Simple Air Evaporative Cooling System

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Simple Air Evaporative Cooling System is similar to the simple cooling system. The only difference between them is the addition of an evaporator between heat exchanger and a cooling turbine. The evaporator gives an extra cooling impact through the vanishing of a refrigerant, for example, water. The evaporative cooling is obtained by using liquor or ammonia. The water, liquor, and alkali have diverse refrigerating impacts at various heights. At 20000 meter height, water boils at 40 C, alcohol at 9 C and ammonia at -70C. The fundamental parts of a simple air refrigeration system are

  1. the primary blower is driven by the gas turbine
  2. a heat exchanger
  3. a cooling turbine
  4. a cooling air fan.
  5. an evaporator
Simple Air Evaporative Air Cooling System Diagram

Simple Air Evaporative Air Cooling System Diagram

Simple Air Evaporative Air Cooling System TS Diagram

Simple Air Evaporative Air Cooling System TS Diagram

Working of Simple Air Evaporative Cooling System

  • Ramming Process- Isentropic ramming of the ambient air from pressure P1 and temperature T1 to pressure P2 and temperature T2.
  •  Actual Ramming Process 1-2` is the actual ramming process because of internal friction due to irreversibilities.
  • Compression Process 2`-3 Isentropic compression of air in the main compressor
  • Actual Compression Process -2-3` Actual compression of air because of internal friction due to irreversibilities.
  • Cooling Process 3`-4 Cooling of the air in the only heat exchanger. We neglect the pressure drop in the heat exchanger.
  • Process 4-4` Cooling of air in the evaporator.
  • Expansion Process 4-5 Isentropic expansion of cooled air in the cooling turbine.
  • Actual Expansion process 4-5` Actual expansion of the cooled air in the cooling turbine, because of internal friction due to irreversibilities.
  • Refrigeration Process 5`-6 Heating of air up to the cabin temperature T6.

C.O.P

= Refrigeration Effect produced/ Work Done = maCp( T6 – T5` ) / maCp( T3` – T2`)

= T6 – T5` / T3` – T2`

Q tonnes of refrigeration is the cooling load in the cabin, then the quantity of air required for the refrigeration purpose will be

ma = 210 Q / Cp ( T6 – T5`)

Unit- Kg/min

Power required for the refrigeration system

P = maCp ( T3` – T2` ) / 60

Unit –  kW

C.O.P  of Refrigeration System

C.O.P = 210 / P x 60


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