BITS Pilani
Hyderabad Campus
Refrigeration cycles-1
Inroduction
❑ Refrigeration is the is the transfer of heat from a lowertemperature region to a higher temperature one
❑ Devices that produce refrigeration are called refrigerators and
the cycles on which they operate are called refrigeration
cycles.
❑ Working fluid of the cycle is called refrigerant
•
•
•
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Vapor Compression Refrigeration Cycle
Gas Refrigeration Cycle
Cascade Refrigeration Cycle
Vapor Absorption Refrigeration Cycle
Analysis of Ideal cycles, Actual cycles Modified Cycle
2
BITS Pilani, Hyderabad Campus
Refrigerator and Heat Pump
Cooling Capacity: Rate of heat removal
from refrigerated space/ object generally
expressed in ton of refrigeration:
The capacity of regurgitation system
that can freeze 1 ton (2000 lbm) 0o C
liquid water into ice at 0o C in 24 h
1 ton of refrigeration (TR) = 211 kJ/min = 3.51 kW
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Ton of Refrigeration (TR)
The capacity of regurgitation system that can freeze 1 ton (2000 lbm) 0o C
liquid water into ice at 0o C in 24 h
ton, is equal to 2,000 pounds (907.18 kg)
Metric Tonne is equal to 1000 kg
1 ton of refrigeration (TR) = 210 kJ/min = 3.50 kW
907.18  333.7
= 3.50 kW
24  60  60
1 tonne of refrigeration = 3.86 kW
1000  333.7
= 3.86 kW
24  60  60
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Reversed Carnot Cycle
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Reversed Carnot & VCRC
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Household Refrigerator
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Ideal Vapor Compression Ref.
Cycle
h1 = hg @ PE or TE
To find h2
Or Use Table
Or Use Ph Chart
T2
s1 = s2 = s2' + c p ln
T2'
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
h1 + wcomp = h2
h3 = h f @ Pc = h4
h4 = h f + x4 h fg @ PE
BITS Pilani, Hyderabad Campus
Ideal Vapor Compression Ref.
Cycle
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
qL = h1 − h4 = RE
COP =
qL
wcomp in
h1 − h4
=
h2 − h1
Capacity in TR
.
m R (h1 − h4 )kW
= TR
3.5
BITS Pilani, Hyderabad Campus
Problem-1
A steady-flow Carnot refrigeration cycle uses refrigerant-134a as the
working fluid. The refrigerant changes from saturated vapor to
saturated liquid at 60°C in the condenser as it rejects heat. The
evaporator pressure is 180 kPa. Show the cycle on a T-s diagram
relative to saturation lines, and determine (a) the coefficient of
performance, (b) the amount of heat absorbed from the refrigerated
space, and (c) the net work input
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Solution
A steady-flow Carnot refrigeration cycle uses refrigerant-134a as the
working fluid. The refrigerant changes from saturated vapor to
saturated liquid at 60°C in the condenser as it rejects heat. The
evaporator pressure is 180 kPa. Show the cycle on a T-s diagram
relative to saturation lines, and determine (a) the coefficient of
performance, (b) the amount of heat absorbed from the refrigerated
space, and (c) the net work input
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Solution
A steady-flow Carnot refrigeration cycle uses refrigerant-134a as the working
fluid. The refrigerant changes from saturated vapor to saturated liquid at
60°C in the condenser as it rejects heat. The evaporator pressure is 180
kPa. Show the cycle on a T-s diagram relative to saturation lines, and
determine (a) the coefficient of performance, (b) the amount of heat
absorbed from the refrigerated space, and (c) the net work input
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Problem-2
A 10-kW cooling load is to be served by operating an ideal
vapor-compression refrigeration cycle with its evaporator
at 400 kPa and its condenser at 800 kPa. Calculate the
refrigerant mass flow rate and the compressor power
requirement when refrigerant-134a is used.
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Solution
A 10-kW cooling load is to be served by operating an ideal vapor-compression
refrigeration cycle with its evaporator at 400 kPa and its condenser at 800
kPa. Calculate the refrigerant mass flow rate and the compressor power
requirement when refrigerant-134a is used.
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Solution
A 10-kW cooling load is to be served by operating an ideal vapor-compression
refrigeration cycle with its evaporator at 400 kPa and its condenser at 800
kPa. Calculate the refrigerant mass flow rate and the compressor power
requirement when refrigerant-134a is used.
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Actual Vapor Compression
Refrigeration Cycle
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Refrigerant
❑ chlorofluorocarbons (CFCs),
❑ hydrofluorocarbons
❑ (HFCs), hydro chlorofluorocarbons (HCFCs),
❑ ammonia,
❑ hydrocarbons (propane, ethane, ethylene, etc.),
❑ carbon dioxide,
❑ air (in the air-conditioning of aircraft),
❑ water (in applications above the freezing point).
The right choice of refrigerant depends on the situation at
hand
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Nomenclature of refrigerants
Inorganic refrigerants:
R(700 + molecular wt )
Ammonia: R717
Water:
R718
Carbon dioxide: R744
Organic refrigerant:
C m H n Fp Cl q
(n + p + q) = 2m + 2
R − (m −1)(n +1)( p)
R12 :CCl2 F2
R 22 :CHClF2
R10 :CCl4
R110 : C2Cl6
R113 : C2Cl3 F3
R134 : C2 H 2 F4
Azeotropes: They are given arbitrary designation
Example:
R410A
CH2F2
(50%)
CHF2CF3
(50%)
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Selection of Refrigerant
Thermodynamic Properties
➢ Evaporator and Condensing temperature
➢ Critical pressure and temperature
➢ Freezing point
➢ Latent heat and specific heat
➢ Liquid and vapour density
Chemical Properties
➢ Inflammability
➢ Toxicity
➢ Solubility in water
➢ Action on material of construction
Physical Properties
➢ Thermal conductivity,
➢ Leak tendency and Viscosity
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Problem
A refrigerator uses refrigerant-134a as the working fluid and
operates on the vapor-compression refrigeration cycle.
The evaporator and condenser pressures are 200 kPa
and 1400 kPa, respectively. The isentropic efficiency of
the compressor is 88 percent. The refrigerant enters the
compressor at a rate of 0.025 kg/s superheated by
10.1°C and leaves the condenser subcooled by 4.4°C.
Determine (a) the rate of cooling provided by the
evaporator, the power input, and the COP.
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Solution
A refrigerator uses refrigerant-134a as the working fluid and operates on
the vapor-compression refrigeration
cycle. The evaporator and
condenser pressures are 200 kPa and 1400 kPa, respectively. The
isentropic efficiency of the compressor is 88 percent. The refrigerant
enters the compressor at a rate of 0.025 kg/s superheated by 10.1°C
and leaves the condenser subcooled by 4.4°C. Determine (a) the rate of
cooling provided by the evaporator, the power input, and the COP.
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Solution
A refrigerator uses refrigerant-134a as the working fluid and operates on the pressures are
200 kPa and 1400 kPa, respectively. The isentropic evapor-compression refrigeration
cycle. The evaporator and condenser fficiency of the compressor is 88 percent. The
refrigerant enters the compressor at a rate of 0.025 kg/s superheated by 10.1°C and
leaves the condenser subcooled by 4.4°C. Determine (a) the rate of cooling provided by
the evaporator, the power input, and the COP.
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Effect of change in operating
conditions
Effect of evaporator pressure
Effect of condenser pressure
Effect of suction vapor superheat
Effect of liquid subcooling
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Suction Superheat and Sub cooling
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Using suction line
regenerative heat exchanger
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Multi stage Compression
VCRS with Flash Chamber
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Problem-4
A two-stage compression refrigeration system operates with refrigerant134a between the pressure limits of 1.4 and 0.10 MPa. The
refrigerant leaves the condenser as a saturated liquid and is
throttled to a flash chamber operating at 0.4 MPa. The refrigerant
leaving the low-pressure compressor at 0.4 MPa is also routed to
the flash chamber. The vapor in the flash chamber is then
compressed to the condenser pressure by the high-pressure
compressor, and the liquid is throttled to the evaporator pressure.
Assuming the refrigerant leaves the evaporator as saturated vapor
and both compressors are isentropic,
Determine (a) the fraction of the refrigerant that evaporates as
it is throttled to the flash chamber, (b) the rate of heat removed
from the refrigerated space for a mass flow rate of 0.25 kg/s
through the condenser, and (c) the coefficient of performance.
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Problem-4
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Solution
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Innovative Vapor-compression
Refrigeration Systems
Cascade Refrigeration Systems
Some industrial applications require moderately low
temperatures,
❑ Temperature range they involve may be too large for a
single vapor compression refrigeration cycle to be
practical.
❑ A large temperature range also means a large pressure
range in the cycle and a poor performance for a
reciprocating compressor.
❑ One way of dealing with such situations is to perform
the refrigeration process in stages, that is, to have two or
more refrigeration cycles that operate in series..
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus
Cascade Refrigeration Systems
MEF217 Applied Thermodynamics Prof. Satish K Dubey, MechE
BITS Pilani, Hyderabad Campus