8-138
8-179 Problem 8-178 is reconsidered. The isentropic efficiencies for the compressor and turbine are to be determined,
and then the effect of varying the compressor efficiency over the range 0.6 to 0.8 and the turbine efficiency over the range
0.7 to 0.95 on the net work for the cycle and the entropy generated for the process is to be investigated. The net work is to
be plotted as a function of the compressor efficiency for turbine efficiencies of 0.7, 0.8, and 0.9.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
“Input Data”
m_dot_air = 10 [kg/s] “air compressor (air) data”
“Conservation of mass for the compressor m_dot_air_in = m_dot_air_out =m_dot_air”
“Conservation of energy for the compressor is:”
E_dot_comp_in – E_dot_comp_out = DELTAE_dot_comp
DELTAE_dot_comp = 0 “Steady flow requirement”
E_dot_comp_in=m_dot_air*(enthalpy(air,T=T_air[1])) + W_dot_comp_in
“Conservation of energy for the turbine is:”
E_dot_turb_in – E_dot_turb_out = DELTAE_dot_turb
DELTAE_dot_turb = 0 “Steady flow requirement”
E_dot_turb_in=m_dot_st*h_st[1]
h_st[1]=enthalpy(steam,T=T_st[1], P=P_st[1])
calculated at P_st[2], h_st[2] and not P_st[2] and x_st[2]”
s_st[2]=entropy(steam,P=P_st[2],h=h_st[2])
T_st[2]=temperature(steam,P=P_st[2], h=h_st[2])
s_st_isen[2]=s_st[1]
“Net work done by the process:”
8-139
“To generate the data for Plot Window 1, Comment out the line ‘ T_air[2]=(700-273) C’
and select values for Eta_comp in the Parmetric Table, then press F3 to solve the table.
8-141
8-181 Air is expanded by an adiabatic turbine with an isentropic efficiency of 85%. The outlet temperature, the work
produced, and the entropy generation are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are
negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Air is an ideal gas with constant specific heats.
Properties The properties of air at the anticipated average temperature of 400
8-143
(b) The total entropy generation during this process is determined by applying the entropy balance on an extended system
that includes both tanks and their immediate surroundings so that the boundary temperature of the extended system is the
8-145
8-184E A rigid tank initially contains saturated R-134a vapor. The tank is connected to a supply line, and is charged until
the tank contains saturated liquid at a specified pressure. The mass of R-134a that entered the tank, the heat transfer with the
surroundings at 100F, and the entropy generated during this process are to be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can
be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies
are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified).
Properties The properties of R-134a are (Tables A-11 through A-13)
RBtu/lbm 0.07934
Btu/lbm 38.17
F80
psia 140
RBtu/lbm 0.07879
Btu/lbm .6237
/lbmft 0.01332
liquid sat.
psia 100
RBtu/lbm 0.22132
Btu/lbm 101.31
/lbmft 0.79462
vapor sat.
psia 06
F80@
F80@
psia 100@2
psia 100@2
3
psia 100@2
2
psia 60@1
psia 60@1
3
psia 60@1
1
fi
fi
i
i
f
f
f
g
g
g
ss
hh
T
P
ss
uu
P
ss
uu
P
vv
vv
Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the
energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and
energy balances for this uniform-flow system can be expressed as
Mass balance:
12systemoutin mmmmmm i
Energy balance:
)0peke (since
1122in
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
WumumhmQ
EEE
ii
140 psia
80F
R-134a
5 ft3
R-134a
Q
100F
8-147
8-186E A steel container that is filled with hot water is allowed to cool to the ambient temperature. The total entropy
generated during this process is to be determined.
Assumptions 1 Both the water and the steel tank are incompressible substances with constant specific heats at room
temperature. 2 The system is stationary and thus the kinetic and potential energy changes are zero. 3 Specific heat of iron
can be used for steel. 4 There are no work interactions involved.
8-149
8-188 One ton of liquid water at 80°C is brought into a room. The final equilibrium temperature in the room and the
entropy change during this process are to be determined.
Assumptions 1 The room is well insulated and well sealed. 2 The thermal properties of water and air are constant at room
temperature. 3 The system is stationary and thus the kinetic and potential energy changes are zero. 4 There are no work
interactions involved.
m 140kPa 100
3
11
P
V
airwater
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin 0 UUUEEE
Tf = 78.4C
(b) Considering that the system is well-insulated and no mass is entering and leaving, the total entropy change during this
KkJ 18.99
K 353
K 351.4
ln KkJ/kg 4.18kg 1000ln
KkJ .7820
K 295
K 351.4
ln KkJ/kg 0.718kg 165.4lnln
1
2
water
0
1
2
1
2
air
/
/
T
T
mcS
mR
T
T
mcS
V
V
v
100 kPa
Water
Heat
8-151
(b) The entropy change of the steam is
kJ/K 0.8013 K 283
K 285.3
ln KkJ/kg 1.005kg 98.5lnln
0
1
2
1
2
air
P
P
mR
T
T
mcSp
airsteamgen SSS
8-152
8-190 An insulated cylinder initially contains a saturated liquid-vapor mixture of water at a specified temperature. The
entire vapor in the cylinder is to be condensed isothermally by adding ice inside the cylinder. The amount of ice added and
the entropy generation are to be determined.
Assumptions 1 Thermal properties of the ice are constant. 2 The cylinder is well-insulated and thus heat transfer is
negligible. 3 There is no stirring by hand or a mechanical device (it will add energy).
fusion of ice at 1 atm are 0C and 333.7 kJ/kg.
Analysis (a) We take the contents of the cylinder (ice and saturated water) as our system, which is a closed system. Noting
that the temperature and thus the pressure remains constant during this phase change process and thus Wb + U = H, the
energy balance for this system can be written as
0
0
waterice
,
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
HH
H
UW
EEE
inb
outin
or
0)]([])C0()C0([ water12iceliquid2solid1 hhmTmcmhTmc if
ice
–18C
Water
0.02 m3
100C
