Department of Mechanical Engineering
ME 322 – Mechanical Engineering
Thermodynamics
Lecture 29
The Vapor Compression Refrigeration
(VCR) Cycle
Refrigerator used for Cooling
Qin
energy sought
E 

 COPC
energy that costs Wcycle
TH
Observation: E may be >1
TL
2
E  100%
The concept of an efficiency being greater than 100%
makes people uneasy. Therefore, the conversion
efficiency for a refrigerator is called the Cooling
Coefficient of Performance (COPC). A refrigeration
system that is used for cooling is called a refrigerator.
Refrigerator used for Heating
Qout
energy sought
E 

 COPH
energy that costs Wcycle
TH
Observation: E may be >1
TL
3
E  100%
The concept of an efficiency being greater than 100%
makes people uneasy. Therefore, the conversion
efficiency for a refrigerator is called the Heating
Coefficient of Performance (COPH). A refrigeration
system that is used for cooling is called a heat pump.
Analysis of the Carnot Refrigerator
For the Refrigeration cycle …
th  COPR 
TH
COPR,Carnot 
 Qout
Qin
Qin
1


Wcycle Qout  Qin Qout / Qin  1
1
1

/ Qin rev  1 TH / TL  1
COPR,Carnot 
TL
TH  TL
For the Heat Pump cycle …
TL
th  COPH 
COPH,Carnot 
4
Qout
Qout
1


Wcycle Qout  Qin 1  Qin / Qout
1
1

1   Qin / Qout rev 1  TL / TH
COPH,Carnot 
TH
TH  TL
Terminology
• Refrigeration cycle
– The cycle is operating in a refrigeration cycle
• The goal is to keep the cold space cold
– Transfer heat from a low-temperature source to the cycle
• Heat pump cycle
– The cycle is operating in a refrigeration cycle
• The goal is to keep the hot space hot
– Transfer heat to a high-temperature sink from the cycle
The words refrigeration or heat pump define the goal of the cycle.
5
The Vapor Compression Refrigeration Cycle
TH
TL
6
The Vapor Compression Refrigeration Cycle
Working fluid = Refrigerant
• Two phase changes
– Boiling (evaporator)
– Condensing (condenser)
• Low temperature boiling fluids
7
Refrigerants
Most refrigerants are halogenated hydrocarbons. The naming
convention adopted by ASHRAE is,
R(a-1)(b+1)d = CaHbClcFd
c = 2(a – 1) – b – d
Example: R22 (R022)
a 1  0  a  1
b 1  2  b  1
d 2
c  2  a  1  b  d
c  2 1  1  1  2  1
8
H
F
C
Cl
F
chlorodifluoromethane
The Pressure-Enthalpy Diagram
9
Vapor Compression Refrigeration Analysis
 m  h 2  h3 
Performance
COPC 
 m  h 2  h1 
h3  h 4
 m  h1  h 4 
10
Qin
Wc
The Ideal VCR Cycle on the P-h Diagram
P
1-2:
2-3:
3-4:
4-1:
qout  Qout / m
3
4
Isentropic compression
Isobaric heat rejection
Isenthalpic expansion
Isobaric heat addition
 m  h 2  h3 
2
 m  h 2  h1 
h3  h 4
1
qin  Qin / m wc  Wc / m
h
11
 m  h1  h 4 
Refrigeration Effect and Capacity
Refrigeration Effect:
qin  h1  h 4
Refrigeration Capacity: Q in  m  h1  h 4 
Refrigeration capacity is often expressed in tons of
refrigeration. Definition …
1 ton of refrigeration is the steady state heat transfer rate
required to melt 1 ton of ice at 32°F in 24 hours.
1 ton = 12,000 Btu/hr = 3.516 kW
12
VCR Cycle Irreversibilities
P
Pressure drop through
the condenser
Pressure drop through
the evaporator
2s
2
3
Isentropic efficiency
of the compressor
4
1
h
13
Practical VCR Cycle
P
T3 SCT
SCT = Saturated Condensing Temperature
DSC = Degrees of Subcooling = SCT – T3
2s 2
Subcooling increases the
refrigeration capacity
3
1
Superheating provides a dry
vapor at the compressor inlet
4
T1
SET
SET = Saturated Evaporating Temperature
DSH = Degrees of Superheat = T1 – SET
h
14
Example
Given: A vapor compression refrigeration cycle is operating
with a saturated evaporating temperature of -20°F and a
saturated condensing temperature of 80°F. The refrigerant,
R22, leaves the condenser as a saturated liquid and enters
the compressor with 5 degrees of superheat. The pressure
drops through the evaporator and condenser can be
considered negligible. The compressor has an isentropic
efficiency of 85%. The cycle has a capacity of 15 tons.
Find: (a) the mass flow rate of the R22 (lbm/hr)
(b) the power requirement of the compressor (hp)
(c) the coefficient of performance of the cycle
15
Example
P
T3  80F
3
T1  SET  5F
4
SCT  80F
SET  20F
2s
2
1
T4  20F
 15 tons
T1  SET  5F
h
16
T3  80F
Example
T1  SET  5F
T4  20F
 15 tons
P
3
4
SCT  80F
SET  20F
2s
2
1
T1  SET  5F
h
17
T3  80F
Example
Strategy: Build the property table
then do the thermodynamics!
T1  SET  5F
The high and low pressures in the cycle
can be found since the saturation
temperatures are given.
T4  20F
 15 tons
P
3
4
SCT  80F
SET  20F
2s
2
1
The pressures at all four states are known!
T1  SET  5F
h
18
T3  80F
Example
T1  SET  5F
T4  20F
 15 tons
P
3
4
SCT  80F
SET  20F
2s
2
1
T1  SET  5F
h
19
T3  80F
Example
The property table is complete!
T1  SET  5F
T4  20F
 15 tons
P
3
4
SCT  80F
SET  20F
2s
2
1
T1  SET  5F
h
20
T3  80F
Example
EES Solution (Key Variables)
T1  SET  5F
T4  20F
 15 tons
P
Comparison (same units) …
3
4
SCT  80F
SET  20F
2s
Q in  15 tons 
2
2545 Btu
Btu
Wc   24.45 hp 
 62, 225
hp-hr
hr
1
T1  SET  5F
h
21
12, 000 Btu
Btu
 180, 000
ton-hr
hr