1. 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]
2. 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
3. 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]
4. 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
5. 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.
6. 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.
7. 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.
8. Which one of the refrigerant is not used in the domestic Electrolux?
a) Ammonia
b) Water
c) Hydrogen
d) Freon – 12
Explanation: The ammonia is used as a refrigerant because it possesses most of the desirable properties. It is toxic, but due to the absence of moving parts, there is very little chance for leakage and the total amount used is also small. The hydrogen, being the lightest gas, is used to increase the rate of evaporation of the liquid ammonia passing through the evaporator.
9. What is water used for in the Domestic Electrolux?
a) Water is used to absorb hydrogen
b) Water is used to absorb ammonia
c) Water is used as a lubricant
d) Water is used to keep the inner surface clean
Explanation: The function of water is used as a solvent as it has the ability to absorb ammonia readily. The hydrogen is also non – corrosive and insoluble in water.
10. The Domestic absorption type refrigerator was invented by?
a) Carl Munters
b) Baltzer Von Platan
c) Carl Munters and Baltzer Von Platan
d) Light Foot
Explanation: The Domestic absorption type refrigerator was invented by two Swedish engineers, Carl Munters and Baltzer Von Platan, in 1925 while they were studying for their undergraduate course of Royal Institute of Technology in Stockholm. Whereas Light foot invented the Bell – Coleman cycle.