1. 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
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.
2. 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
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.
3.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
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.
4. 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
Explanation: Q1 = 2400 kJ/hr
= 2400 / (3600 / 3)
= 2 kW
C.O.P. = Q1 / W
= 2 / 0.25
= 8.
5. 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.
6. Bell-Coleman cycle is also known as __________
a) Carnot cycle
b) Reversed Brayton or Joule’s cycle
c) Rankine cycle
d) Otto cycle
Explanation: Bell-Coleman cycle was developed by Bell-Coleman and Light Foot by reversing the Joule’s air cycle and hence is also known as reversed Brayton or Joule’s cycle. It was one of the easiest types of refrigerators used in ships for carrying frozen meat.
7. In transport aviation, the air conditioning systems are based on ______ cycle.
a) Reversed Carnot cycle
b) Reversed Brayton cycle
c) Reversed Joule’s cycle
d) Otto cycle
Explanation: It is because in vapor-cycle the disadvantage was that due to the leakage loss of fluid it would cause the aircraft to be completely without cooling.
8. Which one of the following is not a true disadvantage of the Bell-Coleman cycle?
a) High running cost
b) Low COP
c) The danger of frosting at the expander valve is more
d) The size of the system is small
Explanation: Mass of air required per ton of refrigeration is large as compared to other systems of refrigeration. Hence the size of the system is large
9. Which one of the following is not a true advantage of the Bell-Coleman cycle?
a) Air is used a refrigerant which is easily available
b) It is safe as air is non-inflammable
c) Air is nontoxic, non-corrosive and stable
d) Weight of air refrigeration equipment per ton of refrigeration is much more in aircraft than other refrigeration systems
Explanation: Weight of air refrigeration equipment per ton of refrigeration is much less in aircraft than other refrigeration systems. Hence it is light. It is because of air compressor is already available in the air-craft.
10. The Bell-Coleman refrigeration cycle uses _____ as refrigerant.
a) Coolant
b) CO2
c) Air
d) H2O
Explanation: The Bell-Coleman refrigeration cycle uses air as refrigerant, as it is easily available, non-inflammable, non-toxic, stable and inert.