9-122
9-154 A vapor-compression refrigeration cycle with refrigerant-22 as the working fluid is considered. The hardware and the
T-s diagram for this air conditioner are to be sketched. The heat absorbed by the refrigerant, the work input to the
compressor and the heat rejected in the condenser are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) In this normal vapor-compression refrigeration cycle, the refrigerant enters the compressor as a saturated vapor
at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure.
2
·
·
(b) The properties as given in the problem statement are
h4 = h3 = hf @ 45C = 101 kJ/kg
1
4
qL
-5C
1
4
s
·
4s
9-124
9-158 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is considered. The power input
to the heat pump is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the
9-125
9-159 A geothermal heat pump is considered. The degrees of subcooling done on the refrigerant in the condenser, the mass
flow rate of the refrigerant, the heating load, the COP of the heat pump, the minimum power input are to be determined.
9-126
9-160 An actual heat pump cycle with R-134a as the refrigerant is considered. The isentropic efficiency of the compressor,
the rate of heat supplied to the heated room, the COP of the heat pump, and the COP and the rate of heat supplied to the
heated room if this heat pump operated on the ideal vapor-compression cycle between the same pressure limits are to be
determined.
28.277
kPa 800
kJ/kg 9507.0
kJ/kg 88.247
C)409.10(
kPa 200
C09.10
kJ/kg 93.87
kJ/kg 93.87
C)306.29(
kPa 750
C06.29
kJ/kg 71.286
C50
kPa 800
2
12
2
1
1
1
1
kPa sat@200
34
3
3
3
kPa sat@7503
2
2
2
s
h
ss
P
s
h
T
P
T
hh
h
T
P
TT
h
T
P
.
QH
750 kPa
Condenser
Evaporator
Compressor
Expansion
valve
800 kPa
50C
QL
Win
1
2
3
4
.
.
9-127
9-161E A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is considered. The power
input to the heat pump and the electric power saved by using a heat pump instead of a resistance heater are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the
throttlingBtu/lbm 79.41
Btu/lbm 79.41
liquid sat.
psia 120
Btu/lbm 64.116
psia 120
RBtu/lbm 22192.0
Btu/lbm 83.108
vapor sat.
psia 50
34
psia 120 @ 3
3
2
12
2
psia 50 @ 1
psia 50 @ 1
1
hh
hh
P
h
ss
P
ss
hh
P
f
g
g
The mass flow rate of the refrigerant and the power input to the
compressor are determined from
lbm/s 0.2227
Btu/lbm 41.79116.64
Btu/s 060,000/360
32
hh
Q
q
Q
mH
H
H
QH
QL
50 psia
1
2
3
4
120 psia
s
T
·
Win
·
·
House
9-128
9-162 A heat pump with refrigerant-134a as the working fluid heats a house by using underground water as the heat source.
The power input to the heat pump, the rate of heat absorption from the water, and the increase in electric power input if an
electric resistance heater is used instead of a heat pump are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
9-129
9-163 Problem 9-162 is reconsidered. The effect of the compressor isentropic efficiency on the power input to the
compressor and the electric power saved by using a heat pump rather than electric resistance heating is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
“Input Data”
“Input Data is supplied in the diagram window”
s[1]=entropy(Fluid$,P=P[1],T=T[1])
h2s=enthalpy(Fluid$,P=P[2],s=s[1]) “Identifies state 2s as isentropic”
h[1]+Wcs=h2s “energy balance on isentropic compressor”
Wc=Wcs/Eta_c“definition of compressor isentropic efficiency”
h[1]+Wc=h[2] “energy balance on real compressor-assumed adiabatic”
“Throttle Valve”
h[4]=h[3] “energy balance on throttle – isenthalpic”
x[4]=quality(Fluid$,h=h[4],P=P[4]) “properties for state 4″
s[4]=entropy(Fluid$,h=h[4],P=P[4])
T[4]=temperature(Fluid$,h=h[4],P=P[4])
9-130
Win [kW]
c
Esaved
3.671
0.6
13
3.249
0.7
13.42
2.914
0.8
13.75
2.641
0.9
14.03
2.415
1
14.25
0,6 0,65 0,7 0,75 0,8 0,85 0,9 0,95 1
2,4
2,6
2,8
3
3,2
3,4
3,6
3,8
c
Win
0,6 0,65 0,7 0,75 0,8 0,85 0,9 0,95 1
12,75
13,05
13,35
13,65
13,95
14,25
c
Esaved [kW]
9-164 A Carnot cycle executed in a closed system uses air as the working fluid. The net work output per cycle is to be
determined.
9-136
9-169E A simple ideal Brayton cycle with air as the working fluid operates between the specified temperature limits. The
net work is to be determined using constant and variable specific heats.
Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential
Analysis (a) Constant specific heats:
R 8.717
12
1
R) 1460(
10.4/1.4
/)1(
34
kk
p
r
TT
)()(
1243
compturbnet
TTcTTc
www
pp
9182.0
Btu/lbm 69.114
R 480
1
1
1
r
P
h
T
Btu/lbm 63.23302.11)9182.0)(12(21
1
2
2 hP
P
P
Prr
Btu/lbm 63.358
3
h
Btu/lbm 63.4
)69.11463.233()32.17663.358(
T
1
2
4
3
qin
qout
1460 R
480 R
9-138
9-172 A spark-ignition engine operates on an ideal Otto cycle with a compression ratio of 11. The maximum temperature
and pressure in the cycle, the net work per cycle and per cylinder, the thermal efficiency, the mean effective pressure, and
the power output for a specified engine speed are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is
an ideal gas with variable specific heats.
1m 000495.000045.0000045.0 dc
kg 0.0004805
K 323K/kgmkPa 0.287
)m 495kPa)(0.000 90(
3
3
1
11
RT
P
m
V
For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of
temperature only whereas entropy is functions of both temperature and pressure.