CANKAYA UNIVERSITY
FACULTY OF ENGINEERING AND ARCHITECTURE
MECHANICAL ENGINEERING DEPARTMENT
ME 211 THERMODYNAMICS I
From Table A-3: p

0 .
8 bar p

0 .
9 bar
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
10
5
Pa p

0 .
834 bar

T

T sat

?

0 .
834 bar
T
T


93 .
50 o
C
96 .
71 o
C
0.9 p
0.834
0.8
93.5 z
96.71
T


0 .
834

0 .
8 z
0 .
9

1

0 .
8
.
0914
 z
96 .
71

93 .
5

T

93 .
5
 z

94 .
59 o
C

2) A rigid tank with a volume of 2.5 m
3
contains 5 kg of saturated liquid-vapor mixture of water at 75 o
C. 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. vapor
T
2
T liquid
75 o
C
1
2 v v = constant
2 .
5 m
3 v
1

V
 
0 .
5 m
3
/ kg m 5 kg
T
1
= 75 o
C
From Table A-2 @ T
1
= 75 o
C: v f
= 1.0259

10
-3
m
3
/kg v g
= 4.131 m
3
/kg v
1
 v f
 x
1 v fg
 x
1

0 .
121
@ State 2: v
2
= v
1
= 0.5 m
3
/kg x
2
= 1 v
2 x
2

v
1

1

0.5
m
3
/kg


 v g

0 .
5 m
3
/kg
From Table A-2 @ v g
= 0.5 m
3
/kg:
0 .
5 m
3
/
0 .
3928 m
3 kg
/

0 .
5089 m
3 kg

0 .
5089 m
/
3 kg
/ kg

T
150
2 o

C
140
 o
140
C o
C

T
2

140 .
7 o
C

3) Consider a two phase mixture at 100 o
C with x
3
= 0.9. Determine the specific volume.
From Table A-2 @100 o
C: 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
 v
3

1 .
0435

10

3 m
3
/ kg

0 .
9 ( 1 .
673

1 .
0435

10

3
) m
3

1 .
506 m
3
/ kg
/ 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 12 o
C is contained in a rigid tank that has a volume of 0.17 m
3
. Find the mass of liquid present. vapor liquid
T
1
= 12 o
C x
1
= 0.4
V = 0.17 m
3 m f
 m
 m g m
 v
 v f
V
, v
 xv fg m g
 xm ,
From Table A-10 @12 o
C: v f v g


0 .
000797 m
3
0 .
046 m
3
/ kg
/ kg v
1
 v f
 v
1
 xv fg
 v
1

0 .
000797 m
3
/ kg

0 .
4 ( 0 .
046

0 .
000797 ) m
3
/ kg

0 .
0189 m
3
/ kg
Total mass m

V v
1

0 .
17 m
0 .
0189 m
3
3
/ kg

9 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 12 o
C. It is specific internal energy 144.58 kJ/kg.
Determine phase description and enthalpy.
From Table A-7 @12 o
C: u u f g


58 .
77 kJ / kg
230 .
38 kJ / kg
So u f
< u < u g
 saturated liquid vapor mixture. x

( u
( u g

 u f u f
)
)

(
(
144
230
.
.
58
38


58
58
.
.
77
77
)
) kJ kJ
/
/ kg kg

0 .
5 h

 h h f
 xh fg

59 .
35

156 .
67 kJ / kg kJ / kg

0 .
5 ( 253 .
99

59 .
35 ) 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) h f
(kJ/kg) h g
(kJ/kg)
5 640.23 2748.7 h > h g
so superheated vapor
From Table A-4 @ p = 5 bar:
T ( o
C)
200 h (kJ/kg)
2855.4
240
2939.9
2939.9 h
2890
2855.4
200 z
240
T


2890

2855 .
4
2939 .
9 z

16 .

37
2855 .
4

240 z

200

T

200
 z

216 .
38 o
C

7) Determine the missing properties and the phase descriptions in the following table for water:
(a)
T, o
C p, kPa
200 h, kJ/kg x
0.7
Phase description
(b) 140
(c) 950
500
800
1800
3161.7
0.0
(d) 80
(e)
(a) p

200 kPa x

0 .
7

2 bar

T h
 h f
 xh fg

T sat

120 .
2 o
C

504 .
7

0 .
7 ( 2201 .
9 )
 h

2046 kJ / kg
 saturated mixture
(b)
T h

140 o
C

1800 kJ / kg


 p h
 h f
 xh fg

3 .
613 bar

1800 kJ / kg

589 .
13 kJ / kg
 x ( 2144 .
7 kJ / kg )
 x

0 .
564
 saturated mixture

361 .
3 kPa
(c) p

950 kPa

9 .
5 bar saturated liquid x

0 p = 950 kPa = 0.95 MPa = 9.5 bar

T = 177.65
o
C h = h f
= 752.82 kJ/kg
(d) p

500 kPa

T

80 o
C
0 .
5 MPa

5 bar


 p
T

T sat
 compressed liquid

5 bar

T sat

151 .
9 o
C

 h
 h f @ T

80 o
C

334 .
91 kJ / kg
(e) p

800 kPa

0 .
8 M Pa

8 bar h

3161 .
7 kJ / kg p

8 bar
 h f

721 .
11 kJ / kg , h g h

 h g superheated vapor p h

8 bar

3161 .
7 kJ / kg 
T

350 .
06 o
C

2769 .
1 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 o
C. Calculate the work done during this process.

9) A piston-cylinder device contains 50 kg of water at 150 kPa and 25 cross-sectional area of the piston is 0.1 m
2 o
C. The
. Heat is now transferred to the water, causing part of it to evaporate and expand. When the volume reaches 0.2 m
3
, 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.
H
2
O
Q = 5876 kJ p
1
= p
2
= 50 kPa m = 2 kg
State 1: p
1

50 kPa
 x
1

0
0 .
5 bar


 h
1
Q – W = U
2
– U
1
 h f

340 .
5 kJ / kg
W
 
2
1 pdV
 p ( V
2

Q

W

( U
2

V
1
)
 p
2
V
2

U
1
)
 p
2
V
2
 p
1
V
1
 p
1
V
1

U
2

U
1

H
2

H
1
 m ( h
2
 h
1
)
 h
2

Q m
 h
1

5876

340 .
5

3278 kJ / kg
2
State 2: p
2

50 kPa
 h
2
0 .

3278 kJ / kg
5 bar
@ 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

T
2
= 400 o
C

12) A vessel having a volume of 5 m
3
contains 0.05 m
3
of saturated liquid water and
4.95 m
3
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
H
2
O liquid

U
2
KE

PE

0 Q
12

W
12

U
1
State 1: p
1

0 .
1 MPa

1 bar
 v f 1 u f 1


0 .
001043 m
3
/ kg ,
417 .
36 kJ / kg , v g 1 u g 1


1 .
694 m
3
/ kg
2506 .
1 kJ / kg m f 1

V v f 1 f
0 .
05 m
3

0 .
001043 m
3
/ kg

47 .
94 kg m g 1

V g v g 1

4 .
95 m
1 .
694 m
3
3
/ kg

2 .
92 kg
U
1

( m f u f
)
1

( m g u g
)
1

( 47 .
94 kg )( 417 .
36 kJ / kg )

( 2 .
92 kg )( 2506 .
1 kJ / kg )

27326 kJ
State 2: m
 m f 1
 m g 1

47 .
94 kg

2 .
92 kg v
2

5 m
3
V m

50 .
86 kg

0 .
09831 m
3
/ kg x
2 v
2

1

0 .
09831 m
3
/ kg



We need to do interpolation v g
(m
3
/kg)
0.09963
0.09831
0.07998 p (bar)
20
?
25

 p

20 .
3 bar ( final pressure ) u g
(kJ/kg)
2600.3
?
2603.1 p (bar)
20
20.3
25
 u
2

2600 .
5 kJ
U
2


Q
12 m
2 u
2

U
2

/ kg
( 50 .
86 kg )( 2600 .
5 kJ

U
1

104935 kJ
/ kg )

132261 kJ

13) A cylinder fitted with a piston has a volume of 0.1 m
3
and contains 0.5 kg of steam at 0.4 MPa. Heat is transferred to the steam until the temperature is 300 o
C while the pressure remains constant. Determine the work and heat transfer for this process. p p
2
= p
1
= 0.4 MPa
1
V
W
12
 
2
1 pdV
 p

2
1 dV
 p ( V
2

V
1
)
 pm ( v
2
 v
1
)
Q
12


Q
12
U
2


U m ( h
2
1

W
12
 h
1
)
State 1:
 m ( u
2
 u
1
)
 mp ( v v
1

V
1 m

0 .
1
0 .
5

0 .
2 m
3
/ kg and
From Table A-3 @ p
1
= 0.4 Mpa = 4 bar:
2
 v
1
)
 m

( u
2
 p
2 v
2
)

( u
1
 p
1 v
1
)
 p
1

0 .
4 MPa v f
= 0.001084 m
3
/kg, v g
= 0.4625 m
3
/kg v f
 v
1
 v g
 liquid
 vapor mixture v
1
 v f
 x
1 v fg
 h
1 x
1
 h
 f
0 .
4311
 x
1 h fg

0 .
2 m
3
/ kg

0 .
001084 m
3
/ kg
 x
1
( 0 .
4625

0 .
001084 )
 h
 h
1

State 2:
1524 .
7 kJ / kg
1

604 .
74 kJ / kg

( 0 .
4311 )( 2738 .
6

604 .
74 ) kJ / kg m
3
/ kg

p
2
= 0.4 Mpa = 4 bar
T
2
= 300 o
C p
2
= 0.4 Mpa = 4 bar

T sat
Using Table A-4: p = 3 bar

143 .
6 o
C ,
T ( o
C)
280
300
320
T

T sat
 superheated vapor h (kJ/kg)
3028.6
?
3110.1
T
320
300
280 z h
3028.6
3110.1
300

320

280
280
 z
( 3110 .
1

3028 .
6 )
 z

40 .
75
 h

3028 .
6
 z

3069 .
35 kJ / kg p = 5 bar
T ( o
C)
280
300
320 h (kJ/kg)
3022.9
?
3105.6
320
300
280
T
3022.9 z
3105.6 h

300

280
320

280
T = 300 o
C
 z
( 3105 .
6

3022 .
9 ) p (bar)
3
4
5
 z

41 .
35
 h

3022 .
9
 z

3064 .
25 kJ / kg h (kJ/kg)
3069.35
?
3064.25
5 p
4
3 z h
3064.25 3069.35
4

3
5

3

( 3069 .
35 z

3064 .
25 )
 z

2 .
55
 h
2

3064 .
25
 z

3066 .
8 kJ
We do similar interpolation for v
2
= 0.6548 m
3
/kg
Q
W
12
12

 m ( mp h
(
2 v
2
 h

1 v
)
1

)

0 .
(
5
0 kg
.
5
( 3066 kg )(
.
8
400

1524 .
7 ) kJ kPa )( 0 .
6548
/ kg


0 .
2 )
771 .
1 kJ m
3 
90 .
96 kJ
/ kg

14) A frictionless piston is used to provide a constant pressure of 400 kPa in a cylinder containing steam originally at 200 o
C with a volume of 2 m
3
. Calculate the final temperature if 3500 kJ of heat added.
Steam
Q = 3500 kJ
State 1: p
1

400 kPa

4

10 5 Pa

4 bar
T
1

200 o
C
V
1

2 m
3
From Table A-3: p
2
= 4 bar

T sat

143 .
6 o
C , T

T sat
 superheated vapor
Using Table A-4 for p
1 v
1

0 .
5342 m
3
/ kg , u
1

4 bar

2647 and kJ /
T
1 kg ,

200 h
1
 o
C :
2860 kJ / kg
W
12
 
2
1 pdV
 p

2
1 dV
 p ( V
2

V
1
)

400 kPa ( V
2

V
1
) m

V
1

2 m
3
0 .
5342 m
3
/ kg

3 .
744 kg v
1
Q


W
3500

( U
2 kJ


U
1
)
 m
400 kPa ( V
2
( u

2

V
1
) u
1
)
 m ( u
2

3500 kJ

400 kPa ( 3 .
744 v
2

2 ) m
3 
 u
1
)
( 3 .
744 kg )( u
2

2647 ) kJ / kg
This problem requires trial error procedure:
1) Guess a value of v
2
2) Calculate u
2
from above equation
3) If this value checks with u
2
from steam tables then the guess is correct one.

A) 1) Let v
2
= 1 m
3
2) 3500 kJ

400
/kg kPa

( 3 .
744 )( 1 )
 u
2

3395 kJ / kg

2
 m
3 
( 3 .
744 kg )( u
2

2647 ) kJ / kg
3) From Table A-4 for: p v
2
2


1
4 bar m
3
/ kg


 u
2

3300 .
2 kJ / kg
These two u values are not the same. So revise value of v.
B) 1) Let v
2
= 1.06 m
3
/kg
2) 3500 kJ

400 kPa

( 3 .
744 )( 1 .
06 )
 u
2

3372 kJ / kg

2
 m
3 
( 3 .
744 kg )( u
2

2647 ) kJ / kg
3) From Table A-4 for: p v
2
2


1
4
.
bar
06 m
3
/ kg


 u
2

3372 kJ / kg
We now accept that u
2
values are close enough. So temperature is T
2

640 o
C.
Second Method:
Q

W

( U
2

U
1
)
W
Q

 
2
1 pdV
W


( U
2 p

2
1
 dV
U
1
)

 p ( V
2 p ( V
2

V
1
)

V
1
)

( U
2

U
1
)

Q

H
2

H
1
 m ( h
2
 h
1
) h
2

Q m
 h
1

3500 kJ
3 .
744 kg

2860 kJ / kg

3795 kJ / kg
From Table A-4 for: p h
2
2


4 bar
3795 kJ / kg
T
2

641 o
C

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 c
P
value at the average temperature (Table A-
20), and (c) the c
P
value at room temperature at 300 K. a) c p    
T
 
T
2
R
From Table A-21:
 
3 .
675 ,
  
1 .
208
 
T
3 

10

3
,
 c p



8 .
314 kJ /
2 .
324

10

6
, kmol .
K
 
  
T
 
T
2
  
0 .
632

10

9
 
T
3

 h


T
2 
T
1 c p dT
 h


8 .
314 kJ / kmol .
K
 


T



8 .
314 kJ
1
2

T
2 
1
3

T
3 
1
4

T
4


T
2
T
1
/ kmol .
K






3 .
675
1
3

2

1000
.
324


600
 

10

6

1000
2
3
1


1 .
208

10

3

1000
2

600
3
 

1
4



600
2
0 .
632

10

9
 

1000
4 
600
4
 



 h

12544 kJ kmol
 h

 h
M

12544 kJ
28 .
01 kg
/ kmol
/ kmol
 kJ
447 .
8 kg b) From Table A-20 at T avg
= 800 K, c p,avg
= 1.121 kJ/kg.K
 h
 c p , avg
( T
2

T
1
)

1 .
121 kJ kg .
K
( 1000

600 ) K
  h

448 .
4 kJ kg c) Taking the c p
at room temperature from Table A-20: c p
= 1.039 kJ/kg.K
 h
 c p
( T
2

T
1
)

1 .
039 kJ kg .
K
( 1000

600 ) K
  h

415 .
6 kJ kg

16) A rigid tank has a volume of 400 cm
3
contains air initially at 22
A paddle wheel stirs the gas until the temperature is 428 o o
C and 100 kPa.
C. 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) c v
= constant


U
U
T avg

Q

W

 mc
22 v

( T
2

428
2
T

1
)
225
From Table A-20: c v o C

498 K

0 .
742 kJ / kg .
K

742 J / kg .
K pV
 mRT

( 28 .
97 kg / pV pVM m
  m

RT R T kmol )( 100 kPa )( 400 cm
3
)
 m
3 m

( 8 .
314 kJ / kmol .
K )( 295 K )
/ 10
6 cm
3


0 .
0004725 kg
W

Q
 
U

Q
 mc v
( T
2

T
1
)
 
742 .
3 J
(b) variable c v
with temperature:

U

Q

W

U
 m ( c v 2
T
2
 c v 1
T
1
)

Q

W
 m ( c v 2
T
2
From Table A-20:
 c v 1
T
1
)

At T
1
= 295 K
T
2
= 701 K c v1
= 0.718 kJ/kg.K = 718 J/kg.K c v2
= 1.098 kJ/kg.K = 1098 J/kg.K pV
 mRT
 m
 pV
 m

RT pVM
R T m

( 28 .
97 kg / kmol )( 100 kPa )( 400 cm
3
)
 m
3
/ 10
6 cm
3


0 .
0004725 kg
( 8 .
314 kJ / kmol .
K )( 295 K )
Q

U
2
W


U
2 mc v 2
T
2

U
1

U
2
U
1
 mc v 1
T
1
W


U
Q

1



U
( 0 .
0004725 kg )( 1098 J /
( 0 .
0004725
2

U
1

 kg )( 718

600 J

J / kg .
K )( 701 K ) kg .
K )( 295 K )

363 .
682 J

100 .
080 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
PV k
= c
2
14
He
2
1
V
Q W
 
U
W

V
2 
V
1 pdV
 m
V
2 
V
1 dV v k
To find T
2
use: k

1
T
2

T
1
( p
2
/ p
1
) k
 m ( p
2 v
2
1

 k p
1 v
1
)
 mR ( T
2
1
 k

T
1
) k
 c p c v

2 .
5
1 .
5

1 .
667

T
2

W m

200 ( 14 / 1 )
1 .
667

1

575 K
1 .
667

( 8 .
314 / 4
1
.
003

)( 575
1 .
667

200 )
 
1167 .
7 kJ
Q
 m ( u
2
 u
1
)

W

Q m
 u
2
 u
1

W m c p
 c v

R
 c v
( T
2

T
1
)

W m c p c v

1

R c v
 k

1

R c v
 c v

Q m
 k
R

1
 k
R

1
( T
2

T
1
)

R ( T
2
1

 k
T
1
)

0

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
T c p c

18

647 .
3 K

R
M

220 .
9
8 .
314 bar
.
02 kJ kg
/
/ kmol .
K kmol

22 .
09 M Pa

0 .
4615 kJ / kg .
K a) ideal gas law: v

RT p

( 0 .
4615 kJ / kg .
K )( 498 K )
1600 kPa

0 .
14364 m
3 kg b) compressibility chart, see Figure A-1, Figure A-2, Figure A-3: p
R
 p

0 .
072
 p c
1 .
6 M Pa
22 .
09 M Pa

T
R

T
T c

498 K
647 .
3 K

0 .
769


Z

0 .
935 v
 error

Z

RT p



0 .
935 ( 0 .
14364 )

0 .
1343

0 .
14364
0 .
14364


100
0

.
13430
7 % m
3
/ kg c) using superheated table: p
T

1 .
6 M Pa

225 o
C

16 bar


 v

0 .
13287

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