Chapter 1

REFRIGERATION
State the need of refrigeration or List the fields of application of refrigeration.
S-17, W-14
a. Domestic applications-Home freezers, fridges, chillers, etc.
b. Commercial applications-Cold storage, ice plant, retail stores, malls, hotels, etc.
c. Industrial applications-Dairy industry, printing industry, textile industry, tool room,
etc.
d. Air conditioning-Comfort air conditioning, Industrial air conditioning
e. Transportation of perishable cargo-Marine, aircraft, railways, vans, trucks, etc.
f. Automotive, Marine & Aircraft applications.
g. Preservation of foodstuff & other perishable items.
h. Chilling & Freezing applications.
Explain the concept of refrigeration. S-17, W-15, 14, 13

SOURCE-
ENCLOSED HEAT LEAKAGE TO
HEAT
REJECTED REF. COOLED SPACE COOLED SPACE
= Q 2 + WR
AT T2 Q2

WORK
REQ. SINK-Surrounding Atmosphere at T1
WR
Refrigeration may be defined as the process of obtaining and maintaining temperature in an
enclosed space lower than the temperature of the surroundings.
The refrigerator pumps heat (Q2) from cooled space at T2 and rejects it to the surrounding
atmosphere at T1. Work (WR) is required to drive the refrigerator. The temperature of the
enclosed space is maintained lower than the surroundings by pumping heat (Q2) equal to the
heat leaking into the enclosed space.

What is a refrigerator? How is the performance of a refrigerator specified? S-19

A refrigerator is a device which maintains an enclosed space at a temperature lower than the
temperature of the surroundings. It works on refrigeration cycle.

Coefficient of Performance of Refrigerator (C.O.Pref): S-19, 18, W-17

The performance of a refrigerating unit is specified in terms of its Coefficient of Performance
(C.O.P.) and is defined as given below:
Definition: Coefficient of Performance (C.O.P.) of a refrigerating unit may be defined as the
ratio of heat extracted (or refrigerating effect produced) to the work done on the refrigerant.
C.O.P ref = Refrigerating effect
Work Required
 C.O.P ref = RE = Q2 .
WR Q1 – Q2

HOT BODY (SINK) at T1

ATMOSPHERE

REFRIGERATOR WORK REQUIRED

WR

COLD BODY (SOURCE) at T2

(COOLED SPACE)
State the fundamental unit of refrigeration. Or Define ton of refrigeration (TR).
S-23, 19, 17, 15, 14, W-18, 17, 14
The fundamental/commercial unit of refrigeration is ton of refrigeration (TR)
Definition: A ton of refrigeration (TR) is defined as the ‘‘refrigerating effect produced by the
uniform melting of one ton (1000 kg) of ice from and at 0°C in 24 hours’’
i.e. 1TR = 1000 x 335 KJ in 24 hours (Latent Heat of fusion of ice =335 KJ/kg)
= (1000 x 335)/(24x60) KJ/min.
= 232.6 KJ/min.
In actual practice, 1TR = 210 KJ/min. = 3.517 KW

Define the following with reference to refrigeration: S-19, 18, 14, W-17
a) Desired Effect or Refrigerating Effect b) COP c) Fundamental Unit of Refrigeration
Desired Effect or Refrigerating Effect or Cooling Capacity:
It is defined as the cooling effect produced per unit time by the refrigerating unit
(COP & Fundamental unit of refrigeration defined above.)
Explain Reversed Carnot Cycle with the help of p-V & T-s Diagrams. S-19, 16
In this refrigeration cycle, air is used as working fluid. It is an ideal refrigeration cycle which
gives the maximum COP when working between the same temperature limits as compared to
other refrigeration cycles.

Processes in Reversed Carnot Cycle:

Process 1-2: Isentropic Compression
Process 2-3: Isothermal or Constant Temperature Heat Rejection
Process 3-4: Isentropic Expansion
Process 4-1: Isothermal or Constant Temperature Heat Addition

Fig. p-V & T-s Diagrams for Reversed Carnot Cycle

Expression for C.O.P of Reversed Carnot Cycle Refrigerator:

Heat Added = Refrigeration Effect = QA = T1 (S1 – S4)

Heat Rejected QR = T2 (S2 – S3)
Net. Work done Wnet = QR – QA = T2 (S2 – S3) - T1 (S1 – S4)
C.O.P = Refrigerating effect = T1 x (S1 – S4)
Work Required T2 (S2 – S3) - T1 (S1 – S4)]

C.O.P Carnot = T1 x (S1 – S4)

T2 (S1– S4) - T1 (S1 – S4)] Since S1 = S2 & S3 = S4
Where T1 = Lower Temperature in K
C.O.P = T1
T2 - T1 T2 = Higher Temperature in K

State why no refrigerator can be built working on Reverse Carnot Cycle. Or State
Limitations of Reversed Carnot Cycle. S-14, 17
No refrigerator can be built working on Reverse Carnot Cycle due to its following limitations:
1) This cycle consists of alternate very fast isentropic process followed by very slow
isothermal process which is not possible in practice.
2) Extremely slow isothermal process is not possible in actual practice.
3) Isentropic expansion & compression processes are not possible in practice due to heat
transfer and frictional losses.
 PROBLEMS ON CARNOT REFRIGERATOR
1.5 kW per ton of refrigeration is required to maintain the temperature of -40°C in the
refrigeration cycle working on Carnot cycle. Determine:
i) C.O.P. of cycle
ii) Temperature of sink (atmosphere)
iii) Heat rejected per ton of refrigeration. S-16

Given data:
Compressor Work W = 1.5 kW
Lower Temperature T1 = -40+273 = 233 K
Refrigeration effect RE = 1 Ton = 3.517 kW (I TR = 3.517 kW)
Cycle; Reversed Carnot Cycle
COP = 2.344
i) COP = R.E = 3.517 = 2.344
W 1.5
ii) COP = T1 = 233 .
T2 -T1 T2 – 233
2.344 = 233
T2 – 233

T2 = 332.40 K Temperature of sink T2 = 332.40 K

iii) Heat Rejected per ton of refrigeration,
QR = W + RE T2 = 332.40 K °K
QR = 1.5 + 3.517
QR = 5.017 kW QR = 5.017 kW

The capacity of a refrigerator is 300 tonnes when working between -4°C and 15°C.
Determine the mass of ice produced per day (24 hours) at 0°C from at water at 15°C Also
find the work done per minute. Assume that the cycle operates on reversed carmot cycle
and latent heat of ice is 335 kJ/kg. Take Cp for water is 4.187 kJ/kgK.

Given data:
Capacity of refrigerator RE = 300 tonnes = 300 x 210 = 63000 kJ/min (1 TR = 210 kJ/min)
Lower Temperature T 1 = - 4°C = - 4+ 273 = 269 K
Higher Temperature T 2 = 15°C =15 +273 288 K
Sp. Heat of Water Cpw = 4.187 kJ/kgK
Cycle Reversed Carnot Cycle
Latent Heat LH= 335 kJ/kg,
Initial Water Temperature T wi =15°C =15 +273 = 288 K
Final Water Temperature (Ice) Twf = 0°C = 273 K

Water Sensible Heat

at Removed = mCpw (Twi - Twf)
15 oC Latent Heat Ice
Water Removed = m * LH
at at
0 oC 0 oC
i) COP = T1 = 269 . = 14.158 C.O.P = 14.158
T2 -T1 288 – 269

ii) Total heat extracted RE = m [Cpw (Twi - Twf) + LH]

63000 = m [4.187 (15-0) + 335]
m = 158.369 kg/min
Mass of ice produced/day m = 158.369 x 60 x 24 m = 228051 kg/day = 228.05 tonnes/day

iii) Work required

COP = R.E
W
14.158 = 63000
W
W = 4449.78 KJ/min W = 4449.78 kJ/min

Explain Bell Coleman Cycle (Air Refrigeration Cycle) or Reversed Brayton Cycle.
Represent it on p-V & T-s Diagrams. S-18, 17, W-18, 17, 16
Bell Coleman Cycle is a modification of the ideal reversed Carnot cycle so as to make it
practicable. The isothermal processes are replaced by constant pressure or isobaric processes.
In this refrigeration cycle air is used as working fluid. As the air is not changing its phase
throughout the cycle the heat carrying capacity or RE per kg of air is very small compared with
vapour compression cycle.

Processes in Bell Coleman Cycle:

Process 1-2: Isentropic Compression
Process 2-3: Constant Pressure or Isobaric Heat Rejection
Process 3-4: Isentropic Expansion
Process 4-1: Constant Pressure or Isobaric Heat Addition

Expression for C.O.P of Bell Coleman cycle Refrigerator:

Heat Added: RE = QA = m Cp (T1 – T4)
Heat Rejected: QR = m Cp (T2 – T3)
Net. Work done: Wnet = QR – QA = m Cp (T2 – T3) - m Cp (T1 – T4)

Coefficient of Performance:
C.O.P = Refrigerating Effect = RE
Work Required Wnet
C.O.P = m Cp (T1 – T4) .
m Cp(T2 – T3) - m Cp (T1– T4)
C.O.P = (T1 – T4)
(T2 –T3) - (T1–T4)
On simplifying,
C.O.P = T1
T2 –T1
C.O.P = 1 .
T2 -1
T1 Where rp = Pressure ratio = P2/P1= P3/P4
C.O.P = 1 .
(rp)ϒ-1/ϒ - 1
 Fig. p-V & T-s Diagrams for Bell Coleman or Reversed Brayton Cycle

Explain construction and working of Air Refrigeration System. W-22, S-23, 18, 17
Explain Bell Coleman Refrigerating Cycle showing the components used.
Construction & Working of Air Refrigeration System:
Air is used as the working fluid in air refrigerating system. It works on Bell Coleman or
Reversed Brayton Cycle involving only sensible heat transfer. It consist of the following
processes & components:
a. Isentropic Compression of air in Centrifugal Air Compressor.
b. Constant Pressure Air Cooling or Heat Rejection in a heat exchanger (Air Cooler)
c. Isentropic Expansion of air in Air Expander (Air Turbine)
d. Constant Pressure Heat Absorption in a heat exchanger (Cold Chamber)

EXPLANATION: Refer p-V & T-s diagram of Bell-Coleman

Compressor: (Isentropic compression 1-2)
The compressor takes the air from the cold chamber and compresses them to high pressure &
temperature isentropically. Work is supplied to compress air.
.
Air Cooler: (Isobaric or Constant Pressure Heat Rejection 2-3)
It is a heat exchanger that is used to remove the heat isobarically from compressed air with the
help of a cooling medium. Heat is rejected and temperature reduces.

3) Air Expander: (Isentropic Expansion 3-4)

It is an expansion device that converts the high-pressure air into low-pressure air isentropically.
It reduces air pressure.& temperature. The work produced during expansion is used to drive
the compressor.

4) Cold Chamber: (Isobaric or Constant Pressure Heat Absorption 4-1)

The low pressure & low-temperature air from the expander enters the cold chamber, where air
absorbs the sensible heat from the cold chamber (which to be cooled) isobarically and provides
cooling effect (RE).

After absorbing heat from the cold chamber, air again flows to the compressor. In such a way,
the cycle continues again & again.
 Components:
i Air Compressor

ii Air Cooler

iii Air expander

iv Cold Chamber

Application of Air Refrigeration or Bell Coleman Cycle: The purpose of air refrigeration is
to absorb heat from the low-temperature cold chamber and reject it to the high-temperature
atmosphere. It is used-
a) In aircrafts for air conditioning & cabin cooling.
b) In aircrafts for pressurization of cabin.
c) For liquefaction of various gases.
d) For the both heating and co oling purpose

Advantages of Air Refrigeration or Bell Coleman Cycle:

a) Air is easily available compared with the other refrigerants.
b) It is cheap.
c) It is simple & light weight in construction.
d) Safe as air is non-flammable & non-toxic in case of leakage.
e) Low weight per TR.
f) Low maintenance cost.

Disadvantages of Air Refrigeration or Bell Coleman Cycle:

a) It is large in size.
b) Mass of air required per TR is very high.
c) Low C.O.P.
d) High running cost.
e) Freezing of moist air.

Explain why air refrigeration cycle is preferred for aircraft cooling.

Even though the COP of air cycle refrigeration is very low compared to vapour compression
refrigeration systems, it is still found to be most suitable for aircraft refrigeration systems due
to following reasons:
a) Air is cheap, safe, non-toxic and non-flammable. Leakage of air is not a problem.
b) Cold air can directly be used for cooling thus eliminating the low temperature heat
exchanger (open systems) leading to lower weight
c) The aircraft engine already consists of a high speed turbo-compressor, hence separate
compressor is not required. This reduces the weight per TR cooling considerably.
d) Simple Design due to low operating pressures.
e) Maintenance required is less.
State the need of aircraft cooling system.
Aircraft cooling is essential for passenger comfort and preserving perishable cargo. Aircraft
cooling system cools cabin and cargo hold as heat is added due to:
i. Internal heat generation due to occupants, equipment etc.
ii. Heat generation due to skin friction caused by the fast moving aircraft
iii. Heat added due to ram effect/cabin pressurization.
iv. Heat added due to solar radiation

Explain simple air-cooling system used in aircrafts with the aid of a neat labelled sketch.
W-22, 17 S-
23, 16
Construction:
Main Components include-
1. Main Compressor
2. Gas Turbine
3. Heat Exchanger
(Air Cooler)
4. Air Cooling Turbine
5. Cooling Air Fan

It makes use of the main

aircraft compressor driven by
gas turbine to compress air. It
operates in open loop.

Schematic of a simple aircraft refrigeration system

Working Principle: Air refrigeration system works on Bell Coleman Cycle using air as
refrigerant in open loop. It consists of following processes:

Isentropic ramming of air: Atmospheric air gets rammed before the main compressor
resulting in pressure & temperature rise.

Isentropic compression in main compressor: This ram air is compressed in the main
compressor resulting in further rise in pressure & temperature. The air required for refrigeration
system is then bled off from this compressor.

Constant pressure heat rejection in heat exchanger: It is then sent to the heat exchanger
(Air cooler) where this high pressure and high temperature air is cooled using ram air.

Isentropic expansion in air cooling turbine: Air is further cooled in the cooling turbine due
to its expansion. The air pressure & temperature reduces. The turbine work drives the cooling
fan which draws ram air through the heat exchanger.

Constant pressure heat addition in cabin: The cooled air from turbine is then sent to the
aircraft cabin. It picks up heat as it flows through the cabin providing useful cooling effect.
Compare Reversed Carnot Cycle & Reversed Brayton/Joule’s Cycle (Bell Coleman
Cycle).

State the similarities & dissimilarities between Reversed Carnot & Bell Coleman Cycle.
S-23
Dissimilarities (Comparison):

Reversed Brayton/ Joule’s or

Parameters Reversed Carnot Cycle
Bell Coleman Cycle
Heat Addition At constant temperature At constant pressure
& Rejection (Isothermal) (Isobaric)

COP Highest Lower

Type Ideal Cycle Real Cycle

Serves as comparison standard
Use Air refrigerator works on this cycle
for other refrigeration cycles

Factors Temperature limits Temperature limits & Pressure Ratio

affecting COP

Similarities between Reversed Carnot & Reversed Joule’s (Bell Coleman Cycle):
a) Both cycle use air as the refrigerant or working fluid.
b) Compression & Expansion are isentropic in both cases.