1. 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
Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)
= \(\frac{120-60}{60}\)
= 1
2. 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
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.
3. 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) \(\frac{1}{8}\)
Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)
= \(\frac{312-271}{273}\)
= \(\frac{1}{7}\).
4. 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
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.
5. 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
Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)
= \(\frac{250-100}{100}\)
= 1.5.
6. Efficiency of the Refrigerator is _________ to the C.O.P of refrigerator
a) inversely proportional
b) equal
c) independent
d) directly proportional
Explanation: Efficiency is the ratio of work done to heat supplied, whereas C.O.P is the ratio of Refrigeration effect to work done. Hence it is totally independent quantity
7. What is the C.O.P. of a refrigeration system if the work input is 40 KJ/kg and work output is 80 KJ/kg and refrigeration effect produced is 130 KJ/kg of refrigerant flowing.
a) 3.00
b) 3.25
c) 2.25
d) 3.75
Explanation: C.O.P. = \(\frac{Refrigeration \,effect}{Work \,Done \,(Output-Input)}\)
= \(\frac{130}{80-40}\)
= 3.25 (unit less).
8. Find the Relative C.O.P. of a refrigeration system if the work input is 100 KJ/kg and refrigeration effect produced is 250 KJ/kg of refrigerant flowing. Also Theoretical C.O.P. is 3.
a) 0.65
b) 0.80
c) 0.83
d) 0.91
Explanation: Actual C.O.P. = \(\frac{Refrigeration \,effect}{Work \,Done}\)
= \(\frac{250}{100}\)
= 2.5 (unit less)
Relative C.O.P. = \(\frac{Actual \,C.O.P.}{Theoretical \,C.O.P.}\)
= \(\frac{2.5}{3}\) = 0.833 i.e. = 83.3 %.
9. If a condenser and evaporator temperatures are 225 K and 100 K respectively, then reverse Carnot C.O.P is ________
a) 0.5
b) 1.5
c) 1.25
d) 1.75
Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)
= \(\frac{225-100}{100}\)
= 1.25.
10. If a condenser and evaporator temperatures are 312 K and X K respectively, and C.O.P. is given as 5 then find the value of X.
a) 52
b) 65
c) 78
d) 82
Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)
5 = \(\frac{312-X}{X}\)
6X = 312
X = 52 K.