Refrigeration Questions and Answers – Coefficient of Performance of Refrigeration – 1

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This set of Refrigeration Multiple Choice Questions & Answers (MCQs) focuses on “Coefficient of Performance of Refrigeration – 1”.

1. What is the term C.O.P. referred in terms of refrigeration?
a) Capacity of Performance
b) Co-efficient of Plant
c) Co-efficient of Performance
d) Cooling for Performance
View Answer

Answer: c
Explanation: Co-efficient of Performance is generally referred as C.O.P. for Refrigeration, which is used to measure the capacity or level up to which the refrigeration will occur.
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2. C.O.P. can be expressed by which equation?
a) \(\frac{Work \,Done}{Refrigeration \,effect}\)
b) \(\frac{Refrigeration \,effect}{Work \,Done}\)
c) \(\frac{Work\, Done}{Heat \,Transfer}\)
d) \(\frac{Heat \,Transfer}{Work \,Done}\)
View Answer

Answer: b
Explanation: Co-efficient of Performance is the ratio of the Refrigeration effect produced to the work done or work supplied to produce the effect.
Whereas the ratio- Work done to Heat transfer is called the efficiency.

3. What is the term relative C.O.P. referred in terms of refrigeration?
a) \(\frac{Actual \,C.O.P.}{Theoretical \,C.O.P.}\)
b) \(\frac{Theoretical \,C.O.P.}{Actual \,C.O.P.}\)
c) \(\frac{Actual \,C.O.P.}{Average \,C.O.P.}\)
d) \(\frac{Average \,C.O.P.}{Theoretical \,C.O.P.}\)
View Answer

Answer: a
Explanation: Relative Co-efficient of Performance is generally referred as the ratio of Actual C.O.P. measured to the Theoretical C.O.P. assumed before calculations. It is in relation with the theoretical C.O.P. and generally expressed in “%” value.
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4. Find the C.O.P. of a refrigeration system if the work input is 40 KJ/kg and refrigeration effect produced is 130 KJ/kg of refrigerant flowing.
a) 3.00
b) 2.25
c) 3.75
d) 3.25
View Answer

Answer: d
Explanation: Here, refrigeration effect = 130 KJ/kg, work input = 40 KJ/kg
C.O.P. = \(\frac{Refrigeration \,effect}{Work \,Done}\)
= \(\frac{130}{40}\)
= 3.25 (unit less).

5. Find the Relative C.O.P. of a refrigeration system if the work input is 60 KJ/kg and refrigeration effect produced is 130 KJ/kg of refrigerant flowing. Also Theoretical C.O.P. is 3.
a) 0.65
b) 0.79
c) 0.72
d) 0.89
View Answer

Answer: c
Explanation: Given, work input = 60 KJ/kg, refrigeration effect = 130 KJ/kg and Theoretical C.O.P. = 3
Actual C.O.P. = \(\frac{Refrigeration \,effect}{Work \,Done}\)
= \(\frac{130}{60}\)
= 2.167 (unit less)
Relative C.O.P. = \(\frac{Actual \,C.O.P.}{Theoretical \,C.O.P.}\)
= 2.167/3 = 0.722 i.e. = 72.2 %.
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6. Find the C.O.P. of a refrigeration system if the work input is 30 KJ/kg and refrigeration effect produced is 120 KJ/kg of refrigerant flowing.
a) 3.00
b) 4.00
c) 0.75
d) 0.25
View Answer

Answer: b
Explanation: Given, work input = 30 KJ/kg and refrigeration effect = 120 KJ/kg
C.O.P. = \(\frac{Refrigeration \,effect}{Work \,Done}\)
= \(\frac{120}{30}\)
= 4 (unit less).

7. Which equation represents efficiency in general?
a) \(\frac{Work \,Done}{Refrigeration \,effect}\)
b) \(\frac{Heat \,Trasfer}{Work \,Done}\)
c) \(\frac{Work\, Done}{Heat \,Transfer}\)
d) \(\frac{Refrigeration \,effect}{Work \,Done}\)
View Answer

Answer: c
Explanation: Work done to Heat transfer is called the efficiency in general sense, which can be applicable to Pumps, Refrigerators, and Turbines etc.
Whereas Co-efficient of Performance is the ratio of Refrigeration effect produced to the work done or work supplied to produce the effect.

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8. The Co-efficient of Performance is always __________
a) greater than 1
b) less than 1
c) equal to 1
d) zero
View Answer

Answer: a
Explanation: C.O.P. is always greater than 1, because in the ratio of C.O.P. the Refrigeration effect is always greater than the work done.

9. In a refrigerating machine, if the lower temperature is fixed, then the C.O.P. of machine can be increased by?
a) Increasing the higher temperature
b) Decreasing the higher temperature
c) Operating the machine at lower speed
d) Operating the machine at higher speed
View Answer

Answer: b
Explanation: By decreasing the higher temperature, the change in temperatures can be reduced. This directly reduces the word done i.e. the denominator in C.O.P. Hence the C.O.P increases.

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10. The reverse Carnot cycle C.O.P. can be expressed as _________ (Where t1 is the lower temperature and t2 is the higher temperature).
a) \(\frac{t1-t2}{t2}\)
b) \(\frac{t2-t1}{t2}\)
c) \(\frac{t2-t1}{t1-t2}\)
d) \(\frac{t2-t1}{t1}\)
View Answer

Answer: d
Explanation: Reverse Carnot C.O.P is the ratio of change in temperature i.e. t2-t1 to the lower temperature i.e. t1 here.

11. If a condenser and evaporator temperatures are 120 K and 60 K respectively, then reverse Carnot C.O.P is _________
a) 0.5
b) 1
c) 3
d) 2
View Answer

Answer: b
Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)
= \(\frac{120-60}{60}\)
= 1.

12. The C.O.P. of reverse Carnot cycle is most strongly dependent on which of the following?
a) Evaporator temperature
b) Condenser temperature
c) Specific heat
d) Refrigerant
View Answer

Answer: a
Explanation: The C.O.P in the reverse Carnot cycle generally depends on the higher of the two temperatures, as it is located in the numerator of C.O.P equation and responsible directly for increase or decrease in C.O.P.

13. If a condenser and evaporator temperatures are 312 K and 273 K respectively, then reverse Carnot C.O.P is _________
a) \(\frac{1}{5}\)
b) \(\frac{1}{6}\)
c) \(\frac{1}{7}\)
d) \(\frac{1}{8}\)
View Answer

Answer: c
Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)
= \(\frac{312-271}{273}\)
= \(\frac{1}{7}\).

14. The C.O.P for reverse Carnot refrigerator is 2. The ratio of lowest temperature to highest temperature will be _____
a) twice
b) half
c) four times
d) three times
View Answer

Answer: d
Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)
2 = \(\frac{X-Y}{Y}\)
2 = \(\frac{X}{Y}\) – 1
Thus, \(\frac{X}{Y}\) = 3 i.e. X=3Y i.e. Higher temperature = 3 times Lower temperature.

15. If a condenser and evaporator temperatures are 250 K and 100 K respectively, then reverse Carnot C.O.P is _________
a) 5.5
b) 1.5
c) 2.5
d) 3.0
View Answer

Answer: b
Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)
= \(\frac{250-100}{100}\)
= 1.5

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter