**preview**has

**blurred**sections. Sign up to view the full version! View Full Document

**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
0
Given
t = 1.8t
32
Find t ()
40
Ans.
1.5 By definition:
P=
F A
F = mass g
Note: Pressures are in gauge pressure.
P
3000bar
D
4mm
A
4
D
2
A
12.566 mm
2
F
PA
g
9.807
m s
2
mass
F g
mass
384.4 kg
Ans.
1.6 By definition:
P 3000atm
P=
D
F A
0.17in
F = mass g
A
4
D
2
A
0.023 in
2
F
PA
g
32.174
ft sec
2
mass
F g
mass
1000.7 lbm
Ans.
1.7 Pabs =
gh
Patm
13.535
gm cm
3
g
9.832
m s
2
h
56.38cm
Patm
101.78kPa
Pabs
gh
Patm
Pabs
176.808 kPa Ans.
1.8
13.535
gm cm
3
g
32.243
ft s
2
h
25.62in
Patm
29.86in_Hg
Pabs
gh
Patm
Pabs
27.22 psia
Ans.
1
1.10 Assume the following:
13.5
gm cm
3
g
9.8
m s
2
P
400bar
h
P g
h
302.3 m
Ans.
1.11
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
0.40kg
g
9.81
m s
2
x
1.08cm
F
mass g
F
3.924 N
Ks
F x
4 10
Ks
3
363.333
N m
On Mars:
x
0.40cm
FMars mass
FMars
Kx
FMars
mK
gMars
gMars
0.01
mK kg
Ans.
1.12 Given:
d P= dz
g
and:
=
MP RT
Substituting:
d P= dz
MP g RT
PDenver
Separating variables and integrating:
Psea
1 dP = P
zDenver
Mg dz RT
0
After integrating:
ln
PDenver Psea
=
Mg zDenver RT
Mg zDenver RT
Taking the exponential of both sides and rearranging: Psea 1atm
PDenver = Psea e gm mol
M
29
g
9.8
m s
2
2
R
82.06
cm atm mol K
3
T
( 10
273.15) K
zDenver
1 mi
Mg zDenver RT
Mg
0.194
PDenver
Psea e
RT
zDenver
PDenver
0.823 atm
Ans.
PDenver 1.13 The same proportionality applies as in Pb. 1.11.
0.834 bar
Ans.
gearth
32.186
ft s
2
gmoon
5.32
ft s
2
lmoon
18.76
learth
lmoon
gearth gmoon
learth
113.498
M
wmoon
learth lbm
M gmoon
M
113.498 lbm
18.767 lbf
Ans.
wmoon
Ans.
1.14
costbulb
hr 5.00dollars 10 day 1000hr
costelec
hr 0.1dollars 70W 10 day kW hr
costbulb
18.262
dollars yr
costelec
25.567
dollars yr
costtotal
costbulb
costelec
costtotal
43.829
dollars yr Ans.
1.15
D
1.25ft
mass
250lbm
g
32.169
ft s
2
3
Patm
30.12in_Hg
A
4
D
2
A
1.227 ft
2
(a) F
Patm A
mass g
F
2.8642
10 lbf
3
Ans.
(b) Pabs
F A
Pabs
16.208 psia
Ans.
(c)
l
1.7ft
Work
F l
Work
4.8691
3 10 ft lbf Ans.
PE
mass g l
PE
424.9 ft lbf
Ans.
1.16
D
0.47m
mass
150kg
g
9.813
m s
2
Patm
101.57kPa
A
4
D
2
A
0.173 m
2
(a) F
(b) Pabs
Patm A
F A
mass g
F
Pabs
1.909
10 N
4
Ans.
110.054 kPa
Ans.
(c)
l
0.83m
Work
F l
Work
15.848 kJ Ans.
EP
mass g l
EP
1.222 kJ
Ans.
1.18
mass
1250kg
u
40
m s
EK
1 2 mass u 2
EK
1000 kJ
Ans.
Work
EK
Work
1000 kJ
Ans.
1.19
Wdot =
mass g h 0.91 0.92 time
Wdot
200W
g
9.8
m s
2
h
50m
4
mdot
Wdot g h 0.91 0.92
25.00 ton
mdot
0.488
kg s
Ans.
1.22 a) cost_coal
MJ 29 kg
2.00 gal
cost_coal
0.95 GJ
1
cost_gasoline
37
GJ m
3
cost_gasoline
14.28 GJ
1
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.
A
Function being fit:
f ( A B C) T
e
B T C
First derivative of the function with respect to parameter A
d f ( A B C) T dA
exp A
B T C
First derivative of the function with respect to parameter B
d f ( A B C) T dB
1 T C
exp A
B T C
First derivative of the function with respect to parameter C
d f ( A B C) T dC
18.5 9.5 0.2 11.8 t 23.1 32.7 44.4 52.1 63.3 75.5
B ( T C)
2
exp A
B T C
3.18 5.48 9.45 16.9 Psat 28.2 41.9 66.6 89.5 129 187
6
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
a1 T
a1 T a2
a2
Guess values of parameters
15 guess 3000 50
F ( a) T
1 exp a0 T a2 a1 T a2
2
a1 T a2 a1 T a2
exp a0
Apply the genfit function
A B C
Compare fit with data.
200
A
genfit ( Psat guess F) T
13.421 2.29 10
3
B C
Ans.
69.053
150
Psat f ( A B C) T
100
50
0 240
260
280
300
T
320
340
360
To find the normal boiling point, find the value of T for which Psat = 1 atm.
7
Psat
1atm
Tnb A
B Psat ln kPa
C K
Tnb
329.154 K
Tnb
1.25 a) t1
C2
273.15K
56.004 degC
dollars gal
dollars gal
Ans.
1970
C1 ( 1 i)
t2
t2 t1
2000
C1
C2
0.35
i
5%
1.513
The increase in price of gasoline over this period kept pace with the rate of inflation. b) t1 1970
Given
C2 C1
t2
= (1
2000
i)
t2 t1
C1
i
16000
dollars yr
i
C2
5.511 %
80000
dollars yr
Find ( i)
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
35 kg
g
9.8
m s
2
z
5m
Work
Mwt g z
Work
1.715 kJ Ans.
(b)
Utotal
Work
Utotal
1.715 kJ
Ans.
(c) By Eqs. (2.14) and (2.21):
dU
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
Utotal
Q
1.715 kJ
Ans.
(e)
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.
i 9.7amp
E
110V
Wdotmech
1.25hp
Wdotelect
i E
Wdotelect
1.067
10 W
3
Qdot
Wdotelect
Wdotmech
Qdot
134.875 W
Ans.
2.5
Eq. (2.3):
U = Q
t
W
Step 1 to 2:
Ut12
200J
W12
6000J
Q12
Ut12
W12
Q12
5.8
10 J
3
Ans.
Step 3 to 4:
Q34
800J
W34
300J
Ut34
Q34
W34
t
Ut34
500 J
t
Ans.
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
U values for all of the steps must sum to zero.
t
Ut41
4700J
Ut23
Ut12
Ut34
Ut41
Ut23 Step 2 to 3:
4000 J
Ans.
Ut23
4
10 J
3
Q23
3800J
W23
Ut23
Q23
W23
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
W12
W23
W34
W41
4.5
10 J
3
Ans.
Step 4 to 1:
Ut41
4700J
W41
4.5
10 J
3
Q41
Ut41
W41
Q41
200 J
Ans.
Note:
Q12341 = W12341
2.11 The enthalpy change of the water = work done.
M
20 kg
CP
4.18
kJ kg degC
t
10 degC
Wdot
0.25 kW
M CP t Wdot
0.929 hr
Ans.
2.12 Q
7.5 kJ
U
12 kJ
W
U
Q
W
U 12 kJ
19.5 kJ
12 kJ
Ans.
Ans.
Q
U
Q
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
40 kg
mt
5 kg
Cc
0.50
kJ kg degC
Ct
0.5
kJ kg degC
Cw
4.18
kJ kg degC
tc
500 degC
t1
25 degC
t2
30 degC
(guess)
Given
mc Cc t2
tc = mw Cw
mt Ct
t2
t1
t2
Find t2
11
t2
27.78 degC
Ans.
2.15
mass
1 kg
CV
4.18
kJ kg K
(a)
T
1K
Ut
mass CV T
Ut
4.18 kJ
Ans.
(b)
g
9.8
m s
2
EP
Ut
z
EP mass g
z
426.531 m Ans.
(c)
EK
Ut
u
EK 1 mass 2
u
91.433
m s
Ans.
2.17
z
50m
1000
kg m
3
u
5
m s
2
D
mdot
2m
uA
A
mdot
4
D
2
A
4 kg
3.142 m
1.571 10
s
3
Wdot
mdot g z Wdot
7.697
10 kW
Ans.
2.18 (a)
U1
kJ 762.0 kg
P1
1002.7 kPa
V1
cm 1.128 gm
3
H1
U1
P1 V 1
H1
763.131
kJ kg
Ans.
(b)
U2
kJ 2784.4 kg
P2
1500 kPa
V2
cm 169.7 gm
3
H2
U2
P2 V 2
U
U2
U1
H
H2
H1
U
2022.4
kJ kg
Ans.
H
2275.8
kJ kg
Ans.
12
2.22
D1
2.5cm
u1
2
m s
D2
5cm
(a)
For an incompressible fluid, =constant. By a mass balance, mdot = constant = u 1A1 = u2A2
u2
(b)
u1
1 2
D1 D2
u2
2
2
u2
0.5
m s
Ans.
EK
1 2
u1
2
EK
1.875
J kg
Ans.
2.23 Energy balance:
mdot3 H3
mdot1 H1
mdot2 H2 = Qdot
Mass balance:
mdot3
mdot1
mdot2 = 0
Therefore:
mdot1 H3
mdot Cp T3
H1
T1
mdot2 H3
H2 = Qdot
T2 = Qdot
or
mdot2 CP T3
T3 CP mdot1
mdot2 = Qdot
mdot1 CP T1
mdot2 CP T2
mdot1
1.0
kg s
T1
25degC
mdot2
0.8
kg s
T2
75degC
Qdot
30
kJ s
CP
4.18
kJ kg K
T3
Qdot
mdot1 CP T1 mdot1
mdot2 CP T2
mdot2 CP
T3
43.235 degC
Ans.
2.25By Eq. (2.32a):
H
By continuity, incompressibility
u2 = u1
u = 0 2 A1
2
H = CP T
A2
CP
4.18
kJ kg degC
13
2 u = u1
2
A1 A2
2
1
2 u = u1
2
D1 D2
4
1
SI units:
u1
2
14
m s
D1
2.5 cm
D2
3.8 cm
T
D2
u1
2 CP
7.5cm
1
D1 D2
4
T
0.019 degC
Ans.
T
u1
2
2 CP
1
D1 D2
4
T
0.023 degC
Ans.
Maximum T change occurrs for infinite D2:
D2
cm
T
u1
2
2 CP
1
D1 D2
4
T
0.023 degC
Ans.
2.26 T1
300K
T2
520K
u1
50
10
kmol hr
m s
u2
3.5
m s
molwt
29
kg kmol
Wsdot
98.8kW
ndot
CP
7 R 2
H
CP T2
T1
H
6.402
10
3 kJ
kmol
By Eq. (2.30):
Qdot
H
u2 2
2
u1 2
2
molwt ndot
2
Wsdot Qdot
9.904 kW
Ans.
2.27By Eq. (2.32b):
H=
u
2 gc
also
V2 V1
=
2
T 2 P1 T 1 P2
2
By continunity, constant area
u2 = u1
V2 V1
14
u2 = u1
T 2 P1 T 1 P2
u = u2
u1
2
2 u = u1
2
T 2 P1 T 1 P2
2
1
H = CP T =
ft s
7 R T2 2
T1
T1
P1
100 psi
P2
20 psi
u1
20
579.67 rankine
R
3.407
ft lbf mol rankine
molwt
28
gm mol
T2
578 rankine
(guess)
Given
T2
7 2
R T2
T1 =
T2
u1 2
2
T 2 P1 T 1 P2
2
1 molwt
Find T2
578.9 rankine
Ans.
( 119.15 degF)
2.28 u1
3
m s
u2
200
m s
H1
2
334.9
2
kJ kg
H2
2726.5
kJ kg
By Eq. (2.32a):
Q
H2
H1
u2
u1 2
Q
2411.6
kJ kg
Ans.
2.29 u1
30
u2
m s m 500 s
H1
3112.5
kJ kg
H2
2945.7
kJ kg
(guess)
By Eq. (2.32a):
Given
H2
H1 =
u1
2
u2 2
2
u2
Find u2
u2
578.36
m s
Ans.
3
D1
5 cm
V1
388.61
cm
gm
15
V2
667.75
cm
3
gm
Continuity:
D2
D1
u1 V2 u2 V1
D2
1.493 cm
Ans.
2.30 (a)
t1
30 degC
t2
250 degC
n
3 mol
CV
20.8
J mol degC
By Eq. (2.19):
Q
n CV t2
t1
Q
13.728 kJ
Ans.
Take into account the heat capacity of the vessel; then
mv
100 kg
cv
0.5
kJ kg degC
Q
mv cv
n CV
t2
t1
Q
11014 kJ
Ans.
(b)
t1
200 degC
t2
40 degC
n
4 mol
CP
29.1
joule mol degC
Q n CP t2 t1
By Eq. (2.23):
Q
18.62 kJ
Ans.
2.31 (a) t1
70 degF
t2
350 degF
n
3 mol
CV
5
BTU mol degF
By Eq. (2.19):
Q
n CV t2
t1
Q
4200 BTU
Ans.
Take account of the heat capacity of the vessel:
mv
200 lbm
cv
0.12
BTU lbm degF
Q
mv cv
n CV
t2
t1
Q
10920 BTU
Ans.
(b) t1
400 degF
t2
150 degF
n
4 mol
16
CP
7
BTU mol degF
By Eq. (2.23):
Q
n CP t2
t1
Q
7000 BTU
Ans.
2.33
H1
1322.6
BTU lbm
H2
1148.6
BTU lbm
u1
10
ft s
V1
3.058
ft
3
lbm
2
V2
78.14
ft
3
lbm
4 lb
D1
3 in
D2
10 in
mdot
4
D 1 u1 V1
mdot
3.463
10
sec
u2
mdot 4
V2
u2
2
22.997
D2
ft sec
2
Eq. (2.32a): Ws
Wdot Ws mdot
H2
H1
Wdot
u2
2
u1 2
Ws
Ans.
173.99
BTU lb
39.52 hp
2.34
H1
307
BTU lbm
H2
330
BTU lbm
u1
20
ft s
molwt
44
gm mol
V1
9.25
ft
3
lbm
V2
0.28
ft
3
lbm
D1
4 in
D2
1 in
mdot
4
D1 u1 V1
2
mdot
679.263
lb hr
u2
mdot 4
V2
u2
2
9.686
D2
ft sec
Ws
5360
BTU lbmol
Eq. (2.32a): Q
H2
H1
u2
2
u1 2
17
2
Ws molwt
Q
98.82
BTU lbm
Qdot
mdot Q
Qdot
67128
BTU hr
Ans.
2.36
T1
300 K
P
1 bar
n
1 kg 28.9 gm mol
3
n
34.602 mol
V1
3 bar cm T1 83.14 mol K P
V1
cm 24942 mol
V2
W= n
V1
P dV = n P V 1
V2 = n P V1
3 V1
Whence
W
n P 2 V1
W
172.61 kJ
Ans.
Given:
T2 = T1
V2 V1
= T1 3
Whence
T2
H 17.4
3 T1
kJ mol
CP
29
joule mol K
H
CP T2
T1
Ans.
Q
n H
Q
602.08 kJ
Ans.
U
Q n
W
U
12.41
kJ mol
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
J mol K
T1
293.15 K
T2
333.15 K
P1
1000 kPa
P2
100 kPa
(a) Cool at const V1 to P2 (b) Heat at const P2 to T2
CP
7 R 2
CV
5 R 2
Ta2
T1
P2 P1
Ta2
29.315 K
18
Tb
T2
Ta2
Tb
303.835 K
Ta
Ta2
T1
Ta
263.835 K
Hb
Ua
V1
Ha
C P Tb
CV Ta
R T1 P1
Ua
Hb
Ua
2.437
8.841
5.484
10
3 J
mol
3 J
10
mol
V2 R T2 P2
10
V1
V 1 P2
10
3 3 m
mol
Ha
V2
3 J
0.028
m
3
mol
P1
7.677
mol
Ub
U
Hb
Ua
Ha
kg m
2 5 2 5
3
P2 V 2
Ub
Hb
V1
U 0.831 kJ
Ub
6.315
Ans.
10
3 J
mol
mol
kJ mol
Ans.
H
H
1.164
2.39
996
9.0 10
1
4 kg
ms
D
0.0001 Note: D = /D in this solution
D
cm
u
1 m 5 s 5
22133
Re
D
u
Re
55333 110667 276667
19
0.00635 fF 0.3305 ln 0.27 D 7 Re
0.9 2
fF
0.00517 0.00452 0.0039
0.313
mdot
u
4
D
2
mdot
1.956 kg 1.565 s 9.778
Ans.
0.632
P L 2 fF D
u
2
P L
kPa 11.254 m
0.206
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
0.573
Wdot
1.009
10 kW
3
Wdot
kW
Cost
799924 dollars Ans.
20
Chapter 3 - Section A - Mathcad Solutions
3.1
=
1 d
dT
=
1
d dP
T
P
At constant T, the 2nd equation can be written:
d
=
dP
ln ( ) 1.01
ln
2 1
=
P
44.1810
6
bar
1
2 = 1.01
1
P
P
225.2 bar
3
P2 = 226.2 bar
Ans.
3.4 b
2700 bar
c
cm 0.125 gm
V2
P1
1 bar
P2
500 bar
Since
Work =
V1
P2
P dV
a bit of algebra leads to
Work
c
P1
P P b
dP
Work
0.516
J gm
Ans.
Alternatively, formal integration leads to
Work
c P2
P1
b ln
P2 P1
b b
Work
0.516
J gm
Ans.
3.5
= a
bP
a
3.9 10
6
atm
1
b
0.1 10
9
atm
2
P1
1 atm
P2
3000 atm
V
1 ft
3
(assume const.)
Combine Eqs. (1.3) and (3.3) for const. T:
P2
Work
V
P1
( a
b P)P dP
21
Work
16.65 atm ft
3
Ans.
3.6
1.2 10
3
degC
3
1
CP
0.84
kJ kg degC
M
5 kg
V1
1
m
1590 kg
P
1 bar
t1
0 degC
t2
20 degC
With beta independent of T and with P=constant,
dV = V
Vtotal
dT
M V
V2
Vtotal
V1 exp
7.638
t2
10
5
t1
m
3
V
V2
V1
Ans.
Work
P Vtotal
(Const. P)
Work
7.638 joule
Ans.
Q
M CP t2
t1
Q
84 kJ
Ans.
Htotal
Q
Htotal
84 kJ
Ans.
Utotal
Q
Work
Utotal
83.99 kJ
7 R 2
Ans.
5 2
3.8
P1
8 bar
P2
1 bar
T1
600 K
CP
CV
R
(a) Constant V:
W= 0
and
U = Q = CV T
T2
T1
P2 P1
T
T2
T1
T
525 K
U
CV T
Q and
U
10.91
kJ mol
Ans.
H
CP T
H
15.28
kJ mol
Ans.
(b) Constant T:
U=
H= 0
Work
and
10.37
Q= W
kJ mol
Work
R T1 ln
P2 P1
Q
and
Ans.
(c) Adiabatic:
Q= 0
22
and
U = W = CV T
1
CP CV
U
T2
T1
P2 P1
T2 331.227 K
T
T2
T1
CV T
H
CP T
W
and
U
5.586
kJ mol
Ans.
H
7.821
kJ mol
Ans.
3.9 P4
2bar
CP
7 R 2
CV
5 R 2
P1
10bar
T1
600K
V1
R T1 P1
R CP
V1
4.988
10
3 3 m
mol
Step 41: Adiabatic
T4
T1
P4 P1
T4
378.831 K
3 J
U41
CV T1
T4
U41
4.597
10
H41
CP T1
T4
H41
6.436
10
mol 3 J
mol
Q41
J 0 mol
Q41
J 0 mol
W41
U41
W41
4.597
10
3
3 J
mol
P2
3bar
T2
600K
V2
R T2 P2
V2
m 0.017 mol
Step 12: Isothermal
U12
0
J mol
U12
0
J mol
H12
0
J mol
H12
0
J mol
23
Q12
W12
R T1 ln
Q12
P2 P1
Q12
6.006
10
3 J
mol
3 J
W12
6.006
10
mol
P3
2bar
V3
V2
T3
P3 V 3 R
T3
400 K
3 J
Step 23: Isochoric
U23
CV T3
T2
H23
CP T3
T2
Q23
CV T3
T2
W23
0
J mol
mol 3 J H23 5.82 10 mol 3 J Q23 4.157 10 mol J W23 0 mol
U23
4.157
10
P4
2 bar
T4
378.831 K
V4
CV T4
R T4 P4
T3
V4
U34
m 0.016 mol
439.997 J mol J
3
Step 34: Isobaric
U34
H34
CP T4
T3
H34
615.996
mol
Q34
CP T4
T3
Q34
615.996
W34
R T4
T3
W34
175.999
J mol J
mol
3.10 For all parts of this problem: T2 = T1
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
12 bar
24
V1
12 m
3
V2
1m
3
(a)
Work = n R T ln
P2
Work
P1
Work
P1 V1 ln
P2 P1 Ans.
2982 kJ
(b) Step 1: adiabatic compression to P2
1
5 3
Vi
V1
P1 P2
(intermediate V)
Vi
2.702 m
3
W1
P2 V i
P1 V 1 1
W1
3063 kJ
Step 2: cool at const P2 to V2
W2
P2 V 2
Vi
W2
2042 kJ
Work
W1
W2
Work
5106 kJ
Ans.
(c) Step 1: adiabatic compression to V2
Pi P1 V1 V2 (intermediate P)
Pi
62.898 bar
W1
Pi V 2
P1 V 1 1
W1
7635 kJ
Step 2: No work.
Work
W1
Work
7635 kJ
Ans.
(d) Step 1: heat at const V1 to P2
W1 = 0
Step 2: cool at const P2 to V2
W2
P2 V 2
V1
Work
W2
Work
13200 kJ
Ans.
(e) Step 1: cool at const P1 to V2
W1
P1 V 2
V1
25
W1
1100 kJ
Step 2: heat at const V2 to P2
W2 = 0
Work
W1
Work
1100 kJ
Ans.
3.17(a)
No work is done; no heat is transferred.
U =
t
T= 0
T2 = T1 = 100 degC
Not reversible
(b)
The gas is returned to its initial state by isothermal compression.
Work = n R T ln
3
V1 V2
V2 4 3
but
n R T = P2 V 2
V1
4m
m
3
P2
6 bar
Work
P2 V2 ln
V1 V2
P2 500 kPa
Work
878.9 kJ Ans.
3.18 (a) P1
100 kPa
T1
303.15 K
CP
7 2
R
CV
5 R 2
CP CV
1
Adiabatic compression from point 1 to point 2:
Q12
U12
0
kJ mol
CV T2 T1
U12 = W12 = CV T12
H12 CP T2 T1
T2
W12
T1
P2 P1
U12
U12
3.679
kJ mol
H12
5.15
kJ mol
W12
3.679
kJ mol
Ans.
Cool at P2 from point 2 to point 3:
T3
T1
H23
CP T3
T2
Q23
H23
U23
CV T3
T2
W23
U23
Q23
26
H23
5.15
kJ mol
U23
3.679
kJ mol
Ans.
Q23
5.15
kJ mol
W23
1.471
kJ mol
Ans.
Isothermal expansion from point 3 to point 1:
U31 =
Q31
H31 = 0
W31
P3
P2
W31
R T3 ln
P1 P3
W31
4.056
kJ mol
Q31
4.056
kJ mol
Ans.
FOR THE CYCLE:
U=
H= 0
Q
Q12
Q23
Q31
Work
Work
W12
1.094
W23
kJ mol
W31
Q
1.094
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
W12 0.8
W12
4.598
kJ mol
Q12
U12
W12
Q12
0.92
kJ mol
Step 23:
W23
W23 0.8
W23
1.839
kJ mol
Q23
U23
W23
Q23
5.518
kJ mol
Step 31:
W31
W31 0.8
W31
3.245
kJ mol
Q31
W31
Q31
27
3.245
kJ mol
FOR THE CYCLE:
Q
Q12
Q23
Q31
Work
W12
W23
W31
Q
3.192
kJ mol
Work
3.192
kJ mol
3.19Here, V represents total volume.
P1
1000 kPa
V1
1m
3
V2
5 V1
T1
600 K
CP
21
joule mol K
CV
CP
R
CP CV
(a) Isothermal:
T2 T1
Work = n R T1 ln
T2 600 K
V1 V2
200 kPa
P2
P1
V1 V2
P2
Work
Ans.
Work
P1 V1 ln
V1 V2
V1 V2
1609 kJ
Ans.
(b) Adiabatic:
P2
P1
T2
T1
P2 V 2 P1 V 1
T2
208.96 K
P2
69.65 kPa
Ans.
Work
P2 V 2
P1 V 1 1
Work
994.4 kJ Ans,
(c) Restrained adiabatic:
Work =
U = Pext V
Pext
100 kPa
Work
Pext V2
V1
Work
400 kJ Ans.
n
P1 V 1 R T1
U = n CV T
T2
Work n CV
T1
T2
442.71 K
Ans.
P2
P1
V 1 T2 V 2 T1
28
P2
147.57 kPa
Ans.
3.20
T1
423.15 K
P1
8bar
P3
3 bar
CP
7 2
R
CV
5 2
R
T2
T1
T3
323.15 K
Step 12:
H12
0
kJ mol
U12
0
kJ mol
If
r=
V1 V2
=
V1 V3
Then
r
T 1 P3 T 3 P1
W12
R T1 ln r ()
W12
2.502
kJ mol
W23 0
Q12
kJ mol
W12
Q12
2.502
kJ mol
Step 23:
U23
CV T3
T2
Q23
U23
H23
CP T3
T2
Q23
2.079
kJ mol
U23
2.079
kJ mol
H23
2.91
kJ mol
Process:
Work
W12
W23
Work
2.502
kJ mol
Ans.
Q
Q12
Q23
Q
0.424
kJ mol
Ans.
H
H12
H23
H
2.91
kJ mol
Ans.
U
U12
U23
U
2.079
kJ mol
Ans.
29
3.21 By Eq. (2.32a), unit-mass basis:
molwt
28
gm mol
H
1 2
u = 0
2
But
H = CP T
Whence
T=
u2
2
u1
2
2 CP
CP
R 7 2 molwt
u1
2.5
m s
u2
2 2
50
m s
t1
150 degC
t2
t1
u2
u1
2 CP
t2
148.8 degC
Ans.
3.22
CP
P1
7 R 2
1 bar
CV
P3
5 R 2
10 bar
T1
303.15 K
T3
403.15 K
U
CV T3
T1
H
CP T3
T1
U
2.079
kJ mol
Ans.
H
2.91
kJ mol
Ans.
Each part consists of two steps, 12 & 23. (a) T2
T3
P2
P1
T2 T1
Work W23
W23
R T2 ln
P3 P2
Work
6.762
kJ mol
Ans.
Q
U
Work
Q
4.684
kJ mol
Ans.
30
(b) P2
P1
T2
T3
U12
CV T2
T1
H12
CP T2
T1
Q12
H12
W12
U12
Q12
W12
W23
0.831
kJ mol
W23
R T2 ln
P3 P2
W23
7.718
kJ mol
Work
W12
Work
6.886
kJ mol
Ans.
Q
U
Work
Q
4.808
kJ mol
Ans.
(c)
T2
H23 U23
T1
CP T3 CV T3
P2
T2 T2
P3
W12
Q23
R T1 ln
H23
P2 P1
W23
U23
Q23
Work
W12
W23
Work
4.972
kJ mol
Ans.
Q
U
Work
Q
2.894
kJ mol
Ans.
For the second set of heat-capacity values, answers are (kJ/mol):
U = 1.247
U = 2.079
(a)
Work = 6.762
Q = 5.515
(b)
Work = 6.886
Q = 5.639
(c)
Work = 4.972
Q = 3.725
31
3.23
T1
303.15 K
T2
T1
T3
393.15 K
P1
1 bar
P3
12 bar
CP
7 R 2
CV
5 R 2
For the process:
U
CV T3
T1
H
CP T3
T1
U
1.871
kJ mol
H
2.619
kJ mol
Ans.
Step 12:
P2
kJ mol
P3
T1 T3
W12
W12
R T1 ln
P2 P1
W12
5.608
Q12
Q12
5.608
kJ mol
Step 23:
For the process:
W23
Work
0
kJ mol
W23
Q23
U
W12
Q
Q12
Q23
Work
5.608
kJ mol
Q
3.737
kJ mol
Ans.
3.24
W12 = 0
Work = W23 = P2 V3
V 2 = R T3
T2
But
T3 = T1
So...
Work = R T2
Therefore
T1
Also
W = R T1 ln
P P1
350 K
ln
T2 T1 P = P1 T1
T2
T1
800 K
P1
4 bar
P
P1 exp
T2 T1
T1
P
2.279 bar
Ans.
32
3.25
VA
256 cm
3
Define:
P P1
= r
r
0.0639
Assume ideal gas; let V represent total volume:
P1 V B = P 2 V A
VB
From this one finds:
P P1
=
VA VA VB
VB
VA ( 1) r r
VB
3750.3 cm
3
Ans.
3.26
T1
300 K
P1
1 atm
CP
7 R 2
CV
CP
R
CP CV
The process occurring in section B is a reversible, adiabatic compression. Let
P ( )= P2 final
nA = nB
TA ( )= TA final
TB ( )= TB final
Since the total volume is constant,
nA R TA 2 nA R T1 = P2 P1
TB
or
TA TB 2 T1 = P2 P1
1
(1)
(a)
P2
TA
1.25 atm
P2 P1
TB
TB
T1
P2 P1
UA UB
(2)
2 T1
Q = nA
Define
q=
Q nA
q
CV TA
TB
2 T1
(3)
TB
319.75 K
TA
430.25 K
q
3.118
kJ mol
Ans.
33
(b)
Combine Eqs. (1) & (2) to eliminate the ratio of pressures:
TA
425 K
(guess)
1
TB
300 K
Given
TB = T1
TA
TB
2 T1
TB
Find TB
TB
319.02 K
1.24 atm
Ans.
Ans.
P2
q
P1
TA
TB
2 T1
TB 2 T1
(1)
P2
CV TA
q
2.993
kJ mol
Ans.
(c)
TB
325 K
1
By Eq. (2),
P2
P1
TB T1
P2 P1
P2
1.323 atm
Ans.
TA
2 T1
TB
(1)
TA
469 K
Ans.
q
CV TA
TB
2 T1
q
4.032
kJ mol
Ans.
(d)
Eliminate
TA
TB
from Eqs. (1) & (3):
q
3
kJ mol
P2
1
q P1 2 T1 C V
P1
P2
1.241 atm
Ans.
TB
T1
P2 P1
P2 P1
(2)
TB
319.06 K
Ans.
TA
2 T1
TB
(1)
34
TA
425.28 K
Ans.
3.30 B
242.5
cm
3
C
25200
cm
6
T
2
373.15 K
mol
mol
P1
1 bar
P2
55 bar
B'
B RT
B'
7.817
10
3 1
bar
C'
C
2
B
2
R T
2
C'
3.492
10
5 1
bar
2
(a) Solve virial eqn. for initial V. Guess:
V1
RT P1
Given
P1 V 1 = 1 RT
B V1
C V1
2
V1
Find V1
V1
cm 30780 mol
3
Solve virial eqn. for final V. Guess:
V2
RT P2
Given
P2 V 2 RT
= 1
B V2
C
2 V2
V2
Find V2
V2
cm 241.33 mol
3
Eliminate P from Eq. (1.3) by the virial equation:
V2
Work
RT
V1
1
B V
C V
2
1 dV V
Work
12.62
kJ mol
Ans.
(b)
Eliminate dV from Eq. (1.3) by the virial equation in P:
P2
dV = R T P
1
2
C' dP
W
RT
P1
1 P
C' P dP
W
35
12.596
kJ mol
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
Tr
T Tc
Tr
1.056
Pc
50.4 bar
P
12 bar
Pr
P Pc
Pr
0.238
0.087 (a)
(guess)
B
140
cm
3
mol
C
7200
cm
6 2
V
mol
RT P
V
2066
cm
3
mol
Given
PV = 1 RT
B V
C V
2
V
Find ( ) V
V
0.422 Tr
1.6
cm 1919 mol
3
Z
PV RT
Z
0.929
Ans.
(b)
B0
0.083
B0
0.304
B1
0.139
0.172 Tr
4.2
B1
2.262
10
3
Z
1
B0
B1
Pr Tr
Z
0.932
V
ZRT P
V
cm Ans. 1924 mol
3
(c)
For Redlich/Kwong EOS:
1
0.5
0
0.08664
0.42748
Table 3.1
( ) Tr
Tr
Table 3.1
q Tr
Tr Tr
Eq. (3.54)
T r Pr
Pr Tr
Eq. (3.53)
36
Calculate Z
Guess:
Z
0.9
Given
Eq. (3.52)
Z= 1
T r Pr
q Tr
T r Pr
Z Z T r Pr
T r Pr Z T r Pr
Z
Find Z) (
Z
0.928
V
Z RT P
V
1916.5
cm
3
mol
Ans.
(d)
For SRK EOS:
1
0
0.08664
1
2
0.42748
Table 3.1
Tr
q Tr
1
Tr
0.480
1.574
0.176
2
1
Tr
2
Table 3.1
Eq. (3.54)
Guess:
Tr Z
T r Pr
0.9
Pr Tr
Eq. (3.53)
Calculate Z
Given
Eq. (3.52)
Z= 1
T r Pr
q Tr
T r Pr
Z Z T r Pr
T r Pr Z T r Pr
Z
Find Z) (
Z
0.928
V
Z RT P
V
1918
cm
3
mol
Ans.
(e)
For Peng/Robinson EOS:
1
2
1
2
0.07779
0.45724
1
2
Table 3.1
Tr
1
0.37464
Tr Tr
1.54226
0.26992
2
1
Tr
Pr Tr
2
Table 3.1
q Tr
Eq. (3.54)
37
T r Pr
Eq. (3.53)
Calculate Z
Guess:
Z
0.9
Given
Eq. (3.52)
Z= 1
T r Pr
q Tr
T r Pr
Z Z T r Pr
T r Pr Z T r Pr
Z
Find ( Z)
Z
0.92
V
ZRT P
V
1900.6
cm
3
mol
Ans.
3.33 Tc
305.3 K
T
323.15 K
Tr
T Tc
Tr
1.058
Pc
48.72 bar
P
15 bar
Pr
P Pc
Pr
0.308
0.100 (a)
(guess)
B
cm 156.7 mol
PV = 1 RT
3
C
C V
2
9650
cm
6 2
V
mol
RT P
V
cm 1791 mol
3
Given
B V
V
Find ( V)
V
cm 1625 mol
3
Z
PV RT
Z
0.907
Ans.
(b)
B0
0.083
0.422 Tr
1.6
B0
0.302
B1
0.139
0.172 Tr
4.2
B1
3.517
10
3
Z
1
B0
B1
Pr Tr
Z
0.912
V
ZRT P
V
cm Ans. 1634 mol
3
(c)
For Redlich/Kwong EOS:
1
0
0.08664
38
0.42748
Table 3.1
( ) Tr
Tr
0.5
Table 3.1
q Tr
Tr Tr
Eq. (3.54)
T r Pr
Pr Tr
Eq. (3.53)
Calculate Z
Guess:
Z
0.9
Given
Eq. (3.52)
Z= 1
T r Pr
q Tr
T r Pr
Z Z T r Pr
T r Pr Z T r Pr
Z
Find Z) (
Z
0.906
V
Z RT P
V
cm 1622.7 mol
3
Ans.
(d)
For SRK EOS:
1
0
0.08664
1
2
0.42748
Table 3.1
Tr q Tr
1 Tr
0.480
1.574
0.176
2
1
Tr
2
Table 3.1
Eq. (3.54)
Tr
T r Pr
Pr Tr
Eq. (3.53)
Calculate Z
Guess:
Z
0.9
Given
Eq. (3.52)
Z= 1
T r Pr
q Tr
T r Pr
Z Z T r Pr
T r Pr Z T r Pr
Z
Find Z) (
Z
0.907
V
Z RT P
V
1624.8
cm
3
mol
Ans.
(e)
For Peng/Robinson EOS:
1
2
1
2
39
0.07779
0.45724
Table 3.1
1
2
Tr
1
0.37464
Tr Tr
1.54226
0.26992
2
1
Tr
Pr Tr
2
Table 3.1
q Tr
Eq. (3.54)
T r Pr
Eq. (3.53)
Calculate Z
Guess:
Z
0.9
Given
Eq. (3.52)
Z= 1
T r Pr
q Tr
T r Pr
Z Z T r Pr
T r Pr Z T r Pr
Z
Find ( Z)
Z
0.896
V
ZRT P
V
cm 1605.5 mol
3
Ans.
3.34 Tc
Pc
318.7 K
T
348.15 K
Tr
Pr
T Tc
P Pc
Tr
Pr
1.092
37.6 bar
P
15 bar
0.399
0.286 (guess) (a)
3
B
194
cm
mol
C
15300
cm
6 2
V
mol
RT P
V
1930
cm
3
mol
Given
PV = 1 RT
B V
C V
2
V
Find ( V)
V
cm 1722 mol
3
Z
PV RT
Z
0.893
Ans.
(b)
B0
0.083
0.422 Tr
1.6
B0
0.283
40
B1
0.139
0.172 Tr
4.2
B1
0.02
3
Z
1
B0
B1
Pr Tr
Z
0.899
V
Z RT P
V
1734
cm
mol
Ans.
(c)
For Redlich/Kwong EOS:
1
0
0.08664
0.42748
Table 3.1
( ) Tr
Tr
0.5
Table 3.1
q Tr
Tr Tr
Eq. (3.54)
T r Pr
Pr Tr
Eq. (3.53) Guess:
Z 0.9
Calculate Z
Given Eq. (3.52)
Z= 1
T r Pr
q Tr
T r Pr
Z Z T r Pr
T r Pr Z T r Pr
Z
Find Z) (
Z
0.888
V
Z RT P
V
cm 1714.1 mol
3
Ans.
(d)
For SRK EOS:
1
0
0.08664
1
2
0.42748
Table 3.1 Table 3.1
Tr
q Tr
1
Tr
0.480
1.574
0.176
2
1
Tr
2
Eq. (3.54)
Tr
T r Pr
Pr Tr
Eq. (3.53)
41
Calculate Z Guess: Eq. (3.52) Given
Z
0.9
Z= 1
T r Pr
q Tr
T r Pr
Z Z T r Pr
T r Pr Z T r Pr
Z
Find ( Z)
Z
0.895
V
ZRT P
V
1726.9
cm
3
mol
Ans.
(e)
For Peng/Robinson EOS:
1
2
1
2
0.07779
0.45724
1
2
Table 3.1
Tr
1
0.37464
Tr Tr
1.54226
0.26992
2
1
Tr
Pr Tr
2
Table 3.1
q Tr
Eq. (3.54)
Guess:
T r Pr
Z 0.9
Eq. (3.53)
Calculate Z
Given Eq. (3.52)
Z= 1
T r Pr
q Tr
T r Pr
Z Z T r Pr
T r Pr Z T r Pr
Z
Find ( Z)
Z
0.882
V
ZRT P
V
cm 1701.5 mol
3
Ans.
3.35
T
523.15 K
P
3
1800 kPa
(a)
B
cm 152.5 mol
C
5800
cm
6 2
V
mol
RT (guess) P
Given
PV = 1 RT
B V
C V
2
V
Find ( V)
Z
PV RT
V
2250
42
cm
3
mol
Z
0.931
Ans.
(b) Tc
647.1 K
Pc
220.55 bar
0.345
Tr
T Tc
Pr
P Pc
B0
0.083
0.422 Tr
1.6
Tr
0.808
Pr
0.082
B0
0.51
B1
0.139
0.172 Tr
4.2
B1
0.281
Z
1
B0
3
B1
Pr Tr
V
Z RT P
Z
0.939
V
cm 2268 mol
Ans.
3
(c) Table F.2:
molwt
gm 18.015 mol
V
cm molwt 124.99 gm
cm 2252 mol
3
or
V
6 2
Ans.
3.37
B
cm 53.4 mol
3
C
2620
cm
D
5000
cm
9 3
n
mol
mol
mol
T
273.15 K
Given
PV RT
= 1
B V
10
C V
2
D V
3
fP V) (
Find V) (
i
0 10
Pi
10
20 i bar
Vi
RT Pi
(guess)
Zi
fPi Vi Pi RT
Eq. (3.12)
Z1i
1
B Pi RT
Eq. (3.38)
Z2i
1 2
1 4
B Pi RT
Eq. (3.39)
43
110 -10 20 40 60 80
Zi
1 0.953 0.906 0.861
Z1i
1 0.953 0.906 0.859 0.812 0.765 0.718 0.671 0.624 0.577 0.53
Z2i
1 0.951 0.895 0.83 0.749 0.622 0.5+0.179i 0.5+0.281i 0.5+0.355i 0.5+0.416i 0.5+0.469i
Pi
100 120 140 160 180 200
bar
0.819 0.784 0.757 0.74 0.733 0.735 0.743
Note that values of Z from Eq. (3.39) are not physically meaningful for pressures above 100 bar.
1
0.9
Zi Z1 i Z2 i
0.8
0.7
0.6
0.5
0
50
100
Pi bar
1
150
200
44
3.38 (a) Propane:
Tc
369.8 K
Pc
42.48 bar
0.152
T
313.15 K
P
13.71 bar
Tr
T Tc
Tr
0.847
Pr
P Pc
Pr
0.323
For Redlich/Kwong EOS:
1
0
0.08664
0.42748
Table 3.1
( ) Tr
Tr
0.5
Table 3.1
Eq. (3.53)
q Tr
Tr Tr
Eq. (3.54)
T r Pr
Pr Tr
Calculate Z for liquid by Eq. (3.56) Guess:
Z
0.01
Given
Z=
T r Pr
Z
T r Pr
Z
T r Pr
1 q Tr
T r Pr T r Pr
Z
Z
Find Z) Z (
0.057
V
Z RT P
V
108.1
cm
3
mol
Ans.
Calculate Z for vapor by Eq. (3.52)
Guess:
Z
0.9
Given
Z= 1
T r Pr
q Tr
T r Pr
Z Z Z
T r Pr T r Pr
Z
Find Z) (
Z
0.789
V
Z RT P
V
cm 1499.2 mol
3
Ans.
45
Rackett equation for saturated liquid:
cm
3
Tr
T Tc
Tr
0.847
Vc
200.0
mol
Zc
0.276
V
V c Zc
1 Tr
0.2857
V
94.17
cm
3
mol
Ans.
For saturated vapor, use Pitzer correlation:
0.422 Tr
1.6
B0
0.083
B0
0.468
B1
0.139
0.172 Tr
4.2
B1
0.207
V
RT P
R B0
B1
Tc Pc
V
1.538
10
3 3 cm
mol
Ans.
46
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 (c) 122.7 (d) 133.6 (e) 148.9 (f) 158.3 (g) 170.4 (h) 187.1 (i) 153.2 (j) 164.2 (k) 179.1 (l) 201.4 (m) (n) (o) (p) (q) (r) (s) (t) 61.7 64.1 66.9 70.3 64.4 67.4 70.8 74.8 1174.7 920.3 717.0 1516.2 1216.1 971.1 768.8 1330.3 1057.9 835.3 645.8 1252.5 1006.9 814.5 661.2 1318.7 1046.6 835.6 669.5 98.1 102.8 109.0 125.4 130.7 137.4 146.4 133.9 140.3 148.6 160.6 53.5 55.1 57.0 59.1 54.6 56.3 58.3 60.6 1228.7 990.4 805.0 1577.0 1296.8 1074.0 896.0 1405.7 1154.3 955.4 795.8 1276.9 1038.5 853.4 707.8 1319.0 1057.2 856.4 700.5
47
3.39 (a) Propane
Tc
369.8 K
Pc
42.48 bar
0.152
T
( 40
273.15)K
T
313.15 K
P
13.71 bar
Tr
T Tc
Tr
0.847
Pr
P Pc
Pr
0.323
From Table 3.1 for SRK:
1
0
0.08664
0.42748
1
2
Tr
q Tr
1
Tr Tr
0.480
1.574
0.176
2
1
Tr
2
Eq. (3.54)
T r Pr
Z
Pr Tr
0.01
Eq. (3.53)
Calculate Z for liquid by Eq. (3.56) Guess:
Given
Z=
T r Pr
Z
T r Pr
Z
T r Pr
1 q Tr
T r Pr T r Pr
3
Z
Z
Find Z) (
Z
0.055
V
Z RT P
V
cm 104.7 mol
Ans.
Calculate Z for vapor by Eq. (3.52)
Guess:
Z
0.9
Given
Z= 1
T r Pr
q Tr
T r Pr
Z Z T r Pr
T r Pr Z T r Pr
3
Z
Find Z) (
Z
0.78
V
Z RT P
V
1480.7
cm
mol
Ans.
48
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 (c) 118.2 (d) 128.5 (e) 142.1 (f) 150.7 (g) 161.8 (h) 177.1 (i) 146.7 (j) 156.9 (k) 170.7 (l) 191.3 (m) (n) (o) (p) (q) (r) (s) (t) 61.2 63.5 66.3 69.5 61.4 63.9 66.9 70.5 1157.8 904.9 703.3 1487.1 1189.9 947.8 747.8 1305.3 1035.2 815.1 628.5 1248.9 1003.2 810.7 657.4 1296.8 1026.3 817.0 652.5 98.1 102.8 109.0 125.4 130.7 137.4 146.4 133.9 140.3 148.6 160.6 53.5 55.1 57.0 59.1 54.6 56.3 58.3 60.6 1228.7 990.4 805.0 1577.0 1296.8 1074.0 896.0 1405.7 1154.3 955.4 795.8 1276.9 1038.5 853.4 707.8 1319.0 1057.2 856.4 700.5
49
3.40 (a) Propane
Tc
369.8 K
Pc
42.48 bar
0.152
T
( 40
273.15)K
T
313.15 K
P
13.71 bar
Tr
T Tc
Tr
0.847
Pr
P Pc
Pr
0.323
From Table 3.1 for PR:
1
2
Tr
1 2
1
0.37464
1 2
1.54226
0.26992
2
1
Tr
2
0.07779
0.45724
q Tr
Tr Tr
Eq. (3.54)
T r Pr
Z
Pr Tr
0.01
Eq. (3.53)
Calculate Z for liquid by Eq. (3.56) Guess:
Given
Z=
T r Pr
Z
T r Pr
Z
T r Pr
1 q Tr
T r Pr T r Pr
3
Z
Z
Find Z) (
Z
0.049
V
Z RT P
V
cm 92.2 mol
Ans.
Calculate Z for vapor by Eq. (3.52)
Guess:
Z
0.6
Given
Z= 1
T r Pr
q Tr
T r Pr
Z Z T r Pr
T r Pr Z T r Pr
Z
Find Z) (
Z
0.766
V
Z RT P
V
1454.5
cm
3
mol
Ans.
50
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 1131.8 879.2 678.1 1453.5 1156.3 915.0 715.8 1271.9 1002.3 782.8 597.3 1233.0 987.3 794.8 641.6 1280.2 1009.7 800.5 636.1 98.1 102.8 109.0 125.4 130.7 137.4 146.4 133.9 140.3 148.6 160.6 53.5 55.1 57.0 59.1 54.6 56.3 58.3 60.6 1228.7 990.4 805.0 1577.0 1296.8 1074.0 896.0 1405.7 1154.3 955.4 795.8 1276.9 1038.5 853.4 707.8 1319.0 1057.2 856.4 700.5
51
(c) 104.4 (d) 113.7 (e) 125.2 (f) 132.9 (g) 143.0 (h) 157.1 (i) 129.4 (j) 138.6 (k) 151.2 (l) 170.2 (m) (n) (o) (p) (q) (r) (s) (t) 54.0 56.0 58.4 61.4 54.1 56.3 58.9 62.2
3.41 (a) For ethylene,
molwt
28.054
gm Tc mol
282.3 K
Pc
50.40 bar
0.087
T
328.15 K
P
35 bar
Tr
T Tc
Pr
P Pc
Tr
1.162
Pr
0.694
From Tables E.1 & E.2:
Z0
0.838
Z1
0.033
Z
Z0
Z1
Z
0.841
n
18 kg molwt
Vtotal
ZnRT P
Vtotal
0.421 m
3
Ans.
(b) T
323.15 K
P
115 bar
Vtotal
0.25 m
3
Tr
T Tc
Tr
1.145
0.482
Pr
Z1
P Pc
0.126
Pr
2.282
From Tables E.3 & E.4: Z0
Z
Z0
Z1
Z
0.493
n
P Vtotal ZRT
n
2171 mol
mass
n molwt
mass
60.898 kg
Ans.
3.42 Assume validity of Eq. (3.38).
P1
1bar
T1
300K
V1
cm 23000 mol
3
Z1
P1 V 1 R T1
Z1
0.922
B
R T1 Z1 P1
1
B
1.942 10
3 3 cm
mol
With this B, recalculate at P2
P2
5bar
3 3 cm
Z2
1
B P2 R T1
Z2
0.611
V2
52
R T1 Z2 P2
V2
3.046
10
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
1.6
B0
0.146
B1
0.139
0.172 Tr
4.2
B1
0.104
V
RT P
B0
B1 R
Tc Pc
V
cm 989 mol
3
Ans.
3
For an ideal gas:
V
RT P
Tc
3
V
cm 1044 mol
Pc
3.44 T
320 K
0.152
P
16 bar
369.8 K
42.48 bar
Vc
cm 200 mol
Zc
0.276
molwt
44.097
gm mol
Tr
T Tc
1 Tr
Tr
0.2857
0.865
Pr
P Pc
Pr
0.377
3
Vliq
V c Zc
Vliq
96.769
cm
mol
Vtank
0.35 m
3
mliq
0.8 Vtank
Vliq molwt
mliq
127.594 kg
Ans.
B0
0.083
0.422 Tr
1.6
B0
0.449
B1
0.139
0.172 Tr
4.2
B1
0.177
53
Vvap
RT P
B0
B1 R
Tc Pc
Vvap
1.318
10
3 3 cm
mol
mvap
0.2 Vtank
Vvap molwt
mvap
2.341 kg
Ans.
3.45 T
298.15 K
Tc
425.1 K
Tr
T Tc P
Tr
0.701
P
2.43 bar
Pc
37.96 bar
3
Pr
Pc
Pr
0.064
0.200
Vvap
16 m
molwt
58.123
gm mol
B0
0.083
0.422 Tr
1.6
B0
0.661
B1
0.139
0.172 Tr
4.2
B1
0.624
3 3 cm
V
RT P
B0
B1 R
Tc Pc
V
9.469
10
mol
mvap
Vvap
V molwt
mvap
98.213 kg
Ans.
3.46 (a) T
333.15 K
Tc
305.3 K
Tr
T Tc
Tr
1.091
P
14000 kPa
Pc
48.72 bar
Pr
3
P Pc
Pr
2.874
0.100
Vtotal
0.15 m
molwt
30.07
gm mol
From tables E.3 & E.4: Z0
0.463
54
Z1
0.037
Z
Z0
Z1
Z
0.459
V
Z RT P
V
90.87
cm
3
mol
methane
Vtotal
V molwt
methane
49.64 kg
Ans.
(b)
V
Vtotal 40 kg
P
20000 kPa
P V = Z R T = Z R Tr Tc
or
Tr =
Z
where
PV R Tc
29.548
mol kg
Whence
Tr =
0.889 Z
at
Pr
P Pc
Pr
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
1.283
T
and
Tr Tc
Z
0.693
Whence
T
391.7 K
or
118.5 degC Ans.
3.47 Vtotal
0.15 m
3
T
298.15 K
Tc
282.3 K
Pc
50.40 bar
P V = P r Pc V = Z R T
0.087
molwt
28.054
gm mol
V
Vtotal 40 kg molwt
Pr = Z
or
where
RT Pc V
4.675
Whence
Pr = 4.675 Z
at
Tr
T Tc
Tr
1.056
55
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
1.582
and
Z
0.338
P
Pc Pr
P
79.73 bar
Ans.
3.48 mwater
15 kg
Vtotal
0.4 m
3
V
Vtotal mwater
V
26.667
cm
3
gm
Interpolate in Table F.2 at 400 degC to find:
P = 9920 kPa
Ans.
3.49 T1
298.15 K
Tc
305.3 K
Tr1
Pr1
T1 Tc
P1 Pc
Tr1
Pr1
0.977
P1
Vtotal
2200 kPa
0.35 m
3
Pc
48.72 bar
0.100
0.452
From Tables E.1 & E.2: Z0
.8105
Z1
0.0479
Z
Z0
Z1
Z
0.806
V1
Z R T1 P1
V1
cm 908 mol
3
T2
493.15 K
Tr2
T2 Tc
Tr2
1.615
Assume Eq. (3.38) applies at the final state.
B0
0.083
0.422 Tr2
1.6
B0
0.113
B1
0.139
0.172 Tr2
4.2
B1
0.116
P2 V1 B0
R T2 B1 R Tc Pc
56
P2
42.68 bar
Ans.
3.50 T
303.15 K
Tc
3
304.2 K
Tr
T Tc
Tr
0.997
Vtotal
0.5 m
Pc
73.83 bar
0.224
molwt
44.01
gm mol
B0
0.083
0.422 Tr
1.6
B0
0.341
B1
0.139
0.172 Tr
4.2
B1
0.036
3 3 cm
V
Vtotal 10 kg molwt
RT V B0 B1 R Tc Pc
V 2.2
10
mol
P
P
10.863 bar
Ans.
3.51 Basis: 1 mole of LIQUID nitrogen
Tn
77.3 K
Tc
126.2 K
Tr
Tn Tc
Tr
0.613
P
1 atm
Pc
34.0 bar
Pr
P Pc
Pr
3
0.03
0.038
molwt
28.014
gm mol
Vliq
34.7 cm
B0
0.083
0.422 Tr
1.6
B0
0.842
B1
0.139
0.172 Tr
4.2
B1
1.209
Z
1
B0
B1
Pr Tr
Z
0.957
57
nvapor
P Vliq Z R Tn
nvapor
5.718
10
3
mol
Final conditions:
ntotal
1 mol
nvapor
V
2 Vliq ntotal
V
69.005
cm
3
mol
T
298.15 K
Tr
T Tc
Tr
2.363
Pig
RT V
Pig
359.2 bar
Use Redlich/Kwong at so high a P.
0.08664
0.42748
2 2
( ) Tr
Tr
.5
Tr
0.651
a
Tr R Tc Pc
Eq. (3.42)
b
R Tc Pc
Eq. (3.43)
a
3 3 bar cm 0.901 m 2
mol
b
cm 26.737 mol
450.1 bar
3
P
RT V b
a V ( b) V
Eq. (3.44)
P
Ans.
3.52 For isobutane:
Tc
408.1 K
Pc
36.48 bar
V1
1.824
cm
3
gm
T1
300 K
P1
4 bar
T2
415 K
P2
75 bar
Tr1
T1 Tc
Pr1
P1 Pc
Tr2
T2 Tc
Pr2
P2 Pc
Tr1
0.735
Pr1
0.11
Tr2
1.017
Pr2
2.056
58
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 Z RT
with Z from Eq. (3.57) and Tables E.3 and E.4. Thus
Vc
262.7
cm
3
mol
0.181
Z0
r2
0.3356
Z1
r2
0.0756
Z
Z0
Z1
Z
0.322
r1 r2
P2 V c Z R T2
1.774
Eq. (3.75):
V2
V1
V2
cm 2.519 gm
3
Ans.
3.53 For n-pentane:
Tc
469.7 K
Pc
33.7 bar
1
0.63
gm cm
3
T1
291.15 K
P1
Pr1
1 bar
P1 Pc
T2
Tr2
413.15 K
T2 Tc
P2
Pr2
120 bar
P2 Pc
Tr1
Tr1
T1 Tc
0.62
Pr1
r1
0.03
Tr2
r2
0.88
Pr2
3.561
From Fig. (3.16):
2.69
r2
2.27
By Eq. (3.75),
2
1 r1
2
0.532
gm cm
3
Ans.
3.54 For ethanol: Tc
513.9 K
T
453.15 K
Tr
T Tc
Tr
0.882
Pc
61.48 bar
P
200 bar
Pr
P Pc
Pr
3.253
Vc
167
cm
3
mol
molwt
59
46.069
gm mol
From Fig. 3.16:
r
2.28
r Vc molwt
=
r
c=
r
Vc
0.629
gm cm
3
Ans.
3.55 For ammonia:
Tc
405.7 K
T
293.15 K
Tr
T Tc
Tr
0.723
Pc
112.8 bar
cm 72.5 mol
3
P
857 kPa
Pr
P Pc
0.253
Pr
0.076
Vc
Zc
0.242
0.2857
Eq. (3.72):
Vliquid
0.422 Tr
1.6
V c Zc
1 Tr
Vliquid
cm 27.11 mol
3
B0
0.083
B0
0.627
B1
0.139
0.172 Tr
4.2
B1
0.534
Vvapor
RT P
B0
B1 R
Tc Pc
Vvapor
cm 2616 mol
3
3
V
Vvapor
Vliquid
V
cm 2589 mol
Ans.
60
Alternatively, use Tables E.1 & E.2 to get the vapor volume:
Z0
0.929
Z1
0.071
Z
Z0
Z1
Z
3
0.911
Vvapor
ZRT P
Vvapor
2591
cm
mol
3
V
Vvapor
Vliquid
V
2564
cm
mol
Ans.
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:
R
ft atm 0.7302 lbmol rankine
3
T
519.67 rankine
P
1 atm
V
1400 ft
3
n
PV RT
T Tc
n
3.689 lbmol
For methane at 3000 psi and 60 degF:
Tc
190.6 1.8 rankine
T
519.67 rankine
Tr
Tr
1.515
Pc
45.99 bar
P
3000 psi
Pr
P Pc
Pr
4.498
0.012 From Tables E.3 & E.4:
Z0
0.819
ZnRT P
Z1
0.234
Z
3
Z0
Z1
Ans.
Z
0.822
Vtank
Vtank
5.636 ft
61
3.59 T
25K
P
3.213bar
Calculate the effective critical parameters for hydrogen by equations (3.58) and (3.56)
Tc 1
43.6 21.8K 2.016T
K
Tc
30.435 K
Pc 1
20.5 44.2K 2.016T
bar
Pc
10.922 bar
0
Pr
P Pc
Pr
0.294
Tr
T Tc
Tr
0.821
3
Initial guess of volume:
V
RT P
V
cm 646.903 mol
Use the generalized Pitzer correlation
B0
Z 1
0.083
0.422 Tr
1.6
B0
0.495
B1
0.139
0.172 Tr
4.2
B1
0.254
B0
B1
Pr Tr
Z
0.823
Ans.
Experimental: Z = 0.7757
For Redlich/Kwong EOS:
1
0.5
0
0.08664
0.42748
Table 3.1
Tr
Tr
Table 3.1
q Tr
Tr Tr
Eq. (3.54)
T r Pr
Pr Tr
Eq. (3.53)
62
Calculate Z
Guess:
Z
0.9
Given
Eq. (3.52)
Z= 1
T r Pr
q Tr
T r Pr
Z Z Z
T r Pr T r Pr
Z
Find Z) (
Z
0.791
Ans.
Experimental: Z = 0.7757
3.61For methane:
0.012
Tc
190.6K
Pc
45.99bar
T 288.706 K
At standard condition:
Pitzer correlations:
T
( 60
32)
5 9
273.15 K
P
1atm
Tr
T Tc
Tr
1.515
Pr
B1
P Pc
0.139 0.172 Tr
4.2
Pr
B1
0.022
B0
Z0
0.083
0.422 Tr Pr
Tr
1.6
B0
0.134
0.109
1
B0
Z0
0.998
Z1
B1
Pr Tr
Z1
0.00158
Z
Z0
Z1
Z
0.998
V1
Z RT P
V1
m 0.024 mol
P 300psi
3
(a) At actual condition:
Pitzer correlations:
T
T
( 50
32)
5 9
273.15 K
283.15 K
Tr
T Tc
1.6
Tr
1.486
Pr
P Pc
Pr
0.45
B0
0.083
0.422 Tr
B0
0.141
B1
0.139
0.172 Tr
4.2
B1
0.106
63
Z0
1
B0
Pr Tr
Z0
0.957
Z1
B1
Pr Tr
Z1
0.0322
3
Z
Z0
Z1
3
Z
0.958
V2
ZRT P
V2
0.00109
3
m
mol
q1
150 10
6 ft
day
q2
q1
V2 V1
3 kmol
q2
6.915
10
6 ft
day
Ans.
(b) n1
q1 V1
n1 7.485 10
hr
Ans.
(c) D
22.624in
q2 A
A
4
D
2
A
0.259 m
2
u
u
8.738
m s
Ans.
64
3.62
0.012 0.087 0.1 0.140 0.152 0.181 0.187 0.19 0.191 0.194 0.196 0.2 0.205 0.21 0.21 0.212 0.218 0.23 0.235 0.252 0.262 0.28 0.297 0.301 0.302 0.303 0.31 0.322 0.326
ZC
0.286 0.281 0.279 0.289 0.276 0.282 0.271 0.267 0.277 0.275 0.273 0.274 0.273 0.273 0.271 0.272 0.275 0.272 0.269 0.27 0.264 0.265 0.256 0.266 0.266 0.263 0.263 0.26 0.261
65
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.
m
slope
ZC
( 0.091)
b
intercept
ZC
( 0.291)
2
r
corr
ZC
( 0.878) r
0.771
0.29
ZC m
0.28
b 0.27
0.26 0.25 0 0.1 0.2 0.3 0.4
The equation of the line is: Zc = 0.291 0.091
Ans.
3.65 Cp
7 R 2
Cv
5 2
R
Cp Cv
1.4
T1
298.15K
P1
1bar
P2
5bar
T3
T1
P3
5bar
Step 1->2 Adiabatic compression
1
T2
U12
T1
P2 P1
Cv T2 T1
T2
472.216 K
U12
3.618
H12
Cp T2
T1
H12
5.065
kJ mol kJ
Ans.
mol
Ans.
Q12
0
kJ mol
Q12
W12
0
kJ mol
kJ mol
Ans.
Ans.
W12
U12
3.618
Step 2->3 Isobaric cooling
U23
Cv T3
T2
U23
3.618
kJ mol kJ
Ans.
H23
Cp T3
T2
H23
5.065
Q23
H23
Q23
5.065
mol kJ
Ans.
W23
R T3
T2
W23
1.447
mol kJ
Ans.
mol
Ans.
Step 3->1 Isothermal expansion
U31
Cv T1
T3
U31
0
H31
Cp T1
T3
66
H31
0
kJ mol kJ
Ans. Ans.
mol
Q31
W31
R T3 ln
Q31
P1 P3
Q31
3.99
W31
kJ mol kJ 3.99 mol
Ans.
Ans.
For the cycle
Qcycle
Q12
Q23
Q31
Qcycle
1.076
kJ Ans. mol
Wcycle
W12
W23
W31
Wcycle
1.076
kJ Ans. mol
Now assume that each step is irreversible with efficiency:
80%
Step 1->2 Adiabatic compression
W12
W12
W12
4.522
kJ mol
Ans.
Q12
U12
W12
Q12
0.904
kJ mol
Ans.
Step 2->3 Isobaric cooling
W23
W23
W23
1.809
kJ mol
Ans.
Q23
U23
W23
Q23
5.427
kJ mol
Ans.
Step 3->1 Isothermal expansion
W31
W31
W31
3.192
Q31
U31
W31
Q31
3.192
kJ mol kJ
Ans.
mol
Ans.
For the cycle
Qcycle
Q12
Q23
Q31
Qcycle
Wcycle
W12
W23
W31
Wcycle
kJ Ans. mol kJ 3.14 mol Ans. 3.14
67
3.67 a) PV data are taken from Table F.2 at pressures above 1atm.
125 150 175 P 200 225 250 275 300
Z PVM RT
kPa
2109.7 1757.0 1505.1 V 1316.2 1169.2 1051.6 955.45 875.29
cm
3
gm
T
( 300
273.15) K
gm mol
M
18.01
1 VM
i
0 7
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
A
Zi
i
1
Xi
i
B
intercept ( Y) X
6 5 cm 10 2
B
cm 128.42 mol
3
Ans.
slope ( Y) X
mol cm
3
A
1.567
5 mol 3
mol
X
0
10
5 mol 3
8 10
cm
cm
Below is a plot of the data along with the linear fit and the extrapolation to the y-intercept.
68
115
120
(Z-1)/p
125
130
0 2 10
(Z-1)/p Linear fit
5
4 10
p
5
6 10
5
8 10
5
b) Repeat part a) for T = 350 C PV data are taken from Table F.2 at pressures above 1atm.
125 150 175 P 200 225 250 275 300
Z PVM RT
kPa
2295.6 1912.2 1638.3 V 1432.8 1273.1 1145.2 1040.7 953.52
cm gm
3
T
( 350
273.15)K
gm mol
M
18.01
1 VM
i
0 7
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
i
69
1
Xi
i
B
intercept ( X Y)
B
105.899
cm
3
mol
Ans.
A
slope ( X Y)
A
5 mol 3
1.784
5 mol 3
6 5 cm 10 2
mol
X
0
mol cm
3
10
8 10
cm
cm
Below is a plot of the data along with the linear fit and the extrapolation to the y-intercept.
90
95
(Z-1)/p
100
105
110
0 2 10
(Z-1)/p Linear fit
5
4 10
p
5
6 10
5
8 10
5
c) Repeat part a) for T = 400 C PV data are taken from Table F.2 at pressures above 1atm.
125 150 175 P 200 225 250 275 300
kPa
2481.2 2066.9 1771.1 V 1549.2 1376.6 1238.5 1125.5 1031.4
70
cm gm
3
T
( 400
273.15)K
M
18.01
gm mol
Z
PV M RT
1 VM
i
0 7
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
i
1
Xi
i
B
intercept ( X Y)
6 5 cm 10 2
B
89.902
cm
3
mol
Ans.
A
slope ( X Y)
A
2.044
5 mol 3
mol
X
0
mol cm
3
10
5 mol 3
8 10
cm
cm
Below is a plot of the data along with the linear fit and the extrapolation to the y-intercept.
70
75
(Z-1)/p
80
85
90
0 2 10
(Z-1)/p Linear fit
5
4 10
p
5
6 10
5
8 10
5
71
3.70
Create a plot of
(Z
1) Z Tr Pr
vs
Pr Z Tr
Data from Appendix E at Tr = 1 0.01 0.05 0.10 Pr 0.20 0.40 0.60 0.80 Pr Z Tr
0.9967 0.9832 0.9659 Z 0.9300 0.8509 0.7574 0.6355
Tr 1
X
Y
(Z
1) Z Tr Pr
Create a linear fit of Y vs X Slope
Intercept
Rsquare
slope ( X Y)
intercept ( X Y)
corr ( X Y)
0.28
Slope
Intercept
Rsquare
0.033
0.332
0.9965
Y Slope X Intercept
0.3
0.32
0.34
0
0.2
0.4
0.6
X
0.8
1
1.2
1.4
The second virial coefficient (Bhat) is the value when X -> 0 Bhat Intercept
B0 0.083 0.422 Tr
72
Bhat
1.6
0.332
0.339
Ans.
Ans.
By Eqns. (3.65) and (3.66)
These values differ by 2%.
B0
3.71 Use the SRK equation to calculate Z
Tc
150.9 K
T
( 30
T 273.15)Kr
T Tc P
Tr
2.009
Pc
48.98 bar
P
300 bar
Pr
Pc
Pr
6.125
0.0
1
0
0.08664
1
0.42748 Table 3.1
2
Tr
q Tr
Calculate Z
1
Tr Tr
0.480
1.574
0.176
2
1
Tr
2
Table 3.1
Eq. (3.54)
Guess:
T r Pr
Z 0.9
Pr Tr
Eq. (3.53)
Given
Eq. (3.52)
Z= 1
T r Pr
q Tr
T r Pr
Z Z T r Pr
T r Pr Z T r Pr
Z
Find Z) (
Z
1.025
V
Z RT P
V
cm Ans. 86.1 mol
3
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
Vt
0.4 1800 cm
3
cm 5 mol 33.0 mol
3
Vt
555 cm
3
V
Vt n
73
V
111
cm
3
mol
Use SRK equation to calculate pressure.
Tc
308.3 K
T
( 125
273.15) K
Tr
T Tc
Tr
1.291
Pc
61.39 bar
0.0
1
0
0.08664
1
2
0.42748 Table 3.1
Tr
q Tr
1
Tr Tr
0.480
1.574
0.176
2
1
Tr
2
Table 3.1
Eq. (3.54)
a
Tr Pc
R Tc
2
2
Eq. (3.45)
b
R Tc Pc
Eq. (3.46)
3
a
3 3 bar cm 3.995 m 2
mol
b
cm 36.175 mol
P
RT V b
a V (V b)
T ( 10 273.15)K
P
197.8 bar
Ans.
3.73 mass
35000kg
0.152
Tc
369.8K
Pc
3
42.48bar
M
n
44.097
7.937
gm mol
5
Zc
0.276
Vc
200.0
cm
mol
n
mass M
10 mol
a) Estimate the volume of gas using the truncated virial equation
Tr
T Tc
Tr
0.766
P
1atm
Pr
P Pc
B0
0.083
0.422 Tr
1.6
Eq. (3-65)
B1
0.139
0.172 Tr
4.2
Eq. (3-66)
B0
0.564
74
B1
0.389
Z
1
B0
B1
Pr Tr
Z
0.981
7 3 3
Vt
ZnRT P
Vt
3
2.379
6
10 m m
This would require a very large tank. If the D tank were spherical the diameter would be:
0.2857
Vt
D
32.565 m
b) Calculate the molar volume of the liquid with the Rackett equation(3.72) Vliq
P
V c Zc
6.294atm
1 Tr
Vliq
Pr
B1 Pr Tr
Z
85.444
0.15
cm
3
mol
P Pc
Pr
Z
1
B0
ZRT P
Vtank
90%
0.878
3 3 cm
Vvap
Guess:
Given
Vvap
90% Vliq n
10% Vtank Vvap
3.24 10
mol
Vtank Vliq
= n
Vtank
Vtank
Find Vtank
75.133 m
6
3
This would require a small tank. If the tank D were spherical, the diameter would be:
3
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
473.15 K
T
1373.15 K
n
10 mol
For SO2: A
5.699
B
0.801 10
3
C
0.0
D
1.015 10
5
H
R ICPH T0 T A B C D
H
47.007
kJ mol
Q
n H
Q (b) T0
470.073 kJ
Ans.
523.15 K
T
1473.15 K
n
3
12 mol
For propane:A
1.213
B
28.785 10
C
8.824 10
6
D
0
H
R ICPH T0 T A B C 0.0
H
161.834
kJ mol
Q
n H
Q
1.942
10 kJ Ans.
3
4.2 (a) T0
473.15 K
n
10 mol
Q
3
800 kJ
For ethylene:
A
1.424
B
14.394 10 K
C
4.392 10 K
2
6
2 (guess)
Given
Q= nR
A T0
1
B 2 T0 2
T
2
1
C 3 T0 3
3
1
Find
2.905
T0
T
1374.5 K
Ans.
(b) T0
533.15 K
n
15 mol
Q
3
2500 kJ
For 1-butene:
A
1.967
B
31.630 10 K
76
C
9.873 10 K
2
6
3
(guess)
Given
Q= nR
Find
A T0
1
B 2 T0 2
2.652
2
1
T0
C 3 T0 3
3
1
T
T
1413.8 K
Ans.
6
(c) T0
500 degF
n
40 lbmol
Q
10 BTU
Values converted to SI units
T0
533.15K
n
1.814
10 mol
4
Q
3
1.055 10 kJ
6
For ethylene:
A
1.424
B
14.394 10 K
2
C
4.392 10 K
2
6
2 (guess)
Given
Q= nR
Find
A T0
1
B 2 T0 2
T
1
C 3 T0 3
3
1
2.256
T0
T
1202.8 K Ans.
T = 1705.4degF 4.3 Assume air at the given conditions an ideal gas. Basis of calculation is 1 second.
P 1 atm
T0
122 degF
V
250 ft
3
T
3
932 degF
Convert given values to SI units
V
7.079 m
T
( T
32degF) 273.15K
T0
T0
32degF
273.15K
T
773.15 K
T0
323.15 K
n
PV R T0
n
266.985 mol
For air:
A
3.355
B
0.575 10
3
C
0.0
D
0.016 10
5
H
R ICPH T0 T A B C D
77
H
13.707
kJ mol
Q
n H
Q 4.4 molwt
3.469
3 10 BTU Ans.
100.1
gm mol
T0
323.15 K
4
T
1153.15 K
n
10000 kg molwt
n
9.99
10 mol
For CaCO3: A
12.572
B
2.637 10
3
C
0.0
D
3.120 10
5
H
R ICPH T0 T A B C D
H
9.441
10
4 J
mol
Q
n H
Q
9.4315 10 kJ
6
Ans.
4.7 Let step 12 represent the initial reversible adiabatic expansion, and step 23 the final constant-volume heating.
T1
P2
298.15 K
101.3 kPa
T3
P3
298.15 K
104.0 kPa
P1
T2
121.3 kPa
T3 P2 P3
CP
30
J mol K
(guess)
R CP
T2
290.41 K
Given
T2 = T1
P2 P1
CP
Find CP
CP
56.95
J Ans. mol K
4.9a) Acetone: Tc
508.2K
Pc
47.01bar
Tn
329.4K
Hn
29.10
kJ mol
Trn
Tn Tc
Trn
0.648
Use Eq. (4.12) to calculate H at Tn ( Hncalc)
1.092 ln Hncalc R Tn
Pc bar
Trn
1.013
0.930
Hncalc
30.108
kJ mol
Ans.
78
To compare with the value listed in Table B.2, calculate the % error.
%error
Hncalc Hn
Hn
%error
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 % 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%
79
b)
469.7 Tc 507.6 562.2 560.4
K
33.70 Pc 30.25 48.98 43.50 366.3
273.15 K bar
25.79 Hn 28.85
kJ 30.72 mol
29.97 72.150
J
36.0 Tn 68.7 80.0 80.7 H25
366.1
433.3 gm 392.5
M
86.177 gm 78.114 mol 82.145
Tr1
Tn Tc
0.658 0.673 0.628 0.631
Tr2
( 25
273.15) K Tc
H2
H25 M
26.429 H2 31.549
H1
Hn
Tr1
kJ 33.847 mol
32.242 1 1 Tr2 Tr1
0.38
H2calc
H1
Eq. (4.13) %error
H2calc H2
0.072
H2
26.448 H2calc 31.533
kJ Ans. 33.571 mol
26.429 H2 31.549
kJ 33.847 mol
%error
0.052 0.814 1.781
%
32.816
32.242
The values calculated with Eq. (4.13) are within 2% of the handbook values.
4.10 The ln P vs. 1/T relation over a short range is very nearly linear. Our procedure is therefore to take 5 points, including the point at the temperature of interest and two points on either side, and to do a linear least-squares fit, from which the required derivative in Eq. (4.11) can be found. Temperatures are in rankines, pressures in psia, volumes in cu ft/lbm, and enthalpies in Btu/lbm. The molar mass M of tetrafluoroethane is 102.04. The factor 5.4039 converts energy units from (psia)(cu ft) to Btu.
80
(a) T
459.67
5
V
1.934
5 0
0.012
i
1 5
18.787 21.162
Data:
P
23.767 26.617 29.726
t
5 10 15
xi
ti
1 459.67
yi
ln Pi
slope
slope ( x y) slope
4952
dPdT
( P) 3 T
2
slopedPdT
0.545
H
T V dPdT 5.4039
H
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
H = 85.817
( 90.111)
( 85.834)
(b)
(c)
H = 81.034
( 81.136)
(d)
H = 76.007
( 75.902)
(e)
H = 69.863
( 69.969)
4.11
119.377 M 32.042 153.822
gm mol
536.4 Tc 512.6 K Pc 556.4
Hexp is the given value at the normal boiling point.
81
54.72 80.97 bar Tn 45.60
334.3 337.9 K 349.8 Tn Tc
H is the value at 0 degC.
Tr1
273.15K Tc
Tr2
270.9 H 1189.5 217.8
J gm
246.9 Hexp 1099.5 194.2
J gm
0.509 Tr1 0.533 0.491 1 1 Tr2 Tr1
0.38
0.623 Tr2 0.659 0.629
(a) By Eq. (4.13)
Hn
H
PCE
Hn
Hexp Hexp
100%
This is the % error
245 Hn
J 1055.2 gm 193.2
0.77 PCE 4.03 % 0.52
(b) By Eq. (4.12):
Hn
R Tn M
1.092 ln
Pc bar
Tr2
1.013
0.930
PCE
Hn
Hexp Hexp
100%
247.7 Hn 1195.3 192.3
J gm
0.34 PCE 8.72 0.96
%
4.12 Acetone 0.307
Vc
Tr
Tc
3
508.2K
329.4K
0.648
Pc
P
Pr
47.01bar
1atm
P Pc
Zc
Hn
Pr
0.233
29.1 kJ mol
cm 209 mol Tn
Tc
Tn
Tr
0.022
82
Generalized Correlations to estimate volumes Vapor Volume
B0 0.083 0.422 Tr
B1 0.139
1.6
B0
0.762
Eq. (3.65)
0.172 Tr
4.2
B1
Pr Tr
0.924
Eq. (3.66)
Z
1
B0
Pr Tr
B1
Z
0.965
(Pg. 102)
3 4 cm
V
Z R Tn P
2
V
2.609
10
mol
Liquid Volume
1 Tr
7
Vsat
V c Zc
Eq. (3.72)
Vsat
H= T V
cm 70.917 mol
d Psat dT
3
Combining the Clapyeron equation (4.11)
B T C
A
with Antoine's Equation
Psat = e
gives
H= T V ( T
3 4 cm
B C)
2
A
B ( C) T
e
V
A
V
Vsat
V
B
2.602
2756.22
10
mol
C 228.060
14.3145
83
A
B Tn 273.15K
Hcalc
Tn V Tn K
B 273.15K
2
C
e C
K
kPa K
Hcalc
29.662
kJ Ans. mol
%error
Hcalc Hn
Hn
%error
1.9 %
The table below shows the values for other components in Table B.2. Values agree within 5% except for acetic acid.
Acetone Acetic Acid Acetonitrile 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
348.15 K
P
100 kPa
(guess)
5622.7 K T
Given
ln
P = 48.157543 kPa
P 5622.7 K T
2
4.70504 ln
T K 0.029 bar K
3
P
Find ( ) dPdT P
4.70504 T
dPdT
P
87.396 kPa
H
joule 31600 mol
Vliq
cm 96.49 mol
Clapeyron equation:
dPdT =
H T V Vliq
V = vapor molar volume. V
Vliq
H T dPdT
cm 1369.5 mol
3
Eq. (3.39)
B
PV V RT
1
B
Ans.
4.14 (a) Methanol: Tc
512.6K
Pc
80.97bar
Tn
3
337.9K
AL
13.431
BL
51.28 10
CL
131.13 10
6
CPL ( ) T
AL
BL K
BV
85
T
CL K
2
T
2
R
AV
2.211
12.216 10
3
CV
3.450 10
6
CPV ( ) T
AV
BV K
T
CV K
2
T
2
R
P
3bar
Tsat
368.0K
T1
300K
T2
500K
Estimate Hv using Riedel equation (4.12) and Watson correction (4.13)
Trn
Tn Tc
Trn
0.659
Trsat
Tsat Tc
Trsat
0.718
1.092 ln Hn
Pc bar
Trn
1.013
0.930
R Tn
Hn
38.301
kJ mol
kJ mol
Hv
Hn
Tsat
1 1
Trsat Trn
0.38
Hv
T2
35.645
H
T1
CPL ( ) T T d
Hv
T sat
CPV ( ) T T d
H
49.38
kJ mol
Ans.
n
100
kmol hr
Q
n H
Q
1.372
10 kW
3
(b) Benzene:
Hv = 28.273
kJ mol
H = 55.296
kJ mol
Q = 1.536 10 kW
3
(c) Toluene
Hv = 30.625
kJ mol
H = 65.586
kJ mol
Q = 1.822 10 kW
3
4.15 Benzene
Tc
562.2K
Pc
48.98bar
Tn
353.2K
T1sat
451.7K
T2sat
358.7K
Cp
162
J mol K
86
Estimate Hv using Riedel equation (4.12) and Watson correction (4.13)
Trn
Tn Tc
Trn
0.628
Tr2sat
T2sat Tc
Tr2sat
0.638
1.092 ln Hn
Pc bar
Trn
1.013
0.930
R Tn
Hn
30.588
kJ mol
Hv
Hn
1 1
Tr2sat Trn
0.38
Hv
30.28
kJ mol
Assume the throttling process is adiabatic and isenthalpic. Guess vapor fraction (x): x
0.5
Given
Cp T1sat
T2sat = x Hv
x
Find ( x)
x
0.498
Ans.
4.16 (a) For acetylene:
Tc
308.3 K
Pc
61.39 bar
Tn
189.4 K
T
298.15 K
Trn
Tn Tc
Trn
0.614
Tr
T Tc
Tr
0.967
ln Hn R Tn 1.092
1 1 Tr Trn
J mol
Pc bar
0.930
0.38
1.013 Trn
Hn
16.91
kJ mol
kJ mol
Hv
Hn
Hv
6.638
Hf
227480
H298
Hf
Hv
H298
220.8
kJ mol
Ans.
87
(b) For 1,3-butadiene:
H298 = 88.5
kJ mol
(c) For ethylbenzene:
H298 = 12.3
kJ mol
(d) For n-hexane:
H298 = 198.6
kJ mol
(e) For styrene:
H298 = 103.9
kJ mol
4.17
1st law:
dQ = dU
dW = CV dT
P dV
(A)
Ideal gas:
PV= RT
and
P dV
V dP = R dT
Whence
Since
V dP = R dT
P V = const then
P dV
P V
1
(B)
dV = V dP
from which
V dP = P
dV
Combines with (B) to yield:
P dV =
R dT 1
Combines with (A) to give:
dQ = CV dT
R dT 1
or
dQ = CP dT
R dT
R dT 1
which reduces to
dQ = CP dT
R dT
1
(C)
R dT
or
dQ =
CP R
1
Since CP is linear in T, the mean heat capacity is the value of CP at the arithmetic mean temperature. Thus Tam
675
88
CPm
R 3.85
0.57 10
3
Tam
950 K
T1 400 K
1.55
Integrate (C):
T2
R T2
Q
CPm R
1
T1
Q
6477.5
J mol
Ans.
P1
1 bar
P2
P1
T2 T1
1
P2
11.45 bar
Ans.
4.18
For the combustion of methanol: CH3OH(g) + (3/2)O2(g) = CO2(g) + 2H2O(g) H298
H298
393509
676485
2 ( 241818) ( 200660)
For 6 MeOH:
H298 = 4 058 910 J
Ans.
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 2 n 2 11.286
i 1 4 A
3.639 5.457 B 3.470 3.280
0.506 1.045 1.450 0.593
10 K
3
0.227 1.157 D 0.121 0.040
10 K
5 2
A
i
ni Ai B
i
ni Bi
D
i
ni D i
A
54.872
B
0.012
T
1 K
D
1.621 10 K
5 2
For the products,
HP = R
T0
CP R
dT
T0
298.15K
The integral is given by Eq. (4.7). Moreover, by an energy balance,
H298
HP = 0
2
(guess)
Given
H298 = R A T0
8.497
1
B 2 T0 2
T
2
1
D T0
1
Find
T
T0
2533.5 K
Ans.
Parts (b), (c), and (d) are worked the same way, the only change being in the numbers of moles of products. (b)
nO = 0.75
2
nn = 14.107
2
T = 2198.6 K
Ans.
(c)
nO = 1.5
2
nn = 16.929
2
T = 1950.9 K
T = 1609.2 K
Ans.
Ans.
(d)
nO = 3.0
2
nn = 22.571
2
90
(e) 50% xs air preheated to 500 degC. For this process, Hair H298 HP = 0
773.15)
Hair = MCPH ( 298.15
For one mole of air:
MCPH 773.15 298.15 3.355 0.575 10
For 4.5/0.21 = 21.429 moles of air: Hair = n R MCPH T
Hair
Hair
3
0.0
0.016 10
5
= 3.65606
21.429 8.314 3.65606 ( 298.15
309399 J mol
H298 0.506 B 1.045 1.450 0.593
773.15)
J mol
The energy balance here gives: 1.5 n 2 2 16.929
Hair
HP = 0
3.639 A 5.457 3.470 3.280
0.227
10 K
3
D
1.157 0.121 0.040
10 K
5
2
A
i
ni Ai
78.84
B
i
ni Bi
1 0.016 K
D
i
ni D i
1.735 10 K
5 2
A
B
D
2 (guess)
Given
H298
Hair = R A T0
1
B 2 T0 2
2
1
D T0
Find
7.656
91
1
T
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 (b) -905,468 J (c) -71,660 J (d) -61,980 J (e) -367,582 J (f) -2,732,016 J (g) -105,140 J (h) -38,292 J (i) 164,647 J (j) -48,969 J (k) -149,728 J (l) -1,036,036 J (m) 207,436 J
(n) 180,500 J (o) 178,321 J (p) -132,439 J (q) -44,370 J (r) -68,910 J (s) -492,640 J (t) 109,780 J (u) 235,030 J (v) -132,038 J (w) -1,807,968 J (x) 42,720 J (y) 117,440 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.
Ho298
T/ K
A
(a) (b ) (f ) (i ) (j ) (l ) (m) (n ) (o ) (r ) (t ) (u ) (v ) (w) (x ) (y )
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
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
4.23
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
150 10
6 ft
3
day
T
( 60
5 32) K 9
273.15K
T
288.71 K
P
1atm
The higher heating value is the negative of the heat of combustion with water as liquid product. Calculate methane standard heat of combustion with water as liquid product: CH4 + 2O2 --> CO2 +2H2O Standard Heats of Formation:
HfCH4
74520
J mol
HfO2
0
J mol
HfCO2
393509
J mol
HfH2Oliq
285830
J mol
Hc
HfCO2
2 HfH2Oliq
HfCH4
2 HfO2
HigherHeatingValue
Hc
Hc
8.906 10
5 J
mol
Assuming methane is an ideal gas at standard conditions:
n q P RT
n 1.793 10
5dollar GJ
8 mol
day
5 dollar
n HigherHeatingValue
4.25
7.985 10
day
Ans.
Calculate methane standard heat of combustion with water as liquid product Standard Heats of Formation: CH4 + 2O2 --> CO2 +2H2O
HfCH4
HfCO2
HcCH4
HcCH4
74520
J mol
J mol
HfO2
0
J mol
285830 J mol
393509
HfCO2
890649
HfH2Oliq
HfCH4
2 HfH2Oliq
J mol
94
2 HfO2
Calculate ethane standard heat of combustion with water as liquid product: Standard Heats of Formation:C2H6 + 7/2O2 --> 2CO2 +3H2O
HfC2H6
83820
J mol
HcC2H6
2 HfCO2
3 HfH2Oliq
HfC2H6
7 2
HfO2
HcC2H6
1560688
J mol
Calculate propane standard heat of combustion with water as liquid product
Standard Heats of Formation:C3H8 + 5O2 --> 3CO2 +4H2O
HfC3H8
104680
J mol
HcC3H8
3 HfCO2
4 HfH2Oliq
HfC3H8
5 HfO2
HcC3H8
2219.167
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
c) 0.85 HcCH4
0.05 HcC2H6
0.07 HcC2H6
0.03 HcC3H8
0.03 HcC3H8
946.194
kJ mol
kJ mol
932.875
Gas b) has the highest standard heat of combustion. 4.26 2H2 + O2 = 2H2O(l)
Ans.
Hf1
2 ( 285830) J
C + O2 = CO2(g)
Hf2
393509 J
N2(g)+2H2O(l)+CO2(g)=(NH2)2CO(s)+3/2O2
H
631660 J
. N2(g)+2H2(g)+C(s)+1/2O2(g)=(NH2)2CO(s)
H298
Hf1
Hf2
H
95
H298
333509 J
Ans.
4.28
On the basis of 1 mole of C10H18 (molar mass = 162.27) Q 43960 162.27 J
Q 7.133 10 J
6
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
H
U=
H
( )= PV
H
ngas
H
R T ngas
( 10 14.5)mol
6
298.15 K
Q R T ngas
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 Moles oxygen Moles nitrogen n1
n2
n3
96
1
2 1.3
2.6 79 21
n2
n3
2.6
9.781
Total moles of dry gases entering
n
n1
n2
n3
n
13.381
At 30 degC the vapor pressure of water is 4.241 kPa. Moles of water vapor entering:
n4
4.241 13.381 101.325 4.241
n4 (1) (2) (3) (4)
0.585
Leaving:
CO2 -- 1 mol H2O -- 2.585 mol O2 -- 2.6 - 2 = 0.6 mol N2 -- 9.781 mol
By an energy balance on the furnace:
Q=
H=
H298
HP
For evaluation of
HP we number species as above.
5.457 1.045 B 1.450 0.506 0.593
10
3
1 n 2.585 0.6 9.781
i 1 4
R
1.157 D 0.121 0.227 0.040
10
5
A
3.470 3.639 3.280
8.314
J mol K
A
i
ni Ai B
i
ni Bi
3
D
i
ni Di
A
48.692
B
10.896983 10
C
0
D
5.892 10
4
The TOTAL value for MCPH of the product stream:
HP
R M CPH ( 303.15K 1773.15K A B C D) ( 1773.15
303.15)K
HP
732.013
kJ mol
From Example 4.7:
H298
802625
J mol
Q
HP
H298
Q = 70 612 J
97
Ans.
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
n
1.578
2.585 1.578
Moles water condensing:
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)( 323.15 1773.15) K n H50
Ans.
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
ENERGY BALANCE:
H=
HR
H298
HP = 0
REACTANTS: 1=NH3; 2=O2; 3=N2
4 n 6.5 24.45
i 1 3 A
3.578 3.639 3.280 B
3.020 0.506 10 0.593
3
0.186 D 0.227 10 0.040
5
A
i
ni Ai
B
i
ni Bi
D
i
ni Di
A
118.161
B
0.02987
C
0.0
D
1.242 10
5
TOTAL mean heat capacity of reactant stream:
HR
R MCPH ( 348.15K 298.15K A B C D) ( 298.15K
348.15K)
HR
52.635
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 2.5 n 3.2 4.8 24.45
i 1 5 A
3.578 3.639 3.387 3.470 3.280 B
3.020 0.506 0.629 1.450 0.593
10 K
3
0.186 0.227 D 0.014 0.121 0.040
10 K
5 2
A
i
ni Ai
B
i
ni Bi
D
i
ni Di
A
119.65
B
1 0.027 K
D
8.873 10 K
4
2
By the energy balance and Eq. (4.7), we can write: (guess) 2 T0 298.15K
99
Given
H298
H R = R A T0
1
B 2 T0 2
2
1
D T0
Find
1
3.283
T
T0
T
978.9 K
Ans.
4.31
C2H4(g) + H2O(g) = C2H5OH(l) BASIS: 1 mole ethanol produced Energy balance:
n
1mol
H= Q=
HR
H298
H298
[ 277690
( 52510
241818) ]
J mol
H298
8.838 10
4 J
mol
Reactant stream consists of 1 mole each of C2H4 and H2O.
i
1 2 n
1.424 3.470
1 1
B 14.394 1.450 10
3
A
C
4.392 0.0
10
6
D
0.0 0.121
10
5
A
i
ni Ai B
i
ni Bi
C
i
ni Ci
D
i
ni Di
A
4.894
B
0.01584
C
4.392 10
6
D
1.21 10
4
HR
R M CP ( H 298.15K 593.15K A B C D)( 298.15K
593.15K)
HR
2.727 10
4 J
mol
Q
HR
H298 1mol
Q
115653 J
Ans.
100
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
H298
4.003 10
4 J
mol
Q=
HR
H298
HP
REACTANTS: 1=CH4; 2=H2O
1.702 3.470
9.081 1.450
i
1 2
2.164 0.0
n
0.2 0.4
10
6
A
B
10
3
C
D
0.0 0.121
10
5
A
i
ni Ai B
i
ni Bi
C
i
ni Ci
D
i
ni Di
A
1.728
B
2.396 10
3
C
4.328 10
7
D
4.84 10
3
HR
RI CPH ( 773.15K 298.15K A B C D)
HR
1.145 10
4 J
mol
PRODUCTS: 1=CO2; 2=CO; 3=H2O; 4=H2
0.0275 n 0.1725 0.1725 0.6275
A
5.457 3.376 3.470 3.249 B
1.045 0.557 1.450 0.422
10
3
1.157 D 0.031 0.121 0.083
10
5
101
i
1 4
A
i
ni Ai
B
i
ni Bi
4
D
i
ni Di
A
3.37
B
6.397 10
C
0.0
D
3.579 10
3
HP
HP
RI CPH ( 298.15K 1123.15K A B C D) 4 J 2.63 10 mol
Q
HR
H298
HP mol
Q
54881 J
Ans.
4.33 CH4 + 2O2 = CO2 + 2H2O(g) C2H6 + 3.5O2 = 2CO2 + 3H2O(g)
H298a
802625
H298b
1428652
BASIS: 1 mole fuel (0.75 mol CH4; 0.25 mol C2H6) burned completely with 80% xs. air. O2 in = 1.8[(0.75)(2) + (0.25)(3.5)] = 4.275 mol N2 in = 4.275(79/21) = 16.082 mol Product gases: CO2 = 0.75 + 2(0.25) = 1.25 mol
H2O = 2(0.75) + 3(0.25) = 2.25 mol O2 = (0.8/1.8)(4.275) = 1.9 mol N2 = 16.082 mol
H298
0.75 H298a
0.25 H298b
J mol
Q
8 10
5
J mol
Energy balance: Q = H = H298 HP PRODUCTS: 1=CO2; 2=H2O; 3=O2; 4=N2
HP = Q
H298
1.157
1.25 n 2.25 1.9 16.082
i 1 4 A
5.457 3.470 3.639 3.280 B
1.045 1.450 0.506 0.593
10 K
3
D
0.121 0.227 0.040
10 K
5
2
A
i
ni Ai B
i
ni Bi
D
i
ni Di
4 2
A
74.292
B
0.015
102
1 K
C
0.0
D
9.62 10 K
By the energy balance and Eq. (4.7), we can write: T0
Given
303.15K
2
(guess)
Q
H298 = R A T0
1
B 2 T0 2
2
1
Find
D T0
1.788
T T0
1
Ans.
T
542.2 K
4.34
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 0.129 A
5.699 3.639 8.060 B
0.801 0.506 10 1.056
3
1.015 D 0.227 10 2.028
D
i
5
i
1 3
A
i
ni Ai B
i
ni Bi
2.58 10
103
ni Di
D 1.16 10
4
A
0.06985
B
7
C
0
Hnet
R M CPH ( 298.15K 773.15K A B C D)( 773.15K
298.15K)
Hnet
77.617
J mol
H773
H298
Hnet
H773
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: Energy balance: moles CO = moles H2O = 0.2 moles CO2 = moles H2 = 0.3
Q=
H=
HR
H298
HP
H298
0.3 [ 393509
( 110525
J 214818) ] mol
0.557 1.450
H298
2.045 10
4 J
mol
Reactants: 1: CO 2: H2O
n
i
0.5 0.5
1 2
A
3.376 3.470
B
10
3
D
0.031 0.121
10
5
A
i
ni Ai B
i
ni Bi
D
i
ni Di
A
3.423
B
1.004 10
3
C
0 D
4.5 10
3
HR
R M CPH ( 298.15K 398.15K A B C D)( 298.15K
398.15K)
HR
3.168 10
3 J
mol
Products:
1: CO 2: H2O 3: CO2 4: H2
3.376 A 3.470 5.457 3.249
104
0.2 n 0.2 0.3 0.3
0.557 B 1.450 1.045 0.422
10
3
0.031 D 0.121 1.157 0.083
10
5
i
1 4 A
i
ni Ai
3.981
B
i
ni Bi
8.415 10
4
D
i
ni Di
3.042 10
4
A
B
C
0 D
HP
HP
Q
R MCPH ( 298.15K 698.15K A B C D) ( 698.15K
1.415 10
HR
4 J
298.15K)
mol
HP mol
Q 9470 J Ans.
H298
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,
15.124 x 100.175
0.21 =
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:
BTU 209.13 lbm lbm
6
Q
0.3 19000
Q
1.192 10 BTU
Reaction upon which net heating value is based:
106
OIL + (21.024)O2 = (14.8)CO2 + (12.448 + 0.116)H2O(g) + (0.149)N2 H298a 19000 209.13 BTU
H298a 3.973 10 BTU
6
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 393509) 0.42993 BTU
Guess: y 50
(y)H2O(l) = (y)H2O(g)
H298c ( y)
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
3 11.8 n y) ( 5.913 79.278 15.797 y A
r
1.986
T
1.045 0.557 B
400
459.67 1.8
T
477.594
5.457 3.376 3.639 3.280 3.470
1.157 0.031
3 5
0.506 10 0.593 1.450
D
0.227 10 0.040 0.121
i
1 5 A ( y)
i
n y) i Ai B ( y) (
i
n y) i Bi D ( y) (
i
n y) i Di (
D ( y) T0
2
T T0
1.602
CP ( y)
r A ( y)
B ( y) T0 2
1
107
Given
CP ()( y 400
77)BTU = Q
H298 () y
y
Find y ()
y Whence
49.782
(lbmol H2O evaporated)
y 18.015 209.13
4.288
(lb H2O evap. per lb oil burned) Ans.
4.37 BASIS: One mole of product gas containing 0.242 mol HCN, and (1-0.242)/2 = 0.379 mol each of N2 and C2H2. The energy balance is
Q=
H=
H298
HP
H298
( 135100 2
227480)
0.242 J 2
H298
5.169 10 J
3
Products:
0.242 n 0.379 0.379
i 1 3 A
i
4.736 A 3.280 6.132 ni Ai B
1.359 0.593 10 1.952
3
0.725 D 0.040 1.299
10
5
B
i
ni Bi
D
i
ni Di
A
4.7133
B
1.2934 10
3
C
0 D
6.526 10
4
HP
R M CPH ( 298.15K 873.15K A B C D)( 873.15K
298.15K)mol
HP
2.495 10 J
4
HP
2.495 10 J
4
Q
H298
HP
Q
30124 J
Ans.
4.38 BASIS: 1 mole gas entering reactor, containing 0.6 mol HCl, 0.36 mol O2, and 0.04 mol N2. HCl reacted = (0.6)(0.75) = 0.45 mol 4HCl(g) + O2(g) = 2H2O(g) + 2Cl2(g)
108
For this reaction,
H298
[ ( 241818) 2
4 ( 92307) ]
J mol
H298
1.144 10
5 J
mol
Evaluate
H823
by Eq. (4.21) with
T0
298.15K
T
823.15K
1: H2O 2: Cl2 3: HCl 4=O2
2 n 2 4 1
i 1 4 A
3.470 4.442 3.156 3.639 B
1.45 0.089 0.623 0.506
10
3
0.121 D 0.344 0.151 0.227
10
5
A
i
ni Ai
B
i
ni Bi
D
i
ni Di
8.23 10
4
A
H823 H298
0.439
MCPH T0 T
B
A
8 10
B
5
C
0
D
C
D R T
T0
H823
117592
J mol
Heat transferred per mol of entering gas mixture:
Q
H823 4
0.45 mol
Q
13229 J
Ans.
4.39 CO2 + C = 2CO 2C + O2 = 2CO Eq. (4.21) applies to each reaction:
H298a
172459
H298b
J (a) mol J (b) 221050 mol
For (a):
2 n 1 1
A
3.376 1.771 5.457
109
0.557 B 0.771 10 1.045
3
0.031 D 0.867 10 1.157
5
i
1 3
A
i
ni Ai
B
i
ni Bi
4
D
i
ni Di
A
H1148a
0.476
B
7.02 10
C
0
D
1.962 10
5
H298a R M CP 298.15K 1148.15K H
A
B
C
D ( 1148.15K
298.15K)
H1148a
1.696 10
5 J
mol
For (b):
2 n 1 2
i 1 3 A
3.376 3.639 1.771 B
0.557 0.506 10 0.771
3
0.031 D 0.227 10 0.867
5
A
i
ni Ai
B
B
i
4
ni Bi
C 0
D
i
ni Di
1.899 10
5
A
H1148b
0.429
9.34 10
D
H298b R M CP 298.15K 1148.15K H
5 J
A
B
C
D ( 1148.15K
298.15K)
H1148b
2.249 10
mol
The combined heats of reaction must be zero: nCO
2
H1148a
nO
2
H1148b = 0
2
nCO
Define:
r=
nO
r
H1148b H1148a
r
1.327
2
110
For 100 mol flue gas and x mol air, moles are: Flue gas Air Feed mix CO2 CO O2 N2 12.8 3.7 5.4 78.1 0 0 0.21x 0.79x 12.8 3.7 5.4 + 0.21x 78.1 + 0.79x
Whence in the feed mix:
r=
12.8 5.4 0.21 x
x
12.5 5.4 r mol 0.21
x
19.155 mol
Flue gas to air ratio =
100 19.155
5.221
Ans.
Product composition:
nCO
3.7
2 ( 12.8
5.4
0.21 19.155)
nCO
nN
2
48.145
93.232
nN
2
78.1
0.79 19.155
nCO nCO nN
2
Mole % CO =
100
34.054 Ans.
Mole % N2 =
100
34.054
65.946
4.40 CH4 + 2O2 = CO2 + 2H2O(g) CH4 + (3/2)O2 = CO + 2H2O(g)
H298a
802625
J mol
H298b
519641
J mol
BASIS: 1 mole of fuel gas consisting of 0.94 mol CH4 and 0.06 mol N2 Air entering contains:
1.35 2 0.94
2.538
mol O2
2.538
79 21
9.548
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
0.282 H298b
H298
6.747 10
5 J
mol
Moles H2O formed by reaction =
0.94 2.0
1.88
Moles O2 consumed by reaction =
2 0.658
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 0.282 n 1.880 0.799 9.608
i 1 5 A
i
5.457 3.376 A 3.470 3.639 3.280 B
1.045 0.557 1.450 10 0.506 0.593
3
1.157 0.031 D 0.121 0.227 0.040
10
5
ni Ai
45.4881
B
B
i
ni Bi
3
D
i
ni Di
3.396 10
298.15K)
4
A
HP
HP
9.6725 10
C
0 D
R M CPH ( 298.15K 483.15K A B C D)( 483.15K
7.541 10
4 J
mol
Hrx H298 HP
Hrx 599.252 kJ mol kg 34.0 sec
Energy balance:
HH2O mdotH2O
Hrx ndotfuel = 0
mdotH2O
112
From Table C.1:
HH2O
( 398.0
104.8)
kJ kg
ndotfuel
HH2O mdotH2O Hrx
ndotfuel
16.635
mol sec
Volumetric flow rate of fuel, assuming ideal gas:
V
ndotfuel R 298.15 K 101325 Pa
V
m 0.407 sec
3
Ans.
4.41 C4H8(g) = C4H6(g) + H2(g)
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
1 n 1 1
2.734 A 3.249 1.967 i 1 3 B
T0
298.15 K
T
798.15 K
by Eq. (4.21): Evaluate H798 1: C4H6 2: H2 3: C4H8 26.786 0.422 31.630
10
3
8.882 C 0.0 9.873
10
6
0.0 D 0.083 10 0.0
5
A
i
ni Ai B
i
ni Bi
3
C
i
ni Ci
7
D
i
ni Di
3
A
4.016
B
4.422 10
C
9.91 10
D
8.3 10
H798
H298
MCPH 298.15K 798.15K
5 J
A
B
C
D R T
T0
H798
1.179 10
mol
Q
0.33 mol H798
Q
38896 J
113
Ans.
4.42Assume Ideal Gas and P = 1 atm
P
1atm
R
7.88 10
3 BTU
mol K
a) T0
( 70
459.67) rankine
T
T0
20rankine
Q
12
BTU sec
T0
294.261 K
3
T
305.372 K
ICPH T0 T 3.355 0.575 10
0
0.016 10
5
38.995 K
ndot
Q R ICPH T0 T 3.355 0.575 10
3
0
3
0.016 10
5
ndot
39.051
3
mol s
Vdot
ndot R T0 P
Vdot
m 0.943 s
T T0
Vdot
ft 33.298 sec
Q 12 kJ s
Ans.
b) T0
R
( 24
273.15) K
3
13K
8.314 10
kJ mol K
3
ICPH T0 T 3.355 0.575 10
0
0.016 10
5
45.659 K
ndot
Q R ICPH T0 T 3.355 0.575 10
3
0
0.016 10
5
ndot
31.611
3
mol s
Vdot
ndot R T0 P
Vdot
m 0.7707 s
Ans.
4.43Assume Ideal Gas and P = 1 atm
P
1atm
a) T0
( 94
459.67) rankine
3
T
( 68
459.67) rankine
R
1.61 10
atm ft mol rankine
3
Vdot
ft 50 sec
3
ndot
P Vdot R T0
114
ndot
56.097
mol s
T0
307.594 K
3
T
0
293.15 K
0.016 10
5
ICPH T0 T 3.355 0.575 10
R 7.88 10
3 BTU
50.7 K
mol K
3
Q
R ICPH T0 T 3.355 0.575 10
0
0.016 10
5
ndot
22.4121 BTU Ans. sec
Q
b) T0
( 35
273.15)K
3 5 atm m
T
( 25
273.15)K
R
8.205 10
m 1.5 sec
3
mol K
ndot
3
Vdot
P Vdot R T0
0 0.016 10
5
ndot
59.325
mol s
ICPH T0 T 3.355 0.575 10
R
Q
35.119 K
8.314 10
3
kJ mol K
3
R ICPH T0 T 3.355 0.575 10
0
0.016 10
5
ndot
17.3216 kJ Ans. s
Q
4.44 First calculate the standard heat of combustion of propane C3H8 + 5O2 = 3CO2(g) + 4H2O (g)
H298
H298
3
393509
J mol
4
241818
J mol
104680
J mol
2.043 10
dollars gal
6 J
mol
80%
Cost
2.20
115
Estimate the density of propane using the Rackett equation Tc
T
369.8K
( 25 273.15) K
1 Tr
Zc
0.276
Tr T Tc
Vc
Tr
cm 200.0 mol
0.806
3
0.2857
Vsat
V c Zc
Vsat
cm 89.373 mol
3
Heating_cost
Vsat Cost H298
Heating_cost
Heating_cost
0.032
dollars MJ
dollars 10 BTU
6
Ans.
33.528
4.45 T0
( 25
273.15) K
Q
Q
T
( 500
273.15) K
3
a) Acetylene
R ICPH T0 T 6.132 1.952 10
2.612 10
4 J
0
1.299 10
5
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
Q
( 25
30000
273.15)K
J mol
T
( 500
273.15)K
a) Acetylene
T
Given Q = R ICPH T0 T 6.132 1.952 10
Find ( T)
T 835.369 K
T
3
0
1.299 10
5
273.15K
562.2 degC
The calculations are repeated and the answers are in the following table:
T( K) 835.4 964.0 534.4 932.9 1248.0 690.2 1298.4 1277.0 877.3 1230.2 1259.7 959.4 927.2 1209.9 636.3 T( C) 562.3 690.9 261.3 659.8 974.9 417.1 1025.3 1003.9 604.2 957.1 986.6 686.3 654.1 936.8 363.2
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
( 250
y
273.15) K
0.5
Q
11500
J mol
a) Guess mole fraction of methane:
Given
y ICPH T0 T 1.702 9.081 10 (1
y
3
2.164 10
3
6
0 R
6
= Q
y) ICPH T0 T 1.131 19.225 10
Find ( y)
y 0.637
Ans.
117
5.561 10
0 R
b) T0
( 100
273.15) K
T
( 400
y
273.15)K
0.5
Q
54000
J mol
Guess mole fraction of benzene
Given
y ICPH T0 T ( 1
y
c) T0
0.206 39.064 10
3
13.301 10
3
6
0 R
6
= Q
y)ICPH T0 T
Find y ()
( 150
3.876 63.249 10
Ans.
T ( 250
y
20.928 10
0 R
y
0.245
273.15) K
273.15)K
0.5
Q
17500
J mol
Guess mole fraction of toluene
Given
y ICPH T0 T 0.290 47.052 10 ( 1
y
3
15.716 10
3
6
0 R
6
= Q
y)ICPH T0 T 1.124 55.380 10
Find y ()
y 0.512
Ans.
18.476 10
0 R
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.
I e m e a e Pi h nt r di t nc Se ton I ci
TH1
Pi h a End nc t
TH1
Se ton I ci I Se ton I ci THi T THi Se ton I ci I
TC1 TCi
TH2 TC1 TCi TC2
TH2 T TC2
118
To solve the problem, apply an energy balance around each section of the exchanger.
T H1
Section I balance:
mdotC HC1
HCi = ndotH
THi
CP dT
T Hi
Section II balance:
mdotC HCi
HC2 = ndotH
TH2
CP dT
If the pinch is intermediate, then THi = TCi + T. If the pinch is at the end,
then TH2 = TC2 + T.
a) TH1
1000degC
TC1
100degC
TCi
100degC
TC2
25degC
T
10degC
HC1
2676.0
kJ kg
HCi
419.1
3
kJ kg
HC2
104.8
kJ kg
5
For air from Table C.1:A
3.355 B
0.575 10
1
C
0 D
0.016 10
Assume as a basis ndot = 1 mol/s.
ndotH
kmol s
Assume pinch at end:
TH2
TC2
T
Guess:
mdotC
1
kg s
THi
110degC
Given
mdotC HC1
mdotC HCi
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
kg s
mdotC THi
mdotC ndotH
Find mdotC THi THi
170.261 degC mdotC
11.255
0.011
kg mol
Ans.
THi
TCi
70.261 degC
TH2
119
TC2
10 degC
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
500degC
TC1
100degC
TCi
100degC
TC2
25degC
T
10degC
HC1
2676.0
kJ kg
HCi
419.1
kJ kg
HC2
104.8
kJ kg
Assume as a basis ndot = 1 mol/s.
ndotH
1
kmol s
Assume pinch is intermediate:
THi
TCi
T
Guess:
mdotC
1
kg s
TH2
110degC
Given
mdotC HC1 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
3 kg
mdotC HCi
mdotC TH2
48.695 degC
mdotC
5.03
kg s
mdotC ndotH
5.03 10
mol
Ans.
THi
TCi
10 degC
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
393.509
kJ mol
H0f4
285.830
kJ mol
M3
44
gm mol
H0r
6 H0f3
6 H0f4
H0f1
6 H0f2
H0r
2801.634
kJ mol
Ans.
b) energy_per_kg
150
kJ kg
mass_person
57kg
mass_glucose
mass_person energy_per_kg H0r
M1
mass_glucose
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
2.216 10 kg
8
Ans.
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
H0f1
74.520
kJ mol
H0f2
83.820
kJ mol
H0f3
0
kJ mol
H0f4
393.509
kJ mol
H0f5
241.818
kJ mol
a) H0c
1.05 H0f4
2 H0f5
0.85 H0f1
0.10 H0f2
1.05 H0f3
H0c
825.096
kJ mol
Ans.
b) For complete combustion of 1 mole of fuel and 50% excess air, the exit gas will contain the following numbers of moles:
n3
0.5 2.05mol
121
n3
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
( 25
273.15) K
T2
( 600
273.15) K
A
n3 3.639
n4 6.311
n5 3.470
n6 3.280
3
mol
B
n3 0.506
n4 0.805
n5 1.450 mol
n6 0.593 10
C
n3 0
n4 0
n5 0 mol
n6 0 10
6
D
n3 ( 0.227) n4 ( 0.906) n5 0.121
mol
n6 0.040 10
5
Q
H0c
ICPH T1 T2 A B C D R
Q
529.889
kJ mol
Ans.
122
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
= 1
TC TH
TC
323.15 K
TH
798.15 K
148.78 kJ s
QH
250
kJ s
Work
QH 1
TC TH
or
Work
Work
QH
148.78 kW which is the power. Ans. Work
By Eq. (5.1),
QC
QC
101.22
kJ s
Ans.
5.3 (a) Let symbols Q and Work represent rates in kJ/s
TH
750 K
TC
300 K
Work
95000 kW
By Eq. (5.8):
Work QH
1
TC TH
QH
0.6
Work
5 1.583 10 kW Ans.
But
=
Whence
QH
QC
(b)
QH QH QC
Work
QC
6.333
4 10 kW Ans.
0.35
Work QH
Work
QH
QC
5 2.714 10 kW Ans.
1.764
5 10 kW Ans.
5.4 (a) TC
303.15 K
TH
623.15 K
Carnot
1
TC TH
0.55
123
Carnot
0.282
Ans.
(b)
0.35
Carnot
0.55
Carnot
0.636
By Eq. (5.8),
TH
TC 1
Carnot
TH
833.66 K Ans.
5.7 Let the symbols represent rates where appropriate. Calculate mass rate of LNG evaporation:
V
9000
m s
3
P
1.0133 bar
T
298.15 K
molwt
17
gm mol
mLNG
PV molwt RT
mLNG
6254
kg s
Maximum power is generated by a Carnot engine, for which
Work
QC
=
QH QC
QC
=
QH QC
1=
TH TC
1
TH
303.15 K
TC
QC
113.7 K
6
QC
Work
512
kJ mLNG kg
TH TC
3.202
10 kW
6
QC
1
Work
5.336
10 kW
Ans.
QH
QC
Work
QH
8.538
10 kW
6
Ans.
5.8
Take the heat capacity of water to be constant at the valueCP
4.184
(a) T1
273.15 K
T2
373.15 K
Q
CP T2
kJ kg K
T1
Q
418.4
kJ kg K kJ
kg
SH2O
Sres
CP ln
Q T2
T2 T1
SH2O
1.305
Sres
124
1.121
kJ kg K
Ans.
Stotal
SH2O
Sres
Stotal
0.184
kJ kgK
Ans.
(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.
Sres
Q 2
1 323.15 K
1 373.15 K
Sres
1.208
kJ kgK
Stotal
Sres
SH2O
Stotal
0.097
kJ kgK
Ans.
(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.
3
5.9
P1
1 bar
T1
500 K
V
0.06m
n
P1 V R T1
n
1.443 mol
CV
5 2
R
Q
15000 J
(a) Const.-V heating;
U= Q
W = Q = n CV T2
T1
T2
T1
Q n CV
T2
1
10 K
3
By Eq. (5.18),
P2 P1
S = n CP ln
T2 T1
T2 T1
R ln
P2 P1
But
=
Whence
S
n CV ln
T2 T1
S
20.794
J K
Ans.
(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
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:
7 R 2
SA
SA
CP ln
8.726
463.15 343.15
J mol K
SB
CP ln
473.15 593.15
SB
6.577
J mol K
Ans.
(b) For both cases:
Stotal
SA
SB
Stotal
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
CP ln
8.726
463.15 343.15
J mol K
SB
CP ln
353.15 473.15
SB
8.512
J mol K
Ans.
Stotal
SA
SB
Stotal
0.214
J mol K
Ans.
5.16 By Eq. (5.8),
dW dQ
= 1
T T
dW = dQ
T
dQ T
Since dQ/T = dS,
dW = dQ
T dS
Integration gives the required result.
T1
600 K
T2
400 K
T
300 K
Q
CP T2
T1
Q
5.82
10
3 J
mol
126
S
CP ln
Q
T2 T1
T S
S
11.799
J mol K
Work
Work
2280
J mol
Ans.
Q
Q
Work
Q
3540
J mol
Ans.
Sreservoir
Q T
Sreservoir
11.8
J mol K
Ans.
S
Sreservoir
0
J mol K
Process is reversible.
5.17 TH1
600 K
TC1
300 K
TH2
300 K
TH1
TC2
TC1
250 K
For the Carnot engine, use Eq. (5.8):
W
QH1
=
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
300K
P1
1.2bar
T2
450K
P2
6bar
Cp
7 R 2
H
C p T2
T1
R ln
3 J
H
P2 P1
4.365
10
3 J
mol
Ans.
S
Cp ln
T2 T1
S
1.582
J mol K
Ans.
(b)
H = 5.82 10
mol
S = 1.484
J mol K
127
(c)
H = 3.118 10
3 J
mol
3 J
S = 4.953
J mol K
(d)
H = 3.741 10
mol
3 J
S = 2.618
J mol K
(e)
H = 6.651 10
mol
S = 3.607
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
T3 V 3
1
= T4 V 4
1
For constant-volume step 4 to 1:
For isobaric step 2 to 3:
V1 = V4
P2 T2
=
P3 T3
Solving these 4 equations for T4 yields: T4 = T1
7 R 2
T2 T3
Cp
Cv
5 R 2
Cp Cv
1.4
T1
( 200
273.15) K
T2
T4
( 1000
273.15) K
T3
( 1700
273.15) K
T4
T1
T2 T3
1
873.759 K
Eq. (A) p. 306
1
T4 T3
T1 T2
0.591
Ans.
128
5.21 CV
CP
R
P1
2 bar
P2
7 bar
T1
298.15 K
CP CV
1.4
With the reversible work given by Eq. (3.34), we get for the actual W:
1
Work
1.35
R T1 1
P2
1 Work 3.6 10
3 J
P1
U = C V T2 T1 Whence
mol
But Q = 0, and W =
T2
T1
Work CV
T2
471.374 K
S
CP ln
T2 T1
R ln
P2 P1
S
2.914
J mol K
Ans.
5.25 P
4
T
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
T
dP = dT P
(A)
PV= RT
P dV
V dP = R dT
P dV = R dT
V dP = R dT
RT
dP P
129
In combination with (A) this becomes P dV = R dT
Moreover,
R dT = 2 R dT
P3 = P 1 T1 T3
= P1
T1 T2
V1
W31 =
V3
P dV = 2 R T 1
U31 W31 = CV T1
2R T1
T3 = 2 R T1
T3 2 R T1
R T1
T2
Q31 =
T3
T2
Q31 = CV
T3 = C P
W23
Q31
=
Wnet
Qin
=
W12
W31
CP
7 R 2
T1
P1
700 K
T2
P3 W12 2.91
350 K
T1 T2
3 J
1.5 bar
P1 10
W12
R T2
T1
mol
3 J
W23
W31
Q31
R T2 ln
P3 P1
T2
T1 T2
W23
W31
Q31
2.017
10
mol
3 J
2 R T1
CP
W12
5.82
10
mol
R
W23
Q31
1.309
10
4 J
mol
W31
0.068
Ans.
130
5.26 T
403.15 K
P1
2.5 bar
P2
6.5 bar
Tres
J mol K
298.15 K
By Eq. (5.18),
S
R ln
P2 P1
S
7.944
Ans.
With the reversible work given by Eq. (3.27), we get for the actual W:
Work
1.3 R T ln
P2 P1
(Isothermal compresion) Work
4.163
10
3 J
mol
Q
Work
Q here is with respect to the system.
So for the heat reservoir, we have
Sres
Stotal
Q Tres
S Sres
Sres
Stotal
13.96
J mol K
J mol K
Ans.
6.02
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
3
0.0
1.015 10
5
S
536.1
J K
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
3
8.824 10
6
0.0
S
2018.7
J K
Ans.
131
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
S
S
10 mol
n R ICPS 473.15K 1374.5K 1.424 14.394 10
900.86 J K
Ans.
3
4.392 10
6
0.0
(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
S
S
15 mol
n R ICPS 533.15K 1413.8K 1.967 31.630 10
2657.5 J K
Ans.
3
9.873 10
6
0.0
(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
S
S
18140 mol
n R ICPS 533.15K 1202.9K 1.424 14.394 10
1.2436 10
6 J
3
4.392 10
6
0.0
K
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
T1
T2
298.15 K
248.15 K
348.15 K
T0 ( 1
Temperature of entering air
Temperature of chilled air
Temperature of warm air x)CP T2 T0 = 0
x C P T1
x 0.3
(guess)
132
Given
x 1 x
=
T2 T1
T0 T0
x
Find x ()
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
R
P0
5 bar
P
T2 T0
1 bar
P P0
Stotal
x CP ln
T1 T0
( 1
x)CP ln
R ln
Stotal
12.97
J mol K
Ans.
Since this is positive, there is no violation of the second law.
5.30
T1
523.15 K
T2
353.15 K
P1
3 bar
P2
1 bar
Tres
CV
Sres
303.15 K
Work
1800
J mol
CP
Q
Q
7 2
R
CP
R
Q
Q=
Sres
U
Work
5.718 J mol K
C V T2
T1
3 J
Work
Tres
1.733 10
mol
S
CP ln
T2 T1
R ln
P2 P1
S
2.301
J mol K
Stotal
S
Sres
Stotal
3.42
J PROCESS IS POSSIBLE. mol K
133
5.33 For the process of cooling the brine:
CP
3.5
kJ kg K
T
40 K
mdot
20
kg sec
t
0.27
T1
( 273.15
25) K
T1
298.15 K
T2
( 273.15
15) K
T2
258.15 K
T
( 273.15
30) K
T
303.15 K
H
CP T
H
140
kJ kg
kJ kg K
S
CP ln
T2 T1
Wdotideal
Wdot
S
0.504
Eq. (5.26):
By Eq. (5.28):
mdot
H
T
S Wdotideal Wdot
256.938 kW
Wdotideal
t
951.6 kW
Ans.
5.34 E
110 volt
i
9.7 amp
T
300 K
Wdotmech
1.25 hp
Wdotelect
iE
Wdotelect
1.067
10 W
3
At steady state: Qdot
Wdotelect
Wdotmech =
d t U = 0 dt
Qdot T
SdotG =
d t S = 0 dt
Qdot
Wdotelect
Wdotmech
Qdot
134.875 W
SdotG
Qdot T
SdotG
0.45
W K
Ans.
134
5.35
25 ohm
i
2
10 amp
T
300 K
Wdotelect
i
Wdotelect
2.5
10 W
3
At steady state:
Qdot
Wdotelect =
d t U = 0 dt
Qdot
Wdotelect
Qdot T
SdotG =
d t S = 0 dt
SdotG
Qdot T
Qdot
2.5
10 watt
3
SdotG
8.333
watt K
Ans.
5.38 mdot
10
kmol hr
T1
( 25
273.15) K
P1
10bar
P2
1.2bar
Cp
7 R 2
Cv
Cp
R
Cp Cv
7 5
(a) Assuming an isenthalpic process:
T2
T2
T1
T2
298.15 K
Ans.
S (b) = R
T1
Cp 1 R T
dT
ln
P2 P1
P2 P1
Eq. (5.14)
S
T2 7 R ln T1 2
R ln
S
17.628
J mol K
Ans.
(c) SdotG
mdot S
SdotG
48.966
W K
Ans.
3 J
(d) T
( 20
273.15) K
Wlost
T
S
Wlost
5.168
10
mol
Ans.
5.39(a) T1
500K
P1
6bar
T2
371K
P2
1.2bar
Cp
7 R 2
T
300K
Basis: 1 mol
n
1mol
H
n C p T2
T1
Ws
135
H
Ws
3753.8 J
Ans.
S
n Cp ln
T2 T1
Wideal
R ln
P2 P1
S
4.698
J K
Ans.
Eq. (5.27)
Eq. (5.30)
H
T
S
Wideal
5163 J
Wlost
SG
Wideal
Wlost T
Wideal
5163J
2953.9J
4193.7J
Ws
Wlost
SG
Wlost
1409.3J
493J
1130J
1409.3 J
4.698
SG
4.698
1.643
3.767
Ans.
Ans.
Eq. (5.39)
Ws
(a)
(b)
(c)
J K
3753.8J
2460.9J
3063.7J
J K
J K
J K
(d)
3853.5J
4952.4J
1098.8J
3.663
J K J
(e)
3055.4J
4119.2J
1063.8J
3.546
K
5.41
P1
2500kPa
P2 S
SdotG
150kPa 0.023 kJ mol K
T
300K
mdot
20
S
SdotG
R ln
P2 P1
mol sec
mdot S
T SdotG
W
0.468
kJ sec K
Ans.
Ans.
TH
TC
Wdotlost
5.42 QH 1kJ
Wdotlost
TH
TC
140.344 kW
( 250
( 25
0.45kJ
273.15)K
273.15)K
523.15 K
298.15 K
W
actual
QH
actual
0.45
136
max
1
TC TH
max
0.43
Since
actual> max,
the process is impossible.
5.43 QH
150 kJ
Q1
50 kJ
Q2
100 kJ
TH
550 K
T1
350 K
T2
250 K
T
300 K
(a)
SG
QH TH
Q1 T1
Q2 T2
SG
0.27
kJ K
Ans.
(b)
Wlost
T SG
Wlost
81.039 kJ
Ans.
5.44
Wdot
750 MW
TH
( 315
273.15)K
TC
TC
( 20
273.15)K
TH
(a)
max
588.15 K
0.502 Ans.
293.15 K
1
TC TH
max
QdotH
Wdot
max
QdotC
QdotC
QdotH
Wdot
(minimum value)
745.297 MW
(b)
0.6
max
QdotH Wdot
Wdot
QdotH
10 MW
3
2.492
10 W
9
QdotC
QdotH
QdotC
1.742
3
(actual value)
River temperature rise: Vdot
m 165 s
1
gm cm
3
Cp
1
cal gm K
T
QdotC Vdot Cp
T
2.522 K
Ans.
137
5.46 T1
P1
( 20
5bar
273.15)K
T2
P2
( 27
1atm
273.15) K
T3
( 22
273.15)K
First check the First Law using Eqn. (2.33) neglect changes in kinetic and potential energy.
H 6 7
R ICPH T1 T2 3.355 0.575 10 1 7
3
0
0.016 10
5
ICPH T1 T3 3.355 0.575 10
10
4 kJ
3
0
0.016 10
5
R
H
8.797
mol
H is essentially zero so the first law is satisfied.
Calculate the rate of entropy generation using Eqn. (5.23)
SG
6 7
R ICPS T1 T2 3.355 0.575 10 1 7
3
0
3
0.016 10
5
R ICPS T1 T3 3.355 0.575 10
kJ mol K
Since SG
0
0.016 10
5
R ln
P2 P1
SG
0.013
0, this process is possible.
5.47 a) Vdot
P
ft 100000 hr
1atm
3
T1
T
( 70
( 70
459.67)rankine T2
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
3
0
0.016 10
3
5 5
R ICPS T1 T2 3.355 0.575 10
0
0.016 10
Wideal
1.776 hp
138
b) Vdot
P
3000
1atm
m
3
hr
T1
T
( 25
( 25
273.15) K
273.15) K
T2
(8
273.15) K
Assume air is an Ideal Gas
R T1 Calculate ideal work using Eqn. (5.26) ndot P Vdot
ndot 34.064 mol s
Wideal
ndot R ICPH T1 T2 3.355 0.575 10 T
3
0
0.016 10
3
5 5
R ICPS T1 T2 3.355 0.575 10
0
0.016 10
Wideal
5.48 T1
1.952 kW
459.67) rankine
T2 ( 300 459.67) rankine
( 2000
Cp ( ) T
T ( 70
3.83
0.000306
T rankine
R
Hv
Tsteam
970
BTU lbm
M
29
gm mol
459.67) rankine
( 212
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
T1
Cp ( ) T T d
T2
mdotsteam Hv = 0
Cp ( ) T T d mdotndot
T1
Hv
mdotndot
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
Ssteam
=
mdotsteam ndotgas
Hv Tsteam
T2
Ssteam
Sgas
BTU lb rankine
3 kg
Ssteam
1.444
Sgas
T1
C p ( T) T
dT
Sgas
Sgas
SdotG
9.969
10
BTU
mol lb rankine
SdotG
mdotndot Ssteam
11.756
BTU lbmol rankine
Calculate lost work by Eq. (5.34) Wlost SdotG T
Hv Tsteam
Wlost
6227
BTU Ans. lbmol
1.444 BTU lb rankine
BTU lb
b) Hsteam
Hv
Ssteam
Ssteam
Wideal
Hsteam
T
Ssteam
Wideal
205.071
Calculate lbs of steam generated per lbmol of gas cooled.
T2
C p ( T ) dT mn
T1
Hv
mn
15.043
lb lbmol
Use ratio to calculate ideal work of steam per lbmol of gas
Wideal mn 3.085
T2
10
3 BTU
lbmol
Ans.
c)
Hgas
T1
Cp ( T) dT
Hgas T Sgas
Wideal 9.312 10
3 BTU
Wideal
lbmol
Ans.
140
5.49 T1
( 1100
273.15) K
T2
( 150
273.15) K
Cp ( ) T
T ( 25
3.83
0.000551
T R K
Hv
Tsteam
2256.9
( 100
kJ kg
M
29
gm mol
273.15) K
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
ndotgas
T1
Cp ( ) T T d
T2
mdotsteam Hv = 0
Cp ( ) T T d mdotndot
T1
Hv
mdotndot
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
Hv Tsteam
T2
Ssteam
Sgas
3
Ssteam
6.048
10
J kg K
Sgas
T1
Cp ( ) T dT T
Sgas
Sgas
SdotG
41.835
J mol K
SdotG
mdotndot Ssteam
49.708
J mol K
Ans.
Calculate lost work by Eq. (5.34) Wlost SdotG T
Wlost 14.8 kJ mol
141
b) Hsteam
Hv
Ssteam
Hv Tsteam
Ssteam
6.048
10
kJ kg
3
J kg K
Wideal
Hsteam
T
Ssteam
Wideal
453.618
Calculate lbs of steam generated per lbmol of gas cooled.
T2
Cp ( T) dT mn
T1
Hv
mn
15.135
gm mol
Use ratio to calculate ideal work of steam per lbmol of gas
Wideal mn 6.866
T2
kJ mol
Ans.
c)
Hgas
T1
Cp ( T) dT
Hgas T Sgas
Wideal 21.686 kJ mol
T
4.392 10
6
Wideal
Ans.
5.50 T1
a)
( 830
273.15)K
T2
( 35
273.15)K
3
( 25
0
273.15)K
Sethylene
Sethylene
Qethylene
Qethylene
Wlost T
R ICPS T1 T2 1.424 14.394 10
0.09 kJ mol K
R ICPH T1 T2 1.424 14.394 10
60.563 kJ mol
Qethylene
3
4.392 10
6
0
Sethylene
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
T
Sethylene
QC
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 =
V dP
and
dH = 1
T V dP
For an estimate, assume properties independent of pressure.
T
270 K
P1
3 3 m
381 kPa
P2
1200 kPa
3
V
1.551 10
kg
2.095 10
K
1
S
V P2
P1
H
1
T V P2
P1
S
2.661
J kg K
Tc
Ans.
H
551.7
J kg
Ans.
6.8
Isobutane:
408.1 K
Zc
0.282
CP
2.78
P1
4000 kPa
J gm K
3
P2
2000 kPa
molwt
gm 58.123 mol
Vc
cm 262.7 mol
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 360 K 361
Tr T Tc
0.88 Tr 0.882 0.885
(The elements are denoted by subscripts 1, 2, & 3
2 7
131.604 V 132.138 132.683
V
V c Zc
1 Tr
cm mol
3
Assume that changes in T and V are negligible during throtling. Then Eq. (6.8) is integrated to yield:
144
H= T S
V P
but
H= 0
Then at 360 K,
S
V 1 P2 T1
P1
S
0.733
J mol K
Ans.
We use the additional values of T and V to estimate the volume expansivity:
V
V3
V1
V
1.079
cm
3
mol
T
T3
T1
T
2K
1 V1
V T
4.098835
10
3
K
1
Assuming properties independent of pressure, Eq. (6.29) may be integrated to give
S = CP
T T
V P
P
P2
P1
P
2
10 kPa
3
Whence
T
T1 CP
S
V1 P molwt
T
0.768 K
Ans.
6.9
T
298.15 K
250 10
6
P1
K
1
1 bar
45 10
6
P2
1500 bar
bar
1
V1
cm 1003 kg
3
By Eq. (3.5),
V1 2 V2
V2 Vave
V1 exp 970.287 cm
P2
3
P1
V2
cm 937.574 kg
3
Vave
kg
By Eqs. (6.28) & (6.29),
H
Vave 1
T
P2
P1
U
H
P2 V 2
P1 V 1
H
134.6
kJ kg
Ans.
U
5.93
kJ kg
Ans.
S
Vave P2
P1
Q
T S
Work
U
Q
S
0.03636
kJ kg K
Ans. Q
10.84
145
kJ Ans. Work kg
4.91
kJ kg
Ans.
6.10
For a constant-volume change, by Eq. (3.5), T2 T1
5
P2
1
P1 = 0
T1
298.15 K
5 1
T2
P1
323.15 K
1 bar
36.2 10
T2
K
4.42 10
bar
P2
6.14 --- 6.16 300 175 575 500 325 175 575 T 650 300 400 150 575 375 475
K
T1
P1
P2
205.75 bar
Ans. for Parts (a) through (n): 61.39 48.98 48.98 37.96 73.83 34.99 45.60 .187 .000 .210 .200 .224 .048 .193
bar
Vectors containing T, P, Tc, Pc, and
40 75 30 50 60 60 35 P 50 35 70 50 15 25 75
bar Tc
308.3 150.9 562.2 425.1 304.2 132.9 556.4
K Pc
553.6 282.3 373.5 126.2 568.7 369.8 365.6
40.73 50.40 89.63 34.00 24.90 42.48 46.65
.210 .087 .094 .038 .400 .152 .140
Tr
T Tc
Pr
P Pc
146
6.14
Redlich/Kwong equation:
Pr Tr
0.08664
0.42748
Eq. (3.53) q
Tr
z 1
q z z z
1.5
Eq. (3.54)
Guess:
Given
Z
i
z= 1
q Find z ()
Eq. (3.52)
1 14
Ii
ln
Z Z
i qi i qi
i
Eq. (6.65b)
HRi
SRi
R Ti
Z
i qi
i qi
1
i
1.5 qi Ii Eq. (6.67) The derivative in these
0.5 qi Ii Eq. (6.68) equations equals -0.5
SRi
R ln Z
Z
i qi
0.695 0.605 0.772 0.685 0.729 0.75 0.709 0.706 0.771 0.744 0.663 0.766 0.775 0.75
HRi
-2.302103 -2.068103 -3.319103 -4.503103 -2.3103 -1.362103 -4.316103 -5.381103 -1.764103 -2.659103 -1.488103 -3.39103 -2.122103 -3.623103
J mol
-5.461 -8.767 -4.026 -6.542 -5.024 -5.648 -5.346 -5.978 -4.12 -4.698 -7.257 -4.115 -3.939 -5.523
J mol K
Ans.
147
6.15
Soave/Redlich/Kwong equation: 0.08664
0.5
2
0.42748 Pr Tr
c
0.480
1.574
0.176
2
1
c 1
Tr
Eq. (3.53) q
Eq. (3.54)
Tr
Guess:
Given
z
z= 1
1
q z z z
Eq. (3.52) Z q Find z ()
The derivative in the following equations equals: ci
i 1 14 Ii ln Z Z Tri
i i qi i qi
0.5
Tri
i
0.5
i
Eq. (6.65b)
HRi
R Ti Z
i qi
1
ci
1 qi Ii
0.5
Eq. (6.67)
SRi
R ln Z
i qi
i
ci
Tri
i
qi Ii
Eq. (6.68)
Z
i qi
0.691 0.606 0.774 0.722 0.741 0.768 0.715 0.741 0.774 0.749 0.673 0.769 0.776 0.787
HRi
-2.595103 -2.099103 -3.751103 -4.821103 -2.585103 -1.406103 -4.816103 -5.806103 -1.857103 -2.807103 -1.527103 -4.244103 -2.323103 -3.776103
SRi
J mol
-6.412 -8.947 -4.795 -7.408 -5.974 -6.02 -6.246 -6.849 -4.451 -5.098 -7.581 -5.618 -4.482 -6.103
J mol K
Ans.
148
6.16 Peng/Robinson equation:
1
2
1
2
0.07779
0.5
0.45724
2
c
Pr Tr
0.37464
1.54226
0.26992
2
1
Guess:
c 1 z
Tr
Eq. (3.53) q
Eq. (3.54)
Tr
1
Given
z= 1
q
z z z
Eq. (3.52) Z
q
0.5
Find ( z)
The derivative in the following equations equals: ci
i 1 14 Ii 1 2
i qi
Tri
i
ln
2 1
Z Z
ci
i qi i qi
i i
Eq. (6.65b)
HRi
R Ti Z
Tri
i
0.5
1 qi Ii
0.5
Eq. (6.67)
SRi
Z
i qi
0.667 0.572 0.754 0.691 0.716 0.732 0.69 0.71 0.752 0.725 0.64 0.748 0.756 0.753
R ln Z
HRi
i qi
i
ci
Tri
i
qi Ii
Eq. (6.68)
SRi
-2.655103 -2.146103 -3.861103 -4.985103 -2.665103 -1.468103 -4.95103 -6.014103 -1.917103 -2.896103 -1.573103 -4.357103 -2.39103 -3.947103
J mol
-6.41 -8.846 -4.804 -7.422 -5.993 -6.016 -6.256 -6.872 -4.452 -5.099 -7.539 -5.631 -4.484 -6.126
J mol K
Ans.
149
Lee/Kesler Correlation --- By linear interpolation in Tables E.1--E.12: h0 equals
( ) HR RTc
( ) SR R
0
0
h1 equals
( ) HR RTc
( ) SR R
1
1
h equals
HR RTc
SR R
s0 equals
s1 equals
s equals
.686 .590 .774 .675 .725 .744 Z0 .705 .699 .770 .742 .651 .767 .776 .746 Z1
.093 .155 .024 .118 .008 .165 .019 .102 .001 .007 .144 .034 .032 .154 h0
.950 1.709 .705 1.319 .993 1.265 .962 1.200 .770 .875 1.466 .723 .701 1.216 h1
1.003 .471 .591 .437 .635 .184 .751 .444 .550 .598 .405 .631 .604 .211
Z
Z0
Z1 Eq. (3.57)
h
h0
h1
(6.85)
HR
( Tc R) h
150
.711 1.110 .497 .829 .631 .710 s0 .674 .750 .517 .587 .917 .511 .491 .688
.961 .492 .549 .443 .590 .276 s1 .700 .441 .509 .555 .429 .589 .563 .287 HRi
-0.891 -1.11 -0.612 -0.918 -0.763 -0.723 -0.809 -0.843 -0.561 -0.639 -0.933 -0.747 -0.577 -0.728
-2.916103 -2.144103 -3.875103 -4.971103 -2.871103 -1.407103 -5.121103 -5.952103 -1.92103 -2.892103 -1.554103 -4.612103 -2.438103 -3.786103
s
s0
s1
SR
( s R)
Eq. (6.86)
Zi
0.669 0.59 0.769 0.699 0.727 0.752 0.701 0.72 0.77 0.743 0.656 0.753 0.771 0.768
hi
-1.138 -1.709 -0.829 -1.406 -1.135 -1.274 -1.107 -1.293 -0.818 -0.931 -1.481 -0.975 -0.793 -1.246
si
SRi
J mol
-7.405 -9.229 -5.091 -7.629 -6.345 -6.013 -6.727 -7.005 -4.667 -5.314 -7.759 -6.207 -4.794 -6.054
J mol K
Ans.
151
6.17
t 50 273.15 K The pressure is the vapor pressure given by the Antoine equation:
T 323.15 K
t
T
P () t
d dt
exp 13.8858
2788.51 t 220.79
P 36.166 kPa
P ( ) 36.166 50
P () 1.375 t
dPdt
1.375
kPa 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:
0.210
Tc
562.2 K
Pc
48.98 bar
Zc
0.271
Vc
259
cm
3
mol
Tr
T Tc
Tr
0.575
Pr
P Pc
Pr
0.007
By Eqs. (3.65), (3.66), (3.61), & (3.63)
B0
0.083
0.422 Tr
1.6
B0
0.941
B1
0.139
0.172 Tr
4.2
B1
3 4 cm
1.621
Vvap
RT P
1
B0
B1
Pr Tr
1 Tr
2/7
Vvap
7.306
10
mol cm
3
By Eq. (3.72),
Vliq
V c Zc
Vliq
93.151
mol
Solve Eq. (6.72) for the latent heat and divide by T to get the entropy change of vaporization:
S
dPdt Vvap
Vliq
S
100.34
J mol K
Ans.
(b) Here for the entropy change of vaporization:
S
RT dPdt P
152
S
102.14
J mol K
Ans.
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
x Hfusion = 0
CP
4.226
J gm K
t
6K
Hfusion
333.4
joule gm
x
CP t Hfusion
x
0.076
Ans.
The entropy change for the two steps is:
T2
273.15 K
T1
x Hfusion T2
( 273.15
6) K
3
S
CP ln
T2 T1
S
1.034709
10
J Ans. gm K
The freezing process itself is irreversible, because it does not occur at the equilibrium temperature of 0 degC. 6.21 Data, Table F.4:
H1 1156.3 BTU lbm
H2
1533.4
BTU lbm
S1
1.7320
BTU S2 lbm rankine
1.9977
BTU lbm rankine
H
H2
H1
S
S2
S1
H
377.1
BTU lbm
S
0.266
BTU Ans. lbm rankine
For steam as an ideal gas, apply Eqs. (4.9) and (5.18). [t in degF]
T1
( 227.96
459.67)rankine
T2
( 1000
459.67)rankine
P1
20 psi
P2
50 psi
T1
382.017 K
T2
810.928 K
153
molwt
18
lb lbmol
3 5
H
H
R MCPH T1 T2 3.470 1.450 10 molwt
372.536 BTU lbm
Ans.
0.0 0.121 10
T2
T1
R MCPS T1 T2 3.470 1.450 10 S
BTU lbm rankine
3
0.0 0.121 10
5
ln
T2 T1
ln
P2 P1
molwt
S
0.259
Ans.
6.22 Data, Table F.2 at 8000 kPa:
Vliq cm 1.384 gm
3
Hliq
3
1317.1
J gm
J gm
Sliq
3.2076
J gm K
J gm K
Vvap
cm 23.525 gm
6
Hvap
2759.9
Svap
0.15 10 3 cm 2 Vvap
6
5.7471
mliq
mliq
Htotal
Stotal
0.15 10 3 cm 2 Vliq
54.191 kg
mliq Hliq
mliq Sliq
mvap
mvap
3.188 kg
Htotal
Stotal
mvap Hvap
mvap Svap
80173.5 kJ
192.145 kJ K
Ans.
Ans.
154
6.23
Data, Table F.2 at 1000 kPa:
Vliq
1.127
cm
3
gm
Hliq
3
762.605
J gm
Sliq
2.1382
J gm K
Vvap
194.29
cm
gm
Hvap
2776.2
J gm
Svap
6.5828
J gm K
Let x = fraction of mass that is vapor (quality)
x
0.5
(Guess)
Given
x Vvap ( 1 x)Vliq
=
70 30
x
Find x ()
x
0.013
H
( 1
x)Hliq
x Hvap
S
S
( 1
x)Sliq
J gm K
x Svap
Ans.
H
789.495
J gm
2.198
6.24 Data, Table F.3 at 350 degF:
Vliq
ft 0.01799 lbm
3
Vvap
ft 3.342 lbm
3
Hliq
321.76
BTU lbm
Hvap
1192.3
BTU lbm
mliq
mvap = 3 lbm mvap Vvap = 50 mliq Vliq mliq
50 mliq Vliq = 3 lbm Vvap
mliq 1
3 lbm 50 Vliq Vvap
mliq
2.364 lb
mvap
3 lbm
mliq
mvap
0.636 lb
Htotal
mliq Hliq
mvap Hvap
Htotal
1519.1 BTU
Ans.
155
6.25
V
1 cm 0.025 gm
3
Data, Table F.1 at 230 degC:
3
Vliq
1.209
cm
gm
Hliq
3
990.3
J gm
Sliq
2.6102
J gm K
Vvap
71.45
cm
gm
Hvap
2802.0
J gm
Svap
6.2107
J gm K
V= ( 1
x)Vliq
x Vvap
x
V Vvap
Vliq Vliq
H
( 1
x)Hliq
x Hvap
S
( 1
x)Sliq
x Svap
x
0.552
H
1991
J gm
S
4.599
J gm K
Ans.
6.26
Vtotal = mtotal Vliq
mvap Vlv Table F.1, 150 degC:
Vtotal
0.15 m
3
Vvap
3
cm 392.4 gm
3
Table F.1, 30 degC:
Vliq
cm 1.004 gm
Vlv
cm 32930 gm
3
mtotal
Vtotal Vvap
mvap
Vtotal
mtotal Vliq Vlv
mtotal
0.382 kg
mvap
4.543
10
3
kg
mliq
mtotal
mvap
Vtot.liq
mliq Vliq
mliq
377.72 gm
Vtot.liq
379.23 cm
3
Ans.
156
6.27
Table F.2, 1100 kPa:
Hliq
781.124
J gm
Hvap
2779.7
Interpolate @101.325 kPa & 105 degC:
H2
2686.1
J gm J
gm
Const.-H throttling:
H2 = Hliq
x Hvap
Hliq
x
H2 Hvap
Hliq Hliq
x
0.953
Ans.
6.28
Data, Table F.2 at 2100 kPa and 260 degC, by interpolation:
H1
2923.5
J gm J
S1
6.5640
J gm K
molwt
18.015
gm mol
H2
2923.5
gm
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
S1
S
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
BTU lbm
S1
1.5703
BTU lbm rankine
H2
1257.7
BTU lbm
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
300 psi
P2
20 psi
molwt
18
lb lbmol
R ln S
P2 P1
S 0.299 BTU lbm rankine
Ans.
molwt
6.30
Data, Table F.2 at 500 kPa and 300 degC
S1
7.4614
J gm K
The final state is at this entropy and a pressure of 50 kPa. This is a state of wet steam, for which
Sliq
1.0912
J gm K
Svap
7.5947
J gm K
Hliq
340.564
J gm
Hvap
2646.9
J gm
S2 = S1 = Sliq
x Svap
Sliq
x
S1 Svap
Sliq Sliq
J gm
x
0.98
H2
Hliq
x Hvap
Hliq
H2
2599.6
Ans.
6.31
Vapor pressures of water from Table F.1: At 25 degC:
Psat
3.166 kPa
P
101.33 kPa
xwater
Psat P
xwater
0.031
Ans.
At 50 degC:
Psat
12.34 kPa
xwater
Psat P
xwater
0.122
Ans.
158
6.32 Process occurs at constant total volume:
Vtotal
( 0.014
0.021)m
3
Data, Table F.1 at 100 degC: Uliq
419.0
J gm
Uvap
3
2506.5
J gm
Vliq
1.044
cm
3
3
gm
Vvap
1673.0
cm
3
gm
mliq
0.021 m Vliq
mvap
0.014 m Vvap
4
mass
mliq
mvap
x
mvap mass
x
4.158 10
(initial quality) This state is first reached as saturated liquid at 349.83 degC
V2
Vtotal mass
V2
cm 1.739 gm
3
For this state, P = 16,500.1 kPa, and
U2
Q
1641.7
J gm
U1
Uliq
x Uvap
Uliq
U1
419.868
J gm
U2
U1
3
Q
1221.8
J gm
Ans.
6.33 Vtotal
0.25 m
Data, Table F.2, sat. vapor at 1500 kPa:
V1
cm 131.66 gm
3
U1
2592.4
J gm
mass
Vtotal V1
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:
x
0.75
V2 = V1 = Vliq
x Vvap
Vliq
Whence
x=
V1 Vvap
Vliq Vliq
(A)
Find P for which (A) yields the value x = 0.75 for wet steam
159
Since the liquid volume is much smaller than the vapor volume, we make a preliminary calculation to estimate:
Vvap
V1 x
Vvap
175.547
cm
3
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
782.41
J gm
Uvap
2584.9
J gm
U2
Uliq
x Uvap
Uliq
U2
2134.3
J gm
Q
mass U2
U1
Q
869.9 kJ
Ans.
6.34 Table F.2,101.325 kPa:
Vliq
cm 1.044 gm
3
Vvap
cm 1673.0 gm
0.02 m Vliq
mvap mtotal
3
3
Uliq
mvap
J 418.959 gm
1.98 m Vvap
Vliq
Uliq
3
Uvap
mtotal
J 2506.5 gm
mliq mvap
cm
3
mliq
x
V1
U1
x Vvap
x Uvap
Vliq
Uliq
V1
U1
98.326
gm
J gm
x
0.058
540.421
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 U1
Q 41860.5 kJ
Ans.
160
6.35 Data, Table F.2 at 800 kPa and 350 degC:
V1
354.34
cm
3
gm
U1
2878.9
J gm
Vtotal
0.4 m
3
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
2638.7
J gm
Q
Vtotal V1
U2
U1
Q
271.15 kJ
Ans.
6.36 Data, Table F.2 at 800 kPa and 200 degC:
U1
2629.9
J gm
S1
6.8148
J gm K
mass
1 kg
(a) Isothermal expansion to 150 kPa and 200 degC
U2
Q
2656.3
J gm
S1
S2
Q
7.6439
J gm K
Ans.
T
473.15 K
mass T S2
392.29 kJ
Also:
Work
mass U2
U1
Q
Work
365.89 kJ
(b) Constant-entropy expansion to 150 kPa. The final state is wet steam:
Sliq
1.4336
J gm K
Svap
7.2234
J gm K
Uliq
444.224
J gm
Uvap
2513.4
J gm
x
S1 Svap
Sliq Sliq
x
0.929
3 J
U2
Uliq
x Uvap
Uliq
U2
2.367
10
gm
W
mass U2
U1
W
262.527 kJ
Ans.
161
6.37 Data, Table F.2 at 2000 kPa:
x
0.94
Hvap
2797.2
J gm
Hliq
908.589
J gm
H1
Hliq
x Hvap
Hliq
H1
2.684
10
3 J
gm
mass
1 kg
For superheated vapor at 2000 kPa and 575 degC, by interpolation:
H2
3633.4
J gm
Q
mass H2
H1
Q
949.52 kJ
Ans.
6.38 First step:
Q12 = 0
W12 = U2
U1
Second step:
For process:
W23 = 0
Q = U3 U2
Q23 = U3
W = U2
U2
U1
Table F.2, 2700 kPa:
Uliq
977.968
J gm
Uvap
2601.8
J gm
Sliq
2.5924
J gm K
Svap
6.2244
J gm K
3 J
x1
0.9
U1
Uliq
x1 Uvap
Uliq U1
2.439
10
gm
S1
Sliq
x1 Svap
Sliq
S1
5.861
2 3 m 10 2
s K
Table F.2, 400 kPa:
Sliq
1.7764
J gm K
Svap
6.8943
J gm K
Uliq
604.237
J gm
3
Uvap
2552.7
J gm
Vliq
cm 1.084 gm
Vvap
462.22
cm
3
gm
162
Since step 1 is isentropic,
S2 = S1 = Sliq
x2 Svap
Sliq
x2
S1 Svap
Sliq Sliq
3 J
x2
0.798
U2
Uliq
x2 Uvap
Uliq
U2
2.159
10
gm
3
V2
Vliq
x2 Vvap
Vliq
V2
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
2560.7
J gm
Whence
Q
U3
U2
Work
U2
U1
Q
401.317
J gm
Ans.
Work
280.034
J gm
Ans.
6.39 Table F.2, 400 kPa & 175 degC:
Table F.1,sat. vapor, 175 degC
U1
2605.8
J gm
J gm
S1
7.0548
J gm K
U2
2578.8
S2
6.6221
J gm K
mass
4 kg
T
( 175
273.15)K
Q
mass T S2
S1
W
mass U2
U1
Q
Q
775.66 kJ
Ans.
W
667.66 kJ
Ans.
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
J gm K
163
H
H2
H1
H
600.1
J gm
Ans.
S
S2
S1
S
0.115
J gm K
Ans.
(b) T1
( 450
273.15)K
T2
( 140
273.15)K
T1
723.15 K
T2
413.15 K
P1
3000 kPa
P2
235 kPa
Eqs. (6.95) & (6.96) for an ideal gas:
Hig
gm mol 3 5 R ICPH T1 T2 3.470 1.450 10 0.0 0.121 10 molwt 18
molwt
R ICPS T1 T2 3.470 1.450 10 Sig
Hig
(c) Tc molwt
3
0.0 0.121 10
5
ln
P2 P1
620.6
647.1 K
J gm Pc 220.55 bar
Sig
0.0605
0.345
J gm K
Ans.
Tr1
T1 Tc
Pr1
P1 Pc
Tr2
T2 Tc
Pr2
P2 Pc
Tr1
1.11752
Pr1
0.13602
Tr2
0.63846
Pr2
0.01066
The generalized virial-coefficient correlation is suitable here
H
H
HRB Tr1 Pr1 R Tc HRB Tr2 Pr2 molwt J Ans. 593.95 gm Hig
S
Sig
R SRB Tr2 Pr2
SRB Tr1 Pr1 molwt
S
0.078
J gm K
Ans.
164
6.41
Data, Table F.2 superheated steam at 550 kPa and 200 degC:
V1
385.19
cm
3
gm
U1
2640.6
J gm
S1
7.0108
J gm K
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
U2
U1
Step 2--3: Isentropic expansion to initial T.
Q12
322.5
J gm
Q23 = 0
S3 = S 2
S3
7.5782
J gm K
Step 3--1: Constant-T compression to initial P.
T
473.15 K
Q31
T S1
S3
Q31
268.465
J gm
For the cycle, the internal energy change = 0.
Wcycle = Qcycle = Q12 Q31
=
Wcycle Q12
1
Q31 Q12
0.1675
Ans.
6.42 Table F.4, sat.vapor, 300(psi):
T1
( 417.35
459.67) rankine H1
1202.9
BTU lbm
T1
877.02 rankine
S1
1.5105
BTU lbm rankine
Superheated steam at 300(psi) & 900 degF
H2
1473.6
BTU lbm
S2
1.7591
BTU lbm rankine
S3
S2
Q12
H2
H1
Q31
T 1 S1
S3
Q31
218.027
BTU lbm
165
For the cycle, the internal energy change = 0.
Wcycle = Qcycle = Q12
Q31
=
Wcycle Q12
Whence
1
Q31 Q12
0.1946
Ans.
6.43
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 0.95
S2 = Sliq
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
J gm K
Svap
7.0520
J gm K
S2
Sliq
x Svap
Sliq
S2
6.7798
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
559.16
(degC)
H1
3598.0
kJ kg
Superheat:
t
( 559.16
212.37)K
t
346.79 K
Wdot
Ans.
(b) mdot
5
kg sec
Wdot
mdot H2
H1
4760 kW Ans.
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
3205.4
kJ kg
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:
Sliq 0.6493 kJ kg K
Svap
8.1511
kJ kg K
Hliq
191.832
kJ kg
Hvap
2584.8
kJ kg
x2
S2 Svap
Sliq Sliq
x2
0.879
H' Hliq
x2 Hvap
Hliq
H' 2.294
10
3 kJ
H2
H1
kg
H' H1
0.681
Ans.
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
kJ kg K
Hliq
317.16
kJ kg
Hvap
2636.9
kJ kg
x2
S2 Svap
Sliq Sliq
x2
0.95
H' Hliq
x2 Hvap
Hliq
H2
H1
H' 2.522
10
3 kJ
kg
H' H1
0.78
Ans.
6.47 Table F.2 at 1600 kPa and 225 degC:
P
1600 kPa
V
cm 132.85 gm
3
H
2856.3
J gm
S
6.5503
J gm K
Table F.2 (ideal-gas values, 1 kPa and 225 degC)
Hig
2928.7
J gm
Sig
10.0681
J gm K
P0
VR
1 kPa
T R molwt P
T
( 225
273.15)K
T
498.15 K
V
The enthalpy of an ideal gas is independent of pressure, but the entropy DOES depend on P:
HR
H
Hig
cm
3
Sig
R molwt 72.4 J gm
ln
P P0
SR
S
Sig
Sig
VR
10.96
gm
HR
SR
0.11
J Ans. gm K
Reduced conditions:
0.345
Tc
647.1 K
Pc
220.55 bar
Tr
T Tc
Tr
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
4.2
B1
0.377
168
By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40)
Z
1
B0
B1
Pr Tr
Z
0.935
VR
RT ( Z 1) P molwt
HR
R Tc molwt
HRB Tr Pr
3
SR
R molwt
SRB Tr Pr
VR
9.33
cm
gm
HR
53.4
J gm
SR
0.077
J gm K
Ans.
6.48
P
1000 kPa
T
( 179.88
273.15) K
T
453.03 K
(Table F.2)
molwt
3
18.015
gm mol
3
Vl
cm 1.127 gm
Vv
cm 194.29 gm
Vlv
Hlv
Vv
Vl
Hl
Sl
762.605
J gm
Hv
Sv
2776.2
J gm
J gm K
Hv
Hl
2.1382
J gm K
3
6.5828
Slv
Sv
Sl
Vlv
cm 193.163 gm
Hlv
2.014
10
3 J
gm
Slv
4.445
J gm K
(a) Gl
Hl
T Sl G l
206.06
J gm
Gv
Hv
T Sv G v
206.01
J gm
(b)
Slv
4.445
J gm K
r
Hlv T
r
3
4.445
J gm K
(c)
VR
Vv
T R molwt P
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
Hig
2841.1
J gm
Sig
9.8834
J gm K
P0
1 kPa
The enthalpy of an ideal gas is independent of pressure; the entropy DOES depend on P:
HR
SR
Hv
Sv
Hig
Sig Sig
Sig
HR
P R ln P0 molwt
64.9 J gm
Sig
3.188
J gm K
Ans. SR
0.1126
J gm K
Ans.
(d) Assume ln P vs. 1/T linear and fit three data pts @ 975, 1000, & 1050 kPa.
975
Data:
178.79 t 179.88 182.02
(degC)
pp
1000 kPa 1050
i
1 3
xi
dPdT
ti
1 yi 273.15
P T
2
ln
ppi kPa
Slope
slope ( y) Slope x
4717
Slope K
dPdT
22.984
kPa K
Slv
Vlv dPdT
Slv
4.44
J gm K
Ans.
Reduced conditions:
0.345
Tc
647.1 K
Pc
220.55 bar
Tr
T Tc
Tr
0.7001
Pr
P Pc
Pr
0.0453
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
0.943
VR
RT ( 1) Z P molwt
170
HR
R Tc molwt
HRB Tr Pr
SR
R SRB Tr Pr molwt
VR
11.93
cm
3
gm
HR
43.18
J gm
SR
0.069
J gm K
Ans.
6.49 T
( 358.43
459.67) rankine
T
818.1 rankine
P
150 psi
(Table F.4)
molwt
3
18.015
gm mol
Vl
ft 0.0181 lbm
Vv
ft 3.014 lbm
3
Vlv
Vv
Vl
Hl
330.65
BTU lbm
Hv
1194.1
BTU lbm
Hlv
Slv
Hv
Hl
Sl
0.5141
BTU lbm rankine
3
Sv
1.5695
BTU lbm rankine
Sv
Sl
Vlv
ft 2.996 lbm
Hlv
863.45
BTU lbm
(a) Gl
Hl
T Sl
Gv
Hv
T Sv
Gl
89.94
BTU lbm
Gv
89.91
BTU lbm
(b)
Slv
1.055
BTU lbm rankine
r
Hlv T
r
1.055
BTU lbm rankine
(c)
VR
Vv
T R molwt P
VR
0.235
ft
3
lbm
Ans.
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
Hig
HR
Sig
28.5
BTU lbm
Ans.
Sig
P R ln P0 molwt
Sv Sig Sig
0.552
BTU lbm rankine
SR
SR
0.0274
BTU lbm rankine
Ans.
(d) Assume ln P vs. 1/T linear and fit threedata points (@ 145, 150, & 155 psia)
145
Data:
355.77 t 358.43 361.02
(degF)
pp
150 psi 155
i
1 3
xi
ti
1 459.67
yi
ln
ppi psi
Slope
Slope
slope ( y) x
8.501 10
3
dPdT
P T
2
Slope rankine
dPdT
1.905
psi rankine
Slv
Vlv dPdT
Slv
1.056
BTU Ans. lbm rankine
Reduced conditions:
0.345
Tc
647.1 K
Pc
220.55 bar
Tr
T Tc
Tr
0.7024
Pr
P Pc
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
4.2
B1
0.62
172
By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40)
Z
1
B0
B1
Pr Tr
Z
0.942
VR
RT P molwt
(Z
1)
HR
R
Tc molwt
HRB Tr Pr
3
SR
R SRB Tr Pr molwt
VR
0.1894
ft
lbm
HR
19.024
BTU lbm
SR
0.0168
BTU lbm rankine
Ans.
6.50
For propane:
Tc
369.8 K
Pc
42.48 bar
0.152
T
( 195
273.15) K
T
468.15 K
P
135 bar
P0
Pr 3.178
1 bar
Tr
T Tc
Tr
1.266
Pr
P Pc
Use the Lee/Kesler correlation; by interpolation,
Z0
0.6141
Z1
0.1636
Z
3
Z0
Z1
Z
0.639
V
ZRT P
V
cm 184.2 mol
Ans.
HR0
2.496 R Tc
HR1
0.586 R Tc
HR0
7.674
10
3 J
mol
HR1
1.802
10
3 J
mol
SR0
1.463 R
SR1
0.717 R
SR0
12.163
J mol K
SR1
5.961
J mol K
HR
HR0
HR1
3 J
SR
SR0
SR1
HR
7.948 10
mol
SR
3
13.069
J mol K
6
H
R ICPH 308.15K T 1.213 28.785 10
173
8.824 10
0.0
HR
S
R ICPS 308.15K T 1.213 28.785 10
J mol
3
8.824 10 J mol K
6
0.0
ln
P P0
SR
H
6734.9
Ans.
S
15.9
Ans.
6.51 For propane:
Tc
369.8 K
Pc
42.48 bar
0.152
T
( 70
273.15)K
T
343.15 K
P0
101.33 kPa P
1500 kPa
Tr
T Tc
Tr
0.92793
Pr
P Pc
Pr
0.35311
Assume propane an ideal gas at the initial conditions. Use generalized virial correlation at final conditions.
H
R Tc HRB Tr Pr
H
1431.3
J mol
J mol K
Ans.
S
R SRB Tr Pr
ln
P P0
S
25.287
Ans.
6.52 For propane:
0.152
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:
3
P
1 bar
A
6.72219
B
1.33236
C
2.13868
D 1.38551 Given
() T
1
T Tc
1.5
Guess:
3
T
6
200 K
P = Pc exp
T
A () B T
() T
C () T
() T
D
() T
1
T
Find T) (
230.703 K
174
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)
Pc exp
230.703 K
A ( T)
B
( T)
1.5
C ( T)
kPa K
( T)
3
D
( T)
6
1
d P ( T) dT
Pr P Pc
B0
T
4.428
dPdT
T Tc
0.172 Tr
4.2
4.428124
kPa K
P
1 bar
Pr
0.024
Tr
Tr
0.624
B0
0.083
0.422 Tr
1.6
0.815
B1
0.139
B1
2 7
1.109
Vvap
RT P
1
B0
3 4 cm
B1
Pr Tr
Vliq
V c Zc
1 Tr
Vvap
1.847
10
mol
Vliq dPdT
Vliq
Hlv
75.546
cm
3
mol
10
4 J
Hlv
T Vvap
1.879
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
0
R Tc
and
175
r1 =
H
R
1
R Tc
T1
370 K
P1
200 bar
Tr
T1 Tc
Tr
1.001
Pr
P1 Pc
Pr
4.708
By interpolation, find:
r0
3.773
r1
3.568
By Eq. (6.85)
H1
R Tc r0
r1
H1
1.327
10
4 J
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
8.824 10
6
0.0
H2
1.048 10
4 J
mol
For Step (3) the enthalpy change is given by Eq. (6.87), for which
Tr
230.703 K Tc
Tr
0.6239
Pr
1 bar Pc
Pr
0.0235
H3
H3
R Tc HRB Tr Pr
232.729 J mol
For Step (4),
H4 = x Hlv
For the process,
H1
H2
H3
x Hlv = 0
x
H1
H2 Hlv
H3
x
0.136
Ans.
6.53 For 1,3-butadiene:
0.190
Tc
425.2 K
Pc
42.77 bar
Zc
0.267
Vc
cm 220.4 mol
3
Tn
268.7 K
T
380 K
P
1919.4 kPa
T0
273.15 K
P0
101.33 kPa
Tr
T Tc
Tr
0.894
Pr
P Pc
Pr
0.449
176
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
Vvap
HR0
HR0
0.7442
ZRT P
0.689 R Tc
2.436 10
Z1
0.1366
Vvap
HR1
HR1
Z
Z0
cm
3
Z1
Ans.
Z
0.718
1182.2
mol
0.892 R Tc
3.153 10
3 J
3 J
mol
mol
SR0
SR0
HR
HR
Hvap
0.540 R
4.49
HR0
3.035
SR1
SR1
SR
SR
0.888 R
7.383
SR0
5.892
3
J mol K
HR1
10
3 J
J mol K
SR1
mol
J mol K
8.882 10
6
6
R ICPH T0 T 2.734 26.786 10
3
0.0
HR
Svap
R ICPS T0 T 2.734 26.786 10
Hvap 6315.9 J mol
8.882 10
0.0
ln
P P0
SR
Ans.
Svap
1.624
J mol K
Ans.
For saturated vapor, by Eqs. (3.63) & (4.12)
2
Vliq
V c Zc
1 Tr
7
Vliq
cm 109.89 mol
3
Ans.
177
1.092 ln Hn R Tn
Pc bar
Tn Tc
1.013
Hn
22449
0.930
J mol
By Eq. (4.13)
H
Hn
1 1
Tr Tn Tc
0.38
H
14003
J mol
Hliq
Hvap
H
Hliq
7687.4
J mol
Ans.
Sliq
Svap
H T
Sliq
38.475
J mol K
Ans.
6.54 For n-butane:
0.200
Tc
425.1 K
Pc
37.96 bar
Zc
P
0.274
1435 kPa
Vc
T0
cm 255 mol
3
Tn
P0
272.7 K
101.33 kPa
T
Tr
370 K
T Tc
273.15 K
Tr
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
0.7692
Z1
0.1372
Z
Z0
Z1
3
Z
0.742
V
ZRT P
V
cm 1590.1 mol
Ans.
HR0
0.607 R Tc
HR1
0.831 R Tc
HR0
2.145
10
3 J
mol
178
HR1
2.937
10
3 J
mol
SR0
SR0
HR
HR
Hvap
0.485 R
4.032
HR0
2.733
SR1
J
SR1
SR
SR
3
0.835 R
6.942
SR0
5.421
mol K
HR1
10
3 J
J mol K
SR1
mol
J mol K
6
R ICPH T0 T 1.935 36.915 10
3
11.402 10
6
0.0
HR
Svap
R ICPS T0 T 1.935 36.915 10
J mol
11.402 10
0.0
ln
P P0
SR
Hvap
7427.4
Ans.
Svap
4.197
J mol K
Ans.
For saturated vapor, by Eqs. (3.72) & (4.12)
1 Tr
2/7
Vliq
V c Zc
Vliq
cm 123.86 mol
3
Ans.
1.092 ln Hn R Tn
Pc bar
Tn Tc
1.013
Hn
22514
0.930
J mol
By Eq. (4.13)
H
Hn
1 1
Tr Tn Tc
0.38
H
15295.2
J mol
Hliq
Sliq
Hvap
Svap
H
H T
Hliq
Sliq
7867.8
J mol
J mol K
Ans.
37.141
Ans.
179
6.55 Under the stated conditions the worst possible cycling of demand can be represented as follows:
10, kg/ 000 hr
Dem and ( hr kg/ ) 2/ hr 3 1/ hr 3
1 hr
6, 000
tm e i
4, kg/ 000 hr
netst age or ofst eam
netdepl i eton ofst eam
This situation is also represented by the equation: 4000
where
10000 1
= 6000
= time of storage liquid 2 Solution gives hr 3 The steam stored during this leg is: mprime mprime
6000
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 m2 =
Hprime Hf2 Vf2
H1
Vtank P2
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
P1
Hfg2 Vfg2
m2 m1
= Hprime
H1 Eq. (A)
However M1 v1 = m2 V2 = Vtank
and therefore
=
V1 V2
180
Making this substitution and rearranging we get
Hprime Hf2 Vf2 Hfg2 Vfg2
V2
P2
P1
Hfg2 Vfg2
=
Hprime V1
H1
In this equation we can determine from the given information everything except Hprime and Vprime. These quantities are expressed by H1 = Hf1 x1 Hfg1
and V1 = Vf1 x1 Vfg1
Therefore our equation becomes (with Hprime = Hg2)
Hg2
Hf2
V2
Vf2
Hfg2 Vfg2
P2 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
therefore
x2 Vf2
19 = 1
0.05V2 = x2 Vg2
or
x2 = Vf2 19Vg2 Vf2
x2 Vf2 x2 Vg2
Then
V2 =
Vg2 0.05
Vf2 19Vg2 Vf2
=
20 19 Vf2
1 Vg2
Eq. (C)
Now we need property values: Initial state in accumulator is wet steam at 700 kPa.
We find from the steam tables
Hf1 697.061 kJ Hg1 kg
2762.0 kJ kg
Hfg1 Hg1 Hf1 Hfg1
2064.939 kJ kg
P1
700kPa
181
Vf1
1.108
cm
3
gm
Vg1
272.68
cm
3
gm
Vfg1
Vg1
Vf1 Vfg1
271.572
cm
3
gm
Final state in accumulator is wet steam at 1000 kPa. From the steam tables
P2
1000kPa
Hf2
762.605
kJ kg
3
Hg2
2776.2
kJ kg
Hfg2
3
Hg2
Hf2 Hfg2
2013.595
kJ kg
3
Vf2
1.127
cm
gm
Vg2
194.29
cm
gm
Vfg2
Vg2
Vf2 Vfg2
193.163
cm
gm
Solve Eq. (C) for V2
V2
Vg2 0.05
Vf2 19Vg2 Vf2
V2
1.18595
10
3 3m
kg
Next solve Eq. (B) for x1
Given
Guess: x1
0.1
Hg2
Hf2
V2
Vf2
Hfg2 Vfg2 P2 P1 Hfg2 Vfg2
=
Hg2
Hf1
Vf1
x1 Hfg1 x1 Vfg1
x1
Find x1
x1
4.279
10
4
Thus
V1
Vf1
x1 Vfg1
V1
1.22419
cm
3
Eq. (A) gives
V1 m2 = V2 m1
gm
and
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
m2
m2
m1 = 2667lb
3.873
182
m1 m2
Find m1 m2
10 kg
10 kg
4
Finally, find the tank volume Vtank
m2 V2
Vtank
45.9 m
3
Ans.
Note that just to store 1333.3 kg of saturated vapor at 1000 kPa would require a volume of: 1333.3kg Vg2 259 m
3
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
=
U1 U2
Hprime
Hg2
=
Hprime Hprime
Uf1 Uf2
2.776
=
Hprime Hprime
3 kJ
Hf1 Hf2
Hprime
Given
m2 m1
Hprime
10
kg
=
Hprime Hprime
Hf1 Hf2
m1
m2 m1 = 2667lb
m2
Find m1 m2
m2
V
3.837
m2 Vf2 0.95
10 kg
3
4
V
45.5 m
0.140
Ans.
Tc
P
6.56 Propylene:
T
365.6 K
38 bar
Pc
P0
46.65 bar
1 bar
400.15 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:
183
Tr
T Tc
Tr
1.095
Pr
P Pc
Pr
0.815
Step (1): Use the Lee/Kesler correlation, interpolate.
H0
0.863 R Tc
H1
0.534 R Tc
HR
H0
H1
H0
2.623 10
3 J
mol
H1
1.623
10
3 J
mol
HR
2.85
10
3 J
mol
S0
0.565 R
S1
0.496 R
SR
S0
S1
S0
4.697
J mol K
S1
4.124
J mol K
SR
5.275
J mol K
Step (2): For the heat capacity of propylene,
A 1.637
B
22.706 10 K
3
C
6.915 10 K
2
6
Solve energy balance for final T. See Eq. (4.7).
1
(guess)
Given
HR = R
Find
AT
1
B 2 T 2
0.908
2
1
Tf
3
C 3 T 3
3
1
T
Tf
6
363.27 K Ans.
Sig
Sig
R ICPS T Tf 1.637 22.706 10
22.774 J mol K
6.915 10
0.0
ln
P0 P
S
SR
Sig
S
28.048
J mol K
Ans.
184
6.57 Propane:
0.152
Tc
369.8 K
Pc
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
1.144
Pr
P Pc
3 J
Pr
0.518
Step (1): Use the generalized virial correlation
HR
R Tc HRB Tr Pr
HR
1.366
10
mol
SR
R SRB Tr Pr
SR
2.284
J mol K
6
Step (2): For the heat capacity of propane,
A 1.213
B
28.785 10 K
3
C
8.824 10 K
2
Solve energy balance for final T. See Eq. (4.7).
1 (guess)
Given
HR = R
Find
AT
1
B 2
T
2
2
1
Tf
3
C 3 T 3
3
1
0.967
T
Tf
408.91 K
Ans.
Sig
Sig
R ICPS T Tf 1.213 28.785 10
22.415 J mol K
8.824 10
6
0.0
ln
P0 P
S
SR
Sig
S
24.699
185
J mol K
Ans.
6.58 For propane:
Tc
369.8 K
Pc
42.48 bar
0.152
T
( 100
273.15)K
T
373.15 K
P0
1 bar
P
10 bar
Tr
T Tc
Tr
1.009
Pr
P Pc
Pr
0.235
Assume ideal gas at initial conditions. Use virial correlation at final conditions.
H
R Tc HRB Tr Pr
H
801.9
J mol
J mol K
Ans.
S
R SRB Tr Pr
ln
P P0
S
20.639
Ans.
6.59 H2S:
0.094
Tc
373.5 K
Pc
89.63 bar
T1
400 K
P1
5 bar
T2
600 K
P2
25 bar
Tr1
T1 Tc
Pr1
P1 Pc
Tr2
T2 Tc
Pr2
P2 Pc
Tr1
1.071
Pr1
0.056
Tr2
1.606
Pr2
0.279
Use generalized virial-coefficient correlation for both sets of conditions. Eqs. (6.91) & (6.92) are written
H
R ICPH T1 T2 3.931 1.490 10 0.0 0.232 10 R Tc HRB Tr2 Pr2 HRB Tr1 Pr1
3
5
S
R ICPS T1 T2 3.931 1.490 10
R SRB Tr2 Pr2
3
0.0
0.232 10
5
ln
P2 P1
SRB Tr1 Pr1
H
7407.3
J mol
S
1.828
J mol K
Ans.
186
6.60 Carbon dioxide:
0.224
Tc
304.2 K
Pc
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:
Tr
T Tc
Tr
1.046
Pr
P Pc
Pr
0.217
Step (1): Use the generalized virial correlation
HR
R Tc HRB Tr Pr
HR
587.999
J mol
SR
R SRB Tr Pr
SR
1.313
J mol K
Step (2): For the heat capacity of carbon dioxide,
A 5.457
B
1.045 10 K
3
D
1.157 10 K
5
2
Solve energy balance for final T. See Eq. (4.7). Given 1 (guess)
HR = R A T
Find
1
B 2 T 2
0.951
2
1
Tf
D T
T
3
1
Tf
302.71 K
5
Ans.
Sig
R ICPS T Tf 5.457 1.045 10
J mol K
0.0
1.157 10
ln
P0 P
Sig
21.047
S
SR
Sig
S
22.36
J mol K
Ans.
187
6.61
T0
523.15 K
P0
3800 kPa
P
120 kPa
S
0
J mol K
For the heat capacity of ethylene:
A
1.424
B
14.394 10 K
3
C
4.392 10
2
6
K (a) For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with D = 0:
0.4
(guess)
Given
S = R A ln
Find
B T0
0.589
C T0
2
1 2
1
ln
P P0
Tf
T0
3
Tf
308.19 K
6
Ans.
Hig
R ICPH T0 Tf 1.424 14.394 10
4.392 10
0.0
Hig
Ws
1.185
10
4 J
mol
Ws 11852 J mol
Hig
Ans.
(b) Ethylene:
0.087
Tc
282.3 K
Pc
50.40 bar
Tr0
T0 Tc
Tr0
1.85317
Pr0
P0 Pc
Pr0
0.75397
At final conditions as calculated in (a)
Tr
T Tc
Tr
1.12699
Pr
P Pc
Pr
0.02381
Use virial-coefficient correlation. The entropy change is now given by Eq. (6.92):
0.5
(guess)
Given
188
S = R A ln
SRB
B T0
T0 Tc
C T0
2
1 2
1
ln
P P0
Pr
SRB Tr0 Pr0
Find
T
T0
T
303.11 K
Ans.
Tr
T Tc
Tr
1.074
The work is given by Eq. (6.91):
Hig
R ICPH T0 T 1.424 14.394 10
3
4.392 10
6
0.0
Hig
1.208
10
4 J
mol
Ws
Hig
R Tc HRB Tr Pr
HRB Tr0 Pr0
Ws
11567
J mol
Ans.
6.62 T0
S
493.15 K
0 J mol K
P0
30 bar
P
2.6 bar
For the heat capacity of ethane:
A
1.131
B
K (a) For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with D = 0: (guess) Given 0.4
19.225 10 K
3
C
5.561 10
2
6
S = R A ln
Find
B T0
C T0
2
1 2
1
ln
P P0
0.745
T
T0
3
T
367.59 K
6
Ans.
Hig
R ICPH T0 T 1.131 19.225 10
189
5.561 10
0.0
Hig
8.735
10
3 J
mol
Ws
Hig
Ws
8735
J mol
Ans.
(b) Ethane:
0.100
Tc
305.3 K
Pc
48.72 bar
Tr0
T0 Tc
Tr0
1.6153
Pr0
P0 Pc
Pr0
0.61576
At final conditions as calculated in (a)
Tr ( ) T
T Tc
Tr ( ) 1.20404 T
Pr
P Pc
Pr
0.05337
Use virial-coefficient correlation. The entropy change is now given by Eq. (6.83):
0.5
(guess)
Given
S = R A ln
SRB
B T0
T0 Tc
C T0
2
1 2
1
ln
P P0
Pr
SRB Tr0 Pr0
Tr
Find T
T
T0
T
362.73 K
Ans.
Tc
Tr
1.188
The work is given by Eq. (6.91):
Hig
R ICPH T0 T 1.131 19.225 10
3
5.561 10
6
0.0
Hig
9.034
10
3 J
mol
Ws
Hig
R Tc HRB Tr Pr
HRB Tr0 Pr0
Ws
8476
J mol
Ans.
190
6.63
n-Butane:
0.200
Tc
425.1 K
Pc
37.96 bar
T0
323.15 K
P0
1 bar
P
7.8 bar
S
0
J mol K
For the heat capacity of n-butane:
A
1.935
B
36.915 10 K
3
C
11.402 10 K
2
6
Tr0
T0 Tc
Tr0
0.76017
Pr0
P0 Pc P
Pr0
0.02634
Pr
Pc
HRB0
Pr
0.205
0.05679
HRB Tr0 Pr0
= 0.05679
The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: (guess)
0.4
Given
S = R A ln
SRB T0 Tc
B T0
Pr
C T0
2
1 2
1
ln
P P0
SRB Tr0 Pr0
Find
1.18
T
T0
T
381.43 K
Ans.
Tr
T Tc
Tr
0.89726
The work is given by Eq. (6.91):
Hig
R ICPH T0 T 1.935 36.915 10
3
11.402 10
6
0.0
Hig
6.551
10
3 J
mol
Ws
Hig
R Tc HRB Tr Pr
HRB Tr0 Pr0
Ws
5680
J mol
Ans.
191
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
3344.6
kJ kg
S1
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
H2
H1
T
S2
S1
Wideal
1224.3
kJ kg
Ans.
6.65
Sat. liquid at 325 K (51.85 degC), Table F.1:
Hliq
kJ 217.0 kg
Sliq
kJ 0.7274 kg K
Vliq
cm 1.013 gm
3
Psat
P1
12.87 kPa
8000 kPa
For the compressed liquid at 325 K and 8000 kPa, apply Eqs. (6.28) and (6.29) with
T
325 K
460 10
6
K
1
H1
Hliq
Vliq 1
T
P1
Psat
H1
223.881
kJ kg
S1
Sliq
Vliq P1
Psat
S1
0.724
kJ kg K
For sat. vapor at 8000 kPa, from Table F.2:
H2
2759.9
kJ kg
S2
5.7471
kJ kg K
T
300 K
Heat added in boiler:
Q
H2
H1
Q
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
Q
Frac
0.4058
Ans.
The heat not converted to work ends up in the surroundings.
SdotG.surr
Q
Wideal T
10
kg sec
SdotG.surr
50.234
kW K
SdotG.system
S1
S2 10
kg sec
SdotG.system
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
Qdot
2536
SdotG
kg sec kg kW Qdot 50.234 K T
kJ
10
T
( 600
273.15)K
T
Ans.
873.15 K
SdotG
21.19
kW K
By Eq. (5.34)
T
300 K
Wdotlost
T SdotG
Wdotlost
6356.9 kW
Ans.
6.67
For sat. liquid water at 20 degC, Table F.1:
H1
83.86
kJ kg
S1
0.2963
kJ kg K
For sat. liquid water at 0 degC, Table F.1:
H0
0.04
kJ kg
S0
0.0000
kJ kg K
For ice at at 0 degC:
H2
H0
333.4
kJ kg
S2
S0
333.4 kJ 273.15 kg K
193
H2
333.44
kJ kg
S2
1.221
kJ kg K
T
293.15 K
mdot
0.5
kg sec
t
0.32
By Eqs. (5.26) and (5.28):
Wdotideal
Wdot
mdot H2
H1
T
S2
S1
Wdot
Wdotideal
13.686 kW
Wdotideal
t
42.77 kW
Ans.
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
S2
2676.0
kJ kg
S1
Q'
7.3554
kJ kg K
kJ kg
H2
T
0.0
kJ kg
0.0
kJ kg K
2000
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
T
Sapparatus.reservoir
Happaratus.reservoir = H2
H1
Q'
Sapparatus.reservoir = S2
S1
Q' T'
Wideal = 0.0
kJ kg
T'
450 K
Given
(Guess) kJ 0 = H2 H1 kg
Find ( ) T'
Q'
T
S2
T'
S1
Q' T'
409.79 K Ans.
T'
(136.64 degC)
194
6.69 From Table F.4 at 200(psi):
H1
1222.6
BTU lbm
S1
1.5737
BTU lbm rankine
(at 420 degF) (Sat. liq. and vapor)
Hliq
BTU 355.51 lbm
Hvap
1198.3
BTU lbm
Sliq
0.5438
BTU lbm rankine
Svap
1.5454
BTU lbm rankine
x
0.96
H2
Hliq
x Hvap
3 BTU
Hliq
S2
Sliq
x Svap
Sliq
H2
1.165 10
lbm
S2
1.505
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
0.5 H1
H Hvap
0.5 H2
H
1193.6
BTU lbm
Ans.
(wet steam)
x
Hliq Hliq
x
0.994
S
Sliq
x Svap
Sliq
S
1.54
BTU lbm rankine
By Eq. (5.22) on the basis of 1 pound mass of exit steam,
SG
S
0.5 S1
0.5 S2
SG
2.895
10
4
BTU lbm rankine
Ans.
6.70
From Table F.3 at 430 degF (sat. liq. and vapor):
Vliq
ft 0.01909 lbm
3
Vvap
ft 1.3496 lbm
3
Vtank
80 ft
3
Uliq
406.70
BTU lbm
Uvap
1118.0
BTU lbm
mliq
4180 lbm
VOLliq
mliq Vliq
195
VOLliq
79.796 ft
3
VOLvap
mvap
U1
Vtank
VOLvap Vvap
VOLliq
VOLvap
mvap
0.204 ft
3
0.151 lbm
406.726 BTU lbm
mliq Uliq mliq
mvap Uvap mvap
U1
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
0
H dm = 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
mvap
Have m = 0
Have
1203.5
m1
BTU lbm
mass
mliq
m2 ( mass)
Property values below are for sat. liq. and vap. at 420 degF
ft 0.01894 lbm
395.81 BTU lbm
Vtank m2 ( mass)
Uliq
3
Vliq
Vvap
ft 1.4997 lbm
1117.4 BTU lbm
3
Uliq
Uvap
V2 ( mass)
U2 ( mass)
mass
x mass) (
Uliq
V2 ( mass) Vliq Vvap Vliq
x mass) Uvap (
50 lbm (Guess)
196
Given
mass =
m1 U1 U2 ( mass) Have U2 ( mass)
mass
Find ( mass)
mass
55.36 lbm
Ans.
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
J gm K
V1
64.721
cm
3
gm
Vtank
50 m
3
S2 = S1 = 6.7093
J gm K
By interpolation in Table F.2 at this entropy and 3500 kPa:
V2
cm 78.726 gm
3
t2 = 362.46 C
Ans.
m1
Vtank V1
m2
Vtank V2
m
m1
m2
m
137.43 kg Ans.
6.72
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,
Whence
mt 1000 kg
mt Vt = Vliq mt
Q = Hliq mt y Hlv
y Vlv = 0
H mt
Required data from Table F.1 are:
H 209.3 kJ kg
Vliq 1.251 cm
3
At 50 degC:
At 250 degC:
Hliq
kJ 1085.8 kg
kJ 1714.7 kg
gm
3
Hlv
y
Q
Vlv
cm 48.79 gm
Vliq mt Vlv
mt Hliq H
y
25.641 kg
Q 832534 kJ
Ans.
kJ kg
y Hlv
6.73
Given:
C 0.43 kJ kg K
Vtank
T1
0.5 m
295 K
3
Hin
mtank
120.8
30 kg
Data for saturated nitrogen vapor:
80 85 90 T 95 100 105 110
K
1.396 2.287 3.600 P 5.398 bar 7.775 10.83 14.67
V
0.1640 0.1017 0.06628 0.04487 0.03126 0.02223 0.01598 m kg
3
198
78.9 82.3 85.0 H 86.8 87.7 87.4 85.6
By Eq. (2.29) multiplied through by dt, d nt Ut H dm = dQ
kJ kg
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
mvap Tvap Vvap Hvap Uvap
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
Also,
Hin mvap = Q = mtank C Tvap
mvap = Vtank Vvap
T1
(A)
(B)
Calculate internal-energy values for saturated vapor nitrogen at the given values of T: 56.006
U
( H
P V)
59.041 61.139 U 62.579 63.395 63.325 62.157
kJ kg
Fit tabulated data with cubic spline: Us lspline ( U) T
interp ( T U t) Us
(guess)
Vs
lspline ( V) T
interp ( T V t) Vs
Uvap () t
Tvap
Vvap () t
100 K
Combining Eqs. (A) & (B) gives:
199
Given
Uvap Tvap
Tvap
mvap
Hin =
mtank C T1
Tvap Vvap Tvap Vtank
Find Tvap
Vtank Vvap Tvap
Tvap
mvap
97.924 K
13.821 kg
Ans.
6.74
The result of Part (a) of Pb. 3.15 applies, with m replacing n: m2 U2
Whence
Also U2 = Uliq.2
V2 = Vliq.2
H
m1 U1
H = Q= 0
m2 H
x2 Ulv.2
x2 Vlv.2
V2 = Vtank m2
U2 = m1 H
U1
Eliminating x2 from these equations gives
Vtank m2 H Uliq.2 m2
Vliq.2
Vlv.2
Ulv.2 = m1 H
U1
which is later solved for m2 Vtank 50 m
3
m1
16000 kg
V1
V1
Vtank m1
3.125 10
3 3m
kg
Data from Table F.1
Vliq.1 cm 1.003 gm
3
@ 25 degC:
Vlv.1 cm 43400 gm
3
Uliq.1
104.8
kJ kg
Ulv.1
2305.1
kJ kg
200
x1
x1
V1
Vliq.1 Vlv.1
U1
5
Uliq.1
104.913
x1 Ulv.1
kJ kg
4.889
10
U1
Data from Table F.2 @ 800 kPa:
Vliq.2 1.115 cm
3
Uliq.2
3
720.043
gm
1.115) cm
Ulv.2
Ulv.2
H
kJ kg
kJ kg
Vlv.2
Vlv.2
( 240.26
m 0.239 kg
3
gm
( 2575.3
720.043)
3 kJ
1.855
kJ kg
10
kg
Data from Table F.2 @ 1500 kPa:
2789.9
m1 H m2
H
U1
Vtank
Ulv.2 Vlv.2
Ulv.2 Vlv.2
m2
2.086
10 kg
4
Uliq.2
Vliq.2
msteam
m2
m1
msteam
4.855
10 kg
3
Ans.
6.75
The result of Part (a) of Pb. 3.15 applies, with
Whence U2 = H
n1 = Q = 0
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
1 100 P2 200 300 400
384.09 384.82 t2 385.57 386.31 387.08
303316 3032.17 V2 1515.61 1010.08 757.34
3
cm
3
gm
i
1 5
Vtank
1.75 m
massi
Vtank V2
i
5.77 mass
10
3
T rises very slowly as P increases
3 2
massi
0.577 1.155 1.733 2.311
kg
1 0
0
200
P2
i
400
6.76
Vtank
Vliq
Hliq
x1
2m
3
3
Data from Table F.2 @ 3000 kPa:
Vvap
Hvap
V1
cm 1.216 gm
1008.4
0.1
cm 66.626 gm
2802.3
Vliq
3
kJ kg
kJ kg
Vliq
m1 Vtank V1
x1 Vvap
3 3m
V1
7.757 10
kg
m1
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 2. Exit of
Hliq
0.6 m1 kg
of liquid from the tank:
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
Vliq
0.6 m1 Vliq = 0
y=
0.6 m1 Vliq Vvap Vliq
Q=
0.6 m1 Vliq Hvap Vvap Vliq
Hliq
0.6 m1 Hliq
0.6 m1 =
H mtank
But
H = Hliq
and
mtank
and therefore the last two terms of the energy equation cancel:
Q 0.6 m1 Vliq Hvap Vvap Vliq
Hliq
Q
5159 kJ
Ans.
6.77
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
mdot2
mdot3
5
kg sec
Whence
mdot2
mdot3 H1 H1 H2
H3
mdot1
mdot3
mdot2
mdot2
0.484
kg sec
Ans.
mdot1
4.516
kg sec
Ans.
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:
Vliq
cm 1.012 gm
3
Hliq
209.3
kJ kg
Sliq
0.7035
kJ kg K
Psat
12.34 kPa
T
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:
V
( 1.015
1.010)
3 3 cm
cm
3
gm
T
10 K
P
3100 kPa
4 1
V
5 10
1 Vliq
V T
gm
4.941
10
K
Apply Eqs. (6.28) & (6.29) at constant T:
H1
Hliq
Vliq 1
T
P
Psat
H1
211.926
S1
Sliq
Vliq P
Psat
204
S1
0.702
kJ kg kJ
kg K
By Eq. (2.30), neglecting kinetic and potential energies and setting the heat and work terms equal to zero:
H3 mdot3 Also
H1 mdot1
H2 mdot2 = 0
mdot2 = mdot3
mdot1
Whence
mdot1
mdot3 H3 H2 H1 H2
mdot1
1.89
kg sec
Ans.
mdot2
mdot3
mdot1
mdot2
13.11
kg sec
For adiabatic conditions, Eq. (5.22) becomes
SdotG
S3 mdot3
S1 mdot1
S2 mdot2
SdotG
1.973
kJ sec K
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
Hliq
H3 mdot3
H1 mdot1
H2 mdot2 = 0
Also
mdot3 = mdot2
mdot1
mdot2
mdot1 H1 H3 H2
H3
mdot2
3.241
kg sec
Ans.
For adiabatic conditions, Eq. (5.22) becomes
205
S2
Sliq
mdot3
mdot2
mdot1
SdotG
S3 mdot3
S1 mdot1
S2 mdot2
SdotG
3.508
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
12 bar
P2
2 bar
n1
1 mol
n2
2.5 mol
CP
7 R 2
CP
29.099
J mol K
1st law:
T
600 K
(guess)
T1 n2 CP T T2 = 0 J
Given
n1 CP T
T
Find ( ) T
T
542.857 K
Ans.
2nd law:
P
5 bar
(guess)
= 0
J K
Given
n1 CP ln
T P R ln T1 P1 T P n2 CP ln R ln T2 P2
P 4.319 bar
P
Find ( ) P
Ans.
6.81
molwt
28.014
lb lbmol
CP
R 7 2 molwt
CP
0.248
BTU lbm rankine
Ms
= steam rate in lbm/sec
Mn
= nitrogen rate in lbm/sec
Mn
40
lbm sec
206
(1) = sat. liq. water @ 212 degF entering (2) = exit steam at 1 atm and 300 degF (3) = nitrogen in at 750 degF
(4) = nitrogen out at 325 degF
BTU lbm
BTU lbm
T3
T4
0.3121 BTU lbm rankine
BTU lbm rankine
1209.67 rankine
784.67 rankine
H1
H2
180.17
S1
S2
(Table F.3)
1192.6
1.8158
(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
Given
Ms
3
lbm sec
Ms H2
(guess)
Q = 60
Mn CP T4
Ms
BTU Ms lbm
Ms
H1
T3 = 60
lbm sec
BTU lbm
Find Ms
3.933
Ans.
Eq. (5.22) here becomes
SdotG = Ms S2 S1 Mn S4 S3 Q T
S4
T
S3 = CP ln
T4 T3
Q
60
BTU Ms lbm
Q
235.967
BTU sec
529.67 rankine
SdotG
SdotG
Ms S2
2.064
S1
Mn CP ln
T4 T3
Q T
BTU sec rankine
Ans.
207
6.82 molwt
28.014
gm mol
CP
R 2 molwt
7
CP
1.039
J gm K
Ms = steam rate in kg/sec Mn= nitrogen rate in kg/sec
Mn
20
kg sec
(1) = sat. liq. water @ 101.33 kPa entering (2) = exit steam at 101.33 kPa and 150 degC (3) = nitrogen in @ 400 degC
T3
673.15 K
(4) = nitrogen out at 170 degC
T4
443.15 K
H1
419.064
kJ kg
S1
1.3069
kJ kg K
(Table F.2)
H2
2776.2
kJ kg
S2
7.6075
kJ kg K
(Table F.2)
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
kg sec
(guess)
Q = 80
kJ Ms kg
Given
Ms H2
H1
Mn CP T4
T3 = 80
kJ Ms kg
Ms
Find Ms
Ms
1.961
kg sec
Ans.
Eq. (5.22) here becomes
SdotG = Ms S2
S1
Mn S4
S3
T
Q T
298.15 K
Q T
S4
S3 = CP ln
Ms S2
4.194
T4 T3
S1 Mn CP ln
Ans.
Q
80
kJ kg
Ms
SdotG
SdotG
T4 T3
kJ sec K
208
6.86
Methane = 1; propane = 2
T
1
363.15 K
P
2
5500 kPa
y1
0.5
y2
1
y1
0.012
0.152
Zc1
0.286
Zc2
0.276
Tc1
190.6 K
Tc2
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
y2 Tc2
Ppc
y1 Pc1
y2 Pc2
Tpc
280.2 K
Ppc
44.235 bar
Tpr
T Tpc
Tpr
1.296
Ppr
P Ppc
Ppr
1.243
By interpolation in Tables E.3 and E.4:
Z0
0.8010
Z1
0.1100
y1
1
y2
2
0.082
Z
Z0
Z1
Z
0.81
For the molar mass of the mixture, we have: gm molwt molwt y1 16.043 y2 44.097 mol
30.07
gm mol
V
ZRT P molwt
V
cm 14.788 gm
3
mdot
3 4 cm
1.4
kg sec
u
30
m sec
2
Vdot
V mdot
Vdot
2.07
10
sec
A
Vdot u
A
6.901 cm
D
4A
D
2.964 cm
Ans.
209
6.87
Vectors containing T, P, Tc, and Pc for the calculation of Tr and Pr:
500 400 450 600 620 T 250 150 500 450 400
1.176 1.315 0.815 0.971
425.2 304.2 552.0 617.7 617.2 Tc 190.6 154.6 469.7 430.8 374.2
0.468 2.709 0.759 0.948
20 200 60 20 20 P 90 20 10 35 15 Pc
42.77 73.83 79.00 21.10 36.06 45.99 50.43 33.70 78.84 40.60 Tr T Tc
Pr
P Pc
Tr
1.005 1.312 0.97 1.065 1.045 1.069
Pr
0.555 1.957 0.397 0.297 0.444 0.369
Parts (a), (g), (h), (i), and (j) --- By virial equation:
500 150 T 500 K P 450 400 T Tc
20 20 10 bar Tc 35 15
P Pc
425.2 154.6 469.7 K Pc 430.8 374.2
42.77 50.43 33.70 bar 78.84 40.6
.190 .022 .252 .245 .327
Tr
Pr
210
1.176 0.97 Tr 1.065 1.045 1.069 0.422 Tr
1.6
0.468 0.397 Pr 0.297 0.444 0.369 0.172 Tr
4.2
B0
0.073
Eq. (3.65) B1
0.139
Eq. (3.66)
DB0
0.675 Tr
2.6
Eq. (6.89)
DB1
0.722 Tr
5.2
Eq. (6.90)
0.253 0.37 B0 0.309 0.321 0.306 B1
0.052 0.056 6.718 4.217 9.009 10 10 10
3 3 3
0.443 0.73 DB0 0.574 0.603 0.568 DB1
0.311 0.845 0.522 0.576 0.51
Combine Eqs. (3.61) + (3.62), (3.63), and (6.40) and the definitions of Tr and Pr to get:
VR
R
Tc B0 Pc
B1
HR
R Tc Pr B0
Tr DB0
( B1
Tr DB1)
Eq. (6.87)
SR
R Pr DB0
DB1
Eq. (6.88)
211
200.647 94.593 VR 355.907 146.1 232.454
cm
3
1.377 10 559.501 HR 1.226 10 1.746 10 1.251 10
3
1.952 2.469
J mol J mol K
3 3 3
SR
1.74 2.745 2.256
mol
Parts (b), (c), (d), (e), and (f) --- By Lee/Kesler correlation: By linear interpolation in Tables E.1--E.12: DEFINE: h0 equals
( ) HR RTc
( ) SR R
0
0
( ) HR h1 equals RTc
s1 equals
( ) SR R
2.008 4.445 h0 3.049 0.671 1.486
1
1
h equals
HR RTc
SR R
s0 equals
.663 .124 Z0 .278 .783 .707 1.137 4.381 s0 2.675 0.473 0.824 400 450 T 600 K 620 250 P s1 Z1
s equals
0.208 .050 .088 .036 0.138 0.405 5.274 2.910 0.557 0.289 200 60 20 20 90
bar
0.233 5.121 h1 2.970 0.596 0.169
304.2 552.0 Tc 617.7 K 617.2 190.6
.224 .111 .492 .303 .012
212
Z s HR
Z0 s0
Z1 Eq. (3.57) s1
(6.86)
h
h0
h1
Eq. (6.85)
( h Tc R)
SR
( s R)
0.71 0.118 Z 0.235 0.772 0.709 HR
5.21 2.301
10
3
10.207
4 4
10
41.291
J mol
2.316 10 4.37 10
SR
34.143 5.336 6.88
J mol K
3 3
2.358 10
48.289
VR
T (Z R P
549.691
1)
VR
1.909 10 587.396 67.284
3
cm mol
3
And.
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)
650 300 600 T 350 400 200 450 250
K
60 100 100 P 75 150 75 80 100
bar
562.2 304.2 304.2 Tc1 305.3 373.5 190.6 190.6 126.2
K Tc2
553.6 132.9 568.7 282.3 190.6 126.2 469.7 154.6 K
213
48.98 73.83 73.83 Pc1 48.72 89.63 45.99 45.99 34.00 bar
40.73 34.99 24.90 Pc2 50.40 45.99 34.00 33.70 50.43 .5 Tc2) Ppc P Ppc 44.855 54.41 49.365
K bar
.210 .224 .224 1 .100 .094 .012 .012 .038 .5 Pc2) .5 1 2
.210 .048 .400 .087 .012 .038 .252 .022 .5 2
Tpc
( Tc1 .5 T Tpc
( Pc1 .5
Tpr
Ppr
557.9 218.55 436.45 Tpc 293.8 282.05 158.4 330.15 140.4 1.165 1.373 1.375 Tpr 1.191 1.418 1.263 1.363 1.781 Ppr Ppc
0.21 0.136 0.312
bar
49.56 67.81 39.995 39.845 42.215 1.338 1.838 2.026 1.513 2.212 1.875 2.008 2.369
214
0.094 0.053 0.025 0.132 0.03
Lee/Kesler Correlation --- By linear interpolation in Tables E.1--E.12:
.6543 .7706 .7527 Z0 .6434 .7744 .6631 .7436 .9168 .890 .658 .729 s0 .944 .704 .965 .750 .361 Z1
.1219 .1749 .1929 .1501 .1990 .1853 .1933 .1839 h0
1.395 1.217 1.346 1.510 1.340 1.623 1.372 0.820 h1
.461 .116 .097 .400 .049 .254 .110 0.172
.466 .235 .242 s1 .430 .224 .348 .250 .095
0
h0 equals
( HR) RTpc
( SR) R
h1 equals
( HR) RTpc
( SR) R
1
h equals
HR RTpc
SR R
0
1
s0 equals
s1 equals
s equals
Z
Z0
Z1 Eq. (3.57)
h
h0
h1
Eq. (6.85)
s
s0
s1
Eq. (6.86)
215
HR
( hTpc R)
0.68 0.794 0.813
SR
( R) s
6919.583 2239.984 4993.974 HR 3779.762
J
8.213 5.736 6.689 SR 8.183
J
Z
0.657 0.785 0.668 0.769 0.922
3148.341 mol 2145.752 3805.813 951.151
5.952 mol K Ans. 8.095 6.51 3.025
6.95 Tc
647.1K
Pc
220.55bar
At Tr = 0.7:
T
0.7 Tc
T
452.97 K
Find Psat in the Saturated Steam Tables at T = 452.97 K
T1
451.15K
P1
957.36kPa
T2
453.15K
P2
1002.7kPa
Psat
P2 T2
Psat Pc
P1 ( T T1
T1) P1
Psat
998.619 kPa
Psat
9.986 bar
Psatr
Psatr
0.045
1
log Psatr
0.344
Ans.
This is very close to the value reported in Table B.1 ( = 0.345).
6.96 Tc
374.2K
Pc
40.60bar
At Tr = 0.7:
T
0.7 Tc
T
471.492 rankine
T
T
459.67rankine
T
11.822 degF
Find Psat in Table 9.1 at T = 11.822 F
T1
10degF
P1
26.617psi
T2
15degF
P2
29.726psi
216
Psat
P2 T2
P1 (T T1
T1)
P1
Psat
27.75 psi
Psat
1.913 bar
Psatr
Psat Pc
Psatr
0.047
1
log Psatr
0.327
Ans.
This is exactly the same as the value reported in Table B.1.
6.101 For benzene a)
0.210
Tc
562.2K
Pc
48.98bar
Zc
0.271
Tn
353.2K
Trn
Tn Tc
Trn
0.628
Psatrn
1
atm Pc
Psatrn
0.021
lnPr0 ( Tr)
5.92714
6.09648 Tr
15.6875 Tr
1.28862 ln ( Tr)
0.169347 Tr
6
6
Eqn. (6.79)
lnPr1 ( Tr)
15.2518
13.4721 ln ( Tr)
0.43577 Tr
Eqn. (6.80)
ln Psatrn
lnPr0 Trn
lnPr1 Trn
Eqn. (6.81).
0.207
lnPsatr ( Tr)
lnPr0 ( Tr)
lnPr1 ( Tr)
2 7
Eqn. (6.78)
Zsatliq
Psatrn Trn
0.083
Zc
1 1 T rn
Eqn. (3.73)
Zsatliq
0.00334
B0
0.422 Trn
1.6
Eqn. (3.65)
Z0
1
B0
Psatrn Trn
Eqn. (3.64)
B0
0.805
Z0
0.974
B1
0.139
0.172 Trn
4.2
Eqn. (3.66)
Z1
B1
Psatrn Trn
Equation following Eqn. (3.64)
B1
1.073
217
Z1
0.035
Zsatvap
Z0
Z1
Eqn. (3.57)
Zsatvap
0.966
Zlv
Zsatvap
Zsatliq
2
Zlv
0.963
Hhatlv
d dTrn
lnPsatr Trn Trn
Zlv
Hhatlv
6.59
Hlv
R Tc Hhatlv
Hlv
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:
0.224
Tc
Tt
304.2K
216.55K
Pc
Pt
73.83bar
5.170bar
At the triple point:
a) At Tr = 0.7
T
0.7Tc
T
212.94 K
Ttr
Tt Tc
Ttr
0.712
Ptr
Pt Pc
Ptr
6
0.07
lnPr0 ( ) Tr
5.92714
6.09648 Tr
1.28862 ln ( ) 0.169347 Tr Tr
6
Eqn. (6.79)
lnPr1 ( ) Tr
15.2518
15.6875 Tr
13.4721 ln ( ) 0.43577 Tr Tr
Eqn. (6.80)
ln Ptr
lnPr0 Ttr
lnPr1 Ttr
Eqn. (6.81).
218
0.224
Ans.
This is exactly the same value as given in Table B.1 b) Psatr
1atm Pc
Psatr
0.014
Guess: Trn
0.7
Given
ln Psatr = lnPr0 Trn
lnPr1 Trn
Trn
Find Trn
Trn
0.609
Tn
Trn Tc
Tn
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 sec
R
8.314
J mol K
molwt
28.9
gm CP mol
7
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
u2 2
2
= 0
But
H = CP T
Whence
T
u2
2
2 CP
T
52.45 K
Ans.
7.4
From Table F.2 at 800 kPa and 280 degC:
H1
3014.9
kJ kg
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
V2
cm 531.21 gm
3
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
u2 2
2
= 0
Whence
u2
2 H2
H1
u2
565.2
m sec
2
Ans.
By Eq. (2.27),
A2
mdot V2 u2
A2
7.05 cm
Ans.
7.5
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 kg
S1
7.1595
kJ kg K
S2 = S 1
Interpolations in Table F.2 at several pressures and at the given entropy yield the following values:
400 425 P 450 kPa 475 500
kg sec
2855.2 2868.2 H2 2880.7 2892.5 2903.9
kJ kg
531.21 507.12 V2 485.45 465.69 447.72
mdot V2 u2
cm
3
gm
mdot
0.75 565.2 541.7
u2
2 H2
H1
A2
7.05 7.022 A2 7.028 cm 7.059 7.127
2
u2
m 518.1 sec 494.8
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
s
pmin
Given
pmin
1 5
cspline P A2
400 kPa
pi
Pi
A( ) P
(guess)
2
a2
i
A2
i
interp s p a2 P
cm d A pmin = 0 kPa dpmin
431.78 kPa
Ans.
pmin
A pmin
Find pmin
2
7.021 cm
Ans.
221
Show spline fit graphically:
7.13
p
400 kPa 401 kPa 500 kPa
7.11
A2 cm
i
7.09 7.07
2
A( ) p cm
2
7.05
7.03
7.01 400
420
440 Pi
460
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.
800 775 P 750 kPa 725 700
H2
2956.0 2948.5 2940.8 2932.8 2924.9 V2 u2
kJ kg
294.81 302.12 V2 309.82 317.97 326.69
cm gm
3
u2
2 H2
H1
A2 =
mdot
222
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
V2 u2
5.553 5.552 5.557 5.577
cm sec kg
2
At the throat,
mdot A2 u2 3 V2
3
A2
mdot
6 cm
2
1.081
kg sec
Ans.
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 1.4098
S1 Svap Sliq Sliq
ft sec
kJ kg K
Svap
7.2479
kJ kg K
x
x
0.966
7.10
u1
230
u2
2000
ft sec
From Table F.4 at 130(psi) and 420 degF:
H1 1233.6 Btu lbm
S1 1.6310
u1
2
Btu lbm rankine
u2
2
By Eq. (2.32a),
H
H2
2
1154.8 Btu lbm
Btu lbm
Btu lbm rankine
H
78.8
Btu lbm
H2
H1
H
From Table F.4 at 35(psi), we see that the final state is wet steam:
Hliq
Sliq
228.03
Btu lbm
Btu lbm rankine
Hvap
Svap
1167.1
0.3809
1.6872
223
x
H2 Hvap
Hliq Hliq
x
0.987
(quality)
S2
Sliq
x Svap
Sliq
S2
1.67
BTU lbm rankine
SdotG
S2
S1
SdotG
0.039
Btu lbm rankine
Ans.
7.11
u2
580
m T2 sec
( 273.15
15)K molwt
28.9
gm CP mol
R 2 molwt
7
By Eq. (2.32a),
H=
u1
2
u2 2
2
=
u2 2
2
But
u2
2
H = CP T
T
Whence
T
2 CP
167.05 K
Ans.
Initial t = 15 + 167.05 = 182.05 degC Ans.
7.12
Values from the steam tables for saturated-liquid water: At 15 degC: V
cm 1.001 gm
3
T
288.15 K
Enthalpy difference for saturated liquid for a temperature change from 14 to 15 degC:
H
( 67.13
58.75)
4
J gm
t
2K
Cp
H t
1.5 10 K
P
4 atm
Cp
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
V 1 Cp
T
P
1
joule
9.86923 cm3 atm
T
0.093 K
The entropy change for this process is given by Eq. (7.26):
S
Cp ln
T T
T
V P
S
293.15 K
J gm
or
1.408 10
3
J gm K
Apply Eq. (5.36) with Q=0:
Wlost T S
Wlost
T
0.413
Wlost
0.413
kJ kg
Ans.
7.13--7.15 350 T1 350 250 400 304.2 Tc 282.3 126.2 369.8
K K
P2
1.2bar 80
P1
60 60 20
bar
73.83 Pc 50.40 34.00 42.48
bar
.224 .087 .038 .152
5.457 A 1.424 3.280 1.213
0.0 C 4.392 0.0 8.824
10 K
2 6
1.045 B 14.394 .593 28.785
1.157 D 0.0 0.040 0.0 10 K
5 2
10 K
3
225
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 1.084 Pr P1 Pc Pr 1.19 1.765 0.471 0.08664
Tr
T1 Tc
Tr
1.24 1.981 1.082
7.13
Redlich/Kwong equation:
0.42748
Pr Tr
Guess:
Eq. (3.53)
q
Tr z 1
1.5
Eq. (3.54)
Given Z
i
z= 1
Find z ()
q
z z z
Eq. (3.52)
q
1 4
Ii
ln
Z Z
i qi i qi
i
Eq. (6.65b)
HRi
SRi
R T1i
R ln Z
Z
i qi
i qi
1
i
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 302 232 385 K
226
Z
i qi
0.721 0.773 0.956 0.862
2.681 HR 2.253
kJ 0.521 mol
5.177 SR 4.346 1.59 2.33
J mol K
1.396
T2 T1
Cp
R A
B T1 2
1
C 2 T1 3
2
1
D T1
2
T2
HR Cp
T1
S
Cp ln 31.545
T2 T1
R ln
P2 P1
SR
279.971 T2 302.026 232.062 384.941
7.14 Soave/Redlich/Kwong equation:
2
K
Ans.
S
29.947
J 31.953 mol K
Ans.
22.163 0.08664
0.42748
0.5 2
c
0.480
Pr Tr
1.574
0.176
1
c 1
Tr
Eq. (3.53)
q
Eq. (3.54)
Tr
Guess:
z
1
Given
z= 1
q
z z z
i
Eq. (3.52) Z
q
Find ( z)
i
1 4
Ii
ln
Z Z
i qi i qi
Eq. (6.65b)
HRi
R T1i Z
i qi
1
ci
Tri
i
0.5
1 qi Ii
Eq. (6.67)
227
SRi
R ln Z
i qi
i
ci
Tri
i
0.5
qi Ii
Eq. (6.68)
0.5
The derivative in these equations equals: Now iterate for T2:
ci
Tri
i
273
Guesses
T2
300 232 384
K
Z
i qi
0.75 0.79 0.975 0.866
2.936 HR 2.356
kJ 0.526 mol
6.126 SR 4.769
J 1.789 mol K
1.523
2.679
T2 T1
Cp
R A
B T1 2
1
C 2 T1 3
2
1
D T1
2
272.757 T2 HR Cp
T1
T2
299.741 231.873 383.554
K
Ans.
31.565
S
Cp ln
T2 T1
R ln
P2 P1
SR
S
30.028
J
32.128 mol K 22.18
Ans.
228
7.15
Peng/Robinson equation:
1 2
1 2
0.07779
0.45724
2
c
0.37464
1.54226
0.26992
2
1
c 1
Tr
0.5
Pr Tr
Eq. (3.53)
q
Eq. (3.54)
Tr
Guess:
Given
Z
i
z
z= 1
1
q z z z
Eq. (3.52)
q
1 4
Find z ()
Ii 1 2 2
ln
Z Z
i qi i qi
0.5
i i
Eq. (6.65b)
HRi
R T1i Z
i qi
1
ci
Tri
i
1 qi Ii
0.5
Eq. (6.67)
SRi
R ln Z
i qi
i
ci
Tri
i
qi Ii
Eq. (6.68)
The derivative in these equations equals: Now iterate for T2: 270 Guesses T2 297 229 383 K
ci
Tri
i
0.5
229
Z
i qi
0.722 0.76 0.95 0.85
3.041 HR 2.459 0.6 1.581
kJ mol
6.152 SR 4.784
J 1.847 mol K
2.689
T2 T1
Cp
R A
B T1 2
1
C 2 T1 3
2
1
D T1
2
269.735 T2 HR Cp
T1
T2
297.366 229.32 382.911
K
Ans.
31.2
S Cp ln T2 T1
R ln P2 P1 SR
S
29.694
J 31.865 mol K
Ans.
22.04
7.18
Wdot
H1
3500 kW
3462.9 kJ kg
Data from Table F.2:
H2 2609.9 kJ kg
S1 7.3439 kJ kg K
By Eq. (7.13),
mdot Wdot H2 H1
mdot 4.103 kg sec
Ans.
For isentropic expansion, exhaust is wet steam:
Sliq 0.8321
S2 Svap Sliq Sliq
kJ kg K
Svap
7.9094
kJ kg K
S2
S1
x
x
0.92
(quality)
230
Hliq
251.453
kJ kg
Hvap
2609.9
kJ kg
3 kJ
H'2
Hliq
x Hvap
Hliq
H'2
2.421
10
kg
H2 H'2
H1 H1
0.819
Ans.
7.19
The following vectors contain values for Parts (a) through (g). For intake conditions:
3274.3
kJ kg
6.5597
kJ kg K
3509.8
kJ kg
6.8143
kJ kg K
0.80 0.77 0.82 0.75 0.75 0.80 0.75
kJ 3634.5 kg
H1
kJ 6.9813 kg K
S1
kJ 3161.2 kg
6.4536
kJ kg K
kJ 2801.4 kg
kJ 6.4941 kg K
1444.7
Btu lbm
1.6000
Btu lbm rankine
1389.6
Btu lbm
1.5677
Btu lbm rankine
231
For discharge conditions:
0.9441
kJ kg K
7.7695
kJ kg K
0.8321
kJ kg K
7.9094
kJ kg K
0.6493
kJ kg K
8.1511
kJ kg K
Sliq
1.0912
kJ kg K
Svap
7.5947
kJ kg K
S'2 = S1
1.5301
kJ kg K
7.1268
kJ kg K
0.1750
Btu lbm rankine
1.9200
Btu lbm rankine
0.2200
Btu lbm rankine
kJ kg
1.8625
Btu lbm rankine
80 kg sec
289.302
2625.4
kJ kg
251.453
kJ kg
2609.9
kJ kg
90
kg sec
191.832
kJ kg
2584.8
kJ kg
70
kg sec
Hliq
340.564
kJ kg
Hvap
2646.0
kJ kg
mdot
65
kg sec
504.701
kJ kg
2706.3
kJ kg
50
kg sec
94.03
Btu lbm
1116.1
Btu lbm
150
lbm sec
120.99
Btu lbm
1127.3
Btu lbm
100
lbm sec
232
x'2
S1 Svap
Sliq Sliq
H'2
Hliq
x'2 Hvap
Hliq
H
H2 Hvap
H'2
Hliq
H1
H2
H1
H
Wdot
H mdot
x2
Hliq
S2
Sliq
x2 Svap
Sliq
H2 H2 H2 H2 H2 H2 H2
1 2 3 4 5 6 7
S2
2423.9 2535.9 kJ 2467.8 kg 2471.4 2543.4
1 2 3 4 5
7.1808 7.6873 7.7842 7.1022 6.7127 kJ kg K
S2 S2 S2 S2
S2 S2
Ans.
1031.9 Btu 1057.4 lbm
6 7
1.7762
Btu 1.7484 lbm rankine
68030 87653 81672 Wdot 44836 kW 12900 65333 35048 Wdot
91230 117544 109523 60126 17299 87613 46999
hp Ans.
233
7.20
T
423.15 K
P0
8.5 bar
P
1 bar
For isentropic expansion,
S
0
J mol K
For the heat capacity of nitrogen:
A
3.280
B
0.593 10 K
3
D
0.040 10 K
5
2
For the entropy change of an ideal gas, combine Eqs. (5.14) & (5.15) with C = 0. Substitute:
0.5
(guess)
Given
S = R A ln
B
T
D T
2
1
1 ln
2
T
P P0
Find
T0
T0
762.42 K
Ans.
Thus the initial temperature is 489.27 degC
7.21
T1
1223.15 K
P1
10 bar
P2
1.5 bar
CP
32
J mol K
0.77
Eqs. (7.18) and (7.19) derived for isentropic compression apply equally well for isentropic expansion. They combine to give:
R CP
W's
C P T1
P2 P1
1
W's
15231
J mol
Ws
W's
H
Ws
234
Ws
11728
J mol
Ans.
Eq. (7.21) also applies to expansion:
T2
T1
H CP
T2
856.64 K
Ans.
7.22
Isobutane:
Tc
408.1 K
Pc
36.48 bar
0.181
T0
523.15 K
P0
5000 kPa
P
500 kPa
S
0
J mol K
For the heat capacity of isobutane:
A
T0 Tc
1.677
B
37.853 10 K
1.282
3
C
P0 Pc P
11.945 10 K
2
6
Tr0
Tr0
Pr0
Pr0
1.3706
Pr
Pc
Pr
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
C T0
2
1 2
1
ln
P P0
Pr
SRB Tr0 Pr0
Find
T
T0
T
445.71 K
Tr
T Tc
Tr
1.092
235
The enthalpy change is given by Eq. (6.91): Hig
Hig
H'
R ICPH T0 T 1.677 37.853 10
11.078 kJ mol
3
11.945 10
6
0.0
Hig
8331.4
R Tc HRB Tr Pr
J mol
HRB Tr0 Pr0
H'
The actual enthalpy change from Eq. (7.16): 0.8
Wdot
ndot
ndot H
700
mol sec
Wdot
H
H'
4665.6 kW
H
Ans.
6665.1
J mol
The actual final temperature is now found from Eq. (6.91) combined with Eq (4.7), written: 0.7
Given (guess)
H = R A T0
1
Tc HRB
Find
B 2 T0 2 T0 Pr Tc
0.875
T
2
1
C 3
T0
3
3
1
HRB Tr0 Pr0
T0
T
457.8 K
Ans.
7.23
From Table F.2 @ 1700 kPa & 225 degC:
H1 2851.0 kJ kg
S1
x2
6.5138
kJ kg K
Sliq 0.6493 kJ kg K
At 10 kPa:
0.95
236
Hliq
191.832
kJ kg
Hvap
2584.8
kJ kg
Svap
8.1511
kJ kg K
mdot
0.5
kg sec
Wdot
180 kW
H2
Hliq
x2 Hvap
Hliq
H
H2
H1
H2
2.465
10
3 kJ
kg
H
385.848
kJ kg
(a)
Qdot
mdot H
Wdot
Qdot
12.92
kJ sec
Ans.
(b)
For isentropic expansion to 10 kPa, producing wet steam:
x'2
S1 Svap
Sliq Sliq
H'2
Hliq
x'2 Hvap
Hliq
x'2
0.782
H'2
2.063
10
3 kJ
kg
Ans.
Wdot'
mdot H'2
H1
Wdot'
394.2 kW
7.24
T0
673.15 K
P0
8 bar
P
1 bar
For isentropic expansion, For the heat capacity of carbon dioxide:
S
0
J mol K
5 2
A
5.457
B
1.045 10 K
3
D
1.157 10 K
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
D T0
2
1 2
T' T0
1
ln
P P0
Find
0.693
237
T'
466.46 K
H'
R ICPH T0 T' 5.457 1.045 10
9.768 kJ mol
Work
H
3
0.0
1.157 10
5
H'
0.75
H Work
H'
7.326 kJ mol
Work
7.326
kJ mol
Ans.
For the enthalpy change of an ideal gas, combine Eqs. (4.2) and (4.7) with C = 0: Given
H = R A T0
1
B 2 T0 2
0.772
2
1
D T0
T0
1
Find
T
T
519.9 K
Ans.
Thus the final temperature is 246.75 degC
7.25
Vectors containing data for Parts (a) through (e):
500 450 T1 525 475 550 H [ ( Cp T2 P1
6 5 10 7 4 T1) ]
R Cp
371 376 T2 458 372 403 P2
1.2 2.0 3.0 1.5 1.2 Cp
3.5 4.0 5.5 R 4.5 2.5
Ideal gases with constant heat capacities
HS
Cp T1
P2 P1
1
Eq. (7.22) Applies to expanders as well as to compressors
238
0.7 0.803
H HS
0.649 0.748 0.699
7.26
Cp
7 2
R
ndot
175
mol sec
T1
550K
P1
6bar
P2
1.2bar
Guesses:
Given
0.75
Wdot
600kW
Wdot =
Wdot
0.065
.08 ln
Wdot kW
ndot Cp T1
P2 P1
R Cp
1
Find ( Wdot) Wdot kW
Wdot
594.716 kW
Ans.
0.065
0.08 ln
0.576
Ans.
For an expander operating with an ideal gas with constant Cp, one can show that:
R Cp
T2
T1 1
P2 P1
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
10
3
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) A second relation follows from Eq. (7.16), written: [x is quality]
H = Hvap - 3207.1 = (
HS) = (0.75)[ (x)(Hvap) + (1-x)(Hliq) - 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
Sl
503.7
1.5276
Hv
Sv
2706.0
7.1293
The two equations for x are:
xH Hv 801.7 .75 Hl .75 ( Hv Hl)
xS 6.7093 Sl Sv Sl
0.924
xS 0.925
The trial values given produce: xH
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:
cm
3
V
1001
kg
Cp
4.190
kJ kg degC
H
kJ kg
T
T
Eqs. (7.16) and (7.24) combine to give:
Ws H
(7.14)
Ws 0.223
V ( P2
P1)
Eq. (7.25) with =0 is solved for T:
H
V ( P2 Cp
P1)
0.044 degC
Ans.
7.30
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):
753.15 673.15 T0 773.15 K 723.15 755.37 200 150 ndot 175 100 0.5 453.59
241
6 bar 5 bar P0 7 bar 8 bar 95 psi 0.80 0.75
mol sec
1 bar 1 bar P 1 bar 2 bar 15 psi
J mol K
S
i
0
1 5
0.78 0.85 0.80
For the heat capacity of nitrogen: A
0.5
Given
3.280
(guess)
B
0.593 10 K
3
D
0.040 10 K
5
2
S = R A ln
B T0
D T0
2 2
1 2
1
ln
P P0
Tau T0 P0 P
Find
460.67 431.36
i
Tau T0 P0 Pi i i
Ti
T0
i
i
T
453.48 K 494.54 455.14
H'i
R ICPH T0 Ti 3.280 0.593 10
i
3
0.0 0.040 10
5
8879.2 7279.8 H' 9714.4 6941.7 9112.1
J mol
7103.4 5459.8
H H'
J mol
H
7577.2 5900.5 7289.7
0.5
Given
(guess)
H = R A T0
Tau T0 H
1
Find
B 2 T0 2
2
1
D T0
i
1
i
Tau T0
Hi
Ti
T0
i
i
242
520.2 492.62 T 525.14 K 529.34 516.28
Ans.
1421 819 Wdot ndot H Wdot 1326 kW Ans. 590 1653
7.31
Property values and data from Example 7.6:
H1
3391.6
kJ kg
S1
6.6858
kJ kg K
mdot
59.02
kg sec
H2
2436.0
kJ kg
S2
7.6846
kJ kg K
Wdot
56400 kW
T
300 K
By Eq. (5.26)
Wdotideal
t
mdot H2
H1
T
t
S2
0.761
S1
Ans.
Wdotideal
74084 kW
Wdot Wdotideal
The process is adiabatic; Eq. (5.33) becomes:
SdotG
mdot S2
S1
SdotG
58.949
kW K
Ans.
Wdotlost
T SdotG
Wdotlost
17685 kW
Ans.
7.32
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
kJ kg K
243
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
kJ kg K
kJ kg K
Sliq
0.8932
Slv
6.9391
0.72
For isentropic expansion of steam in the turbine: S'3
S'3
H23
H23
x3
x3
S2
6.519 kJ kg K
H'3
437.996
H3 Hliq Hlv
0.883
x'3
x'3
H2
kJ kg
S'3
Sliq Slv
H'3
H'3
Hliq
2.174
x'3 Hlv
10
3 kJ
0.811
H3
H3
S3
S3
kg
H2
2.345
H23
10
3 kJ
kg
Sliq
7.023
x3 Slv
kJ kg K
mol sec
197.96)K
For the exhaust gases:
T1
T1
ndot 125
T2
T2
( 273.15
673.15 K
18
400)K
( 273.15
471.11 K
molwt
gm mol
3
Hgas
Sgas
R MCPH T1 T2 3.34 1.12 10
R MCPS T1 T2 3.34 1.12 10
0.0 0.0
T2
T2 T1
T1
3
0.0 0.0 ln
244
Hgas
6.687
10
3 kJ
kmol
Sgas
11.791
kJ kmol K
Energy balance on boiler:
mdot
ndot Hgas H2 H1
mdot
0.30971
kg sec
(a) Wdot
mdot H3
H2
Wdot
135.65 kW Ans.
(b) By Eq. (5.25):
T
293.15 K
Wdotideal
ndot Hgas mdot H3 H1 T ndot Sgas mdot S3
314.302 kW
t
S1
t
Wdotideal
Wdot Wdotideal
0.4316
Ans.
(c) For both the boiler and the turbine, Eq. (5.33) applies with Q = 0. For the boiler:
SdotG Boiler: ndot Sgas mdot S2 S1
SdotG
0.4534
kW K
Ans.
For the turbine: SdotG
mdot S3
S2
Turbine:
SdotG
0.156
kW K
Ans.
(d) Wdotlost.boiler
0.4534
kW K
T
Wdotlost.boiler
132.914 kW
Wdotlost.turbine
0.1560
kW T K
Wdotlost.turbine
Fractionboiler
45.731 kW
Fractionboiler
Wdotlost.boiler Wdotideal
0.4229
Ans.
245
Fractionturbine
Note that:
Wdotlost.turbine Wdotideal
t
Fractionturbine
Fractionturbine 1
0.1455 Ans.
Fractionboiler
7.34
From Table F.2 for sat. vap. at 125 kPa:
H1 2685.2 kJ kg
S1 7.2847 kJ kg K
kJ kg K
For isentropic expansion, S'2 = S1 = 7.2847
Interpolation in Table F.2 at 700 kPa for the enthalpy of steam with this entropy gives
H'2
H2
3051.3
kJ kg
0.78
H
kJ kg
H'2
H1
H
469.359
kJ kg
H1
H
H2
3154.6
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
Ans.
mdot H
Wdot
1173.4 kW
Ans.
246
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):
298.15 353.15 T0 303.15 373.15 299.82 338.71 100 100 ndot 150 50 0.5 453.59 0.5 453.59
mol sec K
101.33 kPa 375 kPa P0 100 kPa 500 kPa 14.7 psi 55 psi
375 kPa 1000 kPa P 500 kPa 1300 kPa 55 psi 135 psi
0.75 0.70 0.80 0.75 0.75 0.70
i
S 0 J mol K
1 6
For the heat capacity of air: A
0.5
Given
3.355
(guess)
B
0.575 10 K
3
D
0.016 10 K
5
2
S = R A ln
B T0
D T0
2 2
1 2
1
ln
P P0
Tau T0 P0 P
Find
247
i
Tau T0 P0 Pi i i
431.06 464.5
Ti T0
i
i
476.19 T 486.87 434.74 435.71
K
H'i
R ICPH T0 Ti 3.355 0.575 10
i
3
0.0
0.016 10
5
3925.2 3314.6 5133.2 H'
J 3397.5 mol
3986.4 2876.6 5233.6 4735.1
H
H'
H
6416.5 4530 5315.2 4109.4
J mol
1.5
(guess)
Given
Tau T0
H = R A T0
H Find
1
B 2 T0 2
i
2
1
D T0
Hi
1
Tau T0
i
Ti
T0
i
i
Wdot
ndot H
248
474.68 511.58 T 518.66 524.3 479.01 476.79
K
702 635 Wdot 1291 304 1617 1250
hp
523 474 Wdot 962 227 1205 932
kW Ans.
7.36
Ammonia:
Tc
405.7 K
Pc
112.8 bar
0.253
T0
294.15 K
P0
200 kPa
P
1000 kPa
S
0
J mol K
For the heat capacity of ammonia:
A
3.578
B
3.020 10 K
0.725
3
D
P0 Pc P
0.186 10 K
5
2
Tr0
T0 Tc
Tr0
Pr0
Pr0
0.0177
Pr
Pc
Pr
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 Tc
D T0
2
1 2
1
ln
P P0
SRB
Pr
SRB Tr0 Pr0
Find
1.437
T
T0
T
422.818 K
Tr
249
T Tc
Tr
1.042
Hig
Hig
H'
H'
R ICPH T0 T 3.578 3.020 10
4.826
Hig
4652
3
0.0
0.186 10
5
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: 1.4
Given (guess)
H = R A T0
1
Tc HRB
B 2 T0 2 T0 Pr Tc
1.521
2
1
D T0
1
HRB Tr0 Pr0
Find
T
Tr
T0
T Tc
T
Tr
447.47 K
1.103
Ans.
S
R A ln
B T0
D
2 2 T0 SRB Tr0 Pr0
1
1
ln
P P0
SRB Tr Pr
S
2.347
J mol K
Ans.
250
7.37
Propylene:
Tc
365.6 K
Pc
46.65 bar
0.140
T0
303.15 K
P0
11.5 bar
P
18 bar
S
0
J mol K
For the heat capacity of propylene:
A
1.637
B
22.706 10 K
3
C
6.915 10 K
2
6
Tr0
T0 Tc
Tr0
0.8292
Pr0
P0 Pc P
Pr0
Pr
0.2465
0.386
Pr
Use generalized second-virial correlation:
Pc
The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0:
1.1
(guess)
Given
S = R A ln
SRB
B T0
T0 Tc
C T0
2
1 2
1
ln
P P0
Pr
SRB Tr0 Pr0
Find
1.069
T
T0
T
324.128 K
Tr
T Tc
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
3
6.915 10
6
0.0
Hig
1.409
10
3 J
mol
H'
Hig
R Tc HRB Tr Pr
251
HRB Tr0 Pr0
H'
964.1
J mol
The actual enthalpy change from Eq. (7.17):
0.80
H
H'
H
1205.2
J mol
ndot
1000
mol sec
Wdot
ndot H
Wdot
1205.2 kW
Ans.
The actual final temperature is now found from Eq. (6.91) combined with Eq (4.7), written:
1.1 (guess)
Given
H = R A T0
1
Tc HRB
Find
7.38 Methane:
B 2 T0 2 T0 Pr Tc T
2
1
C 3 T0 3
3
1
HRB Tr0 Pr0
1.079
Tc 190.6 K
T0
Pc
T
327.15 K
Ans.
0.012
45.99 bar
T0
308.15 K
P0
3500 kPa
P
5500 kPa
S
0
J mol K
For the heat capacity of methane:
A
1.702
B
9.081 10 K
3
C
2.164 10 K
2
6
Tr0
T0 Tc
Tr0
1.6167
Pr0
P0 Pc P
Pr0
0.761
Pr
Pc
Pr
1.196
252
Use generalized second-virial correlation: The entropy change is given by Eq. (6.92) combined with Eq. (5.15) with D = 0: 1.1
Given (guess)
S = R A ln
SRB
B T0
T0 Tc
C T0
2
1 2
1
ln
P P0
Pr
SRB Tr0 Pr0
Find
1.114
T
Tr
T0
T Tc
T
Tr
343.379 K
1.802
The enthalpy change for the final T is given by Eq. (6.91), with HRB for this T: Hig
Hig
H'
H'
R ICPH T0 T 1.702 9.081 10
1.298 10
Hig
1158.8
3 J
3
2.164 10
6
0.0
mol
HRB Tr0 Pr0
R Tc HRB Tr Pr
J mol
The actual enthalpy change from Eq. (7.17): 0.78
mol sec
H
H'
H
1485.6
J mol
Ans.
ndot
1500
Wdot
ndot H
Wdot
2228.4 kW
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
1
Tc HRB
Find
B 2 T0 2 T0 Pr Tc
T
2
1
C 3 T0 3
3
1
HRB Tr0 Pr0
1.14
T0
T
351.18 K
Ans.
7.39
From the data and results of Example 7.9,
T1
293.15 K
T2
428.65 K
P1
140 kPa
P2
560 kPa
Work
5288.3
J T mol
293.15 K
3 6
H
H
R ICPH T1 T2 1.702 9.081 10
5288.2 J mol
2.164 10
0.0
S
R ICPS T1 T2 1.702 9.081 10
J mol K
3
2.164 10
6
0.0
ln
P2 P1
S
3.201
Since the process is adiabatic: SG
S
SG
3.2012
J mol K
Ans.
Wideal
H
T
S
Wideal
4349.8
J mol
Ans.
Wlost
T
S
Wlost
938.4
J mol
Ans.
Wideal
t
Work
t
0.823
Ans.
254
7.42
P1
1atm
T1
( 35
273.15) K
T1
308.15 K
P2
50atm
T2
( 200
273.15) K
T2
473.15 K
0.65
Vdot
0.5
m
3
sec
Cp
3.5 R
V
R T1 P1
ndot
Vdot V
ndot
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:
1 N
r=
P2 P1
(where r is the pressure ratio in each stage and N is the number of stages.)
( T2 T1) T1
Eq. (7.23) may be solved for T2prime: T'2 T'2 415.4 K
R1 N Cp
Eq. (7.18) written for a single stage is:
T'2 = T1
P2 P1
Put in logarithmic form and solve for N:
N
R Cp
ln
ln
P2 P1
T'2 T1
N
3.743
(a) Although any number of stages greater than this would serve, design for 4 stages.
1 N
(b) Calculate r for 4 stages: N
4
r
P2 P1
r
2.659
Power requirement per stage follows from Eq. (7.22). In kW/stage:
R Cp
Wdotr
ndot Cp T1 r
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
mdotw
7.44
Qdotr
Hw
mdotw
1.052
kg sec
Ans.
(in each interchanger)
300 290 T1 295 K 300 305 464 547 T2 455 K 505 496 P2 P1
2.0 1.5 1.2 bar 1.1 1.5 6 5 6 bar 8 7 Cp 3.5 2.5 4.5 R 5.5 4.0
H
[ ( Cp T2
T1) ]
R Cp
Ideal gases with constant heat capacities
HS
Cp T1
P2 P1
1
256
(7.22)
3.219 3.729 HS 4.745 5.959 4.765
kJ mol
0.675 HS H 0.698 0.793 0.636 0.75
Ans.
7.47
The following vectors contain values for Parts (a) through (e). Intake conditions first:
298.15 363.15 T1 333.15 K 294.26 366.48 2000 kPa 5000 kPa P2 5000 kPa 20 atm 1500 psi
100 kPa 200 kPa P1 20 kPa 1 atm 15 psi 0.75 0.70 0.75 0.70 0.75
mdot
20 kg 30 kg 15 kg 50 lb 80 lb
1 sec
257.2 696.2 523.1 217.3 714.3
10 K
6
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 1.002 1.038
By Eq. (7.24) cm gm
3
4.15 4.20 CP 4.20 4.185 4.20 HS V P2 P1
H HS kJ kg K
257
1.906 4.973 HS 5.065 1.929 10.628
kJ kg
2.541 7.104 H 6.753 2.756 14.17 0.188
kJ kg
By Eq. (7.25)
T
H
V 1
T1 CP
P2
P1
0.807 T 0.612 K 0.227 1.506
50.82 213.12 Wdot H mdot Wdot 101.29 kW Wdot 62.5 514.21 298.338 363.957 T2 T1 T T2 333.762 K 294.487 367.986 t2 t2 T2 K
1 2 3
68.15 285.8 135.84 hp 83.81 689.56
Ans.
25.19 90.81 60.61 degC
273.15
t2 t2 t2
t2
T2 K
1.8
459.67
4 5
70.41 202.7
degF
t2
258
7.48 Results from Example 7.10:
H
11.57
kJ kg
W
11.57
kJ kg
S
0.0090
kJ kg K
T
300 K
Wideal
H
T
S
Wideal
t
W
Wideal
8.87
kJ kg
Ans.
t
0.767
Ans.
Since the process is adiabatic.
SG
S
SG
9
10
3 kJ
kg K
Ans.
Wlost
T
S
Wlost
2.7
kJ kg
Ans.
7.53
T1
T3
( 25
( 200
273.15) K
273.15) K
P1
P3
1.2bar
5bar
P2
5bar
Cpv
105
J mol K
Hlv
30.72
kJ mol
0.7
Estimate the specific molar volume of liquid benzene using the Rackett equation (3.72). From Table B.1 for benzene: Tc
562.2K
Zc
0.271 Vc
259
cm
3
mol
From Table B.2 for benzene: Tn
( 80.0
2 7
273.15) K
Trn
Tn Tc
Assume
Vliq
=
Vsat:
V
V c Zc
1 T rn
Eq. (3.72)
V
cm 96.802 mol
3
Calculate pump power
Ws
V P2
P1
Ws
0.053
kJ mol
Ans.
259
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: 13.7819
B 2726.81
C 217.572
Tsat A
B ln P2 kPa
C degC Tsat
142.77 degC
Tsat 273.15K
Tsat 415.9 K
Tsat
Estimate the heat of vaporization at Tsat using Watson's method From Table B.2 Hlv At 80 C:
30.72 kJ mol
Tsat Tc
Tr1
( 80
273.15) K Tc
Tr1
0.38
0.628
Tr2
Tr2
0.74
Hlv2
Hlv
1 1
Tr2 Tr1
Eq. (4.13)
Hlv2
26.822
kJ mol
Calculate the heat exchanger heat duty. Q R ICPH T2 Tsat 0.747 67.96 10 Hlv2 Cpv T3 Tsat
51.1 kJ mol
Ans.
3
37.78 10
6
0
Q
260
7.54
T1
( 25
273.15)K
P1
1.2bar
P2
1.2bar
T3
( 200
273.15)K
P3
5bar
Cpv
105
J mol K
0.75
Calculate the compressor inlet temperature. Combining equations (7.17), (7.21) and (7.22) yields:
T2 1
T3
R Cpv
T2
408.06 K
1
P3 P2
1
T2
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:
13.7819
B
2726.81
C
217.572
Tsat A
B ln P1 kPa
C degC Tsat
85.595 degC
Tsat
Tsat
273.15K
Tsat
358.7 K
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
Tr1
( 80
273.15) K Tc
Tr1
0.38
0.628
Tr2
Tsat Tc
Tr2
0.638
Hlv2
Q
Hlv
1 1
Tr2 Tr1
Eq. (4.13)
3
Hlv2
30.405
6
kJ mol
R ICPH T1 Tsat 0.747 67.96 10 Hlv2 Cpv T2 Tsat
37.78 10
0
Q
44.393
kJ mol
Ans.
7.57 ndot
100
kmol hr
P1
1.2bar
T1
300K
P2
6bar
Cp
50.6
J mol K
0.70
Assume the compressor is adaiabatic.
R Cp
T2
Wdots
Wdote
P2 P1
T1
(Pg. 77)
T2
Wdots
Wdote
390.812 K
ndot Cp T2
Wdots
T1
127.641 kW
182.345 kW
C_compressor
3040dollars
Wdots kW
0.952
C_compressor
307452 dollars Ans.
C_motor
380dollars
Wdote kW
0.855
C_motor
32572 dollars Ans.
262
7.59
T1
375K
P1
18bar
P2
1.2bar
For ethylene:
0.087
Tc
282.3K
Pc
50.40bar
Tr1
T1 Tc
Tr1
1.328
Pr1
P1 Pc
Pr1
0.357
Pr2
P2 Pc
Pr2
6
0.024
A
1.424
B
14.394 10
3
C
4.392 10
D
0
a) For throttling process, assume the process is adiabatic. Find T2 such that H = 0.
H = Cpmig T2 T1 HR2 HR1 Eq. (6-93)
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
= 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
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
T2
T1 R SRB Tr1 Pr1
R ln
P2 P1
Eq. (6-94)
S
22.128
J mol K
Ans.
263
b) For expansion process.
70%
First find T2 for isentropic expansion. Solve Eq. (6-94) with S = 0. Guess:
Given
T2
T1
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 Tc
Tr2
Eq. (6-94)
0.779
Now calculate the isentropic enthalpy change, HS. HR2 HRB Tr2 Pr2 R Tc
HS
R MCPH T1 T2 A B C D T2 T1 HRB Tr2 Pr2 R Tc HRB Tr1 Pr1
6.423 10
3 J
R Tc
HS
mol
Calculate actual enthalpy change using the expander efficiency. H HS
H 4.496 10
3 J
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
Ans.
264
Now recalculate S at calculated T2
S
R MCPS T1 T2 A B C D ln
R SRB Tr2 Pr2
T2
T1 R SRB Tr1 Pr1
R ln
P2 P1
Eq. (6-94)
S
7.77
J mol K
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
Cpgas
200 J mol K
150
J mol K
Hlv 35000 J mol
Light oil:
Exit stream:
b)
T2
T3
25degC Cpoil
200degC
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
Solving for D/F gives:
DF
Cpgas T3
Hlv
T1
T2
Cpoil T3
DF
0.643
Ans.
c)
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
S2
6.9636
At point 4: Table F.1,
H4
209.3
At point 1:
H1
H4
At point 3: Table F.1,
Hliq
H4
Hlv
2382.9
x3
0.96
H3
Hliq
x3 Hlv
H3
2496.9
Sliq
For isentropic expansion,
0.7035
S'3 S2
Slv
7.3241
x'3
S'3
Sliq Slv
x'3
0.855
H'3
Hliq
x'3 Hlv
H'3
2246
H3
turbine
H2 H2
turbine
H'3
0.805
Ans.
Ws
H3
H2
QH
H2
H1
Ws
1.035
10
3
QH
3.322
10
3
Ws
cycle
QH
cycle
0.311
Ans.
266
8.2
mdot
1.0 (kg/s)
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
0.741
x4
S1 Sliq Svap Sliq
3
x4
0.241
(c) The rate of heat addition, Step 1--2:
Qdot12
mdot ( H2
H1)
Qdot12
1.931
10
(kJ/s)
(d) The rate of heat rejection, Step 3--4:
H3
Hliq
x3 ( Hvap Hliq)
H4
Hliq
x4 ( Hvap Hliq)
H3
1.919
10
3
H4
699.083
Qdot34
mdot ( H4
H3)
Qdot34
1.22
10
3
(kJ/s)
(e) Wdot12
0
Wdot34
0
Wdot23
mdot ( H3
H2)
Wdot23
873.222
Wdot41
mdot ( H1
H4)
Wdot41
0.368
161.617
(f)
Wdot23 Wdot41 Qdot12
Note that the first law is satisfied:
Q
Qdot12
Qdot34
W
Wdot23
Wdot41
Q
W
0
267
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
kJ kg
6.9013
kJ kg K
3529.6
kJ kg
6.9485
kJ kg K
3635.4
H2
kJ kg
S2
6.9875
kJ kg K
kJ 3475.6 kg
6.9145
kJ kg K
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
kJ kg
2584.8
kJ kg
251.453
kJ kg
2609.9
kJ kg
191.832
Hliq
kJ kg
Hvap
2584.8
kJ kg
kJ 419.064 kg
2676.0
kJ 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
0.6493
Sliq
kJ kg K
8.1511
Svap
kJ kg K
1.3069
kJ kg K
7.3554
kJ kg K
0.3121
BTU lbm rankine
1.7568
BTU lbm rankine
0.1326
BTU lbm rankine
1.9781
BTU lbm rankine
cm 1.010 gm
3
cm 1.017 gm
3
0.80 0.75 0.80
turbine pump
0.75 0.75 0.80 0.75 0.75 0.75
1.010
Vliq
cm gm
3
cm 1.044 gm
3
0.78 0.78 0.80
0.0167
ft
3
lbm
3
ft 0.0161 lbm
269
80 100 Wdot 70 50 50 80
10 kW
3
10000 kPa 7000 kPa P1 8500 kPa 6500 kPa 950 psi 1125 psi P4
10 kPa 20 kPa 10 kPa 101.33 kPa 14.7 psi 1 psi
Wpump
Vliq P1
pump
P4
H4 Hliq
H1 H4 Wpump
S'3 = S2
x'3
S2 Svap
Sliq Sliq
H'3
Hliq
x'3 Hvap
Hliq
H3
H2
turbine H'3
H2
Wturbine
H3
H2
mdot
Wdot Wturbine Wpump
QdotH
QdotC
H2
QdotH
H1 mdot
Wdot
Answers follow:
mdot1 mdot2 mdot3 mdot4
QdotH
70.43 108.64 62.13 67.29 kg sec
1 2 3 4
240705 355111 kJ 213277 sec 205061
QdotH QdotH QdotH
QdotH QdotH
mdot5 mdot6
145.733 lbm 153.598 sec
5 6
192801 BTU 228033 sec
270
QdotC QdotC
1 2 3 4
0.332
160705 255111 kJ
0.282
Wdot
QdotH
0.328 0.244 0.246 0.333
QdotC QdotC
143277 sec 155061
QdotC QdotC
8.4
5 6
145410 BTU 152208 sec
Subscripts refer to Fig. 8.3. Saturated liquid at 50 kPa (point 4)
V4 cm 1.030 gm
3
H4
340.564
kJ kg
P4
P1
3300 kPa
50 kPa
Saturated liquid and vapor at 50 kPa: Hliq H4
kJ kg K
Hvap
2646.0
kJ kg
kJ kg K
P1
Wpump 3.348 kJ kg
Sliq
1.0912
Svap
Wpump
H1
7.5947
By Eq. (7.24),
V 4 P4
kJ kg
H1
H4
Wpump
343.911
The following vectors give values for temperatures of 450, 550, and 650 degC:
3340.6 H2 3565.3 3792.9
kJ kg S2
7.0373 7.3282 7.5891 kJ kg K
271
S'3
H'3
QH
S2
Hliq
H2 0.914
x'3
x'3 Hvap
H1
S'3 Svap
Sliq Sliq
H'3 H2
Wpump
Hliq
Wturbine
Wturbine
QH
0.297 0.314 0.332
Ans.
x'3
0.959 0.999
8.5
Subscripts refer to Fig. 8.3. Saturated liquid at 30 kPa (point 4)
V4 cm 1.022 gm
3
H4
289.302
kJ kg
P1
30 kPa
Saturated liquid and vapor at 30 kPa: Hliq H4 kJ kg K
Hvap 2625.4 kJ kg 7.7695 kJ kg K
5000 P4 7500 10000
kPa
Sliq
0.9441
Svap
By Eq. (7.24),
Wpump
V 4 P4
P1
kJ kg
294.381 H1 H4 Wpump H1 296.936 299.491
The following vectors give values for pressures of 5000, 7500, and 10000 kPa at 600 degC
3664.5 H2 3643.7 3622.7
272
7.2578 kJ kg S2 7.0526 6.9013 kJ kg K
S'3
H'3
QH
S2
Hliq
H2 0.925
x'3
x'3 Hvap
H1
S'3 Svap
Sliq Sliq
H'3 H2
Wpump
Hliq
Wturbine
Wturbine
QH
0.359 0.375 0.386
Ans.
x'3
0.895 0.873
8.6
From Table F.2 at 7000 kPa and 640 degC:
H1 3766.4 kJ kg
S1 7.2200 kJ kg K
S'2 S1
For sat. liq. and sat. vap. at 20 kPa:
Hliq
Sliq
251.453
kJ kg
Hvap
Svap
2609.9
kJ kg
kJ kg K
0.8321
kJ kg K
7.9094
The following enthalpies are interpolated in Table F.2 at four values for intermediate pressure P2:
725 P2 750 775 800
kPa
3023.9 H'2 3032.5 3040.9 3049.0
W12 H'2 H1
3187.3 H2
kJ 3200.5 kg kJ kg
0.78
579.15 W12
H2
H1
7.4939
W12
572.442 kJ 565.89 kg 559.572
3194
S2
7.4898 7.4851 7.4797
kJ kg K
3206.8
273
where the entropy values are by interpolation in Table F.2 at P2.
x'3 S2 Svap Sliq Sliq
H'3
W12
H'3
Hliq
x'3 Hvap
Hliq
W23
W
H2
W
20.817 7.811 5.073 17.723
kJ kg
W23
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
765.16 kPa
(P2)
Also needed are values of H2 and S2 at this pressure. Again we do linear interpolations:
linterp P2 H2 765.16 kPa
linterp P2 S2 765.16 kPa
3197.9
kJ kg
kJ kg K
H2
S2
3197.9
kJ kg
kJ kg K
7.4869
7.4869
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
H'3
H'3
H2
H1
kJ kg
Hliq
x'3
x'3
W23
W23
S2 Svap
0.94
Sliq Sliq
568.5
Hliq
2.469
x'3 Hvap
10
3 kJ
H'3
568.46
H2
kJ kg
kg
274
Work
W12
W23
Work
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
Sliq Sliq
H'3
Hliq
x'3 Hvap
Hliq
x'3
0.903
H'3
2.38
10
3 kJ
kg
W'
H'3
H1
W'
1386.2
kJ kg
Whence the overall efficiency is:
Work
overall
W'
overall
0.8202
Ans.
275
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
H3
2770.6
H2
kJ kg
H3
0.78
2.918 10
3 kJ
WI
kg
WI
H'3
521.586
H2
kJ kg
WI
Isentropic expansion to 20 kPa: S'4
Hliq
Sliq
S2
251.453 kJ kg
kJ
Exhaust is wet: for sat. liq. & vap.:
Hvap
Svap
2609.9
kJ kg
kJ kg K
0.8321
kg K
7.9094
276
x'4
x'4
H4
S'4 Svap
0.876
H2
Sliq Sliq
H'4
H'4
Hliq
2.317
x'4 Hvap
10
3 kJ
Hliq
kg
H'4
H2
3
H4
2.564
10
3 kJ
kg
H5
Hliq
V5
V 5 P6
1.017
P5
cm
gm
H6
H6
P5
20 kPa
P6
4500 kPa
Wpump
Wpump
H5
Wpump
kJ kg
5.841
kJ kg
257.294
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:
Hsat.liq kJ 558.5 kg
Psat 294.26 kPa
Vsat.liq 1.073 cm
3
gm
Also by approximation, the definition of the volume expansivity yields:
1 Vsat.liq
9.32 10
1.083 20
4 1
1.063
cm gm K
3
P1
P6
K
277
By Eq. (7.25),
H1
Hsat.liq
Vsat.liq 1
T1
P1
Psat
H1
561.305
kJ kg
By an energy balance on the feedwater heater:
mass
H1 H3
H6 H7
kg
mass
0.13028 kg
Ans.
Work in 2nd section of turbine:
WII
( kg 1
mass) H4
H3
WII
307.567 kJ
Wnet
WI
Wpump 1 kg
WII
Wnet
823.3 kJ
QH
H2
H1 1 kg
QH
2878 kJ
Wnet
QH
0.2861
Ans.
8.8
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
BTU lbm
0.78
WI
H'3
H2
H3
H2
WI
H3
1242.2
BTU lbm
WI
219.024
BTU lbm
S3
1.7431
BTU lbm rankine
278
Isentropic expansion to 1(psia):
S'4
S2
Exhaust is wet: for sat. liq. & vap.:
Hliq
69.73
BTU lbm
Hvap
1105.8
BTU lbm
Sliq
0.1326
BTU lbm rankine
Svap
1.9781
BTU lbm rankine
x'4
S'4 Svap
Sliq Sliq
H'4
Hliq
x'4 Hvap
Hliq
x'4
0.831
H'4
931.204
BTU lbm
H4
H2
H'4
H2
H4
1047.8
BTU lbm
x4
H4 Hvap
Hliq Hliq
S4
Sliq
x4 Svap
Sliq
x4
P5
0.944
S4
1.8748
BTU lbm rankine
V5 ft 0.0161 lbm
3
1 psi
H5
Hliq
Wpump
V 5 P6
P5
Wpump
2.489
BTU lbm
P6
650 psi
H6
H5
Wpump
H6
72.219
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
459.67 rankine
t1
270.01
279
At this temperature, 270.01 degF, interpolation in Table F.3 gives:
Psat
Hsat.liq
41.87 psi
238.96 BTU lbm
Vsat.liq
0.1717
ft
3
lbm
BTU lbm rankine
Ssat.liq
0.3960
The definition of the volume expansivity yields:
1 Vsat.liq
4.95 10
0.01726 20
5
0.01709
ft
3
lbm rankine
P1
P6
1 rankine
BTU lbm
BTU lbm rankine
By Eq. (7.25) and (7.26),
H1
S1
Hsat.liq
Ssat.liq
Vsat.liq 1
Vsat.liq P1
T1
Psat
P1
Psat
H1
S1
257.6
0.397
By an energy balance on the feedwater heater:
mass
H1 H3
H6 H7
lbm
mass
0.18687 lbm Ans.
Work in 2nd section of turbine:
WII
Wnet
QH
1 lbm
WI
H2
Wnet
QH
mass
H4
H3
WII
WII
Wnet
QH
158.051 BTU
374.586 BTU
1.204 10 BTU
3
Wpump 1 lbm
H1 1 lbm
0.3112
Ans.
280
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
H'3
H3
S2
3142.6
H2
By double interpolation in Table F.2,
kJ kg
H3
H10
0.80
3.244 10
3 kJ
WI
WI
H'3
407.6
H2
kJ kg
WI
kg
From Table F.1:
829.9
kJ kg
281
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
Sliq
kJ 251.453 kg
0.8321 kJ kg K
Hvap
Svap
2609.9
kJ kg
kJ kg K
Vliq
1.017
cm
3
gm
7.9094
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
H6
60.09
Hliq
V 6 P2 P6
Tsat
V6
tsat
Vliq
273.15 K
P6 20 kPa
Wpump
Wpump
[Eq. (7.24)]
8.238
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 20
10
4 1
1.012
cm
3
gm K
CP
272.0 10
4.18
230.2
kJ kg K
K
CP
kJ kg K
Solving Eq. (7.25) for delta T gives:
T67
H67
Vliq 1 CP
Tsat
P2
P6
T67
0.678 K
t7
tsat
T67 K
t9
190 2
282
t7
t7
t8
t9
5
t7
60.768
t8
130.38
From Table F.1:
H8
547.9
kJ kg
H7
Hliq
H67
t9
125.38
T9
273.15
t9 K
H7
259.691
kJ kg
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
Vsat.9
cm 1.065 gm
3
Psat.9
234.9 kPa
Hsat.1
V9
kJ 807.5 kg
( 1.075
Vsat.1
3
cm 1.142 gm
3
Psat.1
( 1.156
1255.1 kPa
3
cm 1.056) gm
V1
cm 1.128) gm
T
20 K
1
9
V9 T
1
1 Vsat.1
V1 T
Vsat.9
9
8.92
10
4 1
K
1
1.226
10
3 1
K
H9
Hsat.9
Vsat.9 1
9 T9
P2
Psat.9
H9
530.9
kJ kg
T1
( 273.15
190)K
T1
1 T1
463.15 K
H1
Hsat.1
Vsat.1 1
P2
Psat.1
H1
810.089
kJ kg
Now we can make an energy balance on feedwater heater I to find the mass of steam condensed:
mI
H1 H3
H9 H10
kg
283
mI
0.11563 kg
Ans.
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 kg
Then
H4
H4
H2
2.941
H'4
10
3 kJ
H2
kg
We can now make an energy balance on feedwater heater II to find the mass of steam condensed:
mII
H9
H7 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
Then
S2 Svap
0.889
Sliq Sliq
H'5
H'5
Hliq
2.349
x'5 Hvap
10
3 kJ
Hliq
kg
H5 2609.4 kJ kg
H5
H2
H'5
H2
The work of the turbine is:
Wturbine
WI 1 kg 1 kg mI H4 H3 1 kg mI mII H5 H4
936.2 kJ
QH H2 H1 1 kg
QH 2.842 10 kJ
3
Wturbine
Wturbine
Wpump 1 kg
QH
0.3265
Ans.
284
8.10
Isobutane:
Tc
408.1 K
Pc
36.48 bar
0.181
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
533.15 K
P0
4800 kPa
P
450 kPa
S
0
J mol K
For the heat capacity of isobutane:
A
1.677
B
37.853 10 K
3
C
11.945 10 K
2
6
Tr0
T0 Tc
Tr0
1.3064
Pr0
Pr
P0 Pc
P Pc
Pr0
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
C T0
2
1 2
1
ln
P P0
Pr
SRB Tr0 Pr0
Find
0.852
T
T0
T
454.49 K
Tr
T Tc
Tr
1.114
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
3
11.945 10
6
0.0
Hig
1.141
10
4 J
mol
285
Hturbine
Hturbine
Hig
8850.6
R Tc HRB Tr Pr
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
Avp
450 kPa
14.57100
Bvp
Cvp
2606.775
Cvp
274.068
tsat Avp
Bvp VP ln kPa
3
tsat
34
Tsat
Tsat
tsat
273.15 K
307.15 K
Vc
cm 262.7 mol
Zc
2
0.282
Trsat
Tsat Tc
3
Trsat
0.753
Vliq
Wpump
V c Zc
1 T rsat
7
Vliq
P
Wpump
cm 112.362 mol
488.8
Vliq P0
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
Ha
Ha
1.756
Hig
18082
10
4 J
mol
HRB Tr Pr
R Tc HRB Trsat Pr
J mol
For the condensation process, we estimate the latent heat by Eqs. (4.12) and (4.13):
Tn
261.4 K
Pc bar
Trn
0.38
Trn
1.013
Tn Tc
Trn
0.641
R Tn 1.092 ln Hn
0.930
Hn
2.118
10
4 J
mol
Hb
Qdotout
Hn
1 1
Trsat Trn
Ha
Hb
Hb
Qdotout
18378
J mol
mdot
Qdotin
Qdotout
Wturbine
4360 kW
Wpump mdot
Qdotin
1000 kW Qdotin
0.187
Ans.
5360 kW
8.11 Isobutane: Tc
408.1 K
Pc
36.48 bar
0.181
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 J mol K
P0
3400 kPa
P
450 kPa
molwt
58.123
gm mol
287
For the heat capacity of isobutane: A 1.677
T0 Tc
B
37.853 10 K
1.0124
3
C
11.945 10 K
2
6
Tr0
Tr0
Pr0
Pr
P0 Pc
P Pc
Pr0
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: 0.8
Given (guess)
S = R A ln
SRB T0 Tc
B T0
Pr
C T0
2
1 2
1
ln
P P0
SRLK0
Find
0.809
T
Tr
T0
T Tc
T
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
Hig
Hturbine
Hturbine
R ICPH T0 T 1.677 37.853 10
9.3 10
3 J
3
11.945 10
6
0.0
mol
R Tc HRB Tr Pr
J mol
288
Hig
4852.6
HRLK0
Wturbine 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:
Vliq 112.36 cm
3
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
5834 kW
Wpump
Ans.
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
Hig
307.15K
Trsat
Tsat Tc
3
Trsat
11.945 10
0.753
6
R ICPH T Tsat 1.677 37.853 10
2.817 kJ mol
R Tc HRB Trsat Pr
J mol
0.0
Ha
Ha
Hig
2975
HRB Tr Pr
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
33387 kW Ans.
Ans.
Wdot Qdotin
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 0.8
mdot W'turbine W'pump
W'pump
Wdot
414.3
J mol
Ans.
4475 kW
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
28805 kW
Wturbine
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 W'pump Wpump mdot
0.134
Ans.
Qdotin
33280 kW
Ans.
Wdot Qdotin
290
8.13 Refer to Fig. 8.10.
CP
7 R 2
PC
1 bar
TC
293.15 K
PD
5 bar
1.4
By Eq. (3.30c):
1
PC V C = PD V D
1
VC VD
=
PD PC
or
r
1
PD PC
r
3.157
Ans.
Eq. (3.30b):
TD
TC
PD PC
TA QDA CP
QDA
1500
J mol
QDA = CP TA
TD
R TC
TD
TA
515.845 K
re =
VB VA
=
VC VA
=
PC
R TA PA
PA
PD
re
T C PA T A PC
re 8.14
2.841
Ans.
3 Ratio 5 7 9
Eq. (8.12) now becomes:
1
Ratio =
PB PA
1.35
0.248 0.341 0.396 0.434
Ans.
1
1 Ratio
291
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)
R CP
2
WAB = CP TA
PB PA
1 = CP TA cr
7
1
WCD = CP TC
PD PC
R CP
2
1 = CP TC er
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
2 7
er
0.5 (guess)
2 7
Given
TC er
1 = TA cr
1
er
er
Find er) (
0.552
292
By Eq. (7.18):
TD = TC
PD PC
2 7
R CP
This may be written:
TD
TC er
1
By Eq. (7.11)
uE
2
uD =
2
2
PD V D 1
1
PE PD
(A)
We note the following:
er =
PD PC
cr =
PB PA
=
PC PE
cr er =
PD PE
The following substitutions are made in (A):
uD = 0
1
=
2 R = 7 CP
PD V D = R T D
molwt 29
PE PD
gm mol
=
1 cr er
Then
2 7
uE
2
R 7 TD 1 2 molwt
1 cr er
uE
843.4
m sec
Ans.
PE
1 bar
PD
cr er PE
PD
3.589 bar
Ans.
8.17 TA
305 K
PA
1.05bar
PB
7.5bar
0.8
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
3
0.0
0.016 10
5
R
Cpmair
29.921
J mol K
293
Compressor:
R Cpmair
Wsair
TB
Cpmair TA
Wsair Cpmair
PB PA
1
Wsair
8.292
10
3 J
mol
TA
TB
582.126 K
Combustion:
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
HR
HR
1000K
N
57.638 (This is the final value after iteration)
582.03)K 4.217 R ( 298.15 300)K
Cpmair N ( 298.15
4.896 10
5 J
mol
The product stream contains: 1 mol CO2, 2mol H2O, 0.79N mol N2, and (0.21N-2) mol O2
1 n 2 .79 N .21 N
i 1 4
5.457 A 2 3.470 3.280 3.639 B
1.045 1.450 0.593 0.506
10
3
1.157 D 0.121 0.040 0.227
10
5
294
ni
i
58.638
A
i
ni Ai
B
i
ni Bi
D
i
ni D i
5
A
CpmP
HP
198.517
B
0.036
D
1.387
5
10
R
MCPH 298.15K 1000.K 198.517 0.0361 0.0
CpmP TC 298.15K
HP 1.292
J mol
1.3872 10
10
6 J
mol
From Ex. 4.7:
H298
HP 136.223
802625
HR
H298
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
1.3872 10
5
R
J mol K
For 58.638 moles of combustion product:
R Cpm
Ws
58.638 Cpm TC
PD PC
1
Ws
1.214 10
6 J
mol
TD
TC
Ws Cpm
TD
343.123 K
295
(Final result of iteration.) Ans.
Wsnet
Ws
Wsair N
Wsnet
7.364
10
5 J
mol
Ans.
(J per mole of methane) Parts (b) and (c) are solved in exactly the same way, with the following results: (b) TC
1200
N
37.48
Wsnet
7.365 10
5
TD
5
343.123
(c)
TC
1500
N
24.07
Wsnet
5.7519 10
TD
598.94
8.18
tm
0.35
me
0.95
line_losses
20% Cost_fuel
4.00
dollars GJ
Cost_electricity
tm
Cost_fuel 1 me ( line_losses)
Cost_electricity
0.05
cents kW hr
Ans.
This is about 1/2 to 1/3 of the typical cost charged to residential customers.
8.19 TC
111.4K
TH
300K
Hnlv
HE
8.206
Carnot
1
TC TH
Carnot
kJ mol
HE
0.629
0.6
Carnot
0.377
Assume as a basis:
W
1kJ
QH
QC Hnlv
W
HE
QH
2.651 kJ
QC
QH 1
HE
QC
1.651 kJ
W
0.201
mol kJ
Ans.
296
8.20 TH
( 27
273.15)K
TC
(6
273.15)K
a) Carnot
1
TC TH
Carnot
0.07
Ans.
b) actual
Carnot 0.6
2 3
actual
0.028
Ans.
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
273.15) K
TH
293.15 K
TC
( 20
273.15) K
TC
253.15 K
QdotC
Carnot
125000
kJ day
TC TH TC
(9.3)
0.6
Carnot
3.797
Wdot
QdotC
(9.2)
Wdot
0.381 kW
Cost
0.08 kW hr
Wdot
Cost
267.183
dollars yr
Ans.
9.4
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
0.21868
P1
P2
138.83
138.83
State 2, Sat. Vapor at TH: H2
State 3, Wet Vapor at TC: Hliq
116.166 S2
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 S2 Sliq Svap Sliq
x3
0.971
x4
S1 Sliq Svap Sliq
x4
0.302
(c) Heat addition, Step 4--3:
H3
Hliq
x3 ( Hvap Hliq)
H4
Hliq
x4 ( Hvap Hliq)
H3
101.888
H4
42.118
Q43
( H3
H4)
298
Q43
59.77
(Btu/lbm)
(d) Heat rejection, Step 2--1:
Q21
( H1
H2)
Q21
71.223
(Btu/lbm)
(e)
W21
0
W43
0
W32
( H2
H3)
W32
14.278
W14
( H4
H1)
W14
2.825
(f)
Q43 W14 W32
5.219
Note that the first law is satisfied:
Q
Q21
Q43
W
W32
W14
Q
W
0
9.7
TC
298.15 K
TH
523.15 K
(Engine)
(Refrigerator)
T'C
273.15 K
T'H
298.15 K
1 TC TH
By Eq. (5.8):
Carnot
Carnot
0.43
By Eq. (9.3):
T'C
Carnot
T'H
T'C
Carnot
10.926
By definition:
=
Wengine
QH
=
Q'C
Q'C Wrefrig
35 kJ sec
But
Wengine = Wrefrig
QH Q'C
Carnot Carnot
Whence
QH
7.448
kJ sec
Ans.
Given that:
0.6
Carnot
0.6
Carnot
6.556
QH
Q'C
QH
20.689
kJ sec
Ans.
299
9.8
(a) QC
4
kJ sec
W
1.5 kW
QC W
2.667
Ans.
(b) QH
QC
W
QH
5.5
kJ sec
Ans.
(c)
=
TC TH TC
TH
( 40
273.15)K
TH
313.15 K
TC
TH
1
TC
227.75 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.
489.67 479.67 T2 469.67 rankine 459.67 449.67
0.79 0.78 0.77 0.76 0.75 QdotC
600 500 400 300 200
Btu sec
107.320 105.907 H2 104.471 103.015 101.542
Btu lbm
0.22244 0.22325 S2 0.22418 0.22525 0.22647
H4 37.978 Btu lbm
Btu lbm rankine
T4
539.67 rankine
From Table 9.1 for sat. liquid
S'3 = S2
(isentropic compression)
300
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 H'3 116.5 117.2 117.9
Btu lbm
H23
H'3
H2
H3
H1
H2
H4
H23
24.084 30.098
kJ kg
273.711 276.438 H3 279.336 283.026 286.918 8.653
kJ kg
H1
88.337
kJ kg
H23
36.337 43.414 50.732
mdot
QdotC H2 H1
7.361 mdot 6.016 4.613 3.146
lbm sec
Ans.
689.6 595.2
Btu sec
QdotH
mdot H4
H3
QdotH
494 386.1 268.6 94.5 100.5
Ans.
Wdot
mdot H23
Wdot
99.2 90.8 72.4
kW
Ans.
301
6.697
QdotC Wdot
5.25 4.256 3.485 2.914
Ans.
TC
T2
TC
TH
T4
9.793 7.995
Carnot
TH
TC
Carnot
6.71 5.746 4.996
Ans.
9.10
Subscripts in the following refer to Fig. 9.1. All property values come from Tables F.1 and F.2. T2 ( 4 273.15)K
1200
kJ kg
T4
H2
( 34
273.15)K
kJ kg
S2
0.76
9.0526 kJ kg K
QdotC
H4
kJ sec
2508.9
142.4
S'2 = S2
(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
kJ kg
H1
H4
402.368
302
H3
mdot
H2
H23
QdotC
H3
mdot
2.911
10
3 kJ
kg
H2
H1
H3
0.507
kg Ans. sec
kJ Ans. sec
Ans.
QdotH
Wdot
mdot H4
mdot H23
QdotH
Wdot
1404
204 kW
Ans.
QdotC Wdot
T2
Carnot
5.881
T4
T2
Carnot
9.238
Ans.
9.11
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
0.01733
Btu lbm rankine
Svap
0.22714
Btu lbm rankine
For sat. liquid at Point 4 (80 degF):
H4
37.978
Btu lbm
S4
0.07892
Btu lbm rankine
(a) Isenthalpic expansion:
H1
H4
QdotC
5
Btu sec
mdot
QdotC H2 H1
mdot
0.0796
lbm sec
Ans.
(b) Isentropic expansion:
S1
S4
x1
S1 Svap
Sliq Sliq
H1
Hliq
x1 Hvap
Hliq
H1
34.892
BTU lbm
mdot
QdotC H2 H1
mdot
0.0759
lbm sec
Ans.
(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
117.5
Btu lbm
S2A
0.262
Btu lbm rankine
mdot
QdotC H2A H4
mdot
0.0629
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 H2
a
304
H3
3.5896
Ans.
In the second case H1 has its last calculated value [Part (b)]:
H2
b
H1 H2
b
H3
3.7659
Ans.
In Part (c), compression is at constant entropy of 0.262 to the final pressure. Again from Fig. G.2:
H3 138 Btu lbm
Wdot
Wdot
H3
1.289
H2A mdot
BTU sec
(Last calculated value of mdot)
QdotC
c
Wdot
c
3.8791
Ans.
9.12
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
Btu lbm
S2A
0.2435
Btu lbm rankine
Btu lbm R
For sat. liquid at Point 4 (80 degF):
H4 37.978
S4 0.07892
Energy balance, heat exchanger: H1 H4 H2A H2
H1 27.885 BTU lbm
mdot 25.634 lbm sec
QdotC
2000
Btu sec
mdot
QdotC H2 H1
305
For compression at constant entropy of 0.2435 to the final pressure of 101.37(psia), by Fig. G.2:
H'3 127 Btu lbm
0.75
Hcomp H'3 H2A
Wdot
mdot Hcomp
25.634 lbm sec
Wdot 396.66 kW
Hcomp
Ans.
14.667
Btu lbm
mdot
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 QdotC H2 H4
H'3 116 Btu lbm
Hcomp
Hcomp
H'3
H2
Wdot
mdot Hcomp
lbm sec
13.457
Btu lbm
mdot
29.443
Wdot
418.032 kW
Ans.
9.13
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
Btu lbm
113.3 H'3 116.5 119.3
Btu lbm H1 H4
306
(a)
By Eq. (9.4):
H2 H'3
H1 H2
8.294 5.528 4.014
H2
Since H = H3 H2
Ans.
(b)
H
H'3
0.75
Eq. (9.4) now becomes
H2
H1 H
6.221 4.146 3.011
Ans.
9.14
WINTER
TH
293.15
Wdot
1.5
TC
QdotH = 0.75 TH
TH TC Wdot = QdotH TH
TC
Given
250
(Guess)
TH TC Wdot = TH 0.75 TH TC
TC
TC
Find TC
268.94 K
Ans.
Minimum t = -4.21 degC
307
SUMMER
TC
QdotC
298.15
0.75 TH TC
TH TC Wdot = TC QdotC
TH
300
(Guess)
Given
Wdot 0.75 TH TC
=
TH
TC
TC
TH
Find TH
TH
322.57 K
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
H9
284.7
kJ kg
H15
1186.7 1056.4
kJ kg
By Eq. (9.8):
z
H4 H9
H15 H15
z
0.17 0.351
Ans.
9.17
Advertized combination unit:
TH
( 150
459.67)rankine
TC
( 30
459.67)rankine
TH
609.67 rankine
TC
489.67 rankine
QC
50000
Btu hr
WCarnot
QC
308
TH
TC
TC
WCarnot
12253
Btu hr
WI
1.5 WCarnot
WI
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
QC
QH
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
TH
TC
TC
WCarnot
9190
Btu hr
Work
1.5 WCarnot
Work
13785
Btu hr
The total power required is
WII
QH
Work
WII
82165
Btu hr
NO CONTEST
9.18
TC
210
T'H
260
T'C
255
TH
305
By Eq. (9.3):
TC TH TC
I
0.65
TC T'H TC
II
0.65
T'C TH T'C
WCarnot =
QC
WI =
QC
I
WII =
QC
II
Define r as the ratio of the actual work, WI + WII, to the r Carnot work:
9.19
1
I
1
II
r
1.477
Ans.
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
290K
TC
250K
Carnot
Ws
0.40kW
TC
Carnot
TH
TC
6.25
3 2
65% Carnot
4.063 Ans.
QC
9.23
Ws
QC
1.625
-3 Q 10 kgm sec H
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
273.15) K
T4
( 30
273.15) K
T2
T1
Calculate the high and low operating pressures using the given vapor pressure equation Guess:
PL
1bar
PH
4104.67
T1 K
2bar
PL
Given ln
PL bar
= 45.327
5.146 ln
T1 K
615.0
bar
T1 K
2
PL
Find PL
PL
6.196 bar
PH
Given ln
PH bar
= 45.327
4104.67
T4 K
5.146 ln
T4 K
615.0
bar
T4 K
2
PH
Find PH
PH
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
ndottoluene R ICPH T1 T2 15.133 6.79 10
3
16.35 10
6
0
QdotC
177.536 kW
310
Since the throttling process is adiabatic:
H4 = H1
But:
Hliq4 = Hliq1
x1 Hlv1 so:
Hliq4
T4
Hliq1 = x1 Hlv
and:
Hliq4
Hliq1 = Vliq P4
P1
T1
Cpliq ( T) dT
Estimate Vliq using the Rackett Eqn.
0.253
Tc
405.7K
Pc
3
112.80bar
Zc
0.242
Vc
cm 72.5 mol
Tn
239.7K
Hlvn
23.34
kJ mol
Tr
( 20
273.15)K Tc
2
Tr
0.723
3
Vliq
V c Zc
1 Tr
7
Vliq
cm 27.112 mol
Estimate Hlv at 10C using Watson correlation T1 Tn Tr1 Trn 0.591 Trn Tc Tc
Tr1
kJ mol
0.698
Hlv
Hliq41
Hlvn
Vliq PH
1 1
PL
Tr1 Trn
0.38
Hlv
20.798
R ICPH T1 T4 22.626
100.75 10
3
192.71 10
6
0
Hliq41
1.621
kJ mol
x1
Hliq41 Hlv
x1
0.078
For the evaporator
H12 = H2
H1 = H1vap
H1liq
x1 Hlv = 1
x1
Hlv
H12
ndot
1
x1
Hlv
H12
ndot
311
19.177
kJ mol
Ans.
QdotC H12
9.258
mol sec
Chapter 10 - Section A - Mathcad Solutions
10.1 Benzene: Toluene: A1 := 13.7819 A2 := 13.9320
A1 B1 T degC +C1
B1 := 2726.81 B2 := 3056.96
A2 B2 T degC
C1 := 217.572 C2 := 217.625
Psat1 ( T) := e
kPa
Psat2 ( T) := e Guess:
+C2
kPa
(a) Given: x1 := 0.33 Given
T := 100 degC
y1 := 0.5
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
y1 = 0.545
Ans.
P = 109.303 kPa
Ans.
T := 100 degC
Guess:
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
Ans.
P = 92.156 kPa
Ans.
P := 120 kPa
Guess:
y1 := 0.5
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 Given
P := 120 kPa
Guess: x1 := 0.33
T := 100 degC
x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P 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 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P x1 Psat1 ( T) = y1 P
x1 := 0.33
y1 := 0.5
x1 := Find ( x1 , y1) y1
(f) z1 := 0.33 Guess: Given
x1 = 0.282
Ans.
y1 = 0.484
Ans.
x1 = 0.282 L := 0.5 z1 = L x1 + V y1 L+V = 1
y1 = 0.484 V := 0.5
L := Find ( L , V) V
(g)
Vapor Fraction: Liquid Fraction:
V = 0.238 L = 0.762
Ans. 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 A2 := 13.9726 Psat1 ( T) := exp A1 B1 := 2726.81 B2 := 3259.93 C1 := 217.572 C2 := 212.300
B2 Psat2 ( T) := exp A2 T + C2
P-x-y diagram: T := 90 y1 ( x1) := x1 Psat1 ( T) P ( x1 ) P ( x1) := x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) T-x-y diagram: P' := 90
B1 T + C1
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) ) + ( 1 x1) Psat2 ( T ( x1) ) x1 Psat1 ( T ( x1) )
x1 := 0 , 0.05 .. 1.0
150 140 130 P ( x1 ) P ( x1 ) 50 100 120 T ( x1 ) 110 100 T ( x1 ) 90 80 70 0 0 x1 , y1 ( x1 ) 0.5 1 60 0 x1 , y'1 ( x1 ) 0.5 1
314
(b) Antoine coefficients: 1-Chlorobutane=1; Chlorobenzene=2 A1 := 13.7965 A2 := 13.8635 Psat1 ( T) := exp A1 B1 := 2723.73 B2 := 3174.78 B1 T + C1 C1 := 218.265 C2 := 211.700 Psat2 ( T) := exp A2
B2 T + C2
P-x-y diagram:
T := 90 y1 ( x1) := x1 Psat1 ( T) P ( x1 )
P ( x1) := x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) T-x-y diagram: 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) ) + ( 1 x1) Psat2 ( T ( x1) ) x1 Psat1 ( T ( x1) )
x1 := 0 , 0.05 .. 1.0
160 130 122.5 115 113.33 P ( x1 ) P ( x1 ) 66.67 107.5 T ( x1 ) 100 T ( x1 ) 92.5 85 77.5 20 0 0.5 1 70 0 0.5 1
x1 , y1 ( x1 )
x1 , y'1 ( x1 )
315
10.3
Pressures in kPa; temperatures in degC (a) Antoine coefficinets: n-Pentane=1; n-Heptane=2 A1 := 13.7667 A2 := 13.8622 Psat1 ( T) := exp A1 B1 := 2451.88 B2 := 2911.26 C1 := 232.014 C2 := 216.432 Psat2 ( T) := exp A2
B1 T + C1
B2 T + C2
T := 55
P :=
Psat1 ( T) + Psat2 ( T) 2
x1 Psat1 ( T) P
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 := y1 = 0.89
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 z1 := x1 , x1 + 0.01 .. y1 V is obviously linear in z1:
1 x1 y1
V ( z1) :=
z1 x1 y1 x1
V ( z1 ) 0.5
0 0.45
0.5
0.55
0.6
0.65 z1
0.7
0.75
0.8
0.85
316
(b) At fixed T and z1, calculate x1, y1 and P as functions of fraction vapor (V). z1 := 0.5 Guess: Given x := 0.5 y := 0.5
p :=
Psat1 ( T) + Psat2 ( T) 2
Three equations relate x1, y1, & P for given V: 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 P ( V) := f ( V) 3 V := 0 , 0.1 .. 1.0
1
Plot P, x1 and y1 vs. vapor fraction (V)
150
100 P ( V) 50
x1 ( V) y1 ( V) 0.5
0
0
0.5 V
1
0
0
0.5 V
1
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. Benzene: A1 := 13.7819 B1 := 2726.81 B2 := 3259.93
A2
10.7
C1 := 217.572 C2 := 212.300
B2 T degC +C2
Ethylbenzene A2 := 13.9726
A1 B1 T degC +C1
Psat1 ( T) := e
kPa
317
Psat2 ( T) := e
kPa
(a) Given: x1 := 0.35 Given
y1 := 0.70 Guess: T := 116 degC P := 132 kPa
x1 Psat1 ( T) + ( 1 x1) Psat2 ( T) = P x1 Psat1 ( T) = y1 P
T := Find ( T , P) P
T = 134.1 degC
Ans.
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 (c) T = 91.44 deg_C P = 118.72 kPa P = 66.38 kPa P = 36.02 kPa
(d) T = 72.43 deg_C
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
2726.81 B := 3056.96 3259.93
217.572 C := 217.625 212.300
zi := 1 n
i := 1 .. n
Bi T degC +Ci
T := 110 degC
P := 90 kPa
Psat ( i , T) := e
kPa
ki :=
Psat ( i , T) P
Guess:
V := 0.5
318
Given
i=1
n
1 + V ( ki 1) V = 0.836
zi ki
= 1
Eq. (10.17)
V := Find ( V)
Ans.
yi :=
1 + V ( ki 1) yi P Psat ( i , T)
zi ki
Eq. (10.16)
0.371 y = 0.339 0.29 0.142 x = 0.306 0.552 0.188 x = 0.334 0.478 0.238 x = 0.345 0.417 0.293 x = 0.342 0.366
Ans.
xi :=
Ans.
(b) T = 110 deg_C P = 100 kPa (c) T = 110 deg_C P = 110 kPa
V = 0.575
0.441 y = 0.333 0.226 0.508 y = 0.312 0.18 0.572 y = 0.284 0.144
V = 0.352
(d) T = 110 deg_C P = 120 kPa
V = 0.146
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) z1 := 0.75 z2 := 1 z1
Ai Bi T degC +Ci
i := 1 .. n T := ( 340 273.15) degC P := 115 kPa
Psat ( i , T) := e Guess: Given
kPa
ki :=
Psat ( i , T) P
V := 0.5
i=1
n
1 + V ( ki 1) V = 0.656 yi :=
zi ki
= 1 Ans.
Eq. (10.17)
V := Find ( V) Eq. (10.16)
1 + V ( ki 1) yi P Psat ( i , T) y1 V z1 y1 = 0.678 y1 = 0.320 y1 = 0.682
zi ki
y1 = 0.805
Ans.
xi :=
x1 = 0.644
Ans.
r := (b) (c) (d) x1 = 0.285 x1 = 0.183 x1 = 0.340
r = 0.705 V = 0.547 V = 0.487 V = 0.469
Ans. r = 0.741 r = 0.624 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 x1 + x2 = 1 Solve for x1 and y1: x1 := P Psat2 H1 Psat2
3
y2 P = x2 Psat2 P = H1 x1 + ( 1 x1) Psat2
P = y1 P + y2 P
y1 :=
H1 x1 P Ans.
x1 = 4.502 10
y1 = 0.9
10.16 Pressures in kPa Psat1 := 32.27 Psat2 := 73.14 A := 0.67 2 ( x1 , x2) := exp A x1 z1 := 0.65 1 ( x1 , x2) := exp A x2
(
2
)
(
2
)
P ( x1 , x2) := x1 1 ( x1 , x2) Psat1 + x2 2 ( x1 , x2) Psat2 (a) BUBL P calculation: Pbubl := P ( x1 , x2) DEW P calculation: Guess: Given x1 := 0.5 x1 := z1 Pbubl = 56.745 y1 := z1 P' := x2 := 1 x1 Ans. y2 := 1 y1
Psat1 + Psat2 2
y1 P' = x1 1 ( x1 , 1 x1) Psat1 P' = x1 1 ( x1 , 1 x1) Psat1 ... + ( 1 x1) 2 ( x1 , 1 x1) Psat2
x1 := Find ( x1 , P') Pdew
321
Pdew = 43.864
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) := P ( x1 , 1 x1) x1 := 0.75 x2 := 1 x1
x1 1 ( x1 , 1 x1) Psat1
The fraction vapor, by material balance is: V := y1 ( x1) x1 z1 x1 V = 0.379 P ( x1 , x2) = 51.892 Ans.
(c) See Example 10.3(e). 12.0 := 1 ( 0 , 1) Psat1 Psat2 12.1 := Psat1 2 ( 1 , 0) Psat2
12.0 = 0.862
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 A := 0.95 2 ( x1 , x2) := exp A x1
1 ( x1 , x2) := exp A x2
(
2
)
(
2
)
P ( x1 , x2) := x1 1 ( x1 , x2) Psat1 + x2 2 ( x1 , x2) Psat2 y1 ( x1) := x1 1 ( x1 , 1 x1) Psat1 P ( x1 , 1 x1) x1 := 0.05 Pbubl = 47.971 y1 ( x1) = 0.196 (b) DEW P calculation: Guess: x1 := 0.1 y1 := 0.05 P' := x2 := 1 x1 Ans. y2 := 1 y1 Psat1 + Psat2 2
(a) BUBL P calculation: Pbubl := P ( x1 , x2)
322
Given
y1 P' = x1 1 ( x1 , 1 x1) Psat1 P' = x1 1 ( x1 , 1 x1) Psat1 ... + ( 1 x1) 2 ( x1 , 1 x1) Psat2
x1 := Find ( x1 , P') Pdew
(c) Azeotrope Calculation: Guess: Given y1 = x1 := 0.8
Pdew = 42.191 x1 = 0.0104
Ans.
y1 := x1
P := x1 0
Psat1 + Psat2 2 x1 1 x1 = y1
x1 1 ( x1 , 1 x1) Psat1 P
P = x1 1 ( x1 , 1 x1) Psat1 + ( 1 x1) 2 ( x1 , 1 x1) Psat2
xaz1 yaz := Find ( x , y , P) 1 1 1 Paz
10.18 Psat1 := 75.20 kPa At the azeotrope: Therefore ln1 = A x2
2
xaz1 0.857 yaz = 0.857 1 Paz 81.366
Ans.
Psat2 := 31.66 kPa y1 = x1 2 1 = Psat1 Psat2
2
and x1 := 0.294 ln
i =
P Psati x2 := 1 x1
ln2 = A x1 ln
2 1
= A x1 x2
(
2
2
)
Psat1 Psat2
2 2
Whence For
A := x1 := 0.6
A = 2.0998
x2 x1
x2 := 1 x1
323
1 := exp A x2 y1 :=
(
2
)
2 := exp A x1 P = 90.104 kPa Psat1 := 1.24 x1 := 0.65
(
2
)
P := x1 1 Psat1 + x2 2 Psat2 y1 = 0.701 Psat2 := 0.89 x2 := 1 x1 2 := exp A x1 y1 := Ans.
x1 1 Psat1 P
10.19 Pressures in bars: A := 1.8 1 := exp A x2
(
2
)
P = 1.671
(
2
)
P := x1 1 Psat1 + x2 2 Psat2 y1 = 0.6013 By a material balance, V= (c) Guess: z1 x1 y1 x1 For 0V 1
x1 1 Psat1 P
Answer to Part (b)
0.6013 z1 0.65
Ans. (a)
Azeotrope calculation: x1 := 0.6 y1 := x1 P := Psat1 + Psat2 2
2 1 ( x1) := exp A ( 1 x1)
2 ( x1) := exp A x1
(
2
)
Given
P = x1 1 ( x1) Psat1 + ( 1 x1) 2 ( x1) Psat2 x1 0 x1 1 x1 = y1
y1 =
x1 1 ( x1) Psat1 P
x1 y1 := Find ( x1 , y1 , P) P
x1 0.592 y1 = 0.592 P 1.673
Ans.
324
10.20 Antoine coefficients: Acetone(1): Methanol(2): A1 := 14.3145 A2 := 16.5785
P in kPa; T in degC B1 := 2756.22 B2 := 3638.27 C1 := 228.060 C2 := 239.500
P1sat ( T) := exp A1
B1 T + C1
P2sat ( T) := exp A2
B2 T + C2
A := 0.64
x1 := 0.175
z1 := 0.25
p := 100 (kPa)
1 ( x1 , x2) := exp A x2
(
2
)
2 ( x1 , x2) := exp A x1
(
2
)
P ( x1 , T) := x1 1 ( x1 , 1 x1) P1sat ( T) ... + ( 1 x1) 2 ( x1 , 1 x1) P2sat ( T) y1 ( x1 , T) := Guesses: Given F= L+V z1 F = x1 L + y1 ( x1 , T) V p = P ( x1 , T) x1 1 ( x1 , 1 x1) P1sat ( T) P ( x1 , T )
F := 1 T := 100
V := 0.5
L := 0.5
L V := Find ( L , V , T) T
T = 59.531 (degC)
L 0.431 V = 0.569 T 59.531
y1 ( x1 , T) = 0.307 Ans.
325
10.22 x1 := 0.002 A1 := 10.08
A1
y1 := 0.95 B1 := 2572.0
B1
Guess:
T := 300 K B2 := 6254.0
A2 B2
A2 := 11.63
Psat1 ( T) := e x2 := 1 x1 Given
T K bar
y2 := 1 y1 = x2 2 y1 x1 1 y2
Psat2 ( T) := e 1 := e
0.93 x2
2
T K bar
2 := e
0.93 x1
2
Psat1 ( T) Psat2 ( T)
T := Find ( T)
T = 376.453 K
Ans.
P :=
x1 1 Psat1 ( T) y1
P = 0.137 bar
Ans.
326
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) 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
Component methane ethylene ethane
xi 0.100 0.500 0.400
b) DEW P
T=-60 F (-51.11 C) 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
Component methane ethylene ethane
yi 0.500 0.250 0.250
c) BUBL T
P=250 psia (17.24 bar) 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 P=250 psia (17.24 bar) 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
Component methane ethylene ethane
d) DEW T
Component methane ethylene ethane
yi 0.430 0.360 0.210
327
Problem 10.26 a) BUBL P T=60 C (140 F) 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
Component ethane propane isobutane isopentane
xi 0.10 0.20 0.30 0.40
b) DEW P
T=60 C (140 F) 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
Component ethane propane isobutane isopentane
yi 0.48 0.25 0.15 0.12
c) BUBL T
P=15 bar (217.56 psia) 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
Component ethane propane isobutane isopentane
xi 0.14 0.13 0.25 0.48
d) DEW T
P=15 bar (217.56 psia) 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
Component ethane propane isobutane isopentane
yi 0.42 0.30 0.15 0.13
328
Problem 10.27 FLASH T=80 F (14.81 C) 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
Component methane ethane propane n-butane
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
Problem 10.28 First calculate equilibrium composition T=95 C (203 F) 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
Component n-butane n-hexane
xi 0.25 0.75
Now calculate liquid fraction from mole balances z1= x1= y1= L= 0.5 0.25 0.633 0.347
ANSWER Problem 10.29 FLASH
P = 2.00 atm (29.39 psia) T = 200 F (93.3 C) Fraction condensed L= 0.73 ANSWER xi=yi/Ki 0.191 0.455 0.354 SUM = 1.000
Component n-pentane n-hexane n-heptane
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
329
Problem 10.30 FLASH T=40 C (104 F) 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
Component ethane propane n-butane
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
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
Problem 10.31 FLASH T=70 F (21.11 C) 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
Component ethane propane i-butane n-butane
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
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
330
Problem 10.32 FLASH T=-15 C (5 F) 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 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 Target: y1=0.8
Component methane ethane propane n-butane
zi 0.30 0.10 0.30 0.30
L= 0.8145 xi=yi/Ki 0.162 0.103 0.352 0.364 SUM = 0.982
Component methane ethane propane n-butane
zi 0.30 0.10 0.30 0.30
L= 0.6850 xi=yi/Ki 0.073 0.088 0.376 0.424 SUM = 0.960
Component methane ethane propane n-butane
zi 0.30 0.10 0.30 0.30
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 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
Component n-butane n-pentane
xi 0.50 0.50
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 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
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
10.35 a) The equation from NIST is: Mi = ki yi P
Eq. (1)
The equation for Henry's Law is:i Hi = yi P x
Eq. (2)
Solving to eliminate P gives:
Hi
By definition:
Mi
=
ni ns Ms
Eq. (3) ki xi where M is the molar mass and the subscript s refers to the solvent. =
Mi
Dividing by the toal number of moles gives: Mi =
xi xs Ms
Eq. (4)
Combining Eqs. (3) and (4) gives: Hi =
1 xs Ms ki
If xi is small, then x s is approximately equal to 1 and: Hi =
b) For water as solvent: Ms
18.015
For CO2 in H2O:
ki
0.034
gm mol mol
1 Ms ki
Eq. (5)
kg bar
Hi
By Eq. (5):
Hi
1 Ms ki
1633 bar Ans.
The value is Table 10.1 is 1670 bar. The values agree within about 2%.
10.36 Acetone:
14.3145
2756.22 T degC
228.060
Psat1 ( ) T
e
14.8950
kPa
3413.10 T degC
250.523
Acetonitrile
Psat2 ( ) T
e
kPa
a) Find BUBL P and DEW P values
T
50degC
x1
0.5
y1
0.5
333
BUBLP
x1 Psat1 ( T)
1
x1 Psat2 ( T)
BUBLP
0.573 atm Ans.
DEWP
1 y1 Psat1 ( T)
1
y1
DEWP
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
x1
0.5
y1
0.5
Guess:
T
50degC
Given
x1 Psat1 ( T)
1
x1 Psat2 ( T) = P
BUBLT
Find ( T)
x1 Psat1 ( T) = y1 P Find x1 T
BUBLT
1
46.316 degC
Ans.
y1 P
Given
x1 DEWT
x1 Psat2 ( T) = 1 51.238 degC Ans.
DEWT
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 2726.81 T degC
217.572
Benzene:
Psat1 ( T)
e
13.9320
kPa
3056.96 T degC
217.625
Toluene:
Psat2 ( T)
e
kPa
a) Calculate the equilibrium composition of the liquid and vapor at the flash T and P
T
90degC
P
75kPa
Guess:
x1
0.5
y1
0.5
334
Given
x1 Psat1 ( )= y1 P T
Find x1 y1
1
x1 Psat2 ( )= 1 T
y1 0.458
y1 P
x1 y1
x1 0.252
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
0.1604
y1
0.2919
x2
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 x1 Psat2 ( )= 1 T Find y2 y3 y1 y3 P
y1
y2
y3 = 1
y2 y3
y2
0.608
y3
0.1
Ans.
Conclusion: An air leak is consistent with the measured compositions.
10.38 yO21
0.0387
yN21
0.7288
yCO21
0.0775 yH2O1
0.1550
ndot
10
kmol hr
T1
16.3872
100degC
3885.70 T degC
T2
25degC
P
1atm
230.170
PsatH2O ( ) T
e
kPa
335
Calculate the mole fraction of water in the exit gas if the exit gas is saturated with water.
yH2O2 PsatH2O T2 P
yH2O2 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. Guess:
ndotvap
yO22
Given
ndot 2
0.0387
ndotliq
yN22
ndotvap
ndot 2
0.7288
yCO22 0.0775
ndot = ndotliq
Overall balance
O2 balance
N2 balance
CO2 balance
Summation equation
ndot yO21 = ndotvap yO22
ndot yN21 = ndotvap yN22
ndot yCO21 = ndotvap yCO22
yO22 yN22 yCO22 yH2O2 = 1
ndotliq ndotvap yO22 yN22 yCO22
ndotliq
yO22
Find ndotliq ndotvap yO22 yN22 yCO22
1.276
0.044
kmol hr
yN22
ndotvap
0.835
336
8.724
kmol hr
0.089
yH2O2 0.031
yCO22
Apply an energy balance around the cooler to calculate heat transfer rate.
HlvH2O
Qdot
40.66
kJ mol
T1
T1
273.15K
T2
3
T2
273.15K
5 5 5
ndotvap yO22 R ICPH T1 T2 3.639 0.506 10
0
3 3 3
0.227 10
ndotvap yN22 R ICPH T1 T2 3.280 0.539 10 ndotvap yCO22 R ICPH T1 T2 5.457 1.045 10 ndotvap yH2O2 R ICPH T1 T2 3.470 1.450 10 HlvH2O ndotliq
Qdot 19.895 kW
Ans.
0 0.040 10 0
1.157 10
5
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
xC4
xC5
P
18psia
Ans.
0.05
0.85
0.10
KC3
KC4
KC5
3.9
0.925
0.23
The vapor mole fractions must sum to 1. xC3 KC3 xC4 KC4 xC5 KC5 1.004
337
10.40 H2S + 3/2 O2 -> H2O + SO2 By a stoichiometric balance, calculate the following total molar flow rates
Feed: Products
ndotH2S
10
kmol hr
ndotO2
3 ndotH2S 2
ndotSO2
ndotH2S
ndotH2O
ndotH2S
16.3872 3885.70 T degC
Exit conditions:
230.170
P
1atm
T2
70degC
PsatH2O ( ) T
e
kPa
a) Calculate the mole fraction of H2O and SO2 in the exiting vapor stream assuming vapor is saturated with H2O
yH2Ovap
ySO2 1
PsatH2O T2 P
yH2Ovap
yH2Ovap
0.308
Ans.
ySO2
0.692
Ans.
b) Calculate the vapor stream molar flow rate using balance on SO 2
ndotvap
ndotSO2 ySO2
ndotvap
14.461
kmol hr
Ans.
Calculate the liquid H2O flow rate using balance on H2O
ndotH2Ovap
ndotvap yH2Ovap
ndotH2Ovap
4.461
kmol hr
ndotH2Oliq
ndotH2O
ndotH2Ovap
ndotH2Oliq
5.539
kmol Ans. hr
338
10.41 NCL
0.01
kg kg
MH2O
18.01
gm mol
Mair
29
gm mol
a)
YH2O
NCL
Mair MH2O
YH2O
0.0161
yH2O
YH2O 1 YH2O
yH2O
0.0158
Ans.
b)
P
1atm
ppH2O
16.3872
yH2O P
ppH2O
1.606 kPa Ans.
3885.70 T degC
230.170
c)
PsatH2O ( ) T
e
kPa
Guess:
T
Find ( ) T
20degC
Given
Tdp 14.004 degC
yH2O P = PsatH2O ( ) Tdp T
Tdp Tdp 32degF
Tdp
57.207 degF Ans.
10.42 ndot1
50
kmol hr
16.3872
Tdp1
20degC
Tdp2
10degC
P
1atm
3885.70 T degC
230.170
MH2O
18.01
PsatH2O ( ) T
e
kPa
gm mol
y1
PsatH2O Tdp1 P
y1
0.023
y2
PsatH2O Tdp2 P
y2
0.012
By a mole balances on the process Guess: ndot2liq
ndot1 ndot2vap
ndot1
339
Given
ndot1 y1 = ndot2vap y2
ndot2liq
H2O balance
ndot1 = ndot2vap
ndot2liq
Overall balance
ndot2liq ndot2vap
ndot2vap
Find ndot2liq ndot2vap
49.441
kmol hr
ndot2liq
0.559
kmol hr
mdot2liq
ndot2liq MH2O
mdot2liq
10.074
kg hr
Ans.
10.43Benzene:
Cyclohexane:
A1
A2
13.7819
13.6568
B1
B2
kPa
2726.81
2723.44
C1
C2
217.572
220.618
Psat1 ( T)
exp A1
B1 T degC
C1
Psat2 ( T)
exp A2
B2 T degC
kPa
C2
Guess: T
66degC
Given
Psat1 ( T) = Psat2 ( T)
T
Find ( T)
The Bancroft point for this system is:
Psat1 ( T)
39.591 kPa
T
52.321 degC
Ans.
Com ponent1 Benzene 2- t Bu anol Acet ti oni le r
Com ponent2 Cyclohexane W at er Et hanol
T (C) 52.3 8 .7 7 65.8
P ( Pa) k 39.6 64 .2 60 .6
340
Chapter 11 - Section A - Mathcad Solutions
11.1 For an ideal gas mole fraction = volume fraction CO2 (1):
x1
0.7
V1
0.7m
3
N2 (2):
x2
0.3
V2
0.3m
3
i
1 2
P
1bar
T
( 25
273.15) K
P n
i
Vi RT
n
40.342 mol
S
nR
i
xi ln xi
S
204.885
J K
Ans.
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
nAr
4 mol
2.5 mol
TN2
TAr
[75 (
( 130
273.15)K]
273.15)K
PN2
PAr
30 bar
20 bar
TN2
348.15 K
TAr
403.15 K
i
1 2
ntotal
nN2
nAr
x1
nN2 ntotal
x2
nAr ntotal
x1
0.615
x2
0.385
CvAr
3 R 2
CvN2
5 R 2
CpAr
CvAr
R
CpN2
CvN2
R
Find T after mixing by energy balance:
T
TN2 2
TAr
(guess)
Given
nN2 CvN2 T
TN2 = nAr CvAr TAr
341
T
T
Find T) (
T
273.15 K
90 degC
Find P after mixing:
P PN2 2 PAr
(guess)
Given
nN2 nAr R T P
=
nN2 R TN2 PN2
nAr R TAr 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
nN2 CpN2 ln
T TN2 T TAr
R ln
P PN2 P PAr
SN2
11.806
J K J K
SAr
nAr CpAr ln
R ln
SAr
9.547
Smix
ntotal
R
i
xi ln xi
J K
Smix
36.006
J K
S
SN2
SAr
Smix
S
38.27
Ans.
11.3
mdotN2
molwtN2
2
kg sec
gm mol
mdotH2
molwtH2
0.5
kg sec
gm mol
i 1 2
28.014
2.016
molarflowN2
molarflowtotal
mdotN2 molwtN2
molarflowN2
molarflowH2
mdotH2 molwtH2
319.409 mol sec
molarflowH2 molarflowtotal
342
y1
molarflowN2 molarflowtotal
y1
0.224
y2
molarflowH2 molarflowtotal
y2
0.776
S
R molarflowtotal
i
yi ln yi
S
1411
J secK
Ans.
11.4
T1
448.15 K
T2
308.15 K
P1
3
3 bar
P2
6
1 bar
For methane:
MCPHm
MCPH T1 T2 1.702 9.081 10
2.164 10
0.0
MCPSm For ethane:
MCPS T1 T2 1.702 9.081 10
3
2.164 10
6
0.0
MCPHe
MCPH T1 T2 1.131 19.225 10
3
5.561 10
6
0.0
MCPSe
MCPS T1 T2 1.131 19.225 10
3
5.561 10
6
0.0
MCPHmix
0.5 MCPHm
0.5 MCPHe
MCPHmix
MCPSmix
6.21
6.161
MCPSmix
H
0.5 MCPSm
0.5 MCPSe
T1
R ln P2 P1
R MCPHmix T2
H
7228
J mol
S
R MCPSmix ln
T2 T1
R 2 0.5 ln ( ) 0.5
The last term is the entropy change of UNmixing J T 300 K S 15.813 mol K
Wideal
H
T
S
Wideal
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
y2 ln y2
S
4.273
J mol K
3 J
By Eq. (5.27):
Wideal
T
S
Wideal
1.282
10
mol
By Eq. (5.28):
Work
Wideal
t
Work
25638
J mol
Ans.
11.16
0 10 20 40 60 P 80 100 200 300 400 500
bar
1.000 0.985 0.970 0.942 0.913 Z 0.885 0.869 0.765 0.762 0.824 0.910
Fi Zi Pi 1
ln 1
0
1
1
end
rows ( P)
i
2 end
Fi is a well behaved function; use the trapezoidal rule to integrate Eq. (11.35) numerically.
Ai
i
Fi
Fi 1 Pi 2
Pi 1
ln i
ln i 1
i Pi
Ai
exp ln i
fi
Generalized correlation for fugacity coefficient: For CO2:
Tc
304.2 K
Pc
73.83 bar
0.224
T
P G ( P)
( 150
273.15) K
Tr
T Tc
Tr
1.391
exp
Pc
Tr
B0 Tr
B1 Tr
344
fG ( P)
G ( P) P
Pi bar
Calculate values:
10 20 40 60 80 100 200 300 400 500
fi
i
0.993 0.978 0.949 0.922 0.896 0.872 0.77 0.698 0.656 0.636
bar
9.925 19.555 37.973 55.332 71.676 87.167 153.964 209.299 262.377 317.96
400
fi bar
300 200
0.8
i G Pi
f G Pi
0.6
bar
100 0
0.4
0
200
Pi
400
600
0
200
Pi bar
400
600
bar
Agreement looks good up to about 200 bar (Pr=2.7 @ Tr=1.39)
11.17 For SO2:
Tc
T
430.8 K
600 K
1.393
Pc
P
Pr
78.84 bar
300 bar
P Pc
Pr
0.245
Tr
T Tc
Tr
3.805
For the given conditions, we see from Fig. 3.14 that the Lee/Kesler correlation is appropriate.
345
Data from Tables E.15 & E.16 and by Eq. (11.67):
0
0.672
1
1.354
0 1
0.724
f
P
GRRT
ln
f
217.14 bar
GRRT
0.323
Ans.
11.18 Isobutylene:
Tc
417.9 K
Pc
40.00 bar
0.194
a) At 280 degC and 20 bar:
T
( 280
273.15)K
P
20 bar
Tr ( ) T
T Tc
Tr ( ) 1.3236 T
Pr ( ) P
P Pc
Pr ( ) 0.5 P
At these conditions use the generalized virial-coeffieicnt correlation.
f
PHIB Tr ( )Pr ( ) T P
P
T ( 280
f
18.76 bar
273.15)K
Ans.
P 100 bar
b) At 280 degC and 100 bar:
Tr ( ) 1.3236 T
Pr ( ) 2.5 P
At these conditions use the Lee/Kesler correlation, Tables E.15 & E.16 and Eq. (11.67).
0
0.7025
1
1.2335
0 1
f
P
0.732
f
73.169 bar
Ans.
11.19 The following vectors contain data for Parts (a) and (b): (a) = Cyclopentane; (b) = 1-butene
Tc
511.8 420.0
0.273 0.277 383.15 393.15
K
Pc
45.02 40.43 258 239.3 275 34
346
bar cm mol
3
0.196 0.191
Zc
Vc
Tn
322.4 266.9
5.267 25.83
K
T
K
P
bar
Psat
bar
Tr
T Tc
Tr
0.7486 0.9361
Psatr
Psat Pc
Psatr
0.117 0.6389
Calculate the fugacity coefficient at the vapor pressure by Eq. (11.68): (a)
PHIB Tr Psatr
1
1
1
0.900
(b)
PHIB Tr Psatr
2
2
2
0.76
Eq. (3.72), the Rackett equation:
Tr
T Tc
Tr
0.749 0.936
Eq. (11.44):
2
Vsat
V c Zc
1 Tr
7
Vsat
107.546 cm3 133.299 mol
f
PHIB Tr Psatr
11.78 20.29
Psat exp
Vsat ( Psat) P RT
f
bar
Ans.
11.21
Table F.1, 150 degC:
Psat
476.00 kPa
molwt
18
gm mol
Vsat
1.091
cm molwt gm
3
3
T
( 150
273.15)K
P
150 bar
Vsat
cm 19.638 mol
T
423.15 K
Equation Eq. (11.44) with
satPsat = fsat
r
exp
Vsat P RT
Psat
r
1.084
r=
f fsat
= 1.084
Ans.
347
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 3121.2
273.15) K
459.67) rankine J gm
6.2915 S1
J gm K
H1
Btu 1389.6 lbm
1.5677
Btu lbm rankine
T1
Table F.2: (a) 300 kPa & 400 degC; (b) 50(psia) & 800 degF: T2 3275.2 H2 J gm S2 8.0338 J gm K
Btu 1431.7 lbm
1.9227
Btu lbm rankine
Eq. (A) on page 399 may be recast for this problem as:
r
exp
molwt R
H2 T1
H1
S2
S1
f2 f1
r
0.0377 0.0542
Ans.
(a)
r=
f2 f1
= 0.0377
(b)
r=
= 0.0542
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 420.0 Pc
33.70 40.0 40.43 313.0 Vc 238.9 239.3
348
0.252
bar
0.194 0.191
0.270 Zc 0.275 0.277
cm
3
309.2 Tn 266.3 266.9 K
mol
200 P 300 bar 150
0.6583 Tr 0.6372 0.6355 Psat
1.01325 1.01325 bar 1.01325 0.0301 Pr Psat Pc Pr 0.0253 0.0251
Tr
Tn Tc
Calculate the fugacity coefficient at the nbp by Eq. (11.68): (a)
(b)
PHIB Tr Pr 1 1 PHIB Tr Pr
2
1
0.9572
2
2
0.9618
(c)
PHIB Tr Pr 3 3
3
0.9620
1 Tr
0.2857
Eq. (3.72):
Vsat
V c Zc
Eq. (11.44):
f
PHIB Tr Pr 2.445 f 3.326 bar 1.801
Psat exp
Vsat ( Psat) P R Tn
Ans.
11.24 (a) Chloroform: Tc
536.4 K
Pc
3
54.72 bar
0.222
Zc
0.293
Vc
cm 239.0 mol
Tn
334.3 K
Psat
22.27 bar
T
473.15 K
Tr
T Tc
Tr
0.882
2
Trn
Tn Tc
Trn
0.623
3
Eq. (3.72):
Vsat
V c Zc
1 Trn
7
Vsat
cm 94.41 mol
349
Calculate fugacity coefficients by Eqs. (11.68):
P Pc if P Psat Pr ( P ) Tr
Pr ( P )
( P)
exp
B0 Tr
B1 Tr
f ( P)
( P) P
( Psat) Psat exp
Vsat ( P Psat) RT
( P)
P
if P
Psat
( P)
( Psat)
Psat P
exp
Vsat ( P RT
Psat)
0 bar 0.5 bar 40 bar
40 Psat bar
( P) Psat bar
f ( P) bar
30
0.8
P bar
20
0.6
10
0
0
20
40
0.4
0
20
P bar
40
P P bar bar
(b) Isobutane
Zc
T
Tc
Vc
Tr
408.1 K
262.7
T Tc
Vsat
Pc
3
36.48 bar
261.4 K
Tn Tc
Vsat
0.181
Psat 5.28 bar
0.282
cm
mol
Tr
Tn
0.767
2
313.15 K
Trn
Trn
0.641
3
Eq. (3.72):
V c Zc
1 Trn
7
cm 102.107 mol
350
Calculate fugacity coefficients by Eq. (11.68):
Pr ( ) P P Pc if P Psat ( )P P () P exp Pr ( ) P Tr
B0 Tr
B1 Tr
fP) (
( )Psat exp Psat
Vsat ( Psat) P RT
() P
P
if P
Psat
() ( ) P Psat
Psat P
exp
Vsat ( P RT
Psat)
0 bar 0.5 bar 10 bar
10 Psat bar
0.8
Psat bar
fP) ( bar
P bar
5
() P
0.6
0
0
5
10
0.4
0
5 P bar
10
P P bar bar
11.25 Ethylene = species 1; Propylene = species 2
Tc
282.3 365.6
K
Pc
50.40 46.65
bar cm mol
3
w
0.087 0.140
Zc
0.281 0.289
Vc
P
i
131.0 188.4
y1
T
n
423.15 K
2
30 bar
1 n
351
0.35
y2
k
1
1 n
y1
j 1 n
By Eqs. (11.70) through (11.74)
wi
i j
wj 2
1 3
Tc
i j
Tci Tc j
Zc
Zci 2
Zc j
i j
Vc
Vci
Vc j
1 3 3
Zc Pc
i j
i j
i j
2
T
R Tc i j Vc
i j
Tr
i j
Tc
Tr
1.499 1.317 1.317 1.157
50.345 48.189 48.189 46.627
K Zc bar
i j
Vc
157.966 cm3 157.966 188.4 mol 131 0.087 0.114 0.114 0.14 Tc 282.3 321.261
Pc 321.261 365.6
0.281 0.285 0.285 0.289
By Eqs. (3.65) and (3.66): B0i j
B0
B0 Tr
0.138 0.189
i j
B1i j
B1 Tr
i j
0.189 0.251
i j B1i j
B1
0.108 0.085 0.085 0.046
59.892 99.181 99.181 cm3 159.43 mol
Bi j
R Tc i j B0i j Pc
i j
B
By Eq. (11.64):
i j
2 Bi j
Bi i P RT
Bj j
20.96 cm3 mol 20.96 0 0
hatk
exp
Bk k
1 2
i j
yi y j 2 i k
i j
fhatk
hatk yk P
hat
0.957 0.875
352
fhat
10.053 17.059
bar
Ans.
For an ideal solution , id = pure species
Pr
P
k
Pck
Pr
0.595 0.643
Pr idk exp Tr
k
B0k k
k k B1k k
k k
fhatid
k
idk yk P
id
0.95 0.873
fhatid
9.978 17.022
bar
Ans.
Alternatively,
Pr
P
i j
Pr idk exp
k k k k
Pc
i j
Tr
B0k k
k k B1k k
id
0.95 0.873
11.27 Methane = species 1 Ethane = species 2 Propane = species 3 0.21 y 0.43 0.36
T
373.15 K 0.012
P
35 bar
0.286
w
0.100 0.152 45.99
Zc
0.279 0.276 98.6
cm mol
3
190.6 Tc 305.3 K 369.8 Pc
48.72 bar 42.48
Vc
145.5 200.0
n
3
i
1 n
j
1 n
k
1 n
By Eqs. (11.70) through (11.74)
wi
i j
wj 2
1 3
Tc
i j
Tci Tc j
Zc
Zci 2
Zc j
i j
Vc
Vci
Vc j
1 3 3
Zc Pc
i j
i j
i j
2
R Tc i j Vc
i j
353
1.958 1.547 1.406
Tr T
i j
Tc
Tr
1.547 1.222 1.111 1.406 1.111 1.009
i j
98.6 Vc 120.533
120.533 143.378 145.5 171.308 200
cm
3
mol
143.378 171.308
45.964 47.005 43.259 Pc 47.005 48.672 45.253 bar 43.259 45.253 42.428 190.6 Tc 241.226 241.226 265.488 305.3 336.006 K 369.8
0.012 0.056 0.082 0.056 0.1 0.126
0.082 0.126 0.152
0.286 0.282 0.281 Zc 0.282 0.279 0.278 0.281 0.278 0.276
265.488 336.006
By Eqs. (3.65) and (3.66):
B0i j B0 Tr
B1i j
i j
B1 Tr i j
Bi j
R Tc i j B0i j Pc
i j
i j B1i j
By Eq. (11.64):
i j
0
30.442 107.809 0 23.482 0
2 Bi j
Bi i
Bj j
30.442
cm
3
mol
107.809 23.482
hatk
exp
P RT
Bk k
1 2
i j
yi y j 2 i k
i j
fhatk
hatk yk P hat
1.019 0.881 0.775
354
7.491 fhat 13.254 bar 9.764
Ans.
For an ideal solution , id = pure species
0.761 Prk P Pck Pr 0.718 0.824 0.977 fhatid
k
idk
exp
Prk Tr
k k
B0k k
k k B1k k
7.182 fhatid 13.251 bar 9.569
Ans.
idk yk P
id
0.88 0.759
11.28 Given:
(a)
GE RT
=
2.6 x1
1.8 x2 x1 x2
Substitute x2 = 1 - x1: .8 x1 1.8 x1 1 x1 = 1.8 x1 x1
2
GE = RT
0.8 x1
3
Apply Eqs. (11.15) & (11.16) for M = GE/RT:
ln 1 =
GE RT
d 1 x1
GE RT
dx1
ln 2 =
GE RT
d x1
GE RT
dx1
d
GE RT
dx1
= 1.8
2 x1
2.4 x1
2
2
ln 1 = 1.8
ln 2 =
2 x1
2 x1
1.4 x1
1.6 x1
3
3 1.6 x1
Ans.
(b) Apply Eq. (11.100):
GE = x1 RT
1.8 1 x1
2 x1 x1
2
1.4 x1
2 3
1.6 x1
3
1.6 x1
This reduces to the initial condition:
355
(c) Divide Gibbs/Duhem eqn. (11.100) by dx1:
d ln 1 dx1
x1
x2
d ln 2 dx1
= 0
Differentiate answers to Part (a):
d ln 1 = 2 dx1
x1
2
2.8 x1
4.8 x1
2
d ln 2 = 2 x1 dx1
3
4.8 x1
2
d ln 1 = 2 x1 dx1
d ln 1 = 1 dx1
2.8 x1
4.8 x1
x2
x1
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
When x1 = 0, we see from the 3rd eq. of Part (c) that
(e) DEFINE: g = GE/RT g x1
ln 1 x1
ln 2 x1
ln 1 () 0
d ln 1 dx1
d ln 2 dx1
= 0
Q.E.D.
= 0
Q.E.D.
1.8 x1
1.8
x1
1.8
2
x1
2
0.8 x1
3
2
2 x1
1.6 x1
1.4 x1
3
1.6 x1
3
ln 2 () 1
2.6
x1
0 0.1 1.0
356
0
g x1 ln 1 x1 ln 2 x1
1
0 ln 1 ( )
2
ln 2 ( ) 1
3
0
H H1bar H2bar
0.2
0.4
0.6
0.8
x1
0.02715
11.32
87.5 265.6 417.4 534.5 531.7 421.1 347.1 VE 321.7 276.4 252.9 190.7 178.1 138.4 98.4 37.6 10.0
357
0.09329 0.17490 0.32760 0.40244 0.56689 0.63128 x1 0.66233 0.69984 0.72792 0.77514 0.79243 0.82954 0.86835 0.93287 0.98233
253
n
rows x1
i 1 n
x1
0 0.01 1
(a) Guess:
x1 1 F x1 x1
2
a
x1 x1 x1 a b c
3000
b
3000 c
a
250
3.448 10
3 3
1
linfitx1 VE F
b c
Ans.
3.202 10 244.615
3 x1 1
600
VEi x1 ( 1 x1) a b x1 c ( x1)
2
400
200
0
0
0.2
0.4
i
0.6
0.8
x1 x1
By definition of the excess properties
V = x1 x2 a
E
b x1
c x1
3 (c
2
d 3 E V = 4 c x1 dx1
Vbar1
E
b) x1
2
2 (b
2
a) x1
a
= x2 = x1
2
a
2 b x1
3 c x1
Vbar2
E
2
a
b
2 (b
c) x1
3 c x1
2
(b) To find the maximum, set dVE/dx1 = 0 and solve for x1. Then use x1 to find VEmax. Guess:
Given
4 c ( x1)
x1
3
x1
0.5
3 (c
b) ( x1)
x1
2
2 (b
a) x1
a= 0
Find ( x1)
0.353
Ans.
358
VEmax
x1 ( 1
x1) a
b x1
2
c x1
2
VEmax 3c( ) x1
2
536.294
Ans.
(c) VEbar1 ( ) x1
( 1
x1) a
2
2 b x1
VEbar2 ( ) x1
x1 0 0.01 1
4000
( ) a x1
b
2( b
c)x1
3c( ) x1
2
2000 VEbar 1 ( ) x1
x1 VEbar 2 ( )
0
2000
0
0.2
0.4
x1 x1
0.6
0.8
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
466 809
cm mol
3
P
n
2 bar
2
y1
i
0.5
1 n
y2
j
1
1 n
y1
276 466
By Eq. (11.61):
B
i
359
yi y j Bi j
j
B
504.25
cm
3
mol
Use a spline fit of B as a function of T to find derivatives:
331 b11 276 235
cm
3
980 b22 809 684
cm
3
558 b12 466 399
cm
3
mol
mol
mol
50 t 75 100
vs11 lspline ( t b11) B11 ( T)
273.15 K
323.15 t 348.15 K 373.15
3
interp ( vs11 t b11 T)
B11 ( T)
cm 276 mol
vs22
lspline ( t b22) B22 ( T)
interp ( vs22 t b22 T)
B22 ( T)
cm 809 mol
3
vs12
lspline ( t b12) B12 ( T)
interp ( vs12 t b12 T)
B12 ( T)
cm 466 mol
3
dBdT
d d B12 ( T) B11 ( T) dT dT
d dT
B12 ( T)
dBdT
1.92 3.18
d B22 ( T) dT
cm 3.18 5.92 mol K
3
3
Differentiate Eq. (11.61): dBdT
i j
yi y j dBdTi dBdT j
cm 3.55 mol K
By Eq. (3.38): Z
1
BP RT
Z
0.965
V
ZR T P
HRRT R T
By Eq. (6.55): HRRT
By Eq. (6.56):
P R
B T
dBdT HRRT
SRR
0.12
HR
SRR
P dBdT R
0.085
SR
SRR R
V
13968
cm
3
mol
HR
348.037
360
J mol
SR
0.71
J mol K
Ans.
11.34 Propane = 1; n-Pentane = 2 T
B
( 75
273.15)K
466 809
cm
3
P
2 bar
n
y1
2
0.5
y2
1
y1
276 466
i 1 n
mol
j 1 n
ij
2 Bi j Bii B j j
By Eqs. (11.63a) and (11.63b):
hat1 ( ) y1
exp
P B1 1 RT
( 1
2
y1) 1 2
2
hat2 ( ) y1
y1
exp
P B2 2 RT
y1
1 2
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 y1
361
0.6
0.8
0.0426
11.36
23.3 45.7 66.5 86.6 118.2 144.6 176.6 HE 195.7 204.2 191.7 174.1 141.0 116.8 85.6 43.5 22.6
500
a b c
0
0.0817 0.1177 0.1510 0.2107 0.2624 0.3472 x1 0.4158 0.5163 0.6156 0.6810 0.7621 0.8181 0.8650 0.9276 0.9624
n
x1
rows x1
i
1 n
0 0.01 1
(a) Guess:
x1 1 F x1 x1
2 3
a
x1 x1 x1
b
100
c
0.01 a
539.653 1.011 10 913.122
3
1 1
linfit x1 HE F
b c
Ans.
x1
HEi x1 ( 1 x1) a b x1 c ( x1)
2
100
200
300
0
0.2
0.4
i
0.6
0.8
x1 x1
362
By definition of the excess properties
H = x1 x2 a
d dx1
E
E
b x1
3
c x1
3( c
2
H = 4 c x1
E
b) x1
2
2( b
2
a)x1
a
Hbar1
= x2 = x1
2
a
2 b x1
3 c x1
Hbar2
E
2
a
b
2( b
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:
Given
x1
2
x1
0.5
3
HE ( ) x1
3( c
0.512
b x1
x1 ( 1
2
x1) a
a)x1
b x1
a= 0
c x1
4c( ) x1
Find ( ) x1
x1 ( 1
b)( ) x1
Ans. c x1
2
2( b
x1
x1) a
HEmin
(c)
HEmin
204.401
Ans.
HEbar1 ( ) x1
HE ( ) ( x1 1
d x1) HE ( ) x1 dx1
HEbar2 ( ) x1
x1 0 0.01 1
500 0 500
HE ( ) x1 x1
d HE ( ) x1 dx1
HEbar 1 ( ) x1 x1 HEbar 2 ( )
1000
0
0.2
0.4 x1
363
0.6
0.8
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) (1) = Acetone (2) = 1,3-butadiene
y1
0.28
y2
Tc
1
y1
508.2 425.2
T
K Zc
( 60
273.15) K
Vc
P
209
170 kPa
cm mol
3
w
n 2
0.307 0.190
0.233 0.267
220.4
i
1 n
wi wj 2
j
1 n
ki j
0
0.307 0.2485 0.082
Eq. (11.70)
i j
0.2485 0.082 508.2
0.19
0.126
0.126 0.152 464.851 425.2 0
K
Eq. (11.71) Tci j
Tc Tc
i
j
1
ki j
Tc
464.851 369.8
Zc
Eq. (11.73) Zci j
i
Zc 2
1
0.233 0.25
j
Zc
1 3
0.25 0.267 0.276
209
0
214.65
3
Vc
Eq. (11.74) Vci j
3
i
Vc
3
j
2
Vc
214.65 220.4 200 0
cm
mol
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
T Tci j
0.656 0.717 0.717 0.784 B0i j B0 Tri j
Pri j
P Pci j
0.036 0.038 Pr 0.038 0.04 0.824 0
Tr
Eq. (3.65)
0.74636 B0 0.6361 0.16178
Eq. (3.66) B1i j B1 Tri j
0.6361 0.5405 0.27382
0.16178 0.27382 0.33295
0.874 B1 0.558 0.098
0.558 0.098 0.34 0.028 0.028 0.027
Eq. (11.69a) + (11.69b)
Bi j
R Tci j B0i j Pci j
i j B1i j
B
n
910.278 665.188
n
665.188 cm3 499.527 mol
cm
3
Eq. (11.61)
B
i 1 j 1
yi y j Bi j
B
598.524
mol
Eq. (3.38)
Z
V
1
BP RT
Z
V
0.675 Tri j
2.6
0.963
1.5694 10
3 4 cm
RT Z P
mol
Ans.
0.722 Tri j
5.2
Eq. (6.89) dB0dTri j
Eq. (6.90)
dB1dTri j
365
Differentiating Eq. (11.61) and using Eq. (11.69a) + (11.69b)
n n
dBdT
i 1 j 1
yi y j
R dB0dTri j Pci j
i j dB1dTri j
Eq. (6.55) HR
PT
B T
dBdT
HR
344.051
Eq. (6.56) SR
P dBdT
SR
0.727
J mol J
Ans.
Eq. (6.54) GR
BP
3
GR
101.7
mol K J
Ans.
mol
Ans.
(b)
cm V = 15694 mol
SR = 1.006
HR = 450.322
GR = 125.1
J mol
J mol K
3
J mol
(c)
V = 24255
cm
HR = 175.666
mol
J mol
J SR = 0.41 mol K
GR = 53.3
J mol
(d)
V = 80972
cm
3
mol
HR = 36.48
J mol
SR = 0.097
J mol K
3
GR = 8.1
J mol
(e)
cm V = 56991 mol
HR = 277.96
J mol
SR = 0.647
J mol K
GR = 85.2
J mol
366
Data for Problems 11.38 - 11.40
325 200 575 T 350 300 525 225 200 P
15 100 40 35 50 10 25 75 1.054 1.325 1.023 Tc
308.3 150.9 562.2 304.2 282.3 507.6 190.6 126.2 Pc
61.39 48.98 48.98 73.83 50.40 30.25 45.99 34.00
.187 .000 .210 .224 .087 .301 .012 .038
0.244 2.042 0.817
Pr P Pc
Tr
T Tc
Tr
1.151 1.063 1.034 1.18 1.585
Pr
0.474 0.992 0.331 0.544 2.206
11.38 Redlich/Kwong Equation:
0.02 0.133 0.069 Pr Tr
Eq. (3.53)
0.08664
0.42748 4.559 3.234 4.77
0.036 0.081 0.028 0.04 0.121
q
Tr
1.5
Eq. (3.54) q
3.998 4.504 4.691 3.847 2.473
Guess:
z
1
367
Given
z= 1
q
z z z
Eq. (3.52)
Z
q
Find z ()
i
1 8
Ii
i qi
ln
Z Z
i qi i qi
i qi
i
Eq. (6.65)
i
exp Z
i Pi
i qi
0.925 0.722 0.668 0.887 0.639 0.891 0.881 0.859
1
ln Z
i
qi Ii Eq. (11.37)
fi
Z
i
0.93 0.744 0.749 0.896 0.73 0.9 0.89 0.85
fi
13.944 74.352 29.952 31.362 36.504 8.998 22.254 63.743
11.39 Soave/Redlich/Kwong Equation
2
0.08664
0.5
0.42748
2
c
0.480
1.574
0.176 0.02 0.133 0.069
1
c 1
Tr
4.49 3.202 4.737 q
Eq. (3.54) q
Pr Tr
Eq. (3.53)
0.036 0.081 0.028 0.04 0.121
3.79 4.468 4.62 3.827 2.304
Tr
Guess:
z
1
368
Given
z= 1
q
z z z
Eq. (3.52)
Z
q
Find z ()
i
1 8
Ii
ln
Z Z
ln Z
i qi i qi
i qi
i
Eq. (6.65)
i
exp Z
i Pi
i qi
0.927 0.729 0.673 0.896 0.646 0.893 0.882 0.881
i qi
1
i
qi Ii Eq. (11.37)
fi
Z
i
0.931 0.748 0.751 0.903 0.733 0.902 0.891 0.869
fi
13.965 74.753 30.05 31.618 36.66 9.018 22.274 65.155
11.40 Peng/Robinson Equation
1 2
1 2
2
0.07779
0.45724
2
c
0.37464
1.54226
0.26992 0.018 0.12 0.062
1
c 1
Tr
0.5
5.383 3.946 5.658
Pr Tr
Eq.(3.53)
0.032 0.073 0.025 0.036 0.108
q
Tr
Eq.(3.54) q
4.598 5.359 5.527 4.646 2.924
369
Guess:
z
1
q z z z
Eq. (3.52) Z q Find ( z)
Given z = 1
i
1 8
Ii
1 2 2
ln
Z Z
i qi i qi
i i
Eq. (6.65)
i
exp Z
i Pi
i qi
1
i qi
0.918 0.69 0.647 0.882 0.617 0.881 0.865 0.845
ln Z
i qi
i
i
qi Ii Eq. (11.37)
fi
fi
Z
0.923 0.711 0.73 0.89 0.709 0.891 0.876 0.832
13.842 71.113 29.197 31.142 35.465 8.91 21.895 62.363
BY GENERALIZED CORRELATIONS Parts (a), (d), (f), and (g) --- Virial equation: 325 T 350 525 225 T Tc Tc 308.3 304.2 507.6 190.6 P 15 35 10 25 Pc 61.39 73.83 30.25 45.99 .187 .224 .301 .012
Tr
Pr
P Pc
Evaluation of :
B0
B0 ( Tr)
Eq. (3.65)
B1
B1 ( Tr)
Eq. (3.66)
370
DB0
0.675 Tr
2.6
Eq. (6.89)
DB1
0.722 Tr
5.2
Eq. (6.90)
0.932 Pr B0 exp Tr
B1
Eq. (11.60)
0.904 0.903 0.895
(a) (d) (f) (g)
Parts (b), (c), (e), and (h) --- Lee/Kesler correlation: Interpolate in Tables E.13 - E.16:
.7454 0 .7517 .7316 .8554 1
1.1842 0.9634 0.9883 1.2071
0.000 0.210 0.087 0.038
0.745
0 1
Eq. (11.67):
0.746 0.731 0.862
(b) (c) (e) (h)
11.43 ndot1
x1
2
kmol hr
ndot2
x1
4
kmol hr
x2
ndot3
1 x1
ndot1
x2
ndot2
0.667
ndot1 ndot3
0.333
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) St R x1 ln x1 x2 ln x2 ndot3
St 8.82 W K
Ans.
371
11.44 For air entering the process:
xO21
0.21
xN21
0.79
For the enhanced air leaving the process: xO22
0.5
xN22
0.5
ndot2
50
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 O 2
xN21 ndotair = xN22 ndot2
ndotair ndotO2 Find ndotair ndotO2
Mole balance on N2
ndotair
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
R xO21 ln xO21
xN21 ln xN21
Entropy change of mixing
S23
R xO22 ln xO22
xN22 ln xN22
Total rate of entropy generation: SdotG
ndotair S12
ndot2 S23
SdotG
152.919
W K
Ans.
372
10
11.50 T
544.0
273.15K
932.1
J mol
30 K 50
GE
513.0 494.2
HE
893.4 845.9
J mol
Assume Cp is constant. Then HE is of the form:
Find a and c using the given HE and T values. a
c
HE = c
aT
slope ( HE) T
intercept ( HE) T
a
c
2.155
1.544
J mol K
10
3 J
mol
T bT c
GE is of the form: GE = a T ln
T K
Rearrange to find b using estimated a and c values along with GE and T data.
GE B
a T ln
T
T K
T
c
13.543 B 13.559 13.545
J mol K
Use averaged b value
3
Bi b
i 1
3
b
13.549
J mol K
Now calculate HE, GE and T*SE at 25 C using a, b and c values. HE ( ) T aT c
J Ans. mol
HE [25 (
273.15) ] 901.242 K
GE ( ) T
a T ln
T K
T
bT
( c GE [25
273.15) ] 522.394 K
J Ans. mol
J Ans. mol
TSE ( ) T
HE ( ) GE ( ) T T
TSE [25 (
273.15) ] 378.848 K
373
Chapter 12 - Section A - Mathcad Solutions
12.1 Methanol(1)/Water(2)-- VLE data: 39.223 42.984 48.852 52.784 P 56.652 60.614 63.998 67.924 70.229 72.832
kPa
T
333.15 K
0.1686 0.2167 0.3039 0.3681 x1 0.4461 0.5282 0.6044 0.6804 0.7255 0.7776
0.5714 0.6268 0.6943 0.7345 y1 0.7742 0.8085 0.8383 0.8733 0.8922 0.9141
n 10
i 1 n
Number of data points:
Calculate x2 and y2:
n
x2
rows ( ) P
1 x1
y2
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
y2 P
2
x1 Psat1
x2 Psat2
GERT
x1 ln 1
x2 ln 2
374
i
1 2 3 4 5 6 7 8 9 10
1
i
1.572 1.47 1.32 1.246 1.163 1.097 1.05 1.031 1.021 1.012
2
i
1.013 1.026 1.075 1.112 1.157 1.233 1.311 1.35 1.382 1.41
ln 1 i
0.452 0.385 0.278 0.22 0.151 0.093 0.049 0.031 0.021 0.012
ln 2 i
0.013 0.026 0.073 0.106 0.146 0.209 0.271 0.3 0.324 0.343
GERTi
0.087 0.104 0.135 0.148 0.148 0.148 0.136 0.117 0.104 0.086
0.5
0.4
ln 1 i ln 2 i GERT i
0.3
0.2
0.1
0
0
0.2
0.4
0.6
0.8
x1
i
(a) Fit GE/RT data to Margules eqn. by linear least squares: VXi x1
i
VYi
GERTi x1 x2 i i
intercept ( VX VY)
0.683
A12
Ans.
Slope
Slope
A12
A12
slope ( VX VY)
0.208
Intercept
0.683
Intercept
Intercept
A21
A21
Slope
0.475
375
The following equations give CALCULATED values:
1 ( x2) x1 2 ( x2) x1 j 1 101
exp x2 exp x1
2 2
A12 A21
2 A21 2 A12
A12 x1 A21 x2
X1
j
.01 j
.01
X2
j
1
X1
j
pcalc
j
X1
j
1 X1 X2 Psat1 j j
j
X2
j
2 X1 X2 Psat2 j j
X1 Y1calc
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 60 50
j
Pi kPa
pcalc kPa
40
j
pcalc kPa
30 20 10
0
0.2
0.4
i i j
0.6
j
0.8
x1 y1 X1 Y1calc
P-x data P-y data P-x calculated P-y calculated
376
Pcalc
i
x1 1 x1 x2 Psat1 i i i
x1 1 x1 x2 Psat1 i i i Pcalc
i
x2 2 x1 x2 Psat2 i i i
y1calc
i
RMS deviation in P:
Pi RMS
i
Pcalc n
2
i
RMS
0.399 kPa
(b) Fit GE/RT data to van Laar eqn. by linear least squares: VXi
Slope
Slope
a12
a12
x1
i
VYi
x1 x2 i i GERTi
intercept ( VX VY)
1.418
slope ( VX VY)
0.641
1 Intercept
0.705
Intercept
Intercept
a21
a21 a12 x1 a21 x2
2
1 ( Slope Intercept)
0.485
2
Ans.
1 ( x1 x2)
exp a12 1
2 ( x1 x2)
exp a21 1
X1
X2
a21 x2 a12 x1
j
1 101
j
j
.01 j
1 X1
.00999
j
(To avoid singularities)
377
pcalc
j
X1
j
1 X1 X2 Psat1 j j
X2
j
2 X1 X2 Psat2 j j
Pcalc
i
x1 1 x1 x2 Psat1 i i i
x2 2 x1 x2 Psat2 i i i
X1 Y1calc
j
j
1 X1 X2 Psat1 j j pcalc
j
y1calc
i
x1 1 x1 x2 Psat1 i i i Pcalc
i
P-x,y Diagram: van Laar eqn. fit to GE/RT data.
90 80
Pi kPa
70 60 50
j
Pi kPa
pcalc kPa
40
j
pcalc kPa
30 20 10
0
0.2
i
0.4
i j
0.6
j
0.8
1
x1 y1 X1 Y1calc
P-x data P-y data P-x calculated P-y calculated
RMS deviation in P:
Pi RMS
i
Pcalc n
2
i
RMS
0.454 kPa
378
(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
0.5
GERTi
21
1.0 x2
2
12
21 i
x1 ln x1
i
i
i i
12 21
x2 ln x2 i i
x1
12 21
Minimize SSE
12
12 21
0.476 1.026
Ans.
21 12
exp x2 1 ( x2) x1
21
12
x1
x2
x1
x2
12
x1
21
x2
12
exp 2 ( x2) x1
x1
21
12
x1
x2
x2
x2
21
x1
21
x1
j
1 101
X1
X1
j
.01 j
.01
X2
X2
j
1
X1
j
pcalc
Pcalc
j
j
1 X1 X2 Psat1 j j
j
2 X1 X2 Psat2 j j
i
x1 1 x1 x2 Psat1 i i i
X1
x2 2 x1 x2 Psat2 i i i
x1 1 x1 x2 Psat1 i i i Pcalc
i
Y1calc
j
1 X1 X2 Psat1 j j j pcalc
j
y1calc
i
379
P-x,y diagram: Wilson eqn. fit to GE/RT data.
90 80
Pi kPa
70 60 50
j
Pi kPa
pcalc kPa
40
j
pcalc kPa
30 20 10 0 0.2
i
0.4
i j
0.6
j
0.8
1
x1 y1 X1 Y1calc
P-x data P-y data P-x calculated P-y calculated
RMS deviation in P:
Pi RMS
i
Pcalc n
2
i
RMS
0.48 kPa
(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
2 x1 x2 A 12 A 21
exp ( ) A12 x2
exp ( ) A21 x1
2
2
2 A21
2 A12
A12 x1
A21 x2
380
Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses:
SSE A12 A21
i
A12
0.5
Pi x1
A21
1.0
2
x1 x2 A12 A21 Psat1 i 1 i i
i
x2
2 x1i x2i A 12 A 21 Psat2
A12 A21
A12
Minimize SSE A12 A21
0.758 0.435
Ans.
A21
pcalc
j
X1
j j
1 X 1 j X 2 j A 12 A 21 Psat1 2 X 1 j X 2 j A 12 A 21 Psat2
1 X 1 j X 2 j A 12 A 21 Psat1
X2
X1 Y1calc
j
j
pcalc
j
Pcalc
i
x1
i i
1 x1i x2i A 12 A 21 Psat1 2 x1i x2i A 12 A 21 Psat2
1 x1i x2i A 12 A 21 Psat1
x2
x1 y1calc
i
i
Pcalc
i
RMS deviation in P:
Pi RMS
i
Pcalc n
2
i
RMS
0.167 kPa
381
P-x-y diagram, Margules eqn. by Barker's method
90
Pi kPa
80 70 60 50
j
Pi kPa
pcalc kPa
40
j
pcalc kPa
30 20 10 0 0.2
i
0.4
i j
0.6
j
0.8
1
x1 y1 X1 Y1calc
P-x data P-y data P-x calculated P-y calculated
Residuals in P and y1
1
Pi Pcalc kPa
i
0.5
y1 y1calc 100
i i
0
0.5
0
0.2
0.4
0.6
0.8
x1
Pressure residuals y1 residuals
i
382
(e) BARKER'S METHOD by non-linear least squares. van Laar equation. Guesses for parameters: answers to Part (b).
j 1 101
X1
j
.01 j
.00999
a12 x1 a21 x2
2
X2
j
1
X1
j
1 x1 x2 a12 a21
exp a12 1
2 x1 x2 a12 a21
exp a21 1
a21 x2 a12 x1
2
Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses:
SSE a12 a21
i
a12
0.5
Pi x1
a21
1.0
2
x1 x2 a12 a21 Psat1 i 1 i i
i
x2
2 x1i x2i a12 a21 Psat2
a12 a21
Minimize SSE a12 a21
a12 a21
0.83 0.468
Ans.
pcalc
j
X1
j j
1 X 1 j X 2 j a12 a21 Psat1 2 X 1 j X 2 j a12 a21 Psat2
1 X 1 j X 2 j a12 a21 Psat1
X2
X1 Y1calc
Pcalc
j
j
pcalc
x1
i i
j
i
1 x1i x2i a12 a21 Psat1 2 x1i x2i a12 a21 Psat2
1 x1i x2i a12 a21 Psat1
x2
x1 y1calc
i
i
Pcalc
i
383
RMS deviation in P:
Pi RMS
i
Pcalc n
2
i
RMS
0.286 kPa
P-x,y diagram, van Laar Equation by Barker's Method
90
80
70
Pi kPa
60
Pi kPa
pcalc kPa
50
j
pcalc kPa
j
40
30
20
10
0
0.2
i
0.4
i j
0.6
j
0.8
1
x1 y1 X1 Y1calc
P-x data P-y data P-x calculated P-y calculated
384
Residuals in P and y1.
1
Pi Pcalc kPa
i
0.5
y1 y1calc 100
i i
0
0.5
0
0.2
0.4
0.6
0.8
x1
Pressure residuals y1 residuals
i
(f)
BARKER'S METHOD by non-linear least squares. Wilson equation. Guesses for parameters: answers to Part (c).
j
1 101
1 x1 x2 12 21
X1
j
.01 j
exp
.01
x2 x2
12 12
X2
j
1
X1
j
ln x1 x2 x1
21
12
x2
x1
21
2 x1 x2
12
21
exp
ln x2 x1 x1
x1 x2
21 12 12
21
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
21 i
Pi
x1
x1 x2 i 1 i i
i
12 12
21 Psat1 21 Psat2
2
x2
2 x1i x2i
12 21
Minimize SSE
12
12 21
0.348 1.198
Ans.
21
pcalc
j
X1
j j
1 X1 j X2 j 2 X1 j X2 j
12 12
21 Psat1 21 Psat2
X2
X1 Y1calc
j
j
1 X1 j X2 j
12
j
21 Psat1
pcalc
Pcalc
i
x1
i i
1 x1i x2i 2 x1i x2i
12 12
21 Psat1 21 Psat2
x2
x1 y1calc
i
i
1 x1i x2i
12
i
21 Psat1
Pcalc
RMS deviation in P:
2
i
Pi RMS
i
Pcalc n
RMS
0.305kPa
386
P-x,y diagram, Wilson Equation by Barker's Method
90
Pi kPa
80 70 60 50
j
Pi kPa
pcalc kPa
40
j
pcalc kPa
30 20 10 0 0.2
i
0.4
i j
0.6
j
0.8
1
x1 y1 X1 Y1calc
P-x data P-y data P-x calculated P-y calculated
Residuals in P and y1.
1
Pi Pcalc i kPa
0.5
y1 y1calc 100
i i
0
0.5
0
0.2
0.4
0.6
0.8
x1
Pressure residuals y1 residuals
i
387
12.3
Acetone(1)/Methanol(2)-- VLE data: 72.278 75.279 77.524 78.951 82.528 86.762 90.088 93.206 95.017 P 96.365 97.646 98.462 99.811 99.950 100.278 100.467 100.999 101.059 99.877 99.799
kPa
T
328.15 K
0.0287 0.0570 0.0858 0.1046 0.1452 0.2173 0.2787 0.3579 0.4050 x1 0.4480 0.5052 0.5432 0.6332 0.6605 0.6945 0.7327 0.7752 0.7922 0.9080 0.9448
0.0647 0.1295 0.1848 0.2190 0.2694 0.3633 0.4184 0.4779 0.5135 y1 0.5512 0.5844 0.6174 0.6772 0.6926 0.7124 0.7383 0.7729 0.7876 0.8959 0.9336
n y2 20 1 y1
i 1 n
Number of data points:
Calculate x2 and y2:
n
x2
rowsP) (
1 x1
Vapor Pressures from equilibrium data: Psat1 96.885 kPa
Psat2 68.728 kPa
388
Calculate EXPERIMENTAL values of activity coefficients and excess Gibbs energy.
y1 P
1
y2 P
2
x1 Psat1
1
1 2 3 4 5 6 7 8 9
x2 Psat2
2
i
1.013 1.011 1.006 1.002 1.026 1.027 1.057 1.103 1.13 1.14 1.193 1.2 1.278 1.317 1.374 1.431 1.485 1.503 1.644 1.747
GERT
x1 ln 1
x2 ln 2
i
i
1.682 1.765 1.723 1.706 1.58 1.497 1.396 1.285 1.243 1.224 1.166 1.155 1.102 1.082 1.062 1.045 1.039 1.037 1.017 1.018
ln 1 i
0.52 0.568 0.544 0.534 0.458 0.404 0.334 0.25 0.218 0.202 0.153 0.144 0.097 0.079 0.06 0.044 0.039 0.036 0.017 0.018
ln 2 i
0.013 0.011 5.81510-3 1.97510-3 0.026 0.027 0.055 0.098 0.123 0.131 0.177 0.182 0.245 0.275 0.317 0.358 0.395 0.407 0.497 0.558
GERTi
0.027 0.043 0.052 0.058 0.089 0.108 0.133 0.152 0.161 0.163 0.165 0.162 0.151 0.145 0.139 0.128 0.119 0.113 0.061 0.048
10 11 12 13 14 15 16 17 18 19 20
389
0.6
ln 1 i ln 2 i GERTi
0.4
0.2
0
0
0.2
0.4 x1
i
0.6
0.8
(a) Fit GE/RT data to Margules eqn. by linear least squares: VXi
Slope
Slope
A12
A12
x1
i
VYi
GERTi x1 x2 i i
intercept ( VX VY)
0.708
A12
Ans.
slope ( VX VY)
0.018
Intercept
0.708
Intercept
Intercept
A21
A21
Slope
0.69
The following equations give CALCULATED values:
1 ( x2) x1
exp x2
2
A12
2 A21
A12 x1
2 ( x2) x1
j 1 101
X1
exp x1
2
A21
X1
j
2 A12
.01 j .01
X2
A21 x2
X2
j
1
X1
j
pcalc
j
j
1 X1 X2 Psat1 j j
j
j
2 X1 X2 Psat2 j j
X1 Y1calc
j
1 X1 X2
j
j
Psat1
pcalc
j
390
P-x,y Diagram: Margules eqn. fit to GE/RT data.
105 100
Pi kPa
95 90 85
j
Pi kPa
pcalc kPa
80
j
pcalc kPa
75 70 65
0
0.2
0.4
i i j
0.6
j
0.8
x1 y1 X1 Y1calc
P-x data P-y data P-x calculated P-y calculated
Pcalc
i
x1 1 x1 x2 Psat1
i i i
x2 2 x1 x2 Psat2
i i i
y1calc
i
x1 1 x1 x2 Psat1 i i i Pcalc
i
RMS deviation in P:
Pi RMS
i
Pcalc n
2
i
RMS
0.851 kPa
391
(b) Fit GE/RT data to van Laar eqn. by linear least squares:
x1 x2
VXi
Slope
Slope
a12
a12
x1
i
VYi
i
i
GERTi
intercept ( VX VY)
1.442
1 ( Slope
0.686
2
slope ( VX VY)
0.015
1 Intercept
0.693
Intercept
Intercept
a21
a21 a12 x1 a21 x2
2
Intercept)
Ans.
1 ( x2) x1
exp a12 1
2 ( x2) x1
exp a21 1
X1
X2
a21 x2 a12 x1
j
1 101
j
j
.01 j
1 X1
.00999
j
(To avoid singularities)
pcalc
Pcalc
j
X1
j
1 X1 X2
j
j
Psat1
X2
j
2 X1 X2
j
j
Psat2
i
x1 1 x1 x2 Psat1
i i i
x2 2 x1 x2 Psat2
i i i
X1 Y1calc
j
j
1 X1 X2 Psat1 j j pcalc
j
y1calc
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 kPa
95 90 85
j
Pi kPa
pcalc kPa
80
j
pcalc kPa
75 70 65
0
0.2
i
0.4
i j
0.6
j
0.8
1
x1 y1 X1 Y1calc
P-x data P-y data P-x calculated P-y calculated
RMS deviation in P:
Pi RMS
i
Pcalc n
2
i
RMS
0.701 kPa
(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:
12
0.5
21
1.0
SSE
12
21 i
GERTi
x1 ln x1 i i x2 ln x2
i i
x2
2
i i
12 21
x1
393
12 21
Minimize SSE
12
12 21
0.71 0.681
Ans.
21 12
exp x2 1 ( x2) x1
21
12
x1
x2
x1
x2
12
x1
21
x2
12
exp 2 ( x2) x1
x1
21
12
x1
x2
x2
x2
21
x1
21
x1
j
1 101
X1
j
.01 j
.01
X2
j
1
X1
j
pcalc
j
X1
j
1 X1 X2 Psat1 j j
X2
j
2 X1 X2 Psat2 j j
Pcalc
i
x1 1 x1 x2 Psat1 i i i
x2 2 x1 x2 Psat2 i i i
X1 Y1calc
j
1 X1 X2 Psat1 j j j pcalc
j
y1calc
i
x1 1 x1 x2 Psat1 i i i Pcalc
i
394
P-x,y diagram: Wilson eqn. fit to GE/RT data.
105 100
Pi kPa
95 90 85
j
Pi kPa
pcalc kPa
80
j
pcalc kPa
75 70 65 0 0.2
i
0.4
i j
0.6
j
0.8
1
x1 y1 X1 Y1calc
P-x data P-y data P-x calculated P-y calculated
RMS deviation in P:
Pi RMS
i
Pcalc n
2
i
RMS
0.361 kPa
(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
2 x1 x2 A 12 A 21
exp ( x2)
exp ( x1)
2
2
A12
A21
2 A21
2 A12
A12 x1
A21 x2
395
Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses:
SSE A12 A21
i
A12
0.5
Pi x1
A21
1.0
2
x1 x2 A12 A21 Psat1 i 1 i i
i
x2
2 x1i x2i A 12 A 21 Psat2
A12 A21
A12
Minimize SSE A12 A21
0.644 0.672
Ans.
A21
pcalc
j
X1
j j
1 X 1 j X 2 j A 12 A 21 Psat1 2 X 1 j X 2 j A 12 A 21 Psat2
1 X 1 j X 2 j A 12 A 21 Psat1
X2
X1 Y1calc
j
j
pcalc
j
Pcalc
i
x1
i i
1 x1i x2i A 12 A 21 Psat1 2 x1i x2i A 12 A 21 Psat2
1 x1i x2i A 12 A 21 Psat1
x2
x1 y1calc
i
i
Pcalc
i
RMS deviation in P:
Pi RMS
i
Pcalc n
2
i
RMS
0.365 kPa
396
P-x-y diagram, Margules eqn. by Barker's method
105
Pi kPa
100 95 90 85
j
Pi kPa
pcalc kPa
80
j
pcalc kPa
75 70 65 0 0.2
i
0.4
i j
0.6
j
0.8
1
x1 y1 X1 Y1calc
P-x data P-y data P-x calculated P-y calculated
Residuals in P and y1
2
Pi Pcalc kPa
i
1
y1 y1calc 100
i i
0
1
0
0.2
0.4
x1
i
0.6
0.8
Pressure residuals y1 residuals
397
(e) BARKER'S METHOD by non-linear least squares. van Laar equation. Guesses for parameters: answers to Part (b).
j 1 101
X1
j
.01 j
.00999
a12 x1 a21 x2
2
X2
j
1
X1
j
1 x1 x2 a12 a21
exp a12 1
2 x1 x2 a12 a21
exp a21 1
a21 x2 a12 x1
2
Minimize the sum of the squared errors using the Mathcad Minimize function. Guesses:
SSE a12 a21
i
a12
0.5
Pi x1
a21
1.0
2
x1 x2 a12 a21 Psat1 i 1 i i
i
x2
2 x1i x2i a12 a21 Psat2
a12 a21
Minimize SSE a12 a21
a12 a21
0.644 0.672
Ans.
pcalc
j
X1
j j
1 X 1 j X 2 j a12 a21 Psat1 2 X 1 j X 2 j a12 a21 Psat2
1 X 1 j X 2 j a12 a21 Psat1
X2
X1 Y1calc
Pcalc
j
j
pcalc
x1
i i
j
i
1 x1i x2i a12 a21 Psat1 2 x1i x2i a12 a21 Psat2
1 x1i x2i a12 a21 Psat1
x2
x1 y1calc
i
i
Pcalc
i
398
RMS deviation in P:
Pi RMS
i
Pcalc n
2
i
RMS
0.364 kPa
P-x,y diagram, van Laar Equation by Barker's Method
105
100
95
Pi kPa
90
Pi kPa
pcalc kPa
85
j
pcalc kPa
j
80
75
70
65
0
0.2
i
0.4
i j
0.6
j
0.8
1
x1 y1 X1 Y1calc
P-x data P-y data P-x calculated P-y calculated
399
Residuals in P and y1.
1.5 1
Pi Pcalc kPa
i
0.5 0 0.5
1
y1 y1calc 100
i i
0
0.2
0.4
x1
i
0.6
0.8
Pressure residuals y1 residuals
(f)
BARKER'S METHOD by non-linear least squares. Wilson equation. Guesses for parameters: answers to Part (c). 1 101
X1
12 21
j
j
.01 j
.01
x2 x2
12 12
X2
j
1
X1
j
1 x1 x2
exp ln x1 x2 x1
21
12
x2
x1
21
2 x1 x2
12
21
exp ln x2 x1 x1
x1 x2
21 12 12
21
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
21 i
Pi
x1
x1 x2 i 1 i i
i
12 12
21 Psat1 21 Psat2
2
x2
2 x1i x2i
12 21
Minimize SSE
12
12 21
0.732 0.663
Ans.
21
pcalc
j
X1
j j
1 X1 j X2 j 2 X1 j X2 j
12 12
21 Psat1 21 Psat2
X2
X1 Y1calc
j
j
1 X1 j X2 j
12
j
21 Psat1
pcalc
Pcalc
i
x1
i i
1 x1i x2i 2 x1i x2i
12 12
21 Psat1 21 Psat2
x2
x1 y1calc
i
i
1 x1i x2i
12
i
21 Psat1
Pcalc
RMS deviation in P:
Pi RMS
i
Pcalc n
2
i
RMS
0.35 kPa
401
P-x,y diagram, Wilson Equation by Barker's Method
105
Pi kPa
100 95 90 85
j
Pi kPa
pcalc kPa
80
j
pcalc kPa
75 70 65 0 0.2
i
0.4
i j
0.6
j
0.8
1
x1 y1 X1 Y1calc
P-x data P-y data P-x calculated P-y calculated
Residuals in P and y1.
2
Pi Pcalc kPa
i
1
y1 y1calc 100
i i
0
1
0
0.2
0.4
x1
i
0.6
0.8
Pressure residuals y1 residuals
402
12.6
Methyl t-butyl ether(1)/Dichloromethane--VLE data: T 83.402 82.202 80.481 76.719 72.442 68.005 P 65.096 59.651 56.833 53.689 51.620 50.455 49.926 49.720 x2 1 x1
kPa
308.15 K 0.0141 0.0253 0.0416 0.0804 0.1314 0.1975 y1 0.2457 0.3686 0.4564 0.5882 0.7176 0.8238 0.9002 0.9502
0.0330 0.0579 0.0924 0.1665 0.2482 0.3322 x1 0.3880 0.5036 0.5749 0.6736 0.7676 0.8476 0.9093 0.9529
y2
Psat2
1
y1
85.265 kPa
Psat1
49.624 kPa
Calculate EXPERIMENTAL values of activity coefficients and excess Gibbs energy.
y1 P
1
y2 P
2
x1 Psat1
x2 Psat2
GERT
x1 ln 1
x2 ln 2
GERTx1x2
GERT x1 x2
n
rows ( ) P
n
14
i
1 n
403
(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
A21 x1
i
C A12 x2
i
0.2 C x1 x2
i i
SSE A12 A21 C
i
GERTi
x1 x2
i
2
i
A12 A21 C
Minimize SSE A12 A21 C
A12 A21 C
0.336 0.535 0.195
Ans.
(b) Plot data and fit GeRTx1x2 ( x2) x1
GeRT ( x2) x1 ln 1 ( x2) x1
A21 x1
A12 x2
C x1 x2
GeRTx1x2 ( x2)x1 x2 x1 x2
2
A12
2 A21
A12
C x1
3 C x1
2
ln 2 ( x2) x1
j 1 101
x1
2
A21
2 A12
X1
j
A21
C x2
3 C x2
X2
j
2
.01 j
.01
1
X1
j
GERTx1x2 i GeRTx1x2 X1 X2
j j
0 0.1
0.2
ln 1 i ln 1 X1 X2
j j
0.3
0.4
ln 2 i ln 2 X1 X2
j j
0.5
0.6
0
0.2
i
0.4
j i
0.6
j i
0.8
j
x1 X1 x1 X1 x1 X1
404
(c) Plot Pxy diagram with fit and data
1 ( x1 x2)
exp ln 1 ( x1 x2)
2 ( x1 x2)
exp ln 2 ( x1 x2)
j
Pcalc
j
X1
1 X1 X2 Psat1 j j
j
X2
j
2 X1 X2 Psat2 j j
X1 y1calc
j
1 X1 X2 Psat1 j j Pcalc
j
P-x,y Diagram from Margules Equation fit to GE/RT data.
90
Pi kPa
80
Pi kPa
70
j
Pcalc kPa
60
j
Pcalc kPa
50
40
0
0.2
0.4
i i j
0.6
j
0.8
x1 y1 X1 y1calc
P-x data P-y data P-x calculated P-y calculated
(d) Consistency Test:
GERTi
1i 2
i
GeRT x1 x2 i i
GERTi
ln 1 2i
ln
1 x1 x2 i i 2 x1 x2 i i
ln
405
0.004
0
GERTi
0
ln 1 2i
0.025
0.004
0
0.5
1
0.05
0
0.5
1
x1
i
x1
i
Calculate mean absolute deviation of residuals
4
mean
GERT
9.391
10
mean
ln 1 2
0.021
(e) Barker's Method by non-linear least squares: Margules Equation
1 x1 x2 A 12 A 21 C
exp ( ) A12 x2
2
2 A21
2
A12
C x1
3 C x1
2 x1 x2 A 12 A 21 C
exp ( ) A21 x1
2
2 A12
2
A21
C x2
3 C x2
Minimize sum of the squared errors using the Mathcad Minimize function. Guesses: A12 0.3
Pi
i
A21
x1
0.5
C
0.2
2
SSE A12 A21 C
x1 x2 A12 A21 C Psat1 i 1 i i
i
x2
2 x1i x2i A 12 A 21 C Psat2
A12 A21 C
Minimize SSE A12 A21 C
A12 A21 C
0.364 0.521 0.23 Ans.
406
Plot P-x,y diagram for Margules Equation with parameters from Barker's Method.
Pcalc
j
X1
j j
1 X 1 j X 2 j A 12 A 21 C Psat1 2 X 1 j X 2 j A 12 A 21 C Psat2
X2
X1 y1calc
j
j
1 X 1 j X 2 j A 12 A 21 C Psat1
Pcalc
90
j
Pi kPa
80
Pi kPa
70
j
Pcalc kPa
60
j
Pcalc kPa
50
40
0
0.2
0.4
i i j
0.6
j
0.8
x1 y1 X1 y1calc
P-x data P-y data P-x calculated P-y calculated
Pcalc
i
x1
i i
1 x1i x2i A 12 A 21 C Psat1 2 x1i x2i A 12 A 21 C Psat2
x2
x1 y1calc
i
i
1 x1i x2i A 12 A 21 C Psat1
Pcalc
i
407
Plot of P and y1 residuals.
0.8
0.6
Pi Pcalc i kPa
0.4
y1 y1calc 100 0.2 i i 0
0.2
0
0.5
1
x1
i
Pressure residuals y1 residuals
RMS deviations in P:
Pi RMS
i
Pcalc n
2
i
RMS
0.068 kPa
408
12.8
(a)
Data:
0.0523 0.1299 0.2233 0.2764 0.3482 0.4187 x1 0.5001 0.5637 0.6469 0.7832 0.8576 0.9388 0.9813
1
1.202 1.307 1.295 1.228 1.234 1.180 1.129 1.120 1.076 1.032 1.016 1.001 1.003
i 1 n
n 13
2
1.002 1.004 1.006 1.024 1.022 1.049 1.092 1.102 1.170 1.298 1.393 1.600 1.404
n
rows x1
x1 ln 1 i i
x2
i
1
x1
i
GERTi
x2 ln 2 i i
(b) Fit GE/RT data to Margules eqn. by linear least-squares procedure: Xi
Slope
Slope
A12
A12
x1
i
Yi
slope ( Y) X
GERTi x1 x2 i i
intercept ( Y) X
0.286
A12
Ans.
Intercept
Intercept
A21
A21
0.247
Intercept
0.286
exp x2 exp x1
2 2
Slope
0.534
1 ( x2) x1 2 ( x2) x1
A12 A21
2 A21 2 A12
A12 x1 A21 x2
GeRT ( x2) x1
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 ln 1 i ln 1 x1 x2 i i ln 2 i ln 2 x1 x2 i i
0.4
0.3
0.2
0.1
0
0
0.2
0.4
x1
i
0.6
0.8
(c) Calculate and plot residuals for consistency test:
GERTi
GeRT x1 x2 i i
GERTi
1i 2
i
ln 1 2i
ln
1 x1 x2 i i 2 x1 x2
i i
ln
410
GERTi
3.31410-3 -2.26410-3 -3.1410-3 -2.99810-3 -2.87410-3 -2.2210-3 -2.17410-3 -1.55310-3 -8.74210-4 2.94410-4 5.96210-5 9.02510-5 4.23610-4
ln 1 2i
0.098 -9.15310-5 -0.021 0.026 -0.019 5.93410-3 0.028 -9.5910-3 9.13910-3 -5.61710-4 -0.011 0.028 -0.168
0.1
0.05 ln 1 2i 0
0
0.5 x1
i
1
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 0.1056 0.1818 0.2783 0.3607 0.4274
31.957 33.553 35.285 36.457 36.996 P 37.068 36.978 36.778 35.792 34.372 32.331 30.038
kPa
0.0455 0.0940 0.1829 0.2909 0.3980 x1 0.5069 0.5458 0.5946 0.7206 0.8145 0.8972 0.9573
411
y1
0.4885 0.5098 0.5375 0.6157 0.6913 0.7869 0.8916
x2
Psat1
1
x1
27.778 kPa
y2
1
y1
Psat2
29.819 kPa
Calculate EXPERIMENTAL values of activity coefficients and excess Gibbs energy.
y1 P
1
y2 P
2
x1 Psat1
x2 Psat2
GERT
x1 ln 1
x2 ln 2
GERTx1x2
GERT x1 x2
n
rows ( ) P
n
12
i
1 n
(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
i
GERTi
A21 x1 i
A12 x2 i
C x1 x2
i
i
x1 x2 i i
2
A12 A21 C
Minimize SSE A12 A21 C
A12 A21 C
1.128 1.155 0.53
Ans.
(b) Plot data and fit GeRTx1x2 ( x2) x1
GeRT ( x2) x1 ln 1 ( x2) x1
A21 x1
A12 x2
C x1 x2
GeRTx1x2 ( x2)x1 x2 x1 x2
2
A12
2 A21
A12
C x1
3 C x1
2
ln 2 ( x2) x1
j 1 101
x1
2
A21
2 A12
X1
j
A21
C x2
3 C x2
X2
j
2
.01 j
.01
1
X1
j
412
1.2
GERTx1x2 i GeRTx1x2 X1 X2
j j
1
0.8 0.6
ln 1 i ln 1 X1 X2
j j
ln 2 i ln 2 X1 X2
j j
0.4
0.2
0
0
0.2
i
0.4
j i
0.6
j i
0.8
j
x1 X1 x1 X1 x1 X1
(c) Plot Pxy diagram with fit and data
1 ( x1 x2)
exp ln 1 ( x1 x2)
2 ( x1 x2)
exp ln 2 ( x1 x2)
j
Pcalc
j
X1
1 X1 X2
j
j
Psat1
X2
j
2 X1 X2
j
j
Psat2
X1 y1calc
j
j
1 X1 X2 Psat1 j j Pcalc
j
413
P-x,y Diagram from Margules Equation fit to GE/RT data.
38
Pi kPa
36
Pi kPa
34
Pcalc kPa
32
j
30
Pcalc kPa
j
28
26
0
0.2
0.4
i i j
0.6
j
0.8
x1 y1 X1 y1calc
P-x data P-y data P-x calculated P-y calculated
(d) Consistency Test:
GERTi
1i 2
i
GeRT x1 x2
i
i
GERTi
ln 1 2i
ln
1 x1 x2 i i 2 x1 x2 i i
ln
0.004
0
GERTi
0
ln 1 2i
0.025
0.004
0
0.5
1
0.05
0
0.5
1
x1
i
x1
i
414
Calculate mean absolute deviation of residuals
4
mean
GERT
6.237
10
mean
ln 1 2
0.025
(e) Barker's Method by non-linear least squares: Margules Equation
1 x1 x2 A 12 A 21 C
exp ( ) A12 x2
2
2 A21
2
A12
C x1
3 C x1
2 x1 x2 A 12 A 21 C
exp ( ) A21 x1
2
2 A12
2
A21
C x2
3 C x2
Minimize sum of the squared errors using the Mathcad Minimize function. Guesses: A12 0.3
Pi
i
A21
x1
0.5
C
0.2
2
SSE A12 A21 C
x1 x2 A12 A21 C Psat1 i 1 i i
i
x2
2 x1i x2i A 12 A 21 C Psat2
A12 A21 C
Minimize SSE A12 A21 C
A12 A21 C
1.114 1.098 0.387
Ans.
Plot P-x,y diagram for Margules Equation with parameters from Barker's Method. Pcalc
j
X1
j j
1 X 1 j X 2 j A 12 A 21 C Psat1 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
X2
X1 y1calc
j
j
Pcalc
j
415
38
Pi kPa
36
Pi kPa
34
Pcalc kPa
32
j
Pcalc kPa
30
j
28
26
0
0.2
0.4
i i j
0.6
j
0.8
x1 y1 X1 y1calc
P-x data P-y data P-x calculated P-y calculated
Pcalc
i
x1
i i
1 x1i x2i A 12 A 21 C Psat1 2 x1i x2i A 12 A 21 C Psat2
x2
x1 y1calc
i
i
1 x1i x2i A 12 A 21 C Psat1
Pcalc
i
416
Plot of P and y1 residuals.
0.6 0.4
Pi Pcalc i kPa
0.2 0 0.2
0.4
y1 y1calc 100
i i
0
0.5
1
x1
i
Pressure residuals y1 residuals
RMS deviations in P:
Pi RMS
i
Pcalc n
2
i
RMS
0.04 kPa
417
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:
Water:
A1
A2
16.1154
16.3872
B1
B2
3483.67 K
3885.70 K
C1
C2
205.807 K
230.170 K
Psat1 ( T)
exp A1
(T
B1 273.15 K)
C1
kPa
Psat2 ( ) T
exp A2
(T
B2 273.15 K)
C2
kPa
Parameters for the Wilson equation:
V1 cm 75.14 mol
775.48
3
V2
cm 18.07 mol
1351.90
3
a12
cal mol
a21
cal mol
12 ( ) T
a12 V2 exp RT V1 12 ( ) T x1
21 ( ) T
a21 V1 exp RT V2 21 ( ) T x1 21 ( ) T
exp x2 1 ( x2 T) x1
x2 12 ( ) x2 T
x1 x2 12 ( ) T
exp 2 ( x2 T) x1
x1
12 ( ) T x1 x2 12 ( ) x2 T x2 x1
418
21 ( ) T x1 21 ( ) T
21 ( ) T
P-x,y diagram at
T
( 60
273.15) K
Guess:
P
70 kPa
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) (
x1 1 ( x1 1
x1 T) Psat1 ( T)
Peq x1) (
x
0 0.05 1.0
x
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
yeq x) (
0 0.315 0.363 0.383 0.395 0.404 0.413 0.421 0.431 0.441 0.453 0.466 0.483 0.502 0.526 0.556 0.594 0.646 0.718 0.825 1
Peq x) ( kPa
20.007 28.324 30.009 30.639 30.97 31.182 31.331 31.435 31.496 31.51 31.467 31.353 31.148 30.827 30.355 29.686 28.759 27.491 25.769 23.437 20.275
419
P,x,y Diagram at T
32
333.15 K
30
28
Peq x) ( kPa
Peq x) ( kPa
26
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
Water:
Psat1 ( ) T
16.1154
16.3872
B1
B2
3483.67 K
3885.70 K
kPa
C1
C2
205.807 K
230.170 K
A2
exp A1
( T
B1 273.15 K) C1
B2 273.15 K) C2
420
Psat2 ( ) T
exp A2
( T
kPa
Parameters for the Wilson equation:
V1 75.14 cm
3
mol
cal mol
V2
18.07
cm
3
mol
cal mol
a12
775.48
a21
1351.90
12 ( T)
a12 V2 exp RT V1 exp x2 12 ( T) x1
21 ( T)
a21 V1 exp RT V2 21 ( T)
1 ( x1 x2 T)
x2 12 ( T)
x1 x2
x2
12 ( T)
x1
21 ( T)
exp 2 ( x1 x2 T)
x1
12 ( T) x1 x2 12 ( T)
x2 x1
21 ( T) x2 x1 21 ( T)
21 ( T)
T-x,y diagram at P
Guess:
Given
101.33 kPa
273.15) K
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
Teq ( x1)
yeq ( x1)
x
0 0.05 1.0
421
x
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
yeq x ()
0 0.304 0.358 0.381 0.395 0.407 0.418 0.429 0.44 0.453 0.468 0.484 0.504 0.527 0.555 0.589 0.631 0.686 0.759 0.858 1
Teq x () K
373.149 364.159 362.476 361.836 361.49 361.264 361.101 360.985 360.911 360.881 360.904 360.99 361.154 361.418 361.809 362.364 363.136 364.195 365.644 367.626 370.349
T,x,y Diagram at P
375
101.33 kPa
Teq( ) 370 x K
Teq( ) x K
365
360
0
0.2
0.4
0.6
0.8
1
x yeq x) (
422
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
Water:
Psat1 ( T)
16.1154
16.3872
B1
B2
B1 273.15 K)
3483.67 K
3885.70 K
kPa
C1
C2
205.807 K
230.170 K
A2
exp A1
(T
C1
Psat2 ( T)
exp A2
(T
B2 273.15 K)
C2
kPa
Parameters for the NRTL equation:
b12 500.40
b12 RT
exp 12 ( T)
cal mol
b21
1636.57
cal mol
b21 RT
exp
2
0.5081
12 ( T)
G12 ( T)
21 ( T)
G21 ( T)
21 ( T)
1 ( x1 x2 T)
exp x2
2
21 ( T)
G21 ( T) x1 x2 G21 ( T) G12 ( T) 12 ( T)
x1 G12 ( T) )
2
( x2
2 ( x1 x2 T)
exp x1
2
12 ( T)
G12 ( T)
2
x2 x1 G12 ( T) G21 ( T) 21 ( T) x2 G21 ( T) )
423
( x1
2
P-x,y diagram at
T
( 60
273.15)K
Guess:
P
70 kPa
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) (
x1 1 ( 1 x1
x1 T)Psat1 ( ) T
Peq x1) (
x
0 0.05 1.0
x
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
yeq x ()
0 0.33 0.373 0.382 0.386 0.39 0.395 0.404 0.414 0.427 0.442 0.459 0.479 0.503 0.531 0.564 0.606 0.659 0.732 0.836 1
Peq x () kPa
20.007 28.892 30.48 30.783 30.876 30.959 31.048 31.127 31.172 31.163 31.085 30.922 30.657 30.271 29.74 29.03 28.095 26.868 25.256 23.124 20.275
424
P,x,y Diagram at T
35
333.15 K
Peq ( x) kPa
30
Peq ( x) kPa
25
20
0
0.2
0.4
x yeq ( x)
0.6
0.8
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
Water:
Psat1 ( T)
B1
B2
3483.67 K
3885.70 K
kPa
C1
C2
205.807 K
230.170 K
A2
exp A1
16.3872
(T
B1 273.15 K)
C1
Psat2 ( T)
exp A2
(T
B2 273.15 K)
C2
kPa
Parameters for the NRTL equation:
b12 500.40 cal mol
b21 1636.57
425
cal mol
0.5081
12 ( ) T
b12 RT
21 ( ) T
b21 RT
G12 ( ) T
exp
12 ( ) T
G21 ( ) T
exp
2
21 ( ) T
1 ( x2 T) x1
exp x2
2
G21 ( ) T 21 ( ) T x1 x2 G21 ( ) T G12 ( ) 12 ( ) T T
( x2 x1 G12 ( ) T)
2
2 ( x2 T) x1
exp x1
2
G12 ( ) T 12 ( ) T x2 x1 G12 ( ) T G21 ( ) 21 ( ) T T
( x1 x2 G21 ( ) T)
2
2
T-x,y diagram at
P
101.33 kPa
Guess:
T
( 90
273.15)K
Given
P = x1 1 ( 1 x1 T)Psat1 ( ) x1 T ( x1) 2 ( 1 x1 T)Psat2 ( ) 1 x1 T
Teq x1) (
Find T) (
yeq x1) (
x1 1 ( 1 x1
x1 Teq x1) Psat1 ( ( ) ( ) Teq x1 ) P
426
x
0 0.05 1.0
x
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
yeq ( x)
0 0.32 0.377 0.394 0.402 0.408 0.415 0.424 0.434 0.447 0.462 0.48 0.5 0.524 0.552 0.586 0.629 0.682 0.754 0.853 1
Teq ( x) K
373.149 363.606 361.745 361.253 361.066 360.946 360.843 360.757 360.697 360.676 360.709 360.807 360.985 361.262 361.66 362.215 362.974 364.012 365.442 367.449 370.349
T,x,y Diagram at P
375
101.33 kPa
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:
Water:
Psat1 ( ) T
A1
A2
exp A1
16.1154
16.3872
B1
B2
3483.67 K
3885.70 K
kPa
C1
C2
205.807 K
230.170 K
( T
B1 273.15 K) C1
Psat2 ( ) T
exp A2
( T
B2 273.15 K) C2
kPa
Parameters for the Wilson equation:
V1 cm 75.14 mol
775.48
3
V2
cm 18.07 mol
1351.90
3
a12
cal mol
a21
cal mol
12 ( ) T
a12 V2 exp RT V1
21 ( ) T
a21 V1 exp RT V2
exp x2 1 ( x2 T) x1
12 ( ) T x1 x2 12 ( ) x2 T
x1 x2 12 ( ) T
21 ( ) T x1 21 ( ) T
exp 2 ( x2 T) x1
x1
12 ( ) T x1 x2 12 ( ) x2 T x2 x1
428
21 ( ) T x1 21 ( ) T
21 ( ) T
(a) BUBL P:
T
( 60
273.15) K
x1
0.3
x2
1
x1
Guess:
P
101.33 kPa
y1
0.4
y2
1
y1
Given
y1 P = x1 1 ( x1 x2 T) Psat1 ( T)
y1
y2 P = x2 2 ( x1 x2 T) Psat2 ( T)
y2 = 1
Pbubl y1 y2
Pbubl 31.33 kPa
Find ( P y1 y2)
y1
0.413
y2
0.587
Ans.
(b) DEW P:
T
( 60
273.15) K
y1
0.3
y2
1
y1
Guess:
Given
P
101.33 kPa
x1
0.1
x1
x2
x2 = 1
1
x1
y1 P = x1 1 ( x1 x2 T) Psat1 ( T)
y2 P = x2 2 ( x1 x2 T) Psat2 ( T)
Pdew x1 x2
Pdew 27.79 kPa
Find ( P x1 x2)
x1
0.042
x2
0.958
Ans.
(c) P,T-flash Calculation
P
Pdew 2
Pbubl
T
( 60
273.15) K
z1
0.3
Guess:
V
0.5
x1 y1
0.1 0.1
x2 y2
1 1
y1 x1
Given
y1 =
x1 1 ( x1 x2 T) Psat1 ( T) P
x1
x2 = 1
y2 =
x2 2 ( x1 x2 T) Psat2 ( T) P
429
y1
y2 = 1
x1 ( 1
V) y1 V = z1
Eq. (10.15)
x1 x2 y1 y2 V
x1 0.08
Find x1 x2 y1 y2 V) (
x2
0.92
y1
0.351
y2
0.649
V
0.813
(d) Azeotrope Calculation Test for azeotrope at:
T
( 60
273.15)K
1 ( 1 T) 21.296 0
2 ( 0 T) 4.683 1
120
1 ( 1 T)Psat1 ( ) 0 T Psat2 ( ) T
120
21.581
121
Psat1 ( ) T 2 ( 0 T)Psat2 ( ) 1 T
121
0.216
Since one of these values is >1 and the other is <1, an azeotrope exists. See Ex. 10.3(e)
Guess:
P
101.33 kPa
x1 y1
0.3 0.3
x2 y2
1 1
y1 x1
Given
y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T
y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T
x1
x2 = 1
y1
y2 = 1
x1 = y1
x1 x2 y1 y2 Paz
Paz 31.511 kPa
Find x1 x2 y1 y2 P) (
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
Water:
Psat1 ( T)
16.1154
16.3872
B1
B2
B1 273.15 K)
3483.67 K
3885.70 K
kPa
C1
C2
205.807 K
230.170 K
A2
exp A1
(T
C1
Psat2 ( T)
exp A2
(T
B2 273.15 K)
C2
kPa
Parameters for the NRTL equation:
b12 500.40
b12 RT
exp 12 ( T)
cal mol
b21
1636.57
cal mol
b21 RT
exp
2
0.5081
12 ( T)
21 ( T)
G21 ( T)
G12 ( T)
21 ( T)
1 ( x1 x2 T)
exp x2
2
21 ( T)
G21 ( T) x1 x2 G21 ( T) G12 ( T) 12 ( T)
x1 G12 ( T) )
2
( x2
2 ( x1 x2 T)
exp x1
2
12 ( T)
G12 ( T)
2
x2 x1 G12 ( T) G21 ( T) 21 ( T) x2 G21 ( T) )
431
( x1
2
(a) BUBL P: T
( 60
273.15)K
x1
0.3
x2
1
x1
Guess:
P
101.33 kPa
y1
0.4
y2
1
y1
Given
y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T
y1
y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T
y2 = 1
Pbubl y1 y2
Pbubl 31.05 kPa
Find ( y1 y2) P
y1
0.395
y2
0.605
Ans.
(b) DEW P:
T
( 60
273.15)K
y1
0.3
y2
1
y1
Guess:
P
101.33 kPa
x1
0.1
x2
1
x1
Given
y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T
x1
y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T
x2 = 1
Pdew x1 x2
Pdew 27.81 kPa
Find ( x1 x2) P
x1
0.037
x2
0.963
Ans.
(c) P,T-flash Calculation
P
Pdew 2
Pbubl
T
( 60
273.15)K
z1
0.3
Guess:
V
0.5
x1 y1
0.1 0.1
x2 y2
1 1
y1 x1
Given
y1 =
x1 1 ( x2 T)Psat1 ( ) x1 T P
x1
x2 = 1
y2 =
x2 2 ( x2 T)Psat2 ( ) x1 T P
y1
y2 = 1
x1 ( 1
V) y1 V = z1
Eq. (10.15)
432
x1 x2 y1 y2 V
x1 0.06
Find ( x1 x2 y1 y2 V)
x2
0.94
y1
0.345
y2
0.655
V
0.843
(d) Azeotrope Calculation Test for azeotrope at: T
( 60
273.15) K
1 ( 0 1 T)
19.863
2 ( 1 0 T)
4.307
120
1 ( 0 1 T) Psat1 ( T) Psat2 ( T)
120
20.129
121
Psat1 ( T) 2 ( 1 0 T) Psat2 ( T)
121
0.235
Since one of these values is >1 and the other is <1, an azeotrope exists. See Ex. 10.3(e). Guess:
P 101.33 kPa
x1 y1
0.3 0.3
x2 y2
1 1
x1 x1
Given
y1 P = x1 1 ( x1 x2 T) Psat1 ( T)
y2 P = x2 2 ( x1 x2 T) Psat2 ( T)
x1
x2 = 1
y1
y2 = 1
x1 = y1
x1 x2 y1 y2 Paz
Paz 31.18 kPa
Find ( x1 x2 y1 y2 P)
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:
Water:
Psat1 ( ) T
A1
A2
exp A1
16.1154
16.3872
B1
B2
3483.67 K
3885.70 K
kPa
C1
C2
205.807 K
230.170 K
( T
B1 273.15 K) C1
Psat2 ( ) T
exp A2
( T
B2 273.15 K) C2
kPa
Parameters for the Wilson equation:
V1 cm 75.14 mol
775.48
3
V2
cm 18.07 mol
1351.90
3
a12
cal mol
a21
cal mol
12 ( ) T
a12 V2 exp RT V1
21 ( ) T
a21 V1 exp RT V2 21 ( ) T x1 21 ( ) T
exp x2 1 ( x2 T) x1
12 ( ) T x1 x2 12 ( ) x2 T
x1 x2 12 ( ) T
exp 2 ( x2 T) x1
x1
12 ( ) T x1 x2 12 ( ) x2 T x2 x1 21 ( ) T
21 ( ) T x1 21 ( ) T
434
(a) BUBL T:
P
101.33 kPa
x1
0.3
x2
1
x1
Guess:
T
( 60
273.15) K
y1
0.3
y2
1
y1
Given
y1 P = x1 1 ( x1 x2 T) Psat1 ( T)
y1
y2 P = x2 2 ( x1 x2 T) Psat2 ( T)
y2 = 1
Tbubl y1 y2
Tbubl 361.1 K
Find ( T y1 y2)
y1
0.418
y2
0.582
Ans.
(b) DEW T:
P
101.33 kPa
y1
0.3
y2
1
x1
Guess:
T
( 60
273.15) K
x1
0.1
x1
x2
x2 = 1
1
y1
Given
y1 P = x1 1 ( x1 x2 T) Psat1 ( T)
y2 P = x2 2 ( x1 x2 T) Psat2 ( T)
Tdew x1 x2
Tdew 364.28 K
Find ( T x1 x2)
x1
0.048
x2
0.952
Ans.
(c) P,T-flash Calculation
T
Tdew 2
Tbubl
P
101.33 kPa
z1
0.3
Guess:
V
0.5
x1 y1
0.1 0.1
x2 y2
1 1
y1 x1
Given
y1 =
x1 1 ( x1 x2 T) Psat1 ( T) P
x1
x2 = 1
y2 =
x2 2 ( x1 x2 T) Psat2 ( T) P
y1
y2 = 1
x1 ( 1
V)
y1 V = z1
Eq. (10.15)
435
x1 x2 y1 y2 V
x1 0.09
Find x1 x2 y1 y2 V) (
x2
0.91
y1
0.35
y2
0.65
V
0.807
(d) Azeotrope Calculation Test for azeotrope at: P
101.33 kPa
C1
273.15 K
Tb1 A1
B1 P ln kPa
B2 A2 P ln kPa
Tb1
370.349 K
Tb2
C2
273.15 K
Tb2
373.149 K
1 ( 1 Tb2) 16.459 0
2 ( 0 Tb1) 3.779 1
120
1 ( 1 T)Psat1 ( ) 0 Tb2 P
120
121
19.506
121
P 2 ( 0 T)Psat2 ( ) 1 Tb1
0.281
Since one of these values is >1 and the other is <1, an azeotrope exists. See Ex. 10.3(e). Guesses: T ( 60 273.15)K
x1 0.4 x2
1 y1 y1
x2 = 1
y2 = 1
x1 = y1
0.4
y2
1
x1
Given
y1 P = x1 1 ( x2 T)Psat1 ( ) x1 x1 T
y2 P = x2 2 ( x2 T)Psat2 ( ) y1 x1 T
436
x1 x2 y1 y2 Taz
Taz 360.881 K
x1 0.4546
y1 0.4546
Ans.
Find ( x1 x2 y1 y2 T)
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
Water:
Psat1 ( T)
16.1154
16.3872
B1
B2
B1 273.15 K)
3483.67 K
3885.70 K
kPa
C1
C2
205.807 K
230.170 K
A2
exp A1
(T
C1
Psat2 ( T)
exp A2
(T
B2 273.15 K)
C2
kPa
Parameters for the NRTL equation:
b12
12 ( T)
G12 ( T)
500.40
b12 RT
exp
cal mol
b21
1636.57
cal mol
b21 RT
exp
0.5081
21 ( T)
12 ( T)
G21 ( T)
21 ( T)
437
1 ( x2 T) x1
exp x2
2
21 ( ) T
G21 ( ) T x1 x2 G21 ( ) T G12 ( ) 12 ( ) T T
x1 G12 ( ) T)
2
2
( x2
2 ( x2 T) x1
exp x1
2
12 ( ) T
G12 ( ) T x2 x1 G12 ( ) T G21 ( ) 21 ( ) T T
x2 G21 ( ) T)
2
2
( x1 (a) BUBL T:
P
101.33 kPa
x1
0.3
x2
1
x1
Guess:
T
( 60
273.15)K
y1
0.3
y2
1
y1
Given
y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T
y1
y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T
y2 = 1
Tbubl y1 y2
Tbubl 360.84 K P
Find ( y1 y2) T
y1
0.415
y2
y1
0.585
0.3
Ans.
(b) DEW T:
101.33 kPa
y2
1
x1
Guess:
T
( 90
273.15)K
x1
0.05
x2
1
y1
Given
y1 P = x1 1 ( x2 T)Psat1 ( ) x1 T
y2 P = x2 2 ( x2 T)Psat2 ( ) x1 T
x1
x2 = 1
Tdew x1 x2
Tdew 364.27 K
Find ( x1 x2) T
x1
0.042
x2
0.958
Ans.
438
(c) P,T-flash Calculation
T
Tdew 2
Tbubl
P
101.33 kPa
z1
0.3
Guess:
V
0.5
x1 y1
0.1 0.1
x2 y2
1 1
y1 x1
Given
y1 =
x1 1 ( x1 x2 T) Psat1 ( T) P
x1
x2 = 1
y2 =
x2 2 ( x1 x2 T) Psat2 ( T) P
y1
y2 = 1
x1 ( 1
x1 x2 y1 y2 V x1 0.069
V)
y1 V = z1 Eq. (10.15)
Find ( x1 x2 y1 y2 V)
x2
0.931
y1
0.352
y2
0.648
V
0.816
(d) Azeotrope Calculation Test for azeotrope at: P
101.33 kPa
Tb1 A1
B1 P ln kPa
B2 A2 P ln kPa 14.699
C1
273.15 K
Tb1
370.349 K
Tb2
C2
273.15 K
Tb2
373.149 K
1 ( 0 1 Tb2)
2 ( 1 0 Tb1)
4.05
439
120
1 ( 1 T)Psat1 ( ) 0 Tb2 P
120
17.578
121
P 2 ( 0 T)Psat2 ( ) 1 Tb1
121
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
273.15)K
x1
0.4 x2
1
y1 y1
0.4
y2
1
x1
Given
y1 P = x1 1 ( x2 T)Psat1 ( ) x1 x1 T
x2 = 1
y2 P = x2 2 ( x2 T)Psat2 ( ) y1 x1 T
x1 x2 y1 y2 Taz
Taz 360.676 K
x1 0.4461
y1
Find x1 x2 y1 y2 T) (
y2 = 1
x1 = y1
0.4461
Ans.
12.20 Molar volumes & Antoine coefficients:
74.05 V 40.73 18.07 A
14.3145 16.5785 16.3872 B
2756.22 3638.27 3885.70
kPa Ci 0
228.060 C 239.500 230.170
Psat ( T) i
exp Ai
Bi T K
T
( 65
273.15) K
273.15
161.88 291.27 0 107.38 0 cal mol
Wilson parameters:
a
583.11
1448.01 469.55
440
( i j T)
Vj Vi
exp
ai j RT
i
1 3
j
1 3
p
1 3
(a)
BUBL P calculation: No iteration required.
x1
0.3
x2
0.4
x3
1
x1
x2
( i x T)
exp 1
ln
j
xj
xp
p j
( i j T)
( p i T) ( p j T)
xj
Pbubl
i
xi ( i x T) Psat ( i T)
yi
xi ( i x T) Psat ( i T) Pbubl
0.527 y 0.367 0.106
(b) DEW P calculation: Pbubl 117.1 kPa Ans.
y1
Guess:
Given
0.3
0.05
y2
x2
0.4
0.2
y3
x3
1
1
y1
x1
y2
x2
P Pbubl
x1
P y1 = x1 ( 1 x T) Psat ( 1 T)
P y3 = x3 ( 3 x T) Psat ( 3 T)
x1 x2 x3 Pdew
441
P y2 = x2 ( 2 x T) Psat ( 2 T)
xi = 1
i
Find x1 x2 x3 P
0.035 x 0.19 0.775
(c) P,T-flash calculation:
z1
Guess:
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 (
xi = 1
i
i
Pdew
69.14 kPa
Ans.
P
z3
Pdew 2
1
Pbubl
T
338.15 K
0.3
V
z2
0.5
0.4
z1
z2
Use x from DEW P and y from BUBL P as initial guess.
x1 ( 1
x2 ( 1
x3 ( 1
V) y1 V = z1
V) y2 V = z2
V) y3 V = z3
yi = 1
x1 x2 x3 y1 y2 y3 V 0.109 x 0.345 0.546
Find x1 x2 x3 y1 y2 y3 V
0.391 y 0.426 0.183
442
V
0.677
Ans.
12.21 Molar volumes & Antoine coefficients: Antoine coefficients:
74.05 V 40.73 18.07 T ( 65 273.15)K A
14.3145 16.5785 16.3872 Psat ( i T) B
2756.22 3638.27 3885.70 exp Ai Bi T K
228.060 C 239.500 230.170
kPa Ci
273.15
NRTL parameters:
0 0.3084
0.3084 0.5343 0 0.2994 0
0 b 222.64
184.70 631.05 0 253.88 0
cal mol
0.5343 0.2994
1197.41 845.21
bi j RT
i
1 3
j
1 3
i j
Gi j l 1 3 k 1 3 (a) BUBL P calculation: No iteration required.
x1 0.3
x2 0.4
x3 1 x1 x2
exp
i j i j
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
j l
k i j l
Gl j xl
G l j xl
Pbubl
i
xi ( i x T) Psat ( i T)
0.525
yi
xi ( i x T) Psat ( i T) Pbubl
y
0.37 0.105
Pbubl
115.3 kPa
Ans.
443
(b) DEW P calculation:
y1
0.3
y2
0.4
y3
1
y1
y2
Guess:
x1
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 (
P y3 = x3 ( x T)Psat3 T) 3 (
xi = 1
i
x1 x2 x3 Pdew
0.038 x 0.192 0.77
(c) P,T-flash calculation: Find x1 x2 x3 P
Pdew
68.9 kPa
Ans.
P
Pdew 2
Pbubl
T
338.15 K
z1
0.3
z2
0.4
z3
1
z1
z2
Guess:
V
0.5
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
xi = 1
i
yi = 1
i
444
x1 x2 x3 y1 y2 y3 V 0.118 x 0.347 0.534
Find x1 x2 x3 y1 y2 y3 V
0.391 y 0.426 0.183
V 0.667
Ans.
12.22 Molar volumes & Antoine coefficients:
74.05 V 40.73 18.07 A
14.3145 16.5785 16.3872 B
2756.22 3638.27 3885.70
kPa Ci
228.060 C 239.500 230.170
Psat ( i T)
exp Ai
Bi T K
P
101.33kPa
273.15
0 Wilson parameters: a 583.11
161.88 291.27 0 107.38 0
cal mol
1448.01 469.55
( i j T)
Vj Vi
exp
ai j RT
i
1 3
j
1 3
p
1 3
(a) x1
BUBL T calculation: 0.3
x2 0.4
x3
445
1
x1
x2
( x T) i
exp 1
ln
j
xj
xp
p j
( j T) i
( i T) p ( j T) p
xj
Guess:
Given
T
300K
y1
0.3
y2
0.3
y3
1
y1
y2
P y1 = x1 ( x T)Psat1 T) 1 (
P y3 = x3 ( x T)Psat3 T) 3 (
y1 y2 y3 Tbubl
Find y1 y2 y3 T
P y2 = x2 ( x T)Psat2 T) 2 (
P=
i
xi ( x T)Psati T) i (
0.536 y 0.361 0.102
(b) DEW T calculation: y1
Guess:
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 (
i
Tbubl
334.08K
Ans.
0.3
0.05
y2
x2
0.4
0.2
y3
x3
1
1
y1
x1
y2
x2
T Tbubl
x1
xi = 1
446
x1 x2 x3 Tdew
Find x1 x2 x3 T
0.043 x 0.204 0.753
(c) P,T-flash calculation:
z1
Guess:
Tdew
347.4 K
Ans.
T
z3
Tdew 2
1
Tbubl
T
340.75 K
0.3
V
z2
0.5
0.2
z1
z2
Use x from DEW P and y from BUBL P as initial guess.
x1 ( 1
x2 ( 1
x3 ( 1
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)
xi = 1
i
i
V)
y1 V = z1
y2 V = z2
y3 V = z3
V)
V)
yi = 1
x1 x2 x3 y1 y2 y3 V
447
Find x1 x2 x3 y1 y2 y3 V
0.125 x 0.17 0.705
0.536 y 0.241 0.223
V 0.426
Ans.
12.23 Molar volumes & Antoine coefficients: Antoine coefficients:
74.05 V 40.73 18.07 P 101.33kPa A
14.3145 16.5785 16.3872 Psati T) ( B
2756.22 3638.27 3885.70 exp Ai Bi T K
228.060 C 239.500 230.170
kPa Ci
273.15
NRTL parameters:
0 0.3084
0.3084 0.5343 0 0.2994 0
0 b 222.64
184.70 631.05 0 253.88 0
bi j RT
cal mol
0.5343 0.2994
1197.41 845.21
l
exp
i
k
1 3
1 3
j
1 3
1 3
( j T) i
i j
G ( j T) i
( j T) i
(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 l
( j T) i
k
G ( j T)xl l
l
G ( j T)xl l
448
Guess:
T
300K
y1
0.3
y2
0.3
y3
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=
i
xi ( i x T) Psat ( i T)
y1 y2 y3 Tbubl
0.533 y 0.365 0.102 (b) DEW T calculation: Tbubl 334.6 K
Find y1 y2 y3 T
Ans.
y1
Guess:
0.3 0.05
y2
0.4
y3
1
y1
y2
x1
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)
P y3 = x3 ( 3 x T) Psat ( 3 T)
xi = 1
i
x1 x2 x3 Tdew
0.046 x 0.205 0.749
449
Find x1 x2 x3 T
Tdew
347.5 K
Ans.
(c) P,T-flash calculation:
z1
Guess:
T
z3
Tdew 2
1
Tbubl
T
341.011 K
0.3
V
z2
0.5
0.2
z1
z2
Use x from DEW P and y from BUBL P as initial guess.
x1 ( 1
x2 ( 1
x3 ( 1
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 (
xi = 1
i
i
V) y1 V = z1
V) y2 V = z2
V) y3 V = z3
yi = 1
x1 x2 x3 y1 y2 y3 V 0.133 x 0.173 0.694
Find x1 x2 x3 y1 y2 y3 V
0.537 y 0.238 0.225
V 0.414
Ans.
450
12.26
x1
0.4
x2
1
x1
V1
110
cm
3
mol
V2
90
cm
3
mol
VE x1 x2
x1 x2 45 x1
25 x2
cm
3
mol
VE x1 x2
7.92
cm
3
mol
By Eq. (12.27): V x1 x2
VE x1 x2
x1 V1
x2 V2
V x1 x2
105.92
cm
3
mol
By Eqs. (11.15) & (11.16):
Vbar1
V x 1 x2
d x2 V x1 x2 dx1
Vbar1
cm 190.28 mol
49.68 cm mol
3
3
Ans.
Vbar2
Vbar2
V x1 x2
x1
d dx1
V x1 x2
Check by Eq. (11.11):
V
x1 Vbar1
3
x2 Vbar2
V
3
105.92
cm mol
3
OK
12.27 V1
cm 58.63 mol
V2
3
cm 118.46 mol
moles1
750 cm V1
moles2
1500 cm V2
3
moles
moles1
moles2
moles
25.455 mol
x1
moles1 moles
x1
0.503
cm mol
3
x2
1
x1
cm 0.256 mol
3
VE
x1 x2
1.026
0.220 x1
x2
VE
By Eq. (12.27),
V
VE x1 V1
451
x2 V 2
V
88.136
cm
3
mol
Vtotal
V moles
Vtotal
2243 cm
3
Ans.
For an ideal solution, Eq. (11.81) applies:
Vtotal
x1 V 1
x2 V2 moles
Vtotal
2250 cm
3
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
H3
441579 J
2 ( 285830 J)
(Pg. 457)
(Table C.4)
H
H1
H2
H3
H
589 J
(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
53.55 J
Ans.
12.29
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
H2
H
39.2 kJ
Ans.
452
12.30 Calculate moles of LiCl and H2O in original solution:
nLiCl
nLiCl
0.1 125 42.39
kmol
nH2O
nH2O
n'LiCl
nH2O nLiCl
0.9 125 18.015
kmol
3
0.295 kmol
6.245 10 mol
20 kmol 42.39
21.18
Moles of LiCl added:
n'LiCl
0.472 kmol
Mole ratio, original solution:
Mole ratio, final solution:
nH2O nLiCl n'LiCl
8.15
nLiCl
n'LiCl
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
nLiCl 35
kJ mol
(Fig. 12.14, n=21.18)
H2
nLiCl
H1
n'LiCl
Q
32
kJ mol
(Fig. 12.14, n=8.15)
Q
H2
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
H1
104.8
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 25.5 kJ mol
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
H
x
0.8 mol
10.5
M1
0.2 mol
H3
33.16
kJ mol
M1 H1
0.061 kJ
0.2 mol M1 1 mol
0.2 mol H2
0.2 mol H3
Close enough
x 0.087
Ans.
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
H2
H3
H
104.8
kJ kg
kJ mol
21.01
kJ gm 18.015 kg mol
H1
1.509
kJ mol
20.756
From p. 457 ( H LiCl(s) - H LiCl in 3 mol H2O)
From Figure 12.14 H3 0.2 mol
454
25.5
H1
kJ mol
H2
H
646.905 J
Ans.
12.33
(a) LiCl + 4 H2O -----> LiCl(4H2O) H
25.5
kJ From Figure 12.14 mol
0.2 mol H
5.1 kJ
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
kJ mol From p. 457 ( H LiCl(s) - H LiCl in 3 mol H2O)
H2
25.5
kJ From Figure 12.14 mol
H
0.2 mol
H1
H2
H
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
H2
H3
H4
H
(d)
712.58
kJ mol
kJ mol
kJ mol
From p. 457 for LiCl.H2O
From Table C.4 Hf H2O(l) From p. 457 for LiCl
285.83
408.61
25.5
0.2 mol
kJ mol
H1
From Figure 12.14 H2 H3 H4
H 1.472 kJ
Ans.
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
H1
H2
H
(e)
25.5
4 9
kJ mol
kJ mol
H1
From Figure 12.14
( ) 32.4
From Figure 12.14
0.2 mol
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
H2
H3
kJ 5 ( 20.756) mol 6
kJ 1 ( ) 32.4 mol 6
25.5 kJ mol
H1
From p. 457 ( H LiCl(s) - H LiCl in 3 mol H2O)
From Figure 12.14
From Figure 12.14
H
(f)
0.2 mol
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)
kJ 5 ( 712.58) mol 8
5 8
H1
H2
From p. 457 for LiCl.H2O
From Table C.4 Hf H2O(l)
( 285.83)
kJ mol
456
H3
kJ 3 ( 32.4) mol 8
From Figure 12.14
H4
5 8
( 408.61)
kJ mol
From p. 457 for LiCl
H5
25.5
kJ mol
From Figure 12.14
H
0.2 mol
H1
H2
H3
H4
H5
H
0.403 kJ
Ans.
12.34
BASIS: 1 second, during which the following are mixed: (1) 12 kg hydrated (6 H2O) copper nitrate (2) 15 kg H2O
n1
12 kmol 295.61 sec
0.041 kmol sec
n2
15 kmol 18.015 sec
n1
n2
6 n1
0.833
n2
kmol sec
26.51
Mole ratio, final solution:
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
6 ( 285.83 kJ)
(Table C.4)
H2
302.9 kJ
H3
( 2110.8 kJ)
H4
47.84 kJ
H
H1
H2
H3
H4
H
45.08 kJ
457
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
3 ( 285.83 kJ)
(Table C.4)
H3
2 ( 436.805 kJ)
H4
( 439.288 kJ)
(Pg. 457)
H
H1
H2
H3
H4
Ans.
H
19.488 kJ
Q
H
Q
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)
H2
2 ( 285.83 kJ)
H3
1012.65 kJ
(Table C.4)
Since the process is isothermal,
H=
H1
H2
H3
Since it is also adiabatic,
H= 0
Therefore,
H1
H2
H3
H1
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
0.1012
Ans.
458
12.37 Data:
10 15 20 25 n 50 100 300 500 1000
862.74 867.85 870.06 871.07 Hf 872.91 kJ 873.82 874.79 875.13 875.54
Hf
HfCaCl2
Htilde
795.8 kJ
Ca + Cl2 + n H2O ---> CaCl2(n H2O)
CaCl2(s) ---> Ca + Cl2
-------------------------------------------CaCl2(s) + n H2O ---> CaCl2(n H2O)
From Table C.4:
i 1 rows ( n)
65
HfCaCl2
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)
H1
795.8 kJ
(Table C.4)
H2
2 ( 865.295 kJ)
H3
871.07 kJ
H
H1
H2
H3
Q
H
Q
63.72 kJ
Ans.
12.39
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
2607.9 kJ
H3
6 ( 285.83 kJ)
(Table C.4)
H2
871.8 kJ
(Pb. 12.37)
H298
H1
H2
H3
for reaction at 25 degC
H298
21.12 kJ
msoln
( 110.986
34.911 18.015)gm
msoln
739.908 gm
460
CP
3.28
kJ kg degC
H298
CP T = 0
T
H298 msoln CP
T
8.702 degC
T
25 degC
T
T
16.298 degC Ans.
12.43 m1
150 lb (H2SO4)
m2
350 lb
(25% soln.)
H1
8
BTU lbm
H2
23
BTU lbm
(Fig. 12.17)
100 % m1 m1
25 % m2 m2
47.5 %
(Final soln.)
m3
m1
m2
H3
90
BTU lbm
(Fig. 12.17)
Q
m3 H3
m1 H1
m2 H2
Q
38150 BTU
Ans.
12.44 Enthalpies from Fig. 12.17.
x1 0.5
x2
1
x1
H
69
BTU lbm
(50 % soln)
H1
20
BTU lbm
(pure H2SO4)
H2
108
BTU lbm
(pure H2O)
HE
H
x1 H1
x2 H2
HE
133
BTU Ans. lbm
12.45 (a) m1
400 lbm
(35% soln. at 130 degF)
m2
175 lbm
(10% soln. at 200 degF)
H1
100
BTU lbm
H2
152
BTU lbm
(Fig. 12.19)
35 % m1 m1
10 % m2 m2
27.39 %
(Final soln)
m3
m1
m2
H3
41
BTU lbm
(Fig. 12.19)
461
Q
m3 H3
m1 H1
m2 H2
Q
43025 BTU Ans.
(b) Adiabatic process, Q = 0.
H3
m1 H1
m2 H2
m3
H3
115.826
BTU lbm
From Fig. 12.19 the final soln. with this enthalpy has a temperature of about 165 degF.
12.46 m1
25
lbm sec
(feed rate)
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
BTU lbm
(Table F.4, 1.5(psia) & 217 degF]
m2
x1 m1 x2
m2 H2
m2
7.143
lbm sec
m3
m1
m2
m3
17.857
lbm sec
Q
m3 H3
m1 H1
Q
20880
BTU sec
Ans.
12.47 Mix m1 lbm NaOH with m2 lbm 10% soln. @ 68 degF. BASIS: m2
1 lbm
x3
0.35
x2
0.1
m1
1 lbm
(guess)
m3
m1
m2
Given
m1
m2 = m3
m1
m1
x2 m2 = x3 m3
0.385 lbm
m1 m3
Find m1 m3
m3
1.385 lbm
462
From Example 12.8 and Fig. 12.19
H1 478.7 BTU lbm
m2 H2
H2
43
BTU lbm
BTU lbm
H3
m1 H1
m3
H3
164
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
mH2SO4 0.5
8.712
10 J
4
Mix 1 mol or 98.08 gm H2SO4(l) with m gm H2O to form a 50% solution. mH2SO4
mH2O
98.08 gm
msoln mH2SO4
msoln
Data from Fig. 12.17:
HH2SO4
HH2O
Hsoln
Hmix
Hmix
Q
0
BTU lbm
[pure acid @ 77 degF (25 degC)]
45
BTU lbm
BTU lbm
[pure water @ 77 degF (25 degC)]
70
[50% soln. @ 140 degF (40 deg C)] mH2SO4 HH2SO4 mH2O HH2O
msoln Hsoln
18.145 kg
H298 msoln
BTU lbm
Q 283 BTU lbm
Ans.
Hmix
463
12.49
m1
140 lbm
x1
0.15
m2
230 lbm
x2
0.8
H1
65
BTU lb
(Fig. 12.17 at 160 degF)
H2
102
BTU lb
(Fig. 12.17 at 100 degF)
m3
m1
m2
x3
m1 x1
m2 x2
m3
x3
55.4 %
Q
20000 BTU
H3
Q
m1 H1 m3
m2 H2
H3
92.9
BTU lbm
From Fig. 12.17 find temperature about 118 degF
12.50 Initial solution (1) at 60 degF; Fig. 12.17:
m1 1500 lbm
x1
0.40
H1
98
BTU lbm
Saturated steam at 1(atm); Table F.4:
m3 m2
m1
m2
H2
1150.5
BTU lbm
x3 m2
x1 m1 m1 m2
H3 m2
m1 H1
m2 H2
m3 m2
m2
125 lbm
x3 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%
H3 m2
5.5
BTU lbm
This is about as good a result as we can get.
464
12.51 Initial solution (1) at 80 degF; Fig. 12.17:
m1
1 lbm
x1
0.45
H1
95
BTU lbm
Saturated steam at 40(psia); Table F.4:
m3 m2
m1
m2
H2
1169.8
BTU lbm
x3 m2
x1 m1 m1 m2
H3 m2
m1 H1
m2 H2
m3 m2
m2
0.05 lbm
x3 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
1 lbm
x1
0.40
H1
77
BTU lbm
Saturated steam at 35(psia); Table F.4:
H2
1161.1
BTU lbm
x3
0.38
m2
x1 m1 x3
m1
m3
m1
m2
m3
1.053 lbm
m2
0.053 lbm
H3
m1 H1 m2 H2 m3
H3
131.2
BTU 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:
H
56
BTU lbm
H1
8
BTU lbm
H2
68
BTU lbm
x1
0.35
x2
1
x1
H
H
x1 H 1
x2 H2
H
103
BTU lbm
Ans.
12.54 BASIS: 1(lbm) of soln. Read values for H1 & H2 from Fig. 12.17 at 80 degF:
H1
Q=
4
BTU lbm
x1 H 1
H2
48
BTU lbm
x1
0.4
x2
1
x1
H= H
x2 H2 = 0
H
x1 H 1
x2 H2
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:
x1
2 98.08 2 98.08 15 18.015
x1
0.421
Final solution:
x2
3 98.08 3 98.08 14 18.015
x2
0.538
Data from Fig. 12.17 at 100 degF:
HH2O
68
BTU lbm
HH2SO4
9
BTU lbm
H1
75
BTU lbm
H2
101
BTU lbm
466
Unmix the initial solution:
Hunmix
Hunmix
x1 HH2SO4
118.185 BTU lbm
1
x1 HH2O
H1
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)
J mol
J mol
HfH2O HfSO3
Hrx 1.324 10
5 J
HfSO3
HfH2SO4
Hrx
395720
HfH2O
285830
J mol
813989
HfH2SO4
mol
Finally, mix the constituents to form the final solution:
Hmix
Q
H2
x2 HH2SO4
1
x2 HH2O
Hmix
137.231
BTU lbm
Hunmix ( 98.08 2 1 lbmol Hrx Hmix ( 98.08 3
15 18.015)lb 14 18.015)lb
Q
76809 BTU
Ans.
12.56 Read values for H(x1=0.65), H1, & H2 from Fig. 12.17 at 77 degF:
H 125 BTU lbm
x2
H1
1
0
BTU lbm
H2
H
45
BTU lbm
x1 H 1
BTU lbm
x1
0.65
x1
H
x2 H2
Ans.
H
140.8
467
From the intercepts of a tangent line drawn to the 77 degF curve of Fig. 12.17 at 65%, find the approximate values:
Hbar1
136
BTU lbm
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
42 %
m1
1 lb
By a mass balance:
x3 =
0.75 m1 m1 m2
m2
0.75 m1 x3
m1
m2
0.786 lbm
Ans.
12.58 (a) m1
x1
25 lbm
0
m2
x2
40 lbm
1
m3
75 lbm
x3
0.25
Enthalpy data from Fig. 12.17 at 120 degF:
H1
88
BTU lbm
H2
14
BTU lbm
H3
7
BTU lbm
m4
m1
m2
m3
m4
140 lbm
x4
x1 m1
x2 m2 m4
x3 m3
x4
0.42
H4
63
BTU lbm
(Fig. 12.17)
Q
m4 H4
m1 H1
m2 H2
m3 H3
Q
11055 BTU Ans.
468
(b) First step: m1
40 lb
x1
1
H1
14
BTU lbm
m2
75 lb
x2
0.25
H2
7
BTU lbm
m3
m1
m2
x3
x1 m1
x2 m2
m3
H3
Q
m1 H1 m3
m2 H2
x3
0.511
H3
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 285830
5
( 425609
92307) ]J
H298
1.791
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
44.50 kJ
H3
68.50 kJ
H4
3.88 kJ
H
H1
H2
H3
H4
Q
H
Q
62187 J
Ans.
469
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:
x1
HH2O
Hsoln
1 40.00 1 40.00 9 18.015
45 BTU lbm
BTU lbm
BTU lbm
x1
19.789 %
(Table F.3, sat. liq. at 77 degF)
35
(Fig. 12.19 at x1 and 77 degF)
HNaOH
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
molwt
3
40.00
gm mol
Cp
R molwt
0.121
16.316 10 K
T
Cp
0.245
BTU lbm rankine
BTU lbm
HNaOH
HNaOH
Cp ( 77
68)rankine
HNaOH
480.91
H
H
Hsoln
0.224
x1 HNaOH
kJ gm
1
x1 HH2O
This is for 1 gm of SOLUTION.
However, for 1 mol of NaOH, it becomes:
H H x1
molwt
H
45.259
kJ mol
470
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
H2
H3
H4
H
179067 J
74.5 kJ
45.259 kJ
3.88 kJ
H1 H2
(Pb. 12.59)
(Fig. 12.14 with sign change)
(See above; note sign change)
(given) H3 H3
Q H
Q 14049 J Ans.
12.61 Note: The derivation of the equations in part a) can be found in Section B of this manual.
0.1 0.2 0.3 0.4 0.5 x1 0.6 0.7 0.8 0.85 0.9 0.95
HE
73.27 144.21 208.64 262.83 302.84 323.31 320.98 279.58 237.25 178.87 100.71
kJ kg HE x1 x2
x2
1
x1
H
471
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 n a b c d
2 3
H
w w n
regress x1 kJ kg
3 3 3 735.28 824.518 195.199 914.579
3
a b c d kJ kg
H
x1
a
b x1
c x1
d x1
Using the equations given in the problem statement and taking the derivatives of the polynomial analytically:
HEbar1 x1
1
x1
2
H
x1
x1
b
2 c x1
3 d x1
2
kJ kg
kJ kg
HEbar2 x1
x1
2
H
x1
1
x1
b
2 c x1
3 d x1
2
472
0
500
1000
(kJ/kg)
1500
2000
2500
0
0.2
H/x1x2 HEbar1 HEbar2
0.4
x1
0.6
0.8
12.62 Note: This problem uses data from problem 12.61
0.1 0.2 0.3 0.4 0.5 x1 0.6 0.7 0.8 0.85 0.9 0.95 HE
73.27 144.21 208.64 262.83 302.84 323.31 320.98 279.58 237.25 178.87 100.71
kJ kg
x2
1
x1
H
HE x1 x2
473
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 n a b c d
regress x1
H
w w n
kJ kg
3 3 3 735.28 824.518 195.199 914.579
3
a b c d
By the equations given in problem 12.61
H
x1
a
H
b x1
c x1
2
d x1
3
kJ kg
H x1 Hbar1 x1
x1 x1 1 1 x1
2
2
x1 x1 x1 b 2 c x1 3 d x1
2
H
kJ kg
Hbar2 x1
x1
H
x1
1
x1
b
2 c x1
3 d x1
2
kJ 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
m) H
Solving for m:
m x1
( 4000kg)x1 0.9 x1
Eq. (A)
Ht = ( 4000
4000H2
m H3
474
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
Define quantities as a function of x 1
H( ) 0.9
H3
H
178.737
kJ kg
kJ kg
H( ) 0.5
303.265
Q x1
m x1
Qtx1
4000kg
( 4000kg) 1 x 0.9 x1
m x1
H x1
m x1 H3
m ( ) 5000 kg 0.5
6
4000kg
m x1
H
m x1 H3
Qt0.5) (
1.836
10 kJ
Since the heat transfer rate q is constant:
q Qtx1
Q x1 q
and
x1
Eq. (C)
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
1200
x1
0 0.01 0.5
1100
1000
r x1
kg hr
900
800
700
600
0
1
2
3
x1 hr
4
5
6
12.64 mdot1
20000
lb hr
x1
0.8
T1
T2
T3
120degF
H1
H2
H3
92
BTU lb
Enthalpies from Fig. 12.17 x2
x3
0.0
40degF
7
BTU lb
70 BTU lb
0.5
140degF
a) Use mass balances to find feed rate of cold water and product rate. Guess:
Given
mdot2
mdot1
mdot1
mdot2 = mdot3
mdot3
2mdot1
Total balance
H2SO4 balance
mdot1 x1
mdot2 x2 = mdot3 x3
476
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
H3 54.875 BTU lb
mdot3
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:
L
dx1 dt
dx1 dt
dx1 dt
dx1 dt
dx1
x1
dL = Vdot y1 dt
Substituting -Vdot for dL/dt:
L
x1 ( Vdot)= y1 Vdot
Rearranging this equation gives:
L
= x1
= y1
=
dL L
y1 Vdot
x1 dL dt
Substituting -dL/dt for Vdot:
L
Eliminating dt and rearranging:
y1
477
x1
At low concentrations y1 and x1 can be related by:
y1 =
Psat1
inf1
P
x1 = K1 x1
dx1 K1 1 x1
where:
dL L
K1 = inf1
Psat1 P
Substituting gives:
=
Integrating this equation yields:
ln
Lf L0
=
1 K1 1
ln
x1f x10
where L0 and x10 are the initial conditions of the system For this problem the following values apply: L0
T
1mol
x10
P
600 10
6
x1f
50 10
6
130degC
1atm
inf1
5.8
Psat1
exp 16.3872
3885.70 T degC
K1
kPa
230.170
Psat1
270.071 kPa
K1
Psat1
inf1
P
15.459
Lf
norg0
norg0
L0 exp
L0 1
1 K1
x10
1
ln
x1f x10
Lf
norgf
norgf
0.842 mole
Lf 1 x1f
0.999 mole
norg0 norgf
0.842 mole
%lossorg
norg0
%lossorg
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
cal mol
a21
cal 583.11 mol
0.708
V1
74.05
V1
cm
3
mol
a21 RT
V2
40.73
cm
3
mol
V2
12
V1
exp
a12 RT
ln inf1
12
21
V2
21
exp
21
0.733
ln
12
1
ln inf1
0.613
Ans.
From p. 445
ln inf2
ln
21
1
12
ln inf2
0.603
Ans.
From Fig. 12.9(b)
For NRTL equation
ln inf1 = 0.62
ln inf2 = 0.61
b12
12
184.70
cal mol
b21
12
222.64
cal mol
0.3048
b12 RT
0.288
b21
21
RT
21
0.347
G12
exp
12
G12
21
0.916
12 exp
G21
12
exp
21
G21
0.9
From p. 446
ln inf1
ln inf1
0.611
ln inf2
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
183.4kPa
Psat2
96.7kPa
x1
0.253
y1
0.456
P
139.1kPa
Check whether or not the system is ideal using Raoult's Law (RL)
PRL
x1 Psat1
1
1
x1 Psat2
2
PRL
118.635 kPa
Since PRL<P,
and
are not equal to 1. Therefore, we need a model for
GE/RT. A two parameter model will work. From Margules Equation:
GE = x1 x2 A21 x1 RT
2
A12 x2
Eq. (12.10a) Eq. (12.10b)
ln 1 = x2
ln 2 = x1
Find
1
A12
A21
2 A21
2 A12
A12 x1
A21 x2
2
and
2
at x1=0.253 from the given data.
1
y1 P
1
x1 Psat1
1
1.367
1
2
y1 P x1 Psat2
2
1
1.048
Use the values of find A12 and A21.
and
2
at x1=0.253 and Eqs. (12.10a) and (12.10b) to
Guess:
A12
0.5
A21
0.5
Given
ln 1 = 1
ln 2 = x1
2
x1
2
A12
2 A21
A21
A12 x1
1 x1
Eq. (12.10a) Eq. (12.10b)
A21
2 A12 A12
A12 A21
1 x1
Find A12 A21
0.644
A21
0.478
exp
1
2
x1
2
A12
2 A21
A21
A12 x1
1 x1
2 x1
exp x1
A21
2 A12
480
a) x1
0.5
y1
x1 1 x1 Psat1 P
y1
0.743
Ans.
P
x1 1 x1 Psat1
1
x1
2 x1 Psat2
P
160.148 kPa
Ans.
b) 1inf
exp A12
1inf Psat1
1inf
1.904
2inf
exp A21
2inf
1.614
120
Since
Psat2
12
120
3.612
121
Psat1
2inf Psat2
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
108.6kPa
x1
0.389
T
( 35
273.15) K
Psat1
120.2kPa
Psat2
73.9kPa
Check whether or not the system is ideal using Raoult's Law (RL)
PRL
x1 Psat1
1
1
x1 Psat2
2
PRL
91.911 kPa
Since PRL < P,
and
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
P = x1 exp A 1
A
x1
2
Psat1
1
x1 exp A x1
2
Psat2
A
Find A) (
0.677
1 x1
exp A 1
x1 1 x1 Psat1 P
x1
2
2 x1
exp A x1
Ans.
P
2
a) y1
b) P
c) 1inf
120
y1
1
1inf
0.554
2 x1 Psat2
x1 1 x1 Psat1
exp ( A)
1inf Psat1
x1
110.228 kPa
2inf
Ans.
1.968
1.968
2inf
exp ( A)
Psat1
2inf Psat2
Psat2
120
3.201
121
121
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 i=
1
1
1
1= 0
1 2
n0 = 1
1= 2
By Eq. (13.5).
yH = yCO =
2 2
yH2O = yCO =
2
By Eq. (A) and with data from Example 13.13 at 1000 K: T 1000 kelvin
G
1 2
( 395790)
RT 2
1 2
ln
J mol 1
2
( 192420
2
200240)
J mol
2
2
ln
2
Guess:
Given
d d e
e
0.5
e = 0
G
J mol
e
Find e
e
0.45308
0.3 0.31 0.6
2.082
2.084
G 10
5
2.086
2.088
0.2
0.3
0.4
0.5
0.6
483
13.5
(a) H2(g) + CO2(g) = H2O(g) + CO(g) =
i i=
1
1
1
1= 0
1 2
n0 = 1
1= 2
By Eq. (13.5).
yH = yCO =
2 2
yH2O = yCO =
2
By Eq. (A) and with data from Example 13.13 at 1100 K: T 1100 kelvin
G
1 2
( 395960)
RT 2
1 2
ln
J mol 1
2
( 187000
2
209110)
J mol
2
2
ln
2
Guess:
Given
d d e
e
0.5
e = 0
G
J mol
e
Find e
e
0.502
Ans.
0.35 0.36 0.65
2.102
2.103
2.104
G 10
5
2.105
2.106
2.107
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
484
(b) =
H2(g) + CO2(g) = H2O(g) + CO(g)
i= i
1
1
1
1= 0
1 2
n0 = 1
1= 2
By Eq. (13.5),
yH = yCO = 2 2
yH2O = yCO =
2
By Eq. (A) and with data from Example 13.13 at 1200 K: T 1200 kelvin
G
1 2
( 396020)
J
RT 2
1 2
ln
mol 1
2
( 181380
2
217830)
J mol
2
2
ln
2
Guess:
Given
d d e
e
0.1
e = 0
G
J mol
e
Find e
e
0.53988
Ans.
0.4 0.41 0.7
2.121
2.122
2.123
G 10
5 2.124
2.125
2.126
2.127 0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
485
(c) =
H2(g) + CO2(g) = H2O(g) + CO(g)
i= i
1
1
1
1= 0
1 2
n0 = 1
1= 2
By Eq, (13.5),
yH = yCO =
2 2
yH2O = yCO =
2
By Eq. (A) and with data from Example 13.13 at 1300 K: T 1300 kelvin
G
1 2
( 396080)
RT 2
1 2
ln
J mol 1
2
( 175720
2
226530)
J mol
2
2
ln
2
Guess:
d d e
e
0.6
J mol
Given
G
e = 0
e
Find e
e
0.57088
Ans.
0.4 0.41 0.7
2.14
2.142
G 10
5
2.144
2.146
2.148 0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
486
13.6
H2(g) + CO2(g) = H2O(g) + CO(g) =
i i=
1
1
1
1= 0
1 2
n0 = 1
1= 2
By Eq, (13.5),
yH = yCO =
2 2
yH2O = yCO =
2
With data from Example 13.13, the following vectors represent values for Parts (a) through (d):
1000 T 1100 1200 1300
kelvin
3130 G 150 3190 6170
J mol
Combining Eqs. (13.5), (13.11a), and (13.28) gives
2
2
1 2
2 1 2
=
1
2
= K = exp
G RT
0.4531
exp G RT
0.5021
1
Ans.
0.5399 0.5709
13.11 4HCl(g) + O2(g) = 2H2O(g) + 2Cl(g) = 1
H298
n0 = 6
114408 J mol
T
773.15 kelvin
J mol
T0
298.15 kelvin
G298
75948
487
The following vectors represent the species of the reaction in the order in which they appear:
4 1 2 2
end rowsA) (
i Ai i
3.156 A 3.639 3.470 4.442 B
0.623 0.506 1.450 0.089
10
3
0.151 D 0.227 0.121 0.344
10
5
i
1 end
i Bi i
A
B
D
i
i Di
A
0.439
B
8
10
5
C
0
D
8.23
10
4
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
G
1.267
10
4 J
mol
K 7.18041
K
exp
G RT
By Eq. (13.5)
1 6
yHCl = yH2O =
5 6 2 6
4
yO2 =
yCl2 =
2 6
Apply Eq. (13.28);
0.5
4
(guess)
Find
Given
5 6
2 5
4
6 1
4
= 2K
0.793
yHCl
yO2
1 6
yH2O
2 6
yCl2
2 6
yHCl
0.3508
yO2
0.0397
yH2O
488
0.3048
yCl2
0.3048
Ans.
13.12 N2(g) + C2H2(g) = 2HCN(g)
= 0
n0 = 2
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
42720
J mol
G298
39430
J mol
A
0.060
B
0.173 10
3
C
0
D
0.191 10
5
T
923.15 kelvin
T0
298.15 kelvin
G
H298
H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
T
G
3.242
10
4 J
mol
K
exp
G RT
K
0.01464
By Eq. (13.5),
yN2 = 1 2
yC2H4 =
1 2
yHCN =
2e 2
=
By Eq. (13.28),
0.5
(guess)
Given
1 2
2 1
2
= K
Find
0.057
yN2
yC2H4
1 2
yHCN
yN2
0.4715
yC2H4
0.4715
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)
= 1
n0 = 2.5
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
68910
J mol
G298
39630
J mol
6
A
1.424
B
1.601 10
3
C
0.156 10
D
0.083 10
5
T
623.15 kelvin
T0
298.15 kelvin
G
H298
H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
T
G
6.787
10
3 J
mol
1 2.5
K
exp
G RT
1.5 2.5
K
3.7064
By Eq. (13.5), yCH3CHO =
yH2 =
yC2H5OH =
2.5
By Eq. (13.28),
Given
0.5
2.5
(guess)
Find
1
1.5
= 3K
0.818
yCH3CHO
1 2.5
yH2
1.5 2.5
yC2H5OH
2.5
yCH3CHO
0.108
yH2
0.4053
yC2H5OH
0.4867 Ans.
If the pressure is reduced to 1 bar,
Given
2.5 1 1.5
= 1K
Find
0.633
yCH3CHO
1 2.5
yH2
1.5 2.5
yC2H5OH
2.5
yCH3CHO
0.1968
yH2
0.4645
490
yC2H5OH
0.3387 Ans.
13.14 C6H5CH:CH2(g) + H2(g) = C6H5.C2H5(g)
= 1
n0 = 2.5
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
117440
J mol
G298
3
83010
J mol
6
A
4.175
B
4.766 10
C
1.814 10
D
0.083 10
5
T
923.15 kelvin
T0
298.15 kelvin
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
G
2.398
10
3 J
mol
K
exp
G RT
1 2.5
K
1.36672
By Eq. (13.5),
yC6H5CHCH2 =
yH2 =
1.5 2.5
yC6H5C2H5 =
2.5
By Eq. (13.28),
0.5
(guess)
Given
2.5 1 1.5
= 1.0133 K
Find
0.418
yC6H5CHCH2
1 2.5
yH2
1.5 2.5
yC6H5C2H5
2.5
yC6H5CHCH2
0.2794
yH2
0.5196
yC6H5C2H5
0.201
Ans.
491
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
= 0.5
n0 = 1
By Eq. (13.5),
ySO2 =
0.15 1 0.5
yO2 =
0.20 1
0.5 0.5
ySO3 =
1
0.5
From data in Table C.4,
H298
98890
J mol
G298
70866
J mol
The following vectors represent the species of the reaction in the order in which they appear:
1 0.5 1
end rowsA) (
i Ai i
5.699 A 3.639 8.060 B
0.801 0.506 10 1.056
3
1.015 D 0.227 10 2.028
5
i
B
i
1 end
i Bi
A
D
i
i Di
A
0.5415
B
2
10
6
C
0
D
8.995
10
4
T
753.15 kelvin
T0
298.15 kelvin
G
H298
H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
T
G
2.804
10
4 J
mol
K
exp
0.1
G RT
(guess)
K
88.03675
By Eq. (13.28),
Given
1 0.15
0.5 0.2
0.5
0.5
0.5
492
= K
Find
0.1455
By Eq. (13.4),
nSO3 =
= 0.1455
By Eq. (4.18),
H753
H298
R IDCPH T0 T
A
B
C
D
H753
98353
J mol
Q
H753
Q
14314
J Ans. mol
13.16 C3H8(g) = C2H4(g) + CH4(g)
= 1
Basis: 1 mole C3H8 feed. By Eq. (13.4)
nC3H8 = 1
Fractional conversion of C3H8 =
n0
nC3H8 n0
=
1
1 1
=
By Eq. (13.5),
yC3H8 =
1 1
yC2H4 =
1
yCH4 =
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 1
end rowsA) (
i Ai i
1.213 A 1.424 1.702 B
28.785 14.394 10 9.081
3
8.824 C 4.392 10 2.164
6
i
1 end
i Bi i
A
B
C
i
i Ci
A
1.913
B
5.31 10
3
C
2.268
10
6
D
0
(a) T
625 kelvin
T0
298.15 kelvin
493
G
H298
H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
T
G
2187.9
J mol
K
exp
G RT
(guess)
K
1.52356
By Eq. (13.28),
2
0.5
Given
1
1
= K
Find
0.777
This value of epsilon IS the fractional conversion. Ans.
2
(b)
0.85
K
1
4972.3
1
J mol
K
2.604
G
R T ln ( K)
G
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.
n0 = 1.5
By Eq. (13.5),
yC2H6 =
1 1.5
yH =
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 1
A
1.131 3.249 1.424 5.561 C 0.0 4.392
10
6
19.225 B 0.422 14.394 0.0 D 0.083 10 0.0
5
10
3
end
rowsA) (
i Ai i
i
1 end
i Bi
A
B
i
C
i
i Ci
D
i
i Di
A
3.542
B
4.409 10
3
C
1.169
10
6
D
8.3
10
3
T
1100 kelvin
T0
298.15 kelvin
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
G
5.429
10
3 J
mol
0.5
2
K
exp
G RT
K
1.81048
By Eq. (13.28),
(guess)
Given
1.5
1
= K
Find
0.83505
By Eq. (13.4), nC2H6 = 1
nH2 = nC2H4 =
n= 1
yC2H6
1 1
yH2
1
yC2H4
1
yC2H6
0.0899
yC2H4
0.4551
yH2
0.4551
Ans.
495
13.18 C2H5CH:CH2(g) = CH2:CHCH:CH2(g) + H2(g)
(1) (2) (3)
= 1
Number the species as shown. Basis is 1 mol species 1 + x mol steam. n0 = 1 x
y1 = 1 1 x
y2 = y3 = 1 x
By Eq. (13.5),
= 0.10
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 1 A
1.967 2.734 3.249 9.873 C 8.882 10 0.0
6
31.630 B 26.786 10 0.422
3
0.0 D 0.0 0.083
10
5
end
A
rows ( A)
i Ai i
i
B
i
1 end
i Bi
C
i
i Ci
D
i
i Di
A
T
G
4.016
950 kelvin
H298
B
4.422 10
T0
3
C
9.91
10
7
D
8.3
10
3
298.15 kelvin
H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
T
496
G
4.896
10
3 J
mol
K
exp
G RT
x
K
0.53802
By Eq. (13.28),
( )( ) 1 0.1 0.1 1
= K
Since
0.10 1
x
=
K K 0.10
0.843
x
0.10
1
x
6.5894
(a)
y1
1 1 x
yH2O
1
0.2
y1
y1
0.0186
yH2O
ysteam
0.7814
0.8682
Ans.
Ans.
(b)
ysteam
6.5894 7.5894
13.19
C4H10(g) = CH2:CHCH:CH2(g) + 2H2(g) (1) (2) (3)
= 2
Number the species as shown. Basis is 1 mol species 1 + x mol steam entering.
n0 = 1
x
By Eq. (13.5),
y1 =
1 1 x 2
y2 =
1
x
2
= 0.12
From data in Table C.4,
y3 = 2 y2 = 0.24
H298
235030
J mol
G298
166365
J mol
The following vectors represent the species of the reaction in the order in which they appear:
1 1 2
A
1.935 2.734 3.249 B
36.915 26.786 10 0.422
3
497
11.402 C 8.882 0.0
end rows ( A)
i Ai i
0.0
10
6
D
0.0 0.083
10
5
i
1 end
i Bi i
A
B
C
i
i Ci
D
i
i Di
A
7.297
B
9.285 10
3
C
2.52 10
6
D
1.66
10
4
T
925 kelvin
T0
298.15 kelvin
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
G
9.242
10
3 J
mol
K
exp
G RT
2
K 1 x 2
0.30066
By Eq. (13.28),
( 0.12) ( 0.24) 1
= K
Because
0.12 1
x
2
=
K K ( 0.24)
2
x
0.12
1
2
x
4.3151
0.839
(a) y1
1 1 x 2
yH2O
1
0.36
y1
y1
0.023
yH2O
0.617
Ans.
(b) ysteam
4.3151 5.3151
ysteam
0.812
Ans.
498
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
46110
J mol
G298
16450
3
J mol
A
2.9355
B
2.0905 10
C
0
D
0.3305 10
5
(a)
T
300 kelvin
T0
298.15 kelvin
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
G
P 1
1.627
10
4 J
mol
1
K
exp
G RT
K
679.57
P0
From Pb. 13.9 for ideal gases:
1
yNH3
1
1.299 K
P P0
0.5
0.9664
2
yNH3
0.9349
Ans.
(b) For
yNH3 = 0.5
by the preceding equation
2 3
Solving the next-to-last equation for K with P = P0 gives:
1 K 1
2
1
1.299
K
6.1586
499
Find by trial the value of T for which this is correct. It turns out to be
T = 399.5 kelvin
Ans.
(c) For P = 100, the preceding equation becomes
1 K 1
2
1
129.9
K
0.06159
Another solution by trial for T yields
T = 577.6 kelvin Ans.
(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
100bar
For NH3(1): Tc1
405.7kelvin
Pc1
112.8bar
Pr1 P Pc1
1
0.253
0.887
Tr1
For N2(2):
T Tc1
Tr1
1.437
Pc2
Pr1
2
Tc2
126.2kelvin
34.0bar
0.038
Tr2
583 126.2
Tr2
4.62
Pr2
100 34.0
Pr2
2.941
For H2(3), estimate critical constants using Eqns. (3.58) and (3.59)
Tc3 1
43.6 21.8 2.016 T kelvin
kelvin
Tc3
42.806 K
Tr3
T Tc3
Tr3
13.62
Pc3 1
20.5 44.2 T 2.016 kelvin
bar
Pc3
19.757 bar
Pr3
P Pc3
Pr3
5.061
3
0
500
Therefore,
i
1 3
1 2 3
PHIB Tr1 Pr1 PHIB Tr2 Pr2 PHIB Tr3 Pr3
0.924 1.034 1.029
1
i
0.5 1.5
i
i
1.184
The expression used for K in Part (c) now becomes:
1 K 1
2
1
K
129.9 1.184
0.07292
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)
Basis: 1 mol CO, 2 mol H2 feed
From the data of Table C.4,
= 2
n0 = 3
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
10.815 10
T0
3
C
3.45 10
6
D
0.135 10
5
300 kelvin
298.15 kelvin
501
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
G
P 1
2.439
10
4 J
mol
1
K
exp
G RT
K
1.762
10
4
P0
By Eq. (13.5), with the species numbered in the order in which they appear in the reaction,
y1 =
1 3 2
y2 =
2 3
2 2
y3 =
3
2
By Eq. (13.28),
0.8
(guess)
Given
3 4 1
2
2 3
=
P P0
2
K
Find
0.9752
y3
(b)
3
y3
2
0.5
3 y3 2 y3 1
y3
0.9291 Ans.
By the preceding equation
0.75
Solution of the equilibrium equation for K gives
K 3 4 1 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
K
3 4 1
2
2
100
3
2
K
2.7
10
3
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
528kelvin
P
100bar
For CO(1): Tc1
132.9kelvin
Pc1
34.99bar
1
0.048
Tr1
T Tc1
Tc3
Tr1
3.973
Pr1
P Pc1
Pr1
2.858
For CH3OH(3):
512.6kelvin
Pc3
80.97bar
3
0.564
Tr
T Tc3
Tr
1.03
Pr
P Pc3
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
1 3
1
PHIB Tr1 Pr1 1.0 0.612
1 2 1
1.032 1 0.612
i
i i
503
0.5933
The expression used for K in Part (c) now becomes:
3 4 1 2
2
K
100
3
2
0.593
K
1.6011
10
3
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 178321 J mol
G298 130401 J mol
The following vectors represent the species of the reaction in the order in which they appear:
1 1 1 A
12.572 6.104 5.457 B
2.637 0.443 10 1.045
i Ai
3
3.120 D 1.047 10 1.157
i Bi
5
i
1 3
A
i
B
i
D
i
i Di
A
T
1.011
1151.83 kelvin
B
1.149 10
T0
3
C
0
D
9.16
10
4
298.15 kelvin
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
504
G
126.324
J mol
K
exp
G RT
K
1.0133
Thus
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 91121 J mol
The following vectors represent the species of the reaction in the order in which they appear:
1 1 1 A
5.939 3.578 3.156 B
16.105 3.020 0.623
i Ai
0.0
10
3
D
0.186 10 0.151
5
i
1 3
A
i
B
i
i Bi
D
i
i Di
3
A
T
0.795
623.97 kelvin
B
0.012462
T0 298.15 kelvin
C
0
D
3.5
10
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
505
G
Thus
2.986
10
3 J
mol
K
exp
G RT
K
0.5624
T = 623.97 K
Ans.
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)
= 0.5
yNO2 yNO yO2
0.5
=
yNO2 yNO ( ) 0.21
0.5
= K
T
298.15 kelvin
From the data of Table C.4,
G298
K 1.493
(guesses)
35240
J mol
K
yNO
exp
10
G298 RT
12
10
6
yNO2
0.5
10
6
Given
yNO yNO2
yNO2 = ( ) 0.21
K yNO
yNO2 yNO
yNO = 5 10 7.307 10
6
Find yNO yNO2
12
This is about
7 10
6
ppm
(a negligible concentration) Ans.
13.26 C2H4(g) + (1/2)O2(g) = <(CH2)2>O(g)
= 0.5
See Example 13.9, Pg. 508-510 From Table C.4,
H298
105140
J mol
G298
81470
J mol
Basis: 1 mol C2H4 entering reactor. Moles O2 entering: nO2
1.25 0.5
Moles N2 entering:
nN2
nO2
79 21
506
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: 1 n nO2 0.5 0.8
1.424 A 3.639 0.385 3.280
0.0
10
6 2
14.394 B 0.506 23.463 0.593
10 kelvin
3
nN2 4.392 C 0.0 9.296 0.0 D
1
10 kelvin
5 2
0.227 0.0 0.040
0.5 1 0
kelvin
i
1 4
A
i
n
i Ai
B
i
n
i Bi
C
i
n
i Ci
D
i
n
i Di
y
n n0 0.5
K
i
y
i
i
K
15.947
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
Guess:
3.629
0.114 10 kelvin
3 A T0 1
3 3 5 2
B
T0
8.816
10
3
kelvin
C
4.904
10
6 2
kelvin
298.15 kelvin
idcph
B 2
T0
2
2
1
C 3
T0
1
D T0
1
idcps
A ln
B T0
C T0 D
2
1 2
2
1
T0
idcph
Given H298 = R A T0 1
3 3
130.182 kelvin
idcps
0.417
B 2
T0
2
2
1
C 3
H298
T0
1
D T0
H298
idcps
1
K
= exp
G298
1 T0
R T0
R T0
0.88244 3.18374
idcph
Find
508
0.0333 y 0.88244) ( 0.052 0.2496 0.6651
T T0
T 949.23 kelvin Ans.
Ans.
13.27
CH4(g) = C(s) + 2H2(g)
= 1
(gases only)
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 mol
G298 50460 J mol
The following vectors represent the species of the reaction in the order in which they appear:
1 1 2 A
1.702 1.771 3.249 2.164
6
9.081 B 0.771 10 0.422
3
0.0
10
i
1 3
C
0.0 0.0
D
0.867 10 0.083
5
A
i
i Ai
B
i
i Bi
C
i
i Ci
D
i
i Di
A
T
6.567
B
7.466 10
T0
3
C
2.164
10
6
D
7.01 10
4
923.15 kelvin
298.15 kelvin
509
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
G
1.109
10
4 J
mol
n0 = 1
K
exp
G RT
1 1
2
K
4.2392
By Eq. (13.5),
yCH4 =
yH2 =
2 2
2 1
(a)
By Eq. (13.28),
2 1
1
=
4 1
= K
K 4 K
0.7173
(fraction decomposed)
yCH4
1 1
yH2
2 1
yCH4
0.1646
yH2
(b) For a feed of 1 mol CH4 and 1 mol N2,
0.8354
n0 = 2
Ans.
By Eq. (13.28),
.8
(guess)
Given
2 2
2
1
= K
Find
0.7893
(fraction decomposed)
yCH4
1 2
yH2
2 2
yN2
1
yCH4
yH2
yH2
0.5659
yCH4
0.0756
yN2
0.3585
Ans.
510
13.28 1/2N2(g) + 1/2O2(g) = NO(g)
= 0
(1)
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
90250
J mol
G298
86550
3
J mol
5
A
0.0725
B
0.0795 10
C
0
D
0.1075 10
T
2000 kelvin
T0
298.15 kelvin
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
4 J
G
6.501 10
mol
K1
exp
G RT
= 0.5
K1
0.02004
(2)
1/2N2(g) + O2(g) = NO2(g)
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 1 1
i 1 3 A
3.280 3.639 4.982 B
0.593 0.506 10 1.195
3
0.040 D 0.227 10 0.792
i Bi
5
A
i
i Ai
B
i
D
i
i Di
A
0.297
B
3.925
10
4
C
0
D
5.85
10
4
T
2000 kelvin
T0
298.15 kelvin
511
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
10
5 J
G
1.592
mol
K2
exp
G RT
K2
6.9373 10
5
With the assumption of ideal gases, we apply Eq. (13.28): (1)
yN2 yNO
0.5
yO2
0.5
=
yNO ( ) 0.7
0.5
( ) 0.05
0.5
= K1
yNO
(2)
K1 ( ) 0.7
P0 1
0.5
( ) 0.05
200
0.5
yNO
3.74962
10
3
Ans.
P
yNO2 yN2
0.5
=
yO2
yNO2 ( ) 0.7
0.5
=
( ) 0.05
0.5
P P0
0.5
K2
yNO2
P P0
0.5
K2 ( ) 0.7
( ) 0.05
yNO2
4.104
10
5
Ans.
13.29
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 89830 J mol
512
The following vectors represent the species of the reaction in the order in which they appear:
2 1 3 2
i 1 4 A
3.931 5.699 4.114 3.470 B
1.490 0.801 1.728 1.450
10
3
0.232 D 1.015 0.783 0.121
10
5
A
i
i Ai
B
i
i Bi
D
i
i Di
A
5.721
B
6.065 10
3
C
0
D
6.28
10
4
T
723.15 kelvin
T0
298.15 kelvin
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
4 J
G
1.538 10
mol
K
exp
G RT
(basis)
K
12.9169
By Eq. (13.5), gases only:
n0 = 3
yH2S =
2 3
2
ySO2 =
1 3
yH2O =
2 3
By Eq. (13.28),
0.5
2
(guess)
Given
2 2 2
3
2
= 8K
1
Find
0.767
Percent conversion of reactants = PC
PC =
ni0 ni0
ni
100 =
i
ni0
100
[By Eq. (13.4)]
513
Since the reactants are present in the stoichiometric proportions, for each reactant,
ni0 =
i
Whence
PC
100
PC
76.667
Ans.
13.30 N2O4(g) = 2NO2(g) (a) (b)
= 1
Data from Tables C.4 and C.1 provide the following values:
H298
57200
J mol
G298
5080
J mol
T0
298.15 kelvin
T
350 kelvin
3
A
1.696
B
0.133 10
C
0
D
1.203 10
5
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
G
3.968
10
3 J
mol
K
exp
G RT
K
3.911
Basis: 1 mol species (a) initially. Then
ya =
1 1
yb =
2 1
2 1
2
1
K 4P K
=
P P0
1
K
(a)
P
5
P0
1
0.4044
ya
1 1
ya
0.4241
Ans.
(b)
P
1
P0
1
K 4P K
0.7031
514
By Eq. (4.18), at 350 K:
H
H298
R IDCPH T0 T
A
B
C
D
H
56984
J mol
This is Q per mol of reaction, which is
0.7031
0.4044
0.299
Whence
Q
H
Q
17021
J mol
Ans.
13.31 By Eq. (13.32),
2
2
K=
xB B xA A
1
=
2
xA xA A
B
ln a = 0.1 xB
ln b = 0.1 xA
1
=
Whence
2 2
K=
1
xA xA
exp 0.1 xA
xA xA
2 exp 0.1 xB
exp 0.1 xA
xB
K=
1
xA xA
exp 0.1 2 xA J mol
1
K = exp
G RT
G
1000
T
298.15 kelvin
xA
.5
(guess)
Given
xA
1
xA xA
exp 0.1 2 xA
Ans.
1
= exp
G RT
xA
Find xA
0.3955
For an ideal solution, the exponential term is unity:
Given
1
xA xA
= exp
G RT
xA
Find xA
xA
0.4005
This result is high by 0.0050. Ans.
515
13.32 H2O(g) + CO(g) = H2(g) + CO2(g)
= 0
From the the data of Table C.4,
H298
41166
J mol
G298
28618
J mol
T0
298.15 kelvin
T
800 kelvin
3
A
1.860
B
0.540 10
C
0
D
1.164 10
5
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
3 J
G
9.668
10
mol
= 0
K
exp
G RT
K
4.27837
(a) No. Since
, 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),
nCO = 1
nH2 = 1
NCO2 =
1 1 1
=
1 2
= 0.02
0.96 1.02
0.941
Let z = w/2 = moles H2O/mole "Water gas". By Eq. (13.5),
yH2O =
2z w = 2 2z 2 w
yCO =
1 2 2z
yH2 =
1 2 2z
516
yCO2 =
2
2z
By Eq. (13.28)
z
2
(guess)
Given
1 2z 1
= K
z
Find z ()
z
4.1
Ans.
(d)
2CO(g) = CO2(g) + C(s)
= 1 (gases)
Data from Tables C.4 and C.1:
H298
172459
J mol
G298
120021
J mol
D 1.962 10
5
A
G
0.476
H298
B
0.702 10
3
C
0
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
G
3.074
10
4 J
mol
K
exp
G RT
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
2
2z
yCO
1 2 2z
yCO2
0.092
yCO
5.767
10
3
RATIO
yCO2 yCO
2
RATIO
2.775
10
3
No carbon can deposit from the equilibrium mixture.
517
13.33 CO(g) + 2H2(g) = CH3OH(g)
= 2
(1)
This is the reaction of Pb. 13.21, where the following parameter values are given:
H298
90135
J mol
G298
24791
J mol
T
550 kelvin
T0
298.15 kelvin
A
7.663
B
10.815 10
3
C
3.45 10
6
D
0.135 10
5
G
H298
H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
T
G
3.339
10
4 J
mol
K1
exp
G RT
= 0
K1
6.749
(2)
10
4
H2(g) + CO2(g) = CO(g) + H2O(g)
From the the data of Table C.4,
H298
41166
J mol
G298
28618
J mol
T
550 kelvin
T0
298.15 kelvin
The following vectors represent the species of the reaction in the order in which they appear:
1 1 1 1
i 1 4 A
3.249 5.457 3.376 3.470 B
0.422 1.045 0.557 1.450
i Ai
0.083
10
3
D
1.157 0.031 0.121
10
5
A
i
B
i
i Bi
D
i
i Di
A
1.86
B
5.4
10
4
518
C
0
D
1.164
10
5
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
4 J
G
1.856 10
mol
K2
exp
G RT
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.j
i= j 1 2
H2
CO
CO2
CH3OH
H2O
_______________________________________________
-2 -1
-1 1
0 -1
1 0
0 1
By Eq. (13.7)
0.75 yH2 = 1
2 1 2 1
2
2
0.15 yCO = 1
1
2
2 1
1
0.05 yCO2 = 1
2 1
yCH3OH =
1
2 1
yH2O =
2
1
2 1
P
100
P0
1
1
By Eq. (13.40),
0.1
2
0.1
(guesses)
Given
1 1
2 1
2
2
=
1 2
0.75
2 1
0.15
1
2
0.15
P P0
2
K1
2 2
2 2
0.75
2 1
0.05
= K2
1 2
Find 1 2
519
1
0.1186
2
8.8812
10
3
0.75 yH2 1
2 1 2 1
2
2
0.15 yCO 1
1
2
2 1
1
0.05 yCO2 1
2 1
yCH3OH
1
2 1
yH2O
2
1
2 1
yN2
1
yH2
yCO
yCO2
yCH3OH
yH2O
yH2
0.6606
yCO
0.0528
yCO2
0.0539
Ans.
yCH3OH
0.1555
yH2O
0.0116
yN2
0.0655
13.34
CH4(g) + H2O(g) = CO(g) + 3H2(g)
From the the data of Table C.4,
= 2
(1)
H298
205813
J mol
G298
141863
J mol
The following vectors represent the species of the reaction in the order in which they appear:
1 1 1 3
2.164 C 0.0 0.0 0.0
10
6
1.702 A 3.470 3.376 3.249 0.0 D 0.121 0.031 0.083
10
5
9.081 B 1.450 0.557 0.422
10
3
i
1 4
A
i
i Ai
B
i
i Bi
C
i
i Ci
D
i
i Di
A
7.951
B
8.708 10
3
C
520
2.164
10
6
D
9.7
10
3
T
G
1300 kelvin
H298
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
10
5 J
G
1.031
mol
K1
exp
G RT
K1
13845
= 0 (2) This is the reaction of Pb. 13.32, where parameter values are given:
H298
A
G
H2O(g) + CO(g) = H2(g) + CO2(g)
41166
J mol
0.540 10
G298
3
28618
J mol
D 1.164 10
5
1.860
H298
B
C
0.0
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
10
3 J
G
5.892
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 5
yCO2 =
5
yH2 =
3 5
By Eq. (13.28),
0.5
(guess)
Given
3 1
2
= K2
Find
0.1375
Ratio =
yH2 yCO
Ratio
3 1
Ratio
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
G298
120021
J mol
For T = 1300 K,
T
1300 kelvin
T0
298.15 kelvin
A
0.476
B
0.702 10
3
C
0.0
D
1.962 10
5
G
H298
T H298 G298 T0 R IDCPH T0 T A B C D R T IDCPS T0 T A B C D
522
G
5.673
10
4 J
mol
K
exp
G RT
K
5.255685
10
3
As explained in Problem 13.32(d), the question of carbon deposition depends on:
RATIO =
yCO2 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:
Ratio 1
5
2
Ratio
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
CO + H2O = CO2 + H2
(1)
(2)
There are alternative equivalent pairs, but for these: Stoichiometric numbers,
i=
j 1 2 -1 0 -1 -1 1 -1 0 1
523
i.j
CH4
H2O
CO
CO2
H2
j
_________________________________________________
3 1
2 0
For initial amounts: 2 mol CH4 and 3 mol H2O, n0 = 5, and by Eq. (13.7):
2 yCH4 =
yCO2 =
1
3 yH2O =
yH2 =
1
2
5
2 1
2
5
2 1
2
yCO =
1
2
5
2 1
3 1 5
5
2 1
2 1
By Eq. (13.40), with P = P0 = 1 bar
yCO yH2 = k1 yCH4 yH2O
3
yCO2 yH2 yCO yH2O
= k2
From the data given in Example 13.14,
G1 27540 J mol
G2 3130 J mol
T 1000 kelvin
K1
exp
G1 RT
K2
K2
exp
1.457
G2 RT
K1
1
27.453
1.5
2
1
2
(guesses)
3 1
1
2 1 2
Given
2
1
3 2
1
3
2
5
2 1
2
= K1
2 3 1 1 2
3
= K2
1 2
Find 1 2
1
1.8304
2
0.3211
2 yCH4 5
1
3 yH2O 5
524
1
2
2 1
2 1
yCO
1
2
5
2 1
yCO2
2
5 2 1
yH2
3 1
2
5 2 1
yCH4
0.0196
yH2O
0.098
yCO
0.1743
yCO2
0.0371
yH2
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
101.33 kPa
x1 =
x2 =
y1 P 415kPa
y2 P Psat2
Water(2):
x2 Psat2 = y2 P
Psat2
3.166 kPa
(steam tables) Ethylene glycol(3):
Psat3 = 0.0
y3 = 0.0
Therefore, y2 = 1
y1
and
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,
k= y1
3 x3
P P0
=
2 x2
x3 y1 x2
T
298.15kelvin
Data from Table C.4:
G298
k 6.018
72941
10
12
J mol
Ans.
k
exp
G298 RT
525
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 a) Stoichiometric coefficients:
1
Initial numbers of n0 moles
j
50 50 kmol hr 0 0
1 0 1 0 1 1
Number of components: i
1 4
Number of reactions:
n0 n0i
i
1 2
100 kmol hr
vj
i
i j
v
yA
1 1
0.05
n0
Given values:
Guess:
yB
0.10
yC
0.4
yD
0.4
1
1
kmol hr
2
1
kmol hr
Given
yA =
n01 n0
1 1
2 2
yB =
n02 n0
1
1 2
Eqn. (13.7)
yC =
n03 n0
1 1
2 2
yD =
n04 n0
1
2 2
yC yD
Find yC yD 1 2
1 2
1
44.737
kmol hr
2
2.632
kmol hr
526
(i) nA
n01
1
2
nA
2.632
nB
n02
1
nB
2
5.263
kmol hr kmol
nC
n03
1
nC
42.105
hr kmol
nD
n04
2
nD
2.632
hr kmol
n
nA
nB
nC
nD
n
52.632
hr kmol
hr
Ans.
(ii)
yC
0.8
yD
0.05
Ans.
40
1
b) Stoichiometric coefficients:
1 2 0 1
Number of reactions:
1 1 0
Initial numbers of n0 moles
40 kmol hr 0 0 j 1 2
Number of components:
i
1 4
vj
i
i j
v
1 2
n0
n0i
i
n0
80
kmol hr
Given values:
yC
0.52
yD
0.04
Guess:
yA
0.4
yB
0.4
1
1
kmol hr
2
1
kmol hr
Given
yA =
n01 n0
1
1
2
2 2
1
yB =
n02 n0
1 1
2 2 2 2
2
Eqn. (13.7)
yC =
n03 n0
1
2 2
yD =
n04 n0
1
2 2
527
yA yB
1 2
Find yA yB 1 2
1
26
kmol hr
2
2
kmol hr
yA
0.24
yB
0.2
nA
n01
1
2
nA
12
nB
n02
1
2 2
nB
10
kmol hr kmol
nC
n03
1
nC
26
hr kmol
Ans.
nD
n04
2
nD
2
hr kmol
hr
1
c) Stoichiometric coefficients:
1 1 0 1
Number of reactions:
100
1 1 0
Initial numbers of n0 moles
0 0 0 j
kmol hr
Number of components:
i
1 4
1 2
vj
i
i j
v
1 1
n0
n0i
i
n0
100
kmol hr
Given values:
yC
0.3
yD
0.1
Guess:
yA
0.4
yB
0.4
1
1
kmol hr
2
1
kmol hr
Given
yA =
n01 n0
1 1
2 2
0
yB =
n02 n0
1 1
2 2
Eqn. (13.7)
yC =
n03 n0
1
1 2
yD =
n04 n0
1
2 2
528
yA yB
1 2
Find yA yB 1 2
1
37.5
kmol hr
2
12.5
kmol hr
yA
0.4
yB
0.2
nA
n01
1
2
nA
50
nB
n02
1
2
nB
25
kmol hr kmol
nC
n03
1
nC
37.5
hr kmol
Ans.
nD
n04
2
nD
12.5
hr kmol
hr
1 1
d) Stoichiometric coefficients:
1 1 0 1 1
Number of reactions:
40
1 0 0
Initial numbers of n0 moles
60 0 0 0 kmol hr
Number of components:
i
1 5
j
1 2
kmol hr
vj
i
i j
v
1 0
0.25
n0
i
n0i
n0
100
Given values: yC
yD
0.20
Guess:
yA
0.2
yB
0.4
yE
0.1
1
1
kmol hr
2
1
kmol hr
529
Given
yA =
n01 n0
1 1
2
yB =
n02 n0
1 1
2
Eqn. (13.7)
yC =
n03 n0
1 1
yD =
n04 n0
2 1
yE =
n05 n0
2 1
yA yB yE
1 2
Find yA yB yE
1
2
1
20
kmol hr
2
16
kmol hr
(i)
(ii)
nA
n01
1
2
nA
nB
n02
1
2
nB
nC
nD
n03
n04
1
nC
nD
kmol hr kmol 24 hr kmol 20 hr 4
16 kmol hr kmol
yA
0.05
yB
0.3
Ans.
yC
yD
0.25
0.2
2
nE
n05
2
nE
16
hr
yE
0.2
13.45
C2H4(g) + H2O(g) -> C2H5OH(g)
T0
298.15kelvin
P0
1bar
T
400kelvin
P
2bar
1 = C2H4(g)
H0f1
52500
J mol
G0f1
68460
J mol
2 = H2O(g)
H0f2
241818
J mol J
G0f2
228572
3 = C2H5OH(g)
H0f3
235100
mol
G0f3
168490
J mol J
mol
530
H0
H0f1
H0f2
H0f3
H0
45.782
G0
G0f1
G0f2
G0f3
G0
8.378
kJ mol kJ
mol
A
( 1.424) ( 3.470) ( 3.518)
A
3
1.376
B
[( 14.394) ( 1.450) ( 20.001) ]10
B
4.157
10
3
C
[ ( 4.392) () ( 6.002) 0 ]10
6
C
1.61 10
6
D
[ () ( 0 0.121) () 0 ]10
5
D
K0
1.21 10
29.366
4
a) K0
exp
G0 R T0
H0 1 R T0
Eqn. (13.21) K298
K298
Ans.
b) K1
exp
T0 T
Eqn. (13.22)
K1
9.07
10
3
K2
exp
1 IDCPH T0 T A B C D T IDCPS T0 T A B C D K0 K1 K2 Eqn. (13.20)
K2
0.989
Eqn. (13.23)
K400
K400
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
e e
y3 =
e
2
e
Assuming ideal gas behavior
y3 y1 y2
= K
P P0
e
Substituting results in the following expression:
2
e
1 2
e 1 e 2
= K400
e e
P P0
531
Solve for
e using
a Mathcad solve block.
P P0
e
Guess:
e
0.5
e
Given
2
e
1 2
1
e 1 e 2
e e
= K400
e e
Find e
e
0.191
1 y2
y2
y1
y1
e e
2
0.447
2
0.447
y3
y3
e
2
0.105
e
Ans.
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
136.1064
kJ mol
T
298.15kelvin P
1bar
130.680
J mol kelvin
J mol kelvin
S0O2
S0O2
205.152
J mol kelvin
S0H2O2
232.95
S0fH2O2
G0f
S0H2
H0fH2O2
S0H2O2
S0fH2O2
G0f
102.882
kJ mol
J mol kelvin
Ans.
T S0fH2O2
105.432
532
13.48
C3H8(g) -> C3H6(g) + H2(g) (I) C3H8(g) -> C2H4(g) + CH4(g) (II)
T0
298.15kelvin
P0
1bar
T
750kelvin
P
1.2bar
1 = C3H8 (g)
H0f1
104680
J mol
G0f1
24290
J mol
2 = C3H6 (g)
H0f2
19710
J mol
G0f2
62205
J mol
3 = H2 (g)
H0f3
0
J mol
G0f3
0
J mol
4 = C2H4 (g)
H0f4
52510
J mol
G0f4
68460
J mol
5 = CH4 (g)
H0f5
74520
J mol
G0f5
50460
J mol
Calculate equilibrium constant for reaction I:
H0I H0f1 H0f2 H0f3
H0I
124.39
G0I
G0f1
G0f2
G0f3
G0I
86.495
kJ mol kJ
mol
AI
( 1.213) ( 1.637) ( 3.249)
AI
3
3.673
BI
[( 28.785) ( 22.706) ( 0.422) ]10
BI
5.657
10
3
CI
[ ( 8.824) ( 6.915) () 0 ]10
6
CI
1.909
10
6
DI
[ () () ( 0 0 0.083) ]10
5
DI
KI0 KI1 KI2
8.3
0
10
3
KI0
KI1 KI2
exp
G0I R T0
H0I 1 R T0 1
Eqn. (13.21)
exp
T0 T
Eqn. (13.22)
1.348
10
13
exp
IDCPH T0 T AI BI CI DI T IDCPS T0 T AI BI CI DI
533
1.714
Eqn. (13.23)
KI
KI0 KI1 KI2
Eqn. (13.20)
KI
0.016
Calculate equilibrium constant for reaction II:
H0II
H0f1
H0f4
H0f5
H0II
82.67
kJ mol kJ
G0II
G0f1
G0f4
G0f5
G0II
42.29
mol
AII
( 1.213)
( 1.424)
( 1.702)
AII
3
1.913
BII
[ ( 28.785)
( 14.394)
( 9.081) ] 10
BII
6
5.31
10
3
CII
[ ( 8.824)
( 4.392)
( 2.164) ] 10
CII
2.268
10
6
DII
[ ( 0)
( 0)
( 0) ] 10
5
DII
KII0 KII1
0
3.897 5.322 10 10
8 8
KII0
KII1 KII2
exp
G0II R T0
H0II 1 R T0
Eqn. (13.21)
exp
T0 T
Eqn. (13.22)
exp
1 IDCPH T0 T AII BII CII DII T IDCPS T0 T AII BII CII DII
KII2
1.028
Eqn. (13.23)
KII
KII0 KII1 KII2 Eqn. (13.20)
KII
21.328
Assume an ideal gas and 1 mol of C3H8 initially.
1 y1 = 1
I I
II II
y2 =
I
1
y3 =
II
I
I
1
I
II
y4 =
II
1
y5 =
II
II
Eqn. (13.7)
II
I
1
I
The equilibrium relationships are:
y2 y3 P0 = KI P y1
y4 y5 y1
= KII
P0 P
Eqn. (13.28)
534
Substitution yields the following equations:
I
I
1
I
II
1
I I II II
I
II
1 1
= KI
P0 P
II
II
1
I
II
1
I I II II
I
II
1 1
= KII
P0 P
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
II 1
1
= KI
II
P0 P
P0 P
1 1 1 1
I I I I
II II II II
I
II
I
II
1
I
I
II 1
= KII
II
I
Find I
II
II
I
0.026
II
0.948
1 y1 1
I I
II
II II
y2
I
1
y3
II
II
I
I
1
I
II
y4
y1
1
y5
II
I
1
I
II
0.01298 y2
0.0132 y3
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 7 K 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
298.15kelvin
P0
1bar
T
425kelvin
P
15bar
1 = n-C4H10(g)
H0f1
125790
2 = iso-C4H10(g)
H0f2
134180
J mol J
G0f1
16570
mol
G0f2
20760
J mol J
mol
H0
H0f1
H0f2
H0
8.39
kJ mol kJ
G0
G0f1
G0f2
G0
4.19
mol
A 0.258
A
B
( 1.935) ( 1.677)
[( 36.915) ( 37.853) ]10
3
B
9.38 10
4
C
[11.402) ( 11.945) ( ]10
6
C
5.43 10
7
D
[ () () 0 0 ]10
5
D
K0
0
5.421 Ans.
a) K0
exp
G0 R T0
H0 1 R T0
Eqn. (13.21)
b) K1
exp
T0 T
Eqn. (13.22)
K1
0.364
K2
exp
1 IDCPH T0 T A B C D T IDCPS T0 T A B C D
536
K2
1
Eqn. (13.23)
Ke
K0 K1 K2
Eqn. (13.20)
Ke
1.974
Ans.
Assume as a basis there is initially 1 mol of n-C4H10(g)
y1 = 1
e
y2 = e
a) Assuming ideal gas behavior
y2 y1
= Ke
e
Substitution results in the following expression:
1
= Ke
e
e
Solving for Ke yields the following analytical expression for
1
e
1
Ke
e
e
0.336
e
y1
1
y1
0.664
y2
y2
0.336
Ans.
b) Assume the gas is an ideal solution. In this case Eqn. (13.27) applies.
yi i =
i
P P0
K
Eqn. (13.27)
Substituting for yi yields:
1
e e 1
2
= K
e= 2 2
This can be solved analytically for
e
to get:
Ke 1
Calculate i for each pure component using the PHIB function.
For n-C4H10:
1
0.200
Tc1
425.1kelvin
Pc1
37.96bar
Tr1
1
T Tc1
Tr1
1
1
Pr1
1
P Pc1
Pr1
0.395
PHIB Tr1 Pr1
0.872
For iso-C4H10:
2
0.181
Tc2
408.1kelvin
Pc2
36.48bar
Tr2
T Tc2
Tr2
1.041
Pr2
537
P Pc2
Pr2
0.411
2
PHIB Tr2 Pr2
e
2
2
0.884
e
Solving for
yields:
2 e 2
Ke 1
0.339
y1
1
e
y1
0.661
y2
e
y2
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
2
T := ( 55 + 273.15) K
Margules equations: 1 ( x1) := exp ( 1 x1) A12 + 2 ( A21 A12) x1 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 1 ( x1) Psat1 + ( 1 x1) 2 ( x1) Psat2 y1 ( x1) := x1 := 0.25 x1 := 0.50 x1 := 0.75 x1 1 ( x1) Psat1 Pbubl ( x1) Pbubl ( x1) = 64.533 kPa Pbubl ( x1) = 80.357 kPa Pbubl ( x1) = 85.701 kPa y1 ( x1) = 0.562 y1 ( x1) = 0.731 y1 ( x1) = 0.808
(b)
BUBL P calculations with virial coefficients: B11 := 963 cm
3
mol
B22 := 1523
cm
3
mol
B12 := 52
cm
3
mol
12 := 2 B12 B11 B22
B11 ( P Psat1) + P y22 12 1 ( P , T , y1 , y2) := exp R T B22 ( P Psat2) + P y12 12 2 ( P , T , y1 , y2) := exp R T
539
Guess: x1 := 0.25
P :=
Psat1 + Psat2 2 Given
y1 := 0.5
y2 := 1 y1
y1 1 ( P , T , y1 , y2) P = x1 1 ( x1) Psat1 y2 2 ( P , T , y1 , y2) P = ( 1 x1) 2 ( x1) Psat2 y2 = 1 y1
y1 y2 := Find ( y1 , y2 , P) P
x1 := 0.50 Given
y1 0.558 y2 = 0.442 P 63.757 kPa
y1 1 ( P , T , y1 , y2) P = x1 1 ( x1) Psat1 y2 2 ( P , T , y1 , y2) P = ( 1 x1) 2 ( x1) Psat2 y2 = 1 y1
y1 y2 := Find ( y1 , y2 , P) P
x1 := 0.75 Given
y1 0.733 y2 = 0.267 P 79.621 kPa
y1 1 ( P , T , y1 , y2) P = x1 1 ( x1) Psat1 y2 2 ( P , T , y1 , y2) P = ( 1 x1) 2 ( x1) Psat2 y2 = 1 y1
y1 y2 := Find ( y1 , y2 , P) P
y1 0.812 y2 = 0.188 P 85.14 kPa
540
14.3
T := 200 K H1 := 200 bar
P := 30 bar B := 105 cm
3
y1 := 0.95
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: fhat1 = H1 x1 By Eq. (11.36):
l
fhat1 = y1 1 P 1 := exp B P R T 1 = 0.827
v
Equate the liquid- and vapor-phase fugacities and solve for x1: x1 := y1 1 P H1 x1 = 0.118 Ans.
14.4
Pressures in kPa Data:
0.000 0.0895 0.1981 0.3193 x1 := 0.4232 0.5119 0.6096 0.7135
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
i := 2 .. rows ( P) (a)
It follows immediately from Eq. (12.10a) that:
ln 1 = A12
Combining this with Eq. (12.10a) yields the required expression
541
(b)
(c)
Henry's constant will be found as part of the solution to Part (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. 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1
2 2
2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2 Guesses: H1 := 50 A21 := 0.2 A12 := 0.4
Mininize the sums of the squared errors by setting sums of derivatives equal to zero. Given 0=
i
2 d H1 Pi x1 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dA12 + x2i 2 x1i , x2i , A12 , A21 Psat2
( (
( (
) )
) )
0=
i
2 d H1 Pi x1 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dA21 + x2i 2 x1i , x2i , A12 , A21 Psat2 2 d H1 Pi x1 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dH1 + x2i 2 x1i , x2i , A12 , A21 Psat2
0=
i
(
(
)
)
A12 A21 := Find ( A12 , A21 , H1) H 1
A12 0.348 A21 = 0.178 H 1 51.337
Ans.
542
(d) 1 ( x1 , x2) := exp x22 A12 + 2 ( A21 A12) x1 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2
2
Pcalc := x1 1 x1 , x2
i i i i
(
) exp( A12) + x2i 2 ( x1i , x2i) Psat2
H1
y1calc :=
i
H1 x1 1 x1 , x2 i i i exp ( A 12) Pcalc
i
(
)
0.2 0 PiPcalc
i
(y1iy1calci) 100
0.2
0.4
0.6
0
0.2
0.4 x1
i
0.6
0.8
Pressure residuals y1 residuals
Fit GE/RT data to Margules eqn. by least squares: i := 2 .. rows ( P) Given 0= 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
i
... A21 x1 ... x1 x2 i i i + A12 x2i
2
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
... A21 x1 ... x1 x2 i i i + A12 x2i
2
2
0=
i
... A21 x1 ... x1 x2 i i i + A12 x2i
A12 A21 := Find ( A12 , A21 , H1) H 1
2 2
A12 0.375 A21 = 0.148 H 1 53.078
Ans.
1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 H1 Pcalc := x1 1 x1 , x2 + x2 2 x1 , x2 Psat2 i i i i exp ( A 12) i i i
(
)
(
)
y1calc :=
i
H1 x1 1 x1 , x2 i i i exp ( A 12) Pcalc
i
(
)
544
0
0.2 PiPcalc
i
(y1iy1calci) 100
0.4
0.6
0.8
0
0.2
0.4 x1
i
0.6
0.8
Pressure residuals y1 residuals
14.5
Pressures in kPa
Data:
0.3193 0.4232 0.5119 0.6096 x1 := 0.7135 0.7934 0.9102 1.000
x2 := 1 x1
21.63 24.01 25.92 27.96 P := 30.12 31.75 34.15 36.09
0.5934 0.6815 0.7440 0.8050 y1 := 0.8639 0.9048 0.9590 1.000
Psat1 := P8
i := 1 .. 7
(a) It follows immediately from Eq. (12.10a) that:
ln 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. 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1
2 2
2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2 Guesses: H2 := 14 A21 := 0.148 A12 := 0.375
Mininize the sums of the squared errors by setting sums of derivatives equal to zero. Given 0 =
i
d 2 dA Pi x1i 1 x1i , x2i , A12 , A21 Psat1 ... 12 H2 + x2i 2 x1i , x2i , A12 , A21 exp ( A21)
(
)
( (
) )
0=
dA21 Pi x1 1 ( x1 , x2 , A12 , A21) Psat1 ...
i
d
H2 + x2 2 x1 , x2 , A12 , A21 i i i exp ( A21)
i
i
i
2
0=
i
d 2 dH Pi x1i 1 x1i , x2i , A12 , A21 Psat1 ... 2 H2 + x2 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
Ans.
(d) 1 ( x1 , x2) := exp x22 A12 + 2 ( A21 A12) x1 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2
2
546
H2 Pcalc := x1 1 x1 , x2 Psat1 + x2 2 x1 , x2 i i i i i i i exp ( A 21) x1 1 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 2 3 4
(y1iy1calci) 100
0.2
0.4
0.6 x1
i
0.8
Pressure residuals y1 residuals
Fit GE/RT data to Margules eqn. by least squares: i := 1 .. 7 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
A21 x1 ... x1 x2 i i i + A12 x2i
2
547
0=
i
y1 Pi d x1 ln i ... dA21 i x1i Psat1 y2 Pi i + x ln 2i H2 x2i exp ( A ) 21
A21 x1 ... x1 x2 i i i + A12 x2i
2
2
0=
i
y1 Pi d x1 ln i ... dH2 i x1i Psat1 y2 Pi i + x ln 2i H2 x2 i exp ( A ) 21
A21 x1 ... x1 x2 i i i + A12 x2i
A12 A21 := Find ( A12 , A21 , H2) H 2
2 2
A12 0.37 A21 = 0.204 H 2 15.065
Ans.
1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 H2 Pcalc := x1 1 x1 , x2 Psat1 + x2 2 x1 , x2 i i i i i i i exp ( A 21) y1calc :=
i
(
)
(
)
x1 1 x1 , x2 Psat1
i i i
(
)
Pcalc
i
548
0
PiPcalc
0.2
i
(y1iy1calci) 100
0.4 0.6
0.3
0.4
0.5
0.6 x1
i
0.7
0.8
0.9
1
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
0.0 0.0932 0.1248 0.1757 0.2000 x1 := 0.2626 0.3615 0.4750 0.5555 0.6718
0.0 0.1794 0.2383 0.3302 0.3691 y1 := 0.4628 0.6184 0.7552 0.8378 0.9137
i := 2 .. rows ( P) (a)
x2 := 1 x1
Psat2 := P1
It follows immediately from Eq. (12.10a) that:
ln 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. 1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1
2 2
2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2 Guesses: H1 := 35 A21 := 1.27 A12 := 0.70
Mininize the sums of the squared errors by setting sums of derivatives equal to zero. Given 0=
i
2 d H1 Pi x1 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dA12 + x2i 2 x1i , x2i , A12 , A21 Psat2
(
(
)
)
0=
i
2 d H1 Pi x1 1 x1 , x2 , A12 , A21 ... i i exp ( A12) dA21 i + x2i 2 x1i , x2i , A12 , A21 Psat2
(
(
)
)
0=
i
2 d H1 Pi x1 1 x1 , x2 , A12 , A21 ... i i i exp ( A12) dH1 + x2i 2 x1i , x2i , A12 , A21 Psat2
(
(
)
)
A12 A21 := Find ( A12 , A21 , H1) H 1
550
A12 0.731 A21 = 1.187 H 1 32.065
Ans.
(d)
1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1
2
2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2
2
Pcalc := x1 1 x1 , x2
i i i i
(
) exp( A12) + x2i 2 ( x1i , x2i) Psat2
H1
y1calc :=
i
H1 x1 1 x1 , x2 i i i exp ( A 12) Pcalc
i
(
)
0.5 0 PiPcalc
i
0.5 1 1.5 2
(y1iy1calci) 100
0
0.1
0.2
0.3 x1
i
0.4
0.5
0.6
0.7
Pressure residuals y1 residuals
Fit GE/RT data to Margules eqn. by least squares: i := 2 .. rows ( P) Given 0= y2 := 1 y1
i
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
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
... A21 x1 ... x1 x2 i i i + A12 x2i
2
2
0=
i
... A21 x1 ... x1 x2 i i i + A12 x2i
A12 A21 := Find ( A12 , A21 , H1) H 1
2 2
A12 0.707 A21 = 1.192 H 1 33.356
Ans.
1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 H1 Pcalc := x1 1 x1 , x2 + x2 2 x1 , x2 Psat2 i i i i exp ( A 12) i i i x1 1 x1 , x2 y1calc :=
i i i i
(
)
(
)
(
) exp( A12)
H1
i
Pcalc
552
0 0.5 PiPcalc
i
1 1.5 2 2.5
(y1iy1calci) 100
0
0.1
0.2
0.3 x1
i
0.4
0.5
0.6
0.7
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
x2 := 1 x1
19.30 19.89 21.37 24.95 29.82 P := 34.80 42.10 60.38 65.39 69.36
0.3302 0.3691 0.4628 0.6184 0.7552 y1 := 0.8378 0.9137 0.9860 0.9945 1.0000
Psat1 := P10
i := 1 .. 9 (a)
It follows immediately from Eq. (12.10a) that:
ln 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.
1 ( x1 , x2 , A12 , A21) := exp ( x2) A12 + 2 ( A21 A12) x1
2 2
2 ( x1 , x2 , A12 , A21) := exp ( x1) A21 + 2 ( A12 A21) x2 Guesses: H2 := 4 A21 := 1.37 A12 := 0.68
Mininize the sums of the squared errors by setting sums of derivatives equal to zero. Given 0=
dA12 Pi x1 1 ( x1 , x2 , A12 , A21) Psat1 ...
i
d
H2 + x2 2 x1 , x2 , A12 , A21 i i i exp ( A21)
i
i
i
(
)
2
0=
i
d 2 dA Pi x1i 1 x1i , x2i , A12 , A21 Psat1 ... 21 H2 + x2i 2 x1i , x2i , A12 , A21 exp ( A21)
(
)
(
)
0=
dH2 Pi x1 1 ( x1 , x2 , A12 , A21) Psat1 ...
i
d
H2 + x2 2 x1 , x2 , A12 , A21 i i i exp ( A21)
i
i
i
(
)
2
A12 A21 := Find ( A12 , A21 , H2) H 2
A12 0.679 A21 = 1.367 H 2 3.969
Ans.
554
(d) 1 ( x1 , x2) := exp x22 A12 + 2 ( A21 A12) x1 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2
2
H2 Pcalc := x1 1 x1 , x2 Psat1 + x2 2 x1 , x2 i i i i i i i exp ( A 21) y1calc :=
i
(
)
(
)
x1 1 x1 , x2 Psat1
i i i
(
)
Pcalc
1
i
PiPcalc
0
i
(y1iy1calci) 100
1 2
0
0.2
0.4 x1
i
0.6
0.8
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
A21 x1 ... x1 x2 i i i + A12 x2i
2
555
0=
i
y1 Pi d x1 ln i ... dA21 i x1i Psat1 y2 Pi i + x ln 2i H2 x2 i exp ( A ) 21 y1 Pi d x1 ln i ... dH2 i x1i Psat1 y2 Pi i + x ln 2i H2 x2i exp ( A ) 21
A21 x1 ... x1 x2 i i i + A12 x2i A21 x1 ... x1 x2 i i i + A12 x2i
2
2
0=
i
A12 A21 := Find ( A12 , A21 , H2) H 2
2 2
A12 0.845 A21 = 1.229 H 2 4.703
Ans.
1 ( x1 , x2) := exp x2 A12 + 2 ( A21 A12) x1 2 ( x1 , x2) := exp x1 A21 + 2 ( A12 A21) x2 H2 Pcalc := x1 1 x1 , x2 Psat1 + x2 2 x1 , x2 i i i i i i i exp ( A 21) y1calc :=
i
(
)
(
)
x1 1 x1 , x2 Psat1
i i i
(
)
Pcalc
i
556
1
PiPcalc
0
i
(y1iy1calci) 100
1 2
0
0.2
0.4 x1
i
0.6
0.8
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 1 := 0.8050 0.8639 0.9048 0.9590
i
1.304 1.188 1.114 1.071 1.044 2 := 1.023 1.010 1.003 0.997
i i
1.009 1.026 1.050 1.078 1.105 1.135 1.163 1.189 1.268
i
i := 1 .. n
x2 := 1 x1
y2 := 1 y1
Psat2 := 12.30kPa
T := ( 50 + 273.15)K
Data reduction with the Margules equation and Eq. (10.5): 1 :=
i
y1 Pi
i
x1 Psat1
i
2 :=
i
y2 Pi
i
x2 Psat2
i
557
i := 1 .. n
GERTi := x1 ln 1 + x2 ln 2 i i i i A12 := 0.1 A21 := 0.3
( )
(
( )
)
Guess: f ( A12 , A21) :=
i=1
n
2 GERTi A21 x1i + A12 x2i x1i x2i
A12 := Minimize ( f , A12 , A21) A21
A12 = 0.374
A21 = 0.197
Ans.
RMS Error: RMS := RMS = 1.033 10
3
i=1
n
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
i
0.6
0.8
x1 , x1
Data reduction with the Margules equation and Eq. (14.1): cm B11 := 1840 mol
3
cm B22 := 1800 mol
3
cm B12 := 1150 mol
3
12 := 2 B12 B11 B22
B11 ( Pi Psat1) + Pi y2 2 12 i 1 := exp i R T
558
( )
1 :=
i
y1 1 Pi
i i
x1 Psat1
i
B22 ( Pi Psat2) + Pi y1 2 12 i 2 := exp i R T
i := 1 .. n GERTi := x1 ln 1 + x2 ln 2 i i i i A12 := 0.1 A21 := 0.3
( )
2 :=
i
y2 2 Pi
i i
x2 Psat2
i
( )
(
( )
)
Guess: f ( A12 , A21) :=
i=1
n
2 GERTi A21 x1i + A12 x2i x1i x2i
A12 := Minimize ( f , A12 , A21) A21
n
A12 = 0.379
A21 = 0.216
Ans.
RMS Error:
RMS :=
4
i=1
2 GERTi A21 x1i + A12 x2i x1i x2i
(
)
n
RMS = 9.187 10
x1 := 0 , 0.01 .. 1
0.1
GERT i
A21 x1+A12 ( 1x1) x1 ( 1x1)
0.05
0
0
0.5 x1 , x1
i
1
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: T := Tn
Tc := 308.3K Tr := T Tc
Pc := 61.39bar Tr = 0.614
Tn := 189.4K
For Redlich/Kwong EOS: := 1 := 0
1
:= 0.08664
:= 0.42748 ( Tr ) R Tc
2 2
Table 3.1
( Tr) := Tr q ( Tr) :=
2
Table 3.1
a ( Tr) := ( Tr , Pr) := Guess:
Eq. (3.45)
Pc Pr Tr zv := 0.9 zv ( Tr , Pr) Eq. (3.53)
( Tr ) Tr
Eq. (3.54)
Define Z for the vapor (Zv) Given Eq. (3.52)
zv = 1 + ( Tr , Pr) q ( Tr) ( Tr , Pr) Zv ( Tr , Pr) := Find ( zv) Define Z for the liquid (Zl) Given Eq. (3.56)
( zv + ( Tr , Pr) ) ( zv + ( Tr , Pr) )
zl := 0.01
Guess:
zl = ( Tr , Pr) + zl + ( Tr , Pr) zl + ( Tr , Pr)
(
)(
)
1 + ( Tr , Pr) zl q ( T r ) ( 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) := 1 1 ln
Zl ( Tr , Pr) + ( Tr , Pr) Zl ( Tr , Pr) + ( Tr , Pr) Zv ( Tr , Pr) + ( Tr , Pr) Zv ( Tr , Pr) + ( Tr , Pr)
560
Iv ( Tr , Pr) :=
ln
Eq. (6.65b)
lnl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln Zl ( Tr , Pr) ( Tr , Pr) q ( Tr) Il ( Tr , Pr) Eq. (11.37) lnv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln Zv ( Tr , Pr) ( Tr , Pr) q ( Tr) Iv ( Tr , Pr) Guess Psat: Psatr := Given 1bar Pc
(
)
(
)
lnl ( Tr , Psatr) = lnv ( Tr , Psatr) Zl ( Tr , Psatr) = 4.742 10 Psat = 1.6 bar Ans.
3
Psatr := Find ( Psatr) Zv ( Tr , Psatr) = 0.965
Psatr = 0.026 Psat := Psatr Pc
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: := 0.187 T := Tn
Tc := 308.3K
Pc := 61.39bar
Tn := 189.4K Tr := T Tc
Note: For solution at 0.85T c, set T := 0.85Tc.
For SRK EOS: := 1 := 0 := 0.08664
561
Tr = 0.614 := 0.42748 Table 3.1
1 2 2 ( Tr , ) := 1 + ( 0.480 + 1.574 0.176 ) 1 Tr
2
Table 3.1
Tr , R Tc a ( Tr) := Pc q ( Tr) := Tr , Tr
(
)
2
2
Eq. (3.45) ( Tr , Pr) := Pr Tr
(
)
Eq. (3.54)
Eq. (3.53)
Define Z for the vapor (Zv) Given Eq. (3.52)
Guess:
zv := 0.9 zv ( Tr , Pr)
zv = 1 + ( Tr , Pr) q ( Tr) ( Tr , Pr) Zv ( Tr , Pr) := Find ( zv) Define Z for the liquid (Zl) Given Eq. (3.56)
( zv + ( Tr , Pr) ) ( zv + ( Tr , Pr) )
Guess:
zl := 0.01
zl = ( Tr , Pr) + zl + ( Tr , Pr) zl + ( Tr , Pr)
(
)(
)
1 + ( Tr , Pr) zl q ( T r ) ( 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) := 1 1 ln
Zl ( Tr , Pr) + ( Tr , Pr) Zl ( Tr , Pr) + ( Tr , Pr) Zv ( Tr , Pr) + ( Tr , Pr) Zv ( Tr , Pr) + ( Tr , Pr)
Eq. (6.65b)
Iv ( Tr , Pr) :=
ln
562
lnl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln Zl ( Tr , Pr) ( Tr , Pr) q ( Tr) Il ( Tr , Pr) Eq. (11.37) lnv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln Zv ( Tr , Pr) ( Tr , Pr) q ( Tr) Iv ( Tr , Pr) Guess Psat: Psatr := Given 2bar Pc
(
)
(
)
lnl ( Tr , Psatr) = lnv ( Tr , Psatr) Zl ( Tr , Psatr) = 3.108 10 Psat = 1.073 bar Ans.
3
Psatr := Find ( Psatr) Zv ( Tr , Psatr) = 0.975
Psatr = 0.017 Psat := Psatr Pc
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%.
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%
Acetylene Argon Benzene n-Butane Carbon Monoxide n-Decane Ethylene n-Heptane Methane Nitrogen
14.10
(b) Acetylene: := 0.187 T := Tn
Tc := 308.3K
Pc := 61.39bar
Tn := 189.4K Tr := T Tc
Note: For solution at 0.85T c, set T := 0.85Tc.
For PR EOS: := 1 + 2 := 1 2 := 0.07779
563
Tr = 0.614 := 0.45724 Table 3.1
1 2 2 Table 3.1 ( Tr , ) := 1 + ( 0.37464 + 1.54226 0.26992 ) 1 Tr
2
a ( Tr) := q ( Tr) :=
Tr , R Tc Pc
(
)
2
2
Eq. (3.45) ( Tr , Pr) := Pr Tr
Tr , Tr
(
)
Eq. (3.54)
Eq. (3.53)
Define Z for the vapor (Zv) Given Eq. (3.52)
Guess:
zv := 0.9 zv ( Tr , Pr)
zv = 1 + ( Tr , Pr) q ( Tr) ( Tr , Pr) Zv ( Tr , Pr) := Find ( zv) Define Z for the liquid (Zl) Given Eq. (3.56)
( zv + ( Tr , Pr) ) ( zv + ( Tr , Pr) )
Guess:
zl := 0.01
zl = ( Tr , Pr) + zl + ( Tr , Pr) zl + ( Tr , Pr)
(
)(
)
1 + ( Tr , Pr) zl q ( T r ) ( 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) := 1 1 ln
Zl ( Tr , Pr) + ( Tr , Pr) Zl ( Tr , Pr) + ( Tr , Pr) Zv ( Tr , Pr) + ( Tr , Pr) Zv ( Tr , Pr) + ( Tr , Pr)
Eq. (6.65b)
Iv ( Tr , Pr) :=
ln
564
lnl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln Zl ( Tr , Pr) ( Tr , Pr) q ( Tr) Il ( Tr , Pr) Eq. (11.37) lnv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln Zv ( Tr , Pr) ( Tr , Pr) q ( Tr) Iv ( Tr , Pr) Guess Psat: Given Psatr = 0.018 Psat := Psatr Pc Psatr := 2bar Pc
(
)
(
)
lnl ( Tr , Psatr) = lnv ( Tr , Psatr) Zl ( Tr , Psatr) = 2.795 10 Psat = 1.09 bar Ans.
3
Psatr := Find ( Psatr) Zv ( Tr , Psatr) = 0.974
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%.
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%
Acetylene Argon Benzene n-Butane Carbon Monoxide n-Decane Ethylene n-Heptane Methane Nitrogen
14.12
(a)
van der Waals Eqn. := 0 ( Tr) Tr :=
Tr := 0.7 1 8 := Pr Tr 27 64 ( Tr) := 1
:= 0 q ( Tr) :=
( Tr , Pr) :=
zv := 0.9 (guess) zv ( Tr , Pr) ( zv)
2
Given
zv = 1 + ( Tr , Pr) q ( Tr) ( Tr , Pr)
Eq. (3.52)
565
Zv ( Tr , Pr) := Find ( zv) zl := .01 Given (guess) zl = ( Tr , Pr) + ( zl)
2 1 + ( Tr , Pr) zl
q ( Tr) ( Tr , Pr)
Eq. (3.56)
zl < 0.2
Zl ( Tr , Pr) := Find ( zl) Iv ( Tr , Pr) := ( Tr , Pr) Zv ( Tr , Pr) Il ( Tr , Pr) := ( Tr , Pr) Zl ( Tr , Pr) Case II, pg. 218.
By Eq. (11.39): lnv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln ( Zv ( Tr , Pr) ( Tr , Pr) ) q ( Tr) Iv ( Tr , Pr) lnl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln ( Zl ( Tr , Pr) ( Tr , Pr) ) q ( Tr) Il ( Tr , Pr) Psatr := .1 Given lnl ( Tr , Psatr) lnv ( Tr , Psatr) = 0 Zl ( Tr , Psatr) = 0.05 Psatr := Find ( Psatr) Psatr = 0.2
Zv ( Tr , Psatr) = 0.839 lnl ( Tr , Psatr) = 0.148 := 1 log ( Psatr) (b)
lnv ( Tr , Psatr) = 0.148 ( Tr , Psatr) = 0.036 = 0.302 Ans.
Redlich/Kwong Eqn.Tr := 0.7 := 0
.5
:= 1 ( Tr) := Tr q ( Tr) :=
:= 0.08664
:= 0.42748
( Tr) Tr
( Tr , Pr) :=
Pr Tr
Guess:
zv := 0.9
Given zv = 1 + ( Tr , Pr) q ( Tr) ( Tr , Pr) Zv ( Tr , Pr) := Find ( zv) Guess: zl := .01
zv ( zv + ( Tr , Pr) )
zv ( Tr , Pr)
Eq. (3.52)
566
Given zl < 0.2
zl = ( Tr , Pr) + zl ( zl + ( Tr , Pr) ) Zl ( Tr , Pr) := Find ( zl)
1 + ( Tr , Pr) zl q ( Tr) ( Tr , Pr)
Eq. (3.55)
Iv ( Tr , Pr) := ln
Zv ( Tr , Pr) + ( Tr , Pr) Il ( Tr , Pr) := ln Zl ( Tr , Pr) + ( Tr , Pr) Zv ( Tr , Pr) Zl ( Tr , Pr)
By Eq. (11.39): lnv ( Tr , Pr) := Zv ( Tr , Pr) 1 ln ( Zv ( Tr , Pr) ( Tr , Pr) ) q ( Tr) Iv ( Tr , Pr) lnl ( Tr , Pr) := Zl ( Tr , Pr) 1 ln ( Zl ( Tr , Pr) ( Tr , Pr) ) q ( Tr) Il ( Tr , Pr) Psatr := .1 Given lnl ( Tr , Psatr) = lnv ( Tr , Psatr) Zl ( Tr , Psatr) = 0.015 Psatr := Find ( Psatr) Psatr = 0.087 ( Tr , Psatr) = 0.011
Zv ( Tr , Psatr) = 0.913
lnv ( Tr , Psatr) = 0.083 lnl ( Tr , Psatr) = 0.083 := 1 log ( Psatr) = 0.058 Ans.
14.15 (a) x1 := 0.1 Guess:
x2 := 1 x1 A12 := 2
2
x1 := 0.9 A21 := 2
x2 := 1 x1
1 ( A21 , A12) := exp x2 A12 + 2 ( A21 A12) x1 1( A21 , A12) := exp x2 A12 + 2 ( A21 A12) x1
2
2 ( A21 , A12) := exp x1 A21 + 2 ( A12 A21) x2
2
2( A21 , A12) := exp x1 A21 + 2 ( A12 A21) x2
2
567
Given
x1 1 ( A21 , A12) = x1 1( A21 , A12) x2 2 ( A21 , A12) = x2 2( A21 , A12)
A12 := Find ( A12 , A21) A21
(b) x1 := 0.2 Guess:
A21 = 2.747
A12 = 2.747
Ans.
x2 := 1 x1 A12 := 2
2
x1 := 0.9 A21 := 2
x2 := 1 x1
1 ( A21 , A12) := exp x2 A12 + 2 ( A21 A12) x1 1( A21 , A12) := exp x2 A12 + 2 ( A21 A12) x1
2
2 ( A21 , A12) := exp x1 A21 + 2 ( A12 A21) x2
2
2( A21 , A12) := exp x1 A21 + 2 ( A12 A21) x2
2
Given
x1 1 ( A21 , A12) = x1 1( A21 , A12) x2 2 ( A21 , A12) = x2 2( A21 , A12)
A12 := Find ( A12 , A21) A21
(c) x1 := 0.1 Guess:
A12 = 2.148
A21 = 2.781
Ans.
x2 := 1 x1 A12 := 2
2
x1 := 0.8 A21 := 2
x2 := 1 x1
1 ( A21 , A12) := exp x2 A12 + 2 ( A21 A12) x1 1( A21 , A12) := exp x2 A12 + 2 ( A21 A12) x1
2
2 ( A21 , A12) := exp x1 A21 + 2 ( A12 A21) x2
2
2( A21 , A12) := exp x1 A21 + 2 ( A12 A21) x2
2
568
Given
x1 1 ( A21 , A12) = x1 1( A21 , A12) x2 2 ( A21 , A12) = x2 2( A21 , A12)
A12 := Find ( A12 , A21) A21
14.16 (a) x1 := 0.1 Guess: Given
A12 = 2.781
A21 = 2.148
Ans.
x2 := 1 x1 a12 := 2
x1 := 0.9 a21 := 2
x2 := 1 x1
2 2 a12 x1 a12 x1 exp a12 1 + x1 = exp a12 1 + x1 a21 x2 a21 x2 2 2 a21 x2 a21 x2 exp a21 1 + x2 = exp a21 1 + x2 a12 x1 a12 x1
a12 := Find ( a12 , a21) a21
(b) x1 := 0.2 Guess: Given
a12 = 2.747
a21 = 2.747
Ans.
x2 := 1 x1 a12 := 2
x1 := 0.9 a21 := 2
x2 := 1 x1
2 2 a12 x1 a12 x1 exp a12 1 + x1 = exp a12 1 + x1 a21 x2 a21 x2 2 2 a21 x2 a21 x2 exp a21 1 + x2 = exp a21 1 + x2 a12 x1 a12 x1
a12 := Find ( a12 , a21) a21
a12 = 2.199
a21 = 2.81
Ans.
569
(c)
x1 := 0.1 Guess:
x2 := 1 x1 a12 := 2
x1 := 0.8 a21 := 2
x2 := 1 x1
Given
2 2 a12 x1 a12 x1 exp a12 1 + x1 = exp a12 1 + x1 a21 x2 a21 x2 2 2 a21 x2 a21 x2 exp a21 1 + x2 = exp a21 1 + x2 a12 x1 a12 x1
a12 := Find ( a12 , a21) a21
a12 = 2.81
a21 = 2.199
Ans.
14.18
(a) a := 975 T := 250 .. 450
2.1
b := 18.4 A ( T) :=
c := 3
a + b c ln ( T) T
A ( T)
2
1.9 250
300
350 T
400
450
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 A ( T) ( 1 2 x) = ln x 0.5 1 x x Eq. (E), Ex. 14.5 x2 ( T) := 1 x1 ( T)
x1 ( T) := Find ( x)
570
UCST := 300 (guess) Given A ( UCST) = 2 UCST := Find ( UCST) UCST = 272.93
LCST := 400 (guess) Given A ( LCST) = 2 LCST := Find ( LCST) LCST = 391.21
Plot phase diagram as a function of T T1 := 225 , 225.1 .. UCST T2 := LCST .. 450
500
T1 T1 T2 T2
400
300
200 0.2
0.3
0.4
0.5
0.6
0.7
0.8
x1 ( T1 ) , x2 ( T1 ) , x1 ( T2 ) , x2 ( T2 )
(b) a := 540 T := 250 .. 450
2.5
b := 17.1 A ( T) :=
c := 3
a + b c ln ( T) T
A ( T)
2
1.5 250
300
350 T
400
450
Parameter A = 2 at a single temperature. It is a LCST, because A decreases to 2 as T decreases.
571
Guess: Given x0
x := 0.25 A ( T) ( 1 2 x) = ln x 0.5 1 x x Eq. (E), Ex. 14.5
x1 ( T) := Find ( x)
LCST := 350 (guess) Given A ( LCST) = 2 LCST := Find ( LCST) LCST = 346
Plot phase diagram as a function of T
450
T := LCST .. 450
T T
400
350
300 0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
x1 ( T ) , 1x1 ( T)
(c)
a := 1500 T := 250 .. 450
b := 19.9 A ( T) :=
c := 3
a + b c ln ( T) T
3 2.5 A ( T) 2 1.5 250
300
350 T
400
450
Parameter A = 2 at a single temperature. It is an UCST, because A decreases to 2 as T increases.
572
Guess:
x := 0.25 A ( T) ( 1 2 x) = ln x 0.5 1 x x Eq. (E), Ex. 14.5
Given x0
x1 ( T) := Find ( x)
UCST := 350 (guess) Given A ( UCST) = 2 UCST := Find ( UCST) UCST = 339.66
Plot phase diagram as a function of T
350
T := UCST .. 250
T T 300
250
0
0.2
0.4
0.6
0.8
x1 ( T ) , 1x1 ( T)
14.20 Guess: Given
x1 := 0.5
x1 := 0.5
Write Eq. (14.74) for species 1:
2 2 x1 exp 0.4 ( 1 x1) = x1 exp 0.8 ( 1 x1)
x1 1 x1
+
x1 1 x1
= 1
(Material balance)
x1 := Find ( x1 , x1) x1
x1 = 0.371
x1 = 0.291
Ans.
573
14.22 Temperatures in kelvins; pressures in kPa. P1sat ( T) := exp 19.1478
5363.7 T 2048.97 T
water P := 1600 SF6
P2sat ( T) := exp 14.6511
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
6
400
500
6
600
700
y1II ( TII) 10 , y1II ( TII) 10
574
288
286 TI Tstar 284
282
280 650
700
750
800
850
6
900
950
6
1000
1050
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 Toluene P := 101.33 Water
3885.70 P2sat ( T) := exp 16.3872 T + 230.170 Find the three-phase equilibrium T and y: Guess: Given y1star := T := 25 P = P1sat ( T) + P2sat ( T) P1sat ( Tstar) P
Tstar := Find ( T) y1star = 0.444
Tstar = 84.3
For z1 < y1*, first liquid is pure species 2. y1 := 0.2 Given Guess: Tdew := Tstar Tdew := Find ( Tdew) Tdew = 93.855 Ans.
y1 = 1
P2sat ( Tdew) P
For z1 > y1*, first liquid is pure species 1.
575
y1 := 0.7 Given
Guess: y1 =
Tdew := Tstar Tdew := Find ( Tdew) Tdew = 98.494 Ans.
P1sat ( Tdew) P
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 2910.26 T + 216.432 3885.70 T + 230.170 n-heptane P := 101.33 water
P2sat ( T) := exp 16.3872
Find the three-phase equilibrium T and y: Guess: Given y1star := T := 50 P = P1sat ( T) + P2sat ( T) P1sat ( Tstar) P Tstar := Find ( T) y1star = 0.548 Tstar = 79.15
Since 0.35<y1*, first liquid is pure species 2. y1 ( T) := 1 P2sat ( T) P Tdew := Tstar Tdew := Find ( Tdew) Tdew = 88.34
Find temperature of initial condensation at y1=0.35: y10 := 0.35 Given Guess:
y1 ( Tdew) = y10
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 90 T 85 80 75 Tstar Tdew
0.3
0.35
0.4
0.45
0.5
0.55
y1path ( T )
14.26 Pressures in kPa.
P1sat := 75
P2sat := 110
A := 2.25
2 1 ( x1) := exp A ( 1 x1)
2 ( x1) := exp A x1
(
2
)
Find the solubility limits: Guess: Given x1 = 0.224 x1 := 0.1 A ( 1 2 x1) = ln
1 x1 x1
x1 := Find ( x1) x1 = 0.776
x1 := 1 x1
Find the conditions for VLLE:
Guess:
Pstar := P1sat
y1star := 0.5
Given
Pstar = x1 1 ( x1) P1sat + ( 1 x1) 2 ( x1) P2sat y1star Pstar = x1 1 ( x1) P1sat
Pstar := Find ( Pstar , y1star) y1star
Pstar = 160.699
y1star = 0.405
Calculate VLE in two-phase region. Modified Raoult's law; vapor an ideal gas. Guess: x1 := 0.1 P := 50
577
Given
P = x1 1 ( x1) P1sat + ( 1 x1) 2 ( x1) P2sat y1 ( x1) := x1 1 ( x1) P1sat P ( x1)
P ( x1) := Find ( P) 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 PL ( x1) 150 PV ( x1) Pliq Pliq 125 100 75 50 0 0.2 0.4 0.6 0.8 1 Pstar
x1 , y1 ( x1) , x1 , x1
x1 := 0 , 0.05 .. 0.2 x1 =
0 0.05 0.1 0.15 0.2
PL ( x1) =
110 133.66 147.658 155.523 159.598
y1 ( x1) =
0 0.214 0.314 0.368 0.397
578
x1 := 1 , 0.95 .. 0.8 x1 =
1 0.95 0.9 0.85 0.8
PL ( x1) =
75 113.556 137.096 150.907 158.506
y1 ( x1) =
1 0.631 0.504 0.444 0.414
x1 = 0.224
x1 = 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 2451.88 T + 232.014 2910.26 T + 216.432 z3 := 1 z1 z2
n-Pentane:
P2sat ( T) := exp 13.7667 P3sat ( T) := exp 13.8622 z1 := 0.45 z2 := 0.30
n-Heptane: P := 101.33 (a)
Calculate dew point T and liquid composition assuming the hydrocarbon layer forms first: Tdew1 := 100 x2 := z2 x3 := 1 x2
Guess: Given
P = x2 P2sat ( Tdew1) + x3 P3sat ( Tdew1) z3 P = x3 P3sat ( Tdew1) x2 + x3 = 1
x2 x3 := Find ( x2 , x3 , Tdew1) Tdew1
Tdew1 = 66.602 x3 = 0.706
579
x2 = 0.294
Calculate dew point temperature assuming the water layer forms first: x1 := 1 Given Guess: Tdew2 := 100 Tdew2 := Find ( Tdew2) Tdew2 = 79.021 Since Tdew2 > Tdew1, the water layer forms first (b) Calculate the temperature at which the second layer forms: Guess: Tdew3 := 100 y1 := z1 Given x2 := z2 y2 := z2 x3 := 1 x2 y3 := z3
x1 P1sat ( Tdew2) = z1 P
P = P1sat ( Tdew3) + x2 P2sat ( Tdew3) + x3 P3sat ( Tdew3) y1 P = P1sat ( Tdew3) y2 z2 = y3 z3 y1 + y2 + y3 = 1 x2 + x3 = 1
y2 P = x2 P2sat ( Tdew3)
y1 y2 y3 := Find ( y1 , y2 , y3 , Tdew3 , x2 , x3) Tdew3 x2 x3
y1 = 0.288 Tdew3 = 68.437 (c) y2 = 0.388 x2 = 0.1446 y3 = 0.324 x3 = 0.8554
Calculate the bubble point given the total molar composition of the two phases x2 := z2 z2 + z3 x3 := z3 z2 + z3
Tbubble := Tdew3
x2 = 0.545
x3 = 0.455
580
Given P = P1sat ( Tbubble) + x2 P2sat ( Tbubble) + x3 P3sat ( Tbubble) Tbubble := Find ( Tbubble) P1sat ( Tbubble) P x2 P2sat ( Tbubble) y2 := P x3 P3sat ( Tbubble) y3 := P y1 := Tbubble = 48.113 y1 = 0.111 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 2451.88 T + 232.014 2910.26 T + 216.432 z3 := 1 z1 z2
n-Pentane:
P2sat ( T) := exp 13.7667 P3sat ( T) := exp 13.8622 z1 := 0.32 z2 := 0.45
n-Heptane: P := 101.33 (a) Guess: Given
Calculate dew point T and liquid composition assuming the hydrocarbon layer forms first: Tdew1 := 70 x2 := z2 x3 := 1 x2
P = x2 P2sat ( Tdew1) + x3 P3sat ( Tdew1) z3 P = x3 P3sat ( Tdew1) x2 + x3 = 1
x2 x3 := Find ( x2 , x3 , Tdew1) Tdew1
Tdew1 = 65.122 x3 = 0.686
581
x2 = 0.314
Calculate dew point temperature assuming the water layer forms first: x1 := 1 Given Guess: Tdew2 := 70 Tdew2 := Find ( Tdew2) Tdew2 = 70.854 Since Tdew1>Tdew2, a hydrocarbon layer forms first (b) Calculate the temperature at which the second layer forms: Guess: Tdew3 := 100 y1 := z1 Given x2 := z2 y2 := z2 x3 := 1 x2 y3 := z3
x1 P1sat ( Tdew2) = z1 P
P = P1sat ( Tdew3) + x2 P2sat ( Tdew3) + x3 P3sat ( Tdew3) y1 P = P1sat ( Tdew3) y2 z2 = y3 z3 y1 + y2 + y3 = 1 x2 + x3 = 1
y2 P = x2 P2sat ( Tdew3)
y1 y2 y3 := Find ( y1 , y2 , y3 , Tdew3 , x2 , x3) Tdew3 x2 x3
y1 = 0.24 Tdew3 = 64.298 (c) y2 = 0.503 x2 = 0.2099 y3 = 0.257 x3 = 0.7901
Calculate the bubble point given the total molar composition of the two phases x2 := z2 z2 + z3 x3 := z3 z2 + z3
Tbubble := Tdew3
x2 = 0.662
582
x3 = 0.338
Given
P = P1sat ( Tbubble) + x2 P2sat ( Tbubble) + x3 P3sat ( Tbubble) Tbubble = 43.939 y1 = 0.09 y2 = 0.861 y3 = 0.049
Tbubble := Find ( Tbubble) P1sat ( Tbubble) P x2 P2sat ( Tbubble) y2 := P x3 P3sat ( Tbubble) y3 := P y1 := 14.32 :=
0.302 0.224
Tc :=
748.4 K 304.2
Pc :=
40.51 bar 73.83
P := 10bar , 20bar .. 300bar T := 353.15K Use SRK EOS From Table 3.1, p. 98 of text: := 1 := 0 := 0.08664 := 0.42748 T Tr := Tc
:= 1 + 0.480 + 1.574 0.176 1 Tr
2
(
)(
0.5
) 2
Eq. (14.32)
2 2 R Tc Eq. (14.31) a := Pc
R Tc b := Pc
6.842 kg m5 a= 0.325 s2 mol2
2 ( P) := z2 := 1 b2 P R T Eq. (14.33)
1.331 10 4 m3 b= 2.968 10 5 mol
q2 := a2 b2 R T Eq. (14.34)
(guess)
583
Given z2 = 1 + 2 ( P) q2 2 ( P) Z2 ( P) := Find ( z2) I2 ( P) := ln
(
z2 2 ( P) z2 + 2 ( P) z2 + 2 ( P)
)(
)
Eq. (14.36)
Z 2 ( P) + 2 ( P)
Z 2 ( P)
Eq. (6.65b)
For simplicity, let 1 represent the infinite-dilution value of the fugacity coefficient of species 1 in solution. Eq. (14.103): 1 ( P) := exp l12 := 0.088
b1 ( Z2 ( P) 1) ln ( Z2 ( P) 2 ( P) ) ... b2 0.5 b1 a1 + q2 2 ( 1 l12) I2 ( P) b2 a2
V1 := 124.5 cm
3
Psat1 := 0.0102bar
mol
Eqs. (14.98) and (14.99), with sat1 = 1 and (P - Psat1) = P, combine to give: y1 ( P) := Psat1 P 1 ( P) exp
P V 1 R T
0.1
0.01 y1 ( P) 1 .10 1 .10
3
4
0
50
100
150 P bar
200
250
300
584
14.33 :=
0.302 0.038
Tc :=
748.4 K 126.2
Pc :=
40.51 bar 34.00
P := 10bar , 20bar .. 300bar T := 308.15K (K) Use SRK EOS From Table 3.1, p. 98 of text: := 1 := 0 := 0.08664 := 0.42748 T Tr := Tc
2 0.5 := 1 + 0.480 + 1.574 0.176 1 Tr
(
)(
)
2
2 2 R Tc a := Pc
Eq. (14.31)
R Tc b := Pc
Eq. (14.32)
7.298 kg m5 a= 0.067 s2 mol2
2 ( P) := z2 := 1 Given z2 = 1 + 2 ( P) q2 2 ( P) Z2 ( P) := Find ( z2) I2 ( P) := ln b2 P R T Eq. (14.33)
1.331 10 4 m3 b= 2.674 10 5 mol
q2 := a2 b2 R T Eq. (14.34)
(guess)
(
z2 2 ( P) z2 + 2 ( P) z2 + 2 ( P)
)(
)
Eq. (14.36)
Z 2 ( P) + 2 ( P)
Z 2 ( P)
Eq. (6.65b)
For simplicity, let 1 represent the infinite-dilution value of the fugacity coefficient of species 1 in solution.
585
l12 := 0.0
Eq. (14.103):
1 ( P) := exp
b1 ( Z2 ( P) 1) ln ( Z2 ( P) 2 ( P) ) ... b2 0.5 b1 a1 + q2 2 ( 1 l12) I2 ( P) b2 a2
4
Psat1 := 2.9 10
bar
V1 := 125
cm
3
mol
Eqs. (14.98) and (14.99), with sat1 = 1 and (P - Psat1) = P, combine to give: y1 ( P) := Psat1 P 1 ( P) exp
P V 1 R T
10
y1 ( P) 10
5
1
0
50
100
150 P bar
200
250
300
Note: y axis is log scale.
586
14.45 A labeled diagram of the process is given below. The feed stream is taken as the phase and the solvent stream is taken as the phase.
F nF xF1 = 0.99 xF2 = 0.01 S nS xS3 = 1.0
Feed Mixer/ Settler Solvent
R nR x1 x2 = 0.001 E nE x2 x3
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 x2 := 0.001 xF2 := 0.01 x1 := 1 x2 xS3 := 1
Apply mole balances around the process as well as an equilibrium relationship From p. 585 A12 := 1.5 A23 := 0.8
2 2 ( x2) := exp A12 ( 1 x2)
2 2 ( x2) := exp A23 ( 1 x2)
Material Balances nS + nF = nE + nR nS = x3 nE xF1 nF = x1 nR (Total) (Species 3) (Species 1)
Substituting the species balances into the total balance yields xF1 1 nS + nF = nS + nF x3 x1 Solving for the ratio of solvent to feed (nS/nF) gives nS nF =
x1 xF1 x3 1 x3 x1
587
We need x3. Assume exiting streams are at equilibrium. Here, the only distributing species is 2. Then x2 2 = x2 2 Substituting for 2 and 2
2 2 x2 exp A12 1 x2 = x2 exp A23 1 x2
(
)
(
)
Solve for x2 using Mathcad Solve Block Guess: Given
2 2 x2 exp A12 1 x2 = x2 exp A23 1 x2
x2 := 0.5
(
)
(
)
x2 := Find x2
( )
x2 = 0.00979
x3 := 1 x2
x3 = 0.9902
From above, the equation for the ratio nS/nF is: nSnF :=
x1 xF1 x3 1 x3 x1
Ans. Ans.
a) nSnF = 0.9112 b) x2 = 0.00979
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 1 and 2. x1 := 520 10
6
x2 := 1 x1
x2 :=
2 10
6
x1 := 1 x2
588
1 := 2 :=
x1 x1 1 x1 1 x1
1 = 1.923 10 2 = 4.997 10
3
Ans.
5
Ans.
14.50 1 - butanenitrile Psat1 := 0.07287bar 2- benzene Psat2 := 0.29871bar cm
3
V1 := 90 V2 := 92
3
cm
3
mol cm
3
mol cm
3
B1 , 1 := 7993 T := 318.15K i := 1 .. 2
mol
B2 , 2 := 1247
cm
mol
B1 , 2 := 2089 x1 := 0.4819 x2 := 1 x1
mol
B2 , 1 := B1 , 2 y1 := 0.1813 y2 := 1 y1
P := 0.20941bar k := 1 .. 2
j := 1 .. 2 yi P xi Psati
Term A is calculated using the given data. term_Ai :=
Term B is calculated using Eqns. (14.4) and (14.5) j , i := 2 B j , i B j , j Bi , i hati := exp P 1 Bi , i + 2 R T
sati := exp
y jyk( 2 j , i j , k)
j
k
Bi , i Psati R T
term_Bi :=
hati sati
Term C is calculated using Eqn. (11.44) fsati := sati Psati term_A =
Vi ( P Psati) fi := sati Psati exp R T
term_B =
589
term_Ci :=
fsati fi
1.081 1.108
0.986 1.006
term_C =
1 1
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 = 1 and 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) := xr := 0.7 1 1 A ( T) 2 + 2 2 A ( T) xl := 0.3 xspl1 ( T) := 1 1 A ( T) 2 2 2 A ( T) x1 = 1 2 1 2 2 A 2 A
From Eq. (E) in Example 14.5, the solubility curves are solved using a Solve Block:
590
Given
A ( T) ( 1 2xr) = ln A ( T) ( 1 2xl) = ln
1 xr xr
xr > 0.5
xr1 ( T) := Find ( xr)
Given
1 xl xl
xl < 0.5
xl1 ( T) := Find ( xl)
Find the temperature of the upper consolute point. T := 300K Given A ( T) = 2 Tu := Find ( T) Tu = 345.998 K
T := 250K .. 346K
360 340 320 300 280 260 240
0.1
0.2
xl1 xr1 xspl1 xpr1
0.3
0.4
0.5
0.6
0.7
0.8
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 1 := 0.193 A1 := 14.0572 Psat1 ( T) := exp A1 Tc1 := 556.4K B1 := 2914.23 Pc1 := 45.60bar C1 := 232.148
kPa T 273.15 + C 1 K
B1
591
2 - n-heptane 2 := 0.350 A2 := 13.8622 Psat2 ( T) := exp A2 Tc2 := 540.2K B2 := 2910.26 Pc2 := 27.40bar C2 := 216.432
kPa T 273.15 + C 2 K
B2
T := ( 100 + 273.15)K Tr1 := Tr2 := T Tc1 T Tc2 Tr1 = 0.671 Tr2 = 0.691 Psat1r := Psat2r := Psat1 ( T) Pc1 Psat2 ( T) Pc2 Psat1r = 0.043 Psat2r = 0.039
Using Wilson's equation
12 := 1.5410
21 := 0.5197
1 ( x1) := exp ln x1 + ( 1 x1) 12 ... 12 21 + 1 x ( 1) x + 1 x 1 x + x ( 1) 1 21 1) 12 1 ( 2 ( x1) := exp ln ( 1 x1) + x1 21 ... 12 21 + x ( 1) x + 1 x 1 x + x ( 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 1 ( x1) Psat1 ( T) + ( 1 x1) 2 ( x1) Psat2 ( T) yi1 ( x1) := x1 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 hat and sat. sat1 := PHIB Tr1 , Psat1r , 1
( (
) )
sat1 = 0.946 1 ( P) := hat1 ( P) sat1
P hat1 ( P) := PHIB Tr1 , , 1 P
c1
sat2 := PHIB Tr2 , Psat2r , 2
sat2 = 0.95 2 ( P) := hat2 ( P) sat2
P hat2 ( P) := PHIB Tr2 , , 2 P
c2
Solve Eqn. (14.1) for y1 and P given x1. Guess: Given y1 1 ( P) P = x1 1 ( x1) Psat1 ( T) Eqn. (14.1) y1 := 0.5 P := 1bar
( 1 y1) 2 (P) P = ( 1 x1) 2 ( x1) Psat2 (T)
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 ) bar Pi ( x1 ) bar Pii ( x1 ) bar Pii ( x1 ) bar 1.5 1.4 1.3 1.2 1.1 1 1.6 1.7
0
0.2
P-x Raoult's P-y Raoult's P-x Gamma/Phi P-y Gamma/Phi
x1 , yi1 ( x1 ) , x1 , yii1 ( x1 )
0.4
0.6
0.8
594
Chapter 15 - Section A - Mathcad Solutions
15.1 Initial state: Liquid water at 70 degF.
H1
38.05
BTU lbm
S1
0.0745
BTU lbm rankine
(Table F.3)
Final state: Ice at 32 degF.
H2
T
(a)
( 0.02
( 70
143.3)
BTU lbm
S2
0.0
143.3 491.67
BTU lbm rankine
459.67)rankine
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
H1
T
S2
S1
Wideal
12.466
BTU lbm
mdot
1
lbm sec
Wdotideal
mdot Wideal
Wdotideal
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
491.67 rankine
TH
TH
TC
T
QC
H2
Work
H1
QC
BTU lbm
181.37
BTU lbm
Work
QC
TC
14.018
Wdot
t
mdot Work
Wdot
t
14.79 kW
Ans.
Wdotideal Wdot
0.889
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:
H2 mdot H1 1 lbm sec
mdot 2.484 lbm sec
Ans.
Ans.
HA
mdot HB
Wdotideal
HD
HA
Wdot
t
Wdot
t
16.77 kW
0.784
Wdot
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 BTU lbm
BTU lbm rankine
Hvap
106.48
BTU lbm
BTU lbm rankine
HA
Hvap
Sliq
0.0433
Svap
0.2229
SA
Svap
597
For sat. liquid at 98 degF, P=134.75(psia):
HC 44.24 BTU lbm
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 118 BTU lbm
BTU lbm
BTU lbm rankine
xD Svap Sliq
HB
HA
H'B
HA
HB
121.84
The entropy at this H is read from Fig. G.2 at P=134.75(psia)
HD HC
xD HD Hvap
0.094
SB
0.228
Hliq Hliq
xD
0.284
SD
Sliq
SD
BTU lbm rankine
Refrigerent circulation rate:
lbm sec
mdot 2.914 lbm sec
Ans.
H2 mdot
H1 1
HA
mdot HB
Wdotideal
HD
HA
Wdot
Wdot
47.22 kW
t
Wdot
T
SA
t
0.279
( 70
Ans. 459.67)rankine
THERMODYNAMIC ANALYSIS
Wdotlost.compressor
Qdotcondenser
mdot T
SB
HB
SC
598
mdot HC
mdot T
Wdotlost.condenser
SB
Qdotcondenser
Wdotlost.throttle
mdot T
SD
SC
Wdotlost.evaporator
T
mdot SA SD lbm S 2 S1 1 sec
27.85%
17.59%
30.02%
14.02%
10.52%
Wdotideal
Wdotlost.compressor
Wdotlost.condenser
Wdotlost.throttle
Wdotlost.evaporator
13.152 kW
8.305 kW
14.178 kW
6.621 kW
4.968 kW
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
257190 J
J K
298.15 K
86.513
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
nCO2 1 mol
nN2 1.881 mol
599
(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
H298
nN2 nair
For unmixing the air, define
y1 0.79
y2 1 y1
By Eq. (12.35) with no minus sign: Sunmixing
Sunmixing
nair R y1 ln y1
10.174 J K
y2 ln y2
For mixing the products of reaction, define
y1 nCO2 nN2 nCO2
nCO2
y1
0.347
y2
y2 ln y2
1
y1
15.465 J K
Smixing
nN2 R y1 ln y1
Smixing
S
T
Sunmixing
300 K
S298
Wideal
Smixing
H
600
S
S
Wideal
81.223
J K
Ans.
T
259 kJ
(b) Adiabatic combustion:
Heat-capacity data for the product gases from Table C.1:
A nCO2 5.457 mol
B nCO2 1.045 mol
D nCO2 1.157 mol
T
nN2 3.280
A
3
11.627
3
nN2 0.593
10
B
2.16
10
nN2 0.040
10
5
D
1.082
10
5
For the products,
HP = R
T0
CP R
dT
T0
298.15 K
The integral is given by Eq. (4.7). Moreover, by an energy balance, H298
Guess
Given
HP = 0
2
A 11.627
B 2.160 10 K
2
3
.
D 1.082 10 K
5 2
H298 = R mol A T0
Find
8.796
1
B 2 T0 2
T
601
1
T
D T0
1
T0
2622.603 K
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
H
H
t
3
0.0
1.082 10
5
= 29.701
246.934 J K
29.701
H298
2.83 10 J
5
S
R mol ICPS
H T
S
S
Wideal.cooling
Wideal.cooling
t
208904 J
Ans.
Ans.
Wideal.cooling Wideal
0.8078
The surroundings increase in entropy in the amount: Q H298 Wideal.cooling
S Q T
S 246.93 J K
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
For sat. liquid and vapor at 1000 kPa, Table F.2:
Hliq
Hvap
762.6
kJ kg
kJ kg
Sliq
Svap
2.1382
kJ kg K
kJ kg K
Tsat 453.03K
2776.2
6.5828
(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
H3
Given
H3 mdot3
S3 mdot3
0.1
H1 2
kg s
mdot2
S3
mdot1
Sliq
mdot3
H3 Hliq
mdot1
mdot2
H2
Tsat
H1 mdot1
S1 mdot1
H2 mdot2 = 0
S2 mdot2 = 0
H3
kJ s
kJ sK
Hliq mdot3 = 300 kJ s
mdot3 = mdot1
S3 = Sliq
mdot1 mdot2 mdot3 H3 S3
mdot2
Hliq
H3
Tsat
Find mdot1 mdot2 mdot3 H3 S3
603
mdot1
H3
0.086
kg s
3 kJ
mdot2
S3
0.064
kg s
mdot3
Ans.
0.15
kg s
2.767
10
kg
6.563
kJ kg K
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
x'turb
turb
S1 Svap
0.919
0.78
Sliq Sliq
H'turb
H'turb
Hturb
Hturb
Hliq
2.614
H1
2.655
x'turb Hvap
10
3 kJ
Hliq
kg
H1
turb H'turb
10
3 kJ
kg
xturb
xturb
Hturb Hvap
0.94
Hliq Hliq
Sturb
Sturb
Sliq
6.316
xturb Svap
kJ kg K
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 2993.5 kJ kg
comp
0.75
Hcomp
H2
H'comp
comp
H2
Hcomp
kJ kg K
3084.4
kJ kg
By interpolation:
Scomp
7.1803
604
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
Given
Hcomp
H3 mdot3
0.15
kg s
H3
2770.
kJ kg
H2 mdot2 =
H1 mdot1
mdot2
Hturb
H1 mdot1
kJ s
Hliq mdot3 = 300 kJ s
H2 mdot2 = 0
H3
mdot3 = mdot1
mdot1 mdot2 mdot3 H3
kg s
kg s
Find mdot1 mdot2 mdot3 H3
mdot1
mdot3
0.10608
mdot2
H3
0.04274
kg s
Ans.
0.14882
2.77844
3 kJ 10 kg
Steam at Point 3 is slightly superheated. By interpolation,
S3 6.5876 kJ kg K
300K
(assumed)
THERMODYNAMIC ANALYSIS
T
By Eq. (5.25), with the enthalpy term equal to zero: Wdotideal T mdot3 S3 mdot1 S1
605
mdot2 S2
Wdotideal
Wdotlost.turb
6.014 kW
T mdot1 Sturb S1
S2
Wdotlost.comp
T mdot2 Scomp
Wdotlost.mixing
Wdotlost.turb
Wdotlost.comp
Wdotlost.mixing
T
mdot3 S3
mdot1 Sturb
mdot2 Scomp
2.9034 kW
2.054 kW
1.0561 kW
48.2815%
34.1565%
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:
Hliq 12.090 BTU lbm
BTU lbm rankine
Hvap
103.015
BTU lbm
Svap
0.22525
Sliq
0.02744
BTU lbm rankine
For sat. liquid at the condenser outlet temperature of 80 degF:
H4 37.978 BTU lbm
S2
Hliq Hliq
S4
0.07892
BTU lbm rankine
H2
x1
x1
Hvap
H1 Hvap
0.285
Svap
H1
S1
S1
606
H4
Sliq
0.084
x1 Svap
BTU lbm rankine
Sliq
From Example 9.1(b) for the compression step:
H 17.48 BTU lbm
H3 H2 H
H3 120.5 BTU lbm
From Fig. G.2 at H3 and P = 101.37(psia):
S3 0.231 BTU lbm rankine
lbm hr
10
4 BTU
mdot
1845.1
Wdot
mdot H
Wdot
3.225
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
QdotC
( 70
459.67)rankine
120000 BTU hr
TH TC
TH
TC
T
( 10 459.67)rankine
4 BTU
Wdotideal
Wdotlost.comp
Qdot H4
QdotC
TC
Wdotideal
S2
Qdot
1.533
10
hr
T mdot S3
H3 mdot
T mdot S4
T mdot S1
1.523 10
5 BTU
hr
Wdotlost.cond
Wdotlost.throttle
S3
S4
607
Qdot
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).
Wdotideal
Wdotlost.comp
Wdotlost.cond
Wdotlost.throttle
Wdotlost.evap
Wdotlost.evap
15329.9
BTU hr
47.53%
5619.4
BTU hr
BTU hr
BTU hr
BTU hr
17.42%
3625.2
11.24%
4730.2
14.67%
2947.6
9.14%
The percent values above express each quantity as a percentage of the actual work, to which they sum:
Wdot 32252.3 BTU hr
15.5
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 20 10 0
608
600 500 QdotC 400 300 200 BTU sec
TC
tC
459.67 rankine
Wdotideal
QdotC
TH
TC
TC
For sat. liquid and vapor at the evaporator temperature, Table 9.1:
21.486 18.318 Hliq 15.187 12.090 9.026 0.04715 0.04065 Sliq 0.03408 0.02744 0.02073
BTU Svap lbm rankine BTU lbm
107.320 105.907 Hvap 104.471 103.015 101.542
BTU lbm
H2
Hvap
0.22244 0.22325 0.22418 0.22525 0.22647
BTU lbm rankine
S2
Svap
For sat. liquid at the condenser temperature:
H4 37.978 BTU lbm
S4 0.07892 BTU lbm rankine
H1 H4
x1
H1 Hvap
Hliq Hliq
S1
Sliq
x1 Svap
Sliq
From the results of Pb. 9.9, we find:
117.7 118.9 H3 120.1 121.7 123.4 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:
609
0.227 0.229 S3 0.231 0.234 0.237 BTU lbm rankine
8.653 7.361 mdot 6.016 4.613 3.146 T mdot S3
H4 H3 mdot S3 S4 S1
mdot Qdot lbm sec
Wdotlost.comp
Qdot
S2
Wdotlost.cond Wdotlost.throttle Wdotlost.evap
T mdot S4 T mdot S1 T mdot S2
T H1 H2
TC
The final term accounts for the entropy change of the refrigerated space (an internal heat reservoir).
Wdot
mdot H3 36.024 40.844
H2
20.9 22.419 Wdotlost.comp 21.732 21.379 17.547
BTU sec
Wdotideal
BTU 41.695 sec 38.325
30.457
610
11.149 10.52 Wdotlost.cond 9.444 7.292 5.322 12.991 11.268 Wdotlost.evap 9.406 7.369 5.122
BTU sec BTU sec
8.754 10.589 Wdotlost.throttle 11.744 11.826 10.322
BTU sec
89.818 95.641 Wdot 94.024 86.194 68.765
BTU sec
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
TC
( 70
( 30
459.67)rankine
459.67)rankine
TH TC
TH
QdotC
T
2000 BTU sec
Wdotideal
QdotC
TC
Wdotideal
163.375
BTU sec
For sat. liquid and vapor at the evaporator temperature, Table 9.1:
Hliq 18.318 BTU lbm
BTU lbm
Sliq
0.04065
BTU lbm rankine
BTU lbm rankine
Hvap
H2
105.907
Hvap
Svap
S2
0.22325
Svap
611
For sat. liquid at the condenser temperature:
H4 37.978 BTU lbm
S4 0.07892 BTU lbm rankine
From Problem 9.12,
H2A 116. BTU lbm
BTU lbm
S2A
0.2435
BTU lbm rankine
H3
H2A
14.667
H3
130.67
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
H1 Hvap
0.109
H2A
Hliq Hliq
H2
H1
S1
S1
27.885
BTU lbm
Sliq
Sliq
0.061
x1 Svap
BTU lbm rankine
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
lbm sec
From Problem 9.12:
mdot
25.634
612
Wdotlost.comp
Qdot H4
T mdot S3
H3 mdot
T mdot S4
T mdot S1
S2A
Wdotlost.cond
Wdotlost.throttle
S3
S4A
Qdot
Wdotlost.evap
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
Wdot mdot H3
Wdotideal
Wdotlost.comp
Wdotlost.cond
Wdotlost.throttle
Wdotlost.evap
Wdotlost.exchanger
BTU sec
T mdot S2A
H2A
163.38 BTU sec
S2
S4A
S4
43.45%
54.31
BTU sec
BTU sec
14.45%
87.08
23.16%
9.98
BTU sec
BTU sec
BTU sec
2.65%
45.07
11.99%
16.16
4.30%
Wdot
375.97
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
comp
0.75
419.1 kJ kg
kJ kg
H1
H2
S1
S2
1.3069
kJ kg K
kJ kg K
(sat. liquid)
2676.0
7.3554
(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 2757.3 kJ kg
comp
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 500.4 kJ kg
614
This enthalpy is a bit larger than that of sat. liquid at 110 degC; find quality and then the entropy:
Hliq
Slv
461.3
kJ kg
kJ kg K
Hlv
x4
2230.0
H4
kJ kg
Sliq
x4
1.4185
kJ kg K
5.8203
Hliq Hlv
0.018
kg sec
S4
T
Sliq
x4 Slv
S4
1.5206
kJ kg K
mdot
0.5
300 K
Wdotideal
Wdotlost.evap
mdot H4
mdot T
mdot T
H2
H1
S4
S3
T
S3
S2
S4
S2
S1
S1
Wdotlost.comp
Wdot
mdot H3
Wdotideal
Wdotlost.evap
Wdotlost.comp
Wdot
8.606 kW
24.651 kW
7.41 kW
21.16%
60.62%
18.22%
40.667 kW
The figures on the right are percentages of the actual work, to which the terms sum.
15.8
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 H2 H3 H4 H5 H7
QH H2
257.6 1461.2 1242.2 1047.8 69.7 250.2 BTU lbm
S1 S2 S3 S4 S5 S7
Wideal
0.3970 1.6671 1.7431 1.8748 0.1326 0.4112 BTU lbm rankine
H1 1 lbm
QH 1
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 0.18688 lbm
T
S2
S1 1 lbm
QH TH
(From Pb. 8.8)
Wlost.turbine
T
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
1 lbm S5
1 lbm
m S4
m S7
Q
Wlost.pump.heater
T
1 lbm S1
S5
m S7
S3
The absolute value of the actual work comes from Pb. 8.8: Wabs.value = 374.61 BTU
Wlost.boiler.reservoir
Wlost.turbine
Wlost.cond.valve
Wlost.pump.heater
Wideal 742.82 BTU
50.43%
30.24%
13.30%
4.90%
1.13%
224.66 BTU
98.81 BTU
36.44 BTU
8.36 BTU
(absolute value)
The numbers on the right are percentages of the absolute value of the ideal work, to which they sum.
15.9
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
H5
H7
H9
H10
1140.0
kJ kg
kJ kg
S4
S5
S7
S9
S10
9.359
kJ kg K
kJ kg K
kJ kg K
kJ kg K
kJ kg K
1009.7
8.894
719.8
kJ kg
kJ kg
kJ kg
7.544
285.4
4.928
796.9
9.521
617
H14
1042.1
kJ kg
S14
11.015
kJ kg K
H15
1188.9
kJ kg
S15
11.589
kJ kg K
H6
H5
S6
S5
H11
H5 S11
S5
H12
H10 S12
S10
H13
H10 S13
S10
T
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
0.113
x
( m) H fs
0.25
T ( m) SG = S fs
Wideal =
Q Wlost = T SG T ______________________________________________________________ ( m) S fs
Wideal
Wideal
H15 ( 1
489.001
z) H9 z
kJ kg
H4
T
S15 ( 1
z) S9 z
S4
Wout
H12
H11 x
Wout
kJ kg
(a) Heat Exchanger I: SG.a
SG.a 0.044 kJ kg K
S5
S4
S15
S14 ( 1
z)
kJ kg
Wlost.a
T SG.a
Wlost.a
13.021
(b) Heat Exchanger II: SG.b
SG.b 0.313 kJ kg K
S7
S6 ( 1
x)
Wlost.b
S14
S13 ( 1
kJ kg
z)
Wlost.b
T SG.b
92.24
618
(c) Expander:
SG.c
S12
S11 x
SG.c
0.157
kJ kg K
Wlost.c
T SG.c
Wlost.c
46.241
kJ kg
(d) Throttle:
SG.d 0.964
SG.d
kJ kg K
S9 z
Wlost.d
S10 ( 1
T SG.d
z
x) S7 ( 1
Wlost.d
x)
284.304 kJ kg
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
1.478
65.22%
100.00%
Work analysis, Eq. (15.3): kJ/kg Percent of
10.88%
Wout
Wlost.a
53.20
13.02
2.66%
Wlost.b
92.24
18.86%
Wlost.c
46.24
9.46%
Wlost.d
284.30
58.14%
489.00 Note that:
= Wideal
100.00%
619
Chapter 16 - Section A - Mathcad Solutions
16.10(Planck's constant)
h
6.626 10
34
Js
(Boltzmann's constant) 23 J k 1.381 10 K
(Avagodro's number)
NA
6.023 10 mol
23
1
P
1bar
T
298.15K
V
RT P
V
0.025
m
3
mol
39.948
a) For Argon:
gm mol
M
NA
3 2
Sig
R ln
2
MkT h
2
Ve NA
5 2
Sig
154.84
J mol K J mol K
Ans.
83.800
b) For Krypton: M
gm mol
NIST value: 154.84
NA
3
Sig
R ln
2
MkT h
2
2
Ve NA
5 2
Sig
164.08
J mol K
Ans.
NIST value: 164.05
J mol K
131.30
c) For Xenon
M
gm mol
NA
3
Sig
R ln
2
MkT h
2
2
Ve NA
5 2
Sig
164.08
J mol K
J mol K
Ans.
NIST value: 169.68
620
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,
kgm2 Nm energy [=] [=] s3 s time (b) Electric current is by denition the time rate of transfer of electrical charge. Thus
Power [=]
Charge [=] (electric current)(time) [=] As (c) Since power is given by the product of current and electric potential, then
kgm2 power [=] current As3 (d) Since (by Ohms Law) current is electric potential divided by resistance,
Electric potential [=]
kgm2 electric potential [=] 3 current A2 s (e) Since electric potential is electric charge divided by electric capacitance,
Resistance [=]
A2 s charge [=] Capacitance [=] electric potential kgm2
4
1.3 The following are general: ln x = ln 10 log10 x P sat /kPa = P sat /torr 100 kPa 750.061 torr (A)
(B) (C)
t/ C = T /K 273.15 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
+ ln
100 750.061
= 2.3026 a
b T /K 273.15 + c
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 massvelocity2 . Its fundamental units are therefore: E K [=] kgm2 s2 [=] Nm [=] J Potential energy as given by Eq. (1.7) has units of masslengthacceleration. Its fundamental units are therefore: E P [=] kgmms2 [=] Nm [=] 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 Hr [=] 3144.96 Second 1 Dy [=] 31449.6 Second 1 Wk [=] 314496. Second 1 Mo [=] 3144960 Second 1 Second [=] 0.031797 Sc 1 Minute [=] 60 Second [=] 0.19078 Mn 1 Hour [=] 3600 Second [=] 1.14469 Hr 1 Day [=] (24)(3600) Second [=] 2.74725 Dy 1 Week [=] (7)(24)(3600) Second [=] 1.92308 Wk 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 + N . According to the laboratory report a pure material (N = 1) is in 4-phase ( = 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 + 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 + 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
or
t t Tw0 Cw + T0 C t = Tw Cw + T C t
(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 ) d
(C)
In combination with Eq. (A) this becomes: Ct
Ct dT = K Tw0 t (T T0 ) T Cw d
or
dT =K d
T T0 Tw0 T t t Cw C
= T K
1 1 + t t Cw C
+K
T0 Tw0 + t t Cw C
Dene:
K
1 1 + t Cw Ct
K
T0 Tw0 + t Cw Ct
where both and are constants. The preceding equation may now be written: dT = T d
Rearrangement yields:
1 d( T ) dT = d = T T
Integration from T0 to T and from 0 to gives:
T 1 ln T0
=
624
which may be written:
T = exp( ) T0
When solved for T and rearranged, this becomes:
T =
+ T0
exp( )
where by the denitions of and ,
Tw C t + T0 C t = 0 tw Cw + C t
When = 0, the preceding equation reduces to T = T0 , as it should. When = , it reduces to T = /. Another form of the equation for / is found when the numerator on the right is replaced by Eq. (B): t Tw Cw + T C t = t Cw + C t
By inspection, T = / 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 2 2
625
2.21 We reformulate the denition of Reynolds number, with mass owrate m replacing velocity u: m = u A = u 2 D 4
Solution for u gives:
u=
4 m D2
Whence,
Re
4 m 4 m D u D = = 2 D D
(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
and
d(mU ) +Hm =0 dt
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
Multiply by dt and rearrange:
dm dU = m H U
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: P= RT V b
V2 V1
By Eq. (1.3),
W =
P dV =
V2 V1
RT d(V b) V b
Whence,
W = RT ln
V1 b V2 b
2.35 Recall: Whence,
d(P V ) = P d V + V d P d W = V d P d(P V ) and
and
d W = P d V
W =
V dP
(P V )
By Eq. (2.4),
d Q = dU d W and dU = d H P d V V d P the preceding equation becomes d Q = d H V d P
By Eq. (2.11), U = H P V With d W = P d V Whence,
Q=
H
V dP
626
. . 2.38 (a) By Eq. (2.24a), m = u A With m, A, and 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, 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
But a material balance yields
Then
dn . n = dt . dU dn +n Q = (H U ) dt dt
or
. dU dn +n Q = P V dt dt
2.44 (a) By Eq. (2.32a):
By Eq. (2.24a):
H2 H1 + 1 (u 2 u 2 ) = 0 1 2 2 . . 4 m m = u= D2 A
Then
u2 2
u2 1
=
4
2
. m2 2
1 1 4 4 D1 D2
and given
H2 H1 =
1 (P2 P1 )
1 1 (P2 P1 ) + 2
4
2
. m2 2
2
4 4 D1 D2 4 4 D1 D2
=0
1/2
. Solve for m:
. m = 2(P1 P2 ) 4
4 4 D1 D2 4 4 D1 D2
. (b) Proceed as in part (a) with an extra term, Here solution for m yields: . m = 2 (P1 P2 ) 2 C(T2 T1 ) 4
2 4 4 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 : P
=
T
1 V2
V P
T
V T
+
P
1 V
2V P T
= +
2V P T
T
=
P
1 V2
V T
P
V P
T
1 V
2V T P
=
2V P T
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: V P
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]
3.7 (a) For constant T , Eq. (3.4) becomes:
dV = dP V Integration from the initial state (P1 , V1 ) to an intermediate state (P, V ) for constant
gives:
ln
V = (P P1 ) V1
Whence,
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
d V = A(T ) exp( P)d P
P2 P1
P d V = A(T )
P exp( P)d P
A(T )
[( P1 + 1) exp( P1 ) ( P2 + 1) exp( P2 )] 628
With V1 = A(T ) exp( P1 ) and V2 = A(T ) exp( P2 ), this becomes: W = 1 [( P1 + 1)V1 ( P2 + 1)V2 ]
or
W = P1 V1 P2 V2 +
V1 V2
3.11 Differentiate Eq. (3.35c) with respect to T : T 1
P [(1)/]1
dT dP =T + P (1)/ dz dz
1
dT P (1)/ d P =0 + P (1)/ dz dz P
Algebraic reduction and substitution for d P/dz by the given equation yields: T P
dT 1 =0 (Mg) + dz
For an ideal gas T/P = 1/R. This substitution reduces the preceding equation to:
Mg dT = R dz
1
3.12 Example 2.13 shows that U2 = H . If the gas is ideal, H = U + P V = U + RT For constant C V , Whence, U2 U = C V (T2 T ) and and U2 U = RT C V (T2 T ) = RT
C P CV R T2 T = = CV CV T
When C P /C V is set equal to , this reduces to:
T2 = T
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 ( = 0). Here, Eqs. (3.36) and (3.37) reduce to: W = RT1 (1 1) and Q= RT1 (1 1) 1
Both are indeterminate. The easiest resolution is to write Eq. (3.36) and (3.37) in the alternative but equivalent forms: W = RT1 1
T2 1 T1
and
Q=
( )RT1 ( 1)( 1)
T2 1 T1
from which we nd immediately for = 0 that: W = R(T2 T1 ) and Q= R (T2 T1 ) = C P (T2 T1 ) 1
629
Isothermal case ( = 1). Equations (3.36) and (3.37) are both indeterminate of form 0/0. Application of lH pitals rule yields the appropriate results: o W = RT1 ln P2 P1
and
Q = RT1 ln
P2 P1
Note that if
y
P2 P1
(1)/
then
1 dy = 2 d
P2 P1
(1)/
ln
P2 P1
Adiabatic case ( = ). In this case simple substitution yields: W = RT1 1
P2 P1
( 1)/
1
and
Q=0
Isochoric case ( = ). Here, simple substitution yields: W =0 and Q= RT1 1
RT1 P2 1 = 1 P1
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
P1 = 101.33 kPa
P2 = 1500 kPa
630
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
or
Q = 172.71 kJ
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 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
Note also:
n2 =
P2 Vtank RT2
n1 =
P1 Vtank RT1
(c) If n 1 = 0,
n 2 (C V T2 C P T ) = Q
(d) If in addition Q = 0,
C V T2 = C P T
and
T2 =
CP CV
T
Whence,
T2 = T
(e) 1. Apply the result of Part (d), with = 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
P2 = 3 bar Appropriate values for R are therefore:
Vtank = 4 106 cm3
R(denominator) = 83.14 bar cm3 mol1 K1 Numerically,
R(numerator) = 8.314 J mol1 K1
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 dn = 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, Thus, or dU = C V dT or n dU (H U )dn = d Q d Q = mC dT
H U = P V = RT nC V dT RT dn = mC dT
where m is the mass of the tank, and C is its specic heat.
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
Integration yields:
ln
T2 T1
=
n 2 C V + mC R ln n 1 C V + mC CV
or
T2 = T1
n 2 C V + mC n 1 C V + mC
R/C V
In addition,
n1 =
P1 Vtank RT1
and
n2 =
P2 Vtank RT2
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, Whence, U = CV T U= P V = RT CV R (P V ) = R T
(P V )
But
1 CV CV = = 1 C P CV R
Therefore :
U=
1 (P V ) 1
3.28 Since Z = P V /RT the given equation can be written:
V =
RT + B RT P
Differentiate at constant T :
dV =
RT dP P2
V2
The isothermal work is then:
W =
V1
P d V = RT
P2 P1
1 dP P
Whence,
W = RT ln
P2 P1
Compared with Eq. (3.27)
3.29 Solve the given equation of state for V :
V =
RT +b RT P
633
Whence,
V P
=
T
RT P2
By denition [Eq. (3.3)]:
1 V
V P
T
Substitution for both V and the derivative yields:
=
P2
RT RT +b RT P
Solve the given equation of state for P:
P=
RT
V b+
Differentiate:
P T
=
V
R
V b+ RT
+
RT d dT T
V b+
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
d 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 + ( + )bV + b2 V b (V + b)(V + b)
Substitute V = 1/:
Z=
1 a(T ) 1 RT 1 + ( + )b + (b)2 1 b
Expressed in series form, the rst term on the right becomes:
1 = 1 + b + (b)2 + 1 b
The nal fraction of the second term becomes: 1 = 1 ( + )b + [( + )2 ](b)2 + 1 + ( + )b + (b)2
Combining the last three equations gives, after reduction: Z =1+ b
( + )a(T )b 2 a(T ) + + b2 + RT RT
Equation (3.12) may be written: Comparison shows:
Z = 1 + B + C 2 +
B =b
a(T ) RT
and
C = b2 +
( + )ba(T ) RT
634
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= 0.08664RTc Pc
0.42748RTc a(T ) = Tr1.5 Pc RT
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 + 3D P 2 +
T
Whence,
Z P
=B
T,P=0
Equation (3.12) with V = 1/: Differentiate:
Z = 1 + B + C 2 + D 3 + Z
= B + 2C + 3D 2 +
T
Whence,
Z
=B
T,=0
3.56 The compressibility factor is related to the measured quantities by: Z=
M PV t PV t = m RT n RT
(A)
By Eq. (3.39),
B = (Z 1)V =
(Z 1)M V t m
(B)
(a) By Eq. (A),
dT dm dV t dP dM dZ + t + = T m V P M Z
(C)
Thus
Max |% Z | |% M| + |% P| + |% V t | + |% m| + |% 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),
dm dM dV t Z dZ dB + t + = m M V Z 1 Z B
By Eq. (C),
Z dB = Z 1 B
dT dP T P
+
2Z 1 Z 1
dm dM dV t + t m M V
Therefore 635
Max |% B|
Z Z 1
|% P| + |% T |
+
2Z 1 Z 1
|% V t | + |% M| + |% 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
=
T
a(V b)2 b RT 3/2 (V + b)2 RT 3/2 (V b)2 (V + b)2
(A)
P V
=
T
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)
In addition, we have the mathematical relation: Z P
=
T
( Z / V )T ( P/ V )T
(C)
Combining these three equations gives Z P
=
T
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)
For P 0, V , and Eq. (D) becomes:
P0
lim
Z P
=
T
b a/RT 3/2 RT
For P , V b, and Eq. (D) becomes:
P
lim
Z P
=
T
b RT
3.60 (a) Differentiation of Eq. (3.11) gives: Z P
= B + 2C P + 3D P 2 +
T
whence
P0
lim
Z P
=B
T
If the limiting value of the derivative is zero, then B = 0, and
B = B RT = 0
(b) For simple uids, = 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 ( P/ T )V = Constant. (a) By Eq. (3.4) applied to a constant-V process: P T R V P T =
V
P T =
=
V
(b) For an ideal gas P V = RT , and
V
(c) Because a and b are constants, differentiation of Eq. (3.42) yields:
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 , 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 Ps. Inappropriate where changes in V are required. (f ) Rackett equation: Saturated liquids; a corresponding states application. (g) Constant , 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/ substituted everywhere for V . Subtract 1 from each side of the equation and divide by . Take the limit as 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 )
2
= 3Z c
2 = 3Z c
( + ) (
2
+ 1) +
(
+ 1) +
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 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 = 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 = 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): Eliminate :
= 0.077796 where
= 0.457236 = P/Z RT
Z = 1 + B + C 2 + . . . Z =1+
C P2 BP + 2 2 2 + ... Z R T Z RT
Z =1+
2 P2 C P2 B Pc Pr Pr + C Pr + . . . + 2 c2 2 r 2 + . . . = 1 + B Z 2 Tr2 Z Tr R Tc Z Tr RTc Z Tr
Rearrange:
(Z 1)Z Tr Pr + . . . = B +C Z Tr Pr
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: P=
a RT 2 V b V
Z=
Set V = 1/:
Z=
a b a 1 =1+ RT 1 b RT 1 b
a V V b V RT
whence
Z rep =
b 1 b
Z attr =
a RT
638
3.76 Write each modication in Z -form, (a) Z=
The required behavior is: (b) Z=
a V RT V b
V
lim Z = 1
a RT
V
lim Z = 1
The required behavior is: (c) Z=
a V 2 RT (V b)
V
lim Z =
a RT
V
lim Z = 1
The required behavior is: (d) Z =1
a 1 V b V RT
V
lim Z = 0 lim Z = 1
a a =1 RT V RT
V
Although lim Z = 1 as required, the equation makes Z linear in ; i.e., a 2-term virial EOS in . Such an equation is quite inappropriate at higher densities.
V
3.77 Refer to Pb. 2.43, where the general equation was developed;
. dU dn +n Q = P V dt dt
For an ideal gas,
n=
PV t RT
and
dn = dt
PV t RT 2
dT dt
Note that P V t /R = const.
Also for an ideal gas,
dU = C V dT
whence
dT dU = CV dt dt
. PV t Q = RT RT 2
P V t dT dT PV t dT = CP CV + RT dt dt RT dt
t2 t1
Integration yields:
ln
R T2 = CP PV t T1
. Q dt
3.78 By Eq. (3.4),
dV = dT d P V
where and are average values
Integrate:
ln
D2 Vt V2 = ln 2t = ln 2 = ln 2 V1 V1 D1
D1 + D D1
2
= ln 1 +
D D1
2
= (T2 T1 ) (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 ( + 1) + T12 ( 2 + + 1) 3 2
where 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 ( + 1) T1 + T1 T2 + T1 = = 2 2 2
and
2 Tam =
T12 2 ( + 2 + 1) 4
Whence,
C B T1 ( + 1) + T12 ( 2 + 2 + 1) 4 2 Dene as the difference between the two heat capacities:
C Pam = A +
C P C Pam = C T12
2 + + 1 2 + 2 + 1 4 3
This readily reduces to:
C T12 ( 1)2 12 Making the substitution = T2 /T1 yields the required answer. =
4.6 For consistency with the problem statement, we rewrite Eq. (4.8) as CP = A +
D B T1 ( + 1) + 2 T12
where 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 = T1 ( + 1) 2
and
2 Tam =
T12 2 ( + 2 + 1) 4
Whence,
C Pam = A +
4D B T1 ( + 1) + 2 2 2 T1 ( + 2 + 1)
Dene as the difference between the two heat capacities: C P C Pam = D T12
4 1 2 + 2 + 1
This readily reduces to:
=
D T12
1 +1
2
Making the substitution = 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) 2H2 (g) + O2 (g) 2H2 O(g) CH4 (g) C(s) + 2H2 (g)
H298 = 393,509 H298 = (2)(241,818) H298 = 74,520
CH4 (g) + 2O2 (g) CO2 (g) + 2H2 O(g)
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) 11 H2 (g) + 5 1 O2 (g) 11 H2 O(g) 2
H298 = (10)(393,509)
H298 = (11)(241,818) H298 = 249,700
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 = 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
H298 = 6,345,388 H298 = (11)(44,012)
11 H2 O(g) 11 H2 O(l)
C10 H22 (l) + 15 1 O2 (g) 10 CO2 (g) + 11 H2 O(l) 2
H298 = 6,829,520 J (HHV)
4.49 Saturated because the large
H lv overwhelms the sensible heat associated with superheat. H lv .
Water because it is cheap, available, non-toxic, and has a large
The lower energy content is a result of the decrease in H lv 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 U = 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 = U t + E K + 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 Stotal = S t + Ssurr
Just as U t for the egg is zero, so is S t . Therefore,
t Stotal = Ssurr =
Q Q surr = T T
Since Q is negative, Stotal is positive, and the process is irreversible. 5.6 By Eq. (5.8) the thermal efciency of a Carnot engine is: T = 1 TC H
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 S = C P ln P1 T1
(a) If P2 = P1 ,
S P = C P ln
T2 T1
If V2 = V1 ,
T2 P2 = T1 P1
Whence,
SV = C P ln
T2 T1
R ln
T2 T1
= C V ln
T2 T1
642
Since C P > C V , this demonstrates that (b) If T2 = T1 , ST = R ln P2 P1
SP >
SV .
If
V2 = V1 ,
P2 T2 = P1 T1
Whence,
SV = C P ln
P2 P1
R ln
P2 P1
= C V ln
P2 P1
This demonstrates that the signs for
ST and
SV are opposite.
5.12 Start with the equation just preceding Eq. (5.14) on p. 170:
ig dP C dT C dT dS d ln P = P = P P R T R T R
ig
For an ideal gas P V = RT , and ln P + ln V = ln R + ln T . Therefore,
dT dV dP = + T V P
ig
or
ig
dV dT dP = V T P
Whence,
ig
dV dT C dT dS = + = P V T R T R
ig
dT CP + d ln V 1 T R
Because (C P /R) 1 = C V /R, this reduces to:
C dT dS + d ln V = V R T R
ig
Integration yields:
S = R
T T0
V C V dT + ln V0 R T
ig
********************** 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 :
C dV C dP dP C dV C dP dS + P = V + P = P R V R P P R V R P R
ig
ig
ig
ig
Integration yields:
V C C P S + P ln = V ln V0 R P0 R R
ig
ig
5.13
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
or
C t dTH dTC = H t C C TH TC
Whence,
d ln TC = d ln TH
where
t CH t CC
Integration from TH0 and TC0 to TH and TC yields: TC = TC0
TH TH0
or
TC = TC0
TH TH0
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
Integration yields:
t t W = C H (TH TH0 ) + CC (TC TC0 )
Eliminate TC by the boxed equation of Part (a) and rearrange slightly:
W =
t C H TH0
TH t 1 + CC TC0 TH0
TH TH0
1
(c) For innite time, TH = TC T , and the boxed equation of Part (a) becomes: T = TC0 T TH0
+1
= TC0
TH0 T
From which: T = (TC0 )1/(
+1)
T (TH0 )
/( +1)
= TC0 (TH0 ) and T = (TC0 )1/( TH0
+1)
(TH0 )
/( +1)1
Because
/(
+ 1) 1 = 1/( T = TH0
+ 1), then:
TC0 TH0
1/( +1)
and
T TH0
=
TC0 TH0
/( +1)
Because TH = T , substitution of these quantities in the boxed equation of Part (b) yields:
W =
t C H TH0
TC0 TH0
1/( +1)
1 +
t CC TC0
TC0 TH0
/( +1)
1
5.14
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 (a) With d Q C = CC dTC , Eq. (B) becomes:
TH d QH = t TC CC dTC
or
t d Q H = CC
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
dTC t + CC dTC TC
TC t + CC (TC TC0 ) TC0
or
t W = CC TH ln
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:
where
r
TC TH
. |W |
. r 1 1 = |W | 1r 1r
= k Ar 4 (TH )4
or
A=
. |W | k(TH )4
1 (1 r )r 3
Differentiate, noting that the quantity in square brackets is constant: . . |W | 1 3 |W | dA = + = k(TH )4 (1 r )r 4 (1 r )2r 3 k(TH )4 dr
4r 3 (1 r )2r 4
Equating this equation to zero, leads immediately to:
4r = 3
or
r = 0.75
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 m 1 T1 + m 2 T2 m1 + m2
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 T =
(1)
If m 1 = m 2 , 645
T = (T1 + T2 )/2
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
(2)
S t = mC P ln
T2 T1 T2
or
S t = 2mC P ln
T T1 T2
Because T = (T1 + T2 )/2 >
T1 T2 ,
S t is positive.
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. 5.24 By denition,
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.
CP
H
=
T T0
C P dT
=
T0 T
Similarly,
CP
S
dT dT T0 T CP T T = = ln(T0 /T ) ln(T /T0 )
T T0
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 lH 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 | |Q | Heat transfer to the house at temperature T Heat transfer from the furnace at TF Heat transfer from the surroundings at T
The rst and second laws provide the two equations: |Q| = |Q F | + |Q | and
|Q | |Q| |Q F | =0 T TF T
646
Combine these equations to eliminate |Q |, and solve for |Q F |:
|Q F | = |Q|
T T TF T
TF T
With
T = 295 K
TF = 810 K
T = 265 K
and |Q| = 1000 kJ
The result is:
|Q F | = 151.14 kJ
Shown to the right is a scheme designed to accomplish this result. A Carnot heat engine operates with the furnace as heat source and the house as heat sink. The work produced by the engine drives a Carnot refrigerator (reverse Carnot engine) which extracts heat from the surroundings and discharges heat to the house. Thus the heat rejected by the Carnot engine (|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 | Equation (5.7) may be applied to both the engine and the refrigerator: |W |refrig = |Q 2 | |Q |
T |Q | TF |Q F | = = T |Q 2 | T |Q 1 | Combine the two pairs of equations:
|W |engine = |Q 1 |
TF T TF 1 = |Q 1 | T T
|W |refrig = |Q 2 | 1
T T
= |Q 2 |
T T T
Since these two quantities are equal, |Q 1 |
T T TF T = |Q 2 | T T
or
|Q 2 | = |Q 1 |
Because the total heat transferred to the house is |Q| = |Q 1 | + |Q 2 |, |Q| = |Q 1 | + |Q 1 |
TF T T T
But
|Q 1 | = |Q F |
T TF
TF T TF T = |Q 1 | 1 + T T T T
= |Q 1 |
whence
|Q| = |Q F |
T TF
TF T T T
TF T T T
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
|Q| |Q | |Q |
Heat transfer from the tank at temperature T Heat transfer from the house at T Heat transfer to the surroundings at T
The rst and second laws provide the two equations: |Q| + |Q | = |Q | and
|Q | |Q | |Q| =0 T T T
Combine these equations to eliminate |Q |, and solve for |Q|:
|Q| = |Q |
T T T T
T T
With
T = 448.15 K
T = 297.15 K
T = 306.15 K
and |Q | = 1500 kJ
The result is:
|Q| = 143.38 kJ
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 1 | Equation (5.7) may be applied to both the engine and the refrigerator: |W |refrig = |Q 2 | |Q |
Combine the two pairs of equations: |W |engine = |Q| 1 T T
T |Q 1 | = T |Q|
T |Q 2 | = T |Q |
= |Q|
T T T
|W |refrig = |Q |
T T
= |Q |
T t T
Since these two quantities are equal, |Q|
T T T T = |Q | T T
or
|Q| = |Q |
T T T T
T T
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 + T, j dt j
648
. 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 T, 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 T
(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: SG = |Q|
1 1 T1 T2
= |Q|
T1 T2 T1 T2
For the completely reversible process the net work produced is Wideal : |W1 | = |Q| T1 T T1
and
|W2 | = |Q|
T2 T T2
Wideal = |W1 | |W2 | = T |Q|
T1 T2 T1 T2
This is the work that is lost, Wlost , in the direct, irreversible transfer of heat |Q|. Therefore,
Wlost = T |Q|
T1 T2 = T 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 =
Tirrev T0
CP
ig
P dT ln P T
Srev =
Tirrev Trev ig
Trev T0
CP
ig
P dT ln P T
By difference, with
Srev = 0:
Sirrev =
CP
dT T
Since
Sirrev must be greater than zero, Tirrev must be greater than Trev .
650
Chapter 6 - Section B - Non-Numerical Solutions
6.1 By Eq. (6.8),
H S
=T
P
and isobars have positive slope
Differentiate the preceding equation:
2 H S2
=
P
T S
P
Combine with Eq. (6.17):
2 H S2
=
P
T CP
and isobars have positive curvature.
6.2 (a) Application of Eq. (6.12) to Eq. (6.20) yields: C P P
=
T
{V T ( V / T ) P } T
P
or
C P P
=
T
V T
T
P
2V T 2
P
V T
P
Whence,
C P P
= T
T
2V T 2
P
For an ideal gas:
V T
=
P
R P
and
2V T 2
=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
P T
dV +
V
V T
dP
P
Restrict to constant P:
C P = CV + T
P T
V
V T
P
By Eqs. (3.2) and (6.34):
V T
= V
P
and
P T
=
V
Combine with the boxed equation:
C P C V = T V
6.3 By the denition of H , U = H P V . Differentiate: U T
=
P
H T
P
P
V T
or
P
U T
= CP P
P
V T
P
651
Substitute for the nal derivative by Eq. (3.2), the denition of :
U T
= CP PV
P
Divide Eq. (6.32) by dT and restrict to constant P. The immediate result is:
U T
= CV + T
P
P T
P
V
V T
P
Solve for the two derivatives by Eqs. (6.34) and (3.2); substitution gives:
U T
= CV +
P
(T P)V
6.4 (a) In general,
dU = C V dT + T
P T
P dV
V
(6.32)
By the equation of state,
P=
RT V b
whence
P T
=
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 Then, H T
= 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 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
or
R dT d ln(V b) = CV T But from part (b), R/C V = (C P C V )/C V = 1. Then
d ln T = ( 1)d ln(V b) From which:
or
d ln T + d ln(V b) 1 = 0
T (V b) 1 = const.
Substitution for T by the equation of state gives P(V b)(V b) 1 = const. R
or
P(V b) = const.
652
6.5 It follows immediately from Eq. (6.10) that: V = G P
and
T
S=
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 These become: H = (T ) T d (T ) dT and U = H P V = H RT
and
U = (T ) T
d (T ) RT dT
By Eqs. (2.16) and (2.20), CP = H T
and
P
CV =
U T
V
Because
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
T
S=
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 These become: H = F(T ) + K P T d F(T ) dT and U = H PV = H PK
and
U = F(T ) T
d F(T ) dT
By Eqs. (2.16) and (2.20), CP = H T
and
P
CV =
U T
V
653
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 = = 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:
Z
PV RT
V =
1
a q b RT
Then,
Z=
1 exp(qb) 1 b
With the denition, b, this becomes: Z= 1 exp(q ) 1
(A)
Because = P/Z RT ,
=
bP Z RT
Given T and P, these two equations may be solved iteratively for Z and . Because b is a constant, Eqs. (6.58) and (6.59) may be rewritten as: 654
GR = RT
0
(Z 1)
0
d + Z 1 ln Z
(B)
HR = RT
Z T
d + Z 1
(C)
In these equations, Z is given by Eq. (A), from which is also obtained: ln Z = ln(1 ) q and Z T
=
q exp(q ) T (1 )
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:
0
(Z 1)
d = exp(q){E[q(1 )] E(q)} E(q ) ln(q )
where is Eulers constant, equal to 0.57721566. . . .
and
T
0
Z T
s = q exp(q){E[q(1 )] E(q)}
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. 6.19 Write Eq. (6.75) in log10 form: log P sat = A B T
(A)
Apply at the critical point:
log Pc = A
B Tc
(B)
655
By difference,
log Pr sat = B
1 1 T Tc
=B
Tr 1 T
(C)
If P sat is in (atm), then application of (A) at the normal boiling point yields: log 1 = A B Tn
or
A=
B Tn
With Tn /Tc , Eq. (B) can now be written: log Pc = B
1 1 Tc Tn
=B
Tc Tn Tn Tc
=B
1 Tn
Whence,
B=
Equation (C) then becomes: log Pr sat = Tn 1
Tn 1
log Pc
Tr 1 log Pc = T
1
Tr 1 log Pc Tr
Apply at Tr = 0.7:
log(Pr sat )Tr =0.7 =
3 7
1
log Pc
By Eq. (3.48), Whence,
= 1.0 log(Pr sat )Tr =0.7
=
3 7
1
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
and
T S
=
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
P
T C2 P
C P S
=
P
T T 2 2 CP CP
C P T
P
T S
=
P
T T 1 2 CP CP
C P T
P
With C P = a + bT ,
C P T
=b
P
and
1
T CP
C P T
=1
P
a bT = a + bT a + bT
Because this quantity is positive, so then is the curvature of an isobar. 6.84 Division of Eq. (6.8) by d S and restriction to constant T yields: H S
=T +V
T
P S
By Eq. (6.25),
T
P S
=
T
1 V
Therefore,
H S
=T
T
1 1 = (T 1)
656
Also,
2 H S2
=
T
1 2
S
=
T
1 2
P
T
P S
=
T
1 2
P
T
1 V
Whence,
2 H S2
=
T
1
3V
P
T
By Eqs. (3.2) and (3.38):
=
1 V
V T
and
P
V =
RT +B P
Whence,
V T
=
P
dB R + dT P
and
=
1 V
dB R + dT P
Differentiation of the second preceding equation yields: P
=
T
R V P2
dB R + dT P
1 V2
V P
=
T
1 R (V ) 2 V V P2
V P
T
From the equation of state,
V P
=
T
RT P2
Whence,
P
=
T
R RT R (T 1) = + 2 2 V P2 V P VP
Clearly, the signs of quantity (T 1) and the derivative on the left are the same. The sign is determined from the relation of and V to B and d B/dT : T V
T 1 =
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 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 (). By Eqs. (6.54) and (6.56), By Eqs. (3.38) and (6.40), GR = BP V R = B, and S R = P(d B/dT ) 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 (). and V R is (). Combine the equations above for G R , S R , and H R : HR = P B T dB dT
Whence,
HR T
=P
P
dB d2 B dB T 2 dT dT dT
= P T
d2 B dT 2
R Therefore, C P
HR T
is (+).
P
(See Fig. 6.5.)
657
6.89 By Eq. (3.5) at constant T :
P =
V 1 P1 ln V1
(A)
(a) Work
d W = P d V =
1 1 V 1 P1 d V = ln V d V P1 + ln V1 d V ln V1
V2 V1
W =
1
ln V d V P1 +
1 ln V1 (V2 V1 )
W =
1 1 [(V2 ln V2 V2 ) (V1 ln V1 V1 )] P1 (V2 V1 ) (V2 ln V1 V1 ln V1 )
=
V2 1 + V1 V2 P1 (V2 V1 ) V2 ln V1
By Eq. (3.5),
ln
V2 = (P2 P1 ) V1
whence
W = P1 V1 P2 V2
V2 V1
(b) Entropy By Eq. (A),
By Eq. (6.29), P =
d S = V d P
ln V1 ln V P1
and
d P =
1 d ln V
dS =
V d ln V = d V
and
S=
(V2 V1 )
(c) Enthalpy Substitute for d P:
By Eq. (6.28),
d H = (1 T )V d P
d H = (1 T )V
1 T 1 dV d ln V =
H=
1 T (V1 V2 )
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 H = 134.55 kJ kg1
6.90 The given equation will be true if and only if
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
Neither C P
ig
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.
T P
V
H ig P
=
S
H ig P
+
T
H ig T
P
T P
S
= CP
ig
T P
S
658
T P
=
S
T S ig
P
S ig P
=
T
T
CP
T
ig
S ig P
T
H ig P
ig
=T
S
S ig P
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 ): dS = S P
dP +
V
S V
dV
P
By the chain rule for partial derivatives, dS = S T
V
T P
dP +
V
S T
P
T V
dV
P
With Eqs. (6.30) and (6.17), this becomes:
dS =
CV T
T P
dP +
V
CP T
T V
dV
P
6.93 By Eq. (6.31),
P=T
P T
V
U V
T
(a) For an ideal gas,
P=
RT V
and
P T
=
V
R V
Therefore
RT RT = V V
U V
and
T
U V
=0
T
(b) For a van der Waals gas,
P=
a RT 2 V b V
and
P T
=
V
R V b
Therefore
RT a RT 2 = V b V b V
U V
and
T
U V
=
T
a V2
(c) Similarly, for a Redlich/Kwong uid nd:
U V
=
T
T 1/2 V (V
(3/2)A + b)
where
A = a(Tc ) Tc
1 2
6.94 (a) The derivatives of G with respect to T and P follow from Eq, (6.10): S = G T
and
P
V =
G P
T
Combining the denition of Z with the second of these gives:
Z
P PV = RT RT
G P
T
659
Combining Eqs. (2.11) and (3.63) and solving for U gives U = G + T S P V . Replacing S and V by their derivatives gives:
U =GT
G T
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
= T
P
2G T 2
2
P
P
2G T P
U P
= T
T
G T P
P
2G P2
T
From the denition of C V and an equation relating partial derivatives: CV U T
=
V
U T
+
P
U P
T
P T
V
Combining the three equations yields: C V = T 2G T 2
P
P
2G T P
T
2G T P
+P
2G P2
T
P T
V
Evaluate ( P/ T )V through use of the chain rule: P T
=
V
P V
T
V 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 P T
and
V P
=
T
2G P2
T
Then
P T
=
V
( 2 G/ T ) P ( 2 G/ P 2 )T
and C V = T
2G T 2
P
P
2G T P
+ T
2G T P
+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 lv . Both decrease monotonically as T increases, becoming zero at the critical point.
H lv and
660
6.98 By the Clapeyron equation:
If the ratio
S sl to
H sl S sl d P sat = = T V sl V sl dT sl V is assumed approximately constant, then
P sat = A + BT If the ratio H sl to 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:
sat d Psv dT
=
t
Pt Htsv Pt Htsv RTt2 RTt2 Z tsv
sat d Plv dT
=
t
Pt Htlv Pt Htlv RTt2 RTt2 Z tlv
sat d Psv
dT
t
sat d Plv dT
t
Pt RTt2
Htsv
Htlv
Because
Htsv
Htlv
=
Htsl
is positive, then so is the left side of the preceding equation.
6.100 By Eq. (6.72):
H lv d P sat = T V lv dT
But
V lv =
RT P sat
Z lv
whence
H lv d ln P sat = RT 2 Z lv dT
lv
(6.73)
H 1 H lv Tc H lv d ln Pr sat = 2 = = 2 Z lv lv 2 Z lv Tr RTc Tr Z RT dTr
6.102 Convert c to reduced conditions: c 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
Tr =1
Thus, c (L/K) = 5.82 for = 0, and increases with increasing molecular complexity as quantied by .
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
T CP
V T
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
=
U
T U
V
U V
T
The two partial derivatives on the right are found from Eqs. (2.16) and (6.31); thus,
T V
=
U
1 CV
PT
P T
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
=
S
P S
V
S V
P
This can also be written: P V
=
S
P T
V
T S
V
S T
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
=
T
CP CV
1 V
662
Substitute into Eq. (A):
c2 =
V CP MC V
or
c=
V CP MC V
(a) For an ideal gas, V = RT /P and = 1/P. Therefore,
cig =
RT C P M CV
(b) For an incompressible liquid, V is constant, and = 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 2 P1 V1 1 +1 1
= P1 V1
2 +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 Jkg1 . Moreover, P1 V1 = RT1 /M; whence Pa m3 kg1 =
u2 throat =
RT1 M
2 +1
With R in units of J(kg mol)1 K1 , RT1 /M has units of Jkg1 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
x y
+
w
x w
y
w y
z
Z T
=
P
Z T
+
Z
T
T
P
Whence,
Z T
=
Z T
P
Z
T
T
P
Because
P = Z RT ,
=
P Z RT
and
T
=
P
P R
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
=
Z
T
T
and
P
T
=
P
P = 2 T Z RT
Combining these two equations yields:
T
Z T
=
Z
T
(a) Equation (3.42) with van der Waals parameters becomes: P=
a RT 2 V b V
Multiply through by V /RT , substitute Z = P V /RT , V = 1/, and rearrange: Z=
a 1 RT 1 b
In accord with Eq. (3.51), dene q a/b RT . In addition, dene b. Then, Z= 1 q 1
(A)
Differentiate:
Z T
=
Z T
=
dq dT
By Eq. (3.54) with (Tr ) = 1 for the van der Waals equation, q = dq = dT
/ Tr . Whence,
1 Tr2
dTr = dT
1
Tr2 Tc
=
q 1 = T T Tr
Then,
Z T
= ( )
q T
=
q T
In addition,
Z
=b
T
Z
=
T
b qb (1 )2
664
Substitute for the two partial derivatives in the boxed equation: T
b q qb = (1 )2 T
or
q =
Whence,
1 =1 2q
q (1 )2
(B)
By Eq. (3.46), Pc = RTc /b. Moreover, P = Z RT . Division of the second equation by the rst gives Pr = ZbT / Tc . Whence Pr = Z 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 , 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=
By Eq. (3.54) with (Tr ) = Tr0.5 for the Redlich/Kwong equation, q =
q 1 1+ 1
(A) / Tr1.5 . This leads to:
1.5 q dq = T dT
and
Z T
=
1.5 q T (1 + )
Moreover,
Z
T
=
bq b 2 (1 + )2 (1 )
2
Substitution of the two derivatives into the boxed equation leads to a new Eq. (B): q= 1+ 1
As in Part (a), for a given Tr , calculate q, and solve Eq. (B) for , 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.
1 2.5 + 1.5
(B)
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 =
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
l 1.8604 1.3027 S1 S2 = 0.0921 = l v 7.3598 1.3027 S2 S2
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),
T2 =
P2 P1
( 1)/ )
Moreover,
n1 =
P1 Vtank RT1
and
n2 =
P2 Vtank RT2
With = 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 H =n
( H )S
By Eq. (6.8), Assume now that RT1 /P1 . Then
( H )S =
V dP = V
P
P is small enough that V , an average value, can be approximated by V1 = ( H )S = RT1 P1
P
and
RT1 Ws = n P1
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 =
P2 P1
R/C P
R/C P
1
For
P sufciently small, the quantity in square brackets becomes: P2 P1
R/C P
1= 1+
P P1
1
1+
R P C P P1
1
The boxed equation is immediately recovered from this result. 666
7.41 The equation immediately preceding Eq. (7.22) page 276 gives T2 = T1 . With this substitution, Eq. (7.23) becomes: 1 T1 T1 = T1 1 + T2 = T1 +
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:
1 CP S ln 1 + = R R
CP ln ln = R
1+
1
Whence,
+ 1 CP SG ln = R R
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 ( 1) where P2 P1
R/C P
This equation is a proper basis, because compressor efciency and owrate n are xed. With all other variables constant, differentiation yields:
d d Ws = T1 ( 1) + C P dC P dC P
From the denition of , ln =
P2 R ln P1 CP
whence
P2 R 1 d d ln ln = = 2 P1 dC P dC P CP
Then,
P2 R d ln = 2 P1 dC P CP
and
R P2 d Ws ln = T1 1 P1 CP dC P
= T1 ( 1 ln )
When = 1, the derivative is zero; for > 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
(e) Rearrange the given equation:
V ( P/ T )V V = = P ( P/ V )T T
T
P T
=
V
V T
P
For the nal equality see footnote on p. 266. This result is the equation of (c).
7.50 From the result of Pb. 7.3: c =
V CP 1 M CV
where
=
1 V
V P
T
With
V =
RT +B P
then
V P
=
T
RT P2
Also, let =
CP CV
Then
c = PV
BP = 1+ = (RT + B P) RT MRT MRT
RT M
c=
B RT + RT M
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: HS =
ig
V ig d P =
RT dP P
(const S)
HS =
V dP =
Z RT dP P
(const S)
HS
HS
ig
=
Z RT d P (const S) P Z RT d P (const S) P
By extension, and with equal turbine efciencies,
. W H = . ig = Z H ig W
7.52 By Eq. (7.16),
H = ( H ) S
For C P = constant,
T2 T1 = [(T2 ) S T1 ] P2 P1
R/C P
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 :
T2 = T1 1 +
P2 P1
R/C P
1
668
From which
=
T2 1 T1
P2 P1
R/C P
Note that < 1 1
Results: For T2 = 318 K, = 1.123; For T2 = 348 K, = 1.004; For T2 = 398 K, = 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 Diesel 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
or
rc 1 = r re
Equation (8.7) can therefore be written: Diesel 1 =1
(rc /r ) (1/r ) rc /r 1/r
1 (1/r ) =1 1/r
1
rc 1 rc 1
or
Diesel = 1
1 r
rc 1 (rc 1)
(b) We wish to show that: rc 1 >1 (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, R g [1 + (x 1)] (x 1)2 2!
(0 < < 1)
Then,
x a = 1 + a (x 1) + 1 a (a 1) [1 + (x 1)]a2 (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)
and
rc 1 >1 (rc 1)
(c) If = 1.4 and r = 8, then by Eq. (8.6): Otto = 1 1 8
0.4
and
0.4
Otto = 0.5647
rc = 2
Desiel = 1
1 8
21.4 1 1.4(2 1)
and
Diesel = 0.4904
rc = 3
Desiel = 1
1 8
0.4
31.4 1 1.4(3 1)
and
Diesel = 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, W AB WC D Q BC
where,
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,
=
TB T A TA TB + TC TD =1 TC TD TC TD
By Eq. (3.30b), TB = T A PB PA
( 1)/
and
TD = TC
PD PC
( 1)/
= TC
PA PB
( 1)/
TA Then, =1
PB PA
( 1)/
1
TC 1
PA PB
( 1)/
Multiplication of numerator and denominator by (PB /PA )( 1)/ gives:
=1
TA TC
PB PA
( 1)/
671
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 /(1) ln P = ln K + ln T 1
dT dP = 1 T P
P dP = 1T dT
(A)
Sign of d P/dT is that of 1, i.e., + = 0 d P/dT = 0 = 1 d P/dT = Constant P Constant T
Special cases By Eq. (A),
d2 P = 2 1 dT
P 1 dP 2 T T dT
=
1 1T
P P T 1T
P d2 P = ( 1)2 T 2 dT 2
(B)
Sign of d 2 P/dT 2 is that of , i.e., +
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 /(1) , this requires = . 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), = . In this case Eqs. (A) and (B) become:
P dP = 1T dT
P d2 P = ( 1)2 T 2 dT 2
We note here that /( 1) and /( 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.
= =1
P
=
=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 T
Figure 1: The Carnot cycle
0 0 T
Figure 2: The Otto cycle
P
P
0 0 T
Figure 3: The Diesel cycle
0 0 T
Figure 4: The Brayton 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 |Q H | |W |
For a Carnot heat pump, =
TH |Q H | = TH TC |Q H | |Q C |
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 TH ln
TC TH 1 + TH TC
9.5 Differentiation of Eq. (9.3) yields: TC
=
TH
TH TC 1 = + 2 (TH TC )2 (TH TC ) TH TC
and
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, is given by Eq. (9.3). Write this equation for the two cases: = TC TH TC
and
=
TC T H TC
Because the directions of heat transfer require that TH > T H and TC < TC , a comparison shows that < and therefore that 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
Carnot
= 0.6
TC TH TC
If
< 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
and
m air =
Mair pair V t RT
(a) By denition, h
m H2 O m air
or
h=
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:
h pc = h rel
P PHsat (h rel /100) 2O
P PHsat 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 1 and ln 2 ,
2 y1 P = x1 exp(Ax2 )P1sat
and
2 y2 P = x2 exp(Ax1 )P2sat
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 K 1 x1 + K 2 (1 x1 ) = 1
x1 =
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
or for xi 0
Hi =
1 Ms k i
For water, Ms = 18.015 g mol1 Thus,
or
0.018015 kg mol1 .
Hi =
1 = 1633 bar (0.018015)(0.034)
This is in comparison with the value of 1670 bar in Table 10.1.
678
Chapter 11 - Section B - Non-Numerical Solutions
11.6 Apply Eq. (11.7): Ti (nT ) n i
=T
P,T,n j
n n i
=T
T,P,n j
Pi
(n P) n i
=P
P,T,n j
n n i
=P
T,P,n j
11.7 (a) Let m be the mass of the solution, and dene the partial molar mass by: m i
m n i
T,P,n j
Let Mk be the molar mass of species k. Then m= m n i
nk Mk = ni Mi + n j M j
k j
( j = i)
and
=
T,P,n j
(n i Mi ) n i
= Mi
T,P,n j
Whence,
m i = Mi
(b) Dene a partial specic property as: Mi
M t m i
=
T,P,m j
M t n i
T,P,m j
n i m i
T,P,m j
If Mi is the molar mass of species i,
ni =
mi Mi
and
n i m i
=
T,P,m j
1 Mi
Because constant m j implies constant n j , the initial equation may be written:
Mi Mi = Mi
11.8 By Eqs. (10.15) and (10.16),
dV V1 = V + x2 d x1
and
dV V2 = V x1 d x1
Because
V = 1
then
1 d dV = 2 d x1 d x1
whence
1 x2 d 1 = V1 = 2 d x1
1
x2 d d x1
=
d 1 x2 2 d x1
1 x1 d 1 = V2 = + 2 d x1
1+
x1 d d x1
=
d 1 + x1 2 d x1
With
2 = a0 + a1 x 1 + a2 x 1
and
d = a1 + 2a2 x1 d x1
these become:
1 2 V1 = 2 [a0 a1 + 2(a1 a2 )x1 + 3a2 x1 ]
and
1 2 V2 = 2 (a0 + 2a1 x1 + 3a2 x1 )
679
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
For M1 ,
(n M) n 1
T,P,n 2 ,n 3
= M1 + n 2 n 3 C
2n 1 1 3 2 n n
n n 1
T,P,n 2 ,n 3
Because n = n 1 + n 2 + n 3 ,
n n 1
=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
i
For the mixture, P = Z RT /V . These two equations combine to give Z = 11.11 The general principle is simple enough:
yi Z i .
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
or
3 H2 = 600 + 40 x1
680
(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
=
x1 =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
and
3 2 V2 = 70 + x1 + 14 x1
681
(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 )
Whence,
3 2 V = 70 + 58 x1 x1 7 x1
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
and
d V2 2 = 2 x1 + 42 x1 d x1
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
=
x1 =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
dH H2 = H x1 d x1
682
Given that: Then, after simplication,
H = x1 (a1 + b1 x1 ) + x2 (a2 + b2 x2 ) dH = a1 + 2b1 x1 (a2 + 2b2 x2 ) d x1
Combining these equations gives after reduction: H1 = a1 + b1 x1 + x2 (x1 b1 x2 b2 ) and H2 = a2 + b2 x2 x1 (x1 b1 x2 b2 )
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
=
z
x y
+
w
x w
y
w y
z
(n M) n i
=
T,P,n j
(n M) n i
+
T,V,n j
(n M) V
T,n
V n i
T,P,n j
Whence, M Mi = Mi + n V
T,n
V n i
or
T,P,n j
M Mi = Mi n V
T,n
V n i
T,P,n j
By denition, (nV ) Vi n i
=n
T,P,n j
V n i
+V
T,P,n j
or
n
V n i
= Vi V
T,P,n j
Therefore,
M Mi = Mi + (V Vi ) V
T,x
11.20 Equation (11.59) demonstrates that ln i is a partial property with respect to G R /RT . Thus ln i = G i /RT . The partial-property analogs of Eqs. (11.57) and (11.58) are:
ln i P
=
T,x
Vi R RT
and
ln i T
=
P,x
HiR RT 2
The summability and Gibbs/Duhem equations take on the following forms:
GR = RT
xi ln i
i
and
i
xi d ln i = 0
(const T, P)
683
11.26 For a pressure low enough that Z and ln are given approximately by Eqs. (3.38) and (11.36): Z =1+ BP RT
and
ln =
BP RT
then:
ln Z 1
11.28 (a) Because Eq. (11.96) shows that ln i 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 1 =
d(G E/RT ) GE + x2 d x1 RT
ln 2 =
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
whence
d(G E/RT ) 2 = 1.8 + 2 x1 + 2.4 x1 d x1
Then,
3 2 ln 1 = 1.8 + 2 x1 + 1.4 x1 1.6 x1
and
3 2 ln 2 = x1 1.6 x1
(b) In accord with Eq. (11.11), GE 2 3 2 3 = x1 ln 1 + x2 ln 2 = 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 i and divide by d x1 : x1
d ln 2 d ln 1 =0 + x2 d x1 d x1
Differentiate the the boxed equations of part (a): d ln 1 2 = 2 + 2.8 x1 4.8 x1 d x1
and
d ln 2 2 = 2 x1 4.8 x1 d x1
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 1 d x1
=
x1 =1
d ln 2 d x1
=0
x1 =0
684
(e)
11.29 Combine denitions of the activity coefcient and the fugacity coefcients: i fi /xi P f i /P
or
i =
i i
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
and
SE =
G E T
P,x
E = CP
T T
Working equations are then:
E S1 = E H1E G 1 T1
and
E E E S2 = S1 + C P
T T
E H2E = H1E + C P
T
and
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
I.
E II. For C P = 0
E G1
H1E
E S1
E CP
E S2
H2E
E G2
E S2
H2E
E G2
(a) (b) (c) (d) (e) (f) (g)
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 1 = A12 n 2
1 n1 n n2
+ A13 n 3
1 n1 n n2
A23
n2n3 n2
= A12 x2 (1 x1 ) + A13 x3 (1 x1 ) A23 x2 x3 Similarly, ln 2 = A12 x1 (1 x2 ) A13 x1 x3 + A23 x3 (1 x2 ) ln 3 = A12 x1 x2 + A13 x1 (1 x3 ) + A23 x2 (1 x3 )
(b) Each ln i is multiplied by xi , and the terms are summed. Consider the rst terms on the right of each expression for ln i . 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 . (c) For innite dilution of species 1, For pure species 1, For innite dilution of species 2, For innite dilution of species 3, x1 = 0: x1 = 1: x2 = 0: x3 = 0: ln 1 (x1 = 0) = A12 x2 + A13 x3 A23 x2 x3 ln 1 (x1 = 1) = 0
2 ln 1 (x2 = 0) = A13 x3 2 ln 1 (x3 = 0) = A12 x2
11.35 By Eq. (11.87), written with M G and with x replaced by y: G E = B P y1 B11 P y2 B22 P or
GE = GR
Equations (11.33) and (11.36) together give G iR = Bii P. Then for a binary mixture: G E = P(B y1 B11 y2 B22 )
i
yi G iR
Combine this equation with the last equation on Pg. 402:
G E = 12 P y1 y2
From the last column of Table 11.1 (page 415): S E =
G E T
P,x
Because 12 is a function of T only:
SE =
d12 P y1 y2 dT
By the denition of G E , H E = G E + T S E ; whence,
H E = 12 T
d12 dT
P y1 y2
E Again from the last column of Table 11.1: C P =
HE T
P,x
This equation and the preceding one lead directly to:
E C P = T
d 2 12 P y1 y2 dT 2
686
11.41 From Eq. (11.95):
(G E /RT ) T
=
P
H E RT 2
or
(G E /T ) T
=
P
H E T2
To an excellent approximation, write:
(G E /T ) T
P
H E (G E /T ) 2 Tmean T
From the given data:
0.271 785/323 805/298 (G E /T ) = 0.01084 = = 25 323 298 T
and
1060 H E = 0.01082 = 2 3132 Tmean
The data are evidently thermodynamically consistent. 11.42 By Eq. (11.14), the Gibbs/Duhem equation, x1
d M2 d M1 =0 + x2 d x1 d x1
Given that
M1 = M1 + Ax2
and
M2 = M2 + Ax1
then
d M1 = A d x1
and
d M2 =A d x1
Then
d M2 d M1 = x1 A + x2 A = A(x2 x1 ) = 0 + x2 d x1 d x1 The given expressions cannot be correct.
x1
2 2 M E = Ax1 x2
11.45 (a) For
nd
2 E M1 = Ax1 x2 (2 3x1 )
and
2 E M2 = Ax1 x2 (2 3x2 )
Note that at both x1 = 0 (x2 = 1) and x1 = 1 (x2 = 0), In particular, 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 = 2Ax1 (1 6x1 x2 ) = 2Ax2 (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 = 2A and For x1 = 0 d x1 d x1
For
x1 = 1
E d M1 =0 d x1
and
E d M2 = 2A d x1
(b) For
M E = A sin( x1 ) nd: E M1 = A sin( x1 ) + A x2 cos( x1 ) and E M2 = A sin( x1 ) A x1 cos( x1 )
E E d M2 d M1 = A 2 x1 sin( x1 ) = A 2 x2 sin( 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
Pb. 11.45 (a)
A
MEi
10
A . xi
i
2.
0 .. 100 1
xi MEbar1i
3. 1
.00001
.01 . i xi
2
xi 2
xi . 2
A . xi . 1 xi
2
3 . xi
MEbar2i
2 1.5 MEi 1
A . xi . xi . 1
MEbar1
i
0.5 0 0.5
MEbar2i
0
0.2
0.4
xi
0.6
0.8
1
Pb. 11.45 (b)
MEi
A . sin
p
.x
i
p
(pi prints as bf p) .x .x
i i
MEbar1i
MEbar2i
40
30
MEi
A . sin A . sin
A.p . 1
xi . cos
p
p
.x
i
p
A . p . xi . cos
.x
i
20
MEbar1
i
10
0 10
MEbar2i
0
0.2
0.4
xi
0.6
0.8
1
687A
11.46 By Eq. (11.7),
(n M) Mi = n i
= M +n
T,P,n j
M n i
T,P,n j
At constant T and P,
dM =
k
M xk
d xk
T,P,x j
Divide by dn i with restriction to constant n j ( j = i): M n i
=
T,P,n j k
M xk
T,P,x j
xk n i
nj
With
nk xk = n
xk n i
=
nj
n k n2
(k = i)
M n i
=
T,P,n j
1 n
xk
k=i
M xk
1 ni n n2
(k = i)
T,P,x j
M 1 + (1 xi ) xi n
T,P,x j
=
1 n
M xi
T,P,x j
1 n
xk
k
M xk
T,P,x j
Mi = M +
M xi
T,P,x j k
xk
M xk
T,P,x j
For species 1 of a binary mixture (all derivatives at constant T and P): M1 = M + M x1
x2
x1
M x1
x2
x2
M x2
x1
= M + x2
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: dM = M x1
x2
d x1 +
M x2
x1
d x2
Division by d x1 yields: dM = d x1
M x1
+
x2
M 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 = n 1
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
E M1 = An 2
1 1 + n 1 + Bn 2 n 2 + Bn 1
+ n1
B 1 (n 1 + Bn 2 )2 (n 2 + Bn 1 )2
Conversion back to mole fractions yields: E M1 = Ax2
1 1 + x2 + Bx1 x1 + Bx2
x1
B 1 + 2 (x2 + Bx1 )2 (x1 + Bx2 )
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
Similarly,
2 E M2 = Ax1
B 1 + 2 (x2 + Bx1 )2 (x1 + Bx2 )
(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 M1 = x2 2 d x1 d x1
E d2 M E d M2 = x1 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. 1 V
V T
V id =
P i
xi Vi
id =
i
1
xi Vi
xi
i
Vi T
=
P
1
i
xi Vi
xi Vi i
i
The claim is valid only if all the Vi are equal.
690
Chapter 12 - Section B - Non-Numerical Solutions
12.2 Equation (12.1) may be written: yi P = xi i Pi sat . Summing for i = 1, 2 gives: P = x1 1 P1sat + x2 2 P2sat . Differentiate at constant T :
d2 d1 dP 2 + 1 + P2sat x2 = P1sat x1 d x1 d x1 d x1
Apply this equation to the limiting conditions: For x1 = 0 : x2 = 1 1 = 1 2 = 1 d2 =0 d x1 d1 =0 d x1
For x1 = 1 :
x2 = 0
1 = 1
2 = 2
Then,
dP d x1
x1 =0
= P1sat 1 P2sat
or
dP d x1
x1 =0
+ P2sat = P1sat 1
dP d x1
x1 =1
= P1sat P2sat 2
or
dP d x1
x1 =1
P1sat = P2sat 2
Since both Pi
sat
and
i
are always positive denite, it follows that:
dP d x1
x1 =0
P2sat
and
dP d x1
x1 =1
P1sat
12.4 By Eqs. (12.15), Therefore, ln
2 ln 1 = Ax2
and
2 ln 2 = Ax1
1 2 2 = A(x2 x1 ) = A(x2 x1 ) = A(1 2x1 ) 2
By Eq. (12.1),
y1 x2 P2sat 1 = = y2 x1 P1sat 2
y1 /x1 y2 /x2
P2sat P1sat
= 12 r
Whence,
ln(12 r ) = A(1 2x1 )
az x1
If an azeotrope exists, 12 = 1 at 0
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 1 is accompanied by the opposite extremum in ln 2 . Thus the difference ln 1 ln 2 is also an extremum, and Eq. (12.8) becomes useful: d(G E/RT 1 = ln 1 ln 2 = ln d x1 2
Thus, given an expression for G E/RT = g(x1 ), we locate an extremum through:
d ln(1 /2 ) 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
where
A A12 x1 + A21 x2
Moreover,
dA = A12 A21 d x1
and
d 2A =0 2 d x1
Then,
d(G E/RT ) = A12 A21 d x1
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
= A12 A21
2x1 x2 2(x2 x1 ) d A 2 + 2 A3 d x1 A A
dA d x1
2
=
dA dA 2A12 A21 + x1 x2 A2 (x2 x1 )A 3 d x1 d x1 A
2
=
2A12 A21 A3
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(1 /2 ). The Margules equation is given by Eq. (12.9b), here written: GE = Ax1 x2 RT
where
A = A21 x1 + A12 x2
dA = A21 A12 d x1
d 2A =0 2 d x1
Then,
dA d(G E/RT ) = A(x2 x1 ) + x1 x2 d x1 d x1
d 2A dA dA d 2 (G E/RT ) + x1 x2 2 + (x2 x1 ) = 2A + (x2 x1 ) 2 d x1 d x1 d x1 d x1
= 2A + 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 2A12 3(A21 A12 )
or
x1 =
(r 2) 3(r 1)
r
A21 A12
692
When r = 2, x1 = 0, and the extrema in ln 1 and ln 2 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 2 curves at x1 = 1 are larger than the intercepts of the ln 1 curves at x1 = 0. When r = 1/2, x1 = 1, and the extrema in ln 1 and ln 2 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 1 curves at x1 = 0 are larger than the intercepts of the ln 2 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 1 =
d(G E/RT ) GE + x2 d x1 RT
and
ln 2 =
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
Then,
x1 x2 x2 x1 d(G E/RT ) 2 (A12 A21 ) = A12 A21 D D d x1
and ln 1 = 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
D A21 x2
2
= A12
A12 x1 + A21 x2 A21 x2
2
ln 1 = A12 1 +
A12 x1 A21 x2
2
The equation for ln 2 is derived in analogous fashion. (b) With the understanding that T and P are constant, ln 1 = (nG E/RT ) n 1
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
Differentiation in accord with the rst equation gives: ln 1 = A12 A21 n 2
n1 1 n D (n D)2
(n D) n 1
n2
ln 1 =
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
and
1 GE (equimolar) = (A12 + A21 ) 8 RT
But [see page 438, just below Eq. (12.10b)]:
A12 = ln 1
A21 = ln 2
1 GE (equimolar) = (ln 1 + ln 2 ) 8 RT
or
1 GE (equimolar) = ln(1 2 ) 8 RT
12.24 (a) By Eq. (12.6):
GE = x1 ln 1 + x2 ln 2 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 1 = 0.273 x2
and
2 ln 2 = 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 2 d ln 1 =0 + x2 d x1 d x1
whence
x1 d ln 1 x1 d ln 1 d ln 2 = = x2 d x2 x2 d x1 d x1
694
Integrate, recalling that ln 2 = 1 for x1 = 0: ln 2 = ln(1) +
2 (a) For ln 1 = Ax2 ,
x1 0
x1 d ln 1 d x1 = x2 d x2
x1 0
x1 d ln 1 d x1 x2 d x2
d ln 1 = 2Ax2 d x2
Whence
ln 2 = 2A
x1 0
x1 d x1
or
2 ln 2 = Ax1
By Eq. (12.6),
2 (b) For ln 1 = x2 (A + Bx2 ),
GE = Ax1 x2 RT
d ln 1 2 2 = 2x2 (A + Bx2 ) + x2 B = 2Ax2 + 3Bx2 = 2Ax2 + 3Bx2 (1 x1 ) d x2
Whence
2 ln 2 = Ax1 +
ln 2 = 2A 3B 2 3 x Bx1 2 1
x1 0
x1 d x1 + 3B
x1 0
x1 d x1 3B
x1 0
2 x1 d x1
or
2 ln 2 = x1 A +
B 3B 2 Bx1 = x1 A + (1 + 2x2 ) 2 2
Apply Eq. (12.6):
3B GE 2 2 Bx1 ) = x1 x2 (A + Bx2 ) + x2 x1 (A + 2 RT
Algebraic reduction can lead to various forms of this equation; e.g.,
B GE = x1 x2 A + (1 + x2 ) 2 RT
2 2 (c) For ln 1 = x2 (A + Bx2 + C x2 ),
d ln 1 2 2 2 3 = 2x2 (A + Bx2 + C x2 ) + x2 (B + 2C x2 ) = 2Ax2 + 3Bx2 + 4C x2 d x2 = 2Ax2 + 3Bx2 (1 x1 ) + 4C x2 (1 x1 )2
Whence or
ln 2 = 2A
x1 0
x1 d x1 + 3B
x1 0
x1 0
x1 (1 x1 )d x1 + 4C
x1 0
x1 0
x1 (1 x1 )2 d x1
x1 0 3 x1 d x1
ln 2 = (2A + 3B + 4C) ln 2 =
x1 d x1 (3B + 8C)
2 x1
2 x1 d x1 + 4C
2A + 3B + 4C 2
3B + 8C 3
3 4 x1 + C x1
2 ln 2 = x1 A +
8C 3B + 2C B + 3 2
2 x1 + C x1
695
or
2 ln 2 = 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
and H= H x1
H=
H (1 + n)
Eliminating 1 + n gives:
(A)
Differentiation yields:
H d x1 1 d H d H = 2 = x1 d n dn x1 d n
H 1 d H 2 x1 d x1 x1
d x1 dn
where
1 d x1 2 = x1 = (1 + n)2 dn
Whence,
d H = dn
H x1
dHE d H = H E x1 d x1 d x1
Comparison with Eq. (11.16) written with M H E ,
dH H2E = H E x1 d x1
E
shows that
d H = H2E dn
(b) By geometry, with reference to the following gure,
d H = dn
HI n
Combining this with the result of Part (a) gives:
H2E =
HI n
From which, Substitute: H=
I =
H n H2E
HE H = x1 x1
and
n=
x2 x1
696
Whence,
I =
H E x2 H2E x2 HE H2E = x1 x1 x1
However, by the summability equation, H E x2 H2E = x1 H1E Then,
I = H1E
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= n 1+n
A12 n A21 + 1+n 1+n
(a) It follows immediately from the preceding equation that:
n0
lim
H =0
(b) Because n/(1 + n) 1 for n , it follows that:
n
lim
H = A12
(c) Analogous to Eq. (12.10b), page 438, we write: Eliminate the mole fractions in favor of n: H2E = 1 1+n
2
2 H2E = x1 [A21 + 2(A12 A21 )x2 ]
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
n0
lim
d H = A21 dn
12.42 By Eq. (12.29) with M H ,
H=H
i
xi Hi . Differentiate:
H t
=
P,x
H t
P,x i
xi
Hi t
P,x
With
H t
CP,
P,x
this becomes
t t0
H t
= CP
P,x
i
xi C Pi =
t t0
CP
Therefore,
H H0
d( H ) =
C P dt
H=
H0 +
C P dt
697
12.61 (a) From the denition of M: Differentiate:
M E = x1 x2 M
(A)
dM dME = M(x2 x1 ) + x1 x2 d x1 d x1
(B)
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
N! N1 ! N2 ! N3 !
Combining the last three equations gives:
S t = k ln
N! N1 ! N2 ! N3 !
From which:
1 N! 1 St St S = (ln N ! ln = = = N N1 ! N2 ! N3 ! N kN R(N /N A ) R
ln Ni !)
i
ln N ! N ln N N
and
ln Ni ! Ni ln Ni Ni
1 S (N ln N N N R
Ni ln Ni +
i i
Ni ) =
1 (N ln N N
xi N ln xi N )
i
=
1 (N ln N N
xi N ln xi
i i
xi N ln N ) =
i
xi ln x1
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: dG E G 1 = G E + x2 d x1
E
and
dG E G 2 = G E x1 d x1
E
Divide through by RT ;
dene G
GE ; RT
note by Eq. (11.91) that
G iE = ln i RT
Then
ln 1 = G + x2
dG d x1
and
ln 2 = G x1
dG d x1
Given:
GE = A1/k x! x2 RT
with 698
A x 1 Ak + x 2 Ak 21 12
Whence:
G = x1 x2 A1/k
and
d A1/k dG + A1/k (x2 x1 ) = x1 x2 d x1 d x1
1 A1/k k dA 1 d A1/k (A21 Ak ) = = A(1/k)1 12 k A d x1 k d x1
and
A1/k k dG (A21 Ak )+A1/k (x2 x1 ) = x1 x2 12 kA d x1
Finally,
2 ln 1 = x2 A1/k
(Ak Ak )x1 21 12 +1 kA
Similarly,
2 ln 2 = x1 A1/k 1
(Ak Ak )x2 21 12 kA
(b) Appropriate substitition in the preceding equations of x1 = 1 and x1 = 0 yields: ln 1 = A1/k = (Ak )1/k = A12 12 (c) Let g ln 2 = A1/k = (Ak )1/k = A21 21
If k = 1, If k = 1,
GE = A1/k = (x1 Ak + x2 Ak )1/k 12 21 x1 x2 RT g = x1 A21 + x2 A12 (Margules equation)
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
Apply lH 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 , Whence For k +, Whence
k
lim (A12 /A21 )k 0 g = A21
and
k
lim ln g = ln A21
except at x1 = 0 where g = A12 and
k
k
lim (A12 /A21 )k g = A12
lim ln g = ln A12
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 e Pesat = xe e Pesat = ye P e EtOH
Because P is low, we have assumed ideal gases, and for small xe let e e . For volume fraction e in the vapor, the ideal-gas assumption provides ev ye , and for the liquid phase, with xe small
l e =
xe Vel xe Vel xe Vel Vb xb Vb xe Vel + xb Vb
b blood
Then
Vb l sat P ev P Ve e e e
Ve P volume % EtOH in blood Vb e Pesat volume % EtOH in gas
12.70 By Eq. (11.95),
HE = T RT
E
(G E /RT ) T
P,x
G = x1 ln(x1 + x2 RT
12 )
x2 ln(x2 + x1
21 )
(12.18)
(G E /RT ) T
E
x
d 21 d 12 x2 x1 dT dT = x2 + x1 21 x1 + x2 12
x1 x2
d 12 H dT = x1 x2 T x1 + x2 RT
ij
12
d 21 dT + x2 + x1
21
=
ai j Vj exp RT Vi
(i = j)
(12.24)
Vj d ij = Vi dT
exp
ai j RT
12 a12
ai j = RT 2
ij
ai j RT 2
H E = x1 x2
x1 + x2
+
12
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 (x2 + x1 21 )2 (x1 + x2 12 )2 R
Because
12
and
21
E must always be positive numbers, C P must always be positive.
700
Chapter 13 - Section B - Non-Numerical Solutions
13.1 (a) =
i
4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g) i = 4 5 + 4 + 6 = 1 n0 =
i0
=2+5=7
By Eq. (13.5), yNH3 = 2 4 7+
yO2 =
5 5 7+
yNO =
4 7+
yH2 O =
6 7+
(b) =
2H2 S(g) + 3O2 (g) 2H2 O(g) + 2SO2 (g) i = 2 3 + 2 + 2 = 1
i
n0 =
i0
=3+5=8
By Eq. (13.5), yH2 S = 3 2 8
yO2 =
5 3 8
yH2 O =
2 8
ySO2 =
2 8
(c) =
i
6NO2 (g) + 8NH3 (g) 7N2 (g) + 12H2 O(g) i = 6 8 + 7 + 12 = 5 n0 =
i0
=3+4+1=8
By Eq. (13.5), yNO2 = 3 6 8 + 5
yNH3 =
4 8 8 + 5
yN2 =
1 + 7 8 + 5
yH2 O =
12 8 + 5
13.2
C2 H4 (g) + 1 O2 (g) (CH2 )2 O(g) 2
(1) (2)
C2 H4 (g) + 3O2 (g) 2CO2 (g) + 2H2 O(g) The stoichiometric numbers i, j are as follows:
i=
C2 H4
O2
(CH2 )2 O
CO2
H2 O
j
j
1
1
1 2
1
0
0
1 2
2
1
3
0
2
2
0
n0 =
i0
=2+3=5
By Eq. (13.7), yC2 H4 = 2 1 2 5 1 1 2
yO2 =
3 1 1 32 2
5
1 2 1
y (CH2 )2
O
=
1 5 1 1 2
yCO2 =
22 5 1 1 2
yH2 O =
22 5 1 1 2
701
13.3
CO2 (g) + 3H2 (g) CH3 OH(g) + H2 O(g) CO2 (g) + H2 (g) CO(g) + H2 O(g) The stoichiometric numbers i, j are as follows:
(1) (2)
i=
CO2
H2
CH3 OH
CO
H2 O
j
j
1 2
1 1
3 1
1 0
0 1
1 1
2 0
n0 =
i0
=2+5+1=8
By Eq. (13.7), yCO2 = 2 1 2 8 21
yH2 =
5 31 2 8 21
yCH3 OH =
1 8 21
yCO =
1 + 2 8 21
yH2 O =
1 + 2 8 21
13.7 The equation for
G , appearing just above Eq. (13.18) is:
H0
G =
T ( H0 T0
G ) + R 0
T T0
CP dT RT R
T T0
C P dT R T
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: Ky = P P
K
(a) Differentiate this equation with respect to T and combine with Eq. (13.14): Ky T
=
P
P P
Ky H d ln K Ky d K dK = = Ky = RT 2 dT K dT dT
Substitute into the given equation for (e / T ) P :
e T
=
P
K y de RT 2 d K y
H
(b) The derivative of K y with respect to P is: Ky P
=
T
P P
1
1 K = K P
P P
P P
1
K y 1 = P P
702
Substitute into the given equation for (e / P)T :
e P
=
T
K y de () P d Ky
(c) With K y
i
(yi )i , ln K y =
i
i ln yi . Differentiation then yields: 1 d Ky = K y de
i
i dyi yi de
(A)
Because yi = n i /n,
1 n i dn 1 dn i dyi = 2 = n n de n de de
dn dn i yi de de
But Whence,
n i = n i0 + i e dn i = i de
and
n = n 0 + e
and
dn = de
Therefore,
i yi dyi = n 0 + e de
Substitution into Eq. (A) gives 1 d Ky K y de
=
i
i yi
i yi n 0 + e
m i=1
=
1 n 0 + e
m
i
i2 i yi
=
1 n 0 + e
i2 i yi
k
k=1
In this equation, both K y and n 0 + e (= n) are positive. It remains to show that the summation term is positive. If m = 2, this term becomes
2 (y2 1 y1 2 )2 2 1 1 (1 + 2 ) + 2 2 (1 + 2 ) = 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 i=1
i2 i yi
m
m
m k
k
k=1
=
i
(yk i yi k )2 yi yk
(i < k)
All quantities in the sum are of course positive. 13.9
1 N (g) 2 2
+ 3 H2 (g) NH3 (g) 2
For the given reaction, = 1, and for the given amounts of reactants, n 0 = 2. By Eq. (13.5), yN2 =
1 (1 2
e ) 2 e
yH2 =
3 (1 2
e ) 2 e
yNH3 =
e 2 e
By Eq. (13.28),
yNH3 1/2 3/2 yN2 yH2
=
[ 1 (1 2
P e (2 e ) =K 3 P e )]1/2 [ 2 (1 e )]3/2
703
Whence,
e (2 e ) = (1 e )2
1 2
1/2
3 2
3/2
K
P P = 1.299K P P
This may be written: where,
r e 2 2 r e + (r 1) = 0 r 1 + 1.299K P P
The roots of the quadratic are:
e = 1
1 = 1 r 1/2 r 1/2
Because e < 1, e = 1 r 1/2 ,
e = 1 1 + 1.299K
P P
1/2
13.10 The reactions are written: Mary: 2NH3 + 3NO 3H2 O + 5 N2 2
(A) (B)
Paul: Peter:
4NH3 + 6NO 6H2 O + 5N2 3H2 O + 5 N2 2NH3 + 3NO 2
(C)
Each applied Eqs. (13.11b) and (13.25), here written: ln K = G /RT For reaction (A), G = 3 G fH A and K = (P )
i
( fi )i
2O
2 G fNH 3 G fNO
3
For Marys reaction = 1 , and: 2
K A = (P )
1 2
ff3H
2O
5/2 ffN
2
ff2NH ff3NO
3
and
ln K A =
G A RT
For Pauls reaction = 1, and K B = (P )
1
ff6H
2O
ff5N
2
ff4NH ff6NO
3
and
ln K B =
2 G A RT
For Peters reaction = 1 , and: 2
K C = (P )
1 2
ff2NH ff3NO
3
ff3H
2O
5/2 ffN
2
and
ln K C =
G A RT
In each case the two equations are combined: Mary: (P )
1 2
ff3H
2O
5/2 ffN
2
ff2NH ff3NO
3
= exp
G A RT
704
Paul:
(P )
1
ff6H
2O
ff5N
2
ff4NH ff6NO
3
G A = exp RT
2
Taking the square root yields Marys equation. Peter: (P ) 2
1
ff2NH ff3NO
3
ff3H
2O
5/2 ffN
2
= exp
G A RT
1
Taking the reciprocal yields Marys equation. 13.24 Formation reactions:
1 N 2 2
+ 3 H2 NH3 2
(1)
1 N 2 2
+ 1 O2 NO 2
(2)
1 N 2 2
+ 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 = 2 + N r s = 21+520
F =4
13.35 (a) Equation (13.28) here becomes:
yB = yA
P P
0
K =K
Whence,
yB = K (T ) 1 yB
(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 the pressure dependence, which in the general case appears through the i 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
1 yA = K (T ) yA
705
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 (a) Application of Eq. (13.36) yields: and (1 xA )PBsat (T ) = (1 yA )P
F = 2 + N r = 22+21 = 1
(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
KI = KI
yPX = yOX
P P
0
K II = K II
yEB = yOX
P P
0
K III = K III
(b) These equation equations lead to the following set:
yMX = K I yOX
(1)
yPX = K II yOX
(2)
yEB = K III yOX
(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 =
1 1 + K I + K II + K III
(4)
(c) With the assumption that combine to give:
C P = 0 and therefore that K 2 = 1, Eqs. (13.20), (13.21), and (13.22)
K = K 0 K 1 = exp
G 298 exp RT0
H298 1
H298 T0 1 T RT0
Whence,
K = exp
298.15 500 (8.314)(298.15)
G 298
The data provided lead to the following property changes of reaction and equilibrium constants at 500 K: Reaction
H298
G 298
K
I II III
1,750 1,040 10,920
3,300 1,000 8,690
2.8470 1.2637 0.1778
706
(d) Substitution of numerical values into Eqs. (1), (2), (3), and (4) yields the following values for the mole fractions:
yOX = 0.1891
yMX = 0.5383
yPX = 0.2390
yEB = 0.0336
13.40 For the given owrates,
n A0 = 10
and n B0 = 15, nA nB nC nD n
with n A0 the limiting reactant without (II)
= n A0 I II = n B0 I = I II = II = n 0 I II
Use given values of YC and SC/D to nd I and II : YC = I II n A0
and
SC/D =
I II II
Solve for I and II : I = SC/D + 1 n A0 YC = SC/D
2+1 2
10 0.40 = 6
II =
10 0.40 n A0 YC =2 = 2 SC/D
nA nB nC nD n
= 10 6 2 = 15 6 =62 =2 = 17
=2 =9 =4 =2
yA yB yC yD
= 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
i
i G i G P
and from Eq. (6.10), (G i / P)T = Vi
=
T i
i
G i P
=
T
i
i Vi
For the ideal-gas standard state, G P
Vi
= RT /P . Therefore
=
T i
i
RT P
=
RT P
and
G (P2 )
G (P1 ) = RT ln
P2 P1
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
xA =
1 Kl + 1
707
The preceding equation now becomes, P = 1
Kl
1 +1
PBsat +
Kl
1 +1
PAsat
P=
Kl Kl + 1
PBsat +
Kl
1 +1
PAsat
(A) P = PBsat
For K l = 0 (b) Given Raoults law:
P = PAsat
For K l =
1 = x A + x B = yA
P P =P sat + y B PBsat PA
yB yA sat + PBsat PA
P=
y A /PAsat
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=
(K v + 1)PAsat PBsat K v PAsat + PBsat
(B) P = PBsat
For K v = 0
P = PAsat
For K v =
(c) Equations (A) and (B) must yield the same P. Therefore Kl Kl + 1
PBsat +
1 l +1 K
PAsat =
(K v + 1)PAsat PBsat K v PAsat + PBsat
Some algebra reduces this to:
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 r = 2 + 2 2 1 = 1 Thus xing T should sufce.
708
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: ln i =
ln Z (nG R/RT ) (n Z ) +1 +n n i n i n i
where the partial derivatives written here and in the following development without subscripts are understood to be at constant T , n/ (or /n), and n j . Equation (6.61) after multiplication by n can be written: 2 3 nG R n ln Z + n 2 (nC) = 2n(n B) n 2 n RT Differentiate:
3 (nG R/RT ) (n B + n Bi ) + =2 2 n n n i
2
(2n 2 C + n 2 Ci ) n
ln Z ln Z n i
or
ln Z 3 (nG R/RT ) ln Z = 2(B + Bi ) + 2 (2C + Ci ) n n i 2 n i
By denition,
Bi
(n B) n i
and
T,n j
Ci
(nC) n i
T,n j
The equation of state, Eq. (3.40), can be written: Z = 1 + B + C 2 or n Z = n + n(n B)
2
+ n 2 (nC) n n
2
Differentiate:
(n Z ) (n B + n Bi ) + =1+ n n n i
(2n 2 C + n 2 Ci )
or
(n Z ) = 1 + (B + Bi ) + 2 (2C + Ci ) n i
When combined with the two underlined equations, the initial equation reduces to:
ln i = 1 + (B + Bi ) + 1 2 (2C + Ci ) 2
The two mixing rules are:
2 2 B = y1 B11 + 2y1 y2 B12 + y2 B22 3 2 2 3 C = y1 C111 + 3y1 y2 C112 + 3y1 y2 C122 + y2 C222
Application of the denitions of Bi and Ci to these mixing rules yields:
2 2 B1 = y1 (2 y1 )B11 + 2y2 B12 y2 B22 2 2 2 3 C1 = y1 (3 2y1 )C111 + 6y1 y2 C112 + 3y2 (1 2y1 )C122 2y2 C222 2 2 B2 = y1 B11 + 2y1 B12 + y2 (2 y2 )B22
3 2 2 2 C2 = 2y1 C111 + 3y1 (1 2y2 )C112 + 6y1 y2 C122 + 2y2 (3 2y2 )C222
709
In combination with the mixing rules, these give: B + B1 = 2(y1 B11 + y2 B12 ) 2 2 2C + C1 = 3(y1 C111 + 2y1 y2 C112 + y2 C122 ) B + B2 = 2(y2 B22 + y1 B12 )
2 2 2C + C2 = 3(y2 C222 + 2y1 y2 C122 + y1 C112 )
In combination with the boxed equation these expressions along with Eq. (3.40) allow calculation of ln 1 and ln 2 . 14.11 For the case described, Eqs. (14.1) and (14.2) combine to give: yi P = xi Pi sat isat i
If the vapor phase is assumed an ideal solution, i = i , and
yi P = xi Pi sat
isat i
When Eq. (3.38) is valid, the fugacity coefcient of pure species i is given by Eq. (11.36): ln i = Bii P RT
and
isat =
Bii Pi sat RT
Therefore,
ln
Bii (Pi sat P) Bii P Bii Pi sat isat = = ln isat ln i = RT RT RT i
For small values of the nal term, this becomes approximately:
Bii (Pi sat P) isat =1+ RT i
Whence,
yi P = xi Pi sat 1 +
Bii (Pi sat P) RT
or
yi P xi Pi sat =
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 i , Whence, ln i = ln fi ln xi ln f i
1 1 d fi 1 d ln fi d ln i = = i d xi xi xi d xi d xi f
Combination of this expression with Eq. (14.71) yields:
1 d fi >0 fi d xi
Because fi 0,
d fi >0 d xi
(const T, P)
By Eq. (11.46), the denition of fi ,
RT d fi d ln fi di = = RT d xi d xi fi d xi
Combination with Eq. (14.72) yields:
di >0 d xi
(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
i
G = 0, in which
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
i
1 ln N = RT ln N N
For the special case of a binary solution, N = 2, and
G E (max) = RT ln 2
711
14.17 According to Pb. 11.35,
G E = 12 P y1 y2
or
12 P GE y1 y2 = RT RT
This equation has the form:
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: 12 P/RT > 2. Suppose T = 300 K and P = 5 bar. The preceding condition then requires: 12 > 9977 cm3 mol1 for vapor-phase immiscibility. Such large positive values for 12 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.) 14.19 Consider a quadratic mixture, described by: GE = Ax1 x2 RT
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: 1 1 G E > 2 RT = 0.5RT 2 2 This is a model-dependent result. Many liquid mixtures are known which are stable as single phases, even though G E > 0.5RT 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 n 1 n 1
ln(1 )m = C ln 1
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 1 )m ln x1 = C ln(x1 1 ) C ln x1 or Differentiate: ln(x1 1 )m = C ln(x1 1 ) (C 1) ln x1
d ln(x1 1 ) C 1 d ln(x1 1 )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 1 )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):
Hilv 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 Hilv P sat dT ln i = 2 P Tbi RT
Combination with Eq. (10.1) yields: yi = xi exp
T Tbi
Hilv 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: xi i = xi i (i = 1, 2)
If phase is pure species 1 and phase is pure species 2, then x1 = 1 = 1 and x2 = 2 = 1.
Hence,
x1 1 = x1 1 = 1
and
x2 2 = x2 2 = 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 = 1 = exp
sl H1 RTm 1
T Tm 1 T
(a) Differentiate:
sl H1 d x1 = 1 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: xA = exp xB
H sl (Tm B Tm A ) RTm A Tm B
sl HB
where by assumption,
sl HA =
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
sl HA < sl H B , in
Notice that Tm > T (see Fig. 14.21b). Then x A /x B > 1 if and only if accord with the rule of thumb.
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
F1 P
+
y1
F1 y1
P
dy1 dP
=
y1 + y1 P
ln F1 P
+
y1
ln F1 y1
P
dy1 dP
Whence,
dy1 = dP
y1
ln F1 P
y1
1 P
1 y1
ln F1 y1
(A)
P
This is a general result. An expression for F1 is given by Eq. (14.99): F1
sat V s (P P1sat ) 1 exp 1 RT 1
From this, after some reduction: ln F1 P
=
y1
ln 1 P
+
y1
V1s RT
and
ln F1 y1
=
P
ln 1 y1
P
Whence, by Eq. (A),
dy1 = dP
ln 1 y1 P
y1
1 + y1
ln 1 y1
1 V1s + P RT
P
(B)
714
This too is a general result. If the two-term virial equation in pressure applies, then ln 1 is given by Eq. (11.63a), from which: ln 1 P
=
y1
1 2 (B11 + y2 12 ) RT
and
ln 1 y1
=
P
2y2 12 P RT
Whence, by Eq. (B),
dy1 = dP
y1
2 1 V1s B11 y2 12 P RT 2y1 y2 12 P 1 RT
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 lH pitals rule gives: o n=
s0
lim n = lim
m s0 2
Pes Pes + c + Pes c + Pes
=
m 2
P P + c+P c+P
or
lims0 n =
mP c+P
which is the Langmuir isotherm. (b) Henrys constant, by denition: k lim dn dP
P0
Differentiate Eq. (A):
m dn = 2s dP
es es c + Pes c + Pes
Whence,
k=
m 2s
es es c c
=
m cs
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
P0
m dn sinh s = cs dP
715
m d 2n = 2 sinh 2s P0 d P 2 c s 2m d 3n = 3 sinh 3s lim 3 P0 d P c s
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
With z = 1 + Bn + Cn + , this becomes: 3 ln(P/n) = ln k + 2Bn + Cn 2 + 2 Thus a plot of ln(P/n) vs. n produces ln k as the intercept and 2B as the limiting slope (for n 0). Alternatively, a polynomial curve t of ln(P/n) in n yields ln k and 2B as the rst two coefcients.
2
14.37 For species i in a real-gas mixture, Eqs. (11.46) and (11.52) give: i = At constant temperature,
g g g i (T )
+ RT ln yi i P
di = RT d ln yi i P
With di = di , Eq. (14.105) then becomes: a d + d ln P + xi d ln yi i = 0 RT i
(const T )
For pure-gas adsorption, this simplies to: a d = d ln P + d ln (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 = (Z 1)
n = k P exp
P 0
(Z 1)
dP exp P
n 0
(1 z)
dn +1z n
This equation is the real-gas analog of Eq. (14.109). 716
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
(A)
For bn sufciently small,
exp
bn 1 bn
1
bn 1 bn
Whence, by Eq. (A),
n k P(1 2bn)
or
n
kP 1 + 2bk P
which is the Langmuir isotherm. With z = 1 + n, the adsorption isotherm is: n = k P exp(2n) from which, for n sufciently small, the Langmuir isotherm is again recovered. 14.41 By Eq. (14.107) with a = A/n,
dP Ad =n P RT
The denition of and its derivative are: A RT
and
d =
Ad RT
Whence,
d = n
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, d
i
xi = P d
i
yi + Pi
i
yi dP Pi
(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), yi = Pi
i
xi
i
P
=
1 P
With P xed, Eq. (C) can now be written in the simple but approximate form: d
i
xi =
dP P
Equation (A) then becomes: d = n d
i
xi
or
= n
i
xi
i
where we have replaced differentials by deviations. The deviation in value must be unity. Therefore, yi 1 xi = P Pi i i
xi is known, since the true
717
By Eq. (14.132),
n=
i
1
(xi /n i )
Combine the three preceding equations: P =
i
i
yi 1 Pi
(xi /n i )
When xi = yi P/Pi , the Raoults law analogy, is substituted the required equation is reproduced:
P =
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: ln 1 = A12 n 2
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
Introduce solute-free mole fractions: x2 x2 = x2 1 x1 x2 + x3
and
x3 =
x3 1 x1
Whence, For x1 0,
ln 1 = A12 x2 (1 x1 )2 + A13 x3 (1 x1 )2 A23 x2 x3 (1 x1 )2 ln 1 = 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 1,2 = A12
Also
ln 1,3 = A13
Whence,
ln 1 = x2 ln 1,2 + x3 ln 1,3 A23 x2 x3
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 1 Whence, or
ln H1,2 = ln f 1 + ln 1,2 ln H1,3 = ln f 1 + ln 1,3
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 to , probably one on either side of the minimum in curve II. 14.44 By Eq. (14.136) with V2 = V2 : V2 = ln(x2 2 ) RT
Represent ln 2 by a Taylor series: ln 2 = ln 2 |x1 =0 + d ln 2 d x1
x1 =0
x1 +
1 d 2 ln 2 2 2 d x1
x1 =0
2 x1 +
But at x1 = 0 (x2 = 1), both ln 2 and its rst derivative are zero. Therefore, ln 2 = 1 2
d 2 ln 2 2 d x1
x1 =0
2 x1 +
Also,
ln x2 = ln(1 x1 ) = x1
2 x4 x3 x1 1 1 4 3 2
Therefore,
ln(x2 2 ) = + ln x2 + ln 2 = x1
1 2
1
1 2
d 2 ln 2 2 d x1
x1 =0
2 x1
and
1 1 V2 1 =1+ 2 2 x1 RT
d 2 ln 2 2 d x1
x1 +
x1 =0
Comparison with the given equation shows that:
B=
1 1 1 2 2
d 2 ln 2 2 d x1
x1 =0
14.47 Equation (11.95) applies:
E
(G E/RT ) T
=
P,x
HE RT 2
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), yi i P = xi i Pi sat isat
Ki
i Pi sat isat yi = P xi i
12
1 P1sat 1sat 2 K1 sat = 2 P2sat 1 2 K2
(b)
12 (x1 = 0) =
P1sat 1 (P sat ) 1 P1sat 1 (P1sat ) 2 (P2sat ) = 1 sat 1 sat sat sat P2 P2sat 1 (P2 ) 1 (P2 ) 2 (P2 )
12 (x1 = 1) =
(P sat ) P sat 1 (P1sat ) 2 (P1sat ) P1sat = 1 sat 2 1 sat sat sat 2 (P2sat ) 2 P2 1 (P1 ) 2 (P2 ) 2 P2
The nal fractions represent corrections to modied Raoults law for vapor nonidealities. 719
(c) If the vapor phase is an ideal solution of gases, then i = i for all compositions. 14.49 Equation (11.98) applies: ln i T
=
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 i (x, T ) = i R i (x, Tk )
T Tk
HE dT = i R T2
1 1 Tk T
=
HiE RT
T 1 Tk
i (x, T ) = i (x, Tk ) exp
HiE RT
T 1 Tk
14.52 (a) From Table 11.1, p. 415, nd:
G E T
= S E = 0
P,x
and
G E is independent of T .
Therefore
FR (x) GE = RT RT
(b) By Eq. (11.95),
(G E /RT ) T
=
P,x
HE =0 RT 2
GE = FA (x) RT
(c) For solutions exhibiting LLE, G E /RT is generally positive and large. Thus and 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: =2 RTU
TU =
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 = N A d 3 1 + (K 3 1) 1 e/kT (l 3 K 3 ) e /kT 1 3
From this expression,
1 dB (K 3 1) e/kT + (l 3 K 3 ) e = kT 2 dT
/kT
according to which d B/dT = 0 for T
and also for an intermediate temperature Tm :
Tm =
k ln
+ 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)
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
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
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 (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, i j 2 Bi j
1 2 E E E
, HE
, and S E
.
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) i j > 0. If unlike interactions are stronger than like interactions, |Bi j | > 1 |Bii + B j j | 2
Hence i j < 0. For identical interactions of all molecular pairs, Bi j = Bii = B j j , and i j = 0 The rationalizations of signs for H E of binary liquid mixtures presented in Sec. 16.7 apply approximately to the signs of 12 for binary gas mixtures. Thus, positive 12 is the norm for NP/NP, NA/NP, and AS/NP mixtures, whereas 12 is usually negative for NA/NA mixtures comprising solvating species. One expects 12 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 i than would otherwise obtain, the values of Hi are in the observed order. 16.9 By Eq. (6.70), H = T S . For the same temperaature and pressure, less structure or order means larger S. Consequently, S sl , S lv , and S sv are all positive, and so therefore are H sl , H lv , and H sv . 16.11 At the normal boiling point: Therefore H R,l = H R,v H lv H v H l = (H v H ig ) (H l H ig ) = H R,v H R,l H lv
At 1(atm), H R,v should be negligible relative to H lv . Then H R,l H lv . Because the normal boiling point is a representative T for typical liquid behavior, and because H R reects intermolecular forces, H lv has the stated feature. H lv (H2 O) is much larger than H lv (CH4 ) because of the strong hydrogen bonding in liquid water.
R,l R,l 16.12 By denition, write C lP = C P +C P , where C P is the residual heat capacity for the liquid phase. ig R,l Also by denition, C P = ( H R,l / T ) P . By assumption (modest pressure levels) C P C v . P ig
Thus,
R,l
C lP C v + P
H R,l T
P
For liquids, H is highly negative, becoming less so as T increases, owing to diminution of interR,l molecular forces (see, e.g., Fig. 6.5 or Tables E.5 and E.6). Thus C P 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: RT 1/3 V t 1/3 = NA P N
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.
1/3
723
...

View Full Document