1. What is not one of the advantages of using closed Air Refrigeration system?
a) Lighter in weight
b) Compact in construction
c) Environmental Friendly
d) Lower co-efficient of performance
Explanation: Air refrigeration system is used mainly due to light in weight, smaller and environment friendly due to use of air as refrigerant. Also it has higher co-efficient of performance.
2. Reduction in operating pressure in the Air refrigeration cycle results in ________
a) increase in C.O.P.
b) decrease in C.O.P.
c) no change in C.O.P.
d) always decreases
Explanation: As the operating pressure reduces, the change in work done decreases. Hence the C.O.P of refrigeration increases.
3. How is the refrigerant used in the Air refrigeration cycle?
a) In the condenser
b) In the compressor
c) Directly in contact
d) Not used at all
Explanation: Refrigerant used in Air refrigeration cycle is pure air, and it’s used directly in contact with the area of refrigeration. Whereas in Expansion refrigeration, refrigerants are used in condensers etc.
4. 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
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).
5. 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
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.
6. 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
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.
7.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
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.
8. C.O.P. of the heat pump is always _____
a) one
b) less than One
c) greater than One
d) zero
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.
9. 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.
10. 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
Explanation: C.O.P. of reversed Carnot cycle is given by,
C.O.P. = T1 / (T2 – T1)
= 364 / (364 – 294)
= 5.2