250+ TOP MCQs on Two/Multi Stage Compression with Water Intercooling and Liquid Subcooling – 2 and Answers

Refrigeration Question Bank on “Two/Multi Stage Compression with Water Intercooling and Liquid Subcooling – 2”.

1. Why multi-stage or compound VCR with water intercooler and liquid sub-cooler is used?
a) To improve C.O.P.
b) To decrease the refrigeration effect
c) To increase work
d) To increase leakage loss
Answer: a
Clarification: By de-superheating or intercooling the refrigerant before the second stage compression but not till the saturated vapor line, can decrease the total work required. By using the liquid sub-cooler, subcooling is done, and an increase in the refrigeration effect is achieved. So, the increase in R.E. and decrease in work improves the C.O.P. of the system.

2. What is the purpose of using intercooler?
a) To increase the cost of the compressor
b) To reduce the volumetric efficiency
c) To re-heat superheated refrigerant
d) To cool the superheated refrigerant
Answer: d
Clarification: Intercooler is used to cool the superheated refrigerant by some amount but in this case, not till the saturated vapour line. The superheated refrigerant leaving the first compressor is cooled by suitable method before entering the second compressor. This type of cooling is known as intercooling.

3. What is the purpose of using liquid sub-cooler?
a) To increase the cost of compressor
b) To reduce volumetric efficiency
c) To re-heat superheated refrigerant
d) To increase the refrigeration effect
Answer: d
Clarification: Liquid sub-cooler is used to sub-cool the liquid refrigerant coming from the condenser beyond the saturated liquid line. Sub-cooling results in an increase in refrigeration effect and thus increasing the C.O.P.

4. What does process 1-2 and 3-4 represent?

a) Evaporation
b) Compression
c) Condensation
d) Expansion
Answer: b
Clarification: Process 1-2 and 3-4 represent compression. This type of compression is isentropic compression, which is carried out in two stages. The first compression is carried out in low-pressure compressor and then compressed in the high-pressure compressor.

5. What does process 2-3 represent?

a) Evaporation
b) Compression
c) Intercooling
d) Condensation
Answer: d
Clarification: Process 2-3 represent intercooling or de-superheating. Intercooling is carried out to some extent but not till the saturated vapour line before the second stage of compression. Intercooling results in effective compression in the high-pressure compressor resulting in the reduction of the overall work done.

6. What do P_C and P_E represent?

a) Evaporator and expansion pressure
b) Compression and expansion pressure
c) Condenser and evaporator pressure
d) Intercooling pressure
Answer: c
Clarification: P_C is for condenser pressure. The pressure at which latent heat is rejected from the refrigerant and converting vapour form to liquid form. P_E is for evaporator pressure. The pressure at which latent heat of vaporization is absorbed from the medium and phase change from liquid to vapour is carried out. Attaining and maintaining these pressures is essential for the smooth operation of the cycle.

7. What is the mass of refrigerant passing through the evaporator in terms of enthalpies?

a) 210 Q / (h₁ – h₂)
b) 210 Q / (h₂ – h₃)
c) 210 Q / (h₁ – h₄)
d) 210 Q / (h₁ – h₇)
Answer: d
Clarification: The mass of refrigerant flowing in this type of arrangements is m, and the same amount of refrigerant flows through every equipment of the arrangement. As the refrigeration effect is the mass flowing through the evaporator times the enthalpy change across it.
Hence, m = R.E. / delta (h)
R.E. = TR = 210 Q kJ/min = 3.5 Q kW
Delta (h) = h₁ – h_f6 = h₁ – h₇
m = 210 Q / (h₁ – h₇).

8. What is the value of C.O.P. for the following arrangement?

a) (h₁ – h_f6) / (h₂ – h₁) – (h₄ – h₃)
b) (h₁ – h_f6) / (h₂ – h₁) + (h₄ – h₃)
c) (h₁ – h_f6) / (h₂ – h₁)
d) (h₁ – h_f6) / (h₄ – h₃)
Answer: b
Clarification: C.O.P. = Refrigeration effect / work
Hence, by taking the help of above diagrams,
R.E. = Heat absorbed in the evaporator
= mass flowing through the evaporator x enthalpy change
= m₁ (h₁ – h_f6) as, h_f6 = h₇
Work = Work done by low-pressure compressor + Work done by high-pressure compressor
= mass of refrigerant flowing through the low-pressure compressor x enthalpy change + mass of refrigerant flowing through the high-pressure compressor x enthalpy change
Work = m₁ (h₂ – h₁) + m₂ (h₄ – h₃) but here m₁ = m₂
Hence, C.O.P. = m₁ (h₁ – h_f5) / m₁ (h₂ – h₁) + m₂ (h₄ – h₃)
= (h₁ – h_f6) / (h₂ – h₁) + (h₄ – h₃).

9. In a compound vapour compression system, the mass flow rate of 2 kg/min is observed. The enthalpies related to point 1, 2, 3 and 4 in the following figure are 1350, 1580, 1540 and 1650 kJ/kg respectively. What is the amount of power required to run this system?

a) 11.33 kW
b) 22.66 kW
c) 680 kW
d) 5.67 kW
Answer: a
Clarification: Given: m = 2 kg/min
h₁ = 1350 kJ/kg
h₂ = 1580 kJ/kg
h₃ = 1540 kJ/kg
h₄ = 1650 kJ/kg
As, work = m [(h₂ – h₁) + (h₄ – h₃)]
= 2 [(1580 – 1350) + (1650 – 1540)]
= 2 [230 + 110]
= 680 kJ/min
Power required = Work / 60
= 680 / 60
= 11.33 kW.

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