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**Unformatted text preview: **Chapter 1 - Section A - Mathcad Solutions
1.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve this
equation by setting t(F) = t(C).
Guess solution:
t
t = 1.8t
Given
0
Find t
()
32
Ans.
40
P=
1.5 By definition:
F
A
F = mass g
A
P
3000bar
D
4mm
F
PA
g
9.807
m
D
4
F
mass
2
g
s
P=
1.6 By definition:
F
A
A
3000atm
D
0.17in
F
PA
g
32.174
ft
4
D
mass
2
sec
gh
13.535
1.8
gm
3
101.78kPa
13.535
g
9.832
mass
2
384.4 kg
Ans.
2
F
g
A
mass
2
0.023 in
1000.7 lbm
Ans.
gm
3
29.86in_Hg
m
h
2
56.38cm
s
Pabs
g
gh
32.243
Patm
ft
Pabs
h
2
176.808 kPa Ans.
25.62in
s
cm
Patm
12.566 mm
Patm
cm
Patm
A
F = mass g
P
1.7 Pabs =
2
Note: Pressures are in
gauge pressure.
Pabs
gh
1
Patm
Pabs
27.22 psia
Ans.
1.10 Assume the following:
13.5
gm
g
3
m
9.8
2
s
cm
P
1.11
P
h
400bar
h
g
Ans.
302.3 m
The force on a spring is described by: F = Ks x where Ks is the spring
constant. First calculate K based on the earth measurement then gMars
based on spring measurement on Mars.
On Earth:
F = mass g = K x
mass
g
0.40kg
9.81
m
x
2
1.08cm
s
F
F
mass g
F
x
Ks
3.924 N
Ks
363.333
N
m
On Mars:
x
0.40cm
gMars
1.12 Given:
FMars
gMars
FMars
mass
d
P=
dz
g
0.01
FMars
Kx
and:
mK
=
4
MP
RT
Substituting:
Separating variables and integrating:
1
dP =
P
Psea
1atm
ln
M
29
2
mK
PDenver
Psea
=
PDenver = Psea e
gm
mol
d
P=
dz
zDenver
Mg
zDenver
RT
Mg
zDenver
RT
g
9.8
MP
g
RT
Mg
dz
RT
0
Psea
Taking the exponential of both sides
and rearranging:
3
Ans.
kg
PDenver
After integrating:
10
m
2
s
3
R
82.06
cm atm
T
mol K
Mg
zDenver
RT
Mg
RT
(
10
zDenver
273.15)
K
0.194
zDenver
Psea e
PDenver
0.823 atm
Ans.
PDenver
PDenver
1 mi
0.834 bar
Ans.
1.13 The same proportionality applies as in Pb. 1.11.
gearth
ft
32.186
gmoon
2
5.32
M
lmoon
1.14
gearth
gmoon
costbulb
Ans.
18.767 lbf
Ans.
hr
0.1dollars
70W
10
day
kW hr
hr
5.00dollars
10
day
1000hr
costelec
dollars
yr
costelec
25.567
dollars
yr
costtotal
43.829
dollars
yr Ans.
18.262
costtotal
costbulb
D
18.76
113.498
113.498 lbm
wmoon
M gmoon
costbulb
1.15
learth
M
learth lbm
wmoon
lmoon
2
s
s
learth
ft
1.25ft
costelec
mass
250lbm
g
32.169
ft
2
s
3
Patm
(a) F
30.12in_Hg
Patm A
l
Work
1.7ft
D
D
2
3
10 lbf
Work
PE
mass g l
mass
1.227 ft
2
Ans.
Ans.
16.208 psia
Fl
0.47m
A
2.8642
Pabs
PE
1.16
4
F
mass g
F
A
(b) Pabs
(c)
A
3
10 ft lbf Ans.
4.8691
Ans.
424.9 ft lbf
g
150kg
9.813
m
2
s
Patm
(a) F
Patm A
l
0.83m
EP
1.18
mass
EK
2
A
4
110.054 kPa
Work
EP
mass g l
u
1250kg
1
2
mass u
2
40
EK
m
s
Work
EK
Wdot =
200W
Ans.
1000 kJ
mass g h
0.91 0.92
time
Wdot
Ans.
1000 kJ
g
9.8
m
2
s
4
h
2
0.173 m
Ans.
10 N
Fl
Work
1.19
1.909
Pabs
Work
D
4
F
mass g
F
A
(b) Pabs
(c)
A
101.57kPa
50m
Ans.
15.848 kJ Ans.
1.222 kJ
Ans.
Wdot
mdot
1.22
a) cost_coal
mdot
g h 0.91 0.92
0.488
25.00
ton
cost_coal
MJ
29
kg
cost_gasoline
0.95 GJ
kg
s
1
2.00
gal
37
GJ
cost_gasoline
Ans.
14.28 GJ
1
3
m
cost_electricity
0.1000
kW hr
cost_electricity
27.778 GJ
1
b) The electrical energy can directly be converted to other forms of energy
whereas the coal and gasoline would typically need to be converted to heat
and then into some other form of energy before being useful.
The obvious advantage of coal is that it is cheap if it is used as a heat
source. Otherwise it is messy to handle and bulky for tranport and
storage.
Gasoline is an important transportation fuel. It is more convenient to
transport and store than coal. It can be used to generate electricity by
burning it but the efficiency is limited. However, fuel cells are currently
being developed which will allow for the conversion of gasoline to electricity
by chemical means, a more efficient process.
Electricity has the most uses though it is expensive. It is easy to transport
but expensive to store. As a transportation fuel it is clean but batteries to
store it on-board have limited capacity and are heavy.
5
1.24 Use the Matcad genfit function to fit the data to Antoine's equation.
The genfit function requires the first derivatives of the function with
respect to the parameters being fitted.
B
TC
A
Function being fit:
f ( A B C)
T
e
First derivative of the function with respect to parameter A
d
f ( A B C)
T
dA
B
exp A
T
C
First derivative of the function with respect to parameter B
d
f ( A B C)
T
dB
1
T
C
B
exp A
T
C
First derivative of the function with respect to parameter C
d
f ( A B C)
T
dC
B
(
T
2
exp A
C)
18.5
3.18
9.5
5.48
0.2
11.8
t
9.45
16.9
23.1
32.7
Psat
28.2
41.9
44.4
66.6
52.1
89.5
63.3
129
75.5
187
6
B
T
C
T
t
273.15
lnPsat
ln ( )
Psat
Array of functions used by Mathcad. In this case, a0 = A, a1 = B and a2 = C.
exp a0
exp a0
F ( a)
T
a1
T
a2
a1
T
1
exp a0
T a2
a1
T
a2
2
Guess values of parameters
15
a2
guess
a1
T
exp a0
3000
50
a2
a1
T
a2
Apply the genfit function
A
13.421
A
B
B
genfit ( Psat guess F)
T
C
2.29
C
3
10
Ans.
69.053
Compare fit with data.
200
150
Psat
f ( A B C)
T
100
50
0
240
260
280
300
320
340
360
T
To find the normal boiling point, find the value of T for which Psat = 1 atm.
7
Psat
1atm
B
Tnb
A
Tnb
1.25
a) t1
1970
C2
C1 ( 1
t2
i)
273.15K
2000
dollars
gal
C1
0.35
1.513
329.154 K
Ans.
56.004 degC
C2
t2 t1
Tnb
CK
Psat
ln
kPa
i
5%
dollars
gal
The increase in price of gasoline over this period kept pace with the rate of
inflation.
b) t1
1970
Given
t2
C2
C1
= (1
2000
i)
C1
t2 t1
i
16000
dollars
yr
Find ( i)
i
C2
80000
dollars
yr
5.511 %
The salary of a Ph. D. engineer over this period increased at a rate of 5.5%,
slightly higher than the rate of inflation.
c) This is an open-ended problem. The strategy depends on age of the child,
and on such unpredictable items as possible financial aid, monies earned
by the child, and length of time spent in earning a degree.
8
Chapter 2 - Section A - Mathcad Solutions
2.1 (a)
Mwt
g
35 kg
9.8
m
z
2
5m
s
Work
(b)
Work
Mwt g z
Utotal
Utotal
Work
(c) By Eqs. (2.14) and (2.21):
dU
1.715 kJ Ans.
1.715 kJ
Ans.
d ( )= CP dT
PV
Since P is constant, this can be written:
MH2O CP dT = MH2O dU
MH2O P dV
Take Cp and V constant and integrate: MH2O CP t2 t1 = Utotal
kJ
MH2O 30 kg
CP
4.18
t1 20 degC
kg degC
t2
t1
Utotal
MH2O CP
t2
20.014 degC Ans.
(d) For the restoration process, the change in internal energy is equal but of
opposite sign to that of the initial process. Thus
Q
(e)
Utotal
Q
1.715 kJ
Ans.
Ans.
In all cases the total internal energy change of the universe is zero.
2.2 Similar to Pb. 2.1 with mass of water = 30 kg.
Answers are:
(a) W = 1.715 kJ
(b) Internal energy change of
the water = 1.429 kJ
(c) Final temp. = 20.014 deg C
(d) Q = -1.715 kJ
9
2.4
The electric power supplied to the motor must equal the work done by the
motor plus the heat generated by the motor.
E
i 9.7amp
Wdotelect
Qdot
2.5
i
E
Wdotelect
t
Eq. (2.3):
U =Q
Step 1 to 2:
Ut12
Wdotelect
1.067
3
134.875 W
Ans.
W
W12
200J
Ut12
Q34
Q12
800J
Ut34
1.25hp
10 W
Qdot
Wdotmech
Q12
Step 3 to 4:
Wdotmech
110V
W12
6000J
W34
Q34
Ut34
W34
3
5.8
Ans.
10 J
300J
Ans.
500 J
t
t
Step 1 to 2 to 3 to 4 to 1: Since U is a state function, U for a series of steps
that leads back to the initial state must be zero. Therefore, the sum of the
t
U values for all of the steps must sum to zero.
Ut41
4700J
Ut23
Step 2 to 3:
Ut23
W23
Ut34
Ut41
Ans.
4000 J
Ut23
Ut12
4
Ut23
3
Q23
Q23
3800J
W23
10 J
200 J
Ans.
For a series of steps, the total work done is the sum of the work done for each
step.
W12341
1400J
10
W41
W12341
Step 4 to 1:
W12
W23
Ut41
Ut41
Ans.
10 J
3
4.5
Q41
W41
3
4.5
W41
4700J
Q41
Note:
W41
W34
10 J
200 J
Ans.
Q12341 = W12341
2.11 The enthalpy change of the water = work done.
M
Wdot
2.12 Q
CP
20 kg
4.18
kJ
kg degC
t
10 degC
M CP t
0.25 kW
0.929 hr
Wdot
Ans.
U
Q
12 kJ
W
U
19.5 kJ
Ans.
Q
12 kJ
U
W
U
7.5 kJ
Q
12 kJ
Ans.
2.13Subscripts: c, casting; w, water; t, tank. Then
mc Uc
mw Uw
mt Ut = 0
Let C represent specific heat,
C = CP = CV
Then by Eq. (2.18)
mc Cc tc
mw Cw tw
mt Ct tt = 0
mc
2 kg
mw
Cc
0.50
kJ
kg degC
Ct
mt
40 kg
tc
500 degC
Given
t1
mc Cc t2
t2
0.5
5 kg
kJ
kg degC
Cw
25 degC
t2
30 degC
tc = mw Cw
mt Ct
t2
Find t2
11
t2
4.18
kJ
kg degC
(guess)
t1
27.78 degC
Ans.
2.15
mass
(a)
(b)
T
g
CV
1 kg
Ut
1K
9.8
m
Ut
mass CV T
EP
2
kJ
kg K
4.18
Ans.
4.18 kJ
Ut
s
z
(c)
2.17
EP
EK
z
z
mass g
426.531 m Ans.
EK
u
Ut
1
mass
2
1000
50m
u
kg
5
A
3
A
2m
mdot
Wdot
uA
mdot
mdot g z Wdot
4
D
2
(b)
2
4 kg
1.571 10
7.697
s
3
10 kW
U1
P1
1002.7 kPa
U1
H1
763.131
U2
kJ
2784.4
kg
H2
U2
P1 V 1
kJ
kg
kJ
kg
cm
1.128
gm
V1
Ans.
3
P2
U
P2 V 2
2022.4
Ans.
3
kJ
762.0
kg
U
Ans.
3.142 m
H1
2.18 (a)
m
s
m
s
u
m
D
91.433
Ans.
1500 kPa
U2
U1
H
12
cm
169.7
gm
V2
H
2275.8
kJ
kg
H2
H1
Ans.
2.22
D1
(a)
u1
2.5cm
m
s
D2
5cm
For an incompressible fluid, =constant. By a mass balance,
mdot = constant = u 1A1 = u2A2
u2
(b)
D1
u1
u2
D2
1
EK
2
2
2
u2
1
2
2.23 Energy balance:
Mass balance:
u1
2
mdot3 H3
mdot1
1.0
Qdot
30
Qdot
H1
T1
CP
4.18
s
mdot1
mdot2 H2 = Qdot
H2 = Qdot
mdot2 CP T3
25degC
kJ
Ans.
kg
mdot2 H3
T1
kg
s
J
mdot2 = 0
mdot2 = Qdot
mdot1 CP T1
Ans.
1.875
mdot1
mdot Cp T3
T3 CP mdot1
m
s
mdot1 H1
mdot1 H3
or
0.5
EK
mdot3
Therefore:
T3
2
T2 = Qdot
mdot1 CP T1
mdot2 CP T2
mdot2
0.8
kg
s
T2
75degC
kJ
kg K
mdot2 CP T2
mdot2 CP
T3
43.235 degC
2
2.25By Eq. (2.32a):
By continuity,
incompressibility
H
u
=0
2
A1
u2 = u1
13
A2
H = CP T
CP
4.18
kJ
kg degC
Ans.
2
u = u1
u1
u1
2
2 CP
D2
2
u = u1
1
m
s
14
D1
D1
2
1
A2
SI units:
T
2
A1
2
4
1
D2
D1
D2
2.5 cm
3.8 cm
4
T
0.019 degC
Ans.
T
D2
0.023 degC
Ans.
7.5cm
u1
T
2
2 CP
D1
1
4
D2
Maximum T change occurrs for infinite D2:
D2
cm
u1
T
2.26 T1
2 CP
300K
Wsdot
H
2
D1
1
T
D2
T2
u1
520K
ndot
98.8kW
CP T2
4
50
H
T1
10
0.023 degC
m
s
u2
kmol
hr
3.5
10
molwt
29
kg
kmol
7
R
2
CP
6.402
m
s
Ans.
3 kJ
kmol
By Eq. (2.30):
Qdot
H
2
u2
2
2
u1
2
molwt ndot
Wsdot Qdot
2
2.27By Eq. (2.32b):
By continunity,
constant area
H=
u2 = u1
u
2 gc
V2
u2 = u1
V1
14
also
T 2 P1
T 1 P2
Ans.
9.904 kW
V2
V1
=
2
T 2 P1
T 1 P2
2
u = u2
u1
2
2
u = u1
P1
R
T2
T2
P2
molwt
mol rankine
7
2
u1
20 psi
ft lbf
28
20
7
R T2
2
ft
s
T1
T1
579.67 rankine
H2
2726.5
kJ
kg
kJ
kg
Ans.
gm
mol
(guess)
578 rankine
Given
H = CP T =
1
T 1 P2
100 psi
3.407
2
T 2 P1
2
2
R T2
T1 =
T2
Find T2
u1
2
2
T 2 P1
T 1 P2
1 molwt
Ans.
578.9 rankine
(
119.15 degF)
2.28 u1
3
m
s
u2
200
m
s
H1
2
By Eq. (2.32a):
2.29 u1
u2
m
s
m
500
s
30
By Eq. (2.32a):
Q
H2
H1
3112.5
u1
u2
H1
334.9
kJ
kg
2
Q
2
kJ
H2
kg
2411.6
2945.7
(guess)
2
H2
Given
u2
578.36
m
s
H1 =
u2
u1
2
5 cm
V1
388.61
u2
2
Ans.
3
3
D1
kJ
kg
cm
gm
15
V2
667.75
cm
gm
Find u2
Continuity:
2.30 (a)
t1
D2
t2
30 degC
CV
u1 V2
u2 V1
D1
D2
Ans.
1.493 cm
n
250 degC
3 mol
J
mol degC
20.8
By Eq. (2.19):
Q
n CV t2
Q
t1
Ans.
13.728 kJ
Take into account the heat capacity of the vessel; then
cv
mv
Q
(b)
100 kg
mv cv
t1
200 degC
CP
29.1
n CV
CV
Q
Q
Ans.
11014 kJ
n
40 degC
4 mol
joule
mol degC
Q
n CP t2
t2
70 degF
5
kJ
kg degC
t1
t2
By Eq. (2.23):
2.31 (a) t1
t2
0.5
BTU
mol degF
n CV t2
Q
t1
18.62 kJ
n
350 degF
3 mol
By Eq. (2.19):
Q
t1
4200 BTU
Ans.
Take account of the heat capacity of the vessel:
mv
Q
mv cv
cv
200 lbm
(b) t1
400 degF
n CV
t2
0.12
BTU
lbm degF
Q
t1
t2
150 degF
16
10920 BTU
n
Ans.
4 mol
Ans.
CP
7
BTU
mol degF
Q
H1
2.33
n CP t2
1322.6
V1
Q
t1
BTU
BTU
lbm
1148.6
3
V2
lbm
78.14
ft
4
mdot
mdot
V2
10
22.997
u2
Wdot
2.34
H1
V1
H2
307
9.25
BTU
lbm
ft
H2
330
V2
u1
lbm
ft
0.28
173.99
BTU
lb
Ans.
39.52 hp
BTU
3
lbm
Ws
2
Wdot
Ws mdot
u1
u2
H1
20
ft
s
molwt
44
3
D1
lbm
D2
4 in
1 in
2
mdot
u2
4
D1 u1
mdot
V1
V2
mdot
u2
679.263
9.686
2
4
D2
2
Eq. (2.32a): Q
H2
H1
10 in
sec
2
D2
Eq. (2.32a): Ws
D2
ft
sec
2
4
3 in
4 lb
3.463
V1
mdot
10
D1
lbm
2
u2
ft
s
u1
3
2
D 1 u1
Ans.
7000 BTU
H2
lbm
ft
3.058
By Eq. (2.23):
ft
sec
u1
u2
2
17
lb
hr
2
Ws
Ws
molwt
5360
Q
BTU
lbmol
98.82
BTU
lbm
gm
mol
Qdot
2.36
T1
Qdot
mdot Q
P
300 K
67128
BTU
hr
Ans.
1 kg
n
1 bar
28.9
V1
n
gm
34.602 mol
mol
3
3
bar cm T1
83.14
mol K P
cm
24942
mol
V1
V2
P dV = n P V 1
W= n
V2 = n P V1
3 V1
V1
Whence
W
Given:
V2
V1
joule
29
Q
nH
U
mol K
Q
= T1 3
Whence
T1
Q
W
U
n
CP T2
602.08 kJ
12.41
kJ
mol
Ans.
172.61 kJ
H
T2 = T1
CP
W
n P 2 V1
T2
H
3 T1
17.4
kJ
mol
Ans.
Ans.
Ans.
2.37 Work exactly like Ex. 2.10: 2 steps, (a) & (b). A value is required for PV/T,
namely R.
R
8.314
T1
(a) Cool at const V1 to P2
(b) Heat at const P2 to T2
Ta2
T1
P2
P1
Ta2
293.15 K
T2
333.15 K
P1
J
mol K
1000 kPa
P2
100 kPa
7
R
2
CP
29.315 K
18
CV
5
R
2
Tb
T2
Hb
C P Tb
Hb
Ua
CV Ta
Ua
R T1
V1
P1
Ta2
Tb
V1
303.835 K
2.437
8.841
Ta
10
5.484
10
mol
Ha
Ua
V 1 P2
P1
Ha
Ub
Hb
P2 V 2
V1
Ub
U
Ub
U
0.831
H
2.39
Ua
Ha
Hb
H
1.164
996
kg
9.0 10
3
D
5
2
kJ
mol
ms
1
cm
u
5
1m
5s
5
22133
Re
D
u
Re
Ta
263.835 K
mol
3J
mol
V2
3
R T2
V2
P2
7.677
6.315
0.028
m
mol
3J
10
mol
3J
10
mol
Ans.
mol
4 kg
m
2
kJ
T1
3J
3
3m
10
Ta2
55333
110667
276667
19
Ans.
D
0.0001 Note: D = /D
in this solution
0.00635
fF
0.3305 ln 0.27 D
7
Re
0.9
2
0.00517
fF
0.00452
0.0039
0.313
mdot
u
4
D
2
mdot
1.956 kg
1.565 s
Ans.
9.778
0.632
PL
2
fF
D
2
PL
u
0.206
kPa
11.254 m
Ans.
3.88
2.42 mdot
4.5
kg
s
H1
761.1
kJ
kg
H2
536.9
kJ
kg
Assume that the compressor is adiabatic (Qdot = 0). Neglect changes in
KE and PE.
Wdot
Cost
mdot H2
15200
H1
Wdot
Wdot
1.009
3
10 kW
0.573
Cost
kW
20
799924 dollars Ans.
Chapter 3 - Section A - Mathcad Solutions
3.1
1d
=
1
=
dT
d
dP
P
T
At constant T, the 2nd equation can be written:
d
=
2
ln
dP
6
44.1810
P
=
1
bar
2 = 1.01
1
ln ( )
1.01
P
P
Ans.
P2 = 226.2 bar
225.2 bar
3
3.4 b
cm
0.125
gm
c
2700 bar
P1
P2
1 bar
500 bar
V2
Since
Work =
a bit of algebra leads to
P dV
V1
P2
Work
c
P
P
b
dP
Work
0.516
Work
0.516
P1
J
Ans.
gm
Alternatively, formal integration leads to
Work
3.5
=a
P1
c P2
P1
a
bP
P2
1 atm
b ln
P2
b
P1
b
3.9 10
6
atm
1
b
V
3000 atm
0.1 10
1 ft
3
9
J
gm
Ans.
atm
2
(assume const.)
Combine Eqs. (1.3) and (3.3) for const. T:
P2
Work
(
a
V
Work
b P)P dP
P1
21
16.65 atm ft
3
Ans.
1
3.6
1.2 10
V1
3
degC
1
CP
0.84
kJ
kg degC
M
5 kg
t2
20 degC
3
m
1
P
1590 kg
t1
1 bar
0 degC
With beta independent of T and with P=constant,
dV
=
V
V2
dT
Vtotal
Vtotal
MV
Work
V1 exp
P Vtotal
M CP t2
P2
(a) Constant V:
T2
T1
U
H
R T1 ln
(c) Adiabatic:
T1
P1
Ans.
84 kJ
84 kJ
Ans.
83.99 kJ
Ans.
T
T1
U
10.91
15.28
H=0
Q=0
22
kJ
mol
kJ
mol
Work
and
CV
U = Q = CV T
T2
and
7
R
2
CP
600 K
and
Q
Ans.
Utotal
U=
P2
7.638 joule
Work
H
CP T
(b) Constant T:
Work
Q
1 bar
V1
Ans.
Q
Q and
CV T
m
t1
T
P1
3
V2
Htotal
W= 0
P2
10
Q
Utotal
8 bar
5
V
Work
7.638
Htotal
P1
t1
(Const. P)
Q
3.8
t2
525 K
Ans.
Ans.
and
10.37
Q=W
kJ
mol
Ans.
U = W = CV T
5
2
R
1
CP
T2
CV
P2
T1
U
and
U
5.586
3.9 P4
2bar
CP
H
kJ
7
R
2
10bar
T1
Step 41: Adiabatic
T
331.227 K
Ans.
mol
T4
R T1
P1
V1
P4
T1
kJ
mol
Ans.
5
R
2
CV
600K
7.821
T2
CP T
H
CV T
W
P1
T2
P1
V1
4.988
T4
3
3m
10
378.831 K
mol
R
CP
P1
3J
U41
CV T1
T4
U41
4.597
10
H41
CP T1
T4
H41
6.436
10
Q41
3bar
T2
Step 12: Isothermal
Q41
U41
W41
4.597
mol
J
0
mol
W41
P2
J
0
mol
mol
3J
V2
600K
U12
0
J
mol
H12
0
J
mol
23
R T2
P2
3J
10
3
V2
m
0.017
mol
U12
0
J
mol
H12
0
J
mol
mol
T1
Q12
W12
P3
2bar
V3
Q12
Q12
P1
W12
P3 V 3
T3
V2
Step 23: Isochoric
P2
R T1 ln
T3
R
U23
CV T3
H23
CP T3
T2
Q23
2 bar
T4
CV T3
W23
P4
0
T2
R T4
V4
3J
10
mol
3J
6.006
10
400 K
3J
4.157
10
P4
3
V4
m
0.016
mol
U34
CV T4
T3
U34
439.997
H34
CP T4
T3
H34
615.996
Q34
CP T4
T3
Q34
W34
R T4
T3
W34
3.10 For all parts of this problem: T2 = T1
mol
mol
3J
H23
5.82 10
mol
3J
Q23
4.157 10
mol
J
W23 0
mol
J
mol
378.831 K
Step 34: Isobaric
U23
T2
6.006
615.996
175.999
J
mol
J
mol
J
mol
J
mol
and
Also
Q = Work and all that remains is
U= H=0
to calculate Work. Symbol V is used for total volume in this problem.
P1
1 bar
P2
V1
12 bar
24
3
12 m
V2
3
1m
(a)
P2
Work = n R T ln
Work
P1
Work
P 1 V 1 ln
P2
P1
Ans.
2982 kJ
(b) Step 1: adiabatic compression to P2
1
5
3
Vi
P2 V i
W1
V1
P1
P2
(intermediate V)
P1 V 1
Vi
W1
3063 kJ
W2
1
3
2.702 m
2042 kJ
Step 2: cool at const P2 to V2
W2
P2 V 2
Work
W1
Vi
W2
Work
5106 kJ
Ans.
(c) Step 1: adiabatic compression to V2
Pi
W1
P1
V1
(intermediate P)
V2
Pi V 2
1
Step 2: No work.
Work
(d) Step 1: heat at const V1 to P2
62.898 bar
W1
P1 V 1
Pi
7635 kJ
W1
Work
7635 kJ
Ans.
W2
Work
13200 kJ
Ans.
W1 = 0
Step 2: cool at const P2 to V2
W2
P2 V 2
V1
Work
(e) Step 1: cool at const P1 to V2
W1
P1 V 2
W1
V1
25
1100 kJ
Step 2: heat at const V2 to P2
Work
3.17(a)
W2 = 0
Work
W1
No work is done; no heat is transferred.
t
U=
(b)
T=0
Not reversible
T2 = T1 = 100 degC
The gas is returned to its initial state by isothermal compression.
Work = n R T ln
3
V1
3.18 (a) P1
P2 V2 ln
7
2
but
V2
4
3
3
P2
V1
6 bar
878.9 kJ Ans.
Work
V2
T1
500 kPa
303.15 K
CP
5
R
2
CV
R
n R T = P2 V 2
m
P2
100 kPa
CP
V1
V2
4m
Work
CV
Adiabatic compression from point 1 to point 2:
Q12
0
kJ
mol
1
U12 = W12 = CV T12
U12
CV T2
U12
3.679
T1
kJ
mol
T2
H12
CP T2
W12
H12
5.15
T1
kJ
W12
mol
T1
3.679
P2
P1
U12
kJ
mol
Cool at P2 from point 2 to point 3:
T3
Ans.
1100 kJ
H23
T1
U23
CV T3
CP T3
W23
T2
26
Q23
T2
U23
Q23
H23
Ans.
H23
5.15
Q23
5.15
kJ
U23
mol
kJ
W23
mol
kJ
mol
3.679
Ans.
kJ
1.471
Ans.
mol
Isothermal expansion from point 3 to point 1:
U31 =
H31 = 0
Q31
4.056
W31
P2
P1
R T3 ln
P3
W31
W31
P3
kJ
mol
FOR THE CYCLE:
Q
Q
Q12
1.094
Q23
Q31
U=
4.056
kJ
mol
Ans.
H=0
Work
kJ
mol
W12
Work
Q31
1.094
W23
W31
kJ
mol
(b) If each step that is 80% efficient accomplishes the same change of state,
all property values are unchanged, and the delta H and delta U values
are the same as in part (a). However, the Q and W values change.
Step 12:
W12
Q12
Step 23:
W23
Q23
Step 31:
W31
Q31
W12
W12
W23
0.8
U23
4.598
Q12
0.92
kJ
mol
W23
0.8
U12
kJ
W12
1.839
kJ
mol
mol
W31 0.8
W31
Q23
5.518
kJ
mol
W31
W23
3.245
kJ
mol
Q31
27
3.245
kJ
mol
FOR THE CYCLE:
Q
Q12
Q
Q23
3.192
Work
kJ
mol
W12
Work
Q31
3.192
W23
W31
kJ
mol
3.19Here, V represents total volume.
P1
CP
21
CV
joule
mol K
CP
T2
208.96 K
P1 V 1
n
P1 V 1
T2
Work
n CV
P2
P1
V2
Ans.
T1
P2 V 2
P1 V 1
69.65 kPa
Ans.
Work
Work =
Work
Pext V2
994.4 kJ Ans,
U = Pext V
V1
Work
U = n CV T
R T1
V1
Ans.
1609 kJ
V2
P2
P1
200 kPa
T2
1
100 kPa
V2
V1
P1
(c) Restrained adiabatic:
Pext
P2
Work
P2
P2 V 2
V1
P2
600 K
T2
Work
CV
V2
(b) Adiabatic:
600 K
CP
V1
P 1 V 1 ln
T1
5 V1
R
Work = n R T1 ln
T1
Work
V2
1m
(a) Isothermal:
T2
3
V1
1000 kPa
T2
V 1 T2
V 2 T1
28
442.71 K
Ans.
P2
T1
147.57 kPa
Ans.
400 kJ Ans.
3.20
T1
CP
P1
423.15 K
7
2
Step 12:
If
V1
V2
=
2.502
Step 23:
V1
V3
kJ
mol
0
Process:
2.079
U12
mol
mol
R T1 ln r
()
Q12
2.502
CV T3
CP T3
T2
2.079
W23
kJ
mol
H23
Work
Q
Q23
2.502
0.424
2.91
kJ
mol
kJ
mol
kJ
mol
H
H12
H23
H
2.91
U
U12
U23
U
2.079
29
kJ
mol
T2
H23
W12
Q12
kJ
U23
U23
Q
323.15 K
W12
W12
U23
Work
0
T 3 P1
kJ
mol
kJ
mol
T3
T 1 P3
r
Q12
W23
T1
kJ
Then
Q23
Q23
0
3 bar
T2
R
2
H12
r=
W12
5
CV
R
P3
8bar
kJ
mol
kJ
mol
Ans.
Ans.
Ans.
Ans.
3.21 By Eq. (2.32a), unit-mass basis:
But
CP
u1
2.5
t2
3.22
28
Whence
H = CP T
R
7
2 molwt
molwt
u2
1
2
H
u2
T=
m
s
u2
t1
gm
mol
2
2
u1
2 CP
50
m
s
t1
u1
t2
2 CP
7
R
2
CV
5
R
2
P1
1 bar
P3
148.8 degC
Ans.
10 bar
CV T3
U
2.079
T1
mol
CP T3
H
Ans.
T3
303.15 K
H
T1
kJ
2.91
T1
kJ
mol
Each part consists of two steps, 12 & 23.
(a) T2
W23
T3
R T 2 ln
P2
P3
P2
P1
150 degC
2
2
CP
U
2
u =0
T2
T1
Work
W23
Work
6.762
Q
U
Q
4.684
30
kJ
mol
Ans.
Work
kJ
mol
Ans.
Ans.
403.15 K
(b) P2
T2
P1
H12
CP T2
W12
U12
W23
R T 2 ln
Work
W12
Q
(c)
U
T2
U12
T3
CV T2
Q12
Q12
P3
H12
W12
T1
0.831
W23
P2
Work
W23
Q
Work
P2
T1
P3
T1
W12
kJ
mol
7.718
6.886
4.808
kJ
mol
kJ
mol
kJ
mol
CP T3
T2
Q23
CV T3
T2
W23
U23
Work
4.972
P1
H23
U23
Ans.
P2
R T 1 ln
H23
Ans.
Work
Q
W12
U
W23
Q
Work
2.894
Q23
kJ
mol
kJ
mol
For the second set of heat-capacity values, answers are (kJ/mol):
U = 2.079
U = 1.247
(a)
Work = 6.762
Q = 5.515
(b)
Work = 6.886
Q = 5.639
(c)
Work = 4.972
Q = 3.725
31
Ans.
Ans.
3.23
T1
303.15 K
T2
T1
T3
393.15 K
P1
1 bar
P3
12 bar
CP
7
R
2
For the process:
U
U
Step 12:
W12
CV T3
1.871
P2
5.608
kJ
mol
mol
For the process:
Work
W12
Q
3.24
Work
5.608
W12 = 0
Work = W23 = P2 V3
But
T3 = T1
Also
W = R T 1 ln
ln
P
T2 T1
P
=
P1
T1
P1 exp
T2
T2
R T1 ln
5.608
P
P1
Ans.
P2
U
P1
kJ
mol
W23
kJ
Q
mol
3.737
V 2 = R T3
kJ
mol
Ans.
T2
Work = R T2
So...
mol
Q23
kJ
mol
W23
Q23
kJ
Q12
W12
Step 23:
Q12
2.619
T1
W12
T3
0
CP T3
H
kJ
T1
P3
Q12
H
T1
5
R
2
CV
T1
Therefore
T1
T1
32
T1
800 K
P
350 K
2.279 bar
P1
Ans.
4 bar
3.25
3
VA
P
Define:
256 cm
P1
r
0.0639
VB
3750.3 cm
CV
=r
CP
Assume ideal gas; let V represent total volume:
P1 V B = P 2 V A
P
P1
3.26
T1
=
VA
VA
VA ( 1)
r
VB
VB
300 K
From this one finds:
VB
P1
r
1 atm
CP
7
R
2
3
Ans.
CP
R
CV
The process occurring in section B is a reversible, adiabatic compression. Let
TA ( )= TA
final
P ( )= P2
final
TB ( )= TB
final
Since the total volume is constant,
nA = nB
nA R TA
2 nA R T1
=
P2
P1
TB
TA TB
2 T1
=
P2
P1
or
(1)
1
(a)
P2
TA
2 T1
TB
1.25 atm
P2
P1
Q = nA
TB
Define
q=
TB
TA
319.75 K
T1
Q
nA
q
430.25 K
33
P2
(2)
P1
UA
CV TA
q
UB
TB
2 T1
3.118
kJ
mol
(3)
Ans.
(b)
Combine Eqs. (1) & (2) to eliminate the ratio of pressures:
(guess)
425 K
TB
300 K
TB
Find TB
TB
TA
319.02 K
Ans.
P2
1.24 atm
Ans.
kJ
mol
Ans.
1
TB = T1
Given
P2
P1
q
(c)
TA
P1
TA
2 T1
P2
P2
(1)
TB
P1
kJ
mol
2.993
1
T1
CV TA
3
q
2 T1
By Eq. (2),
TB
Eliminate
q
(1)
TB
TB
P2
(d)
2 T1
TB
325 K
q
TB
2 T1
CV TA
TB
TA
Ans.
TA
469 K
Ans.
q
2 T1
TA
1.323 atm
TB
q P1
2 T1 C V
4.032
kJ
mol
Ans.
from Eqs. (1) & (3):
P2
1.241 atm
Ans.
(2)
P2
TB
319.06 K
Ans.
(1)
TA
425.28 K
Ans.
P1
1
TB
T1
TA
2 T1
P2
P1
P2
P1
TB
34
6
3
3.30 B
P1
B'
cm
242.5
C
25200
mol
mol
P2
7.817
T
B
C
10
31
bar
B
2
373.15 K
2
55 bar
B'
1 bar
RT
C'
cm
RT
2
C'
2
3.492
10
51
2
bar
(a) Solve virial eqn. for initial V.
RT
P1
Guess:
V1
Given
P1 V 1
=1
RT
B
V1
3
C
V1
2
V1
V2
Find V1
cm
30780
mol
cm
241.33
mol
V1
Solve virial eqn. for final V.
Guess:
RT
P2
V2
P2 V 2
Given
RT
=1
B
V2
3
C
V2
2
V2
Find V2
Eliminate P from Eq. (1.3) by the virial equation:
V2
Work
RT
1
B
V
C
V
2
1
dV
V
Work
12.62
kJ
mol
V1
(b)
Eliminate dV from Eq. (1.3) by the virial equation in P:
P2
1
dV = R T
P
C' dP
W
1
P
RT
2
P1
W
35
12.596
kJ
mol
Ans.
C' P dP
Ans.
Note: The answers to (a) & (b) differ because the relations between the two
sets of parameters are exact only for infinite series.
3.32 Tc
282.3 K
T
298.15 K
Pc
50.4 bar
P
12 bar
T
Tc
Tr
Pr
Tr
Pr
P
Pc
6
3
B
Given
cm
140
C
mol
PV
=1
RT
7200
cm
mol
B
V
V
(b)
B0
V
Find ( )
V
Tr
V
Tr
Z
1
(c)
B0
cm
1919
mol
B1
B1
4.2
Tr
Z
PV
RT
Z
B0
1.6
Pr
3
V
2066
cm
mol
2
0.172
0.139
RT
P
C
0.422
0.083
V
2
3
B1
0.238
(guess)
0.087
(a)
1.056
Ans.
Z
0.929
V
cm
Ans.
1924
mol
0.304
2.262
0.932
10
V
3
3
ZRT
P
For Redlich/Kwong EOS:
0
1
( r)
T
T r Pr
Tr
0.5
Pr
Tr
Table 3.1
Eq. (3.53)
36
Table 3.1
0.42748
0.08664
q Tr
Tr
Tr
Eq. (3.54)
Calculate Z
Guess:
Given
T r Pr
0.9
Eq. (3.52)
Z=1
Z
Z
Z
Find Z)
(
(d)
q Tr
Z
T r Pr
Z
T r Pr
Z
T r Pr
3
Z RT
V
0.928
T r Pr
V
P
1916.5
cm
mol
Ans.
For SRK EOS:
2
1
Tr
1
0.480
Tr
q Tr
(e)
1
Tr
Guess:
Z
Table 3.1
2
Pr
T r Pr
Tr
T r Pr
q Tr
Z
Find Z)
(
0.9
Z
T r Pr
Z
V
0.928
T r Pr
T r Pr
Z
T r Pr
3
Z RT
V
P
1918
1
2
0.45724
0.07779
2
1
q Tr
cm
mol
Ans.
For Peng/Robinson EOS:
1
Tr
Eq. (3.53)
Tr
Eq. (3.52)
Given
Z
0.176
Eq. (3.54)
Calculate Z
Z=1
1.574
2
Table 3.1
0.42748
0.08664
0
1
1
0.37464
Tr
1.54226
0.26992
Eq. (3.54)
Tr
37
2
T r Pr
1
Tr
Pr
Tr
2
Table 3.1
2
Table 3.1
Eq. (3.53)
Calculate Z
Guess:
0.9
Eq. (3.52)
Given
Z=1
T r Pr
Z
Z
Find ( Z)
q Tr
Z
Z
T r Pr
Z
T r Pr
3
V
1900.6
cm
mol
305.3 K
Pc
T
323.15 K
Tr
T
Tc
Tr
P
15 bar
Pr
P
Pc
Pr
0.308
(guess)
0.100
6
3
(a)
cm
156.7
mol
B
Given
PV
=1
RT
B
V
C
9650
cm
mol
V
(b)
B0
V
Find ( V)
Tr
B1
V
Tr
(c)
B0
cm
1625
mol
B1
B1
4.2
Tr
Z
PV
Z
B0
1.6
Pr
3
V
cm
1791
mol
2
0.172
0.139
RT
P
C
0.422
0.083
V
2
3
1
Ans.
1.058
48.72 bar
3.33 Tc
Z
Z
T r Pr
ZRT
P
V
0.92
T r Pr
Ans.
Z
RT
0.907
V
cm
Ans.
1634
mol
0.302
3.517
0.912
V
10
3
ZRT
P
3
For Redlich/Kwong EOS:
1
0
0.08664
38
0.42748
Table 3.1
( r)
T
0.5
Tr
Table 3.1
Pr
T r Pr
Tr
q Tr
Eq. (3.54)
Tr
Eq. (3.53)
Tr
Calculate Z
Guess:
Given
T r Pr
0.9
Eq. (3.52)
Z=1
Z
Z
Z
Find Z)
(
(d)
q Tr
Z
T r Pr
Z
Z
T r Pr
cm
1622.7
mol
V
P
Ans.
For SRK EOS:
0.42748
0.08664
0
1
Tr
1
0.480
Tr
q Tr
0.176
Eq. (3.54)
Guess:
2
1
Tr
T r Pr
Tr
Calculate Z
Z
Table 3.1
2
Pr
Eq. (3.53)
Tr
0.9
Eq. (3.52)
Given
Z=1
Z
1.574
Table 3.1
2
1
(e)
T r Pr
3
Z RT
V
0.906
T r Pr
T r Pr
Find Z)
(
q Tr
Z
T r Pr
Z
Z
V
0.907
T r Pr
Z RT
P
T r Pr
Z
T r Pr
3
V
1624.8
cm
mol
Ans.
For Peng/Robinson EOS:
1
2
1
0.07779
2
39
0.45724
Table 3.1
1
Tr
1
0.37464
Tr
q Tr
1.54226
0.26992
Eq. (3.54)
1
Guess:
Z
Tr
Table 3.1
2
Pr
T r Pr
Tr
Calculate Z
2
2
Eq. (3.53)
Tr
0.9
Eq. (3.52)
Given
Z=1
T r Pr
q Tr
Z
Z
T r Pr
Z
T r Pr
Z
T r Pr
3
ZRT
V
0.896
T r Pr
cm
1605.5
mol
V
Z
Find ( Z)
3.34 Tc
318.7 K
T
348.15 K
Tr
T
Tc
Tr
Pc
37.6 bar
P
15 bar
Pr
P
Pc
Pr
Ans.
1.092
0.399
P
0.286
(guess)
6
3
(a)
B
Given
194
cm
C
mol
PV
=1
RT
15300
cm
mol
B
V
V
cm
1722
mol
Find ( V)
(b)
B0
0.083
V
1930
mol
2
3
V
3
cm
C
V
V
2
RT
P
0.422
Tr
1.6
B0
40
Z
0.283
PV
RT
Z
0.893
Ans.
B1
0.172
0.139
Tr
Z
1
B0
(c)
B1
4.2
Pr
B1
Z
Tr
0.02
V
0.899
3
Z RT
V
P
1734
cm
mol
Ans.
For Redlich/Kwong EOS:
( r)
T
Tr
0.5
Table 3.1
Pr
T r Pr
Table 3.1
0.42748
0.08664
0
1
Tr
q Tr
Eq. (3.54)
Tr
Eq. (3.53)
Tr
Calculate Z
Guess:
Given
T r Pr
0.9
Eq. (3.52)
Z=1
Z
Z
Find Z)
(
(d)
q Tr
Z
Z
T r Pr
Z
T r Pr
Z
T r Pr
3
Z RT
V
0.888
T r Pr
cm
1714.1
mol
V
P
Ans.
For SRK EOS:
0.42748
0.08664
0
1
1
Tr
q Tr
1
0.480
Tr
1.574
0.176
Eq. (3.54)
Tr
41
2
1
T r Pr
Tr
Table 3.1
2
Table 3.1
2
Pr
Tr
Eq. (3.53)
Calculate Z
Guess:
Eq. (3.52)
Given
Z=1
T r Pr
Z
(e)
q Tr
Z
Find ( Z)
Z
0.9
Z
T r Pr
Z
Z
T r Pr
T r Pr
3
ZRT
P
V
0.895
T r Pr
V
1726.9
Ans.
mol
For Peng/Robinson EOS:
1
1
2
0.45724
0.07779
2
1
Tr
1
0.37464
Tr
q Tr
T r Pr
Guess:
Z
T r Pr
q Tr
Z
Find ( Z)
Tr
2
Table 3.1
2
Pr
Eq. (3.53)
Tr
0.9
523.15 K
Z
T r Pr
Z
ZRT
P
V
0.882
P
T r Pr
cm
152.5
mol
B
Given
Z
PV
RT
T r Pr
Z
T r Pr
3
cm
1701.5
mol
V
Ans.
1800 kPa
6
3
(a)
1
Table 3.1
Eq. (3.52)
Z=1
T
2
0.26992
Tr
Given
Z
1.54226
Eq. (3.54)
Calculate Z
3.35
cm
C
mol
C
B
V
PV
=1
RT
V
2
2
V
RT
(guess)
P
V
5800
cm
Find ( V)
3
V
2250
42
cm
mol
Z
0.931
Ans.
(b) Tc
Tr
Tr
B1
647.1 K
Pc
T
Tc
Pr
0.139
0.172
Tr
(c) Table F.2:
B0
Pc
4.2
0.422
0.083
Tr
B0
0.082
B1
0.51
Z
0.281
1.6
1
B0
B1
Pr
Tr
3
Z RT
P
V
P
Pr
0.808
0.345
220.55 bar
Z
molwt
V
V
cm
molwt
124.99
gm
V
0.939
cm
2268
mol
cm
2252
mol
Ans.
3
gm
18.015
mol
3
or
cm
53.4
mol
B
T
Zi
Z1i
C
2620
cm
mol
2
D
5000
cm
mol
3
n
mol
273.15 K
PV
Given
i
9
6
3
3.37
Ans.
RT
0 10
Pi
fPi Vi Pi
RT
1
B Pi
RT
=1
10
B
V
10
C
V
2
D
V
fP V)
(
3
RT
Pi
Vi
20 i bar
Find V)
(
(guess)
Eq. (3.12)
Eq. (3.38)
43
Z2i
1
2
1
4
B Pi
RT
Eq. (3.39)
1·10 -10
Z2i
Z1i
Zi
20
1
1
1
40
0.953
0.953
0.951
60
0.906
0.906
0.895
0.861
0.859
0.83
0.749
80
Pi
100
bar
0.819
0.812
120
0.784
0.765
0.622
140
0.757
0.718
0.5+0.179i
160
0.74
0.671
0.5+0.281i
180
0.733
0.624
0.5+0.355i
200
0.735
0.577
0.5+0.416i
0.743
0.53
0.5+0.469i
Note that values of Z from Eq. (3.39) are not physically meaningful for
pressures above 100 bar.
1
0.9
Zi
0.8
Z1 i
Z2 i
0.7
0.6
0.5
0
50
100
Pi bar
44
150
1
200
3.38 (a) Propane:
Tc
Tc
313.15 K
P
Tr
T
Pc
T
Tr
369.8 K
0.847
Pr
0.152
42.48 bar
13.71 bar
P
Pc
Pr
0.323
For Redlich/Kwong EOS:
( r)
T
Tr
0.5
Table 3.1
Pr
T r Pr
Table 3.1
0.42748
0.08664
0
1
Tr
q Tr
Eq. (3.54)
Tr
Eq. (3.53)
Tr
Calculate Z for liquid by Eq. (3.56) Guess:
Z
0.01
Given
Z=
Z
T r Pr
Z
Find Z) Z
(
T r Pr
0.057
Z
Calculate Z for vapor by Eq. (3.52)
Z
T r Pr
q Tr
T r Pr
3
Z RT
P
V
1
T r Pr
V
Guess:
108.1
cm
Ans.
mol
Z
0.9
V
cm
1499.2
mol
Given
Z=1
Z
T r Pr
Find Z)
(
q Tr
Z
T r Pr
Z
ZZ
V
0.789
45
T r Pr
T r Pr
Z RT
P
3
Ans.
Rackett equation for saturated liquid:
T
Tc
Tr
Tr
0.847
3
Vc
V
200.0
V c Zc
cm
Zc
mol
1 Tr
0.276
3
0.2857
V
94.17
cm
Ans.
mol
For saturated vapor, use Pitzer correlation:
B0
0.083
0.422
Tr
B1
0.139
V
0.468
B1
0.207
0.172
Tr
RT
P
B0
1.6
4.2
R B0
B1
Tc
V
Pc
46
1.538
3
3 cm
10
mol
Ans.
Parts (b) through (t) are worked exactly the same way. All results are
summarized as follows. Volume units are cu.cm./mole.
R/K, Liq. R/K, Vap. Rackett Pitzer
(a) 108.1
1499.2
94.2 1537.8
(b) 114.5
1174.7
98.1
1228.7
(c) 122.7
920.3
102.8
990.4
(d) 133.6
717.0
109.0
805.0
(e) 148.9
1516.2
125.4
1577.0
(f) 158.3
1216.1
130.7
1296.8
(g) 170.4
971.1
137.4
1074.0
(h) 187.1
768.8
146.4
896.0
(i) 153.2
1330.3
133.9
1405.7
(j) 164.2
1057.9
140.3
1154.3
(k) 179.1
835.3
148.6
955.4
(l) 201.4
645.8
160.6
795.8
(m)
61.7
1252.5
53.5
1276.9
(n)
64.1
1006.9
55.1
1038.5
(o)
66.9
814.5
57.0
853.4
(p)
70.3
661.2
59.1
707.8
(q)
64.4
1318.7
54.6
1319.0
(r)
67.4
1046.6
56.3
1057.2
(s)
70.8
835.6
58.3
856.4
(t)
74.8
669.5
60.6
700.5
47
3.39 (a) Propane
T
Tc
273.15)K
T
Tr
(
40
T
Tc
Tr
Pc
369.8 K
Pr
0.847
0.152
42.48 bar
P
P
Pc
13.71 bar
Pr
313.15 K
0.323
From Table 3.1 for SRK:
0.42748
0.08664
0
1
2
1
Tr
1
0.480
Tr
q Tr
1.574
0.176
Eq. (3.54)
2
1
2
Tr
Pr
T r Pr
Tr
Calculate Z for liquid by Eq. (3.56) Guess:
Eq. (3.53)
Tr
Z
0.01
Given
Z=
Z
T r Pr
Find Z)
(
Z
Z
T r Pr
0.055
Z
Calculate Z for vapor by Eq. (3.52)
Z
T r Pr
q Tr
T r Pr
3
Z RT
P
V
1
T r Pr
cm
104.7
mol
V
Guess:
Z
Ans.
0.9
Given
Z=1
Z
T r Pr
Find Z)
(
q Tr
Z
0.78
Z
T r Pr
Z
V
48
T r Pr
Z
T r Pr
Z RT
P
T r Pr
3
V
1480.7
cm
mol
Ans.
Parts (b) through (t) are worked exactly the same way. All results are
summarized as follows. Volume units are cu.cm./mole.
SRK, Liq. SRK, Vap. Rackett Pitzer
(a) 104.7
1480.7
94.2 1537.8
(b) 110.6
1157.8
98.1
1228.7
(c) 118.2
904.9
102.8
990.4
(d) 128.5
703.3
109.0
805.0
(e) 142.1
1487.1
125.4
1577.0
(f) 150.7
1189.9
130.7
1296.8
(g) 161.8
947.8
137.4
1074.0
(h) 177.1
747.8
146.4
896.0
(i) 146.7
1305.3
133.9
1405.7
(j) 156.9
1035.2
140.3
1154.3
(k) 170.7
815.1
148.6
955.4
(l) 191.3
628.5
160.6
795.8
(m)
61.2
1248.9
53.5
1276.9
(n)
63.5
1003.2
55.1
1038.5
(o)
66.3
810.7
57.0
853.4
(p)
69.5
657.4
59.1
707.8
(q)
61.4
1296.8
54.6
1319.0
(r)
63.9
1026.3
56.3
1057.2
(s)
66.9
817.0
58.3
856.4
(t)
70.5
652.5
60.6
700.5
49
3.40 (a) Propane
T
(
40
Tr
Tc
T
Tc
Pc
369.8 K
T
273.15)K
Tr
P
P
Pc
13.71 bar
Pr
313.15 K
Pr
0.847
0.152
42.48 bar
0.323
From Table 3.1 for PR:
2
1
Tr
1
1
0.37464
1
2
0.26992
2
Eq. (3.54)
1
2
Pr
T r Pr
Tr
Tr
0.45724
0.07779
2
Tr
q Tr
1.54226
Calculate Z for liquid by Eq. (3.56) Guess:
Eq. (3.53)
Tr
Z
0.01
Given
Z=
Z
T r Pr
Find Z)
(
Z
Z
T r Pr
0.049
Z
Calculate Z for vapor by Eq. (3.52)
Z
T r Pr
q Tr
T r Pr
3
Z RT
P
V
1
T r Pr
cm
92.2
mol
V
Guess:
Z
Ans.
0.6
Given
Z=1
Z
T r Pr
Find Z)
(
q Tr
Z
0.766
Z
T r Pr
Z
V
50
T r Pr
Z
T r Pr
Z RT
P
T r Pr
3
V
1454.5
cm
mol
Ans.
Parts (b) through (t) are worked exactly the same way. All results are
summarized as follows. Volume units are cu.cm./mole.
PR, Liq. PR, Vap. Rackett Pitzer
(a) 92.2
1454.5
94.2 1537.8
(b)
97.6
1131.8
98.1
1228.7
(c) 104.4
879.2
102.8
990.4
(d) 113.7
678.1
109.0
805.0
(e) 125.2
1453.5
125.4
1577.0
(f) 132.9
1156.3
130.7
1296.8
(g) 143.0
915.0
137.4
1074.0
(h) 157.1
715.8
146.4
896.0
(i) 129.4
1271.9
133.9
1405.7
(j) 138.6
1002.3
140.3
1154.3
(k) 151.2
782.8
148.6
955.4
(l) 170.2
597.3
160.6
795.8
(m)
54.0
1233.0
(n)
56.0
987.3
55.1
1038.5
(o)
58.4
794.8
57.0
853.4
(p)
61.4
641.6
59.1
707.8
(q)
54.1
1280.2
54.6
1319.0
(r)
56.3
1009.7
56.3
1057.2
(s)
58.9
800.5
58.3
856.4
(t)
62.2
636.1
60.6
700.5
53.5
1276.9
51
3.41 (a) For ethylene,
molwt
28.054
gm
Tc
mol
282.3 K
Pc
50.40 bar
0.087
T
Tr
Tc
328.15 K
P
35 bar
P
Pc
Pr
T
Tr
1.162
Pr
0.694
From Tables E.1 & E.2:
Z
Z0
Z1
Z
Z0
Z1
0.838
0.033
0.841
Vtotal
ZnRT
P
Vtotal
0.421 m
P
115 bar
Vtotal
0.25 m
Tr
1.145
Pr
P
Pc
From Tables E.3 & E.4: Z0
0.482
Z1
0.126
18 kg
n
molwt
(b) T
323.15 K
T
Tc
Tr
Z
Z0
Z
Z1
mass
Pr
ZRT
mass
n molwt
Ans.
3
P Vtotal
n
0.493
3
n
2.282
2171 mol
Ans.
60.898 kg
3.42 Assume validity of Eq. (3.38).
3
P1
1bar
Z1
P1 V 1
R T1
T1
Z1
300K
0.922
cm
23000
mol
V1
R T1
Z1
P1
B
With this B, recalculate at P2
Z2
1
B P2
R T1
Z2
0.611
P2
V2
52
R T1 Z2
P2
B
1
3
3 cm
1.942 10
mol
5bar
V2
3.046
10
3
3 cm
mol
Ans.
3.43 T
753.15 K
Tc
513.9 K
Tr
T
Tc
Tr
1.466
P
6000 kPa
Pc
61.48 bar
Pr
P
Pc
Pr
0.976
0.645
B0
0.083
0.422
Tr
B1
0.172
0.139
Tr
RT
P
V
B0
For an ideal gas:
3.44 T
320 K
0.104
3
V
cm
989
mol
V
Pc
cm
1044
mol
Ans.
3
RT
P
V
P
0.146
B1
4.2
Tc
B1 R
B0
1.6
Tc
369.8 K
Pc
Zc
16 bar
42.48 bar
0.276
molwt
3
Vc
0.152
Tr
Vliq
Vtank
cm
200
mol
T
Tc
Tr
V c Zc
1 Tr
3
B0
0.083
Tr
B1
0.139
1.6
0.172
Tr
4.2
Pr
gm
mol
0.377
3
cm
Vliq
0.8 Vtank
Vliq
molwt
0.422
P
Pc
0.2857
mliq
0.35 m
Pr
0.865
44.097
B0
0.449
B1
0.177
53
96.769
mliq
127.594 kg
mol
Ans.
RT
P
Vvap
B0
B1 R
Tc
3
3 cm
Vvap
1.318
mvap
Pc
10
2.341 kg
mol
0.2 Vtank
mvap
Vvap
Ans.
molwt
3.45 T
B0
Tc
425.1 K
2.43 bar
Pc
37.96 bar
Pr
Vvap
0.083
0.422
Tr
B1
0.139
RT
P
1.6
0.172
Tr
V
T
Tr
0.200
P
298.15 K
4.2
B0
3
gm
mol
0.624
Tc
B1 R
V
Pc
Vvap
mvap
0.064
0.661
B1
58.123
0.701
Pr
Pc
molwt
16 m
B0
Tr
Tc
P
mvap
V
molwt
9.469
98.213 kg
3
3 cm
10
mol
Ans.
P
333.15 K
Tc
305.3 K
Tr
T
Tc
Tr
1.091
14000 kPa
Pc
48.72 bar
Pr
P
Pc
Pr
2.874
0.100
3.46 (a) T
Vtotal
From tables E.3 & E.4: Z0
3
molwt
0.15 m
0.463
54
Z1
0.037
30.07
gm
mol
Z
Z0
Vtotal
methane
(b)
Z
Z1
methane
V
molwt
Vtotal
V
P
40 kg
or
Tr =
Whence
V
0.459
Z
Tr =
PV
29.548
R Tc
at
90.87
mol
P V = Z R T = Z R Tr Tc
20000 kPa
0.889
Z
V
P
cm
Ans.
49.64 kg
where
3
Z RT
Pr
P
Pc
Pr
mol
kg
4.105
This equation giving Tr as a function of Z and Eq. (3.57) in conjunction with
Tables E.3 & E.4 are two relations in the same variables which must be
satisfied at the given reduced pressure. The intersection of these two
relations can be found by one means or another to occur at about:
Tr
Whence
T
V
or
T
or
3
0.15 m
Vtotal
T
298.15 K
molwt
0.087
50.40 bar
P V = P r Pc V = Z R T
40 kg
molwt
Whence
0.693
118.5 degC Ans.
Pc
282.3 K
Pr =
Z
Tr Tc
391.7 K
3.47 Vtotal
Tc
and
1.283
Z
RT
Pc V
where
Pr = 4.675 Z
at
Tr
55
T
Tc
4.675
Tr
1.056
28.054
gm
mol
This equation giving Pr as a function of Z and Eq. (3.57) in conjunction with
Tables E.3 & E.4 are two relations in the same variables which must be
satisfied at the given reduced temperature. The intersection of these two
relations can be found by one means or another to occur at about:
Pr
and
1.582
3.48 mwater
Z
P
Pc Pr
3
Vtotal
15 kg
P
0.338
V
0.4 m
Interpolate in Table F.2 at 400 degC to find:
298.15 K
Tc
305.3 K
Tr1
P1
2200 kPa
Pc
48.72 bar
Pr1
3
T2
Z0
mwater
T1
Tc
P1
Pc
3
V
26.667
Z
Z1
Z1
0.422
Tr2
B1
0.139
T2
Tc
Z R T1
P1
Tr2
0.806
B0
1.6
0.172
Tr2
Tr1
0.977
Pr1
0.452
1.615
3
V1
B1
4.2
0.113
0.116
R T2
P2
V1
B0
gm
0.0479
V1
.8105
Tr2
493.15 K
0.083
cm
Ans.
Assume Eq. (3.38) applies at the final state.
B0
Ans.
0.100
0.35 m
From Tables E.1 & E.2: Z0
Z
Vtotal
P = 9920 kPa
3.49 T1
Vtotal
79.73 bar
B1 R
P2
Tc
Pc
56
42.68 bar
Ans.
cm
908
mol
3.50 T
B0
0.5 m
0.083
0.139
73.83 bar
0.422
Tr
B1
304.2 K
Pc
3
Vtotal
B0
4.2
V
10 kg
molwt
0.224
0.997
molwt
44.01
0.036
Vtotal
V
Tr
Tc
0.341
B1
1.6
0.172
Tr
T
Tr
Tc
303.15 K
3
3 cm
2.2
10
mol
RT
P
V
B0
B1 R
P
Tc
10.863 bar
Ans.
Pc
3.51 Basis: 1 mole of LIQUID nitrogen
P
B0
Tc
126.2 K
Tr
1 atm
Pc
34.0 bar
Pr
0.038
Tn
molwt
77.3 K
0.083
0.139
1
B0
4.2
B1
Pr
Tr
0.842
B1
0.172
Tr
Z
mol
B0
1.6
1.209
Z
0.957
57
Tr
P
Pc
Vliq
0.613
Pr
Tc
gm
0.422
Tr
B1
28.014
Tn
0.03
3
34.7 cm
gm
mol
P Vliq
nvapor
nvapor
Z R Tn
5.718
10
3
mol
Final conditions:
ntotal
T
1 mol
V
nvapor
RT
Pig
Pig
V
V
T
Tc
69.005
Tr
ntotal
Tr
298.15 K
3
2 Vliq
cm
2.363
mol
359.2 bar
Use Redlich/Kwong at so high a P.
2
2
a
Tr R Tc
Pc
a
3
3 bar cm
0.901 m
2
b
Eq. (3.42)
RT
Vb
Tr
.5
R Tc
Tr
0.651
Eq. (3.43)
Pc
3
b
mol
P
( r)
T
0.42748
0.08664
a
V ( b)
V
Eq. (3.44)
cm
26.737
mol
P
450.1 bar
Ans.
3
3.52 For isobutane:
T1
Tr1
Tr1
300 K
T1
Tc
0.735
Tc
408.1 K
Pc
36.48 bar
V1
1.824
P1
4 bar
T2
415 K
P2
cm
75 bar
Pr1
Pr1
P1
Tr2
Pc
Tr2
0.11
58
T2
Tc
1.017
Pr2
Pr2
P2
Pc
2.056
gm
From Fig. (3.17):
r1
2.45
The final T > Tc, and Fig. 3.16 probably should not be used. One can easily
show that
r=
P Vc
with Z from Eq. (3.57) and
Tables E.3 and E.4. Thus
Z RT
3
Vc
Z
262.7
Z0
cm
Z0
0.181
mol
Z
Z1
Eq. (3.75):
V1
r2
Tc
3
r1
cm
2.519
gm
V2
Pc
469.7 K
1.774
r2
Z R T2
r2
3.53 For n-pentane:
0.0756
P2 V c
0.322
V2
Z1
0.3356
33.7 bar
Ans.
0.63
1
gm
3
cm
T1
Tr1
Tr1
P1
291.15 K
T1
Pr1
Tc
T2
1 bar
P1
Tr2
Pc
T2
0.03
Tr2
2.69
r2
r2
2
2
1
0.532
r1
3.54 For ethanol: Tc
Pc
Pc
3.561
gm
3
Ans.
cm
513.9 K
T
453.15 K
Tr
T
Tc
Tr
0.882
61.48 bar
P
200 bar
Pr
P
Pc
Pr
3.253
3
Vc
P2
2.27
From Fig. (3.16):
By Eq. (3.75),
Pr2
0.88
r1
120 bar
Pr2
Tc
Pr1
0.62
P2
413.15 K
167
cm
mol
molwt
59
46.069
gm
mol
From Fig. 3.16:
r
2.28
=
r
r
c=
0.629
Vc
r
Vc
gm
Ans.
3
cm
molwt
3.55 For ammonia:
Tc
405.7 K
T
293.15 K
Tr
T
Tc
Tr
0.723
Pc
112.8 bar
P
857 kPa
Pr
P
Pc
Pr
0.076
Vc
cm
72.5
mol
Zc
0.242
3
Eq. (3.72):
B0
Vliquid
0.083
0.139
Vvapor
cm
27.11
mol
Vliquid
B0
4.2
B0
B1 R
0.627
B1
1.6
0.172
Tr
RT
P
3
0.2857
0.422
Tr
B1
V c Zc
1 Tr
0.253
0.534
Tc
Pc
3
Vvapor
cm
2616
mol
3
V
Vvapor
Vliquid
V
60
cm
2589
mol
Ans.
Alternatively, use Tables E.1 & E.2 to get the vapor volume:
Z0
Z1
0.929
Vvapor
Z
0.071
Z0
Z
Z1
0.911
3
ZRT
P
Vvapor
cm
2591
mol
3
V
Vvapor
V
Vliquid
cm
2564
Ans.
mol
3.5810 gal. of gasoline is equivalent to 1400 cu ft. of methane at 60 degF and 1
atm. Assume at these conditions that methane is an ideal gas:
3
R
ft atm
0.7302
lbmol rankine
V
1400 ft
T
P
1 atm
n
3
519.67 rankine
PV
RT
n
3.689 lbmol
For methane at 3000 psi and 60 degF:
Tc
190.6 1.8 rankine
T
519.67 rankine
Tr
T
Tc
Tr
1.515
Pc
45.99 bar
P
3000 psi
Pr
P
Pc
Pr
4.498
Z
0.822
0.012
From Tables E.3 & E.4:
Z0
0.819
Vtank
ZnRT
P
Z1
Z
0.234
Vtank
61
5.636 ft
Z0
3
Z1
Ans.
3.59 T
P
25K
3.213bar
Calculate the effective critical parameters for hydrogen by equations (3.58)
and (3.56)
43.6
Tc
1
K
1
30.435 K
Pc
10.922 bar
2.016T
20.5
Pc
Tc
bar
21.8K
44.2K
2.016T
0
P
Pc
Pr
Pr
Initial guess of volume:
T
Tr
0.294
Tr
cm
646.903
mol
3
RT
P
V
0.821
V
Tc
Use the generalized Pitzer correlation
B0
0.083
0.422
Tr
Z
1
B0
1.6
B0
B1
0.495
B1
0.139
0.172
Tr
Pr
Tr
Z
0.823
Ans.
4.2
B1
0.254
Experimental: Z = 0.7757
For Redlich/Kwong EOS:
0
1
Tr
T r Pr
Tr
0.5
Pr
Tr
Table 3.1
Eq. (3.53)
62
Table 3.1
0.42748
0.08664
q Tr
Tr
Tr
Eq. (3.54)
Calculate Z
Guess:
Given
T r Pr
0.9
Eq. (3.52)
Z=1
Z
Z
q Tr
Z
Find Z)
(
3.61For methane:
ZZ
Tc
0.012
T
T r Pr
T r Pr
Ans.
0.791
At standard condition:
Z
T r Pr
Experimental: Z = 0.7757
Pc
190.6K
(
60
32)
5
9
45.99bar
273.15 K
T
P
Tr
T
Tc
B0
0.083
Z0
Z
1
Tr
0.422
Tr
Pr
B0
Z0
1.6
Pitzer correlations:
T
Tc
Tr
0.083
0.422
Tr
B1
0.139
1.6
0.172
Tr
4.2
0.134
B1
0.139
0.172
T
T
Tr
Z RT
P
(
50
32)
5
9
0.109
Tr
0.00158
3
V1
273.15 K
m
0.024
mol
P
300psi
Pr
0.45
283.15 K
1.486
B0
B1
B1
V1
0.998
B1
Pr
Z1
0.998
4.2
0.022
Z1
P
Pc
Tr
Z
Z1
(a) At actual condition:
B0
B0
Z0
Tr
Pr
1.515
1atm
Pr
Pitzer correlations:
288.706 K
0.141
0.106
63
Pr
P
Pc
Z0
Z
1
B0
Z0
Pr
Tr
Z1
6 ft
q1
150 10
(b) n1
(c) D
u
Z0
0.957
Z
Z1
0.958
q1
n1
7.485
q2
V1
q1
V1
22.624in
q2
A
A
u
4
D
8.738
2
m
s
64
3 kmol
10
hr
A
Ans.
0.0322
3
V2
P
V2
q2
day
Z1
Tr
ZRT
V2
3
Pr
B1
6.915
Ans.
2
0.259 m
0.00109
10
6 ft
m
mol
3
day
Ans.
3.62
0.012
0.286
0.087
0.281
0.1
0.279
0.140
0.289
0.152
0.276
0.181
0.282
0.187
0.271
0.19
0.267
0.191
0.277
0.194
0.275
0.196
0.273
0.2
0.274
0.205
0.273
0.21
0.273
0.21
ZC
Use the first 29 components in Table B.1
sorted so that values are in ascending
order. This is required for the Mathcad
slope and intercept functions.
0.271
m
slope
b
intercept
r
corr
0.212
0.272
0.218
0.275
0.23
0.272
0.235
0.269
0.252
0.264
0.28
0.265
ZC
0.297
0.256
m
0.301
0.266
0.302
0.266
0.303
0.263
0.31
0.263
0.322
0.26
0.326
0.261
( 0.091)
0.27
0.262
ZC
ZC
ZC
( .291)
0
2
( 0.878) r
0.771
0.29
0.28
b 0.27
0.26
0.25
0
0.1
0.2
The equation of the line is:
Zc = 0.291 0.091
65
0.3
0.4
Ans.
5
7
R
2
Cv
298.15K
P1
P2
T1
T3
5bar
P3
Cp
1bar
5bar
3.65 Cp
T1
2
R
1.4
Cv
Step 1->2 Adiabatic compression
1
T2
T1
P2
T2
P1
472.216 K
U12
Cv T2
T1
U12
3.618
H12
Cp T2
T1
H12
5.065
Q12
W12
0
kJ
mol
kJ
mol
kJ
mol
kJ
mol
Q12
0
U12
W12
3.618
Ans.
Ans.
Ans.
kJ
mol
Ans.
Step 2->3 Isobaric cooling
U23
Cv T3
T2
U23
3.618
H23
Cp T3
T2
H23
5.065
Q23
W23
R T3
Q23
H23
W23
T2
5.065
1.447
kJ
mol
kJ
mol
kJ
mol
kJ
mol
Ans.
Ans.
Ans.
Ans.
Step 3->1 Isothermal expansion
U31
Cv T1
T3
U31
0
H31
Cp T1
T3
H31
0
66
kJ
mol
kJ
mol
Ans.
Ans.
Q31
R T 3 ln
W31
P1
Q31
Q31
P3
W31
kJ
mol
kJ
3.99
mol
3.99
Ans.
Ans.
For the cycle
Qcycle
Q12
Wcycle
Q23
W12
Qcycle
Q31
W23
Wcycle
W31
1.076
1.076
kJ
Ans.
mol
kJ
Ans.
mol
Now assume that each step is irreversible with efficiency:
80%
Step 1->2 Adiabatic compression
W12
Q12
W12
U12
W12
Q12
W12
4.522
kJ
mol
kJ
mol
0.904
Ans.
Ans.
Step 2->3 Isobaric cooling
W23
Q23
W23
U23
W23
1.809
kJ
mol
kJ
mol
Q23
5.427
W31
W23
3.192
Ans.
Ans.
Step 3->1 Isothermal expansion
W31
Q31
W31
U31
Q31
W31
3.192
kJ
mol
kJ
mol
Ans.
Ans.
For the cycle
Qcycle
Q12
Wcycle
W12
Q23
W23
Qcycle
Q31
Wcycle
W31
67
kJ
Ans.
mol
kJ
3.14
mol Ans.
3.14
3.67 a) PV data are taken from Table F.2 at pressures above 1atm.
125
150
1757.0
175
P
2109.7
1505.1
200
1316.2
cm
1169.2
gm
V
kPa
225
250
300
Z
955.45
T
(
300
273.15)
K
M
1051.6
275
3
18.01
gm
mol
875.29
1
VM
PVM
RT
i
07
If a linear equation is fit to the points then the value of B is the y-intercept.
Use the Mathcad intercept function to find the y-intercept and hence, the
value of B
Yi
Zi
1
3
Xi
B
intercept ( Y)
X
A
i
slope ( Y)
X
Ans.
6
5 cm
10
2
i
A
cm
128.42
mol
1.567
B
mol
X
0
mol
3
cm
10
5 mol
3
cm
8 10
5 mol
3
cm
Below is a plot of the data along with the linear fit and the extrapolation to
the y-intercept.
68
115
(Z-1)/p
120
125
130
0
2 10
5
4 10
5
6 10
p
5
8 10
5
(Z-1)/p
Linear fit
b) Repeat part a) for T = 350 C
PV data are taken from Table F.2 at pressures above 1atm.
125
150
1912.2
175
P
2295.6
1638.3
200
1432.8
225
V
kPa
1273.1
250
300
Z
1040.7
T
( 350
273.15)K
M
1145.2
275
3
cm
gm
18.01
gm
mol
953.52
1
VM
PVM
RT
i
07
If a linear equation is fit to the points then the value of B is the y-intercept.
Use the Mathcad intercept function to find the y-intercept and hence, the
value of B
Yi
Zi
1
3
Xi
i
B
intercept ( X Y)
i
69
B
105.899
cm
mol
Ans.
A
A
slope ( X Y)
6
5 cm
10
2
1.784
mol
X
0
mol
3
10
cm
5 mol
3
8 10
5 mol
3
cm
cm
Below is a plot of the data along with the linear fit and the extrapolation to
the y-intercept.
90
(Z-1)/p
95
100
105
110
0
2 10
5
4 10
5
6 10
p
5
8 10
5
(Z-1)/p
Linear fit
c) Repeat part a) for T = 400 C
PV data are taken from Table F.2 at pressures above 1atm.
125
150
2066.9
175
P
2481.2
1771.1
200
1549.2
225
kPa
V
1376.6
250
1238.5
275
1125.5
300
1031.4
70
3
cm
gm
T
( 400
273.15)K
M
18.01
gm
mol
1
VM
PV M
RT
Z
i
07
If a linear equation is fit to the points then the value of B is the
y-intercept.
Use the Mathcad intercept function to find the y-intercept and hence,
the value of B
Yi
Zi
1
3
Xi
B
i
intercept ( X Y)
B
i
A
A
slope ( X Y)
89.902
cm
mol
Ans.
6
5 cm
10
2
2.044
mol
X
0
mol
3
cm
10
5 mol
3
8 10
5 mol
3
cm
cm
Below is a plot of the data along with the linear fit and the extrapolation to
the y-intercept.
70
(Z-1)/p
75
80
85
90
0
2 10
5
4 10
p
(Z-1)/p
Linear fit
71
5
6 10
5
8 10
5
3.70
Create a plot of
(Z
1) Z Tr
vs
Pr
Pr
Z Tr
Data from Appendix E at Tr = 1
0.01
0.9967
0.05
0.9832
0.10
0.9659
Z
0.20
Pr
0.9300
0.40
X
0.7574
0.80
1
0.8509
0.60
Tr
0.6355
Pr
(Z
Y
Z Tr
1) Z Tr
Pr
Create a linear fit of Y vs X
Slope
slope ( X Y)
Intercept
Rsquare
Slope
intercept ( X Y)
0.033
Intercept
corr ( X Y)
Rsquare
0.332
0.9965
0.28
0.3
Y
Slope X Intercept
0.32
0.34
0
0.2
0.4
0.6
0.8
1
1.2
1.4
X
The second virial coefficient (Bhat) is the value when X -> 0
Bhat
Intercept
By Eqns. (3.65) and (3.66)
Bhat
B0
0.083
0.422
Tr
These values differ by 2%.
72
1.6
B0
0.332
0.339
Ans.
Ans.
3.71 Use the SRK equation to calculate Z
T
T
273.15)Kr
Tc
150.9 K
T
(
30
Pc
48.98 bar
P
300 bar
Tr
Pr
Pc
2.009
Pr
Tc
P
6.125
0.0
0.42748 Table 3.1
0.08664
0
1
2
1
Tr
1
0.480
Tr
q Tr
0.176
Eq. (3.54)
1
Tr
Table 3.1
2
Pr
T r Pr
Tr
Calculate Z
Guess:
Z
Eq. (3.53)
Tr
0.9
Eq. (3.52)
Given
Z=1
Z
1.574
2
T r Pr
q Tr
Z
Find Z)
(
Z
T r Pr
Z
V
1.025
T r Pr
T r Pr
Z
T r Pr
3
Z RT
P
cm
Ans.
86.1
mol
V
This volume is within 2.5% of the ideal gas value.
3.72 After the reaction is complete, there will be 5 moles of C 2H2 and 5 moles of
Ca(OH)2.
First calculate the volume available for the gas.
n
5mol
V
Vt
3
0.4 1800 cm
3
cm
5 mol 33.0
mol
Vt
n
Vt
3
555 cm
3
V
73
111
cm
mol
Use SRK equation to calculate pressure.
T
Tc
308.3 K
Pc
( 125
61.39 bar
T
Tc
Tr
273.15) K
Tr
0.0
0.42748 Table 3.1
0.08664
0
1
2
1
Tr
1
1.574
0.176
2
2
2
Pc
Eq. (3.45)
3
3 bar cm
3.995 m
2
RT
Vb
3.73 mass
b
Tc
0.152
( 10
Vc
Pc
369.8K
200.0
197.8 bar
Ans.
273.15)K
3
0.276
cm
36.175
mol
P
b)
T
35000kg
Eq. (3.46)
Pc
3
a
V (V
R Tc
b
mol
Zc
Tr
Eq. (3.54)
R Tc
Tr
P
1
Table 3.1
2
Tr
a
a
0.480
Tr
q Tr
1.291
cm
mol
n
M
42.48bar
mass
M
n
44.097
7.937
gm
mol
5
10 mol
a) Estimate the volume of gas using the truncated virial equation
Tr
B0
T
Tr
Tc
0.083
0.422
Tr
B0
1.6
0.766
P
Eq. (3-65)
B1
Pr
1atm
0.139
0.172
Tr
B1
0.564
74
0.389
4.2
P
Pc
Eq. (3-66)
Z
1
B0
B1
Pr
Z
Tr
ZnRT
Vt
0.981
Vt
P
3
This would require a very large tank. If the
D
tank were spherical the diameter would be:
73
2.379
6
Vt
3
10 m m
D
32.565 m
b) Calculate the molar volume of the liquid with the Rackett equation(3.72)
Vliq
V c Zc
P
1
0.2857
Vvap
3
Vliq
6.294atm
Z
1 Tr
P
Pc
Pr
B0
B1
85.444
Pr
0.878
Pr
Tr
ZRT
P
Vvap
Guess:
Vtank
Given
90%
mol
0.15
Z
cm
3
3 cm
3.24 10
mol
90% Vliq n
Vtank
10%
Vtank
=n
Vvap
Vtank
Find Vtank
Vtank
Vliq
75.133 m
This would require a small tank. If the tank
D
were spherical, the diameter would be:
3
3
6
Vtank
D
5.235 m
Although the tank is smaller, it would need to accomodate a pressure of 6.294
atm (92.5 psi). Also, refrigeration would be required to liquify the gaseous
propane stream.
75
Chapter 4 - Section A - Mathcad Solutions
4.1 (a) T0
T
473.15 K
For SO2: A
B
5.699
n
1373.15 K
0.801 10
H
47.007
C
5
D
0.0
1.015 10
R ICPH T0 T A B C D
H
3
10 mol
kJ
T
523.15 K
For propane:A
28.785 10
H
161.834
3
Ans.
12 mol
C
6
8.824 10
D
0
R ICPH T0 T A B C 0.0
H
470.073 kJ
n
1473.15 K
B
1.213
nH
Q
(b) T0
Q
mol
kJ
mol
n
473.15 K
For ethylene:
A
2 (guess)
Q= nR
A T0
3
10 kJ Ans.
800 kJ
3
14.394 10
K
B
B
2
T0
2
1
n
A
1.967
2
T
2.905
533.15 K
For 1-butene:
1.424
1.942
6
4.392 10
C
K
2
Given
Find
(b) T0
Q
10 mol
nH
Q
4.2 (a) T0
Q
C
3
T0
3
1
T
T0
15 mol
Q
31.630 10
K
B
76
3
1
Ans.
1374.5 K
2500 kJ
3
C
9.873 10
K
2
6
(guess)
3
Q= nR
Given
A T0
2
n
500 degF
C
3
T0
3
1
T
2.652
Find
(c) T0
B
2
T0
2
1
3
T
T0
1
Ans.
1413.8 K
6
Q
40 lbmol
10 BTU
Values converted to SI units
T0
n
533.15K
For ethylene:
A
Q= nR
Q
10 mol
B
14.394 10
K
1.424
2 (guess)
4
1.814
6
1.055 10 kJ
3
4.392 10
C
K
Given
A T0
1
Find
B
2
T0
2
2
T
2.256
C
3
T0
3
1
T
T0
3
6
2
1
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
T0
1 atm
(
T
T
PV
R T0
T0
773.15 K
n
For air:
H
32degF) 273.15K
T0
n
A
T
250 ft
V
Convert given values to SI units
T
3
V
122 degF
932 degF
7.079 m
3
T0
32degF
273.15K
323.15 K
266.985 mol
3.355
B
0.575 10
R ICPH T0 T A B C D
77
3
C
0.0
D
0.016 10
5
kJ
13.707
4.4 molwt
Q
gm
mol
100.1
T0
10000 kg
molwt
n
n
For CaCO3: A
323.15 K
B
10 mol
2.637 10
H
9.441
3
C
D
0.0
3.120 10
5
R ICPH T0 T A B C D
H
1153.15 K
3
10 BTU Ans.
4
9.99
12.572
3.469
T
mol
nH
Q
H
4J
10
Q
mol
nH
6
Q
Ans.
9.4315 10 kJ
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
T2
290.41 K
Given
J
mol K
T2 = T1
4.9a) Acetone: Tc
Hn
(guess)
29.10
P2
P2
P3
R
CP
CP
P1
Pc
508.2K
kJ
47.01bar
Tn
Trn
mol
CP
Find CP
Tc
56.95
Tn
329.4K
Trn
J
Ans.
mol K
0.648
Use Eq. (4.12) to calculate H at Tn ( Hncalc)
1.092 ln
Hncalc
R Tn
Pc
1.013
bar
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
Hn
3.464 %
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
Benz
ene
iso-Butane
n-Butane
1-Butanol
Carbon tetrachloride
Chlorobenz
ene
Chloroform
Cyclohexane
Cyclopentane
n-Decane
Dichloromethane
Diethyl ether
Ethanol
Ethylbenz
ene
Ethylene glycol
n-Heptane
n-Hexane
M ethanol
M ethyl acetate
M ethyl ethyl k
etone
Nitromethane
n-Nonane
iso-Octane
n-Octane
n-Pentane
Phenol
1-Propanol
2-Propanol
Toluene
W ater
o-Xylene
m-Xylene
p-Xylene
Hn (k mol)
J/
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)
33.70
469.7
Tc
507.6
562.2
30.25
Pc
K
48.98
36.0
68.7
H25
273.15 K
80.0
366.1
72.150
J
M
433.3 gm
392.5
Tn
Tr2
Tc
(
25
273.15)
K
Tc
H2
H25 M
0.673
H2
0.628
H1
31.533
Hn
31.549
kJ
33.847 mol
32.242
1
Tr2
1
0.38
Tr1
Eq. (4.13) %error
26.448
H2calc
H1
26.429
0.631
H2calc
86.177 gm
78.114 mol
82.145
0.658
Tr1
kJ
30.72 mol
29.97
366.3
80.7
Tr1
28.85
Hn
bar
43.50
560.4
Tn
25.79
kJ
Ans.
33.571 mol
32.816
31.549
kJ
33.847 mol
32.242
H2
H2
0.072
26.429
H2
H2calc
%error
0.052
0.814
%
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
V
5
1.934
18.787
P
0
23.767
t
5
26.617
dPdT
T
H
2
ti
1
459.67
yi
ln Pi
15
slope ( x y) slope
( P) 3
xi
10
29.726
slope
15
5
21.162
Data:
i
0.012
4952
slopedPdT
T V dPdT
5.4039
H
0.545
90.078
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 ( ):
(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
4.11
M
32.042
153.822
H is the value at
0 degC.
536.4
mol
Tc
334.3
512.6 K Pc
80.97 bar Tn
337.9 K
556.4
gm
54.72
45.60
349.8
Hexp is the given
value at the normal
boiling point.
81
Tr1
273.15K
Tc
Tr2
Tn
Tc
270.9
H
1189.5
0.509
246.9
J
gm
Hexp
217.8
J
gm
1099.5
Hn
Hn
Hexp
Hexp
Hn
PCE
0.38
Tr1
0.52
Hexp
1195.3
1.092 ln
R Tn
Hn
Hexp
4.03 %
M
Pc
1.013
bar
0.930
Tr2
100%
0.34
247.7
Hn
1
H
Tr2
0.77
J
1055.2
gm
193.2
(b) By Eq. (4.12):
PCE
1
0.629
This is the % error
100%
245
Hn
0.659
Tr2
0.491
194.2
(a) By Eq. (4.13)
PCE
0.533
Tr1
0.623
J
gm
PCE
8.72
%
0.96
192.3
4.12 Acetone
0.307
Tc
508.2K
Pc
Tn
329.4K
P
1atm
Pr
P
Pc
47.01bar
Zc
0.233
3
Vc
Tr
cm
209
mol
Tn
Tc
Tr
0.648
82
Hn
29.1
Pr
kJ
mol
0.022
Generalized Correlations to estimate volumes
Vapor Volume
B0
0.422
0.083
Tr
B1
Tr
4.2
Pr
Z
1
V
B1
Pr
Z R Tn
P
Tr
Tr
0.762
Eq. (3.65)
B1
0.172
0.139
B0
B0
1.6
0.924
Eq. (3.66)
Z
V
(Pg. 102)
0.965
3
4 cm
2.609
10
mol
Liquid Volume
2
Vsat
V c Zc
1 Tr
3
7
Eq. (3.72)
Vsat
Combining the Clapyeron equation (4.11)
gives
A
Vsat
14.3145
V
B
2.602
A
B
H=T V
V
2
e
C)
3
4 cm
10
d
Psat
dT
TC
Psat = e
(
T
V
H=T V
B
A
with Antoine's Equation
cm
70.917
mol
mol
2756.22
C
83
228.060
B
( C)
T
B
A
Tn 273.15K
Hcalc
B
Tn V
Tn
C
K
e
2
273.15K
kPa
K
C
K
Hcalc
29.662
kJ
Ans.
mol
%error
Hcalc
Hn
%error
1.9 %
Hn
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
Benz
ene
isoButane
nButane
1Butanol
Carbon tetrachloride
Chlorobenz
ene
Chloroform
Cyclohexane
Cyclopentane
nDecane
Dichloromethane
Diethyl ether
Ethanol
Ethylbenz
ene
Ethylene glycol
nHeptane
nHexane
M ethanol
M ethyl acetate
M ethyl ethyl k
etone
Nitromethane
nNonane
isoOctane
nOctane
nPentane
Phenol
1- panol
Pro
Hn ( J/
k 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
W ater
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%
4.13 Let P represent the vapor pressure.
T
ln
Given
P
P
348.15 K
P
= 48.157543
kPa
Find ( ) dPdT
P
P
5622.7 K
T
P
Clapeyron equation:
B
dPdT =
Pc
AL
13.431
BL
2.211
0.029
bar
K
3
cm
96.49
mol
Vliq
H
T dPdT
1
512.6K
AV
dPdT
3
4.14 (a) Methanol: Tc
AL
4.70504
T
T
K
4.70504 ln
H
T V Vliq
Vliq
PV
V
RT
CPL ( )
T
5622.7 K
T
joule
31600
mol
V = vapor molar volume. V
Eq. (3.39)
2
H
87.396 kPa
(guess)
100 kPa
BL
K
BV
85
cm
1369.5
mol
B
Tn
80.97bar
51.28 10
T
CL
K
2
T
2
3
Ans.
337.9K
CL
131.13 10
CV
3.450 10
6
R
12.216 10
3
6
CPV ( )
T
P
BV
AV
CV
T
K
K
Tsat
3bar
2
T
2
R
T1
368.0K
T2
300K
500K
Estimate Hv using Riedel equation (4.12) and Watson correction (4.13)
Tn
Trn
Trn
Tc
1.092 ln
Hn
Pc
Hv
Trsat
1
Trn
kJ
mol
35.645
kJ
mol
T2
CPL ( ) T
Td
100
38.301
Hv
Hv
CPV ( ) T
Td
T1
n
0.718
0.38
Tsat
H
Trsat
Hn
Tc
R Tn
Trn
1
Hn
Tsat
1.013
bar
0.930
Trsat
0.659
kmol
hr
H
49.38
T sat
Q
nH
Q
1.372
3
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
T1sat
562.2K
451.7K
Pc
48.98bar
T2sat
86
358.7K
3
3
Tn
353.2K
Cp
162
J
mol K
Estimate Hv using Riedel equation (4.12) and Watson correction (4.13)
Trn
Tn
Trn
Tc
Hn
Hv
Hn
30.588
30.28
R Tn
Trn
Tr2sat
1
0.638
Hn
Tc
1.013
bar
0.930
1
Tr2sat
Hv
Pc
1.092 ln
T2sat
Tr2sat
0.628
0.38
Trn
kJ
mol
kJ
mol
Assume the throttling process is adiabatic and isenthalpic.
Guess vapor fraction (x): x
Given
Cp T1sat
0.5
4.16 (a) For acetylene:
Tc
Tn
ln
Hn
R Tn 1.092
1
Tr
1
Hn
Hf
227480
0.498
Tn
61.39 bar
Tr
0.614
Pc
bar
J
mol
T
Tc
Tr
1.013
Trn
16.91
Hv
0.930
Hn
6.638
0.38
Trn
Hv
x
Find ( x)
Ans.
189.4 K
298.15 K
Trn
Tc
Pc
308.3 K
T
Trn
x
T2sat = x Hv
H298
Hf
87
Hv
0.967
kJ
mol
kJ
mol
H298
220.8
kJ
mol
Ans.
kJ
mol
(b) For 1,3-butadiene:
H298 = 88.5
(c) For ethylbenzene:
H298 = 12.3
(d) For n-hexane:
H298 = 198.6
(e) For styrene:
H298 = 103.9
4.17
1st law:
dQ = dU
Ideal gas:
Since
mol
kJ
mol
and
V dP = R dT
V dP = P
P
(B)
1
V
R dT
which reduces to
or
dQ =
P dV =
R
1
R dT
1
dQ = CV dT
R dT
1
R dT
1
dQ = CP dT
CP
dV = V dP
dV
Combines with (A) to give:
dQ = CP dT
V dP = R dT
P dV
Combines with (B) to yield:
or
(A)
P dV
P dV
then
P V = const
from which
kJ
dW = CV dT
PV= RT
Whence
kJ
mol
R dT
1
R dT
(C)
Since CP is linear in T, the mean heat capacity is the value of
CP at the arithmetic mean temperature. Thus
Tam
88
675
CPm
R 3.85
0.57 10
Integrate (C):
P1
4.18
R
950 K
P2
P1
T1
400 K
1.55
J
Q
6477.5
P2
11.45 bar
H298 = 4 058 910 J
R T2
1
1 bar
Tam
T2
CPm
Q
3
Ans.
Ans.
T1
mol
1
T2
T1
Ans.
For the combustion of methanol:
CH3OH(g) + (3/2)O2(g) = CO2(g) + 2H2O(g)
H298
393509
H298
676485
For 6 MeOH:
2 ( 241818) ( 200660)
For the combustion of 1-hexene:
C6H12(g) + 9O2(g) = 6CO2(g) + 6H2O(g)
H298
H298
6 ( 393509) 6 ( 241818) ( 41950)
3770012
H298 = 3 770 012 J
Ans.
Comparison is on the basis of equal numbers of C atoms.
4.19
C2H4 + 3O2 = 2CO2 + 2H2O(g)
H298
[ ( 241818) 2 ( 393509) 52510]
2
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
Index the product species with the numbers:
1 = oxygen
2 = carbon dioxide
3 = water (g)
4 = nitrogen
(a) For the product species, no excess air:
0
3.639
0.506
2
5.457
1.045
n
A
10
B
2
3.470
3.280
3
1.157
10 K
0.593
ni Ai B
A
14
A
0.121
0.040
D
ni Bi
ni D i
i
i
i
B
54.872
52
D
K
1.450
11.286
i
0.227
0.012
1
K
52
D
1.621 10 K
T0
298.15K
T
For the products,
CP
HP = R
dT
R
T0
The integral is given by Eq. (4.7). Moreover, by an energy balance,
H298
HP = 0
(guess)
2
H298 = R A T0
Given
8.497
Find
1
T
T0
B
2
T0
2
2
T
1
D
T0
2533.5 K
1
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
(e) 50% xs air preheated to 500 degC. For this process,
Hair
H298
HP = 0
Hair = MCPH (
298.15
773.15)
For one mole of air:
3
MCPH 773.15 298.15 3.355 0.575 10
0.0
0.016 10
5
= 3.65606
For 4.5/0.21 = 21.429 moles of air:
Hair = n R MCPH T
Hair
Hair
21.429 8.314 3.65606 ( 98.15
2
309399
J
mol
J
mol
The energy balance here gives:
H298
1.5
n
773.15)
3.639
5.457
HP = 0
0.227
0.506
2
Hair
1.045
A
2
B
3.470
1.450
3.280
16.929
A
3
ni Ai
B
B
5
10 K
0.121
2
0.040
ni Bi
D
i
78.84
1.157
D
0.593
i
A
10
K
ni D i
i
1
0.016
K
D
52
1.735
10 K
2 (guess)
H298
Find
Hair = R A T0
1
D
T0
Given
B
2
T0
2
2
1
1
7.656
T
91
T0 K
T
2282.5 K K
Ans.
4.20
n-C5H12 + 8O2 = 5CO2 + 6H2O(l)
By Eq. (4.15) with data from Table C.4:
H298
5 ( 393509) 6 ( 285830) ( 146760)
H298 = 3 535 765 J
4.21
Ans.
The following answers are found by application of Eq. (4.15) with
data from Table C.4.
(a) -92,220 J
(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
The solution to each of these problems is exactly like that shown in
Example 4.6. In each case the value of Ho298 is calculated in Problem
4.21. Results are given in the following table. In the first column the
letter in ( ) indicates the part of problem 4.21 appropriate to the
value.
T/
K
(a)
(b
)
(f
)
(i
)
(j
)
(l
)
(m)
(n
)
(o
)
(r
)
(t
)
(u
)
(v
)
(w)
(x
)
(y
)
4.23
A
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
-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
Ho298
J
HoT/
-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
This is a simple application of a combination of Eqs. (4.18) & (4.19) with
evaluated parameters. In each case the value of Ho298 is calculated in Pb.
4.21. The values of A, B, C and D 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.
A
103 B 106 C 10-5 D
(e)
-7.425 20.778
0.000
3.737
(g)
-3.629
8.816 -4.904
0.114
(h)
-9.987 20.061 -9.296
1.178
(k)
1.704 -3.997
1.573
0.234
(z)
-3.858 -1.042
0.180
0.919
93
4.24 q
6 ft
150 10
3
T
day
(
60
5
32) K
9
T
273.15K
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:
J
mol
HfCH4
74520
HfCO2
393509
Hc
HfCO2
HfO2
J
mol
J
mol
HfH2Oliq
2 HfH2Oliq
HigherHeatingValue
0
285830
HfCH4
J
mol
2 HfO2
5J
Hc
Hc
8.906 10
mol
Assuming methane is an ideal gas at standard conditions:
n
q
P
RT
n HigherHeatingValue
4.25
8 mol
n
1.793 10
5dollar
GJ
day
5 dollar
7.985 10
day
Ans.
Calculate methane standard heat of combustion with water as liquid product
Standard Heats of Formation: CH4 + 2O2 --> CO2 +2H2O
J
mol
HfCH4
74520
HfCO2
393509
HcCH4
HfCO2
HcCH4
890649
J
mol
HfO2
0
J
mol
HfH2Oliq
2 HfH2Oliq
J
mol
94
HfCH4
285830
J
mol
2 HfO2
Calculate ethane standard heat of combustion with water as liquid product:
Standard Heats of Formation:C2H6 + 7/2O2 --> 2CO2 +3H2O
J
mol
HfC2H6
83820
HcC2H6
2 HfCO2
HcC2H6
1560688
3 HfH2Oliq
HfC2H6
7
2
HfO2
J
mol
Calculate propane standard heat of combustion with water as liquid product
Standard Heats of Formation:C3H8 + 5O2 --> 3CO2 +4H2O
J
mol
HfC3H8
104680
HcC3H8
3 HfCO2
HcC3H8
2219.167
4 HfH2Oliq
HfC3H8
5 HfO2
kJ
mol
Calculate the standard heat of combustion for the mixtures
a) 0.95 HcCH4
0.02 HcC2H6
0.02 HcC3H8
921.714
kJ
mol
b) 0.90 HcCH4
0.05 HcC2H6
0.03 HcC3H8
946.194
kJ
mol
c) 0.85 HcCH4
0.07 HcC2H6
0.03 HcC3H8
932.875
kJ
mol
Gas b) has the highest standard heat of combustion.
4.26
2H2 + O2 = 2H2O(l)
Hf1
C + O2 = CO2(g)
Hf2
N2(g)+2H2O(l)+CO2(g)=(NH2)2CO(s)+3/2O2
H
Ans.
2 ( 285830) J
393509 J
631660 J
.
N2(g)+2H2(g)+C(s)+1/2O2(g)=(NH2)2CO(s)
H298
Hf1
Hf2
H298
H
95
333509 J
Ans.
4.28
On the basis of 1 mole of C10H18
(molar mass = 162.27)
Q
43960 162.27 J
6
Q
7.133 10 J
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,
Q=
T
U=
H
( V)=
P
298.15 K
H
Q
H
R T ngas
ngas
R T ngas
(
10
14.5)mol
6
H
7.145 10 J
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)
H
9H2O(l) = 9H2O(g)
Hvap 9 44012 J
___________________________________________________
C10H18(l) + 14.5O2(g) = 10CO2(g) + 9H2O(g)
H298
4.29
H
Hvap
H298
6748436 J
Ans.
FURNACE: Basis is 1 mole of methane burned with 30% excess air.
CH4 + 2O2 = CO2 + 2H2O(g)
Entering:
Moles methane
n1
1
Moles oxygen
n2
2 1.3
Moles nitrogen
n3
2.6
96
79
21
n2
2.6
n3
9.781
Total moles of dry gases entering
n
n1
n2
n
n3
13.381
At 30 degC the vapor pressure of water is
4.241 kPa. Moles of water vapor entering:
n4
Leaving:
4.241
13.381
101.325 4.241
n4
CO2 -- 1 mol
H2O -- 2.585 mol
O2 -- 2.6 - 2 = 0.6 mol
N2 -- 9.781 mol
0.585
(1)
(2)
(3)
(4)
By an energy balance on the furnace:
Q=
H=
H298
For evaluation of
1
A
0.6
9.781
i
HP we number species as above.
5.457
2.585
n
HP
3.470
3.639
1.045
1.450
B
R
8.314
B
48.692
0.121
D
0.227
0.040
D
ni Bi
i
i
10.896983 10
3
ni Di
C
D
0
4
5.892 10
The TOTAL value for MCPH of the product stream:
HP
R M CPH ( 303.15K 1773.15K A B C D) ( 1773.15
HP
732.013
kJ
mol
From Example 4.7:
Q
HP
H298
5
10
J
mol K
i
A
3
0.593
ni Ai B
A
10
0.506
3.280
14
1.157
H298
802625
Q = 70 612 J
97
J
mol
Ans.
303.15)K
HEAT EXCHANGER: Flue gases cool from 1500 degC to
50 degC. The partial pressure of the water in the flue gases leaving the
furnace (in kPa) is
pp
n2
n1
n2
n3
n4
101.325
pp
18.754
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:
n
n1
n3
n4
n
11.381
Moles of water vapor leaving the heat exchanger:
n2
12.34
n
101.325 12.34
n2
Moles water condensing:
1.578
n
2.585
1.578
Latent heat of water at 50 degC in J/mol:
H50
2382.918.015
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)( 23.15
3
1773.15)
K
Q = 766 677 J
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
98
n H50
Ans.
ENERGY BALANCE:
H=
HR
H298
HP = 0
REACTANTS: 1=NH3; 2=O2; 3=N2
4
n
3.578
6.5
A
3.639
24.45
i
3.020
B
0.506 10
3.280
A
13
118.161
ni Ai
B
3
5
D
0.227 10
0.593
B
0.040
D
ni Bi
ni Di
i
i
i
A
0.186
C
0.02987
5
D
0.0
1.242 10
TOTAL mean heat capacity of reactant stream:
HR
HR
R MCPH ( 348.15K 298.15K A B C D) ( 298.15K
52.635
348.15K)
kJ
mol
The result of Pb. 4.21(b) is used to get
J
H298 0.8 ( 905468)
mol
PRODUCTS: =NH3; 2=O2; 3=NO; 4=H2O; 5=N2
1
0.8
3.020
2.5
n
3.578
3.639
0.506
3.2
A
3.387
B
0.629
0.186
10
K
0.227
3
D
0.014
4.8
1.450
3.280
0.593
2
0.121
24.45
i
3.470
5
10 K
0.040
15
A
ni Ai
B
119.65
D
B
1
0.027
K
By the energy balance and Eq. (4.7), we can write:
(guess)
2
T0 298.15K
99
ni Di
i
i
i
A
ni Bi
D
4
8.873 10 K
2
H298
B
2
T0
2
1
2
1
1
T
3.283
Find
4.31
H R = R A T0
D
T0
Given
T
T0
Ans.
978.9 K
C2H4(g) + H2O(g) = C2H5OH(l)
n
BASIS: 1 mole ethanol produced
H=Q=
Energy balance:
H298
[ 277690
HR
( 2510
5
1mol
H298
241818)
J
mol
4J
H298
8.838 10
mol
Reactant stream consists of 1 mole each of C2H4 and H2O.
12n
i
1.424
A
3.470
1
1
14.394
B
1.450
ni Ai B
A
B
HR
Q
4.392
C
10
0.0
C
6
D
C
0.01584
0.0
ni Di
i
4.392 10
6
D
R M CP (
H 298.15K 593.15K A B C D)( 98.15K
2
4
1.21 10
593.15K)
4J
2.727 10
HR
mol
Q
H298 1mol
100
5
10
0.121
D
ni Ci
i
i
4.894
HR
3
ni Bi
i
A
10
115653 J
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
CH4 + 2H2O = CO2 + 4H2
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
Q=
HR
H298
1.702
3.470
9.081
B
1.450
ni Ai B
A
HR
HR
10
i
3
B
2.164
C
0.0
0.4
10
6
D
2.396 10
3
C
0.0
5
10
0.121
D
ni Ci
ni Di
i
i
i
1.728
mol
0.2
n
12
C
ni Bi
i
A
4.003 10
HP
REACTANTS: 1=CH4; 2=H2O
A
4J
H298
4.328 10
7
D
3
4.84 10
RI
CPH ( 773.15K 298.15K A B C D)
4J
1.145 10
mol
PRODUCTS: 1=CO2; 2=CO; 3=H2O; 4=H2
0.0275
n
0.1725
0.1725
0.6275
5.457
A
3.376
3.470
1.045
B
3.249
0.557
1.450
0.422
101
1.157
10
3
D
0.031
0.121
0.083
5
10
i
A
14
B
ni Ai
HP
B
3.37
ni Di
i
i
i
A
D
ni Bi
6.397 10
4
C
3
D
0.0
3.579 10
RI
CPH (
298.15K 1123.15K A B C D)
4J
2.63 10
mol
HP
Q
HR
H298
Q
HP mol
4.33 CH4 + 2O2 = CO2 + 2H2O(g)
Ans.
54881 J
H298a
H298b
C2H6 + 3.5O2 = 2CO2 + 3H2O(g)
802625
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
H298
0.75 H298a
0.25 H298b
J
mol
Energy balance:
Q = H = H298
HP
PRODUCTS: 1=CO2; 2=H2O; 3=O2; 4=N2
1.25
n
5.457
3.470
1.450
A
1.9
14
B
3.639
16.082
i
8 10
HP = Q
1.045
2.25
5
Q
3.280
74.292
10
K
3
D
B
102
0.227
D
ni Bi
0.015
0.121
5
10 K
0.040
ni Di
i
i
i
H298
1.157
0.593
ni Ai B
A
A
0.506
J
mol
1
K
C
0.0
D
42
9.62 10 K
2
By the energy balance and Eq. (4.7), we can write:
T0
303.15K
Q
2
(guess)
H298 = R A T0
2
1
1
1.788
4.34
B
2
T0
2
1
D
T0
Given
T
T0
T
Find
Ans.
542.2 K
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%
SO2 reacted = SO3 formed = (0.15)(0.86) = 0.129 mol
O2 reacted = (0.5)(0.129) = 0.0645 mol
Energy balance:
H773 =
HR
H298
HP
Since HR and HP cancel for the gas that passes through the converter
unreacted, we need consider only those species that react or are formed.
Moreover, the reactants and products experience the same temperature
change, and can therefore be considered together. We simply take the
number of moles of reactants as being negative. The energy balance is
then written: H773 = H298
Hnet
H298
[ 395720
( 296830) ]0.129
J
mol
1: SO2; 2: O2; 3: SO3
0.129
n
0.0645
5.699
A
0.129
i
13
3.639
B
8.060
A
A
0.06985
0.506 10
3
5
D
0.227 10
2.028
1.056
ni Ai B
i
1.015
0.801
ni Bi
D
ni Di
i
B
2.58 10
103
i
7
C
0
D
4
1.16 10
Hnet
R M CPH (
298.15K 773.15K A B C D)( 73.15K
7
Hnet
77.617
H773
298.15K)
J
mol
H298
H773
Hnet
12679
J
mol
Ans.
4.35 CO(g) + H2O(g) = CO2(g) + H2(g)
BASIS: 1 mole of feed consisting of 0.5 mol CO and 0.5 mol H2O.
Moles CO reacted = moles H2O reacted = moles CO2 formed = moles H2
formed = (0.6)(0.5) = 0.3
Product stream:
moles CO = moles H2O = 0.2
moles CO2 = moles H2 = 0.3
Energy balance:
Q=
H298
0.3 [ 393509
H=
HR
H298
J
214818)
mol
( 110525
HP
4J
H298
2.045 10
mol
Reactants: 1: CO 2: H2O
0.5
n
i
0.5
3.376
A
3.470
1.450
ni Ai B
A
12
0.557
B
HR
HR
0.031
D
B
ni Di
i
i
3.423
5
10
0.121
D
1.004 10
3
0D
C
R M CPH (
298.15K 398.15K A B C D)( 98.15K
2
3
4.5 10
398.15K)
3J
3.168 10
Products:
0.2
0.3
0.3
mol
1: CO 2: H2O 3: CO2 4: H2
0.2
n
3
ni Bi
i
A
10
3.376
A
3.470
5.457
0.557
B
3.249
1.450
1.045
0.422
104
0.031
10
3
D
0.121
1.157
0.083
5
10
i
14A
ni Ai
B
i
A
ni Bi
D
i
3.981
B
8.415 10
i
4
C
4
0D
HP
R MCPH ( 298.15K 698.15K A B C D) ( 698.15K
HP
1.415 10
Q
ni Di
3.042 10
298.15K)
4J
HR
mol
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
0.85
209.133 lbm
The oil also contains H2O:
209.133 0.01
lbmol 0.116 lbmol
18.015
Also H2O is formed by combustion of H2 in the oil in the amount
209.133 0.12
lbmol
2.016
12.448 lbmol
Find amount of air entering by N2 & O2 balances.
N2 entering in oil:
209.133 0.02
lbmol
28.013
0.149 lbmol
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)
Total dry air = N2 in air + O2 in air = 85.051 - x + 15.124 + x = 100.175 lbmol
105
Since air is 21 mol % O2,
0.21 =
15.124 x
100.175
x
(
0.21 100.175
15.124)lbmol
x
5.913 lbmol
O2 in air = 15.124 + x = 21.037 lbmols
N2 in air = 85.051 - x = 79.138 lbmoles
N2 in flue gas = 79.138 + 0.149 = 79.287 lbmols
[CHECK: Total dry flue gas
= 3.00 + 11.80 + 5.913 + 79.287
= 100.00 lbmol]
Humidity of entering air, sat. at 77 degF in lbmol H2O/lbmol dry air,
P(sat)=0.4594(psia)
0.4594
14.696 0.4594
0.03227
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 degF. The "products" at 400 degF 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=
H=
H298
HP
where Q = 30% of net heating value of the oil:
Q
0.3 19000
BTU
209.13 lbm
lbm
Reaction upon which net heating value is based:
106
Q
6
1.192 10 BTU
OIL + (21.024)O2 = (14.8)CO2 + (12.448 + 0.116)H2O(g) + (0.149)N2
H298a
19000 209.13 BTU
6
H298a
3.973 10 BTU
To get the "reaction" in the drier, we add to this the following:
(11.8)CO2 = (11.8)CO + (5.9)O2
H298b
11.8 ( 110525
(y)H2O(l) = (y)H2O(g)
H298c ( y)
393509) 0.42993 BTU
Guess: y
50
44012 0.42993 y BTU
[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:
H298 ( y)
H298a
H298b
H298c ( y)
For the product stream we need MCPH:
1: CO2 2: CO 3:O2 4: N2 5: H2O
T0
298.15
r
1.986
400
T
459.67
1.8
T
477.594
3
1.045
1.157
11.8
n y)
(
5.457
3.376
0.557
0.031
5.913
A
i
B
3
5
D
0.227 10
0.593
0.040
3.470
y
1 5 A ( y)
1.450
0.121
n y) i Ai B ( y)
(
i
T
T0
0.506 10
3.280
79.278
15.797
3.639
1.602
n y) i Bi D ( y)
(
n y) i Di
(
i
CP ( y)
107
r A ( y)
i
B ( y)
T0
2
1
D ( y)
T0
2
CP ()(
y 400
Given
y
y
H298 ()
y
Find y
()
(lbmol H2O evaporated)
49.782
y 18.015
209.13
Whence
77)BTU = Q
(lb H2O evap. per lb oil burned)
Ans.
4.288
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
H298
HP
( 135100
2
227480)
0.242
J
2
3
H298
5.169 10 J
Products:
0.242
n
4.736
0.379
A
3.280
0.379
ni Ai
0.725
3
0.593 10
D
0.040
1.952
B
4.7133
B
D
ni Bi
1.2934 10
ni Di
i
3
4
0D
C
HP
R M CPH (
298.15K 873.15K A B C D)( 73.15K
8
HP
6.526 10
2.495 10 J
Q
298.15K)mol
4
HP
H298
Q
HP
4
2.495 10 J
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
5
10
1.299
i
i
A
B
6.132
1 3A
i
1.359
For this reaction,
H298
[ ( 241818)
2
Evaluate
T0
4 ( 92307) ]
5J
H298
1.144 10
mol
by Eq. (4.21) with
H823
T
298.15K
J
mol
823.15K
1: H2O 2: Cl2 3: HCl 4=O2
2
3.470
2
n
4.442
A
4
0.089
B
3.156
1
i
1.45
0.623
3.639
A
14
ni Ai
10
3
D
0.344
0.151
0.506
B
D
ni Di
i
i
0.439
H823
H298
MCPH T0 T
H823
117592
B
5
8 10
A
B
5
10
0.227
ni Bi
i
A
0.121
C
0
DRT
4
D
T0
C
8.23 10
J
mol
Heat transferred per mol of entering gas mixture:
Q
H823
4
Q
0.45 mol
4.39 CO2 + C = 2CO
Ans.
J
(a)
mol
J
(b)
221050
mol
H298a
2C + O2 = 2CO
13229 J
172459
H298b
Eq. (4.21) applies to each reaction:
For (a):
2
n
1
1
3.376
A
1.771
0.557
B
0.771 10
5.457
1.045
109
0.031
3
D
5
0.867 10
1.157
i
A
13
B
ni Ai
H1148a
B
0.476
i
4
7.02 10
H298a
R M CP 298.15K 1148.15K
H
H1148a
ni Di
i
i
A
D
ni Bi
A
C
B
C
D
0
5
1.962 10
D ( 148.15K
1
298.15K)
5J
1.696 10
mol
For (b):
2
n
3.376
1
A
3.639
2
i
13
0.557
B
H1148b
ni Ai
5
D
0.227 10
B
0.867
ni Bi
D
ni Di
i
0.429
B
9.34 10
H298b
R M CP 298.15K 1148.15K
H
H1148b
3
0.771
i
A
0.506 10
1.771
A
0.031
i
4
C
A
B
0
C
D
D ( 148.15K
1
5J
2.249 10
mol
The combined heats of reaction must be zero:
nCO
2
H1148a
nO
H1148b = 0
2
nCO
Define:
r=
2
nO
H1148b
r
H1148a
2
110
r
5
1.899 10
1.327
298.15K)
For 100 mol flue gas and x mol air, moles are:
Flue gas
Air
Feed mix
CO2
12.8
0
12.8
CO
3.7
0
3.7
O2
5.4
0.21x
5.4 + 0.21x
N2
78.1
0.79x
78.1 + 0.79x
Whence in the feed mix:
r=
12.5
5.4
r
mol
0.21
x
12.8
5.4 0.21 x
x
19.155 mol
100
19.155
Flue gas to air ratio =
Ans.
5.221
Product composition:
nCO
3.7
nN
78.1
2
2 ( 12.8
0.21 19.155)
0.79 19.155
nCO
Mole % N2 =
100
48.145
93.232
2
nCO
Mole % CO =
nCO
nN
5.4
nN
100
34.054
Ans.
2
34.054
65.946
H298a
802625
J
mol
H298b
4.40 CH4 + 2O2 = CO2 + 2H2O(g)
519641
J
mol
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
79
21
2.538
9.548
mol O2
mol N2
111
Moles CO2 formed by reaction =
0.94 0.7
0.658
Moles CO formed by reaction =
0.94 0.3
0.282
H298
0.658 H298a
5J
H298
0.282 H298b
Moles H2O formed by reaction =
0.94 2.0
Moles O2 consumed by reaction =
2 0.658
6.747 10
mol
1.88
3
0.282
2
1.739
Product gases contain the following numbers of moles:
(1)
(2)
(3)
(4)
(5)
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
1.045
1.157
0.282
n
5.457
3.376
0.557
0.031
1.880
A
3.470
B
1.450 10
0.799
3.280
D
0.593
0.121
0.506
9.608
i
3.639
3
1 5A
ni Ai
B
A
0.227
0.040
ni Bi
i
D
B
9.6725 10
i
3
C
R M CPH (
298.15K 483.15K A B C D)( 83.15K
4
HP
7.541 10
HH2O mdotH2O
4
0D
HP
Energy balance:
ni Di
i
45.4881
5
10
3.396 10
298.15K)
4J
mol
Hrx
H298
Hrx ndotfuel = 0
112
HP
Hrx
kJ
mol
kg
34.0
sec
599.252
mdotH2O
From Table C.1:
HH2O
( 398.0
104.8)
kJ
kg
HH2O mdotH2O
ndotfuel
ndotfuel
Hrx
16.635
mol
sec
Volumetric flow rate of fuel, assuming ideal gas:
ndotfuel R 298.15 K
V
3
m
0.407
sec
V
101325 Pa
4.41 C4H8(g) = C4H6(g) + H2(g)
Ans.
H298
109780
J
mol
BASIS: 1 mole C4H8 entering, of which 33% reacts.
The unreacted C4H8 and the diluent H2O pass throught the reactor
unchanged, and need not be included in the energy balance. Thus
T
T0
1
n
by Eq. (4.21):
Evaluate
H798
1: C4H6 2: H2 3: C4H8
1
2.734
A
3.249
26.786
B
0.422
1.967
i
298.15 K
798.15 K
1
8.882
10
3
C
0.0
0.0
31.630
6
10
D
9.873
5
0.083 10
0.0
13
ni Ai B
A
4.016
B
H298
H798
1.179 10
Q
3
C
9.91 10
MCPH 298.15K 798.15K
5J
0.33 mol H798
mol
Q
38896 J
113
Ans.
ni Di
i
i
4.422 10
H798
D
ni Ci
i
i
A
C
ni Bi
A
B
7
D
C
3
8.3 10
DRT
T0
4.42Assume Ideal Gas and P = 1 atm
P
R
1atm
a) T0
(
70
T0
3 BTU
7.88 10
mol K
T
294.261 K
ICPH T0 T 3.355 0.575 10
3
T0
T
459.67)
rankine
BTU
sec
305.372 K
5
0
0.016 10
R ICPH T0 T 3.355 0.575 10
Vdot
b) T0
38.995 K
(
24
3
5
0
0.016 10
T
T0
ndot
39.051
Vdot
ft
33.298
sec
Q
13K
12
kJ
s
3
5
0
0.016 10
45.659 K
Q
R ICPH T0 T 3.355 0.575 10
3
0
5
0.016 10
Vdot
4.43Assume Ideal Gas and P = 1 atm
(
94
459.67)
rankine
1.61 10
3
Vdot
31.611
P
T
(
68
m
0.7707
s
1atm
459.67)
rankine
3
atm ft
mol rankine
3
ft
50
sec
ndot
mol
s
3
ndot R T0
P
a) T0
Ans.
kJ
mol K
ndot
Vdot
mol
s
3
m
0.943
s
Vdot
273.15)
K
8.314 10
3
3
ndot R T0
P
ICPH T0 T 3.355 0.575 10
R
12
Q
ndot
R
Q
20rankine
ndot
P Vdot
R T0
114
ndot
56.097
mol
s
Ans.
T0
307.594 K
T
ICPH T0 T 3.355 0.575 10
3
293.15 K
5
0
0.016 10
50.7 K
3 BTU
R
7.88 10
Q
R ICPH T0 T 3.355 0.575 10
mol K
3
0
5
0.016 10
ndot
Q
b) T0
R
( 35
273.15)K
T
273.15)K
mol K
3
m
1.5
sec
3
P Vdot
R T0
ndot
ICPH T0 T 3.355 0.575 10
3
0
ndot
5
0.016 10
59.325
8.314 10
Q
R ICPH T0 T 3.355 0.575 10
35.119 K
3
0
5
0.016 10
Q
ndot
17.3216
4.44 First calculate the standard heat of combustion of propane
C3H8 + 5O2 = 3CO2(g) + 4H2O (g)
H298
Cost
3
393509
J
mol
4
241818
6J
2.043 10
2.20
mol
s
kJ
mol K
R
H298
BTU
Ans.
sec
3
5 atm m
8.205 10
Vdot
( 25
22.4121
dollars
gal
mol
80%
115
J
mol
104680
J
mol
kJ
Ans.
s
Estimate the density of propane using the Rackett equation
3
Tc
T
Vsat
369.8K
(
25
Zc
273.15)
K
V c Zc
Heating_cost
1 Tr
0.276
Tr
Vc
T
Tc
Tr
cm
200.0
mol
0.806
3
0.2857
Vsat
Vsat Cost
cm
89.373
mol
Heating_cost
H298
Heating_cost
0.032
dollars
MJ
33.528
Ans.
dollars
6
10 BTU
4.45 T0
(
25
a) Acetylene
273.15)
K
T
(
500
273.15)
K
Q
R ICPH T0 T 6.132 1.952 10
Q
2.612 10
3
0
5
1.299 10
4J
mol
The calculations are repeated and the answers are in the following table:
J/mol
a) Acetylene
26,
120
b) Ammonia
20,
200
c) nbutane
71,
964
d) Carbon diox
ide
21,
779
e) Carbon monox
ide
14,
457
f) Ethane
3 420
8,
g) Hydrogen
13866
,
h) Hydrogen chloride
14,
040
i) M ethane
233
,18
j Nitric ox
)
ide
14, 0
73
k) Nitrogen
14,
276
l) Nitrogen diox
ide
20,
846
m) Nitrous ox
ide
22,
019
n) Ox
ygen
15,
052
o) Propylene
46,
147
116
4.46 T0
( 25
Q
273.15)K
30000
a) Acetylene
T
( 500
273.15)K
J
mol
Given Q = R ICPH T0 T 6.132 1.952 10
T
Find ( T)
T
835.369 K
T
3
273.15K
0
5
1.299 10
562.2 degC
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
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
( 250
a) Guess mole fraction of methane:
y
273.15) K
Q
11500
0.5
Given
y ICPH T0 T 1.702 9.081 10
(1
y
3
2.164 10
y) ICPH T0 T 1.131 19.225 10
Find ( y)
y
0.637
Ans.
117
3
6
=Q
0R
5.561 10
6
0R
J
mol
b) T0
(
100
273.15)
K
T
(
400
Guess mole fraction of benzene
273.15)K
y
Q
54000
J
mol
0.5
Given
y ICPH T0 T
(
1
3
0.206 39.064 10
y)ICPH T0 T
y
Find y
()
y
c) T0
(
150
3
3.876 63.249 10
273.15)
K
6
13.301 10
=Q
0R
6
20.928 10
0R
Ans.
0.245
T
(
250
Guess mole fraction of toluene
273.15)K
y
Q
17500
J
mol
0.5
Given
y ICPH T0 T 0.290 47.052 10
(
1
y
3
15.716 10
y)ICPH T0 T 1.124 55.380 10
Find y
()
y
0.512
3
6
=Q
0R
18.476 10
6
0R
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.
Pi h a End
nc t
I e m e a e Pi h
nt r di t nc
TH1
Se ton I
ci I
Se ton I
ci
TH1
Se ton I
ci
THi
Se ton I
ci I
THi
T
TH2 TC1
TC1
TCi
TCi
TC2
118
TH2
T
TC2
To solve the problem, apply an energy balance around each section of the
exchanger.
T H1
Section I balance:
mdotC HC1
HCi = ndotH
CP dT
THi
T Hi
Section II balance:
HC2 = ndotH
mdotC HCi
CP dT
TH2
If the pinch is intermediate, then THi = TCi + T. If the pinch is at the end,
then TH2 = TC2 + T.
a) TH1
T
10degC
TC1
2676.0
TCi
For air from Table C.1:A
kJ
kg
3.355 B
100degC
TC2
25degC
HCi
100degC
HC1
1000degC
419.1
kJ
kg
HC2
104.8
0.575 10
3
C
0D
kJ
kg
5
0.016 10
kmol
s
Assume as a basis ndot = 1 mol/s.
ndotH
Assume pinch at end:
TH2
TC2
THi
110degC
Guess:
mdotC
1
kg
s
1
T
Given
mdotC HC1
mdotC HCi
mdotC
THi
mdotC
ndotH
THi
TCi
HCi = ndotH R ICPH THi TH1 A B C D Energy balances
on Section I and
HC2 = ndotH R ICPH TH2 THi A B C D II
170.261 degC mdotC
Find mdotC THi THi
0.011
kg
mol
70.261 degC
Ans.
TH2
119
TC2
10 degC
11.255
kg
s
Since the intermediate temperature difference, THi - TCi is greater than
the temperature difference at the end point, TH2 - TC2, the assumption of a
pinch at the end is correct.
b) TH1
TC1
T
10degC
2676.0
TCi
kJ
kg
100degC
TC2
25degC
HCi
100degC
HC1
500degC
419.1
kJ
kg
HC2
104.8
1
kmol
s
THi
TCi
T
TH2
110degC
Assume as a basis ndot = 1 mol/s.
ndotH
Assume pinch is intermediate:
Guess:
mdotC
1
kg
s
kJ
kg
Given
mdotC HC1
mdotC HCi
mdotC
TH2
mdotC
ndotH
THi
TCi
HCi = ndotH R ICPH THi TH1 A B C D Energy balances
on Section I and
HC2 = ndotH R ICPH TH2 THi A B C D II
Find mdotC TH2 TH2
5.03 10
3 kg
10 degC
mol
48.695 degC
mdotC
5.03
kg
s
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.
4.50a) 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
120
0
kJ
mol
M1
180
gm
mol
H0f3
H0r
393.509
6 H0f3
kJ
mol
6 H0f4
b) energy_per_kg
mass_glucose
H0f4
150
kJ
kg
H0f1
285.830
mass_person energy_per_kg
H0r
M3
H0 r
6 H0f2
mass_person
kJ
mol
44
2801.634
gm
mol
kJ
mol
Ans.
57kg
mass_glucose
M1
0.549 kg Ans.
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
8
Ans.
2.216 10 kg
4.51 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)
1 = CH4, 2 = C2H6, 3 = O2, 4 = CO2, 5 = H2O 6 = N2
kJ
mol
H0f1
74.520
H0f4
393.509
a) H0c
H0c
825.096
kJ
mol
2 H0f5
83.820
H0f5
kJ
mol
1.05 H0f4
kJ
mol
H0f2
241.818
0.85 H0f1
H0f3
0
kJ
mol
kJ
mol
0.10 H0f2
1.05 H0f3
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
n3
0.5 2.05mol
121
1.025 mol
Excess O2
n4
1.05mol
n5
2mol
n6
0.05mol
79
1.5 2.05mol
21
n6
11.618 mol
Total N2
Air and fuel enter at 25 C and combustion products leave at 600 C.
T1
A
B
C
D
Q
(
25
T2
273.15)
K
n3 3.639
n4 6.311
(
600
n5 3.470
273.15)
K
n6 3.280
mol
n3 0.506
n4 0.805
n5 1.450
n6 0.593 10
3
mol
n3 0
n4 0
n5 0
n6 0 10
6
mol
n3 ( 0.227) n4 ( 0.906) n5 0.121
5
n6 0.040 10
mol
H0c
ICPH T1 T2 A B C D R
122
Q
529.889
kJ
mol
Ans.
Chapter 5 - Section A - Mathcad Solutions
5.2 Let the symbols Q and Work represent rates in kJ/s. Then by Eq. (5.8)
Work
=
QH
TC
TC
=1
TH
TH
323.15 K
Work
TC
QH 1
250
kJ
s
kJ
s
Work
or
148.78
Work
TH
By Eq. (5.1),
QH
798.15 K
148.78 kW which is the power. Ans.
QC
QH
QC
Work
101.22
kJ
s
Ans.
5.3 (a) Let symbols Q and Work represent rates in kJ/s
TH
TC
750 K
By Eq. (5.8):
But
TC
1
Work
=
QH
Work
300 K
0.6
TH
Whence
95000 kW
QH
Work
QH
QC
(b)
QH
QH
0.35
QC
5.4 (a) TC
303.15 K
Carnot
1
TC
TH
QC
Work
QH
TH
Work
6.333
QH
Work
5
1.583 10 kW Ans.
5
2.714 10 kW Ans.
QC
1.764
4
10 kW Ans.
5
10 kW Ans.
623.15 K
0.55
123
Carnot
0.282
Ans.
(b)
0.35
Carnot
By Eq. (5.8),
Carnot
0.55
TC
TH
1
0.636
833.66 K Ans.
TH
Carnot
5.7 Let the symbols represent rates where appropriate. Calculate mass rate of
LNG evaporation:
3
V
9000
molwt
m
P
s
17
gm
mLNG
mol
T
1.0133 bar
PV
molwt
RT
298.15 K
mLNG
6254
kg
s
Maximum power is generated by a Carnot engine, for which
Work
QH
=
QC
QC
TH
512
Work
QH
QC
TH
TC
QH
1=
QC
TH
TC
1
1
113.7 K
QC
kJ
mLNG
kg
QC
=
TC
303.15 K
QC
5.8
QC
3.202
Work
QH
Work
6
10 kW
6
10 kW
8.538
Ans.
6
5.336
Ans.
10 kW
Take the heat capacity of water to be constant at the valueCP
(a) T1
273.15 K
SH2O
Sres
CP ln
Q
T2
T2
T2
T1
373.15 K
Q
CP T2
SH2O
1.305
1.121
kJ
kg K
Q
kJ
kg K
Sres
T1
124
Ans.
4.184
418.4
kJ
kg K
kJ
kg
Stotal
SH2O
Sres
Stotal
0.184
kJ
Ans.
kgK
(b) The entropy change of the water is the same as in (a), and the total
heat transfer is the same, but divided into two halves.
Q
Sres
2
Stotal
1
373.15 K
1
323.15 K
Sres
Sres
Stotal
SH2O
0.097
1.208
kJ
kJ
kgK
Ans.
kgK
(c) The reversible heating of the water requires an infinite number of heat
reservoirs covering the range of temperatures from 273.15 to 373.15 K,
each one exchanging an infinitesimal quantity of heat with the water and
raising its temperature by a differential increment.
5.9
P1
T1
P1 V
n
R T1
(a) Const.-V heating;
T2
T1
Q
n CV
But
P2
P1
=
1.443 mol
U= Q
T2
By Eq. (5.18),
1
T1
Whence
3
0.06m
5
CV
2
R
W = Q = n CV T2
Q
15000 J
T1
3
10 K
S = n CP ln
T2
V
500 K
n
1 bar
S
T2
T1
R ln
n C V ln
T2
T1
P2
P1
S
20.794
(b) The entropy change of the gas is the same as in (a). The entropy
change of the surroundings is zero. Whence
Stotal = 10.794
J
K
Ans.
The stirring process is irreversible.
125
J
K
Ans.
5.10 (a) The temperature drop of the second stream (B) in either
case is the same as the temperature rise of the first stream CP
(A), i.e., 120 degC. The exit temperature of the second
stream is therefore 200 degC. In both cases we therefore
have:
SA
SA
8.726
463.15
CP ln
SB
343.15
J
mol K
473.15
593.15
CP ln
SB
7
R
2
6.577
J
Ans.
mol K
(b) For both cases:
Stotal
SA
Stotal
SB
2.149
J
mol K
Ans.
(c) In this case the final temperature of steam B is 80 degC, i.e., there is
a 10-degC driving force for heat transfer throughout the exchanger.
Now
SA
SA
8.726
463.15
343.15
CP ln
Stotal
SB
J
mol K
SA
SB
SB
dQ
Since dQ/T = dS,
8.512
Stotal
dW
5.16 By Eq. (5.8),
Ans.
J
mol K
dW = dQ
T
dW = dQ
J
mol K
0.214
T
=1
353.15
473.15
CP ln
Ans.
T
dQ
T
T dS
Integration gives the required result.
T1
Q
CP T2
T2
600 K
T1
Q
5.82
T
400 K
126
3J
10
mol
300 K
S
CP ln
Work
Q
Q
T2
T1
T
Q
5.17 TH1
11.799
Work
S
Q
Work
Q
Sreservoir
S
S
2280
3540
Sreservoir
T
J
mol K
Sreservoir
0
600 K
TC1
J
mol K
J
mol
Ans.
J
mol
11.8
Ans.
J
Ans.
mol K
Process is reversible.
300 K
TH2
For the Carnot engine, use Eq. (5.8):
W
QH1
TC2
300 K
=
TH1
250 K
TC1
TH1
The Carnot refrigerator is a reverse Carnot engine. W
TH2 TC2
=
Combine Eqs. (5.8) & (5.7) to get:
TC2
QC2
Equate the two work quantities and solve for the required ratio of the heat
quantities:
TC2 TH1 TC1
Ans.
r
r 2.5
TH1 TH2 TC2
5.18 (a) T1
H
S
(b)
C p T2
Cp ln
T2
T1
T2
1.2bar
T1
R ln
P2
mol
4.365
S
P1
3J
H = 5.82 10
450K
H
P1
300K
1.582
S = 1.484
127
P2
6bar
3J
10
mol
J
mol K
J
mol K
Ans.
Ans.
Cp
7
R
2
3J
(c)
H = 3.118 10
(d)
H = 3.741 10
(e)
H = 6.651 10
S = 4.953
mol
3J
S = 2.618
mol
3J
J
mol K
J
mol K
S = 3.607
mol
J
mol K
5.19This cycle is the same as is shown in Fig. 8.10 on p. 305. The equivalent states
are A=3, B=4, C=1, and D=2. The efficiency is given by Eq. (A) on p. 305.
Temperature T4 is not given and must be calaculated. The following equations
are used to derive and expression for T4.
For adiabatic steps 1 to 2 and 3 to 4:
T1 V 1
1
= T2 V 2
1
1
T3 V 3
For constant-volume step 4 to 1:
P2
T2
=
P3
T3
Solving these 4 equations for T4 yields: T4 = T1
7
R
2
T1
(
200
T4
T1
T3
Eq. (A) p. 306
1
(
1000
T4
T2
T3
1.4
Cv
T2
273.15)
K
T2
Cp
5
R
2
Cv
1
V1 = V4
For isobaric step 2 to 3:
Cp
= T4 V 4
T3
(
1700
873.759 K
1
T4
T1
T3
T2
128
273.15)
K
0.591
Ans.
273.15)
K
5.21 CV
CP
P1
R
P2
2 bar
CP
7 bar
T1
298.15 K
1.4
CV
With the reversible work given by Eq. (3.34), we get for the actual W:
1
Work
1.35
P2
R T1
1
3J
Work
3.6 10
P1
1
But Q = 0, and W =
mol
Whence
S
5.25 P
T2
CP ln
T1
T
4
R ln
T1
Work
P2
S
2.914
P1
T2
T1
T2
U = C V T2
471.374 K
J
mol K
CV
Ans.
800
Step 1-2: Volume decreases at constant P.
Heat flows out of the system. Work is done on the system.
W12 =
P V2
V1
=
R T2
T1
Step 2-3: Isothermal compression. Work is done on the system. Heat flows
out of the system.
W23 = R T2 ln
P3
P2
= R T2 ln
P3
P1
Step 3-1: Expansion process that produces work. Heat flows into the
system. Since the PT product is constant,
P dT
T dP = 0
PV= RT
P dV = R dT
T
dP
= dT
P
P dV
V dP = R dT
(A)
V dP = R dT
RT
dP
P
129
In combination with (A) this becomes
P dV = R dT
R dT = 2 R dT
Moreover,
P3 = P 1
T1
T3
= P1
T1
T2
V1
P dV = 2 R T 1
W31 =
T3 = 2 R T1
T2
V3
Q31 =
U31
Q31 = CV
=
CP
W31 = CV T1
2R
Wnet
W12
=
Qin
T1
T3
2 R T1
T3 = C P
W23
R
T1
T3
T2
W31
Q31
7
R
2
W23
W31
Q31
R T2
R T2 ln
W12
T2
350 K
1.5 bar
P3
P1
T1
W23
T2
3J
T1
T2
2.017
10
W31
T2
2.91
W23
P1
R
T1
W12
P3
2 R T1
CP
700 K
P1
W12
T1
5.82
10
Q31
1.309
10
W31
0.068
Q31
130
10
mol
3J
mol
3J
mol
4J
mol
Ans.
5.26 T
403.15 K
P1
By Eq. (5.18),
S
P2
2.5 bar
R ln
P2
P1
S
7.944
Tres
6.5 bar
298.15 K
J
mol K
Ans.
With the reversible work given by Eq. (3.27), we get for the actual W:
Work
Q
1.3 R T ln
Work
P2
P1
(Isothermal compresion) Work
4.163
10
3J
mol
Q here is with respect to the system.
So for the heat reservoir, we have
Sres
Q
Sres
Tres
Stotal
S
Stotal
Sres
13.96
6.02
J
Ans.
mol K
J
mol K
Ans.
5.27 (a) By Eq. (5.14) with P = const. and Eq. (5.15), we get for the entropy change
of 10 moles
n
10 mol
S
n R ICPS 473.15K 1373.15K 5.699 0.640 10
S
536.1
J
K
3
0.0
5
1.015 10
Ans.
(b) By Eq. (5.14) with P = const. and Eq. (5.15), we get for the entropy
change of 12 moles
n
12 mol
S
n R ICPS 523.15K 1473.15K 1.213 28.785 10
S
2018.7
J
K
Ans.
131
3
8.824 10
6
0.0
5.28 (a) The final temperature for this process was found in Pb. 4.2a to be 1374.5 K.
The entropy change for 10 moles is then found as follows
n
10 mol
S
n R ICPS 473.15K 1374.5K 1.424 14.394 10
S
900.86
J
K
3
4.392 10
6
0.0
Ans.
(b) The final temperature for this process was found in Pb. 4.2b to be 1413.8 K.
The entropy change for 15 moles is then found as follows:
n
15 mol
S
n R ICPS 533.15K 1413.8K 1.967 31.630 10
S
2657.5
J
3
9.873 10
6
0.0
Ans.
K
(c) The final temperature for this process was found in Pb. 4.2c to be 1202.9 K.
The entropy change for 18.14 kg moles is then found as follows
n
18140 mol
S
S
n R ICPS 533.15K 1202.9K 1.424 14.394 10
6J
1.2436 10
K
3
4.392 10
6
0.0
Ans.
5.29 The relative amounts of the two streams are determined by an energy
balance. Since Q = W = 0, the enthalpy changes of the two streams must
cancel. Take a basis of 1 mole of air entering, and let x = moles of chilled air.
Then 1 - x = the moles of warm air.
T0
298.15 K
Temperature of entering air
T1
248.15 K
Temperature of chilled air
T2
348.15 K
Temperature of warm air
x C P T1
x
0.3
T0
(
1
x)CP T2
T0 = 0
(guess)
132
x
Given
1
x
=
T2
T0
T1
T0
x
x
Find x
()
0.5
Thus x = 0.5, and the process produces equal amounts of chilled and warmed
air. The only remaining question is whether the process violates the second
law. On the basis of 1 mole of entering air, the total entropy change is as
follows.
CP
7
2
P0
R
Stotal
x CP ln
Stotal
12.97
P
5 bar
T1
(
1
T0
J
mol K
1 bar
T2
x)CP ln
R ln
T0
P
P0
Ans.
Since this is positive, there is no violation of the second law.
5.30
T1
Tres
303.15 K
CV
CP
R
Q=
Sres
Tres
CP ln
Stotal
S
T2
T1
R ln
Sres
P1
353.15 K
Work
Q
Sres
S
T2
523.15 K
1800
U
J
mol
Work
5.718
Q
J
mol K
P2
CP
Q
S
P1
Stotal
133
3.42
3 bar
7
2
P2
1 bar
R
C V T2
T1
Work
3J
1.733 10
2.301
mol
J
mol K
J
PROCESS IS POSSIBLE.
mol K
5.33 For the process of cooling the brine:
kJ
T
mdot
40 K
CP
3.5
T1
( 273.15
25) K
T1
( 273.15
15) K
T2
( 273.15
30) K
T
0.27
258.15 K
T
kg
sec
298.15 K
T2
20
303.15 K
kg K
H
S
CP ln
kJ
kg
H
T2
T1
Eq. (5.26):
Wdotideal
By Eq. (5.28):
140
S
CP T
0.504
mdot
H
kJ
kg K
T
S Wdotideal
Wdotideal
Wdot
t
Wdot
256.938 kW
951.6 kW
Ans.
t
5.34 E
i
110 volt
Wdotmech
1.25 hp
Wdotelect
At steady state: Qdot
Wdotelect
Qdot
T
SdotG =
Qdot
SdotG
Wdotelect
T
9.7 amp
300 K
Wdotelect
iE
Wdotmech =
1.067
dt
U =0
dt
dt
S =0
dt
Qdot
Qdot
T
134
134.875 W
SdotG
Wdotmech
0.45
W
K
Ans.
3
10 W
5.35
i
25 ohm
2
Wdotelect
T
10 amp
Wdotelect
i
At steady state:
300 K
3
2.5
10 W
dt
U =0
dt
SdotG
8.333
kmol
hr
T1
(
25
Cp
R
T2
Cp 1
RT
dT
ln
watt
K
P2
10bar
P2
1.2bar
7
5
Cv
T2
Qdot
T
Ans.
Cp
(a) Assuming an isenthalpic process:
S
(b)
=
R
Wdotelect
SdotG
P1
273.15)
K
Cv
10
Qdot
dt
S =0
dt
10 watt
7
R
2
Cp
SdotG =
3
2.5
5.38 mdot
Wdotelect =
Qdot
T
Qdot
Qdot
T2
T1
298.15 K
Ans.
Eq. (5.14)
P1
T1
T2
7
R ln
T1
2
S
(c) SdotG
(d) T
5.39(a) T1
T
H
R ln
mdot S
(0
2
273.15)
K
P2
P1
SdotG
48.966
Wlost
T
500K
P1
300K
Basis: 1 mol
n C p T2
T1
S
6bar
T2
n
Ws
135
17.628
W
K
S
371K
J
mol K
Ans.
Ans.
Wlost
P2
5.168
3J
10
1.2bar
mol
Cp
Ans.
7
R
2
1mol
H
Ws
3753.8 J
Ans.
S
n Cp ln
T2
T1
P2
R ln
S
P1
Eq. (5.27)
Wideal
H
Eq. (5.30)
Wlost
Wideal
Eq. (5.39)
SG
T
S
Ws
SG
T
K
Ans.
1409.3 J
Wlost
Wlost
J
5163 J
Ans.
J
K
Ans.
Wideal
Wideal
Ws
4.698
4.698
Wlost
SG
(a)
5163J
1409.3J
4.698
J
K
(b)
2460.9J
2953.9J
493J
1.643
J
K
(c)
3063.7J
4193.7J
1130J
3.767
J
K
(d)
3853.5J
4952.4J
1098.8J
3.663
(e)
5.41
3753.8J
3055.4J
4119.2J
1063.8J
3.546
P1
S
R ln
SdotG
QH
P2
S
P1
mdot S
Wdotlost
5.42
P2
2500kPa
1kJ
W
0.023
SdotG
T SdotG
W
QH
actual
K
mdot
20
mol
sec
mol K
kJ
sec K
Ans.
140.344 kW
Ans.
TH
( 250
TC
actual
300K
K
J
kJ
0.468
Wdotlost
0.45kJ
T
150kPa
J
( 25
0.45
136
273.15)K
273.15)K
TH
523.15 K
TC
298.15 K
TC
1
max
actual> max,
Since
5.43 QH
the process is impossible.
50 kJ
Q2
100 kJ
T1
350 K
T2
250 K
QH
Q1
Q2
TH
T1
T2
SG
0.27
550 K
(a)
SG
(b)
Wlost
5.44
0.43
Q1
150 kJ
TH
max
TH
Wlost
T SG
(a)
TH
750 MW
1
max
Ans.
Ans.
81.039 kJ
TC
(
20
TC
273.15)K
273.15)K
293.15 K
Ans.
0.502
max
TH
300 K
kJ
K
588.15 K
TC
Wdot
QdotH
(
315
TH
Wdot
T
QdotC
QdotH
Wdot
QdotC
745.297 MW
max
(b)
0.6
QdotC
QdotH
Wdot
QdotH
max
Wdot
QdotC
1.742
(minimum value)
QdotH
3
Cp
1
cal
gm K
T
m
165
s
QdotC
Vdot
137
Cp
1
9
10 W
(actual value)
10 MW
3
River temperature rise: Vdot
2.492
gm
3
cm
T
2.522 K
Ans.
5.46 T1
( 20
P1
273.15)K
T2
P2
5bar
( 27
273.15) K
1atm
T3
( 22
273.15)K
First check the First Law using Eqn. (2.33) neglect changes in kinetic and
potential energy.
6
H
7
1
7
H
3
R ICPH T1 T2 3.355 0.575 10
ICPH T1 T3 3.355 0.575 10
8.797
10
4 kJ
mol
3
5
0
0
0.016 10
0.016 10
5
R
H is essentially zero so the first law is satisfied.
Calculate the rate of entropy generation using Eqn. (5.23)
6
SG
R ICPS T1 T2 3.355 0.575 10
7
1
7
SG
P
R ICPS T1 T3 3.355 0.575 10
0.013
5.47
a) Vdot
3
kJ
mol K
Since SG
5
0
3
0.016 10
5
0
0.016 10
R ln
P2
P1
0, this process is possible.
3
ft
100000
hr
T1
459.67)rankine T2
T
1atm
( 70
( 70
459.67)rankine
( 20
459.67)rankine
Assume air is an Ideal Gas
ndot
P Vdot
R T1
ndot
258.555
lbmol
hr
Calculate ideal work using Eqn. (5.26)
Wideal
ndot R ICPH T1 T2 3.355 0.575 10
T
Wideal
3
0
R ICPS T1 T2 3.355 0.575 10
1.776 hp
138
5
0.016 10
3
0
5
0.016 10
3
b) Vdot
P
3000
m
T1
1atm
(
25
273.15)
K
T
hr
(5
2
273.15)
K
T2
(8
273.15)
K
0
0.016 10
Assume air is an Ideal Gas
P Vdot
ndot
ndot
34.064
mol
s
R T1
Calculate ideal work using Eqn. (5.26)
Wideal
ndot R ICPH T1 T2 3.355 0.575 10
T
Wideal
5.48 T1
R ICPS T1 T2 3.355 0.575 10
3
5
5
0
0.016 10
1.952 kW
(
2000
Cp ( )
T
T
3
459.67)
rankine
3.83
(0
7
0.000306
T2
T
rankine
(
300
Hv
R
459.67)
rankine
Tsteam
970
459.67)
rankine
BTU
M
lbm
(
212
29
gm
mol
459.67)
rankine
a) First apply an energy balance on the boiler to get the ratio of steam flow
rate to gas flow rate.:
T2
ndotgas
Cp ( ) T
Td
mdotsteam Hv = 0
T1
T2
Cp ( ) T
Td
mdotndot
T1
mdotndot
Hv
15.043
lb
lbmol
Calculate the rate of entropy generation in the boiler. This is the sum of the
entropy generation of the steam and the gas.
SdotG = SdotGsteam
SdotGgas
139
Calculate entropy generation per lbmol of gas:
SdotG
ndotgas
mdotsteam
=
Ssteam
ndotgas
Sgas
Hv
Ssteam
Ssteam
Tsteam
T2
Sgas
C p ( T)
dT
1.444
Sgas
9.969
SdotG
11.756
Wlost
T
6227
BTU
lb rankine
10
3 kg
mol lb rankine
T1
SdotG
mdotndot Ssteam
Sgas
BTU
BTU
lbmol rankine
Calculate lost work by Eq. (5.34)
Wlost
SdotG T
b) Hsteam
Hv
Wideal
Hsteam
Hv
Ssteam
T
Ssteam
Tsteam
Ssteam
BTU
Ans.
lbmol
Wideal
1.444
BTU
lb rankine
205.071
BTU
lb
Calculate lbs of steam generated per lbmol of gas cooled.
T2
C p ( T ) dT
mn
T1
mn
Hv
15.043
lb
lbmol
Use ratio to calculate ideal work of steam per lbmol of gas
Wideal mn
3.085
10
3 BTU
lbmol
Ans.
T2
c)
Cp ( T) dT
Hgas
T1
Wideal
Hgas
T
Sgas
Wideal
140
9.312
3 BTU
10
lbmol
Ans.
5.49 T1
(
1100
Cp ( )
T
T
273.15)
K
3.83
(5
2
T2
0.000551
T
R
K
(
150
Hv
273.15)
K
273.15)
K
2256.9
Tsteam
(
100
kJ
kg
M
29
gm
mol
273.15)
K
a) First apply an energy balance on the boiler to get the ratio of steam flow rate to
gas flow rate.:
T2
Cp ( ) T
Td
ndotgas
mdotsteam Hv = 0
T1
T2
Cp ( ) T
Td
T1
mdotndot
mdotndot
Hv
15.135
gm
mol
Calculate the rate of entropy generation in the boiler. This is the sum of the
entropy generation of the steam and the gas.
SdotG = SdotGsteam
SdotGgas
Calculate entropy generation per lbmol of gas:
SdotG
ndotgas
Ssteam
=
mdotsteam
ndotgas
Ssteam
Sgas
Hv
Ssteam
Tsteam
T2
Sgas
Cp ( )
T
dT
T
Sgas
6.048
41.835
T1
SdotG
mdotndot Ssteam
Sgas
SdotG
49.708
Wlost
14.8
3
10
J
kg K
J
mol K
J
mol K
Calculate lost work by Eq. (5.34)
Wlost
SdotG T
141
kJ
mol
Ans.
b) Hsteam
Hv
Wideal
Hsteam
Hv
Ssteam
T
Ssteam
Tsteam
Ssteam
3
6.048
Wideal
10
453.618
J
kg K
kJ
kg
Calculate lbs of steam generated per lbmol of gas cooled.
T2
Cp ( T) dT
T1
mn
mn
Hv
15.135
gm
mol
Use ratio to calculate ideal work of steam per lbmol of gas
Wideal mn
6.866
kJ
mol
Ans.
T2
c)
Cp ( T) dT
Hgas
T1
Wideal
5.50 T1
a)
Hgas
( 830
Sethylene
Sethylene
Qethylene
T
Sgas
273.15)K
T2
( 35
21.686
0.09
kJ
mol
273.15)K
R ICPS T1 T2 1.424 14.394 10
3
Ans.
T
4.392 10
( 25
6
273.15)K
0
kJ
mol K
R ICPH T1 T2 1.424 14.394 10
3
4.392 10
6
0
kJ
mol
Qethylene
60.563
Wlost
Sethylene
T
Wideal
Qethylene
Wlost
33.803
kJ
mol
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 Stotal = 0.
142
Sethylene
QC
T
=0
Solving for QC gives:
QC
QC
T
Sethylene
26.76
kJ
mol
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.
QH
Qethylene
WHE
QH
QC
WHE
33.803
kJ
mol
The lost work is exactly equal to the work that could be produced by the heat
engine
143
Chapter 6 - Section A - Mathcad Solutions
6.7
At constant temperature Eqs. (6.25) and (6.26) can be written:
dS =
and
V dP
dH = 1
T V dP
For an estimate, assume properties independent
of pressure.
T
V
S
S
6.8
P1
270 K
1.551 10
3
3m
2.661
P1
P2
1200 kPa
3
K
H
Ans.
Tc
551.7
Zc
408.1 K
1
1
H
P1
J
kg K
Isobutane:
2.095 10
kg
V P2
P2
381 kPa
T V P2
J
kg
Ans.
0.282
CP
2.78
Vc
cm
262.7
mol
4000 kPa
molwt
2000 kPa
P1
gm
58.123
mol
J
gm K
3
Eq. (3.63) for volume of a saturated liquid may be used for the volume of a
compressed liquid if the effect of pressure on liquid volume is neglected.
359
T
0.88
360 K
Tr
361
T
Tc
Tr
0.882
0.885
(The elements are denoted by subscripts 1, 2, & 3
2
V
V c Zc
1 Tr
131.604
7
V
132.138
3
cm
mol
132.683
Assume that changes in T and V are negligible during throtling. Then Eq.
(6.8) is integrated to yield:
144
H=T S
V 1 P2
T1
S
but
VP
Then at 360 K,
H=0
P1
S
0.733
J
Ans.
mol K
We use the additional values of T and V to estimate the volume expansivity:
3
V
V3
V
V1
1
V
V1
T
1.079
cm
T
mol
4.098835
3
10
T3
T
T1
2K
1
K
Assuming properties independent of pressure,
Eq. (6.29) may be integrated to give
S = CP
Whence
6.9
T
T
T
S
T1
T
6
P
P1
V1 P
P1
K
P2
T
molwt
CP
298.15 K
250 10
P
VP
1
1 bar
P2
6
45 10
bar
3
2
10 kPa
Ans.
0.768 K
1500 bar
3
1
cm
1003
kg
V1
3
By Eq. (3.5),
Vave
V2
V1
Vave
2
Vave 1
H
134.6
S
S
T
kJ
kg
P2
970.287
cm
P1
V2
U
kJ
kg K
Ans. Q
Q
H
U
P1
P1
By Eqs. (6.28) & (6.29),
kg
Ans.
Vave P2
0.03636
P2
3
V2
H
V1 exp
cm
937.574
kg
5.93
TS
10.84
145
P2 V 2
kJ
kg
P1 V 1
Ans.
Work
kJ
Ans. Work
kg
4.91
U
kJ
kg
Q
Ans.
6.10
For a constant-volume change, by Eq. (3.5),
T2
T1
36.2 10
T2
P2
6.14 --- 6.16
P2
5
K
T1
P1 = 0
T1
1
298.15 K
4.42 10
P1
P2
5
205.75 bar
Vectors containing T, P, Tc, Pc, and
T2
bar
P1
1
323.15 K
1 bar
Ans.
for Parts (a) through (n):
300
40
308.3
61.39
.187
175
75
150.9
48.98
.000
575
30
562.2
48.98
.210
500
50
425.1
37.96
.200
325
60
304.2
73.83
.224
175
60
132.9
34.99
.048
575
T
35
K
P
556.4
bar Tc
45.60
K Pc
.193
bar
650
50
553.6
40.73
.210
300
35
282.3
50.40
.087
400
70
373.5
89.63
.094
150
50
126.2
34.00
.038
575
15
568.7
24.90
.400
375
25
369.8
42.48
.152
475
75
365.6
46.65
.140
Tr
T
Tc
Pr
P
Pc
146
6.14
Redlich/Kwong equation:
Pr
Tr
Eq. (3.53) q
z
Given
z= 1
i
q
z
q
Eq. (3.52)
zz
Find z
()
Ii
HRi
R Ti
SRi
R ln Z
i qi
Eq. (3.54)
1.5
1
1 14
Z
0.42748
Tr
Guess:
Z
0.08664
Z
ln
Z
i qi
Z
i
Eq. (6.65b)
i qi
1
i qi
1.5 qi Ii Eq. (6.67) The derivative in these
i
i qi
0.5 qi Ii Eq. (6.68) equations equals -0.5
HRi
SRi
-2.302·103
J
0.695
-5.461
0.605
-2.068·103
mol
-8.767
0.772
-3.319·103
-4.026
0.685
-4.503·103
-6.542
0.729
-2.3·103
-5.024
0.75
-1.362·103
-5.648
0.709
-4.316·103
-5.346
0.706
-5.381·103
-5.978
0.771
-1.764·103
-4.12
0.744
-2.659·103
-4.698
0.663
-1.488·103
-7.257
0.766
-3.39·103
-4.115
0.775
-2.122·103
-3.939
0.75
-3.623·103
-5.523
147
J
mol K
Ans.
6.15
Soave/Redlich/Kwong equation:
0.08664
1
0.42748
2
0.5
c1
z
z= 1
0.480
1.574
0.176
z
q
Eq. (3.52)
Tr
Z
zz
The derivative in the following equations equals: ci
q
Tri
Find z
()
0.5
i
i
1 14
HRi
R Ti Z
Ii
i qi
ln
1
Z
i qi
Z
Eq. (6.65b)
i qi
0.5
Tri
ci
i
1 qi Ii
Eq. (6.67)
i
SRi
R ln Z
i qi
i
ci
Tri
0.5
qi Ii
Eq. (6.68)
i
Z
i qi
HRi
SRi
J
mol
0.691
-2.595·103
0.606
-2.099·103
0.774
-3.751·103
0.722
-7.408
0.741
-2.585·103
-5.974
0.768
-1.406·103
-6.02
0.715
-4.816·103
-6.246
0.741
-5.806·103
-6.849
0.774
-1.857·103
-4.451
0.749
-2.807·103
-5.098
0.673
-1.527·103
-7.581
0.769
-4.244·103
-5.618
0.776
-2.323·103
-4.482
0.787
-3.776·103
-6.103
J
mol K
-4.795
-4.821·103
148
-6.412
-8.947
2
Eq. (3.54)
Eq. (3.53) q
1
Given
Pr
Tr
Tr
Guess:
c
Ans.
6.16 Peng/Robinson equation:
Guess:
Given
c
2
0.5
c1
0.37464
Pr
Tr
Tr
z
1
2
0.45724
0.07779
1
1
2
1.54226
0.26992
Eq. (3.53) q
Eq. (3.54)
Tr
1
z= 1
z
q
z
Eq. (3.52) Z
z
The derivative in the following equations equals: ci
Tri
q
Find ( z)
0.5
i
i
1 14
HRi
1
Ii
R Ti Z
2
i qi
Z
2
1
i qi
i
Z
ln
i qi
i
ci
R ln Z
i qi
i
ci
Eq. (6.65b)
0.5
Tri
1 qi Ii
i
SRi
Eq. (6.67)
0.5
Tri
Eq. (6.68)
qi Ii
i
Z
i qi
2
HRi
SRi
0.667
-2.655·103
J
-6.41
0.572
-2.146·103
mol
-8.846
0.754
-3.861·103
-4.804
0.691
-4.985·103
-7.422
0.716
-2.665·103
-5.993
0.732
-1.468·103
-6.016
0.69
-4.95·103
-6.256
0.71
-6.014·103
-6.872
0.752
-1.917·103
-4.452
0.725
-2.896·103
-5.099
0.64
-1.573·103
-7.539
0.748
-4.357·103
-5.631
0.756
-2.39·103
-4.484
0.753
-3.947·103
-6.126
149
J
mol K
Ans.
Lee/Kesler Correlation --- By linear interpolation in Tables E.1--E.12:
0
h0 equals
()
HR
RTc
1
h1 equals
0
s0 equals
()
SR
R
.686
()
HR
RTc
()
SR
SR
R
s equals
R
.950
.471
.705
.024
1.003
1.709
.155
.774
RTc
1
s1 equals
.093
.590
HR
h equals
.591
.675
1.319
.437
.725
.008
.993
.635
.744
Z0
.118
.165
1.265
.184
.705
.699
.770
Z1
.019
.962
h0
.102
1.200
.001
h1
.751
.444
.770
.550
.742
.007
.875
.598
.651
.144
1.466
.405
.767
.034
.723
.631
.776
.032
.701
.604
1.216
.211
.746
Z
Z0
.154
Z1 Eq. (3.57)
h
150
h0
h1
(6.85)
HR
( Tc R)
h
.711
1.110
.492
.497
.549
.829
.443
.631
.590
.710
s0
.961
.276
.674
.750
.700
s1
.441
.517
.287
0.669
-1.138
SR i
HRi
si
hi
Zi
Eq. (6.86)
.563
.688
( s R)
.589
.491
SR
.429
.511
s1
.555
.917
s0
.509
.587
s
-0.891
-2.916·103
J
mol
-7.405
0.59
-1.709
-1.11
-2.144·103
0.769
-0.829
-0.612
-3.875·103
-5.091
0.699
-1.406
-0.918
-4.971·103
-7.629
0.727
-1.135
-0.763
-2.871·103
-6.345
0.752
-1.274
-0.723
-1.407·103
-6.013
0.701
-1.107
-0.809
-5.121·103
-6.727
0.72
-1.293
-0.843
-5.952·103
-7.005
0.77
-0.818
-0.561
-1.92·103
-4.667
0.743
-0.931
-0.639
-2.892·103
-5.314
0.656
-1.481
-0.933
-1.554·103
-7.759
0.753
-0.975
-0.747
-4.612·103
-6.207
0.771
-0.793
-0.577
-2.438·103
-4.794
0.768
-1.246
-0.728
-3.786·103
-6.054
J
mol K
151
-9.229
Ans.
6.17
T
t 50
273.15
K
The pressure is the vapor pressure given by the Antoine equation:
T
P ()
t
d
t
323.15 K
2788.51
t 220.79
exp 13.8858
P ( ) 36.166
50
P
P () 1.375
t
dPdt
36.166 kPa
1.375
kPa
dt
K
(a) The entropy change of vaporization is equal to the latent heat divided by
the temperature. For the Clapeyron equation, Eq. (6.69), we need the
volume change of vaporization. For this we estimate the liquid volume by
Eq. (3.63) and the vapor volume by the generalized virial correlation. For
benzene:
Tc
T
Tc
3
Vc
259
cm
mol
Pc
562.2 K
Tr
0.210
Tr
Zc
48.98 bar
P
Pc
Pr
0.575
Pr
0.271
0.007
By Eqs. (3.65), (3.66), (3.61), & (3.63)
B0
0.422
0.083
Tr
Vvap
RT
P
1.6
1
By Eq. (3.72),
B0
B1
0.941
0.139
0.172
Tr
B0
Vliq
B1
Pr
Vvap
Tr
V c Zc
1 Tr
B1
4.2
1.621
3
4 cm
7.306
10
mol
3
2/7
Vliq
93.151
cm
mol
Solve Eq. (6.72) for the latent heat and divide by T to get the entropy change
of vaporization:
S
dPdt Vvap
S
Vliq
100.34
J
mol K
Ans.
102.14
J
mol K
Ans.
(b) Here for the entropy change of vaporization:
S
RT
dPdt
P
S
152
6.20 The process may be assumed to occur adiabatically and at constant
pressure. It is therefore isenthalpic, and may for calculational purposes be
considered to occur in two steps:
(1) Heating of the water from -6 degC to the final equilibrium temperature
of 0 degC.
(2) Freezing of a fraction x of the water at the equilibrium T.
Enthalpy changes for these two steps sum to zero:
CP t
CP
x Hfusion = 0
Hfusion
333.4
joule
gm
J
gm K
4.226
t
CP t
x
x
Hfusion
6K
0.076
Ans.
The entropy change for the two steps is:
T2
S
T1
273.15 K
CP ln
( 273.15
x Hfusion
T2
T2
T1
6) K
S
1.034709
10
3
J
Ans.
gm K
The freezing process itself is irreversible, because it does not occur at the
equilibrium temperature of 0 degC.
H1
1156.3
BTU
lbm
H2
1.7320
BTU
S2
lbm rankine
1.9977
BTU
1533.4
S1
6.21 Data, Table F.4:
H
H2
H1
S
S2
H
377.1
BTU
lbm
S
0.266
lbm
BTU
lbm rankine
S1
BTU
Ans.
lbm rankine
For steam as an ideal gas, apply Eqs. (4.9) and (5.18). [t in degF]
T1
( 227.96
P1
20 psi
T1
382.017 K
T2
459.67)rankine
P2
50 psi
T2
810.928 K
153
( 1000
459.67)rankine
molwt
18
lb
lbmol
R MCPH T1 T2 3.470 1.450 10
molwt
H
H
372.536
BTU
lbm
3
5
0.0 0.121 10
T2
T1
Ans.
R MCPS T1 T2 3.470 1.450 10
S
3
5
0.0 0.121 10
ln
T2
T1
ln
P2
P1
molwt
S
0.259
BTU
lbm rankine
Ans.
6.22 Data, Table F.2 at 8000 kPa:
3
Vliq
cm
1.384
gm
Hliq
1317.1
J
gm
3
Vvap
cm
23.525
gm
Hvap
2759.9
6
mliq
mliq
mvap
54.191 kg
mliq Hliq
Stotal
mliq Sliq
J
gm
Svap
J
gm K
5.7471
J
gm K
6
0.15 10
3
cm
2
Vliq
Htotal
3.2076
Sliq
mvap
mvap Hvap
0.15 10
3
cm
2
Vvap
3.188 kg
Htotal
154
Ans.
Stotal
mvap Svap
80173.5 kJ
192.145
kJ
K
Ans.
6.23
Data, Table F.2 at 1000 kPa:
3
Vliq
1.127
Hliq
gm
762.605
J
gm
Sliq
Hvap
cm
2776.2
J
gm
Svap
3
Vvap
194.29
cm
gm
Let x = fraction of mass that is vapor (quality)
x Vvap
70
30
x
2.1382
6.5828
0.5
(
1
H
(
1
x)Hliq
H
789.495
J
gm
(
1
2.198
J
gm K
(Guess)
0.013
S
x Hvap
gm K
Find x
()
S
x)Vliq
=
x
x
Given
J
x)Sliq
x Svap
J
gm K
Ans.
6.24 Data, Table F.3 at 350 degF:
3
3
Vliq
Hliq
321.76
Vvap
mliq
3 lbm
1
Htotal
lbm
1192.3
BTU
lbm
mvap = 3 lbm mvap Vvap = 50 mliq Vliq mliq
mliq
mvap
BTU
ft
3.342
lbm
Hvap
ft
0.01799
lbm
mliq
50 Vliq
50 mliq Vliq
= 3 lbm
Vvap
2.364 lb
Vvap
3 lbm
mvap
mliq
mliq Hliq
mvap Hvap
Htotal
155
0.636 lb
1519.1 BTU
Ans.
3
6.25
1 cm
0.025 gm
V
Data, Table F.1 at 230 degC:
3
Vliq
1.209
cm
Hliq
gm
990.3
J
gm
3
Vvap
71.45
cm
Hvap
gm
J
gm
2802.0
x Vvap
x
H
(
1
x)Hliq
x Hvap
S
x
6.26
x)Vliq
0.552
Vtotal = mtotal Vliq
Vtotal
3
0.15 m
mtotal
(
1
Vliq
x)Sliq
S
4.599
x Svap
J
gm K
0.382 kg
mliq
mtotal
mvap
mliq
377.72 gm
Ans.
3
Table F.1,
150 degC:
Vvap
Vtotal
Vlv
mtotal Vliq
Vlv
mvap
4.543
3
10
Vtot.liq
Vtot.liq
379.23 cm
kg
mliq Vliq
156
cm
392.4
gm
3
cm
1.004
gm
mvap
Vvap
J
gm K
mvap Vlv
Vliq
Vtotal
J
gm K
Vliq
Vvap
J
gm
6.2107
3
Table F.1,
30 degC:
mtotal
1991
2.6102
Svap
V
V=(
1
H
Sliq
3
Ans.
cm
32930
gm
6.27
Table F.2, 1100 kPa:
Hliq
781.124
J
gm
Hvap
Interpolate @101.325 kPa & 105 degC:
Const.-H throttling:
H2 = Hliq
H2
x
6.28
Hvap
H2
x Hvap
2779.7
2686.1
J
gm
J
gm
Hliq
Hliq
x
Hliq
Ans.
0.953
Data, Table F.2 at 2100 kPa and 260 degC, by interpolation:
H1
H2
J
2923.5
2923.5
gm
J
gm
S1
6.5640
J
gm K
molwt
18.015
gm
mol
Final state is at this enthalpy and a pressure of 125
kPa.
By interpolation at these conditions, the final temperature is 224.80 degC and
S2
7.8316
J
gm K
S
S2
S
S1
1.268
J
gm K
Ans.
For steam as an ideal gas, there would be no temperature change and the
entropy change would be given by:
P1
2100 kPa
P2
125 kPa
S
P2
R
ln
P1
molwt
S
1.302
J
gm K
Ans.
6.29 Data, Table F.4 at 300(psia) and 500 degF:
H1
1257.7
H2
1257.7
BTU
lbm
BTU
lbm
S1
1.5703
BTU
lbm rankine
Final state is at this enthalpy and a pressure of
20(psia).
By interpolation at these conditions, the final temperature is 438.87 degF and
S2
1.8606
BTU
lbm rankine
S
157
S2
S1
S
0.29
BTU
lbm rankine
For steam as an ideal gas, there would be no temperature change and the
entropy change would be given by:
P1
P2
300 psi
6.30
lb
lbmol
P1
S
molwt
0.299
BTU
lbm rankine
Ans.
Data, Table F.2 at 500 kPa and 300 degC
S1
7.4614
J
gm K
1.0912
Hliq
340.564
S2 = S1 = Sliq
H2
Hliq
The final state is at this entropy and a pressure of
50 kPa. This is a state of wet steam, for which
J
gm K
Sliq
6.31
18
P2
R ln
S
molwt
20 psi
Svap
Hvap
J
gm
x Svap
x Hvap
7.5947
J
gm K
2646.9
Sliq
x
H2
Hliq
J
gm
S1
Svap
Sliq
Sliq
x
0.98
J
gm
Ans.
xwater
0.031
Ans.
xwater
0.122
Ans.
2599.6
Vapor pressures of water from Table F.1:
At 25 degC:
Psat
P
xwater
101.33 kPa
At 50 degC:
Psat
xwater
3.166 kPa
Psat
P
12.34 kPa
Psat
P
158
6.32 Process occurs at constant total volume:
Vtotal
3
(
0.014
0.021)m
Data, Table F.1 at 100 degC: Uliq
419.0
J
gm
1.044
cm
1673.0
mvap
0.014 m
Vvap
Vliq
mvap
x
mass
V2
4
4.158 10
3
Vtotal
V2
mass
mliq
gm
3
3
x
mass
gm
cm
Vvap
gm
0.021 m
mliq
J
2506.5
3
3
Vliq
Uvap
cm
1.739
gm
mvap
(initial quality)
This state is first reached as
saturated liquid at 349.83 degC
For this state, P = 16,500.1 kPa, and
U2
Q
1641.7
U2
J
gm
Uliq
Q
U1
U1
1221.8
x Uvap
J
gm
Uliq
U1
419.868
J
gm
Ans.
3
6.33 Vtotal
0.25 m
Data, Table F.2, sat. vapor at 1500 kPa:
3
V1
cm
131.66
gm
U1
2592.4
J
gm
Of this total mass, 25% condenses making the quality 0.75
Since the total volume and mass don't change,
we have for the final state:
V2 = V1 = Vliq
x=
V1
Vvap
Vliq
Vliq
x Vvap
(A)
Vtotal
mass
V1
x
Whence
Vliq
Find P for which (A) yields the value
x = 0.75 for wet steam
159
0.75
Since the liquid volume is much smaller than the vapor volume, we make a
preliminary calculation to estimate:
Vvap
V1
3
Vvap
x
175.547
cm
gm
This value occurs at a pressure a bit above 1100 kPa. Evaluate x at 1100
and 1150 kPa by (A). Interpolate on x to find P = 1114.5 kPa and
Uliq
U2
Q
782.41
Uliq
J
gm
x Uvap
mass U2
Uvap
2584.9
J
gm
U2
2134.3
Q
Uliq
869.9 kJ
U1
J
gm
Ans.
3
3
Vliq
Uvap
6.34 Table F.2,101.325 kPa:
Uliq
J
418.959
gm
Vvap
J
2506.5
gm
cm
1673.0
gm
mliq
cm
1.044
gm
0.02 m
Vliq
3
3
mvap
1.98 m
Vvap
mtotal
mliq
mvap
x
mvap
mtotal
3
V1
Vliq
x Vvap
Vliq
U1
Uliq
x Uvap
Uliq
cm
V1
98.326
U1
540.421
x
gm
0.058
J
gm
Since the total volume and the total mass do not change during the process,
the initial and final specific volumes are the same. The final state is
therefore the state for which the specific volume of saturated vapor is
98.326 cu cm/gm. By interpolation in Table F.1, we find t = 213.0 degC and
U2
2598.4
J
gm
Q
mtotal U2
160
U1
Q
41860.5 kJ
Ans.
6.35 Data, Table F.2 at 800 kPa and 350 degC:
3
V1
cm
354.34
U1
gm
2878.9
J
gm
Vtotal
3
0.4 m
The final state at 200 degC has the same specific volume as the initial
state, and this occurs for superheated steam at a pressure between 575 and
600 kPa. By interpolation, we find P = 596.4 kPa and
U2
J
gm
2638.7
Q
Vtotal
V1
U2
Q
U1
271.15 kJ
Ans.
6.36 Data, Table F.2 at 800 kPa and 200 degC:
U1
J
gm
2629.9
S1
6.8148
J
gm K
mass
1 kg
(a) Isothermal expansion to 150 kPa and 200 degC
U2
J
gm
2656.3
Q
mass T S2
Also:
J
gm K
S2
Q
S1
Work
7.6439
392.29 kJ
mass U2
U1
T
473.15 K
Ans.
Work
Q
365.89 kJ
(b) Constant-entropy expansion to 150 kPa. The final state is wet steam:
J
gm K
Svap
7.2234
J
gm K
J
gm
Uvap
2513.4
J
gm
Sliq
1.4336
Uliq
444.224
x
S1
Svap
Sliq
x
Sliq
U2
Uliq
x Uvap
W
mass U2
U2
Uliq
W
U1
161
0.929
2.367
3J
10
262.527 kJ
gm
Ans.
6.37 Data, Table F.2 at 2000 kPa:
x
H1
Hvap
0.94
Hliq
x Hvap
2797.2
H1
Hliq
J
Hliq
gm
2.684
10
908.589
3J
mass
gm
J
gm
1 kg
For superheated vapor at 2000 kPa and 575 degC, by interpolation:
H2
3633.4
J
gm
6.38 First step:
Q
mass H2
Q
H1
Ans.
949.52 kJ
Q12 = 0
W12 = U2
U1
Second step:
W23 = 0
Q23 = U3
U2
For process:
Q = U3
Table F.2,
2700 kPa:
Uliq
977.968
Sliq
2.5924
x1
0.9
U1
S1
Uliq
Sliq
W = U2
U2
U1
J
gm
Uvap
2601.8
J
gm
J
gm K
Svap
6.2244
J
gm K
x1 Uvap
x1 Svap
Uliq U1
Sliq
S1
3J
2.439
10
5.861
2
3m
10
2
gm
sK
Table F.2, 400 kPa:
J
gm K
Svap
6.8943
J
gm K
J
gm
Uvap
2552.7
J
gm
Sliq
1.7764
Uliq
604.237
Vliq
cm
1.084
gm
3
3
162
Vvap
462.22
cm
gm
Since step 1 is isentropic,
S2 = S1 = Sliq
U2
Uliq
x2 Svap
x2 Uvap
S1
x2
Sliq
Svap
x2
Sliq
2.159
U2
Uliq
Sliq
10
0.798
3J
gm
3
V2
Vliq
x2 Vvap
V2
Vliq
369.135
cm
gm
V3 = V2
and the final state is sat. vapor with this specific volume.
Interpolate to find that this V occurs at T = 509.23 degC and
U3
J
gm
Q
2560.7
401.317
Whence
J
gm
Q
Ans.
U3
Work
U2
Work
280.034
U2
J
gm
Ans.
U1
2605.8
J
gm
S1
7.0548
Table F.1,sat. vapor,
175 degC
U2
2578.8
J
gm
S2
6.6221
mass
T
6.39 Table F.2, 400 kPa &
175 degC:
4 kg
Q
mass T S2
Q
S1
775.66 kJ
Ans.
(
175
273.15)K
W
mass U2
W
U1
667.66 kJ
6.40 (a)Table F.2, 3000 kPa and 450 degC:
H1
3344.6
J
gm
S1
7.0854
J
gm K
Table F.2, interpolate 235 kPa and 140 degC:
H2
2744.5
J
gm
S2
7.2003
163
J
gm K
Ans.
U1
Q
J
gm K
J
gm K
H
H2
S
S2
(b) T1
H
H1
S1
(
450
S
J
gm
Ans.
J
gm K
Ans.
600.1
0.115
T2
273.15)K
(
140
273.15)K
T1
723.15 K
T2
413.15 K
P1
3000 kPa
P2
235 kPa
gm
mol
3
5
R ICPH T1 T2 3.470 1.450 10
0.0 0.121 10
Eqs. (6.95) & (6.96) for an ideal gas:
Hig
molwt
molwt
3
R ICPS T1 T2 3.470 1.450 10
Sig
Tr1
Tr1
5
ln
0.0 0.121 10
620.6
J
gm
647.1 K
T1
Sig
Pc
Pr1
H
S
S
P1
Tc
1.11752
Pr1
Tr2
Pc
0.13602
J
gm K
Tr2
T2
Tc
0.63846
HRB Tr1 Pr1
R Tc HRB Tr2 Pr2
molwt
J
Ans.
593.95
gm
Hig
Sig
0.078
R SRB Tr2 Pr2
J
gm K
SRB Tr1 Pr1
molwt
Ans.
164
Ans.
0.345
220.55 bar
P1
0.0605
Pr2
Pr2
The generalized virial-coefficient correlation is suitable here
H
P2
molwt
Hig
(c) Tc
18
P2
Pc
0.01066
6.41
Data, Table F.2 superheated steam at 550 kPa and 200 degC:
3
V1
385.19
cm
U1
gm
2640.6
J
gm
S1
J
gm K
7.0108
Step 1--2: Const.-V heating to 800 kPa. At the initial specific volume
and this P, interpolation gives t = 401.74 degC, and
U2
2963.1
J
gm
S2
7.5782
J
gm K
Q12
S3
S3 = S 2
Q23 = 0
7.5782
U1
Q12
Step 2--3: Isentropic expansion to initial T.
U2
322.5
J
gm
J
gm K
Step 3--1: Constant-T compression to initial P.
T
473.15 K
Q31
T S1
Q31
S3
268.465
J
gm
For the cycle, the internal energy change = 0.
Wcycle = Qcycle = Q12
1
Q31
Q31
=
0.1675
Q12
Wcycle
Q12
Ans.
6.42 Table F.4, sat.vapor, 300(psi):
T1
T1
459.67) rankine H1
( 417.35
1202.9
877.02 rankine
S1
1.5105
BTU
lbm
BTU
lbm rankine
Superheated steam at 300(psi) & 900 degF
H2
Q12
1473.6
H2
BTU
lbm
H1
S2
Q31
1.7591
BTU
lbm rankine
T 1 S1
165
S3
S3
Q31
S2
218.027
BTU
lbm
For the cycle, the internal energy change = 0.
Wcycle = Qcycle = Q12
1
6.43
Q31
Q31
Q12
Whence
Ans.
0.1946
Q12
Wcycle
=
Data, Table F.2, superheated steam at 4000 kPa and 400 degC:
S1
6.7733
J
gm K
For both parts of the problem:
S2
S1
(a)So we are looking for the pressure at which saturated vapor has the given
entropy. This occurs at a pressure just below 575 kPa. By interpolation,
P2 = 572.83 kPa
Ans.
(b)For the wet vapor the entropy is given by
x
S2 = Sliq
0.95
x Svap
Sliq
So we must find the presure for which this equation is satisfied. This
occurs at a pressure just above 250 kPa. At 250 kPa:
Sliq
1.6071
S2
Sliq
S2
6.7798
J
gm K
x Svap
Svap
7.0520
J
gm K
Sliq
J
gm K
Slightly > 6.7733
By interpolation
P2 = 250.16 kPa
Ans.
6.44 (a) Table F.2 at the final conditions of saturated vapor at 50 kPa:
S2
7.5947
kJ
kg K
H2
2646.0
kJ
kg
S1
S2
Find the temperature of superheated vapor at 2000 kPa with this
entropy. It occurs between 550 and 600 degC. By interpolation
166
t1
(degC)
559.16
Superheat:
(b) mdot
5
t
kg
sec
H1
(
559.16
t
212.37)K
Wdot
kJ
kg
3598.0
mdot H2
H1
Ans.
346.79 K
4760 kW Ans.
Wdot
6.45 Table F.2 for superheated vapor at the initial conditions, 1350 kPa and 375
degC, and for the final condition of sat. vapor at 10 kPa:
H1
kJ
kg
3205.4
S1
7.2410
kJ
kg K
H2
2584.8
kJ
kg
If the turbine were to operate isentropically, the final entropy would be
S2
S1
Table F.2 for sat. liquid and vapor at 10 kPa:
kJ
kg K
Svap
8.1511
kJ
kg K
kJ
kg
Hvap
2584.8
kJ
kg
Sliq
0.6493
Hliq
191.832
x2
S2
Svap
Sliq
Sliq
x2
H' Hliq
0.879
x2 Hvap
H' 2.294
H2
H1
3 kJ
10
kg
Ans.
0.681
H' H1
Hliq
6.46 Table F.2 for superheated vapor at the initial conditions, 1300 kPa and 400
degC, and for the final condition of 40 kPa and 100 degC:
H1
3259.7
kJ
kg
S1
7.3404
kJ
kg K
H2
2683.8
kJ
kg
If the turbine were to operate isentropically, the final entropy would be
S2
S1
Table F.2 for sat. liquid and vapor at 40 kPa:
167
Sliq
1.0261
kJ
kg K
Svap
7.6709
Hliq
317.16
kJ
kg
Hvap
2636.9
S2
x2
Sliq
Svap
H2
x2
Sliq
kJ
kg K
kJ
kg
H' Hliq
0.95
H' 2.522
H1
6.47 Table F.2 at 1600 kPa and 225 degC:
H
3 kJ
kg
P
1600 kPa
S
6.5503
J
gm K
P0
1 kPa
498.15 K
VR
V
3
cm
132.85
gm
10
Hliq
Ans.
0.78
H' H1
V
x2 Hvap
J
gm
2856.3
J
gm K
Table F.2 (ideal-gas values, 1 kPa and 225 degC)
Hig
T
J
gm
2928.7
(
225
Sig
10.0681
T
273.15)K
T
R
molwt P
The enthalpy of an ideal gas is independent of pressure, but the entropy
DOES depend on P:
HR
H
Sig
Hig
R
molwt
3
VR
10.96
cm
gm
HR
Reduced conditions:
Tr
T
Tc
72.4
ln
P
P0
SR
J
gm
S
SR
Sig
0.11
Sig
J
Ans.
gm K
0.345
Tr
Tc
647.1 K
Pc
220.55 bar
0.76982
Pr
P
Pc
Pr
0.072546
The generalized virial-coefficient correlation is suitable here
B0
0.083
0.422
Tr
1.6
B0
0.558
B1
0.139
0.172
Tr
168
4.2
B1
0.377
By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40)
Z
1
B0
B1
R Tc
HR
Pr
Z
Tr
molwt
R
SR
HRB Tr Pr
molwt
3
VR
6.48
9.33
P
cm
HR
gm
53.4
T
1000 kPa
(Table F.2)
SR
273.15) K
18.015
J
gm
Hv
2776.2
J
gm K
Sv
6.5828
762.605
Sl
2.1382
(b)
(c)
Slv
VR
T Sl G l
4.445
Vv
453.03 K
J
gm K
T
R
molwt P
Hlv
206.06
gm
r
10
Gv
3J
Hv
Hl
Sv
Slv
gm
Hv
Vl
Slv
J
gm K
Vv
Hlv
gm
2.014
J
Vlv
J
3
cm
193.163
gm
Hl
Ans.
gm K
gm
mol
Vv
Hl
J
T
cm
194.29
gm
Vl
(a) Gl
0.077
3
3
cm
1.127
gm
Vlv
SRB Tr Pr
J
gm
( 179.88
molwt
RT
( Z 1)
P molwt
VR
0.935
4.445
T Sv G v
Hlv
r
T
Sl
J
gm K
206.01
4.445
J
gm K
3
VR
cm
14.785
gm
Ans.
For enthalpy and entropy, assume that steam at 179.88 degC and 1 kPa
is an ideal gas. By interpolation in Table F.2 at 1 kPa:
169
J
gm
Hig
2841.1
J
gm
Sig
9.8834
J
P0
gm K
1 kPa
The enthalpy of an ideal gas is independent of pressure; the entropy
DOES depend on P:
HR
Hv
SR
Sv
Sig
P
R
ln
P0
molwt
Sig
Hig
HR
Sig
64.9
J
gm
Sig
3.188
Ans. SR
0.1126
J
gm K
J
gm K
Ans.
(d) Assume ln P vs. 1/T linear and fit three data pts @ 975, 1000, & 1050 kPa.
975
Data:
pp
178.79
1000 kPa
t
1050
xi
1
yi
273.15
ti
P
dPdT
T
Slv
2
ln
ppi
Slope
kPa
dPdT
Slope K
13
slope ( y) Slope
x
22.984
4717
kPa
K
J
gm K
Ans.
Slv
4.44
0.345
Reduced conditions:
T
Tc
i
182.02
Tc
647.1 K
Pc
220.55 bar
0.7001
Pr
P
Pc
Pr
0.0453
Vlv dPdT
Tr
(degC)
179.88
Tr
The generalized virial-coefficient correlation is suitable here
B0
0.083
0.422
Tr
1.6
B0
0.664
B1
0.139
0.172
Tr
4.2
B1
0.63
By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40)
Z
1
B0
B1
Pr
Tr
Z
170
0.943
VR
RT
( 1)
Z
P molwt
R Tc
HR
molwt
R
SRB Tr Pr
molwt
SR
HRB Tr Pr
3
VR
cm
11.93
6.49 T
HR
gm
( 358.43
43.18
J
SR
gm
T
459.67) rankine
(Table F.4)
Vv
BTU
lbm
Hv
1194.1
BTU
lbm rankine
Sv
1.5695
330.65
Sl
0.5141
Gl
(b)
(c)
Slv
VR
Gv
Hv
Gv
BTU
lbm rankine
T
R
molwt P
r
T
VR
Hl
Sv
Sl
T Sv
89.91
Hlv
r
Hv
BTU
lbm
863.45
BTU
lbm
1.055
Vv
Hlv
Vl
Slv
lbm rankine
Vv
Hlv
BTU
T Sl
89.94
150 psi
Vlv
BTU
lbm
3
ft
2.996
lbm
Hl
P
gm
mol
ft
3.014
lbm
Hl
(a) Gl
18.015
Ans.
gm K
3
3
Vl
J
818.1 rankine
molwt
ft
0.0181
lbm
Vlv
0.069
0.235
ft
BTU
lbm
1.055
BTU
lbm rankine
3
Ans.
lbm
For enthalpy and entropy, assume that steam at 358.43 degF and 1 psi is
an ideal gas. By interpolation in Table F.4 at 1 psi:
Hig
1222.6
BTU
lbm
Sig
2.1492
171
BTU
lbm rankine
P0
1 psi
The enthalpy of an ideal gas is independent of pressure; the entropy DOES
depend on P:
HR
Hv
P
R
ln
P0
molwt
Sig
SR
HR
Hig
Sv
Sig
28.5
Sig
0.0274
Ans.
BTU
lbm rankine
0.552
SR
Sig
BTU
lbm
BTU
lbm rankine
Ans.
(d) Assume ln P vs. 1/T linear and fit threedata points (@ 145, 150, & 155
psia)
145
Data:
pp
355.77
150 psi
t
155
xi
ti
1
459.67
(degF)
358.43
P
T
Slv
2
yi
ln
ppi
Slope
psi
slope ( y)
x
Slv
Vlv dPdT
T
Tc
Tr
0.345
Tc
1.905
1.056
10
psi
rankine
BTU
Ans.
lbm rankine
Pc
647.1 K
Pr
0.7024
3
8.501
dPdT
Slope rankine
Reduced conditions:
Tr
13
361.02
Slope
dPdT
i
P
Pc
220.55 bar
Pr
0.0469
The generalized virial-coefficient correlation is suitable here
B0
0.083
0.422
Tr
1.6
B0
0.66
B1
0.139
0.172
Tr
172
4.2
B1
0.62
By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40)
Z
1
HR
VR
6.50
B0
R
Tc
ft
Z
Tr
HR
lbm
19.024
For propane:
Tc
T
( 195
T
Tr
T
Tc
273.15) K
Tr
BTU
lbm
SR
(Z
BTU
lbm rankine
1)
P0
135 bar
P
Pc
Pr
Ans.
0.152
42.48 bar
P
468.15 K
1.266
0.0168
Pc
369.8 K
P molwt
R
SRB Tr Pr
molwt
SR
3
RT
VR
0.942
HRB Tr Pr
molwt
0.1894
Pr
B1
Pr
1 bar
3.178
Use the Lee/Kesler correlation; by interpolation,
Z0
V
Z
Z1
ZRT
P
0.1636
V
0.6141
cm
184.2
mol
0.586 R Tc
HR1
1.802
SR1
0.717 R
J
mol K
SR1
5.961
HR1
SR
HR0
7.674
SR0
1.463 R
SR0
12.163
HR
H
0.639
Ans.
HR1
2.496 R Tc
HR0
Z
Z1
3
HR0
HR
Z0
3J
10
mol
3J
7.948 10
SR0
SR
mol
R ICPH 308.15K T 1.213 28.785 10
173
13.069
3
10
3J
mol
J
mol K
SR1
J
mol K
8.824 10
6
0.0
HR
S
R ICPS 308.15K T 1.213 28.785 10
H
6734.9
J
mol
Ans.
6.51 For propane:
T
(
70
Tr
T
Tc
S
Tr
6
8.824 10
15.9
J
mol K
0.0
ln
P
P0
SR
Ans.
369.8 K
Pc
42.48 bar
0.152
343.15 K
P0
101.33 kPa P
1500 kPa
Tc
T
273.15)K
3
P
Pc
Pr
0.92793
Pr
0.35311
Assume propane an ideal gas at the initial conditions. Use
generalized virial correlation at final conditions.
H
S
H
R SRB Tr Pr
P
P0
ln
6.52 For propane:
1431.3
J
mol
Ans.
S
R Tc HRB Tr Pr
25.287
J
mol K
Ans.
0.152
3
cm
Vc 200.0
Zc 0.276
Pc 42.48 bar
Tc 369.8 K
mol
If the final state is a two-phase mixture, it must exist at its saturation
temperature at 1 bar. This temperature is found from the vapor pressure
equation:
P
A
1 bar
D
1.38551
Given
P = Pc exp
T
()
T
A () B
T
Find T)
(
B
6.72219
T
Tc
1
()
T
1.5
C
Guess:
C
1
T
1.33236
()
T
230.703 K
174
()
T
3
D
T
()
T
6
2.13868
200 K
The latent heat of vaporization at the final conditions will be needed for an
energy balance. It is found by the Clapeyron equation. We proceed
exactly as in Pb. 6.17.
P ( T)
A ( T)
Pc exp
P
d
P ( T)
dT
230.703 K
1 bar
Tr
Vvap
Vvap
Hlv
3
( T)
D
kPa
K
Pr
0.815
B1
6
( T)
dPdT
0.024
B1
0.139
Pr
Vliq
Tr
4.428124
T
Tr
Tc
0.172
Tr
B0
4.2
V c Zc
3
4 cm
10
T Vvap
B0
1.6
1
1.847
C
4.428
Pc
0.422
RT
P
1.5
( T)
P
Pr
0.083
( T)
1
T
B0
B
Tr
kPa
K
0.624
B1
1.109
2
1 Tr
7
3
Vliq
mol
Vliq dPdT
Hlv
75.546
1.879
cm
mol
4J
10
mol
ENERGY BALANCE: For the throttling process there is no enthalpy
change. The calculational path from the initial state to the final is made up
of the following steps:
(1) Transform the initial gas into an ideal gas at the initial T & P.
(2) Carry out the temperature and pressure changes to the final T & P in
the ideal-gas state.
(3) Transform the ideal gas into a real gas at the final T & P.
(4) Partially condense the gas at the final T & P.
The sum of the enthalpy changes for these steps is set equal to zero, and
the resulting equation is solved for the fraction of the stream that is liquid.
For Step (1), use the generalized correlation of Tables E.7 & E.8, and let
r0 =
H
R
R Tc
0
and
r1 =
175
H
R
R Tc
1
T1
Tr
P1
370 K
T1
Tr
Tc
By Eq. (6.85)
r0
H1
P1
Pr
1.001
By interpolation, find:
200 bar
r1
3.773
R Tc r0
Pr
Pc
3.568
H1
r1
4.708
4J
1.327
10
mol
For Step (2) the enthalpy change is given by Eq. (6.95), for which
H2
R ICPH T1 T 1.213 28.785 10
3
6
8.824 10
0.0
4J
H2
1.048 10
mol
For Step (3) the enthalpy change is given by Eq. (6.87), for which
Tr
H3
230.703 K
Tc
Tr
R Tc HRB Tr Pr
H3
232.729
H1
Pr
For Step (4),
0.0235
H4 = x Hlv
J
mol
For the process,
x
1 bar
Pc
Pr
0.6239
H2
H1
H3
x
H2
0.136
H3
x Hlv = 0
Ans.
Hlv
6.53 For 1,3-butadiene:
Tc
425.2 K
Vc
0.190
cm
220.4
mol
Tn
268.7 K
101.33 kPa
3
Pc
T
Tr
42.77 bar
380 K
T
Tc
Zc
0.267
P
1919.4 kPa
T0
273.15 K
P0
Tr
0.894
Pr
P
Pc
Pr
176
0.449
Use Lee/Kesler correlation. HOWEVER, the values for a saturated vapor lie
on the very edge of the vapor region, and some adjacent numbers are for the
liquid phase. These must NOT be used for interpolation. Rather,
EXTRAPOLATIONS must be made from the vapor side. There may be
some choice in how this is done, but the following values are as good as any:
Z0
0.7442
Z1
0.1366
Z
Z0
Z1
0.718
3
Vvap
ZRT
P
Vvap
HR0
0.689 R Tc
HR1
0.892 R Tc
HR0
2.436
HR1
3.153
SR0
0.540 R
SR1
0.888 R
SR0
4.49
SR1
7.383
HR
HR
Hvap
Svap
Z
10
3J
mol
J
mol K
HR0
3.035
HR1
SR
3J
10
SR0
SR
mol
6315.9
J
mol
Ans.
3
Svap
Ans.
mol
3J
10
mol
J
mol K
SR1
5.892
R ICPH T0 T 2.734 26.786 10
R ICPS T0 T 2.734 26.786 10
Hvap
1182.2
cm
J
mol K
3
6
8.882 10
8.882 10
1.624
6
0.0
J
mol K
0.0
ln
HR
P
P0
SR
Ans.
For saturated vapor, by Eqs. (3.63) & (4.12)
2
Vliq
V c Zc
1 Tr
3
7
Vliq
177
cm
109.89
mol
Ans.
1.092 ln
Hn
R Tn
Pc
Hn
Tn
0.930
By Eq. (4.13)
1.013
bar
Tc
H
Hn
1
Tr
Hvap
Sliq
Svap
0.38
H
Tn
1
Hliq
14003
J
mol
Tc
H
Hliq
7687.4
H
T
Sliq
38.475
6.54 For n-butane:
J
mol
22449
J
Ans.
mol
J
mol K
Ans.
Tc
425.1 K
Vc
0.200
cm
255
mol
Tn
272.7 K
101.33 kPa
3
Pc
T
Tr
Zc
37.96 bar
370 K
0.274
P
T0
273.15 K
P0
Tr
T
Tc
1435 kPa
0.87
Pr
P
Pc
Pr
0.378
Use Lee/Kesler correlation. HOWEVER, the values for a saturated vapor lie
on the very edge of the vapor region, and some adjacent numbers are for the
liquid phase. These must NOT be used for interpolation. Rather,
EXTRAPOLATIONS must be made from the vapor side. There may be
some choice in how this is done, but the following values are as good as any:
Z0
V
Z
0.1372
ZRT
P
Z0
V
Z1
0.7692
Z
Z1
cm
1590.1
mol
3
HR0
2.145
HR1
10
3J
mol
178
0.831 R Tc
HR1
0.607 R Tc
HR0
Ans.
2.937
3J
10
mol
0.742
SR0
0.485 R
SR0
4.032
HR
HR
Hvap
Svap
SR1
SR1
J
mol K
HR0
2.733
HR1
mol
R ICPH T0 T 1.935 36.915 10
7427.4
J
mol
SR0
SR
3
R ICPS T0 T 1.935 36.915 10
Hvap
6.942
SR
3J
10
0.835 R
Ans.
SR1
5.421
3
J
mol K
6
11.402 10
11.402 10
Svap
J
mol K
6
4.197
0.0
0.0
HR
P
ln
SR
P0
J
mol K
Ans.
For saturated vapor, by Eqs. (3.72) & (4.12)
Vliq
V c Zc
1 Tr
3
2/7
Vliq
1.092 ln
Hn
R Tn
Pc
By Eq. (4.13)
H
Hvap
Sliq
Svap
J
mol
Hn
22514
H
Tn
15295.2
Tc
1
Hn
1
Hliq
Ans.
1.013
bar
0.930
cm
123.86
mol
Tr
0.38
Tn
Tc
H
Hliq
7867.8
J
mol
Ans.
H
T
Sliq
37.141
J
mol K
Ans.
179
J
mol
6.55 Under the stated conditions the worst possible cycling of demand can be
represented as follows:
10, kg/
000
hr
1/ hr
3
2/ hr
3
Dem and
( hr
kg/ )
1 hr
tm e
i
6,
000
4, kg/
000
hr
netst age
or
ofst
eam
netdepl i
eton
ofst
eam
This situation is also represented by the equation:
4000
= 6000
10000 1
= time of storage liquid
2
Solution gives
hr
3
The steam stored during this leg is: mprime
where
6000
mprime
kg
hr
4000
kg
hr
1333.3 kg
We consider this storage leg, and for this process of steam addition to
a tank the equation developed in Problem 6-74 is applicable:
m1 Hprime
H1
Vtank P2
m2 =
Hprime
Hf2
Vf2
P1
Hfg2
Vfg2
Hfg2
Vfg2
We can replace Vtank by m2V2, and rearrange to get
m2
m1
Hprime
Hf2
Vf2
Hfg2
Vfg2
V 2 P2
However M1 v1 = m2 V2 = Vtank
P1
Hfg2
and therefore
180
= Hprime
Vfg2
m2
m1
=
V1
V2
H1
Eq. (A)
Making this substitution and rearranging we get
Hprime
Hf2
Vf2
Hfg2
Vfg2
P2
V2
Hfg2
P1
Vfg2
Hprime
=
H1
V1
In this equation we can determine from the given information everything
except Hprime and Vprime. These quantities are expressed by
H1 = Hf1
and
x1 Hfg1
V1 = Vf1
x1 Vfg1
Therefore our equation becomes (with Hprime = Hg2)
Hg2
Hf2
Hfg2
Vf2
Vfg2
P2
V2
P1
Hfg2
Vfg2
=
Hg2
Hf1
Vf1
x1 Hfg1
Eq. (B)
x1 Vfg1
In this equation only x1 is unknown and we can solve for it as follows. First
we need V2:
From the given information we can write:
0.95V2 = 1
x2 Vf2
therefore
19 =
Then
V2 =
1
0.05V2 = x2 Vg2
x2 Vf2
x2 Vg2
Vf2
Vg2
0.05
or
19Vg2
Vf2
Vf2
x2 =
=
19Vg2
20
19
Vf2
Vf2
Eq. (C)
1
Vg2
Now we need property values:
Initial state in accumulator is wet steam at 700 kPa.
P1
700kPa
We find from the steam tables
Hf1
697.061
kJ
Hg1
kg
2762.0
kJ
kg
181
Hfg1
Hg1
Hf1 Hfg1
2064.939
kJ
kg
Vf1
1.108
3
3
3
cm
Vg1
gm
272.68
cm
Vfg1
gm
Vf1 Vfg1
Vg1
Final state in accumulator is wet steam at 1000 kPa.
From the steam tables
Hf2
762.605
kJ
kg
Hg2
2776.2
1.127
Hfg2
Hf2 Hfg2
Hg2
cm
Vg2
gm
194.29
gm
1000kPa
2013.595
kJ
kg
3
3
3
Vf2
kJ
kg
P2
271.572
cm
cm
Vfg2
gm
Vf2 Vfg2
Vg2
193.163
cm
gm
Solve Eq. (C) for V2
Vf2
Vg2
V2
19Vg2
0.05
V2
Vf2
Next solve Eq. (B) for x1
1.18595
Guess: x1
10
3
3m
kg
0.1
Given
Hg2
Hf2
Hfg2
Vf2
Vfg2
P2
V2
x1
x1
Find x1
4.279
Hfg2
P1
10
Vfg2
=
Hg2
Hf1
Vf1
x1 Hfg1
x1 Vfg1
4
3
Thus
V1
Vf1
V1
x1 Vfg1
V1
m2
=
V2
m1
Eq. (A) gives
1.22419
and
cm
gm
mprime = m2
m1 = 2667kg
Solve for m1 and m2 using a Mathcad Solve Block:
mprime
Guess: m1
m2 m1
2
Given
m1
m2
m1
3.752
=
4
V1
V2
10 kg
m2
m1 = 2667lb
m2
3.873
182
4
10 kg
m1
m2
Find m1 m2
Finally, find the tank volume Vtank
m2 V2
3
Vtank
45.9 m
Ans.
Note that just to store 1333.3 kg of saturated vapor at 1000 kPa would
require a volume of:
3
1333.3kg Vg2
259 m
One can work this problem very simply and almost correctly by ignoring the
vapor present. By first equation of problem 3-15
m2
m1
Hprime
=
Hprime
Hprime
U1
U2
Hg2
=
Hprime
Uf1
Hprime
Uf2
Hprime
2.776
=
Hprime
Hf1
Hprime
Hf2
3 kJ
10
kg
Given
m2
m1
=
Hprime
Hprime
m1
Hf1
Hf2
m2
3.837
V
m2
m1 = 2667lb
m2
Find m1 m2
4
10 kg
m2 Vf2
0.95
V
6.56 Propylene:
3
45.5 m
Ans.
0.140
T
Tc
400.15 K
P
365.6 K
38 bar
Pc
46.65 bar
P0
1 bar
The throttling process, occurring at constant enthalpy, may be split into two
steps:
(1) Transform into an ideal gas at the initial conditions, evaluating property
changes from a generalized correlation.
(2) Change T and P in the ideal-gas state to the
final conditions, evaluating property changes by equations for an ideal gas.
Property changes for the two steps sum to the property change for the
process. For the initial conditions:
183
T
Tc
Tr
Tr
P
Pc
Pr
1.095
Pr
0.815
Step (1): Use the Lee/Kesler correlation, interpolate.
H0
H0
2.623 10
S0
4.697
mol
J
mol K
H1
1.623
0.496 R
S1
3J
4.124
HR
0.534 R Tc
S1
0.565 R
S0
H1
0.863 R Tc
3J
10
mol
HR
SR
J
mol K
H0
H1
3J
2.85
S0
SR
10
mol
S1
5.275
J
mol K
Step (2): For the heat capacity of propylene,
A
B
1.637
22.706 10
K
3
6.915 10
C
K
6
2
Solve energy balance for final T. See Eq. (4.7).
(guess)
1
HR = R
AT
Given
1
B2
T
2
2
Tf
0.908
Find
C3
T
3
1
3
Tf
T
Sig
R ICPS T Tf 1.637 22.706 10
Sig
22.774
SR
Sig
S
184
6.915 10
6
363.27 K Ans.
J
mol K
S
3
1
28.048
J
mol K
0.0
ln
Ans.
P0
P
6.57 Propane:
Tc
0.152
Pc
369.8 K
42.48 bar
P0
1 bar
P
22 bar
T
423 K
The throttling process, occurring at constant enthalpy, may be split into
two steps:
(1) Transform into an ideal gas at the initial conditions, evaluating property
changes from a generalized correlation.
(2) Change T and P in the ideal-gas state to the
final conditions, evaluating property changes by equations for an ideal gas.
Property changes for the two steps sum to the property change for the
process. For the initial conditions:
Tr
T
Tc
Tr
P
Pc
Pr
1.144
Pr
0.518
Step (1): Use the generalized virial correlation
3J
HR
R Tc HRB Tr Pr
HR
1.366
10
SR
R SRB Tr Pr
SR
2.284
J
mol K
mol
Step (2): For the heat capacity of propane,
A
B
1.213
3
28.785 10
K
8.824 10
C
K
6
2
Solve energy balance for final T. See Eq. (4.7).
1 (guess)
HR = R
AT
Given
1
B
2
T
0.967
Find
2
2
C3
T
3
1
Tf
Sig
R ICPS T Tf 1.213 28.785 10
Sig
22.415
J
mol K
S
SR
Sig
S
24.699
185
J
mol K
1
Tf
T
3
3
8.824 10
Ans.
6
Ans.
408.91 K
0.0
ln
P0
P
6.58 For propane:
T
(
100
T
Tc
Pc
42.48 bar
T
273.15)K
Tr
0.152
369.8 K
Tc
P0
1 bar
P
10 bar
Pr
P
Pc
Pr
0.235
373.15 K
Tr
1.009
Assume ideal gas at initial conditions. Use virial correlation at final conditions.
H
S
R SRB Tr Pr
6.59 H2S:
T1
Tr1
Tr1
J
mol
Ans.
J
mol K
Ans.
H
ln
P
P0
801.9
S
R Tc HRB Tr Pr
20.639
0.094
P1
400 K
T1
Tc
373.5 K
Pc
89.63 bar
5 bar
T2
600 K
P2
25 bar
P1
Pr1
Tc
Pr1
1.071
T2
Tr2
Pc
Tc
Tr2
0.056
Pr2
Pr2
1.606
P2
Pc
0.279
Use generalized virial-coefficient correlation for both sets of conditions.
Eqs. (6.91) & (6.92) are written
3
5
H
R ICPH T1 T2 3.931 1.490 10
0.0 0.232 10
R Tc HRB Tr2 Pr2
HRB Tr1 Pr1
S
R ICPS T1 T2 3.931 1.490 10
R SRB Tr2 Pr2
H
7407.3
J
mol
3
0.0
5
0.232 10
ln
P2
P1
SRB Tr1 Pr1
S
186
1.828
J
mol K
Ans.
6.60 Carbon dioxide:
Tc
0.224
Pc
304.2 K
73.83 bar
P0
101.33 kPa
P
1600 kPa
T
318.15 K
Throttling process, constant enthalpy, may be split into two steps:
(1) Transform to ideal gas at initial conditions, generalized correlation
for property changes.
(2) Change T and P of ideal gas to final T & P.
Property changes by equations for an ideal gas.
Assume ideal gas at final T & P. Sum property changes for the process.
For the initial T & P:
T
Tc
Tr
Tr
P
Pr
1.046
Pr
Pc
0.217
Step (1): Use the generalized virial correlation
R Tc HRB Tr Pr
SR
HR
R SRB Tr Pr
587.999
SR
HR
1.313
J
mol
J
mol K
Step (2): For the heat capacity of carbon dioxide,
A
1.045 10
K
B
5.457
3
5
D
1.157 10 K
2
Solve energy balance for final T. See Eq. (4.7).
Given
1 (guess)
HR = R A T
1
2
1
R ICPS T Tf 5.457 1.045 10
Sig
21.047
SR
Sig
T
Tf
T
3
0.0
S
22.36
J
mol K
187
302.71 K
5
1.157 10
J
mol K
S
1
D
Tf
0.951
Find
Sig
B2
T
2
Ans.
ln
P0
P
Ans.
6.61
T0
P0
523.15 K
S
J
mol K
0
A
P
3800 kPa
120 kPa
For the heat capacity of ethylene:
14.394 10
B
1.424
3
C
K
4.392 10
6
2
K
(a) For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15)
with D = 0:
(guess)
0.4
Given
S = R A ln
Hig
Tf
1.185
Ws
10
3
308.19 K
4.392 10
6
Ans.
0.0
mol
Ws
(b) Ethylene:
11852
Tc
0.087
Tr0
Tc
Tf
T0
ln
4J
Hig
T0
P
P0
1
2
R ICPH T0 Tf 1.424 14.394 10
Hig
Tr0
C T0
0.589
Find
1
2
B T0
1.85317
Pr0
J
mol
Ans.
282.3 K
P0
Pc
Pc
50.40 bar
Pr0
0.75397
At final conditions as calculated in (a)
Tr
T
Tc
Tr
1.12699
Pr
P
Pc
Use virial-coefficient correlation.
The entropy change is now given by Eq. (6.92):
0.5
(guess)
Given
188
Pr
0.02381
S = R A ln
SRB
C T0
Pr
Tc
Tc
2
T
T0
Tr
T
1
ln
P
P0
SRB Tr0 Pr0
T
Find
Tr
T0
1
2
B T0
303.11 K
Ans.
1.074
The work is given by Eq. (6.91):
Hig
R ICPH T0 T 1.424 14.394 10
Hig
Ws
Hig
Ws
6.62 T0
S
A
1.208
11567
0.0
HRB Tr0 Pr0
Ans.
P0
P
30 bar
2.6 bar
For the heat capacity of ethane:
B
1.131
6
mol
J
mol
J
mol K
4.392 10
4J
R Tc HRB Tr Pr
493.15 K
0
10
3
19.225 10
K
3
C
5.561 10
6
2
K
(a) For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15)
with D = 0:
(guess)
Given
0.4
S = R A ln
Find
Hig
B T0
1
2
C T0
2
T
0.745
R ICPH T0 T 1.131 19.225 10
189
T0
3
P
P0
1
ln
T
367.59 K
5.561 10
6
0.0
Ans.
Hig
8.735
Ws
3J
10
mol
Ws
Hig
(b) Ethane:
J
mol
Tr0
Tc
Ans.
Tc
0.100
T0
Tr0
8735
Pc
48.72 bar
Pr0
P0
Pr0
1.6153
Pc
305.3 K
0.61576
At final conditions as calculated in (a)
T
Tc
Tr ( )
T
P
Pc
Pr
Tr ( ) 1.20404
T
Pr
0.05337
Use virial-coefficient correlation.
The entropy change is now given by Eq. (6.83):
(guess)
0.5
S = R A ln
B T0
SRB
Tr
Given
T0
Tc
Pr
2
T
T0
Tr
Tc
ln
1
P
P0
SRB Tr0 Pr0
T
Find
T
1
2
C T0
362.73 K
1.188
The work is given by Eq. (6.91):
Hig
Hig
R ICPH T0 T 1.131 19.225 10
9.034
Ws
Hig
Ws
8476
10
5.561 10
3J
mol
R Tc HRB Tr Pr
J
mol
3
Ans.
190
HRB Tr0 Pr0
6
0.0
Ans.
6.63
n-Butane:
T0
S
A
425.1 K
Pc
P0
1 bar
37.96 bar
P
7.8 bar
For the heat capacity of n-butane:
36.915 10
K
B
1.935
T0
Tr0
323.15 K
J
mol K
0
Tc
0.200
Tr0
Tc
3
C
K
Pr0
Pc
P
Pr
HRB Tr0 Pr0
2
P0
Pr0
0.76017
6
11.402 10
Pr
Pc
= 0.05679
0.02634
0.205
HRB0
0.05679
The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0:
(guess)
0.4
S = R A ln
Given
B T0
SRB
T0
Tc
Find
T
Tc
Tr
Pr
Tr
1
2
ln
T
T0
381.43 K
0.89726
The work is given by Eq. (6.91):
Hig
R ICPH T0 T 1.935 36.915 10
Hig
6.551
Ws
Hig
Ws
5680
3
11.402 10
3J
10
mol
R Tc HRB Tr Pr
J
mol
P
P0
SRB Tr0 Pr0
T
1.18
1
2
C T0
HRB Tr0 Pr0
Ans.
191
6
0.0
Ans.
6.64
The maximum work results when the 1 kg of steam is reduced in a
completely reversible process to the conditions of the surroundings,
where it is liquid at 300 K (26.85 degC). This is the ideal work.
From Table F.2 for the initial state of superheated steam:
H1
kJ
3344.6
S1
kg
7.0854
kJ
kg K
From Table F.1, the state of sat. liquid at 300 K is essentially correct:
H2
112.5
kJ
kg
S2
0.3928
kJ
kg K
T
300 K
By Eq. (5.27),
Wideal
6.65
H2
H1
T
S2
Wideal
S1
1224.3
kJ
kg
Ans.
Sat. liquid at 325 K (51.85 degC), Table F.1:
Hliq
kJ
217.0
kg
Psat
12.87 kPa
P1
Sliq
325 K
H1
Hliq
S1
Sliq
cm
1.013
gm
Vliq
For the compressed liquid at
325 K and 8000 kPa, apply
Eqs. (6.28) and (6.29) with
8000 kPa
T
3
kJ
0.7274
kg K
460 10
Vliq 1
T
Vliq P1
P1
6
K
1
H1
Psat
223.881
S1
Psat
0.724
kJ
kg
kJ
kg K
For sat. vapor at 8000 kPa, from Table F.2:
H2
2759.9
kJ
kg
S2
Q
Heat added in boiler:
5.7471
H2
kJ
kg K
H1
T
Q
300 K
2536
kJ
kg
Maximum work from steam, by Eq. (5.27):
Wideal
H1
H2
T
S1
S2
192
Wideal
1029
kJ
kg
Work as a fraction of heat added:
Frac
Wideal
Frac
Q
Ans.
0.4058
The heat not converted to work ends up in the surroundings.
Q
SdotG.surr
Wideal
T
SdotG.system
S1
kg
10
SdotG.surr
sec
S 2 10
kg
50.234
SdotG.system
sec
kW
K
50.234
kW
K
Obviously the TOTAL rate of entropy generation is zero. This is because
the ideal work is for a completely reversible process.
6.66
Treat the furnace as a heat reservoir, for which
kg
sec
kg
kW
Qdot
50.234
K
T
Qdot
2536
SdotG
kJ
T
10
(
600
SdotG
273.15)K
21.19
kW
K
T
873.15 K
Ans.
By Eq. (5.34)
T
6.67
Wdotlost
300 K
Wdotlost
T SdotG
For sat. liquid water at 20 degC, Table F.1:
H1
kJ
kg
83.86
S1
0.2963
kJ
kg K
For sat. liquid water at 0 degC, Table F.1:
H0
0.04
kJ
S0
kg
0.0000
kJ
kg K
For ice at at 0 degC:
H2
H0
333.4
kJ
kg
S2
S0
193
333.4 kJ
273.15 kg K
6356.9 kW
Ans.
kJ
H2
333.44
T
S2
293.15 K
kg
1.221
mdot
0.5
kJ
kg K
kg
0.32
t
sec
By Eqs. (5.26) and (5.28):
Wdotideal
mdot H2
H1
T
S2
Wdotideal
Wdot
S1
Wdotideal
Wdot
42.77 kW
13.686 kW
Ans.
t
6.68
This is a variation on Example 5.6., pp. 175-177, where all property values
are given. We approach it here from the point of view that if the process
is completely reversible then the ideal work is zero. We use the notation of
Example 5.6:
H1
2676.0
S2
0.0
kJ
kg
kJ
kg K
S1
Q'
2000
kJ
kg K
7.3554
kJ
kg
H2
T
kJ
kg
0.0
273.15 K
The system consists of two parts: the apparatus and the heat reservoir at
elevated temperature, and in the equation for ideal work, terms must be
included for each part.
Wideal =
Happaratus.reservoir
Happaratus.reservoir = H2
Sapparatus.reservoir = S2
T'
450 K
Given
H1
S1
(Guess)
kJ
0
= H2 H1
kg
T'
T
Sapparatus.reservoir
Q'
Wideal = 0.0
Q'
T'
Q'
T
S2
S1
T'
Find ( )
T'
Q'
T'
409.79 K
(136.64 degC)
194
kJ
kg
Ans.
6.69 From Table F.4 at 200(psi):
H1
BTU
1222.6
S1
lbm
Hliq
BTU
355.51
lbm
Sliq
0.5438
Hliq
H2
1.165 10
BTU
x Hvap
(at 420 degF)
lbm rankine
Hvap
BTU
lbm rankine
H2
1.5737
1198.3
Svap
1.5454
(Sat. liq.
and vapor)
BTU
lbm
BTU
x
lbm rankine
S2
3 BTU
lbm
Sliq
x Svap
S2
Hliq
1.505
0.96
Sliq
BTU
lbm rankine
Neglecting kinetic- and potential-energy changes,
on the basis of 1 pound mass of steam after mixing, Eq. (2.30) yields for
the exit stream:
H
x
S
0.5 H1
H
Hvap
Sliq
H
Hliq
x Svap
Sliq
1.54
BTU
lbm
0.994
S
Hliq
1193.6
x
0.5 H2
(wet steam)
Ans.
BTU
lbm rankine
By Eq. (5.22) on the basis of 1 pound mass of exit steam,
SG
6.70
S
0.5 S1
SG
0.5 S2
2.895
10
4
BTU
lbm rankine
Ans.
From Table F.3 at 430 degF (sat. liq. and vapor):
3
3
Vliq
ft
0.01909
lbm
Vvap
ft
1.3496
lbm
Uliq
406.70
BTU
lbm
Uvap
1118.0
VOLliq
BTU
lbm
VOLliq
mliq Vliq
195
Vtank
mliq
79.796 ft
3
80 ft
3
4180 lbm
VOLvap
Vtank
VOLliq
VOLvap
VOLvap
mvap
mvap
Vvap
mliq Uliq
U1
mvap Uvap
mliq
U1
mvap
0.204 ft
3
0.151 lbm
406.726
BTU
lbm
By Eq. (2.29) multiplied through by dt, we can write,
d mt Ut
H dm = 0
(Subscript t denotes the contents of the tank.
H and m refer to the exit stream.)
m
Integration gives:
m2 U2
m1 U1
H dm = 0
0
From Table F.3 we see that the enthalpy of saturated vapor changes
from 1203.9 to 1203.1(Btu/lb) as the temperature drops from 430 to 420
degF. This change is so small that use of an average value for H of
1203.5(Btu/lb) is fully justified. Then
m2 U2
m1
m1 U1
mliq
Have m = 0
Have
mvap
1203.5
m2 (
mass)
m1
Property values below are for sat. liq. and vap. at 420 degF
3
3
Vliq
ft
0.01894
lbm
Vvap
ft
1.4997
lbm
Uliq
395.81
BTU
lbm
Uvap
1117.4
V2 (
mass)
U2 (
mass)
mass
Vtank
m2 (
mass)
Uliq
x mass)
(
x mass) Uvap
(
50 lbm (Guess)
196
Uliq
BTU
lbm
V2 (
mass) Vliq
Vvap
Vliq
BTU
lbm
mass
m1 U1 U2 ( mass)
Have U2 ( mass)
Given
mass =
mass
Find ( mass)
mass
Ans.
55.36 lbm
6.71 The steam remaining in the tank is assumed to have expanded isentropically.
Data from Table F.2 at 4500 kPa and 400 degC:
S1
6.7093
3
J
gm K
S2 = S1 = 6.7093
J
gm K
V1
64.721
cm
Vtank
gm
3
50 m
By interpolation in Table F.2
at this entropy and 3500 kPa:
3
cm
78.726
gm
V2
Vtank
m1
6.72
m2
V1
Ans.
t2 = 362.46 C
Vtank
V2
m
m1
m2
m
137.43 kg Ans.
This problem is similar to Example 6.8, where it is shown that
Q=
mt Ht
H mt
Here, the symbols with subscript t refer to the contents of the tank,
whereas H refers to the entering stream.
We illustrate here development of a simple expression for the first term on
the right. The1500 kg of liquid initially in the tank is unchanged during the
process. Similarly, the vapor initially in the tank that does NOT condense
is unchanged. The only two enthalpy changes within the tank result from:
1. Addition of 1000 kg of liquid water. This contributes an enthalpy change of
Hliq mt
2. Condensation of y kg of sat. vapor to sat. liq.
This contributes an enthalpy change of
y Hliq
Thus
Hvap = y Hlv
mt Ht = Hliq mt
y Hlv
197
Similarly,
mt Vt = Vliq mt
Whence
mt
Q = Hliq mt
H
At 250 degC:
209.3
Hliq
Vliq mt
6.73
y
Vlv
mt Hliq
H
Given:
C
kJ
kg K
3
Vliq
cm
1.251
gm
3
kJ
1714.7
kg
cm
48.79
gm
Vlv
25.641 kg
y Hlv
Vtank
0.43
kJ
kg
kJ
1085.8
kg
Hlv
Q
H mt
Required data from Table F.1 are:
1000 kg
At 50 degC:
y
y Hlv
y Vlv = 0
T1
Q
Ans.
832534 kJ
3
0.5 m
295 K
kJ
kg
Hin
120.8
mtank
30 kg
Data for saturated nitrogen vapor:
80
0.1640
85
2.287
0.1017
90
T
1.396
3.600
0.06628
95
K
P
5.398 bar
V
0.04487
100
7.775
0.03126
105
10.83
0.02223
110
14.67
0.01598
198
3
m
kg
78.9
At the point when liquid nitrogen starts to
accumulate in the tank, it is filled with saturated
vapor nitrogen at the final temperature and having
properties
82.3
85.0
H
kJ
kg
86.8
87.7
mvap Tvap Vvap Hvap Uvap
87.4
85.6
By Eq. (2.29) multiplied through by dt, d nt Ut
H dm = dQ
Subscript t denotes the contents of the tank; H and m refer to the inlet
stream. Since the tank is initially evacuated, integration gives
mvap Uvap
Hin mvap = Q = mtank C Tvap
Also,
mvap =
(A)
T1
Vtank
(B)
Vvap
Calculate internal-energy values for saturated vapor nitrogen at the given
values of T:
56.006
U
(
H
59.041
P V)
61.139
U
62.579
kJ
kg
63.395
63.325
62.157
Fit tabulated data with cubic spline:
Us
lspline ( U)
T
Uvap ()
t
Tvap
Vs
interp ( T U t)
Us
100 K
lspline ( V)
T
Vvap ()
t
(guess)
Combining Eqs. (A) & (B) gives:
199
interp ( T V t)
Vs
Given
Uvap Tvap
Tvap
mvap
6.74
Hin =
mtank C T1
Tvap Vvap Tvap
Vtank
Find Tvap
Tvap
97.924 K
mvap
13.821 kg
Vtank
Vvap Tvap
Ans.
The result of Part (a) of Pb. 3.15 applies, with m replacing n:
m2 U2
H
m1 U1
H =Q=0
Whence
Also
m2 H
U2 = m1 H
U2 = Uliq.2
x2 Ulv.2
V2 = Vliq.2
x2 Vlv.2
U1
V2 =
Vtank
m2
Eliminating x2 from these equations gives
Vtank
m2 H
m2
Uliq.2
Vliq.2
Vlv.2
Ulv.2 = m1 H
U1
which is later solved for m2
Vtank
3
50 m
m1
16000 kg
V1
V1
Data from Table F.1
@ 25 degC:
3
Vliq.1
cm
1.003
gm
Uliq.1
104.8
kJ
kg
3
Vlv.1
cm
43400
gm
Ulv.1
2305.1
200
kJ
kg
Vtank
m1
3.125
10
3
3m
kg
V1
x1
x1
Vliq.1
U1
4.889
10
5
Uliq.1
x1 Ulv.1
U1
Vlv.1
104.913
kJ
kg
Data from Table F.2 @ 800 kPa:
Vliq.2
1.115
720.043
kJ
kg
Ulv.2
(
2575.3
Ulv.2
3
cm
1.855
Uliq.2
gm
3
Vlv.2
(
240.26
Vlv.2
1.115)
cm
m
0.239
kg
gm
720.043)
3
Data from Table F.2 @ 1500 kPa:
m1 H
U1
Vtank
m2
H
msteam
6.75
Uliq.2
m2
Vliq.2
m1
H
2789.9
3 kJ
10
kg
kJ
kg
Ulv.2
Vlv.2
m2
Ulv.2
2.086
4
10 kg
Vlv.2
msteam
4.855
3
10 kg
The result of Part (a) of Pb. 3.15 applies, with
Whence
kJ
kg
Ans.
n1 = Q = 0
U2 = H
From Table F.2 at 400 kPa and 240 degC
H = 2943.9
kJ
kg
Interpolation in Table F.2 will produce values of t
and V for a given P where U = 2943.9 kJ/kg.
201
1
384.09
303316
100
384.82
3032.17
1515.61
V2
385.57
t2
200
P2
300
387.08
gm
1010.08
400
i
386.31
3
cm
757.34
15
Vtank
3
1.75 m
Vtank
massi
V2
i
5.77
10
3
T rises very slowly as P increases
0.577
mass
1.155
3
kg
1.733
2
massi
2.311
1
0
0
200
P2
6.76
3
Vtank
400
i
Data from Table F.2 @ 3000 kPa:
2m
3
3
Vliq
cm
1.216
gm
Vvap
cm
66.626
gm
Hliq
1008.4
kJ
kg
Hvap
2802.3
x1
0.1
V1
V1
Vliq
kJ
kg
x1 Vvap
7.757 10
3
3m
kg
Vliq
m1
m1
Vtank
V1
257.832 kg
The process is the same as that of Example 6.8, except that the stream
flows out rather than in. The energy balance is the same, except for a sign:
Q=
mt Ht
H mtank
202
where subscript t denotes conditions in the tank, and H is the enthalpy of
the stream flowing out of the tank. The only changes affecting the
enthalpy of the contents of the tank are:
1. Evaporation of y kg of sat. liq.:
y Hvap
Hliq
2. Exit of
of liquid from the tank:
0.6 m1 kg
0.6 m1 Hliq
Thus
mt Ht = y Hvap
Hliq
0.6 m1 Hliq
Similarly, since the volume of the tank is constant, we can write,
mt Vt = y Vvap
Whence
Q=
But
y=
Vliq
0.6 m1 Vliq = 0
0.6 m1 Vliq
Vvap Vliq
0.6 m1 Vliq
Hvap
Vvap Vliq
Hliq
and
H = Hliq
0.6 m1 Hliq
0.6 m1 =
H mtank
mtank
and therefore the last two terms of the energy equation cancel:
Q
6.77
0.6 m1 Vliq
Hvap
Vvap Vliq
Hliq
Q
5159 kJ
Ans.
Data from Table F.1 for sat. liq.:
H1
100.6
kJ
kg
(24 degC)
H3
355.9
kJ
kg
(85 degC)
Data from Table F.2 for sat. vapor @ 400 kPa:
H2
2737.6
kJ
kg
By Eq. (2.30), neglecting kinetic and potential energies and setting the
heat and work terms equal to zero:
H3 mdot3
H1 mdot1
H2 mdot2 = 0
203
Also
mdot1 = mdot3
Whence
mdot3 H1
mdot2
mdot1
mdot3
mdot2
0.484
H1
kg
mdot3
mdot2
5
kg
sec
Ans.
sec
H3
H2
mdot2
kg
sec
Ans.
mdot1
4.516
6.78 Data from Table F.2 for sat. vapor @ 2900 kPa:
H3
2802.2
kJ
kg
S3
6.1969
kJ
kg K
mdot3
15
kg
sec
Table F.2, superheated vap., 3000 kPa, 375 degC:
H2
3175.6
kJ
kg
S2
6.8385
kJ
kg K
Table F.1, sat. liq. @ 50 degC:
3
Vliq
cm
1.012
gm
Hliq
Psat
12.34 kPa
T
209.3
kJ
kg
Sliq
0.7035
kJ
kg K
323.15 K
Find changes in H and S caused by pressure increase from 12.34 to 3100
kPa. First estimate the volume expansivity from sat. liq, data at 45 and 55
degC:
3
V
V
(
1.015
5 10
1.010)
cm
T
gm
3
3 cm
P
10 K
1
Vliq
gm
V
T
3100 kPa
4.941
Apply Eqs. (6.28) & (6.29) at constant T:
H1
Hliq
S1
Sliq
Vliq 1
Vliq P
T
P
kJ
kg
kJ
Psat
204
H1
211.926
S1
Psat
0.702
kg K
10
4
K
1
By Eq. (2.30), neglecting kinetic and potential energies and setting the heat
and work terms equal to zero:
H3 mdot3
H1 mdot1
H2 mdot2 = 0
Also
mdot2 = mdot3
Whence
mdot1
mdot2
mdot3
mdot1
mdot3 H3 H2
H1 H2
kg
sec
mdot1
mdot2
mdot1
1.89
13.11
Ans.
kg
sec
For adiabatic conditions, Eq. (5.22) becomes
SdotG
S3 mdot3
SdotG
1.973
S1 mdot1
kJ
sec K
S2 mdot2
Ans.
The mixing of two streams at different temperatures is irreversible.
6.79
Table F.2, superheated vap. @ 700 kPa, 200 degC:
H3
2844.2
kJ
kg
S3
6.8859
kJ
kg K
Table F.2, superheated vap. @ 700 kPa, 280 degC:
H1
3017.7
kJ
kg
S1
7.2250
kJ
kg K
mdot1
50
kg
sec
Table F.1, sat. liq. @ 40 degC:
Hliq
167.5
kJ
kg
Sliq
0.5721
kJ
kg K
By Eq. (2.30), neglecting kinetic and potential energies and setting the heat
and work terms equal to zero:
H2
Also
mdot2
H3 mdot3
Hliq
mdot3 = mdot2
mdot1 H1
H3
H1 mdot1
H2 mdot2 = 0
mdot1
H3
mdot2
H2
For adiabatic conditions, Eq. (5.22) becomes
205
3.241
kg
sec
Ans.
S2
mdot3
Sliq
SdotG
S3 mdot3
SdotG
3.508
mdot2
S1 mdot1
mdot1
S2 mdot2
kJ
sec K
Ans.
The mixing of two streams at different temperatures is irreversible.
6.80
Basis: 1 mol air at 12 bar and 900 K
(1)
+ 2.5 mol air at 2 bar and 400 K (2)
= 3.5 mol air at T and P.
T1
900 K
T2
400 K
P1
n1
1 mol
n2
2.5 mol
CP
600 K
2 bar
CP
29.099
(guess)
1st law:
T
Given
n1 CP T
T
T1
n2 CP T
P
Given
T
5 bar
T
P
R ln
T1
P1
T
P
n2 CP ln
R ln
T2
P2
P
Find ( )
P
molwt
28.014
lb
lbmol
CP
Ms
542.857 K
4.319 bar
R
7
2 molwt
= nitrogen rate in lbm/sec
=0
Ans.
J
K
Ans.
CP
0.248
BTU
lbm rankine
= steam rate in lbm/sec
Mn
J
mol K
(guess)
n1 CP ln
P
7
R
2
T2 = 0 J
Find ( )
T
2nd law:
6.81
P2
12 bar
Mn
206
40
lbm
sec
(1) = sat. liq. water @ 212 degF entering
(2) = exit steam at 1 atm and 300 degF
(3) = nitrogen in at 750 degF
T3
1209.67 rankine
(4) = nitrogen out at 325 degF
T4
784.67 rankine
H1
180.17
H2
1192.6
BTU
lbm
S1
lbm
BTU
lbm rankine
(Table F.3)
S2
BTU
0.3121
1.8158
BTU
lbm rankine
(Table F.4)
Eq. (2.30) applies with negligible kinetic and potential energies and with the
work term equal to zero and with the heat transfer rate given by
Ms
3
Given
Ms
lbm
(guess)
sec
Ms H2
H1
Q = 60
Mn CP T4
Find Ms
Ms
T 3 = 60
3.933
lbm
sec
BTU
Ms
lbm
BTU
lbm
Ms
Ans.
Eq. (5.22) here becomes
SdotG = Ms S2
S4
S3 = CP ln
T
S1
Mn S4
T4
Q
T3
Q
T
S3
60
BTU
Ms
lbm
529.67 rankine
SdotG
Ms S2
SdotG
2.064
S1
Mn CP ln
BTU
sec rankine
T4
T3
Ans.
207
Q
T
Q
235.967
BTU
sec
6.82 molwt
28.014
gm
CP
mol
R
2 molwt
7
CP
1.039
J
gm K
Ms = steam rate in kg/sec
Mn= nitrogen rate in kg/sec
kg
sec
Mn
20
(3) = nitrogen in @ 400 degC
T3
673.15 K
(4) = nitrogen out at 170 degC
T4
443.15 K
(1) = sat. liq. water @ 101.33 kPa entering
(2) = exit steam at 101.33 kPa and 150 degC
H1
419.064
H2
2776.2
kJ
kg
S1
S2
kJ
kg
1.3069
7.6075
kJ
kg K
(Table F.2)
kJ
(Table F.2)
kg K
By Eq. (2.30), neglecting kinetic and potential energies and setting
the work term to zero and with the heat transfer rate given by
Ms
1
Given
Ms
kg
sec
(guess)
Ms H2
H1
Ms
Find Ms
Q = 80
Mn CP T4
1.961
T 3 = 80
kg
kJ
Ms
kg
kJ
Ms
kg
Ans.
sec
Eq. (5.22) here becomes
SdotG = Ms S2
S4
S3 = CP ln
SdotG
Ms S2
SdotG
4.194
S1
Mn S4
T4
T
T3
S1
kJ
sec K
Q
T
S3
Mn CP ln
Ans.
208
T4
T3
298.15 K
Q
T
Q
80
kJ
kg
Ms
6.86
Methane = 1; propane = 2
T
363.15 K
1
Tc1
0.012
2
190.6 K
y2
Tc2
0.5
1
y1
5500 kPa
P
y1
0.152
Zc1
0.286
Zc2
0.276
369.8 K
Pc1
45.99 bar
Pc2
42.48 bar
The elevated pressure here requires use of either an equation of state or
the Lee/Kesler correlation with pseudocritical parameters. We choose the
latter.
Tpc
y1 Tc1
Tpc
Ppc
Tpr
T
Tpr
Tpc
1.296
44.235 bar
Ppr
280.2 K
y1 Pc1
Ppc
y2 Tc2
y2 Pc2
P
Ppc
Ppr
1.243
By interpolation in Tables E.3 and E.4:
Z0
Z1
0.8010
y1
1
y2
2
0.1100
Z
0.082
Z0
For the molar mass of the mixture, we have:
gm
molwt
molwt
y1 16.043 y2 44.097
mol
V
Vdot
D
ZRT
P molwt
V mdot
4A
30.07
3
V
Vdot
D
cm
14.788
gm
mdot
3
4 cm
2.07
10
2.964 cm
209
sec
Ans.
Z
Z1
1.4
A
Vdot
u
0.81
gm
mol
kg
sec
u
A
30
m
sec
2
6.901 cm
6.87
Vectors containing T, P, Tc, and Pc for the calculation of Tr and Pr:
500
425.2
20
42.77
400
304.2
200
73.83
450
552.0
60
79.00
600
617.7
20
21.10
620
T
617.2
Tc
20
P
36.06
Pc
250
190.6
90
45.99
150
154.6
20
469.7
10
33.70
450
430.8
35
78.84
400
374.2
15
Pr
P
Pc
40.60
1.176
0.468
1.315
2.709
0.815
0.759
0.971
Tr
T
Tc
50.43
500
Tr
0.948
1.005
Pr
0.555
1.312
1.957
0.97
0.397
1.065
0.297
1.045
0.444
1.069
0.369
Parts (a), (g), (h), (i), and (j) --- By virial equation:
500
425.2
42.77
.190
150
T
20
20
154.6
50.43
.022
500 K P
10 bar Tc
469.7 K Pc
33.70 bar
.252
450
35
430.8
78.84
.245
400
15
374.2
40.6
.327
Tr
T
Tc
Pr
P
Pc
210
1.176
0.97
Tr
0.468
0.397
1.065
Pr
0.297
1.045
0.444
1.069
0.369
B0
0.073
0.422
Eq. (3.65) B1
1.6
0.172
0.139
4.2
Tr
DB0
0.675
2.6
Tr
Eq. (6.89)
DB1
Tr
0.052
0.37
B1
0.321
6.718
4.217
0.306
5.2
Eq. (6.90)
0.443
9.009
10
10
10
3
DB0
3
0.311
0.73
0.056
0.309
0.722
Tr
0.253
B0
Eq. (3.66)
0.845
0.574
DB1
0.522
0.603
0.568
3
0.576
0.51
Combine Eqs. (3.61) + (3.62), (3.63), and (6.40) and the definitions of Tr and
Pr to get:
Tc
B0
Pc
B1
VR
R
HR
R Tc Pr B0
Tr DB0
SR
R Pr DB0
DB1
(1
B
Tr DB1)
Eq. (6.88)
211
Eq. (6.87)
3
1.377 10
200.647
94.593
VR
355.907
146.1
1.952
559.501
3
cm
2.469
3
HR
1.226 10
mol
3
J
SR
mol
2.745
1.746 10
2.256
3
232.454
J
mol K
1.74
1.251 10
Parts (b), (c), (d), (e), and (f) --- By Lee/Kesler correlation:
By linear interpolation in Tables E.1--E.12:
0
DEFINE: h0 equals
1
()
HR
()
HR
h1 equals
RTc
RTc
0
s0 equals
h equals
HR
RTc
s equals
SR
R
1
()
SR
R
s1 equals
()
SR
R
.663
2.008
0.233
.124
Z0
0.208
.050
4.445
5.121
.278
Z1
.088
h0
3.049
h1
2.970
.783
.036
0.671
0.596
.707
0.138
1.486
0.169
1.137
4.381
s0
0.405
5.274
2.675
s1
2.910
0.473
0.557
0.824
0.289
400
304.2
.224
450
T
200
60
552.0
.111
617.7 K
.492
600 K
P
20
Tc
bar
620
20
617.2
.303
250
90
190.6
.012
212
Z
s
Z1 Eq. (3.57)
Z0
s0
HR
s1
h0
( h Tc R)
SR
Eq. (6.85)
( s R)
3
5.21
10
2.301
0.118
0.235
h1
(6.86)
0.71
Z
h
10.207
4
10
4
HR
2.316 10
0.772
4.37
41.291
J
mol
SR
J
mol K
5.336
3
10
0.709
34.143
6.88
3
2.358 10
48.289
VR
T
(Z
R
P
549.691
1)
VR
3
3
1.909 10
cm
mol
And.
587.396
67.284
The Lee/Kesler tables indicate that the state in Part (c) is liquid.
6.88 Vectors containing T, P, Tc1, Tc2, Pc1, Pc2, 1, and 2 for Parts (a) through (h)
650
562.2
553.6
300
100
304.2
132.9
600
T
60
100
304.2
568.7
350
400
K
P
75
150
bar
Tc1
305.3
373.5
K
Tc2
282.3
190.6
200
75
190.6
126.2
450
80
190.6
469.7
250
100
126.2
154.6
213
K
48.98
.210
.210
73.83
34.99
.224
.048
73.83
Pc1
40.73
24.90
.224
.400
48.72
bar
89.63
50.40
Pc2
45.99
1
bar
.100
.087
2
.094
.012
45.99
34.00
.012
.038
45.99
33.70
.012
.252
34.00
50.43
.038
.022
Tpc
( Tc1
.5
Tpr
T
Tpc
.5 Tc2) Ppc
( Pc1
.5
.5 Pc2)
.5
P
Ppc
Ppr
557.9
0.21
218.55
54.41
0.136
436.45
Tpc
44.855
49.365
0.312
293.8
282.05
K
Ppc
49.56
67.81
bar
0.094
0.053
158.4
39.995
0.025
330.15
39.845
0.132
140.4
42.215
0.03
1.165
1.373
1.838
1.375
Tpr
1.338
2.026
1.191
1.418
Ppr
1.513
2.212
1.263
1.875
1.363
2.008
1.781
2.369
214
1
.5
2
Lee/Kesler Correlation --- By linear interpolation in Tables E.1--E.12:
.6543
1.395
.461
.7706
.1749
1.217
.116
.7527
Z0
.1219
.1929
1.346
.097
.6434
Z1
.7744
.1501
1.510
h0
.1990
1.340
h1
.400
.049
.6631
.1853
1.623
.254
.7436
.1933
1.372
.110
.9168
.1839
0.820
0.172
.890
.466
.658
.235
.729
.242
.944
s0
.430
s1
.704
.224
.965
.348
.750
.250
.361
.095
0
h0 equals
s0 equals
Z
Z0
s
s0
( HR)
RTpc
( SR)
R
1
h1 equals
0
s1 equals
Z1 Eq. (3.57)
s1
h
Eq. (6.86)
215
( HR)
RTpc
( SR)
h equals
s equals
SR
R
1
R
h0
HR
RTpc
h1
Eq. (6.85)
HR
(
hTpc R)
SR
( R)
s
0.68
8.213
0.794
2239.984
5.736
0.813
Z
6919.583
4993.974
6.689
0.657
3779.762
HR
0.785
J
SR
3148.341 mol
8.183
5.952 mol K Ans.
0.668
2145.752
8.095
0.769
3805.813
6.51
0.922
6.95 Tc
J
951.151
3.025
Pc
647.1K
At Tr = 0.7:
T
220.55bar
T
0.7 Tc
452.97 K
Find Psat in the Saturated Steam Tables at T = 452.97 K
T1
P1
451.15K
P2
T2
Psat
P1
(
T
T1
Psat
Pc
Psatr
T2
957.36kPa
Psat
T1) P1
Psatr
453.15K
998.619 kPa
1
0.045
P2
Psat
1002.7kPa
9.986 bar
0.344
log Psatr
Ans.
This is very close to the value reported in Table B.1 ( = 0.345).
6.96 Tc
At Tr = 0.7:
T
Pc
374.2K
T
T
T
471.492 rankine
T
0.7 Tc
459.67rankine
40.60bar
11.822 degF
Find Psat in Table 9.1 at T = 11.822 F
T1
10degF
P1
T2
26.617psi
216
15degF
P2
29.726psi
P2
T2
Psat
P1
(T
T1
Psat
Pc
Psatr
T1)
Psatr
P1
Psat
Psat
27.75 psi
1
0.047
1.913 bar
Ans.
0.327
log Psatr
This is exactly the same as the value reported in Table B.1.
6.101 For benzene
a)
Tc
Tn
Trn
Tc
562.2K
Pc
Trn
0.210
0.628
Psatrn
lnPr0 ( Tr)
5.92714
lnPr1 ( Tr)
15.2518
6.09648
Tr
15.6875
Tr
lnPr0 ( Tr)
atm
Pc
Psatrn
353.2K
0.021
6
0.169347 Tr
Eqn. (6.79)
13.4721 ln ( Tr)
0.43577 Tr
6
Eqn. (6.80)
Eqn. (6.81).
lnPr1 Trn
lnPsatr ( Tr)
1
Tn
0.271
1.28862 ln ( Tr)
lnPr0 Trn
ln Psatrn
Zc
48.98bar
lnPr1 ( Tr)
0.207
Eqn. (6.78)
2
Zsatliq
B0
Psatrn
Trn
0.083
Zc
0.422
B1
Eqn. (3.73)
Eqn. (3.65)
1.6
0.805
0.139
0.172
Trn
B1
7
Zsatliq
4.2
Eqn. (3.66)
Z0
1
Z0
Trn
B0
1 1 T rn
Z1
B1
217
Psatrn
Eqn. (3.64)
0.974
Z1
1.073
B0
0.00334
Trn
Psatrn
Trn
0.035
Equation
following
Eqn. (3.64)
Zsatvap
Z0
Zlv
Zsatvap
d
Hhatlv
Hlv
Eqn. (3.57)
Z1
dTrn
Zsatvap
0.966
Zlv
Zsatliq
lnPsatr Trn Trn
2
0.963
Hhatlv
Zlv
Hlv
R Tc Hhatlv
6.59
30.802
kJ
mol
Ans.
This compares well with the value in Table B.2 of 30.19 kJ/mol
The results for the other species are given in the table below.
EstimatedValue (kJ/mol) Table B.2 (kJ/mol)
30.80
30.72
Benz
ene
21.39
21.30
isoButane
29.81
29.82
Carbon tetrachlorid
e
30.03
29.97
Cy
clohex
ane
39.97
38.75
nDecane
29.27
28.85
n- ane
Hex
34.70
34.41
nOctane
33.72
33.18
Toluene
37.23
36.24
o- lene
Xy
6.103 For CO2:
Tc
a) At Tr = 0.7
Ttr
Tt
Ttr
Tc
Pc
73.83bar
216.55K
Pt
5.170bar
T
At the triple point:
304.2K
Tt
0.224
0.7Tc
T
Ptr
0.712
212.94 K
Pt
Ptr
Pc
0.07
lnPr0 ( )
Tr
5.92714
6.09648
Tr
1.28862 ln ( ) 0.169347 Tr
Tr
Eqn. (6.79)
lnPr1 ( )
Tr
15.2518
15.6875
Tr
13.4721 ln ( ) 0.43577 Tr
Tr
6
Eqn. (6.80)
ln Ptr
lnPr0 Ttr
lnPr1 Ttr
6
Eqn. (6.81).
218
0.224
Ans.
This is exactly the same value as given in Table B.1
b) Psatr
1atm
Pc
0.014
ln Psatr = lnPr0 Trn
Given
Trn
Psatr
0.609
Tn
Trn Tc
Guess: Trn
lnPr1 Trn
0.7
Trn
Tn
Find Trn
185.3 K
Ans.
This seems reasonable; a Trn of about 0.6 is common for triatomic species.
219
Chapter 7 - Section A - Mathcad Solutions
7.1
u2
325
m
R
sec
8.314
J
mol K
molwt
28.9
7
gm
CP
mol
R
2 molwt
With the heat, work, and potential-energy terms set equal to zero and
with the initial velocity equal to zero, Eq. (2.32a) reduces to
H
2
u2
Whence
7.4
But
=0
2
T
u2
H = CP T
2
T
2 CP
52.45 K
Ans.
From Table F.2 at 800 kPa and 280 degC:
H1
kJ
kg
3014.9
S1
7.1595
kJ
kg K
Interpolation in Table F.2 at P = 525 kPa and S = 7.1595 kJ/(kg*K) yields:
H2
kJ
2855.2
kg
3
V2
cm
531.21
gm
mdot
0.75
kg
sec
With the heat, work, and potential-energy terms set equal to zero and
with the initial velocity equal to zero, Eq. (2.32a) reduces to:
H
2
u2
2
Whence
=0
u2
2 H2
H1
m
sec
u2
By Eq. (2.27),
7.5
A2
mdot V2
u2
565.2
A2
7.05 cm
2
Ans.
Ans.
The calculations of the preceding problem may be carried out for a
series of exit pressures until a minimum cross-sectional area is found.
The corresponding pressure is the minimum obtainable in the converging
nozzle. Initial property values are as in the preceding problem.
220
H1
3014.9
kJ
S1
kg
7.1595
kJ
S2 = S 1
kg K
Interpolations in Table F.2 at several pressures and at the given
entropy yield the following values:
400
425
P
531.21
2855.2
2868.2
450 kPa
507.12
kJ
kg
2880.7
H2
485.45
V2
475
465.69
500
mdot
2892.5
2903.9
447.72
0.75
kg
sec
u2
2 H2
565.2
7.022
m
518.1
sec
494.8
u2
u2
7.05
541.7
mdot V2
A2
H1
A2
2
7.028 cm
7.059
7.127
471.2
Fit the P vs. A2 data with cubic spline and find
the minimum P at the point where the first
derivative of the spline is zero.
i
15
s
cspline P A2
pmin
pi
400 kPa
Pi
a2
A( )
P
i
A2
i
interp s p a2 P
(guess)
2
Given
pmin
cm
d
A pmin = 0
kPa
dpmin
431.78 kPa
Ans.
pmin
A pmin
221
Find pmin
2
7.021 cm
Ans.
3
cm
gm
Show spline fit graphically:
p
400 kPa 401 kPa 500 kPa
7.13
7.11
A2
7.09
i
2
cm
7.07
A( )
p
2
cm
7.05
7.03
7.01
400
420
440
460
Pi
480
500
p
kPa kPa
7.9
From Table F.2 at 1400 kPa and 325 degC:
H1
3096.5
kJ
kg
S1
7.0499
kJ
kg K
S2
S1
Interpolate in Table F.2 at a series of downstream pressures and at S =
7.0499 kJ/(kg*K) to find the minimum cross-sectional area.
800
294.81
775
P
2956.0
2948.5
302.12
750 kPa
H2
2940.8
kJ
kg
V2
309.82
725
317.97
700
u2
2932.8
2924.9
326.69
2 H2
H1
A2 =
222
V2
u2
mdot
3
cm
gm
Since mdot is constant,
the quotient V2/u2 is a
measure of the area. Its
minimum value occurs very
close to the value at
vector index i = 3.
5.561
5.553
V2
5.552
u2
2
cm sec
kg
5.557
5.577
At the throat,
A2
A2 u2
3
V2
mdot
2
6 cm
mdot
1.081
3
kg
Ans.
sec
At the nozzle exit, P = 140 kPa and S = S1, the initial value. From
Table F.2 we see that steam at these conditions is wet. By
interpolation,
Sliq
x
7.10
u1
kJ
kg K
1.4098
S1
Svap
230
Sliq
Svap
x
Sliq
ft
sec
7.2479
kJ
kg K
0.966
u2
2000
ft
sec
From Table F.4 at 130(psi) and 420 degF:
H1
Btu
lbm
1233.6
S1
2
By Eq. (2.32a),
H2
H1
1.6310
H
H
H2
Btu
lbm rankine
u2
u1
2
H
2
1154.8
78.8
Btu
lbm
From Table F.4 at 35(psi), we see that the final state is wet steam:
Hliq
228.03
Sliq
0.3809
Btu
lbm
Btu
lbm rankine
Hvap
1167.1
Svap
1.6872
223
Btu
lbm
Btu
lbm rankine
Btu
lbm
H2
x
Hvap
S2
u2
x Svap
S2
580
x
Sliq
(
273.15
By Eq. (2.32a),
But
T
1.67
BTU
lbm rankine
0.039
15)K molwt
H=
u1
28.9
u2
2
gm
CP
mol
Ans.
R
2 molwt
7
2
2
2
Btu
lbm rankine
=
u2
2
Whence
H = CP T
u2
(quality)
SdotG
S1
m
T2
sec
0.987
S2
Hliq
Sliq
SdotG
7.11
Hliq
2
T
2 CP
Ans.
167.05 K
Initial t = 15 + 167.05 = 182.05 degC Ans.
7.12
Values from the steam tables for saturated-liquid water:
3
At 15 degC: V
cm
1.001
gm
T
288.15 K
Enthalpy difference for saturated liquid for a temperature change from
14 to 15 degC:
H
(
67.13
1.5 10
K
58.75)
J
gm
t
2K
4
P
4 atm
Cp
Cp
H
t
4.19
J
gm K
Apply Eq. (7.25) to the constant-enthalpy throttling process. Assumes
very small temperature change and property values independent of P.
224
T
V1
joule
1
P
T
T
9.86923 cm3 atm
Cp
0.093 K
The entropy change for this process is given by Eq. (7.26):
S
T
Cp ln
T
T
Apply Eq. (5.36) with Q=0:
Wlost
T
S
Wlost
7.13--7.15
P2
350
T1
350
250
T
J
gm
Wlost
80
P1
K
60
60
bar
20
73.83
282.3
126.2
Pc
K
369.8
.224
50.40
.087
bar
34.00
.038
42.48
.152
5.457
1.045
1.424
14.394
B
3.280
1.213
10
4.392
0.0
3
K
.593
28.785
0.0
C
or
1.2bar
304.2
A
1.408 10
1.157
6
10
K
2
3
J
gm K
293.15 K
0.413
400
Tc
S
VP
D
8.824
0.0
0.040
0.0
225
5
10 K
2
0.413
kJ
kg
Ans.
As in Example 7.4, Eq. (6.93) is applied to this constant-enthalpy
process. If the final state at 1.2 bar is assumed an ideal gas, then Eq.
(A) of Example 7.4 (pg. 265) applies. Its use requires expressions for HR
and Cp at the initial conditions.
1.151
T1
Tr
1.24
Tr
Tc
1.084
P1
Pc
Pr
1.981
Pr
1.082
7.13
0.42748
0.08664
Eq. (3.53)
Tr
1.765
0.471
Redlich/Kwong equation:
Pr
1.19
q
Eq. (3.54)
1.5
Tr
Guess:
z
Given
z= 1
Z
q
Find z
()
i
14
1
q
Ii
HRi
R T1i
Z
SRi
R ln Z
i qi
i qi
z
Eq. (3.52)
zz
ln
1
i
Z
i qi
Z
i
Eq. (6.65b)
i qi
1.5 qi Ii Eq. (6.67) The derivative in these
0.5 qi Ii
Eq. (6.68) equations equals -0.5
The simplest procedure here is to iterate by guessing T2, and then
calculating it.
280
Guesses
T2
302
232
K
385
226
Z
i qi
2.681
5.177
0.721
0.773
2.253
kJ
0.521 mol
HR
0.956
0.862
Cp
HR
Cp
2.33
B
T1
2
RA
C
2
T1
3
1
T1
S
Cp ln
302.026
K
232.062
0.480
Guess:
P2
P1
SR
Ans.
0.42748
1
c1
0.5
2
Tr
Eq. (3.54)
q
Tr
1
Given
z= 1
i
Ii
HRi
R ln
0.08664
2
0.176
Eq. (3.53)
z
14
T2
T1
2
22.163
1.574
Pr
Tr
T1
J
31.953 mol K
Soave/Redlich/Kwong equation:
c
D
1
29.947
S
Ans.
384.941
7.14
2
31.545
279.971
T2
J
mol K
1.59
1.396
T2
T1
T2
4.346
SR
R T1i Z
ln
z
q
Z
i qi
Eq. (3.52) Z
zz
i qi
Z
i
q
Eq. (6.65b)
i qi
1
ci
Tri
i
227
0.5
1 qi Ii
Eq. (6.67)
Find ( z)
SRi
R ln Z
i qi
Tri
ci
i
0.5
Eq. (6.68)
qi Ii
i
The derivative in these equations equals:
Tri
ci
0.5
i
Now iterate for T2:
273
Guesses
300
T2
232
K
384
Z
i qi
2.936
0.75
0.79
HR
0.975
0.866
6.126
2.356
kJ
0.526 mol
4.769
J
1.789 mol K
SR
1.523
T2
T1
Cp
RA
2.679
B
T1
2
1
C
2
T1
3
2
1
D
T1
2
272.757
T2
HR
Cp
T1
299.741
T2
231.873
K
Ans.
383.554
31.565
S
Cp ln
T2
T1
R ln
P2
P1
SR
S
30.028
32.128 mol K
22.18
228
J
Ans.
7.15
Peng/Robinson equation:
1
c
2
1
0.37464
1.54226
Pr
Tr
z
q
i
14
HRi
2
1
c1
Tr
z
q
Eq. (3.52)
z
z
Find z
()
1
Ii
R T1i Z
2
ln
Z
i qi
i
Z
i qi
i
2
i qi
1
ci
Tri
Eq. (6.65b)
0.5
1 qi Ii
R ln Z
i qi
i
ci
Tri
Eq. (6.67)
0.5
Eq. (6.68)
qi Ii
i
The derivative in these equations equals:
ci
Tri
i
Now iterate for T2:
270
Guesses
T2
297
229
K
383
229
0.5
0.5
Tr
Eq. (3.54)
q
i
SRi
0.45724
1
z= 1
Z
0.07779
0.26992
Eq. (3.53)
Guess:
Given
2
2
Z
i qi
3.041
6.152
0.722
0.76
HR
0.95
0.85
2.459
0.6
kJ
mol
4.784
J
1.847 mol K
SR
1.581
T2
T1
Cp
RA
2.689
B
T1
2
C
2
T1
3
1
2
D
1
T1
2
269.735
HR
Cp
T2
T1
297.366
T2
Ans.
K
229.32
382.911
31.2
S
T2
T1
Cp ln
R ln
P2
P1
SR
29.694
J
31.865 mol K
S
Ans.
22.04
7.18
Wdot
H1
3500 kW
3462.9
kJ
kg
Data from Table F.2:
2609.9
H2
kJ
kg
S1
4.103
kg
sec
Ans.
7.3439
kJ
kg K
S2
S1
By Eq. (7.13),
Wdot
H2 H1
mdot
mdot
For isentropic expansion, exhaust is wet steam:
Sliq
x
0.8321
S2
Svap
kJ
kg K
Sliq
Sliq
Svap
x
230
7.9094
0.92
kJ
kg K
(quality)
Hliq
H'2
251.453
Hliq
x Hvap
H2
Hvap
Hliq
H'2
H1
2609.9
2.421
kJ
kg
3 kJ
10
kg
H1
H'2
7.19
kJ
kg
0.819
Ans.
The following vectors contain values for Parts (a) through (g). For intake
conditions:
3274.3
kJ
kg
6.5597
3509.8
kJ
kg
6.8143
kJ
3634.5
kg
H1
kJ
3161.2
kg
1389.6
Btu
lbm
Btu
lbm
kg K
kJ
kg K
kJ
6.9813
kg K
6.4536
S1
kJ
2801.4
kg
1444.7
kJ
kJ
kg K
kJ
6.4941
kg K
1.6000
1.5677
231
Btu
lbm rankine
Btu
lbm rankine
0.80
0.77
0.82
0.75
0.75
0.80
0.75
For discharge conditions:
0.9441
0.8321
0.6493
Sliq
1.0912
1.5301
kJ
7.7695
kg K
kJ
kg K
7.9094
kJ
8.1511
kg K
kJ
kg K
kJ
kg K
7.1268
Btu
0.1750
1.9200
lbm rankine
Btu
lbm rankine
0.2200
7.5947
Svap
1.8625
kJ
kg K
kJ
kg K
kJ
kg K
kJ
kg K
S' 2 = S 1
kJ
kg K
Btu
lbm rankine
Btu
lbm rankine
289.302
2625.4
kJ
kg
80
kg
sec
251.453
kJ
kg
2609.9
kJ
kg
90
kg
sec
191.832
Hliq
kJ
kg
kJ
kg
2584.8
kJ
kg
70
kg
sec
340.564
kJ
kg
2646.0
kJ
kg
65
kg
sec
504.701
kJ
kg
2706.3
kJ
kg
50
kg
sec
Btu
lbm
1116.1
Btu
lbm
150
Btu
lbm
1127.3
Btu
lbm
100
94.03
120.99
Hvap
232
mdot
lbm
sec
lbm
sec
x'2
S1
Svap
H
x2
Sliq
H'2
H2
Hvap
H2
H2
H2
H2
H2
H2
H2
1
H'2
Sliq
H1
Hliq
Hliq
Hliq
H2
H1
S2
Sliq
Wdot
x2 Svap
Sliq
3
S2
kJ
2467.8
kg
2471.4
2
7.6873
7.7842
S2
4
2543.4
kJ
kg K
7.1022
3
S2
Ans.
6.7127
5
S2
1031.9 Btu
1057.4 lbm
6
S2
7
1.7762
Btu
1.7484 lbm rankine
68030
91230
87653
117544
81672
Wdot
H mdot
7.1808
S2
5
7
H
1
2423.9
2535.9
6
Hliq
S2
2
4
x'2 Hvap
109523
44836 kW
Wdot
60126
12900
17299
65333
87613
35048
46999
233
hp
Ans.
7.20
T
P0
423.15 K
P
8.5 bar
1 bar
For isentropic expansion,
S
J
0
mol K
For the heat capacity of nitrogen:
A
B
3.280
3
0.593 10
K
5
D
0.040 10 K
2
For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15)
with C = 0. Substitute:
(guess)
0.5
Given
S = R A ln
B
1
D
T
1
T
2
2
T
T0
Find
ln
P
P0
T0
Ans.
762.42 K
Thus the initial temperature is 489.27 degC
7.21
T1
CP
32
P1
1223.15 K
J
mol K
P2
10 bar
1.5 bar
0.77
Eqs. (7.18) and (7.19) derived for isentropic compression apply equally well
for isentropic expansion. They combine to give:
W's
C P T1
Ws
W's
P2
R
CP
W's
1
P1
H
Ws
234
15231
J
mol
Ws
11728
J
mol
Ans.
Eq. (7.21) also applies to expansion:
T2
7.22
H
T1
T2
CP
Isobutane:
Tc
408.1 K
Pc
T0
P0
5000 kPa
P
523.15 K
S
0
J
mol K
A
0.181
36.48 bar
500 kPa
For the heat capacity of isobutane:
37.853 10
K
B
1.677
T0
Tr0
Ans.
856.64 K
Tr0
Tc
3
C
K
Pr0
1.282
11.945 10
Pr
P0
2
Pr0
Pc
P
Pr
Pc
6
1.3706
0.137
The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0:
0.5
(guess)
Given
S = R A ln
SRB
B T0
T0
Tc
Find
T
Tr
Pr
1
2
C T0
2
SRB Tr0 Pr0
T0
T
Tc
T
Tr
235
445.71 K
1.092
1
ln
P
P0
The enthalpy change is given by Eq. (6.91):
Hig
R ICPH T0 T 1.677 37.853 10
Hig
11.078
H'
Hig
H'
8331.4
3
6
11.945 10
0.0
kJ
mol
R Tc HRB Tr Pr
HRB Tr0 Pr0
J
mol
The actual enthalpy change from Eq. (7.16):
0.8
Wdot
ndot
700
mol
sec
ndot H
H
Wdot
H'
H
6665.1
J
mol
Ans.
4665.6 kW
The actual final temperature is now found from Eq. (6.91) combined with Eq
(4.7), written:
0.7
(guess)
Given
H = R A T0
1
Tc HRB
Find
7.23
B
2
T0
2
T0
Pr
Tc
0.875
2
C
1
3
3
3
T0
1
HRB Tr0 Pr0
T
T0
T
457.8 K
Ans.
From Table F.2 @ 1700 kPa & 225 degC:
H1
2851.0
At 10 kPa:
kJ
kg
S1
6.5138
x2
0.95
236
kJ
kg K
Sliq
0.6493
kJ
kg K
Hliq
kJ
191.832
mdot
0.5
Hvap
kg
kg
2584.8
Wdot
sec
Svap
8.1511
H
H1
2.465
10
(a)
Qdot
mdot H
(b)
For isentropic expansion to 10 kPa, producing wet steam:
S1
Svap
x'2
Hliq
H2
Hliq
H2
3 kJ
H
Wdot
Sliq
H'2
T0
mdot H'2
Hliq
H'2
Sliq
2.063
385.848
Qdot
kg
0.782
Wdot'
7.24
kJ
kg K
180 kW
H2
x'2
x2 Hvap
kJ
kg
12.92
x'2 Hvap
P0
673.15 K
8 bar
P
For isentropic expansion,
For the heat capacity of carbon dioxide:
A
5.457
B
Ans.
Hliq
kg
Ans.
394.2 kW
1 bar
S
0
J
mol K
3
1.045 10
K
kJ
sec
3 kJ
10
Wdot'
H1
kJ
kg
5
D
1.157 10 K
2
For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with C = 0:
(guess)
0.5
Given
S = R A ln
B T0
T0
Find
1
D
2
2
T'
0.693
237
1
T0
T'
ln
P
P0
466.46 K
H'
H'
9.768
0.0
1.157 10
5
kJ
mol
0.75
H
3
R ICPH T0 T' 5.457 1.045 10
Work
Work
H'
H
7.326
Work
7.326
kJ
mol
Ans.
kJ
mol
For the enthalpy change of an ideal gas, combine Eqs. (4.2) and (4.7)
with C = 0:
Given
H = R A T0
B
2
T0
2
1
Find
2
0.772
1
T
D
T0
T0
1
T
Ans.
519.9 K
Thus the final temperature is 246.75 degC
7.25
Vectors containing data for Parts (a) through (e):
500
371
1.2
3.5
450
T1
6
5
376
2.0
4.0
525
P1
10
T2
458
P2
3.0
Cp
5.5 R
475
HS
372
1.5
4.5
550
H
7
4
403
1.2
2.5
[(
Cp T2
Cp T1
Ideal gases with constant heat capacities
T1)
P2
P1
R
Cp
Eq. (7.22) Applies to expanders as
well as to compressors
1
238
0.7
0.803
H
0.649
HS
0.748
0.699
7.26
Cp
7
2
R
ndot
Guesses:
mol
175
T1
sec
0.75
Wdot
550K
P1
6bar
P2
1.2bar
600kW
Given
Wdot =
0.065
Wdot
.08 ln
Wdot
kW
Find ( Wdot)
0.065
ndot Cp T1
Wdot
P2
P1
1
Ans.
594.716 kW
Wdot
kW
0.08 ln
R
Cp
0.576
Ans.
For an expander operating with an ideal gas with constant Cp, one can
show that:
T2
P2
P1
T1 1
R
Cp
1
T2
433.213 K
By Eq. (5.14):
S
R
T2
Cp
ln
T1
R
ln
P2
P1
S
6.435
J
mol K
By Eq. (5.37), for adiabatic operation :
SdotG
ndot S
SdotG
1.126
239
3
10
J
K sec
Ans.
7.27
Properties of superheated steam at 4500 kPa and 400 C from Table F.2,
p. 742.
H1
3207.1
S1
6.7093
If the exhaust steam (Point 2, Fig. 7.4) is "dry," i.e., saturated vapor, then
isentropicexpansion to the same pressure (Point 2', Fig. 7.4) must produce
"wet" steam, withentropy:
S2 = S1 = 6.7093 = (x)(Svap) + (1-x)(Sliq)
[x is quality]
A second relation follows from Eq. (7.16), written:
HS) = (0.75)[ (x)(Hvap) + (1-x)(Hliq) - 3207.1]
H = Hvap - 3207.1 = (
Each of these equations may be solved for x. Given a final temperature
and the corresponding vapor pressure, values for Svap, Sliq, Hvap, and
Hliq are found from the table for saturated steam, and substitution into the
equations for x produces two values. The required pressure is the one for
which the two values of x agree. This is clearly a trial process. For a final
trial temperature of 120 degC, the following values of H and S for
saturated liquid and saturated vapor are found in the steam table:
Hl
503.7
Hv
2706.0
Sl
1.5276
Sv
7.1293
The two equations for x are:
xH
Hv 801.7 .75 Hl
.75 (
Hv Hl)
xS
The trial values given produce: xH
6.7093 Sl
Sv Sl
0.924
xS
0.925
These are sufficiently close, and we conclude that:
t=120 degC;
P=198.54 kPa
If were 0.8, the pressure would be higher, because a smaller pressure
drop would be required to produce the same work and H.
240
7.29
P1
5 atm
P2
1 atm
T1
15 degC
0.55
Data in Table F.1 for saturated liquid water at 15 degC give:
3
V
1001
cm
kg
Cp
4.190
kJ
kg degC
Eqs. (7.16) and (7.24) combine to give:
Ws
H
(7.14)
Ws
0.223
Eq. (7.25) with =0 is solved for T:
H
P1)
kJ
kg
T
T
7.30
V ( P2
H
V ( P2
P1)
Cp
Ans.
0.044 degC
Assume nitrogen an ideal gas. First find the temperature after isentropic
expansion from a combination of Eqs. (5.14) & (5.15) with C = 0. Then
find the work (enthalpy change) of isentropic expansion by a combination
of Eqs. (4.2) and (4.7) with C = 0. The actual work (enthalpy change) is
found from Eq. (7.20). From this value, the actual temperature is found by
a second application of the preceding equation, this time solving it for the
temperature. The following vectors contain values for Parts (a) through
(e):
753.15
1 bar
673.15
T0
6 bar
5 bar
1 bar
P0
773.15 K
7 bar
P
1 bar
723.15
8 bar
2 bar
755.37
95 psi
15 psi
200
150
ndot
0.80
0.75
175
mol
sec
0.78
100
0.85
0.5 453.59
0.80
241
S
i
0
15
J
mol K
For the heat capacity of nitrogen:
A
3.280
3
0.593 10
B
5
D
0.040 10 K
2
K
(guess)
0.5
Given
S = R A ln
B T0
1
D
22
2
T0
Tau T0 P0 P
Find
ln
1
P
P0
Tau T0 P0 Pi
i
i
i
460.67
431.36
Ti
T0
i
T
i
453.48 K
494.54
455.14
H'i
3
R ICPH T0 Ti 3.280 0.593 10
5
0.0 0.040 10
i
8879.2
7279.8
H'
7103.4
5459.8
9714.4
J
mol
H
H
H'
7577.2
6941.7
0.5
5900.5
9112.1
J
mol
7289.7
(guess)
Given
H = R A T0
Tau T0
H
1
B
2
T0
2
2
Find
i
242
1
1
D
T0
Tau T0
i
Hi
Ti
T0
i
i
520.2
492.62
T
1421
819
525.14 K
Ans.
Wdot
ndot H
1326 kW Ans.
Wdot
529.34
516.28
7.31
590
1653
Property values and data from Example 7.6:
kg
H1
3391.6
kJ
kg
S1
6.6858
kJ
kg K
mdot
H2
2436.0
kJ
kg
S2
7.6846
kJ
kg K
Wdot
56400 kW
T
300 K
Wdotideal
74084 kW
Wdotideal
t
59.02
sec
By Eq. (5.26)
mdot H2
H1
Wdot
Wdotideal
T
t
S2
S1
Ans.
0.761
The process is adiabatic; Eq. (5.33) becomes:
SdotG
mdot S2
Wdotlost
7.32
SdotG
S1
Wdotlost
T SdotG
58.949
kW
K
17685 kW
Ans.
Ans.
For sat. vapor steam at 1200 kPa, Table F.2:
H2
2782.7
kJ
kg
S2
6.5194
kJ
kg K
The saturation temperature is 187.96 degC.
The exit temperature of the exhaust gas is therefore 197.96 degC, and
the temperature CHANGE of the exhaust gas is -202.04 K.
For the water at 20 degC from Table F.1,
H1
83.86
kJ
kg
S1
0.2963
243
kJ
kg K
The turbine exhaust will be wet vapor steam.
For sat. liquid and sat. vapor at the turbine exhaust pressure of 25 kPa, the
best property values are found from Table F.1 by interpolation between 64
and 65 degC:
kJ
kJ
Hliq
272.0
Hlv
2346.3
kg
kg
Sliq
0.8932
kJ
kg K
Slv
6.9391
kJ
kg K
0.72
For isentropic expansion of steam in the turbine:
S'3
S2
S'3
6.519
x'3
kJ
kg K
x' 3
S'3
Sliq
H'3
0.811
Hliq
H'3
Slv
2.174
H23
H'3
H2
H3
H2
H23
H23
437.996
kJ
kg
H3
2.345
10
S3
Sliq
x3 Slv
S3
7.023
kJ
kg K
For the exhaust gases:
ndot 125
mol
sec
T1
(
273.15
T2
(
273.15
T1
673.15 K
T2
471.11 K
H3
x3
x3
Hliq
Hlv
0.883
molwt
18
400)K
3 kJ
kg
197.96)K
gm
mol
Hgas
R MCPH T1 T2 3.34 1.12 10
Sgas
R MCPS T1 T2 3.34 1.12 10
244
3
3
0.0 0.0
T2
0.0 0.0 ln
T2
T1
T1
x'3 Hlv
3 kJ
10
kg
Hgas
6.687
3 kJ
10
Sgas
kmol
11.791
kJ
kmol K
Energy balance on boiler:
mdot
ndot Hgas
H2 H1
(a) Wdot
mdot
mdot H3
T
314.302 kW
S1
Wdot
Wdotideal
t
sec
293.15 K
ndot Hgas mdot H3 H1
T
ndot Sgas mdot S3
Wdotideal
kg
135.65 kW Ans.
Wdot
H2
(b) By Eq. (5.25):
Wdotideal
0.30971
t
Ans.
0.4316
(c) For both the boiler and the turbine, Eq. (5.33) applies with Q = 0.
For the boiler:
SdotG
ndot Sgas
Boiler:
mdot S2
SdotG
S1
0.4534
kW
K
For the turbine: SdotG
mdot S3
Turbine:
SdotG
0.156
(d) Wdotlost.boiler
0.4534
Wdotlost.turbine
Fractionboiler
kW
0.1560
K
T
kW
T
K
Wdotlost.boiler
Wdotideal
245
kW
K
Ans.
S2
Ans.
Wdotlost.boiler
132.914 kW
Wdotlost.turbine
45.731 kW
Fractionboiler
0.4229
Ans.
Wdotlost.turbine
Fractionturbine
Note that:
7.34
t
0.1455 Ans.
Fractionturbine
Wdotideal
Fractionboiler
Fractionturbine
1
From Table F.2 for sat. vap. at 125 kPa:
H1
kJ
kg
2685.2
S1
7.2847
kJ
kg K
For isentropic expansion, S'2 = S1 = 7.2847
kJ
kg K
Interpolation in Table F.2 at 700 kPa for the enthalpy of steam with this
entropy gives
H'2
3051.3
H2
H1
kJ
kg
H
0.78
H2
H
3154.6
H'2
kJ
kg
H1
H
469.359
kJ
kg
Ans.
Interpolation in Table F.2 at 700 kPa for the entropy of steam with this
enthalpy gives
S2
mdot
2.5
kg
sec
Wdot
7.4586
kJ
kg K
mdot H
246
Ans.
Wdot
1173.4 kW
Ans.
7.35
Assume air an ideal gas. First find the temperature after isentropic
compression from a combination of Eqs. (5.14) & (5.15) with C = 0. Then
find the work (enthalpy change) of isentropic compression by a
combination of Eqs. (4.2) and (4.7) with C = 0. The actual work (enthalpy
change) is found from Eq. (7.20). From this value, the actual temperature
is found by a second application of the preceding equation, this time
solving it for the temperature. The following vectors contain values for
Parts (a) through (f):
298.15
101.33 kPa
375 kPa
353.15
375 kPa
1000 kPa
303.15
T0
373.15
100 kPa
P0
K
500 kPa
P
500 kPa
1300 kPa
299.82
14.7 psi
55 psi
338.71
55 psi
135 psi
100
0.75
100
0.70
150
ndot
0.80
mol
sec
50
S
0.75
0.5 453.59
16
0.75
0.5 453.59
i
J
mol K
0
0.70
For the heat capacity of air:
A
3.355
0.5
0.575 10
K
3
5
2
(guess)
B
D
0.016 10 K
Given
S = R A ln
B T0
22
T0
Tau T0 P0 P
1
D
Find
1
2
i
247
ln
P
P0
Tau T0 P0 Pi
i
i
431.06
464.5
Ti
T0
i
476.19
i
T
K
486.87
434.74
435.71
H'i
3
R ICPH T0 Ti 3.355 0.575 10
5
0.0
0.016 10
i
3925.2
3314.6
5133.2
J
3397.5 mol
H'
3986.4
2876.6
5233.6
4735.1
H'
H
6416.5
H
4530
J
mol
5315.2
4109.4
1.5
(guess)
Given
H = R A T0
Tau T0
H
Wdot
1
B
2
T0
2
ndot H
Find
i
248
2
1
Tau T0
i
D
T0
Hi
1
Ti
T0
i
i
474.68
702
523
511.58
635
474
518.66
T
524.3
1291
Wdot
K
304
962
Wdot
hp
479.01
1617
1205
476.79
7.36
1250
932
Ammonia:
0
Tc
405.7 K
Pc
294.15 K
P0
200 kPa
J
mol K
A
P
1000 kPa
For the heat capacity of ammonia:
3.020 10
K
B
3.578
T0
Tr0
0.253
112.8 bar
T0
S
Ans.
kW
227
Tr0
Tc
3
0.186 10 K
P0
Pr0
0.725
5
D
Pr0
Pc
P
Pr
Pr
Pc
2
0.0177
0.089
Use generalized second-virial correlation:
The entropy change is given by Eq. (6.92) combined with Eq. (5.15); C = 0:
1.4
(guess)
Given
S = R A ln
B T0
T0
SRB
Find
T0
Tc
1
D
2
ln
1
2
Pr
T
P0
SRB Tr0 Pr0
1.437
P
Tr
249
T0
T
Tc
T
Tr
422.818 K
1.042
Hig
Hig
4.826
H'
Hig
H'
4652
3
R ICPH T0 T 3.578 3.020 10
0.0
5
0.186 10
kJ
mol
R Tc HRB Tr Pr
HRB Tr0 Pr0
J
mol
The actual enthalpy change from Eq. (7.17):
0.82
H'
H
H
5673.2
J
mol
The actual final temperature is now found from Eq. (6.91) combined with Eq
(4.7), written:
(guess)
1.4
Given
1
Tc HRB
Find
B
2
T0
2
T0
Pr
Tc
2
1.521
H = R A T0
T
R A ln
B T0
SRB Tr Pr
S
2.347
J
mol K
T0
T
Tc
D
2
2
T0
SRB Tr0 Pr0
Ans.
250
1
HRB Tr0 Pr0
Tr
S
D
T0
1
1
T
447.47 K
Tr
1.103
1
ln
P
P0
Ans.
7.37
Propylene:
0
A
365.6 K
Pc
303.15 K
P0
11.5 bar
J
22.706 10
B
1.637
3
K
P0
Pr0
0.8292
6.915 10
C
K
Tr0
Tc
P
18 bar
For the heat capacity of propylene:
mol K
T0
Tr0
0.140
46.65 bar
T0
S
Tc
Pr
2
Pr0
Pc
P
Pr
Pc
6
0.2465
0.386
Use generalized second-virial correlation:
The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0:
(guess)
1.1
Given
S = R A ln
B T0
T0
1
2
C T0
2
Tc
Find
Pr
T
P
P0
SRB Tr0 Pr0
1.069
SRB
ln
1
T0
T
T
Tc
Tr
324.128 K
Tr
0.887
The enthalpy change for the final T is given by Eq. (6.91), with HRB for
this T:
Hig
R ICPH T0 T 1.637 22.706 10
Hig
1.409
H'
Hig
3
6.915 10
3J
10
mol
R Tc HRB Tr Pr
251
HRB Tr0 Pr0
6
0.0
H'
J
mol
964.1
The actual enthalpy change from Eq. (7.17):
ndot
H'
H
0.80
1000
mol
Wdot
H
Wdot
ndot H
J
mol
1205.2
Ans.
1205.2 kW
sec
The actual final temperature is now found from Eq. (6.91) combined with Eq
(4.7), written:
(guess)
1.1
Given
H = R A T0
B
2
T0
2
T0
Pr
Tc
1
Tc HRB
7.38 Methane:
C
3
T0
3
1
T
T0
0
Tr0
Tc
190.6 K
Pc
T0
Tc
Ans.
308.15 K
P0
0.012
45.99 bar
3500 kPa
J
mol K
A
1
327.15 K
T0
S
3
HRB Tr0 Pr0
T
1.079
Find
2
P
5500 kPa
For the heat capacity of methane:
B
1.702
Tr0
9.081 10
K
1.6167
3
C
K
Pr0
Pr
252
2.164 10
P0
Pc
P
Pc
6
2
Pr0
Pr
0.761
1.196
Use generalized second-virial correlation:
The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0:
(guess)
1.1
Given
S = R A ln
B T0
T0
1
2
C T0
2
Tc
Find
Pr
T
P
P0
SRB Tr0 Pr0
1.114
SRB
ln
1
T0
T
T
Tc
Tr
343.379 K
Tr
1.802
The enthalpy change for the final T is given by Eq. (6.91), with HRB for
this T:
Hig
Hig
1.298 10
H'
Hig
H'
1158.8
3
R ICPH T0 T 1.702 9.081 10
2.164 10
6
0.0
3J
mol
R Tc HRB Tr Pr
HRB Tr0 Pr0
J
mol
The actual enthalpy change from Eq. (7.17):
0.78
ndot
1500
H
mol
sec
Wdot
H'
ndot H
H
Wdot
1485.6
J
mol
2228.4 kW
Ans.
The actual final temperature is now found from Eq. (6.91) combined with Eq
(4.7), written:
1.1
(guess)
253
Given
H = R A T0
B
2
T0
2
T0
Pr
Tc
1
Tc HRB
7.39
C
3
T0
3
1
3
1
HRB Tr0 Pr0
T
1.14
Find
2
T
T0
Ans.
351.18 K
From the data and results of Example 7.9,
T1
Work
T2
293.15 K
5288.3
P1
428.65 K
J
T
mol
R ICPH T1 T2 1.702 9.081 10
H
5288.2
S
3
R ICPS T1 T2 1.702 9.081 10
3.201
560 kPa
293.15 K
H
S
P2
140 kPa
6
2.164 10
0.0
J
mol
3
6
2.164 10
0.0
ln
P2
P1
J
mol K
Since the process is adiabatic: SG
Wideal
Wlost
H
T
SG
S
T
S
S
Wideal
Wlost
Wideal
t
Work
254
3.2012
t
J
mol K
4349.8
938.4
0.823
J
mol
J
mol
Ans.
Ans.
Ans.
Ans.
7.42
P1
1atm
T1
(
35
P2
50atm
T2
(
200
T1
473.15 K
Cp
273.15)
K
308.15 K
T2
273.15)
K
3.5 R
3
0.65
V
Vdot
R T1
P1
0.5
ndot
m
sec
Vdot
ndot
V
19.775
mol
sec
With compression from the same initial conditions (P1,T1) to the same
final conditions (P2,T2) in each stage, the same efficiency in each stage,
and the same power delivered to each stage, the applicable equations are:
P2
P1
r=
1
N
(where r is the pressure ratio in each stage and N is
the number of stages.)
Eq. (7.23) may be solved for T2prime: T'2
T'2
N
R
Cp
P2
P1
ln
ln
T1)
T1
Eq. (7.18) written for a single stage is:
415.4 K
T'2 = T1
(
T2
R1
N Cp
Put in logarithmic form and solve for N:
P2
P1
N
T'2
(a) Although any number of
stages greater than this
would serve, design for 4
stages.
3.743
T1
(b) Calculate r for 4 stages: N
4
r
P2
P1
1
N
r
2.659
Power requirement per stage follows from Eq. (7.22). In kW/stage:
Wdotr
ndot Cp T1 r
R
Cp
1
Wdotr
255
87.944 kW
Ans.
(c) Because the gas (ideal) leaving the intercooler and the gas entering
the compressor are at the same temperature (308.15 K), there is no
enthalpy change for the compressor/interchanger system, and the first law
yields:
Qdotr
Wdotr
Qdotr
87.944 kW
Ans.
Heat duty = 87.94 kW/interchanger
(d) Energy balance on each interchanger (subscript w denotes water):
With data for saturated liquid water from the steam tables:
kJ
kJ
Hw (
188.4 104.8)
Hw 83.6
kg
kg
Qdotr
mdotw
mdotw
Hw
1.052
kg
sec
Ans.
(in each interchanger)
7.44
300
290
T1
2.0
1.5
295 K
P1
1.2 bar
300
305
1.5
464
6
3.5
547
T2
1.1
5
2.5
455 K
P2
6 bar
Cp
4.5 R
505
HS
5.5
496
H
8
7
4.0
[(
Cp T2
Cp T1
Ideal gases with constant heat capacities
T1)
P2
P1
R
Cp
(7.22)
1
256
3.219
0.675
3.729
HS
kJ
mol
4.745
0.698
HS
H
5.959
0.636
4.765
7.47
Ans.
0.793
0.75
The following vectors contain values for Parts (a) through (e). Intake
conditions first:
298.15
20 kg
363.15
T1
100 kPa
200 kPa
30 kg
P1
333.15 K
20 kPa
mdot
15 kg
294.26
1 atm
50 lb
366.48
15 psi
80 lb
2000 kPa
0.75
257.2
5000 kPa
0.70
696.2
5000 kPa
0.75
523.1
20 atm
0.70
0.75
10
K
6
217.3
1500 psi
P2
1
sec
714.3
From the steam tables for sat.liq. water at the initial temperature (heat
capacity calculated from enthalpy values):
1.003
1.036
V
1.017
4.15
4.20
3
cm
gm
4.20
CP
1.002
4.185
1.038
kJ
kg K
4.20
By Eq. (7.24)
HS
V P2
257
P1
H
HS
1.906
4.973
HS
2.541
7.104
kJ
5.065
H
kg
6.753
1.929
kg
2.756
10.628
kJ
14.17
0.188
By Eq. (7.25)
T
H
V1
T1
P2
0.807
P1
T
CP
0.612 K
0.227
1.506
50.82
213.12
Wdot
H mdot
Wdot
68.15
285.8
101.29 kW Wdot
135.84 hp
62.5
83.81
514.21
689.56
298.338
363.957
T2
T1
T
T2
333.762 K
294.487
367.986
t2
t2
T2
K
273.15
t2
t2
t2
T2
K
t2
1.8
459.67
t2
1
25.19
2
90.81
degC
60.61
3
4
5
258
70.41
202.7
degF
Ans.
7.48 Results from Example 7.10:
H
kJ
11.57
W
kg
11.57
Wideal
300 K
H
T
Wideal
T
8.87
kJ
kg
S
kJ
kg
0.0090
kJ
kg K
Wideal
S
t
Ans.
t
W
0.767
Ans.
Since the process is adiabatic.
SG
S
Wlost
7.53
SG
T
T1
(
25
T3
S
(
200
Cpv
105
9
3 kJ
10
Wlost
2.7
kJ
kg
P1
273.15)
K
J
mol K
Ans.
1.2bar
P3
273.15)
K
Ans.
kg K
5bar
Hlv
30.72
P2
5bar
kJ
mol
0.7
Estimate the specific molar volume of liquid benzene using the Rackett
equation (3.72).
3
From Table B.1 for benzene: Tc
562.2K
From Table B.2 for benzene: Tn
(
80.0
Zc
0.271 Vc
273.15)
K
Trn
259
cm
mol
Tn
Tc
2
Assume
Vliq
=
Vsat:
V
V c Zc
1 T rn
3
7
Eq. (3.72)
Calculate pump power
Ws
V P2
P1
Ws
259
0.053
kJ
mol
Ans.
V
cm
96.802
mol
Assume that no temperature change occurs during the liquid compression.
Therefore:
T2
T1
Estimate the saturation temperature at P = 5 bar using the Antoine
Equation and values from Table B.2
For benzene from
A
Table B.2:
B
Tsat
A
ln
13.7819
B
C degC Tsat
P2
kPa
Tsat
2726.81
C
217.572
Tsat
415.9 K
142.77 degC
Tsat
273.15K
Estimate the heat of vaporization at Tsat using Watson's method
From Table B.2
Hlv
At 80 C:
(
80
Tr1
Hlv2
273.15)
K
Tc
Hlv
1
Tr2
1
Tr1
30.72
kJ
mol
Tr1
0.628
Tr2
Tsat
Tr2
Tc
0.38
Eq. (4.13)
Hlv2
26.822
Calculate the heat exchanger heat duty.
Q
R ICPH T2 Tsat 0.747 67.96 10
Hlv2 Cpv T3 Tsat
Q
51.1
kJ
mol
Ans.
260
3
37.78 10
6
0
0.74
kJ
mol
7.54
T1
( 25
T3
( 200
Cpv
273.15)K
273.15)K
105
P1
P3
P2
1.2bar
1.2bar
5bar
J
mol K
0.75
Calculate the compressor inlet temperature.
Combining equations (7.17), (7.21) and (7.22) yields:
T3
T2
1
P3
1
T2
1
P2
408.06 K
T2
R
Cpv
273.15K
134.91 degC
Calculate the compressor power
Ws
Cpv T3
T2
Ws
6.834
kJ
mol
Ans.
Calculate the heat exchanger duty. Note that the exchanger outlet
temperature, T2, is equal to the compressor inlet temperature. The
benzene enters the exchanger as a subcooled liquid. In the exchanger the
liquid is first heated to the saturation temperature at P1, vaporized and
finally the vapor is superheated to temperature T 2.
Estimate the saturation temperature at P = 1.2 bar using the
Antoine Equation and values from Table B.2
For benzene from
A
Table B.2:
B
Tsat
A
ln
P1
B
13.7819
C degC Tsat
kPa
Tsat
2726.81
C
217.572
Tsat
358.7 K
85.595 degC
Tsat
273.15K
Estimate the heat of vaporization at Tsat using Watson's method
From Table B.2
At 25 C:
Hlv
30.72
kJ
mol
261
From Table B.1
for benzene:
Tc
562.2K
(
80
Tr1
273.15)
K
Tr1
Tc
Hlv2
Hlv
1
Tr2
1
Tr1
Eq. (4.13)
Q
44.393
3
R ICPH T1 Tsat 0.747 67.96 10
Hlv2 Cpv T2 Tsat
7.57 ndot
Cp
100
50.6
mol
Tr2
Tc
0.638
0.38
Q
kJ
Tsat
Tr2
0.628
Hlv2
37.78 10
30.405
6
kJ
mol
0
Ans.
kmol
hr
P1
T1
1.2bar
J
mol K
P2
300K
6bar
0.70
Assume the compressor is adaiabatic.
T2
P2
P1
Wdots
Wdote
T1
(Pg. 77)
ndot Cp T2
T2
T1
Wdots
127.641 kW
Wdote
3040dollars
380dollars
390.812 K
Wdots
Wdots
C_compressor
C_motor
R
Cp
182.345 kW
0.952
C_compressor
kW
Wdote
307452 dollars Ans.
0.855
C_motor
kW
262
32572 dollars Ans.
7.59
T1
18bar
For ethylene:
T1
Tr1
Tr1
Tc
P2
1.2bar
0.087
P1
375K
Tc
282.3K
P2
Pr2
A
B
1.424
14.394 10
3
Pc
C
4.392 10
6
0.357
Pr2
Pc
50.40bar
Pr1
P1
Pr1
1.328
Pc
0.024
D
0
a) For throttling process, assume the process is adiabatic. Find T2 such that
H = 0.
H = Cpmig T2
T1
HR2
Eq. (6-93)
HR1
Use the MCPH function to calculate the mean heat capacity and the HRB
function for the residual enthalpy.
Guess:
T2
T1
Given
0
J
mol
T2
= MCPH T1 T2 A B C D R T2 T1
T2
Pr2
R Tc HRB
R Tc HRB Tr1 Pr1
Tc
Find T2
T2
365.474 K
Ans.
Tr2
T2
Tc
Tr2
1.295
Calculate change in entropy using Eq. (6-94) along with MCPS function for
the mean heat capacity and SRB function for the residual entropy.
S
R MCPS T1 T2 A B C D ln
R SRB Tr2 Pr2
S
22.128
J
mol K
T2
T1
R SRB Tr1 Pr1
Ans.
263
R ln
P2
P1
Eq. (6-94)
b) For expansion process.
70%
First find T2 for isentropic expansion. Solve Eq. (6-94) with S = 0.
Guess:
T2
T1
Given
0
T2
P2
J
R ln
= R MCPS T1 T2 A B C D ln
T1
mol K
P1
T2
Pr2
SRB
R
R SRB Tr1 Pr1
Tc
T2
Find T2
T2
219.793 K
Tr2
T2
Eq. (6-94)
Tr2
Tc
Now calculate the isentropic enthalpy change, HS.
HR2
HS
HS
HRB Tr2 Pr2
R Tc
R MCPH T1 T2 A B C D T2 T1
HRB Tr2 Pr2
R Tc HRB Tr1 Pr1
6.423
R Tc
3J
10
mol
Calculate actual enthalpy change using the expander efficiency.
H
HS
H
4.496
3J
10
mol
Find T2 such that H matches the value above.
Given
HS = MCPH T1 T2 A B C D R T2 T1
T2
Pr2
R Tc HRB
R Tc HRB Tr1 Pr1
Tc
T2
Find T2
T2
268.536 K
264
Ans.
0.779
Now recalculate S at calculated T2
S
R MCPS T1 T2 A B C D ln
T1
R SRB Tr1 Pr1
R SRB Tr2 Pr2
S
7.77
J
mol K
T2
R ln
P2
Eq. (6-94)
P1
Ans.
Calculate power produced by expander
kJ
Ans.
P
H
P
3.147
mol
The advantage of the expander is that power can be produced in the
expander which can be used in the plant. The disadvantages are the extra
capital and operating cost of the expander and the low temperature of the
gas leaving the expander compared to the gas leaving the throttle valve.
7.60
Hydrocarbon gas:
T1
500degC
Light oil:
25degC Cpoil
Exit stream:
b)
T2
T3
Cpgas
150
J
mol K
J
mol K
J
mol
200degC
200
Hlv
35000
Assume that the oil vaporizes at 25 C. For an adiabatic column, the overall
energy balance is as follows.
F Cpgas T3
T1
D
Hlv
Coilp T3
T2
=0
DF
0.643
Solving for D/F gives:
DF
c)
Cpgas T3
Hlv
T1
Cpoil T3
T2
Ans.
Using liquid oil to quench the gas stream requires a smaller oil flow rate.
This is because a significant portion of the energy lost by the gas is used
to vaporize the oil.
265
Chapter 8 - Section A - Mathcad Solutions
8.1
With reference to Fig. 8.1, SI units,
At point 2: Table F.2,
H2
3531.5
At point 4: Table F.1,
H4
209.3
At point 1:
H1
H4
At point 3: Table F.1,
Hliq
H4
x3
H3
Hliq
0.96
Sliq
For isentropic expansion,
x'3
H'3
S'3
Slv
x'3 Hlv
H3
turbine
Ws
Ws
H3
Hlv
x3 Hlv
H2
S'3
S2
0.855
H'3
2246
H2
H'3
1.035
3
10
H2
QH
3.322
Ws
cycle
QH
cycle
266
Ans.
0.805
turbine
QH
H2
H3
Slv
0.7035
x' 3
Sliq
Hliq
S2
H1
10
0.311
3
Ans.
6.9636
2382.9
2496.9
7.3241
8.2
1.0 (kg/s)
mdot
The following property values are found by linear interpolation in Table F.1:
State 1, Sat. Liquid at TH: H1
860.7
S1
2.3482
P1
3.533
State 2, Sat. Vapor at TH: H2
2792.0
S2
6.4139
P2
3.533
State 3, Wet Vapor at TC: Hliq
112.5
Hvap
2550.6
P3
1616.0
State 4, Wet Vapor at TC: Sliq
0.3929
Svap
8.5200
P4
1616.0
(a) The pressures in kPa appear above.
(b) Steps 2--3 and 4--1 (Fig. 8.2) are isentropic, for which S3=S2 and S1=S4.
Thus by Eq. 6.82):
x3
S2 Sliq
Svap Sliq
x3
S1 Sliq
Svap Sliq
x4
0.741
x4
0.241
(c) The rate of heat addition, Step 1--2:
Qdot12
mdot (
H2
Qdot12
H1)
1.931
3
10
(kJ/s)
(d) The rate of heat rejection, Step 3--4:
Hliq
x3 (
Hvap Hliq)
H3
1.919
H4
10
Qdot34
3
mdot (
H4
(e) Wdot12
699.083
1.22
873.222
mdot (
H3
H2)
Wdot23
Wdot41
mdot (
H1
H4)
Wdot41
Wdot23 Wdot41
Qdot12
161.617
0.368
Note that the first law is satisfied:
Q
Qdot12
Q
W
Qdot34
W
Wdot23
0
267
Wdot41
3
10
0
Wdot23
(f)
x4 (
Hvap Hliq)
Qdot34
H3)
Wdot34
0
Hliq
H4
H3
(kJ/s)
8.3
The following vectors contain values for Parts (a) through (f).
Enthalpies and entropies for superheated vapor, Tables F.2 and F.4 @ P2
and T2 (see Fig. 8.4):
3622.7
6.9013
kJ
kg K
3529.6
kJ
kg
6.9485
kJ
kg K
3635.4
H2
kJ
kg
kJ
kg
6.9875
kJ
kg K
6.9145
kJ
kg K
S2
kJ
3475.6
kg
1507.0
BTU
lbm
1.6595
BTU
lbm rankine
1558.8
BTU
lbm
1.6759
BTU
lbm rankine
Sat. liq. and sat. vap. values from Tables F.2 and F.4 @ P3 = P4:
191.832
2584.8
kJ
kg
251.453
kJ
kg
2609.9
kJ
kg
191.832
Hliq
kJ
kg
kJ
kg
2584.8
kJ
kg
2676.0
kJ
kg
Hvap
kJ
419.064
kg
180.17
BTU
lbm
1150.5
BTU
lbm
69.73
BTU
lbm
1105.8
BTU
lbm
268
0.6493
kJ
kg K
8.1511
kJ
kg K
0.8321
kJ
kg K
7.9094
kJ
kg K
kJ
0.6493
kg K
Sliq
0.1326
Svap
kJ
1.3069
0.3121
8.1511
7.3554
kg K
BTU
lbm rankine
1.7568
BTU
1.9781
lbm rankine
kJ
kg K
kJ
kg K
BTU
lbm rankine
BTU
lbm rankine
3
cm
1.010
gm
3
cm
1.017
gm
0.80
1.010
cm
gm
Vliq
3
cm
1.044
gm
0.0167
ft
0.75
0.75
3
0.75
0.80
turbine
0.78
0.80
pump
0.75
0.78
0.80
3
lbm
3
ft
0.0161
lbm
269
0.75
0.75
80
10000 kPa
10 kPa
100
7000 kPa
20 kPa
70
Wdot
50
3
8500 kPa
P1
10 kW
10 kPa
P4
6500 kPa
101.33 kPa
50
950 psi
14.7 psi
80
1125 psi
1 psi
Vliq P1
Wpump
P4
H4
pump
S'3 = S2
H3
H2
x'3
S2
Sliq
Svap
turbine H'3
Sliq
H2
Hliq
H1
H'3
Hliq
H4
Wpump
x'3 Hvap
Hliq
H2
QdotH
Wdot
Wturbine Wpump
H3
H2
H1 mdot
QdotC
mdot
Wturbine
QdotH
Wdot
Answers follow:
QdotH
mdot1
mdot2
108.64
mdot3
62.13
mdot4
1
70.43
67.29
QdotH
kg
sec
2
QdotH
3
QdotH
240705
355111
kJ
213277 sec
205061
4
mdot5
mdot6
QdotH
145.733 lbm
153.598 sec
5
QdotH
6
270
192801 BTU
228033 sec
0.332
QdotC
1
160705
0.282
QdotC
255111
kJ
Wdot
0.328
QdotC
143277 sec
QdotH
0.244
2
3
155061
QdotC
0.246
4
0.333
QdotC
145410 BTU
5
QdotC
152208
6
8.4
sec
Subscripts refer to Fig. 8.3.
Saturated liquid at 50 kPa (point 4)
3
V4
cm
1.030
gm
H4
P4
3300 kPa
P1
kJ
kg
340.564
50 kPa
Saturated liquid and vapor at 50 kPa:
Hliq
H4
Sliq
1.0912
Hvap
kJ
kg K
2646.0
Svap
7.5947
By Eq. (7.24),
Wpump
H1
H1
H4
Wpump
kJ
kg
kJ
kg K
V 4 P4
343.911
P1
Wpump
3.348
kJ
kg
kJ
kg
The following vectors give values for temperatures of 450, 550, and 650 degC:
3340.6
H2
3565.3
7.0373
kJ
kg
S2
3792.9
7.3282
7.5891
271
kJ
kg K
S'3
S2
H'3
Hliq
QH
H2
S'3
x' 3
x'3 Hvap
Hliq
Sliq
Svap
Wturbine
Sliq
H'3
Wturbine
H1
H2
Wpump
QH
0.914
8.5
0.959
0.314
0.999
x'3
0.297
0.332
Ans.
Subscripts refer to Fig. 8.3.
Saturated liquid at 30 kPa (point 4)
3
V4
cm
1.022
gm
H4
289.302
kJ
kg
P1
30 kPa
Saturated liquid and vapor at 30 kPa:
Hliq
H4
Hvap
5000
kJ
kg
2625.4
P4
7500
10000
Sliq
0.9441
kJ
kg K
By Eq. (7.24),
Svap
Wpump
7.7695
V 4 P4
kJ
kg K
P1
294.381
H1
H4
Wpump
H1
296.936
kJ
kg
299.491
The following vectors give values for pressures of 5000, 7500, and
10000 kPa at 600 degC
3664.5
H2
3643.7
7.2578
kJ
kg
S2
3622.7
7.0526
6.9013
272
kJ
kg K
kPa
S'3
S2
H'3
Hliq
QH
H2
S'3
x'3
x'3 Hvap
Hliq
Sliq
Svap
Sliq
Wturbine
H'3
H2
Wturbine
H1
Wpump
QH
0.925
8.6
0.895
0.375
0.873
x'3
0.359
0.386
Ans.
From Table F.2 at 7000 kPa and 640 degC:
H1
kJ
3766.4
S1
kg
7.2200
kJ
kg K
S'2
S1
For sat. liq. and sat. vap. at 20 kPa:
kJ
kg
Hvap
2609.9
kJ
kg
kJ
kg K
Svap
7.9094
kJ
kg K
Hliq
251.453
Sliq
0.8321
The following enthalpies are interpolated in Table F.2 at four values for
intermediate pressure P2:
3023.9
725
P2
750
775
H'2
kPa
W12
H'2
579.15
W12
572.442 kJ
565.89 kg
559.572
3040.9
kJ
kg
3049.0
800
0.78
3032.5
H1
H2
3187.3
H2
3194
kJ
3200.5 kg
3206.8
273
H1
W12
7.4939
S2
7.4898
7.4851
7.4797
kJ
kg K
where the entropy values are by interpolation in Table F.2 at P2.
x'3
S2
Svap
W23
W
Sliq
H'3
Sliq
H'3
W12
Hliq
H2
x'3 Hvap
Hliq
20.817
W23
W
7.811
kJ
kg
5.073
17.723
The work difference is essentially linear in P2, and we interpolate linearly to
find the value of P2 for which the work difference is zero:
linterp
W
P2 0.0
kJ
kg
(P2)
765.16 kPa
Also needed are values of H2 and S2 at this pressure. Again we do linear
interpolations:
linterp P2 H2 765.16 kPa
3197.9
kJ
kg
H2
3197.9
kJ
kg
linterp P2 S2 765.16 kPa
7.4869
kJ
kg K
S2
7.4869
kJ
kg K
We can now find the temperature at this state by interplation in Table F.2.
This gives an intermediate steam temperature t2 of 366.6 degC.
The work calculations must be repeated for THIS case:
W12
W12
H2
H1
568.5
H'3
Hliq
H'3
2.469
x'3
kJ
x'3
kg
x'3 Hvap
10
Hliq
W23
3 kJ
W23
kg
274
S2
Svap
Sliq
Sliq
0.94
H'3
568.46
H2
kJ
kg
Work
W12
Work
W23
1137
kJ
kg
For a single isentropic expansion from the initial pressure to the
final pressure, which yields a wet exhaust:
x'3
S1
Svap
x'3
H'3
H'3
Hliq
x'3 Hvap
H'3
Sliq
0.903
W'
Sliq
2.38
10
W'
H1
Hliq
3 kJ
1386.2
kg
kJ
kg
Whence the overall efficiency is:
Work
overall
overall
W'
275
0.8202
Ans.
8.7
From Table F.2 for steam at 4500 kPa and 500 degC:
H2
3439.3
kJ
kg
S2
7.0311
kJ
kg K
S'3
S2
By interpolation at 350 kPa and this entropy,
H'3
2770.6
H3
H2
kJ
0.78
kg
WI
H3
WI
2.918
10
3 kJ
kg
Isentropic expansion to 20 kPa:
S'4
Exhaust is wet: for sat. liq. & vap.:
S2
Hliq
251.453
Sliq
0.8321
kJ
kg
kJ
kg K
Hvap
2609.9
Svap
7.9094
276
kJ
kg
kJ
kg K
H'3
H2
WI
521.586
kJ
kg
x' 4
S'4
Sliq
Svap
x'4
H2
H'4
2.317
10
H4
H2
Hliq
H'4
0.876
H4
H'4
Sliq
x'4 Hvap
2.564
10
Hliq
3 kJ
kg
3 kJ
kg
3
H5
Hliq
V5
V 5 P6
Wpump
Wpump
5.841
cm
1.017
P5
20 kPa
H6
H5
Wpump
H6
257.294
gm
P5
kJ
kg
P6
4500 kPa
kJ
kg
For sat. liq. at 350 kPa (Table F.2):
H7
584.270
kJ
kg
t7
138.87
(degC)
We need the enthalpy of compressed liquid at point 1, where the pressure is
4500 kPa and the temperature is:
t1
138.87
6
T1
t1
273.15 K
t1
132.87
At this temperature, 132.87 degC, interpolation in Table F.1 gives:
3
kJ
558.5
kg
Hsat.liq
Psat
294.26 kPa
Vsat.liq
1.073
Also by approximation, the definition of the volume expansivity yields:
1
1.083
20
Vsat.liq
9.32 10
1.063
3
cm
gm K
41
K
277
P1
P6
cm
gm
By Eq. (7.25),
H1
Hsat.liq
Vsat.liq 1
T1
P1
H1
Psat
561.305
kJ
kg
By an energy balance on the feedwater heater:
mass
H1
H6
H3
H7
mass
kg
Ans.
0.13028 kg
Work in 2nd section of turbine:
WII
( kg
1
Wnet
mass) H4
WI
Wpump 1 kg
QH
QH
8.8
H2
WII
WII
2878 kJ
307.567 kJ
Wnet
H3
823.3 kJ
H 1 1 kg
Wnet
Ans.
0.2861
QH
Refer to figure in preceding problem.
Although entropy values are not needed for most points in the process, they are
recorded here for future use in Problem 15.8.
From Table F.4 for steam at 650(psia) & 900 degF:
H2
1461.2
BTU
lbm
S2
1.6671
BTU
lbm rankine
S'3
S2
By interpolation at 50(psia) and this entropy,
H'3
1180.4
H3
H2
S3
1.7431
BTU
lbm
WI
WI
0.78
H3
1242.2
BTU
lbm rankine
278
BTU
lbm
H'3
WI
219.024
H2
BTU
lbm
S'4
Isentropic expansion to 1(psia):
S2
Exhaust is wet: for sat. liq. & vap.:
69.73
Sliq
0.1326
x'4
S'4
Hvap
BTU
lbm rankine
Sliq
1.9781
BTU
lbm rankine
H2
H'4
H4
Hvap
Hliq
0.944
1047.8
S4
Hliq
931.204
H4
H2
Hliq
Sliq
S4
H4
H'4
Sliq
0.831
x4
BTU
lbm
H'4
Svap
x' 4
x4
1105.8
Svap
BTU
lbm
Hliq
x'4 Hvap
1.8748
Hliq
BTU
lbm
BTU
lbm
x4 Svap
Sliq
BTU
lbm rankine
3
P5
Wpump
P6
H5
1 psi
V 5 P6
P5
Wpump
H6
650 psi
V5
Hliq
H5
2.489
ft
0.0161
lbm
H6
72.219
BTU
lbm
Wpump
BTU
lbm
For sat. liq. at 50(psia) (Table F.4):
H7
250.21
BTU
lbm
t7
281.01
S7
0.4112
BTU
lbm rankine
We need the enthalpy of compressed liquid at point 1, where the pressure is
650(psia) and the temperature is
t1
281.01
11
T1
t1
279
459.67 rankine
t1
270.01
At this temperature, 270.01 degF, interpolation in Table F.3 gives:
Psat
41.87 psi
Hsat.liq
Vsat.liq
238.96
0.1717
Ssat.liq
0.3960
ft
3
lbm
BTU
lbm
BTU
lbm rankine
The definition of the volume expansivity yields:
0.01726
1
ft
20
Vsat.liq
4.95
0.01709
10
5
3
P1
P6
H1
257.6
BTU
lbm
S1
lbm rankine
0.397
BTU
lbm rankine
1
rankine
By Eq. (7.25) and (7.26),
H1
Hsat.liq
S1
Vsat.liq 1
Ssat.liq
Vsat.liq
T1
P1
P1
Psat
Psat
By an energy balance on the feedwater heater:
mass
H1
H6
H3
H7
lbm
mass
0.18687 lbm Ans.
WII
158.051 BTU
Work in 2nd section of turbine:
WII
1 lbm
Wnet
WI
QH
H2
Wnet
QH
mass
H4
H3
Wpump 1 lbm
WII
Wnet
H1 1 lbm
QH
0.3112
280
Ans.
374.586 BTU
1.204
3
10 BTU
8.9
Steam at 6500 kPa & 600 degC (point 2) Table F.2:
H2
3652.1
kJ
kg
S2
7.1258
kJ
kg K
P2
6500 kPa
At point 3 the pressure must be such that the steam has a condensation
temperature in feedwater heater I of 195 degC, 5 deg higher than the
temperature of the feed water to the boiler at point 1. Its saturation pressure,
corresponding to 195 degC, from Table F.1, is 1399.0 kPa. The steam at point 3
is superheated vapor at this pressure, and if expansion from P2 to P3 is
isentropic,
S'3
S2
H'3
3142.6
H3
H2
By double interpolation in Table F.2,
kJ
kg
WI
From Table F.1:
0.80
H3
H10
3.244
WI
829.9
281
3 kJ
10
kJ
kg
kg
H'3
WI
407.6
H2
kJ
kg
Similar calculations are required for feedwater heater II.
At the exhaust conditions of 20 kPa, the properties of sat. liq. and sat.
vap. are:
Hliq
kJ
251.453
kg
Sliq
0.8321
Hvap
Svap
kJ
kg K
2609.9
7.9094
3
kJ
Vliq
kg
1.017
cm
gm
kJ
kg K
If we find t7, then t8 is the mid-temperature between t7 and t1(190 degC), and
that fixes the pressure of stream 4 so that its saturation temperature is 5 degC
higher. At point 6, we have saturated liquid at 20 kPa, and its properties from
Table F.2 are:
tsat
60.09
Tsat
H6
Hliq
V6
Wpump
Wpump
V 6 P2
8.238
P6
tsat
273.15 K
Vliq
P6
20 kPa
[Eq. (7.24)]
kJ
kg
H67
Wpump
We apply Eq. (7.25) for the calculation of the temperature change from point 6
to point 7. For this we need values of the heat capacity and volume expansivity
of water at about 60 degC. They can be estimated from data in Table F.1:
1
Vliq
5.408
1.023
1.012
20
10
3
cm
CP
gm K
41
CP
K
272.0
kJ
kg K
230.2
10
4.18
kJ
kg K
Solving Eq. (7.25) for delta T gives:
H67
T67
t7
Vliq 1
Tsat
P2
P6
190
t7
T67
CP
tsat
T67
K
t9
2
282
t7
t8
0.678 K
t9
5
t7
60.768
t8
130.38
H7
Hliq
H67
H7
259.691
kJ
kg
H8
From Table F.1:
t9
T9
125.38
547.9
kJ
kg
273.15
t9 K
At points 9 and 1, the streams are compressed liquid (P=6500 kPa), and we
find the effect of pressure on the liquid by Eq. (7.25). Values by
interpolation in Table F.1 at saturation temperatures t9 and t1:
Hsat.9
kJ
526.6
kg
Hsat.1
kJ
807.5
kg
3
Vsat.9
cm
1.065
gm
Vsat.1
cm
1.142
gm
Psat.9
234.9 kPa
Psat.1
1255.1 kPa
3
3
3
V9
T
cm
1.056)
gm
(
1.075
V1
1
20 K
9
9
H9
Hsat.9
Vsat.9 1
T1
(
273.15
Hsat.1
Vsat.1 1
1
T
1
V1
Vsat.1
T
41
9 T9
10
1 T1
P2
1
K
P2
Psat.9
Psat.1
1.226
H9
530.9
T1
8.92
190)K
H1
(
1.156
V9
Vsat.9
cm
1.128)
gm
10
463.15 K
H1
810.089
kJ
kg
kJ
kg
Now we can make an energy balance on feedwater heater I to find the
mass of steam condensed:
mI
H1
H9
H3
H10
mI
kg
283
0.11563 kg
Ans.
31
K
The temperature at point 8, t8 = 130.38 (see above) is the saturation temperture in
feedwater heater II. The saturation pressure by interpolation in Table F.1 is
273.28 kPa.
Isentropic expansion of steam from the initial conditions to this pressure results in
a slightly superheated vapor, for which by double interpolation in Table F.2:
H'4
2763.2
kJ
H4
H2
H4
Then
kg
H'4
2.941
H2
3 kJ
10
kg
We can now make an energy balance on feedwater heater II to find the mass of
steam condensed:
mII
H9
H 7 1 kg
H4
mI H10
H8
H8
mII
0.09971 kg
Ans.
The final stage of expansion in the turbine is to 20 kPa, where the exhaust is wet.
For isentropic expansion,
x'5
x'5
S2
Svap
Sliq
H'5
0.889
H5
Then
H2
Hliq
H'5
Sliq
2.349
H'5
x'5 Hvap
Hliq
3 kJ
10
kg
H2
H5
2609.4
kJ
kg
The work of the turbine is:
Wturbine
Wturbine
WI 1 kg
1 kg mI H4 H3
1 kg mI mII H5 H4
936.2 kJ
Wturbine
QH
H2
Wpump 1 kg
H 1 1 kg
0.3265
QH
284
QH
Ans.
3
2.842 10 kJ
8.10
Tc
Isobutane:
Pc
408.1 K
0.181
36.48 bar
For isentropic expansion in the turbine, let the initial state be represented by
symbols with subscript zero and the final state by symbols with no subscript.
Then
T0
P0
533.15 K
S
0
J
mol K
A
37.853 10
K
B
Tr0
Tc
450 kPa
For the heat capacity of isobutane:
1.677
T0
Tr0
P
4800 kPa
3
11.945 10
C
K
Pr0
1.3064
Pr
6
2
P0
Pr0
Pc
P
Pc
Pr
1.3158
0.123
Use generalized second-virial correlation:
The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0:
0.8 (guess)
Given
S = R A ln
SRB
B T0
T0
Tc
Pr
1
2
C T0
2
ln
P
P0
SRB Tr0 Pr0
T
0.852
Find
1
Tr
T
T0
T
Tc
Tr
454.49 K
1.114
The enthalpy change for this final temperature is given by Eq. (6.91), with HRB at
the above T:
Hig
Hig
R ICPH T0 T 1.677 37.853 10
1.141
10
4J
mol
285
3
11.945 10
6
0.0
Hturbine
Hig
R Tc HRB Tr Pr
Hturbine
8850.6
J
mol
HRB Tr0 Pr0
Wturbine
Hturbine
The work of the pump is given by Eq. (7.24), and for this we need an estimate of
the molar volume of isobutane as a saturated liquid at 450 kPa. This is given by
Eq. (3.72). The saturation temperature at 450 kPa is given by the Antoine
equation solved for t degC:
VP
450 kPa
Avp
14.57100
Bvp
Bvp
tsat
Avp
Cvp
VP
ln
kPa
2606.775
tsat
Cvp
34
Tsat
tsat
Tsat
3
Vc
274.068
cm
262.7
mol
Zc
0.282
Trsat
273.15 K
307.15 K
Tsat
Trsat
Tc
0.753
2
Vliq
Wpump
V c Zc
1 T rsat
Vliq P0
3
7
Vliq
P
cm
112.362
mol
Wpump
488.8
J
mol
The flow rate of isobutane can now be found:
mdot
1000 kW
Wturbine Wpump
mdot
119.59
mol
sec
Ans.
The enthalpy change of the isobutane in the cooler/condenser is calculated in
two steps:
a. Cooling of the vapor from 454.48 to 307.15 K
b. Condensation of the vapor at 307.15 K
Enthalpy change of cooling: HRB at the initial state has already been calculated.
For saturated vapor at 307.15 K:
Hig
R ICPH T Tsat 1.677 37.853 10
286
3
11.945 10
6
0.0
Hig
4J
1.756
Ha
Hig
Ha
18082
10
mol
R Tc HRB Trsat Pr
HRB Tr Pr
J
mol
For the condensation process, we estimate the latent heat by Eqs. (4.12) and (4.13):
Tn
261.4 K
R Tn 1.092 ln
Hn
0.930
Hb
Hn
1
Trsat
1
Pc
bar
mdot
Qdotin
Wturbine
0.641
1.013
Hn
Trn
4J
2.118
10
0.38
Hb
Ha
18378
mol
J
mol
Hb
Wpump mdot
4360 kW
8.11 Isobutane: Tc
Trn
Tc
Trn
Qdotout
Qdotout
Tn
Trn
Qdotout
1000 kW
Qdotin
Qdotin
408.1 K
5360 kW
0.187
Pc
36.48 bar
0.181
Ans.
For isentropic expansion in the turbine, let the initial (inlet) state be
represented by symbols with subscript zero and the final (exit) state by
symbols with no subscript. Then
T0
S
413.15 K
0
P0
3400 kPa
J
mol K
287
P
450 kPa
molwt
58.123
gm
mol
For the heat capacity of isobutane:
A
1.677
B
T0
Tr0
37.853 10
11.945 10
C
K
Tr0
Tc
3
K
1.0124
2
P0
Pr0
Pr0
Pc
P
Pc
Pr
6
Pr
0.932
0.123
Use Lee/Kesler correlation for turbine-inlet state, designating values
by HRLK and SRLK:
HRLK0
1.530
SRLK0
1.160
The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0:
(guess)
0.8
Given
S = R A ln
B T0
T0
SRB
Pr
Tc
Find
1
2
C T0
1
2
ln
P
P0
SRLK0
0.809
T
T0
T
T
Tc
Tr
Tr
334.08 K
0.819
The enthalpy change for this final temperature is given by Eq. (6.91), with
HRB at the above T:
Hig
R ICPH T0 T 1.677 37.853 10
Hig
9.3
3
11.945 10
6
0.0
3J
10
mol
Hturbine
Hig
R Tc HRB Tr Pr
Hturbine
4852.6
J
mol
HRLK0
Wturbine
288
Hturbine
The work of the pump is given by Eq. (7.24), and the required value for the
molar volume of saturated-liquid isobutane at 450 kPa (34 degC) is the
value calculated in Problem 8.10:
3
Vliq
112.36
cm
mol
Wpump
Vliq P0
P
Wpump
331.462
J
mol
For the cycle the net power OUTPUT is:
mdot
kg
75
molwt sec
Wdot
Wdot
mdot Wturbine
Wpump
Ans.
5834 kW
The enthalpy change of the isobutane in the cooler/condenser is calculated in
two steps:
a. Cooling of the vapor from 334.07 to 307.15 K
b. Condensation of the vapor at 307.15 K
Enthalpy change of cooling: HRB at the initial state has already been
calculated. For saturated vapor at 307.15 K it was found in Problem 8.10 as:
Tsat
Hig
307.15K
Tsat
Trsat
Trsat
Tc
R ICPH T Tsat 1.677 37.853 10
Hig
2.817
Hig
R Tc HRB Trsat Pr
Ha
2975
11.945 10
6
0.0
kJ
mol
Ha
3
0.753
HRB Tr Pr
J
mol
For the condensation process, the enthalpy change was found in Problem
8.10:
Hb
18378
J
mol
Qdotout
Qdotout
289
mdot
Ha
Hb
27553 kW Ans.
For the heater/boiler:
Qdotin
Wdot
Qdotout
Qdotin
Wdot
Qdotin
33387 kW Ans.
Ans.
0.175
We now recalculate results for a cycle for which the turbine and pump each
have an efficiency of 0.8. The work of the turbine is 80% of the value
calculated above, i.e.,
W'turbine
0.8 Wturbine
W'turbine
3882
J
mol
The work of the pump is:
W'pump
Wdot
Wpump
W'pump
0.8
mdot W'turbine
W'pump
Wdot
414.3
J
mol
4475 kW
Ans.
The decrease in the work output of the turbine shows up as an increase in
the heat transferred out of the cooler condenser. Thus
Qdotout
Qdotout
Qdotout
Wturbine
28805 kW
W'turbine mdot
Ans.
The increase in pump work shows up as a decrease in the heat added in the
heater/boiler. Thus
Qdotin
Qdotin
Wdot
Qdotin
W'pump
Wpump mdot
0.134
290
Ans.
Qdotin
33280 kW
Ans.
CP
PC
1 bar
By Eq. (3.30c):
7
R
2
TC
8.13 Refer to Fig. 8.10.
293.15 K
PD
PC V C = PD V D
1
VC
VD
1
PD
=
1.4
5 bar
or
PD
r
PC
r
PC
Ans.
3.157
1
Eq. (3.30b):
TD
QDA = CP TA
TC
PD
QDA
PC
QDA
TA
TD
CP
TD
1500
TA
J
mol
515.845 K
R TC
re =
VB
VA
=
VC
VA
=
PC
R TA
PA
PD
T C PA
re
T A PC
PA
re
8.14
2.841
Ans.
3
5
Ratio
Ratio =
7
PB
PA
1.35
9
Eq. (8.12) now becomes:
0.248
1
1
0.341
1
Ratio
0.396
0.434
291
Ans.
8.16
Figure shows the air-standard turbojet power plant on a PV diagram.
7
TA 303.15 K
TC 1373.15 K
R
CP
2
By Eq. (7.22)
WAB = CP TA
WCD = CP TC
PB
R
CP
1 = CP TA cr
PA
PD
2
R
CP
1
2
1 = CP TC er
PC
7
7
1
where cr is the compression ratio and er is the expansion ratio. Since the two
work terms are equal but of opposite signs,
cr
6.5
er
0.5 (guess)
2
Given
TC er
7
2
1 = TA cr
7
er
Find er)
(
er
292
1
0.552
By Eq. (7.18):
PD
TD = TC
R
CP
PC
2
This may be written:
TD
TC er
7
1
By Eq. (7.11)
uE
2
2
2
uD =
PD V D
1
PE
1
(A)
PD
We note the following:
er =
PD
cr =
PC
PB
PA
=
PC
cr er =
PE
PD
PE
The following substitutions are made in (A):
1
uD = 0
=
2
R
=
7
CP
Then
molwt
uE
2
PE
8.17 TA
R
7
TD 1
2 molwt
PD
1 bar
305 K
PA
PE
PD V D = R T D
1
cr er
29
2
7
cr er PE
PB
1.05bar
PD
=
1
cr er
gm
mol
uE
843.4
m
sec
Ans.
PD
3.589 bar
Ans.
0.8
7.5bar
Assume air to be an ideal gas with mean heat capacity (final temperature by
iteration):
Cpmair
MCPH 298.15K 582K 3.355 0.575 10
Cpmair
29.921
J
mol K
293
3
0.0
0.016 10
5
R
Compressor:
TB
PB
Cpmair TA
Wsair
R
Cpmair
PA
Wsair
TA
TB
Cpmair
Combustion:
Wsair
1
8.292
10
3J
mol
582.126 K
CH4 + 2O2 = CO2 + 2H2O
Basis: Complete combustion of 1 mol CH4. Reactants are N mol of
air and 1mol CH4.
Because the combustion is adiabatic, the basic equation is:
HR
H298
HP = 0
For H_R, the mean heat capacities for air and methane are required.
The value for air is given above. For methane the temperature change
is very small; use the value given in Table C.1 for 298 K: 4.217*R.
The solution process requires iteration for N. Assume a value for N until
the above energy balance is satisfied.
(a) TC
1000K
HR
57.638 (This is the final value after iteration)
N
Cpmair N ( 98.15
2
HR
4.896
582.03)K
4.217 R ( 98.15
2
300)K
5J
10
mol
The product stream contains:
1 mol CO2, 2mol H2O, 0.79N mol N2, and (0.21N-2) mol O2
1
5.457
2
n
A
.79 N
.21 N
i
2
3.470
3.280
1.045
B
3.639
1.450
0.593
0.506
14
294
1.157
10
3
D
0.121
0.040
0.227
5
10
ni
58.638
A
ni Ai
i
i
A
CpmP
HP
ni Bi
D
i
198.517
B
CpmP TC
298.15K
HP
H298
802625
H298
HP
136.223
ni D i
i
0.036
D
MCPH 298.15K 1000.K 198.517 0.0361 0.0
From Ex. 4.7:
HR
B
1.292
1.387
10
5
1.3872 10
5
R
6J
10
mol
J
mol
J
(This result is sufficiently close to zero.)
mol
Thus, N = 57.638 moles of air per mole of methane fuel. Ans.
Assume expansion of the combustion products in the turbine is to 1(atm),
i.e., to 1.0133 bar:
PD
1.0133bar
PC
7.5bar
The pertinent equations are analogous to those for the compressor. The
mean heat capacity is that of the combustion gases, and depends on the
temperature of the exhaust gases from the turbine, which must therefore be
found by iteration. For an initial calculation use the mean heat capacity
already determined. This calculation yields an exhaust temperature of
about 390 K. Thus iteration starts with this value. Parameters A, B, and D
have the final values determined above.
Cpm
Cpm
MCPH 1000K 343.12K 198.517 0.0361 0.0
1.849
10
3
5
1.3872 10
R
J
For 58.638 moles of combustion product:
mol K
R
Ws
TD
58.638 Cpm TC
PD
Cpm
1
PC
TC
Ws
Cpm
TD
343.123 K
295
Ws
6J
1.214 10
mol
(Final result of iteration.) Ans.
Wsnet
Ws
Wsair N
Wsnet
5J
7.364
10
Ans.
mol
(J per mole of methane)
Parts (b) and (c) are solved in exactly the same way, with the following
results:
5
(b) TC
N
37.48
Wsnet
7.365 10
(c)
8.18
1200
1500
N
24.07
Wsnet
5.7519 10
me
0.95
line_losses
TC
0.35
tm
TD
TD
5
343.123
598.94
20% Cost_fuel
4.00
dollars
GJ
Cost_fuel
Cost_electricity
tm
Cost_electricity
0.05
1
me (
cents
kW hr
line_losses)
Ans.
This is about 1/2 to 1/3 of the typical cost charged to residential customers.
8.19 TC
TH
111.4K
1
Carnot
TC
Carnot
TH
Assume as a basis:
QH
W
W
QH
Hnlv
300K
0.629
QC
2.651 kJ
QC
W
0.201
mol
kJ
0.6
kJ
mol
Carnot
HE
0.377
1kJ
HE
Hnlv
HE
8.206
Ans.
296
QH 1
HE
QC
1.651 kJ
8.20 TH
( 27
a) Carnot
b) actual
TC
273.15)K
1
(6
TC
Carnot
TH
Carnot 0.6
273.15)K
0.07
Ans.
actual
0.028
Ans.
2
3
c) The thermal efficiency is low and high fluid rates are required to generate
reasonable power. This argues for working fluids that are relatively
inexpensive. Candidates that provide reasonable pressures at the
required temperature levels include ammonia, n-butane, and propane.
297
Chapter 9 - Section A - Mathcad Solutions
9.2
TH
(
20
TC
( 20
QdotC
0.08
Cost
9.4
TC
QdotC
Wdot
kW hr
253.15 K
kJ
day
TC
TH
293.15 K
TC
273.15)
K
125000
Carnot
TH
273.15)
K
(9.3)
0.6
(9.2)
Wdot
Wdot
Cost
3.797
Carnot
0.381 kW
267.183
dollars
yr
Ans.
Basis: 1 lbm of tetrafluoroethane
The following property values are found from Table 9.1:
State 1, Sat. Liquid at TH: H1
44.943
S1
0.09142
P1
138.83
State 2, Sat. Vapor at TH: H2
116.166 S2
0.21868
P2
138.83
State 3, Wet Vapor at TC: Hliq
15.187
Hvap
104.471 P3
26.617
State 4, Wet Vapor at TC: Sliq
0.03408 Svap
0.22418 P4
26.617
(a) The pressures in (psia) appear above.
(b) Steps 3--2 and 1--4 (Fig. 8.2) are isentropic, for which S3=S2 and S1=S4.
Thus by Eq. 6.82):
x3
0.971
x4
S1 Sliq
Svap Sliq
x3 (
Hvap Hliq)
H4
Hliq
H4
42.118
Q43
59.77
S2 Sliq
Svap Sliq
x3
x4
0.302
(c) Heat addition, Step 4--3:
H3
Hliq
H3
101.888
Q43
(
H3
H4)
298
x4 (
Hvap Hliq)
(Btu/lbm)
(d) Heat rejection, Step 2--1:
W21
( H2
W14
( H4
0
H3)
W32
14.278
H1)
W14
2.825
H2)
Q43
W14 W32
(f)
(Btu/lbm)
71.223
0
W32
(e)
( H1
Q21
W43
Q21
5.219
Note that the first law is satisfied:
Q
9.7
Q21
W
Q43
W32
Q
W14
W
TC
298.15 K
TH
523.15 K
(Engine)
T'C
273.15 K
T'H
298.15 K
0
(Refrigerator)
By Eq. (5.8):
Carnot
By Eq. (9.3):
Carnot
By definition:
=
But
TH
T'H
T'C
Wengine
0.43
Carnot
10.926
=
QH
Given that:
0.6
QH
Wrefrig
kJ
sec
35
QH
7.448
Carnot
0.6
Carnot
20.689
Q'C
Q'C
Q'C
QH
Carnot
Q'C
Carnot
T'C
Wengine = Wrefrig
Whence
QH
TC
1
kJ
Ans.
sec
299
Carnot
kJ
sec
Ans.
6.556
9.8
(a) QC
kJ
sec
4
W
QC
(c)
QC
TH
TC
(
40
TC
1
5.5
TH
TC
TH
kJ
QH
W
TC
=
Ans.
2.667
W
(b) QH
1.5 kW
227.75 K
Ans.
sec
273.15)K
TH
313.15 K
Ans.
or -45.4 degC
9.9
The following vectors contain data for parts (a) through (e). Subscripts
refer to Fig. 9.1. Values of H2 and S2 for saturated vapor come from
Table 9.1.
489.67
600
479.67
0.78
500
469.67 rankine
0.77
459.67
0.76
300
449.67
T2
0.79
0.75
200
107.320
H2
0.22325
104.471
103.015
Btu
lbm
S2
Btu
lbm rankine
0.22647
539.67 rankine
S'3 = S2
0.22418
0.22525
101.542
T4
400
0.22244
105.907
QdotC
H4
37.978
Btu
lbm
(isentropic compression)
300
From Table 9.1
for sat. liquid
Btu
sec
The saturation pressure at Point 4 from Table 9.1 is 101.37(psia). For
isentropic compression, from Point 2 to Point 3', we must read values for
the enthalpy at Point 3' from Fig. G.2 at this pressure and at the entropy
values S2. This cannot be done with much accuracy. The most
satisfactory procedure is probably to read an enthalpy at S=0.22 (H=114)
and at S=0.24 (H=126) and interpolate linearly for intermediate values of
H. This leads to the following values (rounded to 1 decimal):
115.5
116.0
Btu
lbm
H23
H'3
H2
117.2
117.9
H3
H2
H1
116.5
H'3
H23
H4
88.337
kJ
kg
H23
273.711
30.098
H1
24.084
276.438
36.337
kJ
kg
H3
279.336
43.414
283.026
50.732
kJ
kg
286.918
8.653
mdot
7.361
QdotC
H2
mdot
H1
6.016
lbm
sec
Ans.
4.613
3.146
689.6
595.2
QdotH
mdot H4
H3
QdotH
Btu
sec
494
Ans.
386.1
268.6
94.5
100.5
Wdot
mdot H23
Wdot
99.2
90.8
72.4
301
kW
Ans.
6.697
QdotC
5.25
Wdot
4.256
Ans.
3.485
2.914
TC
T2
TH
T4
9.793
7.995
TC
Carnot
TH
Carnot
TC
Ans.
6.71
5.746
4.996
9.10
Subscripts in the following refer to Fig. 9.1. All property values come from
Tables F.1 and F.2.
T2
(
4
QdotC
H4
273.15)K
1200
142.4
kJ
sec
kJ
kg
T4
(
34
H2
2508.9
S'2 = S2
273.15)K
kJ
kg
0.76
S2
9.0526
kJ
kg K
(isentropic compression)
The saturation pressure at Point 4 from Table F.1 is 5.318 kPa. We must
find in Table F.2 the enthalpy (Point 3') at this pressure and at the
entropy S2. This requires double interpolation. The pressure lies
between entries for pressures of 1 and 10 kPa, and linear interpolation
with P is unsatisfactory. Steam is here very nearly an ideal gas, for
which the entropy is linear in the logarithm of P, and interpolation must
be in accord with this relation. The enthalpy, on the other hand, changes
very little with P and can be interpolated linearly. Linear interpolation
with temperture is satisfactory in either case.
The result of interpolation is
H'3
2814.7
kJ
kg
H23
H23
H'3
H2
402.368
302
kJ
kg
H1
H4
H3
H2
H23
H3
QdotC
mdot
H2
mdot
H1
QdotH
mdot H4
Wdot
mdot H23
QdotC
H3
0.507
9.11
T2
kg
Ans.
sec
kJ
Ans.
sec
204 kW
Ans.
T2
T4
kg
1404
QdotH
Wdot
10
Ans.
5.881
Wdot
Carnot
3 kJ
2.911
Carnot
9.238
Ans.
Parts (a) & (b): subscripts refer to Fig. 9.1
At the conditions of Point 2 [t = -15 degF and
P = 14.667(psia)] for sat. liquid and sat. vapor from Table 9.1:
Hliq
7.505
Btu
lbm
Hvap
303
100.799
Btu
lbm
H2
Hvap
Sliq
Btu
lbm rankine
0.01733
Svap
0.22714
Btu
lbm rankine
For sat. liquid at Point 4 (80 degF):
H4
37.978
Btu
S4
lbm
(a) Isenthalpic expansion:
QdotC
Btu
sec
5
H1
0.07892
H2
mdot
H1
(b) Isentropic expansion:
S1
x1
Sliq
Svap
H1
Sliq
QdotC
mdot
H2
S1
Hliq
x1 Hvap
mdot
H1
lbm rankine
H4
QdotC
mdot
Btu
0.0759
0.0796
sec
Ans.
S4
H1
Hliq
lbm
lbm
34.892
BTU
lbm
Ans.
sec
(c) The sat. vapor from the evaporator is superheated in the heat
exchanger to 70 degF at a pressure of 14.667(psia). Property values
for this state are read (with considerable uncertainty) from Fig. G.2:
H2A
mdot
117.5
Btu
lbm
H4
0.262
mdot
QdotC
H2A
Btu
S2A
0.0629
lbm rankine
lbm
sec
Ans.
(d) For isentropic compression of the sat. vapor at Point 2,
S3
Svap
and from Fig. G.2 at this entropy and P=101.37(psia)
H3
118.3
Btu
lbm
Eq. (9.4) may now be
applied to the two cases:
In the first case H1 has the value of H4:
H2
a
H4
H3
H2
a
304
3.5896
Ans.
In the second case H1 has its last calculated value [Part (b)]:
H2
H1
H3
b
H2
Ans.
3.7659
b
In Part (c), compression is at constant entropy of 0.262 to the
final pressure. Again from Fig. G.2:
H3
138
Btu
H3
Wdot
lbm
Wdot
1.289
QdotC
c
9.12
c
Wdot
H2A mdot
(Last calculated
value of mdot)
BTU
sec
Ans.
3.8791
Subscripts: see figure of the preceding problem.
At the conditions of Point 2 [sat. vapor, t = 20 degF and P = 33.110(psia)]
from Table 9.1:
H2
105.907
Btu
lbm
S2
0.22325
Btu
lbm rankine
At Point 2A we have a superheated vapor at the same pressure and at
70 degF. From Fig. G.2:
H2A
116
Btu
lbm
S2A
Btu
lbm rankine
0.2435
For sat. liquid at Point 4 (80 degF):
H4
37.978
Btu
lbm
S4
0.07892
Btu
lbm R
Energy balance, heat exchanger:
H1
QdotC
H4
H2A
2000
H2
Btu
sec
H1
mdot
305
27.885
BTU
lbm
QdotC
H2
H1
mdot
25.634
lbm
sec
For compression at constant entropy of 0.2435 to the final pressure of
101.37(psia), by Fig. G.2:
H'3
Wdot
mdot
127
Btu
lbm
0.75
mdot Hcomp
25.634
lbm
sec
Hcomp
Wdot
H'3
Hcomp
396.66 kW
H2A
14.667
Btu
lbm
Ans.
If the heat exchanger is omitted, then H1 = H4.
Points 2A & 2 coincide, and compression is at a constant entropy of
0.22325 to P = 101.37(psia).
mdot
Wdot
mdot
9.13
QdotC
H2
H4
H'3
116
Btu
lbm
mdot Hcomp
29.443
lbm
sec
Hcomp
Wdot
H'3
Hcomp
418.032 kW
H2
13.457
Btu
lbm
Ans.
Subscripts refer to Fig. 9.1.
At Point 2 [sat. vapor @ 10 degF] from Table 9.1:
H2
104.471
Btu
lbm
S2
0.22418
Btu
lbm R
S'3
S2
H values for sat. liquid at Point 4 come from Table 9.1 and H values
for Point 3` come from Fig. G.2. The vectors following give values for
condensation temperatures of 60, 80, & 100 degF at pressures of
72.087, 101.37, & 138.83(psia) respectively.
31.239
H4
37.978
44.943
113.3
Btu
lbm
H'3
116.5
119.3
306
Btu
lbm
H1
H4
(a)
By Eq. (9.4):
8.294
H2
H'3
(b)
H1
H2
H
Ans.
5.528
4.014
H'3
H2
0.75
Since
H = H3
H2
Eq. (9.4) now becomes
H2
6.221
H1
9.14
Ans.
4.146
H
3.011
WINTER
TH
Wdot
293.15
1.5
QdotH = 0.75 TH
TC
TH TC
Wdot
=
QdotH
TH
TC
250
(Guess)
Given
TH TC
Wdot
=
TH
0.75 TH TC
TC
Find TC
TC
268.94 K
Ans.
Minimum t = -4.21 degC
307
SUMMER
TC
298.15
QdotC
0.75 TH
TC
TH TC
Wdot
=
TC
QdotC
TH
(Guess)
300
Given
Wdot
0.75 TH
TC
TH
322.57 K
TH
TC
TC
Find TH
TH
=
Ans.
Maximum t = 49.42 degC
9.15 and 9.16 Data in the following vectors for Pbs. 9.15 and 9.16 come from
Perry's Handbook, 7th ed.
H4
1033.5
785.3
kJ
kg
By Eq. (9.8):
9.17
H9
z
284.7
kJ
kg
H4
H15
H9
1186.7
H15
H15
z
1056.4
0.17
kJ
kg
Ans.
0.351
Advertized combination unit:
TH
(
150
TH
50000
Btu
hr
WCarnot
(
30
TC
609.67 rankine
QC
TC
459.67)rankine
489.67 rankine
QC
308
TH
TC
TC
459.67)rankine
WCarnot
12253
Btu
hr
WI
WI
1.5 WCarnot
18380
Btu
hr
This is the TOTAL power requirement for the advertized combination unit.
The amount of heat rejected at the higher temperature of
150 degF is
QH
WI
QH
QC
68380
Btu
hr
For the conventional water heater, this amount of energy must be supplied
by resistance heating, which requires power in this amount.
For the conventional cooling unit,
TH
( 120
459.67) rankine
WCarnot
QC
Work
TH
TC
WCarnot
TC
Work
1.5 WCarnot
9190
Btu
hr
Btu
hr
13785
The total power required is
WII
9.18
QH
TC
210
WII
Work
T'H
82165
T'C
260
Btu
hr
NO CONTEST
TH
255
305
By Eq. (9.3):
TC
TH
WCarnot =
TC
QC
I
0.65
TC
T'H
WI =
II
TC
QC
WII =
0.65
9.19
TH
T'C
QC
II
I
Define r as the ratio of the
actual work, WI + WII, to the r
Carnot work:
T'C
1
1
I
r
1.477
Ans.
II
This problem is just a reworking of Example 9.3 with different values of x.
It could be useful as a group project.
309
9.22 TH
TC
290K
Ws
250K
TC
Carnot
TH
4.063
65% Carnot
6.25
Carnot
TC
0.40kW
Ans.
QC
9.23
QC
Ws
3
-3
Q
10 kgm sec H
1.625
2
Ws
QC
QH
2.025 kW
Follow the notation from Fig. 9.1
With air at 20 C and the specification of a minimum approach T = 10 C:
T1
(
10
T4
273.15)
K
(
30
T2
273.15)
K
T1
Calculate the high and low operating pressures using the given vapor
pressure equation
Guess:
PL
PH
1bar
2bar
PL
PL
Given ln
4104.67
= 45.327
bar
T1
5.146 ln
T1
615.0
K
T1
K
PL
PL
Find PL
bar
2
K
6.196 bar
PH
PH
Given ln
4104.67
= 45.327
bar
T4
5.146 ln
T4
615.0
K
T4
K
PH
Find PH
PH
bar
2
K
11.703 bar
Calculate the heat load
ndottoluene
50
kmol
hr
T1
(
100
273.15)
K
T2
(
20
273.15)
K
Using values from Table C.3
QdotC
QdotC
ndottoluene R ICPH T1 T2 15.133 6.79 10
177.536 kW
310
3
16.35 10
6
0
Since the throttling process is adiabatic:
But:
Hliq4 = Hliq1
H4 = H1
x1 Hlv1 so:
Hliq4
Hliq1 = x1 Hlv
T4
and:
Hliq4
Hliq1 = Vliq P4
P1
Cpliq ( T) dT
T1
Estimate Vliq using the Rackett Eqn.
Pc
112.80bar
Vc
cm
72.5
mol
Tn
239.7K
273.15)K
Tc
Tr
0.723
Tc
0.253
405.7K
3
Zc
0.242
( 20
Tr
Hlvn
23.34
kJ
mol
2
Vliq
1 Tr
V c Zc
3
7
cm
27.112
mol
Vliq
Estimate Hlv at 10C using Watson correlation
T1
Tn
Tr1
Trn 0.591
Trn
Tc
Tc
Hlv
Hliq41
Hlvn
Vliq PH
Hliq41
1.621
1
Tr1
1
0.698
0.38
Trn
PL
Tr1
Hlv
20.798
R ICPH T1 T4 22.626
kJ
mol
100.75 10
Hliq41
x1
kJ
mol
x1
Hlv
3
192.71 10
0.078
For the evaporator
H12 = H2
H1 = H1vap
H12
x1
ndot
1
QdotC
H12
Hlv
H1liq
x1 Hlv = 1
H12
ndot
311
19.177
9.258
x1
Hlv
kJ
mol
mol
sec
Ans.
6
0
Chapter 10 - Section A - Mathcad Solutions
10.1
Benzene:
A1 := 13.7819
B1 := 2726.81
C1 := 217.572
Toluene:
A2 := 13.9320
B2 := 3056.96
C2 := 217.625
B1
A1
Psat1 ( T) := e
T
degC
(a) Given: x1 := 0.33
Given
+C1
A2
kPa
Psat2 ( T) := e
Guess:
T := 100 degC
B2
T
degC
y1 := 0.5
+C2
kPa
P := 100 kPa
x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P
x1 Psat1 ( T) = y1 P
y1
:= Find ( y1 , P)
P
(b) Given: y1 := 0.33
Given
Ans.
y1 = 0.545
P = 109.303 kPa
Guess:
T := 100 degC
Ans.
x1 := 0.33 P := 100 kPa
x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P
x1 Psat1 ( T) = y1 P
x1
:= Find ( x1 , P)
P
(c) Given: x1 := 0.33
Given
x1 = 0.169
P := 120 kPa
Ans.
Guess:
P = 92.156 kPa
y1 := 0.5
Ans.
T := 100 degC
x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P
x1 Psat1 ( T) = y1 P
y1
:= Find ( y1 , T)
T
y1 = 0.542
312
Ans.
T = 103.307 degC Ans.
(d) Given: y1 := 0.33
P := 120 kPa
Guess: x1 := 0.33
T := 100 degC
x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P
Given
x1 Psat1 ( T) = y1 P
x1
:= Find ( x1 , T)
T
(e) Given:
Given
x1 = 0.173
Ans.
T = 109.131 degC Ans.
T := 105 degC P := 120 kPa Guess:
x1 := 0.33
y1 := 0.5
x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P
x1 Psat1 ( T) = y1 P
x1
:= Find ( x1 , y1)
y1
x1 = 0.282
z1 := 0.33
x1 = 0.282
y1 = 0.484
Guess:
L := 0.5
V := 0.5
Given
z1 = L x1 + V y1
(f)
Ans.
y1 = 0.484
Ans.
L+V = 1
L
:= Find ( L , V)
V
(g)
Vapor Fraction:
V = 0.238
Ans.
Liquid Fraction:
L = 0.762
Ans.
Benzene and toluene are both non-polar and similar in shape and
size. Therefore one would expect little chemical interaction
between the components. The temperature is high enough and
pressure low enough to expect ideal behavior.
313
10.2
Pressures in kPa; temperatures in degC
(a) Antoine coefficients: Benzene=1; Ethylbenzene=2
A1 := 13.7819
B1 := 2726.81
C1 := 217.572
A2 := 13.9726
B2 := 3259.93
C2 := 212.300
Psat1 ( T) := exp A1
B1
T + C1
B2
Psat2 ( T) := exp A2
T + C2
P-x-y diagram:
T := 90
P ( x1) := x1 Psat1 ( T) + ( 1 x1) Psat2 ( T)
T-x-y diagram:
y1 ( x1) :=
x1 Psat1 ( T)
P ( x1 )
P' := 90
Guess t for root function: t := 90
T ( x1) := root x1 Psat1 ( t) + ( 1 x1) Psat2 ( t) P' , t
y'1 ( x1) :=
x1 Psat1 ( T ( x1) )
x1 Psat1 ( T ( x1) ) + ( 1 x1) Psat2 ( T ( x1) )
x1 := 0 , 0.05 .. 1.0
150
140
130
P ( x1 )
120
100
T ( x1 ) 110
100
T ( x1 )
90
P ( x1 )
50
80
70
0
0
0.5
x1 , y1 ( x1 )
60
1
314
0
0.5
x1 , y'1 ( x1 )
1
(b) Antoine coefficients: 1-Chlorobutane=1; Chlorobenzene=2
A1 := 13.7965
B1 := 2723.73
C1 := 218.265
A2 := 13.8635
B2 := 3174.78
C2 := 211.700
Psat1 ( T) := exp A1
P-x-y diagram:
B1
T + C1
T := 90
P ( x1) := x1 Psat1 ( T) + ( 1 x1) Psat2 ( T)
T-x-y diagram:
B2
T + C2
Psat2 ( T) := exp A2
y1 ( x1) :=
x1 Psat1 ( T)
P ( x1 )
P' := 90
Guess t for root function: t := 90
T ( x1) := root x1 Psat1 ( t) + ( 1 x1) Psat2 ( t) P' , t
y'1 ( x1) :=
x1 Psat1 ( T ( x1) )
x1 Psat1 ( T ( x1) ) + ( 1 x1) Psat2 ( T ( x1) )
x1 := 0 , 0.05 .. 1.0
130
160
122.5
115
113.33
107.5
T ( x1 )
100
T ( x1 )
P ( x1 )
P ( x1 )
92.5
66.67
85
77.5
20
0
0.5
x1 , y1 ( x1 )
70
1
315
0
0.5
x1 , y'1 ( x1 )
1
10.3
Pressures in kPa; temperatures in degC
(a) Antoine coefficinets: n-Pentane=1; n-Heptane=2
A1 := 13.7667
B1 := 2451.88
C1 := 232.014
A2 := 13.8622
B2 := 2911.26
C2 := 216.432
B1
T + C1
Psat1 ( T) := exp A1
Psat2 ( T) := exp A2
Psat1 ( T) + Psat2 ( T)
2
P :=
T := 55
B2
T + C2
P = 104.349
Since for Raoult's law P is linear in x, at the specified P, x1 must be 0.5:
x1 := 0.5
y1 :=
x1 Psat1 ( T)
y1 = 0.89
P
For a given pressure, z1 ranges from the liquid composition at the bubble
point to the vapor composition at the dew point. Material balance:
z1 = x1 ( 1 V) + y1 V
V ( z1) :=
z1 := x1 , x1 + 0.01 .. y1
z1 x1
y1 x1
V is obviously linear in z1:
1
x1
y1
V ( z1 ) 0.5
0
0.45
0.5
0.55
0.6
0.65
0.7
z1
316
0.75
0.8
0.85
(b) At fixed T and z1, calculate x1, y1 and P as functions of fraction vapor (V).
z1 := 0.5
Guess:
x := 0.5
Psat1 ( T) + Psat2 ( T)
2
p :=
y := 0.5
Three equations relate x1, y1, & P for given V:
Given
p = x Psat1 ( T) + ( 1 x) Psat2 ( T)
y p = x Psat1 ( T)
z1 = ( 1 V) x + V y
f ( V) := Find ( x , y , p)
x1 ( V) := f ( V) 1
y1 ( V) := f ( V) 2
Plot P, x1 and y1 vs. vapor fraction (V)
P ( V) := f ( V) 3
V := 0 , 0.1 .. 1.0
150
1
100
x1 ( V)
P ( V)
50
0
0.5
y1 ( V)
0
0.5
0
1
0
0.5
1
V
V
10.4
Each part of this problem is exactly like Problem 10.3, and is worked in
exactly the same way. All that is involved is a change of numbers. In
fact, the Mathcad solution for Problem 10.3 can be converted into the
solution for any part of this problem simply by changing one number, the
temperature.
10.7
Benzene:
A1 := 13.7819
B1 := 2726.81
C1 := 217.572
Ethylbenzene A2 := 13.9726
B2 := 3259.93
C2 := 212.300
A1
Psat1 ( T) := e
B1
T
degC
+C1
A2
kPa
Psat2 ( T) := e
317
B2
T
degC
+C2
kPa
y1 := 0.70 Guess: T := 116 degC P := 132 kPa
(a) Given: x1 := 0.35
Given
x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P
x1 Psat1 ( T) = y1 P
T
:= Find ( T , P)
P
Ans.
T = 134.1 degC
P = 207.46 kPa
Ans.
For parts (b), (c) and (d) use the same structure. Set the defined variables
and change the variables in the Find statement at the end of the solve
block.
(b) T = 111.88 deg_C
P = 118.72 kPa
(c)
T = 91.44 deg_C
P = 66.38 kPa
(d) T = 72.43 deg_C
P = 36.02 kPa
To calculate the relative amounts of liquid and vapor phases, one must
know the composition of the feed.
10.8 To increase the relative amount of benzene in the vapor phase, the
temperature and pressure of the process must be lowered. For parts (c)
and (d), the process must be operated under vacuum conditions. The
temperatures are well within the bounds of typical steam and cooling water
temperatures.
10.9
13.7819
(1) = benzene
A := 13.9320
(2) = toluene
(3) = ethylbenzene
13.9726
(a) n := rows ( A)
Ai
Psat ( i , T) := e
i := 1 .. n
2726.81
B := 3056.96
3259.93
T := 110 degC
217.572
C := 217.625
212.300
P := 90 kPa
zi :=
1
n
Bi
T
degC
+Ci
kPa
ki :=
318
Psat ( i , T)
P
Guess:
V := 0.5
n
zi ki
1 + V ( ki 1)
V := Find ( V)
V = 0.836
Given
i=1
yi :=
xi :=
zi ki
1 + V ( ki 1)
=1
Ans.
Psat ( i , T)
T = 110 deg_C
Ans.
V = 0.575
0.441
y = 0.333
0.226
V = 0.352
0.238
x = 0.345
0.417
0.508
y = 0.312
0.18
V = 0.146
0.293
x = 0.342
0.366
0.572
y = 0.284
0.144
P = 110 kPa
(d) T = 110 deg_C
Ans.
0.188
x = 0.334
0.478
P = 100 kPa
(c)
0.371
y = 0.339
0.29
0.142
x = 0.306
0.552
Eq. (10.16)
yi P
(b) T = 110 deg_C
Eq. (10.17)
P = 120 kPa
10.10 As the pressure increases, the fraction of vapor phase formed (V)
decreases, the mole fraction of benzene in both phases increases and the
the mole fraction of ethylbenzene in both phases decreases.
319
14.3145
2756.22
228.060
10.11 (a) (1) = acetone
A :=
B :=
C :=
(2) = acetonitrile
14.8950
3413.10
250.523
n := rows ( A)
i := 1 .. n
z1 := 0.75
T := ( 340 273.15) degC
P := 115 kPa
z2 := 1 z1
Bi
Ai
degC
Psat ( i , T) := e
Guess:
T
+Ci
kPa
ki :=
Psat ( i , T)
P
V := 0.5
n
zi ki
Given
i=1
1 + V ( ki 1)
V := Find ( V)
=1
Ans.
V = 0.656
Eq. (10.16)
yi :=
xi :=
r :=
Eq. (10.17)
zi ki
yi P
Psat ( i , T)
y1 V
z1
y1 = 0.805
Ans.
x1 = 0.644
Ans.
r = 0.705
1 + V ( ki 1)
Ans.
(b)
x1 = 0.285
y1 = 0.678
V = 0.547
r = 0.741
(c)
x1 = 0.183
y1 = 0.320
V = 0.487
r = 0.624
(d)
x1 = 0.340
y1 = 0.682
V = 0.469
r = 0.639
320
10.13 H1 := 200 bar
Psat2 := 0.10 bar
P := 1 bar
Assume at 1 bar that the vapor is an ideal gas. The vapor-phase fugacities
are then equal to the partial presures. Assume the Lewis/Randall rule
applies to concentrated species 2 and that Henry's law applies to dilute
species 1. Then:
y1 P = H1 x1
y2 P = x2 Psat2
P = y1 P + y2 P
P = H1 x1 + ( 1 x1) Psat2
x1 + x2 = 1
Solve for x1 and y1:
x1 :=
P Psat2
y1 :=
H1 Psat2
3
x1 = 4.502 × 10
H1 x1
P
Ans.
y1 = 0.9
10.16 Pressures in kPa
Psat1 := 32.27
Psat2 := 73.14
(
γ 1 ( x1 , x2) := exp A x2
2
)
A := 0.67
z1 := 0.65
(
γ 2 ( x1 , x2) := exp A x1
2
)
P ( x1 , x2) := x1 γ 1 ( x1 , x2) Psat1 + x2 γ 2 ( x1 , x2) Psat2
(a) BUBL P calculation:
x1 := z1
x2 := 1 x1
Pbubl := P ( x1 , x2)
Pbubl = 56.745
DEW P calculation:
y1 := z1
Guess:
P' :=
Given
x1 := 0.5
Ans.
y2 := 1 y1
Psat1 + Psat2
2
y1 P' = x1 γ 1 ( x1 , 1 x1) Psat1
P' = x1 γ 1 ( x1 , 1 x1) Psat1 ...
+ ( 1 x1) γ 2 ( x1 , 1 x1) Psat2
x1
:= Find ( x1 , P')
Pdew
Pdew = 43.864
321
Ans.
The pressure range for two phases is from the dewpoint to the
bubblepoint:
From 43.864 to 56.745 kPa
(b)
BUBL P calculation:
y1 ( x1) :=
x1 := 0.75
x2 := 1 x1
x1 γ 1 ( x1 , 1 x1) Psat1
P ( x1 , 1 x1)
The fraction vapor, by material balance is:
V :=
z1 x1
V = 0.379
y1 ( x1) x1
P ( x1 , x2) = 51.892
Ans.
(c) See Example 10.3(e).
γ 1 ( 0 , 1) Psat1
α 12.0 :=
α 12.1 :=
Psat2
α 12.0 = 0.862
Psat1
γ 2 ( 1 , 0) Psat2
α 12.1 = 0.226
Since alpha does not pass through 1.0 for 0<x1<1, there is no
azeotrope.
10.17 Psat1 := 79.8
Psat2 := 40.5
(
γ 1 ( x1 , x2) := exp A x2
2
)
A := 0.95
(
γ 2 ( x1 , x2) := exp A x1
P ( x1 , x2) := x1 γ 1 ( x1 , x2) Psat1 + x2 γ 2 ( x1 , x2) Psat2
y1 ( x1) :=
x1 γ 1 ( x1 , 1 x1) Psat1
P ( x1 , 1 x1)
(a) BUBL P calculation:
Pbubl := P ( x1 , x2)
x1 := 0.05
x2 := 1 x1
Pbubl = 47.971
Ans.
y1 ( x1) = 0.196
(b) DEW P calculation:
Guess:
y1 := 0.05
x1 := 0.1
P' :=
322
y2 := 1 y1
Psat1 + Psat2
2
2
)
y1 P' = x1 γ 1 ( x1 , 1 x1) Psat1
Given
P' = x1 γ 1 ( x1 , 1 x1) Psat1 ...
+ ( 1 x1) γ 2 ( x1 , 1 x1) Psat2
x1
:= Find ( x1 , P')
Pdew
Pdew = 42.191
Ans.
x1 = 0.0104
(c) Azeotrope Calculation:
Guess:
x1 := 0.8
y1 := x1
x1 γ 1 ( x1 , 1 x1) Psat1
Given
y1 =
P
P :=
Psat1 + Psat2
x1 0
2
x1 1
x1 = y1
P = x1 γ 1 ( x1 , 1 x1) Psat1 + ( 1 x1) γ 2 ( x1 , 1 x1) Psat2
xaz1
yaz := Find ( x , y , P)
11
1
Paz
10.18 Psat1 := 75.20 kPa
y1 = x1
γ2
γ1
2
=
Psat1
and
γi =
lnγ2 = A x1
Psat1
γ2
ln
γ1
A :=
Psat2
2
A = 2.0998
2
x2 x1
For
x1 := 0.6
x2 := 1 x1
323
x2 := 1 x1
(
2
2
= A x1 x2
ln
Whence
P
Psati
x1 := 0.294
Psat2
2
lnγ1 = A x2
Ans.
Psat2 := 31.66 kPa
At the azeotrope:
Therefore
xaz1
0.857
yaz = 0.857
1
Paz
81.366
)
(
γ 1 := exp A x2
)
(
2
x1 γ 1 Psat1
y1 :=
)
2
γ 2 := exp A x1
P = 90.104 kPa
P
10.19 Pressures in bars:
P := x1 γ 1 Psat1 + x2 γ 2 Psat2
Ans.
y1 = 0.701
Psat1 := 1.24
x1 := 0.65
A := 1.8
(
γ 1 := exp A x2
Psat2 := 0.89
x2 := 1 x1
)
(
2
γ 2 := exp A x1
P := x1 γ 1 Psat1 + x2 γ 2 Psat2
y1 = 0.6013
y1 :=
)
2
x1 γ 1 Psat1
P
Answer to Part (b)
P = 1.671
By a material balance,
V=
z1 x1
For
y1 x1
(c)
0V1
Ans. (a)
Azeotrope calculation:
Guess:
x1 := 0.6
y1 := x1
2
γ 1 ( x1) := exp A ( 1 x1)
Given
y1 =
0.6013 z1 0.65
P :=
Psat1 + Psat2
2
(
γ 2 ( x1) := exp A x1
)
2
P = x1 γ 1 ( x1) Psat1 + ( 1 x1) γ 2 ( x1) Psat2
x1 γ 1 ( x1) Psat1
P
x1 0
x1
y1 := Find ( x1 , y1 , P)
P
x1 1
x1 = y1
x1 0.592
y1 = 0.592
P 1.673
324
Ans.
10.20 Antoine coefficients:
P in kPa; T in degC
Acetone(1):
A1 := 14.3145
B1 := 2756.22
C1 := 228.060
Methanol(2):
A2 := 16.5785
B2 := 3638.27
C2 := 239.500
P1sat ( T) := exp A1
A := 0.64
x1 := 0.175
(
B1
T + C1
γ 1 ( x1 , x2) := exp A x2
2
z1 := 0.25
)
p := 100 (kPa)
(
γ 2 ( x1 , x2) := exp A x1
P ( x1 , T) := x1 γ 1 ( x1 , 1 x1) P1sat ( T) ...
+ ( 1 x1) γ 2 ( x1 , 1 x1) P2sat ( T)
y1 ( x1 , T) :=
Guesses:
x1 γ 1 ( x1 , 1 x1) P1sat ( T)
P ( x1 , T )
V := 0.5
L := 0.5
F := 1
T := 100
Given
F= L+V
z1 F = x1 L + y1 ( x1 , T) V
L
V := Find ( L , V , T)
T
T = 59.531
(degC)
B2
T + C2
P2sat ( T) := exp A2
p = P ( x1 , T)
L 0.431
V = 0.569
T
59.531
y1 ( x1 , T) = 0.307 Ans.
325
2
)
10.22 x1 := 0.002
y1 := 0.95
B1 := 2572.0
A1 := 10.08
A1
Psat1 ( T) := e
x2 := 1 x1
Given
P :=
Guess:
A2 := 11.63
B1
Psat2 ( T) := e
0.93 x2
y2 := 1 y1
Psat2 ( T)
=
x1 γ 1 Psat1 ( T)
B2 := 6254.0
A2
T
K bar
Psat1 ( T)
y1
T := 300 K
2
γ 1 := e
x2 γ 2 y1
x1 γ 1 y2
T := Find ( T)
P = 0.137 bar
326
Ans.
B2
T
K bar
0.93 x1
2
γ 2 := e
T = 376.453 K
Ans.
Problems 10.25 to 10.34 have been solved using MS-EXCEL 2000
We give the resulting spreadsheets.
Problem 10.25
a) BUBL P
T=-60 F (-51.11 C)
Component
methane
ethylene
ethane
xi
0.100
0.500
0.400
b) DEW P
T=-60 F (-51.11 C)
Component
methane
ethylene
ethane
yi
0.500
0.250
0.250
c) BUBL T
P=250 psia (17.24 bar)
Component
methane
ethylene
ethane
T=-50 F
T=-60 F
T=-57 F (-49.44 C)
ANSWER
xi
Ki
yi=Ki*xi
Ki
yi=Ki*xi
Ki
yi=Ki*xi
0.120 4.900
0.588
4.600
0.552
4.700
0.564
0.400 0.680
0.272
0.570
0.228
0.615
0.246
0.480 0.450
0.216
0.380
0.182
0.405
0.194
SUM = 1.076
SUM = 0.962
SUM = 1.004 close enough
d) DEW T
P=250 psia (17.24 bar)
Component
methane
ethylene
ethane
yi
0.430
0.360
0.210
P=200 psia
P=250 psia
P=215 psia (14.824 bar) ANSWER
Ki
yi=Ki*xi
Ki
yi=Ki*xi
Ki
yi=Ki*xi
5.600
0.560
4.600
0.460
5.150
0.515
0.700
0.350
0.575
0.288
0.650
0.325
0.445
0.178
0.380
0.152
0.420
0.168
SUM = 1.088
SUM = 0.900
SUM = 1.008 close enough
P=190 psia
P=200 psia (13.79 bar)
ANSWER
Ki
xi=yi/Ki
Ki
xi=yi/Ki
5.900
0.085
5.600
0.089
0.730
0.342
0.700
0.357
0.460
0.543
0.445
0.562
SUM = 0.971
SUM = 1.008
close enough
T=-40 F
T = -50 F
T = -45 F (-27.33 C)
ANSWER
Ki
xi=yi/Ki
Ki
xi=yi/Ki
Ki
xi=yi/Ki
5.200
0.083
4.900
0.088
5.050
0.085
0.800
0.450
0.680
0.529
0.740
0.486
0.520
0.404
0.450
0.467
0.485
0.433
SUM = 0.937
SUM = 1.084
SUM = 1.005 close enough
327
Problem 10.26
a) BUBL P
Component
ethane
propane
isobutane
isopentane
b) DEW P
Component
ethane
propane
isobutane
isopentane
c) BUBL T
Component
ethane
propane
isobutane
isopentane
d) DEW T
Component
ethane
propane
isobutane
isopentane
T=60 C (140 F)
xi
0.10
0.20
0.30
0.40
P=200 psia
P=50 psia
P=80 psia (5.516 bar)
ANSWER
Ki
yi=Ki*xi
Ki
yi=Ki*xi
Ki
yi=Ki*xi
2.015
0.202
6.800
0.680
4.950
0.495
0.620
0.124
2.050
0.410
1.475
0.295
0.255
0.077
0.780
0.234
0.560
0.168
0.071
0.028
0.205
0.082
0.12
0.048
SUM = 0.430
SUM = 1.406
SUM = 1.006 close enough
T=60 C (140 F)
yi
0.48
0.25
0.15
0.12
P=80 psia
P=50 psia
P=52 psia (3.585 bar)
ANSWER
Ki
xi=yi/Ki
Ki
xi=yi/Ki
Ki
xi=yi/Ki
4.950
0.097
6.800
0.071
6.600
0.073
1.475
0.169
2.050
0.122
2.000
0.125
0.560
0.268
0.780
0.192
0.760
0.197
0.12
1.000
0.205
0.585
0.195
0.615
SUM = 1.534
SUM = 0.970
SUM = 1.010 close enough
P=15 bar (217.56 psia)
xi
0.14
0.13
0.25
0.48
T=220 F
T=150 F
T=145 F (62.78 C)
ANSWER
Ki
yi=Ki*xi
Ki
yi=Ki*xi
Ki
yi=Ki*xi
5.350
0.749
3.800
0.532
3.700
0.518
2.500
0.325
1.525
0.198
1.475
0.192
1.475
0.369
0.760
0.190
0.720
0.180
0.57
0.274
0.27
0.130
0.25
0.120
SUM = 1.716
SUM = 1.050
SUM = 1.010 close enough
P=15 bar (217.56 psia)
yi
0.42
0.30
0.15
0.13
T=150 F
Ki
xi=yi/Ki
3.800
0.111
1.525
0.197
0.760
0.197
0.27
0.481
SUM = 0.986
T=145 F
T=148 F (64.44 C)
ANSWER
Ki
xi=yi/Ki
Ki
xi=yi/Ki
3.700
0.114
3.800
0.111
1.475
0.203
1.500
0.200
0.720
0.208
0.740
0.203
0.25
0.520
0.26
0.500
SUM = 1.045
SUM = 1.013 close enough
328
Problem 10.27
FLASH
Component
methane
ethane
propane
n-butane
T=80 F (14.81 C)
zi
0.50
0.10
0.20
0.20
V= 0.855
Ki
yi
10.000
0.575
2.075
0.108
0.680
0.187
0.21
0.129
SUM = 1.000
P=250 psia (17.24 bar)
Fraction condensed
L= 0.145
ANSWER
xi=yi/Ki
0.058
0.052
0.275
0.616
SUM = 1.001
Problem 10.28
First calculate equilibrium composition
T=95 C (203 F)
Component
n-butane
n-hexane
xi
0.25
0.75
P=80 psia
P=65 psia
P=69 psia (4.83 bar)
ANSWER
Ki
yi=Ki*xi
Ki
yi=Ki*xi
Ki
yi=Ki*xi
2.25
0.5625
2.7
0.675
2.6
0.633
0.45
0.3375
0.51
0.3825
0.49
0.3675
SUM = 0.9000 SUM = 1.0575 SUM = 1.0005 Close enough
Now calculate liquid fraction from mole balances
ANSWER
z1=
x1=
y1=
L=
0.5
0.25
0.633
0.347
Problem 10.29
FLASH
Component
n-pentane
n-hexane
n-heptane
P = 2.00 atm (29.39 psia)
T = 200 F (93.3 C)
zi
0.25
0.45
0.30
V= 0.266
Ki
yi
2.150
0.412
0.960
0.437
0.430
0.152
SUM = 1.000
Fraction condensed
L= 0.73
ANSWER
xi=yi/Ki
0.191
0.455
0.354
SUM = 1.000
329
Problem 10.30
FLASH
Component
ethane
propane
n-butane
T=40 C (104 F)
V= 0.60
P=110 psia
zi
Ki
yi
0.15
5.400
0.223
0.35
1.900
0.432
0.50
0.610
0.398
SUM = 1.053
Fraction condensed
L= 0.40
P=100 psia
xi=yi/Ki
Ki
yi
0.041
4.900
0.220
0.227
1.700
0.419
0.653
0.540
0.373
0.921
SUM = 1.012
ANSWER
P=120 psia (8.274 bar)
xi=yi/Ki
Ki
yi
xi=yi/Ki
0.045 4.660 0.219
0.047
0.246 1.620 0.413
0.255
0.691 0.525 0.367
0.699
0.982 SUM = 0.999
1.001
Fraction condensed
L= 0.80
P=40 psia
xi=yi/Ki
Ki
yi
0.004
9.300
0.035
0.039
3.000
0.107
0.508
1.150
0.558
0.472
0.810
0.370
1.023
SUM = 1.071
ANSWER
P=44 psia (3.034 bar)
xi=yi/Ki
Ki
yi
xi=yi/Ki
0.004 8.500 0.034
0.004
0.036 2.700 0.101
0.037
0.485 1.060 0.524
0.494
0.457 0.740 0.343
0.464
0.982 SUM = 1.002
1.000
Problem 10.31
FLASH
Component
ethane
propane
i-butane
n-butane
T=70 F (21.11 C)
zi
0.01
0.05
0.50
0.44
V= 0.20
P=50 psia
Ki
yi
7.400
0.032
2.400
0.094
0.925
0.470
0.660
0.312
SUM = 0.907
330
Problem 10.32
FLASH
Component
methane
ethane
propane
n-butane
Component
methane
ethane
propane
n-butane
Component
methane
ethane
propane
n-butane
T=-15 C (5 F)
zi
0.30
0.10
0.30
0.30
zi
0.30
0.10
0.30
0.30
zi
0.30
0.10
0.30
0.30
Target: y1=0.8
P=300 psia
V= 0.1855
Ki
yi
5.600
0.906
0.820
0.085
0.200
0.070
0.047
0.017
SUM = 1.079
L= 0.8145
xi=yi/Ki
0.162
0.103
0.352
0.364
SUM = 0.982
P=150 psia
V= 0.3150
Ki
yi
10.900
0.794
1.420
0.125
0.360
0.135
0.074
0.031
SUM = 1.086
L= 0.6850
xi=yi/Ki
0.073
0.088
0.376
0.424
SUM = 0.960
P=270 psia (18.616 bar)
V= 0.2535
L= 0.7465 ANSWER
Ki
yi
xi=yi/Ki
6.200
0.802
0.129
0.900
0.092
0.103
0.230
0.086
0.373
0.0495
0.020
0.395
SUM = 1.000
SUM = 1.000
331
Problem 10.33
First calculate vapor composition and temperature on top tray
BUBL T:
P=20 psia
Component
n-butane
n-pentane
xi
0.50
0.50
T=70 F
T=60 F
T=69 F (20.56 C)
ANSWER
Ki
yi=Ki*xi
Ki
yi=Ki*xi
Ki
yi=Ki*xi
1.575
0.788
1.350
0.675
1.550
0.775
0.450
0.225
0.360
0.180
0.440
0.220
SUM = 1.013
SUM = 0.855
SUM = 0.995 close enough
Using calculated vapor composition from top tray, calculate composition out of condenser
FLASH
P=20 psia (1.379 bar)
V= 0.50
Component
n-butane
n-pentane
L= 0.50
T=70 F
zi
Ki
yi
0.78
1.575
0.948
0.22
0.450
0.137
SUM = 1.085
xi=yi/Ki
0.602
0.303
0.905
T=60 F (15.56 C)
Ki
yi
1.350
0.890
0.360
0.116
SUM = 1.007
ANSWER
xi=yi/Ki
0.660
0.324
0.983
Problem 10.34
FLASH
T=40 C (104 F)
V= 0.60
Component
methane
n-butane
L= 0.40
P=350 psia
zi
Ki
yi
0.50
7.900
0.768
0.50
0.235
0.217
SUM = 0.986
P=250 psia
xi=yi/Ki
Ki
yi
0.097
11.000
0.786
0.924
0.290
0.253
1.021
SUM = 1.038
332
ANSWER
P=325 psia (7.929 bar)
xi=yi/Ki
Ki
yi
xi=yi/Ki
0.071 8.400 0.772
0.092
0.871 0.245 0.224
0.914
0.943 SUM = 0.996
1.006
close enough
Eq. (1)
10.35 a) The equation from NIST is: Mi = ki yi P
The equation for Henry's Law is:i Hi = yi P
x
Solving to eliminate P gives:
By definition:
Mi
Hi
ni
=
ns Ms
Eq. (2)
Mi
Eq. (3)
ki xi
where M is the molar mass and the
subscript s refers to the solvent.
=
Dividing by the toal number of moles gives: Mi =
Combining Eqs. (3) and (4) gives: Hi =
xi
1
xs Ms ki
If xi is small, then x s is approximately equal to 1 and: Hi =
b) For water as solvent: Ms
For CO2 in H2O:
By Eq. (5):
Hi
ki
Eq. (4)
xs Ms
gm
mol
mol
1
Ms ki
Eq. (5)
18.015
0.034
kg bar
1
Ms ki
Hi
1633 bar Ans.
The value is Table 10.1 is 1670 bar. The values agree within about 2%.
10.36
14.3145
Acetone:
Psat1 ( )
T
Psat2 ( )
T
T
degC
e
14.8950
Acetonitrile
2756.22
T
a) Find BUBL P and DEW P values
T
50degC
x1
0.5
y1
333
kPa
3413.10
degC
e
228.060
0.5
250.523
kPa
BUBLP
x1 Psat1 ( T)
1
x1 Psat2 ( T)
BUBLP
DEWP
1
DEWP
1
y1
Psat1 ( T)
y1
0.573 atm Ans.
0.478 atm Ans.
Psat2 ( T)
At T = 50 C two phases will form between P = 0.478 atm and 0.573 atm
b) Find BUBL T and DEW T values
P
0.5atm
Given
y1
0.5
x1 Psat1 ( T)
BUBLT
Given
x1
1
BUBLT
1
x1 Psat1 ( T) = y1 P
DEWT
T
50degC
x1 Psat2 ( T) = P
Find ( T)
x1
Guess:
0.5
Find x1 T
DEWT
Ans.
46.316 degC
x1 Psat2 ( T) = 1
51.238 degC
y1 P
Ans.
At P = 0.5 atm, two phases will form between T = 46.3 C and 51.2 C
10.37 Calculate x and y at T = 90 C and P = 75 kPa
13.7819
Benzene:
Psat1 ( T)
Psat2 ( T)
T
degC
e
13.9320
Toluene:
2726.81
kPa
3056.96
T
degC
e
217.572
217.625
kPa
a) Calculate the equilibrium composition of the liquid and vapor at the flash T and P
T
90degC
P
75kPa
Guess:
334
x1
0.5
y1
0.5
1
x1 Psat1 ( )= y1 P
T
Given
x1 Psat2 ( )= 1
T
y1 P
x1
x1
Find x1 y1
y1
y1
0.252
0.458
The equilibrium compositions do not agree with the measured values.
b) Assume that the measured values are correct. Since air will not dissolve
in the liquid to any significant extent, the mole fractions of toluene in the
liquid can be calculated.
x1
y1
0.1604
x2
0.2919
1
x1
x2
0.8396
Now calculate the composition of the vapor. y3 represents the mole
fraction of air in the vapor.
Guess:
y2
0.5
y3
1
y2
y1
Given
1
y3
y1
y3 P
Find y2 y3
y2
0.1
y1
y2
y3 = 1
0.608
y3
y2
x1 Psat2 ( )= 1
T
Ans.
Conclusion: An air leak is consistent with the measured compositions.
10.38 yO21
ndot
yN21
0.0387
10
kmol
T1
hr
16.3872
PsatH2O ( )
T
e
yCO21
0.7288
T2
100degC
3885.70
T
degC
230.170
kPa
335
0.0775 yH2O1
25degC
P
1atm
0.1550
Calculate the mole fraction of water in the exit gas if the exit gas is
saturated with water.
PsatH2O T2
yH2O2
yH2O2
P
0.0315
This is less than the mole fraction of water in the feed. Therefore, some
of the water will condense.
Assume that two streams leave the process: a liquid water stream at rate
ndotliq and a vapor stream at rate ndotvap. Apply mole balances around
the cooler to calculate the exit composition of the vapor phase.
ndotvap
Given
ndot
2
yO22
Guess:
0.0387
ndot = ndotliq
ndot
2
ndotliq
yN22
0.7288
yCO22
ndotvap
0.0775
Overall balance
ndot yO21 = ndotvap yO22
O2 balance
ndot yN21 = ndotvap yN22
N2 balance
ndot yCO21 = ndotvap yCO22
CO2 balance
yO22
yN22
yCO22
yH2O2 = 1
Summation equation
ndotliq
ndotvap
yO22
Find ndotliq ndotvap yO22 yN22 yCO22
yN22
yCO22
ndotliq
yO22
1.276
0.044
kmol
hr
yN22
ndotvap
0.835
8.724
yCO22
336
kmol
hr
0.089
yH2O2
0.031
Apply an energy balance around the cooler to calculate heat transfer rate.
HlvH2O
Qdot
40.66
kJ
T1
mol
T1
273.15K
T2
ndotvap yO22 R ICPH T1 T2 3.639 0.506 10
3
ndotvap yN22 R ICPH T1 T2 3.280 0.539 10
3
ndotvap yH2O2 R ICPH T1 T2 3.470 1.450 10
HlvH2O ndotliq
0.227 10
5
0 0.040 10
3
3
0
0 0.121 10
10.39 Assume the liquid is stored at the bubble point at T = 40 F
Taking values from Fig 10.14 at pressure:
xC3
0.05
KC3
3.9
xC4
0.85
KC4
0.925
xC5
0.10
KC5
0.23
The vapor mole fractions must sum to 1.
xC3 KC3
xC4 KC4
xC5 KC5
1.004
337
P
18psia
5
1.157 10
Ans.
19.895 kW
273.15K
5
0
ndotvap yCO22 R ICPH T1 T2 5.457 1.045 10
Qdot
T2
Ans.
5
10.40 H2S + 3/2 O2 -> H2O + SO2
By a stoichiometric balance, calculate the following total molar flow rates
kmol
Feed:
ndotH2S
10
Products
ndotSO2
ndotH2S
hr
3
ndotH2S
2
ndotO2
ndotH2O
ndotH2S
Exit conditions:
P
T2
1atm
3885.70
16.3872
70degC
PsatH2O ( )
T
T
degC
e
230.170
kPa
a)
Calculate the mole fraction of H2O and SO2 in the exiting vapor stream
assuming vapor is saturated with H2O
PsatH2O T2
yH2Ovap
ySO2
P
1
yH2Ovap
yH2Ovap
ySO2
Ans.
0.308
Ans.
0.692
b)
Calculate the vapor stream molar flow rate using balance on SO 2
ndotvap
ndotSO2
ndotvap
ySO2
14.461
kmol
hr
Ans.
Calculate the liquid H2O flow rate using balance on H2O
ndotH2Ovap
ndotH2Oliq
ndotH2Ovap
ndotvap yH2Ovap
ndotH2O
ndotH2Ovap
338
ndotH2Oliq
4.461
5.539
kmol
hr
kmol
Ans.
hr
10.41 NCL
a)
0.01
YH2O
yH2O
b)
P
kg
Mair
MH2O
YH2O
1
YH2O
PsatH2O ( )
T
e
10.42 ndot1
Tdp
kmol
hr
Tdp1
T
degC
e
PsatH2O Tdp1
P
1.606 kPa Ans.
ppH2O
yH2O P
230.170
Guess:
kPa
Tdp
32degF
20degC
Tdp2
Tdp
T
20degC
Find ( )
T
57.207 degF Ans.
P
10degC
y1
MH2O
230.170
18.01
kPa
y2
0.023
By a mole balances on the process
Guess: ndot2liq
mol
1atm
3885.70
16.3872
y1
T
14.004 degC
PsatH2O ( )
T
gm
Ans.
yH2O P = PsatH2O ( ) Tdp
T
50
29
3885.70
degC
Given
Tdp
0.0158
ppH2O
1atm
Mair
0.0161
yH2O
16.3872
c)
18.01
YH2O
kg
NCL
gm
mol
MH2O
ndot1 ndot2vap
ndot1
339
PsatH2O Tdp2
P
y2
gm
mol
0.012
ndot1 y1 = ndot2vap y2
Given
ndot1 = ndot2vap
ndot2liq
kmol
hr
ndot2liq
ndot2vap
49.441
mdot2liq
ndot2liq MH2O
10.43Benzene:
mdot2liq
Given
kmol
10.074
hr
kg
hr
Ans.
13.7819
B1
2726.81
C1
217.572
A2
13.6568
B2
2723.44
C2
220.618
exp A1
B1
T
degC
Guess: T
0.559
A1
Cyclohexane:
Psat2 ( T)
Overall balance
ndot2liq
Find ndot2liq ndot2vap
ndot2vap
Psat1 ( T)
H2O balance
ndot2liq
exp A2
kPa
C1
B2
T
degC
kPa
C2
66degC
Psat1 ( T) = Psat2 ( T)
T
Find ( T)
The Bancroft point for this system is:
Psat1 ( T)
39.591 kPa
Com ponent1
Benzene
2- t
Bu anol
Acet ti
oni le
r
Com ponent2
Cyclohexane
W at
er
Et
hanol
T
52.321 degC
T (C)
52.3
8 .7
7
65.8
340
P ( Pa)
k
39.6
64
.2
60
.6
Ans.
Chapter 11 - Section A - Mathcad Solutions
11.1
For an ideal gas mole fraction = volume fraction
3
CO2 (1):
x1
0.7
V1
0.7m
N2 (2):
x2
0.3
V2
0.3m
i
P
1bar
T
12
P
3
(
25
273.15)
K
Vi
i
n
n
RT
S
nR
40.342 mol
S
xi ln xi
204.885
J
K
Ans.
i
11.2
For a closed, adiabatic, fixed-volume system, U =0. Also, for an ideal
gas, U = Cv T. First calculate the equilibrium T and P.
nN2
4 mol
TN2
[75
(
273.15)K]
PN2
30 bar
nAr
2.5 mol
TAr
(
130
273.15)K
PAr
20 bar
TN2
348.15 K
TAr
403.15 K
ntotal
nN2
x1
nAr
x1
CvAr
CpAr
CvAr
CvN2
3
R
2
R
CvN2
nN2
ntotal
0.615
x2
12
nAr
ntotal
x2
0.385
T
T
5
R
2
CpN2
i
R
Find T after mixing by energy balance:
T
TN2
Given
TAr
2
nN2 CvN2 T
(guess)
TN2 = nAr CvAr TAr
341
Find T)
(
T
273.15 K
90 degC
Find P after mixing:
PN2
P
PAr
(guess)
2
Given
nN2
nAr R T
P
=
nN2 R TN2
nAr R TAr
PN2
PAr
P
Find ( P)
P 24.38 bar
Calculate entropy change by two-step path:
1) Bring individual stream to mixture T and P.
2) Then mix streams at mixture T and P.
SN2
SAr
nAr CpAr ln
T
TN2
nN2 CpN2 ln
Smix
ntotal
T
TAr
R
R ln
SN2
P
PAr
xi ln xi
J
K
SAr
P
PN2
11.806
9.547
J
K
Smix
R ln
36.006
J
K
i
S
11.3
SN2
mdotN2
molwtN2
SAr
2
Smix
kg
sec
molarflowtotal
38.27
mdotH2
28.014
molarflowN2
S
gm
mol
mdotN2
molwtN2
molarflowN2
molwtH2
J
K
0.5
Ans.
kg
sec
2.016
molarflowH2
gm
mol
i
mdotH2
molwtH2
molarflowH2 molarflowtotal
342
12
319.409
mol
sec
y1
molarflowN2
y1
molarflowtotal
S
R molarflowtotal
molarflowH2
y2
0.224
y2
molarflowtotal
S
yi ln yi
J
secK
1411
0.776
Ans.
i
11.4
T1
T2
448.15 K
P1
308.15 K
P2
3 bar
1 bar
For methane:
MCPHm
MCPSm
3
MCPH T1 T2 1.702 9.081 10
MCPS T1 T2 1.702 9.081 10
6
2.164 10
3
6
2.164 10
0.0
0.0
For ethane:
MCPHe
MCPH T1 T2 1.131 19.225 10
MCPSe
MCPS T1 T2 1.131 19.225 10
3
6
5.561 10
3
6
5.561 10
0.0
0.0
MCPHmix
0.5 MCPHm
0.5 MCPHe
MCPHmix
6.21
MCPSmix
0.5 MCPSm
0.5 MCPSe
MCPSmix
6.161
H
R MCPHmix T2
S
R MCPSmix ln
H
T1
T2
T1
R ln
P2
7228
J
mol
R 2 0.5 ln ( )
0.5
P1
The last term is the entropy change of UNmixing
J
T
300 K
S
15.813
mol K
Wideal
H
T
Wideal
S
2484
J
mol
Ans.
11.5 Basis: 1 mole entering air.
y1
0.21
y2
0.79
t
0.05
T
300 K
Assume ideal gases; then
H=0
The entropy change of mixing for ideal gases is given by the equation
following Eq. (11.26). For UNmixing of a binary mixture it becomes:
343
S
R y1 ln y1
S
y2 ln y2
By Eq. (5.27):
Wideal
By Eq. (5.28):
Work
T
4.273
J
mol K
Wideal
Wideal
25638
t
0
40
mol
Ans.
0.942
60
P
0.970
J
mol
0.985
20
10
1.000
10
11.16
3J
1.282
Work
S
0.913
80
Z
bar
ln 1
0
end
1
rows ( P)
1
0.885
100
0.869
200
0.765
300
0.762
400
0.824
500
0.910
i
2 end
Zi
Fi
1
Pi
Fi is a well behaved function; use the trapezoidal rule to integrate Eq.
(11.35) numerically.
Fi
Ai
i
Fi 1
Pi
2
Pi 1
ln i
fi
exp ln i
ln i 1
Ai
i Pi
Generalized correlation for fugacity coefficient:
For CO2:
Tc
Pc
304.2 K
T
( 150
273.15) K
Tr
T
Tc
0.224
73.83 bar
Tr
P
G ( P)
exp
Pc
Tr
B0 Tr
B1 Tr
344
fG ( P)
G ( P) P
1.391
Pi
fi
bar
bar
i
10
0.993
9.925
20
0.978
19.555
40
0.949
37.973
60
0.922
55.332
80
0.896
71.676
100
0.872
87.167
200
0.77
153.964
300
0.698
209.299
400
0.656
262.377
500
0.636
317.96
Calculate values:
400
300
fi
0.8
i
bar
G Pi
f G Pi
200
0.6
0.4
bar
0
200
400
100
0
600
0
200
400
Pi
Pi
bar
600
bar
Agreement looks good up to about 200 bar (Pr=2.7 @ Tr=1.39)
11.17 For SO2:
Tc
430.8 K
Pc
T
Tr
T
Tc
600 K
P
Tr
1.393
Pr
78.84 bar
0.245
300 bar
P
Pc
Pr
For the given conditions, we see from Fig. 3.14 that the Lee/Kesler
correlation is appropriate.
345
3.805
Data from Tables E.15 & E.16 and by Eq. (11.67):
0
0.672
1.354
1
f
GRRT
P
f
417.9 K
a) At 280 degC and 20 bar:
Tr ( )
T
T
Tc
T
ln
GRRT
217.14 bar
Tc
11.18 Isobutylene:
0.724
01
Pc
(
280
0.194
40.00 bar
P
273.15)K
Pr ( )
P
Tr ( ) 1.3236
T
Ans.
0.323
P
Pc
20 bar
Pr ( ) 0.5
P
At these conditions use the generalized virial-coeffieicnt correlation.
f
PHIB Tr ( )Pr ( )
T
P
f
P
b) At 280 degC and 100 bar:
T
(
280
Ans.
18.76 bar
P
273.15)K
100 bar
Pr ( ) 2.5
P
Tr ( ) 1.3236
T
At these conditions use the Lee/Kesler correlation, Tables E.15 & E.16 and
Eq. (11.67).
1
0.7025
01
f
0.732
0
f
1.2335
P
Ans.
73.169 bar
11.19 The following vectors contain data for Parts (a) and (b):
(a) = Cyclopentane; (b) = 1-butene
Tc
Zc
T
511.8
420.0
K
0.273
Vc
0.277
383.15
393.15
Pc
K
P
45.02
40.43
239.3
34
346
0.191
3
258
275
0.196
bar
bar
cm
mol
Tn
Psat
322.4
266.9
5.267
25.83
K
bar
T
Tr
0.7486
Tr
Tc
Psat
Pc
Psatr
0.9361
Psatr
0.117
0.6389
Calculate the fugacity coefficient at the vapor pressure by Eq. (11.68):
(a)
PHIB Tr Psatr
(b)
PHIB Tr Psatr
1
2
1
2
0.900
2
1
0.76
Eq. (3.72), the Rackett equation:
T
Tc
Tr
0.749
Tr
0.936
Eq. (11.44):
2
Vsat
f
f
11.21
V c Zc
1 Tr
PHIB Tr Psatr
11.78
20.29
107.546 cm3
133.299 mol
7
Vsat
Psat exp
Vsat ( Psat)
P
RT
Ans.
bar
Table F.1, 150 degC:
Psat
molwt
476.00 kPa
18
3
cm
molwt
gm
Vsat
1.091
Vsat
cm
19.638
mol
T
(
150
T
P
273.15)K
150 bar
= 1.084
Ans.
423.15 K
3
Equation Eq. (11.44) with
r
exp
Vsat P
RT
Psat
satPsat = fsat
r
1.084
347
r=
f
fsat
gm
mol
11.22 The following vectors contain data for Parts (a) and (b): molwt
18
gm
mol
Table F.2: (a) 9000 kPa & 400 degC; (b) 1000(psia) & 800 degF:
T1
( 400
( 800
459.67) rankine
3121.2
H1
273.15) K
J
gm
6.2915
S1
Btu
1389.6
lbm
1.5677
J
gm K
Btu
lbm rankine
Table F.2: (a) 300 kPa & 400 degC; (b) 50(psia) & 800 degF: T2
3275.2
H2
J
gm
8.0338
S2
Btu
1431.7
lbm
1.9227
J
gm K
Btu
lbm rankine
Eq. (A) on page 399 may be recast for this problem as:
r
(a)
exp
r=
molwt
R
f2
f1
H2
H1
S2
T1
= 0.0377
(b)
r=
0.0377
r
S1
f2
0.0542
= 0.0542
f1
Ans.
11.23 The following vectors contain data for Parts (a), (b), and (c):
(a) = n-pentane
(b) = Isobutylene
(c) = 1-Butene:
469.7
Tc
417.9 K
33.70
Pc
40.0
420.0
Zc
313.0
0.194
bar
40.43
0.270
0.252
0.275
0.277
Vc
238.9
239.3
348
0.191
309.2
3
cm
mol
Tn
266.3
266.9
K
T1
200
P
1.01325
300 bar
Psat
1.01325 bar
150
Tr
1.01325
0.6583
Tn
0.6372
Tr
Tc
0.0301
Psat
Pc
Pr
0.6355
0.0253
Pr
0.0251
Calculate the fugacity coefficient at the nbp by Eq. (11.68):
(a)
(b)
(c)
PHIB Tr Pr
1
1
PHIB Tr Pr
1
0.9572
2
0.9618
PHIB Tr Pr
3
3
3
0.9620
2
Eq. (3.72):
Eq. (11.44):
2
Vsat
f
V c Zc
1 Tr
PHIB Tr Pr
0.2857
Psat exp
Vsat ( Psat)
P
R Tn
2.445
f
Ans.
3.326 bar
1.801
11.24 (a) Chloroform: Tc
Pc
334.3 K
0.882
Trn
0.222
54.72 bar
Tn
536.4 K
3
Zc
T
0.293
473.15 K
Eq. (3.72):
Vc
cm
239.0
mol
Tr
T
Tc
Vsat
Tr
2
V c Zc
1 Trn
349
Psat
Tn
Tc
Trn
22.27 bar
0.623
3
7
Vsat
cm
94.41
mol
Calculate fugacity coefficients by Eqs. (11.68):
Pr ( P )
P
Pc
f ( P)
if P
Psat
( P) P
( P)
if P
Psat
( P)
P
( P)
exp
Pr ( P )
B0 Tr
Tr
( Psat) Psat exp
Psat
( Psat)
P
B1 Tr
Vsat ( P Psat)
RT
Vsat ( P
exp
Psat)
RT
0 bar 0.5 bar 40 bar
40
f ( P)
Psat
bar
30
Psat
bar
0.8
bar
P
20
( P)
0.6
bar
10
0
0
20
0.4
40
0
20
P
P
bar bar
(b) Isobutane
Tc
40
P
bar
408.1 K
Pc
36.48 bar
Tn
261.4 K
0.767
Trn
0.181
3
Zc
T
0.282
313.15 K
Eq. (3.72):
Vc
Tr
262.7
T
Tc
Vsat
cm
mol
Tr
2
V c Zc
1 Trn
350
Psat
Tn
Tc
Trn
5.28 bar
0.641
3
7
Vsat
cm
102.107
mol
Calculate fugacity coefficients by Eq. (11.68):
Pr ( )
P
P
Pc
fP)
(
if P
Psat
( )P
P
()
P
if P
Psat
( ) ( sat)
P
P
P
()
P
exp
Pr ( )
P
P
exp
B1 Tr
Vsat ( Psat)
P
RT
( sat)Psat exp
P
Psat
B0 Tr
Tr
Vsat (
P
Psat)
RT
0 bar 0.5 bar 10 bar
10
Psat
bar
fP)
(
Psat
bar
0.8
bar
5
P
()
P
bar
0.6
0
0
5
0.4
10
0
P
P
bar bar
5
10
P
bar
11.25 Ethylene = species 1; Propylene = species 2
282.3
Tc
365.6
Pc
K
0.281
Zc
Vc
0.289
T
423.15 K
P
n
2
i
30 bar
1n
50.40
46.65
131.0
188.4
y1
0.140
3
cm
mol
0.35
j1n
351
0.087
w
bar
y2
1
k
1n
y1
By Eqs. (11.70) through (11.74)
wi
wj
ij
Tc
ij
2
1
3
Vc i
Vc
Vc j
ij
Tr
Tci Tc j
Zc
ij
1.499 1.317
Tr
Tc
157.966 cm3
157.966 188.4 mol
131
0.087 0.114
1.317 1.157
50.345 48.189
Pc
48.189 46.627
282.3
321.261
321.261
Tc
0.114 0.14
2
R Tc
ij
Vc
ij
ij
Vc
Zc j
ij
Pc
T
ij
ij
13
3
2
Zci
Zc
365.6
bar
0.281 0.285
K Zc
0.285 0.289
By Eqs. (3.65) and (3.66):
B0i j
B0
Bi j
B0 Tr
B1 i j
ij
0.138
0.189
0.189
0.251
B1
R Tc
ij
B0i j
Pc
i j B1i j
B
ij
B1 Tr
ij
0.108 0.085
0.085 0.046
99.181 cm3
159.43 mol
59.892
99.181
By Eq. (11.64):
ij
2 Bi j
hatk
exp
Bi i
P
RT
20.96 cm3
mol
20.96
0
0
Bj j
Bk k
1
2
yi y j 2 i k
i
fhatk
hatk yk P
hat
0.957
0.875
352
ij
j
fhat
10.053
17.059
bar
Ans.
For an ideal solution , id = pure species
Pr
P
k
Pr
Pck
fhatid
idk
0.643
exp
0.95
id
idk yk P
k
Pr
0.595
Tr
k
B0k k
kk
9.978
fhatid
0.873
k k B1k k
Ans.
bar
17.022
Alternatively,
Pr
Pr
P
ij
idk
Pc
exp
ij
kk
Tr
B0 k k
T
0.873
35 bar
0.012
w
0.43
0.286
0.100
Zc
0.279
0.152
0.276
45.99
0.36
98.6
190.6
305.3 K
Pc
48.72 bar
369.8
n
P
373.15 K
0.21
Tc
0.95
id
kk
11.27 Methane = species 1
Ethane = species 2
Propane = species 3
y
k k B1k k
Vc
145.5
42.48
i
3
j
1n
200.0
k
1n
1n
By Eqs. (11.70) through (11.74)
wi
ij
Vc
ij
wj
Tc
ij
2
Vci
1
3
Vc j
Tci Tc j
Zc
Zci
ij
13
3
Pc
ij
2
Zc j
2
Zc
R Tc
ij
Vc
ij
ij
353
3
cm
mol
1.958 1.547 1.406
Tr
T
ij
1.547 1.222 1.111
Tr
Tc
ij
1.406 1.111 1.009
98.6
Vc
120.533 143.378
120.533
145.5
171.308
143.378 171.308
3
cm
mol
200
45.964 47.005 43.259
47.005 48.672 45.253 bar
0.056
43.259 45.253 42.428
Pc
0.012 0.056 0.082
0.082 0.126 0.152
190.6
Tc
305.3
265.488 336.006
0.126
0.286 0.282 0.281
241.226 265.488
241.226
0.1
Zc
336.006 K
0.282 0.279 0.278
0.281 0.278 0.276
369.8
By Eqs. (3.65) and (3.66):
B0i j
Bi j
B0 Tr
B1 i j
B1 Tr
ij
ij
R Tc
ij
B0i j
Pc
i j B1i j
ij
By Eq. (11.64):
ij
2 Bi j
0
Bi i
Bj j
30.442 107.809
30.442
0
23.482
107.809 23.482
hatk
exp
P
RT
Bk k
1
2
yi y j 2 i k
i
fhatk
hatk yk P
0.881
0.775
354
mol
0
ij
j
1.019
hat
3
cm
7.491
fhat
13.254 bar
9.764
Ans.
For an ideal solution , id = pure species
0.761
P
Pck
Prk
Pr
idk
0.718
exp
0.824
Prk
Tr
idk yk P
k
id
7.182
fhatid
0.88
GE
RT
2.6 x1
=
13.251 bar
9.569
0.759
11.28 Given:
k k B1k k
kk
0.977
fhatid
B0k k
1.8 x2 x1 x2
Substitute x2 = 1 - x1:
(a)
GE
=
RT
.8 x1
1.8 x1 1
2
x1 = 1.8 x1
x1
3
0.8 x1
Apply Eqs. (11.15) & (11.16) for M = GE/RT:
ln 1 =
d
GE
RT
dx1
GE
RT
ln 2 =
1
= 1.8
ln 1 = 1.8
2
x1
GE
RT
d
x1
dx1
2 x1
2 x1
ln 2 =
2.4 x1
1.4 x1
2
2
1.6 x1
3
Ans.
3
1.6 x1
(b) Apply Eq. (11.100):
GE
= x1
RT
1.8
1
2 x1
x1
x1
1.4 x1
2
2
1.6 x1
1.6 x1
3
This reduces to the initial condition:
355
3
GE
RT
d
x1
GE
RT
dx1
Ans.
(c) Divide Gibbs/Duhem eqn. (11.100) by dx1:
x1
d ln 1
dx1
x2
d ln 2
=0
dx1
Differentiate answers to Part (a):
d ln 1
=2
dx1
2.8 x1
x1
d ln 1
= 2 x1
dx1
x2
d ln 1
=1
dx1
4.8 x1
4.8 x1
2
3
2
4.8 x1
2.8 x1
x1
d ln 2
= 2 x1
dx1
2
2 x1
4.8 x1
2
These two equations sum to zero in agreement with the
Gibbs/Duhem equation.
(d) When x1 = 1, we see
from the 2nd eq. of
Part (c) that
d ln 1
When x1 = 0, we see
from the 3rd eq. of
Part (c) that
d ln 2
dx1
dx1
=0
Q.E.D.
=0
Q.E.D.
(e) DEFINE: g = GE/RT
g x1
1.8 x1
ln 1 x1
1.8
ln 2 x1
x1
ln 1 ()
0
1.8
2
2
3
x1
0.8 x1
2 x1
1.4 x1
2
1.6 x1
3
1.6 x1
3
ln 2 ()
1
356
2.6
x1
0 0.1 1.0
0
g x1
1
ln 1 x1
ln 2 x1
0
ln 1 ( )
2
ln 2 ( )
1
3
0
0.2
0.4
0.6
0.8
x1
H
H1bar
H2bar
0.02715
0.09329
417.4
0.32760
534.5
0.40244
531.7
0.56689
421.1
0.63128
x1
265.6
0.17490
11.32
87.5
347.1
0.66233
0.69984
253
VE
321.7
276.4
0.72792
252.9
0.77514
190.7
0.79243
178.1
0.82954
138.4
0.86835
98.4
0.93287
37.6
0.98233
10.0
357
n
x1
rows x1
i1n
0 0.01 1
(a) Guess:
a
x1 1
2
x1
F x1
3000
x1
1
3000 c
a
x1
3
x1 1
b
a
b
3.448
3
10
b
3.202 10
c
linfitx1 VE F
c
x1
250
3
Ans.
244.615
600
400
VEi
x1 ( 1 x1) a b x1 c ( x1)
2
200
0
0
0.2
0.4
0.6
0.8
x1 x1
i
By definition of the excess properties
E
V = x1 x2 a
b x1
d
3
E
V = 4 c x1
dx1
Vbar1
Vbar2
E
E
= x2
= x1
2
2
c x1
3 (c
b) x1
a
a
2
2 b x1
b
2
3 c x1
2 (b
c) x1
2 (b
a) x1
a
2
3 c x1
2
(b) To find the maximum, set dVE/dx1 = 0 and solve for x1. Then use x1 to
find VEmax.
Guess:
x1
0.5
Given
4 c ( x1)
x1
3
3 (c
Find ( x1)
b) ( x1)
x1
2
2 (b
0.353
a) x1
Ans.
358
a= 0
VEmax
x1 (
1
x1) a
(c) VEbar1 ( )
x1
(
1
x1
c x1
() a
x1
b
VEmax
2 b x1
2
x1) a
2
VEbar2 ( )
x1
2
b x1
Ans.
3 c ( 1)
x
2(
b
536.294
2
c)x1
2
3 c ( 1)
x
0 0.01 1
4000
2000
VEbar 1 ( )
x1
x1
VEbar 2 ( )
0
2000
0
0.2
0.4
0.6
0.8
x1 x1
Discussion:
a) Partial property for species i goes to zero WITH ZERO SLOPE as xi -> 1.
b) Interior extrema come in pairs: VEbar min for species 1 occurs at the
same x1 as VEbar max for species 2, and both occur at an inflection point on
the VE vs. x1 plot.
c) At the point where the VEbar lines cross, the VE plot shows a maximum.
11.33
Propane = 1; n-Pentane = 2
T
B
(
75
273.15)K
276
466
466
809
By Eq. (11.61):
P
2 bar
y1
n
2
i
0.5
y2
1
1n
j
1n
y1
3
cm
mol
B
yi y j Bi j
i
359
j
3
B
504.25
cm
mol
Use a spline fit of B as a function of T to
find derivatives:
331
b11
980
3
cm
276
b22
mol
809
b12
mol
466
684
235
50
t
558
3
cm
3
cm
mol
399
323.15
75
t
273.15 K
348.15 K
100
373.15
3
lspline ( t b11) B11 ( T)
vs11
interp ( vs11 t b11 T)
B11 ( T)
cm
276
mol
B22 ( T)
cm
809
mol
B12 ( T)
cm
466
mol
1.92 3.18
3
3
lspline ( t b22) B22 ( T)
vs22
interp ( vs22 t b22 T)
3
lspline ( t b12) B12 ( T)
vs12
dBdT
interp ( vs12 t b12 T)
d
d
B12 ( T)
B11 ( T)
dT
dT
d
B12 ( T)
dT
dBdT
d
B22 ( T)
dT
cm
3.18 5.92 mol K
3
Differentiate Eq. (11.61): dBdT
yi y j dBdTi djBdT
i
By Eq. (3.38): Z
1
By Eq. (6.55): HRRT
By Eq. (6.56):
BP
RT
P
R
3
V
13968
cm
mol
HR
348.037
360
ZR T
P
V
0.965
dBdT HRRT
P
dBdT
R
SRR
j
Z
B
T
cm
3.55
mol K
0.12
SRR
0.085
J
mol
SR
HR
SR
0.71
HRRT R T
SRR R
J
mol K
Ans.
11.34 Propane = 1; n-Pentane = 2
T
(
75
273.15)K
P
2 bar
y1
3
276
ij
cm
n
466
B
466
2
809
mol
0.5
y2
j1n
i1n
2 Bi j Bii B j j
By Eqs. (11.63a) and (11.63b):
hat1 ( )
y1
exp
P
B1 1
RT
hat2 ( )
y1
exp
P
B2 2
RT
y1
(
1
2
y1
2
y1) 1 2
12
0 0.1 1.0
1
0.99
0.98
hat1 ( )
y1
0.97
hat2 ( )
y1
0.96
0.95
0.94
0
0.2
0.4
0.6
y1
361
0.8
1
y1
0.0426
0.0817
45.7
0.1177
66.5
0.1510
86.6
0.2107
118.2
0.2624
144.6
0.3472
11.36
23.3
176.6
0.4158
x1
195.7
HE
0.5163
0.6156
85.6
0.9276
43.5
0.9624
22.6
(a) Guess:
a
x1 1
2
x1
3
x1
0 0.01 1
116.8
0.8650
1n
141.0
0.8181
i
174.1
0.7621
rows x1
x1
191.7
0.6810
F x1
n
204.2
x1
1
x1
1
x1
500
b
100
c
0.01
a
b
linfit x1 HE F
c
539.653
b
1.011 10
c
a
913.122
0
100
HEi
x1 ( 1 x1) a b x1 c ( x1)
2
200
300
0
0.2
0.4
0.6
x1 x1
i
362
0.8
3
Ans.
By definition of the excess properties
E
H = x1 x2 a
d
b x1
E
dx1
H = 4 c x1
Hbar1
Hbar2
E
E
2
= x2
3(
c
b) x1
a
2 b x1
a
2
= x1
3
2
c x1
b
2
2(
b
3 c x1
2(
b
a)x1
a
2
c)x1
3 c x1
2
(b) To find the minimum, set dHE/dx1 = 0 and solve for x1. Then use x1 to
find HEmin.
Guess:
x1
HE ( )
x1
3
Given
x1
0.5
4 c ( 1)
x
Find ( )
x1
HEmin
x1
x1 (
1
x1) a
3(
c
x1 (
1
2
b)( 1)
x
x1) a
2(
b
a)x1
x1
2
b x1
c x1
HEmin
HEbar1 ( )
x1
HE ( ) (
x1
1
a= 0
HE ( ) x1
x1
204.401
d
x1)
HE ( )
x1
dx1
d
HE ( )
x1
dx1
0 0.01 1
500
HEbar 1 ( )
x1
x1
HEbar 2 ( )
0
500
1000
0
2
c x1
Ans.
0.512
HEbar2 ( )
x1
(c)
b x1
0.2
0.4
0.6
x1
363
0.8
Ans.
Discussion:
a) Partial property for species i goes to zero WITH ZERO SLOPE as xi -> 1.
b) Interior extrema come in pairs: HEbar min for species 1 occurs at the same
x1 as HEbar max for species 2, and both occur at an inflection point on the H E
vs. x1 plot.
c) At the point where the HEbar lines cross, the HE plot shows a minimum.
11.37 (a)
y1
(1) = Acetone
0.28
0.307
w
n
0.190
2
(2) = 1,3-butadiene
y2
T
1
508.2
Tc
i
y1
j
ki j
170 kPa
209
Vc
0.267
1n
P
273.15) K
0.233
K Zc
425.2
1n
( 60
220.4
3
cm
mol
0
0.307 0.2485 0.082
Eq. (11.70)
wi
wj
0.2485
2
0.19
0.082
ij
0.126
0.126 0.152
508.2
Tc Tc
i
Zc
Eq. (11.73) Zci j
j
Zc
i
1
ki j
Zc
0
K
3
Vc
i
2
209
3
j
0.25 0.267
0.276
13
1
Eq. (11.74) Vci j
425.2
0.233 0.25
j
2
Vc
Tc
464.851
369.8
Eq. (11.71) Tci j
464.851
Vc
0
214.65
214.65 220.4
200
3
cm
mol
0
47.104 45.013
Eq. (11.72) Pci j
Zci j R Tci j
Vci j
Pc
45.013 42.826 bar
42.48
0
Note: the calculated pure species Pc values in the matrix above do not agree
exactly with the values in Table B.1 due to round-off error in the calculations.
364
Tri j
Tr
T
Tci j
0.036 0.038
0.656 0.717
Pr
0.717 0.784
Eq. (3.65)
P
Pci j
Pri j
0.038 0.04
0.824
B0i j
0
B0 Tri j
0.74636
B1i j
0.5405
0.27382
0.16178
Eq. (3.66)
0.16178
0.6361
B0
0.6361
0.27382
0.33295
B1 Tri j
0.874
0.558
B1
0.558 0.098
0.34
0.098
Eq. (11.69a) + (11.69b)
0.028
910.278
665.188
n
1
V
1j
RT Z
P
Eq. (6.89) dB0dTri j
665.188 cm3
499.527 mol
3
yi y j Bi j
BP
RT
Z
i j B1i j
n
B
i
Eq. (3.38)
0.027
R Tci j
B0i j
Pci j
Bi j
B
Eq. (11.61)
0.028
V
Tri j
2.6
365
598.524
mol
1
Z
0.675
B
cm
0.963
1.5694
Eq. (6.90)
3
4 cm
10
mol
dB1dTri j
Ans.
0.722
Tri j
5.2
Differentiating Eq. (11.61) and using Eq. (11.69a) + (11.69b)
n
n
dBdT
yi y j
i
1j
1
Eq. (6.55) HR
Eq. (6.56) SR
Eq. (6.54) GR
PT
B
T
R
dB0dTri j
Pci j
dBdT
i j dB1dTri j
0.727
GR
BP
344.051
SR
P dBdT
HR
101.7
mol K
J
mol
3
(b)
cm
V = 15694
mol
SR = 1.006
HR = 450.322
J
mol K
GR = 125.1
3
(c)
V = 24255
cm
mol
GR = 53.3
V = 80972
cm
HR = 36.48
mol
SR = 0.097
J
mol K
GR = 8.1
J
mol
J
mol
3
(e)
cm
V = 56991
mol
SR = 0.647
HR = 277.96
J
GR = 85.2
mol K
366
J
mol
J
mol
3
(d)
J
mol
J
mol
HR = 175.666
J
SR = 0.41
mol K
J
mol
J
J
mol
J
mol
Ans.
Ans.
Ans.
Data for Problems 11.38 - 11.40
325
15
308.3
61.39
.187
200
100
150.9
48.98
.000
575
40
562.2
48.98
.210
73.83
.224
50.40
.087
350
T
300
P
35
304.2
Tc
50
Pc
282.3
525
10
507.6
30.25
.301
225
25
190.6
45.99
.012
200
75
126.2
34.00
.038
1.054
1.325
T
Tc
Tr
2.042
1.023
Tr
0.244
0.817
1.151
Pr
1.063
P
Pc
Pr
0.474
0.992
1.034
0.331
1.18
0.544
1.585
2.206
11.38 Redlich/Kwong Equation:
0.08664
0.42748
0.02
0.133
Eq. (3.53)
3.234
0.069
Pr
Tr
4.559
4.77
0.036
0.081
q
1.5
Tr
Eq. (3.54) q
3.998
4.504
0.028
0.04
z
3.847
0.121
Guess:
4.691
2.473
1
367
Given
i
z= 1
Ii
18
exp Z
i
fi
z
q
Z
ln
i qi
Eq. (3.52)
Z
zz
i qi
Z
1
ln Z
i
q
Find z
()
Eq. (6.65)
i qi
i qi
i
qi Ii Eq. (11.37)
i Pi
Z
i qi
fi
i
0.925
0.93
13.944
0.722
0.744
74.352
0.668
0.749
29.952
0.887
0.896
31.362
0.639
0.73
36.504
0.891
0.9
8.998
0.881
0.89
22.254
0.859
0.85
63.743
11.39 Soave/Redlich/Kwong Equation
c
0.480
1.574
0.176
0.08664
2
1
c1
0.42748
0.5
2
Tr
0.02
0.133
Eq. (3.53)
3.202
0.069
Pr
Tr
4.49
4.737
0.036
0.081
q
Tr
Eq. (3.54) q
3.79
4.468
0.028
0.04
z
3.827
0.121
Guess:
4.62
2.304
1
368
Given
i
z= 1
Ii
18
exp Z
i
fi
Z
z
q
i qi
Eq. (3.52)
zz
ln
Z
i qi
i
i qi
q
Find z
()
Eq. (6.65)
i qi
ln Z
1
Z
Z
i
qi Ii Eq. (11.37)
i Pi
i qi
fi
i
0.927
0.931
13.965
0.729
0.748
74.753
0.673
0.751
30.05
0.896
0.903
31.618
0.646
0.733
36.66
0.893
0.902
9.018
0.882
0.891
22.274
0.881
0.869
65.155
11.40 Peng/Robinson Equation
1
c
2
0.37464
1
1.54226
2
0.07779
0.26992
0.45724
2
1
c1
0.5
Tr
0.018
0.12
Eq.(3.53)
3.946
0.062
Pr
Tr
5.383
5.658
0.032
0.073
q
Tr
Eq.(3.54) q
4.598
5.359
0.025
5.527
0.036
4.646
0.108
2.924
369
2
Guess:
z
1
Given z = 1
i
exp Z
fi
z
i qi
i Pi
Z
2
1
Eq. (3.52) Z
z
1
Ii
18
i
z
q
ln
2
ln Z
Z
i qi
i
Z
i qi
q
Find ( z)
i
i qi
i qi
i
Eq. (6.65)
qi Ii Eq. (11.37)
fi
i
0.918
0.923
13.842
0.69
0.711
71.113
0.647
0.73
29.197
0.882
0.89
31.142
0.617
0.709
35.465
0.881
0.891
8.91
0.865
0.876
21.895
0.845
0.832
62.363
BY GENERALIZED CORRELATIONS
Parts (a), (d), (f), and (g) --- Virial equation:
325
T
350
525
308.3
Tc
225
Tr
T
Tc
304.2
507.6
15
P
190.6
Pr
35
10
25
61.39
73.83
.224
30.25
.301
45.99
Pc
.187
.012
P
Pc
Evaluation of :
B0
B0 ( Tr)
Eq. (3.65)
370
B1
B1 ( Tr)
Eq. (3.66)
0.675
DB0
Eq. (6.89)
2.6
0.722
DB1
5.2
Tr
Eq. (6.90)
Tr
0.932
Pr
B0
exp
Tr
Eq. (11.60)
B1
(a)
(d)
(f)
(g)
0.904
0.903
0.895
Parts (b), (c), (e), and (h) --- Lee/Kesler correlation:
Interpolate in Tables E.13 - E.16:
.7454
0
1.1842
0.9634
0.9883
1
.7316
.8554
0.210
0.087
1.2071
.7517
0.000
0.038
0.745
01
(b)
(c)
(e)
(h)
0.746
Eq. (11.67):
0.731
0.862
11.43 ndot1
x1
2
kmol
hr
ndot1
ndot3
ndot2
x1
4
kmol
hr
0.333
ndot3
x2
1
x1
ndot1
ndot2
x2
0.667
a) Assume an ideal solution since n-octane and iso-octane are non-polar and
very similar in chemical structure. For an ideal solution, there is no heat of
mixing therefore the heat transfer rate is zero.
b) S t
R x1 ln x1
x2 ln x2
ndot3
371
St
8.82
W
K
Ans.
11.44 For air entering the process:
0.21
xN21
0.79
For the enhanced air leaving the process: xO22
0.5
xN22
0.5
ndot2
50
xO21
mol
sec
a) Apply mole balances to find rate of air and O2 fed to process
Guess:
ndotair
40
mol
sec
ndotO2
10
mol
sec
Given
xO21 ndotair
ndotO2 = xO22 ndot2
Mole balance on N2
xN21 ndotair = xN22 ndot2
ndotair
ndotO2
ndotair
Mole balance on O 2
Find ndotair ndotO2
31.646
mol
sec
Ans.
ndotO2
18.354
mol
sec
Ans.
b) Assume ideal gas behavior. For an ideal gas there is no heat of mixing,
therefore, the heat transfer rate is zero.
c) To calculate the entropy change, treat the process in two steps:
1. Demix the air to O2 and N2
2. Mix the N2 and combined O2 to produce the enhanced air
Entropy change of demixing
S12
Entropy change of mixing
S23
R xO21 ln xO21
R xO22 ln xO22
Total rate of entropy generation: SdotG
SdotG
372
ndotair S12
152.919
W
K
xN21 ln xN21
xN22 ln xN22
ndot2 S23
Ans.
10
11.50 T
932.1
544.0
30 K
GE
273.15K
J
mol
513.0
50
HE
893.4
J
mol
845.9
494.2
Assume Cp is constant. Then HE is of the form:
HE = c
aT
Find a and c using the given HE and T values.
a
slope ( HE)
T
a
c
intercept ( HE)
T
c
2.155
mol K
3J
1.544
GE is of the form: GE = a T ln
J
10
mol
T
K
T
bT
c
Rearrange to find b using estimated a and c values along with GE and T data.
GE
a T ln
B
T
K
T
13.543
c
B
T
13.559
J
mol K
13.545
Use averaged b value
3
Bi
b
i
1
b
3
13.549
J
mol K
Now calculate HE, GE and T*SE at 25 C using a, b and c values.
HE ( )
T
aT
HE [25
(
c
T
K
GE ( )
T
a T ln
TSE ( )
T
HE ( ) GE ( )
T
T
T
bT
273.15) ] 901.242
K
J Ans.
mol
(
c GE [25
273.15) ] 522.394
K
J
Ans.
mol
TSE [25
(
373
273.15) ] 378.848
K
J
Ans.
mol
Chapter 12 - Section A - Mathcad Solutions
12.1
Methanol(1)/Water(2)-- VLE data:
T
333.15 K
39.223
0.5714
42.984
0.2167
0.6268
48.852
0.3039
0.6943
52.784
P
0.1686
0.3681
0.7345
56.652
60.614
0.4461
x1
kPa
0.7742
y1
0.5282
0.8085
63.998
0.6044
0.8383
67.924
0.6804
0.8733
70.229
0.7255
0.8922
72.832
0.7776
0.9141
Number of data points:
n
Calculate x2 and y2:
x2
rows ( )
P
1
n
10
y2
x1
i1n
1
y1
Vapor Pressures from equilibrium data:
Psat1
84.562 kPa
Psat2
19.953 kPa
Calculate EXPERIMENTAL values of activity coefficients and
excess Gibbs energy.
y1 P
1
x1 Psat1
y2 P
2
GERT
x2 Psat2
374
x1 ln 1
x2 ln 2
1
i
ln 2
i
ln 1
i
2
i
i
GERTi
1
1.572
1.013
0.452
0.013
0.087
2
1.47
1.026
0.385
0.026
0.104
3
1.32
1.075
0.278
0.073
0.135
4
1.246
1.112
0.22
0.106
0.148
5
1.163
1.157
0.151
0.146
0.148
6
1.097
1.233
0.093
0.209
0.148
7
1.05
1.311
0.049
0.271
0.136
8
1.031
1.35
0.031
0.3
0.117
9
1.021
1.382
0.021
0.324
0.104
10
1.012
1.41
0.012
0.343
0.086
0.2
0.4
0.5
0.4
ln 1
i
0.3
ln 2
i
GERT i
0.2
0.1
0
0
0.6
x1
0.8
i
(a) Fit GE/RT data to Margules eqn. by linear least squares:
VXi
Slope
Slope
x1
i
slope ( VX VY)
0.208
VYi
GERTi
x1 x2
i
i
Intercept
intercept ( VX VY)
Intercept
0.683
A12
Intercept
A21
Slope
A12
0.683
A21
0.475
375
A12
Ans.
The following equations give CALCULATED values:
2
1 ( x2)
x1
2 ( x2)
x1
j
exp x2
exp x1
2
X1
j
j
X1
Y1calc
2 A21
A12 x1
A21
2 A12
A21 x2
X1
1 101
pcalc
A12
j
.01 j
1 X1 X2 Psat1
j
j
j
j
X2
.01
X2
j
2 X1 X2 Psat2
j
j
j
1 X1 X2 Psat1
j
j
pcalc
j
P-x,y Diagram: Margules eqn. fit to GE/RT data.
90
80
Pi
kPa
70
Pi
60
kPa
pcalc
50
j
kPa
pcalc
kPa
40
j
30
20
10
0
0.2
0.4
0.6
x1 y1 X1 Y1calc
i
P-x data
P-y data
P-x calculated
P-y calculated
376
i
j
0.8
j
1
X1
j
Pcalc
x1 1 x1 x2 Psat1
i
i
i
i
y1calc
x2 2 x1 x2 Psat2
i
i
i
x1 1 x1 x2 Psat1
i
i
i
Pcalc
i
i
RMS deviation in P:
Pi
RMS
2
Pcalc
i
RMS
n
0.399 kPa
i
(b) Fit GE/RT data to van Laar eqn. by linear least squares:
VXi
x1
VYi
i
x1 x2
i
i
GERTi
Slope
slope ( VX VY)
Intercept
intercept ( VX VY)
Slope
0.641
Intercept
1.418
a12
a12
1
Intercept
0.705
1 ( x1 x2)
2 ( x1 x2)
j
a21
1 101
a21
a12 x1
exp a12 1
X2
j
j
0.485
Ans.
2
a21 x2
exp a21 1
X1
1
( Slope Intercept)
a21 x2
a12 x1
.01 j
1
2
X1
.00999
j
377
(To avoid singularities)
pcalc
Pcalc
X1
j
1 X1 X2 Psat1
j
j
j
x1 1 x1 x2 Psat1
i
i
i
i
X1
Y1calc
j
j
X2
j
2 X1 X2 Psat2
j
j
x2 2 x1 x2 Psat2
i
i
i
1 X1 X2 Psat1
j
j
pcalc
y1calc
x1 1 x1 x2 Psat1
i
i
i
Pcalc
i
i
j
P-x,y Diagram: van Laar eqn. fit to GE/RT data.
90
80
Pi
70
kPa
Pi
60
kPa
pcalc
50
j
kPa
pcalc
kPa
40
j
30
20
10
0
0.2
0.4
0.6
x1 y1 X1 Y1calc
i
i
j
0.8
j
P-x data
P-y data
P-x calculated
P-y calculated
RMS deviation in P:
Pi
RMS
Pcalc
n
2
i
RMS
i
378
0.454 kPa
1
(c) Fit GE/RT data to Wilson eqn. by non-linear least squares.
Minimize the sum of the squared errors using the Mathcad Minimize function.
Guesses:
SSE
0.5
12
12
GERTi
21
1.0
21
x1 ln x1
i
i
2
x2
i
12
i
x2 ln x2
i
i
x1
Minimize SSE
12
12
x1
exp
Pcalc
X1
j
x1
X1
j
X1
Y1calc
j
x1
21
12
21
x2
x2
12
x1
x1
21
21
.01 j
.01
1 X1 X2 Psat1
j
j
X2
j
x1 1 x1 x2 Psat1
i
i
i
i
x2
x2
1 101
x2
12
12
x1
2 ( x2)
x1
pcalc
x2
x1
Ans.
21
12
1 ( x2)
x1
1.026
21
21
exp x2
0.476
21
12
j
21
i
j
X2
j
1
X1
j
2 X1 X2 Psat2
j
j
x2 2 x1 x2 Psat2
i
i
i
1 X1 X2 Psat1
j
j
j
pcalc
y1calc
i
x1 1 x1 x2 Psat1
i
i
i
Pcalc
i
j
379
P-x,y diagram: Wilson eqn. fit to GE/RT data.
90
80
Pi
kPa
70
Pi
60
kPa
pcalc
50
j
kPa
pcalc
40
j
30
kPa
20
10
0
0.2
0.4
0.6
x1 y1 X1 Y1calc
i
i
j
0.8
j
P-x data
P-y data
P-x calculated
P-y calculated
RMS deviation in P:
Pi
RMS
Pcalc
n
2
i
RMS
0.48 kPa
i
(d) BARKER'S METHOD by non-linear least squares.
Margules equation.
Guesses for parameters: answers to Part (a).
2
1 x1 x2 A 12 A 21
exp ( ) A12
x2
2 x1 x2 A 12 A 21
exp ( ) A21
x1
2
380
2 A21
A12 x1
2 A12
A21 x2
1
Minimize the sum of the squared errors using the Mathcad Minimize function.
Guesses:
A12
SSE A12 A21
0.5
A21
Pi
x1
i
x2
X1
j
X2
X1
Y1calc
A21
2 X 1 j X 2 j A 12 A 21 Psat2
j
1 X 1 j X 2 j A 12 A 21 Psat1
pcalc
j
1 x1i x2i A 12 A 21 Psat1
i
x2
2 x1i x2i A 12 A 21 Psat2
x1
1 x1i x2i A 12 A 21 Psat1
i
y1calc
0.435
j
j
x1
i
0.758
1 X 1 j X 2 j A 12 A 21 Psat1
j
i
i
Pcalc
i
RMS deviation in P:
Pi
RMS
Pcalc
n
2
i
2
2 x1i x2i A 12 A 21 Psat2
i
Minimize SSE A12 A21
A21
Pcalc
x1 x2 A12 A21 Psat1
i1
i
i
A12
A12
pcalc
1.0
RMS
i
381
0.167 kPa
Ans.
P-x-y diagram, Margules eqn. by Barker's method
90
80
Pi
kPa
70
Pi
60
kPa
pcalc
50
j
kPa
pcalc
40
j
30
kPa
20
10
0
0.2
0.4
0.6
x1 y1 X1 Y1calc
i
i
j
0.8
1
j
P-x data
P-y data
P-x calculated
P-y calculated
Residuals in P and y1
1
Pi Pcalc
0.5
i
kPa
y1 y1calc 100
i
i
0
0.5
0
0.2
0.4
x1
Pressure residuals
y1 residuals
382
i
0.6
0.8
(e) BARKER'S METHOD by non-linear least squares.
van Laar equation.
Guesses for parameters: answers to Part (b).
j
X1
1 101
1 x1 x2 a12 a21
.01 j
j
.00999
exp a12 1
2 x1 x2 a12 a21
exp a21 1
X2
j
1
X1
j
2
a12 x1
a21 x2
2
a21 x2
a12 x1
Minimize the sum of the squared errors using the Mathcad Minimize function.
Guesses:
a12
SSE a12 a21
0.5
a21
Pi
x1
i
a12
X1
j
x2
i
X1
Y1calc
j
a12
x1
i
2 X 1 j X 2 j a12 a21 Psat2
j
1 X 1 j X 2 j a12 a21 Psat1
j
1 x1i x2i a12 a21 Psat1
i
x2
2 x1i x2i a12 a21 Psat2
x1
1 x1i x2i a12 a21 Psat1
i
y1calc
i
i
Pcalc
i
383
0.83
a21
j
pcalc
2
2 x1i x2i a12 a21 Psat2
1 X 1 j X 2 j a12 a21 Psat1
j
X2
Pcalc
x1 x2 a12 a21 Psat1
i1
i
i
Minimize SSE a12 a21
a21
pcalc
1.0
0.468
Ans.
RMS deviation in P:
Pi
RMS
Pcalc
2
i
RMS
n
0.286 kPa
i
P-x,y diagram, van Laar Equation by Barker's Method
90
80
70
Pi
kPa
60
Pi
kPa
pcalc
50
j
kPa
pcalc
j
40
kPa
30
20
10
0
0.2
0.4
0.6
x1 y1 X1 Y1calc
i
i
P-x data
P-y data
P-x calculated
P-y calculated
384
j
0.8
j
1
Residuals in P and y1.
1
Pi Pcalc
0.5
i
kPa
y1 y1calc 100
i
i
0
0.5
0
0.2
0.4
x1
0.6
0.8
i
Pressure residuals
y1 residuals
(f)
BARKER'S METHOD by non-linear least squares.
Wilson equation.
Guesses for parameters: answers to Part (c).
j
1 101
1 x1 x2
X1
12
21
.01 j
j
exp
ln x1
x2
2 x1 x2
12
21
exp
.01
x2
x1
x2
x1
X1
j
21
12
x2
x1
21
21
21
12
x1
1
j
12
12
ln x2
x1
X2
x2
12
x2
x1
21
Minimize the sum of the squared errors using the Mathcad Minimize function.
Guesses:
12
0.5
21
385
1.0
SSE
12
Pi
21
x1
i
x1 x2
i1
i
i
x2
2 x1i x2i
i
12
X1
j
1 X1 j X2 j
j
X2
X1
Y1calc
2 X1 j X2 j
12
21 Psat2
j
1 X1 j X2 j
12
21 Psat1
pcalc
x1
1 x1i x2i
i
j
21 Psat1
12
x2
2 x1i x2i
12
21 Psat2
x1
1 x1i x2i
12
21 Psat1
i
y1calc
1.198
21 Psat1
12
j
j
i
0.348
21
i
i
Pcalc
i
RMS deviation in P:
Pi
RMS
Pcalc
2
i
RMS
n
i
386
2
21 Psat2
12
Minimize SSE
21
Pcalc
12
21
12
pcalc
21 Psat1
12
0.305kPa
Ans.
P-x,y diagram, Wilson Equation by Barker's Method
90
80
Pi
kPa
70
Pi
60
kPa
pcalc
50
j
kPa
pcalc
40
j
30
kPa
20
10
0
0.2
0.4
0.6
x1 y1 X1 Y1calc
i
i
j
0.8
1
j
P-x data
P-y data
P-x calculated
P-y calculated
Residuals in P and y1.
1
Pi Pcalc
i
0.5
kPa
y1 y1calc 100
i
i
0
0.5
0
0.2
0.4
x1
Pressure residuals
y1 residuals
387
i
0.6
0.8
12.3
Acetone(1)/Methanol(2)-- VLE data:
T
328.15 K
72.278
0.0647
75.279
0.0570
0.1295
77.524
0.0858
0.1848
78.951
0.1046
0.2190
82.528
0.1452
0.2694
86.762
0.2173
0.3633
90.088
0.2787
0.4184
93.206
0.3579
0.4779
95.017
P
0.0287
0.4050
0.5135
96.365
97.646
kPa
0.4480
x1
y1
0.5052
0.5512
0.5844
98.462
0.5432
0.6174
99.811
0.6332
0.6772
99.950
0.6605
0.6926
100.278
0.6945
0.7124
100.467
0.7327
0.7383
100.999
0.7752
0.7729
101.059
0.7922
0.7876
99.877
0.9080
0.8959
99.799
0.9448
0.9336
Number of data points:
n
Calculate x2 and y2:
x2
rowsP)
(
1
x1
n
y2
Vapor Pressures from equilibrium data:
Psat1
96.885 kPa
Psat2
388
68.728 kPa
20
i1n
1
y1
Calculate EXPERIMENTAL values of activity coefficients and
excess Gibbs energy.
y1 P
1
x1 Psat1
ln 1
i
i
i
GERT
x2 Psat2
2
1
i
y2 P
2
x1 ln 1
ln 2
i
x2 ln 2
GERTi
0.013
0.027
0.568
0.011
0.043
0.544
5.815·10-3
0.052
1.002
0.534
1.975·10-3
0.058
1.58
1.026
0.458
0.026
0.089
6
1.497
1.027
0.404
0.027
0.108
7
1.396
1.057
0.334
0.055
0.133
8
1.285
1.103
0.25
0.098
0.152
9
1.243
1.13
0.218
0.123
0.161
10
1.224
1.14
0.202
0.131
0.163
11
1.166
1.193
0.153
0.177
0.165
12
1.155
1.2
0.144
0.182
0.162
13
1.102
1.278
0.097
0.245
0.151
14
1.082
1.317
0.079
0.275
0.145
15
1.062
1.374
0.06
0.317
0.139
16
1.045
1.431
0.044
0.358
0.128
17
1.039
1.485
0.039
0.395
0.119
18
1.037
1.503
0.036
0.407
0.113
19
1.017
1.644
0.017
0.497
0.061
20
1.018
1.747
0.018
0.558
0.048
1
1.682
1.013
0.52
2
1.765
1.011
3
1.723
1.006
4
1.706
5
389
0.6
0.4
ln 1
i
ln 2
i
GERTi
0.2
0
0
0.2
0.4
0.6
x1
0.8
i
(a) Fit GE/RT data to Margules eqn. by linear least squares:
VXi
x1
Slope
VYi
i
slope (
VX VY)
Slope
GERTi
x1 x2
i
i
Intercept
Intercept
0.018
intercept (
VX VY)
0.708
A12
Intercept
A21
Slope
A12
0.708
A21
0.69
A12
Ans.
The following equations give CALCULATED values:
2
1 ( x2)
x1
2 ( x2)
x1
j
exp x2
exp x1
X1
j
j
A12 x1
2 A12
A21 x2
X1
j
X1
Y1calc
2 A21
A21
2
1 101
pcalc
A12
.01 j
j
1 X1 X2 Psat1
j
j
j
1 X1 X2
j
pcalc
j
.01
X2
j
Psat1
j
390
X2
2 X1 X2 Psat2
j
j
j
1
X1
j
P-x,y Diagram: Margules eqn. fit to GE/RT data.
105
100
Pi
kPa
95
Pi
90
kPa
pcalc
85
j
kPa
pcalc
80
j
75
kPa
70
65
0
0.2
0.4
0.6
x1 y1 X1 Y1calc
i
i
j
0.8
j
P-x data
P-y data
P-x calculated
P-y calculated
Pcalc
x1 1 x1 x2 Psat1
i
y1calc
i
i
i
i
x2 2 x1 x2 Psat2
i
i
i
x1 1 x1 x2 Psat1
i
i
i
Pcalc
i
RMS deviation in P:
Pi
RMS
Pcalc
n
2
i
RMS
i
391
0.851 kPa
(b) Fit GE/RT data to van Laar eqn. by linear least squares:
x1 x2
VXi
x1
i
VYi
i
i
GERTi
Slope
slope (
VX VY)
Intercept
intercept (
VX VY)
Slope
0.015
Intercept
1.442
1
a12
Intercept
a12
0.693
X1
X2
Pcalc
X1
j
j
1 X1 X2
j
j
i
X1
j
i
j
i
2
a12 x1
X1
(To avoid singularities)
.00999
j
Psat1
x1 1 x1 x2 Psat1
i
Y1calc
1
j
2
a21 x2
.01 j
j
Ans.
a21 x2
exp a21 1
1 101
Intercept)
0.686
a12 x1
exp a12 1
2 ( x2)
x1
pcalc
(
Slope
a21
1 ( x2)
x1
j
1
a21
X2
j
2 X1 X2
j
j
Psat2
x2 2 x1 x2 Psat2
i
i
1 X1 X2 Psat1
j
j
pcalc
i
y1calc
j
i
x1 1 x1 x2 Psat1
i
i
i
Pcalc
i
392
P-x,y Diagram: van Laar eqn. fit to GE/RT data.
105
100
Pi
95
kPa
Pi
90
kPa
pcalc
85
j
kPa
80
pcalc
j
kPa
75
70
65
0
0.2
0.4
0.6
0.8
x1 y1 X1 Y1calc
i
i
j
1
j
P-x data
P-y data
P-x calculated
P-y calculated
RMS deviation in P:
Pi
RMS
Pcalc
2
i
n
RMS
0.701 kPa
i
(c) Fit GE/RT data to Wilson eqn. by non-linear least squares.
Minimize the sum of the squared errors using the Mathcad Minimize function.
Guesses:
SSE
12
12
0.5
GERTi
21
i
1.0
21
x1 ln x1
i
i
393
12
i
x2 ln x2
i
2
x2
i
x1
i
21
Minimize SSE
12
12
x1
exp
pcalc
Pcalc
X1
j
x1
X1
j
X1
Y1calc
j
x1
21
12
21
x2
x2
12
x1
.01 j
1 X1 X2 Psat1
j
j
X2
x1
21
21
X2
.01
j
x1 1 x1 x2 Psat1
i
i
i
i
x2
x2
1 101
x2
12
12
x1
2 ( x2)
x1
j
x2
x1
Ans.
21
12
1 ( x2)
x1
0.681
21
21
exp x2
0.71
21
12
j
j
1
X1
j
2 X1 X2 Psat2
j
j
x2 2 x1 x2 Psat2
i
i
i
1 X1 X2 Psat1
j
j
j
pcalc
y1calc
i
x1 1 x1 x2 Psat1
i
i
i
Pcalc
i
j
394
P-x,y diagram: Wilson eqn. fit to GE/RT data.
105
100
Pi
kPa
95
Pi
90
kPa
pcalc
85
j
kPa
pcalc
80
j
75
kPa
70
65
0
0.2
0.4
0.6
x1 y1 X1 Y1calc
i
i
j
0.8
j
P-x data
P-y data
P-x calculated
P-y calculated
RMS deviation in P:
Pi
RMS
Pcalc
2
i
n
RMS
0.361 kPa
i
(d) BARKER'S METHOD by non-linear least squares.
Margules equation.
Guesses for parameters: answers to Part (a).
1 x1 x2 A 12 A 21
exp ( x2)
2 x1 x2 A 12 A 21
exp ( x1)
2
2
A12
2 A21
A12 x1
A21
2 A12
A21 x2
395
1
Minimize the sum of the squared errors using the Mathcad Minimize function.
Guesses:
A12
SSE A12 A21
0.5
A21
Pi
x1
i
x2
X1
j
X2
X1
Y1calc
A21
2 X 1 j X 2 j A 12 A 21 Psat2
j
1 X 1 j X 2 j A 12 A 21 Psat1
pcalc
j
1 x1i x2i A 12 A 21 Psat1
i
x2
2 x1i x2i A 12 A 21 Psat2
x1
1 x1i x2i A 12 A 21 Psat1
i
y1calc
0.672
j
j
x1
i
0.644
1 X 1 j X 2 j A 12 A 21 Psat1
j
i
i
Pcalc
i
RMS deviation in P:
Pi
RMS
Pcalc
n
2
i
2
2 x1i x2i A 12 A 21 Psat2
i
Minimize SSE A12 A21
A21
Pcalc
x1 x2 A12 A21 Psat1
i1
i
i
A12
A12
pcalc
1.0
RMS
i
396
0.365 kPa
Ans.
P-x-y diagram, Margules eqn. by Barker's method
105
100
Pi
kPa
95
Pi
90
kPa
pcalc
85
j
kPa
pcalc
80
j
75
kPa
70
65
0
0.2
0.4
0.6
x1 y1 X1 Y1calc
i
i
j
0.8
j
P-x data
P-y data
P-x calculated
P-y calculated
Residuals in P and y1
2
Pi Pcalc
1
i
kPa
y1 y1calc 100
i
i
0
1
0
0.2
0.4
0.6
x1
Pressure residuals
y1 residuals
397
i
0.8
1
(e) BARKER'S METHOD by non-linear least squares.
van Laar equation.
Guesses for parameters: answers to Part (b).
j
X1
1 101
1 x1 x2 a12 a21
.01 j
j
.00999
exp a12 1
2 x1 x2 a12 a21
exp a21 1
X2
j
1
X1
j
2
a12 x1
a21 x2
2
a21 x2
a12 x1
Minimize the sum of the squared errors using the Mathcad Minimize function.
Guesses:
a12
SSE a12 a21
0.5
a21
Pi
x1
i
a12
X1
j
x2
i
X1
Y1calc
j
a12
x1
i
2 X 1 j X 2 j a12 a21 Psat2
j
1 X 1 j X 2 j a12 a21 Psat1
j
1 x1i x2i a12 a21 Psat1
i
x2
2 x1i x2i a12 a21 Psat2
x1
1 x1i x2i a12 a21 Psat1
i
y1calc
i
i
Pcalc
i
398
0.644
a21
j
pcalc
2
2 x1i x2i a12 a21 Psat2
1 X 1 j X 2 j a12 a21 Psat1
j
X2
Pcalc
x1 x2 a12 a21 Psat1
i1
i
i
Minimize SSE a12 a21
a21
pcalc
1.0
0.672
Ans.
RMS deviation in P:
Pi
RMS
Pcalc
2
i
RMS
n
0.364 kPa
i
P-x,y diagram, van Laar Equation by Barker's Method
105
100
95
Pi
kPa
90
Pi
kPa
pcalc
85
j
kPa
pcalc
j
80
kPa
75
70
65
0
0.2
0.4
0.6
x1 y1 X1 Y1calc
i
i
P-x data
P-y data
P-x calculated
P-y calculated
399
j
0.8
j
1
Residuals in P and y1.
1.5
1
Pi Pcalc
i
0.5
kPa
y1 y1calc 100
i
0
i
0.5
1
0
0.2
0.4
0.6
x1
0.8
i
Pressure residuals
y1 residuals
(f)
j
BARKER'S METHOD by non-linear least squares.
Wilson equation.
Guesses for parameters: answers to Part (c).
1 101
1 x1 x2
X1
12
21
.01 j
j
exp ln x1
x2
2 x1 x2
12
21
.01
x2
x1
x2
x1
X1
j
21
12
x2
x1
21
21
21
12
x1
1
j
12
12
exp ln x2
x1
X2
x2
12
x2
x1
21
Minimize the sum of the squared errors using the Mathcad Minimize function.
Guesses:
12
0.5
21
400
1.0
SSE
12
Pi
21
x1
i
x1 x2
i1
i
i
x2
2 x1i x2i
i
12
X1
j
1 X1 j X2 j
j
X2
X1
Y1calc
RMS
2 X1 j X2 j
12
j
1 X1 j X2 j
12
21 Psat1
1 x1i x2i
i
j
21 Psat1
12
x2
2 x1i x2i
12
21 Psat2
x1
1 x1i x2i
12
21 Psat1
i
y1calc
0.35 kPa
21 Psat2
pcalc
x1
0.663
21 Psat1
12
j
j
i
0.732
21
i
i
Pcalc
i
RMS deviation in P:
Pi
RMS
Pcalc
2
i
n
i
401
2
21 Psat2
12
Minimize SSE
21
Pcalc
12
21
12
pcalc
21 Psat1
12
Ans.
P-x,y diagram, Wilson Equation by Barker's Method
105
100
Pi
kPa
95
Pi
90
kPa
pcalc
85
j
kPa
pcalc
80
j
75
kPa
70
65
0
0.2
0.4
0.6
x1 y1 X1 Y1calc
i
i
j
0.8
j
P-x data
P-y data
P-x calculated
P-y calculated
Residuals in P and y1.
2
Pi Pcalc
1
i
kPa
y1 y1calc 100
i
i
0
1
0
0.2
0.4
0.6
x1
Pressure residuals
y1 residuals
402
i
0.8
1
12.6
Methyl t-butyl ether(1)/Dichloromethane--VLE data: T
308.15 K
83.402
0.0141
82.202
0.0579
0.0253
80.481
0.0924
0.0416
76.719
0.1665
0.0804
72.442
0.2482
0.1314
68.005
P
0.0330
0.3322
0.1975
65.096
59.651
0.3880
x1
kPa
y1
0.5036
0.2457
0.3686
56.833
0.6736
0.5882
51.620
0.7676
0.7176
50.455
0.8476
0.8238
49.926
0.9093
0.9002
49.720
Psat1
0.4564
53.689
x2
0.5749
0.9529
0.9502
1
y2
x1
49.624 kPa
1
Psat2
y1
85.265 kPa
Calculate EXPERIMENTAL values of activity coefficients and excess
Gibbs energy.
y1 P
1
x1 Psat1
GERTx1x2
y2 P
2
GERT
x1 x2
GERT
x2 Psat2
n
rows ( )
P
403
x1 ln 1
n
14
x2 ln 2
i
1n
(a) Fit GE/RT data to Margules eqn. by nonlinear least squares.
Minimize sum of the squared errors using the Mathcad Minimize function.
Guesses:
A12
0.3
SSE A12 A21 C
A21
C
A21 x1
GERTi
0.5
A12 x2
i
0.2
C x1 x2
i
i
x1 x2
i
i
i
A12
A21
Minimize SSE A12 A21 C
0.336
A21
A12
0.535
0.195
C
C
Ans.
(b) Plot data and fit
GeRTx1x2 ( x2)
x1
GeRT ( x2)
x1
A21 x1
C x1 x2
GeRTx1x2 ( x2)x1 x2
x1
2
ln 1 ( x2)
x1
x2
ln 2 ( x2)
x1
x1
j
A12 x2
2
A12
A12
C x1
3 C x1
A21
2
2 A21
2 A12
A21
C x2
3 C x2
1 101
X1
j
.01 j
2
.01
X2
1
j
0
GERTx1x2 i
GeRTx1x2 X1 X2
j
ln 1
i
j
0.1
0.2
ln 1 X1 X2
j
j
ln 2
i
0.3
0.4
ln 2 X1 X2
j
j
0.5
0.6
0
0.2
0.4
0.6
0.8
x1 X1 x1 X1 x1 X1
i
404
j
i
j
i
j
X1
j
i
2
(c) Plot Pxy diagram with fit and data
1 ( x1 x2)
exp ln 1 ( x1 x2)
2 ( x1 x2)
exp ln 2 ( x1 x2)
Pcalc
X1
j
X1
y1calc
1 X1 X2 Psat1
j
j
j
j
j
X2
j
2 X1 X2 Psat2
j
j
1 X1 X2 Psat1
j
j
Pcalc
j
P-x,y Diagram from Margules Equation fit to GE/RT data.
90
Pi
80
kPa
Pi
70
kPa
Pcalc
j
60
kPa
Pcalc
j
kPa
50
40
0
0.2
0.4
0.6
x1 y1 X1 y1calc
i
i
j
0.8
j
P-x data
P-y data
P-x calculated
P-y calculated
(d) Consistency Test:
ln 1 2i
ln
1 x1 x2
i
i
2 x1 x2
i
i
GERTi
ln
1i
2
i
405
GeRT x1 x2
i
i
GERTi
0.004
0
GERTi
0
ln 1 2i
0.025
0.004
0
0.5
x1
0.05
1
0
0.5
x1
i
1
i
Calculate mean absolute deviation of residuals
mean
GERT
9.391
4
10
mean
ln 1 2
0.021
(e) Barker's Method by non-linear least squares:
Margules Equation
2
1 x1 x2 A 12 A 21 C
exp ( ) A12
x2
2 A21
A12
C x1
A21
C x2
2
3 C x1
2
2 x1 x2 A 12 A 21 C
exp ( ) A21
x1
2 A12
2
3 C x2
Minimize sum of the squared errors using the Mathcad Minimize function.
Guesses:
A12
0.3
SSE A12 A21 C
Pi
i
A21
C
0.2
x1 x2 A12 A21 C Psat1
i1
i
i
x1
x2
i
A12
A21
0.5
2 x1i x2i A 12 A 21 C Psat2
A12
Minimize SSE A12 A21 C
C
0.364
A21
0.521
C
406
2
0.23
Ans.
Plot P-x,y diagram for Margules Equation with parameters from Barker's
Method.
Pcalc
X1
j
1 X 1 j X 2 j A 12 A 21 C Psat1
j
X2
X1
y1calc
j
j
j
2 X 1 j X 2 j A 12 A 21 C Psat2
1 X 1 j X 2 j A 12 A 21 C Psat1
Pcalc
j
90
Pi
80
kPa
Pi
70
kPa
Pcalc
j
60
kPa
Pcalc
kPa
j
50
40
0
0.2
0.4
0.6
x1 y1 X1 y1calc
i
i
j
P-x data
P-y data
P-x calculated
P-y calculated
Pcalc
x1
i
1 x1i x2i A 12 A 21 C Psat1
i
x2
2 x1i x2i A 12 A 21 C Psat2
x1
1 x1i x2i A 12 A 21 C Psat1
i
y1calc
i
i
Pcalc
i
407
0.8
j
Plot of P and y1 residuals.
0.8
0.6
Pi Pcalc
i
0.4
kPa
y1 y1calc 100 0.2
i
i
0
0.2
0
0.5
x1
1
i
Pressure residuals
y1 residuals
RMS deviations in P:
Pi
RMS
Pcalc
n
2
i
RMS
i
408
0.068 kPa
12.8
(a)
Data:
0.0523
1.002
0.1299
1.307
1.004
0.2233
1.295
1.006
0.2764
1.228
1.024
0.3482
1.234
1.022
0.4187
x1
1.202
1.180
1.049
0.5001
2
1.129
1
1.092
0.5637
1.102
0.6469
1.076
1.170
0.7832
1.032
1.298
0.8576
1.016
1.393
0.9388
1.001
1.600
0.9813
n
1.120
1.003
1.404
rows x1
GERTi
i
x1 ln 1
i
i
1n
n
13
x2
i
1
x1
i
x2 ln 2
i
i
(b) Fit GE/RT data to Margules eqn. by linear least-squares procedure:
Xi
x1
Yi
i
GERTi
x1 x2
i
i
Slope
slope ( Y)
X
Intercept
intercept ( Y)
X
Slope
0.247
Intercept
0.286
A12
Intercept
A21
Slope
A12
0.286
A21
0.534
2
1 ( x2)
x1
exp x2
2 ( x2)
x1
exp x1
GeRT ( x2)
x1
2
A12
A12
2 A21
A12 x1
A21
2 A12
Ans.
A21 x2
x1 ln 1 ( x2)
x1
x2 ln 2 ( x2)
x1
409
Plot of data and correlation:
0.5
GERT i
GeRT x1 x2
i
i
0.4
ln 1
i
ln 1 x1 x2
i
i
ln 2
i
0.3
0.2
ln 2 x1 x2
i
i
0.1
0
0
0.2
0.4
0.6
x1
i
(c) Calculate and plot residuals for consistency test:
GERTi
ln 1 2i
GeRT x1 x2
i
i
ln
1 x1 x2
i
i
2 x1 x2
i
i
GERTi
ln
1i
2
i
410
0.8
ln 1 2i
GERTi
3.314·10-3
-2.264·10-3
-9.153·10-5
-3.14·10-3
0.1
0.098
-0.021
-2.998·10-3
0.026
-2.874·10-3
-0.019
-2.22·10-3
5.934·10-3
-2.174·10-3
0.028
-1.553·10-3
-9.59·10-3
-8.742·10-4
9.139·10-3
2.944·10-4
-5.617·10-4
5.962·10-5
-0.011
9.025·10-5
ln 1 2i
0
0.028
4.236·10-4
0.05
-0.168
0
0.5
x1
1
i
Calculate mean absolute deviation of residuals:
mean
GERT
1.615 10
3
mean
ln 1 2
0.03
Based on the graph and mean absolute deviations,
the data show a high degree of consistency
12.9 Acetonitrile(1)/Benzene(2)-- VLE data
T
318.15 K
31.957
0.1056
33.553
0.0940
0.1818
35.285
0.1829
0.2783
36.457
0.2909
0.3607
36.996
P
0.0455
0.3980
0.4274
37.068
36.978
kPa
x1
0.5069
0.5458
y1
0.4885
0.5098
36.778
0.5946
0.5375
35.792
0.7206
0.6157
34.372
0.8145
0.6913
32.331
0.8972
0.7869
30.038
0.9573
0.8916
411
x2
1
Psat1
x1
y2
1
Psat2
27.778 kPa
y1
29.819 kPa
Calculate EXPERIMENTAL values of activity coefficients and excess
Gibbs energy.
y1 P
1
y2 P
2
x1 Psat1
GERTx1x2
GERT
x2 Psat2
GERT
x1 x2
n
x1 ln 1
n
rows ( )
P
x2 ln 2
i
12
1n
(a) Fit GE/RT data to Margules eqn. by nonlinear least squares.
Minimize sum of the squared errors using the Mathcad Minimize function.
Guesses:
A12
0.3
A21
0.5
C
0.2
SSE A12 A21 C
GERTi
A21 x1
i
A12 x2
i
C x1 x2
i
x1 x2
i
i
i
i
A12
Minimize SSE A12 A21 C
C
1.155
C
A21
1.128
A21
A12
0.53
Ans.
(b) Plot data and fit
GeRTx1x2 ( x2)
x1
GeRT ( x2)
x1
2
x2
ln 2 ( x2)
x1
x1
1 101
A12 x2
C x1 x2
GeRTx1x2 ( x2)x1 x2
x1
ln 1 ( x2)
x1
j
A21 x1
2
2
A12
2 A21
A12
C x1
3 C x1
A21
2 A12
A21
C x2
3 C x2
X1
j
.01 j
412
.01
2
X2
j
1
X1
j
2
1.2
GERTx1x2 i
1
GeRTx1x2 X1 X2
j
j
0.8
ln 1
i
ln 1 X1 X2
j
0.6
j
ln 2
i
0.4
ln 2 X1 X2
j
j
0.2
0
0
0.2
0.4
0.6
0.8
x1 X1 x1 X1 x1 X1
i
j
i
j
i
(c) Plot Pxy diagram with fit and data
1 ( x1 x2)
exp ln 1 ( x1 x2)
2 ( x1 x2)
exp ln 2 ( x1 x2)
Pcalc
X1
j
X1
y1calc
j
1 X1 X2
j
j
j
j
Psat1
X2
j
1 X1 X2 Psat1
j
j
Pcalc
j
413
2 X1 X2
j
j
Psat2
j
P-x,y Diagram from Margules Equation fit to GE/RT data.
38
Pi
36
kPa
34
Pi
kPa
Pcalc
32
j
kPa
30
Pcalc
j
kPa
28
26
0
0.2
0.4
0.6
x1 y1 X1 y1calc
i
i
j
0.8
j
P-x data
P-y data
P-x calculated
P-y calculated
(d) Consistency Test:
ln 1 2i
ln
1 x1 x2
i
i
2 x1 x2
i
i
GERTi
ln
GeRT x1 x2
i
i
GERTi
1i
2
i
0.004
0
GERTi
0
ln 1 2i
0.025
0.004
0
0.5
x1
0.05
1
0
0.5
x1
i
414
i
1
Calculate mean absolute deviation of residuals
mean
GERT
6.237
4
10
mean
ln 1 2
0.025
(e) Barker's Method by non-linear least squares:
Margules Equation
2
1 x1 x2 A 12 A 21 C
exp ( ) A12
x2
2 A21
A12
C x1
A21
C x2
2
3 C x1
2
2 x1 x2 A 12 A 21 C
exp ( ) A21
x1
2 A12
2
3 C x2
Minimize sum of the squared errors using the Mathcad Minimize function.
Guesses:
A12
0.3
SSE A12 A21 C
A21
0.5
C
0.2
x1 x2 A12 A21 C Psat1
i1
i
i
Pi
x1
i
x2
i
2 x1i x2i A 12 A 21 C Psat2
A12
Minimize SSE A12 A21 C
1.114
A21
1.098
C
A12
A21
2
0.387
C
Ans.
Plot P-x,y diagram for Margules Equation with parameters from Barker's
Method.
Pcalc
X1
j
1 X 1 j X 2 j A 12 A 21 C Psat1
j
X2
X1
y1calc
j
j
j
2 X 1 j X 2 j A 12 A 21 C Psat2
1 X 1 j X 2 j A 12 A 21 C Psat1
Pcalc
j
415
38
36
Pi
kPa
34
Pi
kPa
Pcalc
32
j
kPa
Pcalc
kPa
30
j
28
26
0
0.2
0.4
0.6
x1 y1 X1 y1calc
i
i
j
P-x data
P-y data
P-x calculated
P-y calculated
Pcalc
x1
i
1 x1i x2i A 12 A 21 C Psat1
i
x2
2 x1i x2i A 12 A 21 C Psat2
x1
1 x1i x2i A 12 A 21 C Psat1
i
y1calc
i
i
Pcalc
i
416
0.8
j
Plot of P and y1 residuals.
0.6
0.4
Pi Pcalc
i
0.2
kPa
y1 y1calc 100
i
i
0
0.2
0.4
0
0.5
x1
1
i
Pressure residuals
y1 residuals
RMS deviations in P:
Pi
RMS
Pcalc
n
2
i
RMS
i
417
0.04 kPa
12.12 It is impractical to provide solutions for all of the systems listed in the
table on Page 474 we present as an example only the solution for the
system 1-propanol(1)/water(2). Solutions for the other systems can be
obtained by rerunning the following Mathcad program with the appropriate
parameter values substituted for those given. The file WILSON.mcd
reproduces the table of Wilson parameters on Page 474 and includes the
necessary Antoine coefficients.
Antoine coefficients:
1-Propanol:
A1
16.1154
B1
3483.67 K
C1
205.807 K
Water:
A2
16.3872
B2
3885.70 K
C2
230.170 K
Psat1 ( T)
exp A1
Psat2 ( )
T
exp A2
(T
B1
273.15 K)
C1
(T
B2
273.15 K)
C2
kPa
kPa
Parameters for the Wilson equation:
3
V1
a12
775.48
3
cm
75.14
mol
12 ( )
T
V2
a21
cal
mol
cm
18.07
mol
1351.90
a12
V2
exp
RT
V1
exp x2
1 ( x2 T)
x1
2 ( x2 T)
x1
21 ( )
T
x1
x2 12 ( ) x2
T
x1
x2
x1
418
21 ( )
T
21 ( )
T
x2 12 ( ) x2
T
x2
x1
12 ( )
T
12 ( )
T
x1
a21
V1
exp
RT
V2
21 ( )
T
12 ( )
T
x1
exp
cal
mol
21 ( )
T
x1
21 ( )
T
P-x,y diagram at
T
( 60
273.15) K
Guess:
P
Given
P = x1 1 ( x1 1 x1 T) Psat1 ( T)
( 1 x1) 2 ( x1 1 x1 T) Psat2 ( T)
Peq x1)
(
Find P)
(
yeq x1)
(
70 kPa
x1 1 ( x1 1
Peq x1)
(
yeq x)
(
x
x1 T) Psat1 ( T)
Peq x)
(
kPa
0
0
20.007
0.05
0.315
28.324
0.1
0.363
30.009
0.15
0.383
30.639
0.2
0.395
30.97
0.25
0.404
31.182
0.3
0.413
31.331
0.35
0.421
31.435
0.4
0.431
31.496
0.45
0.441
31.51
0.5
0.453
31.467
0.55
0.466
31.353
0.6
0.483
31.148
0.65
0.502
30.827
0.7
0.526
30.355
0.75
0.556
29.686
0.8
0.594
28.759
0.85
0.646
27.491
0.9
0.718
25.769
0.95
0.825
23.437
1
1
20.275
419
x
0 0.05 1.0
P,x,y Diagram at T
333.15 K
32
30
28
Peq x)
(
kPa
Peq x)
(
26
kPa
24
22
20
0
0.2
0.4
0.6
0.8
x yeq x)
(
12.13 It is impractical to provide solutions for all of the systems listed in the
table on Page 474; we present as an example only the solution for the
system 1-propanol(1)/water(2). Solutions for the other systems can be
obtained by rerunning the following Mathcad program with the
appropriate parameter values substituted for those given. The file
WILSON.mcd reproduces the table of Wilson parameters on Page 474
and includes the necessary Antoine coefficients.
Antoine coefficients:
1-Propanol: A1
16.1154
B1
3483.67 K
C1
205.807 K
Water:
16.3872
B2
3885.70 K
C2
230.170 K
A2
Psat1 ( )
T
exp A1
Psat2 ( )
T
exp A2
(
T
B1
273.15 K) C1
kPa
(
T
B2
273.15 K) C2
kPa
420
Parameters for the Wilson equation:
3
V1
75.14
a12
775.48
12 ( T)
3
cm
cm
V2
cal
mol
18.07
a21
mol
1351.90
a12
V2
exp
RT
V1
21 ( T)
x1
1 ( x1 x2 T)
x1
x2
x2
T-x,y diagram at P
x2
x1 21 ( T)
21 ( T)
101.33 kPa
Guess:
x1
273.15) K
Given
Teq ( x1)
yeq ( x1)
x
T
( 90
P = x1 1 ( x1 1 x1 T) Psat1 ( T)
( 1 x1) 2 ( x1 1 x1 T) Psat2 ( T)
Find T)
(
x1 1 ( x1 1
x1 Teq ( x1) ) Psat1 ( Teq ( x1) )
P
0 0.05 1.0
421
21 ( T)
21 ( T)
x2 12 ( T)
x2
x1
12 ( T)
12 ( T)
x1
a21
V1
exp
RT
V2
21 ( T)
x2 12 ( T)
x1
2 ( x1 x2 T)
cal
mol
12 ( T)
exp x2
exp
mol
Teq x
()
K
yeq x
()
x
0
0
373.149
0.05
0.304
364.159
0.1
0.358
362.476
0.15
0.381
361.836
0.2
0.395
361.49
0.25
0.407
361.264
0.3
0.418
361.101
0.35
0.429
360.985
0.4
0.44
360.911
0.45
0.453
360.881
0.5
0.468
360.904
0.55
0.484
360.99
0.6
0.504
361.154
0.65
0.527
361.418
0.7
0.555
361.809
0.75
0.589
362.364
0.8
0.631
363.136
0.85
0.686
364.195
0.9
0.759
365.644
0.95
0.858
367.626
1
1
370.349
T,x,y Diagram at P
101.33 kPa
375
Teq( ) 370
x
K
Teq( )
x
K
365
360
0
0.2
0.4
0.6
x yeq x)
(
422
0.8
1
12.14 It is impractical to provide solutions for all of the systems listed in the
table on Page 474; we present as an example only the solution for the
system 1-propanol(1)/water(2). Solutions for the other systems can be
obtained by rerunning the following Mathcad program with the
appropriate parameter values substituted for those given. The file
NRTL.mcd reproduces the table of NRTL parameters on Page 474 and
includes the necessary Antoine coefficients.
Antoine coefficients:
1-Propanol: A1
16.1154
B1
3483.67 K
C1
205.807 K
Water:
16.3872
B2
3885.70 K
C2
230.170 K
A2
Psat1 ( T)
exp A1
Psat2 ( T)
exp A2
(T
B1
273.15 K)
C1
(T
B2
273.15 K)
C2
kPa
kPa
Parameters for the NRTL equation:
b12
500.40
12 ( T)
exp
b21
1636.57
b12
RT
G12 ( T)
cal
mol
1 ( x1 x2 T)
0.5081
21 ( T )
2
exp x2
2
exp x1
b21
RT
G21 ( T)
12 ( T)
exp
G21 ( T)
x1 x2 G21 ( T)
G12 ( T) 12 ( T)
21 ( T )
2
21 ( T)
( x2
2 ( x1 x2 T)
cal
mol
x1 G12 ( T) )
2
G12 ( T)
12 ( T)
x2 x1 G12 ( T)
G21 ( T) 21 ( T)
( x1
x2 G21 ( T) )
423
2
2
P-x,y diagram at
T
(
60
273.15)K
Guess:
P
Given
P = x1 1 ( 1 x1 T)Psat1 ( )
x1
T
( x1) 2 ( 1 x1 T)Psat2 ( )
1
x1
T
Peq x1)
(
Find P)
(
yeq x1)
(
70 kPa
x1 1 ( 1
x1
Peq x1)
(
yeq x
()
x
x1 T)Psat1 ( )
T
Peq x
()
kPa
0
20.007
0.05
0.33
28.892
0.1
0.373
30.48
0.15
0.382
30.783
0.2
0.386
30.876
0.25
0.39
30.959
0.3
0.395
31.048
0.35
0.404
31.127
0.4
0.414
31.172
0.45
0.427
31.163
0.5
0.442
31.085
0.55
0.459
30.922
0.6
0.479
30.657
0.65
0.503
30.271
0.7
0.531
29.74
0.75
0.564
29.03
0.8
0.606
28.095
0.85
0.659
26.868
0.9
0.732
25.256
0.95
0.836
23.124
1
1
20.275
0
424
x
0 0.05 1.0
P,x,y Diagram at T
333.15 K
35
Peq ( x)
30
kPa
Peq ( x)
kPa
25
20
0
0.2
0.4
0.6
0.8
x yeq ( x)
12.15 It is impractical to provide solutions for all of the systems listed in the
table on Page 474; we present as an example only the solution for the
system 1-propanol(1)/water(2). Solutions for the other systems can be
obtained by rerunning the following Mathcad program with the
appropriate parameter values substituted for those given. The file
NRTL.mcd reproduces the table of NRTL parameters on Page 474 and
includes the necessary Antoine coefficients.
Antoine coefficients:
1-Propanol: A1
16.1154
B1
3483.67 K
C1
205.807 K
Water:
B2
3885.70 K
C2
230.170 K
A2
Psat1 ( T)
exp A1
Psat2 ( T)
exp A2
16.3872
(T
B1
273.15 K)
C1
(T
B2
273.15 K)
C2
kPa
kPa
Parameters for the NRTL equation:
b12
500.40
cal
mol
b21
1636.57
425
cal
mol
0.5081
12 ( )
T
G12 ( )
T
b12
RT
21 ( )
T
exp
1 ( x2 T)
x1
G21 ( )
T
12 ( )
T
2
exp x2
2 ( x2 T)
x1
exp x1
RT
exp
G21 ( )
T
21 ( )
T
x1 x2 G21 ( )
T
G12 ( ) 12 ( )
T
T
2
2
(2
x
2
b21
x1 G12 ( )
T)
G12 ( )
T
12 ( )
T
x2 x1 G12 ( )
T
G21 ( ) 21 ( )
T
T
2
2
(1
x
x2 G21 ( )
T)
T-x,y diagram at
P
Guess:
T
273.15)K
Given
P = x1 1 ( 1 x1 T)Psat1 ( )
x1
T
( x1) 2 ( 1 x1 T)Psat2 ( )
1
x1
T
Teq x1)
(
yeq x1)
(
(
90
101.33 kPa
Find T)
(
x1 1 ( 1
x1
x1 Teq x1) Psat1 ( ( )
()
Teq x1 )
P
426
21 ( )
T
x
0 0.05 1.0
Teq ( x)
K
yeq ( x)
x
0
373.149
0.05
0.32
363.606
0.1
0.377
361.745
0.15
0.394
361.253
0.2
0.402
361.066
0.25
0.408
360.946
0.3
0.415
360.843
0.35
0.424
360.757
0.4
0.434
360.697
0.45
0.447
360.676
0.5
0.462
360.709
0.55
0.48
360.807
0.6
0.5
360.985
0.65
0.524
361.262
0.7
0.552
361.66
0.75
0.586
362.215
0.8
0.629
362.974
0.85
0.682
364.012
0.9
0.754
365.442
0.95
0.853
367.449
1
1
370.349
0
T,x,y Diagram at P
101.33 kPa
375
Teq ( x) 370
K
Teq ( x)
K
365
360
0
0.2
0.4
x yeq ( x)
427
0.6
0.8
1
12.16 It is impractical to provide solutions for all of the systems listed in the
table on Page 474; we present as an example only the solution for the
system 1-propanol(1)/water(2). Solutions for the other systems can be
obtained by rerunning the following Mathcad program with the appropriate
parameter values substituted for those given. The file WILSON.mcd
reproduces the table of Wilson parameters on Page 474 and includes the
necessary Antoine coefficients.
Antoine coefficients:
1-Propanol:
A1
16.1154
B1
3483.67 K
C1
205.807 K
Water:
A2
16.3872
B2
3885.70 K
C2
230.170 K
Psat1 ( )
T
exp A1
Psat2 ( )
T
exp A2
(
T
B1
273.15 K) C1
kPa
(
T
B2
273.15 K) C2
kPa
Parameters for the Wilson equation:
3
V1
a12
775.48
3
cm
75.14
mol
12 ( )
T
V2
a21
cal
mol
cm
18.07
mol
1351.90
a12
V2
exp
RT
V1
exp x2
1 ( x2 T)
x1
2 ( x2 T)
x1
a21
V1
exp
RT
V2
21 ( )
T
21 ( )
T
12 ( )
T
x2 12 ( ) x2
T
x1
x1
exp
cal
mol
x1
x2
x1
12 ( )
T
21 ( )
T
12 ( )
T
x1
x2 12 ( ) x2
T
x2
x1
428
21 ( )
T
21 ( )
T
x1
21 ( )
T
(a) BUBL P:
T
( 60
Guess:
P
101.33 kPa
Given
x1
0.3
x2
1
x1
y1
273.15) K
0.4
y2
1
y1
y1
y2 = 1
y1 P = x1 1 ( x1 x2 T) Psat1 ( T)
y2 P = x2 2 ( x1 x2 T) Psat2 ( T)
Pbubl
Find ( P y1 y2)
y1
y2
Pbubl
0.413
y2
0.587
Ans.
273.15) K
y1
0.3
y2
1
y1
x1
0.1
x2
1
x1
y1
31.33 kPa
(b) DEW P:
T
( 60
Guess:
P
101.33 kPa
y1 P = x1 1 ( x1 x2 T) Psat1 ( T)
Given
x1
x2 = 1
y2 P = x2 2 ( x1 x2 T) Psat2 ( T)
Pdew
Find ( P x1 x2)
x1
x2
Pdew
x1
27.79 kPa
x2
0.042
Ans.
0.958
(c) P,T-flash Calculation
P
Pdew
Guess:
Pbubl
T
2
V
Given
y1 =
y2 =
0.5
x1
y1
( 60
0.1
0.1
x2
y2
z1
273.15) K
x1 1 ( x1 x2 T) Psat1 ( T)
P
x2 2 ( x1 x2 T) Psat2 ( T)
P
429
1
1
x1
x2 = 1
y1
y2 = 1
y1
x1
0.3
x1 (
1
Eq. (10.15)
V) y1 V = z1
x1
x2
y1
Find x1 x2 y1 y2 V)
(
y2
V
x1
x2
0.08
0.92
y1
y2
0.351
V
0.649
0.813
(d) Azeotrope Calculation
Test for azeotrope at:
T
1 ( 1 T)Psat1 ( )
0
T
120
Psat1 ( )
T
2 ( 0 T)Psat2 ( )
1
T
21.581
121
Psat2 ( )
T
121
273.15)K
2 ( 0 T) 4.683
1
1 ( 1 T) 21.296
0
120
(
60
0.216
Since one of these values is >1 and the other is <1, an azeotrope exists.
See Ex. 10.3(e)
Guess:
Given
P
101.33 kPa
x1
y1
x2
y2
0.3
0.3
1
1
y1
x1
y1 P = x1 1 ( x2 T)Psat1 ( )
x1
T
y2 P = x2 2 ( x2 T)Psat2 ( )
x1
T
x1
y1
x2 = 1
x1 = y1
y2 = 1
x1
x2
y1
Find x1 x2 y1 y2 P)
(
y2
Paz
Paz
31.511 kPa
x1
0.4386
430
y1
0.4386
Ans.
12.17
It is impractical to provide solutions for all of the systems listed in the
table on Page 474; we present as an example only the solution for the
system 1-propanol(1)/water(2). Solutions for the other systems can be
obtained by rerunning the following Mathcad program with the
appropriate parameter values substituted for those given. The file
NRTL.mcd reproduces the table of NRTL parameters on Page 474 and
includes the necessary Antoine coefficients.
Antoine coefficients:
1-Propanol: A1
16.1154
B1
3483.67 K
C1
205.807 K
Water:
16.3872
B2
3885.70 K
C2
230.170 K
A2
Psat1 ( T)
exp A1
Psat2 ( T)
exp A2
(T
B1
273.15 K)
C1
(T
B2
273.15 K)
C2
kPa
kPa
Parameters for the NRTL equation:
b12
500.40
12 ( T)
G12 ( T)
cal
mol
b21
1636.57
b12
RT
exp
1 ( x1 x2 T)
0.5081
b21
21 ( T )
12 ( T)
2
exp x2
2
exp x1
RT
G21 ( T)
exp
G21 ( T)
x1 x2 G21 ( T)
G12 ( T) 12 ( T)
21 ( T )
2
21 ( T)
( x2
2 ( x1 x2 T)
cal
mol
x1 G12 ( T) )
2
G12 ( T)
12 ( T)
x2 x1 G12 ( T)
G21 ( T) 21 ( T)
( x1
x2 G21 ( T) )
431
2
2
(a) BUBL P: T
(
60
Guess:
P
101.33 kPa
Given
y1 P = x1 1 ( x2 T)Psat1 ( )
x1
T
x1
0.3
x2
1
x1
y1
273.15)K
0.4
y2
1
y1
y1
y2 P = x2 2 ( x2 T)Psat2 ( )
x1
T
y2 = 1
Pbubl
Find ( y1 y2)
P
y1
y2
Pbubl
y1
31.05 kPa
y2
0.395
(b) DEW P:
T
(
60
Guess:
P
101.33 kPa
Given
Ans.
0.605
y1
0.3
y2
1
y1
x1
273.15)K
0.1
x2
1
x1
y1 P = x1 1 ( x2 T)Psat1 ( )
x1
T
y2 P = x2 2 ( x2 T)Psat2 ( )
x1
T
x1
x2 = 1
Pdew
Find ( x1 x2)
P
x1
x2
Pdew
x1
27.81 kPa
x2
0.037
0.963
Ans.
(c) P,T-flash Calculation
P
Pdew
Pbubl
Guess:
Given
T
2
V
y1 =
y2 =
x1 (
1
0.5
(
60
x1
y1
273.15)K
0.1
0.1
x1 1 ( x2 T)Psat1 ( )
x1
T
P
x2 2 ( x2 T)Psat2 ( )
x1
T
P
V) y1 V = z1
Eq. (10.15)
432
z1
0.3
x2
y2
1
1
x1
x2 = 1
y1
y2 = 1
y1
x1
x1
x2
y1
Find ( x1 x2 y1 y2 V)
y2
V
x1
x2
0.06
y1
0.94
y2
0.345
V
0.655
0.843
(d) Azeotrope Calculation
Test for azeotrope at: T
1 ( 0 1 T)
( 60
273.15) K
2 ( 1 0 T)
19.863
1 ( 0 1 T) Psat1 ( T)
Psat2 ( T)
120
120
2 ( 1 0 T) Psat2 ( T)
20.129
121
Psat1 ( T)
121
4.307
0.235
Since one of these values is >1 and the other is <1, an azeotrope exists.
See Ex. 10.3(e).
Guess:
Given
P
101.33 kPa
x1
y1
x2
y2
0.3
0.3
1
1
x1
x1
y1 P = x1 1 ( x1 x2 T) Psat1 ( T)
y2 P = x2 2 ( x1 x2 T) Psat2 ( T)
x1
y1
x2 = 1
x1 = y1
y2 = 1
x1
x2
y1
Find ( x1 x2 y1 y2 P)
y2
Paz
Paz
31.18 kPa
x1
0.4187
433
y1
0.4187
Ans.
12.18
It is impractical to provide solutions for all of the systems listed in the
table on Page 474; we present as an example only the solution for the
system 1-propanol(1)/water(2). Solutions for the other systems can be
obtained by rerunning the following Mathcad program with the
appropriate parameter values substituted for those given. The file
WILSON.mcd reproduces the table of Wilson parameters on Page 474
and includes the necessary Antoine coefficients.
Antoine coefficients:
1-Propanol:
A1
16.1154
B1
3483.67 K
C1
205.807 K
Water:
A2
16.3872
B2
3885.70 K
C2
230.170 K
Psat1 ( )
T
exp A1
Psat2 ( )
T
exp A2
(
T
B1
273.15 K) C1
kPa
(
T
B2
273.15 K) C2
kPa
Parameters for the Wilson equation:
3
V1
a12
775.48
3
cm
75.14
mol
12 ( )
T
V2
a21
cal
mol
cm
18.07
mol
1351.90
a12
V2
exp
RT
V1
exp x2
1 ( x2 T)
x1
a21
V1
exp
RT
V2
21 ( )
T
12 ( )
T
x2 12 ( ) x2
T
x1
x1
exp
2 ( x2 T)
x1
21 ( )
T
cal
mol
x1
x2
x1
12 ( )
T
21 ( )
T
12 ( )
T
x1
x2 12 ( ) x2
T
x2
434
x1
21 ( )
T
21 ( )
T
x1
21 ( )
T
(a) BUBL T:
P
x1
Guess:
T
Given
0.3
x2
1
x1
y1
101.33 kPa
0.3
y2
1
y1
y1 P = x1 1 ( x1 x2 T) Psat1 ( T)
( 60
273.15) K
y1
y2 = 1
y2 P = x2 2 ( x1 x2 T) Psat2 ( T)
Tbubl
y1
Find ( T y1 y2)
y2
Tbubl
y1
361.1 K
y2
0.418
Ans.
0.582
(b) DEW T:
P
101.33 kPa
y1
0.3
y2
1
x1
Guess:
T
( 60
x1
0.1
x2
1
y1
x1
x2 = 1
Given
273.15) K
y1 P = x1 1 ( x1 x2 T) Psat1 ( T)
y2 P = x2 2 ( x1 x2 T) Psat2 ( T)
Tdew
x1
Find ( T x1 x2)
x2
Tdew
x1
364.28 K
0.048
x2
Ans.
0.952
(c) P,T-flash Calculation
T
Tbubl
Tdew
Guess:
Given
P
2
V
y1 =
y2 =
x1 ( 1
x1
y1
0.5
101.33 kPa
0.1
0.1
x1 1 ( x1 x2 T) Psat1 ( T)
P
x2 2 ( x1 x2 T) Psat2 ( T)
P
V)
y1 V = z1
Eq. (10.15)
435
z1
0.3
x2
y2
1
1
y1
x1
x1
x2 = 1
y1
y2 = 1
x1
x2
y1
Find x1 x2 y1 y2 V)
(
y2
V
x1
x2
0.09
y1
0.91
y2
0.35
V
0.65
0.807
(d) Azeotrope Calculation
Test for azeotrope at: P
B1
Tb1
A1
C1
A2
P
ln
kPa
273.15 K
Tb1
370.349 K
C2
P
ln
kPa
B2
Tb2
101.33 kPa
273.15 K
Tb2
373.149 K
2 ( 0 Tb1) 3.779
1
1 ( 1 Tb2) 16.459
0
1 ( 1 T)Psat1 ( )
0
Tb2
120
120
P
121
2 ( 0 T)Psat2 ( )
1
Tb1
19.506
121
P
0.281
Since one of these values is >1 and the other is <1, an azeotrope exists.
See Ex. 10.3(e). Guesses:
T
(
60
Given
273.15)K
x1
0.4 x2
1
y1 y1
y1 P = x1 1 ( x2 T)Psat1 ( ) x1
x1
T
y2 = 1
y2
1
x2 = 1
y2 P = x2 2 ( x2 T)Psat2 ( ) y1
x1
T
0.4
436
x1 = y1
x1
x1
x2
y1
Find ( x1 x2 y1 y2 T)
y2
Taz
Taz
360.881 K
x1
0.4546
y1
0.4546
Ans.
12.19 It is impractical to provide solutions for all of the systems listed in the
table on page 474; we present as an example only the solution for the
system 1-propanol(1)/water(2). Solutions for the other systems can be
obtained by rerunning the following Mathcad program with the
appropriate parameter values substituted for those given. The file
NRTL.mcd reproduces the table of NRTL parameters on Page 474 and
includes the necessary Antoine coefficients.
Antoine coefficients:
1-Propanol: A1
16.1154
B1
3483.67 K
C1
205.807 K
Water:
16.3872
B2
3885.70 K
C2
230.170 K
A2
Psat1 ( T)
exp A1
Psat2 ( T)
exp A2
(T
B1
273.15 K)
C1
(T
B2
273.15 K)
C2
kPa
kPa
Parameters for the NRTL equation:
b12
500.40
12 ( T)
G12 ( T)
cal
mol
b21
1636.57
b12
RT
exp
cal
mol
21 ( T)
12 ( T)
G21 ( T)
437
0.5081
b21
RT
exp
21 ( T)
1 ( x2 T)
x1
G21 ( )
T
x1 x2 G21 ( )
T
G12 ( ) 12 ( )
T
T
2
exp x2
2
(2
x
2 ( x2 T)
x1
2
21 ( )
T
x1 G12 ( )
T)
G12 ( )
T
x2 x1 G12 ( )
T
G21 ( ) 21 ( )
T
T
2
exp x1
2
12 ( )
T
2
(1
x
x2 G21 ( )
T)
(a) BUBL T:
P
101.33 kPa
x1
0.3
x2
1
x1
Guess:
T
(
60
y1
0.3
y2
1
y1
Given
273.15)K
y1 P = x1 1 ( x2 T)Psat1 ( )
x1
T
y1
y2 P = x2 2 ( x2 T)Psat2 ( )
x1
T
y2 = 1
Tbubl
y1
Find ( y1 y2)
T
y2
Tbubl
y1
360.84 K
0.415
y2
Ans.
0.585
(b) DEW T:
P
101.33 kPa
y1
0.3
y2
1
x1
Guess:
T
(
90
x1
0.05
x2
1
y1
Given
273.15)K
y1 P = x1 1 ( x2 T)Psat1 ( )
x1
T
x1
y2 P = x2 2 ( x2 T)Psat2 ( )
x1
T
x2 = 1
Tdew
x1
Find ( x1 x2)
T
x2
Tdew
364.27 K
x1
0.042
438
x2
0.958
Ans.
(c) P,T-flash Calculation
Tdew
T
Tbubl
P
2
Guess:
Given
V
x1
y1
0.5
0.1
0.1
x1 1 ( x1 x2 T) Psat1 ( T)
y1 =
P
V)
1
1
y1
x1
x2 = 1
y1
x2 2 ( x1 x2 T) Psat2 ( T)
x1 ( 1
0.3
x1
P
y2 =
z1
x2
y2
101.33 kPa
y2 = 1
y1 V = z1 Eq. (10.15)
x1
x2
y1
Find ( x1 x2 y1 y2 V)
y2
V
x1
0.069
x2
y1
0.931
0.352
y2
0.648
V
0.816
(d) Azeotrope Calculation
Test for azeotrope at: P
B1
Tb1
A1
P
ln
kPa
B2
Tb2
A2
1 ( 0 1 Tb2)
P
ln
kPa
101.33 kPa
C1
273.15 K
Tb1
370.349 K
C2
273.15 K
Tb2
373.149 K
2 ( 1 0 Tb1)
14.699
439
4.05
1 ( 1 T)Psat1 ( )
0
Tb2
120
120
P
121
2 ( 0 T)Psat2 ( )
1
Tb1
17.578
121
P
0.27
Since one of these values is >1 and the other is <1, an azeotrope exists.
See Ex. 10.3(e). Guesses:
T
(
90
x1
273.15)K
0.4 x2
1
y1 y1
y1 P = x1 1 ( x2 T)Psat1 ( ) x1
x1
T
y2 = 1
1
x1
x2 = 1
y2 P = x2 2 ( x2 T)Psat2 ( ) y1
x1
T
Given
y2
0.4
x1 = y1
x1
x2
y1
Find x1 x2 y1 y2 T)
(
y2
Taz
Taz
360.676 K
x1
0.4461
y1
Ans.
0.4461
12.20 Molar volumes & Antoine coefficients:
74.05
V
40.73
14.3145
A
18.07
Psat ( T)
i
2756.22
16.5785
B
16.3872
exp Ai
3638.27
kPa
273.15
239.500
a
230.170
T
(
65
583.11
161.88 291.27
0
1448.01 469.55
440
273.15)
K
Ci
0
Wilson parameters:
C
3885.70
Bi
T
K
228.060
107.38
0
cal
mol
Vj
( i j T)
Vi
exp
ai j
RT
i
j
13
(a)
0.3
p
13
BUBL P calculation: No iteration required.
x1
13
x2
( i x T)
exp 1
x3
0.4
ln
xj
1
x1
x2
( i j T)
j
xp
( p i T)
xj
p
( p j T)
j
Pbubl
xi ( i x T) Psat ( i T)
yi
xi ( i x T) Psat ( i T)
Pbubl
i
0.527
y
0.367
Pbubl
Ans.
117.1 kPa
0.106
(b) DEW P calculation:
y1
Guess:
0.3
y2
0.4
y3
1
y1
y2
x1
0.05
x2
0.2
x3
1
x1
x2
P
Pbubl
Given
P y1 = x1 ( 1 x T) Psat ( 1 T)
P y2 = x2 ( 2 x T) Psat ( 2 T)
P y3 = x3 ( 3 x T) Psat ( 3 T)
i
x1
x2
x3
xi = 1
Find x1 x2 x3 P
Pdew
441
0.035
x
0.19
Pdew
69.14 kPa
Ans.
0.775
(c) P,T-flash calculation:
z1
0.3
Guess:
V
z2
Pdew
P
0.4
Pbubl
T
2
z3
1
z1
338.15 K
z2
Use x from DEW P and y from BUBL P as initial
guess.
0.5
Given
P y1 = x1 ( x T)Psat1 T)
1
(
x1 (
1
V) y1 V = z1
P y2 = x2 ( x T)Psat2 T)
2
(
x2 (
1
V) y2 V = z2
P y3 = x3 ( x T)Psat3 T)
3
(
x3 (
1
V) y3 V = z3
xi = 1
yi = 1
i
i
x1
x2
x3
y1
Find x1 x2 x3 y1 y2 y3 V
y2
y3
V
0.391
0.109
x
0.345
0.546
y
0.426
V
0.183
442
0.677
Ans.
12.21 Molar volumes & Antoine coefficients:
Antoine coefficients:
74.05
V
14.3145
40.73
A
16.5785
18.07
T
( 65
B
Psat ( i T)
0
T
K
i
13
0
b
0.2994
0.5343 0.2994
j
230.170
Bi
exp Ai
0.3084 0.5343
0.3084
0.4
bi j
RT
x3
1
x1
Ci
253.88
1197.41 845.21
13
x2
273.15
0
Gi j
l 13
k
13
(a) BUBL P calculation: No iteration required.
0.3
kPa
184.70 631.05
222.64
0
ij
x1
239.500
3885.70
NRTL parameters:
0
C
3638.27
16.3872
273.15)K
228.060
2756.22
cal
mol
0
exp
ij ij
x2
j i G j i xj
( i x T)
exp
j
Gl i xl
l
xk k j G k j
x j Gi j
k
Gl j xl
j
ij
G l j xl
l
l
Pbubl
xi ( i x T) Psat ( i T)
yi
i
0.525
y
0.37
Pbubl
115.3 kPa
0.105
443
Ans.
xi ( i x T) Psat ( i T)
Pbubl
(b) DEW P calculation:
y1
y2
0.4
y3
1
y1
y2
x1
Guess:
0.3
0.05
x2
0.2
x3
1
x1
x2
P
Pbubl
Given
2
(
P y1 = x1 ( x T)Psat1 T) P y2 = x2 ( x T)Psat2 T)
1
(
xi = 1
P y3 = x3 ( x T)Psat3 T)
3
(
i
x1
x2
Find x1 x2 x3 P
x3
Pdew
0.038
x
0.192
Pdew
68.9 kPa
Ans.
0.77
(c) P,T-flash calculation:
z1
z2
0.3
Guess:
V
0.5
0.4
Pdew
P
Pbubl
2
z3
1
z1
T
338.15 K
z2
Use x from DEW P and y from BUBL P as initial
guess.
1
(
Given P y1 = x1 ( x T)Psat1 T)
x1 (
1
V) y1 V = z1
P y2 = x2 ( x T)Psat2 T)
2
(
x2 (
1
V) y2 V = z2
P y3 = x3 ( x T)Psat3 T)
3
(
x3 (
1
V) y3 V = z3
yi = 1
xi = 1
i
i
444
x1
x2
x3
y1
Find x1 x2 x3 y1 y2 y3 V
y2
y3
V
0.391
0.118
x
y
0.347
0.426
V
Ans.
0.667
0.183
0.534
12.22 Molar volumes & Antoine coefficients:
74.05
V
14.3145
40.73
A
16.5785
18.07
Psat ( i T)
2756.22
B
3638.27
16.3872
T
K
C
273.15
Wilson parameters:
230.170
P
kPa
a
161.88 291.27
583.11
0
1448.01 469.55
Vj
Vi
exp
ai j
RT
(a)
0.3
13
j
BUBL T calculation:
x1
i
x2
0.4
x3
445
101.33kPa
Ci
0
( i j T)
239.500
3885.70
Bi
exp Ai
228.060
1
x1
x2
13
107.38
cal
mol
0
p
13
( x T)
i
exp 1
ln
xj
( j T)
i
j
xp
( i T)
p
xj
p
( j T)
p
j
Guess:
T
300K
y1
0.3
y2
0.3
y3
1
y1
y2
Given
P y1 = x1 ( x T)Psat1 T)
1
(
P y2 = x2 ( x T)Psat2 T)
2
(
P y3 = x3 ( x T)Psat3 T)
3
(
P=
xi ( x T)Psati T)
i
(
i
y1
y2
Find y1 y2 y3 T
y3
Tbubl
0.536
y
0.361
Tbubl
Ans.
334.08K
0.102
(b) DEW T calculation:
y1
Guess:
0.3
y2
0.4
y3
1
y1
y2
x1
0.05
x2
0.2
x3
1
x1
x2
T
Tbubl
Given
P y1 = x1 ( x T)Psat1 T) P y2 = x2 ( x T)Psat2 T)
1
(
2
(
P y3 = x3 ( x T)Psat3 T)
3
(
xi = 1
i
446
x1
x2
Find x1 x2 x3 T
x3
Tdew
0.043
x
0.204
Tdew
Ans.
347.4 K
0.753
(c) P,T-flash calculation:
z1
0.3
Guess:
V
z2
0.5
0.2
Tdew
T
Tbubl
2
z3
1
z1
T
340.75 K
z2
Use x from DEW P and y from BUBL P as initial
guess.
Given P y1 = x1 ( 1 x T) Psat ( 1 T)
x1 ( 1
V)
y1 V = z1
P y2 = x2 ( 2 x T) Psat ( 2 T)
x2 ( 1
V)
y2 V = z2
P y3 = x3 ( 3 x T) Psat ( 3 T)
x3 ( 1
V)
y3 V = z3
xi = 1
i
yi = 1
i
x1
x2
x3
y1
Find x1 x2 x3 y1 y2 y3 V
y2
y3
V
447
0.536
0.125
x
y
0.17
0.241
V
Ans.
0.426
0.223
0.705
12.23 Molar volumes & Antoine coefficients:
Antoine coefficients:
74.05
V
14.3145
40.73
A
16.5785
18.07
P
B
Psati T)
(
0
T
K
i
0
13
j
b
0.2994
0.5343 0.2994
273.15
0
13
G ( j T)
i
13
exp
( j T)
i
ij
cal
mol
0
( j T)
i
k
Ci
253.88
1197.41 845.21
l
kPa
184.70 631.05
222.64
0
13
230.170
Bi
exp Ai
0.3084 0.5343
0.3084
239.500
3885.70
NRTL parameters:
0
C
3638.27
16.3872
101.33kPa
228.060
2756.22
bi j
RT
(a) BUBL T calculation:
x1
0.3
x2
0.4
x3
1
x1
x2
( i T)G ( i T)x j
j
j
( x T)
i
exp
j
G ( i T)xl
l
l
xk ( j T)G ( j T)
k
k
x j G ( j T)
i
( j T)
i
k
G ( j T)xl
l
G ( j T)xl
l
j
l
l
448
Guess:
T
y1
300K
y2
0.3
y3
0.3
1
y1
y2
Given
P y1 = x1 ( 1 x T) Psat ( 1 T)
P y2 = x2 ( 2 x T) Psat ( 2 T)
P y3 = x3 ( 3 x T) Psat ( 3 T)
P=
xi ( i x T) Psat ( i T)
i
y1
y2
Find y1 y2 y3 T
y3
Tbubl
0.533
y
0.365
Tbubl
Ans.
334.6 K
0.102
(b) DEW T calculation:
y1
Guess:
0.3
y2
0.4
y3
1
y1
y2
x1
0.05
x2
0.2
x3
1
x1
x2
T
Tbubl
Given
P y1 = x1 ( 1 x T) Psat ( 1 T)
P y2 = x2 ( 2 x T) Psat ( 2 T)
xi = 1
P y3 = x3 ( 3 x T) Psat ( 3 T)
i
x1
x2
Find x1 x2 x3 T
x3
Tdew
0.046
x
0.205
Tdew
347.5 K
0.749
449
Ans.
(c) P,T-flash calculation:
z1
0.3
Guess:
V
z2
Tdew
T
0.2
Tbubl
2
z3
1
z1
T
341.011 K
z2
Use x from DEW P and y from BUBL P as initial
guess.
0.5
Given P y1 = x1 ( x T)Psat1 T)
1
(
x1 (
1
V) y1 V = z1
P y2 = x2 ( x T)Psat2 T)
2
(
x2 (
1
V) y2 V = z2
P y3 = x3 ( x T)Psat3 T)
3
(
x3 (
1
V) y3 V = z3
xi = 1
yi = 1
i
i
x1
x2
x3
y1
Find x1 x2 x3 y1 y2 y3 V
y2
y3
V
0.537
0.133
x
0.173
0.694
y
0.238
V
0.225
450
0.414
Ans.
3
3
12.26
x1
x2
0.4
1
V1
x1
110
cm
V2
mol
90
x1 x2 45 x1
25 x2
By Eq. (12.27): V x1 x2
cm
VE x1 x2
mol
VE x1 x2
mol
3
3
VE x1 x2
cm
x1 V1
7.92
cm
mol
x2 V2
3
V x1 x2
105.92
cm
mol
By Eqs. (11.15) & (11.16):
Vbar1
V x 1 x2
Vbar2
3
d
x2
V x1 x2
dx1
V x1 x2
x1
cm
190.28
mol
Vbar1
Ans.
3
d
dx1
V x1 x2
Vbar2
49.68
cm
mol
Check by Eq. (11.11):
3
V
x1 Vbar1
x2 Vbar2
V
cm
58.63
mol
moles
V2
3
x1
750 cm
V1
moles1
moles2
moles
moles2
moles1
x1
moles
1500 cm
V2
25.455 mol
x2
0.503
1
x1 x2
1.026
0.220 x1
x2
cm
mol
x1
3
3
VE
OK
cm
118.46
mol
3
moles1
cm
mol
3
3
12.27 V1
105.92
VE
cm
0.256
mol
3
By Eq. (12.27),
V
VE x1 V1
451
x2 V 2
V
88.136
cm
mol
Vtotal
Vtotal
V moles
3
2243 cm
Ans.
For an ideal solution, Eq. (11.81) applies:
Vtotal
x1 V 1
x2 V2 moles
Vtotal
3
2250 cm
Ans.
12.28 LiCl.2H2O ---> Li + 1/2 Cl2 + 2 H2 + O2 (1)
Li + 1/2 Cl2 + 10 H2O ---> LiCl(10 H2O) (2)
2(H2 + 1/2 O2 ---> H2O)
(3)
-------------------------------------------------------------------LiCl.2H2O + 8 H2O(l) ---> LiCl(10 H2O)
H1
( 1012650)J
(Table C.4)
H2
441579 J
(Pg. 457)
H3
H
H1
H
589 J
(Table C.4)
2 ( 285830 J)
H2
H3
(On the basis of 1 mol of solute)
Since there are 11 moles of solution per mole of solute, the result on the
basis of 1 mol of solution is
H
11
12.29
53.55 J
Ans.
2(HCl + 2.25 H2O -----> HCl(2.25 H2O)) (1)
HCl(4.5 H2O) -----> HCl + 4.5 H2O (2)
---------------------------------------------HCl(4.5 H2O) + HCl -----> 2 HCl(2.25 H2O)
H1
2 ( 50.6 kJ)
(Fig. 12.14 @ n=2.25)
H2
62 kJ
(Fig. 12.14 @ n=4.5 with sign change)
H
H1
H
39.2 kJ
H2
Ans.
452
12.30 Calculate moles of LiCl and H2O in original solution:
nLiCl
nLiCl
0.1 125
kmol
42.39
0.295 kmol
3
20
kmol
42.39
nH2O
n'LiCl
0.472 kmol
21.18
nLiCl
nH2O
Mole ratio, final solution:
kmol
6.245 10 mol
n'LiCl
Mole ratio, original solution:
n'LiCl
18.015
nH2O
Moles of LiCl added:
nLiCl
0.9 125
nH2O
nLiCl
n'LiCl
8.15
0.7667 kmol
0.2949(LiCl(21.18 H2O) ---> LiCl + 21.18 H2O) (1)
0.7667(LiCl + 8.15 H2O ---> LiCl(8.15 H2O)) (2)
--------------------------------------------------------------------------------------0.2949 LiCL(21.18 H2O) + 0.4718 LiCl ---> 0.7667 LiCl(8.145 H2O)
H1
H2
Q
nLiCl 35
nLiCl
H1
kJ
mol
(Fig. 12.14, n=21.18)
n'LiCl
H2
32
Q
kJ
mol
(Fig. 12.14, n=8.15)
14213 kJ
Ans.
12.31 Basis: 1 mole of 20% LiCl solution entering the process.
Assume 3 steps in the process:
1. Heat M1 moles of water from 10 C to 25 C
2. Unmix 1 mole (0.8 moles water + 0.2 moles LiCl) of 20 % LiCl solution
3. Mix (M1 + 0.8) moles of water and 0.2 moles of LiCl
453
Step 1: From Steam Tables
H1
104.8
H1
1.132
kJ
kg
41.99
kJ
kg
18.015
kg
kmol
kJ
mol
Step 2: From Fig. 12.14 with n = 4 moles H2O/mole solute:
H2
kJ
mol
25.5
Step 3: Guess M1 and find H3 solution from Figure 12.14. Calculate H
for process. Continue to guess M1 until H =0 for adiabatic process.
M1
1.3 mol
n3
n3
H
M1 H1
0.2 mol H2
H
0.061 kJ
M1
H3
0.2 mol
33.16
kJ
mol
10.5
0.2 mol H3
Close enough
0.2 mol
M1 1 mol
x
0.8 mol
x
Ans.
0.087
12.32 H2O @ 5 C
----->
H2O @ 25 C
(1)
LiCl(3 H2O)
----->
LiCl + 3 H2O (2)
LiCl + 4 H2O
----->
LiCl(4 H2O)
(3)
-------------------------------------------------------------------------H2O @ 5 C + LiCl(3 H2O) -----> LiCl(4 H2O)
H1
104.8
H2
20.756
H3
25.5
H
H1
kJ
kg
kJ
mol
kJ
mol
H2
21.01
kJ
gm
18.015
kg
mol
H1
1.509
kJ
mol
From p. 457 ( H LiCl(s) - H LiCl in 3 mol H2O)
From Figure 12.14
H3 0.2 mol
454
H
646.905 J
Ans.
12.33
(a) LiCl + 4 H2O -----> LiCl(4H2O) H
0.2 mol H
5.1 kJ
25.5
kJ
From Figure 12.14
mol
Ans.
(b) LiCl(3 H2O)
-----> LiCl + 3 H2O (1)
LiCl + 4 H2O
-----> LiCl(4 H2O) (2)
----------------------------------------------------LiCl(3 H2O) + H2O -----> LiCl(4 H2O)
H1
20.756
H2
25.5
H
kJ
mol From p. 457 ( H LiCl(s) - H LiCl in 3 mol H2O)
kJ
From Figure 12.14
mol
0.2 mol
H1
H
H2
0.949 kJ Ans.
(c) LiCl*H2O
----->
Li +1/2 Cl2 + H2 + 1/2 O2 (1)
H2 + 1/2 O2
----->
H2O
(2)
Li + 1/2 Cl2
----->
LiCl
(3)
LiCl + 4 H2O
----->
LiCl(4 H2O)
(4)
---------------------------------------------------------------------LiCl*H2O + 3 H2O -----> LiCl(4 H2O)
H1
712.58
kJ
mol
H2
285.83
kJ
mol
H3
408.61
kJ
mol
H4
25.5
H
(d)
0.2 mol
kJ
mol
H1
From p. 457 for LiCl.H2O
From Table C.4 Hf H2O(l)
From p. 457 for LiCl
From Figure 12.14
H2
H3
H4
H
LiCl + 4 H2O
----->
LiCl(4 H2O) (1)
4/9 (LiCl(9 H2O)
----->
LiCl + 9 H2O) (2)
--------------------------------------------------------------5/9 LiCl + 4/9 LiCl(9 H2O) -----> LiCl(4 H2O)
455
1.472 kJ
Ans.
H1
25.5
4
H2
H
(e)
9
kJ
mol
()
32.4
0.2 mol
From Figure 12.14
kJ
From Figure 12.14
mol
H1
H2
H
2.22 kJ
Ans.
5/6 (LiCl(3 H2O)
----->
LiCl + 3 H2O) (1)
1/6 (LiCl(9 H2O)
----->
LiCl + 9 H2O) (2)
LiCl + 4 H2O
----->
LiCl(4 H2O) (3)
-----------------------------------------------------------------------5/6 LiCl(3 H2O) + 1/6 LiCl(9 H2O) -----> LiCl(4 H2O)
H1
kJ
5
(
20.756)
mol
6
From p. 457 ( H LiCl(s) - H LiCl in 3 mol H2O)
H2
kJ
1
()
32.4
mol
6
From Figure 12.14
H3
H
(f)
kJ
mol
25.5
0.2 mol
H1
From Figure 12.14
H2
H3
H
0.561 kJ
Ans.
5/8 (LiCl*H2O
----->
Li +1/2 Cl2 + H2 + 1/2 O2) (1)
5/8 (H2 + 1/2 O2
----->
H2O)
(2)
3/8 (LiCl(9 H2O)
----->
LiCl + 9 H2O)
(3)
5/8 (Li + 1/2 Cl2
----->
LiCl
(4)
LiCl + 4 H2O
----->
LiCl(4 H2O)
(5)
---------------------------------------------------------------------------------------5/8 LiCl*H2O + 3/8 LiCl(9 H2O) -----> LiCl(4 H2O)
H1
H2
kJ
5
(
712.58)
mol
8
5
8
( 285.83)
kJ
mol
From p. 457 for LiCl.H2O
From Table C.4 Hf H2O(l)
456
kJ
3
( 32.4)
mol
8
H3
5
H4
8
H5
H
12.34
( 408.61)
kJ
mol
From p. 457 for LiCl
kJ
25.5
0.2 mol
From Figure 12.14
From Figure 12.14
mol
H1
H2
H3
H4
H
H5
0.403 kJ
Ans.
BASIS: 1 second, during which the following are mixed:
(1) 12 kg hydrated (6 H2O) copper nitrate
(2) 15 kg H2O
n1
n1
12 kmol
295.61 sec
0.041
15 kmol
18.015 sec
n2
kmol
sec
n2
0.833
6 n1
Mole ratio, final solution:
kmol
sec
n2
26.51
n1
6(H2 + 1/2 O2 ---> H2O(l))
Cu + N2 + 3 O2 ---> Cu(NO3)2
(1)
(2)
Cu(NO3)2.6H2O ---> Cu + N2 + 6 O2 + 6 H2
(3)
Cu(NO3)2 + 20.51 H2O ---> Cu(NO3)2(20.51 H2O) (4)
-----------------------------------------------------------------------------------------------Cu(NO3)2.6H2O + 14.51 H2O(l) ---> Cu(NO3)2(20.51 H2O)
H1
H2
H
(Table C.4)
6 ( 285.83 kJ)
H3
302.9 kJ
H1
H2
H3
H4
( 2110.8 kJ)
H
H4
457
45.08 kJ
47.84 kJ
This value is for 1 mol of the hydrated copper nitrate. On the basis of 1
second,
kJ
H
Ans.
Q 1830
Q
n
1
sec
mol
12.35 LiCl.3H2O ---> Li + 1/2 Cl2 + 3H2 + 3/2 O2 (1)
3(H2 + 1/2 O2 ---> H2O(l))
(2)
2(Li + 1/2 Cl2 + 5 H2O ---> LiCl(5H2O)) (3)
LiCl(7H2O) ---> Li + 1/2 Cl2 + 7 H2O
(4)
------------------------------------------------------------------------------LiCl(7H2O) + LiCl.3H2O ---> 2 LiCl(5H2O)
H1
1311.3 kJ
H2
H3
2 ( 436.805 kJ)
H4
( 439.288 kJ)
H
H1
(Table C.4)
3 ( 285.83 kJ)
Q
H2
Q
H
H3
H
H4
(Pg. 457)
19.488 kJ
Ans.
19.488 kJ
12.36 Li + 1/2 Cl2 + (n+2)H2O ---> LiCl(n+2 H2O) (1)
2(H2 + 1/2 O2 ---> H2O)
(2)
LiCl.2H2O ---> Li + 1/2 Cl2 + 2H2 + O2
(3)
-------------------------------------------------------------------------------------LiCl.2H2O + n H2O ---> LiCl(n+2 H2O)
2 ( 285.83 kJ)
H3
1012.65 kJ
Since the process is isothermal,
H=
H1
H2
Since it is also adiabatic,
Therefore,
H1
H2
H2
(Table C.4)
H3
H=0
H1
H3
440.99 kJ
Interpolation in the table on pg. 457 shows that the LiCl is dissolved in
8.878 mol H2O.
xLiCl
1
9.878
xLiCl
Ans.
0.1012
458
10
867.85
20
870.06
25
n
862.74
15
12.37 Data:
871.07
Hf
50
872.91 kJ
100
873.82
300
874.79
500
875.13
1000
875.54
Ca + Cl2 + n H2O ---> CaCl2(n H2O)
CaCl2(s) ---> Ca + Cl2
HfCaCl2
-------------------------------------------CaCl2(s) + n H2O ---> CaCl2(n H2O)
From Table C.4:
i
Hf
HfCaCl2
Htilde
795.8 kJ
1 rows ( n)
65
70
Hf
i
HfCaCl2
kJ
75
80
10
100
ni
459
1 10
3
12.38
CaCl2 ---> Ca + Cl2
(1)
2(Ca + Cl2 + 12.5 H2O ---> CaCl2(12.5 H2O) (2)
CaCl2(25 H2O) ---> Ca + Cl2 + 25 H2O
(3)
-----------------------------------------------------------------------------------CaCl2(25 H2O) + CaCl2 ---> 2 CaCl2(12.5 H2O)
(Table C.4)
H1
H2
2 ( 865.295 kJ)
H
12.39
795.8 kJ
H1
H2
H3
Q
H3
871.07 kJ
H
Q
63.72 kJ
Ans.
The process may be considered in two steps:
Mix at 25 degC, then heat/cool solution to the final temperature. The two
steps together are adiabatic and the overall enthalpy change is 0.
Calculate moles H2O needed to form solution:
85
18.015
n
15
110.986
n
34.911 Moles of H2O per mol CaCl2 in final solution.
Moles of water added per mole of CaCl2.6H2O:
n
6
28.911
Basis: 1 mol of Cacl2.6H2O dissolved
CaCl2.6H2O(s) ---> Ca + Cl2 + 6 H2 + 3 O2 (1)
Ca + Cl2 + 34.991 H2O --->CaCl2(34.911 H2O) (2)
6(H2 + 1/2 O2 ---> H2O) (3)
--------------------------------------------------------------------------------------CaCl2.6H2O + 28.911 H2O ---> CaCl2(34.911 H2O)
H1
H2
H3
2607.9 kJ
871.8 kJ
H298
H1
H298
21.12 kJ
6 ( 285.83 kJ)
(Table C.4)
(Pb. 12.37)
H2
H3
for reaction at 25 degC
msoln
(
110.986
msoln
739.908 gm
460
34.911 18.015)gm
CP
T
12.43 m1
H1
3.28
kJ
kg degC
8
25 % m2
m2
m1
47.5 %
m2
m3 H3
23
m2 H2
BTU
(Fig. 12.17)
lbm
(Final soln.)
H3
m1 H1
(25% soln.)
350 lb
H2
msoln CP
16.298 degC Ans.
T
T
m2
BTU
lbm
H298
T
CP T = 0
25 degC
150 lb (H2SO4)
m1
Q
T
8.702 degC
100 % m1
m3
H298
90
Q
BTU
lbm
(Fig. 12.17)
38150 BTU
Ans.
12.44 Enthalpies from Fig. 12.17.
x1
H1
20
HE
H
x2
0.5
BTU
1
H
(pure H2SO4)
lbm
x1 H1
69
H2
x1
108
HE
x2 H2
BTU
lbm
BTU
lbm
133
12.45 (a) m1
400 lbm
175 lbm
(10% soln. at 200 degF)
H1
100
35 % m1
BTU
lbm
10 % m2
m1
m3
m2
m1
m2
H2
BTU
152
lbm
H3
41
(Fig. 12.19)
(Final soln)
27.39 %
BTU
lbm
461
(pure H2O)
BTU
Ans.
lbm
(35% soln. at 130 degF)
m2
(50 % soln)
(Fig. 12.19)
Q
m3 H3
m1 H1
Q
43025 BTU Ans.
H3
m2 H2
115.826
(b) Adiabatic process, Q = 0.
m1 H1
H3
m2 H2
m3
BTU
lbm
From Fig. 12.19 the final soln. with this enthalpy has a temperature of
about 165 degF.
12.46 m1
25
lbm
(feed rate)
sec
x1
0.2
H1
24
BTU
lbm
(Fig. 12.17 at 20% & 80 degF)
H2
55
BTU
lbm
(Fig. 12.17 at 70% and 217 degF)
[Slight extrapolation]
x2
0.7
H3
1157.7
x1 m1
m2
Q
x2
m2 H2
BTU
lbm
(Table F.4, 1.5(psia) & 217 degF]
lbm
m2
7.143
m3 H3
m1 H1
m3
m2
m3
Q
sec
m1
17.857
20880
BTU
sec
Ans.
12.47 Mix m1 lbm NaOH with m2 lbm 10% soln. @ 68 degF.
BASIS: m2
1 lbm
x3
m1
1 lbm
(guess)
m2 = m3
m1
x2 m2 = x3 m3
m1
0.385 lbm
Given
m1
0.35
x2
0.1
m3
m1
m2
m1
m3
Find m1 m3
462
m3
1.385 lbm
lbm
sec
From Example 12.8 and Fig. 12.19
H1
BTU
lbm
478.7
H3
m1 H1
H2
m3
BTU
lbm
H3
m2 H2
43
164
BTU
lbm
From Fig. 12.19 at 35% and this enthalpy, we find the temperature to be
about 205 degF.
12.48 First react 1 mol SO3(l) with 1 mol H2O(l) to form 1 mol H2SO4(l):
SO3(l) + H2O(l) ---> H2SO4(l)
With data from Table C.4:
H298
[ 813989
( 441040
285830)J
H298
8.712
4
10 J
Mix 1 mol or 98.08 gm H2SO4(l) with m gm H2O to form a 50% solution.
mH2SO4
mH2O
98.08 gm
msoln
msoln
mH2SO4
0.5
mH2SO4
Data from Fig. 12.17:
HH2SO4
HH2O
Hsoln
0
45
70
BTU
lbm
[pure acid @ 77 degF (25 degC)]
BTU
lbm
[pure water @ 77 degF (25 degC)]
BTU
[50% soln. @ 140 degF (40 deg C)]
lbm
Hmix
msoln Hsoln
Hmix
18.145 kg
Q
H298
msoln
mH2SO4 HH2SO4
mH2O HH2O
BTU
lbm
Hmix
Q
283
463
BTU
lbm
Ans.
12.49
m1
H1
65
x1
140 lbm
H2
BTU
lb
m2
0.15
0.8
(Fig. 12.17 at 160 degF)
BTU
lb
(Fig. 12.17 at 100 degF)
m2
102
x2
230 lbm
x3
m3
m1
Q
20000 BTU
H3
92.9
BTU
lbm
H3
m1 x1
m2 x2
x3
m3
Q
m1 H1
55.4 %
m2 H2
m3
From Fig. 12.17 find temperature about 118 degF
12.50 Initial solution (1) at 60 degF; Fig. 12.17:
m1
x1
1500 lbm
H1
0.40
98
BTU
lbm
Saturated steam at 1(atm); Table F.4:
m3 m2
x3 m2
m2
m1
x1 m1
m1
125 lbm
H2
m2
m2
H3 m2
x3 m2
m1 H1
1150.5
BTU
lbm
m2 H2
m3 m2
36.9 %
H3 m2
2
BTU
lbm
The question now is whether this result is in agreement with the value read
from Fig. 12.17 at 36.9% and 180 degF. It is close, but we make a second
calculation:
m2
120 lbm
x3 m2
37%
This is about as good a result as we can get.
464
H3 m2
5.5
BTU
lbm
12.51 Initial solution (1) at 80 degF; Fig. 12.17:
m1
x1
1 lbm
H1
0.45
95
BTU
lbm
Saturated steam at 40(psia); Table F.4:
m3 m2
m1
H2
m2
x1 m1
x3 m2
m2
m1
m2
0.05 lbm
m1 H1
H3 m2
x3 m2
1169.8
BTU
lbm
m2 H2
m3 m2
42.9 %
H3 m2
34.8
BTU
lbm
The question now is whether this result is in agreement with the value read
from Fig. 12.17 at 36.9% and 180 degF. It is close, but we make a second
calculation:
m2
0.048 lbm
x3 m2
42.9 %
H3 m2
37.1
BTU
lbm
This is about as good a result as we can get.
12.52 Initial solution (1) at 80 degF; Fig. 12.19:
m1
x1
1 lbm
0.40
H1
77
BTU
lbm
Saturated steam at 35(psia); Table F.4:
BTU
lbm
H2
1161.1
m3
m1
H3
x3
0.38
m2
m3
1.053 lbm
m2
x1 m1
m1
m1 H1 m2 H2
m3
H3
131.2
m2
BTU
lbm
x3
0.053 lbm
We see from Fig. 12.19 that for this enthalpy
at 38% the temperature is about 155 degF.
465
12.53 Read values for H, H1, & H2 from Fig. 12.17 at 100 degF:
H1
56
x1
0.35
8
BTU
lbm
x2
BTU
lbm
H
1
x1
H2
68
H
BTU
lbm
H
H
x1 H 1
x2 H2
BTU
lbm
Ans.
103
12.54 BASIS: 1(lbm) of soln.
Read values for H1 & H2 from Fig. 12.17 at 80 degF:
BTU
lbm
H1
4
Q=
H2
H=H
H
x1 H 1
x1 H 1
x2 H2
48
BTU
lbm
x1
0.4
x2
1
x1
x2 H2 = 0
H
30.4
BTU
lbm
From Fig. 12.17, for a 40% soln. to have this enthalpy the temperature is
well above 200 degF, probably about 250 degF.
12.55 Initial solution:
Final solution:
x1
x2
2 98.08
2 98.08 15 18.015
3 98.08
3 98.08
14 18.015
Data from Fig. 12.17 at 100 degF:
HH2O
H1
68
75
BTU
lbm
BTU
lbm
HH2SO4
H2
9
101
466
BTU
lbm
BTU
lbm
x1
0.421
x2
0.538
Unmix the initial solution:
Hunmix
x1 HH2SO4
Hunmix
118.185
1
x1 HH2O
H1
BTU
lbm
React 1 mol SO3(g) with 1 mol H2O(l) to form 1 mol H2SO4(l). We
neglect the effect of Ton the heat of reaction, taking the value at 100 degF
equal to the value at 77 degF (25 degC)
HfSO3
395720
HfH2SO4
Hrx
J
mol
285830
J
mol
J
mol
813989
HfH2SO4
HfH2O
HfH2O
HfSO3
Hrx
1.324
5J
10
mol
Finally, mix the constituents to form the final solution:
Hmix
Q
H2
x2 HH2SO4
1
x2 HH2O
Hunmix ( 98.08
2
1 lbmol Hrx
14 18.015)lb
137.231
BTU
lbm
15 18.015)lb
Hmix ( 98.08
3
Hmix
Q
76809 BTU
Ans.
12.56 Read values for H(x1=0.65), H1, & H2 from Fig. 12.17 at 77 degF:
H
125
x1
0.65
BTU
H1
lbm
x2
0
1
BTU
x1
45
H
lbm
H2
H
H
467
BTU
lbm
x1 H 1
140.8
BTU
lbm
x2 H2
Ans.
From the intercepts of a tangent line drawn to the 77 degF curve of Fig.
12.17 at 65%, find the approximate values:
Hbar1
BTU
lbm
136
Hbar2
103
BTU
lbm
Ans.
12.57 Graphical solution: If the mixing is adiabatic and water is added to bring
the temperature to 140 degF, then the point on the H-x diagram of Fig.
12.17 representing the final solution is the intersection of the 140-degF
isotherm with a straight line between points representing the 75 wt %
solution at 140 degF and pure water at 40 degF. This intersection gives
x3, the wt % of the final solution at 140 degF:
x3
m1
42 %
1 lb
By a mass balance:
x3 =
0.75 m1
m1
12.58 (a) m1
x1
0.75 m1
m2
m2
m1
x3
25 lbm
m2
40 lbm
0
x2
m2
1
Ans.
0.786 lbm
m3
75 lbm
x3
0.25
H3
7
x4
0.42
Enthalpy data from Fig. 12.17 at 120 degF:
H1
88
BTU
lbm
m4
m1
m2
x4
x1 m1
BTU
lbm
H2
m4
m3
x2 m2
14
140 lbm
x3 m3
m4
BTU
lbm
H4
63
Q
m4 H4
BTU
lbm
(Fig. 12.17)
m1 H1
m2 H2
m3 H3
468
Q
11055 BTU Ans.
(b) First step: m1
40 lb
x1
1
H1
14
m2
75 lb
x2
0.25
H2
7
m3
m1
x3
m2
x3
x1 m1
x2 m2
H3
m3
H3
0.511
Q
m1 H1
BTU
lbm
BTU
lbm
m2 H2
m3
95.8
BTU
lbm
From Fig. 12.17 at this enthalpy and wt % the temperature is about 100
degF.
12.59 BASIS: 1 mol NaOH neutralized.
For following reaction; data from Table C.4:
NaOH(s) + HCl(g) ---> NaCl(s) + H2O(l)
H298
[ 411153
H298
1.791
285830
( 425609
92307)J
5
10 J
NaOH(s) + HCl(g) ---> NaCl(s) + H2O(l) (1)
NaOH(inf H2O) ---> NaOH(s) + inf H2O
(2)
HCl(9 H2O) ---> HCl(g) + 9 H2O(l)
(3)
NaCl(s) + inf H2O ---> NaCl(inf H2O)
(4)
---------------------------------------------------------------------------------------NaOH(inf H2O) + HCl(9 H2O) ---> NaCl(inf H2O)
H1
H298
H2
H4
3.88 kJ
H
H1
H3
44.50 kJ
Q
H
Q
62187 J
469
H2
H3
Ans.
H4
68.50 kJ
12.60 First, find heat of solution of 1 mole of NaOH in 9 moles of H2O
at 25 degC (77 degF).
Weight % of 10 mol-% NaOH soln:
1 40.00
1 40.00 9 18.015
x1
x1
19.789 %
HH2O
45
BTU
lbm
(Table F.3, sat. liq. at 77 degF)
Hsoln
35
BTU
lbm
(Fig. 12.19 at x1 and 77 degF)
HNaOH
BTU
lbm
478.7
[Ex. 12.8 (p. 468 at 68 degF]
Correct NaOH enthalpy to 77 degF with heat capacity at 72.5 degF
(295.65 K); Table C.2:
T
295.65 K
Cp
R
molwt
HNaOH
H
H
molwt
0.121
HNaOH
Hsoln
0.224
16.316 10
K
Cp (
77
x1 HNaOH
kJ
gm
40.00
3
T
Cp
68)rankine
1
H
x1
molwt
This is for 1 gm of SOLUTION.
H
45.259
kJ
mol
470
0.245
HNaOH
x1 HH2O
However, for 1 mol of NaOH, it becomes:
H
gm
mol
BTU
lbm rankine
480.91
BTU
lbm
Now, on the BASIS of 1 mol of HCl neutralized:
NaOH(s) + HCl(g) ---> NaCl(s) + H2O(l) (1)
HCl(inf H2O) ---> HCl(g) + inf H2O
(2)
NaOH(9 H2O) ---> NaOH(s) + 9 H2O
(3)
NaCl + inf H2O ---> NaCl(inf H2O)
(4)
--------------------------------------------------------------------------------------HCl(inf H2O) + NaOH(9 H2O) ---> NaCl(inf H2O)
H1
179067 J
(Pb. 12.59)
(Fig. 12.14 with sign change)
H2
74.5 kJ
H3
45.259 kJ
H4
3.88 kJ
H
H1
(See above; note sign change)
(given)
H2
H3
H3
Q
H
14049 J Ans.
Q
12.61 Note: The derivation of the equations in part a) can be found in Section B
of this manual.
0.1
0.2
144.21
0.3
208.64
0.4
262.83
0.5
x1
73.27
302.84
0.6
HE
323.31
0.7
320.98
0.8
279.58
0.85
237.25
0.9
178.87
0.95
100.71
471
kJ
kg
x2
1
x1
H
HE
x1 x2
In order to take the necessary derivatives of H, we will fit the data to a
HE
3
2
third order polynomial of the form H =
= a bx.1 c x1
d x1 .
x1 x2
Use the Mathcad regress function to find the parameters a, b, c and d.
w
w
3
w
w
3
n
3
n
a
H
regress x1
kJ
kg
b
a
735.28
b
3
824.518
c
c
d
d
H
x1
a
b x1
c x1
2
d x1
3
195.199
914.579
kJ
kg
Using the equations given in the problem statement and taking the
derivatives of the polynomial analytically:
HEbar1 x1
1
HEbar2 x1
x1
2
x1
H
2
x1
H
x1
1
x1
x1
472
b
2 c x1
3 d x1
b
2 c x1
3 d x1
2
2
kJ
kg
kJ
kg
0
500
(kJ/kg)
1000
1500
2000
2500
0
0.2
0.4
0.6
0.8
x1
H/x1x2
HEbar1
HEbar2
12.62 Note: This problem uses data from problem 12.61
0.1
0.2
144.21
0.3
208.64
0.4
262.83
0.5
x1
73.27
302.84
0.6
HE
323.31
0.7
320.98
0.8
279.58
0.85
237.25
0.9
178.87
0.95
100.71
473
kJ
kg
x2
1
x1
H
HE
x1 x2
Fit a third order polynomial of the form
HE
x1 x2
=a
bx.1
c x1
2
d x1
3
Use the Mathcad regress function to find the parameters a, b, c and d.
w
w
3
w
w
3
n
n
3
a
H
regress x1
b
a
735.28
b
3
kJ
kg
824.518
c
c
d
d
195.199
914.579
By the equations given in problem 12.61
x1
a
H x1
H
H
b x1
c x1
x1 x1 1
Hbar1 x1
1
Hbar2 x1
x1
x1
2
H
2
d x1
3
kJ
kg
x1
2
H
x1
x1
1
x1
x1
2
kJ
kg
2
kJ
b
2 c x1
3 d x1
b
2 c x1
3 d x1
kg
At time , let:
x1 = mass fraftion of H2SO4 in tank
m = total mass of 90% H2SO4 added up to time
H = enthalpy of H2SO4 solution in tank at 25 C
H2 = enthalpy of pure H2O at 25 C
H1 = enthalpy of pure H2SO4 at 25 C
H3 = enthalpy of 90% H2SO4 at 25 C
Material and energy balances are then written as:
x1 ( 4000
Q=
m) = 0.9m
Ht = ( 4000
m) H
Solving for m:
4000H2
m H3
474
m x1
( 4000kg)x1
0.9
x1
Eq. (A)
.
x1 H1 x2 H2 and since T is constant at 25 C, we set
H1 = H2 = 0 at this T, making H = H. The energy balance then
becomes:
Eq. (B)
Q=(
4000 m) H m H3
Since H = H
Applying these equations to the overall process, for which:
6hr
x1
0.5
H3
H( )
0.9
H3
178.737
H
H( )
0.5
H
303.265
kJ
kg
kJ
kg
Define quantities as a function of x 1
Q x1
m x1
4000kg
m x1
H x1
m x1 H3
( 000kg) 1
4
x
0.9
Qtx1
m ( ) 5000 kg
0.5
x1
4000kg
m x1
H
m x1 H3
Qt0.5)
(
1.836
6
10 kJ
Since the heat transfer rate q is constant:
q
Qtx1
and
x1
Q x1
Eq. (C)
q
The following is probably the most elegant solution to this problem, and it
leads to the direct calculation of the required rates,
dm
r=
d
When 90% acid is added to the tank it undergoes an enthalpy change equal
to: 0.9Hbar1+0.1Hbar2-H3, where Hbar1 and Hbar2 are the partial
enthalpies of H2SO4 and H2O in the solution of mass fraction x1 existing in
the tank at the instant of addition. This enthalpy change equals the heat
required per kg of 90% acid to keep the temperature at 25 C. Thus,
r x1
q
0.9 Hbar1 x1
0.1 Hbar2 x1
475
H3
Plot the rate as a function of time
x1
0 0.01 0.5
1200
1100
1000
r x1
kg
900
hr
800
700
600
0
1
2
3
4
5
6
x1
hr
12.64 mdot1
20000
lb
hr
x1
0.8
T1
120degF
H1
Enthalpies from Fig. 12.17 x2
0.0
T2
40degF
H2
x3
0.5
T3
140degF
H3
92
BTU
lb
7
70
a) Use mass balances to find feed rate of cold water and product rate.
Guess:
mdot2
Given
mdot1
mdot1
mdot3
mdot2 = mdot3
mdot1 x1
mdot2 x2 = mdot3 x3
476
2mdot1
Total balance
H2SO4 balance
BTU
lb
BTU
lb
mdot2
mdot3
Find mdot2 mdot3 mdot2
12000
lb
mdot3
hr
32000
lb
Ans.
hr
b) Apply an energy balance on the mixer
Qdot
mdot3 H3
mdot1 H1
mdot2 H2
Qdot
484000
BTU
hr
Since Qdot is negative, heat is removed from the mixer.
c) For an adiabatic process, Qdot is zero. Solve the energy balance to find H3
H3
mdot1 H1
mdot2 H2
mdot3
H3
54.875
BTU
lb
From Fig. 12.17, this corresponds to a temperature of about 165 F
12.65 Let L = total moles of liquid at any point in time and Vdot = rate at
which liquid boils and leaves the system as vapor.
dL
= Vdot
dt
d L x1
An unsteady state species balance on water yields:
= y1 Vdot
dt
An unsteady state mole balance yields:
Expanding the derivative gives:
Substituting -Vdot for dL/dt:
L
Rearranging this equation gives:
L
Substituting -dL/dt for Vdot:
dx1
L
L
Eliminating dt and rearranging:
x1
dt
dx1
x1 ( Vdot)= y1 Vdot
dt
dx1
dt
dx1
dt
= x1
y1 Vdot
= y1
x1
dx1
y1
477
dL
= Vdot y1
dt
x1
=
dL
L
dL
dt
At low concentrations y1 and x1 can be related by:
y1 =
Psat1
inf1
where:
x1 = K1 x1
P
dx1
Substituting gives:
K1
1 x1
=
Integrating this equation yields:
Psat1
K1 = inf1
P
dL
L
ln
Lf
1
=
L0
K1
x1f
ln
1
x10
where L0 and x10 are the initial conditions of the system
For this problem the following values apply:
L0
1mol
600
x10
10
T
130degC
Psat1
P
exp 16.3872
inf1
Lf
L0 exp
6
T
degC
K1
K1
norg0
L0 1
norg0
1
ln
kPa
Psat1
230.170
270.071 kPa
15.459
x1f
Lf
x10
x10
norg0
5.8
inf1
0.999 mole
%lossorg
6
10
3885.70
P
1
50
1atm
Psat1
K1
x1f
0.842 mole
norgf
norgf
norgf
Lf 1
0.842 mole
%lossorg
norg0
x1f
15.744 %
Ans.
The water can be removed but almost 16% of the organic liquid will
be removed with the water.
478
12.69 1 - Acetone
2- Methanol
T
(
50
273.15)
K
For Wilson equation
a12
161.88
V2
12
V1
exp
cal
mol
a21
a12
RT
V1
74.05
V1
0.708
12
3
3
cal
583.11
mol
21
V2
exp
cm
mol
V2
40.73
a21
21
RT
cm
mol
0.733
ln inf1
ln
12
1
21
ln inf1
0.613
Ans.
ln inf2
ln
21
1
12
ln inf2
0.603
Ans.
From p. 445
From Fig. 12.9(b)
ln inf2 = 0.61
ln inf1 = 0.62
For NRTL equation
b12
184.70
cal
mol
b21
b12
12
G12
cal
mol
12
0.3048
b21
12
RT
exp
222.64
0.288
21
G12
0.916
G21
21
RT
exp
21
0.347
G21
0.9
ln inf1
21
12 exp
12
ln inf1
0.611
ln inf2
From p. 446
12
21 exp
21
ln inf2
0.600
Both estimates are in close agreement with the values from Fig. 12.9 (b)
479
12.71 Psat1
x1
Psat2
183.4kPa
y1
0.253
96.7kPa
P
0.456
139.1kPa
Check whether or not the system is ideal using Raoult's Law (RL)
PRL
x1 Psat1
Since PRL<P,
1
1
and
PRL
x1 Psat2
2
118.635 kPa
are not equal to 1. Therefore, we need a model for
GE/RT. A two parameter model will work.
GE
= x1 x2 A21 x1
RT
From Margules Equation:
ln 1 = x2
ln 2 = x1
Find
1
and
2
Use the values of
find A12 and A21.
Given
A12
1
and
A21
2
A12 x1
Eq. (12.10a)
A21
2 A12
A21 x2
Eq. (12.10b)
A21
ln 1 = 1
x1
2
2
A21
exp
1
2 x1
exp x1
2
x1
2
A21
A12
1
2
x1 Psat2
1.048
A12
0.5
2 A21
2 A12
A12
Find A12 A21
1 x1
y1 P
1
2
at x1=0.253 and Eqs. (12.10a) and (12.10b) to
0.5
ln 2 = x1
A12
2 A21
1.367
1
x1 Psat1
Guess:
2
A12
at x1=0.253 from the given data.
y1 P
1
2
A12 x2
1
A21
480
A12 x1
1
x1
A21
0.644
2 A21
2 A12
A21
A12 x1
x1
Eq. (12.10a)
Eq. (12.10b)
0.478
a) x1
x1 1 x1 Psat1
P
y1
0.5
P
x1 1 x1 Psat1
b) 1inf
1
exp A12
12
1.904
3.612
P
0.743
121
Ans.
160.148 kPa
exp A21
2inf
120
Psat2
Since
2 x1 Psat2
1inf
1inf Psat1
120
x1
y1
Psat1
2inf Psat2
Ans.
2inf
1.614
121
1.175
remains above a value of 1, an azeotrope is unlikely based on the
assumption that the model of GE/RT is reliable.
12.72 P
T
x1
108.6kPa
(
35
0.389
Psat1
273.15)
K
Psat2
120.2kPa
73.9kPa
Check whether or not the system is ideal using Raoult's Law (RL)
PRL
x1 Psat1
Since PRL < P,
1
1
and
x1 Psat2
2
PRL
91.911 kPa
are not equal to 1. Therefore, we need a model for
GE/RT. A one parameter model will work.
Assume a model of the form:
GE
= A x1 x2
RT
2
1 = exp A x2
2 = exp A x.1
2
Since we have no y1 value, we must use the following equation to find A:
P = x1 1 Psat1
x2 2 Psat2
481
Use the data to find the value of A
Guess: A
1
Given
A
Find A)
(
1 x1
a) y1
b) P
c) 1inf
120
P = x1 exp A 1
A
exp A 1
x1 1 x1
exp ( A)
1inf Psat1
2
Psat1
1
x1 exp A x1
2
Psat2
0.677
x1
2
Psat1
2 x1
y1
P
x1 1 x1 Psat1
Psat2
x1
1
1inf
120
x1
0.554
2inf
3.201
2
Ans.
2 x1 Psat2
1.968
exp A x1
121
P
110.228 kPa
exp ( A)
Psat1
2inf Psat2
2inf
121
Ans.
1.968
0.826
Since 12 ranges from less than 1 to greater than 1 an azeotrope is likely
based on the assumption that our model is reliable.
482
Chapter 13 - Section A - Mathcad Solutions
Note: For the following problems the variable kelvin is used for the SI
unit of absolute temperature so as not to conflict with the variable K
used for the equilibrium constant
13.4
H2(g) + CO2(g) = H2O(g) + CO(g)
=
i=
1
1
1
1= 0
n0 = 1
1= 2
i
By Eq. (13.5).
yH = yCO =
2
1
2
yH2O = yCO =
2
2
By Eq. (A) and with data from Example 13.13 at 1000 K:
T
1000 kelvin
1
G
2
RT 2
Guess:
e
d
Given
J
mol
1
( 395790)
G
de
1
2
ln
2
2
( 192420
2
2
ln
200240)
J
mol
2
0.5
e =0
J
e
mol
Find e
e
0.3 0.31 0.6
2.082
2.084
G
5
10
2.086
2.088
0.2
0.3
0.4
483
0.5
0.6
0.45308
13.5
(a) H2(g) + CO2(g) = H2O(g) + CO(g)
=
i=
1
1
1
1= 0
n0 = 1
1= 2
i
By Eq. (13.5).
1
yH = yCO =
2
2
yH2O = yCO =
2
2
By Eq. (A) and with data from Example 13.13 at 1100 K:
T
1100 kelvin
1
G
( 395960)
2
RT 2
Guess:
1
2
d
G
de
2
( 187000
2
2
2
ln
209110)
J
mol
2
0.5
e
Given
ln
J
mol
1
e =0
J
mol
e
Find e
e
0.502
Ans.
0.35 0.36 0.65
2.102
2.103
2.104
G
5
10
2.105
2.106
2.107
0.3
0.35
0.4
0.45
484
0.5
0.55
0.6
0.65
(b)
H2(g) + CO2(g) = H2O(g) + CO(g)
=
i=
1
1
1
1= 0
n0 = 1
1= 2
i
By Eq. (13.5),
1
yH = yCO =
2
2
yH2O = yCO =
2
2
By Eq. (A) and with data from Example 13.13 at 1200 K:
T
1200 kelvin
1
G
( 396020)
2
RT 2
Guess:
e
d
Given
G
de
1
ln
2
J
2
mol
1
( 181380
2
2
2
ln
217830)
J
mol
2
0.1
e =0
J
mol
e
Find e
e
0.53988
0.6
0.65
Ans.
0.4 0.41 0.7
2.121
2.122
2.123
G
5 2.124
10
2.125
2.126
2.127
0.35
0.4
0.45
485
0.5
0.55
0.7
(c)
H2(g) + CO2(g) = H2O(g) + CO(g)
=
i=
1
1
1
1= 0
n0 = 1
1= 2
i
By Eq, (13.5),
yH = yCO =
2
1
2
yH2O = yCO =
2
2
By Eq. (A) and with data from Example 13.13 at 1300 K:
T
1300 kelvin
1
G
( 396080)
2
RT 2
Guess:
e
d
Given
G
de
1
ln
2
J
mol
1
2
( 175720
2
2
2
ln
226530)
J
mol
2
0.6
e =0
J
mol
e
Find e
e
0.57088
Ans.
0.4 0.41 0.7
2.14
2.142
G
5
2.144
10
2.146
2.148
0.35
0.4
0.45
486
0.5
0.55
0.6
0.65
0.7
13.6
H2(g) + CO2(g) = H2O(g) + CO(g)
=
1
i=
1
1
1= 0
n0 = 1
1= 2
i
By Eq, (13.5),
1
yH = yCO =
2
2
yH2O = yCO =
2
2
With data from Example 13.13, the following vectors represent values for
Parts (a) through (d):
1000
3130
1100
T
1200
150
G
kelvin
3190
1300
J
mol
6170
Combining Eqs. (13.5), (13.11a), and (13.28) gives
2
1
2
2
=
1
2
1
2
= K = exp
G
RT
2
0.4531
exp
0.5021
G
RT
Ans.
0.5399
1
0.5709
13.11 4HCl(g) + O2(g) = 2H2O(g) + 2Cl(g)
=1
H298
n0 = 6
114408
J
mol
T
773.15 kelvin
G298
487
75948
T0
J
mol
298.15 kelvin
The following vectors represent the species of the reaction in the order in
which they appear:
4
3.156
1
3.639
A
2
A
3
B
G
1.267
0.121
D
i Bi
i Di
i
8
5
10
C
D
0
4
8.23
10
H298
G
B
5
10
0.344
i
0.439
0.227
1 end
i
A
D
0.089
i
i Ai
10
1.450
4.442
rowsA)
(
0.151
0.506
B
3.470
2
end
0.623
K
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
10
4J
mol
G
RT
exp
K
By Eq. (13.5)
yO2 =
yHCl =
1
yH2O =
6
Apply Eq. (13.28);
yHCl
5
5
4
4
0.3508
6
1
yO2
6
yO2
5
4
6
2
0.0397
yCl2 =
6
2
6
(guess)
0.5
4
2
Given
yHCl
7.18041
1
yH2O
6
yH2O
488
0.793
Find
= 2K
0.3048
2
6
yCl2
yCl2
0.3048
2
6
Ans.
13.12 N2(g) + C2H2(g) = 2HCN(g)
n0 = 2
=0
This is the reaction of Pb. 4.21(x). From the answers for Pbs. 4.21(x),
4.22(x), and 13.7(x), find the following values:
H298
A
T
J
mol
42720
G298
B
0.060
T0
C
J
mol
5
D
0
0.191 10
298.15 kelvin
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
G
3
0.173 10
923.15 kelvin
39430
H298
3.242
4J
10
K
mol
exp
G
RT
K
0.01464
By Eq. (13.5),
yN2 =
1
yC2H4 =
2
By Eq. (13.28),
2
Given
yN2
yN2
0.5
1
1
2
0.4715
1
yHCN =
2
2e
2
=
(guess)
2
yC2H4
yC2H4
0.057
Find
=K
1
2
0.4715
yHCN
yHCN
0.057 Ans.
Given the assumption of ideal gases, P has no effect on the equilibrium
composition.
489
13.13
CH3CHO(g) + H2(g) = C2H5OH(g)
n0 = 2.5
=1
This is the reaction of Pb. 4.21(r). From the answers for Pbs. 4.21(r),
4.22(r), and 13.7(r), find the following values:
H298
A
T
J
68910
mol
B
1.424
G298
3
1.601 10
T0
623.15 kelvin
39630
C
J
mol
0.156 10
6
5
D
0.083 10
298.15 kelvin
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
6.787
10
3J
K
mol
By Eq. (13.5), yCH3CHO =
1
Given
yCH3CHO
yCH3CHO
yH2 =
2.5
By Eq. (13.28),
exp
0.5
2.5
1
1.5
= 3K
1
2.5
yH2
0.108
yH2
1.5
2.5
0.4053
G
RT
K
1.5
3.7064
yC2H5OH =
2.5
2.5
(guess)
0.818
Find
yC2H5OH
yC2H5OH
2.5
0.4867 Ans.
If the pressure is reduced to 1 bar,
Given
yCH3CHO
yCH3CHO
2.5
1
1.5
= 1K
1
2.5
0.1968
yH2
yH2
1.5
2.5
0.4645
490
0.633
Find
yC2H5OH
yC2H5OH
2.5
0.3387 Ans.
13.14 C6H5CH:CH2(g) + H2(g) = C6H5.C2H5(g)
n0 = 2.5
=1
This is the REVERSE reaction of Pb. 4.21(y). From the answers for Pbs.
4.21(y), 4.22(y), and 13.7(y) WITH OPPOSITE SIGNS, find the following
values:
H298
J
117440
mol
G298
B
923.15 kelvin
T
3
4.766 10
T0
A
4.175
J
mol
83010
C
1.814 10
6
D
0.083 10
5
298.15 kelvin
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
2.398
10
3J
mol
By Eq. (13.5),
yH2 =
K
G
1.5
yC6H5C2H5 =
2.5
0.5
2.5
1
yC6H5CHCH2
yC6H5CHCH2
1.5
1
2.5
0.2794
yH2
491
1.36672
1
2.5
2.5
(guess)
= 1.0133 K
yH2
K
RT
yC6H5CHCH2 =
By Eq. (13.28),
Given
exp
1.5
2.5
0.5196
0.418
Find
yC6H5C2H5
yC6H5C2H5
2.5
0.201
Ans.
13.15 Basis: 1 mole of gas entering, containing 0.15 mol SO2, 0.20 mol O2,
and 0.65 mol N2.
SO2 + 0.5O2 = SO3
n0 = 1
= 0.5
By Eq. (13.5),
ySO2 =
0.15
1
0.20
yO2 =
0.5
1
0.5
ySO3 =
0.5
1
0.5
From data in Table C.4,
H298
J
mol
98890
G298
70866
J
mol
The following vectors represent the species of the reaction in the order
in which they appear:
1
5.699
0.5
A
3.639
1
end
A
i Ai
1.015
0.506 10
3
2.028
D
i Bi
2
10
T0
753.15 kelvin
i Di
i
i
B
0.227 10
1 end
B
0.5415
5
D
1.056
i
i
T
B
8.060
rowsA)
(
A
0.801
6
C
0
D
8.995
4
10
298.15 kelvin
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
2.804
10
4J
mol
K
By Eq. (13.28),
0.1
1
Given
0.15
exp
0.5
0.2
G
RT
K
88.03675
(guess)
0.5
0.5
0.5
492
=K
Find
0.1455
By Eq. (13.4),
nSO3 =
= 0.1455
By Eq. (4.18),
H753
H298
J
mol
R IDCPH T0 T
Q
Q
H753
13.16 C3H8(g) = C2H4(g) + CH4(g)
A
B
14314
C
D
J
Ans.
mol
=1
H753
98353
Basis: 1 mole C3H8 feed. By Eq. (13.4)
n0
Fractional conversion of C3H8 =
By Eq. (13.5),
nC3H8
n0
1
yC3H8 =
nC3H8 = 1
yC2H4 =
1
=
1
1
1
=
yCH4 =
1
1
From data in Table C.4,
H298
82670
J
mol
G298
42290
J
mol
The following vectors represent the species of the reaction in the order
in which they appear:
1
1.213
1
A
1
end
1.424
i
rowsA)
(
i Ai
8.824
14.394 10
3
625 kelvin
B
4.392 10
6
2.164
1 end
C
i Bi
i Ci
i
i
1.913
C
9.081
B
i
(a) T
B
1.702
A
A
28.785
5.31 10
T0
493
3
C
2.268
298.15 kelvin
10
6
D
0
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
2187.9
J
mol
K
G
RT
exp
By Eq. (13.28),
K
1.52356
(guess)
0.5
2
Given
1
1
Find
=K
This value of epsilon IS the fractional conversion. Ans.
0.777
2
(b)
K
0.85
G
R T ln ( K)
G
1
K
1
4972.3
J
mol
2.604
Ans.
The problem now is to find the T which generates this value.
It is not difficult to find T by trial. This leads to the value:
T = 646.8 K Ans.
13.17 C2H6(g) = H2(g) + C2H4(g)
=1
Basis: 1 mole entering C2H6 + 0.5 mol H2O.
By Eq. (13.5),
n0 = 1.5
yC2H6 =
1
yH =
1.5
1.5
yC2H4 =
1.5
From data in Table C.4,
H298
136330
J
mol
G298
100315
J
mol
The following vectors represent the species of the reaction in the order in
which they appear:
494
1
1.131
1
A
19.225
3.249
1
B
0.422
1.424
14.394
5.561
C
0.0
0.0
6
10
5
D
0.083 10
4.392
end
i
rowsA)
(
A
i Ai
A
T
0.0
1 end
B
C
i Bi
3.542
B
i Di
i
i
3
4.409 10
T0
1100 kelvin
D
i Ci
i
i
3
10
C
1.169
10
6
D
8.3
3
10
298.15 kelvin
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
5.429
10
3J
K
mol
By Eq. (13.28),
0.5
exp
G
RT
K
1.81048
(guess)
2
Given
1.5
1
By Eq. (13.4), nC2H6 = 1
yC2H6
yC2H6
1
1
0.0899
0.83505
Find
=K
n= 1
nH2 = nC2H4 =
yH2
yC2H4
1
yC2H4
495
0.4551
yH2
1
0.4551
Ans.
13.18 C2H5CH:CH2(g) = CH2:CHCH:CH2(g) + H2(g)
(1)
(2)
=1
(3)
Number the species as shown. Basis is 1 mol species 1 + x mol steam.
n0 = 1
x
By Eq. (13.5),
y1 =
1
1
y2 = y3 =
x
1
= 0.10
x
From data in Table C.4,
H298
109780
J
mol
G298
79455
J
mol
The following vectors represent the species of the reaction in the
order in which they appear:
1
1.967
1
A
31.630
2.734
1
B
26.786 10
3.249
0.422
0.0
9.873
C
8.882 10
6
D
rows ( A)
A
i
i Ai
i
4.016
T
950 kelvin
G
B
B
i Bi
C
D
i
4.422 10
T0
i Ci
3
C
9.91
298.15 kelvin
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
H298
5
10
1 end
i
A
0.0
0.083
0.0
end
3
496
i Di
i
10
7
D
8.3
3
10
G
3J
4.896
10
K
mol
G
exp
( )( ) 1
0.1 0.1
By Eq. (13.28),
K
RT
x
=K
1
Since
0.10 1
x
(a)
y1
13.19
K
=
K
x
1
0.10
1
y1
(b)
x
1
0.843
0.10
6.5894
yH2O
6.5894
7.5894
1
yH2O
x
0.0186
ysteam
0.7814
ysteam
0.2
Number the species as shown. Basis is
1 mol species 1 + x mol steam entering.
y1 =
1
1
x
235030
=2
n0 = 1
x
1
x
2
= 0.12
y3 = 2 y2 = 0.24
From data in Table C.4,
H298
Ans.
0.8682
y2 =
2
y1
Ans.
C4H10(g) = CH2:CHCH:CH2(g) + 2H2(g)
(1)
(2)
(3)
By Eq. (13.5),
0.53802
J
G298
mol
166365
J
mol
The following vectors represent the species of the reaction in the order in
which they appear:
1
1
2
1.935
A
36.915
2.734
B
3.249
26.786 10
0.422
497
3
11.402
C
0.0
8.882
6
10
D
0.0
end
i
rows ( A)
A
A
B
D
i Ci
i
3
T0
C
2.52 10
6
D
1.66
298.15 kelvin
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
3J
9.242
10
K
mol
G
RT
exp
( 0.12) ( 0.24)
By Eq. (13.28),
2
K
1
x
2
Because
0.12 1
x
2
K
=
K
1
0.12
1
(a) y1
y1
x
2
1
x
2
0.023
(b) ysteam
4.3151
5.3151
0.30066
=K
1
x
i Di
i
9.285 10
925 kelvin
G
C
i
7.297
T
1 end
i Bi
i
10
0.083
B
i Ai
5
0.0
0.839
4.3151
yH2O
1
yH2O
0.617
ysteam
498
( 0.24)
0.36
0.812
y1
Ans.
Ans.
2
4
10
13.20
1/2N2(g) + 3/2H2(g) = NH3(g)
=1
Basis: 1/2 mol N2, 3/2 mol H2 feed
n0 = 2
This is the reaction of Pb. 4.21(a) with all stoichiometric coefficients divided
by two. From the answers to Pbs. 4.21(a), 4.22(a), and 13.7(a) ALL
DIVIDED BY 2, find the following values:
H298
J
mol
46110
A
B
2.9355
(a)
T
G298
2.0905 10
T0
300 kelvin
16450
3
J
mol
C
0
D
5
0.3305 10
298.15 kelvin
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
1.627
P
10
4J
P0
1
K
mol
exp
G
RT
K
679.57
1
From Pb. 13.9 for ideal gases:
1
1
yNH3
P
P0
yNH3 = 0.5
2
3
0.5
0.9664
yNH3
2
(b) For
0.9349
Ans.
by the preceding equation
Solving the next-to-last equation for K with P = P0 gives:
1
K
1.299 K
2
1
1.299
1
K
6.1586
499
Find by trial the value of T for which this is correct. It turns out to be
Ans.
T = 399.5 kelvin
(c) For P = 100, the preceding equation becomes
1
2
1
1
K
K
129.9
0.06159
T = 577.6 kelvin Ans.
Another solution by trial for T yields
(d)
Eq. (13.27) applies, and requires fugacity coefficients, which can be
evaluated by the generalized second-virial correlation. Since iteration
will be necessary, we assume a starting T of 583 K for which:
T
583kelvin
P
For NH3(1): Tc1
Pc1
405.7kelvin
T
Tc1
Tr1
For N2(2):
100bar
Tr1
Tc2
Tr2
583
126.2
1.437
Pc2
126.2kelvin
Tr2
4.62
1
P
Pc1
Pr1
34.0bar
Pr2
100
34.0
0.253
Pr1
0.887
2
112.8bar
0.038
Pr2
2.941
For H2(3), estimate critical constants using Eqns. (3.58) and (3.59)
Tc3
1
43.6
21.8
2.016
Pc3
1
3
kelvin
T
kelvin
20.5
44.2
Tc3
Tr3
42.806 K
T
Tc3
Pc3
T
2.016
kelvin
0
500
P
Pc3
13.62
Pr3
5.061
19.757 bar
Pr3
bar
Tr3
Therefore,
i
13
PHIB Tr1 Pr1
1
0.924
PHIB Tr2 Pr2
2
1.034
PHIB Tr3 Pr3
3
1.029
1
i
i
0.5
1.184
i
1.5
The expression used for K in Part (c) now becomes:
1
2
1
K
1
K
0.07292
129.9
1.184
Another solution by trial for T yields T = 568.6 K
Ans.
Of course, the INITIAL assumption made for T was not so close to the
final T as is shown here, and several trials were in fact made, but not
shown here. The trials are made by simply changing numbers in the
given expressions, without reproducing them.
13.21 CO(g) + 2H2(g) = CH3OH(g)
=2
Basis: 1 mol CO, 2 mol H2 feed
n0 = 3
From the data of Table C.4,
H298
90135
J
mol
G298
24791
J
mol
This is the reaction of Ex. 4.6, Pg. 142 from which:
A
(a) T
7.663
B
300 kelvin
10.815 10
T0
3
C
298.15 kelvin
501
3.45 10
6
D
5
0.135 10
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
2.439
P
10
P0
1
4J
K
mol
exp
G
RT
K
1.762
4
10
1
By Eq. (13.5), with the species numbered in the order in which they appear
in the reaction,
y1 =
1
3
y2 =
2
2
2
3
2
By Eq. (13.28),
2
41
y3
(b)
3
y3
2
3
=
y3
2
2
P
P0
2
K
Find
0.9752
0.9291 Ans.
By the preceding equation
0.5
3 y3
2 y3
3
(guess)
0.8
3
Given
y3 =
0.75
1
Solution of the equilibrium equation for K gives
K
3
41
2
2
3
K
27
Find by trial the value of T for which this is correct. It turns out to be:
T = 364.47 kelvin Ans.
502
(c) For P = 100 bar, the preceding equation becomes
3
K
2
2
2
100
K
2.7
10
3
3
41
Another solution by trial for T yields T = 516.48 kelvin Ans.
(d) Eq. (13.27) applies, and requires fugacity coefficients. Since iteration
will be necessary, assume a starting T of 528 K, for which:
T
P
528kelvin
For CO(1): Tc1
T
Tc1
For CH3OH(3):
T
Tc3
Tr
Tc3
Tr1
1.03
P
Pc1
Pr1
2.858
3
0.564
Pr1
Pc3
P
Pc3
Pr
0.048
34.99bar
3.973
512.6kelvin
1
80.97bar
Pc1
132.9kelvin
Tr1
Tr
100bar
Pr
1.235
By Eq. (11.67) and data from Tables E.15 & E.16.
3
0.6206 0.9763
3
3
0.612
For H2(2), the reduced temperature is so large that it may be
assumed ideal:
Therefore:
i
13
PHIB Tr1 Pr1
1.032
1
1.0
1
0.612
0.612
1
i
2
1
i
i
503
0.5933
The expression used for K in Part (c) now becomes:
3
K
2
2
100
2
0.593
K
1.6011
10
3
3
41
Another solution by trial for T yields: T = 528.7 kelvin Ans.
13.22 CaCO3(s) = CaO(s) + CO2(g)
Each species exists PURE as an individual phase, for which the activity is
f/f0. For the two species existing as solid phases, f and f0 are for practical
purposes the same, and the activity is unity. If the pure CO2 is assumed
an ideal gas at 1(atm), then for CO2 the activity is f/f0 = P/P0 = P (in bar).
As a result, Eq. (13.10) becomes K = P = 1.0133, and we must find the T
for which K has this value.
From the data of Table C.4,
H298
J
178321
G298
mol
130401
J
mol
The following vectors represent the species of the reaction in the order in
which they appear:
1
12.572
1
A
1
i
6.104
B
0.443 10
5.457
13
A
T
1.011
i Ai
1151.83 kelvin
G
B
3
5
D
1.047 10
1.157
1.045
B
i Bi
i
A
3.120
2.637
D
i
1.149 10
T0
3
i
C
298.15 kelvin
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
H298
504
i Di
0
D
9.16
4
10
G
J
mol
126.324
Thus
K
exp
G
K
RT
1.0133
T = 1151.83 kelvin Ans.
Although a number of trials were required to reach this result, only the
final trial is shown. A handbook value for this temperature is 1171 K.
13.23 NH4Cl(s) = NH3(g) + HCl(g)
The NH4Cl exists PURE as a solid phase, for which the activity is f/f0.
Since f and f0 are for practical purposes the same, the activity is unity. If
the equimolar mixture of NH3 and HCl is assumed an ideal gas mixture at
1.5 bar, then with f0 = 1 bar the activity of each gas species is its partial
pressure, (0.5)(1.5) = 0.75. As a result, Eq. (13.10) becomes K =
(0.75)(0.75) = 0.5625 , and we must find the T for which K has this value.
From the given data and the data of Table C.4,
H298
176013
J
mol
G298
J
mol
91121
The following vectors represent the species of the reaction in the order in
which they appear:
1
1
5.939
A
3.578
1
i
B
3.156
13
A
3.020
T
0.795
i Ai
623.97 kelvin
G
B
3
10
5
D
0.186 10
0.151
0.623
B
i
A
0.0
16.105
i Bi
D
i
0.012462
T0
C
298.15 kelvin
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
H298
505
i Di
i
0
D
3.5
3
10
G
3J
2.986
Thus
10
K
mol
exp
G
K
RT
0.5624
Ans.
T = 623.97 K
Although a number of trials were required to reach this result, only the
final trial is shown.
13.25
NO(g) + (1/2)O2(g) = NO2(g)
yNO2
yNO yO2
0.5
yNO2
=
0.5
= 0.5
=K
T
From the data of Table C.4,
K
Given
yNO
yNO2
G298
G298
exp
yNO
298.15 kelvin
yNO ( )
0.21
K
RT
10
12
yNO2
0.5
yNO2 = ( )
0.21
10
6
1.493
7 10
6
6
10
yNO2
yNO = 5 10
yNO
K yNO
ppm
7.307
J
mol
12
= 0.5
See Example 13.9, Pg. 508-510 From Table C.4,
105140
10
6
(a negligible concentration) Ans.
13.26 C2H4(g) + (1/2)O2(g) = <(CH2)2>O(g)
H298
J
mol
(guesses)
Find yNO yNO2
This is about
35240
G298
81470
J
mol
Basis: 1 mol C2H4 entering reactor.
Moles O2 entering:
nO2
1.25 0.5
Moles N2 entering:
nO2
nN2
506
79
21
n0 1 nO2 nN2
n0 3.976
Index the product species with the numbers:
1 = ethylene
2 = oxygen
3 = ethylene oxide
4 = nitrogen
The numbers of moles in the product stream are given by Eq. (13.5).
For the product stream, data from Table C.1:
Guess:
0.8
1
nO2
n
1.424
0.5
3.639
A
23.463
3.280
4.392
0.0
10
9.296
6
1
kelvin
0.227
D
2
0.0
5
0.5
2
10 kelvin
1
0
0.040
14
A
n
i Ai
B
n
i
n
i Ci
D
i
n
n0
0.5
i Bi
i
C
y
3
10
kelvin
0.593
0.0
0.0
i
0.506
B
0.385
nN2
C
14.394
n
i Di
i
K
y
i
i
K
15.947
i
The energy balance for the adiabatic reactor is:
H298
HP = 0
For the second term, we combine Eqs. (4.3) & (4.7).
The three equations together provide the energy balance.
507
For the equilibrium state, apply a combination of Eqs. (13.11a) &
(13.18).The reaction considered here is that of Pb. 4.21(g), for which the
following values are given in Pb. 4.23(g):
A
D
3.629
B
5
0.114 10 kelvin
T0
8.816
idcph
A T0
C
B
1
2
3
3
T0
3
C
kelvin
4.904
A ln
T0
2
2
1
1
D
T0
1
1
2
B T0
C T0
D
130.182 kelvin
1
2
2
idcps
0.417
Given
H298 = R A
T0
C
3
3
K
= exp
Find
1
T0
H298
R T0
3
G298
B
2
T0
2
2
1
1
D
1
T0
H298
R T0
0.88244
3.18374
508
idcps
10
6
2
kelvin
298.15 kelvin
T0
idcph
3
3
Guess:
idcps
2
10
1
T0
idcph
0.0333
0.052
y 0.88244)
(
Ans.
0.2496
0.6651
T
13.27
T0
949.23 kelvin Ans.
T
CH4(g) = C(s) + 2H2(g)
(gases only)
=1
The carbon exists PURE as an individual phase, for which the activity is
unity. Thus we leave it out of consideration.
From the data of Table C.4,
H298
74520
J
G298
mol
50460
J
mol
The following vectors represent the species of the reaction in the order in
which they appear:
1
9.081
1.702
1
A
2
B
1.771
0.422
3.249
0.0
2.164
i
13
C
0.0
10
6
5
D
0.867 10
0.083
0.0
A
i Ai
i
A
6.567
T
B
i Bi
C
i
B
923.15 kelvin
7.466 10
T0
3
0.771 10
i Ci
D
i
3
C
2.164
298.15 kelvin
509
i Di
i
10
6
D
4
7.01 10
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
1.109
10
4J
mol
By Eq. (13.5),
(a)
K
exp
G
RT
yCH4 =
n0 = 1
K
1
2
1
K
4
yCH4
1
2
yH2
1
=
1
4
2
yCH4
1
By Eq. (13.28),
2
0.7893
yCH4
yH2
1
2
0.5659
=K
0.1646
Ans.
0.8354
(b) For a feed of 1 mol CH4 and 1 mol N2,
Given
1
2
1
yH2
2
2
(fraction decomposed)
0.7173
K
yH2 =
1
2
By Eq. (13.28),
4.2392
n0 = 2
.8
(guess)
2
Find
=K
1
(fraction decomposed)
yH2
yCH4
2
yN2
2
0.0756
510
1
yCH4
yN2
0.3585
yH2
Ans.
13.28 1/2N2(g) + 1/2O2(g) = NO(g)
(1)
=0
This is the reaction of Pb. 4.21(n) with all stoichiometric coefficients
divided by two. From the answers to Pbs. 4.21(n), 4.22(n), and 13.7(n) ALL
DIVIDED BY 2, find the following values:
H298
A
T
90250
J
mol
G298
B
0.0725
0.0795 10
T0
2000 kelvin
86550
3
J
mol
C
0
D
5
0.1075 10
298.15 kelvin
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
6.501 10
4J
K1
mol
G
RT
exp
1/2N2(g) + O2(g) = NO2(g)
K1
0.02004
= 0.5
(2)
From the data of Table C.4,
H298
33180
J
mol
G298
51310
J
mol
The following vectors represent the species of the reaction in the order in
which they appear:
0.5
3.280
1
A
3.639
1
i
13
0.593
B
0.506 10
4.982
A
T
0.297
2000 kelvin
3
5
D
0.227 10
1.195
B
i Ai
0.792
D
i Bi
B
3.925
T0
10
4
298.15 kelvin
511
i Di
i
i
i
A
0.040
C
0
D
5.85
4
10
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
5J
1.592
10
K2
mol
exp
G
K2
RT
6.9373 10
5
With the assumption of ideal gases, we apply Eq. (13.28):
yNO
(1)
yN2
0.5
yNO
(2)
K1 ( )
0.7
P0
1
yNO2
yN2
0.5
yNO2
13.29
0.5
0.5
0.5
0.5
= K1
()
0.05
yNO
3.74962
3
10
Ans.
200
yNO2
=
0.5
()
0.7
( .05)
0
P
0.5
()
0.7
yO2
P
P0
yO2
yNO
=
0.5
P
P0
=
()
0.05
0.5
K2 ( .7)
0
( .05)
0
0.5
K2
yNO2
4.104
10
5
Ans.
2H2S(g) + SO2(g) = 3S(s) + 2H2O(g)
The sulfur exists PURE as a solid phase, for which the activity is f/f0. Since
f and f0 are for practical purposes the same, the activity is unity, and it is
omitted from the equilibrium equation. Thus for the gases only,
=1
From the given data and the data of Table C.4,
H298
145546
J
mol
G298
512
89830
J
mol
The following vectors represent the species of the reaction in the order in
which they appear:
2
3.931
1
5.699
A
3
A
B
i Ai
T
1.015
D
0.783
D
i Bi
6.065 10
T0
723.15 kelvin
i Di
i
3
C
D
0
6.28
298.15 kelvin
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
1.538 10
4J
K
mol
By Eq. (13.5), gases only:
yH2S =
2
2
n0 = 3
ySO2 =
3
By Eq. (13.28),
Given
2
2
1
3
2
G
RT
K
3
1
Percent conversion of reactants = PC
PC =
ni0
ni0
ni
100 =
i
ni0
[By Eq. (13.4)]
100
513
2
3
(guess)
Find
= 8K
12.9169
(basis)
yH2O =
0.5
2
2
exp
5
10
0.121
i
B
5.721
3
1.450
i
A
10
1.728
3.470
14
0.232
0.801
B
4.114
2
i
1.490
0.767
4
10
Since the reactants are present in the stoichiometric proportions, for each
reactant,
ni0 =
Whence
i
PC
PC
100
13.30 N2O4(g) = 2NO2(g)
(a)
(b)
76.667
Ans.
=1
Data from Tables C.4 and C.1 provide the following values:
H298
T0
57200
J
mol
G298
T
298.15 kelvin
A
G
3.968
J
mol
350 kelvin
0.133 10
3
0
D
G
RT
K
C
5
1.203 10
H298
G
B
1.696
5080
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
10
3J
K
mol
exp
3.911
Basis: 1 mol species (a) initially. Then
ya =
(a)
1
1
P
ya
(b)
yb =
P
5
1
1
P0
ya
1
P0
=
1
K
4P K
1
1
1
2
2
2
0.4241
514
1
K
0.4044
Ans.
K
4P K
1
P
P0
0.7031
By Eq. (4.18), at 350 K:
H
H298
R IDCPH T0 T
A
B
C
H
D
56984
J
mol
This is Q per mol of reaction, which is
0.7031
0.299
0.4044
Whence
Q
Q
H
13.31 By Eq. (13.32),
K=
K=
K=
1
2
exp 0.1 xB
xA
xA
G
exp 0.1 2 xA
1000
xA
xA A
xA
B
xA A
Whence
J
mol
1
=
xA
xA
1
T
2
exp 0.1 xA
K = exp
2
xB
G
RT
298.15 kelvin
(guess)
.5
Given
xA
exp 0.1 xA
xA
=
Ans.
mol
ln b = 0.1 xA
2
xA
1
xB B
J
2
2
ln a = 0.1 xB
1
17021
1
xA
xA
exp 0.1 2 xA
1
= exp
G
RT
xA
Find xA
Ans.
0.3955
For an ideal solution, the exponential term is unity:
Given
1
xA
xA
= exp
G
RT
xA
This result is high by 0.0050. Ans.
515
Find xA
xA
0.4005
13.32 H2O(g) + CO(g) = H2(g) + CO2(g)
=0
From the the data of Table C.4,
H298
T0
A
J
41166
G298
mol
T
298.15 kelvin
B
1.860
28618
J
mol
800 kelvin
3
0.540 10
C
5
D
0
1.164 10
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
9.668
10
3J
mol
(a) No. Since
=0
K
exp
G
RT
K
4.27837
, at low pressures P has no effect
(b) No. K decreases with increasing T. (The standard heat of reaction is
negative.).
(c) Basis: 1 mol CO, 1 mol H2, w mol H2O feed.
From the problem statement,
nCO
nCO
nH2
nCO2
= 0.02
By Eq. (13.4),
1
1
1
nCO = 1
=
1
2
NCO2 =
nH2 = 1
0.96
1.02
= 0.02
0.941
Let z = w/2 = moles H2O/mole "Water gas".
By Eq. (13.5),
yH2O =
2z
w
=
2 2z
2w
yCO =
516
1
2 2z
yH2 =
1
2 2z
yCO2 =
2
By Eq. (13.28)
2z
1
Given
2z
(d)
1
z
z
=K
Find z
()
(guess)
2
z
Ans.
4.1
= 1 (gases)
2CO(g) = CO2(g) + C(s)
Data from Tables C.4 and C.1:
H298
A
172459
J
mol
B
0.476
G298
0.702 10
120021
3
C
J
mol
5
D
0
1.962 10
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
3.074
10
4J
K
mol
G
RT
exp
K
101.7
By Eq. (13.28), gases only, with P = P0 = 1 bar
yCO2
yCO
2
= K = 101.7
for the reaction AT EQUILIBRIUM.
If the ACTUAL value of this ratio is GREATER than this value, the
reaction tries to shift left to reduce the ratio. But if no carbon is present, no
reaction is possible, and certainly no carbon is formed. The actual value of
the ratio in the equilibrium mixture of Part (c) is
yCO2
yCO2
RATIO
2
2z
yCO
0.092
yCO2
yCO
1
2 2z
yCO
2
5.767
RATIO
10
2.775
3
3
10
No carbon can deposit from the equilibrium mixture.
517
13.33 CO(g) + 2H2(g) = CH3OH(g)
(1)
=2
This is the reaction of Pb. 13.21, where the following parameter values are
given:
H298
T
90135
J
mol
G298
T0
550 kelvin
A
G
B
7.663
H298
10.815 10
3
J
mol
24791
298.15 kelvin
C
3.45 10
6
5
D
0.135 10
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
3.339
4J
10
K1
mol
exp
G
RT
K1
H2(g) + CO2(g) = CO(g) + H2O(g)
6.749
10
4
(2)
=0
From the the data of Table C.4,
H298
T
41166
J
mol
G298
T0
550 kelvin
J
mol
28618
298.15 kelvin
The following vectors represent the species of the reaction in the order in
which they appear:
1
1
1
3.249
5.457
A
14
B
3.376
1
i
0.422
3.470
A
1.86
B
i Ai
B
0.557
3
10
1.157
D
0.031
1.450
5.4
10
4
D
i Bi
518
C
5
10
0.121
i Di
i
i
i
A
1.045
0.083
0
D
1.164
5
10
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
1.856 10
4J
K2
mol
G
RT
exp
K2
0.01726
Basis: 1 mole of feed gas containing 0.75 mol H2,
0.15 mol CO, 0.05 mol CO2, and 0.05 mol N2.
Stoichiometric numbers,
i=
H2
CO
i.j
CO2
CH3OH
H2O
_______________________________________________
j
1
2
-2
-1
-1
1
0
-1
1
0
0
1
By Eq. (13.7)
0.75
yH2 =
21
1
1
yCO =
21
0.05
yCO2 =
0.15
2
2
1
yCH3OH =
21
100
P0
1
2
21
1
1
yH2O =
21
0.1
2
1
21
1
By Eq. (13.40),
1
P
2
0.1
(guesses)
Given
11
0.75
0.75
2
21
0.15
21
21
2
2
=
0.15
2
1
2
1
2
0.05
2
= K2
2
P
P0
2
K1
1
2
519
Find 1 2
1
0.1186
0.75
yH2
21
1
1
1
yH2
0.6606
yCH3OH
yCO
yCO2
21
1
2
yH2O
21
0.0528
yH2O
2
1
yCH3OH
yCO
0.1555
1
1
yCH3OH
21
yH2
13.34
yCO
2
yN2
3
10
0.15
2
21
0.05
yCO2
8.8812
2
1
21
yH2O
yCO2
0.0539
Ans.
yN2
0.0116
CH4(g) + H2O(g) = CO(g) + 3H2(g)
0.0655
(1)
=2
From the the data of Table C.4,
H298
J
mol
205813
G298
141863
J
mol
The following vectors represent the species of the reaction in the order in
which they appear:
1
1.702
1
A
1
3
3.470
3.249
0.0
0.0
10
6
D
A
7.951
B
0.422
0.121
0.031
5
i
10
C
i Bi
14
8.708 10
i Ci
D
3
C
520
2.164
i Di
i
i
i
i
0.557
3
10
0.083
B
i Ai
1.450
0.0
0.0
A
B
3.376
2.164
C
9.081
10
6
D
9.7
3
10
T
1300 kelvin
T0
298.15 kelvin
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
G
H298
G
1.031
10
5J
K1
mol
exp
G
K1
RT
13845
H2O(g) + CO(g) = H2(g) + CO2(g)
= 0 (2)
This is the reaction of Pb. 13.32, where parameter values are given:
H298
A
G
G
41166
1.860
J
G298
mol
B
0.540 10
3
28618
C
0.0
J
mol
D
5
1.164 10
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
H298
5.892
3J
10
mol
K2
exp
G
RT
K2
0.5798
(a) No. Primary reaction (1) shifts left with increasing P.
(b) No. Primary reaction (1) shifts left with increasing T.
(c) The value of K1 is so large compared with the value of K2 that for all
practical purposes reaction (1) may be considered to go to
completion. With a feed equimolar in CH4 and H2O, no H2O then
remains for reaction (2). In this event the ratio, moles H2/moles CO
is very nearly equal to 3.0.
521
(d) With H2O present in an amount greater than the stoichiometric
ratio, reaction (2) becomes important. However, reaction (1) for all
practical purposes still goes to completion, and may be considered
to provide the feed for reaction (2). On the basis of 1 mol CH4 and
2 mol H2O initially, what is left as feed for reaction (2) is: 1 mol
H2O, 1 mol CO, and 3 mol H2; n0 = 5. Thus, for reaction (2) at
equilibrium by Eq. (13.5):
yCO = yH2O =
1
yCO2 =
5
By Eq. (13.28),
2
1
Ratio =
5
yH2
yCO
0.1375
Find
= K2
Ratio
5
(guess)
0.5
3
Given
3
yH2 =
3
Ratio
1
3.638
Ans.
(e) One practical way is to add CO2 to the feed. Some H2 then reacts
with the CO2 by reaction (2) to form additional CO and to lower the
H2/CO ratio.
(f) 2CO(g) = CO2(g) + C(s)
= 1 (gases)
This reaction is considered in the preceding problem, Part (d), from
which we get the necessary parameter values:
H298
172459
J
mol
For T = 1300 K,
A
G
0.476
G298
T
B
0.702 10
3
120021
1300 kelvin
C
0.0
T
H298
G298
T0
R IDCPH T0 T A B C D
R T IDCPS T0 T A B C D
H298
522
J
mol
T0
298.15 kelvin
D
5
1.962 10
G
4J
5.673
10
K
mol
exp
G
RT
K
5.255685
10
3
As explained in Problem 13.32(d), the question of carbon deposition
depends on:
yCO2
RATIO =
yCO
2
When for ACTUAL compositions the value of this ratio is greater than the
equilibrium value as given by K, there can be no carbon deposition. Thus
in Part (c), where the CO2 mole fraction approaches zero, there is danger
of carbon deposition. However, in Part (d) there can be no carbon
deposition, because Ratio > K:
5
Ratio
Ratio
2
1
0.924
13.37 Formation reactions:
C + 2H2 = CH4
H2 + (1/2)O2 = H2O
C + (1/2)O2 = CO
C + O2 = CO2
Elimination first of C and then of O2 leads to a pair of reactions:
CH4 + H2O = CO + 3H2
(1)
CO + H2O = CO2 + H2
(2)
There are alternative equivalent pairs, but for these:
Stoichiometric numbers,
i=
CH4
H2O
i.j
CO
CO2
H2
j
_________________________________________________
j
1
2
-1
0
-1
-1
1
-1
0
1
523
3
1
2
0
For initial amounts: 2 mol CH4 and 3 mol H2O, n0 = 5, and by Eq. (13.7):
2
yCH4 =
yCO2 =
3
1
5
yH2O =
21
2
5
yH2 =
21
1
5
1
yCO =
21
31
5
2
5
2
21
2
21
By Eq. (13.40), with P = P0 = 1 bar
3
yCO yH2
= k1
yCH4 yH2O
yCO2 yH2
yCO yH2O
= k2
From the data given in Example 13.14,
G1
J
mol
27540
G2
G1
K1
exp
K1
K2
RT
1.5
2
2
3
1
1
5
2
3
1
Find 1 2
G2
RT
2
2
2
1000 kelvin
3
31
231
1
T
(guesses)
2
1
J
mol
1.457
1
1
Given
exp
K2
27.453
1
3130
21
2
= K1
= K2
2
1
1.8304
2
0.3211
2
2
yCH4
5
1
21
3
yH2O
1
5
524
21
2
yCO
1
5
2
21
2
yCO2
31
yH2
521
yCH4
0.0196
0.0371
yH2
521
yH2O
yCO2
2
yCO
0.098
0.1743
0.6711
These results are in agreement with those of Example 13.14.
13.39Phase-equilibrium equations:
Ethylene oxide(1): p1 = y1 P = 415x1 P
Water(2):
x2 Psat2 = y2 P
x1 =
101.33 kPa
Psat2
3.166 kPa
x2 =
y1 P
415kPa
y2 P
Psat2
(steam tables)
Ethylene glycol(3):
Therefore, y2 = 1
Psat3 = 0.0
y1
and
y3 = 0.0
x3 = 1
x2
x3
For the specified standard states:
(CH2)2O(g) + H2O(l) = CH2OH.CH2OH(l)
By Eq. (13.40) and the stated assumptions,
3 x3
k=
y1
P
P0
=
2 x2
x3
T
y1 x2
Data from Table C.4:
k
exp
298.15kelvin
G298
G298
k
RT
525
6.018
72941
12
10
J
mol
Ans.
So large a value of k requires either y1 or x2 to approach zero. If y1
approaches zero, y2 approaches unity, and the phase-equilibrium
expression for water(2) makes x2 = 32, which is impossible. Thus x2
must approach zero, and the phase-equilibrium equation requires y2 also
to approach zero. This means that for all practical purposes the reaction
goes to completion. For initial amounts of 3 moles of ethylene oxide and
1 mole of water, the water present is entirely reacted along with 1 mole of
the ethylene oxide. Conversion of the oxide is therefore
33.3 %.
1
13.41
1
1
1
0
vj
ij
v
i
Initial
numbers of n0
moles
10
a) Stoichiometric coefficients:
Number of components: i
50
50 kmol
hr
0
1
0
Number of reactions:
14
1
n0
1
Given values:
yA
0.05
yB
yC
0.4
yD
n0
n0i
12
kmol
hr
100
i
0.10
Guess:
j
0.4
1
1
kmol
hr
2
1
kmol
hr
Given
yA =
yC =
n01
1
2
n0
1
2
n03
1
2
n0
1
2
n02
yB =
1
n0
yD =
1
n04
n0
2
Eqn. (13.7)
2
1
2
yC
yD
Find yC yD 1 2
1
1
2
526
44.737
kmol
hr
2
2.632
kmol
hr
(i) nA
n01
1
nB
n02
1
nC
n03
1
nD
n04
2
n
nA
yC
(ii)
nB
nB
42.105
nD
nC
5.263
nC
2
2.632
2.632
n
nD
52.632
yD
0.8
1
ij
v
i
hr
kmol
Ans.
hr
Ans.
40
Initial
numbers of n0
moles
0
0
i
hr
kmol
0.05
2
1
Number of components:
hr
kmol
1
1
b) Stoichiometric coefficients:
vj
kmol
hr
kmol
nA
2
40 kmol
hr
0
1
0
Number of reactions:
14
1
Given values:
yC
0.52
yD
yA
0.4
yB
0.4
n0
80
kmol
hr
2
1
1
1
Given
yA =
yC =
n01
n0
1
1
n03
n0
2
22
1
1
22
yB =
yD =
n02
1
22
n0
1
22
n04
n0
527
2
1
kmol
hr
i
0.04
Guess:
12
n0i
n0
2
j
22
Eqn. (13.7)
kmol
hr
yA
yB
Find yA yB 1 2
1
26
1
kmol
hr
2
2
kmol
hr
2
yA
nA
n01
1
2
nA
12
nB
n02
1
22
nB
10
nC
n03
1
nC
26
nD
n04
2
nD
2
hr
kmol
hr
1
1
vj
ij
100
Initial
numbers of n0
moles
1
0
0
i
Ans.
hr
kmol
1
Number of components:
0
Number of reactions:
14
n0
Given values:
yC
0.3
yD
Guess:
yA
0.4
yB
n0i
j
12
100
kmol
hr
n0
i
0.1
0.4
1
1
kmol
hr
2
Given
yA =
yC =
n01
1
2
n0
1
n02
1
2
n0
1
2
2
n03
n0
1
1
0
yB =
yD =
2
n04
n0
528
kmol
hr
0
1
i
0
1
1
v
0.2
kmol
hr
kmol
1
c) Stoichiometric coefficients:
yB
0.24
2
1
2
Eqn. (13.7)
1
kmol
hr
yA
yB
Find yA yB 1 2
kmol
hr
37.5
2
0.4
12.5
yB
1
yA
1
kmol
0.2
hr
2
kmol
hr
kmol
nA
n01
1
2
nA
50
nB
n02
1
2
nB
25
nC
n03
1
nC
37.5
nD
n04
2
nD
12.5
hr
kmol
hr
kmol
hr
1
1
1
d)
Stoichiometric coefficients:
1
ij
Guess:
yA
1
0.2
kmol
hr
0
0
0
n0
n0i
j
n0
12
100
kmol
hr
i
yD
yB
60
Initial
numbers of n0
moles
Number of reactions:
15
0
0.25
40
1
1
v
i
Given values: yC
i
0
0
vj
1
0
Number of components:
Ans.
0.20
0.4
529
yE
0.1
1
1
kmol
hr
2
1
kmol
hr
Given
n01
yA =
1
n0
n02
yB =
1
n0
1
n0
1
n03
yC =
2
2
n0
1
Eqn. (13.7)
1
n04
yD =
2
n05
2
n0
1
1
yE =
yA
yB
yE
Find yA yB yE
1
1
2
20
kmol
hr
16
2
kmol
hr
1
2
(i)
kmol
hr
kmol
24
hr
kmol
20
hr
nA
n01
1
2
nA
nB
n02
1
2
nB
nC
n03
1
nC
nD
n04
2
nD
16
nE
13.45
(ii)
n05
2
nE
16
yA
hr
yC
0.25
0.2
yE
kmol
hr
kmol
0.3
yD
Ans.
0.05
yB
4
0.2
C2H4(g) + H2O(g) -> C2H5OH(g)
T0
298.15kelvin
P0
T
1bar
J
mol
1 = C2H4(g)
H0f1
2 = H2O(g)
H0f2
241818
3 = C2H5OH(g)
H0f3
235100
52500
400kelvin
G0f1
J
mol
J
mol
530
P
2bar
68460
J
mol
G0f2
228572
G0f3
168490
J
mol
J
mol
H0
H0f1
H0f2
H0f3
H0
45.782
G0
G0f1
G0f2
G0f3
G0
8.378
A
A
B
[(
14.394) ( .450) ( 0.001)
1
210
C
[ ( 4.392) () ( 6.002)
010
D
[ () ( .121) ()
0
0
0 ]10
exp
G0
R T0
b) K1
exp
H0
1
R T0
K2
mol
(
1.424) ( .470) ( .518)
3
3
a) K 0
K0 K1 K2
1.376
B
6
4.157
3
10
6
C
1.61 10
D
5
1.21 10
4
K298
K0
Eqn. (13.22)
K2
Eqn. (13.20)
Ans.
29.366
K1
1
IDCPH T0 T A B C D
T
IDCPS T0 T A B C D
exp
K400
3
Eqn. (13.21) K298
T0
T
kJ
mol
kJ
9.07
3
Eqn. (13.23)
0.989
K400
10
0.263
Ans.
c) Assume as a basis there is initially 1 mol of C2H4 and 1 mol of H2O
1
y1 =
2
e
e
1
y2 =
2
Assuming ideal gas behavior
e
y3 =
e
y3
y1 y2
=K
e
2
e
P
P0
e
2
Substituting results in the following expression:
e
= K400
1
e
2
531
e1
e2
e
P
P0
e using
Solve for
a Mathcad solve block.
Guess:
e
0.5
e
2
Given
e
= K400
1
e1
e
2
e2
y1
Find e
e
P0
0.191
e
e
1
y1
P
1
e
2
y2
2
e
0.447
y2
e
e
y3
2
e
0.447
y3
e
Ans.
0.105
d) Since
a decrease in pressure will cause a shift on the reaction
to the left and the mole fraction of ethanol will decrease.
13.46 H2(g) + O2(g) -> H2O2(g)
H0fH2O2
S0H2
130.680
S0H2O2
S0fH2O2
G0f
136.1064
kJ
mol
T
J
mol kelvin
232.95
S0O2
205.152
1bar
J
mol kelvin
J
mol kelvin
S0H2
H0fH2O2
298.15kelvin P
S0O2
S0H2O2
T S0fH2O2
S0fH2O2
G0 f
532
102.882
105.432
kJ
mol
J
mol kelvin
Ans.
13.48
C3H8(g) -> C3H6(g) + H2(g) (I)
C3H8(g) -> C2H4(g) + CH4(g) (II)
T0
P0
298.15kelvin
T
1bar
1 = C3H8 (g)
H0f1
2 = C3H6 (g)
H0f2
19710
3 = H2 (g)
H0f3
0
4 = C2H4 (g)
H0f4
52510
5 = CH4 (g)
H0f5
104680
P
750kelvin
J
G0f1
mol
J
mol
1.2bar
G0f2
J
mol
74520
J
mol
0
G0f4
mol
62205
G0f3
J
J
24290
68460
G0f5
mol
J
mol
J
mol
J
mol
J
mol
50460
Calculate equilibrium constant for reaction I:
H0I
H0f1
H0f2
H0f3
H0I
124.39
G0I
G0f1
G0f2
G0f3
G0I
86.495
AI
BI
[(
28.785) ( 2.706) ( .422)
2
010
CI
[ ( 8.824) ( 6.915) ()
0 ]10
DI
3
[ () () ( .083)
0
0
010
6
KI1
exp
KI2
exp
3.673
BI
5.657
CI
1.909
DI
5
exp
mol
AI
(
1.213) ( .637) ( .249)
1
3
KI 0
kJ
mol
kJ
8.3
3
10
10
6
3
10
G0I
R T0
H0I
1
R T0
Eqn. (13.21)
T0
T
Eqn. (13.22)
1
IDCPH T0 T AI BI CI DI
T
IDCPS T0 T AI BI CI DI
533
KI0
0
KI1
1.348
KI2
1.714
13
10
Eqn. (13.23)
KI
Eqn. (13.20)
KI0 KI1 KI2
KI
0.016
Calculate equilibrium constant for reaction II:
H0II
H0f1
H0f4
H0f5
H0II
82.67
G0II
G0f1
G0f4
G0f5
G0II
42.29
AII
( 1.213)
( 1.424)
( 1.702)
BII
[ ( 28.785)
( 14.394)
( 9.081) ] 10
CII
[ ( 8.824)
( 4.392)
( 2.164) ] 10
DII
[ ( 0)
kJ
( 0) ] 10
( 0)
KII0
exp
G0II
R T0
KII1
exp
H0II
1
R T0
KII2
exp
KII
mol
kJ
mol
AII
BII
5
Eqn. (13.21)
T0
T
10
2.268
10
DII
6
5.31
CII
3
1.913
0
3.897
10
KII1
5.322
10
1
IDCPH T0 T AII BII CII DII
T
IDCPS T0 T AII BII CII DII
KII0 KII1 KII2 Eqn. (13.20)
KII2
Eqn. (13.23)
21.328
Assume an ideal gas and 1 mol of C3H8 initially.
1
y1 =
y4 =
1
I
II
I
II
II
1
I
y2 =
I
1
y5 =
II
y3 =
I
II
II
1
I
I
1
I
y4 y5
y1
= KII
534
II
Eqn. (13.7)
II
The equilibrium relationships are:
y2 y3
P0
= KI
P
y1
P0
P
8
1.028
KII
Eqn. (13.28)
6
8
KII0
Eqn. (13.22)
3
Substitution yields the following equations:
I
1
I
I
1
II
I
1
I
I
II
II
1
P0
P
= KI
II
1
II
II
I
1
II
I
1
I
I
P0
= KII
P
II
1
II
II
Use a Mathcad solve block to solve these two equations for I and II. Note
that the equations have been rearranged to facilitate the numerical
solution.
Guess:
0.5
0.5
I
II
Given
I
I
1
II 1
I
I
1
II 1
I
1
I
II
P
1
I
II
P0
P
1
I
II
1
I
II
II
II
II
P0
= KI
= KII
I
II
I
Find I
0.026
I
II
0.948
II
II
1
y1
y4
y1
1
I
II
I
II
II
1
y2
y5
I
1
I
II
I
1
I
II
II
II
1
0.01298 y2
0.0132 y3
I
y3
I
II
0.0132 y4
535
0.4803 y5
0.4803
A summary of the values for the other temperatures is given in the table below.
T=
y1
y2
y3
y4
y5
7K
50
0.0130
0.0132
0.0132
0.48
03
0.48
03
1000 K
0.00047
0.034
0.034
0.46
58
0.46
58
1250 K
0.00006
0.059
3
0.059
3
0.4407
0.4407
13.49 n-C4H10(g) -> iso-C4H10(g)
T0
P0
298.15kelvin
T
1bar
1 = n-C4H10(g)
H0f1
125790
2 = iso-C4H10(g)
H0f2
134180
J
mol
J
H0f1
H0f2
H0
8.39
G0
G0f1
G0f2
G0
4.19
C
[11.402) ( 11.945)
(10
D
3
[(
36.915) ( 7.853)
310
[ () ()
0
0 ]10
6
5
exp
G0
R T0
b) K1
exp
H0
1
R T0
K2
exp
20760
T0
T
Eqn. (13.22)
1
IDCPH T0 T A B C D
T
IDCPS T0 T A B C D
536
mol
mol
kJ
mol
0.258
B
9.38 10
C
5.43 10
D
Eqn. (13.21)
J
mol
J
kJ
A
B
a) K0
16570
G0f2
mol
(
1.935) ( .677)
1
15bar
G0f1
H0
A
P
425kelvin
4
7
0
K0
5.421
K1
Ans.
0.364
K2
1
Eqn. (13.23)
Ke
Eqn. (13.20)
K0 K1 K2
Ke
Ans.
1.974
Assume as a basis there is initially 1 mol of n-C4H10(g)
y1 = 1
y2 = e
e
y2
a) Assuming ideal gas behavior
= Ke
y1
e
Substitution results in the following expression:
1
= Ke
e
Solving for Ke yields the following analytical expression for
1
e
1
1
y1
e
e
y1
0.664
0.336
y2
Ke
e
e
y2
0.336
Ans.
b) Assume the gas is an ideal solution. In this case Eqn. (13.27) applies.
yi i =
P
P0
Eqn. (13.27)
K
i
1
Substituting for yi yields:
e
2
=K
e1
This can be solved analytically for
e
to get:
2
e=
2
Ke 1
Calculate i for each pure component using the PHIB function.
For n-C4H10:
Tr1
1
0.200
Tc1
Tr1
1
Pr1
PHIB Tr1 Pr1
1
1
T
Tc1
For iso-C4H10:
Tr2
T
Tc2
1
2
0.181
Tc2
Tr2
1.041
Pr2
537
425.1kelvin
P
Pc1
Pc1
37.96bar
Pr1
0.395
Pc2
36.48bar
Pr2
0.411
0.872
408.1kelvin
P
Pc2
2
PHIB Tr2 Pr2
Solving for
e
2
2
2
yields:
e
2
y1
1
e
0.884
y1
0.661
y2
e
Ke 1
e
y2
0.339
0.339
Ans.
The values of y1 and y2 calculated in parts a) and b) differ by less than 1%.
Therefore, the effects of vapor-phase nonidealities is here minimal.
538
Chapter 14 - Section A - Mathcad Solutions
14.1
A12 := 0.59
A21 := 1.42
T := ( 55 + 273.15) K
Margules equations:
γ 1 ( x1) := exp ( 1 x1) A12 + 2 ( A21 A12) x1
2
γ 2 ( x1) := exp x1 A21 + 2 ( A12 A21) ( 1 x1)
2
Psat1 := 82.37 kPa
(a)
Psat2 := 37.31 kPa
BUBL P calculations based on Eq. (10.5):
Pbubl ( x1) := x1 γ 1 ( x1) Psat1 + ( 1 x1) γ 2 ( x1) Psat2
y1 ( x1) :=
x1 γ 1 ( x1) Psat1
Pbubl ( x1)
x1 := 0.25
y1 ( x1) = 0.562
x1 := 0.50
Pbubl ( x1) = 80.357 kPa
y1 ( x1) = 0.731
x1 := 0.75
(b)
Pbubl ( x1) = 64.533 kPa
Pbubl ( x1) = 85.701 kPa
y1 ( x1) = 0.808
BUBL P calculations with virial coefficients:
3
B11 := 963
cm
mol
3
B22 := 1523
cm
mol
δ 12 := 2 B12 B11 B22
B11 ( P Psat1) + P y22 δ 12
Φ 1 ( P , T , y1 , y2) := exp
R T
B22 ( P Psat2) + P y12 δ 12
Φ 2 ( P , T , y1 , y2) := exp
R T
539
3
B12 := 52
cm
mol
Guess:
x1 := 0.25
P :=
Psat1 + Psat2
y1 := 0.5
2
y2 := 1 y1
Given
y1 Φ 1 ( P , T , y1 , y2) P = x1 γ 1 ( x1) Psat1
y2 Φ 2 ( P , T , y1 , y2) P = ( 1 x1) γ 2 ( x1) Psat2
y2 = 1 y1
y1
0.558
y2
= 0.442
P
63.757
kPa
y1
y2 := Find ( y1 , y2 , P)
P
x1 := 0.50
Given
y1 Φ 1 ( P , T , y1 , y2) P = x1 γ 1 ( x1) Psat1
y2 Φ 2 ( P , T , y1 , y2) P = ( 1 x1) γ 2 ( x1) Psat2
y2 = 1 y1
y1
0.733
y2 = 0.267
P
79.621
kPa
y1
y2 := Find ( y1 , y2 , P)
P
x1 := 0.75
Given
y1 Φ 1 ( P , T , y1 , y2) P = x1 γ 1 ( x1) Psat1
y2 Φ 2 ( P , T , y1 , y2) P = ( 1 x1) γ 2 ( x1) Psat2
y2 = 1 y1
y1
0.812
y2 = 0.188
P
85.14
kPa
y1
y2 := Find ( y1 , y2 , P)
P
540
T := 200 K
P := 30 bar
H1 := 200 bar
14.3
B := 105
y1 := 0.95
3
cm
mol
Assume Henry's law applies to methane(1) in the liquid phase, and that the
Lewis/Randall rule applies to the methane in the vapor:
l
v
fhat1 = H1 x1
fhat1 = y1 φ 1 P
By Eq. (11.36):
φ 1 := exp
B P
R T
φ 1 = 0.827
Equate the liquid- and vapor-phase fugacities and solve for x1:
x1 :=
14.4
y1 φ 1 P
x1 = 0.118
H1
Ans.
Pressures in kPa
Data:
0.000
0.0895
0.1981
0.3193
x1 :=
0.4232
0.5119
0.6096
0.7135
i := 2 .. rows ( P)
(a)
12.30
15.51
18.61
21.63
P :=
24.01
25.92
27.96
30.12
x2 := 1 x1
0.000
0.2716
0.4565
0.5934
y1 :=
0.6815
0.7440
0.8050
0.8639
Psat2 := P1
It follows immediately from Eq. (12.10a) that:
ln γ 1 = A12
Combining this with Eq. (12.10a) yields the required expression
541
(b)
Henry's constant will be found as part of the solution to Part (c)
(c)
BARKER'S METHOD by non-linear least squares.
Margules equation.
The most satisfactory procedure for reduction of this set of data is to find
the value of Henry's constant by regression along with the Margules
parameters.
γ 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1
2
γ 2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2
2
H1 := 50
Guesses:
A21 := 0.2
A12 := 0.4
Mininize the sums of the squared errors by
setting sums of derivatives equal to zero.
Given
0=
i
0=
i
0=
i
2
d
H1
Pi x1 γ 1 x1 , x2 , A12 , A21
...
i
i
i
exp ( A12)
dA12
+ x2i γ 2 x1i , x2i , A12 , A21 Psat2
(
)
(
)
2
d
H1
Pi x1 γ 1 x1 , x2 , A12 , A21
...
i
i
i
exp ( A12)
dA21
+ x2i γ 2 x1i , x2i , A12 , A21 Psat2
(
)
(
)
2
d
H1
Pi x1 γ 1 x1 , x2 , A12 , A21
...
i
i
i
exp ( A12)
dH1
+ x2i γ 2 x1i , x2i , A12 , A21 Psat2
(
(
A12
A21 := Find ( A12 , A21 , H1)
H
1
)
)
A12 0.348
A21 = 0.178
H
1 51.337
542
Ans.
(d) γ1 ( x1 , x2) := exp x22 A12 + 2 ( A21 A12) x1
γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2
2
(
) exp( A12) + x2i γ2 ( x1i , x2i) Psat2
H1
Pcalc := x1 γ1 x1 , x2
i
i
y1calc :=
i
(
i
)
H1
x1 γ1 x1 , x2
i
i
i exp ( A 12)
i
Pcalc
i
0.2
0
PiPcalc
i
(y1iy1calci) 100
0.2
0.4
0.6
0
0.2
0.4
x1
0.6
0.8
i
Pressure residuals
y1 residuals
Fit GE/RT data to Margules eqn. by least squares:
i := 2 .. rows ( P)
Given
0=
i
y2 := 1 y1
y1 Pi
i
d
x1 ln
H1
dA12 i
x1i exp ( A )
12
y P
+ x ln 2i i
2i x2 Psat2
i
543
... A21 x1 ... x1 x2
i
i
i
+ A12 x2i
2
0=
i
0=
i
y1 Pi
i
d
x1 ln
H1
dA21 i
x1i exp ( A )
12
y P
+ x ln 2i i
2i x2 Psat2
i
y1 Pi
i
d
x1 ln
H1
dH1 i
x1
i exp ( A )
12
y2 Pi
+ x ln i
2i x2 Psat2
i
A12
A21 := Find ( A12 , A21 , H1)
H
1
... A21 x1 ... x1 x2
i
i
i
+ A12 x2i
... A21 x1 ... x1 x2
i
i
i
+ A12 x2i
A12 0.375
A21 = 0.148
H
1 53.078
γ1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1
2
γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2
H1
Pcalc := x1 γ1 x1 , x2
+ x2 γ2 x1 , x2 Psat2
i
i
i
i exp ( A 12)
i
i
i
2
(
y1calc :=
i
)
(
(
)
H1
x1 γ1 x1 , x2
i
i
i exp ( A 12)
Pcalc
i
544
)
Ans.
2
2
0
0.2
PiPcalc
i
(y1iy1calci) 100
0.4
0.6
0.8
0
0.2
0.4
x1
0.6
0.8
i
Pressure residuals
y1 residuals
14.5
Pressures in kPa
Data:
i := 1 .. 7
0.3193
0.4232
0.5119
0.6096
x1 :=
0.7135
0.7934
0.9102
1.000
21.63
24.01
25.92
27.96
P :=
30.12
31.75
34.15
36.09
x2 := 1 x1
0.5934
0.6815
0.7440
0.8050
y1 :=
0.8639
0.9048
0.9590
1.000
Psat1 := P8
(a) It follows immediately from Eq. (12.10a) that:
ln γ 2 = A21
Combining this with Eq. (12.10a) yields the required expression.
(b) Henry's constant will be found as part of the solution to Part (c).
545
(c) BARKER'S METHOD by non-linear least squares.
Margules equation.
The most satisfactory procedure for reduction of this set of data is to find
the value of Henry's constant by regression along with the Margules
parameters.
γ 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1
2
γ 2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2
2
H2 := 14
Guesses:
A21 := 0.148
A12 := 0.375
Mininize the sums of the squared errors by
setting sums of derivatives equal to zero.
Given 0 =
i
0=
(
)
(
)
d
dA21 Pi x1 γ 1 ( x1 , x2 , A12 , A21) Psat1 ...
i
0=
d
2
dA Pi x1i γ 1 x1i , x2i , A12 , A21 Psat1 ...
12
H2
+ x2i γ 2 x1i , x2i , A12 , A21 exp ( A21)
i
i
i
i
H2
+ x2 γ 2 x1 , x2 , A12 , A21
i
i
i
exp ( A21)
(
)
2
d
2
dH Pi x1i γ 1 x1i , x2i , A12 , A21 Psat1 ...
2
H2
+ x2 γ 2 x1 , x2 , A12 , A21
i
i
i
exp ( A21)
(
)
(
A12
A21 := Find ( A12 , A21 , H2)
H
2
)
A12 0.469
A21 = 0.279
H
2 14.87
(d) γ1 ( x1 , x2) := exp x22 A12 + 2 ( A21 A12) x1
γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2
2
546
Ans.
(
)
(
)
H2
Pcalc := x1 γ1 x1 , x2 Psat1 + x2 γ2 x1 , x2
i
i
i
i
i
i
i exp ( A 21)
(
)
x1 γ1 x1 , x2 Psat1
i
i
i
y1calc :=
i
Pcalc
i
The plot of residuals below shows that the procedure used (Barker's
method with regression for H2) is not in this case very satisfactory, no
doubt because the data do not extend close enough to x1 = 0.
1
0
PiPcalc
i
1
(y1iy1calci) 100
2
3
4
0.2
0.4
0.6
x1
0.8
i
Pressure residuals
y1 residuals
Fit GE/RT data to Margules eqn. by least squares:
i := 1 .. 7
y2 := 1 y1
Given
0=
i
y1 Pi
d
x1 ln i
...
dA12 i x1i Psat1
y2 Pi
i
+ x ln
2i
H2
x2
i exp ( A )
21
547
A21 x1 ... x1 x2
i
i
i
+ A12 x2i
2
0=
i
0=
i
y1 Pi
d
x1 ln i
...
dA21 i x1i Psat1
y2 Pi
i
+ x ln
2i
H2
x2i exp ( A )
21
y1 Pi
d
x1 ln i
...
dH2 i x1i Psat1
y2 Pi
i
+ x ln
2i
H2
x2
i exp ( A )
21
A12
A21 := Find ( A12 , A21 , H2)
H
2
A21 x1 ... x1 x2
i
i
i
+ A12 x2i
A21 x1 ... x1 x2
i
i
i
+ A12 x2i
A12 0.37
A21 = 0.204
H
2 15.065
γ1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1
2
γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2
2
(
)
(
)
H2
Pcalc := x1 γ1 x1 , x2 Psat1 + x2 γ2 x1 , x2
i
i
i
i
i
i
i exp ( A 21)
y1calc :=
i
(
)
x1 γ1 x1 , x2 Psat1
i
i
Pcalc
i
i
548
Ans.
2
2
0
PiPcalc
0.2
i
(y1iy1calci) 100
0.4
0.6
0.3
0.4
0.5
0.6
0.7
x1
0.8
0.9
1
i
Pressure residuals
y1 residuals
This result is considerably improved over that obtained with Barker's method.
14.6
Pressures in kPa
Data:
15.79
17.51
18.15
19.30
19.89
P :=
21.37
24.95
29.82
34.80
42.10
i := 2 .. rows ( P)
(a)
0.0
0.0932
0.1248
0.1757
0.2000
x1 :=
0.2626
0.3615
0.4750
0.5555
0.6718
x2 := 1 x1
0.0
0.1794
0.2383
0.3302
0.3691
y1 :=
0.4628
0.6184
0.7552
0.8378
0.9137
Psat2 := P1
It follows immediately from Eq. (12.10a) that:
ln γ 1 = A12
Combining this with Eq. (12.10a) yields the required expression
549
(b) Henry's constant will be found as part of the solution to Part (c)
(c) BARKER'S METHOD by non-linear least squares.
Margules equation.
The most satisfactory procedure for reduction of this set of data is to
find the value of Henry's constant by regression along with the
Margules parameters.
γ 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1
2
γ 2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2
2
H1 := 35
Guesses:
A21 := 1.27
A12 := 0.70
Mininize the sums of the squared errors by
setting sums of derivatives equal to zero.
Given
0=
i
0=
i
0=
i
2
d
H1
Pi x1 γ 1 x1 , x2 , A12 , A21
...
i
i
i
exp ( A12)
dA12
+ x2i γ 2 x1i , x2i , A12 , A21 Psat2
(
)
(
)
2
d
H1
Pi x1 γ 1 x1 , x2 , A12 , A21
...
i
i
exp ( A12)
dA21
i
+ x2i γ 2 x1i , x2i , A12 , A21 Psat2
(
)
(
)
2
d
H1
Pi x1 γ 1 x1 , x2 , A12 , A21
...
i
i
i
exp ( A12)
dH1
+ x2i γ 2 x1i , x2i , A12 , A21 Psat2
(
A12
A21 := Find ( A12 , A21 , H1)
H
1
(
)
)
A12 0.731
A21 = 1.187
H
1 32.065
550
Ans.
(d)
γ1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1
2
γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2
2
(
) exp( A12) + x2i γ2 ( x1i , x2i) Psat2
H1
Pcalc := x1 γ1 x1 , x2
i
i
y1calc :=
i
i
(
)
H1
x1 γ1 x1 , x2
i
i
i exp ( A 12)
i
Pcalc
i
0.5
0
PiPcalc
0.5
i
(y1iy1calci) 100
1
1.5
2
0
0.1
0.2
0.3
0.4
x1
0.5
0.6
0.7
i
Pressure residuals
y1 residuals
Fit GE/RT data to Margules eqn. by least squares:
i := 2 .. rows ( P)
Given
0=
i
y2 := 1 y1
y1 Pi
i
d
x1 ln
H1
dA12 i
x1
i exp ( A )
12
y2 Pi
+ x ln i
2i x2 Psat2
i
551
... A21 x1 ... x1 x2
i
i
i
+ A12 x2i
2
0=
i
0=
i
y1 Pi
i
d
x1 ln
H1
dA21 i
x1
i exp ( A )
12
y2 Pi
+ x ln i
2i x2 Psat2
i
y1 Pi
i
d
x1 ln
H1
dH1 i
x1i exp ( A )
12
y P
+ x ln 2i i
2i x2 Psat2
i
A12
A21 := Find ( A12 , A21 , H1)
H
1
... A21 x1 ... x1 x2
i
i
i
+ A12 x2i
... A21 x1 ... x1 x2
i
i
i
+ A12 x2i
A12 0.707
A21 = 1.192
H
1 33.356
γ1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1
2
γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2
2
(
)
(
)
H1
Pcalc := x1 γ1 x1 , x2
+ x2 γ2 x1 , x2 Psat2
i
i
i
i exp ( A 12)
i
i
i
(
) exp( A12)
x1 γ1 x1 , x2
y1calc :=
i
i
i
i
Pcalc
H1
i
552
Ans.
2
2
0
0.5
PiPcalc
1
i
(y1iy1calci) 100
1.5
2
2.5
0
0.1
0.2
0.3
0.4
x1
0.5
0.6
0.7
i
Pressure residuals
y1 residuals
14.7
Pressures in kPa
Data:
0.1757
0.2000
0.2626
0.3615
0.4750
x1 :=
0.5555
0.6718
0.8780
0.9398
1.0000
i := 1 .. 9
(a)
19.30
19.89
21.37
24.95
29.82
P :=
34.80
42.10
60.38
65.39
69.36
x2 := 1 x1
0.3302
0.3691
0.4628
0.6184
0.7552
y1 :=
0.8378
0.9137
0.9860
0.9945
1.0000
Psat1 := P10
It follows immediately from Eq. (12.10a) that:
ln γ 2 = A21
Combining this with Eq. (12.10a) yields the required expression.
(b)
Henry's constant will be found as part of the solution to Part (c).
553
(c)
BARKER'S METHOD by non-linear least squares.
Margules equation.
The most satisfactory procedure for reduction of this set of data is to
find the value of Henry's constant by regression along with the
Margules parameters.
γ 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1
2
γ 2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2
2
H2 := 4
Guesses:
A21 := 1.37
A12 := 0.68
Mininize the sums of the squared errors by
setting sums of derivatives equal to zero.
Given
0=
d
dA12 Pi x1 γ 1 ( x1 , x2 , A12 , A21) Psat1 ...
i
0=
i
0=
i
i
i
H2
+ x2 γ 2 x1 , x2 , A12 , A21
i
i
i
exp ( A21)
(
)
2
d
2
dA Pi x1i γ 1 x1i , x2i , A12 , A21 Psat1 ...
21
H2
+ x2i γ 2 x1i , x2i , A12 , A21 exp ( A21)
(
)
(
)
d
dH2 Pi x1 γ 1 ( x1 , x2 , A12 , A21) Psat1 ...
i
i
i
i
H2
+ x2 γ 2 x1 , x2 , A12 , A21
i
i
i
exp ( A21)
(
A12
A21 := Find ( A12 , A21 , H2)
H
2
)
A12 0.679
A21 = 1.367
H
2 3.969
554
2
Ans.
(d) γ1 ( x1 , x2) := exp x22 A12 + 2 ( A21 A12) x1
γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2
2
(
)
(
)
H2
Pcalc := x1 γ1 x1 , x2 Psat1 + x2 γ2 x1 , x2
i
i
i
i
i
i
i exp ( A 21)
y1calc :=
(
)
x1 γ1 x1 , x2 Psat1
i
i
i
Pcalc
i
i
1
PiPcalc
0
i
(y1iy1calci) 100
1
2
0
0.2
0.4
0.6
x1
0.8
i
Pressure residuals
y1 residuals
Fit GE/RT data to Margules eqn. by least squares:
i := 1 .. 9
Given
0=
y2 := 1 y1
i
y1 Pi
d
x1 ln i
...
dA12 i x1i Psat1
y2 Pi
i
+ x ln
2i
H2
x2
i exp ( A )
21
555
A21 x1 ... x1 x2
i
i
i
+ A12 x2i
2
0=
i
0=
y1 Pi
d
x1 ln i
...
dA21 i x1i Psat1
y2 Pi
i
+ x ln
2i
H2
x2
i exp ( A )
21
A21 x1 ... x1 x2
i
i
i
+ A12 x2i
2
y1 Pi
d
x1 ln i
...
dH2 i x1i Psat1
y2 Pi
i
+ x ln
2i
H2
x2i exp ( A )
21
2
i
A12
A21 := Find ( A12 , A21 , H2)
H
2
A21 x1 ... x1 x2
i
i
i
+ A12 x2i
A12 0.845
A21 = 1.229
H
2 4.703
γ1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1
2
γ2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2
2
(
)
(
)
H2
Pcalc := x1 γ1 x1 , x2 Psat1 + x2 γ2 x1 , x2
i
i
i
i
i
i
i exp ( A 21)
y1calc :=
i
(
)
x1 γ1 x1 , x2 Psat1
i
i
Pcalc
i
i
556
Ans.
1
PiPcalc
0
i
(y1iy1calci) 100
1
2
0
0.2
0.4
0.6
x1
0.8
i
Pressure residuals
y1 residuals
14.8 (a)
Data from Table 12.1
15.51
18.61
21.63
24.01
P := 25.92 kPa x1 :=
27.96
30.12
31.75
34.15
n := rows ( P)
Psat1 := 36.09kPa
0.0895
0.1981
0.3193
0.4232
0.5119 y1 :=
0.6096
0.7135
0.7934
0.9102
n=9
0.2716
0.4565
0.5934
0.6815
0.7440 γ 1 :=
0.8050
0.8639
0.9048
0.9590
i := 1 .. n
Psat2 := 12.30kPa
1.304
1.188
1.114
1.071
1.044 γ 2 :=
1.023
1.010
1.003
0.997
x2 := 1 x1
i
i
γ 1 :=
i
y1 Pi
i
x1 Psat1
i
γ 2 :=
i
y2 Pi
i
x2 Psat2
i
557
y2 := 1 y1
T := ( 50 + 273.15)K
Data reduction with the Margules equation and Eq. (10.5):
1.009
1.026
1.050
1.078
1.105
1.135
1.163
1.189
1.268
i
i
()
()
i := 1 .. n
GERTi := x1 ln γ 1 + x2 ln γ 2
i
i
i
i
Guess:
A12 := 0.1
f ( A12 , A21) :=
A21 := 0.3
n
i=1
(
)
2
GERTi A21 x1i + A12 x2i x1i x2i
A12
:= Minimize ( f , A12 , A21)
A21
A12 = 0.374
A21 = 0.197
(
)
n
RMS Error: RMS :=
i=1
3
RMS = 1.033 × 10
Ans.
2
GERTi A21 x1i + A12 x2i x1i x2i
n
x1 := 0 , 0.01 .. 1
0.1
GERT i
A21 x1+A12 ( 1x1) x1 ( 1x1)
0.05
0
0
0.2
0.4
0.6
0.8
x1 , x1
i
Data reduction with the Margules equation and Eq. (14.1):
3
cm
B11 := 1840
mol
3
cm
B22 := 1800
mol
3
cm
B12 := 1150
mol
δ 12 := 2 B12 B11 B22
()
B11 ( Pi Psat1) + Pi y2 2 δ 12
i
Φ 1 := exp
i
R T
558
γ 1 :=
i
y1 Φ 1 Pi
i
i
x1 Psat1
i
()
B22 ( Pi Psat2) + Pi y1 2 δ 12
i
Φ 2 := exp
i
R T
i := 1 .. n
()
i
i
i
x2 Psat2
i
()
GERTi := x1 ln γ 1 + x2 ln γ 2
i
i
i
i
Guess:
f ( A12 , A21) :=
A12 := 0.1
A21 := 0.3
n
i=1
(
)
2
GERTi A21 x1i + A12 x2i x1i x2i
A12
:= Minimize ( f , A12 , A21)
A21
RMS :=
A12 = 0.379
(
4
Ans.
)
n
i=1
RMS = 9.187 × 10
A21 = 0.216
2
GERTi A21 x1i + A12 x2i x1i x2i
n
RMS Error:
γ 2 :=
y2 Φ 2 Pi
x1 := 0 , 0.01 .. 1
0.1
GERT i
A21 x1+A12 ( 1x1) x1 ( 1x1)
0.05
0
0
0.5
1
x1 , x1
i
The RMS error with Eqn. (14.1) is about 11% lower than the RMS error
with Eqn. (10.5).
Note: The following problem was solved with the temperature (T) set at
the normal boiling point. To solve for another temperature, simply change
T to the approriate value.
559
14.9
(a) Acetylene:
Tc := 308.3K
T := Tn
Tr :=
T
Tc
Pc := 61.39bar
Tn := 189.4K
Tr = 0.614
For Redlich/Kwong EOS:
σ := 1
ε := 0
:= 0.08664
Ψ := 0.42748
1
α ( Tr) := Tr
q ( Tr) :=
2
Ψ α ( Tr )
Tr
α ( Tr ) R Tc
β ( Tr , Pr) :=
Eq. (3.54)
Define Z for the vapor (Zv)
Given
a ( Tr) := Ψ
Table 3.1
Table 3.1
Guess:
2
2
Eq. (3.45)
Pc
Pr
Eq. (3.53)
Tr
zv := 0.9
Eq. (3.52)
zv = 1 + β ( Tr , Pr) q ( Tr) β ( Tr , Pr)
zv β ( Tr , Pr)
( zv + ε β ( Tr , Pr) ) ( zv + σ β ( Tr , Pr) )
Zv ( Tr , Pr) := Find ( zv)
Define Z for the liquid (Zl)
Guess:
Guess:
zl := 0.01
Given Eq. (3.56)
(
)(
)
1 + β ( Tr , Pr) zl
zl
zl = β ( Tr , Pr) + zl + ε β ( Tr , Pr) zl + σ β ( Tr , Pr)
q ( T r ) β ( T r , Pr )
To find liquid root, restrict search for zl to values less than 0.2, zl < 0.2
Zl ( Tr , Pr) := Find ( zl)
Define I for liquid (Il) and vapor (Iv)
Il ( Tr , Pr) :=
Iv ( Tr , Pr) :=
1
σε
1
σε
Zl ( Tr , Pr) + σ β ( Tr , Pr)
ln
Zl ( Tr , Pr) + ε β ( Tr , Pr)
Zv ( Tr , Pr) + σ β ( Tr , Pr)
ln
Zv ( Tr , Pr) + ε β ( Tr , Pr)
560
Eq. (6.65b)
(
)
lnφl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln Zl ( Tr , Pr) β ( Tr , Pr) q ( Tr) Il ( Tr , Pr)
Eq. (11.37)
(
)
lnφv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln Zv ( Tr , Pr) β ( Tr , Pr) q ( Tr) Iv ( Tr , Pr)
Guess Psat: Psatr :=
Given
1bar
Pc
lnφl ( Tr , Psatr) = lnφv ( Tr , Psatr)
Psatr = 0.026
Zl ( Tr , Psatr) = 4.742 × 10
Psat := Psatr Pc
Psat = 1.6 bar
3
Psatr := Find ( Psatr)
Zv ( Tr , Psatr) = 0.965
Ans.
The following table lists answers for all parts. Literature values are interpolated
from tables in Perry's Chemical Engineers' Handbook, 6th ed. The last column
shows the percent difference between calculated and literature values at 0.85Tc.
These range from 0.1 to 27%. For the normal boiling point (Tn), Psat should be
1.013 bar. Tabulated results for Psat do not agree well with this value.
Differences range from 3 to > 100%.
Tn (K) Psat (bar) 0.85 Tc (K) Psat (bar) Psat (bar) % Difference
@ Tn
@ 0.85 Tc Lit. Values
189.4
1.60
262.1
20.27
19.78
2.5%
Acetylene
87.3
0.68
128.3
20.23
18.70
8.2%
Argon
353.2
1.60
477.9
16.028
15.52
3.2%
Benzene
272.7
1.52
361.3
14.35
12.07
18.9%
n-Butane
0.92
113.0
15.2
12.91
17.7%
Carbon Monoxide 81.7
447.3
2.44
525.0
6.633
5.21
27.3%
n-Decane
169.4
1.03
240.0
17.71
17.69
0.1%
Ethylene
371.6
2.06
459.2
7.691
7.59
1.3%
n-Heptane
111.4
0.71
162.0
19.39
17.33
11.9%
Methane
77.3
0.86
107.3
14.67
12.57
16.7%
Nitrogen
14.10 (a) Acetylene: ω := 0.187
T := Tn
Tc := 308.3K
Pc := 61.39bar
Note: For solution at 0.85T c, set T := 0.85Tc.
ε := 0
Tr :=
T
Tc
Tr = 0.614
For SRK EOS:
σ := 1
Tn := 189.4K
:= 0.08664
561
Ψ := 0.42748
Table 3.1
1
2
2
α ( Tr , ω ) := 1 + ( 0.480 + 1.574ω 0.176ω ) 1 Tr
(
)
(
)
2
Ψ α Tr , ω
Define Z for the vapor (Zv)
Given
Eq. (3.45)
Eq. (3.54)
Tr
Table 3.1
2
α Tr , ω R Tc
a ( Tr) := Ψ
Pc
q ( Tr) :=
2
β ( Tr , Pr) :=
Guess:
Pr
Eq. (3.53)
Tr
zv := 0.9
Eq. (3.52)
zv = 1 + β ( Tr , Pr) q ( Tr) β ( Tr , Pr)
zv β ( Tr , Pr)
( zv + ε β ( Tr , Pr) ) ( zv + σ β ( Tr , Pr) )
Zv ( Tr , Pr) := Find ( zv)
Define Z for the liquid (Zl)
Given
Guess:
zl := 0.01
Eq. (3.56)
(
)(
)
1 + β ( Tr , Pr) zl
zl = β ( Tr , Pr) + zl + ε β ( Tr , Pr) zl + σ β ( Tr , Pr)
q ( T r ) β ( T r , Pr )
To find liquid root, restrict search for zl to values less than 0.2, zl < 0.2
Zl ( Tr , Pr) := Find ( zl)
Define I for liquid (Il) and vapor (Iv)
Il ( Tr , Pr) :=
Iv ( Tr , Pr) :=
1
σε
1
σε
Zl ( Tr , Pr) + σ β ( Tr , Pr)
ln
Zl ( Tr , Pr) + ε β ( Tr , Pr)
Zv ( Tr , Pr) + σ β ( Tr , Pr)
ln
Zv ( Tr , Pr) + ε β ( Tr , Pr)
562
Eq. (6.65b)
(
)
lnφl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln Zl ( Tr , Pr) β ( Tr , Pr) q ( Tr) Il ( Tr , Pr)
Eq. (11.37)
(
)
lnφv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln Zv ( Tr , Pr) β ( Tr , Pr) q ( Tr) Iv ( Tr , Pr)
Guess Psat: Psatr :=
Given
2bar
Pc
lnφl ( Tr , Psatr) = lnφv ( Tr , Psatr)
Psatr = 0.017
Psat := Psatr Pc
Zl ( Tr , Psatr) = 3.108 × 10
3
Psatr := Find ( Psatr)
Zv ( Tr , Psatr) = 0.975
Psat = 1.073 bar Ans.
The following table lists answers for all parts. Literature values are interpolated
from tables in Perry's Chemical Engineers' Handbook, 6th ed. The last column
shows the percent difference between calculated and literature values at 0.85Tc.
These range from less than 0.1 to 2.5%. For the normal boiling point (Tn), Psat
should be 1.013 bar. Tabulated results for Psat agree well with this value.
Differences range from near 0 to 6%.
Acetylene
Argon
Benzene
n-Butane
Carbon Monoxide
n-Decane
Ethylene
n-Heptane
Methane
Nitrogen
14.10
Tn (K) Psat (bar) 0.85 Tc (K) Psat (bar) Psat (bar) % Difference
@ Tn
@ 0.85 Tc Lit. Values
189.4
1.073
262.1
20.016
19.78
1.2%
87.3
0.976
128.3
18.79
18.70
0.5%
353.2
1.007
477.9
15.658
15.52
0.9%
272.7
1.008
361.3
12.239
12.07
1.4%
81.7
1.019
113.0
12.871
12.91
-0.3%
447.3
1.014
525.0
5.324
5.21
2.1%
169.4
1.004
240.0
17.918
17.69
1.3%
371.6
1.011
459.2
7.779
7.59
2.5%
111.4
0.959
162.0
17.46
17.33
0.8%
77.3
0.992
107.3
12.617
12.57
0.3%
(b) Acetylene: ω := 0.187
T := Tn
Tc := 308.3K
Pc := 61.39bar
Note: For solution at 0.85T c, set T := 0.85Tc.
For PR EOS:
σ := 1 +
2 ε := 1
Tn := 189.4K
Tr :=
T
Tc
Tr = 0.614
2 := 0.07779
563
Ψ := 0.45724
Table 3.1
2
1
2
2 Table 3.1
α ( Tr , ω ) := 1 + ( 0.37464 + 1.54226ω 0.26992ω ) 1 Tr
a ( Tr) := Ψ
q ( Tr) :=
(
)
2
Eq. (3.45)
Pc
(
Ψ α Tr , ω
)
Tr
Define Z for the vapor (Zv)
Given
2
α Tr , ω R Tc
Eq. (3.54)
β ( Tr , Pr) :=
Guess:
Pr
Eq. (3.53)
Tr
zv := 0.9
Eq. (3.52)
zv = 1 + β ( Tr , Pr) q ( Tr) β ( Tr , Pr)
zv β ( Tr , Pr)
( zv + ε β ( Tr , Pr) ) ( zv + σ β ( Tr , Pr) )
Zv ( Tr , Pr) := Find ( zv)
Define Z for the liquid (Zl)
Guess:
zl := 0.01
Given Eq. (3.56)
(
)(
)
1 + β ( Tr , Pr) zl
zl = β ( Tr , Pr) + zl + ε β ( Tr , Pr) zl + σ β ( Tr , Pr)
q ( T r ) β ( T r , Pr )
To find liquid root, restrict search for zl to values less than 0.2,zl < 0.2
Zl ( Tr , Pr) := Find ( zl)
Define I for liquid (Il) and vapor (Iv)
Il ( Tr , Pr) :=
Iv ( Tr , Pr) :=
1
σε
1
σε
Zl ( Tr , Pr) + σ β ( Tr , Pr)
ln
Zl ( Tr , Pr) + ε β ( Tr , Pr)
Zv ( Tr , Pr) + σ β ( Tr , Pr)
ln
Zv ( Tr , Pr) + ε β ( Tr , Pr)
564
Eq. (6.65b)
(
)
lnφl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln Zl ( Tr , Pr) β ( Tr , Pr) q ( Tr) Il ( Tr , Pr)
Eq. (11.37)
(
)
lnφv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln Zv ( Tr , Pr) β ( Tr , Pr) q ( Tr) Iv ( Tr , Pr)
2bar
Pc
Guess Psat:
Psatr :=
Given
lnφl ( Tr , Psatr) = lnφv ( Tr , Psatr)
Psatr = 0.018
Zl ( Tr , Psatr) = 2.795 × 10
Psat := Psatr Pc
Psat = 1.09 bar
Psatr := Find ( Psatr)
Zv ( Tr , Psatr) = 0.974
3
Ans.
The following table lists answers for all parts. Literature values are interpolated
from tables in Perry's Chemical Engineers' Handbook, 6th ed. The last column
shows the percent difference between calculated and literature values at 0.85Tc.
These range from less than 0.1 to 1.2%. For the normal boiling point (Tn), Psat
should be 1.013 bar. Tabulated results for Psat agree well with this value.
Differences range from near 0 to 7.6%.
Acetylene
Argon
Benzene
n-Butane
Carbon Monoxide
n-Decane
Ethylene
n-Heptane
Methane
Nitrogen
14.12
(a)
van der Waals Eqn.
σ := 0
q ( Tr) :=
Given
Tn (K) Psat (bar) 0.85 Tc (K) Psat (bar) Psat (bar) % Difference
@ Tn
@ 0.85 Tc Lit. Values
189.4
1.090
262.1
19.768
19.78
-0.1%
87.3
1.015
128.3
18.676
18.70
-0.1%
353.2
1.019
477.9
15.457
15.52
-0.4%
272.7
1.016
361.3
12.084
12.07
0.1%
81.7
1.041
113.0
12.764
12.91
-1.2%
447.3
1.016
525.0
5.259
5.21
0.9%
169.4
1.028
240.0
17.744
17.69
0.3%
371.6
1.012
459.2
7.671
7.59
1.1%
111.4
0.994
162.0
17.342
17.33
0.1%
77.3
1.016
107.3
12.517
12.57
-0.4%
ε := 0
Ψ α ( Tr)
Tr
:=
Tr := 0.7
1
8
Ψ :=
β ( Tr , Pr) :=
Pr
Tr
zv = 1 + β ( Tr , Pr) q ( Tr) β ( Tr , Pr)
α ( Tr) := 1
27
64
zv := 0.9 (guess)
zv β ( Tr , Pr)
( zv)
565
2
Eq. (3.52)
Zv ( Tr , Pr) := Find ( zv)
zl := .01
(guess)
2 1 + β ( Tr , Pr) zl
zl = β ( Tr , Pr) + ( zl)
Given
q ( Tr) β ( Tr , Pr)
Eq. (3.56)
zl < 0.2
Zl ( Tr , Pr) := Find ( zl)
Iv ( Tr , Pr) :=
β ( Tr , Pr)
Zv ( Tr , Pr)
Il ( Tr , Pr) :=
β ( Tr , Pr)
Zl ( Tr , Pr)
Case II, pg. 218.
By Eq. (11.39):
lnφv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln ( Zv ( Tr , Pr) β ( Tr , Pr) ) q ( Tr) Iv ( Tr , Pr)
lnφl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln ( Zl ( Tr , Pr) β ( Tr , Pr) ) q ( Tr) Il ( Tr , Pr)
Psatr := .1
Given
lnφl ( Tr , Psatr) lnφv ( Tr , Psatr) = 0
Psatr := Find ( Psatr)
Zv ( Tr , Psatr) = 0.839
Zl ( Tr , Psatr) = 0.05
lnφl ( Tr , Psatr) = 0.148
lnφv ( Tr , Psatr) = 0.148 β ( Tr , Psatr) = 0.036
ω := 1 log ( Psatr)
ω = 0.302
(b)
Psatr = 0.2
Ans.
Redlich/Kwong Eqn.Tr := 0.7
σ := 1
ε := 0
:= 0.08664
Ψ := 0.42748
.5
α ( Tr) := Tr
q ( Tr) :=
Ψ α ( Tr)
β ( Tr , Pr) :=
Tr
Pr
Tr
Given zv = 1 + β ( Tr , Pr) q ( Tr) β ( Tr , Pr)
Zv ( Tr , Pr) := Find ( zv)
Guess:
zl := .01
566
Guess:
zv := 0.9
zv β ( Tr , Pr)
zv ( zv + β ( Tr , Pr) )
Eq. (3.52)
Given
zl = β ( Tr , Pr) + zl ( zl + β ( Tr , Pr) )
zl < 0.2
1 + β ( Tr , Pr) zl
q ( Tr) β ( Tr , Pr)
Eq. (3.55)
Zl ( Tr , Pr) := Find ( zl)
Zv ( Tr , Pr) + β ( Tr , Pr) Il ( Tr , Pr) := ln Zl ( Tr , Pr) + β ( Tr , Pr)
Zv ( Tr , Pr)
Zl ( Tr , Pr)
Iv ( Tr , Pr) := ln
By Eq. (11.39):
lnφv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln ( Zv ( Tr , Pr) β ( Tr , Pr) ) q ( Tr) Iv ( Tr , Pr)
lnφl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln ( Zl ( Tr , Pr) β ( Tr , Pr) ) q ( Tr) Il ( Tr , Pr)
Psatr := .1
Given
lnφl ( Tr , Psatr) = lnφv ( Tr , Psatr)
Zv ( Tr , Psatr) = 0.913
Psatr := Find ( Psatr)
Zl ( Tr , Psatr) = 0.015
lnφv ( Tr , Psatr) = 0.083 lnφl ( Tr , Psatr) = 0.083
ω := 1 log ( Psatr)
14.15 (a) x1α := 0.1
Guess:
ω = 0.058
x2α := 1 x1α
A12 := 2
Psatr = 0.087
β ( Tr , Psatr) = 0.011
Ans.
x1β := 0.9
x2β := 1 x1β
A21 := 2
γ1α ( A21 , A12) := exp x2α A12 + 2 ( A21 A12) x1α
2
γ1β( A21 , A12) := exp x2β A12 + 2 ( A21 A12) x1β
2
γ2α ( A21 , A12) := exp x1α A21 + 2 ( A12 A21) x2α
2
γ2β( A21 , A12) := exp x1β A21 + 2 ( A12 A21) x2β
2
567
Given
x1α γ1α ( A21 , A12) = x1β γ1β( A21 , A12)
x2α γ2α ( A21 , A12) = x2β γ2β( A21 , A12)
A12
:= Find ( A12 , A21)
A21
(b) x1α := 0.2
Guess:
A21 = 2.747
A12 = 2.747
x2α := 1 x1α
x1β := 0.9
A12 := 2
Ans.
x2β := 1 x1β
A21 := 2
γ1α ( A21 , A12) := exp x2α A12 + 2 ( A21 A12) x1α
2
γ1β( A21 , A12) := exp x2β A12 + 2 ( A21 A12) x1β
2
γ2α ( A21 , A12) := exp x1α A21 + 2 ( A12 A21) x2α
2
γ2β( A21 , A12) := exp x1β A21 + 2 ( A12 A21) x2β
2
Given
x1α γ1α ( A21 , A12) = x1β γ1β( A21 , A12)
x2α γ2α ( A21 , A12) = x2β γ2β( A21 , A12)
A12
:= Find ( A12 , A21)
A21
(c) x1α := 0.1
Guess:
A12 = 2.148
A21 = 2.781
x2α := 1 x1α
x1β := 0.8
A12 := 2
A21 := 2
x2β := 1 x1β
γ1α ( A21 , A12) := exp x2α A12 + 2 ( A21 A12) x1α
2
γ1β( A21 , A12) := exp x2β A12 + 2 ( A21 A12) x1β
2
γ2α ( A21 , A12) := exp x1α A21 + 2 ( A12 A21) x2α
2
γ2β( A21 , A12) := exp x1β A21 + 2 ( A12 A21) x2β
2
568
Ans.
Given
x1α γ1α ( A21 , A12) = x1β γ1β( A21 , A12)
x2α γ2α ( A21 , A12) = x2β γ2β( A21 , A12)
A12
:= Find ( A12 , A21)
A21
14.16 (a) x1α := 0.1
Guess:
Given
A12 = 2.781
A21 = 2.148
x2α := 1 x1α
x1β := 0.9
a12 := 2
Ans.
x2β := 1 x1β
a21 := 2
2
2
a12 x1α
a12 x1β
exp a12 1 +
x1α = exp a12 1 +
x1β
a21 x2α
a21 x2β
2
2
a21 x2α
a21 x2β
exp a21 1 +
x2α = exp a21 1 +
x2β
a12 x1α
a12 x1β
a12
:= Find ( a12 , a21)
a21
(b) x1α := 0.2
Guess:
Given
a12 = 2.747
x2α := 1 x1α
a12 := 2
a21 = 2.747
x1β := 0.9
Ans.
x2β := 1 x1β
a21 := 2
2
2
a12 x1α
a12 x1β
exp a12 1 +
x1α = exp a12 1 +
x1β
a21 x2α
a21 x2β
2
2
a21 x2α
a21 x2β
exp a21 1 +
x2α = exp a21 1 +
x2β
a12 x1α
a12 x1β
a12
:= Find ( a12 , a21)
a21
a12 = 2.199
569
a21 = 2.81
Ans.
x1α := 0.1
x2α := 1 x1α
x1β := 0.8
Guess:
a12 := 2
a21 := 2
(c)
Given
x2β := 1 x1β
2
2
a12 x1α
a12 x1β
exp a12 1 +
x1α = exp a12 1 +
x1β
a21 x2α
a21 x2β
2
2
a21 x2α
a21 x2β
exp a21 1 +
x2α = exp a21 1 +
x2β
a12 x1α
a12 x1β
a12
:= Find ( a12 , a21)
a21
14.18
(a) a := 975
a12 = 2.81
b := 18.4
T := 250 .. 450
A ( T) :=
a21 = 2.199
Ans.
c := 3
a
+ b c ln ( T)
T
2.1
A ( T)
2
1.9
250
300
350
400
450
T
Parameter A = 2 at two temperatures. The lower one is an UCST,
because A decreases to 2 as T increases. The higher one is a LCST,
because A decreases to 2 as T decreases.
Guess:
Given
x0
x := 0.25
1 x
x
A ( T) ( 1 2 x) = ln
x 0.5
x1 ( T) := Find ( x)
570
Eq. (E), Ex. 14.5
x2 ( T) := 1 x1 ( T)
UCST := 300 (guess)
Given
A ( UCST) = 2
UCST := Find ( UCST)
UCST = 272.93
LCST := Find ( LCST)
LCST = 391.21
LCST := 400 (guess)
Given
A ( LCST) = 2
Plot phase diagram as a function of T
T1 := 225 , 225.1 .. UCST
T2 := LCST .. 450
500
T1
400
T1
T2
T2
300
200
0.2
0.3
0.4
0.5
0.6
x1 ( T1 ) , x2 ( T1 ) , x1 ( T2 ) , x2 ( T2 )
(b) a := 540
b := 17.1
T := 250 .. 450
A ( T) :=
c := 3
a
+ b c ln ( T)
T
2.5
A ( T)
2
1.5
250
300
350
400
450
T
Parameter A = 2 at a single temperature. It is
a LCST, because A decreases to 2 as T decreases.
571
0.7
0.8
Guess:
x := 0.25
Given
A ( T) ( 1 2 x) = ln
1 x
x
x0
x 0.5
Eq. (E), Ex. 14.5
x1 ( T) := Find ( x)
LCST := 350 (guess)
LCST := Find ( LCST)
A ( LCST) = 2
Given
Plot phase diagram as a function of T
LCST = 346
T := LCST .. 450
450
400
T
T
350
300
0.1
0.2
0.3
0.4
0.5
0.6
x1 ( T ) , 1x1 ( T)
a := 1500
b := 19.9
T := 250 .. 450
(c)
A ( T) :=
c := 3
a
+ b c ln ( T)
T
3
2.5
A ( T)
2
1.5
250
300
350
400
450
T
Parameter A = 2 at a single temperature. It is
an UCST, because A decreases to 2 as T increases.
572
0.7
0.8
Guess:
x := 0.25
Given
A ( T) ( 1 2 x) = ln
1 x
x
x0
x 0.5
Eq. (E), Ex. 14.5
x1 ( T) := Find ( x)
UCST := 350 (guess)
A ( UCST) = 2
Given
UCST := Find ( UCST)
Plot phase diagram as a function of T
UCST = 339.66
T := UCST .. 250
350
T
300
T
250
0
0.2
0.4
0.6
0.8
x1 ( T ) , 1x1 ( T)
x1α := 0.5
14.20 Guess:
x1β := 0.5
Write Eq. (14.74) for species 1:
Given
2
2
x1α exp 0.4 ( 1 x1α) = x1β exp 0.8 ( 1 x1β)
x1α
1 x1α
+
x1β
1 x1β
=1
x1α
:= Find ( x1α , x1β)
x1β
(Material balance)
x1α = 0.371
573
x1β = 0.291
Ans.
14.22 Temperatures in kelvins; pressures in kPa.
P1sat ( T) := exp 19.1478
P2sat ( T) := exp 14.6511
5363.7
T
water
2048.97
T
SF6
P := 1600
Find 3-phase equilibrium temperature and vapor-phase composition (pp.
594-5 of text):
Guess:
Given
T := 300
P = P1sat ( T) + P2sat ( T)
Tstar := Find ( T)
Tstar = 281.68
P1sat ( Tstar)
6
y1star 10 = 695
P
Find saturation temperatures of pure species 2:
y1star :=
Guess:
Given
T := 300
P2sat ( T) = P
T2 := Find ( T)
T2 = 281.71
P2sat ( T)
P
P1sat ( T)
TI := Tstar , Tstar + 0.01 .. Tstar + 6
y1I ( T) :=
P
Because of the very large difference in scales appropriate to regions I and
II [Fig. 14.21(a)], the txy diagram is presented on the following page in two
parts, showing regions I and II separately.
TII := Tstar , Tstar + 0.0001 .. T2
y1II ( T) := 1
281.7
TII
Tstar
281.69
281.68
0
100
200
300
400
6
500
6
y1II ( TII) 10 , y1II ( TII) 10
574
600
700
288
286
TI
Tstar
284
282
280
650
700
750
800
850
900
6
950
1000
1050
6
y1I ( TI) 10 , y1I ( TI) 10
14.24 Temperatures in deg. C; pressures in kPa
P1sat ( T) := exp 13.9320
3056.96
T + 217.625
3885.70
P2sat ( T) := exp 16.3872
T + 230.170
Toluene
P := 101.33
Water
Find the three-phase equilibrium T and y:
T := 25
Guess:
P = P1sat ( T) + P2sat ( T)
Given
y1star :=
P1sat ( Tstar)
P
Tstar := Find ( T)
y1star = 0.444
For z1 < y1*, first liquid is pure species 2.
y1 := 0.2
Given
Guess:
y1 = 1
Tdew := Tstar
P2sat ( Tdew)
P
Tdew := Find ( Tdew)
Tdew = 93.855
For z1 > y1*, first liquid is pure species 1.
575
Ans.
Tstar = 84.3
y1 := 0.7
Guess:
y1 =
Given
Tdew := Tstar
P1sat ( Tdew)
P
Tdew := Find ( Tdew)
Tdew = 98.494
Ans.
In both cases the bubblepoint temperature is T*, and the mole fraction of
the last vapor is y1*.
14.25 Temperatures in deg. C; pressures in kPa.
P1sat ( T) := exp 13.8622
P2sat ( T) := exp 16.3872
2910.26
T + 216.432
n-heptane
3885.70
T + 230.170
water
P := 101.33
Find the three-phase equilibrium T and y:
Guess:
T := 50
Given
P = P1sat ( T) + P2sat ( T)
y1star :=
P1sat ( Tstar)
P
Tstar := Find ( T)
Tstar = 79.15
y1star = 0.548
Since 0.35<y1*, first liquid is pure species 2.
y1 ( T) := 1
P2sat ( T)
P
Find temperature of initial condensation at y1=0.35:
y10 := 0.35
Given
Guess:
y1 ( Tdew) = y10
Tdew := Tstar
Tdew := Find ( Tdew)
Tdew = 88.34
Define the path of vapor mole fraction above and below the dew point.
y1path ( T) := if ( T > Tdew , y10 , y1 ( T) )
T := 100 , 99.9 .. Tstar
Path of mole fraction heptane in residual vapor as temperature is
decreased. No vapor exists below Tstar.
576
100
95
Tdew
90
T
85
Tstar
80
75
0.3
0.35
0.4
0.45
0.5
0.55
y1path ( T )
P1sat := 75
14.26 Pressures in kPa.
P2sat := 110
A := 2.25
(
2
γ1 ( x1) := exp A ( 1 x1)
γ2 ( x1) := exp A x1
)
2
Find the solubility limits:
Guess:
x1α := 0.1
Given
A ( 1 2 x1α) = ln
1 x1α
x1α
x1α = 0.224
x1β := 1 x1α
x1α := Find ( x1α)
x1β = 0.776
Find the conditions for VLLE:
Guess:
Given
Pstar := P1sat
y1star := 0.5
Pstar = x1β γ1 ( x1β) P1sat + ( 1 x1α) γ2 ( x1α) P2sat
y1star Pstar = x1α γ1 ( x1α) P1sat
Pstar
:= Find ( Pstar , y1star)
y1star
Pstar = 160.699
Calculate VLE in two-phase region.
Modified Raoult's law; vapor an ideal gas.
Guess:
x1 := 0.1
P := 50
577
y1star = 0.405
P = x1 γ1 ( x1) P1sat + ( 1 x1) γ2 ( x1) P2sat
Given
P ( x1) := Find ( P)
y1 ( x1) :=
x1 γ1 ( x1) P1sat
P ( x1)
Plot the phase diagram.
Define liquid equilibrium line:
PL ( x1) := if ( P ( x1) < Pstar , P ( x1) , Pstar)
Define vapor equilibrium line:
PV ( x1) := if ( P ( x1) < Pstar , P ( x1) , Pstar)
Define pressures for liquid phases above Pstar:
Pliq := Pstar .. Pstar + 10
x1 := 0 , 0.01 .. 1
200
175
Pstar
PL ( x1) 150
PV ( x1)
125
Pliq
Pliq
100
75
50
0
0.2
0.4
0.6
x1 , y1 ( x1) , x1α , x1β
x1 := 0 , 0.05 .. 0.2
x1 =
PL ( x1) =
y1 ( x1) =
0
110
0
0.05
133.66
0.214
0.1
147.658
0.314
0.15
155.523
0.368
0.2
159.598
0.397
578
0.8
1
x1 := 1 , 0.95 .. 0.8
x1 =
PL ( x1) =
y1 ( x1) =
1
75
1
0.95
113.556
0.631
0.9
137.096
0.504
0.85
150.907
0.444
0.8
158.506
0.414
x1α = 0.224
x1β = 0.776
y1star = 0.405
14.27 Temperatures in deg. C; pressures in kPa.
Water:
P1sat ( T) := exp 16.3872
3885.70
T + 230.170
n-Pentane:
P2sat ( T) := exp 13.7667
2451.88
T + 232.014
n-Heptane:
P3sat ( T) := exp 13.8622
2910.26
T + 216.432
P := 101.33
z1 := 0.45
(a)
z2 := 0.30
z3 := 1 z1 z2
Calculate dew point T and liquid composition
assuming the hydrocarbon layer forms first:
Guess:
Tdew1 := 100
x2α := z2
x3α := 1 x2α
Given
P = x2α P2sat ( Tdew1) + x3α P3sat ( Tdew1)
z3 P = x3α P3sat ( Tdew1)
x2α + x3α = 1
x2α
x3α := Find ( x2α , x3α , Tdew1)
Tdew1
Tdew1 = 66.602
x3α = 0.706
579
x2α = 0.294
Calculate dew point temperature assuming the water layer forms first:
x1β := 1
Tdew2 := 100
Guess:
x1β P1sat ( Tdew2) = z1 P
Given
Tdew2 := Find ( Tdew2)
Tdew2 = 79.021
Since Tdew2 > Tdew1, the water layer forms first
(b) Calculate the temperature at which the second layer forms:
Tdew3 := 100
Given
x2α := z2
x3α := 1 x2α
y1 := z1
Guess:
y2 := z2
y3 := z3
P = P1sat ( Tdew3) + x2α P2sat ( Tdew3) + x3α P3sat ( Tdew3)
y1 P = P1sat ( Tdew3)
y2
z2
=
y3
z3
y1 + y2 + y3 = 1
y2 P = x2α P2sat ( Tdew3)
x2α + x3α = 1
y1
y2
y3
:= Find ( y1 , y2 , y3 , Tdew3 , x2α , x3α)
Tdew3
x2α
x3α
y1 = 0.288
y2 = 0.388
y3 = 0.324
Tdew3 = 68.437
x2α = 0.1446
x3α = 0.8554
(c)
Calculate the bubble point given the total molar composition of the
two phases
Tbubble := Tdew3
x2α :=
z2
z2 + z3
x2α = 0.545
580
x3α :=
z3
z2 + z3
x3α = 0.455
Given
P = P1sat ( Tbubble) + x2α P2sat ( Tbubble) + x3α P3sat ( Tbubble)
Tbubble := Find ( Tbubble)
Tbubble = 48.113
P1sat ( Tbubble)
P
x2α P2sat ( Tbubble)
y2 :=
P
x3α P3sat ( Tbubble)
y3 :=
P
y1 = 0.111
y1 :=
y2 = 0.81
y3 = 0.078
14.28 Temperatures in deg. C; pressures in kPa.
Water:
P1sat ( T) := exp 16.3872
3885.70
T + 230.170
n-Pentane:
P2sat ( T) := exp 13.7667
2451.88
T + 232.014
n-Heptane:
P3sat ( T) := exp 13.8622
2910.26
T + 216.432
P := 101.33
(a)
Guess:
Given
z1 := 0.32
z2 := 0.45
z3 := 1 z1 z2
Calculate dew point T and liquid composition
assuming the hydrocarbon layer forms first:
Tdew1 := 70
x2α := z2
x3α := 1 x2α
P = x2α P2sat ( Tdew1) + x3α P3sat ( Tdew1)
z3 P = x3α P3sat ( Tdew1)
x2α + x3α = 1
x2α
x3α := Find ( x2α , x3α , Tdew1)
Tdew1
Tdew1 = 65.122
x3α = 0.686
581
x2α = 0.314
Calculate dew point temperature assuming the water layer forms first:
x1β := 1
Tdew2 := 70
Guess:
x1β P1sat ( Tdew2) = z1 P
Given
Tdew2 := Find ( Tdew2)
Tdew2 = 70.854
Since Tdew1>Tdew2, a hydrocarbon layer forms first
(b) Calculate the temperature at which the second layer forms:
Tdew3 := 100
Given
x2α := z2
x3α := 1 x2α
y1 := z1
Guess:
y2 := z2
y3 := z3
P = P1sat ( Tdew3) + x2α P2sat ( Tdew3) + x3α P3sat ( Tdew3)
y1 P = P1sat ( Tdew3)
y2
z2
=
y3
z3
y2 P = x2α P2sat ( Tdew3)
y1 + y2 + y3 = 1
x2α + x3α = 1
y1
y2
y3
:= Find ( y1 , y2 , y3 , Tdew3 , x2α , x3α)
Tdew3
x2α
x3α
y1 = 0.24
y2 = 0.503
y3 = 0.257
Tdew3 = 64.298
x2α = 0.2099
x3α = 0.7901
(c)
Calculate the bubble point given the total
molar composition of the two phases
Tbubble := Tdew3
x2α :=
z2
z2 + z3
x2α = 0.662
582
x3α :=
z3
z2 + z3
x3α = 0.338
Given
P = P1sat ( Tbubble) + x2α P2sat ( Tbubble) + x3α P3sat ( Tbubble)
Tbubble := Find ( Tbubble)
Tbubble = 43.939
P1sat ( Tbubble)
P
x2α P2sat ( Tbubble)
y2 :=
P
x3α P3sat ( Tbubble)
y3 :=
P
y1 = 0.09
y1 :=
0.302
0.224
y2 = 0.861
y3 = 0.049
748.4
K
304.2
14.32 ω :=
40.51
bar
73.83
Tc :=
Pc :=
P := 10bar , 20bar .. 300bar
T
Tr :=
Tc
T := 353.15K
Use SRK EOS
From Table 3.1, p. 98 of text:
σ := 1
ε := 0
:= 0.08664
Ψ := 0.42748
(
)(
α := 1 + 0.480 + 1.574 ω 0.176 ω 1 Tr
2
0.5
) 2
2
2
Ψ α R Tc
Eq. (14.31)
a :=
Pc
R Tc
b :=
Pc
6.842 kg m5
a=
0.325 s2 mol2
1.331 × 10 4 m3
b=
2.968 × 10 5 mol
β 2 ( P) :=
z2 := 1
b2 P
R T
q2 :=
Eq. (14.33)
(guess)
583
a2
b2 R T
Eq. (14.32)
Eq. (14.34)
Given
z2 = 1 + β 2 ( P) q2 β 2 ( P)
(
z2 β 2 ( P)
)(
z2 + ε β 2 ( P) z2 + σ β 2 ( P)
Eq. (14.36)
)
Z2 ( P) := Find ( z2)
Z 2 ( P) + β 2 ( P)
I2 ( P) := ln
Z 2 ( P)
Eq. (6.65b)
For simplicity, let φ1 represent the infinite-dilution value of the fugacity
coefficient of species 1 in solution.
l12 := 0.088
Eq. (14.103):
b1
( Z2 ( P) 1) ln ( Z2 ( P) β 2 ( P) ) ...
b2
0.5
b1
a1
+ q2 2 ( 1 l12)
I2 ( P)
b2
a2
φ 1 ( P) := exp
3
Psat1 := 0.0102bar
V1 := 124.5
cm
mol
Eqs. (14.98) and (14.99), with φsat1 = 1 and (P - Psat1) = P, combine to give:
y1 ( P) :=
Psat1
P φ 1 ( P)
P V 1
R T
exp
0.1
0.01
y1 ( P)
1 .10
3
1 .10
4
0
50
100
150
P
bar
584
200
250
300
0.302
0.038
748.4
K
126.2
14.33 ω :=
40.51
bar
34.00
Tc :=
Pc :=
P := 10bar , 20bar .. 300bar
T
Tr :=
Tc
T := 308.15K (K)
Use SRK EOS
From Table 3.1, p. 98 of text:
σ := 1
ε := 0
:= 0.08664
Ψ := 0.42748
(
)(
)
2
0.5
α := 1 + 0.480 + 1.574 ω 0.176 ω 1 Tr
a :=
2
2
Ψ α R Tc
R Tc
b :=
Pc
Eq. (14.31)
Pc
b2 P
R T
z2 := 1
Eq. (14.32)
1.331 × 10 4 m3
b=
2.674 × 10 5 mol
7.298 kg m5
a=
0.067 s2 mol2
β 2 ( P) :=
2
q2 :=
Eq. (14.33)
a2
b2 R T
Eq. (14.34)
(guess)
Given
z2 = 1 + β 2 ( P) q2 β 2 ( P)
(
z2 β 2 ( P)
)(
z2 + ε β 2 ( P) z2 + σ β 2 ( P)
)
Eq. (14.36)
Z2 ( P) := Find ( z2)
Z 2 ( P) + β 2 ( P)
I2 ( P) := ln
Z 2 ( P)
Eq. (6.65b)
For simplicity, let φ1 represent the infinite-dilution value of the fugacity
coefficient of species 1 in solution.
585
l12 := 0.0
Eq. (14.103):
b1
( Z2 ( P) 1) ln ( Z2 ( P) β 2 ( P) ) ...
b2
0.5
b1
a1
+ q2 2 ( 1 l12)
I2 ( P)
b2
a2
φ 1 ( P) := exp
3
4
Psat1 := 2.9 10
V1 := 125
bar
cm
mol
Eqs. (14.98) and (14.99), with φsat1 = 1 and (P - Psat1) = P, combine to
give:
y1 ( P) :=
Psat1
P φ 1 ( P)
P V 1
R T
exp
10
5
y1 ( P) 10
1
0
50
100
150
P
bar
Note: y axis is log scale.
586
200
250
300
14.45 A labeled diagram of the process is given below. The feed stream is taken as
the α phase and the solvent stream is taken as the β phase.
F
nF
xF1 = 0.99
xF2 = 0.01
R
nR
xα 1
xα2 = 0.001
Feed
Mixer/
Settler
S
nS
xS3 = 1.0
E
nE
xβ 2
xβ 3
Solvent
Define the values given in the problem statement. Assume as a basis a feed
rate nF = 1 mol/s.
nF := 1
mol
s
xF1 := 0.99
xF2 := 0.01
xS3 := 1
xα2 := 0.001
xα1 := 1 xα2
Apply mole balances around the process as well as an equilibrium relationship
A12 := 1.5
From p. 585
2
γα 2 ( x2) := exp A12 ( 1 x2)
A23 := 0.8
2
γβ 2 ( x2) := exp A23 ( 1 x2)
Material Balances
nS + nF = nE + nR
(Total)
nS = xβ3 nE
(Species 3)
xF1 nF = xα1 nR
(Species 1)
Substituting the species balances into the total balance yields
xF1
1
nS + nF =
nS +
nF
xβ3
xα1
Solving for the ratio of solvent to feed (nS/nF) gives
nS
nF
xα1 xF1 xβ3
1 xβ3 xα1
=
587
We need xβ3. Assume exiting streams are at equilibrium. Here, the only
distributing species is 2. Then
xα2 γα 2 = xβ2 γβ 2
Substituting for γα2 and γβ2
(
)
(
)
(
)
2
2
xα2 exp A12 1 xα2 = xβ2 exp A23 1 xβ2
Solve for xβ2 using Mathcad Solve Block
xβ2 := 0.5
Guess:
Given
(
)
2
2
xα2 exp A12 1 xα2 = xβ2 exp A23 1 xβ2
()
xβ2 := Find xβ2
xβ2 = 0.00979
xβ3 := 1 xβ2
xβ3 = 0.9902
From above, the equation for the ratio nS/nF is:
xα1 xF1 xβ3
1 xβ3 xα1
nSnF :=
a) nSnF = 0.9112
Ans.
b) xβ2 = 0.00979
Ans.
c) "Good chemistry" here means that species 2 and 3 "like" each other, as
evidenced by the negative GE23. "Bad chemistry" would be reflected in a
positive GE23, with values less than (essential) but perhaps near to GE12.
14.46 1 - n-hexane
2 - water
Since this is a dilute system in both phases, Eqns. (C) and (D) from Example
14.4 on p. 584 can be used to find γ1α and γ2β.
xα1 :=
520
6
xα2 := 1 xα1
xβ2 :=
10
2
6
10
588
xβ1 := 1 xβ2
γα 1 :=
γβ 2 :=
xβ1
3
Ans.
5
Ans.
γα 1 = 1.923 × 10
xα1
1 xα1
γβ 2 = 4.997 × 10
1 xβ1
3
14.50 1 - butanenitrile Psat1 := 0.07287bar
V1 := 90
cm
Psat2 := 0.29871bar
V2 := 92
cm
mol
3
2- benzene
3
B1 , 1 := 7993
cm
mol
T := 318.15K
i := 1 .. 2
mol
3
B2 , 2 := 1247
cm
mol
3
B1 , 2 := 2089
P := 0.20941bar
j := 1 .. 2
cm
mol
B2 , 1 := B1 , 2
x1 := 0.4819
x2 := 1 x1
k := 1 .. 2
y1 := 0.1813
y2 := 1 y1
Term A is calculated using the given data.
term_Ai :=
yi P
xi Psati
Term B is calculated using Eqns. (14.4) and (14.5)
δ j , i := 2 B j , i B j , j Bi , i
φhati := exp
P
1
Bi , i +
2
R T
Bi , i Psati
R T
φsati := exp
y jyk( 2 δ j , i δ j , k)
j
k
term_Bi :=
φhati
φsati
Term C is calculated using Eqn. (11.44)
fsati := φsati Psati
1.081
1.108
term_A =
Vi ( P Psati)
fi := φsati Psati exp
R T
0.986
1.006
term_B =
589
term_Ci :=
1
1
term_C =
fsati
fi
Ans.
14.51 a) Equivalent to d2(G/RT)/dx12 = 0, use d2(GE/RT)/dx12 = -1/x1x2
For GE/RT = Ax1x2 = A(x1-x12)
d(GE/RT)/dx1 = A(1-2x1)
d2(GE/RT)/dx12 = -2A
Thus, -2A = -1/x1x2 or 2Ax1x2 = 1.
Substituting for x2: x1-x12 = 1/(2A) or x12-x1+1/(2A) = 0.
1+
The solution to this equation yields two roots:
x1 =
2
A
2
1
and
1
x1 =
1
2
A
2
The two roots are symmetrical around x1 = 1/2
Note that for:
A<2: No real roots
A = 2: One root, x1 = 1/3 (consolute point)
A>2: Two real roots, x1 > 0 and x1 <1
b) Plot the spinodal curve along with the solubility curve
540K
T
+ 21.1 3 ln
T
K
Both curves are symmetrical around x1 = 1/2. Create functions to
represent the left and right halves of the curves.
From Fig. 14.15:
A ( T) :=
From above, the equations for the spinodal curves are:
xspr1 ( T) :=
1 1 A ( T) 2
+
22
A ( T)
xr := 0.7
xl := 0.3
xspl1 ( T) :=
1 1 A ( T) 2
22
A ( T)
From Eq. (E) in Example 14.5, the solubility curves are solved using
a Solve Block:
590
1 xr
xr
xr > 0.5
xr1 ( T) := Find ( xr)
1 xl
xl
xl < 0.5
xl1 ( T) := Find ( xl)
Given
A ( T) ( 1 2xr) = ln
Given
A ( T) ( 1 2xl) = ln
Find the temperature of the upper consolute point.
T := 300K
A ( T) = 2
Tu := Find ( T)
0.3
Given
0.5
Tu = 345.998 K
T := 250K .. 346K
360
340
320
300
280
260
240
0.1
0.2
0.4
0.6
0.7
0.8
xl1
xr1
xspl1
xpr1
14.54 The solution is presented for one of the systems given. The solutions for the
other systems follow in the same manner.
f) 1- Carbon tetrachloride
ω 1 := 0.193
Tc1 := 556.4K
Pc1 := 45.60bar
A1 := 14.0572
B1 := 2914.23
C1 := 232.148
Psat1 ( T) := exp A1
kPa
T 273.15 + C
1
K
B1
591
2 - n-heptane
ω 2 := 0.350
Tc2 := 540.2K
Pc2 := 27.40bar
A2 := 13.8622
B2 := 2910.26
C2 := 216.432
Psat2 ( T) := exp A2
kPa
T 273.15 + C
2
K
B2
T := ( 100 + 273.15)K
Tr1 :=
T
Tc1
Tr1 = 0.671
Psat1r :=
Tr2 :=
T
Tc2
Tr2 = 0.691
Psat2r :=
Psat1 ( T)
Pc1
Psat2 ( T)
Pc2
Λ 12 := 1.5410
Using Wilson's equation
Psat1r = 0.043
Psat2r = 0.039
Λ 21 := 0.5197
γ 1 ( x1) := exp ln x1 + ( 1 x1) Λ 12 ...
Λ 12
Λ 21
+ 1 x
( 1) x + 1 x Λ 1 x + x Λ
( 1) 1 21
1) 12
1 (
γ 2 ( x1) := exp ln ( 1 x1) + x1 Λ 21 ...
Λ 12
Λ 21
+ x
( 1) x + 1 x Λ 1 x + x Λ
( 1) 1 21
1) 12
1 (
For part i, use the modified Raoult's Law. Define the pressure and vapor
mole fraction y1 as functions of the liquid mole fraction, x 1.
Pi ( x1) := x1 γ 1 ( x1) Psat1 ( T) + ( 1 x1) γ 2 ( x1) Psat2 ( T)
yi1 ( x1) :=
x1 γ 1 ( x1) Psat1 ( T)
Pi ( x1)
Modified Raoult's Law: Eqn. (10.5)
592
For part ii, assume the vapor phase is an ideal solution. Use Eqn. (11.68)
and the PHIB function to calculate φhat and φsat.
(
φsat1 := PHIB Tr1 , Psat1r , ω 1
)
φsat1 = 0.946
P
φhat1 ( P) := PHIB Tr1 ,
, ω1
P
(
c1
φsat2 := PHIB Tr2 , Psat2r , ω 2
)
φ 1 ( P) :=
c2
φsat1
φsat2 = 0.95
P
φhat2 ( P) := PHIB Tr2 ,
, ω2
P
φhat1 ( P)
φ 2 ( P) :=
φhat2 ( P)
φsat2
Solve Eqn. (14.1) for y1 and P given x1.
Guess:
y1 := 0.5
P := 1bar
Given
y1 φ 1 ( P) P = x1 γ 1 ( x1) Psat1 ( T)
( 1 y1) φ2 (P) P = ( 1 x1) γ 2 ( x1) Psat2 (T)
Eqn. (14.1)
fii ( x1) := Find ( P , y1)
fii is a vector containing the values of P and y 1. Extract the pressure, P and
vapor mole fraction, y1 as functions of the liquid mole fraction.
Pii ( x1) := fii ( x1) 0
yii1 ( x1) := fii ( x1) 1
Plot the results in Mathcad
x1 := 0 , 0.1 .. 1.0
593
2
1.9
1.8
Pi ( x1 )
1.7
bar
Pi ( x1 )
1.6
bar
Pii ( x1 )
bar
Pii ( x1 )
bar
1.5
1.4
1.3
1.2
1.1
1
0
0.2
0.4
0.6
x1 , yi1 ( x1 ) , x1 , yii1 ( x1 )
P-x Raoult's
P-y Raoult's
P-x Gamma/Phi
P-y Gamma/Phi
594
0.8
Chapter 15 - Section A - Mathcad Solutions
15.1
Initial state: Liquid water at 70 degF.
H1
38.05
BTU
lbm
S1
0.0745
BTU
(Table F.3)
lbm rankine
Final state: Ice at 32 degF.
H2
( 0.02
T
(0
7
143.3)
BTU
S2
lbm
0.0
143.3
491.67
BTU
lbm rankine
459.67)rankine
(a)
Point A: sat. vapor at 32 degF.
Point C: sat. liquid at 70 degF. P = 85.79(psia).
Point D: Mix of sat. liq. & sat. vapor at 32 degF with the enthalpy of Point C.
Point B: Superheated vapor at 85.79(psia) and the entropy of Point A.
Data for Points A, C, & D from Table 9.1. Data for Point B from Fig. G.2.
595
Wideal
H2
Wideal
12.466
Wdotideal
H1
T
S2
S1
BTU
mdot
lbm
lbm
1
sec
Wdotideal
mdot Wideal
13.15 kW
Ans.
(b) For the Carnot heat pump, heat equal to the enthalpy change of the
water is extracted from a cold reservoir at 32 degF, with heat
rejection to the surroundings at 70 degF.
TC
Work
Wdot
TH
491.67 rankine
TH
QC
QC
T
TC
H2
QC
mdot Work
14.018
Wdot
TC
14.79 kW
t
Wdot
0.889
181.37
BTU
lbm
BTU
lbm
Work
Wdotideal
t
H1
Ans.
Ans.
The only irreversibility is the transfer of heat from the water as it cools
from 70 to 32 degF to the cold reservoir of the Carnot heat pump at 70
degF.
(c) Conventional refrigeration cycle under ideal conditions of operation:
Isentropic compression, infinite flow rate of cooling water, &
minimum temp. difference for heat transfer = 0.
For sat. liquid and vapor at 32 degF, by interpolation in the table:
HA
107.60
BTU
lbm
SA
0.2223
BTU
lbm rankine
For sat. liquid at 70 degF:
HC
34.58
BTU
lbm
HD
HC
For superheated vapor at 85.79(psia) and S = 0.2223:
HB
114
BTU
lbm
596
Refrigerent circulation rate:
H1 1
H2
mdot
Wdot
HA
lbm
sec
HA
2.484
Wdot
HD
mdot HB
16.77 kW
Wdotideal
t
lbm
mdot
Ans.
Ans.
0.784
t
Wdot
sec
The irreversibilities are in the throttling process and in heat transfer in
both the condenser and evaporator, where there are finite temperature
differences.
(d)
Practical cycle.
0.75
Point A: Sat. vapor at 24 degF.
Point B: Superheated vapor at 134.75(psia).
Point D: Mix of sat. liq. and sat. vapor at 24 degF with H of point C,
Point C: Sat. Liquid at 98 degF.
(Note that minimum temp. diff. is not at end of condenser, but it is not
practical to base design on 8-degF temp. diff. at pinch. See sketch.)
For sat. liquid and vapor at 24 degF:
Hliq
19.58
Sliq
0.0433
BTU
lbm
BTU
lbm rankine
Hvap
106.48
Svap
0.2229
597
BTU
lbm
BTU
lbm rankine
HA
Hvap
SA
Svap
For sat. liquid at 98 degF, P=134.75(psia):
HC
BTU
lbm
44.24
SC
0.0902
BTU
lbm rankine
For isentropic compression, the entropy of Point B is 0.2229 at
P=134.75(psia). From Fig. G.2,
H'B
BTU
lbm
118
HB
121.84
SB
0.228
SD
Sliq
HB
BTU
H'B
HA
HA
The entropy at this H is
read from Fig. G.2 at
P=134.75(psia)
lbm
BTU
lbm rankine
xD Svap
HD
HC
Sliq
xD
SD
HD
Hliq
Hvap
0.094
Hliq
xD
0.284
BTU
lbm rankine
Refrigerent circulation rate:
H2
mdot
Wdot
H1 1
HA
lbm
sec
HA
2.914
Wdot
HD
mdot HB
47.22 kW
Wdotideal
t
t
Wdot
THERMODYNAMIC ANALYSIS
Wdotlost.compressor
Qdotcondenser
mdot T
mdot HC
Wdotlost.condenser
mdot T
lbm
mdot
SB
T
0.279
(0
7
sec
Ans.
459.67)rankine
SA
HB
SC
598
SB
Ans.
Qdotcondenser
Wdotlost.throttle
mdot T
Wdotlost.evaporator
T
SD
SC
mdot SA SD
lbm
S 2 S1
1
sec
13.152 kW
27.85%
Wdotlost.compressor
8.305 kW
17.59%
Wdotlost.condenser
14.178 kW
30.02%
Wdotlost.throttle
6.621 kW
14.02%
Wdotlost.evaporator
4.968 kW
10.52%
Wdotideal
The percent values above express each quantity as a percentage of the
actual work, to which the quantities sum.
15.2
Assume ideal gases. Data from Table C.4
H298
S298
282984 J
H298
G298
G298
S298
298.15 K
257190 J
86.513
J
K
BASIS: 1 mol CO and 1/2 mol O2 entering with
accompanying N2=(1/2)(79/21)=1.881 mol
nCO
1 mol
nair
2.381 mol
599
nCO2
1 mol
nN2
1.881 mol
(a) Isothermal process at 298.15 K:
Since the enthalpy change of mixing for ideal gases is zero, the overall
enthalpy change for the process is
H
y1
For unmixing the air, define
H298
nN2
y1
nair
0.79
y2
1
y1
By Eq. (12.35) with no minus sign:
Sunmixing
nair R y1 ln y1
Sunmixing
10.174
y2 ln y2
J
K
For mixing the products of reaction, define
y1
nCO2
nN2
Smixing
S
T
y1
nCO2
nCO2
Sunmixing
300 K
0.347
y2
nN2 R y1 ln y1
S298
Wideal
600
T
y1
y2 ln y2
S
Smixing
S
Smixing
H
1
81.223
Wideal
15.465
J
K
J
K
259 kJ
Ans.
(b) Adiabatic combustion:
Heat-capacity data for the product gases from Table C.1:
A
nCO2 5.457
nN2 3.280
A
11.627
B
2.16
mol
B
nCO2 1.045
nN2 0.593
3
10
10
3
mol
D
nCO2 1.157
nN2 0.040
5
10
D
1.082
5
10
mol
T
For the products,
CP
HP = R
dT
T0
R
298.15 K
T0
The integral is given by Eq. (4.7). Moreover, by an energy balance,
H298
Guess
HP = 0
2
.
A
11.627
B
3
2.160 10
K
D
5
1.082 10 K
Given
H298 = R mol A T0
Find
8.796
1
B
2
T0
2
T
601
T0
2
D
T0
1
T
1
2622.603 K
2
For the cooling process from this temperature to the final temperature of
298.15 K, the entropy change is calculated by
ICPS 2622.6 298.15 11.627 2.160 10
ICPS
29.701
H
H298
H
2.83
S
5
0.0
1.082 10
R mol ICPS
H
Wideal.cooling
5
10 J
208904 J
0.8078
t
Wideal
= 29.701
S
Wideal.cooling
Wideal.cooling
t
3
T
246.934
J
K
S
Ans.
Ans.
The surroundings increase in entropy in the amount:
Q
H298
Wideal.cooling
S
Q
T
The irreversibility is in the combustion reaction. Ans.
15.3
For the sat. steam at 2700 kPa, Table F.2:
H1
2801.7
kJ
kg
S1
6.2244
kJ
kg K
For the sat. steam at 275 kPa, Table F.2:
H2
2720.7
kJ
kg
S2
7.0201
602
kJ
kg K
S
246.93
J
K
For sat. liquid and vapor at 1000 kPa, Table F.2:
Hliq
762.6
Hvap
kJ
kg
2776.2
Sliq
kJ
kg
Svap
2.1382
6.5828
kJ
kg K
kJ
Tsat
kg K
453.03K
(a) Assume no heat losses, no shaft work, and negligible changes in kinetic
and potential energy. Then by Eqs. (2.30) and (5.22) for a completely
reversible process:
H
fs ( mdot)= 0
S
fs ( mdot)= 0
We can also write a material balance, a quantity requirement, and relation
between H3 and S3 which assumes wet steam at point 3.
The five equations (in 5 unknowns) are as follows:
Guesses: mdot1
0.1
H1
H3
kg
s
mdot2
H2
mdot1
S3
2
Sliq
mdot3
H3
mdot1
Hliq
Tsat
Given
H3 mdot3
H1 mdot1
S3 mdot3
S1 mdot1
mdot3 = mdot1
S3 = Sliq
H3
H2 mdot2 = 0
S2 mdot2 = 0
mdot2
H3
kJ
s
kJ
sK
Hliq mdot3 = 300
Hliq
Tsat
mdot1
mdot2
mdot3
Find mdot1 mdot2 mdot3 H3 S3
H3
S3
603
kJ
s
mdot2
mdot1
H3
0.086
2.767
kg
s
10
mdot2
3 kJ
S3
kg
0.064
6.563
kg
s
kJ
kg K
mdot3
0.15
kg
s
Ans.
Steam at Point 3 is indeed wet.
(b) Turbine: Constant-S expansion of steam from Point 1 to 1000 kPa
results in wet steam of quality
x'turb
S1
Sliq
turb
xturb
xturb
Hliq
0.919
H'turb
2.614
10
0.78
Hturb
H1
turb H'turb
Hturb
x'turb
H'turb
2.655
10
Sturb
Sliq
xturb Svap
Sturb
6.316
kJ
kg K
Svap
Sliq
Hturb
Hliq
Hvap
Hliq
0.94
x'turb Hvap
Hliq
3 kJ
kg
H1
3 kJ
kg
Sliq
Compressor: Constant-S compression of steam from Point 2 to 1000 kPa
results in superheated steam. Interpolation in Table F.2 yields
H'comp
Hcomp
H2
kJ
2993.5
comp
kg
H'comp
H2
Hcomp
comp
By interpolation:
Scomp
7.1803
604
kJ
kg K
0.75
3084.4
kJ
kg
The energy balance, mass balance, and quantity requirement equations of
Part (a) are still valid. In addition, The work output of the turbine equals
the work input of the compressor. Thus we have 4 equations (in 4
unknowns):
kg
kg
Guesses:
mdot1
0.086
mdot2
0.064
s
s
mdot3
0.15
kg
H3
s
2770.
kJ
kg
Given
Hcomp
H2 mdot2 =
H3 mdot3
H1 mdot1
mdot3 = mdot1
Hturb
H1 mdot1
H2 mdot2 = 0
mdot2
kJ
s
H3
Hliq mdot3 = 300
kJ
s
mdot1
mdot2
Find mdot1 mdot2 mdot3 H3
mdot3
H3
mdot1
mdot3
0.10608
0.14882
kg
s
kg
s
mdot2
H3
0.04274
2.77844
kg
s
3 kJ
10
kg
Ans.
Steam at Point 3 is slightly superheated.
By interpolation,
S3
THERMODYNAMIC ANALYSIS
6.5876
T
kJ
kg K
300K
By Eq. (5.25), with the enthalpy term equal to zero:
Wdotideal
T
mdot3 S3
mdot1 S1
605
mdot2 S2
(assumed)
Wdotideal
6.014 kW
Wdotlost.turb
T mdot1 Sturb
Wdotlost.comp
S1
T mdot2 Scomp
Wdotlost.mixing
Wdotlost.turb
T
mdot3 S3
S2
mdot1 Sturb
2.9034 kW
48.2815%
2.054 kW
mdot2 Scomp
34.1565%
Wdotlost.comp
Wdotlost.mixing
1.0561 kW
17.5620%
The percent values above express each quantity as a percentage of the
absolute value of the ideal work, to which the quantities sum.
15.4
Some property values with reference to Fig. 9.1 are given in Example 9.1.
Others come from Table 9.1 or Fig. G.2.
For sat. liquid and vapor at the evaporator temperature of 0 degF:
BTU
lbm
Hliq
12.090
Svap
0.22525
Hvap
BTU
Sliq
lbm rankine
103.015
0.02744
BTU
lbm
BTU
lbm rankine
For sat. liquid at the condenser outlet temperature of 80 degF:
H4
37.978
H2
BTU
lbm
Hvap
x1
x1
H1
Hvap
S4
S2
H1
Hliq
0.285
606
Sliq
S1
Hliq
0.084
BTU
lbm rankine
H4
S1
Svap
0.07892
x1 Svap
BTU
lbm rankine
Sliq
From Example 9.1(b) for the compression step:
H
BTU
lbm
17.48
H3
H2
H
H3
120.5
BTU
lbm
From Fig. G.2 at H3 and P = 101.37(psia):
S3
0.231
Wdot
BTU
lbm rankine
mdot
Wdot
mdot H
1845.1
3.225
lbm
hr
10
4 BTU
hr
The purpose of the condenser is to transfer heat to the surroundings. Thus
the heat transferred in the condenser is Q in the sense of Chapter 15; i.e.,
it is heat transfer to the SURROUNDINGS, taken here to be at a
temperature of 70 degF.
Internal heat transfer (within the system) is not Q. The heat transferred in
the evaporator comes from a space maintained at 10 degF, which is part of
the system, and is treated as an internal heat reservoir.
The ideal work of the process is that of a Carnot engine operating between
the temperature of the refrigerated space and the temperature of the
surroundings.
T
(0
7
QdotC
459.67)rankine
120000
Wdotideal
Qdot
H4
TH
T
TC
BTU
hr
QdotC
Wdotlost.comp
TH
(
10
TC
Wdotideal
TC
T mdot S3
Wdotlost.throttle
T mdot S4
T mdot S1
1.533
4 BTU
10
hr
S2
H3 mdot
Wdotlost.cond
459.67)rankine
Qdot
S3
S4
607
Qdot
5 BTU
1.523 10
hr
Wdotlost.evap
T mdot S2 S1
H1 H2
mdot
T
TC
The final term accounts for the entropy change of the refrigerated space (an
internal heat reservoir).
BTU
47.53%
Wdotideal
15329.9
Wdotlost.comp
5619.4
BTU
hr
17.42%
Wdotlost.cond
3625.2
BTU
hr
11.24%
Wdotlost.throttle
4730.2
BTU
hr
14.67%
Wdotlost.evap
2947.6
BTU
hr
9.14%
hr
The percent values above express each quantity as a percentage of the
actual work, to which they sum:
Wdot
15.5
32252.3
BTU
hr
The discussion at the top of the second page of the solution to the
preceding problem applies equally here. In each case,
T
( 70
459.67) rankine
TH
T
The following vectors refer to Parts (a)-(e):
40
30
tC
600
500
20
QdotC
400
10
300
0
200
608
BTU
sec
TC
tC
Wdotideal
459.67 rankine
QdotC
TH
TC
TC
For sat. liquid and vapor at the evaporator temperature, Table 9.1:
21.486
18.318
Hliq
107.320
105.907
BTU
15.187
lbm
12.090
104.471
Hvap
103.015
9.026
Sliq
Hvap
0.22244
0.04065
H2
lbm
101.542
0.04715
BTU
0.22325
BTU
Svap
lbm rankine
0.03408
0.02744
0.22418
0.22525
BTU
lbm rankine
S2
Svap
0.22647
0.02073
For sat. liquid at the condenser temperature:
H4
x1
37.978
H1
Hvap
BTU
lbm
Hliq
Hliq
S4
0.07892
S1
Sliq
BTU
lbm rankine
x1 Svap
H1
H4
Sliq
From the results of Pb. 9.9, we find:
117.7
118.9
H3
120.1
121.7
BTU
lbm
From these values we must find the
corresponding entropies from Fig. G.2.
They are read at the vapor pressure for
80 degF of 101.37 kPa. The flow rates
come from Problem 9.9:
123.4
609
8.653
0.227
7.361
0.229
0.231
S3
0.234
BTU
lbm rankine
mdot
H4
Wdotlost.cond
S2
H3 mdot
T mdot S4
Wdotlost.throttle
Wdotlost.evap
sec
3.146
T mdot S3
Qdot
lbm
4.613
0.237
Wdotlost.comp
6.016
S3
T mdot S1
S4
T mdot S2
T
H1
S1
H2
TC
Qdot
mdot
The final term accounts for the entropy change of the refrigerated space (an
internal heat reservoir).
Wdot
mdot H3
H2
36.024
40.844
Wdotideal
20.9
22.419
BTU
41.695
sec
38.325
Wdotlost.comp
21.732
21.379
17.547
30.457
610
BTU
sec
11.149
8.754
10.52
9.444
Wdotlost.cond
10.589
BTU
sec
11.744
Wdotlost.throttle
7.292
11.826
5.322
10.322
89.818
12.991
95.641
11.268
9.406
Wdotlost.evap
BTU
sec
BTU
Wdot
sec
BTU
94.024
7.369
5.122
sec
86.194
68.765
In each case the ideal work and the lost work terms sum to give the actual
work, and each term may be expressed as a percentage of the actual work.
15.6
The discussion at the top of the second page of the solution to Problem
15.4 applies equally here.
T
(0
7
459.67)rankine
TH
TC
(
30
459.67)rankine
QdotC
Wdotideal
QdotC
TH
TC
T
Wdotideal
TC
2000
BTU
sec
163.375
BTU
sec
For sat. liquid and vapor at the evaporator temperature, Table 9.1:
Hliq
Hvap
H2
18.318
BTU
lbm
105.907
Hvap
BTU
lbm
Sliq
Svap
S2
611
0.04065
0.22325
Svap
BTU
lbm rankine
BTU
lbm rankine
For sat. liquid at the condenser temperature:
H4
37.978
BTU
lbm
S4
0.07892
BTU
lbm rankine
From Problem 9.12,
H2A
H3
BTU
lbm
116.
H2A
S2A
14.667
BTU
lbm
H3
0.2435
130.67
BTU
lbm rankine
BTU
lbm
From Fig. G.2 at this enthalpy and 33.11(psia):
S3
0.2475
BTU
lbm rankine
Energy balance on heat exchanger:
H1
x1
x1
H4
H2A
H1
Hliq
Hvap
H2
BTU
lbm
H1
S1
0.109
Sliq
x1 Svap
S1
Hliq
27.885
0.061
BTU
lbm rankine
Sliq
Upstream from the throttle (Point 4A) the state is subcooled liquid with
the enthalpy:
H4A
H1
The entropy at this point is essentially that of sat. liquid with this
enthalpy; by interpolation in Table 9.1:
S4A
0.05986
BTU
lbm rankine
From Problem 9.12:
mdot
25.634
612
lbm
sec
Wdotlost.comp
Qdot
H4
T mdot S3
S2A
H3 mdot
Wdotlost.cond
T mdot S4
Wdotlost.throttle
Wdotlost.evap
T mdot S1
S3
Qdot
S4A
T mdot S2 S1
H1 H2
T
mdot
TC
The final term accounts for the entropy change of the refrigerated space (an
internal heat reservoir).
Wdotlost.exchanger
T mdot S2A
Wdot
H2A
mdot H3
54.31
BTU
sec
14.45%
87.08
BTU
sec
S4
43.45%
23.16%
163.38
Wdotlost.comp
Wdotlost.cond
Wdotlost.throttle
9.98
BTU
sec
Wdotlost.evap
45.07
Wdotlost.exchanger
16.16
375.97
S4A
BTU
sec
Wdotideal
Wdot
S2
BTU
sec
2.65%
BTU
sec
11.99%
BTU
4.30%
sec
The figures on the right are percentages of the
actual work, to which the terms sum.
613
15.7
Compression to a pressure at which condensation in coils occurs at
110 degC. Table F.1 gives this sat. pressure as 143.27 kPa
0.75
comp
kJ
kg
H1
419.1
H2
2676.0
S1
kJ
kg K
(sat. liquid)
S2
kJ
kg
1.3069
7.3554
kJ
kg K
(sat. vapor)
For isentropic compression to 143.27 kPa, we find by double interpolation in
Table F.2:
H'3
2737.0
kJ
kg
H3
H2
H'3
H2
H3
comp
2757.3
kJ
kg
By more double interpolation in Table F.2 at 143.27 kPa,
S3
7.4048
kJ
kg K
By an energy balance, assuming the slurry passes through unchanged,
H4
H1
H3
H2
H4
614
500.4
kJ
kg
This enthalpy is a bit larger than that of sat. liquid at 110 degC; find quality
and then the entropy:
Hliq
461.3
Slv
Sliq
T
Hlv
kJ
kg K
x4
5.8203
S4
kJ
kg
kJ
kg K
300 K
x4 Slv
Wdotideal
Wdotlost.comp
Wdot
Wdotideal
Wdotlost.evap
Wdotlost.comp
Wdot
H1
mdot T
S4
mdot T
mdot H3
H4
S3
Hliq
Hlv
S4
mdot H4
Wdotlost.evap
15.8
kJ
kg
2230.0
1.5206
T
S4
S3
S2
kJ
kg K
Sliq
x4
mdot
1.4185
0.018
0.5
kg
sec
S1
S1
S2
H2
8.606 kW
21.16%
24.651 kW
60.62%
7.41 kW
18.22%
40.667 kW
The figures on the right are percentages of the
actual work, to which the terms sum.
A thermodynamic analysis requires an exact definition of the overall
process considered, and in this case we must therefore specify the source
of the heat transferred to the boiler.
Since steam leaves the boiler at 900 degF, the heat source may be
considered a heat reservoir at some higher temperature. We assume in
the following that this temperature is 950 degF.
The assumption of a different temperature would provide a variation in the
solution.
615
The ideal work of the process in this case is given by a Carnot engine
operating between this temperature and that of the surroundings, here
specified to be 80 degF.
We take as a basis 1 lbm of H2O passing through the boiler. Required
property values come from Pb. 8.8.
TH
(
459.67
950)rankine
TC
(
459.67
80)rankine
T
TC
Subscripts below correspond to points on figure of Pb. 8.7.
H1
257.6
S1
0.3970
H2
1461.2
S2
1.6671
H3
1242.2
S3
1.7431
H4
1047.8
S4
1.8748
H5
69.7
S5
0.1326
H7
250.2
S7
0.4112
QH
H2
BTU
lbm
H1 1 lbm
Wideal
QH 1
BTU
lbm rankine
TC
TH
For purposes of thermodynamic analysis, we consider the following 4 parts
of the process:
The boiler/heat reservoir combination
The turbine
The condenser and throttle valve
The pump and feedwater heater
Wlost.boiler.reservoir
m
T
T
S1 1 lbm
QH
TH
(From Pb. 8.8)
0.18688 lbm
Wlost.turbine
S2
m S3
S2
1 lbm
m
S4
S2
The purpose of the condenser is to transfer heat to the surroundings. The
amount of heat is
Q
1 lbm H5
1 lbm m H4 m H7
Q
829.045 BTU
616
Wlost.cond.valve
T
Wlost.pump.heater
T
1 lbm S5
1 lbm
m S4
m S7
S5
m S7
S3
1 lbm S1
Q
The absolute value of the actual work comes from Pb. 8.8:
Wabs.value = 374.61 BTU
Wlost.boiler.reservoir
50.43%
224.66 BTU
30.24%
Wlost.turbine
98.81 BTU
13.30%
Wlost.cond.valve
36.44 BTU
4.90%
8.36 BTU
1.13%
Wlost.pump.heater
Wideal
742.82 BTU
(absolute value)
15.9
The numbers on the right are percentages of the
absolute value of the ideal work, to which they
sum.
Refer to Figure 9.7, page 330 The analysis presented here is for
the liquefaction section to the right of the dashed line. Enthalpy and
entropy values are those given in Ex. 9.3 plus additional values from
the reference cited on page 331 at conditions given in Ex. 9.3.
Property values:
H4
1140.0
kJ
kg
S4
9.359
kJ
kg K
H5
1009.7
kJ
kg
S5
8.894
kJ
kg K
H7
719.8
kJ
kg
S7
7.544
H9
285.4
kJ
kg
S9
4.928
H10
796.9
kJ
kg
S10
617
9.521
kJ
kg K
kJ
kg K
kJ
kg K
H14
1042.1
kJ
kg
S14
11.015
kJ
kg K
H15
1188.9
kJ
kg
S15
11.589
kJ
kg K
H11
H5 S11
S5
H13
H10 S13
S10
H6
S6
H5
S5
H10 S12
H12
T
S10
295K
The basis for all calculations is 1 kg of methane entering at point 4. All
work quantities are in kJ. Results given in Ex. 9.3 on this basis are:
Fraction of entering methane that is liquefied:
Fraction of entering methane passing through the expander:
On this basis also Eq. (5.26) for Ideal Work, Eq. (5.33) for Entropy
Generation,and Eq. (5.34) for Lost Work can be written:
z
x
0.113
0.25
Q
Wlost = T SG
T
______________________________________________________________
Wideal =
( m)
H fs
Wideal
H15 (
1
Wideal
489.001
Wout
H12
T
( m) SG =
S fs
z) H9 z
0.044
H11 x
kJ
kg K
Wout
Wlost.a
(b) Heat Exchanger II: SG.b
SG.b
0.313
kJ
kg K
T
S15 (
1
z) S9 z
S4
S14 (
1
z)
kJ
kg
(a) Heat Exchanger I: SG.a
SG.a
H4
( m)
S fs
Wlost.b
kJ
kg
S5
S4
T SG.a
S7
S6 (
1
T SG.b
618
S15
Wlost.a
x)
Wlost.b
13.021
S14
kJ
kg
S13 (
1
92.24
kJ
kg
z)
(c) Expander:
SG.c
0.157
kJ
kg K
(d) Throttle:
SG.d
0.964
SG.c
S12
Wlost.c
SG.d
kJ
kg K
S9 z
Wlost.d
S11 x
Wlost.c
T SG.c
S10 (
1
T SG.d
z
46.241
x) S7 (
1
x)
Wlost.d
Entropy-generation analysis:
kJ/kg-K
Percent of
S_Ga
0.044
2.98%
S_Gb
0.313
21.18%
S_Gc
0.157
10.62%
S_Gd
0.964
65.22%
1.478
100.00%
Work analysis, Eq. (15.3):
kJ/kg
Percent of
Wout
53.20
10.88%
Wlost.a
13.02
2.66%
Wlost.b
92.24
18.86%
Wlost.c
46.24
9.46%
Wlost.d
284.30
58.14%
489.00
100.00%
Note that:
= Wideal
619
kJ
kg
284.304
kJ
kg
Chapter 16 - Section A - Mathcad Solutions
16.10(Planck's constant)
h
P
34
6.626 10
Js
T
1bar
(Boltzmann's constant)
23 J
k
1.381 10
K
M
6.023 10 mol
3
RT
V
P
0.025
R ln
MkT
5
2
2
Ve
NA
2
h
83.800
b) For Krypton: M
Sig
R ln
MkT
2
131.30
M
R ln
NIST value: 154.84
J
mol K
Sig
164.08
J
mol K
J
mol K
Ans.
Ve
NA
Ans.
gm
mol
NA
3
Sig
J
mol K
5
2
2
2
mol
NA
h
c) For Xenon
154.84
Sig
gm
mol
3
2
m
NA
NIST value: 164.05
Sig
1
mol
3
2
23
NA
gm
39.948
a) For Argon:
V
298.15K
(Avagodro's number)
MkT
2
h
2
5
2
Sig
Ve
NA
164.08
NIST value: 169.68
620
J
mol K
J
mol K
Ans.
Chapter 1 - Section B - Non-Numerical Solutions
1.1 This system of units is the English-system equivalent of SI. Thus,
gc = 1(lbm )(ft)(poundal)1 (s)2
1.2 (a) Power is power, electrical included. Thus,
kg·m2
N·m
energy
[=]
[=]
s3
s
time
(b) Electric current is by denition the time rate of transfer of electrical charge. Thus
Power [=]
Charge [=] (electric current)(time) [=] A·s
(c) Since power is given by the product of current and electric potential, then
kg·m2
power
[=]
current
A·s3
(d) Since (by Ohms Law) current is electric potential divided by resistance,
Electric potential [=]
kg·m2
electric potential
[=]
3
current
A2 ·s
(e) Since electric potential is electric charge divided by electric capacitance,
Resistance [=]
4
A2 ·s
charge
[=]
Capacitance [=]
electric potential
kg·m2
1.3 The following are general:
ln x = ln 10 × log10 x
P sat /kPa = P sat /torr ×
( A)
100 kPa
750.061 torr
(B)
t / C = T /K 273.15
(C )
By Eqs. ( B ) and ( A),
ln P sat /kPa = ln 10 × log10 P sat /torr + ln
100
750.061
The given equation for log10 P sat /torr is:
log10 P sat /torr = a
b
t / C + c
Combining these last two equations with Eq. (C ) gives:
ln P sat /kPa = ln 10 a
b
T /K 273.15 + c
= 2.3026 a
b
T /K 273.15 + c
+ ln
100
750.061
2.0150
Comparing this equation with the given equation for ln P sat /kPa shows that:
A = 2.3026 a 2.0150
B = 2.3026 b
621
C = c 273.15
1.9 Reasons result from the fact that a spherical container has the minimum surface area for a given interior
volume. Therefore:
(a) A minimum quantity of metal is required for tank construction.
(b) The tensile stress within the tank wall is everywhere uniform, with no sites of stress concentration.
Moreover, the maximum stress within the tank wall is kept to a minimum.
(c) The surface area that must be insulated against heat transfer by solar radiation is minimized.
1.17 Kinetic energy as given by Eq. (1.5) has units of mass·velocity2 . Its fundamental units are therefore:
E K [=] kg·m2 ·s2 [=] N·m [=] J
Potential energy as given by Eq. (1.7) has units of mass·length·acceleration. Its fundamental units are
therefore:
E P [=] kg·m·m·s2 [=] N·m [=] J
1.20 See Table A.1, p. 678, of text.
1(atm) 1 bar = 1/0.986923 = 1.01325 bar
1(Btu) 1 kJ = 1/0.947831 = 1.05504 kJ
1(hp) 0.75 kW = 1/1.34102 = 0.745701 kW
1(in) 2.5 cm = 2.54 cm exactly, by denition (see p. 651 of text)
1(lbm ) 0.5 kg = 0.45359237 kg exactly, by denition (see p. 651 of text)
1(mile) 1.6 km = 5280/3280.84 = 1.60934 km
1(quart) 1 liter = 1000/(264.172 × 4) = 0.94635 liter (1 liter 1000 cm3 )
1(yard) 1 m = (0.0254)(36) = 0.9144 m exactly, by denition of the (in) and the (yard)
An additional item could be:
1(mile)(hr)1 0.5 m s1 = (5280/3.28084)(1/3600) = 0.44704 m s1
1.21 One procedure here, which gives results that are internally consistent, though not exact, is to assume:
1 Year [=] 1 Yr [=] 364 Days
This makes 1 Year equivalent to exactly 52 7-Day Weeks. Then the average Month contains 30 1 Days
3
and 4 1 Weeks. With this understanding,
3
1 Year [=] 1 Yr [=] 364 Days [=] (364)(24)(3600) = 31,449,600 Seconds
Whence,
1 Sc [=] 31.4496 Second
1 Mn [=] 314.496 Second
1 Second [=] 0.031797 Sc
1 Minute [=] 60 Second [=] 0.19078 Mn
1 Hr [=] 3144.96 Second
1 Hour [=] 3600 Second [=] 1.14469 Hr
1 Dy [=] 31449.6 Second
1 Day [=] (24)(3600) Second [=] 2.74725 Dy
1 Wk [=] 314496. Second
1 Week [=] (7)(24)(3600) Second [=] 1.92308 Wk
1 Mo [=] 3144960 Second
1 Month [=] (4 1 )(7)(24)(3600) Second[=] 0.83333 Mo
3
The nal item is obviously also the ratio 10/12.
622
Chapter 2 - Section B - Non-Numerical Solutions
2.3 Equation (2.2) is here written:
U t + E P + E K = Q + W
(a) In this equation W does not include work done by the force of gravity on the system. This is
accounted for by the E K term. Thus, W = 0.
(b) Since the elevation of the egg decreases, sign( E P ) is ().
(c) The egg is at rest both in its initial and nal states; whence E K = 0.
(d) Assuming the egg does not get scrambled, its internal energy does not change; thus U t = 0.
(e) The given equation, with U t = E K = W = 0, shows that sign(Q) is (). A detailed examination of the process indicates that the kinetic energy of the egg just before it strikes the surface
appears instantly as internal energy of the egg, thus raising its temperature. Heat transfer to the
surroundings then returns the internal energy of the egg to its initial value.
2.6 If the refrigerator is entirely contained within the kitchen, then the electrical energy entering the refrigerator must inevitably appear in the kitchen. The only mechanism is by heat transfer (from the
condenser of the refrigerator, usually located behind the unit or in its walls). This raises, rather than
lowers, the temperature of the kitchen. The only way to make the refrigerator double as an air conditioner is to place the condenser of the refrigerator outside the kitchen (outdoors).
2.7 According to the phase rule [Eq. (2.7)], F = 2 κ + N . According to the laboratory report a pure
material ( N = 1) is in 4-phase (κ = 4) equilibrium. If this is true, then F = 2 4 + 1 = 1. This is
not possible; the claim is invalid.
2.8 The phase rule [Eq. (2.7)] yields: F = 2 κ + N = 2 2 + 2 = 2. Specication of T and P xes the
intensive state, and thus the phase compositions, of the system. Since the liquid phase is pure species
1, addition of species 2 to the system increases its amount in the vapor phase. If the composition of
the vapor phase is to be unchanged, some of species 1 must evaporate from the liquid phase, thus
decreasing the moles of liquid present.
2.9 The phase rule [Eq. (2.7)] yields: F = 2 κ + N = 2 2 + 3 = 3. With only T and P xed,
one degree of freedom remains. Thus changes in the phase compositions are possible for the given
T and P . If ethanol is added in a quantity that allows T and P to be restored to their initial values,
the ethanol distributes itself between the phases so as to form new equilibrium phase compostions and
altered amounts of the vapor and liquid phases. Nothing remains the same except T and P .
2.10 (a) Since F = 3, xing T and P leaves a single additional phase-rule variable to be chosen.
(b) Adding or removing liquid having the composition of the liquid phase or adding or removing
vapor having the composition of the vapor phase does not change the phase compositions, and
does not alter the intensive state of the system. However, such additions or removals do alter the
overall composition of the system, except for the unusual case where the two phase compositions
are the same. The overall composition, depending on the relative amounts of the two phases, can
range from the composition of the liquid phase to that of the vapor phase.
2.14 If the uid density is constant, then the compression becomes a constant-V process for which the work
is zero. Since the cylinder is insulated, we presume that no heat is transferred. Equation (2.10) then
shows that U = 0 for the compression process.
623
2.16 Electrical and mechanical irreversibilities cause an increase in the internal energy of the motor, manifested by an elevated temperature of the motor. The temperature of the motor rises until a dynamic
equilibrium is established such that heat transfer from the motor to the srroundings exactly compensates for the irreversibilities. Insulating the motor does nothing to decrease the irreversibilities in
the motor and merely causes the temperature of the motor to rise until heat-transfer equilibrium is
reestablished with the surroundings. The motor temperature could rise to a level high enough to cause
damage.
2.19 Let symbols without subscripts refer to the solid and symbols with subscript w refer to the water.
Heat transfer from the solid to the water is manifested by changes in internal energy. Since energy is
t
conserved, U t = Uw . If total heat capacity of the solid is C t (= mC ) and total heat capacity of
t
the water is Cw (= m w Cw ), then:
t
C t (T T0 ) = Cw (Tw Tw0 )
or
Tw = Tw0
Ct
(T T0 )
t
Cw
( A)
This equation relates instantaneous values of Tw and T . It can be written in the alternative form:
t
t
T C t T0 C t = Tw0 Cw Tw Cw
t
t
Tw0 Cw + T0 C t = Tw Cw + T C t
or
(B)
˙
The heat-transfer rate from the solid to the water is given as Q = K (Tw T ). [This equation implies
that the solid is the system.] It may also be written:
Ct
dT
= K (Tw T )
dτ
(C )
In combination with Eq. ( A) this becomes:
Ct
or
dT
=K
dτ
Dene:
Ct
dT
= K Tw0 t (T T0 ) T
Cw
dτ
T T0
Tw0 T
t
t
Cw
C
βK
1
1
+t
t
Cw
C
= T K
1
1
+t
Cw
Ct
αK
+K
T0
Tw0
+t
t
Cw
C
T0
Tw0
+t
Cw
Ct
where both α and β are constants. The preceding equation may now be written:
dT
= α βT
dτ
Rearrangement yields:
1 d (α β T )
dT
= dτ
=
β α βT
α βT
Integration from T0 to T and from 0 to τ gives:
α βT
1
ln
α β T0
β
624
=τ
α βT
= exp(βτ )
α β T0
which may be written:
When solved for T and rearranged, this becomes:
T=
where by the denitions of α and β ,
α
α
+ T0
β
β
exp(βτ )
Tw C t + T0 C t
α
= 0 tw
Cw + C t
β
When τ = 0, the preceding equation reduces to T = T0 , as it should. When τ = , it reduces to
T = α/β . Another form of the equation for α/β is found when the numerator on the right is replaced
by Eq. ( B ):
t
Tw Cw + T C t
α
=
t
Cw + C t
β
By inspection, T = α/β when Tw = T , the expected result.
2.20 The general equation applicable here is Eq. (2.30):
˙
H + 1 u 2 + zg m
2
fs
˙
˙
= Q + Ws
(a) Write this equation for the single stream owing within the pipe, neglect potential- and kineticenergy changes, and set the work term equal to zero. This yields:
˙
( H )m = Q
˙
(b) The equation is here written for the two streams (I and II) owing in the two pipes, again neglecting
any potential- and kinetic-energy changes. There is no work, and the the heat transfer is internal,
˙
between the two streams, making Q = 0. Thus,
( H )I m I + ( H )II m II = 0
˙
˙
(c) For a pump operating on a single liquid stream, the assumption of negligible potential- and kineticenergy changes is reasonable, as is the assumption of negligible heat transfer to the surroundings.
Whence,
˙
( H )m = W
˙
(d) For a properly designed gas compressor the result is the same as in Part (c).
(e) For a properly designed turbine the result is the same as in Part (c).
(f ) The purpose of a throttle is to reduce the pressure on a owing stream. One usually assumes
adiabatic operation with negligible potential- and kinetic-energy changes. Since there is no work,
the equation is:
H =0
(g) The sole purpose of the nozzle is to produce a stream of high velocity. The kinetic-energy change
must therefore be taken into account. However, one usually assumes negligible potential-energy
change. Then, for a single stream, adiabatic operation, and no work:
˙
H + 1 u2 m = 0
2
The usual case is for a negligible inlet velocity. The equation then reduces to:
H + 1 u2 = 0
22
625
2.21 We reformulate the denition of Reynolds number, with mass owrate m replacing velocity u :
˙
m = u Aρ = u
˙
Solution for u gives:
u=
Whence,
Re
π2
Dρ
4
˙
4m
π D2ρ
˙
4m
˙
4 m ρD
uρ D
=
=
2ρ µ
π Dµ
πD
µ
(a) Clearly, an increase in m results in an increase in Re.
˙
(b) Clearly, an increase in D results in a decrease in Re.
2.24 With the tank as control volume, Eqs. (2.25) and (2.29) become:
dm
+m =0
˙
dt
d (mU )
˙
+Hm =0
dt
and
Expanding the derivative in the second equation, and eliminating m by the rst equation yields:
˙
m
dm
dm
dU
=0
H
+U
dt
dt
dt
dm
dU
=
m
H U
Multiply by dt and rearrange:
Substitution of H for H requires the assumption of uniform (though not constant) conditions throughout the tank. This requires the absence of any pressure or temperature gradients in the gas in the tank.
2.32 From the given equation:
By Eq. (1.3),
P=
W =
RT
V b
V2
V1
W = RT ln
Whence,
2.35 Recall:
Whence,
By Eq. (2.4),
d(P V ) = P d V + V d P
d W = V d P d(P V )
V2
V1
RT
d ( V b)
V b
V1 b
V2 b
and
and
dW = P dV
W=
V dP
(PV )
d Q = dU d W
By Eq. (2.11), U = H P V
With d W = P d V
Whence,
P dV =
Q=
and
dU = d H P d V V d P
the preceding equation becomes d Q = d H V d P
H
V dP
626
.
.
2.38 (a) By Eq. (2.24a),
m = u Aρ
With m , A, and ρ all constant, u must also be constant. With
q = u A, q is also constant.
.
.
.
(b) Because mass is conserved, m must be constant. But n = M/m may change, because M may
change. At the very least, ρ depends on T and P . Hence u and q can both change.
2.40 In accord with the phase rule, the system has 2 degrees of freedom. Once T and P are specied, the
intensive state of the system is xed. Provided the two phases are still present, their compositions
cannot change.
2.41 In accord with the phase rule, the system has 6 degrees of freedom. Once T and P are specied, 4
remain. One can add liquid with the liquid-phase composition or vapor with the vapor-phase composition or both. In other words, simply change the quantities of the phases.
.
2.43 Let n represent the moles of air leaving the home. By an energy balance,
.
dn
dU
d (nU )
.
.
+U
=n H +n
Q=nH+
dt
dt
dt
dn
.
n =
dt
.
dU
dn
+n
Q = ( H U )
dt
dt
But a material balance yields
Then
.
dU
dn
+n
Q = PV
dt
dt
or
H2 H1 + 1 (u 2 u 2 ) = 0
1
22
.
.
4m
m
=
u=
π ρ D2
Aρ
2.44 (a) By Eq. (2.32a):
By Eq. (2.24a):
Then
u2
2
u2
1
=
4
π
2
.
m2
ρ2
1
1
4
4
D1
D2
1
1
( P2 P1 ) +
2
ρ
.
Solve for m :
and given
4
π
2
π
.
m = 2ρ( P1 P2 )
4
.
m2
ρ2
2
H2 H1 =
4
4
D1 D2
44
D1 D2
44
D1 D2
4
4
D1 D2
1
( P2 P1 )
ρ
=0
1 /2
.
(b) Proceed as in part (a) with an extra term, Here solution for m yields:
.
m = 2 ρ ( P1 P2 ) ρ 2 C (T2 T1 )
π
4
2
44
D1 D2
4
4
D1 D2
1 /2
Because the quantity in the smaller square brackets is smaller than the leading term of the preceding result, the effect is to decrease the mass owrate.
627
Chapter 3 - Section B - Non-Numerical Solutions
3.2 Differentiate Eq. (3.2) with respect to P and Eq. (3.3) with respect to T :
πξ
πP
π
πT
T
=
P
πV
πP
1
V2
=
1
V2
πV
πT
T
P
πV
πT
πV
πP
+
P
T
π2V
πT π P
1
V
=ξ +
π2V
π PπT
= ξ
π2V
π PπT
π2V
π PπT
1
V
Addition of these two equations leads immediately to the given equation.
One could of course start with Eq. (3.4) and apply the condition for an exact differential, but this topic
is not covered until Chapter 6.
3.3 The Tait equation is given as:
V = V0 1
AP
B+P
where V0 , A, and B are constants. Application of Eq. (3.3), the denition of , requires the derivative
of this equation:
πV
πP
T
= V0
AV0
AP
A
=
+
2
B+P
(B + P)
B+P
1 +
P
B+P
Multiplication by 1/ V in accord with Eq. (3.3), followed by substitution for V0 / V by the Tait equation leads to:
AB
=
( B + P )[ B + (1 A) P ]
dV
= dP
V
Integration from the initial state ( P1 , V1 ) to an intermediate state ( P , V ) for constant
3.7 (a) For constant T , Eq. (3.4) becomes:
ln
Whence,
gives:
V
= ( P P1 )
V1
V = V1 exp[ ( P P1 )] = V1 exp( P ) exp( P1 )
If the given equation applies to the process, it must be valid for the initial state; then, A(T ) =
V1 exp( P1 ), and
V = A(T ) exp( P )
(b) Differentiate the preceding equation:
Therefore,
W =
=
V2
V1
A(T )
d V = A(T ) exp( P )d P
P d V = A(T )
P2
P1
P exp( P )d P
[( P1 + 1) exp( P1 ) ( P2 + 1) exp( P2 )]
628
With V1 = A(T ) exp(κ P1 ) and V2 = A(T ) exp(κ P2 ), this becomes:
W=
1
[(κ P1 + 1)V1 (κ P2 + 1)V2 ]
κ
W = P1 V1 P2 V2 +
or
V1 V2
κ
3.11 Differentiate Eq. (3.35c) with respect to T :
T
1δ
δ
P [(1δ)/δ]1
dT
dP
=T
+ P (1δ)/δ
dz
dz
dT
P (1δ)/δ d P
=0
+ P (1δ)/δ
dz
dz
P
1δ
δ
Algebraic reduction and substitution for d P /dz by the given equation yields:
dT
1δ
=0
(Mρ g ) +
dz
δ
T
P
For an ideal gas T ρ/ P = 1/ R . This substitution reduces the preceding equation to:
δ1
δ
Mg
dT
=
R
dz
3.12 Example 2.13 shows that U2 = H . If the gas is ideal,
H = U + P V = U + RT
For constant C V ,
and
U2 U = RT
U2 U = C V (T2 T )
and
C V (T2 T ) = RT
C P CV
R
T2 T
=
=
CV
CV
T
Whence,
T2 = γ T
When C P / C V is set equal to γ , this reduces to:
This result indicates that the nal temperature is independent of the amount of gas admitted to the
tank, a result strongly conditioned by the assumption of no heat transfer between gas and tank.
3.13 Isobaric case (δ = 0). Here, Eqs. (3.36) and (3.37) reduce to:
W = RT1 (1 1)
and
Q=
γ RT1
(1 1)
γ 1
Both are indeterminate. The easiest resolution is to write Eq. (3.36) and (3.37) in the alternative but
equivalent forms:
W=
RT1
δ1
T2
1
T1
and
Q=
(δ γ ) RT1
(δ 1)(γ 1)
T2
1
T1
from which we nd immediately for δ = 0 that:
W = R (T2 T1 )
and
Q=
629
γR
(T2 T1 ) = C P (T2 T1 )
γ 1
Isothermal case (δ = 1). Equations (3.36) and (3.37) are both indeterminate of form 0/0. Application
of lHˆ pitals rule yields the appropriate results:
o
W = RT1 ln
Note that if
P2
P1
y
P2
P1
and
(δ 1)/δ
Q = RT1 ln
(δ 1)/δ
P2
P1
1
dy
=2
δ
dδ
then
P2
P1
ln
P2
P1
Adiabatic case (δ = γ ). In this case simple substitution yields:
W=
P2
P1
RT1
γ 1
(γ 1)/γ
1
and
Q=0
Isochoric case (δ = ). Here, simple substitution yields:
W =0
and
Q=
RT1
P2
1 =
γ 1
P1
RT1
γ 1
T2
1 = C V (T2 T1 )
T1
3.14 What is needed here is an equation relating the heat transfer to the quantity of air admitted to the tank
and to its temperature change. For an ideal gas in a tank of total volume V t at temperature T ,
n1 =
P1 V t
RT
and
n2 =
P2 V t
RT
The quantity of air admitted to the tank is therefore:
V t ( P2 P1 )
RT
The appropriate energy balance is given by Eq. (2.29), which here becomes:
n=
( A)
d (nU )tank
˙
n H = Q
˙
dt
where the prime ( ) identies the entrance stream of constant properties. Multiplying by dt and integrating over the time of the process yields:
n 2 U2 n 1 U1 n H = Q
With n = n 2 n 1 ,
n 2 (U2 H ) n 1 (U1 H ) = Q
Because U2 = H2 RT and U1 = H1 RT , this becomes:
n 2 ( H2 H RT ) n 1 (U1 H RT ) = Q
or
n 2 [C P (T T ) RT ] n 1 [C P (T T ) RT ] = Q
Because n = n 2 n 1 , this reduces to:
Q = n [C P (T T ) RT ]
Given:
V t = 100, 000 cm3
T = 298.15 K
T = 318.15 K
630
P1 = 101.33 kPa
P2 = 1500 kPa
By Eq. ( A) with R = 8, 314 cm3 kPa mol1 K1 ,
n=
(100, 000)(1500 101.33)
= 56.425 mol
(8, 314)(298.15)
With R = 8.314 J mol1 K1 and C P = (7/2) R , the energy equation gives:
Q = (56.425)(8.314)
7
(298.15 318.15) 298.15 = 172, 705.6 J
2
Q = 172.71 kJ
or
3.15 (a) The appropriate energy balance is given by Eq. (2.29), here written:
d (nU )tank
˙
n H = Q
˙
dt
where the prime ( ) identies the entrance stream of constant properties. Multiplying by dt and
integrating over the time of the process yields:
n 2 U2 n 1 U1 n H = Q
Since n = n 2 n 1 , rearrangement gives:
n 2 (U2 H ) n 1 (U1 H ) = Q
(b) If the gas is ideal,
H = U + P V = U + RT
Whence for an ideal gas with constant heat capacities,
U2 H = U2 U RT = C V (T2 T ) RT
Substitute R = C P C V :
Similarly,
and
Note also:
(c) If n 1 = 0,
(d) If in addition Q = 0,
Whence,
U2 H = C V T2 C V T C P T + C V T = C V T2 C P T
U1 H = C V T1 C P T
n 2 (C V T2 C P T ) n 1 (C V T1 C P T ) = Q
n2 =
P2 Vtank
RT2
n1 =
P1 Vtank
RT1
n 2 (C V T2 C P T ) = Q
C V T2 = C P T
and
T2 =
CP
CV
T
T2 = γ T
(e) 1. Apply the result of Part (d ), with γ = 1.4 and T = 298.15 K:
T2 = (1.4)(298.15) = 417.41 K
631
Then, with R = 83.14 bar cm3 mol1 K1 :
n2 =
(3)(4 × 106 )
P2 Vtank
= 345.8 mol
=
(83.14)(417.41)
RT2
2. Heat transfer between gas and tank is: Q = m tank C (T2 T )
where C is the specic heat of the tank. The equation of Part (c) now becomes:
n 2 (C V T2 C P T ) = m tank C (T2 T )
Moreover
n2 =
P2 Vtank
RT2
These two equations combine to give:
P2 Vtank
(C V T2 C P T ) = m tank C (T2 T )
RT2
With C P = (7/2) R and C V = C P R = (7/2) R R = (5/2) R , this equation becomes:
R
P2 Vtank
(5T2 7T ) = m tank C (T2 T )
2
RT2
Note: R in the denominator has the units of P V ; R in the numerator has energy units.
Given values in the appropriate units are:
m tank = 400 kg
C = 460 J mol1 kg1
T = 298.15 K
Vtank = 4 × 106 cm3
P2 = 3 bar
Appropriate values for R are therefore:
R (denominator) = 83.14 bar cm3 mol1 K1
R (numerator) = 8.314 J mol1 K1
Numerically,
8.314
(3)(4 × 106 )
= (400)(460)(T2 298.15)
[(5)(T2 ) (7)(298.15)]
2
(83.14)(T2 )
Solution for T2 is by trial, by an iteration scheme, or by the solve routine of a software package.
The result is T2 = 304.217 K. Then,
n2 =
(3)(4 × 106 )
P2 Vtank
= 474.45 mol
=
(83.14)(304.217)
RT2
3.16 The assumption made in solving this problem is that the gas is ideal with constant heat capacities.
The appropriate energy balance is given by Eq. (2.29), here written:
d (nU )tank
˙
+Hn =Q
˙
dt
Multiplied by dt it becomes:
d (nU ) + H d n = d Q
632
where n and U refer to the contents of the tank, and H and n refer to the exit stream. Since the stream
bled from the tank is merely throttled, H = H , where H is the enthalpy of the contents of the tank.
By material balance, dn = dn . Thus,
n dU + U dn H dn = Q
Also,
dU = C V dT
or
n dU ( H U )dn = d Q
H U = P V = RT
d Q = mC dT
where m is the mass of the tank, and C is its specic heat.
nC V dT RT dn = mC dT
Thus,
R d (nC V + mC )
R d (nC V )
R
dT
=
dn =
=
C V nC V + mC
C V nC V + mC
nC V + mC
T
or
Integration yields:
ln
T2
T1
=
n 2 C V + mC
R
ln
n 1 C V + mC
CV
T2
=
T1
or
In addition,
n1 =
n 2 C V + mC
n 1 C V + mC
P1 Vtank
RT1
and
R/CV
P2 Vtank
RT2
n2 =
These equations may be solved for T2 and n 2 . If mC >>> nC V , then T2 = T1 . If mC = 0, then we
recover the isentropic expansion formulas.
3.27 For an ideal gas,
U = CV T
P V = RT
Whence,
But
U=
1
CV
CV
=
=
γ 1
C P CV
R
CV
R
The isothermal work is then:
Whence,
W =
W = RT ln
3.29 Solve the given equation of state for V :
U=
Therefore :
dV =
V1
P2
P1
V=
P d V = RT
P2
P1
1
dP
P
Compared with Eq. (3.27)
V=
633
1
(PV )
γ 1
RT
+ B RT
P
RT
dP
P2
V2
T
(PV )
3.28 Since Z = P V / RT the given equation can be written:
Differentiate at constant T :
(PV ) = R
θ
RT
+b
RT
P
V
P
Whence,
By denition [Eq. (3.3)]:
T
V
P
1
V
κ
RT
P2
=
T
Substitution for both V and the derivative yields:
RT
θ
RT
+b
RT
P
κ=
P2
Solve the given equation of state for P :
RT
P=
V b+
P
T
Differentiate:
R
=
+
θ
V b+
RT
V
θ
RT
dθ
θ
dT
T
V b+
θ
RT
2
By the equation of state, the quantity in parentheses is RT / P ; substitution leads to:
P
T
V
P
+
=
T
P
RT
2
dθ
θ
dT
T
3.31 When multiplied by V / RT , Eq. (3.42) becomes:
Z=
a (T )V / RT
V
a (T )V / RT
V
2
=
V b V + ( + σ )bV + σ b2
V b (V + b)(V + σ b)
Substitute V = 1/ρ :
Z=
1
a (T )ρ
1
RT 1 + ( + σ )bρ + σ (bρ)2
1 bρ
Expressed in series form, the rst term on the right becomes:
1
= 1 + bρ + (bρ)2 + · · ·
1 bρ
The nal fraction of the second term becomes:
1
= 1 ( + σ )bρ + [( + σ )2 σ ](bρ)2 + · · ·
1 + ( + σ )bρ + σ (bρ)2
Combining the last three equations gives, after reduction:
Z =1+ b
Equation (3.12) may be written:
Comparison shows:
B =b
( + σ )a (T )b 2
a (T )
ρ + ···
ρ + b2 +
RT
RT
Z = 1 + Bρ + Cρ 2 + · · ·
a (T )
RT
and
634
C = b2 +
( + σ )ba (T )
RT
For the Redlich/Kwong equation, the second equation becomes:
C = b2 +
a (T )
ba (T )
=b b+
RT
RT
Values for a (T ) and b are found from Eqs. (3.45) and (3.46), with numerical values from Table 3.1:
b=
0.42748 RTc
a (T )
=
Tr1.5 Pc
RT
0.08664 RTc
Pc
The numerical comparison is an open-ended problem, the scope of which must be decided by the
instructor.
3.36 Differentiate Eq. (3.11):
Z
P
= B + 2C P + 3 D P 2 + · · ·
T
Z
P
Whence,
=B
T , P =0
Z = 1 + Bρ + Cρ 2 + Dρ 3 + · · ·
Equation (3.12) with V = 1/ρ :
Z
ρ
Differentiate:
= B + 2Cρ + 3 Dρ 2 + · · ·
T
Z
ρ
Whence,
=B
T ,ρ =0
3.56 The compressibility factor is related to the measured quantities by:
Z=
By Eq. (3.39),
(a) By Eq. ( A),
Thus
M PV t
PV t
=
m RT
n RT
B = ( Z 1)V =
( A)
( Z 1) M V t
m
dT
dm
dV t
dP
dM
dZ
+ t
+
=
T
m
V
P
M
Z
(B)
(C )
Max |% δ Z | |% δ M | + |% δ P | + |% δ V t | + |% δ m | + |% δ T |
Assuming approximately equal error in the ve variables, a ±1% maximum error in Z requires
errors in the variables of <0.2%.
(b) By Eq. ( B ),
By Eq. (C ),
dm
dM
dV t
Z dZ
dB
+ t+
=
m
M
V
Z 1 Z
B
Z
dB
=
Z 1
B
dT
dP
T
P
Therefore
635
+
2Z 1
Z 1
dm
dM
dV t
+
t
m
M
V
Max |% δ B |
+
Z
Z 1
|% δ P | + |% δ T |
2Z 1
Z 1
|% δ V t | + |% δ M | + |% δ m |
For Z 0.9 and for approximately equal error in the ve 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 innity.
3.57 The Redlich/Kwong equation has the following equivalent forms, where a and b are constants:
Z=
a
V
3/2 ( V + b)
RT
V b
P=
a
RT
1 /2
V b T V ( V + b)
From these by differentiation,
Z
V
P
V
=
T
a (V b)2 b RT 3/2 (V + b)2
RT 3/2 (V b)2 (V + b)2
( A)
a (2V + b)(V b)2 RT 3/2 V 2 (V + b)2
T 1/2 V 2 (V b)2 (V + b)2
(B)
=
T
In addition, we have the mathematical relation:
( Z / V )T
( P / V )T
(C )
aV 2 (V b)2 b RT 3/2 V 2 (V + b)2
a RT (2V + b)(V b)2 R 2 T 5/2 V 2 (V + b)2
( D)
Z
P
=
T
Combining these three equations gives
Z
P
=
T
lim
For P 0, V , and Eq. ( D ) becomes:
P 0
lim
For P , V b, and Eq. ( D ) becomes:
P
Z
P
Z
P
=
T
b a / RT 3/2
RT
=
T
b
RT
3.60 (a) Differentiation of Eq. (3.11) gives:
Z
P
= B + 2C P + 3 D P 2 + ·
whence
T
If the limiting value of the derivative is zero, then B = 0, and
lim
P 0
Z
P
=B
T
B = B RT = 0
(b) For simple uids, ω = 0, and Eqs. (3.52) and (3.53) combine to give B 0 = B Pc / RTc . If B = 0,
then by Eq. (3.65),
0.422
B 0 = 0.083 1.6 = 0
Tr
636
and
Tr =
0.422
0.083
(1/1.6)
= 2.763
3.63 Linear isochores require that (γ P /γ T )V = Constant.
γP
γT
(a) By Eq. (3.4) applied to a constant-V process:
γP
γT
(b) For an ideal gas P V = RT , and
=
V
=
V
β
κ
R
V
(c) Because a and b are constants, differentiation of Eq. (3.42) yields:
γP
γT
=
V
R
V b
In each case the quantities on the right are constant, and so therefore is the derivative.
3.64 (a) Ideal gas: Low P , or low ρ , or large V and/or high T . See Fig. 3.15 for quantitative guidance.
(b) Two-term virial equation: Low to modest P . See Fig. 3.14 for guidance.
(c) Cubic EOS: Gases at (in principle) any conditions.
(d) Lee/Kesler correlation: Same as (c), but often more accurate. Note that corresponding states
correlations are strictly valid for non-polar uids.
(e) Incompressible liquids: Liquids at normal T s and P s. Inappropriate where changes in V are
required.
(f ) Rackett equation: Saturated liquids; a corresponding states application.
(g) Constant β , κ liquids: Useful where changes in V are required. For absolute values of V , a
reference volume is required.
(h) Lydersen correlation for liquids: a corresponding-states method applicable to liquids at extreme
conditions.
3.66 Write Eq. (3.12) with 1/ρ substituted everywhere for V . Subtract 1 from each side of the equation
and divide by ρ . Take the limit as ρ 0.
3.68 Follow the procedure laid out on p. 93 with respect to the van der Waals equation to obtain from
Eq. (3.42) the following three more-general equations:
1 + (1 σ )
σ
2
( + σ) (
σ
2
(
= 3Zc
+ 1) +
+ 1) +
2
= 3Zc
3
= Zc
where by denition [see Eqs. (3.45) and (3.46)]:
b Pc
RTc
and
ac Pc
R 2 Tc2
For a given EOS, and σ are xed, and the above set represents 3 equations in 3 unknowns, , ,
and Z c . Thus, for a given EOS the value of Z c is preordained, unrelated to experimental values of Z c .
637
PROPRIETARY MATERIAL. © 2005 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers
and educators for course preparation. If you are a student using this Manual, you are using it without permission.
(a, b) For the Redlich/Kwong and Soave/Redlich/Kwong equations, = 0 and σ = 1. Substitution
of these values into the 3-equation set allows their solution to yield:
Zc =
1
3
= 0.086640
= 0.427480
(c) For the Peng/Robinson equation, = 1 2 and σ = 1 + 2. As for the Soave and SRK
equations the 3-equation set can be solved (with considerably greater difculty) to yield:
Z c = 0.30740
3.69 Equation (3.12):
Z = 1 + Bρ + Cρ 2 + . . .
Eliminate ρ :
Z =1+
= 0.077796
Z =1+
= 0.457236
where
ρ = P / Z RT
C P2
BP
+ 2 2 2 + ...
ZRT
Z RT
2
P2
C P2
B Pc Pr
ˆ
ˆ Pr + C · Pr + . . .
+ 2 c2 · 2 r 2 + . . . = 1 + B ·
·
Z 2 Tr2
Z Tr
R Tc Z Tr
RTc Z Tr
( Z 1) Z Tr
ˆ
ˆ Pr + . . .
= B +C ·
Z Tr
Pr
Rearrange:
ˆ
B = lim ( Z 1) Z Tr / Pr
Pr 0
3.74 In a cylinder lled with 1 mole of an ideal gas, the molecules have kinetic energy only, and for a given
T and P occupy a volume V ig .
(a) For 1 mole of a gas with molecules having kinetic energy and purely attractive interactions at the
same T and P , the intermolecular separations are smaller, and V < V ig . In this case Z < 1.
(b) For 1 mole of a gas with molecules having kinetic energy and purely repulsive interactions at the
same T and P , the intermolecular separations are larger, and V > V ig . In this case Z > 1.
(c) If attractive and repulsive interactions are both present, they tend to cancel each other. If in balance, then the average separation is the same as for an ideal gas, and V = V ig . In this case
Z = 1.
3.75 van der Waals EOS:
Set V = 1/ρ :
whence
P=
a
RT
2
V b V
Z=
Z rep =
Z=
a
V
V b V RT
aρ
bρ
aρ
1
=1+
RT
1 bρ
RT
1 bρ
bρ
1 bρ
Z attr =
638
aρ
RT
3.76 Write each modication in Z -form,
(a)
Z=
a
V
RT
V b
lim Z = 1
V
The required behavior is:
(b)
Z=
lim Z = 1
V
a
V
2
RT
( V b)
lim Z =
V
Z=
V
a
1
V b V RT
lim Z = 0
V
lim Z = 1
The required behavior is:
(d )
Z =1
a
RT
lim Z = 1
The required behavior is:
(c)
a
RT
V
aρ
a
=1
RT
V RT
Although lim Z = 1 as required, the equation makes Z linear in ρ ; i.e., a 2-term virial EOS in
V
ρ . Such an equation is quite inappropriate at higher densities.
3.77 Refer to Pb. 2.43, where the general equation was developed;
For an ideal gas,
n=
Also for an ideal gas,
PV t
RT
dn
=
dt
and
dU = C V dT
whence
.
PV t
Q = RT
RT 2
ln
Integration yields:
3.78 By Eq. (3.4),
Integrate:
ln
PV t
RT 2
D2
Vt
V2
= ln 2t = ln 2 = ln
2
V1
V1
D1
Note that P V t / R = const.
dT
dU
= CV
dt
dt
P V t dT
dT
PV t
dT
= CP
CV
+
RT dt
dt
RT
dt
t2
R
T2
=
CP PV t
T1
dV
= β dT κ d P
V
dT
dt
.
dU
dn
+n
Q = PV
dt
dt
t1
.
Q dt
where β and κ are average values
D1 + δ D
D1
2
= ln 1 +
δD
D1
2
= β(T2 T1 ) κ( P2 P1 )
ln(1.0035)2 = 250 × 106 (40 10) 45 × 106 ( P2 6)
Solution for P2 yields:
P2 = 17.4 bar
639
Chapter 4 - Section B - Non-Numerical Solutions
4.5 For consistency with the problem statement, we rewrite Eq. (4.8) as:
CP = A +
C
B
T1 (ν + 1) + T12 (ν 2 + ν + 1)
3
2
where ν T2 / T1 . Dene C Pam as the value of C P evaluated at the arithmetic mean temperature Tam .
Then:
2
C Pam = A + BTam + C Tam
where
Tam
T1 (ν + 1)
T1 ν + T1
T2 + T1
=
=
2
2
2
and
2
Tam =
T12 2
(ν + 2ν + 1)
4
C
B
T1 (ν + 1) + T12 (ν 2 + 2ν + 1)
4
2
Dene ε as the difference between the two heat capacities:
C Pam = A +
Whence,
ν 2 + ν + 1 ν 2 + 2ν + 1
4
3
ε C P C Pam = C T12
C T12
(ν 1)2
12
Making the substitution ν = T2 / T1 yields the required answer.
This readily reduces to:
ε=
4.6 For consistency with the problem statement, we rewrite Eq. (4.8) as
CP = A +
D
B
T1 (ν + 1) +
2
ν T12
where ν T2 / T1 . Dene C Pam as the value of C P evaluated at the arithmetic mean temperature Tam .
Then:
D
C Pam = A + BTam + 2
Tam
As in the preceding problem,
Tam =
Whence,
T1 (ν + 1)
2
C Pam = A +
and
2
Tam =
T12 2
(ν + 2ν + 1)
4
4D
B
T1 (ν + 1) + 2 2
2
T1 (ν + 2ν + 1)
Dene ε as the difference between the two heat capacities:
ε C P C Pam =
This readily reduces to:
ε=
D
T12
D
T12 ν
4
1
2
ν + 2ν + 1
ν
ν 1
ν +1
2
Making the substitution ν = T2 / T1 yields the required answer.
640
4.8 Except for the noble gases [Fig. (4.1)], C P increases with increasing T . Therefore, the estimate is
likely to be low.
4.27 (a) When the water formed as the result of combustion is condensed to a liquid product, the resulting
latent-heat release adds to the heat given off as a result of the combustion reaction, thus yielding a
higher heating value than the lower heating value obtained when the water is not condensed.
(b) Combustion of methane(g) with H2 O(g) as product (LHV):
C(s) + O2 (g) CO2 (g)
H298 = 393,509
2H2 (g) + O2 (g) 2H2 O(g)
H298 = (2)(241,818)
CH4 (g) C(s) + 2H2 (g)
CH4 (g) + 2O2 (g) CO2 (g) + 2H2 O(g)
H298 = 74,520
H298 = 802,625 J (LHV)
Combustion of methane(g) with H2 O(l) as product (HHV):
CH4 (g) + 2O2 (g) CO2 (g) + 2H2 O(g)
2H2 O(g) 2H2 O(l)
H298 = 802,625
H298 = (2)(44,012)
CH4 (g) + 2O2 (g) CO2 (g) + 2H2 O(l)
H298 = 890,649 J (HHV)
(c) Combustion of n-decane(l) with H2 O(g) as product (LHV):
10 C(s) + 10 O2 (g) 10 CO2 (g)
H298 = (10)(393,509)
11 H2 (g) + 5 1 O2 (g) 11 H2 O(g)
2
H298 = (11)(241,818)
C10 H22 (l) 10 C(s) + 11 H2 (g)
C10 H22 (l) + 15 1 O2 (g) 10 CO2 (g) + 11 H2 O(g)
2
H298 = 249,700
H298 = 6,345,388 J (LHV)
Combustion of n-decane(l) with H2 O(l) as product (HHV):
C10 H22 (l) + 15 1 O2 (g) 10 CO2 (g) + 11 H2 O(g)
2
11 H2 O(g) 11 H2 O(l)
C10 H22 (l) + 15 1 O2 (g) 10 CO2 (g) + 11 H2 O(l)
2
4.49 Saturated because the large
H298 = 6,345,388
H298 = (11)(44,012)
H298 = 6,829,520 J (HHV)
H l v overwhelms the sensible heat associated with superheat.
Water because it is cheap, available, non-toxic, and has a large
H lv .
The lower energy content is a result of the decrease in H l v with increasing T , and hence P .
However, higher pressures allow higher temperature levels.
641
Chapter 5 - Section B - Non-Numerical Solutions
5.1
Shown to the right is a P V 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 πU = 0, and
therefore by the rst 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.
Q = πU t + π E K + π E P
5.5 The energy balance for the over-all process is written:
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:
t
π Stotal = π S t + π Ssurr
Just as πU t for the egg is zero, so is π S t . Therefore,
t
π Stotal = π Ssurr =
Q
Q surr
=
Tξ
Tξ
Since Q is negative, π Stotal is positive, and the process is irreversible.
T
= 1 TC
H
5.6 By Eq. (5.8) the thermal efciency of a Carnot engine is:
Differentiate:
TC TH
=
1
TH
and
TH TC
=
TC 1
TC
=
2
TH TH
TH
Since TC /TH is less unity, the efciency changes more rapidly with TC than with TH . So in theory it is
more effective to decrease TC . In practice, however, TC is xed by the environment, and is not subject
to control. The practical way to increase is to increase TH . Of course, there are limits to this too.
5.11 For an ideal gas with constant heat capacities, and for the changes T1 T2 and P1 P2 , Eq. (5.14)
can be rewritten as:
P2
T2
R ln
π S = C P ln
P1
T1
(a) If P2 = P1 ,
Whence,
π S P = C P ln
If V2 = V1 ,
T2
T1
π SV = C P ln
T2
T1
R ln
642
T2
P2
=
T1
P1
T2
T1
T2
T1
= C V ln
Since C P > C V , this demonstrates that
(b) If
T2 = T1 ,
ST = R ln
Whence,
SP >
SV .
P2
P1
If
P2
P1
SV = C P ln
This demonstrates that the signs for
R ln
ST and
V2 = V1 ,
P2
P1
= C V ln
P2
T2
=
P1
T1
P2
P1
SV are opposite.
5.12 Start with the equation just preceding Eq. (5.14) on p. 170:
ig
ig
dP
C dT
C dT
dS
d ln P = P
=P
P
RT
RT
R
For an ideal gas P V = RT , and ln P + ln V = ln R + ln T . Therefore,
dT
dV
dP
=
+
T
V
P
ig
dV
dT
C dT
dS
=
+
=P
V
T
RT
R
Whence,
dV
dT
dP
=
V
T
P
or
ig
dT
CP
+ d ln V
1
T
R
ig
ig
Because (C P / R ) 1 = C V / R , this reduces to:
ig
C dT
dS
+ d ln V
=V
RT
R
S
=
R
Integration yields:
T
T0
ig
V
C V dT
+ ln
V0
RT
**********************
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 = d P / P + d V / V :
ig
ig
ig
ig
C dV
C dP
dP
C dV
C dP
dS
+P
=V
+P
=P
RV
RP
P
RV
RP
R
ig
Integration yields:
5.13
ig
V
C
C
P
S
+ P ln
= V ln
V0
R
P0
R
R
As indicated in the problem statement the
basic differential equations are:
d W d Q H d QC = 0
( A)
TH
d QH
=
TC
d QC
(B)
where Q C and Q H refer to the reservoirs.
643
t
t
(a) With d Q H = C H dTH and d Q C = CC dTC , Eq. ( B ) becomes:
t
TH
C H dTH
=
t
TC
CC dTC
Whence,
C t dTH
dTC
= H
t
C C TH
TC
or
d ln TC = d ln TH
where
t
CH
t
CC
Integration from TH0 and TC0 to TH and TC yields:
TH
TH0
TC
=
TC0
TH
TH0
TC = TC0
or
t
t
(b) With d Q H = C H dTH and d Q C = CC dTC , Eq. ( A) becomes:
t
t
d W = C H dTH + CC dTC
t
t
W = C H (TH TH0 ) + CC (TC TC0 )
Integration yields:
Eliminate TC by the boxed equation of Part (a) and rearrange slightly:
W=
TH
TH0
TH
t
1 + CC TC0
TH0
t
C H TH0
1
(c) For innite time, TH = TC T , and the boxed equation of Part (a) becomes:
T = TC0
From which:
T
T = (TC0 )1/(
Because
/(
+1)
(TH0 )
TC0
TH0
+1
/( +1)
+ 1) 1 = 1/(
T
=
TH0
T
TH0
= TC0
TH0
T
= TC0 (TH0 )
and
T
= (TC0 )1/(
TH0
+1)
(TH0 )
/( +1)1
+ 1), then:
1/( +1)
and
T
TH0
=
TC0
TH0
/( +1)
Because TH = T , substitution of these quantities in the boxed equation of Part (b) yields:
W=
5.14
t
C H TH0
TC0
TH0
1/( +1)
1 +
As indicated in the problem statement the
basic differential equations are:
d W d Q H d QC = 0
( A)
TH
d QH
=
TC
d QC
(B)
where Q C and Q H refer to the reservoirs.
644
t
CC TC0
TC0
TH0
/( +1)
1
t
(a) With d Q C = CC dTC , Eq. ( B ) becomes:
TH
d QH
=
t
TC
CC dTC
t
d Q H = CC
or
TH
dTC
TC
Substitute for d Q H and d Q C in Eq. ( A):
t
d W = CC TH
Integrate from TC0 to TC :
t
W = CC TH ln
TC
t
+ CC (TC TC0 )
TC0
dTC
t
+ CC dTC
TC
t
W = CC TH ln
or
TC0
+ TC TC0
TC
(b) For innite time, TC = TH , and the boxed equation above becomes:
t
W = CC TH ln
TC0
+ TH TC0
TH
.
5.15 Write Eqs. (5.8) and (5.1) in rate form and combine to eliminate | Q H |:
.
.
.
.
|W |
TC
|W |
= |W | + | Q |
= 1r
or
.
. =1
1r
TH
|W | + | Q C |
.
With | Q C | = k A(TC )4 = k A(r TH )4 , this becomes:
.
|W |
.
r
1
1 = |W |
1r
1r
= k Ar 4 (TH )4
or
where
A=
Differentiate, noting that the quantity in square brackets is constant:
.
.
|W |
1
3
|W |
dA
=
+
=
k ( T H )4
(1 r )r 4 (1 r )2r 3
k (TH )4
dr
Equating this equation to zero, leads immediately to:
4r = 3
.
|W |
k (TH )4
r
TC
TH
1
(1 r )r 3
4r 3
(1 r )2r 4
r = 0.75
or
5.20 Because W = 0, Eq. (2.3) here becomes:
Q=
U t = mC V 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).
5.22 An appropriate energy balance here is:
Q=
Ht = 0
Applied to the process described, with T as the nal temperature, this becomes:
m 1 C P (T T1 ) + m 2 C P (T T2 ) = 0
whence
If m 1 = m 2 ,
645
T=
T = (T1 + T2 )/2
m 1 T1 + m 2 T2
m1 + m2
(1)
The total entropy change as a result of temperature changes of the two masses of water:
T
T
+ m 2 C P ln
T2
T1
Equations (1) and (2) represent the general case. If m 1 = m 2 = m ,
S t = m 1 C P ln
S t = mC P ln
Because T = (T1 + T2 )/2 >
T2
T1 T2
S t = 2mC P ln
or
(2)
T
T1 T2
S t is positive.
T1 T2 ,
5.23 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 amd 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
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.
T
T0
C P dT
T0
T
C P dT
T0 T
T T0
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 C P H is positive.
5.24 By denition,
CP
Similarly,
CP
H
=
dT
dT
T0
T CP
T
T=
=
ln(T0 / T )
ln(T / T0 )
T
T0
S
=
CP
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 C P S is positive.
When T = T0 , both the numerators and denominators of the above fractions become zero, and the
fractions are indeterminate. Application of lHˆ pitals rule leads to the result: C P H = C P S = C P .
o
5.31 The process involves three heat reservoirs: the house, a heat sink; the furnace, a heat source; and the
surroundings, a heat source. Notation is as follows:
|Q|
|Q F |
|Qσ |
Heat transfer to the house at temperature T
Heat transfer from the furnace at TF
Heat transfer from the surroundings at Tσ
The rst and second laws provide the two equations:
|Q| = |Q F | + |Qσ |
and
646
|Qσ |
|Q| |Q F |
=0
Tσ
TF
T
Combine these equations to eliminate | Q σ |, and solve for | Q F |:
|Q F | = |Q|
With
T = 295 K
TF = 810 K
T Tσ
TF Tσ
TF
T
Tσ = 265 K
and | Q | = 1000 kJ
| Q F | = 151.14 kJ
The result is:
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 (| Q 1 |) and by
the Carnot refrigerator (| Q 2 |) together provide the
heat | Q | for the house. The energy balances for
the engine and refrigerator are:
|W |engine = | Q F | | Q 1 |
|W |refrig = | Q 2 | | Q σ |
Equation (5.7) may be applied to both the engine
and the refrigerator:
Tσ
|Qσ |
TF
|Q F |
=
=
T
| Q2|
T
| Q1|
Combine the two pairs of equations:
|W |engine = | Q 1 |
TF T
TF
1 = | Q1|
T
T
|W |refrig = | Q 2 | 1
Tσ
T
= | Q2|
T Tσ
T
Since these two quantities are equal,
| Q1|
T Tσ
TF T
= | Q2|
T
T
or
| Q2| = | Q1|
Because the total heat transferred to the house is | Q | = | Q 1 | + | Q 2 |,
| Q| = | Q1| + | Q1|
But
| Q1| = | Q F |
T
TF
TF T
TF T
= | Q1| 1 +
T Tσ
T Tσ
whence
|Q| = |Q F |
= | Q1|
T
TF
TF T
T Tσ
TF Tσ
T Tσ
TF Tσ
T Tσ
Solution for | Q F | yields the same equation obtained more easily by direct application of the two laws
of thermodynamics to the overall result of the process.
5.32 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:
647
Heat transfer from the tank at temperature T
Heat transfer from the house at T
Heat transfer to the surroundings at Tσ
|Q|
|Q |
|Qσ |
The rst and second laws provide the two equations:
|Q| + |Q | = |Qσ |
|Q |
|Qσ | |Q|
=0
T
T
Tσ
and
Combine these equations to eliminate | Q σ |, and solve for | Q |:
Tσ T
T Tσ
|Q| = |Q |
With
T = 448.15 K
T = 297.15 K
T
T
Tσ = 306.15 K
and | Q | = 1500 kJ
| Q | = 143.38 kJ
The result is:
Shown to the right 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:
|W |engine = | Q | | Q σ1 |
|W |refrig = | Q σ2 | | Q |
Equation (5.7) may be applied to both the engine
and the refrigerator:
Tσ
| Q σ1 |
=
T
|Q|
Tσ
| Q σ2 |
=
T
|Q |
|W |engine = | Q | 1
Tσ
T
Combine the two pairs of equations:
= |Q|
|W |refrig = | Q |
T Tσ
T
Tσ
T
= |Q |
|Q| = |Q |
Tσ T
T Tσ
T
T
Since these two quantities are equal,
|Q|
Tσ T
T Tσ
= |Q |
T
T
or
5.36 For a closed system the rst term of Eq. (5.21) is zero, and it becomes:
.
.
Qj
d (m S )cv
= SG 0
+
Tσ, j
dt
j
648
Tσ t
T
.
where Q j is here redened to refer to the system rather than to the surroundings. Nevertheless, the sect
ond term accounts for the entropy changes of the surroundings, and can be written simply as d Ssurr /dt :
t
.
d (m S )cv d Ssurr
= SG 0
dt
dt
or
T
t
.
d Scv d Ssurr
= SG 0
dt
dt
Multiplication by dt and integration over nite time yields:
t
Scv +
t
Ssurr 0
or
Stotal 0
5.37 The general equation applicable here is Eq. (5.22):
.
( S m )fs
j
.
.
Qj
= SG 0
Tσ, j
(a) For a single stream owing within the pipe and with a single heat source in the surroundings, this
becomes:
.
.
Q
.
= SG 0
( S )m
Tσ
(b) The equation is here written for two streams (I and II) owing in two pipes. Heat transfer is
.
internal, between the two streams, making Q = 0. Thus,
.
.
.
( S )I m I + ( S )II m II = SG 0
(c) For a pump operatiing on a single stream and with the assumption of negligible heat transfer to
the surroundings:
.
.
( S )m = SG 0
(d) For an adiabatic gas compressor the result is the same as for Part (c).
(e) For an adiabatic turbine the result is the same as for Part (c).
(f ) For an adiabatic throttle valve the result is the same as for Part (c).
(g) For an adiabatic nozzle the result is the same as for Part (c).
5.40 The gure on the left below indicates the direct, irreversible transfer of heat | Q | from a reservoir at T1
to a reservoir at T2 . The gure on the right depicts a completely reversible process to accomplish the
same changes in the heat reservoirs at T1 and T2 .
649
The entropy generation for the direct heat-transfer process is:
1
1
T1
T2
SG = | Q |
T1 T2
T1 T2
= |Q|
For the completely reversible process the net work produced is Wideal :
T1 Tσ
T1
| W1 | = | Q |
and
| W2 | = | Q |
Wideal = |W1 | |W2 | = Tσ | Q |
T2 Tσ
T2
T1 T2
T1 T2
This is the work that is lost, Wlost , in the direct, irreversible transfer of heat | Q |. Therefore,
Wlost = Tσ | Q |
T1 T2
= Tσ SG
T1 T2
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 .
5.45 Equation (5.14) can be written for both the reversible and irreversible processes:
Sirrev =
By difference, with
Since
Tirrev
T0
ig
CP
Srev = 0:
P
dT
ln
P
T
Sirrev =
Srev =
Tirrev
Trev
ig
CP
Trev
T0
dT
T
Sirrev must be greater than zero, Tirrev must be greater than Trev .
650
ig
CP
P
dT
ln
P
T
Chapter 6 - Section B - Non-Numerical Solutions
νH
νS
6.1 By Eq. (6.8),
and isobars have positive slope
=T
P
ν2 H
ν S2
Differentiate the preceding equation:
ν2 H
ν S2
Combine with Eq. (6.17):
νT
νS
=
P
T
CP
=
P
P
and isobars have positive curvature.
6.2 (a) Application of Eq. (6.12) to Eq. (6.20) yields:
νCP
νP
νCP
νP
or
ν {V T (ν V /ν T ) P }
νT
=
T
νV
νT
=
T
νCP
νP
Whence,
νV
νT
For an ideal gas:
=
P
ν2V
νT 2
T
P
= T
T
R
P
P
νV
νT
P
ν2V
νT 2
P
ν2V
νT 2
and
P
=0
P
(b) Equations (6.21) and (6.33) are both general expressions for d S , and for a given change of state
both must give the same value of d S . They may therefore be equated to yield:
(C P C V )
dT
=
T
νP
νT
νV
νT
By Eqs. (3.2) and (6.34):
V
νP
νT
C P = CV + T
Restrict to constant P :
dV +
= εV
V
νV
νT
dP
P
P
νP
νT
and
P
C P CV = ε T V
Combine with the boxed equation:
νV
νT
=
V
ε
ρ
ε
ρ
6.3 By the denition of H , U = H P V . Differentiate:
νU
νT
=
P
νH
νT
P
P
νV
νT
or
P
651
νU
νT
= CP P
P
νV
νT
P
Substitute for the nal derivative by Eq. (3.2), the denition of β :
U
T
= CP β PV
P
Divide Eq. (6.32) by dT and restrict to constant P . The immediate result is:
U
T
P
T
= CV + T
P
P
V
V
T
P
Solve for the two derivatives by Eqs. (6.34) and (3.2); substitution gives:
U
T
6.4 (a) In general,
= CV +
P
β
(β T κ P )V
κ
P
T
dU = C V dT + T
By the equation of state,
P=
RT
V b
P dV
(6.32)
V
P
T
whence
=
V
P
R
=
T
V b
Substituting this derivative into Eq. (6.32) yields dU = C V dT , indicating that U = f (T )
only.
(b) From the denition of H ,
From the equation of state,
d H = dU + d ( P V )
d ( P V ) = R dT + b d P
Combining these two equations and the denition of part (a) gives:
d H = C V dT + R dT + b d P = (C V + R )dT + b d P
H
T
Then,
= CV + R
P
By denition, this derivative is C P . Therefore C P = C V + R . Given that C V is constant, then
so is C P and so is γ C P / C V .
(c) For a mechanically reversible adiabatic process, dU = d W . Whence, by the equation of state,
C V dT = P d V =
d ( V b)
RT
d V = RT
V b
V b
R
dT
d ln(V b)
=
CV
T
But from part (b), R / C V = (C P C V )/ C V = γ 1. Then
or
d ln T = (γ 1)d ln(V b)
From which:
d ln T + d ln(V b)γ 1 = 0
or
T (V b)γ 1 = const.
Substitution for T by the equation of state gives
P (V b)(V b)γ 1
= const.
R
652
or
P (V b)γ = const.
6.5 It follows immediately from Eq. (6.10) that:
V=
G
P
and
S=
T
G
T
P
Differentation of the given equation of state yields:
V=
RT
P
and
S=
d (T )
R ln P
dT
Once V and S (as well as G ) are known, we can apply the equations:
H = G +TS
and
U = H P V = H RT
These become:
H = (T ) T
d (T )
dT
and
U = (T ) T
d (T )
RT
dT
By Eqs. (2.16) and (2.20),
CP =
Because
H
T
and
CV =
P
U
T
V
is a function of temperature only, these become:
C P = T
d2
dT 2
and
C V = T
d2
R = CP R
dT 2
The equation for V gives the ideal-gas value. The equations for H and U show these properties to
be functions of T only, which conforms to ideal-gas behavior. The equation for S shows its relation
to P to be that of an ideal gas. The equations for C P and C V show these properties to be functions
of T only, which conforms to ideal-gas behavior, as does the result, C P = C V + R . We conclude
that the given equation of state is consistent with the model of ideal-gas behavior.
6.6 It follows immediately from Eq. (6.10) that:
V=
G
P
and
S=
T
G
T
P
Differentation of the given equation of state yields:
V=K
and
S=
d F (T )
dT
Once V and S (as well as G ) are known, we can apply the equations:
H = G +TS
and
U = H PV = H PK
These become:
H = F (T ) + K P T
d F (T )
dT
and
U = F (T ) T
By Eqs. (2.16) and (2.20),
CP =
H
T
and
P
653
CV =
U
T
V
d F (T )
dT
Because F is a function of temperature only, these become:
C P = T
d2 F
dT 2
and
C V = T
d2 F
= CP
dT 2
The equation for V shows it to be constant, independent of both T and P . This is the denition of
an incompressible uid. H is seen to be a function of both T and P , whereas U , S , C P , and C V are
functions of T only. We also have the result that C P = C V . All of this is consistent with the model
of an incompressible uid, as discussed in Ex. 6.2.
6.11 Results for this problem are given in the text on page 217 by Eqs. (6.61), (6.62) and (6.63) for G R ,
H R , and S R respectively.
6.12 Parameter values for the van der Waals equation are given by the rst line of Table 3.1, page 98. At
the bottom of page 215, it is shown that I = / Z . Equation (6.66b) therefore becomes:
q
GR
= Z 1 ln( Z )
Z
RT
For given T and P , Z is found by solution of Eq. (3.52) for a vapor phase or Eq. (3.56) for a liquid
phase with σ = δ = 0. Equations (3.53) and (3.54) for the van der Waals equation are:
=
Pr
8Tr
and
q=
27
8Tr
With appropriate substitutions, Eqs. (6.67) and (6.68) become:
q
HR
= Z 1
Z
RT
and
SR
= ln( Z )
R
6.13 This equation does not fall within the compass of the generic cubic, Eq. (3.42); so we start anew.
First, multiply the given equation of state by V / RT :
a
V
PV
exp
=
V RT
V b
RT
Substitute:
Then,
Z
PV
RT
V=
Z=
1
ρ
a
q
b RT
1
exp(qbρ)
1 bρ
With the denition, ξ bρ , this becomes:
Z=
Because ρ = P / Z RT ,
1
exp(q ξ )
1ξ
ξ=
bP
Z RT
Given T and P , these two equations may be solved iteratively for Z and ξ .
Because b is a constant, Eqs. (6.58) and (6.59) may be rewritten as:
654
( A)
GR
=
RT
ξ
0
( Z 1)
HR
=
RT
ξ
0
dξ
+ Z 1 ln Z
ξ
(B)
dξ
+ Z 1
ξ
(C )
Z
T
ξ
In these equations, Z is given by Eq. ( A), from which is also obtained:
ln Z = ln(1 ξ ) q ξ
Z
T
and
=
ξ
qξ
exp(q ξ )
T (1 ξ )
The integrals in Eqs. ( B ) and (C ) must be evaluated through the exponential integral, E (x ), a special
function whose values are tabulated in handbooks and are also found from such software packages
as MAPLE R . The necessary equations, as found from MAPLE R , are:
ξ
0
( Z 1)
dξ
= exp(q ){ E [q (1 ξ )] E (q )} E (q ξ ) ln(q ξ ) γ
ξ
where γ is Eulers constant, equal to 0.57721566. . . .
ξ
and
T
0
Z
T
ξ
sξ
= q exp(q ){ E [q (1 ξ )] E (q )}
ξ
Once values for G R / RT and H R / RT are known, values for S R / R come from Eq. (6.47). The
difculties of integration here are one reason that cubic equations have found greater favor.
6.18 Assume the validity for purposes of interpolation of Eq. (6.75), and write it for T2 , T , and T1 :
ln P2sat = A
B
T2
( A)
ln P sat = A
B
T
(B)
ln P1sat = A
B
T1
(C )
Subtract (C ) from ( A):
ln
P2sat
=B
P1sat
1
1
T2
T1
=B
(T2 T1 )
T1 T2
Subtract (C ) from ( B ):
ln
P sat
=B
P1sat
1
1
T
T1
=B
(T T1 )
T1 T
The ratio of these two equations, upon rearrangement, yields the required result.
B
T
( A)
B
Tc
(B)
6.19 Write Eq. (6.75) in log10 form:
log P sat = A
Apply at the critical point:
log Pc = A
655
By difference,
1
1
T
Tc
log Pr sat = B
=B
Tr 1
T
(C )
If P sat is in (atm), then application of ( A) at the normal boiling point yields:
B
Tn
log 1 = A
or
A=
=B
Tc Tn
Tn Tc
=B
Tn
1θ
B
Tn
log Pc
With θ Tn / Tc , Eq. ( B ) can now be written:
1
1
Tc
Tn
log Pc = B
Whence,
B=
Equation (C ) then becomes:
Tr 1
log Pc =
T
Tn
1θ
log Pr sat =
log( Pr sat )Tr =0.7 =
Apply at Tr = 0.7:
3
7
1θ
Tn
Tr 1
log Pc
Tr
θ
1θ
θ
1θ
log Pc
By Eq. (3.48),
ω = 1.0 log( Pr sat )Tr =0.7
Whence,
ω=
3
7
θ
1θ
log Pc 1
6.83 The slopes of isobars and isochores on a T S diagram are given by Eqs. (6.17) and (6.30):
T
S
=
P
T
CP
T
S
and
=
V
T
CV
Both slopes are necessarily positive. With C P > C V , isochores are steeper.
An expression for the curvature of isobars results from differentiation of the rst equation above:
2T
S2
=
P
1
CP
T
S
With C P = a + bT ,
P
CP
S
T
C2
P
CP
T
=
P
=b
T
T
2
2
CP CP
and
P
CP
T
1
P
T
CP
T
S
CP
T
=
P
T
T
1
2
CP
CP
=1
P
6.84 Division of Eq. (6.8) by d S and restriction to constant T yields:
Therefore,
=T +V
T
P
S
By Eq. (6.25),
T
H
S
=T
T
656
1
1
= (β T 1)
β
β
P
S
P
a
bT
=
a + bT
a + bT
Because this quantity is positive, so then is the curvature of an isobar.
H
S
CP
T
=
T
1
βV
2 H
S2
Also,
=
T
1
β2
β
S
=
T
1
β2
2 H
S2
Whence,
By Eqs. (3.2) and (3.38):
V
T
Whence,
β=
=
P
1
V
β
P
=
T
V
T
P
S
T
1
=
T
β
P
β3V
1
β2
β
P
T
and
V=
P
dB
R
+
dT
P
and
T
1
βV
β=
1
V
RT
+B
P
dB
R
+
dT
P
Differentiation of the second preceding equation yields:
β
P
=
T
V
P
From the equation of state,
β
P
Whence,
=
T
V
P
1
V2
dB
R
+
dT
P
R
V P2
=
T
=
T
1
R
(β V ) 2
V
V P2
V
P
T
RT
P2
R
β RT
R
(β T 1)
=
+
2
2
V P2
VP
VP
Clearly, the signs of quantity (β T 1) and the derivative on the left are the same. The sign is
determined from the relation of β and V to B and d B /dT :
βT 1 =
T
V
dB
R
+
dT
P
dB
dB
RT
B
T
+T
dT 1 = dT
1= P
RT
RT
+B
+B
P
P
In this equation d B /dT is positive and B is negative. Because RT / P is greater than | B |, the quantity
β T 1 is positive. This makes the derivative in the rst boxed equation positive, and the second
derivative in the second boxed equation negative.
6.85 Since a reduced temperature of Tr = 2.7 is well above normal temperatures for most gases, we
expect on the basis of Fig. 3.10 that B is () and that d B /dT is (+). Moreover, d 2 B /dT 2 is ().
GR = BP
By Eqs. (6.54) and (6.56),
S R = P (d B /dT )
and
Whence, both G R and S R are (). From the denition of G R , H R = G R + T S R , and H R is ().
V R = B,
By Eqs. (3.38) and (6.40),
and V R is ().
Combine the equations above for G R , S R , and H R :
HR = P B T
R
Therefore, C P
dB
dT
HR
T
Whence,
is (+).
HR
T
=P
P
(See Fig. 6.5.)
P
657
dB
d2 B
dB
T
2
dT
dT
dT
= PT
d2 B
dT 2
6.89 By Eq. (3.5) at constant T :
(a) Work
P =
dW = P dV =
W=
W=
=
1
κ
V
1
P1
ln
κ V1
1
1
V
1
P1 d V = ln V d V P1 + ln V1 d V
ln
κ
κ
κ V1
V2
V1
ln V d V P1 +
1
ln V1 (V2 V1 )
κ
1
1
[(V2 ln V2 V2 ) (V1 ln V1 V1 )] P1 (V2 V1 ) (V2 ln V1 V1 ln V1 )
κ
κ
V2
1
+ V1 V2 P1 (V2 V1 )
V2 ln
V1
κ
By Eq. (3.5),
ln
(b) Entropy
V2
= κ( P2 P1 )
V1
ln V1
ln V
P1
κ
κ
dS =
β
βV
d ln V = d V
κ
κ
and
1
d ln V
κ
d P =
S=
and
By Eq. (6.28),
Substitute for d P :
V2 V1
κ
d S = β V d P
P =
(c) Enthalpy
W = P1 V1 P2 V2
whence
By Eq. (6.29),
By Eq. ( A),
( A)
β
(V2 V1 )
κ
d H = (1 β T ) V d P
d H = ( 1 β T ) V ·
H=
1 βT
1
dV
d ln V =
κ
κ
1 βT
(V1 V2 )
κ
These equations are so simple that little is gained through use of an average V . For the conditions
given in Pb. 6.9, calculations give:
W = 4.855 kJ kg1
S = 0.036348 kJ kg1 K1
M
P
6.90 The given equation will be true if and only if
H = 134.55 kJ kg1
dP = 0
T
The two circumstances for which this condition holds are when ( M / P )T = 0 or when d P = 0.
The former is a property feature and the latter is a process feature.
6.91
ig
Neither C P
T
H ig
H ig
H ig
ig
= CP
+
=
T P P V
P T
P V
nor ( T / P )V is in general zero for an ideal gas.
H ig
P
=
S
H ig
P
+
T
658
H ig
T
P
T
P
T
P
ig
S
= CP
V
T
P
S
T
P
T
S ig
=
S
S ig
P
P
H ig
P
=
T
S ig
P
=T
S
S ig
P
T
ig
CP
T
T
ig
Neither T nor ( S / P )T is in general zero for an ideal gas. The difculty here is that the
expression independent of pressure is imprecise.
6.92 For S = S ( P , V ):
S
P
dS =
S
V
dP +
V
dV
P
By the chain rule for partial derivatives,
S
T
dS =
T
P
V
P
V
T
V
T
V
S
T
dP +
P
dV
P
With Eqs. (6.30) and (6.17), this becomes:
dS =
6.93 By Eq. (6.31),
P=T
(a) For an ideal gas,
Therefore
RT
RT
=
V
V
(b) For a van der Waals gas,
Therefore
P
T
RT
V
P=
T
P
CV
T
V
dP +
V
U
V
P
T
and
U
V
P=
CP
T
T
=
V
U
V
T
a
RT
2
V b V
U
V
=0
T
P
T
and
=
V
T
U
V
=
T
A = a (Tc ) · Tc
6.94 (a) The derivatives of G with respect to T and P follow from Eq, (6.10):
G
T
and
V=
P
Combining the denition of Z with the second of these gives:
Z
P
PV
=
RT
RT
659
G
P
T
=
T
(3/2) A
+ b)
T 1 /2 V ( V
1
2
S =
R
V b
U
V
and
(c) Similarly, for a Redlich/Kwong uid nd:
where
R
V
and
RT
a
RT
2=
V b
V b V
dV
G
P
T
a
V2
Combining Eqs. (2.11) and (3.63) and solving for U gives U = G + T S P V .
G
T
U =GT
Replacing S and V by their derivatives gives:
P
P
G
P
T
Developing an equation for C V is much less direct. First differentiate the above equation for U
with respect to T and then with respect to P : The two resulting equations are:
U
T
2G
T 2
= T
P
U
P
P
2
G
T P
= T
T
2G
T P
P
2G
P2
P
T
From the denition of C V and an equation relating partial derivatives:
U
T
CV
U
T
=
V
U
P
+
P
P
T
T
V
Combining the three equations yields:
C V = T
2G
T 2
P
P
2G
T P
T
2G
T P
2G
P2
+P
T
P
T
V
Evaluate ( P / T )V through use of the chain rule:
P
T
=
V
P
V
V
T
T
=
P
( V / T ) P
( V / P )T
The two derivatives of the nal term come from differentiation of V = ( G / P )T :
V
T
=
P
2G
PT
P
T
Then
and C V = T
2G
T 2
P
P
V
P
and
2G
T P
=
V
2G
P2
=
T
T
( 2 G / T ) P
( 2 G / P 2 )T
2G
T P
+T
+P
2G
P2
T
( 2 G / P T )
( 2 G / P 2 )T
Some algebra transforms this equation into a more compact form:
C V = T
2G
T 2
+T
P
( 2 G / T P )2
( 2 G / P 2 )T
(b) The solution here is analogous to that of part (a), but starting with the derivatives inherent in
Eq. (6.9).
6.97 Equation (6.74) is exact:
H lv
d ln P sat
=
R Z lv
d (1/ T )
The right side is approximately constant owing to the qualitatively similar behaviior of
Z l v . Both decrease monotonically as T increases, becoming zero at the critical point.
660
H l v and
H sl
S sl
d P sat
=
=
T V sl
V sl
dT
sl
V is assumed approximately constant, then
6.98 By the Clapeyron equation:
S sl to
If the ratio
P sat = A + BT
H sl to
If the ratio
V sl is assumed approximately constant, then
P sat = A + B ln T
6.99 By Eq, (6.73) and its analog for sv equilibrium:
=
dT
Because
Hts v
=
t
Htsl
t
sat
d Pl v
dT
t
Pt
RTt2
Hts v
Htl v
is positive, then so is the left side of the preceding equation.
H lv
d P sat
=
T V lv
dT
6.100 By Eq. (6.72):
But
Htl v
Pt Htl v
Pt Htl v
RTt2
RTt2 Z tl v
t
sat
d Pl v
dT
d Ps sat
v
Pt Hts v
Pt Hts v
RTt2
RTt2 Z ts v
=
d Ps sat
v
dT
V lv =
RT
P sat
Z lv
H lv
d ln P sat
=
RT 2 Z l v
dT
whence
(6.73)
lv
H
1
H lv
Tc H l v
d ln Pr sat
=
·2
=
=
2 Z lv
lv
2 Z lv
Tr
RTc Tr Z
RT
dTr
6.102 Convert αc to reduced conditions:
αc
d ln P sat
d ln T
=
T =Tc
d ln Prsat
d ln Tr
= Tr
Tr =1
d ln Prsat
dTr
=
Tr =1
d ln Prsat
dTr
Tr =1
From the Lee/Kesler equation, nd that
d ln Prsat
dTr
= 5.8239 + 4.8300 ω
Tr =1
Thus, αc (L/K) = 5.82 for ω = 0, and increases with increasing molecular complexity as quantied
by ω.
661
Chapter 7 - Section B - Non-Numerical Solutions
7.2 (a) Apply the general equation given in the footnote on page 266 to the particular derivative of interest
here:
S
T
T
=
S P P T
P S
The two partial derivatives on the right are found from Eqs. (6.17) and (6.16); thus,
T
P
=
S
V
T
T
CP
P
For gases, this derivative is positive. It applies to reversible adiabatic expansions and compressions
in turbines and compressors.
(b) Application of the same general relation (page 266) yields:
T
V
T
U
=
U
U
V
V
T
The two partial derivatives on the right are found from Eqs. (2.16) and (6.31); thus,
T
V
=
U
1
CV
P
T
PT
V
For gases, this may be positive or negative, depending on conditions. Note that it is zero for an
ideal gas. It applies directly to the Joule expansion, an adiabatic expansion of gas conned in a
portion of a container to ll the entire container.
7.3 The equation giving the thermodynamic sound speed appears in the middle of page 257. As written,
it implicitly requires that V represent specic volume. This is easily conrmed by a dimensional
analysis. If V is to be molar volume, then the right side must be divided by molar mass:
c2 =
V2
M
P
V
( A)
S
Applying the equation given in the footnote on page 266 to the derivative yields:
P
V
P
S
=
S
V
S
V
P
This can also be written:
P
V
=
S
P
T
V
T
S
S
T
V
P
T
V
=
P
T
S
V
S
T
P
P
T
V
T
V
P
Division of Eq. (6.17) by Eq. (6.30) shows that the rst product in square brackets on the far right is
the ratio C P /C V . Reference again to the equation of the footnote on page 266 shows that the second
product in square brackets on the far right is ( P / V )T , which is given by Eq. (3.3).
Therefore,
P
V
=
S
CP
CV
P
V
662
=
T
CP
CV
1
κV
c2 =
Substitute into Eq. ( A):
V CP
MC V κ
V CP
MC V κ
c=
or
(a) For an ideal gas, V = RT / P and κ = 1/ P . Therefore,
cig =
RT CP
M CV
(b) For an incompressible liquid, V is constant, and κ = 0, leading to the result: c = . This of
course leads to the conclusion that the sound speed in liquids is much greater than in gases.
7.6
As P2 decreases from an initial value of P2 = P1 , both u 2 and m steadily increase until the critical˙
pressure ratio is reached. At this value of P2 , u 2 equals the speed of sound in the gas, and further
reduction in P2 does not affect u 2 or m .
˙
7.7 The mass-ow rate m is of course constant throughout the nozzle from entrance to exit.
˙
The velocity u rises monotonically from nozzle entrance ( P / P1 = 1) to nozzle exit as P and P / P1
decrease.
The area ratio decreases from A / A1 = 1 at the nozzle entrance to a minimum value at the throat and
thereafter increases to the nozzle exit.
7.8 Substitution of Eq. (7.12) into (7.11), with u 1 = 0 gives:
u2
throat =
2
2γ P1 V1
1
γ +1
γ 1
= γ P1 V1
2
γ +1
where V1 is specic volume in m3 ·kg1 and P1 is in Pa. The units of u 2
throat are then:
N
· m3 · kg1 = N · m · kg1 = kg · m · s2 · m · kg1 = m2 · s2
m2
With respect to the nal term in the preceding equation, note that P1 V1 has the units of energy per unit
mass. Because 1 N · m = 1 J, equivalent units are J·kg1 . Moreover, P1 V1 = RT1 / M ; whence
Pa · m3 · kg1 =
u2
throat =
γ RT1
M
2
γ +1
With R in units of J·(kg mol)1 ·K1 , RT1 / M has units of J·kg1 or m2 ·s2 .
663
7.16 It is shown at the end of Ex. 7.5 that the Joule/Thomson inversion curve is the locus of states for which
( Z / T ) P = 0. We apply the following general equation of differential calculus:
x
y
z
Z
T
Because
P
P = ρ Z RT ,
ρ
ρ=
w
+
ρ
Z
T
=
x
w
+
Z
T
=
Z
T
Whence,
x
y
=
Z
ρ
P
P
Z RT
Z
ρ
y
w
y
z
T
ρ
T
P
T
ρ
T
P
ρ
T
and
P
R
=
P
1
Z +T
( Z T )2
Z
T
P
Setting ( Z / T ) P = 0 in each of the two preceding equations reduces them to:
Z
T
=
ρ
Z
ρ
T
ρ
T
ρ
T
and
P
=
P
ρ
P
=
2
T
Z RT
Combining these two equations yields:
Z
T
T
=ρ
ρ
Z
ρ
T
(a) Equation (3.42) with van der Waals parameters becomes:
a
RT
2
V b V
P=
Multiply through by V / RT , substitute Z = P V / RT , V = 1/ρ , and rearrange:
Z=
aρ
1
RT
1 bρ
In accord with Eq. (3.51), dene q a /b RT . In addition, dene ξ bρ . Then,
Z=
Z
T
Differentiate:
1
qξ
1ξ
=
ρ
Z
T
( A)
= ξ
ξ
dq
dT
By Eq. (3.54) with α (Tr ) = 1 for the van der Waals equation, q =
dq
=
dT
Z
T
Then,
In addition,
Z
ρ
Tr2 Tc
= (ξ )
ρ
=b
T
1
d Tr
=
dT
1
Tr2
Z
ξ
664
=
T
q
T
=
=
/ Tr . Whence,
q
1
=
T
T Tr
qξ
T
b
qb
(1 ξ )2
Substitute for the two partial derivatives in the boxed equation:
T
bρ
qξ
qbρ
=
(1 ξ )2
T
or
qξ =
1
ξ =1
2q
Whence,
ξ
qξ
(1 ξ )2
(B)
By Eq. (3.46), Pc = R Tc /b. Moreover, P = Zρ RT . Division of the second equation by the
rst gives Pr = Zρ bT / Tc . Whence
Pr =
Z ξ Tr
(C )
These equations allow construction of a Tr vs. Pr inversion curve as in Fig. 7.2. For a given value
of Tr , calculate q . Equation ( B ) then gives ξ , Eq. ( A) gives Z , and Eq. (C ) gives Pr .
(b) Proceed exactly as in Part (a), with exactly the same denitions. This leads to a new Eq. ( A):
Z=
qξ
1
1+ξ
1ξ
( A)
By Eq. (3.54) with α(Tr ) = Tr0.5 for the Redlich/Kwong equation, q =
1.5 q
dq
=
T
dT
Z
ρ
Moreover,
Z
T
and
T
=
ρ
=
/ Tr1.5 . This leads to:
1.5 q ξ
T (1 + ξ )
bq
b
2
(1 + ξ )2
(1 ξ )
Substitution of the two derivatives into the boxed equation leads to a new Eq. ( B ):
q=
1+ξ
1ξ
2
1
2.5 + 1.5 ξ
(B)
As in Part (a), for a given Tr , calculate q , and solve Eq. ( B ) for ξ , by trial or a by a computer
routine. As before, Eq. ( A) then gives Z , and Eq. (C ) of Part (a) gives Pr .
7.17 (a) Equal to.
(b) Less than.
(c) Less than.
(d) Equal to.
(e) Equal to.
7.28 When a saturated liquid is expanded in a turbine some of the liquid vaporizes. A turbine properly
designed for expansion of liquids cannot handle the much larger volumes resulting from the formation
of vapor. For example, if saturated liquid at 5 bar expands isentropically to 1 bar, the fraction of the
original liquid that vaporizes is found as follows:
l
v
v
l
S2 = S2 + x2 ( S2 S2 ) = S1
or
v
x2 =
l
1.8604 1.3027
S1 S2
= 0.0921
=
l
v
7.3598 1.3027
S2 S2
Were the expansion irreversible, the fraction of liquid vaporized would be even greater.
7.33 Apply Eq. (2.29) to this non-steady-state process, with n replacing m , with the tank as control volume,
and with a single inlet stream. Since the process is adiabatic and the only work is shaft work, this
equation may be multiplied by dt to give:
d (nU )tank H dn = d Ws
665
Because the inlet stream has constant properties, integration from beginning to end of the process
yields:
Ws = n 2 U 2 n 1 U 1 n H
where the subscripted quantities refer to the contents of the tank and n and H refer to the inlet stream.
Substitute n = n 2 n 1 and H = U + P V = U + RT :
Ws = n 2U2 n 1U1 (n 2 n 1 )(U + RT ) = n 2 (U2 U RT ) n 1 (U1 U RT )
With
U = C V T for an ideal gas with constant heat capacities, this becomes:
Ws = n 2 [C V (T2 T ) RT ] n 1 [C V (T1 T ) RT ]
However, T = T1 , and therefore:
Ws = n 2 [C V (T2 T1 ) RT1 ] + n 1 RT1
By Eq. (3.30b),
n1 =
Moreover,
(γ 1)/γ )
P2
P1
T2 =
P1 Vtank
RT1
and
n2 =
P2 Vtank
RT2
With γ = 1.4, T2 = 573.47 K. Then, with R = 8.314 m3 kPa kmol1 K1 ,
n1 =
(101.33)(20)
= 0.8176 kmol
(8.314)(298.15)
and
n2 =
(1000)(20)
= 4.1948 kmol
(8.314)(573.47)
Substitution of numerical values into the boxed equation, with R = 8.314 kJ kmol1 K1 , gives:
Ws = 15, 633 kJ
7.40 Combine Eqs. (7.13) and (7.17):
˙
˙
Ws = n
By Eq. (6.8),
Assume now that
RT1 / P1 . Then
H =n
˙
( H )S
η
V dP = V
( H )S =
P
P is small enough that V , an average value, can be approximated by V1 =
( H )S =
RT1
P1
P
RT1
˙
Ws = n
˙
η P1
and
P
Equation (7.22) is the usual equation for isentropic compression of an ideal gas with constant heat
capacities. For irreversible compression it can be rewritten:
nC P T1
˙
˙
Ws =
η
For
P2
P1
R/C P
1
P sufciently small, the quantity in square brackets becomes:
P2
P1
R/C P
1= 1+
P
P1
R/C P
1
The boxed equation is immediately recovered from this result.
666
1+
RP
C P P1
1
7.41 The equation immediately preceding Eq. (7.22) page 276 gives T2 = T1 π . With this substitution,
Eq. (7.23) becomes:
π 1
T1 π T1
= T1 1 +
T2 = T1 +
η
η
The entropy generation SG is simply S for the compression process, for which Eq. (5.14) may be
rewritten:
P2 R / C P
C P T2 C P
P2
C P T2
S
ln
ln
=
ln
ln
=
P1
R
T1
R
P1
T1
R
R
Combine the two preceding equations:
CP
ln
ln π =
R
π 1
CP
S
ln 1 +
=
η
R
R
1+
π 1
η
π
η+π 1
CP
SG
ln
=
ηπ
R
R
Whence,
7.43 The relevant fact here is that C P increases with increasing molecular complexity. Isentropic compression work on a mole basis is given by Eq. (7.22), which can be written:
Ws = C P T1 (π 1)
where
π
P2
P1
R/C P
This equation is a proper basis, because compressor efciency η and owrate n are xed. With all
˙
other variables constant, differentiation yields:
dπ
d Ws
= T1 (π 1) + C P
dC P
dC P
From the denition of π ,
ln π =
Then,
and
P2
R
ln
P1
CP
whence
P2
R
1 dπ
d ln π
ln
=
=
2
P1
π dC P
dC P
CP
P2
πR
dπ
ln
=
2
P1
dC P
CP
π R P2
d Ws
ln
= T1 π 1
P1
CP
dC P
= T1 (π 1 π ln π )
When π = 1, the derivative is zero; for π > 1, the derivative is negative (try some values). Thus, the
work of compression decreases as C P increases and as the molecular complexity of the gas increases.
7.45 The appropriate energy balance can be written: W = H Q . Since Q is negative (heat transfer is
out of the system), the work of non-adiabatic compression is greater than for adiabatic compression.
Note that in order to have the same change in state of the air, i.e., the same H , the irreversibilities of
operation would have to be quite different for the two cases.
7.46 There is in fact no cause for concern, as adiabatic compression sends the steam further into the superheat region.
667
7.49 (a) This result follows immediately from the last equation on page 267 of the text.
(b) This result follows immediately from the middle equation on page 267 of the text.
(c) This result follows immediately from Eq. (6.19) on page 267 of the text.
T
Z
Z
but by (a), this is zero.
=
(d )
T P V P
V P
V
( P / T )V
V
=
=
P
( P / V )T
T
(e) Rearrange the given equation:
P
T
T
=
V
V
T
P
For the nal equality see footnote on p. 266. This result is the equation of (c).
V CP 1
·
·
M CV κ
7.50 From the result of Pb. 7.3: c =
With
Then
V=
RT
+B
P
c = PV
V
P
then
=
T
where
RT
P2
κ=
Also, let γ =
B
γ RT
+
RT
M
T
CP
CV
γ RT
M
BP
γ
γ
= 1+
= ( RT + B P )
RT
M RT
M RT
c=
V
P
1
V
γ RT
·P
M
A value for B at temperature T may be extracted from a linear t of c vs. P .
7.51 (a) On the basis of Eq. (6.8), write:
ig
HS =
HS =
V dP =
RT
dP
P
(const S )
Z RT
dP
P
(const S )
V ig d P =
HS
ig
HS
=
Z RT
d P (const S )
P
Z
RT
d P (const S )
P
By extension, and with equal turbine efciencies,
7.52 By Eq. (7.16),
H = η( H ) S
.
W
H
= . ig = Z
H ig
W
For C P = constant,
T2 T1 = η[(T2 ) S T1 ]
For an ideal gas with constant C P , (T2 ) S is related to T1 by (see p. 77): (T2 ) S = T1
Combine the last two equations, and solve for T2 :
668
T2 = T1 1 + η
P2
P1
R/C P
P2
P1
1
R/C P
From which
η=
T2
1
T1
P2
P1
R/C P
Note that η < 1
1
Results: For T2 = 318 K, η = 1.123; For T2 = 348 K, η = 1.004; For T2 = 398 K, η = 0.805.
Only T2 = 398 K is possible.
7.55 The proposal of Pb. 7.53, i.e., pumping of liquid followed by vaporization. The reason is that pumping
a liquid is much less expensive than vapor compression.
7.56 What is required here is the lowest saturated steam temperature that satises the T constraint. Data
from Tables F.2 and B.2 lead to the following:
Benzene/4.5 bar; n-Decane/17 bar; Ethylene glycol/33 bar; o-Xylene/9 bar
669
Chapter 8 - Section B - Non-Numerical Solutions
8.12 (a) Because Eq. (8.7) for the efciency ηDiesel includes the expansion ratio, re VB / V A , we relate
this quantity to the compression ratio, r VC / VD , and the Diesel cutoff ratio, rc V A / VD . Since
VC = VB , re = VC / V A . Whence,
VA
VC / VD
r
= rc
=
=
VD
VC / V A
re
rc
1
=
r
re
or
Equation (8.7) can therefore be written:
ηDiesel
ηDiesel = 1
or
1 (1/ r )γ
=1
γ 1/ r
(rc / r )γ (1/ r )γ
rc / r 1/ r
1
=1
γ
1
r
γ 1
γ
rc 1
rc 1
γ
rc 1
γ (rc 1)
(b) We wish to show that:
γ
rc 1
>1
γ (rc 1)
or more simply
xa 1
>1
a (x 1)
Taylors theorem with remainder, taken to the 1st derivative, is written:
g = g (1) + g (1) · (x 1) + R
where,
Then,
R
g [1 + θ (x 1)]
· ( x 1)2
2!
(0 < θ < 1)
x a = 1 + a · (x 1) + 1 a · (a 1) · [1 + θ (x 1)]a 2 · (x 1)2
2
Note that the nal term is R. For a > 1 and x > 1, R > 0. Therefore:
x a > 1 + a · ( x 1)
x a 1 > a · (x 1)
γ
rc 1
>1
γ (rc 1)
and
(c) If γ = 1.4 and r = 8, then by Eq. (8.6):
1
8
0.4
ηOtto = 1
ηDesiel = 1
1
8
0.4
rc = 2
ηDesiel = 1
1
8
0.4
rc = 3
ηOtto = 0.5647
and
21.4 1
1.4(2 1)
and
ηDiesel = 0.4904
31.4 1
1.4(3 1)
and
ηDiesel = 0.4317
670
8.15 See the gure below. In the regenerative heat exchanger, the air temperature is raised in step B B ,
while the air temperature decreases in step D D . Heat addition (replacing combustion) is in step
B C .
By denition,
η
where,
W AB WC D
Q BC
W AB = ( H B H A ) = C P (TB TA )
WC D = ( H D HC ) = C P (TD TC )
Q B C = C P (TC TB ) = C P (TC TD )
Whence,
η=
TB T A
TA TB + TC TD
=1
TC TD
TC TD
By Eq. (3.30b),
TB = T A
PB
PA
(γ 1)/γ
and
PB
PA
TA
Then,
PD
PC
TD = TC
η =1
TC 1
(γ 1)/γ
(γ 1)/γ
1
PA
PB
(γ 1)/γ
Multiplication of numerator and denominator by ( PB / PA )(γ 1)/γ gives:
η =1
TA
TC
671
PB
PA
(γ 1)/γ
= TC
PA
PB
(γ 1)/γ
8.21 We give rst a general treatment of paths on a P T diagram for an ideal gas with constant heat capacities
undergoing reversible polytropic processes. Equation (3.35c), p. 78, may be rewritten as
P = K T δ/(δ1)
ln P = ln K +
δP
dP
=
δ1T
dT
δ = 0 d P /dT = 0
δ = 1 d P /dT =
δ
d2 P
=
2
δ1
dT
P
1 dP
2
T
T dT
P
δ
d2 P
=
(δ 1)2 T 2
dT 2
δ dT
dP
=
δ1 T
P
Sign of d P /dT is that of δ 1, i.e., +
( A)
Special cases
By Eq. ( A),
δ
ln T
δ1
=
δ1
δ1T
Constant P
Constant T
P
δP
T
δ1T
Sign of d 2 P /dT 2 is that of δ , i.e., +
(B)
For a constant-V process, P varies with T in accord with the ideal-gas law: P = RT / V or P = K T
With respect to the initial equation, P = K T δ/(δ1) , this requires δ = . Moreover, d P /dT = K
and d 2 P /dT 2 = 0. Thus a constant-V process is represented on a P T diagram as part of a straight
line passing through the origin. The slope K is determined by the initial P T coordinates.
For a reversible adiabatic process (an isentropic process), δ = γ . In this case Eqs. ( A) and ( B ) become:
P
γ
d2 P
=
(γ 1)2 T 2
dT 2
γP
dP
=
γ 1T
dT
We note here that γ /(γ 1) and γ /(γ 1)2 are both > 1. Thus in relation to a constant-V process
the isentropic process is represented by a line of greater slope and greater curvature for the same T
and P . Lines characteristic of the various processes are shown on the following diagram.
δ=γ
δ=1
δ=
P
δ=0
0
0
T
The required sketches appear on the following page. (Courtesy of Prof. Mark T. Swihart, State University of New York at Buffalo.)
672
P
P
0
0
0
0
T
T
Figure 2: The Otto cycle
Figure 1: The Carnot cycle
P
P
0
0
0
0
T
T
Figure 4: The Brayton cycle
Figure 3: The Diesel cycle
8.23 This is a challenging and open-ended problem for which we offer no solution. Problem 8.21 may offer
some insight.
673
Chapter 9 - Section B - Non-Numerical Solutions
9.1 Since the object of doing work |W | on a heat pump is to transfer heat | Q H | to a heat sink, then:
What you get = | Q H |
What you pay for = |W |
Whence ν
For a Carnot heat pump,
ν=
|Q H |
|W |
TH
|Q H |
=
TH TC
| Q H | | QC |
9.3 Because the temperature of the nite cold reservoir (contents of the refrigerator) is a variable, use
differential forms of Carnots equations, Eqs. (5.7) and (5.8):
TH
d QH
=
TC
d QC
and
dW = 1
TC
TH
d QH
In these equations Q C and Q H refer to the reservoirs. With d Q H = C t dTC , the rst of Carnots
equations becomes:
dTC
d Q H = C t TH
TC
Combine this equation with the second of Carnots equations:
d W = C t TH
dTC
+ C t dTC
TC
Integration from TC = TH to TC = TC yields:
W = C t TH ln
TC
+ C t (TC TH )
TH
or
W = C t T H ln
TC
TH
1
+
TH
TC
9.5 Differentiation of Eq. (9.3) yields:
ερ
ε TC
=
TH
TH
TC
1
=
+
2
(TH TC )2
(TH TC )
TH TC
and
ερ
ε TH
=
TC
TC
(TH TC )2
Because TH > TC , the more effective procedure is to increase TC .
For a real refrigeration system, increasing TC is hardly an option if refrigeration is required at a particular value of TC . Decreasing TH is no more realistic, because for all practical purposes, TH is xed by
environmental conditions, and not subject to control.
9.6 For a Carnot refrigerator, ρ is given by Eq. (9.3). Write this equation for the two cases:
ρ=
TC
TH TC
and
ρσ =
TσC
Tσ H TσC
Because the directions of heat transfer require that TH > Tσ H and TC < TσC , a comparison shows that
ρ < ρσ and therefore that ρ is the more conservative value.
674
9.20 On average, the coefcient of performance will increase, thus providing savings on electric casts. On
the other hand, installation casts would be higher. The proposed arrangement would result in cooling of
the kitchen, as the refrigerator would act as an air conditioner. This would be detrimental in the winter,
but benecial in the summer, at least in temperate climates.
9.21
= 0.6
If
Carnot
= 0. 6
TC
TH TC
< 1, then TC < TH /1.6. For TH = 300 K, then TC < 187.5 K, which is most unlikely.
675
Chapter 10 - Section B - Non-Numerical Solutions
10.5 For a binary system, the next equation following Eq. (10.2) shows that P is linear in x1 . Thus no
maximum or minimum can exist in this relation. Since such an extremum is required for the existence
of an azeotrope, no azeotrope is possible.
10.6 (a) Because benzene and toluene are chemically similar and the pressure is only 1(atm), this system
can be modeled by Raoults law to a good approximation.
(b) Although n-hexane and n-heptane are chemically similar, a pressure of 25 bar is too high for
modeling this system by Raoults law.
(c) At 200 K, hydrogen is supercritical, and modeling the hydrogen/propane system at this temperature by Raoults law is out of the question, because no value of P sat for hydrogen is known.
(d) Because isooctane and n-octane are chemically similar and at a temperature (373.15 K) close to
their normal boiling points, this system can be modeled by Raoults law to a good approximation.
(e) Water and n-decane are much too dissimilar to be modeled by Raoults law, and are in fact only
slightly soluble in one another at 300 K.
10.12 For a total volume V t of an ideal gas, P V t = n RT . Multiply both sides by yi , the mole fraction of
species i in the mixture:
yi P V t = n i RT
or
pi V t =
mi
RT
Mi
where m i is the mass of species i , Mi is its molar mass, and pi is its partial pressure, dened as
pi yi P . Solve for m i :
Mi pi V t
mi =
RT
Applied to moist air, considered a binary mixture of air and water vapor, this gives:
m H2 O =
M H 2 O p H2 O V t
RT
(a) By denition,
h
and
m H2 O
m air
or
m air =
h=
Mair pair V t
RT
MH2 O pH2 O
Mair pair
Since the partial pressures must sum to the total pressure, pair = P pH2 O ; whence,
h=
p H2 O
MH2 O
Mair P pH2 O
(b) If air is in equilibrium with liquid water, then the partial pressure of water vapor in the air equals
the vapor pressure of the water, and the preceding equation becomes:
h sat =
PHsat
MH2 O
2O
Mair P PHsat
2O
676
(c) Percentage humidity and relative humidity are dened as follows:
h pc
pH2 O P PHsat
h
2O
(100)
sat = sat
PH2 O P pH2 O
h
and
h rel
p H2 O
(100)
PHsat
2O
Combining these two denitions to eliminate pH2 O gives:
P PHsat
2O
h pc = h rel
P PHsat (h rel /100)
2O
10.14 Because the vapor space above the liquid phase is nearly pure gas, Eq. (10.4) becomes P = xi Hi .
For the same mole fraction of gas dissolved in the liquid phase, P is then proportional to Hi . Values
given in Table 10.1 indicate that were air used rather than CO2 , P would be about 44 times greater,
much too high a pressure to be practical.
10.15 Because Henrys constant for helium is very high, very little of this gas dissolves in the blood streams
of divers at approximately atmospheric pressure.
10.21 By Eq. (10.5) and the given equations for ln γ1 and ln γ2 ,
2
y1 P = x1 exp( Ax2 ) P1sat
2
y2 P = x2 exp( Ax1 ) P2sat
and
These equations sum to give:
2
2
P = x1 exp( Ax2 ) P1sat + x2 exp( Ax1 ) P2sat
Dividing the equation for y1 P by the preceding equation yields:
y1 =
2
x1 exp( Ax2 ) P1sat
2
2
x1 exp( Ax2 ) P1sat + x2 exp( Ax1 ) P2sat
For x1 = x2 this equation obviously reduces to:
P=
P1sat
P1sat + P2sat
10.23 A little reection should convince anyone that there is no other way that BOTH the liquid-phase and
vapor-phase mole fractions can sum to unity.
10.24 By the denition of a K -value, y1 = K 1 x1 and y2 = K 2 x2 . Moreover, y1 + y2 = 1. These equations
combine to yield:
K 1 x1 + K 2 x2 = 1
Solve for x1 :
or
x1 =
K 1 x 1 + K 2 (1 x 1 ) = 1
1 K2
K1 K2
Substitute for x1 in the equation y1 = K 1 x1 :
y1 =
K 1 (1 K 2 )
K1 K2
677
Note that when two phases exist both x1 and y1 are independent of z 1 .
By a material balance on the basis of 1 mole of feed,
x1 L + y1 V = z 1
or
x1 (1 V ) + y1 V = z 1
Substitute for both x1 and y1 by the equations derived above:
K 1 (1 K 2 )
1 K2
V = z1
(1 V ) +
K1 K2
K1 K2
Solve this equation for V :
V=
z 1 ( K 1 K 2 ) (1 K 2 )
( K 1 1)(1 K 2 )
Note that the relative amounts of liquid and vapor phases do depend on z 1 .
10.35 Molality Mi =
xi
ni
=
x s Ms
ms
where subscript s denotes the solvent and Ms is the molar mass of the solvent. The given equation
may therefore be written:
xi
= ki yi P
x s Ms
or
xi
1
x s Ms k i
= yi P
Comparison with Eq. (10.4) shows that
Hi =
1
x s Ms k i
For water, Ms = 18.015 g mol1
Thus,
or
Hi =
or for xi 0
Hi =
0.018015 kg mol1 .
1
= 1633 bar
(0.018015)(0.034)
This is in comparison with the value of 1670 bar in Table 10.1.
678
1
Ms k i
Chapter 11 - Section B - Non-Numerical Solutions
11.6 Apply Eq. (11.7):
ν (nT )
ν ni
¯
Ti
νn
ν ni
=T
P ,T ,n j
=T
ν (n P )
ν ni
¯
Pi
T , P ,n j
=P
P ,T ,n j
νn
ν ni
νm
ν ni
11.7 (a) Let m be the mass of the solution, and dene the partial molar mass by: m i
¯
=P
T , P ,n j
T , P ,n j
Let Mk be the molar mass of species k . Then
m=
ε nk Mk = ni Mi + ε n j M j
k
and
νm
ν ni
=
T , P ,n j
( j = i)
j
ν (n i Mi )
ν ni
Whence,
= Mi
m i = Mi
¯
T , P ,n j
νM t
ν mi
˜
(b) Dene a partial specic property as: Mi
If Mi is the molar mass of species i ,
=
T , P ,m j
mi
Mi
ni =
νM t
ν ni
ν ni
ν mi
and
T , P ,m j
ν ni
ν mi
=
T , P ,m j
1
Mi
Because constant m j implies constant n j , the initial equation may be written:
11.8 By Eqs. (10.15) and (10.16),
Because
V = ρ 1
then
dV
¯
V1 = V + x2
d x1
1 dρ
dV
=2
ρ d x1
d x1
¯
Mi
˜
Mi =
Mi
dV
¯
V2 = V x1
d x1
and
whence
1
x2 dρ
1
¯
=
V1 = 2
ρ
ρ d x1
ρ
1
x2 dρ
ρ d x1
=
dρ
1
ρ x2
2
d x1
ρ
1
x1 dρ
1
¯
=
V2 = + 2
ρ
ρ d x1
ρ
With
T , P ,m j
1+
x1 dρ
ρ d x1
=
dρ
1
ρ + x1
2
d x1
ρ
2
ρ = a0 + a1 x 1 + a2 x 1
and
dρ
= a1 + 2a2 x1
d x1
1
2
¯
V1 = 2 [a0 a1 + 2(a1 a2 )x1 + 3a2 x1 ]
ρ
679
and
these become:
1
2
¯
V2 = 2 (a0 + 2a1 x1 + 3a2 x1 )
ρ
11.9 For application of Eq. (11.7) all mole fractions must be eliminated from the given equation by the
relation xi = n i / n :
n1n2n3
C
n M = n 1 M1 + n 2 M2 + n 3 M3 +
n2
(n M )
n1
¯
For M1 ,
T , P ,n 2 ,n 3
n
n1
Because n = n 1 + n 2 + n 3 ,
n
n1
2n 1
1
3
2
n
n
= M1 + n 2 n 3 C
T , P ,n 2 ,n 3
=1
T , P ,n 2 ,n 3
Whence,
n1
n2n3
¯
C
M1 = M1 + 2 1 2
n
n
and
¯
M1 = M1 + x2 x3 [1 2x1 ]C
Similarly,
¯
M2 = M2 + x1 x3 [1 2x2 ]C
and
¯
M3 = M3 + x1 x2 [1 2x3 ]C
One can readily show that application of Eq. (11.11) regenerates the original equation for M . The
innite dilution values are given by:
¯
Mi = Mi + x j xk C
( j, k = i )
Here x j and xk are mole fractions on an i -free basis.
11.10 With the given equation and the Daltons-law requirement that P =
P=
RT
V
i
pi , then:
yi Z i
i
For the mixture, P = Z RT / V . These two equations combine to give Z =
i
yi Z i .
11.11 The general principle is simple enough:
¯
¯
¯
Given equations that represent partial properties Mi , MiR , or MiE as functions of composition, one may combine them by the summability relation to yield a mixture property.
Application of the dening (or equivalent) equations for partial properties then regenerates
the given equations if and only if the given equations obey the Gibbs/Duhen equation.
11.12 (a) Multiply Eq. ( A) of Ex. 11.4 by n (= n 1 + n 2 ) and eliminate x1 by x1 = n 1 /(n 1 + n 2 ):
n H = 600(n 1 + n 2 ) 180 n 1 20
n3
1
(n 1 + n 2 )2
Form the partial derivative of n H with respect to n 1 at constant n 2 :
¯
H1 = 600 180 20
n3
n2
2n 3
3n 2
1
1
1
1
+ 40
= 420 60
(n 1 + n 2 )3
(n 1 + n 2 )2
(n 1 + n 2 )2 (n 1 + n 2 )3
3
2
¯
H1 = 420 60 x1 + 40 x1
Whence,
Form the partial derivative of n H with respect to n 2 at constant n 1 :
¯
H2 = 600 + 20
2 n3
1
(n 1 + n 2 ) 3
680
or
3
¯
H2 = 600 + 40 x1
(b) In accord with Eq. (11.11),
2
3
3
H = x1 (420 60 x1 + 40 x1 ) + (1 x2 )(600 + 40 x1 )
Whence,
3
H = 600 180 x1 20 x1
(c) Write Eq. (11.14) for a binary system and divide by d x1 : x1
¯
¯
d H2
d H1
=0
+ x2
d x1
d x1
Differentiate the the boxed equations of part (a):
¯
d H1
2
= 120 x1 + 120 x1 = 120 x1 x2
d x1
and
¯
d H2
2
= 120 x1
d x1
Multiply each derivative by the appropriate mole fraction and add:
2
2
120 x1 x2 + 120x1 x2 = 0
(d) Substitute x1 = 1 and x2 = 0 in the rst derivative expression of part (c) and substitute x1 = 0
in the second derivative expression of part (c). The results are:
¯
d H1
d x1
=
x 1 =1
¯
d H2
d x1
=0
x1 =0
(e)
11.13 (a) Substitute x2 = 1 x1 in the given equation for V and reduce:
3
2
V = 70 + 58 x1 x1 7 x1
¯
¯
Apply Eqs. (11.15) and (11.16) to nd expressions for V1 and V2 . First,
dV
2
= 58 2 x1 21 x1
d x1
Then,
3
2
¯
V1 = 128 2 x1 20 x1 + 14 x1
681
and
3
2
¯
V2 = 70 + x1 + 14 x1
(b) In accord with Eq. (11.11),
2
3
2
3
V = x1 (128 2 x1 20 x1 + 14 x1 ) + (1 x1 )(70 + x1 + 14 x1 )
3
2
V = 70 + 58 x1 x1 7 x1
Whence,
which is the rst equation developed in part (a).
(c) Write Eq. (11.14) for a binary system and divide by d x1 : x1
¯
¯
d V2
d V1
=0
+ x2
d x1
d x1
Differentiate the the boxed equations of part (a):
¯
d V1
2
= 2 40 x1 + 42 x1
d x1
¯
d V2
2
= 2 x1 + 42 x1
d x1
and
Multiply each derivative by the appropriate mole fraction and add:
2
2
x1 (2 40 x1 + 42 x1 ) + (1 x1 )(2 x1 + 42 x1 ) = 0
The validity of this equation is readily conrmed.
(d) Substitute x1 = 1 in the rst derivative expression of part (c) and substitute x1 = 0 in the second
derivative expression of part (c). The results are:
¯
d V1
d x1
=
x 1 =1
¯
d V2
d x1
=0
x1 =0
(e)
11.14 By Eqs. (11.15) and (11.16):
dH
¯
H1 = H + x2
d x1
and
682
dH
¯
H2 = H x1
d x1
Given that:
H = x1 (a1 + b1 x1 ) + x2 (a2 + b2 x2 )
dH
= a1 + 2b1 x1 (a2 + 2b2 x2 )
d x1
Then, after simplication,
Combining these equations gives after reduction:
¯
H1 = a1 + b1 x1 + x2 (x1 b1 x2 b2 )
¯
H2 = a2 + b2 x2 x1 (x1 b1 x2 b2 )
and
These clearly are not the same as the suggested expressions, which are therefore not correct. Note
that application of the summability equation to the derived partial-property expressions reproduces
the original equation for H . Note further that differentiation of these same expressions yields results
that satisfy the Gibbs/Duhem equation, Eq. (11.14), written:
x1
¯
¯
d H2
d H1
=0
+ x2
d x1
d x1
The suggested expresions do not obey this equation, further evidence that they cannot be valid.
11.15 Apply the following general equation of differential calculus:
x
y
(n M )
ni
=
T , P ,n j
x
y
=
z
(n M )
ni
+
w
x
w
y
w
y
(n M )
V
+
T , V ,n j
z
V
ni
T ,n
T , P ,n j
Whence,
M
¯
˜
Mi = Mi + n
V
T ,n
V
ni
or
T , P ,n j
M
˜
¯
Mi = Mi n
V
T ,n
V
ni
T , P ,n j
By denition,
(nV )
¯
Vi
ni
=n
T , P ,n j
V
ni
+V
or
T , P ,n j
M
˜
¯
¯
Mi = Mi + (V Vi )
V
Therefore,
n
V
ni
¯
= Vi V
T , P ,n j
T ,x
ˆ
ˆ
11.20 Equation (11.59) demonstrates that ln φi is a partial property with respect to G R / RT . Thus ln φi =
¯
G i / RT . The partial-property analogs of Eqs. (11.57) and (11.58) are:
ˆ
ln φi
P
=
T ,x
¯
Vi R
RT
ˆ
ln φi
T
and
=
P ,x
¯
HiR
RT 2
The summability and Gibbs/Duhem equations take on the following forms:
GR
=
RT
ˆ
xi ln φi
and
i
i
683
ˆ
xi d ln φi = 0
(const T , P )
11.26 For a pressure low enough that Z and ln φ are given approximately by Eqs. (3.38) and (11.36):
Z =1+
BP
RT
and
ln φ =
BP
RT
ln φ Z 1
then:
11.28 (a) Because Eq. (11.96) shows that ln γi is a partial property with respect to G E/ RT , Eqs. (11.15)
and (11.16) may be written for M G E/ RT :
ln γ1 =
d (G E/ RT )
GE
+ x2
d x1
RT
ln γ2 =
d (G E/ RT )
GE
x1
d x1
RT
Substitute x2 = 1 x1 in the given equaiton for G E/ RT and reduce:
GE
2
3
= 1.8 x1 + x1 + 0.8 x1
RT
Then,
d (G E/ RT )
2
= 1.8 + 2 x1 + 2.4 x1
d x1
whence
3
2
ln γ1 = 1.8 + 2 x1 + 1.4 x1 1.6 x1
3
2
ln γ2 = x1 1.6 x1
and
(b) In accord with Eq. (11.11),
GE
2
3
2
3
= x1 ln γ1 + x2 ln γ2 = x1 (1.8 + 2 x1 + 1.4 x1 1.6 x1 ) + (1 x1 )(x1 1.6 x1 )
RT
Whence,
GE
3
2
= 1.8 x1 + x1 + 0.8 x1
RT
which is the rst equation developed in part (a).
¯
(c) Write Eq. (11.14) for a binary system with Mi = ln γi and divide by d x1 :
x1
d ln γ2
d ln γ1
=0
+ x2
d x1
d x1
Differentiate the the boxed equations of part (a):
d ln γ1
2
= 2 + 2.8 x1 4.8 x1
d x1
d ln γ2
2
= 2 x1 4.8 x1
d x1
and
Multiply each derivative by the appropriate mole fraction and add:
2
2
x1 (2 + 2.8 x1 4.8 x1 ) + (1 x1 )(2 x1 4.8 x1 ) = 0
The validity of this equation is readily conrmed.
(d) Substitute x1 = 1 in the rst derivative expression of part (c) and substitute x1 = 0 in the second
derivative expression of part (c). The results are:
d ln γ1
d x1
=
x 1 =1
684
d ln γ2
d x1
=0
x1 =0
(e)
11.29 Combine denitions of the activity coefcient and the fugacity coefcients:
γi
fˆi /xi P
fi / P
or
γi =
ˆ
φi
φi
Note: See Eq. (14.54).
E
11.30 For C P = const., the following equations are readily developed from those given in the last column
of Table 11.1 (page 415):
E
H E = CP
T
SE =
and
GE
T
P ,x
T
T
E
= CP
Working equations are then:
E
S1 =
E
H1E G 1
T1
E
H2E = H1E + C P
T
T
and
and
T
E
E
E
S2 = S1 + C P
E
E
G 2 = H2E T2 S2
For T1 = 298.15, T2 = 328.15, T = 313.15 and
given in the following table:
T = 30, results for all parts of the problem are
E
II. For C P = 0
I.
E
G1
(a)
(b)
(c)
(d)
(e)
(f)
(g)
H1E
E
S1
E
CP
E
S2
H2E
E
G2
E
S2
H2E
E
G2
622
1095
407
632
1445
734
759
1920
1595
984
208
605
416
1465
4.354
1.677
1.935
2.817
2.817
3.857
2.368
4.2
3.3
2.7
23.0
11.0
11.0
8.0
3.951
1.993
1.677
0.614
1.764
2.803
1.602
1794
1694
903
482
935
86
1225
497.4
1039.9
352.8
683.5
1513.7
833.9
699.5
4.354
1.677
1.935
2.817
2.817
3.857
2.368
1920
1595
984
208
605
416
1465
491.4
1044.7
348.9
716.5
1529.5
849.7
688.0
685
11.31 (a) Multiply the given equation by n (= n 1 + n 2 ), and convert remaining mole fractions to ratios of
mole numbers:
n2n3
n1n3
n1n2
nG E
+ A23
+ A13
= A12
n
n
n
RT
Differentiation with respect to n 1 in accord with Eq. (11.96) yields [( n / n 1 )n 2 ,n 3 = 1]:
ln γ1 = A12 n 2
1 n1
n n2
1 n1
n n2
+ A13 n 3
A23
n2n3
n2
= A12 x2 (1 x1 ) + A13 x3 (1 x1 ) A23 x2 x3
Similarly,
ln γ2 = A12 x1 (1 x2 ) A13 x1 x3 + A23 x3 (1 x2 )
ln γ3 = A12 x1 x2 + A13 x1 (1 x3 ) + A23 x2 (1 x3 )
(b) Each ln γi is multiplied by xi , and the terms are summed. Consider the rst terms on the right of
each expression for ln γi . Multiplying each of these terms by the appropriate xi and adding gives:
2
2
A12 (x1 x2 x1 x2 + x2 x1 x2 x1 x1 x2 x3 ) = A12 x1 x2 (1 x1 + 1 x2 x3 )
= A12 x1 x2 [2 (x1 + x2 + x3 )] = A12 x1 x2
An analogous result is obtained for the second and third terms on the right, and adding them
yields the given equation for G E/ RT .
x1 = 0:
ln γ1 (x1 = 0) = A12 x2 + A13 x3 A23 x2 x3
For pure species 1,
x1 = 1:
ln γ1 (x1 = 1) = 0
For innite dilution of species 2,
x2 = 0:
2
ln γ1 (x2 = 0) = A13 x3
For innite dilution of species 3,
x3 = 0:
2
ln γ1 (x3 = 0) = A12 x2
(c) For innite dilution of species 1,
GE = GR
11.35 By Eq. (11.87), written with M G and with x replaced by y :
i
yi G iR
Equations (11.33) and (11.36) together give G iR = Bii P . Then for a binary mixture:
G E = B P y1 B11 P y2 B22 P
G E = P ( B y1 B11 y2 B22 )
or
G E = δ12 P y1 y2
Combine this equation with the last equation on Pg. 402:
GE
T
From the last column of Table 11.1 (page 415): S E =
Because δ12 is a function of T only:
SE =
P ,x
d δ12
P y1 y2
dT
H E = δ12 T
By the denition of G E , H E = G E + T S E ; whence,
E
Again from the last column of Table 11.1: C P =
This equation and the preceding one lead directly to:
686
HE
T
d δ12
dT
P ,x
E
C P = T
d 2 δ12
P y1 y2
dT 2
P y1 y2
(G E / RT )
T
11.41 From Eq. (11.95):
=
P
To an excellent approximation, write:
H E
RT 2
(G E / T )
T
or
(G E / T )
T
=
P
H E
T2
H E
(G E / T )
2
Tmean
T
P
0.271
785/323 805/298
(G E / T )
= 0.01084
=
=
25
323 298
T
From the given data:
1060
H E
= 0.01082
=
2
3132
Tmean
and
The data are evidently thermodynamically consistent.
11.42 By Eq. (11.14), the Gibbs/Duhem equation,
Given that
¯
M1 = M1 + Ax2
and
x1
¯
¯
d M2
d M1
=0
+ x2
d x1
d x1
¯
M2 = M2 + Ax1
then
¯
d M1
= A
d x1
and
¯
d M2
=A
d x1
¯
¯
d M2
d M1
= x 1 A + x 2 A = A( x 2 x 1 ) = 0
+ x2
d x1
d x1
The given expressions cannot be correct.
Then
11.45 (a) For
x1
22
M E = Ax1 x2
nd
2
¯E
M1 = Ax1 x2 (2 3x1 )
and
Note that at both x1 = 0 (x2 = 1) and x1 = 1 (x2 = 0),
In particular,
2
¯E
M2 = Ax1 x2 (2 3x2 )
¯E
¯E
M1 = M2 = 0
¯E
¯E
( M1 ) = ( M2 ) = 0
¯E
¯E
Although M E has the same sign over the whole composition range, both M1 and M2 change
sign, which is unusual behavior. Find also that
¯E
¯E
d M2
d M1
= 2 Ax1 (1 6x1 x2 )
= 2 Ax2 (1 6x1 x2 ) and
d x1
d x1
The two slopes are thus of opposite sign, as required; they also change sign, which is unusual.
¯E
¯E
d M2
d M1
=0
= 2 A and
For x1 = 0
d x1
d x1
For
(b) For
x1 = 1
¯E
d M1
=0
d x1
and
¯E
d M2
= 2 A
d x1
M E = A sin(π x1 ) nd:
¯E
M1 = A sin(π x1 ) + Aπ x2 cos(π x1 )
and
¯E
M2 = A sin(π x1 ) Aπ x1 cos(π x1 )
¯E
¯E
d M2
d M1
= Aπ 2 x1 sin(π x1 )
= Aπ 2 x2 sin(π x1 ) and
d x1
d x1
The two slopes are thus of opposite sign, as required. But note the following, which is unusual:
For
x1 = 0
and
x1 = 1
¯E
d M1
=0
d x1
and
¯E
d M2
=0
d x1
PLOTS OF THE FUNCTIONS ARE SHOWN ON THE FOLLOWING PAGE.
687
10
A
Pb. 11.45 (a)
i
A . xi
MEi
MEbar2i
xi
0 .. 100
2.
xi 2
1
A . xi . xi . 1
MEbar1i
xi . 2
3. 1
.01 . i
.00001
A . xi . 1
xi
xi
2
1.5
MEi
MEbar1
1
i
MEbar2i
0.5
0
0.5
0
MEi
Pb. 11.45 (b)
0.2
0.4
A . sin
p
.x
0.6
xi
0.8
1
(pi prints as bf p)
i
MEbar1i
A . sin
p
.x
i
A.p . 1
xi . cos
MEbar2i
A . sin
p
.x
i
A . p . xi . cos
p
.x
p
.x
i
i
40
30
MEi
MEbar1
20
i
MEbar2i
10
0
10
0
0.2
0.4
xi
687A
0.6
0.8
1
2
2
3 . xi
(n M )
¯
Mi =
ni
11.46 By Eq. (11.7),
At constant T and P ,
M
ni
= M +n
T , P ,n j
M
xk
dM =
k
T , P ,n j
d xk
T , P ,x j
Divide by dn i with restriction to constant n j ( j = i ):
M
ni
M
ni
T , P ,n j
k
xk
ni
nk
xk =
n
With
=
T , P ,n j
=
1
n
M
xk
=
1
n
k =i
M
xi
T , P ,x j
(k = i )
M
1
+ (1 x i )
xi
n
M
xk
xk
k
xk
T , P ,x j
(k = i )
1
ni
n n2
T , P ,x j
1
n
nj
n
k
n2
=
M
xi
¯
Mi = M +
T , P ,x j
nj
M
xk
xk
xk
ni
k
M
xk
T , P ,x j
T , P ,x j
T , P ,x j
For species 1 of a binary mixture (all derivatives at constant T and P ):
¯
M1 = M +
M
x1
x2
M
x1
x1
x2
x2
M
x2
= M + x2
x1
M
x1
x2
M
x2
x1
Because x1 + x2 = 1, the partial derivatives in this equation are physically unrealistic; however, they
do have mathematical signicance. Because M = M(x1 , x2 ), we can quite properly write:
M
x1
dM =
x2
d x1 +
M
x2
x1
d x2
Division by d x1 yields:
dM
=
d x1
M
x1
M
x2
+
x2
x1
d x2
=
d x1
M
x1
x2
M
x2
x1
wherein the physical constraint on the mole fractions is recognized. Therefore
dM
¯
M1 = M + x 2
d x1
¯
The expression for M2 is found similarly.
688
11.47 (a) Apply Eq. (11.7) to species 1:
(n M E )
¯E
M1 =
n1
n2
Multiply the given equation by n and eliminate the mole fractions in favor of mole numbers:
n M E = An 1 n 2
1
1
+
n 1 + Bn 2 n 2 + Bn 1
1
1
+
n 1 + Bn 2 n 2 + Bn 1
¯E
M1 = An 2
+ n1
B
1
(n 1 + Bn 2 )2 (n 2 + Bn 1 )2
x1
B
1
+
2
(x2 + Bx1 )2
(x1 + Bx2 )
Conversion back to mole fractions yields:
1
1
+
x2 + Bx1
x1 + Bx2
¯E
M1 = Ax2
The rst term in the rst parentheses is combined with the rst term in the second parentheses
and the second terms are similarly combined:
¯E
M1 = Ax2
x1
1
1
x1 + Bx2
x1 + Bx2
+
Bx1
1
1
x2 + Bx1
x2 + Bx1
Reduction yields:
2
¯E
M1 = Ax2
1
B
+
(x1 + Bx2 )2 (x2 + Bx1 )2
2
¯E
M2 = Ax1
B
1
+
2
(x2 + Bx1 )2
(x1 + Bx2 )
Similarly,
(b) The excess partial properties should obey the Gibbs/Duhem equation, Eq. (11.14), when written
for excess properties in a binary system at constant T and P :
x1
¯E
¯E
d M2
d M1
=0
+ x2
d x1
d x1
If the answers to part (a) are mathematically correct, this is inevitable, because they were derived
¯
from a proper expression for M E . Furthermore, for each partial property MiE , its value and
derivative with respect to xi become zero at xi = 1.
1
1
¯E
¯E
+1
( M 2 ) = A 1 +
(c)
( M 1 ) = A
B
B
11.48 By Eqs. (11.15) and (11.16), written for excess properties, nd:
¯E
d2 M E
d M2
= x1
2
d x1
d x1
¯E
d2 M E
d M1
= x2
2
d x1
d x1
¯E
At x1 = 1, d M1 /d x1 = 0, and by continuity can only increase or decrease for x1 < 1. Therefore the
2
E
¯ 1 /d x1 is the same as the sign of d 2 M E /d x1 . Similarly, at x1 = 0, d M2 /d x1 = 0, and by
¯E
sign of d M
2
E
2
E
¯
the same argument the sign of d M2 /d x1 is of opposite sign as the sign of d M /d x1 .
689
11.49 The claim is not in general valid.
β
1
V
V
T
1
β id =
i
xi Vi
V id =
P
Vi
T
xi
i
xi Vi
i
The claim is valid only if all the Vi are equal.
690
1
=
P
i
xi Vi
xi Vi βi
i
Chapter 12 - Section B - Non-Numerical Solutions
12.2 Equation (12.1) may be written: yi P = xi πi Pi sat .
Summing for i = 1, 2 gives: P = x1 π1 P1sat + x2 π2 P2sat .
d π2
d π1
dP
π2
+ π1 + P2sat x2
= P1sat x1
d x1
d x1
d x1
Differentiate at constant T :
Apply this equation to the limiting conditions:
For x1 = 0 :
x2 = 1
π1 = π1
π2 = 1
For x1 = 1 :
x2 = 0
π1 = 1
π2 = π2
= P1sat π1 P2sat
or
Then,
dP
d x1
dP
d x1
Since both Pi
sat
and
x1 =0
x1 =1
πi
= P1sat P2sat π2
Whence,
x1 =0
P2sat
and
2
ln π1 = Ax2
12.4 By Eqs. (12.15),
By Eq. (12.1),
dP
d x1
dP
d x1
x 1 =0
x 1 =1
+ P2sat = P1sat π1
P1sat = P2sat π2
are always positive denite, it follows that:
dP
d x1
Therefore,
or
d π2
=0
d x1
d π1
=0
d x1
ln
and
dP
d x1
x1 =1
P1sat
2
ln π2 = Ax1
π1
2
2
= A ( x 2 x 1 ) = A ( x 2 x 1 ) = A (1 2 x 1 )
π2
y1 x2 P2sat
π1
=
=
y2 x1 P1sat
π2
y1 /x1
y2 /x2
P2sat
P1sat
= ξ12 r
ln(ξ12 r ) = A(1 2x1 )
If an azeotrope exists, ξ12 = 1 at 0
az
x1
1. At this value of x1 ,
az
ln r = A(1 2x1 )
The quantity A(1 2x1 ) is linear in x1 , and there are two possible relationships, depending on the
sign of A. An azeotrope exhists whenever | A| | ln r |. NO azeotrope can exist when | A| < | ln r |.
12.5 Perhaps the easiest way to proceed here is to note that an extremum in ln π1 is accompanied by the
opposite extremum in ln π2 . Thus the difference ln π1 ln π2 is also an extremum, and Eq. (12.8)
becomes useful:
d (G E/ RT
π1
=
ln π1 ln π2 = ln
d x1
π2
Thus, given an expression for G E/ RT = g (x1 ), we locate an extremum through:
d ln(π1 /π2 )
d 2 (G E/ RT )
=0
=
2
d x1
d x1
691
For the van Laar equation, write Eq. (12.16), omitting the primes ( ):
x1 x2
GE
= A12 A21
A
RT
dA
= A12 A21
d x1
Moreover,
d 2A
=0
2
d x1
and
d (G E/ RT )
= A12 A21
d x1
Then,
A A12 x1 + A21 x2
where
x1 x2 d A
x2 x1
2
A d x1
A
x2 x1
2x1 x2 d A
dA
x1 x2 d 2A
x2 x1 d A
2
d 2 (G E/ RT )
+
2
= A12 A21
2
2
3 dx
2
A2
A
d x1
A d x1
d x1
A
A
d x1
1
dA
d x1
2
dA
dA
2 A12 A21
+ x1 x2
A2 ( x 2 x 1 ) A
3
d x1
d x1
A
2
=
=
2 A12 A21
A3
= A12 A21
2x1 x2
2( x 2 x 1 ) d A
2
+
2
A3
d x1
A
A
A + x2
dA
d x1
x1
dA
A
d x1
This equation has a zero value if either A12 or A21 is zero. However, this makes G E/ RT everywhere
zero, and no extremum is possible. If either quantity in parentheses is zero, substitution for A and
d A /d x1 reduces the expression to A12 = 0 or A21 = 0, again making G E/ RT everywhere zero. We
conclude that no values of the parameters exist that provide for an extremum in ln(γ1 /γ2 ).
The Margules equation is given by Eq. (12.9b), here written:
GE
= Ax1 x2
RT
where
d 2A
=0
2
d x1
dA
= A21 A12
d x1
A = A21 x1 + A12 x2
dA
d (G E/ RT )
= A( x 2 x 1 ) + x 1 x 2
d x1
d x1
Then,
d 2A
dA
dA
d 2 (G E/ RT )
+ x1 x2 2
+ (x2 x1 )
= 2 A + (x2 x1 )
2
d x1
d x1
d x1
d x1
= 2 A + 2(x2 x1 )
dA
dA
A
= 2 (x1 x2 )
d x1
d x1
This equation has a zero value when the quantity in square brackets is zero. Then:
(x2 x1 )
dA
A = (x2 x1 )( A21 A12 ) A21 x1 A12 x2 = A21 x2 + A12 x1 2( A21 x1 + A12 x2 ) = 0
d x1
Substituting x2 = 1 x1 and solving for x1 yields:
x1 =
A21 2 A12
3( A21 A12 )
or
692
x1 =
(r 2)
3(r 1)
r
A21
A12
When r = 2, x1 = 0, and the extrema in ln γ1 and ln γ2 occur at the left edge of a diagram such
as those of Fig. 12.9. For values of r > 2, the extrema shift to the right, reaching a limiting value for
r = at x1 = 1/3. For positive values of the parameters, in all of these cases A21 > A12 , and the
intercepts of the ln γ2 curves at x1 = 1 are larger than the intercepts of the ln γ1 curves at x1 = 0.
When r = 1/2, x1 = 1, and the extrema in ln γ1 and ln γ2 occur at the right edge of a diagram
such as those of Fig. 12.9. For values of r < 1/2, the extrema shift to the left, reaching a limiting
value for r = 0 at x1 = 2/3. For positive values of the parameters, in all of these cases A21 < A12 ,
and the intercepts of the ln γ1 curves at x1 = 0 are larger than the intercepts of the ln γ2 curves at
x1 = 1.
No extrema exist for values of r between 1/2 and 2.
12.7 Equations (11.15) and (11.16) here become:
ln γ1 =
d (G E/ RT )
GE
+ x2
d x1
RT
and
ln γ2 =
d (G E/ RT )
GE
x1
d x1
RT
(a) For simplicity of notation, omit the primes that appear on the parameters in Eqs. (12.16) and
(12.17), and write Eq. (12.16) as:
x1 x2
GE
= A12 A21
D
RT
where
D A12 x1 + A21 x2
x1 x2
x2 x1
d (G E/ RT )
2 ( A12 A21 )
= A12 A21
D
D
d x1
Then,
and ln γ1 = A12 A21
x1 x2
+ x2
D
x1 x2
x2 x1
2 ( A12 A21 )
D
D
=
2
x1 x2
A12 A21
2
( A12 A21 )
x1 x2 + x2 x1 x2
D
D
=
2
2
A12 A21 x2
A12 A21 x2
( A21 x2 + A21 x1 )
( D A12 x1 + A21 x1 ) =
D2
D2
=
2
A12 A2 x2
21
= A12
D2
A21 x2
D
2
= A12
ln γ1 = A12 1 +
D
A21 x2
2
A12 x1
A21 x2
2
= A12
A12 x1 + A21 x2
A21 x2
The equation for ln γ2 is derived in analogous fashion.
(b) With the understanding that T and P are constant,
ln γ1 =
(nG E/ RT )
n1
n2
and Eq. (12.16) may be written:
A12 A21 n 1 n 2
nG E
=
nD
RT
where
693
n D = A12 n 1 + A21 n 2
2
Differentiation in accord with the rst equation gives:
ln γ1 = A12 A21 n 2
ln γ1 =
=
n1
1
n D ( n D )2
(n D )
n1
n2
A12 x1
A12 A21 x2
n1
A12 A21 n 2
1
A12 =
1
D
D
nD
nD
2
A12 A2 x2
A12 A21 x2
A12 A21 x2
21
A21 x2 =
( D A12 x1 ) =
D2
D2
D2
The remainder of the derivation is the same as in Part (a ).
12.10 This behavior requires positive deviations from Raoults law over part of the composition range and
negative deviations over the remainder. Thus a plot of G E vs. x1 starts and ends with G E = 0 at
x1 = 0 and x1 = 1 and shows positive values over part of the composition range and negative values
over the remainder, with an intermediate crossing of the x1 axis. Because these deviations are usually
quite small, the vapor pressures P1sat and P2sat must not be too different, otherwise the dewpoint and
bubblepoint curves cannot exhibit extrema.
12.11 Assume the Margules equation, Eq. (12.9b), applies:
GE
= x1 x2 ( A21 x1 + A12 x2 )
RT
But [see page 438, just below Eq. (12.10b)]:
A12 = ln γ1
1
GE
(equimolar) = (ln γ1 + ln γ2 )
8
RT
12.24 (a) By Eq. (12.6):
1
GE
(equimolar) = ( A12 + A21 )
8
RT
and
or
A21 = ln γ2
1
GE
(equimolar) = ln(γ1 γ2 )
8
RT
GE
= x1 ln γ1 + x2 ln γ2
RT
2
2
= x1 x2 (0.273 + 0.096 x1 ) + x2 x1 (0.273 0.096 x2 )
= x1 x2 (0.273 x2 + 0.096 x1 x2 + 0.273 x1 0.096 x1 x2 )
= x1 x2 (0.273)(x1 + x2 )
GE
= 0.273 x1 x2
RT
(b) The preceding equation is of the form from which Eqs. (12.15) are derived. From these,
2
ln γ1 = 0.273 x2
and
2
ln γ2 = 0.273 x1
(c) The equations of part (b) are not the reported expressions, which therefore cannot be correct. See
Problem 11.11.
12.25 Write Eq. (11.100) for a binary system, and divide through by d x1 :
x1
d ln γ2
d ln γ1
=0
+ x2
d x1
d x1
whence
694
x1 d ln γ1
x1 d ln γ1
d ln γ2
=
=
x2 d x2
x2 d x1
d x1
Integrate, recalling that ln γ2 = 1 for x1 = 0:
ln γ2 = ln(1) +
x1
0
x1 d ln γ1
d x1 =
x2 d x2
0
x1 d ln γ1
d x1
x2 d x2
d ln γ1
= 2 Ax2
d x2
2
(a) For ln γ1 = Ax2 ,
Whence
x1
x1
ln γ2 = 2 A
0
x1 d x1
2
ln γ2 = Ax1
or
GE
= Ax1 x2
RT
By Eq. (12.6),
2
(b) For ln γ1 = x2 ( A + Bx2 ),
d ln γ1
2
2
= 2x2 ( A + Bx2 ) + x2 B = 2 Ax2 + 3 Bx2 = 2 Ax2 + 3 Bx2 (1 x1 )
d x2
Whence
ln γ2 = 2 A
2
ln γ2 = Ax1 +
3B 2
3
x Bx1
21
x1
0
x1 d x1 + 3 B
x1
0
x1 d x1 3 B
2
ln γ2 = x1 A +
or
x1
0
2
x1 d x1
B
3B
2
Bx1 = x1 A + (1 + 2x2 )
2
2
3B
GE
2
2
Bx1 )
= x1 x2 ( A + Bx2 ) + x2 x1 ( A +
2
RT
Apply Eq. (12.6):
Algebraic reduction can lead to various forms of this equation; e.g.,
B
GE
= x 1 x 2 A + (1 + x 2 )
2
RT
2
2
(c) For ln γ1 = x2 ( A + Bx2 + C x2 ),
d ln γ1
2
2
2
3
= 2x2 ( A + Bx2 + C x2 ) + x2 ( B + 2C x2 ) = 2 Ax2 + 3 Bx2 + 4C x2
d x2
= 2 Ax2 + 3 Bx2 (1 x1 ) + 4C x2 (1 x1 )2
Whence
or
ln γ2 = 2 A
x1
0
x1 d x1 + 3 B
ln γ2 = (2 A + 3 B + 4C )
ln γ2 =
x1
0
x1
0
x1 (1 x1 )d x1 + 4C
x1 d x1 (3 B + 8C )
2 A + 3 B + 4C
2
2
ln γ2 = x1 A +
2
x1
x1
0
695
0
x1 (1 x1 )2 d x1
2
x1 d x1 + 4C
3 B + 8C
3
8C
3B
+ 2C B +
3
2
x1
3
4
x1 + C x1
2
x1 + C x1
x1
0
3
x1 d x1
or
2
ln γ2 = x1 A +
C
B
2
(1 + 2x2 ) + (1 + 2x2 + 3x2 )
3
2
The result of application of Eq. (12.6) reduces to equations of various forms; e.g.:
C
B
GE
2
= x1 x2 A + (1 + x2 ) + (1 + x2 + x2 )
3
2
RT
12.40 (a) As shown on page 458,
x1 =
1
1+n
˜
Eliminating 1 + n gives:
˜
Differentiation yields:
H=
H=
( A)
H
1d H
2
x1 d x1
x1
1
d x1
2
= x1
=
(1 + n ) 2
˜
dn
˜
dH
=
dn
˜
H x1
dHE
dH
= H E x1
d x1
d x1
E
dH
¯
H2E = H E x1
d x1
Comparison with Eq. (11.16) written with M H E ,
dH
¯
= H2E
dn
˜
shows that
(b) By geometry, with reference to the following gure,
Combining this with the result of Part (a) gives:
From which,
Substitute:
H (1 + n )
˜
H
x1
H d x1
1d H
dH
=
2
=
˜
˜
x1 d n
dn
˜
x1 d n
where
Whence,
and
I=
H=
¯
H2E =
HI
n
˜
H n H2E
˜¯
HE
H
=
x1
x1
696
HI
n
˜
dH
=
dn
˜
and
n=
˜
x2
x1
d x1
dn
˜
Whence,
¯
H E x2 H2E
x2 ¯
HE
H2E =
x1
x1
x1
I=
¯
¯
However, by the summability equation, H E x2 H2E = x1 H1E
¯
I = H1E
Then,
12.41 Combine the given equation with Eq. ( A) of the preceding problem:
H = x2 ( A21 x1 + A12 x2 )
With x2 = 1 x1 and x1 = 1/(1 + n ) (page 458): x2 =
˜
n
˜
1+n
˜
The preceding equations combine to give:
H=
˜
A12 n
A21
+
˜
1+n 1+n
˜
n
˜
1+n
˜
lim
(a) It follows immediately from the preceding equation that:
lim
(b) Because n /(1 + n ) 1 for n , it follows that:
˜
˜
˜
H =0
n 0
˜
H = A12
n
˜
2
¯
H2E = x1 [ A21 + 2( A12 A21 )x2 ]
(c) Analogous to Eq. (12.10b), page 438, we write:
Eliminate the mole fractions in favor of n :
˜
1
1+n
˜
¯
H2E =
2
A21 + 2( A12 A21 )
n
˜
1+n
˜
In the limit as n 0, this reduces to A21 . From the result of Part (a) of the preceding problem,
˜
it follows that
lim
n 0
˜
12.42 By Eq. (12.29) with M H ,
H=H
H
t
With
H
t
Therefore,
CP,
=
P ,x
i
dH
= A21
dn
˜
xi Hi . Differentiate:
H
t
H0
d( H ) =
t
t0
C P dt
697
P ,x
i
H
t
this becomes
P ,x
H
xi
Hi
t
P ,x
P ,x
xi C Pi =
= CP
H=
i
H0 +
t
t0
C P dt
CP
M E = x1 x2 M
( A)
dM
dME
= M( x 2 x 1 ) + x 1 x 2
d x1
d x1
(B)
12.61 (a) From the denition of M:
Differentiate:
Substitution of Eqs. ( A) & ( B ) into Eqs. (11.15) & (11.16), written for excess properties, yields
the required result.
(b) The requested plots are found in Section A.
12.63 In this application the microscopic state of a particle is its species identity, i.e., 1, 2, 3, . . . . By
assumption, this label is the only thing distinguishing one particle from another. For mixing,
t
t
t
S t = Smixed Sunmixed = Smixed
i
Sit
where the total emtropies are given by Eq. (5.42). Thus, for an unmixed species i , and for the mixed
system of particles,
Sit = k ln
i
= k ln
Ni !
=0
Ni !
t
Smixed = k ln
S t = k ln
Combining the last three equations gives:
N!
N1 ! N2 ! N3 ! · · ·
N!
N1 ! N2 ! N3 ! · · ·
1
N!
1
St
St
S
= (ln N !
ln
=
=
=
N
N1 ! N2 ! N3 ! · · ·
N
kN
R( N / N A)
R
From which:
ln N ! N ln N N
1
S
( N ln N N
N
R
=
1
( N ln N
N
ln Ni ! Ni ln Ni Ni
and
Ni ) =
Ni ln Ni +
i
i
xi N ln xi
i
ln Ni !)
i
1
( N ln N
N
xi N ln N ) =
i
xi N ln xi N )
i
xi ln x1
i
12.66 Isobaric data reduction is complicated by the fact that both composition and temperature vary from
point to point, whereas for isothermal data composition is the only signicant variable. (The effect
of pressure on liquid-phase properties is assumed negligible.) Because the activity coefcients are
strong functions of both liquid composition and T , which are correlated, it is quite impossible without
additional information to separate the effect of composition from that of T . Moreover, the Pi sat values
depend strongly on T , and one must have accurate vapor-pressure data over a temperature range.
12.67 (a) Written for G E , Eqs. (11.15) and (11.16) become:
E
dG
¯E
G 1 = G E + x2
d x1
Divide through by RT ;
Then
Given:
dene G
ln γ1 = G + x2
and
GE
;
RT
dG
d x1
GE
= A1/ k
x! x2 RT
and
with
698
E
dG
¯E
G 2 = G E x1
d x1
note by Eq. (11.91) that
ln γ2 = G x1
dG
d x1
A x 1 Ak + x 2 Ak
21
12
¯
G iE
= ln γi
RT
G = x 1 x 2 A 1/ k
Whence:
d A 1/ k
dG
+ A 1/ k ( x 2 x 1 )
= x1 x2
d x1
d x1
and
1 A 1/ k k
dA
1
d A 1/ k
( A21 Ak )
=
= A(1/ k )1
12
kA
d x1
k
d x1
2
ln γ1 = x2 A1/ k
Finally,
( Ak Ak ) x 1
21
12
+1
kA
2
ln γ2 = x1 A1/ k 1
Similarly,
A 1/ k k
dG
( A21 Ak )+ A1/ k (x2 x1 )
= x1 x2
12
kA
d x1
and
( Ak Ak ) x 2
21
12
kA
(b) Appropriate substitition in the preceding equations of x1 = 1 and x1 = 0 yields:
ln γ2 = A1/ k = ( Ak )1/ k = A21
21
ln γ1 = A1/ k = ( Ak )1/ k = A12
12
GE
= A1/ k = (x1 Ak + x2 Ak )1/ k
12
21
x1 x2 RT
g = x1 A21 + x2 A12
(Margules equation)
(c) Let
g
If k = 1,
A21 A12
(van Laar equation)
x1 A12 + x2 A21
For k = 0, , +, indeterminate forms appear, most easily resolved by working with the
logarithm:
1
ln g = ln(x1 Ak + x2 Ak )1/ k = ln x1 Ak + x2 Ak
21
12
21
12
k
g = (x1 A1 + x2 A1 )1 =
21
12
If k = 1,
Apply lHˆ pitals rule to the nal term:
o
d ln x1 Ak + x2 Ak
x1 Ak ln A21 + x2 Ak ln A12
21
12
21
12
=
k
dk
x1 A21 + x2 Ak
12
( A)
Consider the limits of the quantity on the right as k approaches several limiting values.
For k 0,
x1
x2
ln g x1 ln A21 + x2 ln A12 = ln A21 + ln A12
and
x1 x2
g = A21 A12
For k ± , Assume A12 / A21 > 1, and rewrite the right member of Eq. ( A) as
x1 ln A21 + x2 ( A12 / A21 )k ln A12
x1 + x2 ( A12 / A21 )k
For k ,
lim ( A12 / A21 )k 0
k
Whence
For k +,
Whence
g = A21
lim ln g = ln A21
k
except at x1 = 0 where g = A12
lim ( A12 / A21 )k
k
g = A12
and
and
lim ln g = ln A12
k
except at x1 = 1 where g = A21
If A12 / A21 < 1 rewrite Eq. ( A) to display A21 / A12 .
699
12.68 Assume that Eq. (12.1) is the appropriate equilibrium relation, written as
xe γe Pesat = xe γe Pesat = ye P
e EtOH
Because P is low, we have assumed ideal gases, and for small xe let γe γe . For volume fraction
ξe in the vapor, the ideal-gas assumption provides ξev ye , and for the liquid phase, with xe small
xe Vel
xe Vel
xe Vel
Vb
xb Vb
xe Vel + xb Vb
l
ξe =
Ve P
volume % EtOH in blood
Vb γe Pesat
volume % EtOH in gas
Vb l sat
ξ γ P ξev P
Ve e e e
Then
12.70 By Eq. (11.95),
κ (G E / RT )
κT
HE
= T
RT
b blood
P ,x
E
G
= x1 ln(x1 + x2
RT
κ (G E / RT )
κT
12 )
x2 ln(x2 + x1
d 21
d 12
x2 x1
dT
dT
=
x 2 + x 1 21
x 1 + x 2 12
x
d 12
H
dT
= x1 x2 T
x1 + x2
RT
ij
=
Vj
d ij
=
Vi
dT
H E = x1 x2
12
d 21
dT
+
x2 + x1
ai j
Vj
exp
RT
Vi
exp
12 a12
x1 + x2
+
12
21
(i = j )
ai j
=
RT 2
ai j
RT
ij
(12.24)
ai j
RT 2
21 a21
x2 + x1
21
E
Because C P = d H E /dT , differentiate the preceding expression and reduce to get:
E
x2 21 (a21 / RT )2
x1 12 (a12 / RT )2
CP
+
= x1 x2
( x 2 + x 1 21 ) 2
(x1 + x2 12 )2
R
Because
12
and
21
(12.18)
x1 x2
E
21 )
E
must always be positive numbers, C P must always be positive.
700
Chapter 13 - Section B - Non-Numerical Solutions
13.1 (a)
4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g)
νi = 4 5 + 4 + 6 = 1
ν=
n0 =
=2+5=7
i
i0
By Eq. (13.5),
yNH3 =
2 4ε
7+ε
yO2 =
5 5ε
7+ε
4ε
7+ε
yNO =
yH2 O =
6ε
7+ε
2H2 S(g) + 3O2 (g) 2H2 O(g) + 2SO2 (g)
(b)
νi = 2 3 + 2 + 2 = 1
ν=
=3+5=8
n0 =
i0
i
By Eq. (13.5),
yH2 S =
3 2ε
8ε
yO2 =
5 3ε
8ε
yH2 O =
2ε
8ε
ySO2 =
2ε
8ε
6NO2 (g) + 8NH3 (g) 7N2 (g) + 12H2 O(g)
(c)
νi = 6 8 + 7 + 12 = 5
ν=
n0 =
=3+4+1=8
i0
i
By Eq. (13.5),
yNO2 =
3 6ε
8 + 5ε
yNH3 =
4 8ε
8 + 5ε
yN2 =
1 + 7ε
8 + 5ε
yH2 O =
12ε
8 + 5ε
C2 H4 (g) + 1 O2 (g) (CH2 )2 O(g)
2
13.2
(1)
C2 H4 (g) + 3O2 (g) 2CO2 (g) + 2H2 O(g)
(2)
The stoichiometric numbers νi, j are as follows:
i=
C2 H4
O2
(CH2 )2 O
CO2
H2 O
j
νj
1
1
1
2
1
0
0
1
2
2
1
3
0
2
2
0
n0 =
=2+3=5
i0
By Eq. (13.7),
yC2 H4 =
2 ε1 ε2
5 1 ε1
2
yCO2 =
yO2 =
3 1 ε1 3ε2
2
2ε2
5 1 ε1
2
701
5
1
ε
21
yH2 O =
y (CH2 )2
2 ε2
5 1 ε1
2
O
=
ε1
5 1 ε1
2
13.3
CO2 (g) + 3H2 (g) CH3 OH(g) + H2 O(g)
(1)
CO2 (g) + H2 (g) CO(g) + H2 O(g)
(2)
The stoichiometric numbers νi, j are as follows:
i=
CO2
H2
CH3 OH
CO
H2 O
j
νj
1
2
1
1
3
1
1
0
n0 =
0
1
1
1
2
0
=2+5+1=8
i0
By Eq. (13.7),
yCO2 =
2 ε1 ε2
8 2ε1
13.7 The equation for
5 3ε1 ε2
8 2ε1
yH2 =
yCH3 OH =
ε1
8 2ε1
yCO =
1 + ε2
8 2ε1
yH2 O =
ε 1 + ε2
8 2ε1
G , appearing just above Eq. (13.18) is:
H0
G =
T
( H0
T0
G ) + R
0
T
CP
dT RT
R
T0
T
C P dT
RT
T0
To calculate values of G , one combines this equation with Eqs. (4.19) and (13.19), and evaluates
parameters. In each case the value of H0 = H298 is tabulated in the solution to Pb. 4.21. In
addition, the values of A, B , C , and D are given in the solutions to Pb. 4.22. The required
values of G = G in J mol1 are:
0
298
(a) 32,900; (f ) 2,919,124; (i) 113,245; (n) 173,100; (r) 39,630; (t) 79,455; (u) 166,365;
(x) 39,430; (y) 83,010
13.8 The relation of K y to P and K is given by Eq. (13.28), which may be concisely written:
P
P
Ky =
ν
K
(a) Differentiate this equation with respect to T and combine with Eq. (13.14):
Ky
T
=
P
P
P
ν
Ky H
d ln K
Ky d K
dK
=
= Ky
=
RT 2
dT
K dT
dT
Substitute into the given equation for (εe / T ) P :
εe
T
=
P
K y d εe
RT 2 d K y
H
(b) The derivative of K y with respect to P is:
Ky
P
= ν
T
P
P
ν 1
1
K = ν K
P
702
P
P
ν
P
P
1
ν K y
1
=
P
P
Substitute into the given equation for (εe / P )T :
εe
P
(c) With K y
i
( yi )νi , ln K y =
K y d εe
(ν)
P d Ky
=
T
νi ln yi . Differentiation then yields:
i
1 d Ky
=
K y d εe
i
νi dyi
yi d εe
( A)
dn
d ni
yi
d εe
d εe
Because yi = n i / n ,
1
n i dn
1 dn i
dyi
=
2
=
n
n d εe
n d εe
d εe
But
n i = n i0 + νi εe
and
n = n 0 + νεe
dn i
= νi
d εe
and
dn
=ν
d εe
Whence,
νi yi ν
dyi
=
n 0 + νεe
d εe
Therefore,
Substitution into Eq. ( A) gives
1 d Ky
K y d εe
νi yi ν
n 0 + νεe
νi
yi
=
i
1
n 0 + νεe
=
m
=
νi2
νi
yi
i =1
1
n 0 + νεe
νi2
νi ν
yi
i
m
νk
k =1
In this equation, both K y and n 0 + νεe (= n ) are positive. It remains to show that the summation
term is positive. If m = 2, this term becomes
2
( y2 ν1 y1 ν2 )2
ν2
ν1
ν1 (ν1 + ν2 ) + 2 ν2 (ν1 + ν2 ) =
y1 y2
y2
y1
where the expression on the right is obtained by straight-forward algebraic manipulation. One
can proceed by induction to nd the general result, which is
m
νi2
νi
yi
i =1
m
m
νk
m
i
k
=
k =1
( yk νi yi νk )2
yi yk
(i < k )
All quantities in the sum are of course positive.
1
N (g)
22
13.9
+ 3 H2 (g) NH3 (g)
2
For the given reaction, ν = 1, and for the given amounts of reactants, n 0 = 2.
By Eq. (13.5),
By Eq. (13.28),
yN2 =
1
(1
2
εe )
2 εe
yNH3
1 /2 3 /2
yN2 yH2
=
yH2 =
[ 1 (1
2
3
(1
2
εe )
2 εe
yNH3 =
εe
2 εe
P
εe (2 εe )
=K
3
P
εe )]1/2 [ 2 (1 εe )]3/2
703
Whence,
1 /2
1
2
εe (2 εe )
=
(1 εe )2
3/2
3
2
K
P
P
= 1.299 K
P
P
r εe 2 2 r εe + (r 1) = 0
This may be written:
where,
r 1 + 1.299 K
The roots of the quadratic are:
1
= 1 ± r 1 /2
r 1 /2
εe = 1 ±
Because εe < 1, εe = 1 r 1/2 ,
P
P
εe = 1 1 + 1.299 K
1/2
P
P
13.10 The reactions are written:
Mary:
2NH3 + 3NO 3H2 O + 5 N2
2
( A)
Paul:
4NH3 + 6NO 6H2 O + 5N2
(B)
Peter:
3H2 O + 5 N2 2NH3 + 3NO
2
(C )
Each applied Eqs. (13.11b) and (13.25), here written:
ln K = G / RT
and
( fˆi )νi
K = ( P )ν
i
For reaction ( A),
G = 3 G fH
A
2O
2 G fNH 3 G fNO
3
For Marys reaction ν = 1 , and:
2
5/2
fˆfN
fˆf3H
1
2
2O
K A = (P )
and
2
fˆf2NH fˆf3NO
G
A
RT
ln K A =
3
For Pauls reaction ν = 1, and
fˆf6H
1
KB = (P )
2O
fˆf5N
2
and
fˆf4NH fˆf6NO
ln K B =
3
2 G
A
RT
For Peters reaction ν = 1 , and:
2
KC = ( P )
1
2
fˆf2NH fˆf3NO
and
3
fˆf3H
2O
5 /2
fˆfN
ln K C =
2
In each case the two equations are combined:
Mary:
1
2
(P )
fˆf3H
2O
5 /2
fˆfN
2
fˆf2NH fˆf3NO
3
704
= exp
G
A
RT
G
A
RT
1
Paul:
(P )
fˆf6H
2O
fˆf5N
G
A
= exp
RT
2
fˆf4NH fˆf6NO
3
2
Taking the square root yields Marys equation.
1
Peter:
( P ) 2
fˆf2NH fˆf3NO
3
fˆf3H
2O
5/2
fˆfN
= exp
2
G
A
RT
1
Taking the reciprocal yields Marys equation.
13.24 Formation reactions:
1
N
22
+ 3 H2 NH3
2
(1)
1
N
22
+ 1 O2 NO
2
(2)
1
N
22
+ O2 NO2
(3)
H 2 + 1 O2 H 2 O
2
(4)
Combine Eq. (3) with Eq. (1) and with Eq. (2) to eliminate N2 :
NO2 + 3 H2 NH3 + O2
2
(5)
NO2 1 O2 + NO
2
(6)
The set now comprises Eqs. (4), (5), and (6); combine Eq. (4) with Eq. (5) to eliminate H2 :
(7)
NO2 + 3 H2 O NH3 + 1 3 O2
4
2
Equations (6) and (7) represent a set of independent reactions for which r = 2. Other equivalent sets
of two reactions may be obtained by different combination procedures. By the phase rule,
F = 2π + N r s = 21+520
13.35 (a) Equation (13.28) here becomes:
yB
=
yA
P
P
F =4
0
K=K
yB
= K (T )
1 yB
Whence,
(b) The preceding equation indicates that the equilibrium composition depends on temperature only.
However, application of the phase rule, Eq. (13.36), yields:
F =2+211=2
This result means in general for single-reaction equilibrium between two species A and B that
two degrees of freedom exist, and that pressure as well as temperature must be specied to x the
equilibrium state of the system. However, here, the specication that the gases are ideal removes
ˆ
the pressure dependence, which in the general case appears through the φi s.
13.36 For the isomerization reaction in the gas phase at low pressure, assume ideal gases. Equation (13.28)
then becomes:
yB
=
yA
P
P
0
K=K
whence
705
1 yA
= K (T )
yA
Assume that vapor/liquid phase equilibrium can be represented by Raoults law, because of the low
pressure and the similarity of the species:
xA PAsat (T ) = yA P
(1 xA ) PBsat (T ) = (1 yA ) P
and
F = 2π + N r = 22+21 = 1
(a) Application of Eq. (13.36) yields:
(b) Given T , the reaction-equilibriuum equation allows solution for yA . The two phase-equilibrium
equations can then be solved for xA and P . The equilibrium state therefore depends solely on T .
13.38 (a) For low pressure and a temperature of 500 K, the system is assumed to be a mixture of ideal
gases, for which Eq. (13.28) is appropriate. Therefore,
yMX
=
yOX
P
P
0
yPX
=
yOX
KI = KI
P
P
0
yEB
=
yOX
K II = K II
P
P
0
K III = K III
(b) These equation equations lead to the following set:
yMX = K I yOX
yEB = K III yOX
(2)
yPX = K II yOX
(1)
(3)
The mole fractions must sum to unity, and therefore:
yOX + K I yOX + K II yOX + K III yOX = yOX (1 + K I + K II + K III ) = 1
yOX =
(c) With the assumption that
combine to give:
(4)
C P = 0 and therefore that K 2 = 1, Eqs. (13.20), (13.21), and (13.22)
K = K 0 K 1 = exp
Whence,
1
1 + K I + K II + K III
K = exp
G
298
exp
RT0
H298
T0
1
T
RT0
298.15
500
(8.314)(298.15)
H298 1
G
298
The data provided lead to the following property changes of reaction and equilibrium constants
at 500 K:
Reaction
I
II
III
H298
G
298
1,750
1,040
10,920
3,300
1,000
8,690
706
K
2.8470
1.2637
0.1778
(d) Substitution of numerical values into Eqs. (1), (2), (3), and (4) yields the following values for the
mole fractions:
yOX = 0.1891
13.40 For the given owrates,
yMX = 0.5383
n A0 = 10
yPX = 0.2390
and n B0 = 15,
nA
nB
nC
nD
n
yEB = 0.0336
with n A0 the limiting reactant without (II)
= n A0 εI εII
= n B0 εI
= εI εII
= εII
= n 0 εI εII
Use given values of YC and SC / D to nd εI and εII :
YC =
εI εII
n A0
and
εI εII
εII
SC / D =
Solve for εI and εII :
εII =
= 10 6 2
= 15 6
=62
=2
= 17
nA
nB
nC
nD
n
2+1
2
SC / D + 1
n A0 YC =
SC / D
εI =
× 10 × 0.40 = 6
10 × 0.40
n A0 YC
=2
=
2
SC / D
yA
yB
yC
yD
=2
=9
=4
=2
= 2/17
= 9/17
= 4/17
= 2/17
=1
= 0.1176
= 0.5295
= 0.2353
= 0.1176
13.42 A compound with large positive G f has a disposition to decompose into its constituent elements.
Moreover, large positive G f often implies large positive H . Thus, if any decomposition product
f
is a gas, high pressures can be generated in a closed system owing to temperature increases resulting
from exothermic decomposition.
13.44 By Eq. (13.12),
G
νi G i
G
P
i
and from Eq. (6.10), ( G i / P )T = Vi
=
For the ideal-gas standard state,
G
P
=
T
νi
i
RT
P
Vi
T
νi
i
G i
P
=
T
i
νi Vi
= RT / P . Therefore
=
ν RT
P
and
G ( P2 )
G ( P1 ) = ν RT ln
13.47 (a) For isomers at low pressure Raoults law should apply:
P = x A PAsat + x B PBsat = PBsat + x A ( PAsat PBsat )
For the given reaction with an ideal solution in the liquid phase, Eq. (13.33) becomes:
Kl =
1 xA
xB
=
xA
xA
from which
707
xA =
1
Kl + 1
P2
P1
The preceding equation now becomes,
P = 1
P=
Kl
Kl
Kl + 1
For K l = 0
1
+1
PBsat +
PBsat +
Kl
Kl
1
+1
1
+1
PAsat
For K l =
P = PAsat
PAsat
( A)
P = PBsat
(b) Given Raoults law:
1 = x A + x B = yA
P=
y A / PAsat
yB
yA
sat +
PBsat
PA
P
P
=P
sat + y B
PBsat
PA
PAsat PBsat
PAsat PBsat
1
sat
sat
sat =
sat =
PA + y A ( PBsat PAsat )
y A PB + y B PA
+ y B / PB
For the given reaction with ideal gases in the vapor phase, Eq. (13.28) becomes:
yB
= Kv
yA
whence
yA =
Kv
1
+1
Elimination of y A from the preceding equation and reduction gives:
P=
For K v = 0
( K v + 1) PAsat PBsat
K v PAsat + PBsat
P = PAsat
(B)
For K v =
P = PBsat
(c) Equations ( A) and ( B ) must yield the same P . Therefore
Kl
Kl + 1
Some algebra reduces this to:
PBsat +
1
l +1
K
PAsat =
( K v + 1) PAsat PBsat
K v PAsat + PBsat
P sat
Kv
= Bsat
PA
Kl
(d) As mentioned already, the species (isomers) are chemically similar, and the low pressure favors
ideal-gas behavior.
(e) F = N + 2 π r = 2 + 2 2 1 = 1
708
Thus xing T should sufce.
Chapter 14 - Section B - Non-Numerical Solutions
14.2 Start with the equation immediately following Eq. (14.49), which can be modied slightly to read:
ˆ
ln νi =
ε ln Z
ε(nG R/ RT ) ε(n Z )
+1
+n
ε ni
ε ni
ε ni
where the partial derivatives written here and in the following development without subscripts are
understood to be at constant T , n /ρ (or ρ/ n ), and n j . Equation (6.61) after multiplication by n can
be written:
ρ2
3
ρ
nG R
n ln Z
+ n 2 (nC )
= 2n (n B )
n
2
n
RT
Differentiate:
3ρ
ρ
ε(nG R/ RT )
¯
(n B + n Bi ) +
=2
2n
n
ε ni
or
By denition,
2
¯
(2n 2 C + n 2 Ci ) n
ε ln Z
ln Z
ε ni
ε ln Z
3
ε(nG R/ RT )
¯
¯
ln Z
= 2ρ( B + Bi ) + ρ 2 (2C + Ci ) n
ε ni
2
ε ni
ε (n B )
ε ni
¯
Bi
and
T ,n j
¯
Ci
ε (nC )
ε ni
T ,n j
The equation of state, Eq. (3.40), can be written:
Z = 1 + Bρ + Cρ 2
Differentiate:
or
n Z = n + n (n B )
ρ
ρ
ε(n Z )
¯
(n B + n Bi ) +
=1+
n
n
ε ni
2
ρ
ρ
+ n 2 (nC )
n
n
2
¯
(2n 2 C + n 2 Ci )
ε(n Z )
¯
¯
= 1 + ρ( B + Bi ) + ρ 2 (2C + Ci )
ε ni
or
When combined with the two underlined equations, the initial equation reduces to:
¯
¯
ˆ
ln νi = 1 + ρ( B + Bi ) + 1 ρ 2 (2C + Ci )
2
The two mixing rules are:
2
2
B = y1 B11 + 2 y1 y2 B12 + y2 B22
3
2
2
3
C = y1 C111 + 3 y1 y2 C112 + 3 y1 y2 C122 + y2 C222
¯
¯
Application of the denitions of Bi and Ci to these mixing rules yields:
2
2
¯
B1 = y1 (2 y1 ) B11 + 2 y2 B12 y2 B22
2
2
2
3
¯
C1 = y1 (3 2 y1 )C111 + 6 y1 y2 C112 + 3 y2 (1 2 y1 )C122 2 y2 C222
2
2
¯
B2 = y1 B11 + 2 y1 B12 + y2 (2 y2 ) B22
3
2
2
2
¯
C2 = 2 y1 C111 + 3 y1 (1 2 y2 )C112 + 6 y1 y2 C122 + 2 y2 (3 2 y2 )C222
709
In combination with the mixing rules, these give:
¯
B + B1 = 2( y1 B11 + y2 B12 )
2
2
¯
2C + C1 = 3( y1 C111 + 2 y1 y2 C112 + y2 C122 )
¯
B + B2 = 2( y2 B22 + y1 B12 )
2
2
¯
2C + C2 = 3( y2 C222 + 2 y1 y2 C122 + y1 C112 )
In combination with the boxed equation these expressions along with Eq. (3.40) allow calculation of
ˆ
ˆ
ln φ1 and ln φ2 .
14.11 For the case described, Eqs. (14.1) and (14.2) combine to give:
yi P = xi Pi sat
φisat
ˆ
φi
ˆ
If the vapor phase is assumed an ideal solution, φi = φi , and
yi P = xi Pi sat
φisat
φi
When Eq. (3.38) is valid, the fugacity coefcient of pure species i is given by Eq. (11.36):
ln φi =
Therefore,
ln
Bii P
RT
and
φisat =
Bii Pi sat
RT
Bii ( Pi sat P )
Bii P
Bii Pi sat
φisat
=
= ln φisat ln φi =
RT
RT
RT
φi
For small values of the nal term, this becomes approximately:
Bii ( Pi sat P )
φisat
=1+
RT
φi
yi P = xi Pi sat 1 +
Whence,
yi P xi Pi sat =
or
Bii ( Pi sat P )
RT
xi Pi sat Bii ( Pi sat P )
RT
Write this equation for species 1 and 2 of a binary mixture, and sum. This yields on the left the
difference between the actual pressure and the pressure given by Raoults law:
P P (RL) =
x1 B11 P1sat ( P1sat P ) + x2 B22 P2sat ( P2sat P )
RT
Because deviations from Raoults law are presumably small, P on the right side may be replaced by
its Raoults-law value. For the two terms,
P1sat P = P1sat x1 P1sat x2 P2sat = P1sat (1 x2 ) P1sat x2 P2sat = x2 ( P1sat P2sat )
P2sat P = P2sat x1 P1sat x2 P2sat = P2sat x1 P1sat (1 x1 ) P2sat = x1 ( P2sat P1sat )
Combine the three preceding equations:
P P (RL) =
=
x1 x2 B11 ( P1sat P2sat ) P1sat x1 x2 B22 ( P1sat P2sat ) P2sat
RT
x1 x2 ( P1sat P2sat )
( B11 P1sat B22 P2sat )
RT
710
Rearrangement yields the following:
P P (RL) =
x1 x2 ( P1sat P2sat )2
RT
B11 P1sat B22 P2sat
P1sat P2sat
=
( B11 B22 ) P2sat
x1 x2 ( P1sat P2sat )2
B11 +
P1sat P2sat
RT
=
B22
x1 x2 ( P1sat P2sat )2
( B11 ) 1 + 1
B11
RT
P2sat
P1sat P2sat
Clearly, when B22 = B11 , the term in square brackets equals 1, and the pressure deviation from the
Raoults-law value has the sign of B11 ; this is normally negative. When the virial coefcients are not
equal, a reasonable assumption is that species 2, taken here as the heavier species (the one with
the smaller vapor pressure) has the more negative second virial coefcient. This has the effect of
making the quantity in parentheses negative and the quantity in square brackets < 1. However, if this
latter quantity remains positive (the most likely case), the sign of B11 still determines the sign of the
deviations.
14.13 By Eq. (11.90), the denition of γi ,
Whence,
ln γi = ln fˆi ln xi ln f i
1
1 d fˆi
1
d ln fˆi
d ln γi
=
=
ˆi d xi
xi
xi
d xi
d xi
f
1 d fˆi
>0
fˆi d xi
Combination of this expression with Eq. (14.71) yields:
Because fˆi 0,
d fˆi
>0
d xi
(const T , P )
RT d fˆi
d ln fˆi
d µi
=
= RT
d xi
d xi
fˆi d xi
By Eq. (11.46), the denition of fˆi ,
d µi
>0
d xi
Combination with Eq. (14.72) yields:
(const T , P )
14.14 Stability requires that
G < 0 (see Pg. 575). The limiting case obtains when
event Eq. (12.30) becomes:
G E = RT
xi ln xi
G = 0, in which
i
For an equimolar solution xi = 1/ N where N is the number of species. Therefore,
G E (max) = RT
i
1
1
= RT
ln
N
N
For the special case of a binary solution, N = 2, and
711
i
1
ln N = RT ln N
N
G E (max) = RT ln 2
G E = δ12 P y1 y2
14.17 According to Pb. 11.35,
This equation has the form:
or
δ12 P
GE
y1 y2
=
RT
RT
GE
= Ax1 x2
RT
for which it is shown in Examples 14.5 and 14.6 that phase-splitting occurs for A > 2. Thus, the
formation of two immiscible vapor phases requires: δ12 P / RT > 2.
Suppose T = 300 K and P = 5 bar. The preceding condition then requires: δ12 > 9977 cm3 mol1
for vapor-phase immiscibility. Such large positive values for δ12 are unknown for real mixtures.
(Examples of gas/gas equilibria are known, but at conditions outside the range of applicability of the
two-term virial EOS.)
GE
= Ax1 x2
RT
14.19 Consider a quadratic mixture, described by:
It is shown in Example 14.5 that phase splitting occurs for such a mixture if A > 2; the value of
A = 2 corresponds to a consolute point, at x1 = x2 = 0.5. Thus, for a quadratic mixture,
phase-splitting obtains if:
11
G E > 2 · · · RT = 0.5 RT
22
This is a model-dependent result. Many liquid mixtures are known which are stable as single phases,
even though G E > 0.5 RT for equimolar composition.
14.21 Comparison of the Wilson equation, Eq. (12.18) with the modied Wilson equation shows that
(G E/ RT )m = C (G E/ RT ), where subscript m distinguishes the modied Wilson equation from
the original Wilson equation. To simplify, dene g (G E/ RT ); then
gm = Cg
ngm = Cng
(ng )
(ngm )
=C
n1
n1
ln(γ1 )m = C ln γ1
where the nal equality follows from Eq. (11.96). Addition and subtraction of ln x1 on the left side
of this equation and of C ln x1 on the right side yields:
ln(x1 γ1 )m ln x1 = C ln(x1 γ1 ) C ln x1
or
Differentiate:
ln(x1 γ1 )m = C ln(x1 γ1 ) (C 1) ln x1
d ln(x1 γ1 ) C 1
d ln(x1 γ1 )m
=C
x1
d x1
d x1
As shown in Example 14.7, the derivative on the right side of this equation is always positive. However, for C sufciently greater than unity, the contribution of the second term on the right can make
d ln(x1 γ1 )M
<0
d x1
over part of the composition range, thus violating the stability condition of Eq. (14.71) and implying
the formation of two liquid phases.
14.23 (a) Refer to the stability requirement of Eq. (14.70). For instability, i.e., for the formation of two
liquid phases,
1
d 2 (G E / RT )
<
2
x1 x2
d x1
712
over part of the composition range. The second derivative of G E must be sufciently negative so
as to satisfy this condition for some range of x1 . Negative curvature is the norm for mixtures for
which G E is positive; see, e.g., the sketches of G E vs. x1 for systems (a), (b), (d), (e), and (f ) in
Fig. 11.4. Such systems are candidates for liquid/liquid phase splitting, although it does not in
fact occur for the cases shown. Rather large values of G E are usually required.
(b) Nothing in principle precludes phase-splitting in mixtures for which G E < 0; one merely requires that the curvature be sufciently negative over part of the composition range. However,
positive curvature is the norm for such mixtures. We know of no examples of liquid/liquid phasesplitting in systems exhibiting negative deviations from ideal-solution behavior.
14.29 The analogy is Raoults law, Eq. (10.1), applied at constant P (see Fig. 10.12): yi P = xi Pi sat
If the vapor phase in VLE is ideal and the liquid molar volumes are negligible (assumptions inherent
in Raoults law), then the Clausius/Clapeyron equation applies (see Ex. 6.5):
Hil v
d ln Pi sat
=
RT 2
dT
Integration from the boiling temperature Tbi at pressure P (where Pi sat = P ) to the actual temperature
T (where Pi sat = Pi sat ) gives:
T
Hil v
P sat
dT
ln i =
2
P
Tbi RT
Combination with Eq. (10.1) yields:
yi = xi exp
T
Tbi
Hil v
dT
RT 2
which is an analog of the Case I SLE equations.
14.30 Consider binary (two-species) equilibrium between two phases of the same kind. Equation (14.74)
applies:
ββ
xiα γiα = xi γi
(i = 1, 2)
β
β
α
If phase β is pure species 1 and phase α is pure species 2, then x1 = γ1 = 1 and x2 = γ2α = 1.
Hence,
β
β
α
x1 γ1α = x1 γ1 = 1
and
β
β
α
x2 γ2α = x2 γ2 = 1
The reasoning applies generally to (degenerate) N -phase equilibrium involving N mutually immiscible species. Whence the cited result for solids.
14.31 The rules of thumb are based on Case II binary SLE behavior. For concreteness, let the solid be pure
species 1 and the solvent be liquid species 2. Then Eqs. (14.93) and (14.92a) apply:
x1 = ψ1 = exp
(a) Differentiate:
sl
H1
RTm 1
T Tm 1
T
sl
H1
d x1
= ψ1 ·
RT 2
dT
Thus d x1 /dT is necessarily positive: the solid solubility x1 increases with increasing T .
(b) Equation (14.92a) contains no information about species 2. Thus, to the extent that Eqs. (14.93)
and (14.92a) are valid, the solid solubility x1 is independent of the identity of species 2.
713
(c) Denote the two solid phases by subscripts A and B . Then, by Eqs. (14.93) and (14.92a), the
solubilities x A and x B are related by:
H sl (Tm B Tm A )
RTm A Tm B
xA
= exp
xB
sl
HA =
where by assumption,
sl
HB
H sl
Accordingly, x A /x B > 1 if and only if TA < TB , thus validating the rule of thumb.
(d) Identify the solid species as in Part (c). Then x A and x B are related by:
sl
sl
( H B H A )(Tm T )
xA
= exp
RTm T
xB
where by assumption,
Tm A = Tm B Tm
Notice that Tm > T (see Fig. 14.21b). Then x A /x B > 1 if and only if
accord with the rule of thumb.
sl
HA <
sl
H B , in
14.34 The shape of the solubility curve is characterized in part by the behavior of the derivative dyi /d P
(constant T ). A general expression is found from Eq. (14.98), y1 = P1sat P / F1 , where the enhancement factor F1 depends (at constant T ) on P and y1 . Thus,
P sat
P sat
dy1
= 1 2 F1 + 1
P
P
dP
=
ln F1
P
y1
+ y1
P
dy1
=
dP
Whence,
F1
P
y1
y1
+
y1
ln F1
P
ln F1
y1
y1
ln F1
y1
1 y1
F1
y1
+
P
P
dy1
dP
dy1
dP
1
P
( A)
P
This is a general result. An expression for F1 is given by Eq. (14.99):
sat
V s ( P P1sat )
φ1
exp 1
ˆ
RT
φ1
F1
From this, after some reduction:
ln F1
P
=
y1
Whence, by Eq. ( A),
ˆ
ln φ1
P
+
y1
dy1
=
dP
V1s
RT
ln F1
y1
and
ˆ
ln φ1
y1
P
1 + y1
714
y1
1
V1s
+
P
RT
ˆ
ln φ1
y1
P
=
P
ˆ
ln φ1
y1
P
(B)
ˆ
This too is a general result. If the two-term virial equation in pressure applies, then ln φ1 is given by
Eq. (11.63a), from which:
ˆ
ln φ1
P
=
y1
1
2
( B11 + y2 δ12 )
RT
dy1
=
dP
Whence, by Eq. ( B ),
and
ˆ
ln φ1
y1
=
P
2 y2 δ12 P
RT
2
1
V1s B11 y2 δ12
P
RT
2 y1 y2 δ12 P
1
RT
y1
The denominator of this equation is positive at any pressure level for which Eq. (3.38) is likely to be
valid. Hence, the sign of dy1 /d P is determined by the sign of the group in parentheses. For very low
pressures the 1/ P term dominates and dy1 /d P is negative. For very high pressures, 1/ P is small,
and dy1 /d P can be positive. If this is the case, then dy1 /d P is zero for some intermediate pressure,
and the solubility y1 exhibits a minimum with respect to pressure. Qualitatively, these features are
consistent with the behavior illustrated by Fig. 14.23. However, the two-term virial equation is only
valid for low to moderate pressures, and is unable to mimic the change in curvature and attening
of the y1 vs. P curve observed for high pressures for the naphthalene/CO2 system.
14.35 (a) Rewrite the UNILAN equation:
m
ln(c + Pes ) ln(c + Pes )
( A)
2s
As s 0, this expression becomes indeterminate. Application of lHˆ pitals rule gives:
o
n=
Pes
P es
+
c + Pes
c + Pes
m
s 0 2
lim n = lim
s 0
=
P
P
+
c+P
c+P
m
2
lims 0 n =
or
mP
c+P
which is the Langmuir isotherm.
(b) Henrys constant, by denition:
Differentiate Eq. ( A):
Whence,
k=
m
2s
k lim
P 0
m
dn
=
2s
dP
es
es
c
c
=
m
cs
dn
dP
e s
es
c + Pes
c + Pes
es es
2
or
k=
m
sinh s
cs
(c) All derivatives of n with respect to P are well-behaved in the zero-pressure limit:
lim
P 0
m
dn
sinh s
=
cs
dP
715
m
d 2n
= 2 sinh 2s
P 0 d P 2
cs
2m
d 3n
= 3 sinh 3s
lim
3
P 0 d P
cs
lim
Etc.
Numerical studies show that the UNILAN equation, although providing excellent overall correlation of adsorption data at low-to-moderate surface coverage, tends to underestimate Henrys
constant.
14.36 Start with Eq. (14.109), written as:
ln( P / n ) = ln k +
n
0
(z 1)
dn
+z1
n
2
With z = 1 + Bn + Cn + · · ·, this becomes:
3
ln( P / n ) = ln k + 2 Bn + Cn 2 + · · ·
2
Thus a plot of ln( P / n ) vs. n produces ln k as the intercept and 2 B as the limiting slope (for
n 0). Alternatively, a polynomial curve t of ln( P / n ) in n yields ln k and 2 B as the rst
two coefcients.
14.37 For species i in a real-gas mixture, Eqs. (11.46) and (11.52) give:
g
i (T )
µi =
ˆ
+ RT ln yi φi P
g
ˆ
d µi = RT d ln yi φi P
At constant temperature,
g
With d µi = d µi , Eq. (14.105) then becomes:
a
ˆ
d + d ln P +
xi d ln yi φi = 0
RT
i
(const T )
For pure-gas adsorption, this simplies to:
a
d = d ln P + d ln φ
(const T )
( A)
RT
which is the real-gas analog of Eq. (14.107). On the left side of Eq. ( A), introduce the adsorbate
compressibility factor z through z a / RT = A / n RT :
dn
a
d = dz + z
n
RT
where n is moles adsorbed. On the right side of Eq. ( A), make the substitution:
(B)
dP
(C )
P
which follows from Eq. (11.35). Combination of Eqs. ( A), ( B ), and (C ) gives on rearrangement (see
Sec. 14.8):
dP
dn
n
dz + ( Z 1)
d ln = (1 z )
P
n
P
which yields on integration and rearrangement:
d ln φ = ( Z 1)
n = k P · exp
P
0
( Z 1)
dP
· exp
P
This equation is the real-gas analog of Eq. (14.109).
716
n
0
(1 z )
dn
+1z
n
14.39 & 14.40 Start with Eq. (14.109). With z = (1 bm )1 , one obtains the isotherm:
n = k P (1 bn ) exp
bn
1 bn
bn
1 bn
bn
1 bn
For bn sufciently small,
Whence, by Eq. ( A),
exp
n k P (1 2bn )
1
or
n
( A)
kP
1 + 2bk P
which is the Langmuir isotherm.
With z = 1 + β n , the adsorption isotherm is:
n = k P exp(2β n )
from which, for β n sufciently small, the Langmuir isotherm is again recovered.
dP
Ad
=n
P
RT
14.41 By Eq. (14.107) with a = A / n ,
The denition of ψ and its derivative are:
ψ
A
RT
and
Whence,
dψ =
dψ = n
Ad
RT
dP
P
( A)
By Eq. (14.128), the Raoults law analogy, xi = yi P / Pi . Summation for given P yields:
yi
xi = P
Pi
i
i
(B)
By general differentiation,
xi = P d
d
i
i
yi
+
Pi
yi
dP
Pi
i
(C )
The equation,
i x i = 1, is an approximation that becomes increasingly accurate as the solution
procedure converges. Thus, by rearrangement of Eq. ( B ),
i
yi
=
Pi
xi
i
P
=
1
P
With P xed, Eq. (C ) can now be written in the simple but approximate form:
xi =
d
i
dP
P
Equation ( A) then becomes:
xi
dψ = n d
or
xi
δψ = n δ
i
i
where we have replaced differentials by deviations. The deviation in
value must be unity. Therefore,
yi
1
δ
xi = P
Pi
i
i
717
i
xi is known, since the true
By Eq. (14.132),
1
n=
(xi / n i )
i
Combine the three preceding equations:
yi
1
Pi
P
i
δψ =
(xi / n i )
i
When xi = yi P / Pi , the Raoults law analogy, is substituted the required equation is reproduced:
P
δψ =
i
P
i
yi
1
Pi
yi
Pi n i
14.42 Multiply the given equation for G E/ RT by n and convert all mole fractions to mole numbers:
n2n3
n1n3
n1n2
nG E
+ A23
+ A13
= A12
n
n
n
RT
Apply Eq. (11.96) for i = 1:
n2n3
1 n1
1 n1
A23 2
+ A13 n 3
n
n n2
n n2
= A12 x2 (1 x1 ) + A13 x3 (1 x1 ) A23 x2 x3
ln γ1 = A12 n 2
Introduce solute-free mole fractions:
x2
x2
=
x2
1 x1
x2 + x3
Whence,
and
x3 =
x3
1 x1
ln γ1 = A12 x2 (1 x1 )2 + A13 x3 (1 x1 )2 A23 x2 x3 (1 x1 )2
For x1 0,
ln γ1 = A12 x2 + A13 x3 A23 x2 x3
Apply this equation to the special case of species 1 innitely dilute in pure solvent 2. In this case,
x2 = 1, x3 = 0, and
ln γ1 = A12
,2
Whence,
Also
ln γ1 = A13
,3
ln γ1 = x2 ln γ1 + x3 ln γ1 A23 x2 x3
,2
,3
In logarithmic form the equation immediately following Eq. (14.24) on page 552 may be applied to
the several innite-dilution cases:
ln H1 = ln f 1 + ln γ1
Whence,
or
ln H1,2 = ln f 1 + ln γ1
,2
ln H1,3 = ln f 1 + ln γ1
,3
ln H1 ln f 1 = x2 (ln H1,2 ln f 1 ) + x3 (ln H1,3 ln f 1 ) A23 x2 x3
ln H1 = x2 ln H1,2 + x3 ln H1,3 A23 x2 x3
718
14.43 For the situation described, Figure 14.12 would have two regions like the one shown from α to β ,
probably one on either side of the minimum in curve II.
V2
= ln(x2 γ2 )
RT
¯
14.44 By Eq. (14.136) with V2 = V2 :
Represent ln γ2 by a Taylor series:
ln γ2 = ln γ2 |x1 =0 +
d ln γ2
d x1
x 1 =0
x1 +
1 d 2 ln γ2
2
2 d x1
x1 =0
2
x1 + · · ·
But at x1 = 0 (x2 = 1), both ln γ2 and its rst derivative are zero. Therefore,
ln γ2 =
Also,
Therefore,
and
1
2
d 2 ln γ2
2
d x1
x1 =0
ln x2 = ln(1 x1 ) = x1
ln(x2 γ2 ) = + ln x2 + ln γ2 = x1
1
1
V2
1
=1+
2
2
x1 RT
Comparison with the given equation shows that:
(G E/ RT )
T
14.47 Equation (11.95) applies:
1
2
2
x4
x3
x1
1 1 ···
4
3
2
1
1
2
d 2 ln γ2
2
d x1
B=
d 2 ln γ2
2
d x1
2
x1 · · ·
x 1 =0
x1 + · · ·
x 1 =0
1
1
1
2
2
=
P ,x
2
x1 + · · ·
d 2 ln γ2
2
d x1
x 1 =0
HE
RT 2
E
For the partially miscible system G / RT is necessarily large, and if it is to decrease with increasing
T , the derivative must be negative. This requires that H E be positive.
14.48 (a) In accord with Eqs. (14.1) and (14.2),
α12
(b)
yi
ˆ
φi
P = xi γi Pi sat
φisat
Ki
γi Pi sat φisat
yi
·
=
ˆ
P
xi
φi
ˆ
γ1 P1sat φ1sat φ2
K1
· sat
·
=
ˆ
γ2 P2sat φ1 φ2
K2
α12 (x1 = 0) =
γ P1sat φ1 ( P sat )
γ1 P1sat φ1 ( P1sat ) φ2 ( P2sat )
= 1 sat · 1 sat
· sat ·
sat
ˆ
ˆ
P2
P2sat
φ1 ( P2 )
φ1 ( P2 ) φ2 ( P2 )
α12 (x1 = 1) =
ˆ
ˆ
φ ( P sat )
P sat
φ1 ( P1sat ) φ2 ( P1sat )
P1sat
= 1 sat · 2 1
·
·
sat
sat
sat
φ2 ( P2sat )
γ2 P2
φ1 ( P1 ) φ2 ( P2 )
γ2 P2
The nal fractions represent corrections to modied Raoults law for vapor nonidealities.
719
ˆ
(c) If the vapor phase is an ideal solution of gases, then φi = φi for all compositions.
ln γi
T
14.49 Equation (11.98) applies:
=
P ,x
¯
HiE
RT 2
¯
Assume that H E and HiE are functions of composition only. Then integration from Tk to T gives:
ln
¯
HE
γi (x , T )
=i
R
γi (x , Tk )
T
Tk
¯
HE
dT
=i
R
T2
1
1
Tk
T
γi (x , T ) = γi (x , Tk ) · exp
14.52 (a) From Table 11.1, p. 415, nd:
GE
T
(G E / RT )
T
¯
HiE
RT
T
1
Tk
T
1
Tk
= S E = 0
and
G E is independent of T .
P ,x
Therefore
(b) By Eq. (11.95),
¯
HiE
RT
=
=
P ,x
FR ( x )
GE
=
RT
RT
HE
=0
RT 2
GE
= FA ( x )
RT
(c) For solutions exhibiting LLE, G E / RT is generally positive and large. Thus α and β are positive
for LLE. For symmetrical behavior, the magic number is A = 2:
A<2
homogeneous;
A=2
consolute point;
A>2
LLE
With respect to Eq. ( A), increasing T makes G E / RT smaller. thus, the consolute point is an upper consolute point. Its value follows from:
α
=2
RTU
TU =
α
2R
The shape of the solubility curve is as shown on Fig. 14.15.
14.53 Why? Because they are both nontoxic, relatively inexpensive, and readily available. For CO2 , its
Tc is near room temperature, making it a suitable solvent for temperature-sensitive materials. It is
considereably more expensive than water, which is probably the cheapest possible solvent. However,
both Tc and Pc for water are high, which increases heating and pumping costs.
720
Chapter 16 - Section B - Non-Numerical Solutions
16.1 The potential is displayed as follows. Note that K is used in place of k as a parameter to avoid
confusion with Boltzmanns constant.
Combination of the potential with Eq. (16.10) yields on piecewise integration the following expression
for B :
2
B = π N A d 3 1 + ( K 3 1) 1 eξ/ kT (l 3 K 3 ) e / kT 1
3
From this expression,
1
dB
( K 3 1)ξ eξ/ kT + (l 3 K 3 ) e
=
kT 2
dT
/ kT
and also for an intermediate temperature Tm :
according to which d B /dT = 0 for T
Tm =
k ln
ξ
+ξ
K3 1
l3 K 3
That Tm corresponds to a maximum is readily shown by examination of the second derivative d 2 B /dT 2 .
16.2 The table is shown below. Here, contributions to U (long range) are found from Eq. (16.3) [for U (el)],
Eq. (16.4) [for U (ind)], and Eq. (16.5) [for U (disp)]. Note the following:
1. As also seen in Table 16.2, the magnitude of the dispersion interaction in all cases is substantial.
2. U (el), hence f (el), is identically zero unless both species in a molecular pair have non-zero
permanent dipole moments.
3. As seen for several of the examples, the fractional contribution of induction forces can be substantial for unlike molecular pairs. Roughly: f (ind) is larger, the greater the difference in polarity
of the interacting species.
721
Molecular Pair
C6 /1078 J m6
f (el)
f (ind)
f (disp)
f (el)/ f (disp)
49.8
34.3
24.9
22.1
161.9
119.1
106.1
95.0
98.3
270.3
0
0
0
0
0
0
0
0.143
0.263
0.806
0
0.008
0.088
0.188
0.008
0.096
0.205
0.087
0.151
0.052
1.000
0.992
0.912
0.812
0.992
0.904
0.795
0.770
0.586
0.142
0
0
0
0
0
0
0
0.186
0.450
5.680
CH4 /C7 H16
CH4 /CHCl3
CH4 /(CH3 )2 CO
CH4 /CH3 CN
C7 H16 /CHCl3
C7 H16 /(CH3 )2 CO
C7 H16 /CH3 CN
CHCl3 /(CH3 )2 CO
CHCl3 /CH3 CN
(CH3 )2 CO/CH3 CN
16.3 Water (H2 O), a highly polar hydrogen donor and acceptor, is the common species for all four systems;
in all four cases, it experiences strong attractive interactions with the second species. Here, interactions
between unlike molecular pairs are stronger than interactions between pairs of molecules of the same
kind, and therefore H is negative. (See the discussion of signs for H E in Sec. 16.7.)
16.4 Of the eight potential combinations of signs, two are forbidden by Eq. (16.25). Suppose that H E is
negative and S E is positive. Then, by Eq. (16.25), G E must be negative: the sign combination G E ,
H E , and S E is outlawed. Similar reasoning shows that the combination G E , H E , and S E
is inconsistent with Eq. (16.25). All other combinations are possible in principle.
16.5 In Series A, hydrogen bonding occurs between the donor hydrogens of CH2 Cl2 and the electron-rich
benzene molecule. In series B, a charge-transfer complex occurs between acetone and the aromatic
benzene molecule. Neither cyclohexane nor n-hexane offers the opportunity for these special solvation
interactions. Hence the mixtures containing benzene have more negative (smaller positive) values of
H E than those containing cyclohexane and n-hexane. (See Secs. 16.5 and 16.6.)
16.6 (a) Acetone/cyclohexane is an NA/NP system; one expects G E , H E , and S E .
(b) Acetone/dichloromethane is a solvating NA/NA mixture. Here, without question, one will see
G E , H E , and S E .
(c) Aniline/cyclohexane is an AS/NP mixture. Here, we expect either Region I or Region II behavior:
G E and H E , with S E or . [At 323 K (50 C), experiment shows that S E is for this
system.]
(d) Benzene/carbon disulde is an NP/NP system. We therefore expect G E , H E , and S E .
(e) Benzene/n-hexane is NP/NP. Hence, G E , H E , and S E .
(f ) Chloroform/1,4-dioxane is a solvating NA/NA mixture. Hence, G E
E
E
, HE
, and S E
.
E
(g) Chloroform/n-hexane is NA/NP. Hence, G , H , and S .
(h) Ethanol/n-nonane is an AS/NP mixture, and ethanol is a very strong associator. Hence, we expect
Region II behavior: G E , H E , and S E .
16.7 By denition,
δi j 2 Bi j
1
2
Bii + B j j
At normal temperature levels, intermolecular attractions prevail, and the second virial coefcients are
negative. (See Sec. 16.2 for a discussion of the connection between intermolecular forces and the
second virial coefcient.) If interactions between unlike molecular pairs are weaker than interactions
between pairs of molecules of the same kind,
| Bi j | < 1 | Bii + B j j |
2
722
and hence (since each B is negative) δi j > 0. If unlike interactions are stronger than like interactions,
| Bi j | > 1 | Bii + B j j |
2
Hence δi j < 0. For identical interactions of all molecular pairs, Bi j = Bii = B j j , and δi j = 0
The rationalizations of signs for H E of binary liquid mixtures presented in Sec. 16.7 apply approximately to the signs of δ12 for binary gas mixtures. Thus, positive δ12 is the norm for NP/NP, NA/NP, and
AS/NP mixtures, whereas δ12 is usually negative for NA/NA mixtures comprising solvating species.
One expects δ12 to be essentially zero for ideal solutions of real gases, e.g., for binary gas mixtures of
the isomeric xylenes.
16.8 The magnitude of Henrys constant Hi is reected through Henrys law in the solubility of solute i in a
liquid solvent: The smaller Hi , the larger the solubility [see Eq. (10.4)]. Hence, molecular factors that
inuence solubility also inuence Hi . In the present case, the triple bond in acetylene and the double
bond in ethylene act as proton acceptors for hydrogen-bond formation with the donor H in water, the
triple bond being the stronger acceptor. No hydrogen bonds form between ethane and water. Because
hydrogen-bond formation between unlike species promotes solubility through smaller values of G E
and γi than would otherwise obtain, the values of Hi are in the observed order.
16.9 By Eq. (6.70),
H α β = T S α β . For the same temperaature and pressure, less structure or order
means larger S . Consequently,
S sl , S l v , and S s v are all positive, and so therefore are
H sl ,
H l v , and H s v .
H l v H v H l = ( H v H ig ) ( H l H ig ) = H R ,v H R ,l
16.11 At the normal boiling point:
Therefore
H R ,l = H R ,v
H lv
At 1(atm), H R ,v should be negligible relative to H l v . Then H R ,l H l v . Because the normal
boiling point is a representative T for typical liquid behavior, and because H R reects intermolecular
forces, H l v has the stated feature. H l v (H2 O) is much larger than H l v (CH4 ) because of the strong
hydrogen bonding in liquid water.
ig
R
R
16.12 By denition, write C lP = C P +C P ,l , where C P ,l is the residual heat capacity for the liquid phase.
ig
R
Also by denition, C P ,l = ( H R ,l / T ) P . By assumption (modest pressure levels) C P C v .
P
C lP C v +
P
Thus,
H R ,l
T
P
R ,l
For liquids, H is highly negative, becoming less so as T increases, owing to diminution of interR
molecular forces (see, e.g., Fig. 6.5 or Tables E.5 and E.6). Thus C P ,l is positive, and C lP > C v .
P
16.13 The ideal-gas equation may be written:
Vt =
N RT
n RT
·
=
NA P
P
RT
Vt
=
NA P
N
The quantity V t / N is the average volume available to a particle, and the average length available is
about:
R T 1 /3
V t 1 /3
=
NA P
N
1 /3
83.14 cm3 bar mol1 K1 × 300 K
V t 1/3
˚
= 34.6 × 1010 m or 34.6 A
=
N
6.023 × 1023 mol1 × 1 bar × 106 cm3 m3
For argon, this is about 10 diameters. See comments on p. 649 with respect to separations at which
attractions become negligible.
723
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