# Refrigeration Questions and Answers – Air Refrigerator Working on Reverse Carnot Cycle – 2

«
»

This set of Refrigeration Questions and Answers for Experienced people focuses on “Air Refrigerator Working on Reverse Carnot Cycle – 2”.

1. If a refrigeration system having T1 and T2 as lower and higher temperatures respectively then, what is the value of C.O.P of the refrigeration system working on the reversed Carnot cycle?
a) T2 / (T2 − T1)
b) (T2 − T1) / T1
c) (T2 − T1) / T2
d) T1 / (T2 − T1)

Explanation: As, C.O.P. = Desired effect / Work done
Here, work-done = Q1 − Q1
The desired temperature is T1. So, the heat delivered to achieve the desired temperature is Q1.
C.O.P. of the heat pump = Q1 / (Q2 − Q1).
According to Carnot’s theorem,
C.O.P. = T1 / (T2 − T1).

2. In a refrigerating machine working on the reversed Carnot cycle, if the lower temperature is fixed, then what can be done to increase the C.O.P.?
a) Increasing higher temperature
b) Operating machine at higher speed
c) Decreasing higher temperature
d) Operating machine at a lower speed

Explanation: As C.O.P. of the refrigerator working on the Carnot cycle is given by,
C.O.P. = T1 / (T2 − T1)
So, If T1 is fixed, so by decreasing the denominator C.O.P. can be increased. If the higher temperature is reduced then by keeping numerator constant, the denominator decreases and leads to an increase in C.O.P.

3. If the condenser and evaporator temperatures are 320 K and 240 K respectively, then what is the value of the C.O.P.?
a) 0.25
b) 4
c) 0.33
d) 3

Explanation: Given: T1 = 240 K
T2 = 320 K
C.O.P. = T1 / (T2 − T1)
= 240 / (320 − 240)
= 240 / (80)
= 3.

4. The efficiency of Carnot heat engine is 40%. What is the value of C.O.P. of a refrigerator operating on reversed Carnot cycle?
a) 2.5
b) 1.5
c) 4
d) 10

Explanation: ηE = 40% = 0.4
C.O.P. of heat pump = 1 / ηE = 1 / 0.4 = 2.5
As we know, (C.O.P.)R = (C.O.P.)P − 1
C.O.P. of refrigerator = 2.5 − 1
= 1.5.

5. The C.O.P. of a reversed Carnot refrigerator is 5. What is the ratio of highest temperature to lower temperature?
a) −1.2
b) 0.8
c) 1.2
d) −0.8

Explanation: As C.O.P. of the refrigerator working on the Carnot cycle is given by,
C.O.P. = T1 / (T2 − T1)
5 = T1 / (T2 − T1)
(T2 − T1) / T1 = 1 / 5
(T2 / T1) − (T1 / T1) = 0.2
T2 / T1 − 1 = 0.2
T2 / T1 = 1.2.

6. For the systems working on reversed Carnot cycle, what is the relation between C.O.P. of Refrigerator i.e. (C.O.P.)R and Heat Pump i.e. (C.O.P.)P?
a) (C.O.P.)R + (C.O.P.)P = 1
b) (C.O.P.)R = (C.O.P.)P
c) (C.O.P.)R = (C.O.P.)P − 1
d) (C.O.P.)R + (C.O.P.)P + 1 = 0

Explanation: If we put the values of C.O.P. for standard system i.e. (C.O.P.)R = T1 / (T2 − T1) and
(C.O.P.)P = T2 / (T2 − T1),
(C.O.P.)P − (C.O.P.)R = 1.
{T2 / (T2 − T1)} − {T1 / (T2 − T1)} = 1.

7. If the reversed Carnot cycle operating as a refrigerator between temperature limits of 405 K and 255 K, then what is the value of C.O.P.?
a) 2.7
b) 0.588
c) 1.7
d) 0.370

Explanation: C.O.P. of reversed Carnot cycle is given by,
C.O.P. = T1 / (T2 − T1)
= 255 / (405 − 255)
= 1.7.

8. A reversed Carnot cycle is operating between temperature limits of 272 K and (+) 49°C. If it acts as a heat engine gives an efficiency of 15.52%. What is the value of C.O.P. of a refrigerator operating under the same conditions?
a) 6.44
b) 0.1838
c) 5.44
d) 2

Explanation: T1 = 272 K, T2 = 49°C = 322 K
Temperature limits are given in the question so; we can calculate C.O.P. using the formula
C.O.P. = T1 / (T2 − T1)
= 272 / (322 − 272)
= 5.44
Alternative approach:
But as the efficiency of the heat engine is given so directly by the relation, we can find out the C.O.P.
C.O.P. of heat pump = 1 / ηE
= 1 / (0.1552) = 6.44
C.O.P. of refrigerator = C.O.P. of heat pump − 1 = 6.44 − 1 = 5.44.

9. If the Coefficient of performance of a heat pump is 5, then what is the value of the Coefficient of performance of the refrigerator operating under the same conditions?
a) 4
b) 6
c) 0.2
d) 3

Explanation: As we know, the equation between Coefficient of performance of Refrigerator and heat pump:
(C.O.P.)R = (C.O.P.)P − 1
Hence, C.O.P. of refrigerator = C.O.P. of heat pump − 1
= 5 − 1
= 4.

10. C.O.P. of the refrigerator is always _____ the C.O.P. of the heat pump when both are working between the same temperature limits.
a) less than
b) greater than
c) equal to
d) inverse of

Explanation: C.O.P. = Desired effect / Work
As the desired effect for the heat pump is higher than the refrigerator. So, numerator value is higher for heat pump keeping denominator constant.
Can also be proved by this equation,
(C.O.P.)R = (C.O.P.)P − 1.

11. Two refrigerators M and N operate in series. The refrigerator M absorbs energy at the rate of 1 kW from a body at temperature 400 K and rejects energy as heat to a body at a temperature T. The refrigerator N absorbs the same quantity of energy which is rejected by the refrigerator N from the body at temperature T, and rejects energy as heat to a body at temperature 800 K. If both the refrigerators have the same C.O.P., what is the value temperature T?
a) 600 K
b) 200 K
c) 565.68 K
d) 459.21 K

Explanation: Given: T1 = 400 K, T2 = 800 K, C.O.P. of M = C.O.P. of N
C.O.P. of refrigerator M = T1 / (T2 − T1)
= 400 / (T − 400)
C.O.P. of refrigerator N = T1 / (T2 − T1)
= T / (800 − T)
Equating both the C.O.P. we get,
400 / (T − 400) = T / (800 − T)
T2 = 400 x 800
T = √320000
T = 565.68 K.

12. A refrigerator works on a reversed Carnot cycle. The Coefficient of performance is 3.5. If the capacity of the refrigerator is 6 TR. Find out the power required to run the refrigerator.
a) 5 kW
b) 3.5 kW
c) 360 kJ/min
d) 520 kJ/min

Explanation: Given: Q = 6 TR = 6 x 3.5 = 21 kW
C.O.P.= 3.5
C.O.P. = Desired effect / Work
Work = Q / C.O.P.
= 21 / 3.5
= 6 kw = 360 kJ/min.

13. A refrigerator works on a reversed Carnot cycle. The Coefficient of performance is 7. If the capacity of the refrigerator is 12 TR. What is the value of heat rejected from the system?
a) 46 kW
b) 50 kW
c) 42 kJ/min
d) 48 kJ/min

Explanation: Given: Q1 = 12 TR = 12 x 3.5 = 42 kW
C.O.P. = 5
C.O.P. = Desired effect / Work
Work = Q1 / C.O.P.
= 42/7
= 6 kW
Heat rejected from the system, Q2 = Q1 + W
= 42 + 6
= 48 kW.

Sanfoundry Global Education & Learning Series – Refrigeration & Air Conditioning.

To practice all areas of Refrigeration for Experienced people, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs! 