8-42
8-74E The entropy difference between the two states of air is to be determined.
Assumptions Air is an ideal gas with constant specific heats.
Properties The specific heat of air at the average temperature of (90+210)/2=150F is cp = 0.241 Btu/lbmR (Table A-2Eb).
The gas constant of air is R = 0.06855 Btu/lbmR (Table A-2Ea).
RBtu/lbm 0.01973
psia 15
)R460(90
8-75 Oxygen gas is compressed from a specified initial state to a specified final state. The entropy change of oxygen during
this process is to be determined for the case of constant specific heats.
Assumptions At specified conditions, oxygen can be treated as an ideal gas.
Properties The gas constant and molar mass of oxygen are R = 0.2598 kJ/kg.K and M = 32 kg/kmol (Table A-1).
O2
0.8 m3/kg
25C
8-44
8-77 An insulated cylinder initially contains air at a specified state. A resistance heater inside the cylinder is turned on, and
air is heated for 15 min at constant pressure. The entropy change of air during this process is to be determined for the cases
of constant and variable specific heats.
Assumptions At specified conditions, air can be treated as an ideal gas.
 
kJ 180s 6015kJ/s 0.2
ine,ine,
1
tWW
The energy balance for this stationary closed system can be expressed as
)()( 1212ine,outb,ine,
kinetic, internal,in Change
sy stem
nsferenergy traNet
outin
TTchhmWUWW
EEE
p
since U + Wb = H during a constant pressure quasi-equilibrium process.
(a) Using a constant cp value at the anticipated average temperature of 450 K, the final temperature becomes
1
P
AIR
0.3 m3
120 kPa
17C
8-48
8-81 Air is accelerated in an nozzle, and some heat is lost in the process. The exit temperature of air and the total entropy
change during the process are to be determined.
Assumptions 1 Air is an ideal gas with variable specific heats. 2 The nozzle operates steadily.
Analysis (a) Assuming variable specific heats, the inlet properties are determined to be,
h
1
8-49
8-82 Problem 8-81 is reconsidered. The effect of varying the surrounding medium temperature from 10°C to 40°C on
the exit temperature and the total entropy change for this process is to be studied, and the results are to be plotted.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
Function HCal(WorkFluid$, Tx, Px)
“Function to calculate the enthalpy of an ideal gas or real gas”
WorkFluid$ = ‘Air’
T[1] = 77 [C]
P[1] = 280 [kPa]
Vel[1] = 50 [m/s]
P[2] = 85 [kPa]
“If we knew the inlet or exit area, we could calculate the mass flow rate. Since we don’t know these areas, we
write the conservation of energy per unit mass.”
“Conservation of mass: m_dot[1]= m_dot[2]”
“Conservation of Energy – SSSF energy balance for neglecting the change in potential energy, no work, but heat
transfer out is:”
8-50
stotal
[kJ/kg-K]
Tsurr
[C]
T2
[C]
0.189
10
24.22
0.1888
15
24.22
0.1886
20
24.22
0.1884
25
24.22
0.1882
30
24.22
0.188
35
24.22
0.1879
40
24.22
10 15 20 25 30 35 40
0.1878
0.188
0.1882
0.1884
0.1886
0.1888
0.189
Tsurr [C]
Dstotal [kJ/kg–K]
8-51
8-83E A fixed mass of helium undergoes a process from one specified state to another specified state. The entropy change
of helium is to be determined for the cases of reversible and irreversible processes.
Assumptions 1 At specified conditions, helium can be treated as an ideal gas. 2 Helium has constant specific heats at room
temperature.
Properties The gas constant of helium is R = 0.4961 Btu/lbm.R (Table A-1E). The
8-53
8-87E Air is compressed in an isentropic compressor. The outlet temperature and the work input are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 The process is adiabatic, and thus there is
(Table A-2Eb).
Analysis There is only one inlet and one exit, and thus
mmm 21
. We take the compressor as the system, which is a
control volume since mass crosses the boundary. The energy balance for this steady–flow system can be expressed in the
rate form as
outin
energies etc. potential,
kinetic, internal,in change of Rate
(steady) 0
sy stem
mass and work,heat,by
nsferenergy tranet of Rate
outin 0
EE
EEE
 
2in1
hmWhm
Air
compressor
200 psia
15 psia
8-55
8-90 Problem 8-89 is reconsidered. The effect of the final pressure on the final mass in the tank is to be investigated
as the pressure varies from 450 kPa to 150 kPa, and the results are to be plotted.
R=0.2081 [kPa-m^3/kg-K]
P_1= 450 [kPa]
T_1 = 30 [C]
m_1 = 4 [kg]
expressed in the rate form as
0
energies etc. potential,
kinetic, internal,in change of Rate
(steady ) 0
sy stem
mass and work,heat,by
nsferenergy tranet of Rate
outin
EE
EEE
  
8-58
8-94 Air is compressed in a piston-cylinder device in a reversible and isothermal manner. The entropy change of air and the
work done are to be determined.
Assumptions 1 At specified conditions, air can be treated as an ideal gas. 2 The process is specified to be reversible.
Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).
Analysis (a) Noting that the temperature remains constant, the entropy change of air is determined from
lnlnln
1
2
1
2
0
1
2
avg,air P
P
R
P
P
R
T
T
cS p
kJ/kg 125.4
outin
qw
8-60
8-96 Air is expanded adiabatically in a piston-cylinder device. The entropy change is to be determined and it is to be
discussed if this process is realistic.
Assumptions 1 Air is an ideal gas with constant specific heats.
(Table A-2a).
Analysis (a) 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
)(
0)=PE=KE (since )(
12out
12out
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
TTmcW
QuumUW
EEE
v
Solving for the final temperature,
K 9.532
K)kJ/kg 718.0(kg) 5(
kJ 600
K) 273427()( out
1221out
v
vmc
W
TTTTmcW
From the entropy change relation of an ideal gas,
KkJ/kg 0.240
kPa 600
kPa 100
K)lnkJ/kg (0.287
K 700
K 9.532
K)lnkJ/kg (1.005
lnln
1
2
1
2
air P
P
R
T
T
cs p
(b) Since the entropy change is positive for this adiabatic process, the process is irreversible and realistic.