250+ TOP MCQs on Theoretical VCR with Dry Saturated Vapour after Compression and Answers

Refrigeration Quiz on “Theoretical VCR with Dry Saturated Vapour after Compression”.

1. Refrigerating effect corresponds to which of the following processes?
a) Condensing process
b) Vaporizing process
c) Compression process
d) Expansion process

Answer: b
Clarification: During evaporation, the refrigerant absorbs its latent heat of vaporization from the medium or system which is to be cooled. So, the heat absorbed by the refrigerant is known as Refrigerating effect carried out in the vaporizing process.

2. Find out the value of C.O.P. for the given values of enthalpies.
h₁ = 1126.25 kJ/kg
h₄ = 250.25 kJ/kg
W = 120 kJ/kg
a) 5.3
b) 6.3
c) 7.3
d) 8.3

Answer: c
Clarification: Refrigerating effect = h₁ − h₄ = 1126.25 − 250.25 = 876 kJ/kg
Work = 120 kJ/kg
C.O.P. = Refrigerating effect / Work done
= 876 / 120
= 7.3.

3. If a process 1 – 2 undergoes a throttling. Which of the following represents the process 1 – 2?
a) T1 = T2
b) S1 = S2
c) P1 = P2
d) H1 = H2

Answer: d
Clarification: In a throttling process, no heat is absorbed or rejected by the liquid refrigerant. It is also called as an isenthalpic process, in which change in internal energy and flow work remains the same. Enthalpies remain same, so H1 = H2 represents throttling.
T1 = T2 is the isothermal process.
S1 = S2 is an isentropic process.
P1 = P2 is an isobaric process.

4. In a vapor compression refrigeration system, the coefficient of performance is observed as 10. If the enthalpy after compression is 1368.28 kJ/kg and after condensation is 598.58 kJ/kg. If the system works on 50% of the C.O.P., then what is the value of enthalpy after evaporation?
a) 1220
b) 1210
c) 1240
d) 1280

Answer: c
Clarification: Given data: Theoretical C.O.P. = 10
h₂ = 1368.28 kJ/kg
h₄ = h₃ = 598.58 kJ/kg
Calculations: Actual C.O.P. = 50 * 10 / 100 = 5
As, C.O.P. = (h₁ − h₄) / (h₂ − h₁)
5 = (h₁ − 598.58) / (1368.28 − h₁)
h₁ = 1239.99 = 1240 kJ/kg.