This set of Refrigeration Multiple Choice Questions & Answers (MCQs) focuses on “Vapour Refrigeration Cycles – Heat Pump”.
1. For a standard system with temperatures T1 and T2, where T1 < Ta < T2 (Ta – Atmospheric Temperature). Q1 is the heat extracted from a body at temperature T1, and Q2 is heat delivered to the body at temperature T2. What is the C.O.P. of the heat pump for given conditions?
a) Q2 / (Q2 − Q1)
b) (Q2 − Q1) / Q1
c) (Q2 − Q1) / Q2
d) Q1 / (Q2 − Q1)
View Answer
Explanation: As, C.O.P. = Desired effect / Work done
Here, work-done = Q2 − Q1
The desired temperature is T2. So, the heat delivered to achieve the desired temperature is Q2.
C.O.P. of the heat pump = Q2 / (Q2 − Q1).
2. What is the difference between Heat Pump and Refrigerator?
a) Heat Pump Gives efficiency and refrigerator gives C.O.P.
b) Both are similar
c) Both are almost similar, just the desired effect is different
d) Work is output in refrigerator and work is input in heat pump
View Answer
Explanation: Heat Pump and Refrigerator work on the same principle. Work needs to be given to get the desired effect. The characteristic which differentiates both of them is the temperature of the desired effect, heat pump desires for higher temperature whereas Refrigerator desires for lower temperature than atmospheric temperature.
3. What is the equation between efficiency of Heat engine and C.O.P. of heat pump?
a) ηE = (C.O.P.)P
b) ηE = 1 / (C.O.P.)P
c) ηE / (C.O.P.)P = 1
d) ηE x (C.O.P.)P = 0
View Answer
Explanation: ηE = W / Q hence for Carnot engine it is equal to (T2 – T1) / T2.
(C.O.P.)P for Carnot cycle is equal to T2 / (T2 – T1) .
So, these terms are related reciprocally.
4. How is the Relative coefficient of performance represented?
a) Theoretical C.O.P. / Actual C.O.P.
b) Actual C.O.P. / Theoretical C.O.P.
c) Theoretical C.O.P. x Actual C.O.P.
d) 1 / Theoretical C.O.P. x Actual C.O.P.
View Answer
Explanation: Relative C.O.P. is the ratio of an actual to the theoretical coefficient of performance. It is used to show the deviation of C.O.P. due to the ideal state and real state conditions.
5. C.O.P. of the heat pump is always _____
a) one
b) less than One
c) greater than One
d) zero
View Answer
Explanation: The second law of Thermodynamics states that a 100% conversion of heat into work is not possible without ideal conditions. So, efficiency will be less than 1. As C.O.P. is the reciprocal of efficiency, it tends to be more than 1.
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
View Answer
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 heat pump between temperature limits of 364 K and 294 K, then what is the value of C.O.P?
a) 4.2
b) 0.19
c) 5.2
d) 0.23
View Answer
Explanation: C.O.P. of reversed Carnot cycle is given by,
C.O.P. = T1 / (T2 – T1)
= 364 / (364 – 294)
= 5.2.
8. A reversed Carnot cycle is operating between temperature limits of (-) 33°C and (+) 27°C. If it acts as a heat engine gives an efficiency of 20%. What is the value of C.O.P. of a heat pump operating under the same conditions?
a) 6.5
b) 8
c) 5
d) 2.5
View Answer
Explanation: Temperature limits are given in the question so, we can calculate C.O.P. using the formula
C.O.P. = T1 / (T2 – T1)
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. = 1 / ηE
= 1 / (0.2) = 5.
9. If the coefficient of performance of the refrigerator is 4.67, then what is the value of the coefficient of performance of the heat pump operating under the same conditions?
a) 3.67
b) 5.67
c) 0.214
d) 9.34
View Answer
Explanation: As we know, the equation between the coefficient of performance of the Refrigerator and heat pump:
(C.O.P.)R = (C.O.P.)P – 1
Hence, C.O.P. of heat pump = C.O.P. of Refrigerator + 1
= 4.67 + 1
= 5.67.
10. A heat pump is used to maintain a hall at 30°C when the atmospheric temperature is 15°C. The heat loss from the hall is 1200 kJ/min. Calculate the power required to run the heat pump if its C.O.P. is 40% of the Carnot machine working between the same temperature limits.
a) 0.495
b) 4.04
c) 0.247
d) 8.08
View Answer
Explanation: Given data: T1 = 30°C = 30 + 273 = 303 K
T2 = 15°C = 15 + 273 = 288 K
Q1 = 1200 kJ/min = 1200/60 = 2 kW
Calculations: C.O.P. of heat pump working on Carnot cycle,
Ideal C.O.P. = T1 / (T1 − T2)
= 303 / (303 − 288)
= 20.2
Actual C.O.P = 0.4 x Ideal C.O.P.
= 0.4 x 20.2 = 8.08
C.O.P. = Q1 / W
Hence, W = Q1 / C.O.P.
W = 2 / 8.08
W = 0.247 kW.
11. A heat pump which runs (1/3)rd of time removes on an average 2400 kJ/hr of heat. If power consumed is 0.25 kW, what is the value of the C.O.P.?
a) 4
b) 2
c) 8
d) 6
View Answer
Explanation: Q1 = 2400 kJ/hr
= 2400 / (3600 / 3)
= 2 kW
C.O.P. = Q1 / W
= 2 / 0.25
= 8.
12. 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
View Answer
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.
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