5-24
5-32 A cylinder is initially filled with steam at a specified state. The steam is cooled at constant pressure. The mass of the
steam, the final temperature, and the amount of heat transfer are to be determined, and the process is to be shown on a T-
v
diagram.
Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work
interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The
compression or expansion process is quasi-equilibrium.
Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The
energy balance for this stationary closed system can be expressed as
energies etc. p otential,
kinetic, internal,in Change
sy stem
mass and work,heat,by
nsferenergy traNet
outin
EEE
5-25
5-33 A cylinder is initially filled with saturated liquid water at a specified pressure. The water is heated electrically as it is
stirred by a paddle-wheel at constant pressure. The voltage of the current source is to be determined, and the process is to be
shown on a P-
v
diagram.
Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The cylinder is well-
insulated and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The
compression or expansion process is quasi-equilibrium.
Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The
energy balance for this stationary closed system can be expressed as
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
EEE
5-26
5-34 A cylinder equipped with an external spring is initially filled with steam at a specified state. Heat is transferred to the
steam, and both the temperature and pressure rise. The final temperature, the boundary work done by the steam, and the
amount of heat transfer are to be determined, and the process is to be shown on a P-
v
diagram.
Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The thermal energy
stored in the cylinder itself is negligible. 3 The compression or expansion process is quasi-equilibrium. 4 The spring is a
linear spring.
Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. Noting
that the spring is not part of the system (it is external), the energy balance for this stationary closed system can be expressed
as
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
EEE
kJ/kg 3312.0
/kgm 1.6207
kPa 250
/kgm 1.6207
kg 0.3702
m 0.6
2
2
3
2
2
3
3
2
2
u
T
P
m
C606
v
V
v
(b) The pressure of the gas changes linearly with volume, and thus the process curve on a P-V diagram will be a straight
line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus,
3
21
kJ 1
kPa0)25(200
PP
v
H2O
Q
5-28
10-3 10-2 10-1 100101102
100
101
102
103
104
105
106
v [m3/kg]
P [kPa]
200°C
606°C
Ste am IAPW S
1
2
Area = Wb
5-30
5-37 A room is heated by an electrical radiator containing heating oil. Heat is lost from the room. The time period during
which the heater is on is to be determined.
Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of –
141C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible,
0ΔpeΔke
. 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 The local atmospheric pressure is 100 kPa. 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.m3/kg.K (Table A-1). Also, c
v
= 0.718 kJ/kg.K for air at room
temperature (Table A-2). Oil properties are given to be ρ = 950 kg/m3 and cp = 2.2 kJ/kg.C.
Analysis We take the air in the room and the oil in the radiator 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
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
EEE
10C
Radiator
Room
5-31
5-38 A saturated water mixture contained in a spring-loaded piston-cylinder device is heated until the pressure and volume
rise to specified values. The heat transfer and the work done are to be determined.
Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work
interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The
compression or expansion process is quasi-equilibrium.
Analysis We take the contents of the cylinder as the system. This is a closed
system since no mass enters or leaves. The energy balance for this stationary
closed system can be expressed as
)(
12ou,in
12ou,in
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
uumWQ
EEE
tb
tb
The initial state is saturated mixture at 75 kPa. The specific volume and
internal energy at this state are (Table A-5),
kJ/kg 30.553)8.2111)(08.0(36.384
/kgm 1783.0)001037.02172.2)(08.0(001037.0
1
3
1
fgf
fgf
xuuu
x
vvv
The mass of water is
/kgm 1783.0
m 2
3
3
1
1
v
V
The final specific volume is
kg 22.11
m 5 3
3
2
V
The final state is now fixed. The internal energy at this specific volume and 225 kPa pressure is (Table A-6)
Since this is a linear process, the work done is equal to the area under the process line 1–2:
kJ 450
3
3
12
21
out,mkPa 1
kJ 1
)m2(5
2
)kPa225(75
)(
2
Area
VV
PP
Wb
Substituting into energy balance equation gives
P
v
2
225 kPa
75 kPa
5-33
5-40 One part of an insulated tank contains compressed liquid while the other side is evacuated. The partition is then
removed, and water is allowed to expand into the entire tank. The final temperature and the volume of the tank are to be
determined.
Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 The tank is insulated and
thus heat transfer is negligible. 3 There are no work interactions.
Analysis We take the entire 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
21
12
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
uu
EEE
The properties of water are (Tables A-4 through A-6)
/kgm 0.001017
kPa 600
3
C60@1
1
f
P
vv
02644.0
4.2245
79.19116.251
kJ/kg 2245.4,79.191
/kgm 14.670,001010.0
)(
kPa 10
2
2
3
12
2
fg
f
fgf
gf
u
uu
x
uu
uu
P
vv
Thus,
H2O
Evacuated
Partition
5-35
Specific Heats, u and h of Ideal Gases
5-43C It can be used for any kind of process of an ideal gas.
5-47C For the constant pressure case. This is because the heat transfer to an ideal gas is mcpT at constant pressure, mc
v
T
at constant volume, and cp is always greater than c
v
.
5-48 The desired result is obtained by multiplying the first relation by the molar mass M,
5-49 The enthalpy change of oxygen is to be determined for two cases of specified temperature changes.
Assumptions At specified conditions, oxygen behaves as an ideal gas.
Properties The constant-pressure specific heat of oxygen at room temperature is cp = 0.918 kJ/kgK (Table A-2a).
kJ/kg 91.8K)150K)(250kJ/kg (0.918Tch p
If we use the same room temperature specific heat value, the enthalpy change will be the same for the second case.
However, if we consider the variation of specific heat with temperature and use the specific heat values from Table A-2b,
we have cp = 0.964 kJ/kgK at 200C (= 473 K) and cp = 0.934 kJ/kgK at 100C ( 373 K). Then,
The two results differ from each other by about 3%. The pressure has no influence on the enthalpy of an ideal gas.
5-37
5-53 The internal energy change of hydrogen gas during a heating process is to be determined using an empirical specific
heat relation, constant specific heat at average temperature, and constant specific heat at room temperature.
Analysis (a) Using the empirical relation for
)(Tc p
from Table A-2c and relating it to
)(Tc
v
,
32
kJ/kg 6110
200)KK)(800kJ/kg (10.183)(
KkJ/kg 10.183
12avg,
K 300@avg,
TTcu
cc
v
vv
5-54 The enthalpy change of nitrogen gas during a heating process is to be determined using an empirical specific heat
relation, constant specific heat at average temperature, and constant specific heat at room temperature.
Analysis (a) Using the empirical relation for
)(Tc p
from Table A-2c,
32 dTcTbTacp
kJ/kg415.6 600)KK)(1000kJ/kg (1.039)(
12avg,
TTch
p
5-38
Closed System Energy Analysis: Ideal Gases
5-55C No, it isn’t. This is because the first law relation Q – W =
U reduces to W = 0 in this case since the system is
5-56 The hydrogen gas in a rigid tank is cooled until its temperature drops to 300 K. The final pressure in the tank and the
of –240C and 1.30 MPa. 2 The tank is stationary, and thus the kinetic and potential energy changes are negligible,
 ke pe 0
.
kPa 159.1kPa) (250
K 550
K 350
1
1
2
2
2
2
1
1P
T
T
P
T
P
T
P
VV
(b) We take the hydrogen in the tank as the system. This is a closed system since no mass enters or leaves. The energy
5-40
5-58E The air in a rigid tank is heated until its pressure doubles. The volume of the tank and the amount of heat transfer are
to be determined.
Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of –
221F and 547 psia. 2 The kinetic and potential energy changes are negligible,
 pe ke 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.
Properties The gas constant of air is R = 0.3704 psia.ft3/lbm.R (Table A-1E).
Analysis (a) The volume of the tank can be determined from the ideal gas relation,
3
R) R)(525/lbmftpsia 4lbm)(0.370 (10 3
1
1
P
mRT
(b) We take the air in the tank as our system. The energy balance for
this stationary closed system can be expressed as
)()( 1212in
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
TTmcuumQ
EEE
v
The final temperature of air is
1
2
2
2
1
1T
P
P
T
P
T
P
VV
The internal energies are (Table A-21E)
kJ/kg
48.89
R 525@1
uu
Air
30 psia
65F
Q