Refrigeration and Air Conditioning Questions and Answers Part-8

1. Which one of the following is not a component of a simple air cooling system
a) Main compressor
b) Cooling fan
c) Heat exchanger
d) Generator

Answer: d
Explanation: The main components of simple air cooling system are the main compressor driven by the gas turbine, a cooling fan, heat exchanger and a cooling turbine.

2. The COP of simple air cooling system is given by?
T6 = Inside temperature of cabin
T5′ = Exit temperature of cooling turbine
T3′ = Temperature at the exit of compressor
T2′ = Stagnation temperature
a) COP = \(\frac{(T_6 – T_{5′})}{(T_3′ – T_{2′})}\)
b) COP = \(\frac{(T_6 + T_{5′})}{(T_3′ – T_{2′})}\)
c) COP = \(\frac{(T_6 + T_{5′})}{(T_3′ + T_{2′})}\)
d) COP = \(\frac{(T_6 – T_{5′})}{(T_3′ + T_{2′})}\)

Answer: a
Explanation: COP = Refrigeration effect produced i.e. (T6 – T5′) divided by Work done i.e. (T3′ – T2′).
Hence COP = \(\frac{(T_6 – T_{5′})}{(T_3′ – T_{2′})}\)

3.The simple air cooling system is good for _____ flight speed.
a) low
b) high
c) moderate
d) any

Answer: a
Explanation: The simple air cooling system is good for low flight speed so as fan can maintain airflow over the air cooler, which is difficult for it while at high speeds

4. What is the main difference between simple air cooling system and simple air evaporative cooling system?
a) Simple air evaporative cooling system has an evaporator
b) Simple air evaporative cooling system has two evaporator
c) Simple air evaporative cooling system has an extra compressor
d) Simple air evaporative cooling system has two extra compressor

Answer: a
Explanation: The difference between the simple air cooling system and simple air evaporative cooling system is that it has an evaporator between the heat exchanger and the cooling turbine

5. A simple evaporative air refrigeration system is used for an airplane to take 20 TR of refrigeration load (Q). The power required for the refrigerating system P is 746 KW. What is its COP?
a) 0.086
b) 0.094
c) 0.079
d) 0.099

Answer: b
Explanation: COP = \(\frac{(210×Q)}{(P×60)}\)
COP = \(\frac{(210×20)}{(746×60)}\)
= 0.094.

6. In Boot-strap air cooling system how many heat exchangers are there?
a) 1
b) 2
c) 3
d) 0

Answer: b
Explanation: In Boot-strap air cooling system there are two heat exchangers and a cooling turbine driving the secondary compressor instead of the cooling fan.

7. What is the difference between Boot-strap air cooling system and Boot-strap evaporative cooling system?
a) Boot strap evaporative system has an evaporator
b) Boot strap evaporative system has two evaporator
c) Boot-strap evaporator eliminates the need for evaporator
d) Boot-strap evaporator system has three evaporators

Answer: a
Explanation: Boot strap evaporative system has an evaporator between the second heat exchanger and the cooling turbine

8. Mass of air per tonne of refrigeration will be _____ in Boot-strap air cooling system than Boot-strap evaporative system.
a) less
b) more
c) equal to
d) can’t say

Answer: b
Explanation: Since the temperature of air leaving the cooling turbine in boot-strap system is lower than the simple boot-strap system, therefore the mass of air per tonne of refrigeration will be more in boot-strap air cooling system.

9. The air cooling system that is used mostly in transport type aircraft is?
a) Boot-strap air cooling system
b) Simple air cooling system
c) Simple evaporative cooling system
d) Regenerative air cooling system

Answer: a
Explanation: Boot-strap air cooling system uses a secondary air compressor along with an after cooler for achieving higher pressures of compression and more cooling effect. The cooling turbine replaces the use of cooling fan as well. Hence the air cooling system that is used mostly in transport type aircraft is Boot-strap air cooling system.

10. A Boot-strap air cooling system is used for a transport aircraft to take 10TR of refrigeration load (Q). The power required for the refrigerating system P is 800 KW. What is its COP?
a) 0.06426
b) 0.04375
c) 0.05435
d) 0.04367

Answer: b
Explanation: COP = \(\frac{210×Q}{P×60}\)
COP = \(\frac{210×10}{800×60}\)
= 0.04375.