SP-18. Methane gas is stored at a pressure of 150 bar and a temperature of -10ºC in a tank with
a volume of 10 m3. Calculate the mass of methane in the tank using
(a)
(b)
(c)
(d)
the ideal gas equation of state
the generalized compressibility chart
the Van der Waals equation of state (iteration required, use ideal gas EOS solution for first
guess)
EES
Solution:
(a) Ideal gas EOS
For methane: M CH 4  16.04
kg
kmol
kJ
kJ
kmol K

 0.5183
kg
kg K
16.04
kmol
8.314
RCH 4
kJ 

150 100 3  10 m3 

p
m 

mIG 

 1100 kg
RT 
kJ 
 0.5183
  10  273.15 K 
kg K 

(b) Generalized compressibility chart
For methane: TC ,CH 4  191 K pC ,CH 4  46.4 bar
For the tank conditions: TR 
T 263.15
p
150

 1.38, pR 

 3.23
TC
191
pC 46.4
From the gen comp chart: Z 
pv
p

 0.71
RT mRT
mGCC
kJ 

150 100 3  10 m3 

p
m 



 1550 kg
ZRT

kJ 
0.71 0.5183
  10  273.15 K 
kg K 

(c) Van der Waals EOS
RT
a
p
 2
v b v
for methane:
bar m6
kPa m 6
kJ m3

229.3

229.3
kmol 2
kmol 2
kmol 2
m3
b  0.0428
 Table A-24 
kmol
a  2.293

kJ 
kJ m3
 8.314
  263.15 K  229.3
kmol K 
kJ
kmol 2
p  150 bar  15, 000 3  

m3
m
v2
v  0.0428
kmol
?
2188
229.3
15, 000 

v  0.0428
v2
From the ideal gas EOS:
kg 
10 m  16.04 kmol


3
 M
v 

n
m
1100 kg
 0.1458
m3
kmol
Set up an iteration table:
v
RHS
0.1458 10456
0.1000 15321
0.1100 13610
0.1020 14920
0.1015 15017
0.1016 14997
From the Van der Waals EOS:
v
 M
m3

 0.1016
n
m
kmol
kg 
10 m  16.04 kmol


3
mVDW
M


v
m3
0.1016
kmol
(d) from EES
p_tank = 15000 [kPa]
V_tank=10.00 [m^3]
T_tank =273.15-10
Ru = 8.314 [kJ/kmol-K]
MWCH4= 16.04 [kg/kmol]
RCH4=Ru/MWCH4
vCH4=volume(methane,T=T_tank, p=p_tank)
massCH4=V_tank/vCH4
From EES:
Result:
mEES  1572 kg
 1579 kg
SP-19. (a) One (1.0) kg of ammonia (NH3) is contained in a piston cylinder device. The
ammonia is present initially as a saturated liquid at a pressure of 2.0 bars and a temperature
of -18.86 ºC. Heat is added to the piston cylinder device and the ammonia undergoes a phase
change from a saturated liquid to a saturated vapor. The specific volume of the saturated
liquid is 1.507 103 m3 / kg . Assume that the saturated vapor is an ideal gas. The
temperature dependence of the ammonia saturation pressure is given by
 dp 
2
Tsat   K ,  psat   kPa

   Tsat   Tsat
 dT  sat
kPa
kPa
where   67.745
,   0.1502 2
K
K
Using the Clapeyron relation, calculate the enthalpy change h2 ? h1 of the ammonia for this
constant pressure, constant temperature vaporization process. Compare your answer with the
enthalpy of vaporization listed in the ammonia saturation table.
Solution
Given: Polynomial fit for psat as a function of Tsat . Vaporizaiton process as shown:
T1 = -18.86°C, p1=2 bar
T2 = -18.86°C, p2=2 bar
x1=0.0
x2=1.0
v1 = v1f = 1.507x10-3 m3/kg
mNH3 = 1 kg
mNH3 = 1 kg
Find:
hvap
Assumptions:
(1) Clapeyron equation applies
(2) Vapor phase is ideal gas
(3) Closed system
Basic equations:


hg  h f
 dp 



 dT 
Tsat  v g  v f 

 sat
psat     Tsat   Tsat2
where   67.745
psat v g  RNH3 Tsat
Tsat   K ,  psat   kPa
kPa
kPa
,   0.1502 2
K
K
Solution
(a)


 dp 
   2  Tsat


 dT 

 sat
Tsat  18.86  273.15  254.29 K
dp 
kPa  
kPa 

  67.745
 2  0.1502 2   254.29 K


dT  sat 
K  
K 


kPa
kJ
 dp 
 8.644
 8.644 3


 dT 
K
m K

 sat






psat v g  RNH3 Tsat
kJ
R
kJ
kmol K
RNH3 

 0.4882
kg
RNH3
kg K
17.03
kmol

kJ 
0.4882

  254.29 K 
RNH3 Tsat 
kg K 
m3
vg 

 0.6207
kJ
psat
kg
200 3
m



hg  h f
dp 
 dp 

 hg  h f  hvap  
Tsat  v g  v f



 dT 
 dT 
Tsat  v g  v f 

 sat

 sat
8.314
hvap

3


kJ 
3 m 
  8.644 3   254.29 K   0.6207  1.507 10

m K
kg 


hvap  h fg  1361
kJ
kg
compared to 1325.51 from Table A-14
(b) Check to make sure that ammonia tables at p = 5.0 bar, T = 80ºC are consistent with Eq.
11.35 from the textbook.
From Table A-15, for T  80C
pressure (bar)
3.5
4.0
4.5
5.0
5.5
6.0
7.0
entropy (kJ/kg-K)
6.0586
5.9897
5.9285
5.8733
5.8230
5.7768
5.6939
From Table A-15, for p  5 bar
temperature (ºC)
30
40
60
80
100
120
140
specifc volume
(m3/kg)
0.28103
0.29227
0.31410
0.33535
0.35621
0.37681
0.39722
Specific Entropy (kJ/kg-K)
Plotting the data and performing polynomial curve fits we obtain:
6.1
T=80oC
6
5.9
5.8
5.7
s = 6.8165 - 0.003039p
+ 2.94x10-6p2 - 1.271x10-9p3
5.6
350 400 450 500 550 600 650 700
Pressure (kPa)
s  6.8165  0.003039 p  2.94  10-6 p 2  1.271 10-9 p 3
 s 
 0.003039  5.88 106 p  3.813 109 p 2
 
 p T 80C
at p  500 kPa
 s 
2
 0.003039  5.88 106  500   3.813 109  500 
 
 p T 80C
 s 
kJ 1
kJ m3
3
3
 1.052 10
 1.052 10
 
kg K kPa
kg K kJ
 p T 80C
 s 
m3
3


1.052

10
 
kg K
 p T 80C
3
Specific Volume (m /kg)
0.4
v = -0.099354+ 0.0014027T
0.38
-7 2
- 4.8654x10 T
0.36
0.34
0.32
0.3
0.28
300
p = 500 kPa
320
340
360
380
400
420
Temperature (K)
v  0.099354  0.0014027 T  4.8654  107 T 2
 v 
 0.0014027  9.7308  107 T


 T  p 5bar
at T  80C  353.15 K
 v 
 0.0014027  9.7308  107  353.15 


 T  p 5bar
m3
 v 
3

1.059

10


kg K
 T  p 5bar
For ammonia vapor at T  80C and p  5 bar ,
 s 
m3
m3
 v 
3
3

1.059

10



1.052

10
 


kg K
kg K
 T  p 
 p T
Eqn. 11.35 holds to within 0.7%.
SP-20. Fifty (50) kg of high pressure butane (C4H10, M = 58 kg/kmol) is compressed
isothermally in a piston-cylinder container. The initial volume of the container is 0.8 m3, and
the final volume is 0.3 m3. Use the van der Waals equation of state to analyze the isothermal
compression process.
p
RT
a
 2
v b v
for butane:
(a)
(b)
(c)
(d)
a  13.86
bar m6
kmol 2
b  0.1162
m3
kmol
 Table A-24 
Find the pressures of the C4H10 in the tank at the initial and final conditions.
Find the change in internal energy for the butane as a result of the isothermal compression.
Find the change in enthalpy for the butane.
Find the change in entropy for the butane.
Solution
Given: As shown
T1 = 450 K
V1 = 0.8 m3
mC4H10 = 50 kg
T2 = 450 K
V2 = 0.3 m3
Find:
(a)
(b)
(c)
(d)
p1 , p2
U 2  U1
H 2  H1
S2  S1
Assumptions:
(1) Van der Waals gas
(2) Quasistatic process
(3) Closed system
Basic equations:
p
RT
a
 2
v b v

  p 

 v  
dh  c p dT   v  T 
du  cv dT  T 
  dp,
  p  dv
 T  p 

  T  v

cp
c
 v 
 p 
ds 
dT  
ds  v dT  
 dp,
 dv
T
T
 T  p
 T  v
Solution
(a)
m
50 kg

 0.862 kmol
M 58 kg
kmol
3
0.8 m
m3
v1 
 0.928
0.862 kmol
kmol
n
kJ

100 3

bar m6  
m
a  13.86
2 
kmol
1
bar



0.3 m3
m3
v2 
 0.348
0.862 kmol
kmol


kJ m3

1386

kmol 2



kJ 
kJ m3
 8.314
  450 K 
1386
kmol K 
RT1
a
kmol 2
p1 
 2 

2
3
v1  b v1

m  
m3 
0.928

0.1162
0.928



kmol  
kmol 

p1  4609  1609  3000
kJ
 3000 kPa
m3

kJ 
kJ m3
 8.314
  450 K 
1386
kmol K 
RT2
a
kmol 2
p2 
 2 

2
3
3
v2  b v2

m  

m
0.348
 0.348  0.1162


kmol  
kmol 

p2  16,140  11, 440  4700
(b)
in specific molar units,

du  cv dT  T

 p 

 
 T  v

p  dv

kJ
 4700 kPa
m3
p
RT
a
 2,
v b v
R
 p 

 
 T  v v  b
RT  RT
a 
a
 p 
T 

 2    2   av 2
  p
v b  v b v 
v
 T  v

2
1
2
 p 
du  u2  u1   T 
 
1
  T  v
2
v2

p  dv   av 2 dv   av 1 
v1
1


 1 1

kJ m3  
1
1
u2  u1  a      1386

3
2 
m
m3
kmol  

 v2 v1 
0.928
 0.348
kmol
kmol

kJ
u2  u1  2490
kmol






kJ 

U 2  U1  n  u2  u1    0.862 kmol   2490
  2150 kJ
kmol 

(c)
h  u  pv 
h2  h2   u2  u1   p2 v2  p1 v1
kJ
kJ  
m3  
kJ  
m3 

  4700 3   0.348

3000
0.928
 


kmol 
m 
kmol  
m3  
kmol 
kJ
 2490  1636  2784  3640
kmol
h2  h2  2490
kJ 

H 2  H1  n  h2  h1    0.862 kmol   3640
  3140 kJ
kmol 

(d)
ds 
 v 
dT  
 dp
T
 T  p
cp
or
ds 
cv
 p 
dT  
 dv
T
 T  v
Choose second relation because we have already evalvuated the partial derivative

2
1
2  p 
2 R 
 v b 
v2
ds  s2  s1   
dv   
dv  R ln  v  b   v  R ln  2



1
1
1
 T  v
 v b 
 v1  b 

kJ   0.348  0.1162 
kJ
s2  s1   8.314
 ln 
  10.42
kmol K   0.928  0.1162 
kmol K


kJ 
kJ
S2  S1  n  s2  s1    0.862 kmol   10.42
  8.98
kmol K 
K
