy Change of Heat Capacity Change of Reaction at To Integral Reaction at T J/(mol K) (dimensionless) J/(mol K) -1.415 Heat Capacity Coefficients Standard Stoichiometric Entropy at T 0 coefficient (J/(mol K)) A B (1/K) -0.5 3.28 5.93E-04 -1.5 3.249 4.22E-04 1 3.578 3.02E-03

1

 C dT  T  A T  T   2 T  T   3 T  T   D  T  T  

A 3.28 3.249 3.578

Temperature of Interest T (K) 604

Updated 4/5/2017

T

1 RT

2

T

1 R

iBi -2.97E-04 -6.33E-04 3.02E-03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iDi -2.00E+03 -1.25E+04 1.86E+04 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

2

T 

1 

D  1

 T dT   A ln  T   B T  T   2 T  T   2  T  T   Cp

C

2

o

2 o

2

o

To

2 o

 T  C 2 D  1 1  o o S rxn T  To2       B T  To   ,T  S rxn ,T0  R  A ln   2 2  T 2 To2    To  

2

2

C (1/K ) 0.00E+00 0.00E+00 0.00E+00

D (K ) iHf,i iAi 4.00E+03 0.00E+00 -1.64E+00 8.30E+03 0.00E+00 -4.87E+00 1.86E+04 0.00E+00 3.58E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

C (1/K ) 0.00E+00

D (K ) 4.15E+03

2

iBi -2.97E-04 -6.33E-04 3.02E-03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

iDi -2.00E+03 -1.25E+04 1.86E+04 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

2

p. 38 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 46110   46110  6 K  K0 K1K 2  6.365 106 exp    0.6584  4.1905 10 exp    8.3145  T   8.3145  T  46110 T  578.0 K  0.06158  8.3145  ln  6   4.1905 10 

 46110   46110  6 K  K0 K1K 2  6.365 106 exp    0.6873  4.3747 10 exp    8.3145  T   8.3145  T  46110 T  580.6 K  0.06158  8.3145  ln  6   4.3747 10 



P P   yi   K  o   K    P   1 bar  i 1 yNH3  P  K  0.5 1.5  1 bar  y  y  N

i

N2



H2

 4   

1  0.5   3  1.5  0.5

1.5

P i yi   K  o   P  i 1 NH3 yNH3



 KP

 P  K   1 bar   P  K  0.5 1.5  1 bar   N 2 y N 2   H 2 y H 2  N

i



  NH3    KP 0.5 1.5 0.5 1.5 1  0.5   3  1.5   N2  H2  

 4   

Updated 4/5/2017

p. 39 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

     NH3 NH3    6.158    KP 2 0.5 1.5 0.5 1.5 1.5     1  0.5 1.333   3  N2  H2    N2  H2   1.333  4  1.333

K

  NH3 6.158   0.5 1.5  P      H2  N2 

  

 

i  exp  Bi0   Bi1

Updated 4/5/2017

 TP  r

r

p. 40 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T (K)

P (bar)

580.6

100

Chapter 14

Tc (K) Pc (bar) 405.7

112.8

0.253

Tr

Pr

B

0

B

1

1.4311

0.8865

-0.1548

0.1008

0.9230

 TP 

B 0  0.083 

0.422

B  0.139 

0.172

Tr1.6

1

T (K)

  exp  B 0   B1 

r

r

Tr4.2

P (bar)

580.6

100

Tc (K) Pc (bar) 126.6

34

0.038

Tr

Pr

B

0

B

1

4.5861

2.9412

0.0461

0.1387

1.0335

 TP 

B 0  0.083 

0.422

B  0.139 

0.172

1

T (K)

Tr1.6

  exp  B 0   B1 

r

r

Tr4.2

P (bar)

580.6

100

Tc (K) Pc (bar) 33.19

13.13

-0.216

Tr

Pr

B

0

B

1

17.4932

7.6161

0.0787

0.1390

1.0214

 TP 

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6

  exp  B 0   B1 

r

r

Tr4.2

K

  NH3 6.158   P   0.5  1.5  H2  N2 

K

 6.158  0.9230    0.06158  0.8795  0.05416 100  1.0335 0.5 1.0214 1.5 

Updated 4/5/2017

p. 41 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T

Chapter 14

46110  588.5 K  0.05416  8.3145  ln  6   4.3747 10 

       Updated 4/5/2017

   p. 42 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

yCH3OH yCO yH2 2

  3  2  4 1   

  3  2  4 1   

3

2

3

Chapter 14

 P K o  P 

2

2

 17632

 1

   3  2 2  3   1    4 17632   

 yCH3OH yCO yH2 2

  3  2  4 1   

3

2

0.75  3  1.5  4 1  0.75 

3

2

 P  K o  P 

2

2

 P  K  o   27 P 

Updated 4/5/2017

p. 43 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Reference Temperature T 0 (K) 298.15

Temperature of Interest T (K) 366

Chapter 14

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T 0 kJ/mol (dimensionless) kJ/mol -90.135 -0.849 -92.717 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -1 -110.53 -2 0.00

Species Name CO H2 CH3OH

1

-200.66

Species Name CO H2 CH3OH

2

0

o o  H rxn H rxn ,T ,T0

 R  A T  

T0 

1 D  T

1 T0

-3.45E-06

0.00E+00 -2.01E+02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

-3.45E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

B (1/K) 1.08E-02

C (1/K2) -3.45E-06

D (K2) -1.35E+04

T

1 R

T 

0.75  3  1.5  4 1  0.75 

3

2

2.21E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1.22E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1 

D  1

 T dT   A ln  T   B T  T   2 T  T   2  T  T   Cp

C

2

o

2 o

2

o

To

2 o

 T  C 2 D  1 1  o o Srxn T  To2       B T  To   ,T  S rxn ,T0  R  A ln   2 2  T 2 To2    To  

C (1/K2) 0.00E+00 0.00E+00

D (K2) iHf,i iAi iBi -3.10E+03 -8.94E+01 -3.38E+00 -5.57E-04 8.30E+03 0.00E+00 -6.50E+00 -8.44E-04

iCi 0.00E+00 0.00E+00

iDi 3.10E+03 -1.66E+04

-3.45E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

D (K ) -1.35E+04

T03

1.22E-02

C (1/K ) -3.45E-06

3

C 3 T 3

2.211

B (1/K) 1.08E-02

4 1   

iDi 3.10E+03 -1.66E+04

A -7.663 Note: Light blue fields are inputs, pink fields are the final output.

T02

iCi 0.00E+00 0.00E+00

0.00E+00 -1.30E+02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

yCO yH2 2

 B 2 T 2

D (K2) iHf,i iAi iBi -3.10E+03 1.11E+02 -3.38E+00 -5.57E-04 8.30E+03 0.00E+00 -6.50E+00 -8.44E-04

-3.45E-06

2

1 

1

3 0

0

1.22E-02

  3  2 

3

To

2.211

yCH3OH

C

2 0

C (1/K2) 0.00E+00 0.00E+00

Entropy Entropy Change of Heat Capacity Change of Reaction at To Integral Reaction at T J/(mol K) (dimensionless) J/(mol K) -219.16 -0.941 -226.986 Heat Capacity Coefficients Standard Stoichiometric Entropy at T 0 coefficient (J/(mol K)) A B (1/K) -1 89.36 3.376 5.57E-04 -2 0.00 3.249 4.22E-04 -129.80

B

p

B (1/K) 5.57E-04 4.22E-04

Temperature of Interest T (K) 366

1

1

 C dT  T  AT  T   2 T  T   3 T  T   D  T  T  

A 3.376 3.249

A -7.663 Note: Light blue fields are inputs, pink fields are the final output. Reference Temperature T 0 (K) 298.15

T

1 RT

2

 P  K o  P 

2.21E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1.22E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

2

2

2

 P  K  o   27 P 

Updated 4/5/2017

p. 44 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Reference Temperature T 0 (K) 298.15

Temperature of Interest T (K) 530

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T 0 kJ/mol (dimensionless) kJ/mol -90.135 -1.696 -97.609 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -1 -110.53 -2 0.00

Species Name CO H2 CH3OH

Chapter 14

1

-200.66

Species Name CO H2 CH3OH

2

0

 R  A T  

T0 

Updated 4/5/2017

1 D  T

1 T0

-3.45E-06

0.00E+00 -2.01E+02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

-3.45E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

B (1/K) 1.08E-02

C (1/K2) -3.45E-06

D (K2) -1.35E+04

T

1 R

T 

Cp

3

CO H  2

2

4 1   

1.22E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

C

2

o

1 

D  1

2 o

2

o

To

2 o

 T  C 2 D  1 1  o o Srxn T  To2       B T  To   ,T  S rxn ,T0  R  A ln   2 2  T 2 To2    To  

3

2

C (1/K2) 0.00E+00 0.00E+00

D (K2) iHf,i iAi iBi -3.10E+03 -8.94E+01 -3.38E+00 -5.57E-04 8.30E+03 0.00E+00 -6.50E+00 -8.44E-04

iCi 0.00E+00 0.00E+00

iDi 3.10E+03 -1.66E+04

-3.45E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

2

  3  2 

CH OH

2.21E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

 T dT   A ln  T   B T  T   2 T  T   2  T  T  

D (K ) -1.35E+04

2

T03

1.22E-02

C (1/K ) -3.45E-06

2

C 3 T 3

2.211

B (1/K) 1.08E-02

iDi 3.10E+03 -1.66E+04

A -7.663 Note: Light blue fields are inputs, pink fields are the final output.

2

T02

iCi 0.00E+00 0.00E+00

0.00E+00 -1.30E+02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

CO yCO H yH 

 B 2 T 2

D (K2) iHf,i iAi iBi -3.10E+03 1.11E+02 -3.38E+00 -5.57E-04 8.30E+03 0.00E+00 -6.50E+00 -8.44E-04

-3.45E-06

3

1 

1

3 0

C (1/K2) 0.00E+00 0.00E+00

1.22E-02

3

3

B (1/K) 5.57E-04 4.22E-04

2.211

CH OH yCH OH

C

2 0

0

o o  H rxn H rxn ,T ,T0

Entropy Entropy Change of Heat Capacity Change of Reaction at To Integral Reaction at T J/(mol K) (dimensionless) J/(mol K) -219.16 -2.284 -238.155 Heat Capacity Coefficients Standard Stoichiometric Entropy at T 0 coefficient (J/(mol K)) A B (1/K) -1 89.36 3.376 5.57E-04 -2 0.00 3.249 4.22E-04 -129.80

B

p

To

Temperature of Interest T (K) 530

1

1

 C dT  T  AT  T   2 T  T   3 T  T   D  T  T  

A 3.376 3.249

A -7.663 Note: Light blue fields are inputs, pink fields are the final output. Reference Temperature T 0 (K) 298.15

T

1 RT

 P   K o  P 

2.21E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1.22E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

2

2

p. 45 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T (K)

P (bar)

516.5

100

Chapter 14

Tc (K) Pc (bar)

132.9

34.99

0.048

Tr

Pr

B

0

B

1

3.8864

2.8580

0.0349

0.1384

1.0310

B 0 0.083

0.422

 TP 

B1 0.139

Tr1.6 0.172

 

  exp  B0   B1

r

r

Tr4.2

T (K)

P (bar)

516.5

100

Tc (K) Pc (bar)

33.19

-0.216

13.13

Tr

Pr

B

0

B

1

15.5619

7.6161

0.0778

0.1390

1.0236

B 0 0.083

0.422

 TP 

B1 0.139

Tr1.6 0.172

Updated 4/5/2017

 

  exp  B0   B1

r

r

Tr4.2

p. 46 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T (K) 516.5

Chapter 14

P (bar) 100

Tc (K) 512.6

Pc (bar) 80.97

Tr 1.00 1.00 1.01 1.01

Pr 1.20 1.50 1.20 1.50

 0.564

Tr 1.0076

Pr 1.2350

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values

0

1

0.5781 0.4875 0.5970 0.5047

0.9204 0.9078 0.9462 0.9333

0.5818

0.9385

Final Value

CH OH yCH OH 3

3

CO yCO H yH  2

2

2

0.5613

  3  2 

CH OH 3

CO H 

2

2

2

4 1   

3

2

 P    K  o   104 K 2 P  1.0402 1.0236  27 *0.5613

 

 CH OH yCH OH 3

3

CO yCO H yH  2

2

2

  3  2 

CH OH 3

CO H 

2

2

4 1   

3

2

2

 P    K  o   104 K 2 P  1.0417 1.0231 27 *0.6268

Updated 4/5/2017

p. 47 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 fˆCO   fˆ   y P   o 2   CaO   COo2  1 i o   N  ˆ  f f P  CO2   CaO  f P K    io      o   P  fˆCaCO  1 i 1  f i   o 3  fCaCO  3  

Updated 4/5/2017

p. 48 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Reference Temperature T 0 (K) 298.15

Species Name CaCO3 CO2 CaO

Temperature of Interest T (K) 1109

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T 0 kJ/mol (dimensionless) kJ/mol 178.321 -1.128 167.923 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -1 -1206.92 1 1

Chapter 14

-393.509 -635.09

Species Name CaCO3 CO2 CaO

C

2 0

3

1 

1

3 0

0

o o  H rxn H rxn ,T ,T0

C (1/K2)

 R  A T  

T0 

 B 2 T 2

T02

C 3 T 3

T03

2

1 D  T

1 T0

D (K ) iHf,i iAi iBi -3.12E+05 1.21E+03 -1.26E+01 -2.64E-03

iCi 0.00E+00

iDi 3.12E+05

0.00E+00 0.00E+00

-1.16E+05 -3.94E+02 -1.05E+05 -6.35E+02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

-1.16E+05 -1.05E+05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

C (1/K2) 0.00E+00

D (K2) 9.16E+04

5.457 6.104

1.05E-03 4.43E-04

B (1/K) -1.15E-03

T

1 R

T 

5.46E+00 6.10E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1.05E-03 4.43E-04 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1 

D  1

 T dT    A ln  T   B T  T   2 T  T   2  T  T   Cp

C

2

o

2 o

2

o

To

2 o

 T  C 2 D  1 1  o o S rxn T  To2   2  2     B T  To   ,T  S rxn,T0  R  A ln    T 2 2 T T  o o   

2

C (1/K ) 0.00E+00

2

iHf,i iAi iBi -3.12E+05 2.62E+02 -1.26E+01 -2.64E-03

iCi 0.00E+00

iDi 3.12E+05

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

-1.16E+05 -1.05E+05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

D (K )

5.457 6.104

1.05E-03 4.43E-04

0.00E+00 0.00E+00

-1.16E+05 2.85E+00 -1.05E+05 -1.04E+02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

A -1.011 Note: Light blue fields are inputs, pink fields are the final output.

B (1/K) -1.15E-03

C (1/K2) 0.00E+00

D (K2) 9.16E+04

Updated 4/5/2017

2.85 -104.18

2

0

0.00E+00

Entropy Entropy Change of Heat Capacity Change of Reaction at To Integral Reaction at T J/(mol K) (dimensionless) J/(mol K) 160.72 -1.782 145.910 Heat Capacity Coefficients Standard Stoichiometric Entropy at T 0 coefficient (J/(mol K)) A B (1/K) -1 -262.05 12.572 2.64E-03

B

p

To

B (1/K) 2.64E-03

Temperature of Interest T (K) 1109

1 1

1

 C dT  T  AT  T   2 T  T   3 T  T   D  T  T  

A 12.572

A -1.011 Note: Light blue fields are inputs, pink fields are the final output. Reference Temperature T 0 (K) 298.15

T

1 RT

5.46E+00 6.10E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1.05E-03 4.43E-04 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

p. 49 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 50 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 51 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 52 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 53 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 54 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 

 

 

  

  Updated 4/5/2017

 

  p. 55 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

 

 

 yNO   2

yO2 yN2

Chapter 14

 21  K  0.70  1   2  0.05  1  2 2  1 2

y    2  1     P  K    200 K P   0.70      0.05    2  y y  2

2

NO2

2

2

2

O2

2

1

N2

2

1

2

o

2

2

1  K1  0.70  1   2  0.05  1  2 2  / 4 2 

200 K 2  0.70  1   2  0.05  1  2 2  4 1   2 

 

2

  

Updated 4/5/2017

p. 56 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 57 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 58 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 59 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

 d  nG E   ln  1      dn1  RT   T , P ,n2

Chapter 14

 d  nG E   ln  2      dn2  RT   T , P ,n1

an n nG E  nax1 x2  1 2 RT n1  n2

 n  n  n2   n1 n2   d  an1 n2    ln  1    a 2 1   2   dn n  n n  n  1 2 2   T , P ,n  1 1   2 ln  1 

an22

 n1  n2 

 ax22  0.1x22

2

 n n  n   n n   d  an1 n2   ln  2    a 1 1 2 2 1 2       n1  n2   dn2  n1  n2   T , P ,n2   ln  2 

an12

 n1  n2 

 ax12  0.1x12

2

i

 fˆ  K    io    i 1  f i  N

N

  x   K i

i i

i 1

 B xB K  A xA 

Updated 4/5/2017

p. 60 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

xB xB   1.497 x A 1  xB xB 

1.497  0.600 2.497

 B xB exp  0.1xA  xB x x   exp 0.1 xA2  xB2  B  exp  0.1 xA  xB  xA  xB   B  K 2  A xA exp  0.1xB  xA xA xA 2

 B xB x  exp  0.11  2 xB   B  1.497  A xA 1  xB

Updated 4/5/2017

p. 61 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 62 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

 

       Updated 4/5/2017

p. 63 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

      

 

 

 

 

  

   

 

 

  

 

 

 H o To   T    90135  550   8 K1,1  exp   1     exp  1      5.878  10   RT  To    8.3145*550  298.15    Updated 4/5/2017

p. 64 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

T  1 T C po  C po K 2  exp    dT   dT   TT R  RT To o  

T 1 (K) 298.15

T 2 (K) 550 T2

ICPH 

R

Cp

A -7.663

dT  A  T2  T1  

T1

T 1 (K) 298.15

T 2 (K) 550 T2

ICPS 

T1

A -7.663

C (1/K2) D (K2) B (1/K) 1.08E-02 -3.45E-06 -1.35E+04

ICPH (K) -9.56E+02

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 

C (1/K2) D (K2) B (1/K) 1.08E-02 -3.45E-06 -1.35E+04

ICPS -2.391

T  C D 2 dT  A ln  2   B T2  T1   T22  T12  T2  T12 RT T 2 2  1

Cp

 956  K 2,1  exp   2.391  0.521  550  

 H o To   T    41166  550   K1,2  exp   1     exp  1      2006   RT T 8.3145*550 298.15    o    T  1 T C po  C po K 2  exp    dT   dT   TT R  RT To o  

Updated 4/5/2017

p. 65 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

T 1 (K) 298.15

T 2 (K) 550 T2

ICPH 

Chapter 14

R

Cp

A -1.95

dT  A  T2  T1  

T1

T 1 (K) 298.15

T 2 (K) 550 T2

ICPS 

T1

C (1/K2) D (K2) B (1/K) 5.40E-04 0.00E+00 1.16E+05

ICPH (K) -2.55E+02

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 

A -1.95

C (1/K2) D (K2) B (1/K) 5.40E-04 0.00E+00 1.16E+05

ICPS -0.596

T  C D 2 dT  A ln  2   B T2  T1   T22  T12  T2  T12 RT T 2 2  1

Cp

 255  K 2,2  exp   0.596   0.876  550   

1 100  21 

2

 75  21   2  15  1   2  2

 6.74

15  1   2   2  0.01703  75  21   2  5   2   

Updated 4/5/2017

p. 66 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

e1 ( 1002 e1 ) 2 eq1 := 6.74 ( 752 e1 e2 ) 2 ( 15e1 e2 ) ( 15e1 e2 ) e2 eq2 := .01703 ( 752 e1 e2 ) ( 5e2 ) { e2 -50.32722105, e1 -31.79049524 }, { e1 39.967893954.768453918 I, e2 .02238269229 .02815501014 I }, { e1 39.967893954.768453918 I, e2 .02238269229 .02815501014 I }, { e2 .8787346421, e1 11.85184823 }, { e1 43.71142266, e2 29.30072336 }, { e1 109.7205911, e2 98.65062054 }

e1 := 11.85184823 e2 := .8787346421

yH2 := .6608127335 yCO := .05277957418 yCH3OH := .1553397436 yCO2 := .05401657966 yH2O := .01151739470 yN2 := .06553397435

Updated 4/5/2017

p. 67 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 68 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 69 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 70 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 71 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 72 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 73 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 74 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 75 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 76 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 77 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 78 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 79 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 80 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 81 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 82 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 83 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 84 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 85 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 14

Updated 4/5/2017

p. 86 of 86

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

xi  i fi  xi  i fi 

xi  i  xi  i

    x exp   x   A  2  A  A  x    x exp   x   A  2  A  A  x   x1 exp  x2   A12  2  A21  A12  x1   x1 exp  x2   A12  2  A21  A12  x1  2

2

1

2

2

21

12

21

2

2

1

2

21

12

21

2

    0.9 exp   0.1  A  2  A  A  0.9    0.1exp   0.9   A  2  A  A  0.1  0.1exp  0.9   A12  2  A21  A12  0.1  0.9 exp  0.1  A12  2  A21  A12  0.9  2

2

2

2

21

Updated 4/5/2017

12

21

21

12

21

p. 1 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

xi  i fi  xi  i fi 

xi  i  xi  i

    x exp   x   A  2  A  A  x    x exp   x   A  2  A  A  x   x1 exp  x2   A12  2  A21  A12  x1   x1 exp  x2   A12  2  A21  A12  x1  2

2

1

2

2

21

12

21

2

2

1

2

21

12

21

2

    0.8exp   0.2   A  2  A  A  0.8    0.1exp   0.9   A  2  A  A  0.1  0.2 exp  0.8  A12  2  A21  A12  0.2   0.9 exp  0.1  A12  2  A21  A12  0.9  2

2

2

2

21

12

21

21

12

21

xi  i fi  xi  i fi 

xi  i  xi  i Updated 4/5/2017

p. 2 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

    x exp   x   A  2  A  A  x    x exp   x   A  2  A  A  x   x1 exp  x2   A12  2  A21  A12  x1   x1 exp  x2   A12  2  A21  A12  x1  2

2

1

2

2

21

12

21

2

2

1

2

21

12

21

2

    0.9 exp   0.1  A  2  A  A  0.9    0.2 exp   0.8   A  2  A  A  0.2   0.1exp  0.9   A12  2  A21  A12  0.1  0.8exp  0.2   A12  2  A21  A12  0.8  2

2

2

2

21

Updated 4/5/2017

12

21

21

12

21

p. 3 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 4 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 5 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 6 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 7 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 8 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

x1  1  x1  1 

GE  ax1 x2 RT

ln  2  ax12

ln  1  ax22 

2

 1  exp 0.8  x3   exp 0.8 1  x1 

2

2

2

 1  exp 0.4  x2   exp 0.4 1  x1  

x1 exp 0.4 1  x1   x1 exp 0.8 1  x1  2

2

 

 

x1 

n1 n1  n1  n2 n1  1 

n1 

x1 1  x1

n1  n1 

2 x1  1 3x1  2

Updated 4/5/2017

n1 n1  n1  n3 n1  1

n1 

x1 1  x1

x1 x1  1 1  x1 1  x1 

x1 

x1 

1  x1 

x1  1 3x1  2 p. 9 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

x1 exp 0.4 1  x1  

 2

  x  1 2  2 x1  1    exp  0.8  1   3x1  2   3x1  2  

eq := xa e

2 ( .4 ( 1 xa ) )



2    .8 ( xa 1 )     2   ( 3 xa 2 ) 

( 2 xa 1 ) e 3 xa 2

.3711378130 

Updated 4/5/2017

p. 10 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

⁡ ⁡

Updated 4/5/2017

p. 11 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 12 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

⁡ ⁡

Updated 4/5/2017

p. 13 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

⁡ ⁡

Updated 4/5/2017

p. 14 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Updated 4/5/2017

Chapter 15

p. 15 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 16 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 17 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

⁡ ⁡ ⁡

Updated 4/5/2017

p. 18 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 19 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

⁡ ⁡ ⁡

Updated 4/5/2017

p. 20 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 21 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 22 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 23 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 24 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 25 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 26 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 27 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 28 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 29 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 30 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


π

SVNAS 8th Edition Annotated Solutions

Chapter 15

400 350 300 250 200 150 100 50 0 0

1

2

3

4

5

n

Updated 4/5/2017

p. 31 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

2.4 2.2

2 z

1.8 1.6 1.4 1.2 1

0

Updated 4/5/2017

1

2

n

3

4

5

p. 32 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 33 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 34 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 35 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 36 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 37 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 38 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 39 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 40 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 41 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 42 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 43 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 15

Updated 4/5/2017

p. 44 of 44

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 1 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 2 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 3 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 4 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 5 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 6 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 7 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 8 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 9 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

𝑚̇3 = 𝑚̇1 + 𝑚̇2 (A)

𝐻3 𝑚̇3 − 𝐻1 𝑚̇1 − 𝐻2 𝑚̇2 = 0 𝑘𝐽 ∙ 𝑠 −1 (B)

𝑆3 𝑚̇3 − 𝑆1 𝑚̇1 − 𝑆2 𝑚̇2 = 0 𝑘𝐽 ∙ 𝑠 −1 ∙ 𝐾 −1 (C)

(𝐻3 − 𝐻𝑙𝑖𝑞 )𝑚̇3 = 300 𝑘𝐽 ∙ 𝑠 −1 (D)

𝑆3 = 𝑆𝑙𝑖𝑞 +

𝐻3 − 𝐻𝑙𝑖𝑞 𝑇𝑠𝑎𝑡

(𝐸)

Defining 𝑥1 = 𝑚̇1 ⁄𝑚̇3 , equations B and C can be re-written:

𝐻3 = 𝐻2 + 𝑥1 (𝐻1 − 𝐻2 ) Updated 4/5/2017

(𝐹)

p. 10 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

𝑆3 = 𝑆2 + 𝑥1 (𝑆1 − 𝑆2 )

Chapter 16

(𝐺)

Rearranging equation E yields: 𝐻3 = 𝐻𝑙𝑖𝑞 + 𝑇𝑠𝑎𝑡 (𝑆3 − 𝑆𝑙𝑖𝑞 )

(𝐻)

Substituting Equation G into H and equation to F yields:

𝐻3 = 𝐻2 + 𝑥1 (𝐻1 − 𝐻2 ) = 𝐻𝑙𝑖𝑞 + 𝑇𝑠𝑎𝑡 [𝑆2 + 𝑥1 (𝑆1 − 𝑆2 ) − 𝑆𝑙𝑖𝑞 ]

𝑥1 =

(𝐼)

(𝐻𝑙𝑖𝑞 − 𝐻2 ) − 𝑇𝑠𝑎𝑡 (𝑆𝑙𝑖𝑞 − 𝑆2 ) = 0.574 (𝐻1 − 𝐻2 ) − 𝑇𝑠𝑎𝑡 (𝑆1 − 𝑆2 )

𝐻3 = 𝐻2 + 𝑥1 (𝐻1 − 𝐻2 ) = 2767.2 𝑘𝐽 ∙ 𝑘𝑔−1

𝑆3 = 𝑆2 + 𝑥1 (𝑆1 − 𝑆2 ) = 6.563 𝑘𝐽 ∙ 𝑘𝑔−1 ∙ 𝐾 −1

𝑚̇3 =

300 𝑘𝐽 ∙ 𝑠 −1 (𝐻3 − 𝐻𝑙𝑖𝑞 )

=

300 𝑘𝐽 ∙ 𝑠 −1 = 0.15 𝑘𝑔 ∙ 𝑠 −1 (2767.2 − 762.6)𝑘𝐽 ∙ 𝑘𝑔−1

𝑚̇1 = 𝑥1 𝑚̇3 = 0.086 𝑘𝑔 ∙ 𝑠 −1 and 𝑚̇2 = 𝑚̇3 − 𝑚̇1 = 0.064 𝑘𝑔 ∙ 𝑠 −1

The steam at Point 3 is wet.

Updated 4/5/2017

p. 11 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 12 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 13 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 14 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 15 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 16 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 17 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 18 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 19 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 20 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 21 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 22 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 23 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 24 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 25 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 26 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 27 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


SVNAS 8th Edition Annotated Solutions

Chapter 16

Updated 4/5/2017

p. 28 of 28

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.


Turn static files into dynamic content formats.

Create a flipbook
Issuu converts static files into: digital portfolios, online yearbooks, online catalogs, digital photo albums and more. Sign up and create your flipbook.