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 O C) 05/04/0 T.Al-Shemmeri
Refrigeration In principle, It is based on the fact that when a liquid is allowed to evaporate, it will absorb heat from its surroundings. 05/04/0 T.Al-Shemmeri
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. 05/04/0 T.Al-Shemmeri 3
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. = Q - Q according to the first law of thermodynamics HOT RESERVOIR Q HEAT ENGINE COLD RESERVOIR Q 05/04/0 T.Al-Shemmeri 4
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 Q = Q - according to the first law of thermodynamics HOT RESERVOIR Q HOT RESERVOIR Q HEAT PUMP HEAT ENGINE COLD RESERVOIR Q COLD RESERVOIR Q 05/04/0 T.Al-Shemmeri 5
The reason this is known as Heat Pump because it is possible to utilise both heat terms Q and Q. HOT RESERVOIR Q HEAT PUMP COLD RESERVOIR Q 05/04/0 T.Al-Shemmeri 6
Both COP are related to each other, According to Energy Conservation : Q Q = Or Q = Q + Divide equation by, to get : Q / = Q / + COP H = COP R + 05/04/0 T.Al-Shemmeri 7
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Simple Refrigeration cycle: 4 processes: - Isentropic compression 3 CONDENSER - 3 Condensation at constant pressure 3-4 Throttling at constant enthalpy EVAPORATOR 4 - Evaporation at constant pressure 4 05/04/0 T.Al-Shemmeri 9
3 CONDENSER P 3 EVAPORATOR 4 4 h 05/04/0 T.Al-Shemmeri 0
Calculations for ideal compression, 00% isentropic: 3 Input ork = m r ( h h ) Refrigeration Effect Q 4 = m r ( h h 4 ) Heating Effect Q 3 = m r (h h 3 ) If the Refrigerant mass is not specified, calculate on kg/s basis. Q COP = 3 H 4 h Q COP = 4 R 05/04/0 T.Al-Shemmeri
Calculations for irreverible compression, less than 00%: 3 if h is the actual enthalpy after compression, and h is the ideal value, at constant entropy curve. Then the compressor s isentropic efficiency is defined as: η c 4 = h h ' h h h 05/04/0 T.Al-Shemmeri
Calculations for irreverible compression, less than 00%: 3 Input ork = m r ( h h ) Refrigeration Effect Q 4 = m r ( h h 4 ) Heating Effect Q 3 = m r (h h 3 ) If the Refrigerant mass is not specified, calculate on kg/s basis. Q COP = 4 R 05/04/0 T.Al-Shemmeri 3 4 h Q COP = 3 H
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Enthalpy h (kj/kg) Pressure P( bar ) X- axis Y-axis Temp T ( C) on inside of the Curve Volume v( m 3 /kg) lines ( curves ) at 0 o to horizontal Entropy s (kj/kgk) lines ( curves ) at 70-80 o to horizontal 05/04/0 T.Al-Shemmeri 5
3 condenser 4 Heat Exchanger 5 6 evaporator compressor p 4 3 5 6 Expansion Valve h 05/04/0 T.Al-Shemmeri 6
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SOLVED EXAMPLE A heat pump is used to heat a factory during winter with average outside temperature being 0 o C and the factory temperature is to be maintained at 0 o C. The building heating load is estimated as 00 k. Determine the minimum power requirement to drive the HP unit for this application. 05/04/0 T.Al-Shemmeri 8
SOLUTION EXAMPLE COP but H COP hence min max = = T / T 73/ 93 = H max = = Q COP Q max L = H 00 4. 65 = 6. 8 k 4. 65 05/04/0 T.Al-Shemmeri 9
SOLVED EXAMPLE A refrigerator using R as the working fluid. The refrigerant enters the compressor as saturated vapour at bar and on leaving the condenser it is saturated liquid at 8 bar (absolute). Assume that the compression process is 00% 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 h=45 kj/kg h=70 kj/kg h3 = h4 = 30 kj/kg Hence: ork of compressor c = h-h=4 kj/kg Refrigeration effect RE = h-h4=6 kj/kg COP R = RE/c=4.83 Note that COP ideal T = T T 60 = 5 77 305 60 =. 05/04/0 T.Al-Shemmeri 0
SOLVED EXAMPLE 3 A refrigerator using R34a as a Retrofit working fluid to replace R in the system of example 7.5. The refrigerant enters the compressor as saturated vapour at bar and on leaving the condenser it is saturated liquid at 8 bar (absolute). Assume that the compression process is 00% 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: h =95 kj/kg h =35 kj/kg h3 = h4 =45 kj/kg Hence: ork of compressor c= h-h=30 kj/kg Refrigeration effect RE=h-h4=50 kj/kg Coefficient of Performance COP R =RE/c= 5.0 which is better than that for R but still not as good as the ideal value of 5.77 05/04/0 T.Al-Shemmeri