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Refrigeration. Introduction to Food Engineering. Ice Box. Ice -> water, latent heat = 333 kJ/kg. Refrigerant : liquid -> vapor. Selection of Refrigerant. 1. Latent heat of vaporization High value Small amount needed per unit time 2. Condensing pressure
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Refrigeration Introduction to Food Engineering
Ice Box Ice -> water, latent heat = 333 kJ/kg
Selection of Refrigerant 1. Latent heat of vaporization • High value • Small amount needed per unit time 2. Condensing pressure • High pressure needs expenses on heavy construction of condenser & piping 3. Freezing temperature • Should be below evaporator temperature
Selection... 4. Critical temperature • Temperature that refrigerant vapor cannot be liquefied, should be high 5. Toxicity, must be non-toxic 6. Flammability –nonflammable 7. Corrosiveness –noncorrosive 8. Chemical stability - stable
Selection... 9. Detection of leaks –easy to detect 10. Cost 11. Environmental impact
Commonly used refrigerants • Ammonia • CFCs • Freon 12 = dichloro difluoromethane • Freon 22 = monochloro difluoromethane • Stable – long life in lower atmostphere • Migrate to upper atmostphere, Cl split off by UV, reacts with ozone -> deplete • More UV
Alternatives • HCFs • Less stable • Hydrofluorocarbons • hydrochlorofluorocarbons
Changes • d saturated liquid condensation temp • Passing through expansion valve • Pressure & temp drop • Some liquid -> gas • Liquid/gas mixture enters evaporator coils at e • Completely vaporize, -> saturated vapor (gain additional heat)
Changes... • Vapors enter compressor • Compressed -> high pressure, temp increase • Superheated refrigerant • Superheated vapor cooled by air or water in condenser • Saturated liquid
Pressure-Enthalpy Charts • Enthalpy H = U + PV H = enthalpy (kJ/kg) U = internal energy (kJ/kg) P = pressure (kPa) V = specific volume (m3/kg)
Evaporator & Condenser • Enthalpy change, pressure constant • Compression : work done • Increase enthalpy, increase pressure • Expansion valve • Constant enthalpy • Flow from high P -> low P
condenser compressor evaporator
Cooling Load • Rate of heat energy removal from a given space or object to lower temp. to a desired level • One ton of refrigeration = latent heat of fusion of 1 ton of ice = 288,000 Btu/24 hr = 303,852 kJ/24 hr = 3.5168 kW
Cooling load calculation must consider heat of respiration, walls, floor, doors, etc.
Example Calculate cooling load caused by heat of evolution of 2000 kg cabbage stored at 5 °C. Given heat of evolution of cabbage at 5 °C = 28 – 63 W/Mg Total heat evolution (2000 kg)(63W/Mg)(1Mg/1000 kg) = 126 W
Calculations • Compression qw = work done on refrigerant (kW)
Condenser • Evaporator Rate of heat exchanged in condenser Rate of heat accepted by refrigerant
Coefficient of performance • Indicate efficiency of the system. • Ratio between heat absorbed by refrigerant in evaporator to heat equivalence of the energy supplied to the compressor
Refrigerant flow rate • Depends on cooling load & refrigeration effect q = total cooling load (kW) m = mass flow rate (kg/s)
Example • Cold storage room (2 °C) uses Freon-12 as refrigerant. Evaporator temp = -5 °C, Condenser temp = 40 °C, refrigeration load = 20 tons, calculate m, compressor power requirement and C.O.P. Assume saturated conditions and compressor efficiency 85 %.
From chart Evaporator pressure = 260 kPa Condenser pressure = 950 kPa H1 = 238 kJ/kg H2 = 350 kJ/kg H3 = 395 kJ/kg (1 ton of refrigeration = 303,852 kJ/24 hr)
qw = m(H3– H2), 85 % efficiency = 18.47 kW = 4.48
Assume vapors leave evaporator 10 °C super-heated, liquid from condenser is subcooled 15 °C. • m = 0.54 kg/s • qw = 15.9 kW • C.O.P. = 5.2 • Slower m, less power, higher C.O.P.