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Chapter 1
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h
1
2
2.1
d C ˜V
dM =
=M IN -M OUT
dt
dt
a) at steady state
g
M IN =1
M OUT C ˜ volumetric flow rate out
min
g
10m3
1 g
g
1
0.1 3
Cu
C
min
min
10 m3
m
dC
dC
m3
m3
V
M C u 10
V
M C 10
b)
dt
min
min
dt
3
V 6 u 6 u 3 108 m
V 6 u 6 u 3 108 m3
3
dC
g
10m
Cu
108m3
1
min
min
dt
dC
g
1
10 1
C
m
3
dt 108 m min
108
or for C in g 3
m
dC
dC
108
10C 1
10.8
dt
dt
C 0.1
10.8 ln
C-0.1
-0.1
t
§ -0.005 ·
10.8 ln ¨
¸
© -0.1 ¹
t = 32.35 minutes
For C = 0.95 * 0.1 = 0.095
108 ln 20 = t
2.2
dC
dt
kC
If Cf = 0.5 Ci
ln
Cf
Ci
kt
ln 0.5 = -kt or k
ln 2
t
for a half life of seven years
k = 0.099 years-1
so that
C
ln f k 25 0.099 u 25 2.475
22
Cf 22 exp (-2.475) 1.85 ppm
2.3
Basis: 2-liter vessel = 2000 cm3 = 2×103 ×10-6
Initial moles of phosphine = 5
m3
= 2 × 10-3 m3
cm3
kmol
mol
×103
×2×10-5 m3 = 10 mol
kmol
m3
t
2
a)
Mass balance from stoichiometry
Species
Initial
Final
PH3
10
10 – 4X = 7
P4
0
X
H2
0
6X
Since 10 - 4X = 7
3 = 4X
X=¾
Ÿ P4 = 0.75 moles
H2 = 6 moles
10-4X moles
moles
b)
CPH3 at any time =
= 5-2X ×10-3
2×10-3 m3
m3
Therefore
dCPH3
=3.715×10-6 CPH3
dt
dX
-2×10-3
=-3.715×10-6 5-2X ×10-3
dt
dX
=3.715×10-6 2.5-X
dt
dX
d(2.5-X )
=
=-3.715×10-6 dt
2.5 - X
2.5-X
ln 2.5-X t -ln 2.5-X 0 =-3.715×10-6 t
2.5-X -2.715u10 t
=e
2.5
-6
2.5-X =2.5 e 2.715u10 t
-6
-6
X =2.5 1-e-2.715×10 t
Therefore
Moles PH 3 =10-4X =10-10 1-e-2.715u10 t = 10 e-2.715u10 t
-6
-6
Moles PH 4 =2.5 1-e-2.715u10 t
-6
Moles H 2 = 15 1- e-2.715u10 t
-6
Time required for 3 moles to react = time for X = 0.75
-6
X =2.5 1-e-2.715×10 t
= 0.75
T = 1.314 × 105 sec = 36.49 hours
2.4
2 NO + O2 ĺ122
a)
0.5 moles NO produces 0.5 moles NO2
b)
Initial
Species
Conc
NO
O2
NO2
1
3
0
dCNO
2
CO2
=-1.4×10-4 CNO
dt
d 1- 2 X
dX
2
=-1.4×10-4 1-2X 3-X =-2
dt
dt
Conc. at
X
1–2X
3–X
6X
4–X
3
Ÿ
dX
2
=0.7×10-4 1-2X 3-X
dt
2X-1
1
1
0.7 × 10-4 t =
+
ln
5 1-2X 25
X-3
Now for 0.5 mole of NO to react Ÿ 1 – 2X = 0.5 X = 0.25
1
1
0.7×10-4 t =
+
ln 0.5/2.75 =0.332
5×0.5 25
0.332
t=
×104 =0.474×104 sec = 1.317 hrs
0.7
3
3.1
(a) By an energy balance, the bicycle stops when final potential energy equals
initial kinetic energy. Therefore 1 2
mvi
2
vi2
mgh f or h f
F 20 km u 1000 m u 1 hr I
H hr
km 3600 sec K
2g
2 u 9.807
2
m
sec2
or h=1.57 m.
(b) The energy balance now is
1 2 1 2
mv f
mvi mghi or v 2f vi2 2 ghi
2
2
3600 sec 2
km 2
m
km
v 2f
20
2 u 9.807
u
u
u
70
m
1000 m
hr
hr
sec2
v f = 134.88 km/hr. Anyone who has bicycled realizes that this number is much
F
H
I
K
F
H
I
K
too high, which demonstrates the importance of air and wind resistance.
3.2
The velocity change due to the 55 m fall is
3600 sec I
km
u
c'v h 2 u 9.807 secm u 55 m u FH 1000
m
hr K
2
2
2
v f = 118.24 km/hr. Now this velocity component is in the vertical direction. The
initial velocity of 8 km/hr was obviously in the horizontal direction. So the final
velocity is
km
v
vx2 v 2y 11851
.
hr
3.3
(a) System: contents of the piston and cylinder
(closed isobaric = constant pressure)
M.B.: M 2 M1 'M 0 Ÿ M 2 M1 M
0
0
E.B.: M 2U 2 M1U1 'M H
Q Ws PdV
z
d i
c
h Q z PdV Q Pz dV Q PaV V f
M cU U h Q PM cV V h
Q M cU U h M c PV PV h M cU PV h cU PV h
M c H H h
M U 2 U1
2
2
1
2
2
1
2
1
2
1
1
1
2
2
1
1
Solutions to Chemical and Engineering Thermodynamics, 5h ed
P
1013
.
bar | 01
. MPa
V
1.6958
1.9364
T 100
T 150
Linear interpolation
1.8161
T 125q C
Final state
P 01
. MPa , V2
3.565
T 500q C
4.028
T 600q C
Linear interpolation
W
U
2506.7
2582.8
Chapter 3
H
2676.2
2776.4
2544.8
2726.3
3
m /kg
.
36322
3488.1
3704.7
Initial state
T2 500
3565
36322
.
.
T2 514.5q C
4.028 3565
.
600 500
514.5 500
.
H 2 34881
H 2 3519.5
600 500
3704.7 34881
.
Q 1 kg 3519.5 2726.3 kJ kg 7932
. kJ
z
PdV
a
1 bar u V2 V1
1 bar u 100,000
18161
m kg
.
.
f 1 bar u 36322
3
Pa
1 kg
1J
u
u
u 18161
m3 kg
.
bar m ˜ s2 ˜ Pa m2 ˜ s2 ˜ kg
1816
. kJ kg
(b) System is closed and constant volume
M.B.: M 2 M1 M
E.B.: M 2U 2 M1U1
Q
c
M U 2 U1
d i 0 Q W 0 – z PdV 0
'M H
h
s
Here final state is P 2 u 1013
.
bar ~ 0.2 MPa ; V2
(since piston-cylinder volume is fixed)
.
P 0.2 MPa ; V2 18161
T(q C)
V
U
500
1.7814
3130.8
600
2.013
3301.4
17814
.
.
18161
.
2.013 17814
T 500
0.0347
600 500 0.2316
T 515q C
V1
18161
m3 kg
.
.
~ 015
.
01498
U 2 31308
.
.
U 2 3156.4 kJ kg
01498
. 31308
.
33014
Q 1 kg u 3156.4 2544.8 kJ kg 6116
. kJ
(c) Steam as an ideal gas—constant pressure
N
PV
PV
Ÿ 1 1
RT
RT1
P2V 2
but V 2
RT2
2V 1 ; P1
P2
Solutions to Chemical and Engineering Thermodynamics, 5h ed
P12V 1
Ÿ T2
T2
PV
1 1
T1
T1
T2
27315
. 125 39815
. K
2 u T1 796.3 K 52315
. qC
760.9 kJ
z
PdV
W
2 u T1
1000 g kg
1 kJ
u 34.4 J mol K u 796.3 39815
. Ku
18 g mol
1000 J
N' H
Q
Chapter 3
P'V
P
FG NRT NRT IJ NRaT T f
H P P K
2
1
2
1
1
1000
.
u 8.314 u 39815
18
. kJ
1839
(d) Ideal gas - constant volume
PV
1 1
RT1
So again
2 P1 ˜V1
; T2
T2
PV
1 1
T1
Q
P2V 2
here V 1 V 2 ; P2
RT2
N'U
2T1
2 P1
796.3 K .
1
1000 g kg
. u
u 34.4 8.314 u 796.3 39815
1000
18 g mol
CP R ; Q
CV
577.0 kJ
3.4
M wU w, f M wU w, i
Mw
M weight
a
1 kg u CP Tf Ti
3.5
M weight u g u 1 m
1 kg
f 1 kg u 9.807 m s u 1 m u m 1kgJ s
1 kg u 4184
.
J g Ku
'T
Ws
2
2
1000 g
u 'T
kg
9.807
K
4184
.
u 1000
2
9.807 J
9.807
2.344 u 103 K
a
f a
f
From Illustration 3.2-3 we have that H T1, P1 H T2 , P2 for a Joule-Thomson
expansion. On the Mollier diagram for steam, Fig. 3.3-1a, the upstream and
downstream conditions are connected by a horizontal line. Thus, graphically,
we find that T ~ 383 K . (Alternatively, one could also use the Steam Tables
of Appendix III.)
Solutions to Chemical and Engineering Thermodynamics, 5h ed
Chapter 3
For the ideal gas, enthalpy is a function of temperature only.
Thus,
becomes
which
implies
that
H T1, P1 H T2 , P2
H T1 H T2 ,
T1 T2 600q C .
a
3.6
f a
af a f
f
System: Contents of Drum (open system)
mass balance: M t2 M t1 'M
energy balance:
MU MU
t2
z
'MH in Q Ws PdV
t1
steam
but Q
0 by problem statement, Ws
and
PdV
P'V
80q C
.
u 103 m3 kg ). Also from the Steam Tables
1029
V T
z
H in
0
(Note V T
is negligible.
H T
300q C, P
. bar
30
.
u 103 m3 kg ,
1003
25q C
300 kPa
3069.3 kJ kg
and recognizing that the internal energy of a liquid does not depend on pressure
gives
U
t1
U T
25q C, 1.013 bar
U sat., T
25q C
104.88 kJ kg
U T
80q C, 1.013 bar
U sat., T
80q C
334.86 kJ kg
and
U
t2
Now using mass balance and energy balances with M t1
100 kg yields
Solutions to Chemical and Engineering Thermodynamics, 5h ed
M t2 u 334.86 kJ 100 u 104.88 kJ
Chapter 3
M t2 100 u 3069.3 kJ
Thus
M t2 3069.3 334.86
M t2
3.7
108.41 kg , and 'M
100 u 3069.3 104.88
M t2 M t1
8.41 kg of steam added.
(a) Consider a change from a given state 1 to a given state 2 in a closed system.
Since initial and final states are fixed, U1 , U 2 , V1 , V2 , P1 , P2 , etc. are all
fixed. The energy balance for the closed system is
U 2 U1
where W
z
Ws PdV
adiabatic. Thus, U 2 U1
z
Q Ws PdV
total work. Also, Q
Q W
0 since the change of state is
W.
Since U1 and U 2 are fixed (that is, the end states are fixed regardless of the
path), it follows that W is the same for all adiabatic paths. This is not in
contradiction with Illustration 3.5-6, which established that the sum Q W is
the same for all paths. If we consider only the subset of paths for which
Q 0 , it follows, from that illustration that W must be path independent.
(b) Consider two different adiabatic paths between the given initial and final
states, and let W * and W ** be the work obtained along each of these paths,
i.e.,
Path 1: U 2 U1 W * ; Path 2: U 2 U1 W **
Now suppose a cycle is constructed in which path 1 is followed from the
initial to the final state, and path 2, in reverse, from the final state (state 2)
back to state 1. The energy balance for this cycle is
U 2 U1 W *
U 2 U1 W **
0 W * W **
a
f
Thus if the work along the two paths is different, i.e., W * z W ** , we have
created energy!
3.8
System contents of tank at any time
mass balance: M 2 M1 'M
energy balance: MU MU
'MH
c h c h
2
1
(a) Tank is initially evacuated Ÿ M1
in
0
Solutions to Chemical and Engineering Thermodynamics, 5h ed
Chapter 3
H in H 5 bar, 370q C 3209.6 kJ kg
interpolation).
Then
U P 5 bar, T ? 3209.6 kJ kg .
2
interpolation, using the Steam Tables (Appendix A.III) T 548q C
Thus
U 2
U
'M , and
M2
V P
V V
Therefore M
5 bar, T
548q C # 0.756 m3 kg
c
.
kg .
h 13228
1 m3 0.756 m3 kg
(by
By
(b) Tank is initially filled with steam at 1 bar and 150qC
and
Ÿ V1 V P 1 bar, T 150q C 194
. m3 kg
U1 2583 kJ kg ,
M V V 1 V 05155
.
kg . Thus, M
05155
.
'M kg . Energy balance
1
2
is M 2U 2 05155
.
u 2583
T2 , using T2 and P2
M 05155
.
u 3209.6 . Solve by guessing value of
5 bar to find V and U in the Steam Tables
2
2
1 m3 V2 are satisfied. By
trial and error: T2 ~ 425q C and M 2 # 1563
.
kg of which 1.323 kg was
present in tank intially. Thus, 'M M 2 M1 0.24 kg .
(Appendix A.III). See if energy balance and M 2
3.9
a) Use kinetic energy = mv2/2 to find velocity.
1 kg u
v 2 m2
2 sec2
1000 J = 1000
kg
so v= 44.72 m/sec
m2 sec2
b) Heat supplied = specific heat capacity u temperature change, so
1 mol
J
u 2510
.
u 'T 1000 J so 'T=2.225 K.
1000g u
55.85g
mol ˜ K
3.10 System resistor
Energy balance: dU dt Ws Q
where W E ˜ I , and since we are interested only in steady state dU dt
s
Thus
Q
Ws
1 amp u 10 volts
and 1 watt 1 volt u 1 amp 1 J s .
10 watt u 1 J s ˜ watt
25q C
ŸT
0.2 J s ˜ K
3.11 System
0.2 u T 25q C J s
. qC
750
gas contained in piston and cylinder (closed)
0
Energy balance: U t2 U t1 Q Ws PdV
(a) V
constant,
z
z
PdV
0.
0 , Q U t2 U t1
d
N U t2 U t1
i NC aT T f
V
2
1
From ideal gas law
N
Thus
PV
RT
114,367 Pa u 0120
m3
.
3
8.314 Pa ˜ m mol ˜ K u 298 K
5539
mol (see note following)
.
Solutions to Chemical and Engineering Thermodynamics, 5h ed
Q
NCV
298 K 298 630
.
3610
. K
T1 T2
Chapter 3
10,500 J
5.539 mol u 30.1 J mol ˜ K
Since N and V are fixed, we have, from the ideal gas law, that
P2
P1
3610
.
u 114.367 kPa
298.0
T2
P1
T1
T2
or P2
T1
1385
u 105 Pa
.
114367
.
u 105 Pa ;
301
. 8.314 38.414 J mol ˜ K
(b) P constant
CP CV R
Energy balance U t2 U t1
a
Q P'V , since P
constant
f Q PaV V f Q N aRT RT f
Ÿ Q NC aT T f
Ÿ NCV T2 T1
P
2
T2
2
1
2
1
1
T1 Q
NCP
298 10,500
5539
u 38.414
.
347.35 K
and
NR'T
P
'V
5539
.
mol u 8.314 Pa ˜ m3 mol ˜ K u 49.35 K
114,367 Pa
012
. 0.0199
V
Note: The initial pressure P
Patm
1013
.
bar
0.01987 m3
01399
.
m3
Patm Pwt of piston
1013
.
u 105 kPa
200 kg 1 Ns2
u
u 9.8 m s2
2
kg
m
˜
0.15 m
kPa
13067
.
Thus, intial pressure 114.367 kPa .
Pwt piston
13,067 N m2
3.12 System contents of storage tank (open system)
Mass balance: M 2 M1 'M
Energy balance:
'M H
MU MU
c h c h
2
1
in
since Q
13,067 Pa
W
0 and steam
entering is of constant properties.
Initially system contains 0.02 m3 of liquid water and 40 0.02 39.98 m3 of
steam.
Since vapor and liquid are in equilibrium at 50qC, from Steam Tables,
Also from the Steam Tables V L 0.001012 m3 kg ,
P 12.349 Pa .
V V 12.03 m3 kg ,
H V 25921
. kJ kg ,
H L 209.33 kJ kg ,
U L
209.32 kJ kg , and U V
24435
. kJ kg .
Solutions to Chemical and Engineering Thermodynamics, 5h ed
0.02 m3
0.001012 m3 kg
M1L
Chapter 3
U|
|V M
||
W
19.76 kg;
M1L M1V
. kg .
2308
39.98 m
. kg;
332
12.03 m3 kg
. u 24435
. 12,248.6 kJ
U1 19.76 u 209.32 332
1
3
M1V
Also
H in
. 010
. u 419.04
0.90 u 26761
2450.4 kJ kg
Possibilities for final state: 1) vapor-liquid mixture, 2) all vapor, and 3) all liquid.
First possibility is most likely, so we will assume V-L mixture. Since
P 1013
.
bar , T must be 100qC. Thus we can find properties of saturated vapor
and saturated liquid in the Steam Tables:
V L 0.001044 m3 kg ,
and
U L 418.94 kJ / kg ,
V V 16729
.
m3 kg ,
H V 26761
. kJ kg ,
U V 25065
. kJ / kg.
1 x 0.001044
V x 16729
.
2
0.001044 16719
.
x m3 kg , where
x quality
U 2 x 25065
. 1 x 418.94 418.94 2087.56 x kJ kg
Substituting into energy balance
M 2 418.94 2087.56 x 12,248.6
. f ˜ 2450.4
a M 2308
2
where
M2
V
V2
40 m3
0.001044 16719
.
x
(quality), M 2 46.36 kg , and
Solving by trial and error yields x 05154
.
'M 2328
. kg . Also the final state is a vapor-liquid mixture, as assumed.
3.13 System = tank and its contents (open system)
(a) Steady state mass balance
dM
M
M
0 M
1
2
3
dt
M
ŸM
M
10 kg min
3
1
2
b
g
.
.
M1
T1
M2
T2
.
M3 , T3
Steady state energy balance
dU
H M
H M
H
0 M
1 1
2 2
3 3
dt
H 3 H exit stream H at temperature of tank contents
Solutions to Chemical and Engineering Thermodynamics, 5h ed
Also T3
T
Chapter 3
temperature of tank contents
H C T T , assuming C is not a function of temperature
a
f
0 m
a fr m
a
5T 5T
1
ŸT
aT T f 65q C
10
2
Now H
0
P
0
P
a
fr m
5 H 0 CP T1 T0 5 H 0 CP T2 T0 10 H 0 CP T T0
1
2
1
fr
2
dM
M
M
(no useful information here)
0 M
1
2
3
dt
dU
H M
H M
H
energy balance:
M
1 1
2 2
3 3
dt
dU
d
dU
dT
dT
since CP | CV
but
MU
M
MCV
~ MCP
dt
dt
dt
dt
dt
dT
liquids. Thus MCP 3 5CPT1 5CPT2 10CPT3 and M 50 kg .
dt
(b) mass balance:
c h
10
dT3
2T3
dt
At t o f , T3
80 50
130 Ÿ T3
Ae t 5 C t
0 , T3
C 65q C
A C 25q C Ÿ A
40q C
So finally T3
65q C 40q Ce t 5 t
minutes
At t
E.B:
M FU F M iU i 'MH in
M FU F M FU F M i U i M i U i
c
L
L
V
V
Also known is that V
h c
60 m3
L
L
V
V
h
minutes
. H L, in 0.9 H V, in
M LF M VF 01
M LFVLF M VFVVF . Ÿ2 equations and 2 unknowns
V M VFVVF
VLF
M LF
FG V M V U M U IJ c M U M U h
K
H V
LMV M V M OP 01. H 0.9 H
Q
N V
F F
F
V V
L
F
L
F F
V V
F
L
F
V
F
V
F
V
i
L
L, in
i
L
i
V
V, in
3.14 Thermodynamic properties of steam from the Steam Tables
Initial conditions:
Specific volume of liquid and of vapor:
m3
m3
; VVi 08857
.
kg
kg
Specific internal energy of liquid and of vapor
kJ
kJ
U Li 3139
.
; U Vi 24759
.
kg
kg
VLi
.
1061
u 103
M.B: M f M i
'M i
for
i
V
Solutions to Chemical and Engineering Thermodynamics, 5h ed
200 liters
= 194.932 kg;
VLi
Mi
M Li M Vi ; M Li
M Vi
60 m3 200 liters
=14.476 kg and so Mi=209.408 kg
VVi
E.B.
c
M f U f M iU i 'MH in
M f U f M f U f M i U i M i U i
L
V
L
V
Chapter 3
h c
L
L
V
V
h
. H L, in 0.9 H V, in
M Lf M Vf 01
Total internal energy of steam + water in the tank
194.932u313.0 + 14.476u2475.9 = 9.686u104 kJ
Properties of steam entering, 90% quality
Specific volume = Vin = 0.1u1.061u10-3+ 0.9u0.8857 = 0.797 m3/kg
Specific enthalpy = H =0.1u504.70 + 0.9 u 2706.7 = 2.486 u103 kJ/kg
in
60 m3
Also have that V
M LfVLf M VfVVf .
This gives two equations, and two unknowns, M Lf and M Vf .
The solution (using MATHCAD) is M Lf = 215.306 kg and M Vf = 67.485 kg.
Therefore, the fraction of the tank contents that is liquid by weight is 0.761.
3.15 System contents of both chambers (closed, adiabatic system of constant volume.
Also Ws 0 ).
Energy balance: U t2 U t1 0 or U t2 U t1
(a) For the ideal gas u is a function of temperature only.
Thus,
U t2 U t1 Ÿ T t2 T t1 500 K . From ideal gas law
af af
af af
af af af af
PV
1 1
PV
2 2
N1RT1
N 2 RT2
but
N1 N 2 since system is closed
T1 T2 see above
and V2 2V1 see problem statement.
1
. MPa Ÿ T2 500 K, P2 05
. MPa
P1 5 bar 05
2
(b) For steam the analysis above leads to U t2 U t1 or, since the system is
closed U t
U t , V t
2V t . From the Steam Tables, Appendix III,
Ÿ P2
af af
af af af af
U at f U T 500 K, P 1 MPa U T 22685
. q C, P 1 MPa
2
1
2
1
1
# 2669.4 kJ kg
V t1 V T 22685
. q C, P
af
af af
af
1 MPa # 0.2204 m3 kg
Therefore U t2 U t1 2669.4 kg kg and
V t
2V t
0.4408 m3 kg . By, essentially, trial and error, find that
af
2
1
T ~ 216.3q C , P ~ 05
. MPa .
Solutions to Chemical and Engineering Thermodynamics, 5h ed
af af
af
af
Chapter 3
af
af
af
(c) Here U t2 U t1 , as before, except that U t1 U I t1 U II t1 , where
superscript denotes chamber.
Also, M t
M I t1 M II t1 {mass balance} and
2V M t
2V M I t M II t
V t
af
2
1
af
2
af
1
1
af
1
For the ideal gas, using mass balance, we have
a f P V P V Ÿ 2P
I
1 1
T1I
P1I P1II
T1I T1II
(1)
U 0 NCV T T0 , and cancel terms, use N
PV RT and get
P2 2V1
T2
Energy balance: N 2U 2
Substitute U
a
II
1 1
T1II
Eqns.
(1)
T2
N1IU 1I N1IIU 1II
f
2 P2
Using
2
and
(2)
P1I P1II
get
P2
(2)
7.5 u 10 Pa
5
0.75 MPa
and
529.4 K 256.25q C .
T2
(d) For steam, solution is similar to (b). Use Steam Table to get M1I and M1II in
terms of V.
Chamber 1: U1I 2669.4 kJ kg ; V1I 0.2204 m3 kg ;
M I V V I 4.537V
1
1
1
a
Chamber
2:
U1II U T 600 K, P
V II 05483
.
m3 kg ; M II 1824
. V V V II
1
1
2V1
Thus, V2
I
M M II
U
M IU I M IIU II
2
c
1
1
1
2V1
4.537V1 1824
. V1
. MPa
05
f 28459. kJ kg ;
1
0.3144 m3 kg ;
h b M M g 2720.0 kJ kg
I
1
II
1
By trial and error: T2 ~ 302q C 575 K and P ~ 0.76 MPa .
3.16 System: contents of the turbine (open, steady state)
(a) adiabatic
dM
M
ŸM
0 M
M
mass balance:
1
2
2
1
dt
0
dU
H M
H Q 0 W P dV
0 M
energy balance:
1 1
2 2
s
dt
dt
Ÿ Ws M1 H1 H2 M1 3450.9 28656
. kJ kg
c
c
h
5853
M
u 10
1 .
But Ws
5
h J kg
7.5 u 105 watt
7.5 u 105 J s
7.50 u 105 J s
.
kg s 4.613 u 103 kg h
1281
.
5853
u 105 J kg
(b) Energy balance is
0
dU
H M
H Q W P dV
0 M
1 1
2 2
s
dt
dt
M
1
Solutions to Chemical and Engineering Thermodynamics, 5h ed
a
Chapter 3
f
60 kJ kg
H 2 H 150q C, 0.3 MPa 27610
. kJ kg
where Q M
1
Thus
Ws 1281
.
kg s 3450.9 27610
. 60 kJ kg 807 kJ s
8.07 u 105 watt
807 kW
3.17 System: 1 kg of water (closed system).
Work of vaporization
PdV P dV
z
z
P'V since P is constant at 1.013 bar.
Also, from Steam Tables
m3 kg ; 'V 16719
.
m3 kg
V L 0.001044 m3 kg ; V V 16729
.
Energy balance for closed system (1 kg):
U U Q PdV Q 1013
.
.
u 105 Pa u 16719
m3 kg
2
z
1
Q 16945
.
u 105 J kg
U 2
U
1
. kJ kg
25065
2.5065 u 106 J kg
418.94 kJ kg
.
u 105 J kg
41894
Thus
Q U 2 U1 W
u 105 16945
u 105
.
.
2.5065 u 106 41894
2.2570 u 106 J kg
z
PdV
W
16945
u 105 J kg .
.
So heat needed to vaporize liquid
2.2570 u 106 J kg of which 016945
.
u 106 is
recovered as work against the atmosphere. The remainder, 2.088 u 106 kJ kg ,
goes to increase internal energy.
3.18 System = Contents of desuperheater (open, steady state)
Superheated steam
T=500ºC
P=3 MPa
Desuperheater
Water
25ºC
M
1
M
2
500 kg hr ; H 1 34565
. kJ kg
H sat’d liq., T 25q C
?; H
2
Mass B: 0
Saturated steam
2.25 MPa
104.89 kJ kg
M
M
M
1
2
3
H M
H M
H Q 0 W
0 P dV
M
1 1
2 2
3 3
s
dt
500 M 2 kg hr ; H3 H sat’d steam, P 2.25 MPa
0
Energy B: 0
M
3
Thus,
b
g
. kJ kg
28017
Solutions to Chemical and Engineering Thermodynamics, 5h ed
b
0 500 u 34565
. M
2
ŸM
2
Chapter 3
g
u 28017
104.89 500 M
.
2
1214
. kg hr
3.19 The process here is identical to that of Illustration 3.5-3, so that we can use the
equation
P2
P1 T1 CV CP P2 P1 Tin
a
T2
2.0 MPa , Tin
20.99 J mol K .
developed in the illustration.
CP
29.3 J mol K , CV
Here, P2
CP R
CP
Tin 5488
. K
CV
Cylinder 1: P1
0 , T2
Cylinder 2: P1
01
. MPa , T1
f
20q C
Cylinder 3: P1
1 MPa , T1
20q C
39315
. K,
27565
. qC
29315
. K
2.0
. 29315
. 20.99 29.3 2.0 01
. 39315
.
01
T2
120q C
29315
. K ; Ÿ T2
252.7q C
. K
52587
38216
. K 109.01q C
3.20 System: Gas contained in the cylinder (closed system)
M piston g
4000 kg
9.8 m s2
(a) P 01013
10133
u 105 u
.
.
MPa A
2.5 m2 1 kgm Ns2
11701
.
u 105 Pa
moles of
N
gas initially
in system
0117
.
MPa
PV
RT
11701
.
u 105 Pa u 25 m3
8.314 Pa ˜ m3 mol K u 29315
. K
1200
kmol
.
.
u 103 mol 1200
(b) Energy balance: U 2 U1 Q PdV
z
'V 3 m u 2.5 m
Final temperature:
2
T2
PV2
NR
U 2 U1
7.5 m ; P'V
Q P'V since P is constant.
11701
.
u 105 Pa u 7.5 m3
3
11701
u 105 Pa u 25 7.5 m3
.
3
. u 10 mol u 8.314 Pam3 mol K
12
a
N U 2 U1
8.7758 u 105 J
3812
. K 108.05q C
f NC aT T f
V
2
1
12
. u 10 mol u 30 8.314 J mol K u 3812
. 29315
. K
3
2.291 u 106 J
(c)
Q
'U P'V
'T of
gas
work
2.291 u 106 8.7758 u 105
work 27.7% of energy absorbed
'T 72.3%
.
3169
u 106 J
.
3169
MJ
(d) System: Gas contained within Piston + Cylinder (open system).
Solutions to Chemical and Engineering Thermodynamics, 5h ed
Chapter 3
[Note: Students tend to assume dT dt 0 . This is true, but not obvious!]
dN
mass balance:
N
dt
dV
d
0
NU
N H out Q
P
energy balance:
dt
dt
Here (1) P is constant, (2) Ideal Gas Law V NRT P , (3) T and P of Gas
Leaving Cylinder T and P of gas in the system. Thus,
d NRT
dN
dN
dU
N
U
P
H
dt
P
dt
dt
dt
dT
dN
d
NCV
NT
Ÿ H U
R
dt
dt
dt
dN
dT
dT
dN
dT
RT
0
NCV
NR
RT
Ÿ N CV R
dt
dt
dt
dt
dt
dT
0 Q. E.D.
Ÿ
dt
Thus T3 T2 3812
. K
Now going back to
F
H
I
K
a
N
dN
dU
U
dt
dt
H
dN
dt
Ÿ H U
f
dT
dU
dV
dN
and using
P
0
dt
dt
dt
dt
dN
P dV
dV
dN
P
RT
or
dt
dt
dt
RT dt
(**)
Since P and T are constants
N3
N2
V3
V2
25 m3
25 7.5 m3
0.7692
Thus N 3 0.7692 u 1200 mol 923 mol ;
'N 277 mol 0.277 kmol
3.21 (a) System: Gas contained within piston-cylinder (closed system) [neglecting the
potential energy change of gas]
energy balance:
d NU
dt
But T
Thus
PV
dT
Ÿ
NR
dt
N
dU
dt
dV
dT
Q P
; NCV
dt
dt
F I PA dh .
H K NR dt
P dV
NR dt
dh
Q PA
dt
Solutions to Chemical and Engineering Thermodynamics, 5h ed
F
H
Chapter 3
I
K
ACV P dh
dh
C
dh PACP dh
PA V 1
AP
R dt
dt
R
dt
R dt
30 J mol K
.
u 11701
u 105 Pa u 2.5 m2 u 0.2 m s
8.314 J mol K
Q
2111
. u 105 J s
(b) System: Gas contained within piston and cylinder (open system). Start from
result of Part (d), Problem 3.20 (see eqn. (**) in that illustration)
dN
dt
P dV
RT dt
PA dh
with P and T constant
RT dt
(See solution to Problem 3.20)
u 105 Pa u 2.5 m2
11701
.
u 0.2 m s
8.314 J mol K u 3812
. K
0.01846 kmol s
a
dN
dt
[check:
18.46 mol sec u 15 sec
Problem 3.20]
f 18.46 mol s
276.9 mol compare with part d of
3.22 System: gas contained in the cylinder (open system)
Important observation . . . gas leaving the system (That is, entering the exit valve
of the cylinder) has same properties as gas in the cylinder.
dN ½ Note that these are
mass balance
N
°°
dt
¾ Eqns. (d) and (e) of
d NU
° Illustration 3.5-5
energy balance
NH
°¿
dt
dN
Note that these are
N
mass balance
dt
Eqns. (d) and (e) of
d NU
energy balance
N H Illustration 2.5- 5
dt
Proceeding as in that illustration we get Eqn. (f)
|VU
|W
FG T t IJ
HT 0 K
CP R
FG P t IJ or T t
HP0K Pt
R CP
320
10a8.314 30f
169.05 K
(1)
where we have used a slightly different notation. Now using the mass balance we
get
dN
dt
F I V d a P T f N
H K R dt
d PV
dt RT
or
a4.5 28f mol s u 8.314 Pa ˜ m mol K
a f NR
8.908 Pa K ˜ s
d PT
dt
3
V
3
. m
015
Solutions to Chemical and Engineering Thermodynamics, 5h ed
Chapter 3
and
P
Tt
P
8.908 u 105 t
Tt 0
Using t 5 minutes
and (2) yields
bar K for P in bar and t in secs.
(2)
300 secs in Eqn. (2) and simultaneously solving Eqns. (1)
T 5 min
152.57 K , P 5 min
0.6907 bar
Computation of rates of change from mass balance
d ln P d ln T
F I 1 F dP Pd ln T I NR
H K T H dt dt K V or dt dt
d P
dt T
NRT
PV
(3)
From energy balance (using 2 eqns. above and eqn. (f) in Illustration (3.5-5))
a f
CV d ln T
R dt
d ln P T
C d ln T
or P
R dt
dt
d ln P
dt
(4)
Now using Eqn. (4) in Eqn. (3). Thus,
CV d ln T CV dT NRT
or
R dt
RT dt
PV
dT
N RT 2
.
K sec
1151
dt t 5 min
PVCV t 5 min
and
CP P dT
RT dt 5 min
dP
dt 5 min
0.0188 bar s
3.23 Consider a fixed mass of gas as the (closed) system for this problem. The energy
balance is:
d NU
dt
N
From the ideal gas law we have P
CV N
dT
dt
dU
dt
NCV
dT
dt
P
dV
dt
NRT V . Thus
NRT dV
C d ln T
Ÿ V
R dt
V
dt
d ln V
dt
or
CV T2
ln
R
T1
FG IJ
H K
V
T
ln 2 Ÿ 2
V1
T1
CV R
FG V IJ
HV K
1
2
(*)
Solutions to Chemical and Engineering Thermodynamics, 5h ed
Chapter 3
or
V2T2 CV R
V1T1CV R
VT CV R
constant
Substituting the ideal gas law gives PV CP CV PV J constant. Note that the
heat capacity must be independent of temperature to do the integration in Eqn. (*)
as indicated.
3.24 System: Contents of the tank (at any time)
(a) Final temperature T 330 K and pressure P
c
h
.
1013
u 105 Pa are known.
Thus, there is no need to use balance equations.
1013
u 105 Pa u 0.3 m3
.
8.314 J mol K ˜ 330 K
PV
RT
Nf
1108
. mol
0.01108 kmol
(b) Assume, as usual, that enthalpy of gas leaving the cylinder is the same as gas
in the cylinder . . . See Illustration 3.5-5. From Eqn. (f) of that illustration we
have
Pf
Pi
Thus
Nf
2136
. mol
FG T IJ
HTK
CP R
f
i
Tf
or
Tf
Ti
FG P IJ
H PK
R CP
f
FG 10133
.
u 10 I
H 10. u 10 JK
5 8.314 29
.
05187
6
i
05187
.
u 330 K 17119
. K,
0.02136 kmol .
Pf
1013
.
bar ,
and
3.25 Except for the fact that the two cylinders have different volumes, this problem is
just like Illustration 3.5-5. Following that illustration we obtain
2 P1i
T1i
2 P1i
2 P1f P2f
f
T1f
T2
2 P1f P2f or P f
for Eqn. (a')
2 i
P1
3
for Eqn. (c')
and again get Eqn. (f)
FG T IJ
HT K
f
1
i
1
Then we obtain P f
1333
. bar , T1f
CP R
FG P IJ
HP K
f
1
i
1
2234
. K , and T2f
328.01 K .
3.26 From problem statement P1f P2f P f and T1f T2f T f .
Mass balance on the composite system of two cylinders
N1f N 2f
N1i or
2 P1f P2f
f
T1f
T2
3P f
Tf
2 Pi
Ti
Solutions to Chemical and Engineering Thermodynamics, 5h ed
Chapter 3
Energy balance on composite system
N1iU 1i
and T f
FI
HK
3P f i
T1
2 P1i
3 2 i
T1
2 3
T1i
2 u 200 bar
3
2 Pi
3
N1f U 1f N 2f U 2f Ÿ P f
1333
. bar (as before)
250 K .
3.27 Even though the second cylinder is not initially evacuated, this problem still bears
many similarities to Illustration 3.5-5). Proceeding as in that illustration, we
obtain
2 P1i P2i
i
T1i
T2
2 P1i P2i
2 P1f P2f
f instead of Eqn (a')
T1f
T2
2 P1f P2f
3 P f instead of Eqn. (c)
2 u 200 1 u 20 / 3 140 bar ] and again recover Eqn. (f) for
[Thus, P f
Cylinder 1
FG T IJ
HT K
f
1
i
1
Solution is P1f
P2f
140 bar , T1f
FG P IJ
HP K
CP R
f
1
i
1
226.47 K , T2f
Eqn. (f)
28651
. K.
3.28 (a) System = Gas contained in room (open system)
dN
N
mass balance:
dt
d NU
dN N H Q H
Q
energy balance:
dt
dt
Thus,
Q
dN
d NU
H
dt
dt
UH
dU
dN
N
dt
dt
F I V d F PI
H K R dt H T K
V d P
NT d F P I
dT
dT
Q RT F I F I NC
RT ˜
NC
H R K dt H T K
P dt H T K
dt
dt
For the ideal gas, H U
PV
RT ;
dN
dt
d PV
dt RT
V
Q
Since P
NRT dP
dT
dT
NR
NCV
dt
dt
P dt
constant,
dP
dt
0 , Q
NCPdT
or
dt
V
Solutions to Chemical and Engineering Thermodynamics, 5h ed
dT
dt
Q RT
CP PV
Chapter 3
15
. u 103 W ˜ 8.314 J mol K ˜ 28315
. K
5
29 J mol K ˜ 10133
u 10 Pa ˜ 35
.
. u 5 u 3 m3
0.0229 K s 137
. K min
(b) System = Gas contained in sealed room (closed system) N
d NU
dU
dT
N
NCv
Q
Energy balance:
dt
dt
dt
Q
NCV
dT
dt sealed
room
CP dT
CV dt unsealed
room
0
29
. K min
u 137
29 8.314
1925
.
K min
3.29 In each case we must do work to get the weights on the piston, either by pushing
the piston down to where it can accept the weights, or by lifting the weights to the
location of the piston. We will consider both alternatives here. First, note that
choosing the gas contained within piston and cylinder as the system, 'U Q W .
Also
But 'U 0 , since the gas is ideal and T constant.
W PdV NRT ln Vf Vi , for the same reasons. Thus, in each case, we have
z
a
f
that the net heat and work flows to the gas are
W work done on gas
u 10
.
FG V IJ 2479 ln 1213
HV K
2.334 u 10
NRT ln
2
f
i
and Q
W
2
1622.5 J
1622.5 J (removed from gas)
If more work is delivered to the piston, the piston will oscillate eventually
dissipating the addition work as heat. Thus, more heat will be removed from the
gas + piston and cylinder than if only the minimum work necessary had been used.
Note that in each case the atmosphere will provide
Watm
P'V
1013
.
u 105 kPa u 2.334 1213
.
u 102 m3
11356
. J
and the change in potential energy of piston
mg'h
5 kg u 9.8 m s2 u
.
2.334 1213
u 102 m3
2
1 u 10 m
54.9 J
The remainder 1622.5 11356
. 54.9 432.0 J must be supplied from other
sources, as a minimum.
(a) One 100 kg weight.
An efficient way of returning the system to its original state is to slowly (i.e.,
at zero velocity) force the piston down by supplying 432.0 J of energy. When
the piston is down to its original location, the 100 kg is slid sideways, onto the
piston, with no energy expenditure.
An inefficient process would be to lift the 100 kg weight up to the present
location of the piston and then put the weight on the piston. In this case we
would supply
Solutions to Chemical and Engineering Thermodynamics, 5h ed
Mg'h
Chapter 3
b
g
2
u 102 m3
.
'V
m 2.334 u 10 1213
100 kg u 9.8 2 u
2
2
A
s
1 u 10 m
2
2
1098.6 kg m s 1098.6 J
Mg
This energy would be transmitted to the gas as the piston moved down. Thus
11356
. J 54.9 J 1098.6 J
W on gas
atmosphere
PE of piston
W J Q J
Efficient
1622.5
Inefficient 2289.1
22891
.
PE of weight
Wcycle Qcycle
1622.5 11905
. 432.0
22891
. 11905
. 1098.6
(b) Two 50 kg weights
In this case we also recover the potential energy of the topmost weight.
mg'h
50 kg u 9.8
1213
u 102 m3
m 1597
.
.
u
2
0.01 m2
s
188.2 J
Thus in an efficient process we need supply only
1622.5 11356
. 54.9 188.2
2438
. J
An efficient process would be to move the lowest weight up to the position of
the piston, by supplying
50 kg u 9.8
u 102 m3
.
m 2.334 1213
u
2
2
1 u 10 m2
s
549.3 J
Slide this weight onto the piston and let go. The total work done in this case
is
11356
. atmosphere
54.9
'PE of piston
2438
.
'PE of weight
549.3
supplied by us
19836
. J
Therefore
W J Q Wcycle Qcycle
1622.5 1378.7 2438
. J
Efficient
1622.5
19836
. 1378.7 604.9 J
Inefficient 1983.6
(c) Four 25 kg weights.
In this case the recovered potential energy of weights is
25 kg u 9.8 m s2 u
FG 1897
1213
1597
1213
1379
1213
.
.
.
.
.
.
u 10 I
JK m
H
1 u 10
302.3 J
Thus in an efficient process we need supply only
2
2
Solutions to Chemical and Engineering Thermodynamics, 5h ed
Chapter 3
1622.5 11356
. 54.9 302.3 129.7 J
An inefficient process would be to raise the lowest weight up to the piston,
expending
25 kg u 9.8 m s2 u
.
u 102 m3
2.334 1213
1 u 102 m
274.6 J
Thus the total work done is
11356
. 54.9 302.3 274.6 1767.4 J
and
W Q Wcycle Qcycle
. 129.5
Efficient
1622.5 1622.5 14930
. 274.4
Inefficient 1767.4 1767.4 14930
(d) Grains of sand
Same analysis as above, except that since one grain of sand has essentially
zero weight W 1622.5 J , Q 1622.5 J , Wcycle Qcycle 0 .
3.30 System = Gas contained in the cylinder (closed system)
d NU
dU
dT
dV
P
N
NCV
energy balance:
dt
dt
dt
dt
ideal gas equation of state}
Since CV and CP are constant
CV 1 dT
R T dt
Ÿ T2
FG IJ FG V IJ
H K HV K
1 dV
T
or 2
V dt
T1
. u
25 27315
P2
NRT dV
V
dt
R CV
1
FG L IJ
HL K
{Using the
R CV
1
2
FG 0.03 m IJ
H 0.03 0.6 u 0.05 m K
3
2
8.314 30 8.314
3
. K 44.58q C and
22857
.
1 22857
V T2
20 u u
7.666 bar
P1 1
.
2 29815
V2 T1
FG IJ FG IJ
H KH K
From the difference (change of state) form of energy balance
'U
Q
0
W
a
NCV T2 T1
f z PdV
PV
20 bar u 0.03 m3 kmol ˜ K
0.0242 kmol
RT 298.15 K u 8.314 u 102 bar ˜ m3
. 22857
. K
Ÿ W 'U 0.0242 kmol u 30 8.314 kJ kmol ˜ K ˜ 29815
3652
. kJ
Where has this work gone?
and N
Solutions to Chemical and Engineering Thermodynamics, 5h ed
Chapter 3
(a) To increase potential energy of piston
(b) To increase kinetic energy of piston
(c) To push back atmosphere so system can expand
(d) Work done against friction (and converted to heat).
To see this, write Newton’s 2nd Law of Motion for the piston
Patm u A
f
Fr
Frictional Force
mg
f
Fr
Pressure of gas (P) u A
a
MA Ÿ PA Patm A mg f fr
f
Thus, P
m dv
mg f fr
Patm A dt
A
A
'U 36,520 J PdV
z
z
PatmdV Now
f m dvdt ; v velocity of piston
1 dV
A dt
dh
dt
v (h
z
z
piston height) and v
36,520 J
z
1
mg
dv
m dv dV
dt dV f fr
dt
A
A
dt
A dt dt
Patm 'V 3000 J
Work against
atmosphere
mv 2
2
since
vinitial 0
dv
dt
(1)
c h
1 d 2
v
2 dt
mg'h
z
f fr vdt
1760 J
Work used to
increase potential
energy of piston
z
mv 2
1760 f fr vdt .
2
(a) If there is no friction f fr 0 then
Thus 36,520 J
3000 36520 3000 1760 J u 2
300 kg
v2
(b) If we assume only sliding friction, f fr
z
f fr vdv
2117
. m2 s2 Ÿ v
14.55 m s
kv
z
k v 2dt Ÿ 36520 3000 1760
z
m 2
v k v 2dt
2
In order to determine the velocity now we need to know the coefficient of
sliding friction k, and then would have to solve the integral equation above (or
integrate successively over small time steps). It is clear, however, that
v with friction v without friction
14.55 m s
System for
part a
25q C,
3.0 u 106 Pa = 3Mpa
125 kg/s
System for
part b
Solutions to Chemical and Engineering Thermodynamics, 5h ed
Chapter 3
3.31 25qC, 30
. u 106 Pa 3 MPa
125 kg s
(a) mass balance (steady-state)
M
0 M
1
2
ŸM
M
125 kg s
1
2
Energy balance (neglecting PE terms)
0
IJ
K
FG
H
FG
H
2
2
H v1 M
H v2
M
1
1
2
2
2
2
IJ
K
UvA mnvA ; U mass density, n molar density,
M
v velocity, A pipe area, m molecular weight.
M
P
vA
m RT
125 kg s
30
. u 106 Pa
u v m s u S u 0.09 m2
Ÿ
16 kg kmol 298.15 K u 8.314 u 103 Pa ˜ m3
Ÿ v 22.83 m s
a f
a
f
mv 2 16 kg kmol u 22.83 m s 2
4170
u 103 J kmol
.
2
2 u 1 kg ˜ m Ns2
Back to energy balance, now on a molar basis
a
mv22 mv12
C p T1 T2
2
2
As a first guess, neglect kinetic energy terms . . .
Cp T1 T2 0 Ÿ T1 T2 25q C
H1 H 2
a
417
. kJ kmol
f
f
Now check this assumption
. u 106 v1
n1v1 Pv
30
1 1
34.24 m s
n2
P2
2.0 u 106
Recalculate including the kinetic energy terms
m 2 2 16
Cp T1 T2
v1 v2
34.242 22.832 5209 J kmol
2
2
5209 J kmol
T2 T1 T1 014
. qC
. J mol u 1000 mol kmol
368
Thus the kinetic energy term makes such a small contribution, we can safely
ignore it.
v2
a
(b)
f
c
h
c
Mass balance on compressor (steady-state) 0
2.0 u 106 Pa
T1=25q C
compressor
h
N 1 N 2
3.0 u 106 Pa
T2= ?
Energy balance on compressor, which is in steady-state operation
Solutions to Chemical and Engineering Thermodynamics, 5h ed
0
Ws Ÿ Ws
N 1 H 1 N 2 H 2 Q
0
Chapter 3
N 1Cp (T2 T1)
25qC
adiabatic compressor
Can compute Ws if T2 is known or vice versa. However, can not compute both
without further information.
2.0 u 106 Pa
T2= ?
Gas cooler
3.0 u 106 Pa
T3=25q C
Analysis as above except that Q z 0 but W
0 N 2 N 3
Here we get Q N 1Cp ( T3 T2 )
R|
S|
T
0.
25q C
Can not compute Q until T2 is known.
See solution to Problem 3.10.
3.32 a) Define the system to be the nitrogen gas. Since a Joule-Thomson expansion is
isenthalpic, H T1, P1 H T2 , P2 . Using the pressure enthalpy diagram for
nitrogen, Figure 3.3-3, we have
H 135 K,20 MPa 153 kJ / kg and then T2 T P2 0.4 MPa, H 153 kJ / kg
a f a
f
c
h
From which we find that T = 90 K, with approximately 55% of the nitrogen as
vapor, and 45% as liquid.
b) Assuming nitrogen to be an ideal gas (poor assumption), then the enthalpy
depends only on temperature. Since a Joule-Thomson expansion is isenthalpic, this
implies that the temperature is unchanged, so that the final state will be all vapor.
3.33 Plant produces 136
. u 109 kwh of energy per year
Ÿ Plant uses 136
. u 109 u 4 544
. u 109 kwh of heat
6
1 kwh 36
. u 10 J
J year
. u 106
u 544
. u 109 kwh 19.584 u 1015 J year
Ÿ Plant uses 36
kwh
'H of rock (total) M ˜ Cp Tf Ti
a
12
10
f
kg u 1 J g K u 1000 g kg u 110 600 K
= 490 u 1015 J
Ÿ 19.58 u 1015 J year u x years 490 u 1015 J
x 2502
. years
3.34
a)
Body temperature is 37oC. From Appendix A.III, ice at 0oC has a volume of
0.0010908 m3/kg and an enthalpy (and internal energy) of -333.43 kJ/kg. For
water at 35oC Û = 146.67 kJ/kg and at 40oC Û = 167.56 kJ/kg, so that at 37oC
Solutions to Chemical and Engineering Thermodynamics, 5h ed
Chapter 3
Û = 155.03 kJ/kg. Also, 1 L = 10-3 m3, so that 1 L of water 10-3 m3/0.0010908
m3/kg = 0.9168 kg. So that the amount of energy needed to melt 1 L of ice is
0.9168 kg ¯(155.03-(-333.43))kJ/kg = 447.8 kJ.
b) 447.8 kJ/(42 kJ/kg fat) = 10.66 g of fat (= 0.023 lbs of fat)
c) For water at 0oC, Û = 0 kJ/kg, and V̂ = 0.001 m3/kg. Therefore, 1 L of
water = 1 kg, and the energy required to warm up the water is
1 kg¯155.03 kJ/kg = 155.03 kJ, and only 155.03/42 = 3.69 g of fat would be
consumed.
3.35
a)
M
2
M
3
M
M
1
2
Mass balance
M
M
10 kg/s M
3
1
2
Energy balance
Stream 120°C (1 atm)
M
3
M
10 kg/s M
1
2
ˆ
H
ˆ
M
1 1 M2H2
10
M
3
10 kq/s
kg ˆ ˆ
ˆ
H
H 3 M1H1 M
2 2
s
10
kq ˆ
H3
s
Ĥ1 = 83.96 J/g
Stream 2450°C, 2.5 MPa Ĥ 2 = 3344.0 J/g
Ĥ 3 = 376.92 J/g
Stream 390°C (1 atm)
M
10-M
u 83.96 10-M
M
1
MB
EB
M
1
M
2
b)
3344 10 x 376.92
9.101 kg/s M
1
9.101 kg/s M
1
9.101 kq/s
0.899 kq/s
M
10 kg/s, T=90°C
10 kg/s, T=20°C
O
N
P
10 kg/s, T=20°C
100 kg/s, quality = 50%
Ĥ
Steady balance:
1
1
2676.1 419.04
2
2
12547.6 J/g
Solutions to Chemical and Engineering Thermodynamics, 5h ed
Chapter 3
H
ˆ
H
ˆ
ˆ
ˆ
M
M
1 1 M2H2
3 3 M4H4
M
M
10 kg/s
But
1
3
M
2
M
4
? kg/s
x3344 = 10 × 376.92 + M
× 1547.6
10 × 83.96 + M
2
2
M ¯ (3344 – 1547.6) = 10 × (376.92 – 83.96)
2
=1.63 kg/s
M
2
3.36 ǻ vap H at 37°C ~ 2412 J/g
Amount evaporated = 4184 u 103
J
hr
1
2412 J
1.735 u 102 g
g
hr
1.735 kg/hr
If only 75% evaporates Ÿ
M
3.37
1.735
0.75
N
1.631 kg/hr of sweat produced
25°C, 0.8 bar
–50°C
0.1 bar
Mass balance – steady state
N
N
O N
N
1
2
2
1
Energy balance – steady state
W
W
H Q
N
H H Q
O N
1
1
2
1
2
W
Q
N
1
H 2 H1
H 2 H1
³ C dT
Cp T2 T1
p
T1
30
W
Q
N1
T2
J
u 75k
mol K
T2
³ C dT
p
2250
J
mol
Cp T2 T1
T1
30
J
u 75k
mol K
PV NRT
100 m3 Ÿ
N
Ÿ
N
1
2250
J
Mol
0.8bar u 100 m3
u 298.15 M
bar m3
8.314 u 105
mol K
3228 mols/min
PV
RT
0.03228 u 105 mol
3228 mols
Solutions to Chemical and Engineering Thermodynamics, 5h ed
W
Q
W
Q
Chapter 3
7.262 u 106 J/min 1.210 u 105 J/s
1.210 u 105 Watts 121.0 kw
$
Cost 0.2
u 121.0 kw 24.2 $/hr
kw ˜ hr
7.262 u 106 J/min 1.210 u 105 J/s
1.210 u 105 Watts 121.0 kw
$
Cost 0.2
u 121.0 kw 24.2 $/hr
kw ˜ hr
3.38 In the folder Aspen for Textbook>Chapter 3>Problem 3.38 (Prob 3.5)
3.39 In the folder Aspen for Textbook>Chapter 3>Problem 3.39 (Prob 3.16a)
3.40 In the folder Aspen for Textbook>Chapter 3>Problem 3.40 (Prob 3.22)
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
4
4.1
Ball (1) + Water (2) Energy balance: M U f M U f M U i M U i
(a)
System
1
c
h
1
2
c
Ÿ M1CV,1 T1 f T1i M 2CV,2 T2f T2i
Tf
M1CV,1T1i M 2CV,2T2i
M1CV,1 M 2CV,2
2
1
1
2
0
2
h 0 ; also T T . Thus
f
f
2
1
. u 75 12 u 103 u 4.2 u 5
5 u 103 u 05
. 12 u 103 u 4.2
5 u 103 u 05
8.31q C
[Note: Since only 'T s are involved, q C were used instead of K)].
dT
(b) For solids and liquids we have (eqn. 4.4-6). That 'S M CP
T
z
which CP is a constant. Thus
Ball: 'S
.
5 u 103 g u 05
RS
T
.
8.31 27315
J
u ln
.
75 27315
g˜K
. s
53161
12 u 103 g u 4.2
Water: 'S
RS
T
T
MCP ln 2 for the case in
T1
J
UV 53161
W . K
.
8.31 27315
J
u ln
.
5 27315
g˜K
UV 596.22 J
K
W
and
J
J
64.61
K
K
Note that the system Ball + Water is isolated. Therefore
J
'S Sgen 64.61
K
Energy balance on the combined system of casting and the oil bath
'S Ball Water
4.2
c
596.22 53161
.
h
c
h 0 since there is a common final temperature.
kJ
kJ
T 450iK 150 kg u 2.6
20 kg u 0.5
d
dT 450iK 0
kg ˜ K
kg ˜ K
M cCV,c T f Tci M oCV,o T f Toi
f
f
This has the solution Tf = 60oC = 313.15 K
Since the final temperature is known, the change in entropy of this system can be calculated
kJ
. 60
. 60
27315
27315
.
. u ln
4135
150 u 2.6 u ln
from 'S 20 u 05
K
. 50
. 450
27315
27315
F
H
4.3
I
K
Closed system energy and entropy balances
dU
dV dS Q ;
Sgen ;
Q Ws P
dt
dt dt T
dS
Thus, in general Q T
TSgen and
dt
F
H
I
K
Solutions to Chemical and Engineering Thermodynamics, 5th ed
dU
dS
dV
T
TSgen P
dt
dt
dt
dU dV
Q P
dt
dt
Ws
c
Reversible work: WsRev
h dUdt T dSdt P dVdt
WsRev Sgen
c
S
W
s gen
(b) System at constant S & P Ÿ
0
dU
dt
(a) System at constant U & V Ÿ
dS
dt
Chapter 4
0 and
dV
dt
h W
0
dS
dt
0
Rev
S
T
0 and
dP
dt
0Ÿ P
dV
dt
d
PV
dt
d
U PV
dt
dH
dt
so that
c
h W
Ws Sgen
rev
S
0
dU d
PV
dt dt
4.4
700 bar, 600oC
10 bar, T = ?
Steady-state balance equations
dM
M
0 M
1
2
dt
0
dU
H M
H
H M
H Q 0 W 0 P dV
M
0 M
1 1
2 2
1 1
2 2
s
dt
dt
or H 1 H 2
Drawing a line of constant enthalpy on Mollier Diagram we find, at P
At 700 bar and 600q C
V 0.003973 m3 kg
H 3063 kJ kg
S 5522
.
kJ kg K
10 bar, T # 308q C
At 10 bar and 308q C
V | 0.2618 m3 kg
H | 3063 kJ kg
S 7145
.
kJ kg K
Also
dS
dt
c
0
S S
M
1 2
1
Ÿ Sgen
h
0
S M
S Q
Sgen 0
M
1 1
2 2
T
Sgen kJ
or
S2 S1 7145
1623
. 5522
.
.
M1
kg ˜ K
4.5
1
System
2
Ws
Energy balance
'U
cU U h cU U h Q
f
2
i
2
f
1
i
1
adiabatic
z
constant
WS PdV volume
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Ws
c
h
c
T2f
Tf Ÿ
MCp T2f T2i MCp T1 f T1i
but T1 f
Ws
MCP
Chapter 4
h MC cT T h cT T h
p
f
2
i
2
f
1
i
1
2T f T1i T2i
Entropy balance
adiabatic
'S
f
2
i
2
f
1
i
1
cS S h cS S h 0
R T T UV 0 ; T T
or lnS
TTT W
Ÿ cT h cT T h or T
f
2
z
cS S h cS S h
i
2
f
1
i
1
f
1
i i
2 1
0
Q
dt Sgen
T
0 for maximum work
T
Tf
MCP ln 2i MCP ln 1 i
T2
T1
f
f
f
2
f 2
1
i i
1 2
f
2
T1i T2i ; but T1 f
f
T1i T2i and
Ws
MCP
4.6
2T f T1i T2i
1 bar
10 bar
290 K
575 K
T2f
Tf
2 T1i T2i T1i T2i
(a) Entropy change per mole of gas
T
P
'S CP ln 2 R ln 2 eqn. (4.4-3)
T1
P1
J
J
J
10
575
Thus 'S 29.3
ln
ln
0.9118
8.314
mol K 290
mol K 1
mol K
(b) System contents of turbine (steady-state system)
dN
Mass balance
0 N 1 N 2 Ÿ N 2 N 1 N
dt
dV 0
dU
0
Energy balance
Ws P
0 N 1 H 1 N 2 H 2 Q
dt
dt
T T
Ws N H 2 H 1 NC
1
P 2
J
Ws
W
CP T2 T1 29.3
u 575 290 K
mol K
N
a
a
f
a
f
f
J
mol
(c) In Illustration 4.5-1, W 7834.8 J mol because of irreversiblitities 'S z 0 , more work is
done on the gas here. What happens to this additional energy input? It appears as an increase of
the internal energy (temperature) of the gas.
83505
.
4.7
Heat loss from metal block
dU
dT
CP
Q
dt
dt
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
RS
T
Q heat out of metal
T T2 Q 1
T
Q heat into heat engine
W
a
T
f W Ÿ z Wdt
C z F1 I dT
H TK
T
T O
L
C MaT T f T ln P
W C aT T f C ˜ T ln
T
TQ
N
LF T I T O
W C T MG1 J ln P
NH T K T Q
F TI
Q z C dT C aT T f C T G1 J
H TK
CP
dT T T2
dt
T
T2
t
2
P
T1
0
2
P
2
1
2
2
P
2
P
1
2
1
1
2
2
1
1
P 2
T2
1
P
P
2
1
P 2
2
T1
Alternate way to solve the problem
T2
System is the metal block + heat engine (closed)
dU
dT
E.B.:
CP
Q W
dt
dt
dS
Q
Sgen
S.B.:
dt T 2
0 for maximum work
Q
W
W
W
4.8
dS dU
;
T2
dt
dt
dU
dS
T2
dt
dt
z
Wdt
z FH
T2
CP
dS dT
W ; dU CPdT ; dS
T
dt
C
T
CPdT T2 P dT CP 1 2 dT
T
T
T2
F
H
CP 1 T1
a
I
K
T2
dT
T
f
T
CP T2 T1 T2CP ln 2
T1
z FH
I
K
I
K
LF T I T O
C T MG1 J ln P
NH T K T Q
T2
CP
T1
1
T2
dT
T
1
2
2
1
P 2
This problem is not well posed since we do not know exactly what is happening. There are several
possibilities:
(1) Water contact is very short so neither stream changes T very much. In this case we have the
Carnot efficiency
K
W
Q
Thigh Tlow
Thigh
22
27 273
22
300
0.0733
7.33%
(2) Both warm surface water (27qC) and cold deep water (5qC) enter work producing device, and
they leave at a common temperature.
TH
TL
TO
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
b
dM
M
M
ŸM
M
0 M
M
H
L
0
0
H
L
dt
dU
H M
H M
H W 0
E.B.:
0 M
H H
L L
0 0
dt
H M
M
H
H M
W M
M.B.:
H
c
H
L
h
L
b
c
H
L
H H M
H H
M
0
0
H
H
L
L
a
f
a
h
g
C T T M
C T T
M
H P 0
H
L P 0
L
S.B.:
g
0
f
0
0
dS
S M
S M S Q
Sgen
0 M
H H
L L
0 0
T
dt
M H SH M L SL M H M L S0 0
b
c h
FG T IJ FG T IJ
HT K HT K
g
c
S S M
S S
M
H H
0
L L
0
M
H
H
H
L
H P
L P
0
M
L
T0 M H M L
f
a
1 or TH M H TL M L
L
0
h 0 Ÿ M C ln TT M C ln TT
0
0
0
M
TH H
b M M gT M b M M g
H
L
L
H
T0
L
From this can calculate T0 . Then
L
a
W
C T T M
C T T
M
H P 0
H
L P 0
L
f
This can be used for any flow rate ratio.
(3) Suppose very large amount of surface water is contacted with a small amount of deep water, i.e.,
!! M
. Then T ~ T
M
H
L
0
H
W
a
a
f
a
f
C T T M
C T T ~ M
C T T
M
H P H
H
L P H
L
L P H
L
f
(4) Suppose very large amount of deep water is contacted with a small amount of surface water, i.e.,
M
, T ~T .
M
H
L
0
L
C T T M
C T T ~ M
C T T
W M
H P
4.9
a
L
H
f
L P
a
L
L
f
H P
a
L
H
f
System
contents of the turbine. This is a steady-state, adiabatic, constant volume system.
dM
M
or M
M
(a) Mass balance
0 M
2
1
1
2
dt
Energy balance
constant
dU
H M
H Q adiabatic W P dV volume
0 M
s
1 1
2 2
dt
dt
Entropy balance
dS
S Q Sgen
S M
0 M
1 1
2 2
T
dt
0, by problem statement
Thus
M
M
4500 kg h
2
1
W M H H
M.B.
S2
S.B.
S
1
S1
c
1
2
h
E.B.
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
State T1 500q C
1
P1 60 bar
Steam
 o
Tables
H 1 3422.2 kJ kg
S1 68803
.
kJ kg
State
2
Steam
 o
Tables
T2 # 240.4q C
P2
10 bar
S2
S1
kJ
kgK
.
68803
H 2 | 29205
. kJ kg
kg
kJ
kJ
2257650
u 29205
. 3422.2
h
kg
h
(b) Same exit pressure P2 10 bar , and still adiabatic
H H .
Ÿ W M
Ws
4500
a
c
s
1
1
2
6271
. kW
f
h
Here, however,
Ws
c
08
. Ws Part a
Ÿ H 2
P
Thus
08
. 2.258 u 106
2
T2 # 286.7 K
S | 7.0677 kJ kg K
Steam
 o
Tables
. kJ kg
30208
10 bar
Sgen
h kJh 4500c H 3422.2h kJh
c
2
kJ
kJ
.
7.0677
8433
.
h 4500 kgh u 68803
kg K
K˜h
S S
M
1 1
2
(c) Flow across valve is a Joule-Thompson (isenthalpic expansion) ... See Illustration 3.4-1.
Thus, H into valve H out of valve , and the inlet conditions to the turbine are
H 1
P1
H out of valve
H into valve
3422.2 kJ kg
30 bar
T1 | 484.8q C
S1 | 71874
.
kJ kg K
Steam
 o
Tables
Flow across turbine is isentropic, as in part (a)
S2
P2
S1 71874
.
kJ kg K
10 bar
Steam
 o
Tables
T2 # 3181
. qC
H | 3090.4 kJ kg
2
kg
kJ
kJ
u 3090.4 3422.2
1493
.
u 106
414.8 kW
h
kg
h
4.10 Since compression is isentropic, and gas is ideal with constant heat capacity, we have
Ws
4500
FG T IJ FG P IJ
HT K H PK
2
2
1
1
FG 3 u 10 IJ
H 2 u 10 K
aT T f
Problem 3.31, that W NC
So that T2
T1
FG P IJ
HPK
6 8.314 36.8
R CP
2
29815
.
6
1
s
R CP
P
2
1
326.75 K . Now using, from solution to
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Ws
125
Chapter 4
kg 1 mol
J
1000 g
u
u 368
u 326.75 29815
.
. Ku
mol K
kg
s
16 g
8.23 u 106 J s
The load on the gas cooler is, from Problem 3.31,
Q
a
T T
NC
p 3
2
f
125 kg s u 1000 g kg
J
.
. 326.75 K
u 368
u 29815
16 g mol
mol K
8.23 u 106 J s
4.11 (a) This is a Joule-Thomson expansion Ÿ H 70 bar, T ? H 10133
.
bar, T 400q C | H 1 bar, T 400q C
3278.2 kJ kg
and T 447q C , S 6.619 kJ kg K
(b) If turbine is adiabatic and reversible S
S
6.619 kJ kg K and P
0 , then S
c
h
gen
out
in
1013
.
bar. This suggests that a two-phase mixture is leaving the turbine
Let x
fraction vapor
S V 7.3594 kJ kg K
.
S L 13026
kJ kg K
Then x 7.3594 1 x 13026
.
fluid leaving turbine is
H
.
u
08788
.
26755
H V sat’d, 1 bar
6.619 kJ kg K or x
.
1 08778
u
.
08778
.
417.46
H L sat’d, 1 bar
2399.6
Energy balance
0
H M
H Q 0 W P dV
M
s
in in
out out
dt
0
but M
M
in
out
Ws
kJ
3278.2 2399.6 878.6
Ÿ
M in
kg
(c) Saturated vapor at 1 bar
S
Ws
M
in Actual
Efficiency %
Sgen
M
in
7.3594 kJ kg K ; H
.
3278.2 26755
602.7 u 100
878.6
7.3594 6.619
. kJ kg
26755
602.7 kJ kg
68.6%
0.740 kJ Kh
Therefore the enthalpy of
kJ
kg
Solutions to Chemical and Engineering Thermodynamics, 5th ed
(d)
M
ŸM
M
1
2
2
0
W
Steam
70 bar
447q C
c
c
M
1
h
H H W Q P dV
M
1
1
2
s
dt
Q
S S S
M
1 1
2
gen
T
0
Water
1 bar
25q C
Chapter 4
0
Q
h
Simplifications to balance equations
dV
Sgen 0 (for maximum work); P
0 (constant volume)
dt
Q Q
where T0 25q C (all heat transfer at ambient temperature)
T T0
kJ kJ
; S sat' d liq, T 25q C 0.3674
H sat' d liq, 25q C 104.89
kg
kg K
Q
Ws
T0 S2 S1 ;
H 1 H 2 T0 S2 S1
H 1 T0 S1 H 2 T0 S2
M
M
c
h
c
h c
h c
h
max
Ws
M
. u 6.619 104.89 29815
. u 0.3674
3278.2 29815
max
1304.75 4.65 1309.4 kJ kg
4.12
Take that portion of the methane initially in the tank that is also in the tank finally to be in the
system. This system is isentropic S f Si .
(a) The ideal gas solution
S i Ÿ Tf
Sf
FP I
TG J
HPK
f
R Cp
i
N=
PV
Ÿ Ni
RT
'N
N f Ni
PV
i
RTi
0.0195
35.90 kg u 1000
28
g
mol
150.2 K
1964.6 mol; N f
Si
m3
, so that mi
kg
Ni
8.314 36
Pf V
RTf
1768.4 mol
(b) Using Figure 2.4-2.
70 bar | 7 MPa, T = 300 K
Vi
F 35. I
H 70 K
300
i
g
kg
505
. kJ kg K
0.7 m3
m3
0.0195
kg
1282 mol
S f
. kg.
3590
196.2 mol
Solutions to Chemical and Engineering Thermodynamics, 5th ed
At 3.5 bar = 0.35 MPa and S f
3
Vf
.
0192
.
3646
kg u 1000
Nf
28
'N
4.13
g
kg
.
3646
kg.
130.2 mol
g
mol
N f Ni
505
. kJ kg K Ÿ T | 138 K. Also,
0.7 m3
m3
.
0192
kg
m
, so that m f
kg
Chapter 4
130.2 1282
11518
. mol
dS
C
dV
dT
eqn. (4.4-1)
R
V
T
'S
z LMN
a R bT cT 2 dT 3 e dT
dV
R
2
T
V
T
f a
2
OP
Q
z
so that
a
S T2 , V 2 S T1, V 1
f a R ln TT baT T f 2c cT T h
2
2
2
1
2
1
1
c
h c
h
V
d 3
e
T2 T13 T22 T12 R ln 2
V1
3
2
Now using
PV
a
f a
S T2 , P2 S T1, P1
RT Ÿ
V2
V1
T2 P1
˜ Ÿ
T1 P2
f a ln TT baT T f 2c cT T h
d
e
P
cT T h cT T h R ln
P
3
2
2
2
2
2
1
2
1
1
3
2
2
2
3
1
2
1
2
1
Finally, eliminating T2 using T2
a
f a
S P2 ,V 2 S P1,V 1
T1 P2 V 2 PV
1 1 yields
f a lnFGH PPVV IJK Rb a P V PV f
c
a P V f a PV f
2R
d
a P V f a PV f
3R
2
2
2
1 1
2
1 1
2
2
2
2
2
3
2
2
1 1
3
3
1 1
eR 2
P
2
P2V 2 2 PV
R ln 2
1 1
P1
2
d
i d
i
Solutions to Chemical and Engineering Thermodynamics, 5th ed
4.14
Chapter 4
System: contents of valve (steady-state, adiabatic, constant volume system)
Mass balance
0
Energy balance
0
N 1 N 2
0
dV
N 1 H 1 N 2 H 2 Q 0 Ws P
dt
Ÿ H1 H 2
0
Q 0
N 1 S 1 N 2 S 2 Sgen T
Sgen
Ÿ 'S S 2 S 1
N
(a) Using the Mollier Diagram for steam (Fig. 3.3-1a) or the Steam Tables
Entropy balance 0
T1
P1
H 1
'S
7 bar
T2 | 293q C
Ÿ . J g
30453
S2 7.277 J g K
H 2 30453
. J g . Thus S1
S S 0.717 J g K
2
H 2 Ÿ T1
a
293q C
p
T2
600 K
f a f C ln TT R ln PP
S T2 , P2 S T1, P1
P
R ln 2
P1
'S
.
65598
J g K ; Texit
1
(b) For the ideal gas, H 1
'S
4.15
600 K P2
35 bar H 2
2
2
1
1
. J mol K Ÿ
1338
0.743 J mol K
From the Steam Tables
P 15538
.
MPa
L
V
0.001157 m3 / kg VV 012736
.
m3 / kg
U L 850.65 kJ / kg
U V 25953
. kJ / kg
At 200oC,
L
L
H
H
852.45 kJ / kg
27932
. kJ / kg
L
V
S
2.3309 kJ / kg ˜ K S
6.4323 kJ / kg ˜ K
.
kJ / kg ˜ K
'H vap 1940.7 kJ / kg 'S vap 41014
(a) Now assuming that there will be a vapor-liquid mixture in the tank at the end, the properties
of the steam and water will be
P 0.4578 MPa
V L 0.001091 m3 / kg VV 0.3928 m3 / kg
o
At 150 C,
U L
H L
63168
. kJ / kg
632.20 kJ / kg
L
S
18418
.
kJ / kg ˜ K
vap
'H
2114.3 kJ / kg
U V
H V
2559.5 kJ / kg
27465
. kJ / kg
V
S
68379
.
kJ / kg ˜ K
vap
'S
4.9960 kJ / kg ˜ K
(b) For simplicity of calculations, assume 1 m3 volume of tank.
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
Then
Mass steam initially =
0.8 m3
0.12736 m3 / kg
6.2814 kg
0.2 m3
172.86 kg
0.001157 m3 / kg
6.2814
Weight fraction of steam initially =
0.03506
179.14
6.2814
Weight fraction of water initially =
0.96494
179.14
The mass, energy and entropy balances on the liquid in the tank (open system) at any time
yields
L L
L L
dM L
L ; dM U
L H V ; and dM S
L SV
M
M
M
dt
dt
dt
L
dU L L dM L
L H V H V dM
ML
M
U
or
dt
dt
dt
L
L
dU
dM
ML
H V U L
dt
dt
Also, in a similar fashion, from the entropy balance be obtain
dS L dM L V L
dM L vap
ML
'S
S S
dt
dt
dt
There are now several ways to proceed. The most correct is to use the steam tables, and to use
either the energy balance or the entropy balance and do the integrals numerically (since the
internal energy, enthalpy, entropy, and the changes on vaporization depend on temperature.
This is the method we will use first. Then a simpler method will be considered.
Using the energy balance, we have
dM L
dU L
, or replacing the derivatives by finite differences
L
V
M
H U L
MiL1 MiL U iL1 U iL
U L U iL
or finally MiL1 MiL 1 i V1
L
V
L
Mi
H i U i
H i U iL
So we can start with the known initial mass of water, then using the Steam Tables and the data
at every 5oC do a finite difference calculation to obtain the results below.
Mass water initially =
c
c
h
h
FG
H
i
1
2
3
4
5
6
7
8
9
10
11
T (oC)
200
195
190
185
180
175
170
165
160
155
150
U iL (kJ/kg K)
850.65
828.37
806.19
784.10
762.09
740.17
718.33
696.56
674.87
653.24
631.68
IJ
K
H iV (kJ/kg K)
2793.2
2790.0
2786.4
2782.4
2778.2
2773.6
2768.7
2763.5
2758.1
2752.4
2746.5
MiL (kg)
172.86
170.88
168.95
167.06
165.22
163.42
161.67
159.95
158.27
156.63
155.02
So the final total mass of water is 155.02 kg; using the specific volume of liquid water at
150oC listed at the beginning of the problem, we have that the water occupies 0.1691 m3
leaving 0.8309 m3 for the steam. Using its specific volume, the final mass of steam is found to
be 2.12 kg. Using these results, we find that the final volume fraction of steam is 83.09%, the
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
final volume fraction of water is 16.91%, and the fraction of the initial steam + water that has
been withdrawn is
(172.86+6.28-155.02-2.12)/(172.86+6.28) = 0.1228 or 12.28%. A total of 22.00 kg of steam
has withdrawn, and 87.7% of the original mass of steam and water remain in the tank.
For comparison, using the entropy balance, we have
dM L
dS L
, or replacing the derivatives by finite differences
L
V
M
S S L
SiL1 SiL
MiL1 MiL SiL1 SiL
L
L
of
finally
1
M
M
1
i
i
L
M
'S vap
'S vap
i
FG
H
i
i
IJ
K
So again we can start with the known initial mass of water, then using the Steam Tables and
the data at every 5oC do a finite difference calculation to obtain the results below.
i
T (oC)
1
2
3
4
5
6
7
8
9
10
11
200
195
190
185
180
175
170
165
160
155
150
SiL (kJ/kg K)
2.3309
2.2835
2.2359
2.1879
2.1396
2.0909
2.0419
1.9925
1.9427
1.8925
1.8418
SiL (kJ/kg K)
6.4323
6.4698
6.5079
6.5465
6.5857
6.6256
6.6663
6.7078
6.7502
6.7935
6.8379
MiL (kg)
172.86
170.86
168.92
167.02
165.17
163.36
161.60
159.87
158.18
156.53
154.91
So the final total mass of water is 154.91 kg; using the specific volume of liquid water at
150oC listed at the beginning of the problem, we have that the water occupies 0.1690 m3
leaving 0.8310 m3 for the steam. Using its specific volume, the final mass of steam is found to
be 2.12 kg. Using these results, we find that the final volume fraction of steam is 83.10%, the
final volume fraction of water is 16.90%, and the fraction of the initial steam + water that has
been withdrawn is
(172.86+6.28-154.91-2.12)/(172.86+6.28) = 0.1234 or 12.34%. A total of 22.11 kg of steam
has withdrawn, and 87.7% of the original mass of steam and water remain in the tank.
These results are similar to that from the energy balance. The differences are the result of
round off errors in the simple finite difference calculation scheme used here (i.e., more
complicated predictor-corrector methods would yield more accurate results.).
A simpler method of doing the calculation, avoiding numerical integration, is to assume that
the heat capacity and change on vaporization of liquid water are independent of temperature.
Since liquid water is a condensed phase and the pressure change is small, we can make the
following assumptions
U L | H L and H V H L 'H vap
dT L
dS L CPL dT L
dU L dH L
|
| CPL
|
; and
dt
dt
T dt
dt
dt
With these substitutions and approximations, we obtain from the energy balance
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
dH L dM L vap
dU L dM L V L
'H
o ML
H U
dt
dt
dt
dt
dT dM L vap
'H
M LCPL
dt
dt
Now using an average value of CPL and 'H vap over the temperature range we obtain
c
h
1 dM L
M L dt
or
ML
CPL dT
'H vap dt
CPL
150 200
'H vap
FG M IJ
HM K
L
f
L
i
ln
and from the entropy balance
dS L dM L vap
C L dT dM L vap
ML
'S
'S
o ML P
dt
dt
T dt
dt
Now using an average value of CPL and 'S vap over the temperature range we obtain
CPL dT
T'S vap dt
F
H
1 dM L
M L dt
150 27315
CPL
.
ln
200 27315
.
'S vap
or
I lnFG M IJ
K HM K
L
f
L
i
From the Steam Table data listed above, we obtain the following estimates:
U (T 200o C) U (T 150o C) 852.45 632.20
kJ
CPL
4.405
kg ˜ K
200o C - 150o C
50
or using the ln mean value (more appropriate for the entropy calculation) based on
FG T IJ SaT f SaT f
HTK
CPL ln
2
2
1
1
S(T
kJ
.
200o C) S(T 150o C) 2.3309 18418
4.3793
.
.
200 27315
47315
kg
˜K
ln
ln
.
.
150 27315
42315
Also, obtaining average values of the property changes on vaporization, yields
1
kJ
1
u 'H vap T 150o C 'H vap T 200o C
u 2114.3 1940.7 2027.5
'H vap
2
2
kg
1
1
kJ
4.5487
u 'S vap T 150o C 'S vap T 200o C
u 4.9960 41014
'S vap
.
2
2
kg ˜ K
With this information, we can now use either the energy of the entropy balance to solve the
problem. To compare the results, we will use both (with the linear average Cp in the energy
balance and the log mean in the entropy balance. First using the energy balance
M fL
CPL
4.405 u 50
ln
150
200
.
010863
L
vap
M
2027.5
'H
CPL
F
H
F
H
I
K
b
g
b
g
I
K
b
b
g
g
FG IJ
H K
i
M
L
f
L
i
exp 010863
.
.
089706
M
Now using the entropy balance
M fL
.
CPL
150 27315
ln
ln
L
vap
.
M
200 27315
'S
FG IJ
. I
. I
42315
F
I 4.3793 lnF 42315
0.9628 lnF
H
K
H
K
H
.
. K
4.5487
47315
47315
H K
M
. I
F 42315
.
089805
H 47315
. K
M
i
L
f
L
i
0.9628
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
Given the approximations, the two results are in quite good agreement. For what follows, the
energy balance result will be used. Therefore, the mass of water finally present (per m3) is
.
. kg
u M L initial 15506
M L final 0897
L
L
o
. u 0.001091 01692
.
occupying V M final u V 150 C 15506
m3
b
g
Therefore, the steam occupies 0.8308 m3 , corresponding to
0.8308 m3
2115
.
kg
m3
0.3928
kg
So the fraction of liquid in the tank by mass at the end is 155.06/(155.06+2.12) = 0.9865,
though the fraction by volume is 0.1692. Similarly the fraction of the tank volume that is
steam is 0.8308, though steam is only 2.12/(155.06+2.12) = 0.0135 of the mass in the tank.
0.8308 m3
V V 150o C
M V final
b
g
(c) Initially there was 6.28 + 172.86 = 179.14 kg of combined steam and water, and finally from
the simpler calculation above there is 155.06 + 2.12 = 157.18 kg. Therefore, 87.7% of the
total amount of steam + water initially in the tank are there finally, or 12.3% has been
withdrawn. This corresponds to 21.96 kg being withdrawn. This is in excellent agreement
with the more rigorous finite difference calculations done above.
4.16 (a)
dN
dt
0
N 1 N 2 ;
dU
dt
0
dV
N 1 H 1 N 2 H 2 WS Q P
dt
dS
dt
0
or N 2
N 1
Q
N 1 S 1 N 1 S 2 Sgen
T0
z
Tf
WS
= H 2 - H1 =
CPdT
N 1
298.15K
Sgen
N
W
WS N 1 H 1 N 1 H 2 or S = H 2 - H 1
N
1
S 2 S1
CP
37151
.
1
c
h
CP ˜ Tf 29815
. K
J
mol ˜ K
if the heat capacity is independent of
temperature. First consider the reversible case,
z
Tf
S 2 S1
0 gives
CP
dT
T
T
i
WSrev
N 1
Wact
CP ˜ 49914
. 29815
. K
125
. WSrev
(b)
dP
P
1
7467
c
The
J
.
mol
The
solution
is
499.14K.
actual
work
is
25%
Then
greater
h
J
CP ˜ Tf 29815
. K
mol
549.39 K
9334
The solultion is Tf
z
10
R
Repeat the calculation with a temperature-dependent heat capacity
CP T 22.243 5977
.
˜ 102 T 3499
.
˜ 105 T 2 7.464 ˜ 109 T 3
Assuming reversibility Tf = 479.44K. Repeating the calculations above with the temperaturedependent heat capacity we find Wact = 9191 J, and Tf =520.92K.
So there is a significant difference between the results for the constant heat capacity and variable
heat capacity cases.
Solutions to Chemical and Engineering Thermodynamics, 5th ed
4.17
Chapter 4
Ti = 300 K, Tf = 800 K, and Pi = 1.0 bar
CP (T ) = 29.088 - 0.192 u 10-2 T + 0.4 u 10-5 T 2 - 0.870 u 10-9 T 3
z
T f 800K
J
mol ˜ K
z
Pf
CP (T )
dT
T
T 300K
P
i
dP
P
P 1
i
Calculated final pressure Pf = 3.092 u 106 Pa.
z
T f 800K
Wrev
1458
u 104
.
CP (T )dT
Ti 300K
4.18
J
mol
Stage 1 is as in the previous problem.
Stage 2
Following the same calculation as above.
Stage 2 allowed pressure Pf ,2 = 9.563 u 107 Pa
Wrev = 1.458 u 10-4
J
= Stage 2 work
mol
Stage 3
Following the same calculation method
Pf ,3 = 2.957 u 10-9 Pa = Stage 3 allowed pressure.
J
= Stage 3 work
mol
Question for the student: Why is the calculated work the same for each stage?
Wrev = 1.458 u 104
4.19
The mass, energy and entropy balances are
dM
M
0, M
M
M
1
2
2
1
dt
dU
H M
H Q W ; M
H H W
0 M
s
s
1 1
2 2
1
1
2
dt
H H
Ws M
1
2
1
c
c
dS
dt
0
300q C, 5 bar
05
. MPa
100q C, 1 bar
01
. MPa
c
gen
M
1
2676.2 3064.2
S2 S1
h
0
S S S
M
gen
1 1
2
h
H 1 3064.2 kJ kg
S1 7.4599 kJ kg K
H
2676.2 kJ kg
2
S2
Ws
M
1
S
0;
h
S M
S Q S
M
gen
1 1
2 2
T
S S
Sgen M
1 2
1
c
h
7.3614 kJ kg K
388 kJ kg satisfied the energy balance.
7.3614 7.4599
0.0985 kJ kg K can not be. Therefore the process is impossible.
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
4.19
N
M
0.5 kg/s 100°C, steam
1 kg/s
e
300°C
1 MPa
Streams
1
2
3
0.5 kg/s 100°C, sat’d liq
T
300°C
100°C
100°C
P
1 MPa
Condition
sup heated
Ĥ kJ/kg
3051.2
Ŝ kJ/kg
7.1229
0.10135 MPa
sat’d liq
419.04
1.3069
Energy balance
kg
ˆ 0.5 u 1344.0 W
0
1 u 3051.2 0.5 u H
2
s
1.15 MW 1500 kJ
but W
s
ˆ 0.5 u 419.04 1500
3051.2 0.5 u H
2
Ĥ 2
2683.4
kJ
kg
ˆV
Now at at 0.1 MPa and T 100o C H
2676.1
kJ
kJ
and Sˆ 7.3549
kg
kg ˜ K
So steam is slightly superheated and Quality is 1
Entropy balance (using Ŝ at saturation)
1 u 7.1229 – 0.5 u 1.3069 – 0.5 u 7.3549 + S gen = 0
S
0.5 1.3069 7.3549 7.1229
gen
2.792 kJ / K S
So process is not possible
(using better estimate, by interpolation, for exit entropy of steam would have no effect on
conclusion).
4.20 Steam 20 bar
2 MPa and 300q C
H 30235
. kJ kg
S 6.7664 kJ kg
U 2772.6 kJ kg
Final pressure = 1 bar. For reference saturation conditions are
P 01
. MPa, T 99.63
L
H L 417.46
U
417.36
S L 13026
.
V
V
25061
26755
S V 7.3594
U
.
H
.
(a) Adiabatic expansion valve W 0 and Q 0
(from Steam Tables)
Solutions to Chemical and Engineering Thermodynamics, 5th ed
dM
;
M
M
0; M
M
2
1
1
2
dt
dU
H M
H
E.B.:
0 ; H 2 H 1
M
1 1
2 2
dt
From Steam Tables
T 250q C
.
Ÿ H2 30235
H 2974.3 kJ kg
Chapter 4
M.B.:
P
By interpolation T
kJ kg K
S 81245
.
dS
dt
S
gen
M
gen
300q C
H 3074.3 kJ/kg
275q C gives H = 3023.5 kJ / kg Ÿ all vapor
T
01
. MPa
S 8.0333 kJ kg K
S 8.2158 kJ/kg K
M
S S
MS
gen
1
2 2
0
S2 S1 = 8.1254 6.7664
kJ kg K
1359
.
(b) Well designed, adiabatic turbine
H M
H W
E.B.: M
1 1
2 2
cH H h
0 ; W
2
1
S M
S 0 ; S S ; S 6.7664 kJ kg K
S.B.: M
1 1
2 2
2
2
1
Ÿ Two-phase mixture. Solve for fraction of liquid using entropy balance.
.
6.7664
x ˜ 7.3594 1 x ˜ 13026
x 0.902 not good for turbine!
. 0.098 u 417.46
0.902 u 26755
H
2
W
M
.
2454.2 30235
569.3 kJ kg
W
M
2454.2 kJ kg
569.3 kJ kg
(c) Isorthermal turbine Ÿ superheated vapor
T 300q C
H 3074.3 kJ kg
final state
P 01
. MPa
S 8.2158 kJ kg K
H M
H Q W 0
E.B.: M
OP
Q
1 1
2
2
s
0
S M
S Q Sgen
S.B.: M
0
1 1
2 2
T
Q
S M
S
S S
M
M
1 1
2 2
1 2
1
T
Q
. 8.2158 6.7664 kJ kg K
T S2 S1
300 27315
M
c
c
W
s
M
h
h
830.7 kJ kg
Q
H 1 H 2
M
c
h 830.7 30235. 3074.3 779.9 kJ kg
Ÿ get more work out than in adiabatic case, but have to put in heat.
Solutions to Chemical and Engineering Thermodynamics, 5th ed
4.21
Chapter 4
contents of the compressor (steady-state, constant volume). Also, gas is ideal. (a) Mass balance
0 N N o N
N
System
1
Energy balance
2
2
1
N 1 H 1 N 2 H 2 0
dV
0
Ws P
Q
dt
adiabatic
0
Entropy balance
reversible
0
Q 0 compressor
Sgen Ÿ S 1
N 1 S 1 N 2 S 2 T
0
From the energy balance Ws
T T or Ws
NC
P 2
1
N
a
f
From the entropy balance S 1
S 2 Ÿ T2
T1
FG P IJ
HPK
S2
a
CP T2 T1
f
R CP
2
1
Thus
Ws
N
CPT1
LMF P I
MNGH P JK
R CP
2
OP
PQ
1
1
(b) Two stage compression, with intercooling, so that gas is returned to initial temperature, before
entering 2nd compressor
LF P I
work in stage 1
NC T MG J
MNH P K
LP
NC T MF I
work in stage 2 W
NH P K
L
T MFG P IJ
Total work W W
NC
MNH P K
R CP
WsI
1
II
s
R CP
2
P 1
*
I
s
R CP
II
1
To find P* for minimum work, set d Ws dP
R R F P I aR C f1 1
T |S G J
b g 0 NC
|T C H P K
Ÿ cP h
aP P f
d Ws
dP
P
P 1
P
P1
1
2 R CP
1 2
R CP
2
P 1
s
OP
PQ
O
1P
Q
P
F I
HP K
1 ; where P
P 1
2
0.
F Ia
H K
R P2
CP P
pressure after 1st compressor.
OP W
PQ
s
f F P I U|
R CP 1
H P K V|W
2
2
R CP
or
P
P1 P2
Students should check that this results in minimum, and not maximum work.
4.22
System: nitrogen contained in both tanks (closed, adiabatic, constant volume)
(1)
Mass balance: M1i M1f M 2f
f f
f f
(2)
Energy balance: M U
M U M U
1 1
1
P1 f
1
P2f
2
2
(3)
Final pressure condition:
For the entropy balance, the nitrogen in the first tank that remains in the tank will be taken as
the system. Then
(4)
S1i S1f
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
Equation (1) –(4), together with eqn. of state information of the form S S T , P , U U T , P
and V V T , P which we can get from Fig. 3.3-3 provides 4 eqns. for the 4 unknowns
T1 f , P1 f , T2f and P2f . Procedure to be followed in solution
(iv)
Guess a final pressure P f
Use eqn. (4) + Fig. 3.3-3 to compute T1 f , caluculate U1f
Use Fig. 2.4-3 to get V1 f , compute M1f V f
V1
f
f
f
f
V M
M
M M , and V
(v)
Use P and V2f to get T2f and U 2f
(i)
(ii)
(iii)
2
2
1
1
2
f
(vi) See if energy balance, Eqn. (2), is satisfied. If it is, guessed P f is correct. If it is not,
guess new p f , go back to (ii), and repeat calculation.
Some preliminaries
Figure
H 1i 368 kJ kg
T1i 250 K
 
o
2.4 - 3
V1i | 0.0037 m3 kg
P1i 200 bar
Thus M1i
V / V1i
0.01 m3 / 0.0037 m3 kg
2.703 kg
As a first guess, use ideal gas solution for pressure. (Also try some neighboring pressures.) My
solution is
P1 f
P2f
. bar same as ideal gas solution
1333
c
285 K cideal gas solution: T
T1
f
226 K ideal gas solution: T1 f
T2
f
f
2
h
330.4 Kh
222.8 K
4.23 (a) Set up just as in Problem 3.22 above. Solution after a number of iterations is P1 f
T1 f
2756
. q C and T2f
P2f | 5 bar .
497.7q C .
(b) Since now there is heat exchange between the two chambers we have T1 f T2f . This equation
is used instead of entropy balance. Solution procedure is to guess a final pressure, and then
compute final temperature using first the mass balance
M1f M 2f Ÿ
M1i
V1
V1i
V1 V2
Ÿ Vf
V f
V1 V2
V1 f V2 f
V1 V2 i
V1
V1
(1)
That is, choose T f until eqn (1) is satisfied. Then compute T f from energy balance i.e.
M1iU1i
M1f U1f M 2f U 2f
f
c M M hU Ÿ U
f
1
f
2
f
2
i
1
U f
f
(2)
When guessed P is correct, T computed from eqns. (1) and (2) using the Steam Tables will
be identical. My solution is P f 5 bar and T f # 366q C .
4.24
System
contents of turbine (open, constrant volume, steady-state)
dN
Mass balance:
0 N 1 N 2 Ÿ N 2 N 1
dt
constant
volume
Energy balance:
dN
dt
0
N 1 H 1 N 2 H 2 dV 0
0
Ws P
Q
dt
adiabatic
Solutions to Chemical and Engineering Thermodynamics, 5th ed
a
Ÿ Ws
N 1 H 1 H 2
Chapter 4
f N C aT T f for the ideal gas
1 P
1
2
Q 0 dN
Entropy balance:
Sgen
0 N 1 S 1 N 2 S 2 T
dt
P
T
N 1 CP ln 2 R ln 2
Ÿ Sgen N 1 S 1 S 2
P1
T1
or
R CP
Sgen
P
exp
T2 T1 2
P
N C
a
LM
N
FG IJ
H K
1
RS FG IJ
T H K
f
1 P
FG IJ UV
H KW
OP
Q
(a) For T2 to be a minimum, since Sgen t 0 and N 1 ! 0 , Sgen must be zero. Thus the minimum
outlet temperature occurs in reversible operation.
(b) Ws N 1CP T2 T1 . Since T1 ! T2 , the maximum work occurs when T2 is a minimum. Thus,
W is a maximum (in magnitude) for a reversible process.
a
f
s
4.25 (a) For any system:
LM M S Q S OP
T
Q
N
dS
dt
i i
gen
S t 0 or ¦ M
S d 0 . Also, S
depending 0n the process Q t 0 or Q d 0 and ¦ M
gen
i i
i i
0 or
Sgen ! 0 , depending on whether or not the process is reversible. Thus, dS dt for a system can
be greater than, less than, or equal to zero.
Since, by definition, the universe contains everything, it must be a closed system and
adiabatic, since there is nothing for the universe to exchange mass or heat with. Therefore
dS
dt
dS
0 0 Sgen Ÿ
dt
Sgen t 0
Thus the entropy of the universe can not decrease, and the statement is true.
(b) Consider the change from any state 1 to any state 2 in a closed system. The energy and entropy
balances for this transformation are:
1
2
U2 U2
S2 S1
Q W
Sgen
RS
T
W Since the process is adiabatic
If the transformation is possible, then Sgen t 0 now consider the transformation from state 2 to
state 1. Here
3
4
U1 U1
S1 S2
W
Sgen
Comparing eqns. (1) and (3) we have W W (This is ok).
Comparing eqns. (2) and (4) we have Sgen Sgen
(5)
Separately we have, if the processes are possible, that Sgen t 0 and Sgen t 0 . The only way that
all these three equations for Sgen and Sgen can be satisfied is if Sgen
Sgen
0 , that is, both
processes are reversible. Generally, processes are not reversible. However, eqn. (5) requires
that only one of Sgen and Sgen can be greater than zero. Thus,
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
If Sgen ! 0 1 o 2 is possible, but 2 o 1 is not possible.
If Sgen ! 0 2 o 1 is possible, but 1 o 2 is not possible.
4.26
W
1 § T Tamb · ¨ c
¸Q
2 © TC
¹
Q
q k Tc Tamb
Heat flow into
collector
Convective heat loss from collector
1 § Tamb ·
¨1 ¸ q k Tc Tamb
2 ©
Tc ¹
W
1
1
1 Tamb
1
1 T2
q k Tc Tamb q kTamb k amb
2
2
2 Tc
2
2 Tc
dW
dTc
1
k
2
kTc2
1 Tamb
1 T2
q k amb
2
2 Tc
2 Tc2
2
qTamb kTamb
Tc
4.27
0 (for a maximum)
amb
qT
2
Tamb
k
q
1
kTamb
Tamb
System: contents of the tank at any time (open, adiabatic, constant volume system).
dN
(a) Mass balance:
N
dt
PV
PV
ideal gas law N
; N
where V volumetric flow rate.
RT
RT
d P
P
d PV
PV
Thus,
ŸV
V since V and V are both constant.
dt T
T
dt RT
RT
d
P
V
P
P
Vt
bar
or
or
exp 1082
.
u 103
ln
dt
T
V
T 5 min T 0
V
K
F I
H K
F I
H K
F I
H K
P 5 min
Energy balance:
or N
dU
dt
FG IJ
H K
1082
.
u 103 T 5 min
d
NU
dt
dN
dU
N
dt
dt
dT
NCV
dt
N H Ÿ U
dN
dN
H U Ÿ RT
dt
dt
^P
K`
bar, T
H
dN
dt
FG IJ
H K
RC
P
P
dP P
dT
or T2 T1 2
CP
P1
dt
T
dt
[Note: could have gotten this result from the entropy balance also!]
using N
PV RT yields R
(1)
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Ÿ T 5 min
Chapter 4
F P 5 min I
H 1 bar K
8.314 39
(2)
340
simultaneously solving equations (1) and (2) yields
0.281 bar and T 5 min
P 5 min
259.3 K
S out or T3
(b) Since pump is adiabatic and reversible, S in
T2
FG P IJ
HPK
R CP
3
since P3
P1 . This
2
equation implies that T3
4.28
T1
340 K .
Number of moles of gas in tank initially
N 0
N0
N 0
15 bar u 0.2 m3 u 105 J m3 bar
8.314 J mol K u 22 27315
. K
P 0V
RT 0
122.26 mol
dN
N 4.5 mol min Ÿ N t 122.26 4.5t mol t min
dt
(a) Entropy balance on an element of gas that remains in the tank (see Illustration 3.5-2) yields
S t S 0
0Ÿ
Tt
T0
FG P t IJ
HP0K
R CP
Ÿ T (t )
FPt I
H 15 K
0.3779
29515
.
(1)
From the ideal gas equation of state
V
R
Tt N t
Pt
PtV
N t RT t Ÿ
Tt
Pt
29515
. K
122.26
u
15 bar
122.26 4.5t
T0N 0
P0
Thus
19.68
1 0.03681t
(2)
Now using eqn. (1) in eqn. (2) to solve for P t and T t yields
Pt
15 u 1 0.03681t 1.6075
Tt
. u 1 0.03681t 0.6075
29515
But T t is temperature in the tank. What about temperature of gas leaving the throttling valve?
Gas going thru valve undergoes a Joule-Thomson expansion Ÿ H in H out . Since gas is ideal,
this implies Tin
Tout . Thus, T t out of valve
29515
. u 1 0.03681t 0.6075
(b) If tank is isothermal, then, instead of eqn. (2), we have
Pt
N t
RT
V
P0
ŸPt
N 0
15 1 0.03681t bar
Solutions to Chemical and Engineering Thermodynamics, 5th ed
and T t
constant
Chapter 4
29515
. K
Summary
T K
P bar
0.6075
. 1 0.03681t
15 1 0.03681t 1.6075
Adiabatic 29515
.
15 1 0.03681t
Isothermal 29515
4.29
This is a tough problem!
Subscript 1 denotes properties in initally filled tank
Subscript 2 denotes properties in initially "evacuated" tank
We will use i and f (superscripts) to denote initial and final properties, and we will assume
negligible mass hold-up in engine.
1) Mass balance on closed system consisting of both tanks
N1i N 2i
N1f N 2f Ÿ
P1iV1 P2iV2
i
T1i
T2
P1 f V1 P2f V2
T1 f T2f
but
P1 f
Pf Ÿ Pf
P2 f
FG 0.3 0.75IJ 14.0 u 0.3 0.35 u 0.75
H T T K 97315
29815
.
.
f
f
1
2
bar m3
u 10
5196
.
K
(1)
3
2) Entropy balance on gas contained in tank 1 initially and finally. This is a closed, adiabatic,
reversible system
dS
dt
0 Ÿ S1i
FG T IJ R lnFG P IJ
HT K HP K
(2)
F P IJ
97315
. G
H 14.0K
(3)
S1f Ÿ CP ln
f
f
1
1
i
1
i
1
Thus
FG P IJ
HPK
f
T1
f
T1i
R CP
f
1
27
1
i
1
Equation (2) implies that T1 and P1 are related as follows
d ln T1
dt1
R d ln P1
CP dt
(4)
3) Mass and energy balances on tank 1
or N1CV
dN1
dt
d
N 1 { N ;
dt
dT1
dt
dN1
H1 U 1
dt
a
a N U f N H
1
1
f RT dNdt
1
1
1
RT1 N
U1
dU 1
dN1
N1
dt
dt
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
using eqn. (4) gives
N1
CV d ln P1
CP dt
dN1
dt
N
(5)
4) balances on the engine: adiabatic, reversible (for maximum work) and since no hold-up of
mass, dN1 dt 0
Subscript eng refers to gas leaving engine and going into tank 2.
0 H 1 H eng N Ws
d
i
0 dS S i N
1
eng
Ÿ S1
S eng ;
Also P2
Peng
c
Ÿ Teng
T1 Peng P1
h
a
R CP
f
T1 P2 P1 R CP
(6)
(Note that Teng z T2 )
and
c
Ws
T T
NC
P 1
eng
h
(7)
5) balances on tank 2 [Note, irreversible mixing occurs unless, fortuitously, Teng
T2 at all times
(can this occur?). Thus, Sgen ! 0 , and entropy balance gives no useful information]
N
Mass balance: dN 2 dt
a
f U dN N dU N H dN H But,
dt
dt
dt
H C aT T f
UV where T reference temperature
U C aT T f RT C T C T W
Energy balance:
P
0
V
0
d N 2U 2
dt
2
2
2
2
2
eng
eng
0
0
Ÿ
dN 2
H eng U 2
dt
or
d
V
P 0
i dNdt mC T C T C T C T r N C dTdt
2
2
P eng
P 0
m
N CPTeng CVT2
V 2
P 0
2 V
r N C dTdt
2
(8)
2 V
and
N
IJ PV dT V dP
a f dtd FGH PV
RT dt
RT K
RT dt
d
N2
dt
2 2
2
2
2 2
2
Ÿ
dT2
dt
2
2
2
2
RT22 N T2 dP2
PV
P2 dt2
2 2
(9)
using eqns. (6) and (9) in eqn. (8)
R| F I
S| GH JK
T
P
N CPT1 2
P1
R CP
CVT2
U| N C ˜ RT N N C T dP
V|
PV
P dt
W
2
2
2 V
2
2
2
2
2 V
2 2
PV
T dP
CVT2 N 2 2 ˜ CV 2 2
RT2
P2 dt2
Solutions to Chemical and Engineering Thermodynamics, 5th ed
FG IJ
H K
T P2
NC
P 1
P1
R CP
Chapter 4
CVV2 dP2
R dt
and using eqn. (5)
N
FG IJ
H K
CV
P
CPT1 2
CP
P1
R CP
d ln P1
dt
CV
V2 dP2
R dt
or, finally
FG P IJ
HPK
R CP
dP1
dt
1
dP
V2 P2 R CP 2
dt
V1
2
V2
dP2
dt
V1 P1 R CP
dP1
dt
or
z
dP2
V2
R CP
P
Pi 2
2
z
P1 f
P2f
V1
dP1
R CP
P
Pi 1
1
Ÿ
{c h
2.5 P f
Ÿ Pf
CV CP
0.35 CV CP
} {c P h
3053
.
bar: using Eqn. (3), T1 f
f CV CP
14.0 CV CP
}
629.8 K .
f
Now using eqn. (1), T2
6119
. K
Finally, to get the total work, we do an overall energy balance (system
closed, constant volume).
N f U 1f N 2f U 2f N1i U 1i N 2i U i2 Ws
Ws
4.30
m a
f
two tanks; adiabatic,
r
CV
P f V1 V2 P1iV1 P2iV2
R
5
^3053
.
0.3 0.75 14 u 0.3 0.35 u 0.75`
2
3142
u 105 J 314.2 kJ
bar m3 3142
.
.
Note: be careful about coordinate system. A mass flow in the negative u direct is negative!
L
Heat exchanger is
in steady-state operation
+x
'L
Case I: Concurrent flow
Mass balance on shaded volume
Solutions to Chemical and Engineering Thermodynamics, 5th ed
mass in
element at
time t 't
mass in
element
at time t
Chapter 4
mass in
at face
at L in 't
b
mass out
at face at
L 'L in 't
g
M
M
0
L
L 'L 't
steady - state
M
M
M
L 'L
L
Energy balance on the shaded volume
energy in
energy in
element element
at t 't
at t
energy flow in
by heat flow
in time 't
energy flow in
energy flow out
by mass flow by mass flow at
at L in 't
L 'L in 't
(steady-state)
H 't M
0 M
L L
L 'L H L 'L 't Q'L't
H
H
T
M
MC
T
Q 'L
a
L 'L
L
f
P
a
L 'L
L
f
dividing by 'L , taking limit as 'L o 0 , and using subscript 1 to denote fluid 1
a
C dT1
M
1 P ,1
dC
Q N T2 T1
f
Q heat flow rate per unit length of exchanger.
Similarly, for fluid 2 (other part of exchanger)
C dT2
M
2 P ,2
dL
a
Q
N T2 T1
f
and M
are both + for concurrent flow)
(M
1
2
Adding the 2 equations
C dT1 M
C dT2 0
M
1 P ,1
2 P ,2
dL
dL
From problem statement, M
1
a
and C
M
2
P ,1
f
dT1 dT2
d
T1 T2 0 or T1 T2
dL dL dL
and T1 C T2 ; T2 C T1
now going back to
Ÿ
dT1
dL
and integrating
N
MC
P
CP,2
constant
C
2T f
aT T f N aCMC
1
2
1
P
Solutions to Chemical and Engineering Thermodynamics, 5th ed
FG C 2T IJ 2NL
H C 2T K MC
f
T1 f
15q C , T1i
1
ln
i
1
35q C , T2f
P
5q C , T2i
Chapter 4
MC
P
2N
L
; L0
L0
15q C .
Also, C T1i T2i T1 f T2f 20q C
[ i initial conditions, conditions at L
L* length of exchanger]
Using this in equation above gives
0; f
final conditions, conditions at L * where
F 20 70I ln 5 1609
.
H 20 30K
L*
L0
ln
And, more generally, at any point in the exchanger
FG L IJ Ÿ T L C FG1 expFG L IJ IJ T expFG L IJ
H L KK
H LK
H LK
2H
F LI
T L 10 25 expG J q C
H LK
F LI
T L C T L 10 25 expG J q C
H LK
C 2T1 L
C 2T1i
exp i
1
1
0
0
0
1
0
2
1
0
Now writing an entropy balance
S 't M
M
L L
L 'L S L 'L 't
0
ŸM
dS
dS
dL
Q
T1
a
N T2 T1
f
Q
'L't
T
N 50 exp L L0
.
10 25 exp L L0 27315
a
T1
a
a
f
f
need absolute
T here
FG IJ
a fH K
25 expa L L f
F LI
dG J
C
28315
. 25 expa L L f H L K
28315
. 25 expa L L f U
Ÿ S L S L 0 C lnRS
VW
30815
.
T
f
L
50NL0
exp L L0
d
M 28315
L0
. 25 exp L L0
0
p
0
0
0
p
Case II Countercurrent flow
M
M
2
1
dT
1
C
(1)
M
N T2 T1
1 P
dL
35
15q C
dT2
(2)
M 2CP
N T2 T1
-15
5q C
dL
C dT2 N T T
(3)
M
1 P
2
1
dL
. Subtracting eqn. (1) from eqn. (3) gives
Eqn. (3) comes from eqn. (2) using M
M
2
1
a
f
a
a
f
f
Solutions to Chemical and Engineering Thermodynamics, 5th ed
a
f
d
T1 T2 0 Ÿ T1 T2
dL
T2 T1 30q C
Chapter 4
constant
30q C
C
Thus
a
f
C dT1 N T T
30N
M
1 P
2
1
dL
30NL
T1i
T1 MC
P
30NL
L
L
35 35 15
T2 5 15
MC
L
L
P
0
0
MC
P
and the entropy balance
2N
where L0
dS1
M
dL
dS
Q
T1
N 30
T1
N 30
27315
M (35 30NL MC
. )
P
1
dL
absolute
temperature
needed here
15C
F LI
FG IJ
dG J
HL K
K
H
L
L
.
30815
15
g
L
30NL0
d
30815
M
L
. 30NL MC
0
P
dS1
b
CP
dx
. x
2054
FG
H
L
S 0 CP ln 1 0.048
L0
Ÿ S L
P
0
IJ
K
Summary
Concurrent flow
FG L IJ
H LK
F LI
T 10 25 expG J
H LK
T1
10 25 exp 0
2
0
S L
S L
0 CP ln
. 25 expa L L f U
RS 28315
VW
.
30815
T
0
Countercurrent flow
T1
35 15
T2
5 15
S L
4.31 (a) L
L0
L
L0
FG
H
L
S 0 CP ln 1 0.048
L0
dU
dt
dV
WS Q P
dt
JIK
a
f
dV
dV
WS Q P0
P P0
dt
dt
0
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
and
dS
dt
Q Sgen
T0
Now let
a
f
dV
dV
P P0
WS P0
dt
dt
W
and
Wu
a
dV
W P0
dt
f
dV
dU
Ÿ
WS P P0
dt
dt
dV
Wu Q P0
dt
or
a
f
U 2 U1
Wu Q P0 V2 V1
S2 S1
Q
Sgen Ÿ Q
T0
U 2 U1
Wu T0 S2 T0 S1 T0 Sgen P0 V2 V1
T0 S2 T0S1 T0Sgen
and
Wu
a
f
0 1
0 gen
aU PV T S f aU PV T S f T S
2
0 2
0 2
1
0 1
since T0 Sgen t 0
A 2 A1, where A
Wumax
(b)
0
0
Here WS
U PV
0 T0 S
H M
H Q W
M
S
1
2
Q
S M
S S
M
gen
1
2
T0
Wu
Ÿ Wu
H M
H Q
M
2
1
Ÿ Wu
H T S M
H T S T S
M
0 2
0 1
0 gen
2
1
a
Since T0 Sgen t 0
Wumax
b
f a
g
H M
H MT
S MT
S T S
M
0 1
0 2
0 gen
2
1
B B where B
M
2
1
f
H T0 S
(c) Using the Steam Tables we find
i)
at 30 bar 3 MPa and 600qC
U 32850
. kJ kg , S 7.5085 kJ kg K , V
013243
.
m3 kg
Solutions to Chemical and Engineering Thermodynamics, 5th ed
1
A
Chapter 4
U PV
0 T0 S
32850
bar 013243
m3 kg u 102 kJ bar ˜ m3 29815
. 1013
.
.
. u 7.5085
1059.76 kJ kg
ii)
. MPa and 300qC
at 5 bar 05
U 2802.9 kJ kg , S 7.4599 kJ kg K , V
2
A
05226
.
m3 kg
2802.9 1013
.
u 05226
.
u 102 29815
. u 7.4599
Wu
2A
1
A
(63167
. 1059.76) kJ kg
63167
. kJ kg
428.09 kJ kg
This is the maximum useful work that can be obtained in the transformation with the
environment at 25qC and 1.013 bar. It is now a problem of clever engineering design to
develop a device which will extract this work from the steam in a nonflow process.
bar, any component which
(d) Since the inlet and exit streams are at 25qC and P 1013
.
passes through the power plant unchanged (i.e., the organic matter, nitrogen and excess
oxygen in the air, etc.) does not contribute to the change in availability, or produce any
useful work. Therefore, for each kilogram of coal the net change is:
0.7 kg of carbon
58.33 mol of C
58.33 mol of O2
to produce 58.33 mol CO2
also
015
. kg of water
8.33 mol of H 2O undergoes a phase change
from liquid to vapor
Therefore
B
in
M
¦ b N i Bi gin
i
B
out
M
58.33 u 0 58.33 u 0 8.33 u 68.317 29815
. u 0.039
carbon
oxygen
liquid water
1976 kJ kg coal
58.33 u 94.052 8.33 u 57.8 29815
. u 0.0106
¦ N B
b
i
i
g
i out
carbon dioxide
water vapor
24858 kJ kg coal
Wumax
Wuactual
24858 1976 kJ kg coal
22882 kJ kg coal
2.2 kW - hr kg coal 7920 kJ kg coal
7920 u 100
Efficiency in %
34.6%
22882
Thus a coal-fired electrical power generation plants converts slightly more than 1/3 of the
useful work obtainable from the coal it consumes. This suggests that it would be useful
to look for another method of generating electrical power from coal . . . for example,
using an electro-chemical fuel cell. Considering the amount of coal consumed each year
in power generation, and the consequences (strip mining, acid rain, greenhouse effect,
etc.) the potential economic savings and environmental impact of using only 1/3 as much
coal is enormous.
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
4.32 Three subsystems: unknowns T1 f , P1 f , T2f , P2f , T3f , P3f (6 unknowns)
After process P1 f P2f P3f (2 equations)
Subsystem 1 has undergone a reversible adiabatic expansion
Ÿ S 1f
FG P IJ
HPK
f
S 1i , or T1 f
T1i
R CP
1
(1 equation)
i
1
(#1)
Subsystem 3 has undergone a reversible adiabatic compression
Ÿ S 3f
S i3 , or T3f
FG P IJ
HPK
f
3
i
3
T3i
R CP
FG P IJ
HPK
f
T3i
R CP
(1 equation)
i
3
(#2)
Mass balance subsystems 1 + 2
N1f N 2f
or
Pf
N1i N 2i Ÿ
P1 f V1 P2f V2
T1 f
T2f
P1iV1 P2iV2i
i
T1i
T2
FG 05. V IJ 10 u 05. 1 u 0.25 0.017909 (1 equation)
29315
.
.
H T T K 29315
f
1
f
2
f
2
(#3)
Energy balance on subsystems 1 + 2 + 3
N1f U 1f N 2f U 2f N 3f U 3f
N1i U 1i N 2i U i2 N 3i U i3
P fV
P fV
P1 f V1
CVT1 f 2 f2 CVT2f 3 f3 CVT3f
f
RT3
RT2
RT1
c
P f V1 V2f V3f
but V1 V2f V3f
V1 V2i V3i
h PV PV PV
i
1 1
05
. 0.25 0.25
i
2 2
in eqn. (1) o T1 f
252.45 K
in eqn. (2) o T3
448.93 K
f
P3i T3f
˜
P3f T3i
0.25 u
i
3 3
1 m3
10 u 05
. 1 u 0.25 1 u 0.25
1
Pf
using this result
PiV
PiV
P1iV1
C T i 2 2i CVT2i 3 i3 CVT3i
i V 1
RT3
RT2
RT1
1 448.93
u
. 29315
.
55
V3f
V3i
V2f
0.25 u 2 0.06961 0.4304 m3
55
. bar
0.06961 m3
Now using Eqn. (#3)
Pf
FG 05. V IJ 55. FG 05. 0.4304 IJ 0.017909 Ÿ T
H T T K H 252.45 T K
f
1
f
2
f
2
f
f
2
2
Thus the state of the system is as follows
T1
P1
Initial
293.15 K
10 bar
Final
252.45 K
5.5 bar
337.41 K
Solutions to Chemical and Engineering Thermodynamics, 5th ed
T2
P2
V2
T3
P3
V3
Chapter 4
293.15 K
1 bar
337.41 K
5.5 bar
0.25 m3
293.15 K
0.4304 m3
448.93 K
1
5.5 bar
3
0.25 m
0.0696 m3
Work done on subsystem 3
Energy balance
N 3f U 3f N 3i U i3
W
z
PdV
P3f V3f
P3iV3i
f
CVT3i
C
T
V 3
RT3i
RT3f
W
P3f V3f
CV
C
P3iV3i V
R
R
W
CP R f f
. u 0.0696 1 u 0.25
P3 V3 P3iV3i 3 55
R
0.3984 bar ˜ m3 39.84 kJ
c
h
From simple statics the change in atmospheric pressure dP accompanying a change in height dh is
dP Ugdh
where U is the local mass density and g is the gravitational constant. Assuming a packet of air undergoes an
altitude change relatively rapidly (compared to heat transfer), the entropy change for this process is
CP
R
dS
dT dP 0 since both Q and Sgen equal zero.
T
P
Combining the two equations above we have
CP
R
R
R N
M
dT
dP Ugdh Mgdh gdh
T
P
P
PV
T
Mg
dT
or
dh
CP
K
dT
dT
For dry air
# 9.7
is referred to as the adiabatic lapse rate.
. Note that
km
dh
dh
Also, its value will be less than that above as the humidity increases.
In fact, if the humidity is 100%, so water will condense as the pressure decreases, the adiabatic lapse rate
will be almost zero.
4.33
4.34
W
-10 kW
M
i
300°C
1 MPa
1 kg/s
Mass balance
100°C
Steady State
N
300°C
Sat’d L
Solutions to Chemical and Engineering Thermodynamics, 5th ed
0
Chapter 4
0.5
1 M
1
M
M
M
i
1
2
0 Ÿ M
1
0.5 kg/s
Energy balance
0
assume no heat loss Q
1344.0 kJ/kg
Ĥ 2
10 kw
10
kJ
s
O
3.2534 kJ/kg ˜ K
Sˆ i
1344.0 kJ/kg Sˆ 2
Ĥ i
0 1
W
H
ˆ
ˆ
ˆ
M
i i 0.5 H1 0.5 H 2 W
3.2534 kJ/kg ˜ K
kg
kJ
ˆ 0.5 u 1344.0 10 kJ
u 1344.0
0.5 u H
1
s
kg
s
ˆ
0.5 H
1
1344.0 0.5 u 1344.0 10
Ĥ1 =1324.0 kJ/kg
662
at 100°C
HV = 2676.1
At 100°C HL = 419.04
1324 = x (2676.1) + (1-x) 419.04
1324 – 419.04 = x (2676.1 – 419.04)
x = 0.401
quality
where x = quality
Ŝ1 = 0.401 × 7.3549 + 0.599 × 1.3069 = 3.7322 kJ/kg.K
Entropy balance
0
Sˆ M
Sˆ M
Sˆ S
M
i i
1 1
2 2
gen
where S gen t 0
1u 3.2534 0.5 u 3.7322 0.5 u 3.2534 S gen
0.2394 S gen
O
Ÿ S gen
kJ
kg ˜ K ˜ s
0.2394
0
so device is possible
4.35 a) Basis: 1 mole air
Energy balance
Uf Ui
WQ
0
C v Tf Ti
Entropy balance
Sf Si 0 Ÿ Sf Si
Ÿ S Tf , Vf S Ti , Vf
ln
C*V ln
Tf
V
R ln f
Tf
Vi
0
8.314 § 1 ·
ln ¨ ¸ 1.2743
21
© 25 ¹
Tf
Ti
Tf
Ti e1.2743
W
8.314
303.15e1.2743
1084.1 K
J
u 1084.1 303.15 K
mol K
6493
J
mol
6.493
kJ
mol
6.493 kJ / mol
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
b)
ln
Tf
Ti
8.314 § 1 ·
ln ¨ ¸
21
© 10 ¹
0.9116
Tf
303.15 u e0.9116
754.33
W
8.314 u 754.33 303.15
3751.1
J
mol
3.751
kJ
mol
4.36
N
M
2 MPa, 200°C
200 kPa, 100°C
Find initial number of moles in each compartment
PV=NRT
N
PV
RT
200 kPa u 100 u 103 cm3 u
N1
kPa m3
8.314 u 10
u 373.15 K
mol K
a)
8.314 u 103
6.447 moles
3
2 u 103 kPa u 2 u 105 cm3 u
N2
1 m3
106 cm3
1 m3
106 cm3
kPa m3
u 473.15 K
mol K
Mass balance: Nf – N1 – N2 = 0
101.68 moles
Nf = N1 + N2 = 6.45 + 101.68
Nf = 108.13
Energy balance: Nf Uf – N1 U1 – N2 U2 = 0
6.448 × CV (Tf – 373.15) + 101.68 × CV (Tf – 473.15) = 0
(6.447 + 101.68) Tf = 6.447 × 373.15 + 101.68 × 473.15
6.447 u 373.15 101.68 u 473.15
= 467.19 K
Tf
6.447 101.68
Final Pressure
kPa m3
u 467.19 K
NRT
mol K
P
V
m3
3 u 105 cm3 u 106
cm3
3
2
2
1.0813 u 10 u 8.314 u 10 u 4.6719 u 10
14.0 u 102 kPa 1.4 MPa
3 u 101
108.16 mol u 8.314 u 103
b)
Entropy Change
N f Sf N1S1 N 2 S2
Sgen t 0
­
­
T
P ½
T
P ½
N1 ®CP ln f R ln f ¾ N 2 ®CP ln f R ln f ¾ Sgen
T1
P1 ¿
T2
P2 ¿
¯
¯
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
467.9
1.4 ½ J
467.9
1.4 ½
­
­
8.314 ln
101.68 ®30 ln
8.314 ln
6.45 mol u ®30 ln
¾
¾ Sgen
373.15
0.2 ¿ mol K
473.15
2 ¿
¯
¯
0
6.45 u ^30 u 0.2263 8.314 u 1.9459` 101.68 ^30 u 0.01116 8.314 0.35667 ` =
60.56
J
J
267.47
K
K
Sgen
Q
4.37
206.9
J
t 0! (as it should be)
K
0
Assume steady state operation
N
M
2 bar = 0.2 MPa
20 bar
400°C
W
M
M
0
1
2
2 bar = 0.2 MPa
M
2
M
1
Entropy balance
0 0 for maximum work
Q
ˆ
ˆ
S M
S S
0 M
1 1
2 2
gen
T
Ÿ Sˆ 2 Sˆ 1
Sˆ 1 Sˆ 2 MPa, 400qC 7.1271
Ĥ1
Ŝ2 0.2MPa, T
at T
?
kJ
kg ˜ K
3247.6 kJ / kg
7.1271 kJ / kg ˜ K
200qC Sˆ 7.5066
150qC Sˆ 7.1272
kJ / kg K
ˆ
H
2870.5 kJ / kg
kJ / kg
Ĥ
2768.8 kJ / kg
Ÿ
T2 = 150°C
Energy Balance
0
ˆ
H
ˆ
0 M
1 1 M1 H 2 W Q
W
5000
kg
kJ
9247.6 2768.8
hr
kg
2.394 u 106
kJ
hr
665
kJ
S
665 kW
4.38
20°C
T=?
1 bar
5 bar
Heater
a) Assume adiabatic and reversible operation of compressor
Ÿ Isentropic
S (20°C, 1 bar) – S (T = ?, 5 bar) = 0
C*p ln
T
5
R ln
293.15
1
0
400°C
5 bar
Solutions to Chemical and Engineering Thermodynamics, 5th ed
R
Chapter 4
§ 5 · Cp
T 293.15 ¨ ¸
462.7 K since C*p C*v R
©1¹
E.B. H (20°C, 1 bar) – H (462.7 K, 5 bar) + W = 0
W = H (462.7 K, 5 bar) – H (293.15 K, 1 bar)
Cpx 462.7 293.15 4971.2 J / mol
*
21 8.314
29.314
J
mol ˜ K
b) Heater
H (462.7 K, 5 bar) – H (673.15 K, 5 bar) + Q = 0
Cpx 673.15 462.7 6169.1 J / mol
4.39 From Fig. 3.3-2
ˆ
At 160 K
H
260 kJ / kg
L
ŜL 2.05 kJ / kg K
ˆ # 0.00295 m3 / kg
V
L
If adiabatic and reversible, Sgen
0
Sˆ initial
Sˆ initial
?, P
? 0.25 Sˆ V T
2.05 kJ / kg
0.75 Sˆ L T
Sˆ final
?, P
?
but since it is an equilibrium V-L mixture T & P are connected by the saturation line
ˆ
Sˆ V 5.03 Ÿ Sˆ
Guess T = 120 K SL 1.03
Ŝ 0.75 u 1.03 0.25 5.03 2.03
ˆ
H
L
ˆ
V
L
ˆ
H
V
3
ˆ
0.0025 m / kg VV
105 kJ / kg
ˆ #H
ˆ
U
L
L
ˆ
U
V
ˆ
U
f
0.3 m3 / kg
P
0.2 MPa
105 kJ / kg
ˆ PV
ˆ
H
L
L
595
kJ
Pa
1J
kJ
0.2 MPa u 106
u 3 u 103
kg
MPa m Pa
J
595
kJ
kJ
60
kg
kg
ˆ 0.25 U
ˆ
0.75 U
L
V
ˆ H
ˆ
U
i
i
595 kJ / kg
Close Enough
535
kJ
kg
212.5 kJ / kg
260 kJ / kg
Energy balance on piston-cylinder (closed system)
o
ˆ U
ˆ
U
Q
W 212.5 260 47.5 kJ / kg
f
i
(Negative sign Ÿ system does work on surroundings)
W
4.40
a)
20 bar
550°C
20 bar = 2 MPa = 2,000 kPa
From the superheated steam tables
1 bar
0.1 MPa
Adiabatic and Isentropic
Ÿ Sˆ IN Sˆ OUT
Ÿ Vertical line of Fig. 3.3-1b
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Ĥ (20 bar, 600°C) = 3690.1
Chapter 4
kJ/kg
kJ/kg K
Ŝ (20 bar, 600°C) = 8.6451
Estimate From Fig. 3.3-1b (vertical line down to 1 bar)
ˆ 2820 kJ / kg Also all vapor
T # 172qC
H
Steady state energy balance
W
H
ˆ
ˆ
ˆ
ˆ
0 M
H
IN IN M OUT H OUT W Ÿ OUT H IN
M IN
W
kJ
870.1 kJ u 5000 kg 1327.2 kJ
3690.16 2820 870.1
W
kg
kg
hr
s
M
1327.2 kW
IN
If only 90% of the work generated, W 0.9 u 870.1 783.1 kW
kJ
All vapor
so Ĥ out 3690.1 783.1 2907.0
at 1 bar (0.1 MPa)
kg
By interpolation using the Superheated Steam Tables
kJ
T 216o C and S=7.8983
kg ˜ K
b)
0
Entropy balance
dS
Sˆ M
Sˆ Q S
0 M
in in
out out
gen
dt
T
S gen
kJ
7.8983 7.7024 0.1959
or S gen
kg ˜ K
M in
979.5
kJ
hr ˜ K
4.41 The gas in the tank undergoes a uniform adiabatic expansion, which is isentropic, so that its
temperature and pressure are related by
R / C*p
§P ·
TT,f TT,i ¨ T,f ¸
¨P ¸
© T,i ¹
The gas in the cylinder to the right of the piston undergoes a uniform adiabatic compression, which is also
isentropic, so its temperature and pressure are related by
TCR ,f
§P
·
TCR ,i ¨ CR ,f ¸
¨P ¸
© CD,i ¹
R / C*p
Also PT,f = Pc,f.
An energy balance on the whole system gives
(UT,f + UCL,f + UCR,f) – (UT,I + UCR,i) = 0
NT,f UT,f + NCL,f UCL,f + NCR,f UCR,f = NT,i UT,i + NCR,i UCR,i
Also, the mass balance
NT,f + NCL,f + NCR,f = NT,i + NCR,i
Now using the ideal gas law
V
V
V
M.B.
Pf TT Pf CL,f Pf CR ,f
RTCL,f
RTCR ,f
R Tf
E.B.
PT,i
V
VT
PCR ,i CR ,i
RTT,i
RTCR ,i
PV
PV
Pf VT
CV TT,f t CL C v TCL,f f CR ,f C V TCR ,f
RTT,f
kTCL,f
RTCR ,i
These equations reduce to
PT,i VT
RTT,i
C V TT,i PCR ,i VCR ,i
RTCR ,i
CV TCR ,i
Solutions to Chemical and Engineering Thermodynamics, 5th ed
ªV
V
V º
Pf « T CL,f CR ,f »
¬« TTf TCL,f TCR ,f ¼»
and
Pf ª¬ VT VCL,f VCR ,f º¼
but
VCL,f VCR ,f
Pf
TT,i
PCR ,i VCR ,i
TCR ,i
PT,i VT PCR ,i VCR ,i
0.5 m3
VCR ,i
Pf ª¬ VT 0.5 m3 º¼
PT,i VT
Chapter 4
0.4 VT 0.1u 0.5 m3
0.4VT 0.1u 0.5
VT 0.5
0.4 u 0.25 0.1u 0.5
0.25 0.5
0.2 MPa
8.314
TT,f
§ 0.2 · 29.3
473.15 ¨
¸
© 0.4 ¹
TCR ,f
§ 0.2 ·
288.15 ¨
¸
© 0.1 ¹
8.314
388.7 K
29.3
350.8 K
Mass balance on region to right of piston
P V
PCR ,f VCR ,f
N CR ,i N CR ,f Ÿ CR ,i CR ,i
RTCR ,i
RTCR ,f
Ÿ
0.1u 0.5
288.15
0.2 u VCR ,f
350.8
VCR ,f
0.1u 0.5 350.8
u
288.15
0.2
0.3043 m3
Ÿ VCL,f = 0.5 – 0.3043 = 0.1557 m3
Moles in tank initially
PT,i VT,i
0.4 u 0.25
N T,i
25.42 moles
RTi
8.314 u 106 u 473.15
Moles in tank finally
0.2 u 0.25
N T,f
15.47 moles
8.314 u 106 u 388.7
Ÿ N CL,f 25.42 15.47 9.95 moles
TCL,f
PCL,f VCL,f
R ˜ N CL,f
0.2 u 0.1557
8.314 u 106 u 9.95
376.43
0.1u 0.5
20.87 moles
8.314 u 106 u 288.15
(QWURS\FKDQJH ǻ6of N2 finally in tank ǻ6 of N2 finally left of piston ǻ6of N2 finally right of piston
N CR
388.7
0.2 ·
376.43
0.2 ·
350.8
0.2 ·
§
§
§
15.47 ¨ C*P ln
R ln
9.95 ¨ C*P ln
R ln
N CR ¨ C*P ln
R ln
¸
¸
473.15
0.4 ¹
473.15
0.4 ¹
288.15
0.1 ¸¹
©
©
©
9.26
[Note that the first and third terms should be zero, since the gas in the tank under went a uniform, adiabatic
expansion, and the gas to the right of the piston under went a uniform, adiabatic compression. So only the
middle term is important.]
4.42 Initial
To the left of the piston
TCL,f = 376.43 K VCL,f = 0.1557
P = 0.2 MPa
J
K
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
NCL,f = 9.95 Mols
TCR,f = 350.8 K VCR,f = 0.3043 m3
NCR,f = 20.87 K
Mass balance after conduction of heat
Pf VCL
Pf VCR
N CL,f 9.95
N CR 20.87
RTf
RTf
but Pf is the same on both sides of piston, and Tf is the same on both sides of piston
Ÿ
9.95
20.87
VCL
VCR
VCL
0.5 VCL
0.5 – VCL = 2.097 VCL
0.5 VCL
20.87
VCL
9.95
0.5
3.097
0.1614 m3
VCL
VCR = 0.5 –0.1614 = 0.3386 m3
Energy balance
9.95 UL,2 + 20.87 UR,2 = 9.95 UL,1 + 20.87 UR,1
9.95 (T–376.43) + 20.87 (T–350.8) = 0
30.82 T = 9.95 u376.43 + 20.87 + 350.8
T = 359.1 K
Pf
9.95 moles u 8.314 u 10 5
9.95 u R u T
VCL
bar m3
u 359.1 K
mol K
0.1614 m3
1.840 bar 184.0 kPa
4.43
110 bar
80oC
V=0.1 m3, initial conditions 25oC and 1 bar
a) Mass balance: N f
N i 'N with N i
PVi
RT
1 bar ˜ 0.1m3
bar ˜ m3
8.314 ˜10-5
˜ 298.15 K
mol ˜ K
4.034 mol
Energy balance:
NfU f
Ni U i 'N H in
For simplicity in the calculation of thermal properties, will use 25oC as Tref so that Uref = 0 and
Href=-298.15R, so that
100 bar ˜ 0.1 m3
1 bar ˜ 0.1 m3
˜ CV ˜ T Tref
˜ CV ˜ Tin Tref
R ˜Tf
R ˜ Ti
§ 100 bar ˜ 0.1 m3 1 bar ˜ 0.1 m3 ·
¨
¸ ˜ CP ˜ Tin CV ˜ Tref
¨
¸
R ˜Tf
R ˜ Ti
©
¹
which reduces to
Solutions to Chemical and Engineering Thermodynamics, 5th ed
100
CV ˜ T f CP ˜ Tin
Tf
Chapter 4
1
˜ CV ˜ Ti CP ˜ Tin
Ti
So that
Tf
100 ˜ CP ˜ Tin
489.86 K = 216.71o C
T
100 ˜ CV CV CP ˜ in
Ti
100 bar ˜ 0.1 m3
245.54 mol
3
-5 bar ˜ m
˜ 489.86 K
8.314 ˜10
mol ˜ K
241.51 mol
so that 'N 241.51 mol and the fill time is t
12.075 sec
20 mol/s
c) Final pressure, since volume and number of moles is constant, can be computer from
Pf ' Pf
Pf
100
˜Tf '
˜ 298.15 60.86 bar
so that Pf '
Tf ' Tf
Tf
489.86
b) Number of moles finally N f
4.44
c
1 bar
20°C
10 bar
e
d
Liquid, 85°C
10 bar = 1000 kPa =1 MPa
200°C
­1 bar, 20qC
L®
¯1 bar,85qC
ˆ
H
1
ˆ
H
3
ˆ
V ^10 bar, 200qC H
2
M
M
M
1
2
3
M.B.
E.B.
0
355.90
Sˆ 1
Sˆ
3
1.1343
2827.9
Sˆ 2
6.6940
83.96
0
M
3
0.2966
kg M1 M
2
s
M
1 M
1
2
1
ˆ
H
ˆ
ˆ
M
1 1 M 2 H 2 M3 H3
83.96 M
˜ 2827.9 1 ˜ 355.90
1 M
2
2
2827.9 83.96 355.90
83.96 M
2
M
2
M
1
355.90 83.96
0.0991
2827.9 83.96
1 0.0991 0.9009
S.B.
0
Sˆ M
Sˆ M
Sˆ S
M
1 1
2 2
3 3
gen
0
0.9091u 0.2966 0.0991u 6.6940 1u 1.1343 S gen
0
Solutions to Chemical and Engineering Thermodynamics, 5th ed
S gen
0.2013
Chapter 4
J
K ˜s
Reactor is I, initially evacuated tank is II. Mass balance is
4.45
N I N II , which when using the ideal gas law becomes
Ni
200 bar ˜ V I
673.15 K
P I V I P II V II
TI
T II
200
673.15
4 ·
§ 1
P I ¨ I II ¸
T
T
©
¹
or
§ V I 4V I ·
P I ¨¨ I II ¸¸
T ¹
©T
since P I
P II
The energy balance of the reactor and tank is
N I U I N II U II , which when using the ideal gas law and constant heat capacity becomes
NiU i
200 bar ˜ CV ˜ 673.15 K
673.15 K
or 200=P I 4 P I
P I CVT I 4 ˜ P I CVT II
TI
T II
200
and therefore P I
40 bar
5
The entropy balance on just the contents of the reactor that undergo a uniform expansion leads to
S
constant which leads to
R
T
I
§ P I · CP
Ti ¨¨
¸¸
© Pi ¹
8.314
§ 40 · 73.2
673.15 ¨
¸
© 200 ¹
560.7 K
Using this result in the mass balance gives
200
673.15
4 º
ª 1
II »
40 «
560.7
T ¼
¬
which gives T II
708.7 K
4.46
N
steam
1 bar, 250°C
M
steam
1 bar, 100°C
1 bar, liquid water
e
ˆ
H
2676.1
ˆ
O liquid 1 bar, 100qC O H
ˆ
N steam 1 bar, 250qC N H
419.04
kJ
Sˆ 7.3549
kg ˜ K
ˆS 1.3069
2974.3
Sˆ 8.0333
M steam 1 bar, 100qC M
M.B.
E.B.
M
M
M
0
1
2
3
H
ˆ M
H
ˆ M
H
ˆ
M
1
1
2
2
3
kJ
kg
M
M
2
3
3
0
M
1
1
Solutions to Chemical and Engineering Thermodynamics, 5th ed
M
2676.1 M
2974.3 M
419.04
M
2
3
2
3
Chapter 4
0
2974.3 2676.1 M
419.04 2676.1
M
2
3
M 2 2676.1 419.04
7.5689
2974.3 2676.1
M
3
Ÿ M
2
M
M
2
3
but
M
1 7.5689 M
3
3
1
0.1167
8.5689
Sˆ M
Sˆ M
Sˆ Sˆ
M
0
M
3
S.B.
1 1
2 2
3 3
1
M
2
1 0.1167
0.8833
gen
1u 7.3549 0.8833 u 8.0333 0.1167 u 1.3069 Sˆ gen
0.1066
Ŝgen
7.5689 M
3
0.1066
kJ ˆ
Sgen
K
0
0
kJ
Ÿ process is not possible
K
4.47
4.48
The maximum temperature difference will occur if the process is carried out reversibly.
In this case the energy balance is
1
1
1
H1 H 2 H 3 0 or H1
H 2 H 3 and since CP is constant T1
T2 T3
2
2
2
The entropy balance is
T
P
T
P
0 CP ln 1 R ln 1 CP ln 1 R ln 1
T2
P2
T3
P3
0
29.3ln
273.15 25
4
273.15 25
4
8.314 ln
29.3ln
8.314 ln
T2
T3
1.013
1.013
The solution to these equations using MATHCAD is
T2 298.15
T3 298.15
Given
298.15
§ 1 · ˜ ( T2 T3)
¨ ¸
© 2¹
29.3˜ ln §¨
298.15˜ 298.15 ·
©
T2˜ T3
¸
¹
8.314˜ 2˜ ln §¨
4
·
¸
© 1.013 ¹
z Find ( T2 T3)
z
§ 78.788 ·
¨
¸
© 517.512 ¹
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
T2=78.8 K and T3=517.5 K, a difference of more than 438 K. However, in fact a HilschRanque
vortex tube is very inefficient and very irreversible
4.49 In the folder Aspen for Textbook>Chapter 4>Problems
4.50 In the folder Aspen for Textbook>Chapter 4>Problems
4.51 In the folder Aspen for Textbook>Chapter 4>Problems
4.52 In the folder Aspen for Textbook>Chapter 4>Problems
4.53 In the folder Aspen for Textbook>Chapter 4>Problems
4.54 In the folder Aspen for Textbook>Chapter 4>Problems
4.55 In the folder Aspen for Textbook>Chapter 4>Problems
4.56 In the folder Aspen for Textbook>Chapter 4>Problems
4.57 In the folder Aspen for Textbook>Chapter 4>Problems
4.58 In the folder Aspen for Textbook>Chapter 4>Problems
4.59 Equation 4.6-14 is to be used to solve this problem
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
M ª Uˆ T1 , P1 PambVˆ T1 , P1 Tamb Sˆ T1 , P1 Uˆ T2 , P2 PambVˆ T2 , P2 Tamb Sˆ T2 , P2 º
¬
¼
a) So for part a, assuming the exiting stream is a vapor
WS ,max
ª Uˆ 800o C, 2 MPa (1 bar)Vˆ 800o C, 2 MPa 298.15 K Sˆ 800o C, 2 MPa
WS ,max M «
« Uˆ 298.15 K),1 bar (1 bar)Vˆ 298.15 K),1 bar 298.15 K Sˆ 298.15 K),1 bar
«¬
ª§
kJ
m3
kJ · º
- 298.15 K 8.1765
Ǭ 3657.0 +(1 bar)0.2467
¸»
kg
kg
kg K ¹ »
©
«
100 kg «
3
§
·»
«- ¨104.88 kJ +(1 bar)43.36 m - 298.15 K 8.5580 kJ ¸ »
kg
kg
kg K ¹ »¼
«¬ ©
ª§
kJ
m3
kJ
kJ · º
u100
3657.0
+(1
bar)0.2467
- 298.15 K 8.1765
Ǭ
¸»
3
kg
kg
bar m
kg K ¹ »
©
=100 kg ««
3
§
·»
«- ¨104.88 kJ +(1 bar)43.36 m u100 kJ 3 - 298.15 K 8.5580 kJ ¸ »
kg
kg
bar m
kg K ¹ »¼
«¬ ©
º
»
»
¼»
=100 kJ ª¬ 3657.0+24.7-2437.8 - 104.88+4336-2551.6 º¼
=100 kJ >1243.9-1889.3@ 64540 kJ
b) WS ,max
ª Uˆ 800o C, 2 MPa (1 bar)Vˆ 800o C, 2 MPa 298.15 K Sˆ 800o C, 2 MPa
M«
« Uˆ 673.15 K, 0.6 MPa (1 bar)Vˆ 673.15 K, 0.6 MPa 298.15 K Sˆ 673.15 K, 0.6 MPa
«¬
ª§
kJ
m3
kJ · º
- 298.15 K 8.1765
Ǭ 3657.0 +(1 bar)0.2467
¸ »
kg
kg
kg K ¹ »
©
«
100 kg «
3
§
·»
«- ¨ 2962.1 kJ +(1 bar)0.5137 m - 298.15 K 7.7079 kJ ¸ »
kg
kg
kg K ¹ »¼
«¬ ©
100 kJ ª¬ 3657.0+24.67-2437.8 - 2962.1+51.37-2298.1 º¼
100 kJ >1243.9-715.4@ 52850 kJ
c) The simplest way to calculate this is as
WS ,max
WS ,max part a WS ,max part b
64540 52850 kJ 11690 kJ
4.60 In the first printing, Problem 4.60 inadvertently duplicates Problem 4.59
4.61 In first printing, this problem duplicates parts a and b of Problem 4.59
4.62
º
»
»
¼»
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
The starting point is
Wˆ
S ,max
Hˆ T1 , P1 PambVˆ T1 , P1 Tamb Sˆ T1 , P1
Hˆ Tamb , Pamb PambVˆ Tamb , Pamb Tamb Sˆ Tamb , Pamb
o
For the 800 C and 2 MPa stream
WˆS ,max
Hˆ 800o C, 2 MPa (1 bar)Vˆ 800o C, 2 MPa 298.15 K Sˆ 800o C, 2 MPa
Hˆ 298.15 K),1 bar (1 bar)Vˆ 298.15 K),1 bar 298.15 K Sˆ 298.15 K),1 bar
ª§
kJ
m3
kJ · º
4150.3
+(1
bar)0.2467
- 298.15 K 8.1765
Ǭ
¸»
kg
kg
kg K ¹ »
©
«
=
« §
3
·»
«- ¨104.88 kJ +(1 bar)43.36 m - 298.15 K 8.5580 kJ ¸ »
kg
kg
kg K ¹ »¼
«¬ ©
For the 900o C and 1.4 MPa stream
WˆS ,max
Hˆ 900o C,1.4 MPa (1 bar)Vˆ 900o C,1.4 MPa 298.15 K Sˆ 900o C,1.4 MPa
Hˆ 298.15 K),1 bar (1 bar)Vˆ 298.15 K),1 bar 298.15 K Sˆ 298.15 K),1 bar
ª§
kJ
m3
kJ · º
4391.5
+(1
bar)0.3861
- 298.15 K 8.5556
Ǭ
¸»
kg
kg
kg K ¹ »
©
«
=
« §
3
·»
«- ¨104.88 kJ +(1 bar)43.36 m - 298.15 K 8.5580 kJ ¸ »
kg
kg
kg K ¹ ¼»
¬« ©
4.63
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
The starting point is
WS ,max
Hˆ T1 , P1 PambVˆ T1 , P1 Tamb Sˆ T1 , P1
Hˆ Tamb , Pamb PambVˆ Tamb , Pamb Tamb Sˆ Tamb , Pamb
o
For the 1000 C and 25 bar stream
WS ,max
H 1000o C, 25 bar (1 bar)V 1000o C, 25 bar 298.15 K S 800o C, 25 bar
H 298.15 K),1 bar (1 bar)V 298.15 K),1 bar 298.15 K S 298.15 K),1 bar
For the 900o C and 40 bar stream
WS ,max
H 900o C, 40 bar (1 bar)V 900o C, 40 bar 298.15 K S 800o C, 25 bar
Toevaluate which stream has the potential to do more work, calculate the difference in
available work
'WS ,max
WS ,max 1000o C, 25 bar WS ,max 900o C, 40 bar
= H 1000o C, 25 bar H 900o C, 40 bar 1 bar u V 1000o C, 25 bar V 900o C, 40 bar
298.15 K S 1000o C, 25 bar S 900o C, 40 bar
Now since the fluid is an ideal gas
§ 1273.15 K 1173.15 K ·
CP 1000o C 900o C 1 bar u R ¨
¸
40 bar ¹
© 25 bar
1273.15
25 ·
§
298.15 K ¨ CP ln
R ln ¸
1173.15
40 ¹
©
3
bar m
1273.15
25 ·
§
CP 100 1 bar u 8.314 u 105
R ln ¸
29.329 298.15 K ¨ CP ln
mol K
1173.15
40 ¹
©
'WS ,max
4.64
A U PambV Tamb S
§ wA ·
¨
¸
© wT ¹V
§ wA ·
¨
¸
© wV ¹T
§C ·
§ T ·
§ wU ·
§ wS ·
CV Tamb ¨ V ¸ CV ¨1 amb ¸
¨
¸ Tamb ¨
¸
T ¹
© wT ¹V
© wT ¹V
© T ¹
©
§ wU ·
§ wS ·
§ wS · § wP ·
§ wU ·
¨
¸ Pamb Tamb ¨
¸ ; but ¨
¸ ¨
¸ and ¨
¸
© wV ¹T
© wV ¹T
© wV ¹T © wT ¹V
© wV ¹T
§ wA ·
so ¨
¸
© wV ¹T
§ wP ·
§ wP ·
T¨
¸ P Pamb Tamb ¨
¸
© wT ¹V
© wT ¹V
§ wP ·
T¨
¸ P
© wT ¹V
§ wP ·
T Tamb ¨
¸ Pamb P
© wT ¹V
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Chapter 4
A U PambV Tamb S
§ wA ·
¨
¸
© wT ¹V
§ wA ·
¨
¸
© wV ¹T
§C ·
§ T ·
§ wU ·
§ wS ·
CV Tamb ¨ V ¸ CV ¨1 amb ¸
¨
¸ Tamb ¨
¸
T ¹
© wT ¹V
© wT ¹V
© T ¹
©
§ wU ·
§ wS ·
§ wS · § wP ·
§ wU ·
¨
¸ Pamb Tamb ¨
¸ ; but ¨
¸ ¨
¸ and ¨
¸
© wV ¹T
© wV ¹T
© wV ¹T © wT ¹V
© wV ¹T
§ wA ·
so ¨
¸
© wV ¹T
§ wP ·
§ wP ·
T¨
¸ P Pamb Tamb ¨
¸
© wT ¹V
© wT ¹V
§ wP ·
T¨
¸ P
© wT ¹V
§ wP ·
T Tamb ¨
¸ Pamb P
© wT ¹V
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: methane
Formula: CH4
Molecular weight (g/mol) = 16.043
Normal boiling point (K) = 111.6
Critical temperature (K) = 190.4
Critical pressure (bar)
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Critical volume (cm3/mol) = 99.2
Critical compressibility factor = 0.288
Pitzer's acentric factor = 0.011
Solutions to Chemical and Engineering Thermodynamics, 5th ed.
Isobaric heat capacity of the ideal gas (J/mol.K)
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A = 1.925e+01
B = 5.213e-02
C = 1.197e-05
D = -1.132e-08
-----------------------------------------------------------------** Vapor-Liquid Equilibrium **
Temperature
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Vapor pressure (bar) = 1.0249
Vapor
Compressibility factor
0.96673
Specific volume (m^3/mol)
8.75155e-03
Density (g/cm^3)
1.83316e-03
Fugacity (bar)
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Fugacity coeff
0.96771
Enthalpy departure (J/mol)
-73.98
Enthalpy ideal
(J/mol)
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Enthalpy
(J/mol)
-5735.73
Entropy departure (J/mol K)
-0.39
Entropy ideal
(J/mol K)
-29.21
Entropy
(J/mol K)
-29.60
** Vapor-Liquid Equilibrium **
T (K)
115.0
120.0
125.0
130.0
135.0
140.0
145.0
150.0
155.0
160.0
165.0
170.0
175.0
180.0
185.0
190.0
P (bar)
1.343
1.943
2.726
3.723
4.969
6.496
8.342
10.539
13.128
16.143
19.625
23.612
28.146
33.270
39.030
45.478
Liquid
0.00373
3.37236e-05
4.75720e-01
0.99186
0.96774
-8266.44
-5661.75
-13928.18
-73.80
-29.21
-103.01
Molar volume (m^3/mol)
Density (g/cm^3)
Vapor
Liquid
Vapor
Liquid
6.82605e-03 3.41685e-05
2.35026e-03 4.69525e-01
4.85847e-03 3.48771e-05
3.30207e-03 4.59986e-01
3.54925e-03 3.56597e-05
4.52011e-03 4.49891e-01
2.65044e-03 3.65291e-05
6.05295e-03 4.39184e-01
2.01609e-03 3.75015e-05
7.95749e-03 4.27797e-01
1.55770e-03 3.85977e-05
1.02992e-02 4.15647e-01
1.21889e-03 3.98450e-05
1.31620e-02 4.02636e-01
9.63678e-04 4.12800e-05
1.66477e-02 3.88639e-01
7.67798e-04 4.29535e-05
2.08948e-02 3.73497e-01
6.15040e-04 4.49387e-05
2.60845e-02 3.56998e-01
4.93846e-04 4.73457e-05
3.24858e-02 3.38848e-01
3.96190e-04 5.03529e-05
4.04932e-02 3.18611e-01
3.15943e-04 5.42758e-05
5.07781e-02 2.95583e-01
2.48265e-04 5.97618e-05
6.46204e-02 2.68449e-01
1.88258e-04 6.85514e-05
8.52179e-02 2.34029e-01
1.18996e-04 9.18577e-05
1.34820e-01 1.74651e-01
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1.3299
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1.3186
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1.3074
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98.30
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1.2961
1.0043
116.04
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1.2848
1.0060
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1.2735
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1.2510
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179.74
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1.2285
1.0203
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0.246
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0.263
0.272
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0.287
0.295
0.302
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0.317
0.323
0.330
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d
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d
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a
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Chap e 13
d
13.23
a) U g he De b gh e h d, e ha e
5C + 12H = C5
(1)
5C + 12H = C5
(2)
5C + 12H = e C5
(3)
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C H
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E . (2 a d 3)
ba
C5 = C5
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C5 = e C5
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e
a
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d
1 ba
d be ee .) S
1.0
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ha
1.0 ba
PCH 4 ˜ 0.714
5.75
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0.1 MPa
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5.885 MPa
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0.0238 ˜ 0.714
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9.528 MPa
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1
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d e
b e , ha , a
cha ged.
eac
1
.
a
1 X
1 X
1
X
2
1
X
2
1
he C
1
X
2
1
X
2
a1 2f X a P 1 a f
1 X aP 1 a f
2 Ka , 1
g
a1H 2 a1I 2
2
F a
12 2
. u 102
Ÿ 372
3.72 u 10
­° ' ,2 G ½°
®
¾
¯° RT
¿°
S d ec a
f e
c a
he e
e a h ch he f
b f
d a ea
The ef e, he f
e
c
e he a
c
4.63 ca =-19.37 J
,2 G
'
1 X P
1a
1 2 XP
1a
1 2 XP
1a
a f
a f
a E
1a
Ha
X
2 1 X
Ka,1
0.069248 ( de e de
f
e
e!)
gh ha
ha e c
ec
dea
.
a
ca be
S
C
ca a d E
T
d
H2
h ch
cc
.
N
I2
he c
c
d
0.03462 ;
HI
he f
1
aI 2
I2
2478.3
1
I2 P 1 a
Th , f P ! 0.01181 ba
2NaHCO3
ec
a
ha e a d N H 2 O
P
CO2
f
0.01181 ba .
cc .
H2O
UV
W
b
aNaHCO 3 1
aNaCO 3 1
Ÿ Ka
aH 2 OaCO 2
PH 2 O PCO 2
˜
1 ba 1 ba
he e N
a
aNaCO 3 aCO 2 aH 2 O
aNaHCO 2
N
P
N
N CO2 Ÿ PH 2 O
PCO2
The ef e
d I2
f
1a
Ka , 2 I 2
ŸP
d I2
f
3
P
a
F PI
H1 a K
0.01166 a
a
NaCO3
Ka
b
ec
:
1a
0.03462 u 2478.3
P
093075
.
he
de he ec d eac
Ÿ Ka , 2
Chap e 13
c ,5
f he a
aI 2
13.2 (a)
a
e
d
ha e, N
ga
e
f
ga
1
P.
2
Ka
LM a1 2f P OP
N 1 ba Q
2
a d
LM a1 2f u 0.826 Pa OP 1706
N 100 Pa Q . u 10
a1 2f u 166.97 Pa O 0.697
K 110q C LM
N 100 Pa PQ
2
Ka 30q C
5
2
a
Ka 30q C
10.979;
0.3610
Ka 110q C
N
K a T2
K a T1
K a T2 K a T1
'
R
H§ 1 1·
¨ ¸
© T2 T1 ¹
(1)
S
C
ca a d E
T
d
Th a
e ha ' H
ed da a
he
be
h he
a
c ,5
de e de
ae e
Chap e 13
d
f T, he
a
e ca
a e
Ÿ ' H 128.2 J
(b)
Ka T
Ka
G
g bac
E
. (1) ab e e ha e
§ ' H§1
1 ··
5
¨
¨ ¸ ¸ 1.706 u10 e
R © T 303.15 ¹ ¹
©
15420
15420
10.9788 508659
39.8871 .
T
T
Ka T
30 C e
Ÿ Ka (T )
ŸT
13.2 Reac
U
10
. ba Ÿ PH 2O
PCO2
(c)
LM a1 2f u 2.0 OP
N 1 Q
2
g he
ga
P
PCO2
1 ba .
CH4
CHEMEQ e ha e Ka T
S ec e
I
a (ga )
1000 K
H2
1
1 2 X
CH4
0
X
6
1 X
aCH 4
aCaH2 2
aCH 4
aH2 2
009838
.
F a (ga )
C
Ka
2.0 ba
1
3866
. K 11345
. qC f
C 2H2
10
. ba
§
1 ··
§1
¨ 15420 ¨ ¸¸
© T 303.15 ¹ ¹
©
1 2 X
1 X
X
1 X
F
H
XP
1 X 2 1 ba
˜
P
1 X 1 ba 1 2 X 2
I
K
2
X 1 X
X 1 X 1 ba
1 2 X 2 P
1 2 X 2
h e a
a e X 0.0769 a d X 0.9231 .
a a e f he e
b
c
a , he X 0.0769
0.917 a d CH 4 0.083
e
H2
01004
.
c
The
W h ch a
ec
e. Th
Th
( bab
h e e e a a d ca c a
a e
)e e
a e a he e
b
c
. The ef e, he eac
ce
he
d a ca
ed,
a
a fe
ed. C e e
c ea
e
ba
eb
g he h d ge f
ha e
effec
he
13.2
Reac
(a a a ab e a a Ma hcad
hee )
he
a
he
g he
ce .
S
C
ca a d E
CO H 2
C 2H 2O
CO2 C
CO2 2H 2
2CO
CO H 2O
CO2 H 2
c
, e
e eac
CO H 2
C H 2O
CO2 2H 2
CO2 C
C 2H 2O
2CO
CO2 H 2
CO H 2O
N
eed
Sa b
de
g
f
he
CO
CO
E
2H O
C 2O
H 2O
CO2
eac
ce
H2
H
2H
e O
d
C H 2O
F
N
T
4
che
C 2O
CO2
ce
e e e
ca eac
CO
H 2O
aeO
Chap e 13
d
CO
Ÿ H2 O
ee
e
c ,5
1, 2 a d 4
de e de
ae
1
1
CO2 C
2
2
a
a
c
1
CO2
2
1
H 2 CO2
2
1
C CO
2
1
C H 2O
2
C CO 2
UV Th e e f
C 2H O W de e de eac
C
2H 2 CO 2
.
ge
ee .
2CO
2
.
Add he e
2H 2 C 2CO 2
We
2CO C 2H 2O Ÿ H 2 CO2
CO H 2O
e
H 2 CO 2
2H 2 CO 2
UV a he
C 2H O W de e de eac
CO H 2 O
2
.
S
C
S ec e
ca a d E
I
T
a
d
1 X1 X 2
H2
1
1 X1 2 X 2
CO
0
X1
H 2O
0
X1 2 X 2
C
0
aa
Ka,2
aCO aH 2 O
f
2
2
fa
1
2
1
2
CHEMEQ, I f d he f
2
1
2
g
Ka,2
758.6
48.43
5.950
1.137
0.2974
1
03665
.
u 10
0.1110
0.2493
0.4596
0.7387
X2 0
de
Ka,1
a
f
2
2
Ka,1
600
700
800
900
1000
N ca b
f
a X 2 X f a2 X f 1 ba
a1 X 2 X f a1 X X f P
1 a H2 2 O
C
a
a
aCO 2 aH 2
T K
de
1
1
X 1 X1 2 X 2
1 X1 X 2 1 X1 2 X 2
a H2 2 aCO 2
ga
g he e e
2
1
X1 2 X 2
¦
0
2
1
ga ha e)
¦ 2 X2
Ka,1
S
LM1 X X OP P
N ¦ Q 1 ba
LM1 X 2 X OP P
N ¦ Q 1 ba
LM X OP P
N ¦ Q 1 ba
LM X 2 X OP P
N ¦ Q 1 ba
1 X1 X 2
¦
1 X1 2 X 2
¦
X1
¦
aX f
(
(a)
Chap e 13
d
a
1
g he
c ,5
F a
CO 2
U
a
X 12
a1 X f
1
X 12 ˜ 2 1 ba
1 X1 3 P
a d Ka,2
2
a
f
,If d
TK
600
700
800
900
1000
P ba
1151
. u 104
6111
. u 103
0.126
1.357
9.237
If he
.
e
ef
ag e
e
ea
e
ab
e he
e
e ca c a ed, ca b
S
C
(b)
ca a d E
E ac
T
30% f ca b
Ka,1
a
TK
X1
P ba
a
de
d
a
c ,5
ed X 2
03
.
f
Chap e 13
d
f
a 0.6f 17. 1 ba
a0.7 X fa0.4 X f P
600
700
800
900
1000
0.0157
0.0410
0.0750
0.1104
0.1427
0.0084
0.170
1.972
15.266
85.419
X 1 X 1 0.6
a d Ka,2
0.7 X 1 0.4 X 1
fa
2
1
1
1
2
13.2
The eac
he e g ee
c ce ed ab
T ( ) + S O 2( ) = T O 2( ) + S ( )
Th
e
ae
he f
eac
he
be
ae e
eac
. The ef e
J
J
J
' G
674 (644)
30
30, 000
a d he e
Ka
C
e
b
c
a
f
h
he ec
d
eac
§ ' G ·
§ 30000 ·
¨¨ ¸¸ e ¨
¸ 17.02
R ˜T ¹
© 8.314 ˜1273 ¹
©
e e
, a he e g ee fea , he a
e a e c ac
h c d
de.
e
be affec ed b h gh
13.30
a)
R 8.314
T 80 273.15
1203.828 ·
§
¨ 4.01816
¸
T
53.229 ¹
P a B( T) 10©
1209.299 ·
§
¨ 3.98022
¸
T 49.623 ¹
P a C( T) 10©
B 0.2
P( T) P a C ( T)
0.991
2
ª
(1 ) º
« 100˜
»
R˜ T ¼
¬
ga C( T) e
2
ª
( ) º
« 100˜
»
R˜ T ¼
¬
˜ ga B( T) ˜ P a B ( T) ( 1 ) ˜ ga C( T) ˜ P a C( T)
P( 0.2 353.15)
1
ba
( ˜ ga B( T) ˜ P a B( T) )
P( T)
So b=0.206 and nC=0.794
) The
1.01
C 1 B
ga B( T) e
( T) P a B ( T)
a
ba a ce ab e :
( 0.2 353.15)
0.206
S
C
ca a d E
T
a
0.794
0.206
0
f a
0.794-X
0.206+X
3X
S ec e
C6
be e e
h d ge
a
S
d
a
c ,5
Chap e 13
d
e f ac
(0.794-X)/(1+3X)
(0.206-X)/(1+3X)
3X/(1+3X)
1+3X
ha he e
b
ea
a 3H a B
aC
Ka
3
3
§ H P · § BP ·
¨
¸ ¨
¸
© 1 ba ¹ © 1 ba ¹
§ CP ·
¨
¸
© 1 ba ¹
§ 3X · § 0.206 X ·
¨
¸ ¨
¸
© 1 3X ¹ © 1 3X ¹
§ 0.794 X ·
¨
¸
© 1 3X ¹
From the Vis al Basic chemical eq ilibri m program
Ka 0.064608
X 0.1
3
1
1 º
· ª
§
¨ 3˜ X˜
¸ ˜ «( 0.206 X) ˜
»
1 3˜ X ¹ ¬
1 3˜ X¼
©
G e
Ka
( 0.794 X) ˜
X f d( X)
cC H
X
( 0.799 X)
1
1 3˜ X
0.248
cC 0.316
1 3˜ X
( 3˜ X)
cB ( 0.201 X)
1 3˜ X
cB
0.257
H 0.427
1 3˜ X
c)
Since h drogen is so far abo e its critical point, onl a negligible amo nt of h drogen ill
appear in the liq id phase,.
PP 1
g e
B˜ ga B( B T) ˜ P a B ( T)
0.257˜ PP
cC˜ ga C( B T) ˜ P a C ( T)
0.32˜ PP
B cC
1
f d( B cC PP)
§ 0.44 ·
¨ 0.56 ¸
¨
¸
© 1.747 ¹
So the de point press re is 1.77 bar, and the liq id that forms has a mole fraction
of ben ene of 0.448, a mole fraction of c clohe ane of 0.552, and (b ass mption), a
negligible amo nt of h drogen.
S
C
ca a d E
T
d
13.31 The che ca eac
CH3-CHOH-CH3 = CH3-CO-CH3 + H2
A
g e a
h
e ace e, he
(g e he h gh e e a e a d
e
S ec e
ace e
h d ge
6
B X = .564,
Ka
aace aH
a P
0.8043 c ,5
Chap e 13
d
a ba a ce ab e
e)
ha
ec e a
a
a
1-X
(1-X)/(1+X)
(1-X)u95.9/(1+X) )u100
X
X/(1+X)
Xu95.9/(1+X) )u100
X
X/(1+X)
Xu95.9/(1+X) )u100
1+X
aI-P=0.2673, a d aace = aH =0.3458. The ef e
1
0
0
0.3458 u 0.3458
0.2673
'
a
G
;
RT
'
G
0.4474
§ ' G ·
¨¨ ¸
RT ¸¹
©
e
8.314 u 452.2 u 0.8043 3023.8
J
13.32 The eac
ae
C6H6 + H2 = 1,3-c c he ad e e
C6H6 + 2H2 = c c he e e
C6H6 + 3H2 = c c he a e
The G bb f ee e e g
ff
J
' f G (be e e) 124.5
' f G (c c he a e)
26.9
a
da a eeded
e h
' f G (c c he e e) 106.9
J
be
J
' f G (1, 3 c c he d e e) 178.97
J
The G bb f ee e e g f f
a
f 1,3-c c he ad e e
a a ab e
IV, Pe
The Che ca E g ee Ha db
he Ha db
f Che
Ph c . The a e a f d
g da a
he WWW e
h :// ebb
. .g /che
. Th Web e c a
he Na
a I
S a da d a d Tech
g (NIST) che
da a b
. The a e f d
ea e
J
' f H (1,3 c c he d e e, 298.15 K) 71.41
S (1,3 c c he d e e, 298.15 K) 197.3
S (C,g a h e, 298.15 K)
S (H 2 , 298.15 K) 130.68
5.88
J
J
A
e d
a d
e f
h Web
J
˜K
˜K
˜K
N e ha he e e
e ae
h e ec
he e
c
e a d 0 K. A , he e
cha ge f eac
The ef e
e
a
e f he
a 0K
e f a
e
eac
.
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
' f S (1,3 c c he d e e, 298.15 K)
S (1, 3 c c he d e e, 298.15 K) 6 ˜ S (C, 298.15 K) 4 ˜ S (H 2 , 298.15 K)
197.3 6 ˜ 5.88 4 ˜130.68
360.75
' f G (1,3 c c he d e e, 298.15 K)
J
˜K
' f H T 'f S
178967
J
178.97
71410 298.15 ˜ 360.75
J
The a ba a ce ab e a
g a he ga c a e e e
he
d ha e, a d
ha he h d ge
ee
g ea e ce
ee
a a e
e f ed a 1 ba .
A , ce a he ga c a e
a, e
a
e he f
a dea
d
e, a d ha he e
h d ge
he
d ha e. Ma ba a ce ab e f
d
ha e:
S ec e
I
O
Be e e
1,3-c c he
c c he e e
c c he a e
T a
1
0
0
0
1-X1-X2-X3
X1
X2
X3
1
The e
ea
b
a1,3 c c
Ka,1
abe aH 2
e
Ka,2
( 21973
. )
Ka,3
X2
1 X1 X 2 X 3
(7.1002)
1212.2
ac cha e
X3
1 X1 X 2 X 3
abe aH3 2
e
(39.374)
1-X1-X2-X3
X1
X2
X3
1-X1-X2-X3
X1
X2
X3
e
F 178970 124500 I
H 8.314 u 29815
K
.
e
F 106900 124500 I
H 8.314 u 29815
K
.
e
F 26900 124500 I
H 8.314 u 29815
K
.
2.866 u 1010
ac che e
abe aH2 2
e
ae
X1
1 X1 X 2 X 3
ac
12587
u 1017
.
B e a
g he a e f he e
b
c
a
he e e a
, e ee ha X3 ab
e a
f he e e a
he h d, e ha e
10
X 1 2.866 u 10
2.27 u 1027 | X 1
X 3 12587
.
u 1017
a d b a g he a
f he ec d f he e e a
X2
1212.2
9.63 u 1015 | X 2
17
X 3 12587
u 10
.
,
e d ec b a
. The b a g he a
he h d
Th
gge
ha X3 1, X2
f he de
10-17, a d X1
f de
-27
10 . Th he be e e
eac
f
e e a a c c he a e.
13.33
(a
a a ab e a a Ma hcad
hee ).
g a
f
f he f
S
C
ca a d E
T
G
2400 e
R
8.31451 aA
A J A
AJ A
aC
A J C
1
Ka
e
K
e
T
Chap e 13
d
298.15 K
V
4 e
1
Ka
2.633
a D 0.5 10 Pa a C
A 1
R T
c ,5
e
a D( P )
G
a
e
aB
A J C
d
P
5
10 Pa
Part (a):
A
(initial g ess)
0.5
5
G e
Ka
C
1
A
0.16
a B a A
F d
A
A 1
A
A
C
0.84
Part (b):
Recogni ing that the partial molar Gibb's e cess is in the form of the one constant Marg les
e pression ields:
2
0.3 C
J A e
G e
J C e
1
A
0.132
2
5
a D 0.5 10 Pa a C
Ka
C
0.3 A
a B a A
A
C
0.868
A e
A e
0.3 A
0.3 1
2
A
2
A
F d
A
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
Part (c):
Mass Balance Table:
Species
A
B
C
D
In
1
2
0
0
O t
1-X
2-X
X
X
Ass ming that D is an ideal gas:
P=(n*R*T)/V
P( X )
e X R T a hee
13.3
(a a a ab e a Da Ma hcad
C( X). 1 )
V
F e he S ea TabKae
P a = 12.349
Pa
G
F
CHEMEQ Ka = 171.2 a B a A ( 1 X 1 )
HNH3 (f
be
a e e ) = 384.5 Pa/
e f ac
a
NA
1
X
A
1
X
NA
0.575
A
0.575
V
(initial g ess)
X 0.5
13.34 Need
X
The 0.425
X R T
a
e
e
P ( Pa)
NB
0.10
0.25
0.50
1.00
2.00
4.00N B
6.00
8.00
10.00
12.00
14.00
16.00
18.00
18.40
18.50
19.00
20.00
25.00
30.00
35.00
40.00
50.00
2
X
1.575
X
F d( X)
X NC X
0.0899
0
0.2017 C X
0
0.3467
0
0.5478
0
0.7868
0
1.0322 N C 0.425 0
1.1670
0
1.2559 C 0.425
0
1.3204
0
1.3701
0
1.4099
0
1.4429
0
1.4707
0
1.4757
2.02u10-05
1.4797
0.0114
1.4996
0.0680
1.5339
0.1589
1.6437
0.3989
1.7063
0.5068
1.7488
0.5713
1.7806
0.6162
1.8267
0.6792
ND
PD
X
X R T e
V
ND
0.425
PD
5
2.633 10 Pa
S
C
ca a d E
T
d
60.00
70.00
80.00
90.00
100.00
104.50
a
c ,5
Chap e 13
d
1.8602
1.8869
1.9093
1.9291
1.9472
1.9449
0.7261
0.7661
0.8030
0.8388
0.8745
0.8908
2.5
X and f
2.0
1.5
X, molar e tent of reaction
f, fraction liq id
1.0
0.5
0.0
0
20
40
60
80
100
HCONH 3 R NH 3 CO , f
h ch he
Press re, kPa
13.34
ab e
The che
S ec e
HCONH2
NH3
CO
ca eac
a
0
0
f a
1
X
X
T a
A
, P
NRT
V
a NH3 ˜ a CO
Ka
a HCONH3
=
1-X
X2
ba ˜ 3 1
˜ 8.314 u105
u
˜ K 25
1 X
1 X ˜ 8.314 u 105 ˜ T
25 u 10 3
ba
2
RT
§ X ·
¨
¸ ˜ 1 X
V
© 1+X ¹
1
X
§
·
¨
¸
© 1+X ¹
u
103
3
uT
ba a ce
e f ac
(1-X)/(1+X)
X/(1+X)
X/(1+X)
1+X
1 X RT
1 X P
V
§ NH3 ˜ P · § CO ˜ P ·
¨
¸˜¨
¸
© 1 ba ¹ © 1 ba ¹
§ HCONH3 ˜ P ·
¨
¸
© 1 ba
¹
a
X 2 RT
1 X V
S
X2
1 X
C
ca a d E
',
he e ' =
T
d
' X'; X 2 'X ' 0, a d X
U
g he che
N
ca e
b
Ka
7.8184u105
4.5987u105
2.8598u105
1.8660u105
1.2695u105
8.9571u104
c) I
a
PV
RT
be
c
e
HCONH2
fa
ba ˜
u 10-3
3
A d he
T(K)
P(ba )
e
CO
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
1.022
e a
u 298.15 K
ge a d 0.807
ce he e
b
e
c
f
a
ge .
3.022 u 8.315 u 105 u T
25 u 10 3
ha
480
4.82
500
5.02
NH3=0.331
CO=0.331
O2=0.071
N2=0.267
e
P
NRT
V
400
4.02
P(ba )
2.66
2.794
2.927
3.06
3.193
3.326
3
˜K
The ef e, he e
be 0.215
e f
The c
a a e ea e (
eac
g
c
e
)
1.0
e NH3
1.0
e CO
0.215
e O2
0.807
e N2
NH3
0
0
0
0
0
0
1.013 ba u 25
8.314 u 10 5
Chap e 13
d
' ' 2 4'
2
ga
c
e K a, e b a
a
X
1
1
1
1
1
1
f
c ,5
K a u 25
8.314 u 102
X2
T(K)
400
420
440
460
480
500
a
420
4.22
440
4.42
460
4.62
a ge ha he
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
13.3
(a a a ab e a a Ma hcad
hee ).
F
he S ea Tab e
P a = 12.349 Pa
F
CHEMEQ Ka = 171.2
HNH3 (f
be
a e e ) = 384.5 Pa/
e f ac
The
a a
e
e
P ( Pa)
X
0.10
0.0899
0
0.25
0.2017
0
0.50
0.3467
0
1.00
0.5478
0
2.00
0.7868
0
4.00
1.0322
0
6.00
1.1670
0
8.00
1.2559
0
10.00
1.3204
0
12.00
1.3701
0
14.00
1.4099
0
16.00
1.4429
0
18.00
1.4707
0
18.40
1.4757
2.02u10-05
18.50
1.4797
0.0114
19.00
1.4996
0.0680
20.00
1.5339
0.1589
25.00
1.6437
0.3989
30.00
1.7063
0.5068
35.00
1.7488
0.5713
40.00
1.7806
0.6162
50.00
1.8267
0.6792
60.00
1.8602
0.7261
70.00
1.8869
0.7661
80.00
1.9093
0.8030
90.00
1.9291
0.8388
100.00
1.9472
0.8745
104.50
1.9449
0.8908
2.5
X and f
2.0
1.5
X, molar e tent of reaction
f, fraction liq id
1.0
0.5
0.0
0
20
40
60
Press re, kPa
80
100
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
13.3 a) The eac
2H 2S CH 4 R CS2 4H 2
F
he he che ca e
b
c
a
g a , e ha e he a e be
e
b
c
a (A ,
a e ace, I h
e f he a e ca c a ed
T(K)
Ka
X
773.15
863.15
923.15
1013.15
1073.15
4.7915u10-7
2.9923u10-5
3.0602u10-4
6.0722u10-3
3.3960u10-2
0.023
0.052
0.080
0.137
0.183
b)
S ec e
H2S
CH4
CS2
H2
1
1
0
0
a
CH4
CS2
H2
0.466
0.426
0.388
0.319
0.268
0.477
0.451
0.426
0.379
0.345
0.011
0.025
0.037
0.060
0.077
0.045
0.099
0.149
0.242
0.310
f a
1-2X
1-X
X
4X
T a
(
e f ac
)
(1-2X)/(2+2X)
(1-X)/(2+2X)
X/(2+2X)
4X/(2+2X)
2+2X
A 1 ba
Ka
H2S
f he
a b)
4
4X
X
˜
2 2X 2 2X 4
44 X 5
2
64X5
2
1 2X 1 X 2 2X
§ 1 2X · 1 X
¨
¸
2
2X
2
2X
©
¹
The
f X a d he
e f ac
a eg e
c) A 10 ba (a
g he a
ha e
dea )
2
1 2X
2
he ab e ab
1 X 2 2X
2
e.
4
Ka
5
4X
X
§ 10 ba ·
˜
¨
¸
2 2X 2 2X 4 © 1 ba ¹
2
3
§ 1 2X · 1 X § 10 ba ·
¨
¸
¨
¸
© 2 2X ¹ 2 2X © 1 ba ¹
The e
ae
he ab e be
T(K)
X
773.15
863.15
923.15
1013.15
1073.15
9.611u10-3
0.021
0.033
0.059
0.082
100 ˜ 64 ˜ X5
1 2X
2
1 X 2 2X
H2S
CH4
0.486
0.469
0.452
0.416
0.386
0.490
0.479
0.468
0.444
0.424
S he eac
e ed b he h ghe
LeCha e e
c e a d he eac
ch
he a
ha e).
2
CS2
e
4.76u10-3
0.010
0.016
0.028
0.038
e
H2
0.019
0.042
0.065
0.112
0.151
e, a
d be e ec ed f
( ha c ea e he
be f
e
S
C
ca a d E
T
13.3 (a
a a ab e a a Ma hcad
d
a
Chap e 13
c ,5
d
P2
10 Pa
hee ).
Gi en:
298.15 K
T1
5
10 Pa
P1
e
' G C3H8
24300
' H C3H8
104700 e
e
e
650 K
T2
' G CH4
50500 ' H CH4
74500
e
e
e
e
8.31451 R
6
' G C2H4
68500 ' H C2H4
52500
e
K
e
e
(at 298.15 K)
e
e
Mass Balance Table:
Species
In
O t
C3H8
CH4
C2H4
Total
1
0
0
1-X
X
X
1+X
(1-X)/(1+X)
X/(1+X)
X/(1+X)
Calc lation of mole fractions and acti ities:
C3H8 ( X )
1
X
1
X
CH4( X )
P
aC3H8( X P )
C3H8( X ) aC2H4( X P )
C2H4( X ) X
1
CH4( X ) aCH4( X P )
5
10 Pa
C2H4( X )
X
10 Pa
P
5
10 Pa
' G CH4
' G C3H8
'G
4
4.23 10
e
'H
' H C2H4
' H CH4
' H C3H8
'H
4
8.27 10
e
'G
e
Ka 298.15
R T1
Part (a):
X
G e
Ka 298.15
C3H8( Xa )
1
10
4
3.885 10
1
e
1
e
8
(initial g ess)
aC2H4( X P1 ) aCH4( X P1 )
aC3H8( X P1 )
C2H4( Xa )
1.971 10
4
Xa
F d( X)
CH4( Xa )
1.971 10
Xa
1.971 10
Xb
0.854
4
Part (b):
From eq ation 9.1-22b:
Ka 650
X
Ka 298.15 e
.5
'H
R
1
T2
1
Ka 650
T1
2.704
(initial g ess)
G e
Ka 650
C3H8( Xb )
0.079
X
P
' G C2H4
Ka 298.15
X
1
5
'G
aC2H4( X P1 ) aCH4( X P1 )
aC3H8( X P1 )
C2H4( Xb )
0.461
Xb
CH4( Xb )
F d( X)
0.461
e
4
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
Part (c):
X
(initial g ess)
.5
G e
aC2H4( X P2 ) aCH4( X P2 )
Ka 650
C3H8( Xc )
Xc
aC3H8( X P2 )
0.369
C2H4( Xc )
13.3
(a
R
8.31451 aA
A J A
AJ A
aC
A J C
1
Ka
e
0.316
CH4( Xc )
a a ab e a a Ma hcad
e
K
e
T
V
4 e
P
5
10 Pa
Ka
2.633
a D 0.5 10 Pa a C
A 1
RT
298.15 K
1
a D( P )
G
0.461
0.316
e
aB
A J C
Xc
hee ).
2400 G
e
F d( X)
Part (a):
A
(initial g ess)
0.5
5
G e
Ka
C
1
A
0.16
a B a A
A
A 1
F d
A
A
0.84
C
Part (b):
Recogni ing that the partial molar Gibb's e cess is in the form of the one constant Marg les
e pression ields:
J A e
2
0.3 C
J C e
0.3 A
5
G e
a D 0.5 10 Pa a C
Ka
C
1
A
0.132
a B a A
A
C
0.868
A e
2
A e
0.3 A
0.3 1
2
A
2
A
F d
A
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
Part (c):
Mass Balance Table:
Species
A
B
C
D
In
1
2
0
0
O t
1-X
2-X
X
X
Ass ming that D is an ideal gas:
P=(n*R*T)/V
X
P( X )
X R T
V
(initial g ess)
0.5
G e
Ka
X
0.425
NA
1
X
A
1
X
NA
0.575
A
0.575
aD
X R T V
a Ba A ( 1
NB
NB
2
X
1.575
e a C( X 1 )
X
X 1)
F d( X)
NC
X
ND
X
C
X
PD
NC
0.425
ND
0.425
C
0.425
PD
5
2.633 10 Pa
X R T e
V
S
C
13.3
(a
ca a d E
T
d
a
a a ab e a a Ma hcad
c ,5
Chap e 13
d
hee ).
Witho t dissociation:
a h drogen gas is eq al to the press re of h drogen gas:
Ass
ming
it ofh molec
A
_Adthe acti
bed_W
K1lar
H2
A
_Ad
K1 P H2
bed_W h
With dissociation:
A
_Ad
bed_W h K1 a H2
1
2
K3 a H
Using the eq ilibri m constant for the reaction H2 = 2H, the acti it of atomic h drogen can be
sol ed for in terms of the acti it of molec lar h drogen:
2
G e
K2
aH
F d aH
a H2
Using the positi e root for the acti it of h drogen ields:
A
_Ad
bed_W h K1 a H2
1
2
K3 K2 a H2
Ass ming the acti it of molec lar h drogen gas is eq al to the press re of h drogen gas:
A
_Ad
bed_W h K1 P H2
1
2
K3 K2 P H2
If the amo nt adsorbed aries linearl
ith the partial press re of molec lar h drogen then no
dissociation is occ rring. If the amo nt adsorbed aries as the sq are root of the partial press re,
then dissociation is occ rring.
13.40
F
he da a
he
be
G
'
H T'
S
'
'
The e
b
c
a
§ ' G ·
K e ¨
¨
¸¸
© R ˜ 298.15 ¹
The e
b
c
a
e
ae h
be
ae e
H
'
e ha e ha
G T'
S
148
a 25 C
e
19600
§
·
¨
¸
© 8.314 ˜ 298.15 ¹
a af c
f e
2716
ea
e
b a ed f
E .13.1-22b; he
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
Eq ilibri m constant
5
1 10
4
1 10
3
1 10
K
100
10
260
280
f
he eac
The
a
ba a ce ab e f
S ec e
M
e
D e
I a
M
0
Ka
aD
2
aM
b
h
h ch ca be
2
eac
F a
M(1-2D)
DM
ea
DM
1 2D
320
2A U A2
Th
The e
300
T
D
M
2
1 2D
2
M
e a
1 ·
§
4D 2 ¨ 4 ¸ 1 0
M ˜K ¹
©
a d ha he
2
D
The
1 ·
1 ·
§
§
¨4
¸ ¨4
¸ 16
˜
˜K ¹
M
K
M
©
¹
©
8
f
he f ac
d
e
ed a a f
c
f e
ea
e
g e be
:
S
C
13.41
(a
ca a d E
H N2 N2 e a
aN2 ga
P N2
H N2
A P N2
ge
Process #2:
aN e a
Chap e 13
d
hee ).
here
A
2 Ka 1
H N2
N2(gas) =2*N(metal)
2
H N N e a
aN2 ga
2
P N2
Ka 2 P N2
N ea
%
c ,5
Ka 1 P N2
N2 e a
Ka 2
a
N2(gas) = N2(metal)
aN2 e a
%
d
a a ab e a a Ma hcad
Process #1:
Ka 1
T
HN
ge
B P N2
here
B
Ka 2
HN
The empirical e pression gi en in the problem is s pported b process #2.
13.42 a) F
he che
ca e
b
a CO 2
P
1 ba
CaCO3 R CaO CO2
Ka
1.1395
a CaO a CO 2
a CaCO3
The ef e, he e
b
a d e f ha a d
ba e e a a ed
N
e
e
c a ed. The
c
a
ga
a 1150 K f
he eac
ce a CaO =a CaCO3 =1 a he a e
e
d .
he c
de
1.1395 ba f a CaCO3
ee ,
e f CO2 ha
d be
he ga ha e f 1.1395
1.1395 ba u 10 u10-3 3
0.119
e
3
5 ba ˜
8.314 u10
˜1150 K
˜K
ha 0.801
e f
d CaCO3 e a a 1150 K.
PV
RT
b) If a 100 e e e
ed a 1150 K, 1.19
e f CO2 c d be acc
da ed a a
e
e f 1.1395 ba , h ch
e ha a he ca c
ca b a e
d ha e d
c a ed
d ce 1
e f CO2 a d he e
e
d be
S
C
NRT
V
P
ca a d E
1
T
u 8.314 u 10 5
d
3
ba ˜
10 u10-3
3
˜K
a
c ,5
˜1150 K
Chap e 13
d
0.965 ba
2H2Se g R 2H2 g Se2
The a ba a ce ab e
S ec e
a
f a
H2Se
1
1-2X
1-2X
H2
0
2X
2X
Se2
0
(X)
(X)
13.43 The eac
T a
S
ha K a 1 4X 4X 2
F
each
X
1000
1050
1100
1150
1200
1250
0.239
0.271
0.299
0.324
0.346
0.364
13.44
K a,1
S ec e
N2
O2
N 2O
NO
NO2
a
0.79
0.21
0
0
0
S
e
f Se2 f
ed e
a 2H2Se
f Se2
f
a N 2O
; K a,2
0.5
2
0
e H2Se ha eac = X/2
a N 2 ˜ a O2
a 2NO
; a d K a,3
a N 2 ˜ a O2
f a
0.79-X-Y-0.5Z
0.21-0.5X-Y-Z
X
2Y
Z
4Y 2
0.79-X-Y-0.5Z ˜ 0.21-0.5X-Y-Z
c
1 2X
ed.
a
f
he che
a NO2
0.5
a N2 ˜ a O2
e f ac
(0.79-X-Y-0.5Z)/(1-0.5X-0.5Z)
(0.21-0.5X-Y-Z)/ (1-0.5X-0.5Z)
X/ (1-0.5X-0.5Z)
2Y/ (1-0.5X-0.5Z)
Z/ (1-0.5X-0.5Z)
1
1
1
1
a d K a,3
ca e
b
1
X
§
·§ 1-0.5X-0.5Z · 2 § 1 · 2
¨
¸¨
¸ ¨
¸
© 0.79-X-Y-0.5Z ¹© 0.21-0.5X-Y-Z ¹ © 1.013 ¹
0.79-X-Y-0.5Z § 0.21-0.5X-Y-Z · 2 § 1.013 · 2
˜¨
¸ ˜¨
¸
1-0.5X-0.5Z © 1-0.5X-0.5Z ¹ © 1 ¹
a , e ba
b
2
2X ˜1
0.119
0.135
0.150
0.162
0.173
0.182
X
(1-0.5X-0.5Z)
K a,2
E
e
a 2H2 ˜ a Se2
4X 2 Ÿ 4 K a 1 X 2 4K a X K a
e f H2Se ha eac ,
T(K)
K a,1
Ka
1
1
Z
§ 1-0.5X-0.5Z · 2 § 1 · 2
¨
¸ ¨
¸
0.21-0.5X-Y-Z © 0.79-X-Y-0.5Z ¹ © 1.013 ¹
c
a
ga
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
T(K)
1000
1200
1400
1600
1800
2000
Ka,1
6.7142u10-9
3.6675u10-8
1.2688u10-7
3.3069u10-7
7.1595u10-7
1.3675u10-6
Ka,2
7.5363u10-9
2.8345u10-7
3.7886u10-6
2.6496u10-5
1.2000u10-4
4.0052u10-4
Ka,3
1.0131u10-5
1.9428u10-5
3.1098u10-5
4.4434u10-5
5.8845u10-5
7.3931u10-5
T(K)
1000
1200
1400
1600
1800
2000
X
1.719u10-9
1.336u10-8
4.605u10-8
1.200u10-7
2.588u10-7
4.909u10-7
Y
1.986u10-5
1.084u10-4
1.002u10-4
1.044u10-3
2.216u10-3
4.026u10-3
Z
1.902u10-6
3.648u10-6
5.839u10-6
8.300u10-6
1.092u10-5
1.359u10-5
T(K)
1000
0.79
0.21
1.719u10-9
3.972u10-5
0.79
0.21
1.336u10-8
2.168u10-4
0.79
0.21
4.605u10-8
2.004u10-4
0.789
0.209
1.200u10-7
2.089u10-3
0.788
0.208
2.588u10-7
4.432u10-3
0.786
0.206
4.909u10-7
8.052u10-3
1200
1400
1600
1800
2000
N2
1.902u10-6
3.648u10
-6
5.839u10
-6
8.300u10
-6
1.092u10
-5
1.359u10
-5
N e ha a he e
he a
e
(
de
O2
N2O
f
ge (f e e
) a ge.
efe ed
NO
a NOX) a e a
NO2
a
S
C
13.4
(a
ca a d E
T
d
a a ab e a a Ma hcad
a
c ,5
Chap e 13
d
hee ).
Gi en:
'GAgC 108700˜
'GAg 77110˜
e
'GT C 186020˜
R 8.31451˜
e
e
'GT 32450˜
e
e
K˜
e
'GC 131170˜
e
e
e
e
e
T 298.15K
˜
e
Part (a):
'G
AgC 'GAg 'GC 'GAgC
'G
§ 'G AgC ·
¨
¸
R˜ T
©
¹
KAgC
KAgC e
AgC
4
-1
5.464 u 10
e
e
10
2.676 u 10
The sol bilit prod ct gi en in ill stration 9.3-2 is 1.607E-10. This e perimental al e is of the
same order of magnit de as the theoretical al e calc lated abo e.
Part (b):
'G
T C 'GT 'GC 'GT C
'G
§ 'G T C ·
¨
¸
R˜ T
©
¹
KT C
KT C e
TC
4
2.24 u 10
-1
e
e
4
1.19 u 10
The sol bilit prod ct gi en in ill stration 9.3-2 is 1.116E-2. This e perimental al e is t o orders
of magnit de greater than the theoretical al e calc lated abo e.
C2H4 H2O R C2H5OH
13.4 The eac
N e ha e h e e ha a b
g
f 169.3 K a d a c ca e e a e f 282.4 K.
The ef e, e h e e a ga a 298.15 K, h e a e a d e ha
ae
d.I a ,I
a
e ha a e a d e ha
f
a dea
, ha he e
ae
e ha
he a , a d
eh e e
he
d. The
c ec f he e a
b
ea
.
a C2H5OH
C2 H5OH J C2 H5OH
Ka
1 ba
a C 2 H 4 ˜ a H 2O
˜ H 2 O J H 2O
1.013 ba
a , a 298.15 K, f
he P e
ga
a
a
PH O 0.03171 ba a d PC H OH 0.07889 ba
2
The
S ,a
2
a
ba a ce ab e
S ec e
Wa e
C2H5OH
T a
d
g dea
1
0
5
a
beha
,
f a
1-X
X
1
,
1-X
X
d
e f ac
S
C
ca a d E
Ka
C 2H5OH
1 ba
˜ H 2O
1.013 ba
S he
a e
aef
0.9578 a d
C2H5OH
T
22.89
he
H 2O
d
a
c ,5
Chap e 13
d
X
22.89
, a d X=
1 ba
22.89+1.013
˜ 1 X
1.013 ba
d
e f ac
ae
0.0422.
0.9578
H e e, e
ha e ha
+ ae
e ae
dea . U g he V a Ba c
UNIFAC
ga
e
a e he ac
c eff c e g e
JC2H5OH ( 0.9578) 1.0011 a d J H2O (
0.0422) 2.4802 . The ef e,
Ka
X ˜1.0011
22.89
,
1 ba
˜ 1 X ˜ 2.4802
1.013 ba
22.89
a d X=
0.9824
0.0176
C2 H5OH a d H 2O
22.89+0.4089
Reca c a g he ac
c eff c e
h he UNIFAC
ga g e
JC2H5OH (
0.9824) 1.0002 a d J H2O (
0.0176) 2.5557
a d eca c a
g he e
b
X=0.9830
C2H5OH a d
H 2O
h ch
dea
ea
c
c
g e
0.0170 .
ee
gh
he e
e
ha I
: C2H5OH 0.9578 a d H2O 0.0422.
:
C2 H5OH =0.9830
a d
c
a
e
e a e:
0.0170
H 2O
N
eed c ec f e ha
a d ae
he a
PC 2 H5OH 0.9830 u 1.0002 u 0.07889 0.0776 ba
PH 2O
e a e aga . S f
ha e. The
a a
0.017 u 2.5557 u 0.03171 0.00138 ba
S
he a a e
e f e h e e 1 - 0.0776 - 0.00138 = 0.9210 ba a d
ac
0.9210
a C2 H 4
0.9092 ( 1/1.013 a ha
fa bee a
ed).
1.013
S eca c a g he e
b
(
g he a e f he e
b
c
a
f
a ea
) e ha e
X ˜1.0002
X
K a 22.89
0.4360
,
1 ba
1 X
˜ 1 X ˜ 2.5557 u 0.9092
1.013 ba
22.89
a d X=
0.9811 C2H5OH a d H 2O 0.0189.
22.89+0.44
A h e
e d ffe e f
he e
ea
,I
e a e aga .
13.4
F
he che ca e
K.
The a ba a ce ab e
S ec e
a
f a
SO2
1
1-X
O2
0.5
0.5-0.5X
SO3
0
X
T a
(1.5-0.5X)
b
c
a
ca c a
(
e f ac
)
(1-X)/(1.5-0.5X)
(0.5-0.5X)/(1.5-0.5X)
X/(1.5-0.5X)
he
g a , Ka = 4.5905 a 1000
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
X
P
ISO3
1.5
0.5X
1.013
Ka
0.5
0.5
a SO2 a O
1 X
P
§ 0.5 0.5X P
·
2
ISO2 ˜ ¨
IO 2 ¸
1.5 0.5X 1.013
© 1.5 0.5X 1.013
¹
he e he I e
a e f gac c eff c e . The e a
be
ed f
XISO3
1
4.5905
0.5
1 X ISO2 § 0.5 0.5X P
·
I
¨
O ¸
© 1.5 0.5X 1.013 2 ¹
a)
A P = 1.013 ba , e ca a
e ha he a e f a f gac
,
ha
a SO3
4.5905
X § 1.5 0.5X ·
¨
¸
1 X © 0.5 0.5X ¹
0.304,
SO2
0.5
Ÿ X
0.152 a d
O2
a
aa dba e
c eff c e
ae
0.642
SO3 =0.544.
b)
A P=101.3 ba , he
dea ga ha e beha
( ha , he f gac c eff c e
ca
be eg ec ed. The
ced e I
e
f
e he f gac c eff c e
ca c a e he e
b
c
, e he e c
he Pe g-R b
e a
f ae
ca c a e he f gac c eff c e , eca c a e he e
b
c
h e f gac c eff c e , a d c
e ea g
c e ge ce ach e ed.
·
X § 1.5 0.5X
¨¨
¸¸
1 X © 100 ˜ 0.5 0.5X ¹
4.5905
0.089,
SO2
0.045 a d
O2
0.5
Ÿ X 0.907
SO3 =0.866.
A
g ha he b a
e ac
a a e e f he e
f gac c eff c e a e
ISO2 1.0056, IO2 1.0405 a d ISO3 =1.0090. The ef e
KI
ISO3
1.0090
0.5
ISO2 IO
2
The
X 0.908,
1.0056 u 1.0405
he
SO2
S he cha ge f
e e a e be g
0.088,
O2
he dea ga
h gh, ha
0.5
ga c ac d a e
, he
0.984
0.048 a d
ha e ca e
SO3 =0.867.
h ea
e
g.
a , a ge
a a e
f he
)
,
h
S
C
13.4
ca a d E
(a
T
d
a a ab e a a Ma hcad
a
c ,5
Chap e 13
d
hee ).
Pa a
'G
R T
( Ka )
57.33
d
d T R T
0.17677
'H
R
R T
2
57.33 ' H
R T
2
R T
2
57.33 'H
J
e
Part (b):
'G
( T)
Ka( T )
e
e
57330 'G
176.77 e
e T
eK
R
In
1
0
1
O t
1-X
2*X
1+X
8.31451 K
e
e
( T)
R T
Mass Balance Table:
Species
N2O4
NO2
Total
Calc lation of mole fractions and acti ities:
N2O4( X )
a N2O4( X P )
KaX( X P )
1
X
1
X
NO2( X )
N2O4( X )
a NO2( X P )
2
a N2O4( X P )
P
5
10 Pa
a NO2( X P )
2X
1
X
NO2( X )
P
5
10 Pa
S
C
ca a d E
T
Pa b
X 0.7
(initial g ess)
G e
Ka( T ) KaX( X P )
Xb 0.1
4
X 323.15 K 10 Pa
d
a
c ,5
Chap e 13
d
X( T P )
Xb 1
F d( X)
5
X 323.15 K 10 Pa
Xb 10
6
X 323.15 K 10 Pa
NO2 Xb 0.1
0.91
NO2 Xb 1
0.605
NO2 Xb 10
0.261
N2O4 Xb 0.1
0.09
N2O4 Xb 1
0.395
N2O4 Xb 10
0.739
Part (c):
Xc 0.1
4
X 473.15 K 10 Pa
NO2 Xc 0.1
13.4
N2O4 Xc 0.1
1
1.246 10
4
Xc 1
5
X 473.15 K 10 Pa
NO2 Xc 1
0.999
N2O4 Xc 1
1.243 10
3
Xc 10
6
X 473.15 K 10 Pa
NO2 Xc 10
0.988
N2O4 Xc 10
0.012
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
13.49
Gi en:
Ta
e
58620 'H
138.2 'S
e
298.15 K
Tb
373.15 K
K
e
R
e
4
10 Pa
P1
P2
8.31451 5
10 Pa
K
P3
e
e
6
10 Pa
Mass Balance Table:
Species
In
O t
M
D
Total
2
0
2-2*X
X
2-X
(2-2*X)/(2-X)
X/(2-X)
Acti ities, Eq ilibri m Constant, and Eq ilibri m E pression:
a M( X P )
'G
( T)
KaX( X P )
X
0.999
2
2
2X X
P
a D( X P )
5
10 Pa
T ' S
'H
Ka( T )
X P
5
2 X 10 Pa
e
'G
( T)
R T
a D( X P )
a M( X P )
2
(initial g ess for sol er)
G e
KaX( X P ) Ka( T )
Part (a):
Deg eeOfD e a
( Ta P1 )
0.953
Deg eeOfD e a
( Ta P2 )
0.985
Deg eeOfD e a
( Ta P3 )
0.995
X 1
Deg eeOfD e a
( T P)
F d( X)
S
C
13. 0
(a
ca a d E
T
d
a
a a ab e a a Ma hcad
c ,5
Chap e 13
d
hee ).
Gi en (All Units are SI):
T1
267
T2
255
6
2 10
P1 CH4
Acti ities:
ah d ae
1
a H2O
1
6
1.5 10
P2 CH4
R
8.31451
P CH4
a CH4 P CH4
5
10
Part (a):
ah d a e
Ka P CH4
5.75 a H2O
'G
a CH4 P CH4
'G
T1 P1 CH4
3
6.65 10
'G
T2 P2 CH4
3
5.742 10
R T T P CH4
Ka P CH4
Part (b):
3
'H
10
G e
a
'G
F d 'H
'H
(initial g esses)
'S
10
T1 P1 CH4
'H
'S
T1 ' S
'H
4
1.357 10
'S
a
'G
0
T2 P2 CH4
'S
a
1
75.737
Part (c):
Ka 273
P273 CH4
G e
P273 CH4
'H
Ka P1 CH4 e
6
10
R
1
1
273
T1
Ka 273
0.044
(initial g ess)
Ka P273 CH4
6
2.288 10
Ka 273
P273 CH4
F d P273 CH4
'H
T2 ' S
S
C
ca a d E
13. 1 The eac
The
T
d
a
c ,5
Chap e 13
d
ae
ea
a
B1 œ B
ba a ce
ec e A
a
NA
be
e
O ea
a
f
a
e
a d
A1 B1 œ A B
¦ N ¦¦ N
A
f A1
ba a ce
NB
A e
A1 œ A
F
I
¦ GH N ¦ N JK
AB
A
AB
ec e B
be
f B1
f
¦ N ¦¦ N
B
F
I
¦ GH N ¦ N JK
AB
B
AB
b
0;
GA GA1
GB GB1
c
0 ; GA B G A1 GB1
h 0;
a d
G
A e
b
G
a
¦ N A G A ¦ N B GB ¦ ¦ N A B G A B
a
h e ec
0
each e e
f eac
.
¦ GA dNA ¦ GB dNB ¦ ¦ GA B dNA B
dGtotal
¦ N A dGA ¦ N B dGB ¦ ¦ N A B dGA B
0
0 b the GibbsD hem eq ation
A
,
G
g he e
a
0
0
b
e
a
¦ G A1 N A ¦ GB1 N B ¦ ¦ b G A1 GB1g N A B
G A1 ¦ N A GB1 ¦ N B ¦ G A1¦ N A B ¦ GB1 ¦ N A B
LM F N ¦ N I OP G LM¦ e N ¦ N jOP
JK Q N
N GH
Q
G A1 ¦
A
AB
B1
GA . A
FG w G IJ
H wN K
AB
G A1 N A GB1 N B
FG w G IJ
H wN K
a
A
N BT , P
{ GA
a
B
N A, T, P
{ GB
GB
S
C
ca a d E
T
G A1
The ef e
A
d
eg a
c ,5
Chap e 13
d
GA
, e ha e ha , b def
Th ,
a
GB1
a d
FG w G IJ
H w PK
g be ee a
RT
T
FG w IJ
H wP K
a e , e ha e
GB
(1)
(2)
T
a f a f RT aaTT,, PP ff
2
G T , P2 G T , P1
N
g E
a f ed a a T a d P
. (3)
h E
e ha
A T, P
A e a e
e ha E . (1)
. (1), a d ec g
A1 T , P
A1 P
b
Ÿ
A P
a
0
a d
Po0
A1 P
0
A1 P
ha
13. 2
A P
ee
be
B1 T , P
e
e ha
A P
be
A1
. (1)
P 0 a d he
A P
0
g ha E
B T, P
e c d eg a e E . (2) be ee
be a f ed a a T a d P. Th
A1 P
(3)
1
ePa d
0
(LeCha e e
c
.
A1 A1 œ A2
A1 A2 œ A3
#
A1 A œ A 1
e c.
T a
f A1
N0
NT
NT
T
¦
he e N T
a
¦N
be
f
e
a1
d¦ i
¦
¦
N
a
h ch
N1 2 N 2 3 N 3 " A "
N
¦ N
N0
NT
Ÿ
e
a
¦¦
e ha
¦
1
aa
1
¦
a d
¦¦
a1
2
e
e)
S
(a)
C
a
ca a d E
a1
T
FG N IJ a d
HN K
2
0
1
T
A
K 1
d
a
c ,5
Chap e 13
d
FG N IJ
HN K
K
0
T
a
f
a P 1 ba f
a 1
I 1
1 P 1 ba
a a1
I I1
1
I 1 1 1 ba
I I1 1 P
2
The
I I1P
I 1
F
he
e
e
1K
a
S
f he
1K
1
e ha
I I1P
I 1
de e de
I 1
I I1P
1
1
1K
de
D
a d he
2
1D K
2
1 2D K
3
4
3
4
1 DK
a D Kf
1
h
2
ha h
deed he ca e. The
a D Kf D K
2
1
1
a f
2
3
1 D K
a
a f
#
aD Kf
1
1
#
e c.
NT
The
N0 ¦
A
N
f
he
N0 ¦ KD 1 1 .
¦
1
e e
f ge
f
1
1 T
¦T
0
¦ KD 1 1
1
e c
e ha e
f
¦ KD 1 1
1
f
1
¦ a KD 1f 1
1
f
1
¦ a KD 1f
0
1
1 KD 1
1
(*)
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
ha
1
1 KD 1
1
1 ; KD
1 1 KD
1
1
A
FG ¦ T IJ
K
TH
f
F 1I
T H1T K
f
¦ T 1
0
0
f
1
1 T 2
¦ T 1
1
ha
f
f
¦ KD 1 1
1
1
¦ a KD 1f 1
1
1
1
Ÿ
F
1
NT
N0
1
he a de Waa e
0
1
2
1
FG
IJ
H 1 KD K
2
FG N IJ
HN K
2
1
1
E . (*) ab e
a d
1
a
FG N IJ
HN K
0
T
1 1; a
a1
0
T
1
1
12
2
1 a1
f a e e ha e
RT
a
2
V
V
P
¦ a KD 1f 1
NT
N0
a1 KD f
f
f
1
NRT
N 2a
VN
V
a d
I
B
ZB
B1
ZB
ZB h ch he e bec
I
2¦ a
RTV
e
I
ha
ZB 2¦
a1
B1
2a
ZB ¦
RTV
ZB
RTV
B
2 a1 N 0
ZB ZB
RTV N T
S
C
ca a d E
T
I I1
I 1
d
a
I 1
de e de
RT
V
2a1 N 0
RTV N T
ZB
1
ZB
RT
V 1 N0 NT
a
1
ZB
1
RT P
V
N T RT
N TV N 0 1
f
h ch
de e de
f he de
f he de . We ca
e
a
D
a a
N T RT
NTV N0 1
1 a dD
1 1 DK
B1
1 1 1
B
1
I I1
I 1
PI I1
Chap e 13
d
I I1 I 1
1
(b)
c ,5
ed. N
ha
RT
V N0 NT
a
1
1 1
2
1 1
2
1 1
1
1 1
1
1
aV RTK f r aV RTK f 4 V
1
1
RT
a
2
V
V
2
RT
a
1 21
V 1
V
1
2
2
1
1
21
ha
P
a
RT
a
12
V 1 V
a de Waa e a
.
1
f a e f he
h ch de e d
aV f r dV 4 V i aV f aV f 1
1
2
(***)
E a
(**) a d (***) a e he e h ch f
he e a
a c a g a de Waa f d. N ce ha
e f V e eed 1
ge c b c.
V ; he ef e, he e a
N e ha f he f d
-a c a g, he K 0
h
he
1
21
(**)
A
h ch
1 1
1
(c)
P
RT
V 1
f
FG1 RTK IJ 1 Ÿ aV RTK f V H V K
aV RTK f V 0
aV RTK f V 0
1
ha D
e
1
21
1
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
13. 3 The de c
f HF c a
g
e
de c bed
he a c e C ec
f Pha e E
b
Da a f Se a a
Tech
g b W a Sch e I . E
C
.P
D . D . (1980), 19, 432 439. B a ca ef e a
a
f he de
a d he da a, he
ed ha HF a c a e
he a
ha e acc d g he eac
.
2HF œ HF 2
6HF œ HF 6
a d,
e he e
8HF œ HF 8
e a ge f 195 240 K, he e
ea
K2
2
2
1
K6
6
6
1
K8
8
8
1
b
c
a
ae
LM 6429.4 24.1456OP
N T
Q
21100.9
O
e LM
N T 69.7292PQ
25224.5
e LM
83.4689OP
N T
Q
e
1
f a
,a d
he f gac
f ec e . Ne , Sch e
ed a (
e b Ta
a d W a [C
. E . S (1978), 33, 651]
P f he
I
e , d e , he a e a d
ha he f gac
c eff c e
ca e aea a
a e e a , a d ca be ca c a ed f
he f gac
f
e HF.
Th a e a ed he eed
ec f he
ec a a a e e f he a c a
c
e e .
U g h
de , Sch e b a ed e g d ag ee e f he a c a
fac (de
)
f
e HF a d a HF-F e
e
g a
e e a
f a e ch a Pe gR b
( h ch
be
ed e a e
ce che ca e
b
e
ed
he ha e e
b
ca c a
).
A a e a e,
ead f
g he Ta -W a a
f e a
f f gac
c eff c e
e he
de
P b e 9.52
he e K ha
e
ab e) a g
a d he
ea HF a a che
HF 8 . S
a
, HF +
e :
-a
ca
gc
-a
ca
a
2
ca
eac
gc
a1 a d
1
e
e
h HF, HF 2 , HF 6 a d
d be ea ed a a f e c
e
e + HF, HF 2 , HF 6 a d HF 8 . I each f he e ca e
he c
f he HF c
e cha ge a he e
b
cha ged. The
be
h h
a
ha he a a a e e f
he a c a
c
e e bec e
ea
ca
a ge.
F
e a
e, a8 64 u a1 . C e e
,
c
ee
he e ca -c ec
de f he HF a c a g
e e
, h gh he
de
ea e
bab a fac
f e g ee g ca c a
.
S
C
13. 4
Sa
af c
13.
ca a d E
KW
KW
gf
§ K T2
¨¨
© K T1
a
a ge f e
fa
C
H
1)
c ,5
Chap e 13
d
7.085
25
7.0
30
6.915
40
6.77
50
6.63
60
6.51
g ha '
1
KW
2
' H ( J/
e, a d ha
20
º
H
e ha e ha
H
)
59.22
58.79
56.89
58.83
52.70
54.24
49.46
K
W

o H OH
BOH o B OH a d H 2 O m

ªM - º
¬ OH ¼ W
ª¬ M H+ º¼
W
ª¬ MOH- º¼
W
` ª¬M
º J 2r
OH- ¼ W
M H+
^M
h
ha
º
BOH ª
¬ MOH- ¼ W
` ª¬M
º
OH- ¼ W
J r2
ha
M H
E
ea ed
ea
a
H§ 1
1 ·
¸ a da
¨
R © T2 T1 ¹
7.27
OH- ¼ BOH
M OH a
'
10
ªM º
¬ OH ¼ W
13.
H
7.47
de he eac
S
·
¸¸
¹
d
T( C)
0
M H M OH J r2 b
^ª¬M
T
a
e
M BOH M 2BOH 4K W / J r2
M 2BOH 4K W / J r2 M BOH
2
2
ª
º
¬ M OH ¼ W M BOH
KW
M 2BOH 4K W / J r2 M BOH
2K W
M OH J 2r
M 2BOH 4K W / J r2 M BOH J r2
(15.1-10)
e aced b K a,W
,
e aced b MH J r . B
g M H J r , e
cha ged.
1
2
ce aga
ha e H
|
KW
M BOH J 2r
M 2H J 2r
M HA, M 1
he e e MH
ha
h he def
1
1
K a,HA g M HA,
2
2
ha
,
e
. (15.1-
S
C
ca a d E
13.
The eac
T
d
a
c ,5
Chap e 13
d
ae
K a ,HA
K
a ,W

o H A a d H 2O m

o H OH HA m

M H J r M A J r
h K a,HA
M HA
b
The ef e
M H J r M OH J r
a d K a,W
M HA, M H a d M A K a,HA
ha
M HA ˜ M 1
M 1
2
M H
M 2H J r2
M HA, M H J r ˜ M 1
he e e MH a
ea ed
e
½°
Ka,HA ­°
4M HA,
1¾ a d
® 1
2 °¯
Ka,HA
°¿
e . 15.1-10
cha ged.
M H J r
13.
Wea ba e +
f ae
g ac d, eg ec
e aced b MH J r .
,
g H g M H J r
g H+ a d OH-
d ced f
he
a
K
BOH
ZZZZ
X B OH HA o H A a d BOH YZZZZ
h K a,BOH
MBOH
M B MOH
M BOH M 1
U g D be he
MBOH, D; MB+
(
a
f BOH d
c a ed,
D; a d MOH- D MHA
ce each
e f OH- f
ed b
+
fH f
ed f
he d
ca
D MHA ˜ D
K a,BOH
MBOH, D ˜ M 1
he d
ca
f he ac d f
( h ch
e e
e cha ge f BOH
(ab e I
a
e ca
ed a e
e a
h HA.) S
K a,BOH M HA D
K a,BOH M HA
2
15.1-7) e ce
e
f
e
he
4K a,BOH ˜ M BOH,
2
M OH D M HA
K a,BOH M HA K a,BOH M HA
M H
KW
M OH The ef e
4K a,BOH ˜ M BOH,
2K W
K a,BOH M HA
2
4K a,BOH ˜ M BOH, K a,BOH M HA
Add g e ec
e
dea
M B J r M OH J r
K a,BOH
K a,BOH
Ÿ
M BOH M 1
J r2
ha e e
2
2
a d
13.
f BOH eac
ha
a e .) S ha
he e e ace K a,BOH
M B M OH M BOH M 1
h K a,BOH / J r2
S
D
C
ca a d E
§K
·
¨¨ a,BOH
M HA ¸¸ 2
© Jr
¹
M OH T
d
13. 0
2
§ K a,BOH
·
K
M HA ¸¸ 4 a,BOH
˜ M BOH,
¨¨
2
J 2r
© Jr
¹
2
KW
M OH J 2r
2K W
2
§K
·
§K
·
K
M HA ¸¸
J 2r ¨¨ a,BOH
M HA ¸¸ 4 a,BOH
˜ M BOH, ¨¨ a,BOH
2
2
2
Jr
© Jr
¹
© Jr
¹
Add g e ec
e
K a,1M HA
M A
; J r M A 2
M H J 2r
ST S
ST S
S
13. 1 The e
Chap e 13
d
2
a d
M H
c ,5
§ K a,BOH
·
K
M HA ¸¸ 4 a,BOH
˜ M BOH,
¨¨
2
J 2r
© Jr
¹
2
§K
·
¨¨ a,BOH
M HA ¸¸ 2
© Jr
¹
D M HA
a
K a,1M HA
M H J 2r
dea
K a,2 M A J r
M H J r
K a,1K a,2 M HA
M H
2 3
Jr
ha M A2
K a,1Ka,2 M HA
MH J 3r
ª
º
K
K K
... « a,1 a,1 a,22 ...» M HA
«M »N
M H
¬« H
¼» S
ª K
º
K1 K 2
10 1 10
«
.... « H ....»»
2 3
H 2
M H J r2
10
J
r
M H J r
10
Jr
«¬
»¼
K a,1
b
K a,1K a,2
ea
ae
H ˜ NH3 CH 2 COO
NH3 CH 2COO
H ˜ NH 2CH 2COO
K1 =
; K2 =
; KD =
NH3 CH2COOH
NH3 CH 2 COO
NH 2CH 2COOH
The ef e
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
NH 3 CH 2COO K D ˜ NH 2CH 2COOH=NH 2CH 2COOH ˜10 K D
NH 3 CH 2COOH
H ˜ NH 3 CH 2COO K1
H ˜ NH 2 CH 2 COOH ˜10 KD
K1
NH 2 CH 2 COO
K 2 ˜ NH3 CH 2 COO
H
NH 2 CH 2COOH ˜10 H K 2 K D
NH 2 CH 2 COOH
NH 2 CH 2 COOH
NH 2 CH 2 COOH NH CH 2 COO NH 3 CH 2COOH NH 2CH 2COO 3
1
1 10
KD
10
K1 H K D
10 H K2 KD
10 K D
NH 3 CH 2 COO
NH 3 CH 2 COOH
NH 2 CH 2 COO A
H=3
NH 2 CH 2 COOH
A
1 10 K D 10 K1 H K D 10 H K 2 K D
10 K1 H KD
1 10 KD 10 K1 H K D 10 H K2 KD
10 H K2 KD
1 10 KD 10 K1 H KD 10 H K2 KD
| 0;
NH 3 CH 2COO 0.818;
NH3 CH 2 COOH
;
0.182;
NH 2 CH 2 COO |0
H=10
NH 2 CH 2 COOH
NH 2CH 2COOH ˜10 K1 H K D
| 0;
NH 3 CH 2COO 0424;
NH 3 CH 2COOH
| 0;
NH 2CH 2 COO | 0.576
S
C
ca a d E
13. 2
T
d
a
c ,5
Chap e 13
d
a) The eac
NH 3 + 2O 2 + (7.5238N 2 ) o HNO 3 + H 2O+ (7.5238N 2 )
The
e gh a e HNO 3 =63.01; H 2O= 18.01; NH 3 =17.03; H 2O=18.01
ec a
S 100 g f a 60
ae,
HNO3
=
The
e
%
f HNO 3 c
he f
g
g
a
e f ac
0.9522
e
he
d c
0.9522
=0.0979, H 2 O =0.1285 a d
0.9522 1.25 7.5238
he feed a e
e f ac
N2
f HNO3 a d 1.25
c d g
S
f
ge :
0.773576
1
=0.0950 O2 =0.1900 a d N 2 0.71493
1 2 7.5238
; 'Gf (H 2O, ) 237.1 J/
; 'Gf (HNO 3 , )
'Gf (NH3 ,g) 16.5 J/
NH3
e
=
'G = 111.3 237.1 ( 16.5)
Th
he
a
a
S
effec
he
f
d c
331.9 J/
ha c
e
111.3 J/
a eac ed,
fa
d be b a ed
h
c
de
g
effec .
e NH 3 eac ed
'GP = -RT (1 0.0979+1 0.1285+7.7238 0.7736)=1.710 J/
S
effec
he feed
'Gf = -RT (1 0.0950+2 0.1900+7.7238 0.71493)=2.714 J/
The
S
e a 'G f
he
a
ce ha
50 g=64.7
he
a
ce ='G
f
'GP 'Gf
ha ca be
effec a e eg g b e c
=0.446 MPa
A 50
g=64.7
a =0.446 MPa, I
331.9 1.7 2.7
d ced
a ed
a 25 C H =104.89 J/ g a d S =0.3674 J/ g-K
332.9 J/
he f ee e e g
T 147.9 C, H =2743.9 J/ g a d S =6.8565 J/ g-K
S ce
1
= 2000 b =4410 g,
J/ g
332.9 J/
NH 3
NH 3 eac ed. A
f eac
.
G=104.89-0.3674 u 298.15=-4.65
e 0.45 MPa
'G f ea = -143.03-(-4.65)=-138.4 J/ g
NH 3
ce a e ca be ead f
ea
ab e
G =2743.9-6.8586 u (147.9+273.15)=-143
S
C
ca a d E
13. 3 See f
13. 4
g
T
d
a
c ,5
c ce
a
Chap e 13
d
be
S
gf
he h d ge
a
he
ec e
b
gf
a de
a
g he c
ce
a
f
e ge
1
§
·2
K K a1
ª¬ H º¼ ¨10 14 PCO2 ¸
KH
©
¹
h ch ead
PCO2
0
H=7
PCO2
2.7 u 104
H=4.233
PCO2
3.85 u 104
H=4.156
PCO2
5.5 u 104
H=4.079
13.
Pa a Ma
bala ce able
S b a ce
I
CH3OH
F a
M e f ac
1
1-X
(1-X)/(1+X)
CH2O
0
X
X/(1+X)
H2
0
X
X/(1+X)
T a
1
1+X
Ka
aCH 2O aH 2
CH 2 O H 2
aCH3OH
Ka 1 X
X
a
2
P / 1 ba
X2
1 X 1 X
CH 3OH
2
X2
;
1 X 2
(1 K a ) X 2 ;
X ; Ka
Ka
1 Ka
' G ( J/
) ' H ( J/
CH 2O
-102.5
-108.6
H2
0
0
CH 3OH
-162.0
-200.7
Reac
59.5
92.1
K 298.15 K
e
§ ' G·
¨
¸ e
© RT ¹
)
59000
§
·
¨
¸ e ( 24.0034); K
© 8.314 u 298.15 ¹
3.7625 u1011
S
C
ca a d E
Pa c Ma
T
bala ce able
S b a ce
CH3OH
CH2O
H2
N2
T a
Ka
aCH2 O aH2
I a
1
0
0
0.5
1.5
CH 2 O H 2
aCH3OH
P / 1 ba
CH3OH
K a 1.5 0.5 X X
2 2
S
B
e
a
c ,5
Chap e 13
d
F a
1-X
X
X
0.5
1.5+X
M e f ac
(1-X)/(1.5+X)
X/(1.5+X)
X/(1.5+X)
0.5/(1.5+X)
X2
1.5 X 1 X
X2
;
1.5 0.5 X X 2
X 2 ; 1.5K a 0.5 K a X K a X 2
(1 K a ) X 2 0.5 K a 1.5 K a
13.
13.
d
X 2;
0
he a e a P b e 13.39
, h
he a e a
be
h
ge e ac g h d ge .
13.38,
ee he
ha
be .
S
C
ca a d E
a) Reac
ac
fBa dDa e
§ 'G ·
¨
¸
© RT ¹
Th ha he
a
c ,5
Chap e 13
d
A
he e a
0.5 CJ C
J
A
c) U e N A
S ec e
§ P ·
aC ¨ D ¸
© 1 ba ¹
aA
aC aD
aA aB
ha e a e
0.5 C
0.5 1 A
A
A
ed
e ee
he
=0.166 a d C =0.834.
he
be
d ha e a a
f
e
e
be c
each
e f A eac ed. D
M e
a
NA
NA N
B
NB
NB N
C
NA N
D
N
S
he e
b
K
aC aD
aA
NA N
f A added
a
a . N e ha he
a h
g he c
e f a
A
be
, e ha e he f
g
e f ac
Pa a
be
uN
/ NA
ed
u R uT / V
NA N
, h e a
be
ed f
>Cha
>Cha
>Cha
>Cha
>Cha
>Cha
>Cha
>Cha
e 13>P
e 13>P
e 13>P
e 13>P
e 13>P
e 13>P
e 13>P
e 13>P
N .
13.
13.
h
h
h
h
h
h
h
h
be
be
be
be
be
be
be
be
f
f
f
f
f
f
f
f
de A
de A
de A
de A
de A
de A
de A
de A
e
e
e
e
e
e
e
e
f
f
f
f
f
f
f
f
Te
Te
Te
Te
Te
Te
Te
Te
b
b
b
b
b
b
b
b
e
/ NA
N
ea
f
e f he eac
N
e
be
dC
NA N
ce
a
e f
d
/ NA
A N A , R, T a d V a e a
13. S
13. 0 S
13. 1 S
13. 2 S
13. 3 S
13. 4 S
13. S
13. S
e, he
=0.195 a d C =0.805
be
Th ha he
d ced f
d a d ga e
0.5 C e (0.3(1 C ) 2 )
(1 C ) e (0.3 C2 )
A A
e
g he
. The ef e
2.058
K
K
d
A( )+B( ) o C( )+D(g). S a
e
b) N
T
be
be
be
be
be
be
be
be
13.69
13.70
13.71
13.72
13.73
13.74
13.75
13.76
u R uT / V
e
S
13.
C
ca a d E
T
d
a
c ,5
d
Chap e 13
S
C
13.
M A M H
K1
b
M A
K1M A MA
M H
def e
F
ca a d E
c ,5
Chap e 13
d
M A, M A M A M A 2 M A3
H K1
10
a d def e M
def e
=
M A-
K 2M A
M H
M / M A,
K2
M A, M A M A M A 2 M A3
M H
;
M A
2
K2
M H
10
2
1 M A + M A- M A2- M A3- =
MA
M A,
a
M A, M A M A M A 2 M A3
eac
2
M A M H
M A
d
1 M A + M A- M A 2- M A3-
1 M A+
K2
MA
K1
M H
1
T
H K 2
2 1 M A+
2 1M A+
F
eac
3a d4
M A 2 M H K3
; M A 2
M A
M A 2- =
K 3M A 3M A MH
3 M A- =
3
2 1 M A+
4 M A 2- =
4
3
10
H K 3
M A
a d
M A3h
N
10
3
1 M A+ =
H K3
2 1M A+
a d
10
4
H K4
1 M A+ M A- M A2- M A3-
1+ 1 M A +
1
1 2+
1
2
3+
1 2
3
4
M A+ a d
1
M A+
1 1 1 2 + 1 2 3 + 1 2
The cha ge he c
ed f
ch a g e M A + M A- 2M A2- 3 M A3= M A+ 1
2 M A+ 2 1
2
3
4
3 M A+ 3 1 2
1 1 2 2 1 2 3 3 1 2 3 4
1 1 1 2 + 1 2 3 + 1 2 3 4
3
4 M A+
S
C
ca a d E
1 2.20
h
1 d
2 8.75
0 1 140
R1 10
T
Chap e 13
d
4 13.40
h
h 10
h
R2 10
c ,5
3 9.81
h 0.1˜
1
a
2
R3 10
h
3
R4 10
h
4
1
1 R1 R1 ˜ R2 R1 ˜ R2 ˜ R3 R1 ˜ R2 ˜ R3 ˜ R4
1 R1 ˜ R2 ˜ 1
2 R1 ˜ R2 ˜ R3 ˜ 1
3 R4 ˜ 2
0 1
1 1 2 3
cha ge 1 1 2˜ 2 3˜ 3
1
2
1
0
1
0
cha ge
0.5
2
2
3
4
0
0
5
10
h
15
0
5
10
h
15
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
13.80
part a
1 2.63
RP 10
cha ge
2 14.30
1 h
RN 10
1
1 RN
cha ge
cha ge cha ge
0 1 150
h
h 0.1˜ 1.0
2
RP
1 RP
cha ge
1
e ec c cha ge
0
1
e ec c
0
5
10
15
(
1
2)
2
8.465
20
h
ª 1 10 1 h 10 h 2º
¬
¼
S 16˜
(
1
7
)
(
7
2
)
ª¬ 1 10
º¼
10
3
1 10
S
100
10
5
10
h
15
10
h
2
S
84.249
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
13.81
1 2.55
0 1 70
R1 10
2 h
2 4.37
h
h 0.1˜
1
R2 10
h 10
h
2
1
1 R1 ˜ 2
1 R1 R1 ˜ R2
0 R1 ˜ R2 ˜ 2
cha ge 1
2
1
0.5
0
0
0
2
4
6
8
h
2
cha ge
1
0
0
2
4
h
6
8
2˜ 2 1
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
13.82
1 2.55
2 4.37
0 1 140
R1 10
2 h
3 13.03
h
h 0.1˜
1
h 10
h
R2 10
2
R3 10
1
1 R1 ˜ R2 ˜ R3 ˜ 2
cha ge 2˜ 2 1
2
1
0
0.5
1
0
0
5
10
15
h
2
1
cha ge
0
1
0
3
1 R1 ˜ 2
1 R1 R1 ˜ R2 R1 ˜ R2 ˜ R3
0 R1 ˜ R2 ˜ 2
h
5
10
h
15
1 1
S
C
ca a d E
T
d
a
c ,5
Chap e 13
d
13. 3
1 1.0
0 1 120
R1 10
2 2 2.1
3 8.02
h
h 0.1˜
h
1
R2 10
4 8.71
h 10
h
2
h
R3 10
3
R4 10
1
1 R1 R1 ˜ R2 R1 ˜ R2 ˜ R3 R1 ˜ R2 ˜ R3 ˜ R4
1 R1 ˜ 2
1 R1 ˜ R2 ˜ R3 ˜ 2
2 R1 ˜ R2 ˜ R3 ˜ R4 ˜ 2
0 R1 ˜ R2 ˜ 2
cha ge 2˜ 2 1 1 2˜ 2
1
2
1
0
0.5
1
2
0
0
2
4
6
h
8
10
12
h
4
S
13. 4
C
ca a d E
T
a) C
de he f
g eac
K a1
AH 2 
o AH H f
h ch he e
K a1
c ,5
ea
ae
M A MH
M AH
2
The a e age e cha ge
he
ab e
ec e = M AH M A .
2
The ef e, a he
e ec c
H
M AH M H M A M H ˜
M AH M AH
e e K a1 ˜ K a2
Chap e 13
d
K
a d K a2
M AH a
a2
a d AH 
o A H
b
M AH M H d
M AH = M A ,
2
e ec c
b
e .. F
K
M AH
2
=
M AH M H 2
M AH
2
M H
M AH 2
, 2 H= K a1 K a 2
b) He e he e a e
M A M H
K a2
M A M H
2
ha a he
ha
c
H=
K a1 K a 2
2
de
K
K
a1
a2
a3
AH3 
o AH 2 H , AH 2 
o AH H , a d AH 
o A H
f
h ch he e
K a1
b
M AH M H 2
M AH
ea
ae
M AH M H , K a2
M AH 3
he
ab e
3
e ec c
H
M AH M H M AH M 2
H
˜
M AH M AH e e K a1 ˜ K a2
M A
ec e = 2M AH M AH M A .
The ef e, a he
3
M AH 2
M AH
K a3
a d
2 M H
M AH
2
The a e age e cha ge
a
M A MH
a d K a3
M H
K a2
3
K a1 ˜ K a2
+
. S
M H
Ka 2
2M AH M AH ,
3
e ec c
N e ha d ffe e , b
h ch cha ge ec e
The ec d
b
e
e
ha
3
M AH
+
M H
M AH
K a1 ˜ K a2
3
M AH M H ,
3
+K a1 ˜ M H
2
= M H
3§
K ·
¨ 2 a1 ¸
¨
M H ¸¹
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Chapter 15
Solutions to Chemical and Engineering Thermodynamics, th ed
Section 10.1
10.1
10.1-1
A simpler solution using Mathcad is available as a Mathcad worksheet.
(a) At the bubble point we have
xi Pi vap P where P
yi
5 bar
xET 0.05 ; xP 010
. , xNB 0.40 and xMP 0.45 . Procedure used was
i) Guess T,
ii) Compute each yi , and the sum ¦ yi . If ¦ yi 1 , guessed T is correct; if ¦ yi ! 1 ,
guessed T is too high; if ¦ yi 1 , guessed T is too low. If ¦ yi z 1 , we correct T and
K (bubble point) yET 0.4167 , yP 01730
,
recalculate.
Solution:
T 29366
.
.
and yMP 0.2502 .
yNB 01601
.
(b) The dew point calculation is similar. Here, yET 0.05 , yP 010
. , yNB 0.40 and yMP 0.45 .
P 5 bar, and T and the xi 's are the unknowns. Thus, here we guess the dew point
temperature, compute each of the xi 's from xi
Pyi Pi vap and evaluate ¦ xi . If ¦ xi
guessed temperature is the dew point temperature; if
1 , the
¦ xi ! 1 , guessed T is too low; if
¦ xi 1 , guessed T is too high. Solution (obtained using the computer) T
314.23 K (dew
and xMP 0.4409 .
point) xET 0.0039 , xP 0.0337 , xNB 05215
.
(c) The advantage of the Mathcad worksheet for the isothermal flash calculation is that one can use
the initial flash equations directly, rather than having to make the substitutions below.
For the isothermal flash vaporization calculation, we proceed as in Illustration 8.1-3. First, we
calculate the K factors, i.e.
. 4.402229
10 817.08 30315
,
KET
10185
.
5
and, similarly KP 2.238 , KNB 0546
and KMP 0.743 . Thus, the equations to be solved
.
are:
(1)
xET xP xNB xMP 1
(2)
yET yP yNB yMP 1 Ÿ 10185
. xET 2.238 xP 0546
. xNB 0.743xMP 1
Also,
xET L 1 KET KET 0.05 Ÿ xET 10185
. 9185
. L 0.05
(3)
and, similarly
(4)
xP 2.238 1238
.
L 010
.
(5)
xNB 0546
.
0.454 L 0.40
(6)
xMP 0.743 0.257 L 0.45
Solution procedure I used was to guess L, compute the xi 's from eqns. (3 to 6), and then
ascertain whether eqns. (1) and (2) were satisfied. After a number of iterations, I obtained the
following solution:
a
a
f
f
Solutions to Chemical and Engineering Thermodynamics, th ed
L
xET
xP
xNB
xMP
¦ xi
086667
.
0.0225
0.0858
0.4258
0.4659
.
1000
Section 10.1
V
013333
.
0.2289
01921
.
0.2326
0.3464
1000
.
yET
yP
yNB
yMP
¦ yi
(d) For an adiabatic flush vaporization, shown below, the energy balance must also be satisfied
liquid
liquid
X
vapor
pressure reducing valve or device
This is a (two-phase) Joule-Thomson expansion, so that the energy balance yields H in
H out ,
or
¦ xi H i T , P, x inlet conditions
L ¦ xi H i T , P, x
L
L
c
V ¦ yi H i T , P, y
V
h
outlet liquid conditions
outlet vapor conditions
This equation must be satisfied, together with the mass balances and phase equilibrium
equations of part c. Thus, we have one new unknown here, the outlet temperature, and an
additional equation from which to find that unknown.
10.1-2
(a) We start with eqn. (8.2-12b):
temperature SdT
¦ xi dGi SdT VdP
0.
And note that at constant
0 and
dGi
RTd ln f i
a
a ff
RTd ln xiJ i f i T1 P ,
so that
RT ¦ xi d ln xi RT ¦ xi d ln J i RT ¦ xi d ln fi VdP
0
However, for the pure fluid fugacity, we have, from eqn. (7.2-8a)
RTd ln f i
dGi
V i dP
Thus
b
g
RT ¦ xi d ln xi RT ¦ xi d ln J i ¦ xiV i V dP
Also
0
Solutions to Chemical and Engineering Thermodynamics, th ed
Section 10.1
¦ xiV i V ¦ xiV i ¦ xiV i ¦ xiV i
V ex Ÿ RT ¦ xi d lna xiJ i f V exdP
ex
Now assuming
i) Ideal gas-phase behavior: xiJ i Pi vap
yi P or xiJ i
0
yi P Pi vap and
ii) That PV ex / RT 1 we obtain
ex
a f ¦ x d lnc y P P h PVRT d ln P
¦ xi d ln xiJ i
vap
i
i
i
or
¦ xi d ln yi ¦ xi d ln P ¦ xi d ln Pi vap d PV ex RT id ln P
Now noting that ¦ xid ln Pi vap 0 , since Pi vap is a function of temperature only and
b g d ln P , yields ¦ x d ln y FGH PVRT 1IJK d ln P or
¦ xid ln P
a
y1 y2
x1 1 x1
y1 1 y1
i
ex
1
2
1
1
2
dy2
dy1 .
1
1
1
1
1
1,
f x a1 y f y a1 x f x y
y a1 y f
y a1 y f
a y x f dy d ln P
y a1 y f dx
dx
1
i
FG x x IJ dy FG PV 1IJ d ln P | d ln P
H y y K H RT K
x
x1
dy1 2 dy2
y2
y1
Since
ex
d ln P ¦ xi
1
1
1
1
1
1
y2
1
1
1
Also
1 y ,
and
x2
1 x1 ,
so
1
1
To obtain the x y diagram, I used the equation above in a finite difference form. Using
the argument i to denote the ith data point, the equation above becomes
y1 i x1 i
˜ y1 i y1 i 1
y1 i 1 y1 i
a
f
ln P i ln P i 1
y1 i is unknown, however, P1 i , P1 i 1 , x1 i are known. Also y1 1 is either 0 or 1
depending on which end of the data one starts with. In fact, I started at both ends, in two
separate calculations, to check the results. I solved this problem using the equation above
rewritten as
y1 i
where
B
B r B 2 4C
2
x1 i y1 i 1 ' ln P
and C
1 ' ln P
x1 i y1 i 1
1 ' ln P
and averaged the results from starting at the x1 0 and x1 1 ends.
Once x1 and y1 , were known, the activity coefficients were calculated from
J1
y1 P x1 P1vap and J 2
y2 P x2 P2vap . Results are given below.
Solutions to Chemical and Engineering Thermodynamics, th ed
b.
CCl4 + n-Heptane System
= exprmntl
x-y data
c.
Ethylene bromide + 1-nitropentane System
Azeotrope
= exprmntl
x-y data
Section 10.1
Solutions to Chemical and Engineering Thermodynamics, th ed
Section 10.1
(also available as a Mathcad worksheet)
10.1-3
T
69
p5
10.422
p6
26799
8.314 ( 273.15
p5
exp ( p5 )
p5
2.721
8.314 ( 273.15
T)
p6
exp ( p6 )
p6
1.024
35200
8.314 ( 273.15
T)
p7
exp ( p7 )
p7
0.389
x7
29676
10.456
p7
T)
11.431
x5
0.25
x6
0.45
P
x5 p5
x6 p6
x7 p7
( x5 p5 )
y5
0.3
P
1.258
( x6 p6 )
y6
P
y7
( x7 p7 )
P
Bubble point pressure
P
Bubble point compositions
P
1.258
y5
0.541
y6
0.366
x6
0.6
y7
0.093
Now on to dew point calculation
Initial guesses
z5
0.25
GIVEN
x5 x6
soln
x5
P
z6
1
0.45
x5
z7
x7
0.3
0.3
x5 p5 z5 P
x7 1
0.1
x6 p6 z6 P
x7 p7 z7 P
x7
soln 2
soln 3
P
0.768
FIND( x5 x6 x7 P )
soln 0
x6
soln 1
Dew point pressure
Dew point compositions
x5
0.071
P
x6
0.338
x7
0.592
Solutions to Chemical and Engineering Thermodynamics, th ed
Section 10.1
(also available as a Mathcad worksheet)
10.1-4
Solving for the bubble point pressure
T
P
69
p5( T )
26799
8.314 ( 273.15
exp 10.422
p6( T )
exp 10.456
p7( T )
y5
1.013
exp 11.431
0.33
y6
p5( T )
T)
35200
8.314 ( 273.15
T)
y7
p5( T )
P
T)
29676
8.314 ( 273.15
0.33
K5( T P )
2.721
p6( T )
p6( T )
P
p7( T )
K7( T P )
0.389
p7( T )
P
z5
0.33
K6( T P )
1.024
0.25
z6
0.45
z7
0.3
GIVEN
K6( T P ) z6
K5( T P ) z5
soln
y6 K6( T P ) z6
y5 K5( T P ) z5
K7( T P ) z7 1
y7 K7( T P ) z7
FIND( y5 y6 y7 P )
y5
soln 0
y5
0.541
y6
y7
soln 1
y6
soln 2
0.366
y7
P
soln 3
0.093
P
1.258
This is the bubble-point pressure solution. Now on to the dew-point pressure problem.
x5
0.33
x6
0.33
x7
Note that xi=yi/Ki
0.33
GIVEN
z5
z6
z7
K5( T P )
K6( T P )
K7( T P )
soln
1
x5
soln 2
P
z5
K5( T P )
x6
z6
K6( T P )
x7
z7
K7( T P )
FIND( x5 x6 x7 P )
x5
soln 0
x5
0.071
x6
soln 1
x6
x7
0.338
x7
0.592
soln 3
P
0.768
This is the dew-point pressure solution.
So for a mixture of the composition z5=0.25, z6=0.45 and z7=0.30, at a temperature of 69 C, the
mixture will be all liquid at pressures above 1.258 bar, and all vapor at pressures below 0.768 bar.
Vapor-liquid equilibrium will exist at this temperature only between 0.768 and 1.258 bar, so this is
the pressure range we will examine.
Solutions to Chemical and Engineering Thermodynamics, th ed
T
69
L
0.99
P
1.2
K5( T P ) x5
GIVEN
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
x6
x7
x5
soln 0
Section 10.1
soln 1
K7( T P ) x7
soln
soln 2
L
x7
0.32
0.223
x6
0.456
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
K7( T P ) x7
y5
0.507
y6
0.389
0.104
P
1.1
L
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
x6
x7
x5
soln 0
soln 1
0.362
0.458
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
y5
0.445
y6
0.427
y7
1
L
L
0.906
( x5
x6
x7 ) 0
FIND( x5 x6 x7 L)
soln 3
V
1
L
V 0.264
L
0.736
K7( T P ) x7
0.128
0.60
K5( T P ) x5
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
x6
x7
x5
soln
x7
x6
GIVEN
K7( T P ) x7
L
0.18
L
V
V 0.094
soln 2
x5
1.0
x7 ) 0
0.80
K5( T P ) x5
GIVEN
P
y7
x6
FIND( x5 x6 x7 L)
soln 3
x5
( x5
soln 0
soln 1
K7( T P ) x7
soln
soln 2
L
x7
0.414
soln 3
x5
0.141
x6
0.445
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
K7( T P ) x7
y5
0.383
y6
0.456
0.161
y7
( x5
x6
x7 ) 0
FIND( x5 x6 x7 L)
V
1
V 0.451
L
L
0.549
Solutions to Chemical and Engineering Thermodynamics, th ed
P
0.9
L
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
x6
x7
x5
P
soln 0
soln
soln 2
L
x7
0.481
x6
0.412
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
K7( T P ) x7
y5
0.323
y6
0.469
0.208
L
0.8
y7
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
x6
x7
soln 0
soln 1
K7( T P ) x7
soln
soln 2
L
x7
0.563
x6
0.359
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
K7( T P ) x7
y5
0.267
y6
0.459
0.274
y7
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
x6
x7
soln 0
soln 1
K7( T P ) x7
soln
soln 2
L
x7
0.59
x6
0.339
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
K7( T P ) x7
y5
0.251
y6
0.451
0.298
y7
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
x6
x7
x5
L
L
0.337
x6
x7 ) 0
FIND( x5 x6 x7 L)
V
1
L
L
0.09
( x5
x6
x7 ) 0
FIND( x5 x6 x7 L)
V
1
L
V 0.995
L
5.309 10
0.95
K5( T P ) x5
GIVEN
( x5
soln 3
0.071
L
1
V 0.91
x5
1.25
V
0.10
K5( T P ) x5
x5
FIND( x5 x6 x7 L)
soln 3
0.078
GIVEN
x7 ) 0
V 0.663
x5
L
x6
0.20
K5( T P ) x5
0.77
( x5
soln 3
0.107
x5
P
soln 1
K7( T P ) x7
x5
GIVEN
P
0.40
K5( T P ) x5
GIVEN
Section 10.1
soln 0
soln 1
K7( T P ) x7
soln
soln 2
L
x7
0.303
soln 3
x5
0.246
x6
0.451
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
K7( T P ) x7
y5
0.536
y6
0.369
0.094
y7
( x5
x6
x7 ) 0
FIND( x5 x6 x7 L)
V
1
V 0.013
L
L
0.987
3
Solutions to Chemical and Engineering Thermodynamics, th ed
Section 10.1
0 1 9
i
0.768
0.071
0.338
0.592
0
.77
0.071
0.339
0.590
0.0053
.8
0.078
0.359
0.563
0.090
.9
0.107
0.412
0.481
0.337
1.0
PP
xx5
0.141
xx6
0.445
xx7
0.414
0.549
LL
1.1
0.180
0.458
0.362
0.736
1.2
0.223
0.456
0.320
0.906
1.25
0.246
0.451
0.303
0.987
1.258
0.250
0.450
0.300
1.0
(also available as a Mathcad worksheet)
10.1-5
Solving for the bubble point temperature
T
69
p5( T )
p6( T )
p7( T )
P
1.013
exp 10.422
exp 10.456
exp 11.431
26799
8.314 ( 273.15
p5( T )
K5( T P )
T)
29676
8.314 ( 273.15
2.721
P
p6( T )
1.024
K6( T P )
T)
T)
p6( T )
P
35200
8.314 ( 273.15
p5( T )
p7( T )
0.389
K7( T P )
p7( T )
P
Solutions to Chemical and Engineering Thermodynamics, th ed
y5
0.33
y6
0.33
y7
0.33
z5
Section 10.1
0.25
z6
0.45
z7
0.3
GIVEN
K5( T P ) z5
K6( T P ) z6
K7( T P ) z7 1
y5 K5( T P ) z5
y6 K6( T P ) z6
y7 K7( T P ) z7
FIND( y5 y6 y7 T )
soln
y5
soln 0
y5
0.548
y6
soln 1
y6
y7
soln 2
0.363
T
y7
soln 3
0.088
T
61.788
This is the bubble-point temperature solution. Now on to the dew-point temperature problem.
x5
0.33
x6
0.33
x7
Note that xi=yi/Ki
0.33
GIVEN
z5
z6
z7
K5( T P )
K6( T P )
K7( T P )
1
x5
soln 2
T
z5
x6
K5( T P )
z6
K6( T P )
x7
z7
K7( T P )
FIND( x5 x6 x7 T )
soln
x5
soln 0
x5
0.074
x6
soln 1
x6
x7
0.346
x7
soln 3
0.579
T
77.436
This is the dew-point pressure solution.
So for a mixture of the composition z5=0.25, z6=0.45 and z7=0.30, the mixture will be all liquid
at temperatures below 61.79 C, and all vapor at temperatures above 77.44 C. Vapor-liquid
equilibrium will exist only between 61.79 and 77.44 C, so this is the temperature range we
will examine.
T
62
L
0.99
P
1.013
K5( T P ) x5
GIVEN
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
x6
x7
x5
soln 0
soln 1
K7( T P ) x7
soln
soln 2
L
x7
0.303
soln 3
x5
0.246
x6
0.451
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
K7( T P ) x7
y5
0.543
y6
0.367
0.09
T
65
L
0.80
y7
( x5
x6
x7 ) 0
FIND( x5 x6 x7 L)
V
1
V 0.012
L
L
0.988
Solutions to Chemical and Engineering Thermodynamics, th ed
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
x6
x7
x5
T
soln 0
x7
0.343
0.459
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
y5
0.476
y6
0.41
L
y7
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
x6
x7
soln 0
soln 1
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
y5
0.411
y6
0.444
y7
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
x6
x7
soln 0
( x5
L
L
0.813
x6
x7 ) 0
FIND( x5 x6 x7 L)
soln 3
V
1
L
V 0.365
L
0.635
K7( T P ) x7
0.144
soln 1
K7( T P ) x7
soln
soln 2
L
x7
0.443
0.124
x6
0.432
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
K7( T P ) x7
y5
0.352
y6
0.464
0.183
y7
( x5
x6
x7 ) 0
FIND( x5 x6 x7 L)
soln 3
x5
V
1
L
V 0.551
L
0.20
K5( T P ) x5
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
GIVEN
1
0.40
K5( T P ) x5
L
V
V 0.187
soln
0.389
0.453
74
FIND( x5 x6 x7 L)
soln 3
K7( T P ) x7
x7
x6
x5
x7 ) 0
0.114
L
0.157
GIVEN
x6
K7( T P ) x7
soln 2
x5
L
( x5
0.60
K5( T P ) x5
71
soln
L
x6
68
K7 ( T P ) x7
soln 2
0.198
x5
T
soln 1
x5
GIVEN
T
K6 ( T P ) x6
K5 ( T P ) x5
GIVEN
Section 10.1
K7( T P ) x7
soln
( x5
x6
x7 ) 0
FIND( x5 x6 x7 L)
0.449
Solutions to Chemical and Engineering Thermodynamics, th ed
x5
T
soln
x7
soln 2
L
x7
0.504
soln 3
0.098
x6
0.398
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
K7( T P ) x7
y5
0.301
y6
0.467
0.232
76
L
y7
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
x6
x7
soln 0
soln 1
K7( T P ) x7
soln
soln 2
L
x7
0.548
x6
0.369
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
K7( T P ) x7
y5
0.27
y6
0.46
0.27
L
y7
L
L
0.251
x6
x7 ) 0
FIND( x5 x6 x7 L)
V
1
L
V 0.891
L
0.109
0.10
K5( T P ) x5
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
x6
x7
GIVEN
soln 0
soln 1
K7( T P ) x7
soln
soln 2
L
x7
0.57
0.077
x6
0.353
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
K7( T P ) x7
y5
0.256
y6
0.453
0.291
L
y7
x6
x7 ) 0
FIND( x5 x6 x7 L)
V
1
L
V 0.966
L
0.05
K5( T P ) x5
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
GIVEN
( x5
soln 3
x5
78
( x5
soln 3
0.083
x5
1
V 0.749
x5
77
V
0.10
K5( T P ) x5
x5
T
soln 1
x5
GIVEN
T
x6
0
Section 10.1
K7( T P ) x7
( x5
x6
x7 ) 0
0.034
Solutions to Chemical and Engineering Thermodynamics, th ed
x7 ( L ( 1
x5
T
K7( T P ) ) z7
x6
x7
soln 0
soln 1
FIND( x5 x6 x7 L)
soln
soln 2
L
x7
0.592
soln 3
x5
0.071
x6
0.337
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
K7( T P ) x7
y5
0.242
y6
0.445
0.313
61
L
y7
K6( T P ) x6
x5 ( L ( 1
K5( T P ) )
K5( T P ) ) z5
x6 ( L ( 1
K6( T P ) )
K6( T P ) ) z6
x7 ( L ( 1
K7( T P ) )
K7( T P ) ) z7
x6
x7
x5
V
1
L
V 1.045
L
0.045
0.9
K5( T P ) x5
GIVEN
i
K7( T P ) )
Section 10.1
soln 0
soln 1
K7( T P ) x7
soln
soln 2
L
x7
0.29
0.264
x6
0.446
y5
K5( T P ) x5
y6
K6( T P ) x6 y7
K7( T P ) x7
y5
0.566
y6
0.351
0.083
x6
x7 ) 0
FIND( x5 x6 x7 L)
soln 3
x5
y7
( x5
V
V
1
L
0.047
L
1.047
0 1 9
TT
61.788
0.25
0.45
0.3
1.0
62
0.246
0.451
0.303
0.988
65
0.198
0.459
0.343
0.813
68
0.157
0.453
0.389
0.635
71
xx5
0.124
xx6
0.432
xx7
0.443
LL
0.449
74
0.098
0.398
0.504
0.251
76
0.083
0.369
0.548
0.109
77
0.077
0.353
0.570
0.034
77.436
0.074
0.346
0.570
0.0
Solutions to Chemical and Engineering Thermodynamics, th ed
10.1-6
Section 10.1
Clearly, it is only water that condenses out at the dew point, since O2 and N 2 are far above their
critical temperatures. Thus, at the dew point
PHvap
2O
yH 2 O P
PH 2 O
partial pressure of H 2O in air.
[In writing this expression, all fugacity coefficients have been assumed equal to unity.]
From the data in Problem 5.12 we have, at the dew point,
5432.8
7.8086
26.3026 ln PHvap
2O
27315
. 256
.
and
PHvap
dew point
PH 2 O 24618
. Pa
2O
at the air conditions PHvap
T
2O
256
. qC
3354.9
PH 2 O
PHvap
T
2O
20.6q C
Ÿ relative humidity
PHvap
T
2O
256
. qC
u 100%
7338%
.
Solutions to Chemical and Engineering Thermodynamics, th ed
Section 10.1
10.1-7
Start by determining the pressure if only vapor was present using ideal gas law PV=NRT
5
1 u 8.315˜ 10
P
˜ 298.15
P
3
8.264
bar
3˜ 10
which is much higher than the vapor pressure of either pure component. Therefore, a vapor-liquid
mixture must be present. The molecular weight of butane is 58.124 and its density is 0.575 g/cc,
so that 0.1 moles of liquid butane would occupy 10.1 cc; the molecular weight of hexane is
86.178, and its density is 0.655 g/cc so that 0.9 moles of hexane would occupy 118.4 cc.
Therefore to proceed, I will assume that finally there will be about 2875 cc of vapor and 125 cc of
vapor.
LL 0.9
xb 0.075
P 0.5
yb xb ˜
V 1 LL
2.428
yh xh ˜
P
xh 1 xb
0.2
P
Given
xb ˜ 2.428
yb ˜ P
xh ˜ 0.2
P
equilibrium relation for butane
equilibrium relation for hexane
yh ˜ P
xb ˜ LL yb ˜ V
0.1
xh ˜ LL yh ˜ V
0.9
xb xh
1
yb yh
1
5
V˜ 8.314˜ 10
mass balance on butane
mass balance on hexane
sum of liquid mole fractions = 1
sum of vapor mole fractions = 1
Pressure from ideal gas equation of state
298.15
˜
3
2.875˜ 10
z find ( xb xh yb yh LL V P)
§ 0.081 ·
¨
¸
¨ 0.919 ¸
¨ 0.516 ¸
z ¨ 0.484 ¸
¨
¸
¨ 0.956 ¸
¨ 0.044 ¸
¨ 0.38 ¸
©
¹
xb z
xh z
0
V z
2
yh z
LL z
3
4
P z
5
6
yb
0.516
LL
0.956
Volume of liquid
yb z
1
yh
0.484
VL LL˜ §¨ xb ˜
©
58.124
0.575
xb
xh ˜
0.081
86.178 ·
¸
0.655 ¹
xh
0.919
VL
123.415
Close enough to the guess value of 125 cc that it is not worth iterating.
Solutions to Chemical and Engineering Thermodynamics, th ed
Section 10.1
10.1-8
p5 2.755
p6 1.021
p7 0.39
a) Bubble point pressure
P 0.55˜ p5 0.25˜ p6 0.2˜ p7
y5 0.55˜
y7 0.2˜
p5
P
1.8485
p6
y5
0.81972
y6 0.25˜
y7
0.0422
y5 y6 y7
P
p7
P
y6
P
0.13808
1
b) Dew point pressure
P 2
Initial guess
x5 0.55˜
P
x6 0.25˜
p5
P
x7 0.2˜
p6
P
p7
Given
0.55˜
x5
P
x6
p5
x5 x6 x7
0.25˜
x5 z
x5
0.20854
x6 z
x6
0.25578
x7
0.53569
1
x7 z
2
x5 x6 x7
p6
x7
P
0.2˜
p7
1
z find ( x5 x6 x7 P)
0
P
1
0.20854 ·
¨§
¸
0.25578 ¸
¨
z
¨ 0.53569 ¸
¨ 1.04459 ¸
©
¹
P z
3
P
1.04459
bar
Solutions to Chemical and Engineering Thermodynamics, th ed
Section 10.1
10.1-9
p5 2.755
a)
p6 1.021
p7 0.39
0.55
x5( L p ) x6( L p ) § 1 p5 · ˜ L p5
¨
¸
p
p ¹
©
0.25
p6
§1 ˜L ¨
¸
p
p ¹
©
p6 ·
0.20
x7( L p ) p7
§1 ¨
¸ ˜L p
p ¹
©
p7 ·
y6( L p ) x6( L p ) ˜
y5( L p ) x5( L p ) ˜
p6
y7( L p ) x7( L p ) ˜
p
p 0.9775
Given
L 0.1
x5( L p ) ˜ §¨ 1 0
©
p5
p
p7
p
initial guess for P, fix L
p6 ·
p7 ·
x6( L p ) ˜ §¨ 1 x7( L p ) ˜ §¨ 1 ¸
¸
¸
p ¹
p ¹
p ¹
©
©
p5 ·
p Find ( p )
p
1.129
x5( L p )
0.24
x6( L p )
y5( L p )
0.584
y6( L p )
0.274
0.247
x5( L p ) x6( L p ) x7( L p )
1
y5( L p ) y6( L p ) y7( L p )
1
x7( L p )
0.487
y7( L p )
0.168
b)
p 0.9775
Given
0
L 0.9
x5( L p ) ˜ §¨ 1 ©
initial guess for P, fix L
p6 ·
p7 ·
x7( L p ) ˜ §¨ 1 x6( L p ) ˜ §¨ 1 ¸
¸
¸
p ¹
p ¹
p ¹
©
©
p5 ·
p Find ( p )
L
0.9
p
1.789
x5( L p )
0.522
x6( L p )
y5( L p )
0.804
y6( L p )
0.261
0.149
x7( L p )
y7( L p )
0.217
0.047
Solutions to Chemical and Engineering Thermodynamics, th ed
Section 10.1
10.1-10
p5 2.755
a)
x5( L p ) p6 1.021
p7 0.39
0.55
x6( L p ) § 1 p5 · ˜ L p5
¨
¸
p
p ¹
©
0.25
p6
§1 ˜L ¨
¸
p
p ¹
©
p6 ·
0.20
x7( L p ) § 1 p7 · ˜ L p7
¨
¸
p
p ¹
©
y6( L p ) x6( L p ) ˜
y5( L p ) x5( L p ) ˜
p6
y7( L p ) x7( L p ) ˜
p
p 0.9775
Given
0
L 0.5
x5( L p ) ˜ §¨ 1 ©
p5
p
p7
p
initial guess for P, fix L
p6 ·
p7 ·
x6( L p ) ˜ §¨ 1 x7( L p ) ˜ §¨ 1 ¸
¸
¸
p ¹
p ¹
p ¹
©
©
p5 ·
p Find ( p )
p
b)
x5( L p )
0.386
x6( L p )
y5( L p )
0.714
y6( L p )
y5( L p ) y6( L p ) y7( L p )
1
0
x5( L p ) ˜ §¨ 1 ©
p
0.317
y7( L p )
0.203
1
L 0.27152
x7( L p )
0.297
x5( L p ) x6( L p ) x7( L p )
p 0.9775
Given
10.1-11 a)
1.491
0.083
Guess p, iterate on L
p6 ·
p7 ·
x6( L p ) ˜ §¨ 1 x7( L p ) ˜ §¨ 1 ¸
¸
¸
p ¹
p ¹
p ¹
©
©
p5 ·
1.285
x5( L p )
0.3
x6( L p )
y5( L p )
0.643
y6( L p )
0.294
0.234
x5( L p ) x6( L p ) x7( L p )
1
y5( L p ) y6( L p ) y7( L p )
1
x7( L p )
y7( L p )
0.406
0.123
Number of unknowns
x5, x6, x7, y5, y6, y7, TI, PI, TII, PII, V and L = 12
(4 constraints)
Number of constraints
TI = TII, PI = PII™[i ™\i = 1
VAP
x i Pi
yi P i 5, 6 and 7 (3 constraints)
Mass balances xi L + yi V = zi
(3 constraints) where z is the feed composition
p Find ( p )
Solutions to Chemical and Engineering Thermodynamics, th ed
Section 10.1
(Note summing the three mass balances gives L + V = 1, so that is not an independent
equation)
So the number of degrees of freedom is 12 – 4 – 3 – 3 = 2
Setting T = 69°C fixes one degree of freedom.
Therefore, can not also set one liquid phase composition and one vapor phase composition!
The system would be overspecified.
b) Joe has to release one degree of freedom. Let T vary, then can find a T and P to meet the
composition specifications.
Pivap
10.1-12
Ÿ
yi P 4.35 kPa and ln P (MPa) = 8.347 – 2644/T
T = 191.8 K
10.1-13
xBZ = 0.99
98% of benzene in the feed
xBZ = 0.4
xTMP = 0.6
105 mols/hr
xBZ = ?
Benzene mass balance
TMP mass balance
Product restriction
Also xC6 = 1 - xBZ
0.4 × 105 = 0.99 D + xBZB
0.6 × 105 = 0.01 D + xTMPB
0.98 × 0.4 × 105 = 0.99 D
D = 3.9596 × 104 moles/hr
B = 105–D = 6.0404 × 104 moles/hr
0.6 × 105 = 0.01 × 3.9596 × 104 + xTMP × 6.0404 × 104
6 0.039596 ×104
x TMP
0.98676
6.0404 u 104
x BZ
0.01324
10.1-14
Since the pressure in the smaller vessel is 0.6 bar, that is the partial pressure of the butane
in the vapor phase of the larger vessel. That is
0.6
yB P 0.6 xB PBvap xB u 2.583 so that xB
0.232 and xH 1 xB 0.768
2.583
Now P xB PBvap xH PHvap 0.232 u 2.583 0.768 u 0.218 0.767 bar
so that finally
0.6
0.6
yB
P 0.767
0.782 and yB 1 0.782 0.218
10.1-15 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.15
Solutions to Chemical and Engineering Thermodynamics, th ed
Section 10.1
10.1-16
The Gibbs energy of a mixture consisting of N1 moles of species 1 and N 2 moles of species 2 is
G
N1 G1 RT ln x1 N 2 G 2 RT ln x2
o
o
§ o
§ o
N1 ·
N2 ·
N1 ¨ G1 RT ln
¸
¸ N 2 ¨ G 2 RT ln
N1 N 2 ¹
N1 N 2 ¹
©
©
So the change in Gibbs energy between a state I and a state II (without chemical reaction) is
'G
N1II G1 RT ln x1II N 2II G 2 RT ln x2II N1I G1 RT ln x1I N 2I G 2 RT ln x2I
o
o
o
o
Since there is no chemical reaction, N1II = N1I and N 2II = N 2I so this equation reduces to
'G
N1II RT ln x1II N 2II RT ln x2II N1I RT ln x1I N 2I RT ln x2I
Based on 1 mole of feed that is 0.5 moles of species 1 and 0.5 moles of species going to one stream
of 0.5 moles of a mixture 0.495 moles of species 1 and 0.005 of species 2, we have
'G
RT
0.495ln 0.99 0.005ln 0.01 0.5ln 0.5 0.5ln 0.5
= 0.495 u 0.01005 0.005 u 4.605 0.5 u 0.693 u 2
= 0.0050 0.0230 0.693 0.665
J
kJ
'G 0.665 RT 0.665 u 8.314
u 298.15 K=1.648
mol ˜ K
mol
This increase in free energy must occur as the result of work be supplied to the system.
Therefore, the work required for the separation is 1.648 kJ per mole of feed.
10.1-17 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.17
10.1-18 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.18
10.1-19 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.1
10.1-20 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.2
10.1-21 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.3
10.1-22 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.4
10.1-23 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.5
10.1-24 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.8
10.1-25 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.9
Solutions to Chemical and Engineering Thermodynamics, th ed
10.1-26 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.10
10.1-27 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.13
Section 10.1
Section 10.2
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Section 10.2
Solutions to Chemical and Engineering Thermodynamics,5th ed
F
Ethylene bromide + 1-nitropentane System
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-------------COMPONENT DATA
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Name
Number
Description
Frequency
---- -------------------------------- -----------------------1
carbon tetrachloride
53
CCl4
1
2
n-heptane
1
2
CH3
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Section 10.2
Solutions to Chemical and Engineering Thermodynamics,5th ed
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Name
Number
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isobutane
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ethyl bromide
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2.9105
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390.41
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2.7976
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2.5849
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588.89
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2.4847
1.0484
648.55
0.2500
2.3885
1.0613
704.77
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2.2963
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2.2078
1.0935
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1.1132
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Section 10.2
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Section 10.2
Solutions to Chemical and Engineering Thermodynamics,5th ed
/ŶƚŚĞĨŽůĚĞƌƐƉĞŶĨŽƌdĞdžƚŬхŚĂƉƚĞƌϭϬ͘ϮхWƌŽďůĞŵƐхWƌŽďϭϬ͘ϮͲϯϴ
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Section 10.2
Solutions to Chemical and Engineering Thermodynamics, 5th ed
10.3-1
Section 10.3
Using the critical properties in the text, the program VLMU and the following interaction
parameters kC2 C3
0.001 ; kC2 C4
0.010 ; kC3 C4
0.003
I obtain the following
Bubble point, P bar
1
5
10
15
20
25
30
35
40
42
43
10.3-2
T K
.
23131
280.26
.
30817
.
32712
.
34199
354.47
.
36583
37527
.
384.56
388.31
390.31
yC2
0.3444
0.2222
.
01738
.
01460
.
01261
.
01102
0.0932
0.0835
0.0699
0.0632
0.0588
yC3
.
05907
0.6576
0.6693
0.6691
0.6642
0.6565
0.6434
0.6329
0.6133
0.6011
05919
.
yC4
0.0649
.
01202
.
01569
.
01849
0.2097
0.2333
0.2634
0.2836
0.3168
0.3357
0.3493
Using the same program and information as above, I obtain the following
Dew point, P bar
T K
xC2
xC3
xC4
1
5
10
15
20
25
30
32
35
38
40
42
43
25393
.
299.65
32527
.
342.32
35543
.
36615
.
37515
.
378.44
38300
.
387.09
389.57
39176
.
392.64
0.0038
0.0081
0.0117
0.0148
0.0180
0.0213
0.0257
0.0271
0.0296
0.0331
0.0361
0.0402
0.0434
0.2309
0.3128
0.3595
0.3924
0.4198
0.4446
0.4724
0.4806
0.4936
05103
.
05231
.
05389
.
05499
.
0.7653
0.6791
0.6289
05928
.
05623
.
05341
.
05019
.
0.4923
0.4768
0.4566
0.4408
0.4209
0.4068
See figure on following page.
10.3-3
Again we use the program VLMU and the data from Problem 10.3-1. Also, since, at 20 bar, the
bubble point of the mixture is 341.99 K and the dew point is 355.43 K, we only need to consider
temperatures between these two extremes. The results follow:
Solutions to Chemical and Engineering Thermodynamics, 5th ed
P
20 bar
T K
342.5
344.0
346.0
348.0
350.0
352.0
354.0
3550
.
x1
0.0478
0.0418
0.0352
0.0299
0.0258
0.0224
0.0197
0.0185
x2
.
05670
.
05564
.
05382
.
05170
0.4928
0.4668
0.4397
0.4259
x3
0.3852
0.4019
0.4263
0.4531
0.4814
0.5107
05407
.
05557
.
y1
0.0120
.
01069
0.0914
0.0789
0.0690
0.0609
0.0542
0.0512
Section 10.3
y2
0.6648
0.6641
0.6580
0.6464
0.6304
0.6106
05878
.
05756
.
y3
0.2143
0.2290
0.2506
0.2746
0.3007
0.3285
0.3580
0.3732
V L split
0.0303 0.9697
.
.
01265
08735
0.2635 0.7365
.
0.4095 05905
.
05612
0.4388
0.7175 0.2825
08796
.
01204
.
0.9628 0.0372
Note to instructor:
Re: Problems 10.3-1, 2 and 3
You should take time to discuss how the compositions are changing with pressure in each
of the cases above. For example, in Problem 10.3-1, at low bubble point pressures the vapor
composition is very different than the liquid, with the vapor greatly enriched in the light C2
component. However, as the pressure increases, and the critical point is approached, the vapor
composition becomes similar to that of the liquid. Analogous comments apply to the dew point
case of Problem 10.3-2.
a f
Vapor Liquid Equlibrium
40
30
P_bp
P in bar
i
P_dp
i
Bubble point
curve
20
Dew point
curve
10
0
220
240
260
280
300
320
340
360
380
400
T_bp T_dp
i
i
T in K
10.3-4
See solution to Problem 6.2. The derivation of Eqs. 10.3-8 is identical to the derivation of eqns.
6.4-29 & 30.
10.3-5 (a and b) These algorithms are incorporated into the program VLMU Examine that program to see
the algorithms used.
10.3-6 (a) The equations to be used to solve this problem are the mass balances, the equilibrium condition
(equality of fugacities) and the energy balance. Writing these equations for an open, steadyflow system, we have (for 1 mole of feed of compositions zi ).
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Section 10.3
Mass balances zi xi L yiV i 1, 2, ! , N
1 L V (summing equations above over all species)
phase equilibrium condition f i L f iV
Ÿ xi P
fi L
xi ˜ P
yi P
fi V
yi ˜ P
or
L
xi I i T , P, x
V
i
1, 2, ! , N
c
h
yi I i T , P, y
(1)
(2)
or
Ki
energy balance
L
Ii
yi
xi
Ki T , P, x
V
Ii
i
1, 2, ! , N
0
¦ b N H g ¦ b N H g ¦ b N H g
0
¦ b z H g V ¦ b x H g L¦ b y H g
or
V
i
i
in
i
i
i
out
i
i
in
i
i
Also, we have the summation conditions
¦ xi 1 and ¦ yi
i
L
out
V
i
(3)
L
(4)
i
(5)
1
(b) With the exception of Eq. (4), the other equations are the ones used in the isothermal flash
calculation (see Eqs. 10.1-14 & 15, and 10.3-4). Therefore the easiest algorithm to implement is
to use the one for the isothermal flash with an extra, outer loop which iterates on final
temperature. This is done by adding an enthalpy calculation (see eq. 10.3-8a) to the program,
and calculating the enthalpy of the feed stream, and the liquid and vapor streams. If eq. (4) is
not satisfied, the exit temperature is adjusted, and the calculation repeated.
(c) A Mathcad worksheet (10.3-6.MCD) is available for this calculation. This worksheet is also
available as the Adobe PDF file 10.3-6.pdf.
10.3-7
This problem is probably most easily solved using an equation of state, such as the Peng-Robinson
equation. Using the program VLMU, the critical properties in Table 6.6-1 and k12 0.01 (from
Table 9.4-1) we obtain the following
mole % ethane
Calculated
Measurement
in liquid
7.8
vapor mol %
6.62
P bar
39.68
vapor mol %
6.2
P bar
39.73
22.8
19.35
37.08
19.7
37.07
30.3
2581
.
3579
.
255
.
3560
.
59.0
5213
.
30.94
531
.
3213
.
89.0
8528
.
2589
.
854
.
2545
.
Thus we see, using the program VLMU, we obtain very accurate predictions in a simple manner
(though, of course, much work went into preparing the program). With the exception of the 59
mol % ethane liquid, the compositions are predicted to about 0.004 mole fraction and 0.4 bar
accuracy.
10.3-8
Again the program VLMU will be used with the critical properties in Table 6.6-1 and the value
k12 0.055 given in Table 9.4-1. The only question is how to use the program to get K values. I
have used the isothermal flash, since that is the only option that allows me to specify T and P.
One then has to choose a feed composition that assures that one is in the two-phase region (this
can be checked by a collection of bubble point or similar calculations, if desired). This
corresponds to methane compositions in the range
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Section 10.3
0.0711 d xC1 d 0.9916
In this range we find the following
x
y
K
methane 0.0711 0.9916
1395
.
benzene 0.9289 0.0084 0.009083
A
asked for in problem statement
10.3-9
We use the program VLMU, critical properties in Table 6.6-1, and a value of k12 0.003 given in
Table 9.4-1. The problem must be solved by trial-and-error. The easiest way is to start with the
bubble-point P program. This leads to the result that the bubble-point pressure is 8.47 bar.
Therefore, a lower pressure than 8.47 bar must be used to produce appreciable vapor. Likewise,
using the dew point pressure program gives a dew point pressure of 6.12 bar. Therefore, the
pressure must be between 6.12 bar and 8.47 bar to produce appreciable vapor. By trial-and-error
. 05
. 1 and
. K , P 7.234 bar V L 05
we find at T 31315
C3
nC4
Phase compressibility, Z
Feed
Liq
05
.
0.3699
05
.
0.6301
0.0268
Vapor
K
0.6301 1703
.
0.3699 05871
.
08594
.
10.3-10 Using the program VLMU, critical properties in Table 9.4-1 and the following interaction
parameters kC2 C3 0.001 ; kC2 nC4 0.010 ; kC2 iC 4 0.007 ; kC3 nC4 0.003 ;
kC3 iC4
0.007 and k nC4 iC4
0.0 we obtain the following
Component
ethane
propane
n - butane
i - butane
moles
compressibility
Feed
0.3100
0.3400
0.2100
01400
.
10
.
Liquid
.
01638
0.3372
0.3084
01907
.
0.4676
0.0548
Vapor
K
0.4384 2.677
0.3425 1016
.
01236
.
0.4008
0.0955 05008
.
05324
.
0.7952
Therefore, 53.24 mol % of tank contents will be vapor and the remaining 46.76 mol % will be
liquid. [If we believe the compressibilities, even though the liquid compressibility has some error,
then we have that 5.71% of the volume of the tank is filled with liquid, and the remaining 94.29%
is filled with vapor.]
10.3-11 The analysis is similar to that of Problem 10.3-6 except that the energy balance
0
¦ b z H g L¦ b x H g V ¦ b y H g
L
i
i
in
i
V
i
i
i
used there is now replaced with the entropy balance
0
¦ c z S h L¦ e x S j V ¦ e y S j
i
i
in
i
L
i
i
V
i
A Mathcad worksheet (10-3-11.MCD) is available for this calculation. This worksheet can be
viewed as the Adobe PDF file 10-3-11.pdf.
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Section 10.3
10.3-12 A reasonably good fit is obtained with kij = 0.11
Bubble point pressure at 273.13 K
35
30
25
20
15
10
5
0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
mole fraction of carbon dioxide
Vapor
Liquid
Bubble point pressure calculation using the Peng Robinson equation of state
Temperature = 273.13 K
T (K)
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
P (bar)
1.5559
2.0182
2.4785
2.9347
3.3918
3.8465
4.2990
4.7491
5.1970
5.6425
6.0857
6.5263
6.9645
7.4002
7.8332
8.2635
8.6911
9.1159
9.5379
9.9569
10.3729
10.7860
11.1959
11.6026
12.0062
12.4065
12.8034
13.1970
13.5872
13.9739
14.3570
14.7366
15.1126
15.4849
15.8534
16.2183
16.5794
16.9366
17.2900
17.6396
17.9852
18.3269
x1
0.0000
0.0100
0.0200
0.0300
0.0400
0.0500
0.0600
0.0700
0.0800
0.0900
0.1000
0.1100
0.1200
0.1300
0.1400
0.1500
0.1600
0.1700
0.1800
0.1900
0.2000
0.2100
0.2200
0.2300
0.2400
0.2500
0.2600
0.2700
0.2800
0.2900
0.3000
0.3100
0.3200
0.3300
0.3400
0.3500
0.3600
0.3700
0.3800
0.3900
0.4000
0.4100
y1
PV/RT(liq) PV/RT(vap)
0.0000 0.0064
0.9485
0.2237 0.0083
0.9497
0.3650 0.0102
0.9491
0.4623 0.0120
0.9478
0.5334 0.0138
0.9458
0.5876 0.0156
0.9435
0.6303 0.0173
0.9409
0.6648 0.0190
0.9382
0.6933 0.0207
0.9354
0.7172 0.0224
0.9325
0.7375 0.0240
0.9295
0.7551 0.0257
0.9264
0.7703 0.0273
0.9233
0.7837 0.0288
0.9202
0.7956 0.0304
0.9170
0.8062 0.0319
0.9139
0.8157 0.0334
0.9107
0.8242 0.0349
0.9075
0.8320 0.0363
0.9043
0.8391 0.0377
0.9011
0.8456 0.0391
0.8978
0.8515 0.0404
0.8946
0.8570 0.0418
0.8914
0.8621 0.0431
0.8882
0.8668 0.0444
0.8850
0.8712 0.0456
0.8818
0.8753 0.0469
0.8786
0.8791 0.0481
0.8754
0.8827 0.0492
0.8722
0.8861 0.0504
0.8690
0.8893 0.0515
0.8659
0.8923 0.0526
0.8627
0.8951 0.0537
0.8596
0.8978 0.0547
0.8565
0.9003 0.0557
0.8533
0.9028 0.0567
0.8503
0.9051 0.0577
0.8472
0.9073 0.0586
0.8441
0.9094 0.0596
0.8411
0.9114 0.0604
0.8381
0.9133 0.0613
0.8351
0.9151 0.0621
0.8321
Solutions to Chemical and Engineering Thermodynamics, 5th ed
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
273.13
18.6648
18.9986
19.3286
19.6545
19.9766
20.2947
20.6088
20.9191
21.2255
21.5280
21.8267
22.1216
22.4129
22.7004
22.9843
23.2648
23.5417
23.8153
24.0856
24.3528
24.6169
24.8780
25.1364
25.3921
25.6452
25.8960
26.1446
26.3912
26.6360
26.8791
27.1208
27.3612
27.6007
27.8394
28.0751
28.3156
28.5511
28.7918
29.0282
29.2704
29.5088
29.7537
29.9979
30.2444
30.4934
30.7453
31.0006
31.2596
31.5228
31.7907
32.0637
32.3424
32.6272
32.9189
33.2178
33.5246
33.8398
34.1638
34.4965
0.4200
0.4300
0.4400
0.4500
0.4600
0.4700
0.4800
0.4900
0.5000
0.5100
0.5200
0.5300
0.5400
0.5500
0.5600
0.5700
0.5800
0.5900
0.6000
0.6100
0.6200
0.6300
0.6400
0.6500
0.6600
0.6700
0.6800
0.6900
0.7000
0.7100
0.7200
0.7300
0.7400
0.7500
0.7600
0.7700
0.7800
0.7900
0.8000
0.8100
0.8200
0.8300
0.8400
0.8500
0.8600
0.8700
0.8800
0.8900
0.9000
0.9100
0.9200
0.9300
0.9400
0.9500
0.9600
0.9700
0.9800
0.9900
1.0000
0.9169
0.9186
0.9202
0.9218
0.9234
0.9248
0.9263
0.9276
0.9290
0.9303
0.9316
0.9328
0.9340
0.9352
0.9363
0.9375
0.9386
0.9396
0.9407
0.9418
0.9428
0.9438
0.9448
0.9458
0.9468
0.9478
0.9488
0.9498
0.9508
0.9518
0.9528
0.9538
0.9548
0.9558
0.9568
0.9578
0.9589
0.9600
0.9611
0.9622
0.9634
0.9646
0.9659
0.9672
0.9685
0.9699
0.9714
0.9730
0.9746
0.9764
0.9782
0.9802
0.9823
0.9846
0.9872
0.9899
0.9929
0.9963
1.0000
0.0630
0.0637
0.0645
0.0653
0.0660
0.0667
0.0673
0.0680
0.0686
0.0692
0.0698
0.0703
0.0708
0.0714
0.0718
0.0723
0.0727
0.0732
0.0736
0.0740
0.0743
0.0747
0.0750
0.0753
0.0756
0.0758
0.0761
0.0763
0.0765
0.0767
0.0769
0.0771
0.0772
0.0773
0.0774
0.0775
0.0776
0.0776
0.0777
0.0777
0.0777
0.0777
0.0776
0.0776
0.0775
0.0774
0.0772
0.0771
0.0769
0.0766
0.0764
0.0761
0.0758
0.0754
0.0750
0.0746
0.0742
0.0736
0.0731
Section 10.3
0.8291
0.8262
0.8233
0.8204
0.8175
0.8147
0.8119
0.8091
0.8063
0.8035
0.8008
0.7981
0.7955
0.7928
0.7902
0.7876
0.7850
0.7825
0.7800
0.7775
0.7750
0.7725
0.7701
0.7677
0.7653
0.7629
0.7606
0.7583
0.7559
0.7536
0.7514
0.7491
0.7468
0.7446
0.7424
0.7401
0.7379
0.7357
0.7335
0.7312
0.7291
0.7268
0.7246
0.7224
0.7202
0.7180
0.7158
0.7136
0.7114
0.7092
0.7070
0.7049
0.7027
0.7006
0.6986
0.6966
0.6947
0.6930
0.6915
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Section 10.3
Points are experimental data, and lines are correlation using the Peng-Robinson equation of state
with k12 = 0.11
40
25
0.9
35
0.8
30
20
0.7
0.6
15
20
15
k2
k1
P. bar
25
0.5
0.4
10
0.3
10
0.2
5
5
0.1
0
0
0
0.2
0.4
0.6
0.8
1
0
0
0.2
x 1, y 1
0.4
0.6
0.8
1
0
0.2
0.4
x1
0.6
0.8
1
x1
10.3-13 Clearly, because high pressures and a simple carbon dioxide + hydrocarbon system is involved,
and equation of state, such as the Peng-Robinson equation, should be used. Using the program
VLMU with kCO 2 nC6 011
. (Table 9.4-1) results in no solution at 140 bar and 75qC. However,
trying the bubble point and dew point pressure programs we obtain the following results (at
T 34815
. K ).
bubble point
P, bar
12.96
37.93
64.15
89.14
91.36
93.49
95.52
97.43
99.17
xCO 2
0.1
0.3
0.5
0.7
0.72
0.74
0.76
0.78
0.80
yCO 2
0.001
0.05
0.1
0.3
0.4
0.5
0.6
0.7
0.73
0.75
0.78
0.80
0.82
Program doesn’t
converge at higher
CO2 concentrations
dew point
P, bar
1.21
1.28
1.35
1.76
2.07
2.51
3.17
4.31
4.83
5.25
6.03
6.70
7.56
yCO 2
0.9
0.92
0.93
0.94
0.945
0.947
0.949
0.9495
dew point
P, bar
15.06
20.25
24.64
32.02
38.56
42.62
49.39
52.75
Program doesn’t
converge at higher
CO2 concentrations
120
100
80
bubble curve
60
dew curve
40
20
0
0
0.2
0.4
0.6
0.8
1
Since the program doesn’t converge at higher concentrations of CO2 along either the bubble point
or dew point curves, we have to make an estimate of the CO2 concentration based on the data
Solutions to Chemical and Engineering Thermodynamics, 5th ed
Section 10.3
above. There are two possibilities: (1) The CO2 saturation of the liquid at 140 bar is in the
retrograde region at somewhere between xCO 2 of 0.8 and 0.95 [Note, simple equation of state
programs, such as VLMU typically do not converge in the retrograde region, and more sophisticated
algorithms and numerical methods must be used]; (2) at 140 bar only the vapor exists, that is, all the
hexane vaporizes.
Note the enormous solubility of carbon dioxide in hexane and, indeed, in reservoir crude! That is
why carbon dioxide has been used in enhanced oil recovery (crude oil swells so more is recovered,
and viscosity drops so the trapped oil in the earth matrix flows more easily.)
10.3-14 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-14
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10.3-33 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-33
Solutions to Chemical and Engineering Thermodynamics, 5th ed.
Section 11.1
11.1
11.1-1 (a) H V data are obtained from the Superheated Steam Tables. Also
H L T , P 01
. MPa | H L T , any pressure since, at low and moderate pressures, the liquid
enthalpy is independent of pressure to an excellent approximation. Therefore, H L T , P is
taken along the saturation line, and then extrapolated to high temperature, as shown below. 11.1-2
General: The starting point for solving this problem is
f NL2 f NV2 Ÿ xN 2 J N 2 f NL2 yN 2 P
PN 2
where PN 2 is the partial pressure of nitrogen in the gas phase. Also, f NL2
"liquid" nitrogen, is 1000 bar according to the problem statement.
(a) Ideal solution
J N 2 1 ; also PN 2 1 bar (problem statement)
thus
xN 2
(b) Non-ideal solution
PN 2
f NL2
1
1000
0.001
fugacity of pure
Solutions to Chemical and Engineering Thermodynamics, 5th ed.
c
Here ln J N 2
0526
.
1 xN 2
Section 11.1
h . Since, from the ideal solution case above, x
2
N2
is quite small
(and will be even smaller here), it is reasonable to assume that 1 xN 2 | 1 , ln J N 2
J N2
, and
0526
.
. Thus
16922
.
PN 2
xN 2
J N 2 f NL2
1
.
u 1000
16922
. u 103
0591
The Henry's Law Constant is the constant in the expression
Hi xi
thus
1 bar
xN 2
HN 2
f i V , here, HN 2 xN 2
1 bar
1692 bar.
Also, since HN 2 { J N 2 f NL2 and V ex
0 , we have that the activity coefficient J N 2 is
independent of pressure. Thus, the only pressure variation of the Henry's law constant will be
through the Poynting pressure correction to the "liquid" phase of pure nitrogen.
11.1-3 (a) To compute the ideal solubility, we will use Shair's correlation, Fig. 11.1-1 first we need to
compute the reduced temperature
29815
.
Tr
1564
.
190.6
thus,
fL
# 365
. and f L
PC Fig. 8.3-1
365
. u 46.0 bar
167.9 bar
Therefore at 1 atm partial pressure of methane
PCH 4
ideal
xCH
4
(b) To
compute
J CH 4
L
f CH
4
the
L
1013
.
xCH 4 f CH
1013
.
bar
167.9 bar
activity
4
CH 4 - benzene: J CH 4
CH 4 - C6 H12: J CH 4
6.03 u 103 (all solvents)
coefficients,
we
use
L
xCH 4 J CH 4 f CH
4
1013
.
bar
or
6.03 u 103 xCH 4 The following results are obtained
2.91 CH 4 - CCl4: J CH 4
213
.
CH 4 - C6 H14: J CH 4
211
.
191
.
McDaniel
142
.
Guerry
(c) The regular solution model gives, for the CH 4 - C6 H14 mixture
J CH4
R|V cG G h I
expS
RT
|T
L
CH 4
2
CH 4
C6H14
U| expR52 u 568
. 7.3 U
. ,
V| ST 1987
V 126
u 298.2 W
.
W
2
2
C6H14
Since I C6 H14 | 1 . Note: CH 4 parameters from Table 11.1-1 C6 H14 parameters from Table 9.61.
This result is not in agreement with either set of data, but is distinctly closer to Guerry's
result!
Solutions to Chemical and Engineering Thermodynamics, 5th ed.
f i G Ÿ xiJ i f i L
11.1-4 (a) Start from f i L
yi P
FfI
H PK
Section 11.1
i
Determination of nitrogen properties:
Tr N
From
Fig.
37315
. K
126.2 K
2
2.96 ; Pr N
F f I #1.
H PK
7.4-1
2
75 bar
33.94 bar
2.21
using
Shair's
Now
correlation
(Fig.
11.1-1)
N2
f NL2 P
1013
.
bar
61
. ; however, at 75 bar we have
PC
f NL2 P
f NL2 P
75 bar
PC
1013
.
PC
exp
R|V
S|
T
L
N2
.
P 1013
RT
6.6 Ÿ f NL2 P
61
. exp 0.0784
U|
V|
W
75 bar
224 bar
Determination of benzene properties:
Tr C H
6
6
From Fig. 9.8-1
. K
37315
. K
5621
0.664 and Pr C H
6
6
75 bar
48.96 bar
1532
.
f
# 0.23 and, for the liquid, we need the vapor pressure. From the
P vapor
bar at T 801
Chemical Engineers Handbook we have P vap 1013
.
. q C , and P vap 2.026 bar
A
at T 1038
.
. q C using ln P vap
bar.
B as the interpolating formula, we find P vap 1823
T
f
From Fig. 7.4-1, we have (with a little extrapolation) that
| 0.96 . [Along the saturation
P sat
line]. Thus
F I
H K
f CL6 H 6
PCvap
6H 6
.
UV 2159
R 89 u 73173
F f I expR|SV c P P h U|V 1823
.
bar
H P K |T RT |W . u 0.96 u expST8314
. u 37315
. W
L
C6 H 6
vap
sat
Thus, the equations to be solved are
xN 2 J N 2 224 bar
yN 2 75 bar
xC6 H 6 J C6 H 6 2.159 bar
xN 2 xC6 H 6
1
yN 2 yC6 H 6
1
together with
J N2
and
yC6 H 6 75 bar
R|V cG G h I
expS
RT
|T
L
N2
N2
C6 H 6
2 2
C6 H 6
U|
V|
W
Solutions to Chemical and Engineering Thermodynamics, 5th ed.
Section 11.1
R|V cG G h I U|
expS
V|
RT
|T
W
L
C6 H 6
J C6 H 6
N2
C6 H 6
2 2
N2
where the N 2 parameters are gotten from Table 11.1-1 and the benzene parameters from Table
9.6-1. Because of the nonlinear nature of the equations (due to the composition dependence of
the activity coefficient), this problem is best solved by trial and error. I chose xN 2 0.0 as the
initial guess.
Solution obtained was:
xN 2
0.047 and yN 2
0883
.
xN 2
0.045 and yN 2
0.944
Measured values:
(b) At 100 bar the calculation is similar, however the numbers are a little different. I find
f NL2
f V
| 0175
.
;
| 6.76 and f CL6 H 6 | 2.290
P C6 H 6
PC
Solution obtained was:
xN 2
0.061 and
yN 2
0874
.
Measurement yields
xN2
0.0595 y N 2
0.968
In both cases the liquid compositions are in better agreement with experiment than the vapor
compositions!
Note: I have found that some students try to make a large extrapolation of the vapor pressure,
rather than using Shair's correlation ... it is a large extrapolation here, since the nitrogen critical
temperature is 126.2 K . If we extrapolate the low temperature vapor pressure, and make the
(small) fugacity coefficient correction, we obtain f NL2 1195 bar compared to 224 bar here at
75 bar. This leads to xN 2
0.0082 , compared to 0.047 calculated here and 0.045 observed
experimentally.
Moral: Use Shair's correlation instead of making large extrapolations of the vapor pressure.
11.1-5 (a) Suppose a small amount of liquid, 'N is, vaporized, then there are y'N moles of dissolved
gas in the vapor, and x 'x N 'N
moles of dissolved gas left in the liquid we are
interested in a different distillation, i.e., the case where 'x and 'N will be very small. Thus
x 'x N 'N
xN x'N N'x 'x 'N
| xN x'N N'x
Now writing a mass (mole) balance on the dissolved gas yields
xN
Thus
'x
'N
y'N x 'x N 'N | y'N xN x'N N'x
yx
dx
and taking the limit as 'N o 0 yields
N
dN
yx
N
Solutions to Chemical and Engineering Thermodynamics, 5th ed.
(b) Using y
ax , N f
0
0
z
Hx
P x
dx
Hx
yields
dN
P
and x, N yields
x
x0
d ln x
N
F
H
x HP
N
P
Section 11.1
I or d ln x
K d ln N
z
HP N
d ln N or ln x ln x0
N0
P
HP
. Now integrating between
P
HP
ln N ln N 0
P
which can be rewritten as
x
x0
FG N IJ
HN K
( H P) P
or
0
N
N0
FG x IJ
Hx K
P HP
0
x
N
0.01 1 266 0.983 . Thus, only 1.7% of initial number of moles of
0.01 we have
x0
N0
liquid need be vaporized in a differential distillation (i.e., no violent boiling, otherwise
equilibrium will not be obtained) to remove 99% of the CO2 .
x
N
For
0.0001
0.0001 1 266 0.966
x0
N0
(c) For
Thus only 3.4% of initial number of moles of liquid need be vaporized to remove 99.99% of the
CO2 .
11.1-6
I used the bubble point option of program VLMU treating the liquid mole fraction of CO2 as an
adjustable parameter, until a CO2 partial pressure of 1.013 bar in the vapor phase was obtained.
The results appear below:
xCO 2
0.006
0.020
0.022
0.0221
0.0222
Ptot
0.31
0.96
105
.
105
.
106
.
yCO 2
PCO 2 yCO 2 ˜ Ptot
08680
.
0.269 bar
0.9562
0.918
0.960
1008
.
0.9601
10081
.
0.9603
1018
.
Therefore, the predicted solubility is xCO 2 | 0.02215 .
11.1-7 (a) This problem is treated in the same manner as the previous problem. The results are given
below:
k12 0
xCO 2
Ptot
yCO 2
0.015 141
.
0.6729
0.016 148
.
0.6869
0.0159 147
.
0.6856
PCO 2
yCO 2 ˜ Ptot
0.9488
101662
.
10078
.
Therefore, the predicted solubility is | 0.01595
experimental value of xCO 2
0.00328
xCO 2 . This is considerably higher than the
Solutions to Chemical and Engineering Thermodynamics, 5th ed.
Section 11.1
(b) k12 0.2
xCO 2
Ptot
yCO 2
PCO 2 yCO 2 ˜ Ptot
0.015
4.94 0.9011
4.4514
0.003
136
0.6566
089298
.
.
0.0033 145
0.6775
0.9824
.
0.0034 148
0.6839
10121
.
.
In this case xCO 2 | 0.0034 which is quite close to the experimental value of 0.00328. This
illustrates the importance of the binary interaction parameter k12 .
11.1-8 (a) N
m
. K
u 27315
N G ˜ 8.314 u 105 bar
mol K
3
PV
RT
V
.
1013
bar
D u 1013
.
1 m3U W g m3
N
W
8.314 u 105 u 27315
.
18.015 g mol
NG
LM1 N OP
N NQ
1
1 NNWG
NG
NG NW
x
D
1
W
G
. KO
U O
L U u 8.314 u 10 u 27315
LM1 1244
.
u 10
x M1 P
.
D u 1013
D PQ
N 18.015
Q N
1
5
3
W
1
W
volume gas
; for simplicity, assume 1 m3 liquid
volume liquid
(b) L
N G ˜ 8.314 u 105 T
; NG
1013
.
V
1 m3
L
.
V ˜ 1013
8.314 u 105 T
LM1 N OP LM1 U u 8.314 u 10 T OP
.
Q
N N Q N 18.015 L u 1013
LM1 4.555 u 10 T U OP
L Q
N
1
5
W
x
1013
. L
8.314 u 105 T
1
W
G
6
(c) S
V
11.1-9
. Ku
N G ˜ 8.314 u 105 bar m3 mol K u 27315
106 cm3 m3
1013
.
bar
1013
.
1
; NW
8.314 u 27315
.
18.015
1.013
S u 8314
NG
. u 27315
.
NG
Su
x
NG NW
(d) xKH
1013
Ÿx
.
(e) N G
S0
; NW
MS
NG
NG NW
x
1
W
S u1.013
181.05
. u 27315
.
8314
S u 1013
.
S u 1013
1260599
.
.
1013
.
KH
100
18.015
S0
MS
18.015S0
18.015S0 100 MS
S0
18100
.015
MS
Condition for equilibrium as the solubility limit is f i L
'G eq
0
RT ln
fi L
fi V
S
S 1244
RT ln
a
f RT ln x RT ln
xisol f i L T , P, xi 1
f i V T , P, yi 1
a
f
f i V which implies
sol
i
a
f
f i L T , P, xi 1
f i T , P 1013
.
bar
V
Solutions to Chemical and Engineering Thermodynamics, 5th ed.
Section 11.1
since the fugacity of the species in the vapor phase as this low pressure is given by the Lewis-Randall rule,
and just equal to the fugacity of the pure species at 1.013 bar. Now the last term on the right of this
equation is equal to the Gibbs free energy change of transferring the gas from the ideal gas state to a
solution of unit mole fraction. Therefore
f L T , P, xi 1
RT ln xisol 'G
'G eq 0 RT ln xisol RT ln V i
.
bar
f i T , P 1013
a
f
where 'G is the free energy changed asked for in the problem statement . Therefore 'G
A
ln xisol
w GT
wT
B
C ln T DT ET 2 , then 'G
T
G
H
2 wT
'
H
T
Ÿ
wT
T2
w
B
'H
R A C ln T DT ET 2
wT
T
'H 'G
H G
'S
G H TS Ÿ S
T
T
RS LM
T N
'S
CP
LM
N
RT A OP
Q
RT ln xisol , but
wG
B
C ln T DT ET 2 , also T1
wT
T
OPUVcT h R B CT DT 2 ET
QW
2
2
H or
3
R
B CT DT 2 2 ET 3 AT B CT ln T DT 2 ET 3
T
R
AT CT ln T CT 2 DT 2 3ET 3
T
R A C ln T C 2 DT 3ET 2
F wH I Ÿ 'C F w' H I R C 2 DT 6ET
H wT K
H wT K
2
P
P
11.1-10 a) We expect the solubility of bromine in water to be quite low, so that the molar concentration of
water will be essentially unchanged. Then the change in free energy for the bromine dissolution
process is
aq
'G
aq
f H O( M)
f Br2
M
554
.
RT ln
RT ln aq 2
f Br2 M 554
M 554
.
.
f
( M 0)
H 2O
M
147 M
5555
5555
.
. M
RT ln
RT ln
M 5555
M 5555
28.4
5555
.
.
.
where 55.55 is the molar concentration of water. Assuming M will be very small, the second term
can be neglected and we obtain
'G
M
147 M
RT ln
M 5555
.
28.4
The results are plotted below.
Solutions to Chemical and Engineering Thermodynamics, 5th ed.
Section 11.1
1
0.307734
0
1
G
i
2
3
3.167164
4
0
0.01
b)
0.05
0.1
M
i
0.15
0.2
0.2
At saturation, the fugacity of liquid bromine equals that of bromine in solution. Therefore
aq
f Br2
f BrL2 ; or 147 M
28.4; so M
01932
.
Either by using this value in the equation above, or using the graph above, we see that the
Free energy change is zero if the liquid is saturated with bromine.
Z 0.225 TB 194.7 K
11.1-11 CO2
TC 304.2 K PC 7376
. bar
Toluene
591.7
41.13
0.257
383.8
From Prausnitz-Shair Correlation
fL
Tr 0.98 ;
# 0.6 . Assume very little toluene in vapor yCO 2 ~ 1
PC
xCO 2
a
L
T, P
J CO 2 f CO
2
At 10 bar Tr
0.98 , Pr
a f
1013
.
bar f expn V
10 1013
.
bar RTs
yCO 2 P f P
CO 2
10
7376
.
0136
.
f
Figure 7.4-1 | 0.96
P
L
f CO
0.6 u 73.76 44.26 bar
2
Now calculate the Poynting correction
55 cc mol u 8.987 bar
1 m3
u
exp
298.15 K u 8.314 u 105 bar m3 mol cc 106 cc
LM
N
LM 55. u 8.987 u 10 OP exp 0.02 102
.
N 2.9816 u 10 u 8.314 u 10 Q
1
exp
2
1
So ideal “Prausnitz-Shair” solubility
10 u 0.96
xCO2
0.213
44.26 u 102
.
OP
Q
Solutions to Chemical and Engineering Thermodynamics, 5th ed.
Section 11.1
Now consider solution nonideality
Table 9.6-1
Table 11.1-1
G CO 2 6.0
G T 8.9
VT 107
VCO 2 55
ln J CO2
IT
a55 cc molfI 6.0 8.9 cal cc
2
2
T
0.781I 2T
8.314 J mol K u 29815
. K u 0.239 cal J
VT xT
VT xT VCO2 xCO2
c
107 1 xCO2
c
107 1 xCO2
c
h
h
107 1 xCO2 55xCO2
h
c
107 1 xCO2
107 107 55 xCO2
h
107 52 xCO2
So to find the nonideal solubility must solve
0.213
0.213
xCO2
exp 0.781I 2T
exp 0.781 107 1 xCO 107 52 xCO
c
{
which has the solution xCO2
2
hc
h}
2
2
(using MATHCAD)
0102
.
From Program VLMU with kCO 2 T
0.0
(using flash with equimolar feed)
Pressure
10.05 bar
x
y
0.2062
0.9952
0.7938
0.0048
yCO 2 u P 0.9952 u 10.05 ~ 10 bar
CO2
T
Note PCO 2
kCO2 T
010
.
CO2
T
PCO 2 ~ 10 bar
x
0.1174
0.8826
y
0.9949
0.0051
Which is reasonably close to the Prausnitz-Shair correlation result, especially given the difference in
the methods.
11.1-12 (a)
From the Steam Tables T 25q C , P 3169
.
kPa
(b) 1 atm = 1.-10 kPa
3169
.
yW
0.0313 bar ; remainder is oxygen and nitrogen.
1013
.
Initial partial pressure of N 2 0.79 u 1013
. 32
.
77.5 kPa
Initial partial pressure of O2 0.21 u 1013
. 32
.
20.6 kPa
Mole fraction N 2 in water xN 2 HN 2 PN 2
xN 2
77.5 kPa 100 kPa bar
8.48 u 104 bar mole fraction
0.775
8.48 u 104
0.0914 u 104 914
. u 106
Mole fraction O2 in water
20.6 100 bar
xO 2
4.35 u 104 bar mole fraction
20.6
100 u 4.35 u 104
4.74 u 106
11.1-13 a) Since the vapor pressures are low, we will assume the fugacity coefficients at saturation are
unity. Then the fugacity of methyl acetate (MA) and methanol (M) is
Solutions to Chemical and Engineering Thermodynamics, 5th ed.
2 · vap
2
§ A
x MA exp ¨
1 x MA ¸ PMA
x MA exp 1.06 1 x MA 1.1260 bar
© RT
¹
2·
2
§ A
x M exp ¨
1 x M ¸ PMvap x M exp 1.06 1 x M 0.8465 bar
© RT
¹
vap
x MA J MA x MA PMA
fMA
fM
Section 11.1
x M J M x M PMvap
1.5
1
fugma
fugm
0.5
0
0
0.2
0.4
0.6
0.8
1
xx
methyl acetate mole fraction
b) The partial pressure of species i is gotten from
P
P
fi x i J i x i Pivap yi P Pi so that i J i x i Pivap and lim i
x o0 x i
xi
lim J i x i Pivap
x o0
§ A ·
Pivap exp ¨
¸
© RT ¹
§ A ·
Pivap exp ¨
¸ , so that the hypothetical standard state of a component in a mixture
© RT ¹
§ A ·
that obeys the one0constant Margules equation is Pivap exp ¨
¸ , while the actual pure component
© RT ¹
Therefore, Hi
fugacity is simply Pivap .
xair
11.1-14
xair
Pair
H
Henry’s
law
constant
Ÿ xH 2 O
mole fraction of air in water (liquid)
01333
.
kPa u 103 Pa kPa
# 0.3 u 107
4.3 u 104 bar u 105 Pa bar
1 (Do not have to consider air trapped in water)
At equilibrium
Ÿ xH 2 O PHvap
2O
f HL2 O
f HS2 O
f HV2 O ;
PHsub
2O
yH 2 O P
since xH 2 O
PHvap
2O
1
[Note: Because of the low pressures involved, we have neglected f P terms and Poynting
corrections.]
For comparison, in an air-free measurement, we have
PHvap
PHsub
PTP
2O
2O
where PTP is the true triple point temperature.
Solutions to Chemical and Engineering Thermodynamics, 5th ed.
Since PHvap
2O
Section 11.1
has to be satisfied in both cases, we would obtain the same triple point
PHsub
2O
temperature in both the air-free experiment, and the measurement with air.
In the air-free experiment we measure P PH 2 O and get the triple point pressure. In the experiment
with air we measure P
PH 2 O Pair and, mistakenly, assume this is the triple point pressure;
actually PH 2 O is the triple point pressure. The error, 'P , is equal to the partial pressure of air; here
0.1333 kPa. Thus, we have
% error
'P
u 100
PHTP2 O
01333
.
u 100
0.6113
218%
.
[Note: From the Steam Tables, triple point pressure is 0.6113 kPa.]
11.1-15 The fugacity of a species 1 at infinite dilution in species 2 is, from eq. 9.4-10
b Pº
ª L
Z 1 2 2 »
ª
f1L
b1 L
a2
b1 º « 2
§ L b2P ·
RT
lim x1o0 ln
Z2 1 ln ¨ Z2 »
« 2 » ln «
¸
RT ¹ 2 2b 2 RT ¬
b 2 ¼ « ZL 1 2 b 2 P »
x1P1vap b 2
©
2
¬
RT ¼
so that
fL
Lim x1o0 1
H1
x1
we have that
ª
«
b
P1vap exp « 1
« b2
H1
b P ºº
ª L
Z 1 2 2 »»
ª
a2
b1 º « 2
RT
»»
« 2 » ln «
RT ¹ 2 2b 2 RT ¬
b 2 ¼ « ZL 1 2 b 2 P » »
2
¬
RT ¼ »¼
b P·
§
Z2L 1 ln ¨ Z2L 2 ¸ ©
¬«
11.1-16 We start with eqns. (9.9-11, 12 and 13) and have the following at infinite dilution of species 1
§ wND ·
a1
ln J f
*1
¨
¸
C
© wN1 ¹T,N ,x o0 b1RT
2
1
1 § wN 2Q ·
¨
¸
N ©¨ wN1 ¹¸T,N ,x o0
2 1
D x o0
1
a ·
§
2 ¨ b2 2 ¸
RT
©
¹
a2
b 2 RT
§ wNb ·
¨
¸
© N1 ¹T,N 2 ,x1o0
§
b 2 RT
a ·
b 22
a1
ln J f ·
§
2 ¨ b 2 2 ¸ RT
*1 ¸¸
¨¨1 b 2 RT a 2 ©
RT ¹
b 2 a 2 RT © b1RT
C ¹
2b 2 §
b 22
a RT ln J1f ·
¨¨ RT 1 ¸
b 2 RT a 2 ©
b1
C* ¸¹
Solutions to Chemical and Engineering Thermodynamics, 5th ed.
1 § wN 2a ·
¨
¸
N ¨© wN1 ¸¹T,N ,x o0
2
1
Section 11.1
§
ª a RT ln J1f º
a 2 ª b 2 RT
b 22
a
ln J f · º
2 b 2 RT a 2 RT
«
¨¨1 1 *1 ¸¸ » b 2 « 1 »
b 2 «¬ b 2 RT a 2
b 2 RT a 2 © b1RT
C ¹ »¼
C* ¼
¬ b1
ª a1 RT ln J1f º
§
a2 ª
b 22
a
ln J f · º
« 2b 2 RT RT
¨¨1 1 *1 ¸¸ » b 2 « »
b 2 «¬
b 2 RT a 2 © b1RT
C* ¼
C ¹ »¼
¬ b1
These expressions are substituted into eq. (9.9-11) to obtain
ª 1 § wNb ·
º
b P·
§
ZL2 1 ln ¨ ZL2 2 ¸
« ¨
»
¸
RT ¹
©
« b1 © wN1 ¹T,N 2 ,x1o0
»
«
»
f1
b2P ·»
H1 Lim x1o0
P«
§ L
­ 1 1 § wN 2a ·
½ ¨ Z2 1 2
x1
a2
1 § wNb ·
°
°
«
RT ¸ »
ln ¨
¨
¸
¸»
®
¾
¨
¸
« 2 2b RT a N ¨ wN ¸
b
w
b
N
2P ¸
1 ¹T,N ,x o0
2©
1 ¹T,N 2 ,x1 o0 ° ¨ ZL 1 2
2
°¯ 2 ©
«
»
¿
2 1
2
©
RT ¹ ¼
¬
Solutions to Chemical and Engineering Thermodynamics, 5th ed.
Section 11.1
11.1-17
I will use the Prausnitz-Shair correlation
From Table 9.6-1
From Table 11.3-1
From Table 6.6-1
Tr ( 25 273.15)
Tr
TcO2
From Fig. 11.1-1
) ( x) GBr 11.5
VBr 51
GO2 4.0
VO2 33.0
TcO2 154.6
PcO2 50.46
1.929
fO2 5.6˜ PcO2
f/Pc=5.6 so that
x˜ VO2
)B( x) 1 ) ( x)
x˜ VO2 ( 1 x) ˜ VBr
ª ª¬ VO2˜ ) ( x) 2˜ 4.184˜ GO2 GBr 2º¼ º
»
8.314˜ 298.15
¬
¼
J ( x) exp «
Initial guess
given
x 0.002
x˜ J ( x) ˜ fO2
1
x find ( x)
x
3
3.539 u 10
fO2
282.576
Section 11.2
Solutions to Chemical and Engineering Thermodynamics, 5th ed
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Solutions to Chemical and Engineering Thermodynamics, 5th ed
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Solutions to Chemical and Engineering Thermodynamics, 5th ed
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Section 11.2
Solutions to Chemical and Engineering Thermodynamics, 5th ed
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Section 11.2
Solutions to Chemical and Engineering Thermodynamics, 5th ed
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Section 11.2
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Section 11.2
Solutions to Chemical and Engineering Thermodynamics, 5th ed
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Section 11.2
Solutions to Chemical and Engineering Thermodynamics, 5th ed
U / T U ˜ U T ˜ U T [ I [ / [˜ T
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Section 11.2
Solutions to Chemical and Engineering Thermodynamics, 5th ed
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Section 11.2
Solutions to Chemical and Engineering Thermodynamics, 5th ed
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