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TER 201E THERMODYNAMICS
Problem Hour - 3
Problems and Solutions
Subject: The First Law of Thermodynamics: Energy Analysis of Closed Systems
(Moving Boundary Work + Closed System Energy Analysis + Closed System Energy Analysis: Ideal
Gases )
Prof. Dr. Ayşegül ABUŞOĞLU
2019 - 2020 Spring Term
Problems
1. A piston-cylinder device initially contains 0.07 m
3
of nitrogen gas at 130 kPa and 120ºC. The nitrogen
is now expanded polytropically to a state of 100 kPa and 100ºC. Determine the boundary work done
during this process.
Solution: Properties The gas constant for nitrogen is 0.2968 kJ/kg.K (Table A-2).
Analysis The mass and volume of nitrogen at the initial state are
- 07802 kg ( 0. 2968 kJ/kg.K)(1 20 273 K)
(130 kPa)(0.07m) 3
1
1 1
RT
P
m
V
3
3
2
2 2 0.^08637 m 100 kPa
( 0. 07802 kg)(0.2968kPa.m /kg.K)(100 273 K)
P
mRT
V
The polytropic index is determined from
(130 kPa)(0.07m ) (100kPa)(0.086 37 m) 1. 249
3 3 P 1 (^) 1 = P 2 2 ⎯⎯→ = ⎯⎯→ n =
n n n n
V V
The boundary work is determined from
= 1.86kJ −
(100kPa)(0.086 37 m) (130kPa)(0.07m)
1
3 3 2 2 11
n
P P
Wb
V V
2. A piston-cylinder device with a set of stops initially contains 0.3 kg of steam at 1.0 MPa and 400ºC.
The location of the stops corresponds to 60 percent of the initial volume. Now the steam is cooled.
Determine the compression work if the final state is (a) 1.0 MPa and 250ºC and (b) 500 kPa. (c) Also
determine the temperature at the final state in part (b).
Solution: Analysis (a) The specific volumes for the initial and final states are (Table A-6)
- 23275 m/kg 250 C
1 MPa
- 30661 m/kg 400 C
1 MPa 3 2 2
3 2 1 1
1
v v
T
P
T
P
Noting that pressure is constant during the process, the boundary work is determined from
= ( − )=( 0. 3 kg)(1000kPa)( 0. 30661 − 0. 23275 )m/kg= 22.16kJ 3
Wb mP v 1 v 2
(b) The volume of the cylinder at the final state is 60% of initial volume. Then, the boundary
work becomes
= ( − 0. 60 )=( 0. 3 kg)(1000kPa)( 0. 30661 − 0. 60 0. 30661 )m/kg= 36.79kJ 3
Wb mP v 1 v 1
The temperature at the final state is
= 151.8 C
3 2 2
2
( 0. 60 0. 30661 )m /kg
- 5 MPa T
P
v
(Table A- 5 )
3. A frictionless piston-cylinder device initially contains 200 L of saturated liquid refrigerant-134a. The
piston is free to move, and its mass is such that it maintains a pressure of 900 kPa on the refrigerant.
The refrigerant is now heated until its temperature rises to 70ºC. Calculate the work done during this
process.
Solution: Assumptions The process is quasi-equilibrium.
Steam 0.3 kg 1 MPa 400 C Q
Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are
(Table A-11 through A-13)
0.027413m/kg 70 C
900 kPa
0.0008580m/kg Sat.liquid
900 kPa
3 2 2
2
3 1 @ 900 kPa
1
v
v v
T
P
P
f
Analysis The boundary work is determined from its definition to be
233.1kg 0.0008580m/kg
0.2m 3
3
1
1 = = =
v
V
m
and
= 5571 kJ
3
3
2 1 2 1
2
1 ,out
1 kPa m
1 kJ (233.1kg)(900kPa)(0.027 413 0.0008580)m/kg
Wb Pd V P (V V) mP (v v)
Discussion The positive sign indicates that work is done by the system (work output).
4. A piston-cylinder device contains 0.15 kg of air initially at 2 MPa and 350 ºC. The air is first expanded
isothermally to 500 kPa, then compressed polytropically with a polytropic exponent of 1.2 to the
initial pressure, and finally compressed at the constant pressure to the initial state. Determine the
boundary work for each process and the net work of the cycle.
Solution: A piston-cylinder device contains air gas at a specified state. The air undergoes a cycle with three processes. The
boundary work for each process and the net work of the cycle are to be determined.
Properties The properties of air are R = 0.287 kJ/kg.K , k = 1.4 (Table A-2a).
Analysis For the isothermal expansion process:
3
1
1 0.^01341 m (2000kPa)
( 0. 15 kg)( 0. 287 kJ/kg.K)(3 50 273 K)
P
mRT
V
3
2
2 0.^05364 m (500kPa)
( 0. 15 kg)( 0. 287 kJ/kg.K)(3 50 273 K)
P
mRT
V
= 37.18kJ
3
3 3
1
2 , 1 2 11
01341 m
05364 m ln (2000kPa)(0.013 41 m)ln
V
V
Wb P V
For the polytropic compression process:
3 3
2 3
2
2 V 2 = 3 V 3 ⎯⎯→(500^ kPa)(0.053^64 m ) =(2000kPa)V ⎯⎯→^ V =^0.^01690 m
n n P P
= -34.86kJ −
(2000kPa)(0.016 90 m) (500kPa)(0.053 64 m)
1
3 3 3 3 2 2 , 23 n
P P
Wb
V V
For the constant pressure compression process:
− = − = − = -6.97kJ
3
Wb , 3 1 P 3 (V 1 V 3 ) (2000kPa)(0.013 41 0.01690)m
The net work for the cycle is the sum of the works for each process
W net = Wb , 1 − 2 + Wb , 2 − 3 + Wb , 3 − 1 = 37. 18 +(− 34. 86 )+(− 6. 97 )= -4.65kJ
5. A well-insulated rigid tank contains 5 kg of a saturated liquid-vapor mixture of water at 100 kPa.
Initially, three-quarters of the mass is in the liquid phase. An electric resistor placed in the tank is
connected to a 110-V source, and a current of 8 A flows through the resistor when the switch is
turned on. Determine how long it will take to vaporize all the liquid in the tank. Also, show the
process on a T-v diagram with respect to saturation lines.
Solution: An insulated rigid tank is initially filled with a saturated liquid-vapor mixture of water. An electric heater in the tank is
turned on, and the entire liquid in the tank is vaporized. The length of time the heater was kept on is to be determined, and the process
is to be shown on a P- v diagram.
Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 The device is well-insulated and
thus heat transfer is negligible. 3 The energy stored in the resistance wires, and the heat transferred to the tank itself is negligible.
Analysis We take the contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the
volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be
expressed as
v
P
(kPa)
Air 2 MPa
350 C
7. Two tanks (Tank A and Tank B) are separated by a partition. Initially Tank A contains 2-kg steam at
1 MPa and 300ºC while Tank B contains 3-kg saturated liquid-vapor mixture with a vapor mass
fraction of 50 percent. Now the partition is removed, and the two sides are allowed to mix until the
mechanical and thermal equilibrium are established. If the pressure at the final state is 300 kPa,
determine (a) the temperature and quality of the steam (if mixture) at the final state and (b) the
amount of heat lost from the tanks.
Solution: Assumptions 1 The tank is stationary and thus the kinetic and potential energy
changes are zero. 2 There are no work interactions.
Analysis ( a ) We take the contents of both tanks as the system. This is a closed system since
no mass enters or leaves. Noting that the volume of the system is constant and thus there is
no boundary work, the energy balance for this stationary closed system can be expressed as
out ^ (^21 )^ ^ ( 2 1 )^ (since KE=PE=0)
potential, etc.energies
Changeininternal,kinetic,
system
by heat,work,andmass
Netenergy transfer
in out
Q U U mu u mu u W
E E E
A B A B
The properties of steam in both tanks at the initial state are (Tables A-4 through A-6)
7 kJ/kg
25799 m/kg
300 C
1000 kPa
1 ,
3 1 ,
1 ,
1 ,
=
A
A
A
A
T u
P v
( )
631 .66 ( 0.50 1927. 4 ) 1595. 4 kJ/kg
0.001091 0.50 0.39248 0.001091 0.19679m/kg
66 , 1927. 4 kJ/kg
001091 , 0 .39248m/kg
50
150 C
1 , 1
3 1 , 1
3
1
1 ,
B f fg
B f fg
f fg
B f g
u u x u
x
x u u
T
v v v
v v
The total volume and total mass of the system are
3 2 5 kg
( 2 kg)( 0. 25799 m /kg) ( 3 kg)( 0. 19679 m/kg) 1. 106 m 3 3 3 1 , 1 ,
= + = + =
A B
A B A A B B
m m m
V V V m v m v
Now, the specific volume at the final state may be determined
22127 m/kg 5 kg
106 m 3 3
2 =^ = = m
V v
which fixes the final state and we can determine other properties
561 .11 ( 0.3641 1982. 1 ) 1282. 8 kJ/kg
- 22127 m/kg
300 kPa
2 2
2 2
2 sat@ 300 kPa
3 2
2
f fg
g f
f
u u x u
x
T T
P
133.5C
v v
v v
v
( b ) Substituting,
( 2 kg)( 1282. 8 2793. 7 )kJ/kg ( 3 kg)( 1282. 8 1595. 4 )kJ/kg 3959 kJ
out (^21 ) ( 2 1 )
= − + − = −
A B A B Q U U mu u mu u
or Q out= 3959 kJ
8. A 4-m × 5 - m × 6 - m room is to be heated by a base-board resistance heater. It is desired that the
resistance heater be able to raise the air temperature in the room from 7 to 23ºC within 15 min.
Assuming no heat losses from the room and an atmospheric pressure of 100 kPa, determine the
required power of the resistance heater. Assume constant specific heats at room temperature.
Solution: Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -
141 C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ke pe 0. 3 Constant specific heats at room
temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. 4 Heat losses
from the room are negligible. 5 The room is air-tight so that no air leaks in and out during the process.
Properties The gas constant of air is R = 0.287 kPa.m 3 /kg.K (Table A-1). Also, c v = 0.718 kJ/kg.K for air at room temperature (Table
A-2).
Analysis We take the air in the room to be the system. This is a closed system since no mass crosses the system boundary. The energy
balance for this stationary constant-volume closed system can be expressed as
e,in ,avg(^21 )(since KE PE^0 )
potential, etc.energies
Changeininternal,kinetic,
system
by heat,work,andmass
Netenergy transfer
in out
W U mc T T Q
E E E
v
or,
W^ e ,in t = mc v,avg( T 2 − T 1 )
The mass of air is
149.3kg (0.287kPa m/kg K)(280K)
(100kPa)(120m)
4 5 6 120 m
3
3
1
1
3
RT
P
m
V
V
Substituting, the power rating of the heater becomes
= 1.91kW
15 60 s
(149.3kg)(0.718kJ/kg C)(23 7)C , in
We
Discussion In practice, the pressure in the room will remain constant during this process rather than the volume, and some air will
leak out as the air expands. As a result, the air in the room will undergo a constant pressure expansion process. Therefore, it is more
proper to be conservative and to use H instead of using U in heating and air-conditioning applications_._
9. An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 4 kg
of an ideal gas at 800 kPa and 50º C, and the other part is evacuated. The partition is now removed,
and the gas expands into the entire tank. Determine the final temperature and pressure in the tank.
Solution: Assumptions 1 The kinetic and potential energy changes are negligible, ke pe 0. 2 The tank is insulated and thus
heat transfer is negligible.
Analysis We take the entire tank as the system. This is a closed system since no mass crosses the boundaries of the system. The
energy balance for this system can be expressed as
E E E
U m u u
u u
in −^ out =
Net energy transfer by heat, work, and mass
system Change in internal, kinetic, potential, etc. energies
2 1
Therefore,
T 2 = T 1 = 50 C
Since u = u ( T ) for an ideal gas. Then,
= ⎯⎯→ = = (800kPa)= 400 kPa 2
1 2
1 2 2
22
1
1 1 P P T
P
T
P
V
V V V
10. A piston-cylinder device contains 0.8 kg of nitrogen initially at 100 kPa and 27ºC. The nitrogen is now
compressed slowly in a polytropic process during which PV
1.
= constant until the volume is reduced
by one-half. Determine the work done and the heat transfer for this process.
Solution: Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The N 2 is an ideal
gas with constant specific heats. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion
process is quasi-equilibrium.
Properties The gas constant of N 2 are R = 0.2968 kPa.m^3 /kg.K (Table A-1). The c v value of N 2 at the average temperature
(369+300)/2 = 335 K is 0.744 kJ/kg.K (Table A-2b).
Analysis We take the contents of the cylinder as the system. This is a closed system since no mass crosses the system boundary. The
energy balance for this closed system can be expressed as
4 5 6 m 3
7 C
AIR
We
e
IDEAL GAS
800 kPa
50 C
Evacuated
