Simple Air Cooling System

Published by skhattar36@gmail.com on

A simple air cooling system is utilized in airplanes. The air required for the refrigeration system seeps off the principle compressor. This high pressure and high-temperature air cool in the heat exchanger. This slam air is further used in cooling and then air expands in the cooling turbine. This turbine drives the cooling fan which draws cooling air through the heat exchanger. A simple air refrigeration system is useful for ground surface cooling and low flight speeds. The fundamental parts of a simple air refrigeration system are

  1. the primary blower driven by the gas turbine
  2. a warmth exchanger
  3. a cooling turbine
  4. a cooling air fan.
Simple Air Cooling System Diagram

Simple Air Cooling System Diagram

 

TS Diagram of Simple Air Cooling System

TS Diagram of Simple Air Cooling System

Working of Simple air 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. 
  • 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 the refrigeration system

C.O.P = 210 / P x 60


1 Comment

savita chopra · March 23, 2020 at 5:35 pm

very useful 👍

Leave a Reply

Your email address will not be published. Required fields are marked *