Refrigeration and Air Conditioning Lab Manual

VIDYA PRATISHTHAN’S

COLLEGE OF ENGINEERING, BARAMATI

DEPARTMENT OF MECHANICAL ENGINEERING

LABORATORY MANUAL

SUBJECT: REFRIGERATION AND AIR-CONDITIONING

[SUBJECT CODE: 302051]

CLASS: T.E. MECHANICAL

YEAR: 2011-12

APPROVED BY: H.o.D. [Mech] PRINCIPAL Prof. P. R. Shinde Dr. S. B. Deosarkar

VALIDITY UP TO: ACADEMIC YEAR 2012 – 2013

PRAPARED BY: PROF. P. R. Chitragar, K.M. Jadhav

VIDYA PRATISHTHAN’S

COLLEGE OF ENGINEERING, BARAMATI

DEPARTMENT OF MECHANICAL ENGINEERING

List of Experiments

FACULTY: PROF.K. M. JADHAV SUBJECT: REFRIGERATION AND

AIR-CONDITIONING

YEAR: 20011-12 CLASS: T.E. (MECH)

1. Test on Vapour compression Test Rig.

2. Test on ice plant test rig.

3. Test on air conditioning test rig.

4. Trial on vapour absorption refrigeration system.

5. Study of installation/operatio/maintenance practices for refrigeration systems.

6. Determination of refrigeration load in cold storage (case study/visit).

7. Visit to any refrigeration or air conditioning plant.

8. Thermal analysis of any refrigeration or air-conditioning cycle.

Prof. K M. JADHAV

[SUB-Incharge]

REFRIGERATION AND AIR-CONDITIONING – LABORATORY MANUAL

DEPARTMENT OF MECHANICAL ENGINEERING - VPCOE

EXPERIMENT NO. 1

TITLE: TRIAL ON VAPOUR COMPRESSION TESTS Objective: Trial on vapour compression test rig to find out actual and theoretical

COP. Theory: VAPOUR COMPRESSION REFRIGERATION SYSTEM

A vapour compression system is improved type of air refrigeration system in which

suitable working substance termed as refrigerant is used, it condensates and

evaporates at temperature and pressure close to atmospheric conditions. The refrigerants used for this purpose are ammonia, carbon dioxide, and sulphur dioxide.

The refrigerant used does not leave the system, alternatively condensing and

evaporating. In evaporation, the refrigerant absorbs it’s latent heat from which it is used for circulating it around the cold chamber while cost during it gives latent heat to the circulating water of the cooler. The VCS is thus latent heat pump as it pumps

its latent heat from brine end and delivers it to the cooler. It is generally used for all individual purposes from small domestic refrigerator to a big air conditioning plant.

ADVANTAGES:

1. It has similar size for given capacity of refrigerator. 2. It has less running cost. 3. It is employed for large range of temperature. 4. The COP is quite higher.

DISADVANTAGES:

1. The initial cost is higher. 2. The leakage of refrigerant is difficult to avoid.

ESSENTIAL PARTS OF THE SYSTEM:

1. Compressor: The low pressure and temperature of the vapour refrigerant from evaporator is

drawn into the compressor through IV or suction valve where it is compressed to high pressure and temperature. Refrigerant is discharged into the condenser through delivery or discharge valve.

2. Condenser It consists of coils of pipe in which high pressure and temperature vapour refrigerant gets cooled and condensed. The refrigerant while running through

the condenser gives up its latent heat to the surrounding condensing medium

which is normally air or water.



3. Receiver The condensed liquid refrigerant from the condenser is stored in vessel known as

receiver from where it is supplied to the evaporator through the expansion valve. 4. Expansion Valve: It allows the liquid refrigerant under high pressure and temperature to pass at a

controlled rate after reducing its pressure and temperature. Some of the refrigerants evaporate and pass through expansion valve but their greater portion is vapourised in the evaporator at low pressures and temperatures.

5. Evaporator: It consists of coils of pipe in which the liquid refrigerant at low pressure and

temperature. In evaporation, the liquid vapour refrigerant absorbs its latent heat of vapourisation from the medium which is to be cooled.

ANALYSIS OF VAPOUR COMPRESSION SYSTEM:

It depends upon: 1. Load on refrigerant plant in Tonnes.

2. Temperature of refrigerant. 3. Atmospheric temperature.

The temperature of refrigerant required and atmospheric conditions decide the

temperature of the refrigerant entering into the condenser and automatically it

decides the pressure ratio. Similarly, the working cycle is represented on T-S and P-h diagram, where the

process 3-2 represents superheating and 4-5 represents undercooling. The COP of

this cycle increases as it approaches near to the Carnot cycle. This is explained as

follows: The absorption of heat in evaporation and rejection of heat in the condenser are in the form of the latent heat. This heat transfer is carried out by maintaining the

temperatures differences as low as possible in evaporator and condenser. By comparing with the refrigerant cycle where high temperature cycle is necessary to carry out into compressor isolating it from the atmosphere. Irreversibility only

exists in the throttling process. The VCS approaches Carnot cycle as compared with air refrigeration system and COP of VCS is higher than air refrigeration system.

SPECIFICATIONS:

1. Cooling capacity: 30 kg of ice in 24 hours. 2. Insulation: PUF Insulation 3. Number of ice cans: 12 4. Size of tank: (0.83*0.345*0.345) m3.



5. Height of the tank: 0.27 m 6. Size of the ice can: (0.325*0.116*0.078) m3.

PROCEDURE:

1. Fill the tank in vapour compression test rig with the brine solution.

2. Switch ON the main supply. 3. Start the compressor and stirrer supply button. 4. Take initial energy meter reading of both compressor and stirrer. 5. Wait for about 50 minutes so that the system gets stabilized.

6. After stabilization take all readings at required temperature. 7. Measure evaporator (L.P) and condenser (H.P) pressures.

FIGURES:

VAPOUR COMPRESSION SYSTEM



SIMPLE VCC (T-S Graph) OBSERVATIONS:

1. Inlet temperature of brine at the start = 30°C.

2. Total weight of brine solution = 110.9 kg 3. Specific heat of brine = 23.32 kJ/kg°C 4. Final temperature of the brine at time of loading ice plant = -6°C 5. Final temperature of brine = 0°C

OBSERVATION TABLE:

Sr.

NO.

TIME T1 T2 T3 T4 T5 T6 Energy

meter

reading

(amp)

Energy

Meter

Reading

kW

LP

(bar)

HP

(bar)

CALCULATIONS:

1. Refrigeration Effect = m*Cp*(dT/dt) a. Size of can = 0.325*0.116*0.078



= 0.0294 m3 b. Total size of the can = 12*0.00294 = 0.3528 m3 c. Size of the tank = 0.83*0.395*0.395 = 0.1295 m3

d. Volume occupied by the brine solution =

e. Weight of water =

f. Weight of salt water =

g. Refrigeration Effect =

2. Compressor work =

3. COP (actual)

4. COP (from p-h chart)

1. absolute suction pressure = LP = 2. absolute delivery pressure = HP = W= h2=

h3=h4= h1=

COPth = (h1-h4)/ (h2-h1)



RESULT:

1. Theoretical COP =

2. Actual COP =

p-h diagram for refrigerant 134a



EXPERIMENT NO. 2

TITLE: TEST ON ICE PLANT TEST RIG

Objective: 1. Test on ice formation

2. Calculation of actual and theoretical COP

Apparatus: Ice plant test rig.

Specifications:

1. Compressor: hermetically sealed kirloskar made. 2. Condensor: Forced convection air cooled 3. Expansion valve: thermostatic expansion valve 4. Evaporator: Immerse type direct expansion 5. Cooling capacity: 30 kg of ice in 24 hours 6. Insulation: PUF insulation 7. Number of ice cans: 12 8. Flowmeter: GPs flow make rotameter 9. Size of tank: 0.83*0.395*0.395 m3. 10. Size of ice can: 0.35*0.166*0.078 m3.

Procedure:

1. Place machine in proper position where its level is horizontal and it is well insulated.

2. Machine must have atleast 1.5 m clearance from all sides. 3. Give 230 V, 50 Hz, 1 phase supply. 4. Prepare brine solution of water with approximate 20% of NaCl and mix it

thoroughly with water in the tank.

5. Start the compressor by putting switch ON. 6. Check the solution and discharge pressure and note down energy meter

reading.

7. Allow unit to run for some hours to bring down the temperature of brine in the range of -15°C to -120°C.

8. After attaining the temperature, fill ice can with water in each can and insert the ice can.

9. Now start agitator. Take all the readings of temperature at various point and pressure flow rate.



10. Allow the unit to run between the periods of 4 hours to 10 hours. Take all the readings until ice formation takes place.

11. After formation of ice remove the ice blocks.

Figures:

Ice plant test rig

Observations:

1. Time required to obtain temperature = 2 hours 2. Total weight of water in ice can = 30 kg 3. Initial temperature of water in ice can = 25°C. 4. Final temperature = 0°C. 5. Final temperature of brine when ice is formed = -2°C. 6. Time required = 5 hours



7. Initial temperature of brine start up = 15°C. 8. Total weight of brine solution = 110.9 kg 9. Specific heat of brine = 23.32 kJ/kg°C 10. Final temperature of the brine at time of loading ice plant = -6°C

Observation Table:

Sr.

No.

TIME T1 T2 T3 T4 T5 T6 Energy

Meter

Reading

kW

LP

(bar)

HP

(bar)

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

Where,

T1= Discharge gas temperature

T2= Liquid line temperature condenser output

T3= After expansion Temperature

T4= Suction gas Temperature

T5= Temperature of brine

T6= Ambient Temperature



Conversions:

PSI = 0.06897 bar

P1 = 23*0.6897 bar = 1.5 bar

P2 = 51*0.6897 = 10.25 bar

Calculations:

1. Refrigeration Effect

2. Compressor work =

3. Stirrer work =

4. COP (actual)

5. COP (from p-h chart)

Result:

1. Theoretical COP = 2. Actual COP =



p-h diagram for 134a refrigerant



EXPERIMENT NO. 3

TITLE: TEST ON AIR CONDITIONING TEST RIG

Objective:

1. Study of vapour compression cycle for conditioner.

2. To calculate actual COP.

3. To study different air conditioning processes:

a. Humidification

b. Dehumidification

c. Cooling

d. Heating

e. Heating and Humidification

f. Heating and Dehumidification

g. Cooling and Humidification

h. Cooling and Dehumidification

Apparatus: Air conditioning test rig.

Specifications:

1. Compressor: hermetically sealed Kirloskar made.

2. Condensor: Forced convection air cooled

3. Expansion valve: thermostatic expansion valve

4. Evaporator: Immerse type direct expansion

5. Flowmeter: GPs flow make rotameter

6. Condensor and Evaporator fan: Axial flow type

7. Heater: 1000W

8. Animometer: Digital type

Theory:

Simple Vapour Compression

Figure shows the schematic diagram of simple vapour compression refrigeration

system.

It consists of following parts:

1. Compressor

The low pressure and temperature of refrigerant from the evaporator is drawn into

the compressor through the inlet and suction valve as it is compressed to high



temperature and pressure. This high temperature and pressure vapour refrigerant is

discharged into the condenser through the delivery pipe.

2. Condensor

The condenser or cooler consists of pipe in which the high pressure and temperature

refrigerant is cooled and condensed. The refrigerant while passing through the

condenser gives up heat to the surrounding which consists of condensing medium

which is normally air or water.

3. Receiver

The condenser liquid refrigerant from the condenser is stored in a vessel known as

receiver from where it is supplied to evaporator through expansion valve or

refrigeration control valve.

4. Expansion valve

It is also called as throttle valve and its use is to allow the refrigerant under high

temperature and pressure to pass at controlled rate after reducing its high

temperature and pressure some of the refrigerant evaporates as it passes through

the expansion valve and the graded portion is vapourised in the evaporator at low

temperature and pressure.

5. Evaporator

It consists of coils of pipe in which the liquid vapour refrigerant at low pressure and

temperature is evaporated and changed to the vapour refrigerant at low

temperature and pressure. In evaporating the liquid vapour refrigerant absorbs the

latent heat of vapourisation from the medium which is to be cooled.

Psychrometric Terms:

1. Dry Air: the pure dry air is mixture of number of gases such as Nitrogen,

Oxygen, Carbon Dioxide, Hydrogen, Argon and Helium.

2. Moisture Air: It is the mixture of dry air and water vapour.

3. Saturated Vapour: It is mixture of dry air and water vapour when the air has

diffused the maximum amount of water vapour in it.

4. Degree of Saturation: It is the ratio of actual mass of water vapour in a unit

mass of dry air to the mass of water vapour is same mass of dry air when it is

(saturated) at the same temperature.

5. Humidity: It is the mass of the vapour present in 1 kg of water and generally

expressed in terms of grams per kg of dry air.

6. Absolute Humidity: It is the mass of water vapour present in 1 m3 of dry air

and is generally expressed in terms of gm/ m3.



7. Relative Humidity: It is the ratio of actual mass of water vapour in given

volume of moist air to the mass of water vapour in same volume of saturated

air.

8. Dew Point Temperature: It is the temperature of air recorded by the

thermometer when the moisture present in it begins to condense.

PRECAUTIONS:

1. Before taking the readings ensure that the sensing bulb of wet bulb

thermometer is properly wetted by soft cloth.

2. Do not start the heater while conducting the experiment for calculating

COP.

3. Keep the heater ON during dehumidification process.

Procedure:

4. Keep the test rig on leveled surface.

5. Give 230 V, 50 Hz, and single phase stabilized power supply to the

system and ensure that all the electrical connections are properly made.

6. Start the evaporator fan.

7. Start the compressor.

8. Allow the system to reach the steady state. This can be achieved by

turning it for atleast 15 minutes.

9. Make the provision to collect condensed moisture.

10. Take readings as per observation table.

Figures:

SENSIBLE COOLING



Air conditioning test rig

Sensible Heating



Cooling And Humidification

Heating And Humidification

Cooling And Dehumidification



Heating And Dehumidification

Observations:

1. Air velocity = 3.5 m/s.

2. Inlet air duct area = 0.05335 m2

3. Density of air = 1.1 kg/ m3

Observation Table:

S. NO. TIME INLET

DBT

TEMP

WBT

OUTLET

DBT

TEMP

WBT

LP

PSI

HP

PSI

1

2.

3.

4

5.

S. NO. T1

°C

T2

°C

T3

°C

T4

°C

1.

2.

3.

4.

5.



Energy Meter:

Where,

T1 = compressor outlet temperature

T2 = liquid line temperature

T3 = compressor inlet temperature

T4 = temperature after expansion

S.

NO

PROCES INLET

DBT

INLET

WBT

OUTLET

DBT

OUTLET

WBT

REMARKS

1. Heating Only heater ON

2. Cooling Only lamp ON

3. Humidification Humidification

4. Humidification and

Cooling

Humidification

and lamp ON

5. Heating and

Dehumidification

Humidification

and heater ON

6. Cooling and

Dehumidification

Dehumidification

and Lamp ON

CALCULATIONS:

m= Q*ρ

Now,

Inlet DBT = 92°F = (92-32)/1.8 °C = 33.5°C

Outlet DBT = 58°F = (58-32)/1.8 °C = 14.44°C

Refrigeration Effect = m. δh

From chart,

h1 =

h2 =

Refrigeration Effect =

Now work done by compressor = Final EMR – Initial EMR



So COP (actual) = R.E/W.D

Thus,

W.D =

COP = R.E/W.D

P1 = Absolute suction pressure

= Patm + Pg

P2 = Absolute suction pressure

T1 = T2 = T3 = T4 =

By p-h chart,

h1 = kJ/kg

h2 = kJ/kg

h3 = h4 = kJ/kg

COP (theoretical) = (h1 – h4) / (h1 – h2)

Result:


2. Actual COP =



EXPERIMENT NO. 4

TITLE: VAPOUR ABSORPTION REFRIGERATION SYSTEM

Objective: Trial on vapour absorption refrigeration system (Electrolux

Refrigeration) to find out actual and theoretical COP.

Apparatus: Refrigeration test rig.

Theory:

Electrolux Refrigerator:

This type of refrigeration is usually used for domestic purposes only as it is complex

in the construction and working. This type of refrigerator was developed in 1925 by

Munters and Battzervan when they were studying at Royal Institute of Technology

At Stockholm for their undergraduate course. This type of refrigerator was known

as three fluid refrigeration system. The elimination of aqua pump from the

absorption system with the complete absence of moving parts and work input. The

main purpose of removing the pump was to make the machine noiseless. It uses

refrigerant as a solvent s and an inlet gas for inlet of the system. The inert gas is

continued to the lower side of system only by its system. It is possible to maintain

the uniform pressure throughout the system and after sometime permitting the

refrigerant to evaporator at low temperature corresponding to its partial pressure.

In the high pressure side system (generator and condenser), there exists only the

refrigerant which is subjected to total pressure of the system so that it is

condensed by using normal cooling water as air as it is done in other system.

In lower side of the system, the total pressure is sum of the partial pressure of

hydrogen which is used as an inert gas. The liquid ammonia which comes into the

evaporator as the partial pressure of ammonia is low.

The strong aqua ammonia solution is heated in generator by the application of

external heat source. The water vapour carried with ammonia vapour is removed in

separate form as shown in figure. Then the dry ammonia vapour is passed into the

condenser and it is condensed by using external cooling source. The liquid ammonia

flows under gravity in the evaporator and it evaporates. The mixture of hydrogen

and ammonia vapour is passed into the absorber and the weak solution from aqua

ammonia from the separator is allowed to follow into the absorber, through tray this

weak aqua ammonia solution comes into contact with hydrogen separated. This strong

solution is further passed to the generator and it completes the cycle.



There is no pump to create the pressure differential between condenser and

evaporator or not an expansion valve.

The hydrogen returns to the evaporator having no affinity for the absorbent. The

hydrogen is held in this condition by the V tube due to small pressure difference in

the system. The solution is circulated through absorber and generated by thermal

action alone. The paths are so arranged that the liquid refrigerant flows through

evaporator by gravity, only care is to be taken to keep hydrogen isolated in the

proper part of the system otherwise pressure will be unbalanced and the machine

will stop.

Procedure:

1. Switch on the main supply.

2. Start the bulb inside the cabin.

3. Adjust current and the supply voltage.

4. Take the initial temperature reading.

5. Repeat the procedure after sometime.

6. Take readings until the steady state is reached.

7. When steady state is reached, take this reading for calculation and find the

theoretical and actual COP>

Figures:

VAPOUR ABSORPTION SYSTEM



DOMESTIC TYPE ELECTROLUX REFRIGERATOR

Observation Table: Sr.

NO

TIME

TAKEN

WITH

OUT

LOAD

T1 T2 T3 T4 T5 T6 V (VOLTS)

I (AMPERE)

WATT

(V*I)

1.

2.

3.

4.

5.

Where,

T1= Temperature before generation

T2= Temperature after generation

T3= Temperature of evaporator

T4= Temperature of condenser

T5= Temperature of absorber

T6= Temperature of cabinet



Calculations:

1. COP= T3(T2-T5))/T2(T5-T3)

2. COP= (V*I)/W

Result:


2. Actual COP =

Refrigeration and Air Conditioning Lab Manual

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