05/04/2011 T.Al-Shemmeri 1
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)
05/04/2011 T.Al-Shemmeri 2
Refrigeration In principle, It is based on the fact that when a liquid is allowed to evaporate, it will absorb heat from its surroundings.
05/04/2011 T.Al-Shemmeri 3
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.
05/04/2011 T.Al-Shemmeri 4
∴
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
05/04/2011 T.Al-Shemmeri 5
∴
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
05/04/2011 T.Al-Shemmeri 6
∴
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.
05/04/2011 T.Al-Shemmeri 7
∴
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
05/04/2011 T.Al-Shemmeri 8
∴
05/04/2011 T.Al-Shemmeri 9
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
05/04/2011 T.Al-Shemmeri 10
h
2
1 4
3 P
4
3
CONDENSER
EVAPORATOR
W
2
1
05/04/2011 T.Al-Shemmeri 11
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=
05/04/2011 T.Al-Shemmeri 12
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’
05/04/2011 T.Al-Shemmeri 13
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’
05/04/2011 T.Al-Shemmeri 14
05/04/2011 T.Al-Shemmeri 15
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
05/04/2011 T.Al-Shemmeri 16
∴
p
h
2
1 5
4 3
6
condenser
compressor
evaporator
Heat Exchanger 6
Expansion Valve
4
2 3
1
5
05/04/2011 T.Al-Shemmeri 17
∴
05/04/2011 T.Al-Shemmeri 18
∴
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.
05/04/2011 T.Al-Shemmeri 19
∴
SOLUTION EXAMPLE 1
65142932731
11
1 .//max =
−=
−=
HLH TT
COP
kWCOPQW
henceWQCOP
but
H
866514
100 ..max
min
max
===
=
05/04/2011 T.Al-Shemmeri 20
∴
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
05/04/2011 T.Al-Shemmeri 21
∴
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
