# 250+ TOP MCQs on Simple VCR System – Pressure – Enthalpy Chart and Answers

Refrigeration Multiple Choice Questions on “Simple VCR System – Pressure – Enthalpy Chart”.

1. What is the effect of using a flash chamber in VCR system?
a) C.O.P. increases
b) C.O.P. decreases
c) C.O.P. remains the same
d) Mass of refrigerant flowing through evaporator increases
Clarification: As C.O.P. of the VCR system containing flash chamber is given by,
C.O.P. = h1 − hf3 / h2 − h1
Which is same as the simple VCR cycle. The purest forms of refrigerant enter the respective system but effectively do not give enormous impact on C.O.P., So C.O.P. remains the same.
The mass of refrigerant required to flow through the evaporator decreases as even the smaller amount can produce the same refrigeration effect from its pure liquid form, resulting in a reduction in the size of evaporator.

2. From the following line diagram and p-h chart, what is the refrigeration effect in terms of the mass of mixture i.e., m2?  a) m2 (h1 − hf3)
b) m2 (h2 − h1)
c) m2 (h1 − h2)
d) m2 (h4 − h1)
Clarification: Refrigerating effect is considered in the evaporator i.e. heat absorbed by the evaporator is called as refrigerating effect.
R.E. = m1 (h1 − hf4’)
And also, m1 / m2 = h1 − h4 / h1 − hf4’
Hence, R.E. = m2 (h1 − hf3)

3. What is the value of C.O.P. in the VCR with flash chamber?  a) h1 – h2 / h1 – h4
b) h1 – hf3 / h2 – h1
c) h2 – h1 / h4 – h1
d) h1 – hf4’ / h1 – h4
Clarification: C.O.P. = Refrigerating effect / Work done
Refrigerating effect in terms of m2 = m2 (h1 – hf3)
Work = m2 (h2 – h1)
C.O.P. = m2 (h1 – hf3) / m2 (h2 – h1)
C.O.P. = (h1 – hf3) / (h2 – h1)

4. What is the effect of using accumulator in the VCR system?
a) C.O.P. increases
b) C.O.P. decreases
c) Total dry compression of refrigerant occurs
d) Sub-cooling happens
Clarification: As C.O.P. of the VCR system containing accumulator is given by,
C.O.P. = h1 – hf3 / h2 – h1
Which is same as the simple VCR cycle. The purest forms of refrigerant enter the respective system but effectively do not give enormous impact on C.O.P. So C.O.P. remains the same.
The C.O.P., R.E., and power required are the same, but it protects the liquid refrigerant from entering the compressor. Dry compression is always ensured by using accumulator or pre-cooler.

5. What is the value of C.O.P. in the VCR with a flash chamber?  a) h1 – h2 / h1 – h4
b) h1 – hf3 / h2 – h1
c) h2 – h1 / h4 – h1
d) h1 – hf4’ / h1 – h4
Clarification: C.O.P. = Refrigerating effect / Work done
Refrigerating effect in terms of m2 = m2 (h1 – hf3)
Work = m2 (h2 – h1)
C.O.P. = m2 (h1 – hf3) / m2 (h2 – h1)
C.O.P. = (h1 – hf3) / (h2 – h1)

6. From the following line diagram and p-h chart, what is the power required in terms of the mass of refrigerant flowing in the condenser i.e., m2?  a) m2 (h1 – h2) / 60
b) m2 (h1 – h4) / 60
c) m2 (h2 – h1) / 60
d) m2 (h1’ – hf4’) / 60
Clarification: Work is done by the compressor. So, the enthalpies related to compressor using the diagrams are h1 and h2. The mass of refrigerant flowing through the compressor is m2.
Hence, the power required is the work done per unit time.
Power i.e. P = Work / 60 (kW)
Work = m2 (h2 – h1)
P = m2 (h2 – h1) / 60.

7. What is the effect of using the following arrangement of VCR? a) C.O.P. increases
b) C.O.P. decreases
c) Total dry compression of refrigerant occurs
d) Sub-cooling does not occur
Clarification: Though sub-cooling is carried out in this arrangement. But the actual installations and many other losses affect widely over the sub-cooling. Sub-cooling increases the refrigeration effect and losses increase the compressor work and decreases refrigeration effect in some cases. So, C.O.P. being the ratio of refrigeration effect and work, effectively it reduces.

8. What is the ratio of the mass of the vapor refrigerant and mass of the liquid refrigerant i.e., m1 / m2? a) 1
b) h1 – h4 / h1’ – hf4’
c) 0
d) h1’ – hf4’ / h1 – h4
Clarification: In the case of this kind of heat exchanger, heat transfer takes place between vapor and liquid refrigerant due to the change of temperature and specific heat. As the mass flowing through condenser and evaporator is same. Due to heat transfer, mass does not get affected. Hence,
m1 = m2 or m1 / m2 = 1 = mR = 210 Q / (h1 – hf3’) kg/min.

9. What is the value of excess power required in this arrangement of VCR if P1 is the power required without heat exchanger and P2 is the power required with heat exchanger?  a) Pexcess = P1 / P2
b) Pexcess = P1 x P2
c) Pexcess = P1 + P2
d) Pexcess = P1 – P2
Clarification: Power required to drive the compressor,
P1 = mR (h2’ – h1’) / 60 = 210 Q [h2’ – h1’ / h1 – hf3’] / 60
Power required to drive the compressor without heat exchanger,
P2 = mR (h2 – h1) / 60 = 210 Q [h2 – h1 / h1 – hf3] / 60
Pexcess = P1 – P2
= 210 [(h2’ – h1’ / h1 – hf3’) – (h2 – h1 / h1 – hf3)] / 60 kW.

10. What is the value of C.O.P. in the VCR with a heat exchanger in the following diagram?  a) h1 – h2 / h1 – h4
b) h1 – hf3 / h2 – h1
c) h2 – h1 / h4 – h1
d) h1 – hf4’ / h1 – h4
Clarification: C.O.P. = Refrigerating effect / Work done
Refrigerating effect in terms of m2 = m2 (h1 – hf3)
Work = m2 (h2 – h1)
C.O.P. = m2 (h1 – hf3) / m2 (h2 – h1)
C.O.P. = (h1 – hf3) / (h2 – h1).

11. What is effect of using flash chamber in the following VCR system arrangement?  a) C.O.P. remains the same
b) C.O.P. decreases
c) C.O.P. increases
d) Subcooling affects significantly on C.O.P.