Refrigeration cycles

11.119
A refrigerator with R-12 as the working fluid has a minimum temperature of
−10°C and a maximum pressure of 1 MPa. Assume an ideal refrigeration cycle as
in Fig. 11.24. Find the specific heat transfer from the cold space and that to the
hot space, and the coefficient of performance.
Solution:
Exit evaporator sat. vapor −10°C from B.3.1: h1 = 183.19, s1 = 0.7019 kJ/kgK
Exit condenser sat. liquid 1 MPa from B.3.1: h3 = 76.22 kJ/kg
Compressor: s2 = s1 & P2 from B.3.2 ⇒ h2 ≈ 210.1 kJ/kg
Evaporator: qL = h1 - h4 = h1 - h3 = 183.19 - 76.22 = 107 kJ/kg
Condenser: qH = h2 - h3 = 210.1 - 76.22 = 133.9 kJ/kg
COP: β = qL/wc = qL/(qH - qL) = 3.98

T
Ideal refrigeration cycle 2
Pcond = P3= P2 = 1 MPa
3
Tevap = -10oC = T1

Properties from Table B.3 4 1

s
11.120
Consider an ideal refrigeration cycle that has a condenser temperature of 45°C
and an evaporator temperature of −15°C. Determine the coefficient of
performance of this refrigerator for the working fluids R-12 and R-22.
Solution:
T
Ideal refrigeration cycle
2
Tcond = 45oC = T3
3
Tevap = -15oC = T1

4 1
s

Property for: R-12, B.3 R-22, B.4

Compressor h1, kJ/kg 180.97 244.13
s2 = s1, kJ/kg K 0.7051 0.9505
P2, MPa 1.0843 1.729
T2, oC 54.7 74.4
h2, kJ/kg 212.63 289.26
wC = h2 - h1 31.66 45.13
Exp. valve h3 = h4, kJ/kg 79.71 100.98
Evaporator qL = h1 - h4 101.26 143.15
β = qL/wC 3.198 3.172

The value of h2 is taken from the computer program as it otherwise will be

a double interpolation due to the value of P2.
11.121
The environmentally safe refrigerant R-134a is one of the replacements for R-12
in refrigeration systems. Repeat Problem 11.120 using R-134a and compare the
result with that for R-12.
Consider an ideal refrigeration cycle that has a condenser temperature of 45°C
and an evaporator temperature of −15°C. Determine the coefficient of
performance of this refrigerator for the working fluids R-12 and R-22.
Solution:
T
Ideal refrigeration cycle 2
Tcond = 45oC = T3 3
Tevap = -15oC = T1
4 1
s

Property for: R-12, B.3 R-134a, B.5

Compressor h1, kJ/kg 180.97 389.2
s2 = s1, kJ/kg K 0.7051 1.7354
P2, MPa 1.0843 1.16
T2, oC 54.7 51.8*
h2, kJ/kg 212.63 429.9*
wC = h2 - h1 31.66 40.7
Exp. valve h3 = h4, kJ/kg 79.71 264.11
Evaporator qL = h1 - h4 101.26 125.1
β = qL/wC 3.198 3.07

* To get state 2 an interpolation is needed:

At 1 MPa, s = 1.7354 : T = 45.9 and h = 426.8 kJ/kg
At 1.2 MPa, s = 1.7354 : T = 53.3 and h = 430.7 kJ/kg
make a linear interpolation to get properties at 1.16 MPa
11.122
A refrigerator using R-22 is powered by a small natural gas fired heat engine with
a thermal efficiency of 25%, as shown in Fig.P11.122. The R-22 condenses at
40°C and it evaporates at −20°C and the cycle is standard. Find the two specific
heat transfers in the refrigeration cycle. What is the overall coefficient of
performance as QL/Q1?
Solution:
Evaporator: Inlet State is saturated liq-vap with h4 = h3 =94.27 kJ/kg
The exit state is saturated vapor with h1 = 242.06 kJ/kg
qL = h1 - h4 = h1 - h3 = 147.79 kJ/kg
Compressor: Inlet State 1 and Exit State 2 about 1.6 MPa
wC = h2 - h1 ; s2 = s1 = 0.9593 kJ/kgK
2: T2 ≈ 70°C h2 = 287.2 kJ/kg
wC = h2 - h1 = 45.14 kJ/kg
Condenser: Brings it to saturated liquid at state 3
qH = h2 - h3 = 287.2 - 94.27 = 192.9 kJ/kg
Overall Refrigerator:
β = qL / wC = 147.79 / 45.14 = 3.274
Heat Engine:
. . . .
WHE = ηHEQ1 = WC = QL / β
. .
QL / Q1 = ηβ = 0.25 × 3.274 = 0.819

T
Ideal refrigeration cycle 2
Tcond = 40oC = T3 3
Tevap = -20oC = T1
Properties from Table B.4
4 1
s
11.123
A refrigerator in a meat warehouse must keep a low temperature of -15°C and the
outside temperature is 20°C. It uses R-12 as the refrigerant which must remove 5
kW from the cold space. Find the flow rate of the R-12 needed assuming a
standard vapor compression refrigeration cycle with a condenser at 20°C.
Solution:
Basic refrigeration cycle: T1 = T4 = -15°C, T3 = 20°C
Table B.3: h4 = h3 = 54.87 kJ/kg; h1 = hg = 180.97 kJ/kg
. . .
QL = mR-12 × qL = mR-12(h1 - h4)
qL = 180.97 - 54.87 = 126.1 kJ/kg
.
mR-12 = 5.0 / 126.1 = 0.03965 kg/s

T
Ideal refrigeration cycle 2
Tcond = 20oC 3
Tevap = -15oC = T1

Properties from Table B.3 4 1

s
11.124
A refrigerator with R-12 as the working fluid has a minimum temperature of
−10°C and a maximum pressure of 1 MPa. The actual adiabatic compressor exit
temperature is 60°C. Assume no pressure loss in the heat exchangers. Find the
specific heat transfer from the cold space and that to the hot space, the coefficient
of performance and the isentropic efficiency of the compressor.
Solution:
State 1: Inlet to compressor, sat. vapor -10°C,
h1 = 183.19 kJ/kg, s1 = 0.7019 kJ/kg K
State 2: Actual compressor exit, h2AC = 217.97 kJ/kg
State 3: Exit condenser, sat. liquid 1MPa, h3 = 76.22 kJ/kg
State 4: Exit valve, h4 = h3
C.V. Evaporator: qL = h1 - h4 = h1 - h3 = 107 kJ/kg
C.V. Ideal Compressor: wC,S = h2,S - h1, s2,S = s1
State 2s: 1 MPa, s = 0.7019 kJ/kg K; T2,S = 49.66°C, h2,S = 210.1 kJ/kg
wC,S = h2,S - h1 = 26.91 kJ/kg
C.V. Actual Compressor: wC = h2,AC - h1 = 34.78 kJ/kg
qL
β= = 3.076, ηC = wC,S/wC = 0.774
wC
C.V. Condenser: qH = h2,AC - h3 = 141.75 kJ/kg

Ideal refrigeration cycle T 2ac

with actual compressor 2s
Pcond = P3= P2 = 1 MPa
3
T2 = 60oC
Tevap = -10oC = T1
4 1
Properties from Table B.3 s
11.125
Consider an ideal heat pump that has a condenser temperature of 50°C and an
evaporator temperature of 0°C. Determine the coefficient of performance of this
heat pump for the working fluids R-12, R-22, and ammonia.
Solution:
T
Ideal heat pump 2
Tcond = 50oC = T3 3
Tevap = 0oC = T1
4 1
s

C.V. Property for: R-12 R-22 NH3

From Table: B.3 B.4 B.2
h1, kJ/kg 187.53 249.95 1442.32
Compressor s2 = s1, kJ/kgK 0.6965 0.9269 5.3313
P2, MPa 1.2193 1.9423 2.0333
T2, oC 56.7 72.2 115.6
h2, kJ/kg 211.95 284.25 1672.84
wC = h2 - h1 24.42 34.3 230.52
Exp. valve h3 = h4, kJ/kg 84.94 107.85 421.58
Condenser qH = h2 - h 3 127.01 176.4 1251.26
β′ =qH/wC 5.201 5.143 5.428
11.126
The air conditioner in a car uses R-134a and the compressor power input is 1.5 kW
bringing the R-134a from 201.7 kPa to 1200 kPa by compression. The cold space is
a heat exchanger that cools atmospheric air from the outside 30°C down to 10°C
and blows it into the car. What is the mass flow rate of the R-134a and what is the
low temperature heat transfer rate. How much is the mass flow rate of air at 10°C?
Standard Refrigeration Cycle
Table B.5: h1 = 392.28 kJ/kg; s1 = 1.7319 kJ/kg K; h4 = h3 = 266
C.V. Compressor (assume ideal)
. .
m1 = m2 wC = h2 - h1; s2 = s1 + sgen
P2, s = s1 => h2 = 429.5 kJ/kg => wC = 37.2 kJ/kg
. . .
m wC = WC => m = 1.5 / 37.2 = 0.0403 kg/s
C.V. Evaporator
. .
QL = m(h1 - h4) = 0.0405(392.28 - 266) = 5.21 kW
C.V. Air Cooler
. . .
mair∆hair = QL ≈ mairCp∆T
. .
mair = QL / (Cp∆T) = 5.21 / (1.004×20) = 0.26 kg / s

T
Ideal refrigeration cycle 2
3
Pcond = 1200 kPa = P3
Pevap = 201.7 kPa = P1
4 1
s
11.127
A refrigerator using R-134a is located in a 20°C room. Consider the cycle to be
ideal, except that the compressor is neither adiabatic nor reversible. Saturated
vapor at -20°C enters the compressor, and the R-134a exits the compressor at
50°C. The condenser temperature is 40°C. The mass flow rate of refrigerant
around the cycle is 0.2 kg/s, and the coefficient of performance is measured and
found to be 2.3. Find the power input to the compressor and the rate of entropy
generation in the compressor process.
Solution:
Table B.5: P2 = P3 = Psat 40C = 1017 kPa, h4 = h3 = 256.54 kJ/kg
s2 ≈ 1.7472 kJ/kg K, h2 ≈ 430.87 kJ/kg;
s1 = 1.7395 kJ/kg K, h1 = 386.08 kJ/kg
β = qL / wC -> wC = qL / β = (h1- h4) / β = (386.08 - 256.54) / 2.3 = 56.32
. .
WC = m wC = 11.26 kW
C.V. Compressor h1 + wC + q = h2 ->
qin = h2 - h1 - wC = 430.87 - 386.08 - 56.32 = -11.53 kJ/kg i.e. a heat loss
s1 + ∫ dQ/T + sgen = s2
sgen = s2 - s1 - q / To = 1.7472 - 1.7395 + (11.53 / 293.15) = 0.047 kJ/kg K
. .
Sgen = m sgen = 0.2 × 0.047 = 0.0094 kW / K

Ideal refrigeration cycle T 2ac

with actual compressor 2s
Tcond = 40oC 3
T2 = 50oC
Tevap = -20oC = T1 4 1
Properties from Table B.5 s
11.128
A refrigerator has a steady flow of R-22 as saturated vapor at –20°C into the
adiabatic compressor that brings it to 1000 kPa. After the compressor, the
temperature is measured to be 60°C. Find the actual compressor work and the
actual cycle coefficient of performance.
Solution:
Table B.4.1: h1 = 242.06 kJ/kg, s1 = 0.9593 kJ/kg K
P2 = P3 = 1000 kPa, h4 = h3 = hf = 72.86 kJ/kg
h2 ac = 286.97 kJ/kg
C.V. Compressor (actual)
Energy Eq.: wC ac = h2 ac - h1 = 286.97 – 242.06 = 44.91 kJ/kg
C.V. Evaporator
Energy Eq.: qL = h1- h4 = h1- h3 = 242.06 – 72.86 = 169.2 kJ/kg
qL 169.2
β= = = 3.77
wC ac 44.91

Ideal refrigeration cycle T 2ac

with actual compressor 2s
Tcond = 23.4oC = Tsat 1000 kPa 3
T2 = 60oC
Tevap = -20oC = T1 4 1
Properties from Table B.4 s
11.129
A small heat pump unit is used to heat water for a hot-water supply. Assume that
the unit uses R-22 and operates on the ideal refrigeration cycle. The evaporator
temperature is 15°C and the condenser temperature is 60°C. If the amount of hot
water needed is 0.1 kg/s, determine the amount of energy saved by using the heat
pump instead of directly heating the water from 15 to 60°C.
Solution:

Ideal R-22 heat pump T

2
o o
T1 = 15 C, T3 = 60 C 3
From Table B.4.1
h1 = 255.02 kJ/kg, s2 = s1 = 0.9062 kJ/kg K
P2 = P3 = 2.427 MPa, h3 = 122.18 kJ/kg 4 1
s

Entropy compressor: s2 = s1 => T2 = 78.4oC, h2 = 282.86 kJ/kg

Energy eq. compressor: wC = h2 - h1 = 27.84 kJ/kg
Energy condenser: qH = h2 - h3 = 160.68 kJ/kg
To heat 0.1 kg/s of water from 15oC to 60oC,
. .
QH2O = m(∆h) = 0.1(251.11 - 62.98) = 18.81 kW
Using the heat pump
. .
WIN = QH2O(wC/qH) = 18.81(27.84/160.68) = 3.26 kW
a saving of 15.55 kW
11.130
The refrigerant R-22 is used as the working fluid in a conventional heat pump
cycle. Saturated vapor enters the compressor of this unit at 10°C; its exit
temperature from the compressor is measured and found to be 85°C. If the
compressor exit is at 2 MPa what is the compressor isentropic efficiency and the
cycle COP?
Solution:
R-22 heat pump: T 2
Table B.4 2s
State 1: TEVAP = 10oC, x = 1 3
h1 = 253.42 kJ/kg, s1 = 0.9129 kJ/kg K

State 2: T2, P2: h2 = 295.17 kJ/kg 4 1

s
C.V. Compressor
Energy Eq.: wC ac = h2 - h1 = 295.17 – 253.42 = 41.75 kJ/kg
State 2s: 2 MPa , s2S = s1 = 0.9129 kJ/kg T2S = 69oC, h2S = 280.2 kJ/kg
wC s h2S - h1 280.2 - 253.42
Efficiency: η= = = = 0.6414
wC ac h2 - h1 295.17 - 253.42
C.V. Condenser
Energy Eq.: qH = h2 - h3 = 295.17 – 109.6 = 185.57 kJ/kg
qH 185.57
COP Heat pump: β= = = 4.44
wC ac 41.75
11.131
A refrigerator in a laboratory uses R-22 as the working substance. The high
pressure is 1200 kPa, the low pressure is 201 kPa, and the compressor is
reversible. It should remove 500 W from a specimen currently at –20°C (not
equal to T in the cycle) that is inside the refrigerated space. Find the cycle COP
and the electrical power required.
Solution:
State 1: 201 kPa, x = 1, Table B.4.1: h1 = 239.92 kJ/kg, s1 = 0.9685 kJ/kg K
State 3: 1200 kPa, x = 0, Table B.4.1: h3 = 81.57 kJ/kg
C.V. Compressor
Energy Eq.: wC = h2 - h1
Entropy Eq.: s2 = s1 + sgen = s1
State 2: 1.2 MPa , s2 = s1 = 0.9685 kJ/kg, T2 ≈ 60oC, h2 = 285.21 kJ/kg
wC = h2 - h1 = 285.21 – 239.92 = 45.29 kJ/kg
Energy Eq. evaporator: qL = h1 – h4 = h1 – h3 = 239.92 – 81.57 = 158.35 kJ/kg
qL 158.35
COP Refrigerator: β= = = 3.5
wC 45.29
. .
Power: WIN = QL / β = 500 W/ 3.5 = 142.9 W
11.132
Consider the previous problem and find the two rates of entropy generation in the
process and where they occur.
Solution:
From the basic cycle we know that entropy is generated in the valve as the throttle
process is irreversible.
State 1: 201 kPa, x = 1, Table B.4.1: h1 = 239.92 kJ/kg, s1 = 0.9685 kJ/kg K
State 3: 1200 kPa, x = 0, Table B.4.1: h3 = 81.57 kJ/kg, s3 = 0.30142 kJ/kg K
Energy Eq. evaporator: qL = h1 – h4 = h1 – h3 = 239.92 – 81.57 = 158.35 kJ/kg
. .
Mass flow rate: m = QL / qL = 0.5 / 158.35 = 0.00316 kg/s
C.V. Valve
Energy Eq.: h4 = h3 = 81.57 kJ/kg => x4 = (h4 – hf)/hfg
81.57 - 16.19
x4 = = 0.29223
223.73
s4 = sf + x4 sfg = 0.067 + x4 × 0.9015 = 0.33045 kJ/kg K
Entropy Eq.: sgen = s4 - s3 = 0.33045 – 0.30142 = 0.02903 kJ/kg K
. .
Sgen valve = msgen = 0.00316 × 0.02903 = 0.0917 W/K

There is also entropy generation in the heat transfer process from the specimen at
–20°C to the refrigerant T = -25°C = Tsat (201 kPa).
. . 1 1 1 1
Sgen inside = QL [ – ] = 500 ( – ) = 0.04 W/K
Tspecimen TL 248 253
11.133
In an actual refrigeration cycle using R-12 as the working fluid, the refrigerant flow
rate is 0.05 kg/s. Vapor enters the compressor at 150 kPa, −10°C, and leaves at 1.2
MPa, 75°C. The power input to the compressor is measured and found be 2.4 kW.
The refrigerant enters the expansion valve at 1.15 MPa, 40°C, and leaves the
evaporator at 175 kPa, −15°C. Determine the entropy generation in the compression
process, the refrigeration capacity and the coefficient of performance for this cycle.
Solution:
Actual refrigeration cycle
T 2
1: compressor inlet T1 = -10oC, P1 = 150 kPa
2: compressor exit T2 = 75oC, P2 = 1.2 MPa 3
3: Expansion valve inlet T3 = 40oC 5
P3 = 1.15 MPa 4 1

5: evaporator exit T5 = -15oC, P5 = 175 kPa s

Table B.3 h1 = 184.619, s1 = 0.7318, h2 = 226.543, s2 = 0.7404
CV Compressor: h1 + qCOMP + wCOMP = h2 ; s1 + ∫ dq/T + sgen = s2
. .
wCOMP = WCOMP/m = 2.4/0.05 = 48.0 kJ/kg
qCOMP = h2 - wCOMP - h1 = 226.5 - 48.0 - 184.6 = -6.1 kJ/kg
sgen = s2 - s1 - q / To = 0.7404 - 0.7318 + 6.1/298.15 = 0.029 kJ / kg K
C.V. Evaporator
qL = h5 - h4 = 181.024 - 74.527 = 106.5 kJ/kg
. .
⇒ QL = mqL = 0.05 × 106.5 = 5.325 kW
COP: β = qL/wCOMP = 106.5/48.0 = 2.219