CANKAYA UNIVERSITY
FACULTY OF ENGINEERING AND ARCHITECTURE
MECHANICAL ENGINEERING DEPARTMENT
ME 211 THERMODYNAMICS I
CHAPTER 3 EXAMPLES
SOLUTIONS
1) The average atmospheric pressure in Denver (elevation=1610 m) is 83.4 kPa.
Determine the temperature at which water in an uncovered pan will boil in
Denver.
p  83.4 kPa  83.4  10 3 Pa 
1 bar
 0.834 bar
10 5 Pa
p  0.834 bar  T  Tsat  ?
From Table A-3:
p  0.8 bar
T  93.50 o C
p  0.9 bar
T  96.71 o C
p
0.9
0.834
0.8
z
93.5
0.834  0.8
z

0.9  0.8
96.71  93.5
 z  1.0914

 T  93.5  z  94.59 o C
96.71
T
2) A rigid tank with a volume of 2.5 m3 contains 5 kg of saturated liquid-vapor
mixture of water at 75oC. Now the water is slowly heated. Determine the quality
x at the initial state. Also, determine the temperature at which the liquid in the
tank is completely vaporized.
T
vapor
2
T2
liquid
75oC
1
v
v = constant
V 2.5 m 3
v1  
 0.5 m 3 / kg
m
5 kg
o
T1 = 75 C
From Table A-2 @ T1 = 75oC:
vf = 1.0259  10-3 m3/kg
vg = 4.131 m3/kg
v1  v f  x 1 v fg  x1  0.121
@ State 2:
v2 = v1 = 0.5 m3/kg
x2 = 1
v 2  v1  0.5 m 3 /kg 
3
v g  0.5m /kg
x 2 1

From Table A-2 @ vg = 0.5 m3/kg:
T2  140 o C
0.5m 3 / kg  0.5089m 3 / kg

 T2  140.7 o C
3
3
o
o
0.3928m / kg  0.5089m / kg 150 C  140 C
3) Consider a two phase mixture at 100oC with x3 = 0.9. Determine the specific
volume.
From Table A-2 @100oC:
v f  1.0435  10 3 m 3 / kg
v g  1.673 m 3 / kg
v 3  v f  x(v g  v f )
 v 3  1.0435  10 3 m 3 / kg  0.9(1.673  1.0435  10 3 ) m 3 / kg
 v 3  1.506 m 3 / kg
Note: Once the quality x, is known, it can be applied to calculate v, h or s in the same
manner as above.
4) Refrigerant 134a with a quality of 0.4 and a temperature of 12oC is contained in
a rigid tank that has a volume of 0.17 m3. Find the mass of liquid present.
vapor
liquid
m
V
,
v
m g  xm ,
T1 = 12oC
x1 = 0.4
V = 0.17 m3
mf  m  mg
v  v f  xv fg
From Table A-10 @12oC:
v f  0.000797 m 3 / kg
v g  0.046 m 3 / kg
v1  v f  xv fg  v1  0.000797 m 3 / kg  0.4(0.046  0.000797)m 3 / kg
 v1  0.0189 m 3 / kg
Total mass m 
0.17 m 3
V

 9 kg
v1 0.0189 m 3 / kg
m g  xm  (0.4)(9)  3.6 kg
m f  m  m g  9 kg  3.6 kg  5.4 kg
5) Consider Refrigerant-22 at 12oC. It is specific internal energy 144.58 kJ/kg.
Determine phase description and enthalpy.
From Table A-7 @12oC:
u f  58.77 kJ / kg
u g  230.38 kJ / kg
So uf < u < ug  saturated liquid vapor mixture.
x
(u  u f )
(144.58  58.77) kJ / kg

 0.5
(u g  u f ) (230.38  58.77) kJ / kg
h  h f  xh fg  59.35 kJ / kg  0.5(253.99  59.35) kJ / kg
 h  156.67 kJ / kg
6) Determine the temperature of water at a state of p = 0.5 MPa and h = 2890
kJ/kg.
P = 0.5 MPa = 500 kPa = 5bar
h = 2890 kJ/kg
From Table A-3 for p = 5 bar
p (bar)
hf (kJ/kg)
hg (kJ/kg)
2748.7
640.23
5
h > hg so superheated vapor
From Table A-4 @ p = 5 bar:
T (oC)
h (kJ/kg)
200
2855.4
240
2939.9
h
2939.9
2890
2855.4
z
200
2890  2855.4
z

2939.9  2855.4 240  200
 z  16.37

 T  200  z  216.38 o C
240
T
7) Determine the missing properties and the phase descriptions in the following
table for water:
T, oC
(a)
(b) 140
(c)
(d) 80
(e)
p, kPa
200
h, kJ/kg
x
0.7
Phase description
1800
950
500
800
0.0
3161.7
(a)
p  200 kPa  2 bar 
o
T  Tsat  120.2 C
x  0.7

h  h f  xh fg  504.7  0.7(2201.9)  h  2046 kJ / kg  saturated mixture
(b)

T  140 o C
p  3.613 bar  361.3 kPa
h  1800 kJ / kg 
h  h f  xh fg  1800 kJ / kg  589.13 kJ / kg  x (2144.7 kJ / kg)
 x  0.564
 saturated mixture
(c)
p  950 kPa  9.5 bar 
 saturated liquid
x0

p = 950 kPa = 0.95 MPa = 9.5 bar  T = 177.65oC
h = hf = 752.82 kJ/kg
(d)
p  500 kPa  0.5 MPa  5 bar 
o
p  5 bar  Tsat  151.9 C
o
T  80 C

T  Tsat  compressed liquid
 h  h f @ T 80o C  334.91kJ / kg
(e)
p  800 kPa  0.8 MPa  8 bar 

h  3161.7 kJ / kg

p  8 bar  h f  721.11 kJ / kg, h g  2769.1 kJ / kg
h  hg
 superheated vapor
p  8 bar

o
T  350.06 C
h  3161.7 kJ / kg 
8) A frictionless piston-cylinder device initially contains 200 L of saturated liquid
refrigerant R-134a. The piston is free to move, and its mass is such that it
maintains a pressure of 800 kPa on the refrigerant. The refrigerant is now heated
until its temperature rises to 50 oC. Calculate the work done during this process.
9) A piston-cylinder device contains 50 kg of water at 150 kPa and 25 oC. The
cross-sectional area of the piston is 0.1 m2. Heat is now transferred to the water,
causing part of it to evaporate and expand. When the volume reaches 0.2 m3, the
piston reaches a linear spring whose spring constant is 100kN/m. More heat is
transferred to the water until the piston rises 20 cm more. Determine (a) the
final pressure and temperature and (b) the work done during this process. Also,
show the process on p-v diagram.
10) The radiator of a steam heating system has a volume of 20 L and is filled with
superheated vapor at 300 kPa and 250°C. At this moment both the inlet and
exit valves to the radiator are closed. Determine the amount of heat that will
be transferred to the room when the steam pressure drops to 100 kPa. Also,
show the process on a p-v diagram with respect to saturation lines.
11) Two kilograms of saturated liquid water at 50 kPa are heated slowly at constant
pressure. During the process 5876 kJ of heat are added. Find the final
temperature.
H2O
Q = 5876 kJ
p1 = p2 = 50 kPa
m = 2 kg
State 1:
p1  50 kPa  0.5 bar 
h 1  h f  340.5 kJ / kg
x1  0

Q – W = U2 – U1
2
W   pdV  p(V2  V1 )  p 2 V2  p1V1
1
 Q  W  ( U 2  U1 )  p 2 V2  p1 V1  U 2  U1  H 2  H1  m(h 2  h 1 )
 h2 
Q
5876
 h1 
 340.5  3278 kJ / kg
m
2
State 2:
p 2  50 kPa  0.5 bar 

h 2  3278 kJ / kg

@ p 2  0.5 bar  h f  340.5 kJ / kg, h g  2645.9 kJ / kg
h 2  h g  superheated vapor
From Table A-4  T2 = 400oC
12) A vessel having a volume of 5 m3 contains 0.05 m3 of saturated liquid water and
4.95 m3 of saturated water vapor at 0.1 MPa. Heat is transferred until the vessel
is filled with saturated vapor. Determine the heat transfer for this process.
vapor
H2O
liquid
Q12  W12  U 2  U1
KE  PE  0
State 1:
p1  0.1MPa  1bar  v f 1  0.001043 m 3 / kg, v g1  1.694 m 3 / kg
u f 1  417.36 kJ / kg, u g1  2506.1kJ / kg
Vf
0.05 m 3

 47.94 kg
v f 1 0.001043 m 3 / kg
Vg
4.95 m 3
m g1 

 2.92 kg
v g1 1.694 m 3 / kg
mf1 
U1  (m f u f )1  (m g u g )1  (47.94 kg)(417.36 kJ / kg)  (2.92 kg)(2506.1kJ / kg)  27326 kJ
State 2:
m  m f 1  m g1  47.94 kg  2.92 kg
v2 
5 m3
V

 0.09831 m 3 / kg
m 50.86 kg
x2  1

 We need to do interpolation
v 2  0.09831 m / kg 
3
vg (m3/kg)
0.09963
0.09831
0.07998
p (bar)
20
?
25
p  20.3 bar (final pressure)
ug (kJ/kg)
2600.3
?
2603.1
p (bar)
20
20.3
25
u 2  2600.5 kJ / kg
U 2  m 2 u 2  (50.86 kg)(2600.5 kJ / kg)  132261kJ
 Q12  U 2  U1  104935 kJ
13) A cylinder fitted with a piston has a volume of 0.1 m3 and contains 0.5 kg of
steam at 0.4 MPa. Heat is transferred to the steam until the temperature is 300oC
while the pressure remains constant. Determine the work and heat transfer for
this process.
p
p2 = p1 = 0.4 MPa
1
V
2
2
1
1
W12   pdV  p  dV  p(V2  V1 )  pm( v 2  v1 )
Q12  U 2  U1  W12  m(u 2  u 1 )  mp( v 2  v1 )  m(u 2  p 2 v 2 )  (u 1  p1 v1 )
 Q12  m(h 2  h 1 )
State 1:
v1 
V1 0.1

 0.2 m 3 / kg
m 0.5
and
p1  0.4 MPa
From Table A-3 @ p1 = 0.4 Mpa = 4 bar:
vf = 0.001084 m3/kg, vg = 0.4625 m3/kg
v f  v1  v g  liquid  vapor mixture
v1  v f  x 1 v fg  0.2 m 3 / kg  0.001084 m 3 / kg  x 1 (0.4625  0.001084) m 3 / kg
 x 1  0.4311
h 1  h f  x 1h fg  h 1  604.74 kJ / kg  (0.4311)(2738.6  604.74) kJ / kg
 h 1  1524.7 kJ / kg
State 2:
p2 = 0.4 Mpa = 4 bar
T2 = 300oC
p2 = 0.4 Mpa = 4 bar  Tsat  143.6 o C, T  Tsat  superheated vapor
Using Table A-4:
p = 3 bar
T (oC)
280
300
320
h (kJ/kg)
3028.6
?
3110.1
T
320
300
280
z
3028.6
3110.1
h
300  280
z

 z  40.75  h  3028.6  z  3069.35 kJ / kg
320  280 (3110.1  3028.6)
p = 5 bar
T (oC)
280
300
320
h (kJ/kg)
3022.9
?
3105.6
T
320
300
280
z
3022.9
3105.6
h
300  280
z

 z  41.35  h  3022.9  z  3064.25 kJ / kg
320  280 (3105.6  3022.9)
T = 300 oC
p (bar)
3
4
5
h (kJ/kg)
3069.35
?
3064.25
p
5
4
3
z
3064.25
3069.35
h
43
z

 z  2.55  h 2  3064.25  z  3066.8 kJ / kg
5  3 (3069.35  3064.25)
We do similar interpolation for v2 = 0.6548 m3/kg
Q12  m(h 2  h 1 )  0.5 kg(3066.8  1524.7) kJ / kg  771.1 kJ
W12  mp( v 2  v1 )  (0.5 kg)(400 kPa)(0.6548  0.2) m 3  90.96 kJ
14) A frictionless piston is used to provide a constant pressure of 400 kPa in a
cylinder containing steam originally at 200oC with a volume of 2 m3. Calculate
the final temperature if 3500 kJ of heat added.
Steam
Q = 3500 kJ
State 1:
p1  400 kPa  4  10 5 Pa  4 bar
T1  200 o C
V1  2 m 3
From Table A-3:
p2 = 4 bar  Tsat  143.6 o C, T  Tsat  superheated vapor
Using Table A-4 for p1  4 bar and T1  200 o C :
v1  0.5342 m 3 / kg, u1  2647 kJ / kg, h1  2860 kJ / kg
2
2
1
1
W12   pdV  p  dV  p(V2  V1 )  400 kPa(V2  V1 )
m
V1
2 m3

 3.744 kg
v1 0.5342 m 3 / kg
Q  W  ( U 2  U 1 )  m( u 2  u 1 )
 3500 kJ  400 kPa(V2  V1 )  m(u 2  u 1 )
 3500 kJ  400 kPa(3.744v 2  2) m 3  (3.744 kg)(u 2  2647) kJ / kg
This problem requires trial error procedure:
1) Guess a value of v2
2) Calculate u2 from above equation
3) If this value checks with u2 from steam tables then the guess is correct one.
A) 1) Let v2 = 1 m3/kg
2) 3500 kJ  400 kPa(3.744)(1)  2m 3  (3.744 kg)(u 2  2647) kJ / kg
 u 2  3395 kJ / kg
3) From Table A-4 for:
p 2  4 bar 
u 2  3300.2 kJ / kg
v 2 1 m 3 / kg 
These two u values are not the same. So revise value of v.
B) 1) Let v2 = 1.06 m3/kg
2) 3500 kJ  400 kPa(3.744)(1.06)  2m 3  (3.744 kg)(u 2  2647) kJ / kg
 u 2  3372 kJ / kg
3) From Table A-4 for:
p 2  4 bar

u 2  3372 kJ / kg
3
v 2 1.06 m / kg 
We now accept that u2 values are close enough. So temperature is T2  640oC.
Second Method:
Q  W  (U 2  U1 )
2
2
1
1
W   pdV  p  dV  p(V2  V1 )
Q  W  (U 2  U1 )  p(V2  V1 )  (U 2  U1 )  Q  H 2  H1  m(h 2  h1 )
h2 
3500 kJ
Q
 h1 
 2860 kJ / kg  3795 kJ / kg
m
3.744 kg
From Table A-4 for:
p 2  4 bar

o
T2  641 C
h 2  3795 kJ / kg 
15) Determine the enthalpy change h of nitrogen, in kJ/kg, as it is heated from 600
to 1000 K, using (a) the empirical specific heat equation as a function of
temperature (Table A-21), (b) the cP value at the average temperature (Table A20), and (c) the cP value at room temperature at 300 K.
a)
cp
R

   T  T 2  T 3  c p  8.314 kJ / kmol.K    T  T 2  T 3

From Table A-21:
  3.675,   1.208  10 3 ,   2.324  10 6 ,   0.632  10 9
T
T2
2
1
1
1


h   c p dT  8.314 kJ / kmol.K T  T 2  T 3  T 4 
2
3
4

 T1
T1



1

3
2
2
3.6751000  600  2  1.208  10 1000  600
 h  8.314 kJ / kmol.K 
 1 2.324  10 6 1000 3  600 3  1  0.632  10 9 1000 4  600 4
 3
4


 
kJ
kmol
h 12544 kJ / kmol
kJ
h 

 447.8
M 28.01 kg / kmol
kg
h  12544
b) From Table A-20 at Tavg = 800 K, cp,avg = 1.121 kJ/kg.K
h  c p,avg (T2  T1 )  1.121
kJ
kJ
(1000  600) K  h  448.4
kg.K
kg
c) Taking the cp at room temperature from Table A-20:
cp = 1.039 kJ/kg.K
h  c p (T2  T1 )  1.039
kJ
kJ
(1000  600) K  h  415.6
kg.K
kg







16) A rigid tank has a volume of 400 cm3 contains air initially at 22oC and 100 kPa.
A paddle wheel stirs the gas until the temperature is 428oC. During the process,
600 J of heat are transferred from air to surroundings. Calculate the work done
by the paddlewheel two ways:
(a) assuming constant specific heat
(b) assuming variable specific heat
air
(a) cv = constant
U  Q  W
U  mc v (T2  T1 )
22  428
Tavg 
 225 o C  498 K
2
From Table A-20:
c v  0.742 kJ / kg.K  742 J / kg.K
pV
pVM
m
RT
RT
(28.97 kg / kmol)(100 kPa)(400 cm 3 )m 3 / 10 6 cm 3 
m
 0.0004725 kg
(8.314 kJ / kmol.K)(295 K)
pV  mRT  m 
W  Q  U  Q  mc v (T2  T1 )  742.3 J
(b) variable cv with temperature:
U  Q  W
U  m(c v 2 T2  c v1T1 )  Q  W  m(c v 2 T2  c v1T1 )
From Table A-20:
At
T1 = 295 K
T2 = 701 K
pV  mRT  m 
cv1 = 0.718 kJ/kg.K = 718 J/kg.K
cv2 = 1.098 kJ/kg.K = 1098 J/kg.K
pV
pVM
m
RT
RT


(28.97 kg / kmol)(100 kPa)(400 cm 3 ) m 3 / 10 6 cm 3
m
 0.0004725 kg
(8.314 kJ / kmol.K)(295 K)
Q  W  U 2  U1
U 2  mc v 2 T2  U 2  (0.0004725 kg)(1098 J / kg.K )(701 K )  363.682 J
U1  mc v1T1  U1  (0.0004725 kg)(718 J / kg.K )(295 K )  100.080 J
W  Q  U 2  U1   600 J  (363.682  100.080)  863.602 J
17) A mass of 15 kg of air in a piston-cylinder device is heated from 25 to 77°C by
passing current through a resistance heater inside the cylinder. The pressure
inside the cylinder is held constant at 300 kPa during the process, and a heat
loss of 60 kJ occurs. Determine the electric energy supplied, in kWh.
18) Helium (He) gas initially at 2 bar and 200 K undergoes a polytropic process
with n = k to a final pressure of 14 bar. Determine work and heat transfer for the
process, each in kJ/kg of helium. Assume ideal gas behavior.
p
14
PVk = c
2
He
2
Q - W  U
dV m(p 2 v 2  p1 v1 ) mR (T2  T1 )


k
1 k
1 k
V1 v
V2
V2
W   pdV  m 
V1
To find T2 use:
T2  T1 (p 2 / p1 )
k
cp
cv

k 1
k
2.5
 1.667
1.5
 T2  200(14 / 1)

1.6671
575K
1.667
W (8.314 / 4.003)(575  200)

 1167.7 kJ
m
1  1.667
Q  m(u 2  u 1 )  W 
Q
W
W
 u 2  u1 
 c v (T2  T1 ) 
m
m
m
cp  cv  R
cp
cv
1 
R
R
 k 1 
cv
cv
R
k 1
R (T2  T1 )
Q
R
 
(T2  T1 ) 
0
m k 1
1 k
cv 
1
V
19) Determine the specific volume of superheated water vapor at 1.6 MPa and 225
o
C, using (a) the ideal-gas equation, (b) the generalized compressibility chart,
and (c) the steam tables. Also determine the error involved in the first two cases.
R 8.314 kJ / kmol.K

 0.4615 kJ / kg.K
M
18.02 kg / kmol
Tc  647.3 K
R
p c  220.9 bar  22.09 MPa
a) ideal gas law:
RT (0.4615 kJ / kg.K)(498 K)
m3
v

 0.14364
p
1600 kPa
kg
b) compressibility chart, see Figure A-1, Figure A-2, Figure A-3:
1.6 M Pa
p


 0.072
p c 22.09 M Pa

Z  0.935
498 K
T
TR 

 0.769 

Tc 647.3 K
pR 
 RT 
  0.935(0.14364)  0.13430 m 3 / kg
v  Z
 p 
0.1343  0.14364
error 
 100  7%
0.14364
c) using superheated table:
p  1.6MPa  16 bar 
v  0.13287
T  225 o C

20)
21 )
21) Air is confined to one side of a rigid container divided by a partition, as shown in
Figure below. The other side is initially evacuated. The air is initially at =5 bar , and
=500 K, and
. When the partition is removed, the air expands to fill the
entire chamber. Measurements show that
and
. Assuming the air
behaves as an ideal gas, determine (a) the final temperature, in K, and (b) the heat
transfer, kJ.
solution