1. Isentropic compression of saturated vapour of all fluids leads to superheated vapuor. (T/F)
(2 Mark, 1990)

Ans: T
2. A refrigeration compressor designed to operate with R22 _____ (can/cannot) be operated with R12 because the condensing pressure of R22 at any given temperature is ______ (higher/lower) than that of R12.
(2 Mark, 1992)

Ans: cannot, higher
3. Round the clock cooling of an apartment having a load of 300 MT/day requires an airconditioning plant of capacity about
(a) 1 ton
(b) 5 tons
(c) 10 tons
(d) 100 tons
(2 Mark, 1993)

Ans: a
Explanation:
Refrigeration capacity = \frac { 300\times { 10 }^{ 3 } }{ 24\times 3600 }
=> RC = 3.472 kW \simeq 3.5 kW = 1 ton
4. The use of Refrigerant22 (R22) for temperatures below 30°C is not recommended due to its
(a) Good miscibility with lubricating oil
(b) Poor miscibility with lubricating oil
(c) Low evaporating pressure
(d) High compressor discharge temperature
(2 Mark, 1993)

Ans: d
Explanation:
The discharge temperature of R22 is quite high, hence the degree of superheat in the systems using this refrigerant should be minimum.
5. Clearance volume of a reciprocating compressor is 100 ml, and the volume of the cylinder at bottom dead centre is 1.0 litre. The clearance ratio of the compressor is
(a) 1/11
(b) 1/10
(c) 1/9
(d) 1/12
(2 Mark, 1997)

Ans: c
Explanation:
Clearance volume (V_{c}) = 100 ml
Total volume (V) = 1 lit = 1000 ml
Swept volume (V_{s}) = 1000 – 100 = 900 ml
Clearance ratio = \frac { { V }_{ c } }{ { V }_{ s } } =\frac { 100 }{ 900 } =\frac { 1 }{ 9 }

Ans: A – 2, B – 4, C – 6, D – 1
7. In a vapour compression refrigeration system, liquid to suction heat exchanger is used to
(a) Keep the COP constant
(b) Prevent the liquid refrigerant from entering the compressor
(c) Subcool the liquid refrigerant leaving the condensor
(d) Subcool the vapour refrigerant from the evaporator
(1 Mark, 2000)

Ans: c
Explanation:
Benefits of liquid to suction heat exchanger:
1) It provides subcooling to the liquid refrigerant prior to entering the expansion valve.
2) Eliminate the flash gas forming in the liquid line
8. Global warming is caused by
(a) Ozone
(b) Carbon dioxide
(c) Nitrogen
(d) Carbon monoxide
(1 Mark, 2000)

Ans: b
Data for Q.9 – 10 are given below.
A refrigerator based on ideal vapour compression cycle operates between the temperature limits of –20°C and 40°C. The refrigerant enters the condenser as saturated vapour and leaves as saturated liquid. The enthalpy and entropy values for saturated liquid and vapour at these temperatures are given in the table below.
9. If refrigerant circulation rate is 0.025 kg/s, the refrigeration effect is equal to
(a) 2.1 kW
(b) 2.5 kW
(c) 3.0 kW
(d) 4.0 kW
(2 Mark, 2003)
10. The COP of the refrigerator is
(a) 2.0
(b) 2.33
(c) 5.0
(d) 6.0
(2 Mark, 2003)

Ans: b
Explanation:
COP = \frac { Refrigeration\quad effect }{ Compressor\quad work }
=> COP = \frac { { h }_{ 1 }{ h }_{ 4 } }{ { h }_{ 2 }{ h }_{ 1 } }
=> COP = \frac { 16480 }{ 200164 } = 2.33
11. An industrial heat pump operates between the temperatures of 27°C and 13°C. The rates of heat addition and heat rejection are 750 W and 1000W, respectively. The COP for the heat pump is
(a) 7.5
(b) 6.5
(c) 4.0
(d) 3.0
(1 Mark, 2003)

Ans: c
Explanation:
COP of heat pump = \frac { Heat\quad rejected }{ Work\quad input }
=> COP = \frac { 1000 }{ 1000750 } = 4
12. A heat engine having an efficiency of 70% is used to drive a refrigerator having a coefficient of performance of 5. The energy absorbed from low temperature reservoir by the refrigerator for each kJ of energy absorbed from high temperature source by the engine is
(a) 0.14 kJ
(b) 0.71 kJ
(c) 3.5 kJ
(d) 7.1 kJ
(2 Mark, 2004)
13. A R12 refrigerant reciprocating compressor operates between the condensing temperature of 30^{0}C and evaporator temperature of 20^{0}C. The clearance volume ratio of the compressor is 0.03. Specific heat ratio of the vapour is 1.15 and the specific volume at the suction is 0.1089 m^{3}/kg. Other properties at various states are given in the figure. To realize 2 Tons of refrigeration, the actual volume displacement rate considering the effect of clearance is
(a) 6.35 × 10^{3} m^{3}/s
(b) 63.5 × 10^{3} m^{3}/s
(c) 635 × 10^{3} m^{3}/s
(d) 4.88 × 10^{3} m^{3}/s
(2 Mark, 2004)

Ans: a
Explanation:
Net refrigeration effect = 2 Ton = 2 * 3.5 = 7 kW
Net refrigeration effect = m(h_{1} – h_{4})
=> 7 = m (176 – 65)
=> m = 0.063 kg/s
Volume = mass * Specific volume
=> V = 0.063 * 0.1089 = 6.867 * 10^{3} m^{3}/s
Volumetric efficiency = 1 + C – C{ \left( \frac { { p }_{ 2 } }{ { p }_{ 1 } } \right) }^{ \frac { 1 }{ n } }
=> { \eta }_{ V }=1+0.030.03{ \left( \frac { 7.45 }{ 1.50 } \right) }^{ \frac { 1 }{ 1.15 } } = 0.909
Volume displacement rate considering effect of clearance = 6.867 * 10^{3 }* 0.909 = 6.867 * 10^{3 }m^{3}/s
14. Environment friendly refrigerant R134 is used in the new generation domestic refrigerators. Its chemical formula is
(a) CH Cl F_{2}
(b) C_{2} Cl_{3} F_{3}
(c) C_{2} Cl_{2} F_{4}
(d) C_{2} H_{2} F_{4}
(1 Mark, 2004)

Ans: d
Explanation:
R134:
Here, m – 1 = 1 => m = 2
n + 1 = 3 => n = 2
p = 4
We know: n + p + q = 2m + 2
=> 2 +4 + q = 2 * 2 + 2
=> q = 0
Now, General chemical formula: C_{m}H_{n}F_{p}Cl_{q}
After putting the values of m, n, p, q the chemical formula becomes C_{2}H_{2}F_{4}
15. The vapor compression refrigeration cycle is represented as shown in the figure below, with state 1 being the exit of the evaporator.
The coordinate system used in this figure is
(a) ph
(b) Ts
(c) ps
(d) Th
(2 Mark, 2005)

Ans: d
Explanation:
1 – 2: isentropic process. So, option ‘b’ and ‘c’ is not correct.
2 – 3: Isobaric process. So, option ‘a’ is not correct.
From elimination process, the answer will be option ‘d’.
16. A vapour absorption refrigeration system is a heat pump with three thermal reservoirs as shown in the figure. A refrigeration effect of 100W is required at 250 K when the heat source available is at 400 K. Heat rejection occurs at 300 K. the minimum value of heat required (in W) is
(a) 167
(b) 100
(c) 80
(d) 20
(2 Mark, 2005)

Ans: c
Explanation:
Consider, the temperature of evaporator (T_{E}) = 250 K
The temperature of Condensor and Absorber (T_{A}) = 300 K
The temperature of generator (T_{G}) = 400 K
COP of VARS = \frac { { T }_{ E }\left( { T }_{ G }{ T }_{ A } \right) }{ { T }_{ G }\left( { T }_{ A }{ T }_{ E } \right) }
=> COP = \frac { 250\left( 400300 \right) }{ 400\left( 300250 \right) }
=> \frac { { Q }_{ E } }{ { Q }_{ G } } =\frac { 5 }{ 4 }
=> \frac { 100 }{ { Q }_{ G } } =\frac { 5 }{ 4 }
=> { Q }_{ G } = 80W
So, the minimum value of heat required is 80 W.
17. In an ideal vapour compression refrigeration cycle, the specific enthalpy of refrigerant (in kJ/kg) at the following states is given as:
Inlet of condenser: 283
Exit of condenser: 116
Exit of evaporator: 232
The COP of this cycle is
(a) 2.27
(b) 2.75
(c) 3.27
(d) 3.75
(1 Mark, 2009)
Data for Q.18 – 19 are given below.
A refrigerator operates between 120 kPa and 800 kPa in an ideal vapor compression cycle with R134a as the refrigerant. The refrigerant enters the compressor as saturated vapor and leaves the condenser as saturated liquid. The mass flow rate of the refrigerant is 0.2 kg/s. Properties for R134a are as follows:
18. The rate at which heat is extracted, in kJ/s from the refrigerated space is
(a) 28.3
(b) 42.9
(c) 34.4
(d) 14.6
(2 Mark, 2012)
19. The power required for the compressor in kW is
(a) 5.94
(b) 1.83
(c) 7.9
(d) 39.5
(2 Mark, 2012)
20. Which one of the following is a CFC refrigerant?
(a) R744
(b) R290
(c) R502
(d) R718
(1 Mark, 2014[1])

Ans: c
Explanation:
R744 ——> CO_{2}
R290 ——> C_{3}H_{8}
R718———> H_{2}O
R502——> 48.8% R22 + 51.2% R115
R502—–> 48.8% CHClF_{2} + 51.2% C_{2}ClF_{5}
21. A heat pump with refrigerant R22 is used for space heating between temperature limits of −20°C and 25°C. The heat required is 200 MJ/h. Assume specific heat of vapour at the time of discharge as 0.98 kJ/kg.K. Other relevant properties are given below. The enthalpy (in kJ/kg) of the refrigerant at isentropic compressor discharge is _______.

Ans: 433.33
Explanation:
We know: s_{2} = s_{1}
=> s_{2′ }+ C.ln\frac { { T }_{ 2 } }{ { T }_{ 2' } } = s_{1}
=> 1.7183 + 0.98\times ln\frac { { T }_{ 2 } }{ 298 } = 1.7841
=> T_{2 }= 318.695 K
Again, h_{2} = h_{2’} + C(T_{2} – T_{3})
=> h_{2 }= 413.02 + 0.98 * (318.695 – 298)
=> h_{2 }= 433.3 kJ/kg
22. A reversed Carnot cycle refrigerator maintains a temperature of −5°C. The ambient air temperature is 35°C. The heat gained by the refrigerator at a continuous rate is 2.5 kJ/s. The power (in watt) required to pump this heat out continuously is ______.
(1 Mark, 20014[4])

Ans: 373.13
Explanation:
COP of refrigerator = \frac { { T }_{ 2 } }{ { T }_{ 1 }{ T }_{ 2 } }
=> COP = \frac { 268 }{ 308268 } = 6.7
Again, COP = \frac { { Q }_{ 2 } }{ W }
=> 6.7 = \frac { 2.5\times { 10 }^{ 3 } }{ W }
=> W = 373.13 Watt
23. The COP of a Carnot heat pump operating between 6°C and 37°C is _____.
(1 Mark, 2015[2])

Ans: 10
Explanation:
T_{h }= 37 + 273 = 310 K
T_{l }= 6 + 273 = 279 K
COP = \frac { { T }_{ h } }{ { T }_{ h }{ T }_{ l } } =\frac { 310 }{ 310279 } = 10
24. Refrigerant vapor enters into the compressor of a standard vapor compression cycle at −10°C (h = 402 kJ/kg) and leaves the compressor at 50°C (h = 432 kJ/kg). It leaves the condenser at 30°C (h = 237 kJ/kg). The COP of the cycle is _____.
(2 Mark, 2015[3])
25. A refrigerator uses R134a as its refrigerant and operates on an ideal vapourcompression refrigeration cycle between 0.14 MPa and 0.8 MPa. If the mass flow rate of the refrigerant is 0.05 kg/s, the rate of heat rejection to the environment is ______kW.
Given data:
At P = 0.14 MPa, h = 236.04 kJ/kg, s = 0.9322 kJ/kgk (saturated vapour)
At P = 0.8 MPa, h = 272.05 kJ/kg, s = 0.9322 kJ/kgk (superheated vapour)
At P = 0.8 MPa, h = 93.42 kJ/kg (saturated liquid)
(2 Mark, 2016[2])
26. The heat removal rate from a refrigerated space and the power input to the compressor are 7.2 kW and 1.8 kW, respectively. The coefficient of performance (COP) of the refrigerator is ______.
(1 Mark, 2016[2])

Ans: 4
Explanation:
We know: COP = \frac { Refrigeration\quad Effect }{ Compressor\quad work }
=> COP = \frac { 7.2 }{ 1.8 } = 4
27. In the vapour compression cycle shown in the figure, the evaporating and condensing temperatures are 260 K and 310 K respectively. The compressor takes in liquidvapour mixture (state 1) and isentropically compresses it to a dry saturated vapour condition (state 2). The specific heat of the liquid refrigerant is 4.8 kJ/kgK and may be treated as constant. The enthalpy of evaporation for the refrigerant at 310 K is 1054 kJ/kg.
The difference between the enthalpies at state points 1 and 0 (in kJ/kg) is __.
(2 Mark, 2016[3])

Ans: 1103.51
Explanation:
Assume at some temperature x, the entropy is s_{x}.
s_{3} – s_{x} = cln\frac { { T }_{ 3 } }{ x }
=> s_{3} = s_{x} + 4.8ln\frac { 310 }{ x } …..(1)
Simillarily, s_{0} = s_{x} + 4.8ln\frac { 260 }{ x } …..(2)
Again, s_{2} – s_{3 }= \frac { { h }_{ fg } }{ T }
=> s_{2} = s_{3 }+ \frac { 1054 }{ 310 }……..(3)
putting the value of s_{3 }in equation (3)
s_{2} = s_{x} + 4.8ln\frac { 310 }{ x }_{ }+ \frac { 1054 }{ 310 }…(4)
We know, s_{2} = s_{1}
So, Equation 4 becomes
s_{1} = s_{x} + 4.8ln\frac { 310 }{ x }_{ }+ \frac { 1054 }{ 310 }…..(5)
Between state 0 and 1: h_{1} – h_{0} = T_{1}(s_{1} – s_{0})
=> h_{1} – h_{0} = 260(s_{1} – s_{0})……(6)
By using the Equations (6), (5) and (2)
h_{1} – h_{0 }= 4.8\times \left\{ ln\left( \frac { 310 }{ T } \right) ln\left( \frac { 260 }{ T } \right) \right\} +\frac { 1054 }{ 310 }
=> h_{1} – h_{0 }= 1103.51 kJ/kg
28. In a 3stage air compressor, the inlet pressure is P_{1}, discharge pressure is P_{4} and the intermediate pressures are P_{2 }and P_{3} (P_{2} < P_{3}). The total pressure ratio of the compressor is 10 and the pressure ratios of the stages are equal. If P_{1} = 100 kPa, the value of the pressure P_{3} (in kPa) is _____.
(2 Mark, 2016[3])

Ans: 464.16
Explanation:
Given, Pressure ratios of the stages are equal.
i.e. \frac { { P }_{ 2 } }{ { P }_{ 1 } } =\frac { { P }_{ 3 } }{ { P }_{ 2 } } =\frac { { P }_{ 2 } }{ { P }_{ 1 } } ={ r }_{ p }
Again, \frac { { P }_{ 4 } }{ { P }_{ 1 } } =10
=> \frac { { P }_{ 2 } }{ { P }_{ 1 } } \times \frac { { P }_{ 3 } }{ { P }_{ 2 } } \times \frac { { P }_{ 4 } }{ { P }_{ 3 } } = 10
=> { r }_{ p }\times { r }_{ p }\times { r }_{ p } = 10
=> { r }_{ p }={ 10 }^{ \frac { 1 }{ 3 } }=2.1544
Given, \frac { { P }_{ 4 } }{ { P }_{ 1 } } =10
=> { P }_{ 4 }=10\times 100=1000 kPa
Again, \frac { { P }_{ 4 } }{ { P }_{ 3 } } ={ r }_{ p }
=> { P }_{ 3 }=\frac { { P }_{ 4 } }{ { r }_{ p } } = 464.16 kPa
29. A standard vapour compression refrigeration cycle operating with a condensing temperature of 35^{0}C and an evaporating temperature of 10^{0}C develops 15 kW of cooling. The ph diagram shows the enthalpies at various states. If the isentropic efficiency of the compressor is 0.75, the magnitude of compressor power (in kW) is _________ (correct to two decimal places).
(2 Mark, 2018[2])

Ans: 10
Explanation:
Refrigeration effect = \dot { m } \left( 400250 \right)
=>15 = \dot { m } \times 150
=> \dot { m } = 0.1 kg/s
Isentropic compression work \left( { W }_{ i } \right) =\dot { m } \left( 475400 \right)
=> { W }_{ i }=\dot { m } \left( 475400 \right)
=> { W }_{ i }=0.1\times 75 = 7.5 kW
Actual compressor work \left( { W }_{ c } \right) =\frac { { W }_{ i } }{ \eta } =\frac { 7.5 }{ 0.75 } = 10 kW
30. Consider an ideal vapor compression refrigeration cycle. If the throttling process is replaced by an isentropic expansion process, keeping all the other processes unchanged, which one of the following statements is true for the modified cycle?
(a) Coefficient of performance is the same as that of the original cycle
(b) Coefficient of performance is lower than that of the original cycle
(c) Refrigerating effect is lower than that of the original cycle
(d) Coefficient of performance is higher than that of the original cycle
(1 Mark, 2019[1])
31. The figure shows a heat engine (HE) working between two reservoirs. The amount of heat (Q_{2}) rejected by the heat engine is drawn by a heat pump (HP). The heat pump receives the entire work output (W) of the heat engine. If temperatures, { T }_{ 1 }>{ T }_{ 3 }>{ T }_{ 2 } , then the relation between the efficiency \left( \eta \right) of the heat engine and the coefficient and the coefficient of performance (COP) of the heat pump is
(a) COP = \eta
(b) COP = 1+\eta
(c) COP = { \eta }^{ 1 }
(d) COP = { \eta }^{ 1 }1
(2 Mark, 2019[2])

Ans: c
Explanation:
COP of Heat pump = \frac { { Q }_{ 3 } }{ W }
=> COP = \frac { { W+Q }_{ 2 } }{ W } =\frac { { { Q }_{ 1 }{ Q }_{ 2 }+Q }_{ 2 } }{ W } =\frac { { { Q }_{ 1 } } }{ W }……(1)
Efficiency of heat engine = \left( \eta \right) =\frac { W }{ { Q }_{ 1 } } ……..(2)
From (1), (2): COP = \frac { 1 }{ \eta }