Chapter4 A
Chapter4 A
Chapter4 A
T := 1373.15 K
B := 0.801 10
n := 10 mol
5
C := 0.0
D := 1.015 10
H := R ICPH ( T0 , T , A , B , C , D)
H = 47.007
kJ
mol
Q := n H
Q = 470.073 kJ
(b) T0 := 523.15 K
T := 1473.15 K
Ans.
n := 12 mol
3
B := 28.785 10
C := 8.824 10
D := 0
H := R ICPH ( T0 , T , A , B , C , 0.0)
H = 161.834
kJ
mol
Q := n H
3
Q = 1.942 10 kJ Ans.
n := 10 mol
Q := 800 kJ
3
For ethylene:
:= 2 (guess)
A := 1.424
(b) T0 := 533.15 K
C :=
= 2.905
T := T0
n := 15 mol
A := 1.967
B
2 2
C 3 3
T0 1 + T0 1
2
3
T = 1374.5 K
31.630 10
B :=
K
76
Ans.
Q := 2500 kJ
3
For 1-butene:
4.392 10
Given
Q = n R A T0 ( 1) +
:= Find ( )
B :=
14.394 10
C :=
9.873 10
K
:= 3
(guess)
Given
Q = n R A T0 ( 1) +
:= Find ( )
B
2 2
C 3 3
T0 1 + T0 1
2
3
= 2.652
T := T0
T = 1413.8 K
Ans.
6
n := 40 lbmol
Q := 10 BTU
T0 := 533.15K
n = 1.814 10 mol
Q = 1.055 10 kJ
3
For ethylene:
A := 1.424
:= 2 (guess)
C :=
4.392 10
K
Given
Q = n R A T0 ( 1) +
:= Find ( )
B :=
14.394 10
B
2 2
C 3 3
T0 1 + T0 1
2
3
= 2.256
T := T0
T = 1202.8 K
Ans.
T = 1705.4degF
4.3 Assume air at the given conditions an ideal gas. Basis of calculation is 1 second.
P := 1 atm
T0 := 122 degF
V := 250 ft
T := 932 degF
3
V = 7.079 m
T0 := ( T0 32degF) + 273.15K
T = 773.15 K
T0 = 323.15 K
n :=
P V
R T0
For air:
n = 266.985 mol
A := 3.355
B := 0.575 10
H := R ICPH ( T0 , T , A , B , C , D)
77
C := 0.0
D := 0.016 10
H = 13.707
kJ
mol
Q := n H
3
Q = 3.469 10 BTU Ans.
gm
mol
T0 := 323.15 K
10000 kg
molwt
T := 1153.15 K
n = 9.99 10 mol
3
B := 2.637 10
C := 0.0
D := 3.120 10
H := R ICPH ( T0 , T , A , B , C , D)
H = 9.441 10
4 J
Q := n H
mol
Q = 9.4315 10 kJ
Ans.
4.7 Let step 12 represent the initial reversible adiabatic expansion, and step 23
the final constant-volume heating.
T1 := 298.15 K
T3 := 298.15 K
P1 := 121.3 kPa
P2 := 101.3 kPa
P3 := 104.0 kPa
T2 := T3
CP := 30
Given
J
mol K
P2
T2 = T1
P1
CP := Find ( CP)
Pc := 47.01bar
kJ
mol
P3
T2 = 290.41 K
(guess)
R
CP
P2
Trn :=
Tn
Tc
CP = 56.95
J
Ans.
mol K
Tn := 329.4K
Trn = 0.648
Pc
1.013
bar
1.092 ln
Hncalc := R Tn
0.930 Trn
78
Hncalc = 30.108
kJ
mol
Ans.
To compare with the value listed in Table B.2, calculate the % error.
%error :=
Hncalc Hn
%error = 3.464 %
Hn
Values for other components in Table B.2 are given below. Except for
acetic acid, acetonitrile. methanol and nitromethane, agreement is within
5% of the reported value.
Acetone
Acetic Acid
Acetonitrile
Benzene
iso-Butane
n-Butane
1-Butanol
Carbon tetrachloride
Chlorobenzene
Chloroform
Cyclohexane
Cyclopentane
n-Decane
Dichloromethane
Diethyl ether
Ethanol
Ethylbenzene
Ethylene glycol
n-Heptane
n-Hexane
Methanol
Methyl acetate
Methyl ethyl ketone
Nitromethane
n-Nonane
iso-Octane
n-Octane
n-Pentane
Phenol
1-Propanol
2-Propanol
Toluene
Water
o-Xylene
m-Xylene
p-Xylene
Hn (kJ/mol)
30.1
40.1
33.0
30.6
21.1
22.5
41.7
29.6
35.5
29.6
29.7
27.2
40.1
27.8
26.6
40.2
35.8
51.5
32.0
29.0
38.3
30.6
32.0
36.3
37.2
30.7
34.8
25.9
46.6
41.1
39.8
33.4
42.0
36.9
36.5
36.3
79
% error
3.4%
69.4%
9.3%
-0.5%
-0.7%
0.3%
-3.6%
-0.8%
0.8%
1.1%
-0.9%
-0.2%
3.6%
-1.0%
0.3%
4.3%
0.7%
1.5%
0.7%
0.5%
8.7%
1.1%
2.3%
6.7%
0.8%
-0.2%
1.2%
0.3%
1.0%
-0.9%
-0.1%
0.8%
3.3%
1.9%
2.3%
1.6%
b)
469.7
507.6
Tc :=
K
562.2
560.4
36.0
68.7
Tn :=
+ 273.15 K
80.0
80.7
Tr1 :=
Tn
Tc
Tr2 :=
33.70
30.25
Pc :=
bar
48.98
43.50
366.3
366.1 J
H25 :=
433.3 gm
392.5
( 25 + 273.15)K
Tc
0.658
0.673
Tr1 =
0.628
0.631
H2 := H25 M
H1 := Hn
26.429
31.549 kJ
H2 =
33.847 mol
32.242
0.38
1 Tr2
H2calc := H1
1 Tr1
H2calc
25.79
28.85 kJ
Hn :=
30.72 mol
29.97
72.150
86.177 gm
M :=
78.114 mol
82.145
H2calc H2
H2
26.448
26.429
0.072
31.533 kJ
31.549 kJ
0.052
Ans. H2 =
=
%error =
%
33.571 mol
33.847 mol
0.814
32.816
32.242
1.781
The values calculated with Eq. (4.13) are within 2% of the handbook values.
4.10 The ln P vs. 1/T relation over a short range is very nearly linear. Our
procedure is therefore to take 5 points, including the point at the
temperature of interest and two points on either side, and to do a linear
least-squares fit, from which the required derivative in Eq. (4.11) can be
found. Temperatures are in rankines, pressures in psia, volumes in cu
ft/lbm, and enthalpies in Btu/lbm. The molar mass M of tetrafluoroethane is
102.04. The factor 5.4039 converts energy units from (psia)(cu ft) to Btu.
80
(a) T := 459.67 + 5
V := 1.934 0.012
5
0
t := 5
10
15
18.787
21.162
P := 23.767
26.617
29.726
Data:
xi :=
1
ti + 459.67
i := 1 .. 5
yi := ln ( Pi)
dPdT :=
( P) 3
T
H :=
slopedPdT = 0.545
T V dPdT
H = 90.078
5.4039
Ans.
The remaining parts of the problem are worked in exactly the same
way. All answers are as follows, with the Table 9.1 value in ( ):
4.11
(a) H = 90.078
( 90.111)
(b) H = 85.817
( 85.834)
(c) H = 81.034
( 81.136)
(d) H = 76.007
( 75.902)
(e) H = 69.863
( 69.969)
119.377
gm
M := 32.042
Tc :=
mol
153.822
H is the value at
0 degC.
536.4
512.6 K Pc :=
556.4
54.72
80.97 bar Tn :=
45.60
273.15K
Tr1 :=
Tc
334.3
337.9 K
349.8
Tr2 :=
Tn
Tc
270.9
J
H := 1189.5
Hexp :=
gm
217.8
246.9
0.509
J
Tr1 = 0.533
1099.5
gm
194.2
0.491
0.38
1 Tr2
Hn := H
1 Tr1
Hn Hexp
PCE :=
100%
Hexp
245
J
Hn = 1055.2
193.2 gm
0.623
Tr2 = 0.659
0.629
0.77
PCE = 4.03 %
0.52
c
R Tn 1.092 ln bar 1.013
Hn :=
0.930 Tr2
M
n
exp
PCE :=
100%
Hexp
247.7
J
Hn = 1195.3
192.3 gm
0.34
PCE = 8.72 %
0.96
4.12 Acetone
:= 0.307
Tc := 508.2K
Pc := 47.01bar
Tn := 329.4K
P := 1atm
Tr = 0.648
Pr :=
Zc := 0.233
cm
Vc := 209
mol
Tn
Tr :=
Tc
82
P
Pc
Hn := 29.1
kJ
mol
Pr = 0.022
B0 := 0.083
Tr
0.172
B1 := 0.139
Tr
Z := 1 + B0
V :=
1.6
Pr
Tr
4.2
+ B1
Pr
Tr
Z R Tn
B0 = 0.762
Eq. (3.65)
B1 = 0.924
Eq. (3.66)
Z = 0.965
(Pg. 102)
V = 2.609 10
3
4 cm
mol
Liquid Volume
2
(1Tr) 7
Vsat := Vc Zc
Vsat = 70.917
Eq. (3.72)
cm
mol
gives
Psat = e
B
T +C
H = T V
( T + C)
V := V Vsat
V = 2.602 10
A := 14.3145
B := 2756.22
A B
( T +C)
3
4 cm
mol
C := 228.060
83
B
A
Tn 273.15K
+C
B
K
kPa
Hcalc := Tn V
e
2
K
Tn 273.15K
+C
Hcalc = 29.662
kJ
Ans.
mol
%error :=
Hcalc Hn
Hn
%error = 1.9 %
The table below shows the values for other components in Table B.2. Values
agree within 5% except for acetic acid.
Acetone
Acetic Acid
Acetonitrile
Benzene
iso-Butane
n-Butane
1-Butanol
Carbon tetrachloride
Chlorobenzene
Chloroform
Cyclohexane
Cyclopentane
n-Decane
Dichloromethane
Diethyl ether
Ethanol
Ethylbenzene
Ethylene glycol
n-Heptane
n-Hexane
Methanol
Methyl acetate
Methyl ethyl ketone
Nitromethane
n-Nonane
iso-Octane
n-Octane
n-Pentane
Phenol
1-Propanol
Hn (kJ/mol)
29.7
37.6
31.3
30.8
21.2
22.4
43.5
29.9
35.3
29.3
29.9
27.4
39.6
28.1
26.8
39.6
35.7
53.2
31.9
29.0
36.5
30.4
31.7
34.9
37.2
30.8
34.6
25.9
45.9
41.9
84
% error
1.9%
58.7%
3.5%
0.2%
-0.7%
0.0%
0.6%
0.3%
0.3%
0.1%
-0.1%
0.4%
2.2%
0.2%
0.9%
2.8%
0.5%
4.9%
0.4%
0.4%
3.6%
0.2%
1.3%
2.6%
0.7%
-0.1%
0.6%
0.2%
-0.6%
1.1%
p
2-Propanol
Toluene
Water
o-Xylene
m-Xylene
p-Xylene
40.5
33.3
41.5
36.7
36.2
35.9
1.7%
0.5%
2.0%
1.2%
1.4%
0.8%
P := 100 kPa
(guess)
P
5622.7 K
T
= 48.157543
4.70504 ln
T
kPa
K
Given
ln
P := Find ( P)
dPdT := P
5622.7 K
T
4.70504
T
P = 87.396 kPa
joule
H := 31600
mol
Clapeyron equation:
dPdT =
AL := 13.431
cm
Vliq := 96.49
mol
H
T dPdT
P V 1
B := V
R T
bar
K
H
T ( V Vliq)
Eq. (3.39)
dPdT = 0.029
cm
B = 1369.5
mol
Pc := 80.97bar
Ans.
Tn := 337.9K
3
BL := 51.28 10
CL := 131.13 10
BL
CL 2
CPL ( T) := AL +
T +
T R
2
K
K
AV := 2.211
BV := 12.216 10
85
CV := 3.450 10
BV
CV 2
CPV ( T) := AV +
T +
T R
2
K
K
P := 3bar
Tsat := 368.0K
T1 := 300K
T2 := 500K
Tn
Trn = 0.659
Tc
Trsat :=
Pc
1.013
bar
R T
Tsat
Tc
Trsat = 0.718
1.092 ln
Hn :=
0.930 Trn
1 Trsat
Hv := Hn
1 Trn
kJ
mol
Hv = 35.645
kJ
mol
0.38
sat
2
H :=
CPL ( T) dT + Hv +
CPV ( T) dT
T
T
1
n := 100
Hn = 38.301
H = 49.38
sat
kmol
hr
Q := n H
Q = 1.372 10 kW
kJ
mol
Ans.
(b) Benzene:
Hv = 28.273
kJ
mol
H = 55.296
kJ
mol
Q = 1.536 10 kW
(c) Toluene
Hv = 30.625
kJ
mol
H = 65.586
kJ
mol
Q = 1.822 10 kW
4.15 Benzene
Tc := 562.2K
T1sat := 451.7K
Pc := 48.98bar
T2sat := 358.7K
86
Tn := 353.2K
Cp := 162
J
mol K
Tn
Trn = 0.628
Tc
Tr2sat :=
Pc
1.013
bar
R T
T2sat
Tc
Tr2sat = 0.638
1.092 ln
Hn :=
0.930 Trn
1 Tr2sat
Hv := Hn
1 Trn
kJ
mol
Hn = 30.588
0.38
Hv = 30.28
kJ
mol
Cp ( T1sat T2sat) = x Hv
Tc := 308.3 K
x := Find ( x)
x = 0.498
Pc := 61.39 bar
Ans.
Tn := 189.4 K
T := 298.15 K
Trn :=
Tn
Tc
Trn = 0.614
Tr :=
Pc
1.013
bar
Hn := R Tn 1.092
T
Tc
ln
0.930 Trn
1 Tr
Hv := Hn
1 Trn
Hf := 227480
J
mol
Hn = 16.91
kJ
mol
Hv = 6.638
kJ
mol
0.38
H298 := Hf Hv
87
Tr = 0.967
H298 = 220.8
kJ
mol
Ans.
4.17
kJ
mol
kJ
mol
H298 = 12.3
H298 = 198.6
H298 = 103.9
kJ
mol
dQ = dU dW = CV dT + P dV
1st law:
P V = R T
Ideal gas:
and
P V = const
P dV + V dP = R dT
then
P V
dV = V dP
P dV =
dQ = CP dT R dT +
which reduces to
R dT
1
dQ = CV dT +
or
(B)
V dP = P dV
from which
or
(A)
V dP = R dT P dV
Whence
Since
kJ
mol
R dT
1
R dT
1
dQ = CP dT +
CP
+
R dT
R 1
dQ =
R dT
(C)
88
Tam
T2 := 950 K
Integrate (C):
T1 := 400 K
CPm
+
R ( T2 T1)
1
R
Q :=
Q = 6477.5
:= 1.55
J
mol
Ans.
P1 := 1 bar
4.18
T2
P2 := P1
T1
P2 = 11.45 bar
Ans.
Ans.
Ans.
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.
89
2
n :=
A :=
2
11.286
3.639
5.457 B :=
3.470
3.280
i := 1 .. 4
( niAi)B := ( ni Bi)
A :=
0.506
3
1.045 10
D :=
1.450 K
0.593
D :=
0.227
1.157 105K2
0.121
0.040
( niDi)
A = 54.872
B = 0.012
1
K
5 2
D = 1.621 10 K
CP
HP = R
dT
R
T
T0 := 298.15K
(guess)
Given
H298 = R A T0 ( 1) +
:= Find ( )
= 8.497
B
D 1
2 2
T0 1 +
2
T0
T := T0
T = 2533.5 K
Ans.
Parts (b), (c), and (d) are worked the same way, the only change being in the
numbers of moles of products.
(b)
nO = 0.75
nn = 14.107
T = 2198.6 K
Ans.
(c)
nO = 1.5
nn = 16.929
T = 1950.9 K
Ans.
(d)
nO = 3.0
nn = 22.571
T = 1609.2 K
Ans.
2
2
2
2
2
2
90
) = 3.65606
, 0.0 , 0.016 10
J
mol
H298 + Hair + HP = 0
1.5
2
n :=
2
16.929
3.639
5.457
A :=
3.470
3.280
A :=
J
mol
( niAi)
0.506
1.045 10 3
B :=
1.450 K
0.593
B :=
( ni Bi)
A = 78.84
0.227
1.157 5 2
D :=
10 K
0.121
0.040
D :=
( niDi)
i
1
B = 0.016
K
5 2
D = 1.735 10 K
:= 2 (guess)
Given
:= Find ( )
B
2 2
H298 Hair = R A T0 ( 1) + T0 1 ...
2
D 1
= 7.656
T := T0 K
91
T = 2282.5 K K
Ans.
4.20
4.21
Ans.
(n) 180,500 J
(b) -905,468 J
(o) 178,321 J
(c) -71,660 J
(p) -132,439 J
(d) -61,980 J
(q) -44,370 J
(e) -367,582 J
(r) -68,910 J
(f) -2,732,016 J
(s) -492,640 J
(g) -105,140 J
(t) 109,780 J
(h) -38,292 J
(u) 235,030 J
(i) 164,647 J
(v) -132,038 J
(j) -48,969 J
(w) -1,807,968 J
(k) -149,728 J
(x) 42,720 J
(l) -1,036,036 J
(y) 117,440 J
(m) 207,436 J
(z) 175,305 J
92
4.22
T/K
(a)
(b)
(f)
(i)
(j)
(l)
(m)
(n)
(o)
(r)
(t)
(u)
(v)
(w)
(x)
(y)
4.23
873.15
773.15
923.15
973.15
583.15
683.15
850.00
1350.00
1073.15
723.15
733.15
750.00
900.00
673.15
648.15
1083.15
A
-5.871
1.861
6.048
9.811
-9.523
-0.441
4.575
-0.145
-1.011
-1.424
4.016
7.297
2.418
2.586
0.060
4.175
103 B
4.181
-3.394
-9.779
-9.248
11.355
0.004
-2.323
0.159
-1.149
1.601
-4.422
-9.285
-3.647
-4.189
0.173
-4.766
106 C
0.000
0.000
0.000
2.106
-3.450
0.000
0.000
0.000
0.000
0.156
0.991
2.520
0.991
0.000
0.000
1.814
10-5 D
-0.661
2.661
7.972
-1.067
1.029
-0.643
-0.776
0.215
0.916
-0.083
0.083
0.166
0.235
1.586
-0.191
0.083
IDCPH/J
-17,575
4,729
15,635
25,229
-10,949
-2,416
13,467
345
-9,743
-2,127
7,424
12,172
3,534
4,184
125
12,188
HoT/J
-109,795
-900,739
-2,716,381
189,876
-59,918
-1,038,452
220,903
180,845
168,578
-71,037
117,204
247,202
-128,504
-1,803,784
42,845
129,628
93
ft
day
5
T := ( 60 32) K + 273.15K
9
T = 288.71 K
P := 1atm
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:
HfCH4 := 74520
J
mol
HfCO2 := 393509
HfO2 := 0
J
mol
J
mol
HfH2Oliq := 285830
J
mol
HigherHeatingValue := Hc
Hc = 8.906 10
mol
P
R T
n HigherHeatingValue
4.25
8 mol
n = 1.793 10
day
5dollar
5 dollar
= 7.985 10
GJ
day
Ans.
HfO2 := 0
J
mol
HfH2Oliq := 285830
J
mol
J
mol
94
J
mol
7
HfO2
2
J
mol
J
mol
kJ
mol
Ans.
2H2 + O2 = 2H2O(l)
Hf1 := 2 ( 285830) J
C + O2 = CO2(g)
Hf2 := 393509 J
N2(g)+2H2O(l)+CO2(g)=(NH2)2CO(s)+3/2O2 H := 631660 J
.
N2(g)+2H2(g)+C(s)+1/2O2(g)=(NH2)2CO(s)
H298 := Hf1 + Hf2 + H
H298 = 333509 J
95
Ans.
4.28
Q := 43960 162.27 J
Q = 7.133 10 J
H := Q + R T ngas
H = 7.145 10 J
9H2O(l) = 9H2O(g)
Hvap := 9 44012 J
___________________________________________________
C10H18(l) + 14.5O2(g) = 10CO2(g) + 9H2O(g)
H298 := H + Hvap
4.29
H298 = 6748436 J
Ans.
Moles methane
n1 := 1
Moles oxygen
n2 := 2 1.3
Moles nitrogen
n3 := 2.6
96
79
21
n2 = 2.6
n3 = 9.781
n := n1 + n2 + n3
n = 13.381
4.241
13.381
101.325 4.241
n4 = 0.585
CO2 -- 1 mol
H2O -- 2.585 mol
O2 -- 2.6 - 2 = 0.6 mol
N2 -- 9.781 mol
(1)
(2)
(3)
(4)
2.585
n :=
0.6
9.781
5.457
3.470
A :=
3.639
3.280
i := 1 .. 4
R = 8.314
A :=
1.045
1.450 3
B :=
10
0.506
0.593
J
mol K
( niAi) B := ( ni Bi)
i
A = 48.692
1.157
0.121 5
D :=
10
0.227
0.040
i
3
B = 10.896983 10
D :=
( niDi)
i
4
C := 0
D = 5.892 10
kJ
mol
H298 := 802625
Q := HP + H298
Q = 70 , 612 J
97
J
mol
Ans.
n2
n1 + n2 + n3 + n4
101.325
pp = 18.754
n = 11.381
12.34
n
101.325 12.34
n2 = 1.578
n := 2.585 1.578
J
mol
Sensible heat of cooling the flue gases to 50 degC with all the water as
vapor (we assumed condensation at 50 degC):
Q := R MCPH ( 323.15 K , 1773.15 K , A , B , C , D) ( 323.15 1773.15)K n H50
Q = 766 , 677 J
4.30
Ans.
ENERGY BALANCE:
H = HR + H298 + HP = 0
REACTANTS: 1=NH3; 2=O2; 3=N2
4
n := 6.5
24.45
3.578
A := 3.639
3.280
i := 1 .. 3
A :=
3.020
3
B := 0.506 10
0.593
( niAi)
B :=
( ni Bi)
A = 118.161
0.186
5
D := 0.227 10
0.040
D :=
B = 0.02987
( niDi)
i
5
C := 0.0
D = 1.242 10
kJ
mol
0.8
2.5
n := 3.2 A :=
4.8
24.45
i := 1 .. 5
A :=
3.578
3.639
3.387 B :=
3.470
3.280
3.020
0.506
10 3
0.629
1.450 K
0.593
( niAi)
B :=
A = 119.65
0.186
0.227
5 2
D := 0.014 10 K
0.121
0.040
( ni Bi)
i
1
B = 0.027
K
99
D :=
( niDi)
i
4
D = 8.873 10 K
:= Find ( )
4.31
B
2 2
H298 HR = R A T0 ( 1) + T0 1 ...
2
D 1
Given
= 3.283
T := T0
T = 978.9 K
Ans.
J
mol
4 J
H298 = 8.838 10
mol
1
1
i := 1 .. 2 n :=
1.424
B :=
3.470
A :=
A :=
14.394 3
10 C :=
1.450
( niAi)B := ( ni Bi)
i
C :=
A = 4.894
4.392 6
10
D :=
0.0
0.0 5
10
0.121
( ni Ci)
( niDi)
D :=
i
6
B = 0.01584
C = 4.392 10
D = 1.21 10
HR = 2.727 10
mol
Q := HR + H298 1mol
Q = 115653 J
100
Ans.
4.32 One way to proceed is as in Example 4.8 with the alternative pair of reactions:
CH4 + H2O = CO + 3H2
H298a := 205813
H298b := 164647
BASIS: 1 mole of product gases containing 0.0275 mol CO2; 0.1725 mol CO;
& H2O 0.6275 mol H2
Entering gas, by carbon & oxygen balances:
0.0275 + 0.1725 = 0.2000 mol CH4
0.1725 + 0.1725 + 2(0.0275) = 0.4000 mol H2O
J
H298 := 0.1725 H298a + 0.0275 H298b
mol
The energy balance is written
4 J
H298 = 4.003 10
mol
Q = HR + H298 + HP
i := 1 .. 2
1.702
B :=
3.470
A :=
A :=
9.081 3
10
C :=
1.450
( niAi)B := ( ni Bi)
i
A = 1.728
C :=
0.2
0.4
n :=
2.164 6
10
D :=
0.0
0.0 5
10
0.121
( ni Ci)
( niDi)
D :=
i
3
i
7
B = 2.396 10
C = 4.328 10
D = 4.84 10
HR = 1.145 10
mol
0.0275
0.1725
n :=
A :=
0.1725
0.6275
5.457
3.376
3.470
3.249
1.045
0.557 3
B :=
10
1.450
0.422
101
1.157
0.031 5
D :=
10
0.121
0.083
i := 1 .. 4
A :=
( niAi)
B :=
( ni Bi)
( niDi)
i
4
A = 3.37
D :=
B = 6.397 10
i
3
C := 0.0
D = 3.579 10
Q := HR + H298 + HP mol
Q = 54881 J
Ans.
H298a := 802625
H298b := 1428652
BASIS: 1 mole fuel (0.75 mol CH4; 0.25 mol C2H6) burned completely with
80% xs. air.
O2 in = 1.8[(0.75)(2) + (0.25)(3.5)] = 4.275 mol
N2 in = 4.275(79/21) = 16.082 mol
Product gases: CO2 = 0.75 + 2(0.25) = 1.25 mol
H2O = 2(0.75) + 3(0.25) = 2.25 mol
O2 = (0.8/1.8)(4.275) = 1.9 mol
N2 = 16.082 mol
J
mol
Q := 8 10
Energy balance:
Q = H = H298 + HP
PRODUCTS: 1=CO2; 2=H2O; 3=O2; 4=N2
1.25
2.25
n :=
A :=
1.9
16.082
i := 1 .. 4
A :=
5.457
3.470 B :=
3.639
3.280
HP = Q H298
1.045
3
1.450 10
0.506 K
0.593
( niAi) B := ( ni Bi)
i
A = 74.292
B = 0.015
102
J
mol
1.157
0.121 5 2
D :=
10 K
0.227
0.040
D :=
( niDi)
i
1
K
C := 0.0
4 2
D = 9.62 10 K
:= 2
B
2 2
Q H298 = R A T0 ( 1) + T0 1 ... := Find ( )
2
D 1
= 1.788
4.34
(guess)
T := T0
T = 542.2 K
Ans.
BASIS: 1 mole of entering gases containing 0.15 mol SO2; 0.20 mol
O2; 0.65 mol N2
SO2 + 0.5O2 = SO3
Conversion = 86%
J
mol
0.129
n := 0.0645
0.129
i := 1 .. 3
A :=
5.699
A := 3.639
8.060
0.801
3
B := 0.506 10
D :=
1.056
( niAi) B := ( ni Bi)
i
A = 0.06985
B = 2.58 10
( niDi)
i
103
D :=
1.015
5
0.227 10
2.028
C := 0
D = 1.16 10
J
mol
H773 = 12679
J
mol
Ans.
Energy balance:
Q = H = HR + H298 + HP
J
4 J
H298 = 2.045 10
mol
mol
Reactants: 1: CO 2: H2O
0.5
0.5
3.376
3.470
n :=
A :=
i := 1 .. 2
A :=
0.031 5
10
0.121
0.557 3
10
1.450
B :=
D :=
( niAi) B := ( ni Bi)
i
A = 3.423
D :=
( niDi)
i
B = 1.004 10
C := 0 D = 4.5 10
HR = 3.168 10
Products:
0.2
0.2
n :=
0.3
0.3
mol
1: CO 2: H2O 3: CO2 4: H2
3.376
3.470
A :=
5.457
3.249
0.557
1.450 3
B :=
10
1.045
0.422
104
0.031
0.121 5
D :=
10
1.157
0.083
i := 1 .. 4 A :=
( niAi)
B :=
( ni Bi)
D :=
i
4
A = 3.981
( niDi)
B = 8.415 10
C := 0 D = 3.042 10
HP = 1.415 10
mol
Q := HR + H298 + HP mol
Q = 9470 J Ans.
4.36 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:
14.8
12.011
lbm = 209.133 lbm
0.85
0.21 =
15.124 + x
100.175
x = 5.913 lbmol
Q = H = H298 + HP
BTU
209.13 lbm
lbm
Q = 1.192 10 BTU
Guess: y := 50
11.8
n ( y) := 5.913
79.278
15.797 + y
i := 1 .. 5 A ( y) :=
r := 1.986
5.457
3.376
A := 3.639
3.280
3.470
T :=
1.045
0.557
3
B := 0.506 10
0.593
1.450
T
T0
T = 477.594
1.157
0.031
5
D := 0.227 10
0.040
0.121
:=
400 + 459.67
1.8
= 1.602
CP ( y) := r A ( y) +
107
B ( y)
D ( y)
T0 ( + 1) +
2
2
T0
Given
y = 49.782
y 18.015
= 4.288
209.13
Whence
y := Find ( y)
4.37 BASIS: One mole of product gas containing 0.242 mol HCN, and
(1-0.242)/2 = 0.379 mol each of N2 and C2H2. The energy balance is
Q = H = H298 + HP
H298 := ( 2 135100 227480)
0.242
J
2
H298 = 5.169 10 J
Products:
0.242
n := 0.379
0.379
i := 1 .. 3 A :=
4.736
A := 3.280
6.132
( niAi)
i
B :=
1.359
3
B := 0.593 10
1.952
( ni Bi)
0.725
5
D := 0.040 10
1.299
D :=
A = 4.7133
( niDi)
i
B = 1.2934 10
C := 0 D = 6.526 10
HP = 2.495 10 J
HP = 2.495 10 J
Q := H298 + HP
Q = 30124 J
Ans.
4.38 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)
108
Evaluate
T0 := 298.15K
J
mol
5 J
H298 = 1.144 10
mol
2
n :=
4
3.470
4.442
A :=
3.156
3.639
i := 1 .. 4 A :=
1.45
0.089 3
B :=
10
0.623
0.506
( niAi) B := ( ni Bi)
i
0.121
0.344 5
D :=
10
0.151
0.227
D :=
( niDi)
i
5
A = 0.439
B = 8 10
C := 0 D = 8.23 10
J
mol
H823
4
0.45 mol
Q = 13229 J
Ans.
J
(a)
mol
J
(b)
H298b := 221050
mol
H298a := 172459
2
n := 1
1
3.376
A := 1.771
5.457
0.557
3
B := 0.771 10
1.045
109
0.031
5
D := 0.867 10
1.157
i := 1 .. 3 A :=
( niAi) B := ( ni Bi)
i
D :=
( niDi)
i
4
A = 0.476
B = 7.02 10
C := 0
D = 1.962 10
H1148a = 1.696 10
mol
For (b):
2
n := 1
2
3.376
A := 3.639
1.771
i := 1 .. 3 A :=
0.557
3
B := 0.506 10
0.771
( niAi) B := ( ni Bi)
i
A = 0.429
i
4
B = 9.34 10
0.031
5
D := 0.227 10
0.867
D :=
( niDi)
i
C := 0
D = 1.899 10
H1148b = 2.249 10
mol
Define:
nCO
r=
nO
r :=
110
H1148b
H1148a
r = 1.327
For 100 mol flue gas and x mol air, moles are:
Flue gas
Air
Feed mix
CO2
12.8
12.8
CO
3.7
3.7
O2
5.4
0.21x
5.4 + 0.21x
N2
78.1
0.79x
78.1 + 0.79x
r=
12.5
5.4
r
x :=
mol
0.21
12.8
5.4 + 0.21 x
x = 19.155 mol
100
= 5.221
19.155
Ans.
nCO = 48.145
nN = 93.232
Mole % CO =
nCO
nCO + nN
100 = 34.054
Ans.
Mole % N2 =
H298a := 802625
J
mol
H298b := 519641
J
mol
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
2.538
79
= 9.548
21
mol O2
mol N2
111
5 J
H298 = 6.747 10
2 0.658 +
mol
3
0.282 = 1.739
2
CO2: 0.658
CO: 0.282
H2O: 1.880
O2: 2.538 - 1.739 = 0.799
N2: 9.548 + 0.060 = 9.608
0.658
0.282
n := 1.880 A :=
0.799
9.608
i := 1 .. 5 A :=
5.457
3.376
3.470
B :=
3.639
3.280
( niAi)
1.045
0.557
3
1.450
10
0.506
0.593
B :=
( ni Bi)
i
A = 45.4881
1.157
0.031
5
D := 0.121 10
0.227
0.040
D :=
( niDi)
i
B = 9.6725 10
C := 0 D = 3.396 10
HP = 7.541 10
Energy balance:
mol
Hrx := H298 + HP
112
kJ
mol
kg
mdotH2O := 34.0
sec
Hrx = 599.252
HH2O mdotH2O
ndotfuel :=
kJ
kg
ndotfuel = 16.635
Hrx
mol
sec
ndotfuel R 298.15 K
m
V = 0.407
sec
101325 Pa
Ans.
H298 := 109780
1
n := 1
1
J
mol
T := 798.15 K
Evaluate
by Eq. (4.21):
H798
1: C4H6 2: H2 3: C4H8
2.734
A := 3.249 B :=
1.967
26.786
3
C :=
0.422 10
31.630
8.882
6
D :=
0.0 10
9.873
0.0
5
0.083 10
0.0
i := 1 .. 3
A :=
( niABi) := ( ni Bi)
i
A = 4.016
C :=
( ni Ci)
i
B = 4.422 10
D :=
( niDi)
i
C = 9.91 10
H798 = 1.179 10
mol
Q = 38896 J
113
Ans.
D = 8.3 10
P := 1atm
R = 7.88 10
mol K
a) T0 := ( 70 + 459.67)rankine
T := T0 + 20rankine
T0 = 294.261 K
, 0 , 0.016 10
ndot :=
) = 38.995 K
5
, 0 , 0.016 10
ndot R T0
P
m
s
R = 8.314 10
kJ
mol K
5
, 0 , 0.016 10
ndot :=
ft
sec
Q := 12
kJ
s
Ans.
) = 45.659 K
5
, 0 , 0.016 10
ndot = 31.611
mol
s
ndot R T0
Vdot :=
P
m
Vdot = 0.7707
s
P := 1atm
T := ( 68 + 459.67)rankine
atm ft
mol rankine
ft
Vdot := 50
sec
Vdot = 33.298
T := T0 + 13K
mol
s
ndot = 39.051
Vdot = 0.943
b) T0 := ( 24 + 273.15)K
R = 1.61 10
BTU
sec
T = 305.372 K
Vdot :=
Q := 12
ndot :=
P Vdot
R T0
114
ndot = 56.097
mol
s
Ans.
T0 = 307.594 K
T = 293.15 K
, 0 , 0.016 10
) = 50.7 K
3 BTU
R = 7.88 10
mol K
, 0 , 0.016 10 ndot
Q = 22.4121
b) T0 := ( 35 + 273.15)K
BTU
Ans.
sec
T := ( 25 + 273.15)K
3
5 atm m
R = 8.205 10
mol K
m
Vdot := 1.5
sec
ndot :=
R = 8.314 10
P Vdot
R T0
ndot = 59.325
, 0 , 0.016 10
) = 35.119 K
kJ
mol K
, 0 , 0.016 10 ndot
Q = 17.3216
J
J
J
+ 4 241818
104680
mol
mol
mol
6 J
H298 = 2.043 10
Cost := 2.20
dollars
gal
mol
:= 80%
115
mol
s
kJ
Ans.
s
Tc := 369.8K
Zc := 0.276
T := ( 25 + 273.15)K
Tr :=
Vc := 200.0
T
Tc
Tr = 0.806
(1Tr)0.2857
Vsat := Vc Zc
Heating_cost :=
a) Acetylene
Vsat = 89.373
Vsat Cost
cm
mol
dollars
MJ
Ans.
dollars
Heating_cost = 33.528
6
10 BTU
Heating_cost = 0.032
H298
4.45 T0 := ( 25 + 273.15)K
cm
mol
T := ( 500 + 273.15)K
, 0 , 1.299 10
4 J
Q = 2.612 10
mol
The calculations are repeated and the answers are in the following table:
J/mol
a) Acetylene
26,120
b) Ammonia
20,200
c) n-butane
71,964
d) Carbon dioxide
21,779
e) Carbon monoxide
14,457
f) Ethane
38,420
g) Hydrogen
13,866
h) Hydrogen chloride
14,040
i) Methane
23,318
j) Nitric oxide
14,730
k) Nitrogen
14,276
l) Nitrogen dioxide
20,846
m) Nitrous oxide
22,019
n) Oxygen
15,052
o) Propylene
46,147
116
4.46 T0 := ( 25 + 273.15)K
Q := 30000
T := ( 500 + 273.15)K
J
mol
a) Acetylene
T := Find ( T)
T = 835.369 K
The calculations are repeated and the answers are in the following table:
a) Acetylene
b) Ammonia
c) n-butane
d) Carbon dioxide
e) Carbon monoxide
f) Ethane
g) Hydrogen
h) Hydrogen chloride
i) Methane
j) Nitric oxide
k) Nitrogen
l) Nitrogen dioxide
m) Nitrous oxide
n) Oxygen
o) Propylene
4.47 T0 := ( 25 + 273.15)K
T (K)
835.4
964.0
534.4
932.9
1248.0
690.2
1298.4
1277.0
877.3
1230.2
1259.7
959.4
927.2
1209.9
636.3
T ( C)
562.3
690.9
261.3
659.8
974.9
417.1
1025.3
1003.9
604.2
957.1
986.6
686.3
654.1
936.8
363.2
T := ( 250 + 273.15) K
y = 0.637
Q := 11500
y := 0.5
, 2.164 10
Ans.
117
, 0 R ...
, 5.561 10
, 0 R
, 0 , 1.299 10
= Q
J
mol
b) T0 := ( 100 + 273.15)K
T := ( 400 + 273.15) K
, 13.301 10
3
y = 0.245
c) T0 := ( 150 + 273.15)K
y = 0.512
, 20.928 10
= Q
, 0 R
Q := 17500
y := 0.5
6
, 15.716 10
3
T := ( 250 + 273.15) K
, 0 R ...
Ans.
J
mol
y := 0.5
Q := 54000
, 0 R ...
6
, 18.476 10
J
mol
= Q
, 0 R
Ans.
4.48 Temperature profiles for the air and water are shown in the figures below.
There are two possible situations. In the first case the minimum
temperature difference, or "pinch" point occurs at an intermediate location
in the exchanger. In the second case, the pinch occurs at one end of the
exchanger. There is no way to know a priori which case applies.
Intermediate Pinch
Section I
TH1
Pinch at End
Section II
TH1
Section I
THi
Section II
THi
TC1
T
TCi
TH2 TC1
TCi
TC2
118
TH2
T
TC2
To solve the problem, apply an energy balance around each section of the
exchanger.
T
H1
mdotC ( HC1 HCi) = ndotH
CP dT
T
Section I balance:
Hi
Hi
mdotC ( HCi HC2) = ndotH
CP dT
T
Section II balance:
H2
If the pinch is intermediate, then THi = TCi + T. If the pinch is at the end,
then TH2 = TC2 + T.
a) TH1 := 1000degC
T := 10degC
TC1 := 100degC
HC1 := 2676.0
kJ
kg
TCi := 100degC
HCi := 419.1
kJ
kg
TC2 := 25degC
HC2 := 104.8
kJ
kg
ndotH := 1
TH2 := TC2 + T
Guess:
mdotC := 1
kg
s
THi := 110degC
Given
mdotC ( HC1 HCi) = ndotH R ICPH ( THi , TH1 , A , B , C , D)Energy balances
on Section I and
mdotC ( HCi HC2) = ndotH R ICPH ( TH2 , THi , A , B , C , D)II
mdotC
kg
:= Find ( mdotC , THi) THi = 170.261 degC mdotC = 11.255
s
THi
mdotC
ndotH
= 0.011
kg
mol
Ans.
TH2 TC2 = 10 degC
119
TC1 := 100degC
T := 10degC
HC1 := 2676.0
kJ
kg
TCi := 100degC
HCi := 419.1
ndotH := 1
THi := TCi + T
mdotC := 1
kg
s
HC2 := 104.8
kJ
kg
kmol
s
Guess:
kJ
kg
TC2 := 25degC
TH2 := 110degC
Given
mdotC ( HC1 HCi) = ndotH R ICPH ( THi , TH1 , A , B , C , D)Energy balances
on Section I and
mdotC ( HCi HC2) = ndotH R ICPH ( TH2 , THi , A , B , C , D)II
mdotC
kg
:= Find ( mdotC , TH2) TH2 = 48.695 degC mdotC = 5.03
s
TH2
mdotC
ndotH
3 kg
= 5.03 10
mol
Ans.
TH2 TC2 = 23.695 degC
Since the intermediate temperature difference, THi - TCi is less than the
temperature difference at the end point, TH2 - TC2, the assumption of an
intermediate pinch is correct.
4.50 a) C6H12O6(s) + 6 O2(g)= 6 CO2(g) + 6 H2O(l)
1 = C6H12O6 , 2 = O2 , 3 = CO2 , 4 = H2O
H0f1 := 1274.4
kJ
mol
H0f2 := 0
120
kJ
mol
M1 := 180
gm
mol
H0f3 := 393.509
kJ
mol
H0f4 := 285.830
kJ
mol
M3 := 44
kJ
kg
gm
mol
kJ
mol
Ans.
mass_person := 57kg
mass_person energy_per_kg
H0r
M1
c) 6 moles of CO2 are produced for every mole of glucose consumed. Use
molecular mass to get ratio of mass CO2 produced per mass of glucose.
6
275 10 mass_glucose
6 M3
M1
= 2.216 10 kg
Ans.
kJ
mol
H0f4 := 393.509
H0f2 := 83.820
kJ
mol
kJ
mol
H0f5 := 241.818
H0f3 := 0
kJ
mol
kJ
mol
a) H0c := 1.05 H0f4 + 2 H0f5 0.85 H0f1 0.10 H0f2 1.05 H0f3
H0c = 825.096
kJ
mol
Ans.
b) For complete combustion of 1 mole of fuel and 50% excess air, the exit
gas will contain the following numbers of moles:
n3 := 0.5 2.05mol
n3 = 1.025 mol
121
Excess O2
n4 := 1.05mol
n5 := 2mol
n6 := 0.05mol +
79
1.5 2.05mol
21
n6 = 11.618 mol
Total N2
B :=
C :=
T2 := ( 600 + 273.15)K
( n3 0 + n4 0 + n5 0 + n6 0) 10 6
mol
5
Q := H0c + ICPH ( T1 , T2 , A , B , C , D) R
122
Q = 529.889
kJ
mol
Ans.