# Refrigeration Questions and Answers – Coefficient of Performance of Ideal VAR System – 2

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This set of Refrigeration Assessment Questions and Answers focuses on “Coefficient of Performance of Ideal VAR System – 2”.

1. What is the fundamental equation of the C.O.P. of Ideal VAR system?
a) Heat absorbed in the evaporator / Work done by a pump
b) Heat absorbed in the evaporator / Heat supplied in the generator
c) Heat absorbed in the evaporator / Work done by compressor + Heat supplied in the generator
d) Heat absorbed in the evaporator / Work done by pump + Heat supplied in the generator

Explanation: VAR uses the generator to do work as well as pump to pressurize the liquid to the generator. So, total work done is the summation of work done by a pump and generator. Refrigeration effect is the heat absorbed in the evaporator.
So, C.O.P. = Refrigeration effect / Total work done
= Heat absorbed in the evaporator / Work done by pump + Heat supplied in the generator.

2. What is the relation between QC, QG, and QE if they represent heat given to the refrigerant in the generator, heat discharging to atmosphere from condenser to absorber and heat absorbed by the refrigerant in the evaporator?
a) QC + QG = QE
b) QC = QG + QE
c) QC + QE = QG
d) QC = QG – QE

Explanation: According to the first law of thermodynamics, heat discharged to the atmosphere or cooling water from the condenser to absorber is equal to the heat given to the refrigerant in a generator, heat absorbed by the refrigerant in the evaporator and heat added to the refrigerant due to pump work. Heat added due to pump work is neglected due to minimal effect.
So, QC = QG + QE

3. What is the equation of VAR system in terms of entropy?
a) QC + QE / TC = QG / TG + QE / TE
b) QG + QC / TC = QG / TG + QE / TE
c) QG + QE / TC = QG / TG + QE / TE
d) QG + QE / TC = QC / TC + QE / TE

Explanation: As vapour absorption system is considered as a perfectly reversible system,
Hence the initial entropy of the system should be equal to the entropy of the system after some change in its conditions.
As entropy is heat absorbed divided by the temperature.
From the first law of thermodynamics, QC = QG + QE
So, the equation in terms of entropy QC / TC = QG / TG + QE / TE
Modified as, QG + QE / TC = QG / TG + QE / TE
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4. What is the ratio of QG / QE in terms of temperature?
If TC, TG and TE if they represent temperature at which heat is given to the refrigerant in the generator, temperature at which heat is discharging to atmosphere from condenser to absorber and temperature at which heat is absorbed by refrigerant in the evaporator
a) [TC – TE / TE] [TG / TG – TC]
b) [TC – TE / TE] [TG / TG – TE]
c) [TC – TE / TE] [TC / TG – TC]
d) [TC – TE / TC] [TG / TG – TC]

Explanation: By using the phenomenon of perfectly reversible system,
QG + QE / TC = QG / TG + QE / TE
QG / TG – QG / TC = QE / TC – QE / TE
QG [TC – TG / TC x TG] = QE [TE – TC / TC x TE]
By rearranging, QG / QE = [TC – TE / TE] [TG / TG – TC]

5. What is the value of maximum C.O.P. in terms of heat?
a) QG / QE
b) QE / QG
c) QG / QC
d) QC / QE

Explanation: As VAR uses the generator to do work as well as pump to pressurize the liquid to the generator. So, total work done is the summation of work done by a pump and generator. Refrigeration effect is the heat absorbed in the evaporator.
So, C.O.P. = Refrigeration effect / Total work done
= Heat absorbed in the evaporator / Work done by pump + Heat supplied in the generator
For maximum C.O.P., due to the minimal effect of work done by a pump, QP is neglected.
QE = Heat absorbed in the evaporator and QG = Heat supplied in the generator
C.O.P. maximum = QE / QG

6. What is the value of maximum C.O.P. in terms of temperature?
a) [TC – TE / TE] [TG / TG – TC]
b) [TC – TE / TE] [TG / TG – TC]
c) [TC – TE / TE] [TG / TG – TC]
d) [TG – TC / TG] [TE / TC – TE]

Explanation: As VAR uses the generator to do work as well as pump to pressurize the liquid to the generator. So, total work done is the summation of work done by a pump and generator. Refrigeration effect is the heat absorbed in the evaporator.
So, C.O.P. = Refrigeration effect / Total work done
= Heat absorbed in the evaporator / Work done by pump + Heat supplied in the generator
For maximum C.O.P., due to the minimal effect of work done by a pump, QP is neglected.
QE = Heat absorbed in the evaporator and QG = Heat supplied in generator
C.O.P. maximum = QE / QG
By using the phenomenon of perfectly reversible system,
QG + QE / TC = QG / TG + QE / TE
QG / TG – QG / TC = QE / TC – QE / TE
QG [TC – TG / TC x TG] = QE [TE – TC / TC x TE]
By rearranging, QE / QG = [TG – TC / TG] [TE / TC – TE]
(C.O.P.) max = QE / QG = [TG– TC / TG] [TE / TC – TE]

7. Which of the following is true?
a) (C.O.P.)max = (C.O.P.)Carnot / ηCarnot
b) (C.O.P.)max = (C.O.P.)Carnot + ηCarnot
c) (C.O.P.)max = (C.O.P.)Carnot x ηCarnot
d) (C.O.P.)max = (C.O.P.)Carnot – ηCarnot

Explanation: As, (C.O.P.) max = QE / QG = [TG – TC / TG] [TE / TC – TE]
Where, it may be noted that, [TE / TC – TE] is the coefficient of performance of a Carnot refrigerator working between TE and TC temperatures.
And, [TG – TC / TG] is the efficiency of a Carnot engine working between TG and TC temperatures.
Hence, (C.O.P.)max = (C.O.P.)Carnot x ηCarnot

8. C.O.P. of the VAR system is lower than the C.O.P. of VCR system.
a) True
b) False

Explanation: Though the VAR system has many advantages over VCR, VAR gives lower C.O.P. as compared to VCR. Compressor used in the VCR system given higher pressure difference, which increases the temperature more and thus increasing the heat rejection. Overall impacting the C.O.P. to increase but in case of VAR system, heat supplied is limited so overall C.O.P. is not as high as in the VCR.

9. In a VAR system, heating, cooling, and refrigeration take place at temperatures 200°C, 30°C, and -10°C respectively. What is the value of maximum C.O.P. of the system?
a) 2.202
b) 2.808
c) 3.404
d) 2.404

Explanation: Given: TG = 200°C = 200 + 273 = 473 K
TC = 30°C = 273 + 30 = 303 K
TE = – 10°C = 273 – 10 = 263 K
As, (C.O.P.)max = QE / QG = [TG – TC / TG] [TE / TC – TE]
= [263 / 303 – 263] [473 – 303 / 473]
= [263/ 40] [173 / 473]
= [45499 / 18920]
= 2.404.

10. What is the value of maximum C.O.P. if the VAR system has a capacity of 12 TR and heat given to the refrigerant in the generator is 40 kW?
a) 0.9523
b) 0.4878
c) 1.05
d) 2.05

Explanation: Given: QE = 12 TR = 12 x 3.5 = 42 kW
QG = 40 kW
As, C.O.P. maximum = QE / QG
= 42 / 40
= 1.05.

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