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Refrigeration is a process of controlled removal of heat from a substance to keep it at a temperature below the ambient condition, often below the freezing point of water (0 OC)

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Refrigeration In principle, It is based on the fact that when a liquid is allowed to evaporate, it will absorb heat from its surroundings.

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Domestic: fridges, freezers. Transportation: refrigerated containers Industrial: dairies, brewing, ice cream, Commercial: supermarkets, hotels and

restaurants. Air-conditioning: hospitals, offices, aeroplanes,

trains, cars. Leisure and Sports : ice rinks.

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∴

HOT RESERVOIR

COLD RESERVOIR

HEAT ENGINE W

Q1

Q2

The principle of a Heat Engine in which fuel is burnt to produce work (drive the car), the exhaust rejects lower quality heat to the ambient. W= Q1- Q2 according to the first law of thermodynamics

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∴

HOT RESERVOIR

COLD RESERVOIR

HEAT PUMP

Q1

W

Q2

HOT RESERVOIR

COLD RESERVOIR

HEAT ENGINE W

Q1

Q2

The principle of a Heat Pump in which work (motor), is used to drive a compressor pumping the gas to a higher pressure, then let it through a condenser to loose heat, and finally suddenly drop its pressure allowing to vapourise and hence absorb heat from its surroundings, hence

Q2 = Q1- W according to the first law of thermodynamics

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∴

HOT RESERVOIR

COLD RESERVOIR

HEAT PUMP

Q1

W

Q2

The reason this is known as Heat Pump because it is possible to utilise both heat terms Q2 and Q1.

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∴

Both COP are related to each other, According to Energy Conservation : Q1 – Q2 = W Or Q1 = Q2 + W Divide equation by W, to get : Q1/W = Q2/W + 1

COPH = COPR + 1

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∴

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Simple Refrigeration cycle: 4 processes: 1 - 2 Isentropic compression 2 - 3 Condensation at constant pressure 3 - 4 Throttling at constant enthalpy 4 - 1 Evaporation at constant pressure

4

3

CONDENSER

EVAPORATOR

W

2

1

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h

2

1 4

3 P

4

3

CONDENSER

EVAPORATOR

W

2

1

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Calculations for ideal compression, 100% isentropic: Input Work W12 = mr ( h2 – h1)

Refrigeration Effect Q41 = mr ( h1 – h4)

Heating Effect Q23 = mr (h2 – h3)

If the Refrigerant mass is not specified, calculate on 1 kg/s basis.

2

h

1 4

3

WQCOPH 23=

WQCOPR 41=

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Calculations for irreverible compression, less than 100%: if h2 is the actual enthalpy after compression, and

h2’ is the ideal value, at constant entropy curve.

Then the compressor’s isentropic efficiency is defined as:

2

h

1 4

3

12

1'2

hhhh

c −

−=η

2’

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Calculations for irreverible compression, less than 100%: Input Work W12 = mr ( h2 – h1)

Refrigeration Effect Q41 = mr ( h1 – h4)

Heating Effect Q23 = mr (h2 – h3)

If the Refrigerant mass is not specified, calculate on 1 kg/s basis.

2

h

1 4

3

WQCOPH 23=

WQCOPR 41=

2’

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Enthalpy h (kJ/kg) X- axis

Pressure P( bar ) Y-axis

Temp T ( C) on inside of the Curve

Volume v( m3/kg) lines ( curves ) at 10o to horizontal

Entropy s (kJ/kgK) lines ( curves ) at 70-80o to horizontal

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∴

p

h

2

1 5

4 3

6

condenser

compressor

evaporator

Heat Exchanger 6

Expansion Valve

4

2 3

1

5

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∴

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∴

SOLVED EXAMPLE 1 A heat pump is used to heat a factory during winter with average outside temperature being 0oC and the factory temperature is to be maintained at 20 oC. The building heating load is estimated as 100 kW. Determine the minimum power requirement to drive the HP unit for this application.

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∴

SOLUTION EXAMPLE 1

65142932731

11

1 .//max =

−=

−=

HLH TT

COP

kWCOPQW

henceWQCOP

but

H

866514

100 ..max

min

max

===

=

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∴

SOLVED EXAMPLE 2 A refrigerator using R12 as the working fluid. The refrigerant enters the compressor as

saturated vapour at 2 bar and on leaving the condenser it is saturated liquid at 8 bar (absolute). Assume that the compression process is 100% isentropic, and that there is no superheating or

subcooling effects and that pressure losses are negligible. Calculate using the chart: the work of compression, the refrigerating effect, the coefficient of performance - compare this value with the Carnot performance Solution: Construct the cycle on the P-h chart find

h1=245 kJ/kg h2=270 kJ/kg h3 = h4 = 130 kJ/kg

Hence: Work of compressor Wc = h2-h1=24 kJ/kg Refrigeration effect RE = h1-h4=116 kJ/kg COPR = RE/Wc=4.83 Note that

775

260305260

21

2 .=−

=−

=TT

TCOPideal

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∴

SOLVED EXAMPLE 3 A refrigerator using R134a as a Retrofit working fluid to replace R12 in the system of example

7.5. The refrigerant enters the compressor as saturated vapour at 2 bar and on leaving the condenser it is saturated liquid at 8 bar (absolute). Assume that the compression process is 100% isentropic, and that there is no superheating or subcooling effects and that pressure looses are negligible. Calculate using the chart: the work of compression, the refrigerating effect, the coefficient of performance. Solution: Construct the cycle on the P-h chart, find enthalpies at key state points:

h1 =295 kJ/kg h2 =325 kJ/kg h3 = h4 =145 kJ/kg

Hence: Work of compressor Wc= h2-h1=30 kJ/kg Refrigeration effect RE=h1-h4=150 kJ/kg Coefficient of Performance COPR =RE/Wc= 5.0 which is better than that for R12 but still not as good as the ideal value of 5.77