Bootstrap Air Evaporative Cooling System

Published by skhattar36@gmail.com on

Bootstrap Air Evaporative Cooling System is similar to the bootstrap air cooling system. The only difference between them is the addition of an evaporator between the second heat exchanger and a cooling turbine. The fundamental parts of a simple air refrigeration system are

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

Bootstrap Cooling System

Working Of Bootstrap Air Evaporative Cooling System

  • 1-2  Isentropic ramming of the surrounding air from pressure P1 and temperature T1 to pressure P2 and temperature T2.
  • 1-2` Actual ramming process on account of inner erosion because of irreversibilities.
  • 2′- 3 Isentropic compression of air in the primary compressor.
  • 2-3′ Actual compression of air as a result of inward erosion because of irreversibilities.
  • 3′- 4 Cooling of the air in the primary heat exchanger. During this process, the drop in pressure is neglected.
  • 4-5 Isentropic compression of cooled air, from the main heat exchanger, in the auxiliary compressor.
  • 4-5′ Actual Compression process in light of interior erosion because of irreversibilities.
  • 5`- 5“ Cooling by ram air in the secondary heat exchanger. During this process, the drop in pressure is neglected.
  • 5“-6  Cooling of air in the evaporator.
  • 6-7 Isentropic expansion of cooled air in the cooling turbine up to compartment pressurization.
  • 6-7′ Actual Expansion of the cooled air in the cooling turbine.
  • 7′- 8 Heating of air up to the compartment temperature T8.
Bootstrap Air Evaporative Cooling System TS Diagram

Bootstrap Air Evaporative Cooling System TS Diagram

Q tonnes of refrigeration is the cooling load in the compartment, at that point, the amount of air required for the refrigeration will be

ma = 210 Q / Cp ( T8 – T7`)

Unit- Kg/min

Power required during refrigeration system

P = maCp ( T3` – T2` ) / 60

Unit –  kW

The temperature of the air leaving the cooling turbine in this system is low. Therefore, the mass of air (ma) per tonne of refrigeration will be less in this system.

C.O.P  of Refrigeration System

C.O.P = 210 / P x 60


2 Comments

Mohit Dhyani · April 1, 2020 at 5:09 pm

Thank’s for sharing this sir. Your articles are helping me very much in preparations for my college exam. Please add more articles.
Thank You

Siddharth · April 1, 2020 at 6:30 pm

Thank you for exploring my knowledge by letting me know about the techniques behind refrigeration…
Please upload more article’s…

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