Introduction to Chemical Engineering Thermodynamics 8th Edition Smith Solutions Manual
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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 =
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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
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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)
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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
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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
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Chapter 1
m q 1.25 kg m3 *55880 m3 s1 = 69850 kg s1
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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 / KC
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℃
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Chapter 2
(b)
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Chapter 2
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Chapter 2
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Chapter 2
m'
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Chapter 2
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Chapter 2
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Chapter 2
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Chapter 2
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Chapter 2
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Chapter 2
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Chapter 2
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Chapter 2
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Chapter 2
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SVNAS 8th Edition Annotated Solutions
Chapter 2
∆𝑇
℃
℃
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Chapter 2
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Chapter 2
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NAS 8th Edition Annotated Solutions
Chapter 3
Updated 5/23/2017
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NAS 8th Edition Annotated Solutions
Chapter 3
◦
Updated 5/23/2017
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Chapter 3
k 44.18*106
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Chapter 3
1 V V P T
AP V V0 1 BP 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 BP
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
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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 3.9 106 0.1 109 P dP V V ln 3.9 106 Po P 5 1011 P 2 Po2 Vo
V Vo exp 3.9 106 Po P 5 1011 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 3.9 106 Po P 5 1011 P 2 Po2 6
9
6
o
Updated 5/23/2017
11
o
2
2 o
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Chapter 3
dW 3.9 10 P 0.1 10 P exp 3.9 10 P 5 10
P dP
dW 3.9 106 0.1 109 P P exp 3.9 106 1 P 5 1011 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 1.2 103 K -1 dT V
V ln 2 T2 T1 1.2 103 K -1 T2 T1 V1 V ln 2 1.2 103 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
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Chapter 3
dV dT dP V
V ln T P Vo
V ln (104 K 1 ) *(5 K) (4.8 105 bar -1 ) * 20 bar 5 104 9.6 104 0.0015 Vo
Updated 5/23/2017
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Chapter 3
𝜕 𝜕
Updated 5/23/2017
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Chapter 3
°
Updated 5/23/2017
°
°
1 V V P T
1 dV dP V
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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 1V1 P 1 V PV 1 1 PV
1
Updated 5/23/2017
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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
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Updated 5/23/2017
Chapter 3
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Chapter 3
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Chapter 3
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Chapter 3
T = f(P) ?
dn n dt
dU t nH dt
Updated 5/23/2017
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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
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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
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Chapter 3
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Chapter 3
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Chapter 3
Updated 5/23/2017
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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
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Chapter 3
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Chapter 3
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Chapter 3
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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·mol1·K 1 × (335.947 K - 298.15 K) = 471.366 J·mol1
H = U + W = 471.366 J·mol1 + 471.366 J·mol1 = 942.732 J·mol1
U = CV T = 1.5 × 8.314 J·mol1·K 1 × (573.15 K - 335.947 K) = 2958.16 J·mol1
H = CP T = 2.5 × 8.314 J·mol1·K1 × (573.15 K - 335.947 K) = 4930.26 J·mol1
W = -RT = -8.314 J·mol1·K1 × (573.15 K - 335.947 K) = -1972.11 J·mol1
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·mol1·K 1 × (335.947 K - 298.15 K) = 471.366 J·mol1 Updated 5/23/2017
p. 36 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
H = U + W = 471.366 J·mol1 + 471.366 J·mol1 = 942.732 J·mol1
Q = U = CV T = 1.5 × 8.314 J·mol1·K 1 × (573.15 K - 335.947 K) = 2958.16 J·mol1 H = CP T = 2.5 × 8.314 J·mol1·K1 × (573.15 K - 335.947 K) = 4930.26 J·mol1
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·mol1·K 1 × (335.947 K - 298.15 K) = 471.366 J·mol1
H = U + W = 471.366 J·mol1 + 471.366 J·mol1 = 942.732 J·mol1
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 d2
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 1.05560.5 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
4860171V 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 15.005V 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 1.0580.5 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 1.09190.5 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 22.71V 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 0.66370.5 *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 1.2820.5 *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 BP CP2 DP3
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 1T V sat Vc Zc r
10.8468 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 1T V sat Vc Zc r
10.8739 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 1T V sat Vc Zc r
V
sat
10.9009 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 1T V sat Vc Zc r
10.9279 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 1T V sat Vc Zc r
10.8778 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 1T V sat Vc Zc r
10.9013 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 1T V sat Vc Zc r
10.9248 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 1T V sat Vc Zc r
10.9484 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 1T V sat Vc Zc r
10.8899 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 1T V sat Vc Zc r
10.9144 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 1T V sat Vc Zc r
10.9389 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 1T V sat Vc Zc r
V
sat
10.9632 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 1T V sat Vc Zc r
10.7985 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 1T V sat Vc Zc r
10.8225 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 1T V sat Vc Zc r
10.8465 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 1T V sat Vc Zc r
10.8704 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 1T V sat Vc Zc r
10.8198 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 1T V sat Vc Zc r
10.8430 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 551101V 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
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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
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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
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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 707471V 72.43 3 -1 V 1533.0 cm mol V 30.00 V 174.86
Updated 5/23/2017
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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 1T V sat Vc Zc r 200.0*0.276(1320 / 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
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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·mol1
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·mol1·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·mol1
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·mol1·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·mol1
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·mol1·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·mol1
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·mol1·K 1 × 293 K
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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·mol1
V = 2.4 L = 2400 cm3
Using this,
mass = MWAsH3 ×
Updated 5/23/2017
PV 25 bar × 2400 cm3 = 77.945 g·mol1 × = 260.54 g ZRT 0.7368 × 83.14 bar·cm3·mol1·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·mol1
V = 2.4 L = 2400 cm3
Using this,
mass = MWNF3 ×
Updated 5/23/2017
PV 25 bar × 2400 cm3 = 71 g·mol1 × = 192.33 g ZRT 0.9092 × 83.14 bar·cm3·mol1·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
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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
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Chapter 3
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Chapter 3
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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
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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
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NAS 8th Edition Annotated Solutions
Chapter 3
Z Z0 Z 1 = 0.322
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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
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NAS 8th Edition Annotated Solutions
Chapter 3
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Chapter 3
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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
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Chapter 3
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Chapter 3
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Chapter 3
p. 168 of 182
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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
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Chapter 3
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Chapter 3
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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
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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
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NAS 8th Edition Annotated Solutions
=
1 + 102 + 4 3
Chapter 3
(2)
(1 - 2) 1 + 2 2 + 1 + = Z c3 1 + 102 + 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
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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
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Chapter 3
ˆ 0.339 0* 0.033 0.339 B
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Chapter 3
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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
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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.
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NAS 8th Edition Annotated Solutions
Chapter 3
dV dT dP V
V ln T P Vo
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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 1100C, 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 = 1.95106 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 200C (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 122F 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 122F 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 = 0.57510-3 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 2.88 104 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 = 9.43109 J = 9.43106 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, 25C 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 25C, 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 -16C, we see that the vapor volume is 0.12551 m3kg-1 and the liquid volume is 0.000743 m3kgH 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.K1 210.4kPa.m3 .kg1 210.4kJ.kg1
(b) At 0C, 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 .kg1 198.59kJ.kg1
(c) At 12C, 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 .kg1 190.22kJ.kg1
(d) At 26C, 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 .kg1 176.91kJ.kg1
(e) At 40C, 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 .kg1 163.17kJ.kg1
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
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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
1T 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
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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 212.32 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
1T 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
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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 217.652 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
1T 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
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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
1T 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:
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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)
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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 AT 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·mol1·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 AT 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:
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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 25C. 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 25C. 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
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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
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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 25C (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 = -1.323106 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 = 1.32310 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 = -1.32310 ) 6
and then heat the products with both the heat removed to cool the air to 298K and the heat of reaction 6 (Htot = 1.66310 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
niHf,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 650C 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·mol1 ) + (-31.9 J·mol1 ) + 0 J·mol1 - (-288.3 J·mol1 ) + 0 J·mol1 = 44.2 J·mol1 H = 0.217 mol × H f 298 = 0.217 mol × 44.2 J·mol1 = 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·mol1 ) + 2(-285830 J·mol 1 ) + (-31.9 J·mol1 ) + 0 J·mol1 - -288.3 J·mol 1 + 0 J·mol1 H = -0.217 mol × H f 298 = 0.217 mol × (-1358421.6 J·mol1 ) = -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 25C 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 25C, 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 60F 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
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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
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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 320C to 25C, 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 25C) 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
DT1 T1 2
1
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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 -8105 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 30C to 25C, then react it at 25C, 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
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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
DT1 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) +2Hf(H2O) --- Hf(CH4)) + 0.25*(2Hf(CO2) +3Hf(H2O) --- Hf(C2H6)) Hrxn =1.25*Hf(CO2) +2.25Hf(H2O) ---0.75Hf(CH4)) --- 0.25Hf(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
DT1 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 = 270C 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 400C to 500C and then compute the heat of reaction at 500C, or we could compute the heat of reaction at 400 C and then compute the heat required to heat the products from 400C to 500C. 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.
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SVNAS 8th Edition Annotated Solutions
Chapter 4
Using the handy spreadsheet from the notes, we can compute the heat of reaction at 400C 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) niHf,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 400C 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 400C to 500C. 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 500C. 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 125C to the final temperature of 425C (698.15 K), and then compute the heat of reaction at 425C. 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 AT 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:
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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) niHf,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
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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
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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)
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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:
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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 25C 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 AT 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 25C 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
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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 25C 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 25C 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 25C, the final temperature for a heat input of 30 kJ/mol is 835 K or 562C:
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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 25C, the final temperature for a heat input of 30 kJ/mol is 964 K or 691C: 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 25C, 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
DT1 T1 2
T1
1
(d) For CO2 starting at 25C, 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
DT1 T1 2
1
(e) For carbon monoxide starting at 25C, the final temperature for a heat input of 30 kJ/mol is 1248 K or 975C:
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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 25C, the final temperature for a heat input of 30 kJ/mol is 664 K or 391C: 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 25C, the final temperature for a heat input of 30 kJ/mol is 1298 K or 1025C: 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 AT 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 25C, the final temperature for a heat input of 30 kJ/mol is 1277 K or 1004C: 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
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Chapter 4
H liquid = R × ICPH = 8.314 J·mol1·K 1 × 683.3399 K = 5681.288 J·mol1 H vapor = R × ICPH = 8.314 J·mol1·K 1 × 0 K = 0 J·mol1 H boiling = H latent × mH = 2257 J·g 1 × 18 g·mol1 = 40626 J·mol1 H = 46307.29 J·mol1
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·mol1·K 1 × 683.3399 K = 5681.288 J·mol1 H vapor = R × ICPH = 8.314 J·mol1·K 1 × 0 K = 0 J·mol1 H boiling = H latent × mH = 2257 J·g 1 × 18 g·mol1 = 40626 J·mol1 H = 46307.29 J·mol1
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 2.2210 g/day = 2.2210 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.
⋅
⋅
⋅
⋅
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SVNAS 8th Edition Annotated Solutions
Chapter 4
⋅
⋅
⋅
⋅ ⋅
⋅ ⋅
⋅
⋅
⋅ ⋅
⋅
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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
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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
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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
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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
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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
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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.
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(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:
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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:
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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
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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
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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.
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For adiabatic steps 1 to 2 and 3 to 4:
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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
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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
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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
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Chapter 5
Moreover
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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 T02 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
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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 AT 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 T12 RT T 2 2 1
Cp
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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
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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 T12 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
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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:
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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:
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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:
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|W|engine = |Q| − |Qσ1| |W|refrig = |Qσ2| − |Q’| Equation (5.7) may be applied to both the engine and the refrigerator:
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Q Sg T
Sg
Chapter 5
135 J s1 0.45 J K -1 s-1 300 K
Q Sg T
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Sg
Chapter 5
2500 J s1 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)
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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
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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
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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
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Qj Tj
Sg p. 29 of 39
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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
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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
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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
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SVNAS 8th Edition Annotated Solutions
Sg
Chapter 5
QH QC 1 0.55 6.7 105 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
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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
DT1 T1 2
T1
DH (J/mol) 204.12
1
DH (J/mol) -1218.58
DT1 T1 2
1
p. 34 of 39
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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
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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
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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
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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
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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
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=
−
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G V RT P RT T
G RT T RT P H
2
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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
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dS V dP T dH 1 T V dP T
S V P H 1 T V P
S 2.095 103 K -1 1.551 103 m3 kg -1 819000 Pa = -2.66 Pa m3 kg-1 K -1 = -2.66 J kg-1 K -1
H 1 2.095 103 K -1 270 K 1.551 103 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
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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
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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 45 106 bar -1 1499 bar 937.6 cm3 kg-1 9.376 104 m3 kg-1
P2
W Vave PdP
Vave 2 P2 P12 2
P1
S Vave P H 1 T Vave P
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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 45 106 bar -1 1499 bar 937.5741 cm3 kg-1 9.375741 104 m3 kg-1 6
1
3
-1
2
2
2
3
-1
1
S 250 106 K-1 *970.3 cm3 kg 1 *1499 bar 363.6 bar cm3 kg1 K1 36.4 J kg1 K1
H 1 250 106 K-1 *298.15 K *970.3 cm3 kg 1 *1499 bar 1.346 106 bar cm3 kg 1 134.6 kJ kg 1
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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
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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
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S R H R GR R RT RT q q SR Z 1 Z 1 ln Z R Z Z SR ln Z R
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⋅
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θ≡
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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
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−
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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
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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.
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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
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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
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⋅
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H T V
dP sat -1 818.1* 2.996 *1.888 4628 ft 3psia lb-1 m 856.5 Btu lb m dT
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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
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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
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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
DT1 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
DT1 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 AT 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
DT1 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
DT1 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
DT1 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
DT1 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 mol1 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 mol1 K 1
dT 17.08 J mol1 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 mol1 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.
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⋅
−
⋅
56.0056 59.04121 61.1392 62.579174 63.39535 63.32491 62.15734
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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
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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)
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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.
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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
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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
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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
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=
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≡
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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).
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−
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ω
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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
≡
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∙
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SVNAS 8th Edition Annotated Solutions
Chapter 7
Updated 4/5/2017
p. 1 of 73
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Chapter 7
Ϻ
Ϻ
Ϻ Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 7
∞
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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
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Chapter 7
=
and
or
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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
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Chapter 7
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Chapter 7
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Chapter 7
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Chapter 7
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Chapter 7
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Chapter 7
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Chapter 7
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Chapter 7
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Chapter 7
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Chapter 7
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Chapter 7
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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 AT 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
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Chapter 7
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Chapter 7
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Chapter 7
Δ
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Chapter 7
Δ
Updated 4/5/2017
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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
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Chapter 7
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Chapter 7
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=
−
Chapter 7
=
+
=
+
−
=
Updated 4/5/2017
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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
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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
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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
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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
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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 AT 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
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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 AT 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 AT 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
257.2 106 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* 1 257.2 106 *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* 1 257.2 106 * 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 696.2 106 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* 1 696.2 106 *363.15 *4800 4.976 kJ kg 1
H C p T V 1 T P 7.108 T
7.108 0.001036* 1 696.2 106 *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
523.1 106 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 * 1 523.1 106 *333.15 * 4980 5.065 kJ kg 1
H C p T V 1 T P 6.753 T
6.753 0.001017 * 1 523.1 106 * 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
217.3 106 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* 1 217.3 106 *294.26 *1925 1.927 kJ kg 1
H C p T V 1 T P 2.753 T
2.753 0.001002* 1 217.3 106 *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 * 1 714.3 106 *366.5 *10239 10.64 kJ kg 1
H C p T V 1 T P 14.19 T
14.19 0.001039 * 1 714.3 106 * 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
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Chapter 8
Updated 4/5/2017
p. 14 of 41
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Chapter 8
Updated 4/5/2017
p. 15 of 41
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Chapter 8
Updated 4/5/2017
p. 16 of 41
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Chapter 8
Updated 4/5/2017
p. 17 of 41
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Chapter 8
Updated 4/5/2017
p. 18 of 41
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Chapter 8
Updated 4/5/2017
p. 19 of 41
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Chapter 8
Updated 4/5/2017
p. 20 of 41
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Chapter 8
Updated 4/5/2017
p. 21 of 41
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Chapter 8
Updated 4/5/2017
p. 22 of 41
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Chapter 8
Updated 4/5/2017
p. 23 of 41
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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·mol1 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·mol1
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·mol1·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·mol1 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·mol1 = -18.378 kJ·mol1
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
|
|
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Chapter 9
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SVNAS 8th Edition Annotated Solutions
Updated 4/5/2017
Chapter 9
⋅
−
⋅
−
p. 3 of 18
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SVNAS 8th Edition Annotated Solutions
Chapter 9
ω
> ω < ωσ
Updated 4/5/2017
σ
< σ
ω
p. 4 of 18
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Chapter 9
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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
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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
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Chapter 9
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Chapter 9
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Chapter 9
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Chapter 9
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SVNAS 8th Edition Annotated Solutions
Chapter 9
.
Updated 4/5/2017
p. 12 of 18
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Chapter 9
Updated 4/5/2017
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Chapter 9
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Chapter 9
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Chapter 9
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SVNAS 8th Edition Annotated Solutions
Chapter 9
Δ
Updated 4/5/2017
p. 17 of 18
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SVNAS 8th Edition Annotated Solutions
Chapter 9
so:
Δ
Updated 4/5/2017
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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
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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
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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
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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 AT 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
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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
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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
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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
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SVNAS 8th Edition Annotated Solutions
Chapter 10
nM Mi ni T , P ,n ji
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
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Chapter 10
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Chapter 10
and
nV Vi ni T , P ,n j i Updated 4/5/2017
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Chapter 10
V1 V1 V1 V1 V1
V2 V2 V2 V2 V2
x1V1 x2V2
Updated 4/5/2017
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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
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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 ji
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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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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 yii P
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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 yethyleneethylene 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
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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 y2212 y3213 y2 y3 12 13 23 RT P ln ˆ2 B22 y32 23 y1212 y3 y1 23 12 13 RT P ln ˆ3 B33 y1213 y22 23 y1 y2 13 23 12 RT ln ˆ1
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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 y2212 y3213 y2 y3 12 13 23 RT P ln ˆ2 B22 y32 23 y1212 y3 y1 23 12 13 RT P ln ˆ3 B33 y1213 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 yii 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
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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
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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
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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
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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
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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
≡
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Chapter 10
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has the same sign over the whole composition range, both
Updated 4/5/2017
Chapter 10
and
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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
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Chapter 10
)
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. 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
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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
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Chapter 11
nV Vi ni T , P ,n ji 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
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Chapter 11
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Chapter 11
⋅
Updated 4/5/2017
−
⋅
⋅
⋅
⋅
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Chapter 11
H
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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
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Chapter 11
n 10 Updated 4/5/2017
Hf (kJ) -862.74
ΔHf HfCaCl2 -66.94 p. 11 of 28
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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
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Chapter 11
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Chapter 11
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Chapter 11
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Chapter 11
M: Differentiate:
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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
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Chapter 11
θ θ
−
Updated 4/5/2017
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Chapter 11
3600 3100
r(x1) (kg/h)
2600 2100 1600 1100 600 0
1
2
3
4
5
6
Θ (x1)
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Chapter 12
As the system is closed, extensive coordinates remain unchanged. The intensive variables will change value as temperature changes.
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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1.0000
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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
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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)
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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
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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
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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
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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
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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
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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 1V 2 K 2 1 K1 1V 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
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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
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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
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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
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SVNAS 8th Edition Annotated Solutions
Chapter 13
=
H H
Updated 4/5/2017
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Chapter 13
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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
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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
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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
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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
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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
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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
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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
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Chapter 13
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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
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SVNAS 8th Edition Annotated Solutions
Chapter 13
Using the equations
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 13
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SVNAS 8th Edition Annotated Solutions
Chapter 13
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SVNAS 8th Edition Annotated Solutions
Component 1 Benzene 2-butanol Acentonitrile
Updated 4/5/2017
Chapter 13
𝑃𝑠𝑎𝑡1 (𝑇) = 39.591𝑘𝑃𝐴𝑇 = 52.321𝐶 Component 2 T (deg C) P (kPA) Cyclohexane 52.3 39.6 water 87.7 64.2 Ethanol 65.8 60.6
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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
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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
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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
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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
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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
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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
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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
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Chapter 13
Updated 4/5/2017
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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
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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
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SVNAS 8th Edition Annotated Solutions
Chapter 13
Updated 4/5/2017
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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
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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
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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
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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
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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
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SVNAS 8th Edition Annotated Solutions
Chapter 13
Updated 4/5/2017
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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SVNAS 8th Edition Annotated Solutions
Chapter 13
γ γ
γ −
γ
( ‘)
/
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SVNAS 8th Edition Annotated Solutions
=
Chapter 13
γ
=
γ
> =∞
=
/
> γ
=
=
/ =
Updated 4/5/2017
γ
=
γ < /
= =
γ
/
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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
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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
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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
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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
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Chapter 13
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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
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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
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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
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Chapter 13
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Chapter 13
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Chapter 13
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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
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Chapter 13
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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)
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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
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Wislon Parameters (cal/mol) 1 2 0 -161.88 583.11 0 1448.01 469.55
Chapter 13
3 291.27 107.38 0
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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
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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
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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
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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
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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
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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
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2
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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 =
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=
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Chapter 13
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Chapter 13
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Chapter 13
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Chapter 13
=
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=
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Ʌ
Chapter 13
Ʌ
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Ʌ Ʌ
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Ʌ Ʌ
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Chapter 13
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Chapter 13
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γ
Chapter 13
γ
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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
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P
y1
1 x1 P1sat 1
Chapter 13
2 x2 P2sat 2
1 x1 P1sat 1 P
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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
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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
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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
Tr0.5 R 2Tc2 0.61430.5 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
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Zv 1
Chapter 13
q Zv Zv Zv
Zl Zl Zl
1 Zl q
Z ln Z 1 ln Z q ln Z
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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
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P
Chapter 13
a T RT V b V V b
0.6282 Tr0.5 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
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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
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Chapter 13
0.6415 Tr0.5 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
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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
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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
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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
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P
Chapter 13
a T RT V b V V b
T 0.5 R 2Tc2 0.68790.5 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
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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
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Chapter 13
T 0.5 R 2Tc2 0.61250.5 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
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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
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Chapter 13
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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
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Chapter 13
=
<
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Chapter 13
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Chapter 13
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Chapter 13
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SVNAS 8th Edition Annotated Solutions
Chapter 13
Ʌ
Ʌ
Ʌ
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Ʌ
Ʌ Ʌ
Ʌ p. 156 of 158
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Chapter 13
Ʌ
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Ʌ
Ʌ Ʌ
Ʌ
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Chapter 13
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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
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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
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Chapter 14
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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
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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
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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
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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
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SVNAS 8th Edition Annotated Solutions
Chapter 14
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SVNAS 8th Edition Annotated Solutions
Chapter 14
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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
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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
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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
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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
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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
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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
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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
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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 T12 RT T 2 2 1
Cp
104 K 2 exp 0.176 1.066 923
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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
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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 ) iHf,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 ) iHf,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
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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
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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 3.7011.5 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
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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
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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 AT 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) iHf,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
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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 ) iHf,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
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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
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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) iHf,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 AT 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) iHf,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
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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
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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
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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
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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
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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 3.901 108 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
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SVNAS 8th Edition Annotated Solutions
Chapter 14
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 14
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 14
Updated 4/5/2017
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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
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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
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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 6.365 106 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
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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 6.365 106 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 ) iHf,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 ) iHf,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
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SVNAS 8th Edition Annotated Solutions
Chapter 14
46110 46110 6 K K0 K1K 2 6.365 106 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 6.365 106 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
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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
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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
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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
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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
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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) iHf,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) iHf,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 AT 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
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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) iHf,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) iHf,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 AT 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
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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
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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
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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
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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 ) iHf,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
iHf,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 AT 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
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Chapter 14
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Chapter 14
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Chapter 14
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Chapter 14
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Chapter 14
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Chapter 14
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SVNAS 8th Edition Annotated Solutions
yNO 2
yO2 yN2
Chapter 14
21 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
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Chapter 14
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Chapter 14
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Chapter 14
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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
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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 0.11 2 xB B 1.497 A xA 1 xB
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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
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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 T12 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
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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 T12 RT T 2 2 1
Cp
255 K 2,2 exp 0.596 0.876 550
1 100 21
2
75 21 2 15 1 2 2
6.74
15 1 2 2 0.01703 75 21 2 5 2
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SVNAS 8th Edition Annotated Solutions
Chapter 14
e1 ( 1002 e1 ) 2 eq1 := 6.74 ( 752 e1 e2 ) 2 ( 15e1 e2 ) ( 15e1 e2 ) e2 eq2 := .01703 ( 752 e1 e2 ) ( 5e2 ) { e2 -50.32722105, e1 -31.79049524 }, { e1 39.967893954.768453918 I, e2 .02238269229 .02815501014 I }, { e1 39.967893954.768453918 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
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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
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12
21
21
12
21
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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
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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
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12
21
21
12
21
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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
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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
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π
SVNAS 8th Edition Annotated Solutions
Chapter 15
400 350 300 250 200 150 100 50 0 0
1
2
3
4
5
n
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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
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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)
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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.
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