Refrigeration Questions and Answers for Experienced people 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)
Answer: d
Clarification: 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
Answer: c
Clarification: 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
Answer: d
Clarification: 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
Answer: b
Clarification: 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
Answer: c
Clarification: 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
Answer: c
Clarification: 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
Answer: c
Clarification: 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