# 250+ TOP MCQs on Compound VCR System – Cascade Systems – 2 and Answers

Refrigeration Questions and Answers for Campus interviews on “Compound VCR System – Cascade Systems – 2”.

1. Multiple refrigerants can be used in the cascade refrigeration system.
a) True
b) False
Clarification: Multiple refrigerants can be used in the cascade refrigeration system. It helps to attain lower temperatures, and the refrigeration effect can be enormously increased.

2. How is the cascade system achieved?
a) VCR system in a parallel combination
b) VAR system in a series combination
c) VAR system in a parallel combination
d) VCR system in a series combination
Clarification: The cascade refrigeration system is structured by connecting two or more vapor compression refrigeration systems in series which use different refrigerants.

3. Cascade refrigeration system reduces the C.O.P.
a) True
b) False
Clarification: Due to the usage of multiple refrigerants and temperature overlap phenomenon, the refrigeration effect is increased, and overall work is reduced, leading to enhance the C.O.P. of the system.

4. What is the value of m1 in the following diagram? a) 210 Q / (h1 – h7) kg/min
b) 210 Q / (h1 – h4) kg/s
c) 210 Q / (h1 – h4) kg/min
d) 210 Q / (h1 – h7) kg/s
Clarification: As m1 flows through the low-temperature cascade system. So, when it passes through the evaporator i.e., between point 1 and point 4, there is a change in enthalpy, which produces the refrigeration effect.
Hence, R.E. = m x delta h
210 Q = m1 x (h1 – h4)
m1 = 210 Q / (h1 – h1) kg/min.

5. What is the value of m2 / m1 in the following diagram? a) (h2 – h4) / (h5 – h8)
b) (h6 – h4) / (h7 – h8)
c) (h1 – h4) / (h6 – h8)
d) (h2 – h1) / (h1 – h8)
Clarification: The cascade condenser is an intermediate between low temperature and high temperature cascade system. The heat between both systems is balanced hence, taking the thermal equilibrium of condenser.
Heat absorbed in the high temperature cascade system = Heat rejected in the low temperature cascade system
m2 h5 + m1 h4 = m1 h2 + m2 h8
m1 (h2 – h4) = m2 (h5 – h8)
m2 / m1 = (h2 – h4) / (h5 – h8).

6. What is the value of work done in the following diagram? a) m1 (h2 – h4) + m2 (h5 – h6)
b) m1 (h2 – h1) + m2 (h6 – h5)
c) m1 (h2 – h8) + m2 (h5 – h8)
d) m1 (h2 – h5) + m2 (h4 – h8)
Clarification: Work done is across the compressors.
Work = mass of refrigerant flowing through the compressor x change of enthalpy
Work done by low-temperature cascade system = m1 (h2 – h1)
Work done by high-temperature cascade system = m2 (h6 – h5)
Total work = Work done by low-temperature cascade system + Work done by high-temperature cascade system
Total work done = m1 (h2 – h1) + m2 (h6 – h5).

7. What is the value of C.O.P. in the following diagram? a) 210 Q / m1 (h2 – h4) + m2 (h5 – h6)
b) 210 Q / m1 (h2 – h7) + m2 (h6 – h8)
c) 210 Q / m1 (h2 – h8) + m2 (h5 – h8)
d) 210 Q / m1 (h2 – h1) + m2 (h6 – h5)
Clarification: As m1 flows through the low-temperature cascade system. So, when it passes through the evaporator i.e. between point 1 and point 4, there is change in enthalpy which produces the refrigeration effect.
Hence, R.E. = m x delta h
= 210 Q where, Q is the load on the low-temperature cascade system in tones of refrigeration
Work = mass of refrigerant flowing through the compressor x change of enthalpy
Work done by low-temperature cascade system = m1 (h2 – h1)
Work done by high-temperature cascade system = m2 (h6 – h5)
Total work = Work done by low-temperature cascade system + Work done by high-temperature cascade system
Total work done = m1 (h2 – h1) + m2 (h6 – h5)
As, C.O.P. = Refrigeration effect / Total work done
= 210 Q / m1 (h2 – h1) + m2 (h6 – h5).

8. What is the value of power required to drive the systems in the following diagram? a) m1 (h2 – h4) + m2 (h5 – h6) kW
b) m1 (h2 – h1) + m2 (h8 – h1) kJ/s
c) m1 (h2 – h1) + m2 (h6 – h5) kW
d) m1 (h2 – h1) + m2 (h6 – h5) kJ/min
Clarification: Power required to drive the system is the power required to do the work using compressors.
Work = mass of refrigerant flowing through the compressor x change of enthalpy
Work done by low-temperature cascade system = m1 (h2 – h1)
Work done by high-temperature cascade system = m2 (h6 – h5)
Total work = Work done by low-temperature cascade system + Work done by high-temperature cascade system
Total work done = m1 (h2 – h1) + m2 (h6 – h5) kJ/min
Power required = Work done / 60
= m1 (h2 – h1) + m2 (h6 – h5) kW.

9. What is the value of C.O.P. of low-temperature cascade system in terms of intermediate temperature in the following diagram? If TEL and TCL = Evaporator and condenser temperatures for low-temperature cascade system
TEH and TCH = Evaporator and condenser temperatures for high-temperature cascade system
a) TEL / TCL – TCL
b) TCL / TCL – TEL
c) TCL / TI – TCL
d) TEL / TI – TCL
Clarification: Temperature of the condenser is higher than the temperature of the evaporator. The desired effect is obtained in the evaporator. So, by using the Carnot’s theorem to low-temperature cascade system,
C.O.P. = Temperature of the evaporator / Temperature of the condenser – Temperature of the evaporator
= TEL / TCL – TEL
As, when low-temperature cascade condenser temperature is equal to high-temperature cascade evaporator temperature.
TCL = TEH = TI
Hence, C.O.P. = TEL / TI – TCL.

10. What is the value of C.O.P. of high-temperature cascade system in terms of intermediate temperature in the following diagram? If TEL and TCL = Evaporator and condenser temperatures for low-temperature cascade system
TEH and TCH = Evaporator and condenser temperatures for high-temperature cascade system
a) TEH / TCH – TCH
b) TI / TCH – TI
c) TCH / TI – TCH
d) TEH / TI – TCH
Clarification: Temperature of the condenser is higher than the temperature of the evaporator. The desired effect is obtained in the evaporator. So, by using the Carnot’s theorem to high-temperature cascade system,
C.O.P. = Temperature of the evaporator / Temperature of the condenser – Temperature of the evaporator
= TEH / TCH – TEH
As, when low-temperature cascade condenser temperature is equal to high-temperature cascade evaporator temperature.
TCL = TEH = TI
Hence, C.O.P. = TI / TCH – TI.

11. What is the value of pressure ratio if the pressure at point 1 and 2 is 1.809 and 3.467 bar, respectively? a) 1.916
b) 1.916 bar
c) 0.521
d) 0.521 bar