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UPPSC Lec’21 ME: Daily Practices Quiz. 10-Dec-2021

Quiz: Mechanical Engineering
Exam: UPPSC AE/Lecturer
Topic: RAC

Each question carries 3 marks
Negative marking: 1 mark
Time: 8 Minutes

Q1. A heat pump working on reversed Carnot cycle has a COP of 7. What is the ratio of maximum to minimum absolute temperatures?
(a) 7/8
(b) 1/7
(c) 7/6
(d) 6/7

Q2. A Carnot refrigerator has a COP of 6. What is the COP of heat pump?
(a) 7/6
(b) 5
(c) 7
(d) 1/6

Q3. In case of vapour-compression refrigeration system, the enthalpies at various points in the cycle are:
1. Enthalpy at inlet to the compressor (saturated vapour), h₁ = 280kJ/kg
2. Enthalpy at outlet of the compressor (after isentropic compression), h₂ = 300 kJ/kg
3. Enthalpy at exit of the condenser (saturated liquid), h_3 = 180 kJ/kg
What is the COP of the plant?
(a) 3
(b) 4
(c) 5
(d) 6

Q4. Ammonia used as refrigerant is non-corrosive to
(a) iron and steel
(b) copper and copper alloys
(c) Both (a) and (b)
(d) Neither (a) nor (b)

Q5. In a VCR plant, the refrigerant leaves the compressor and evaporator with enthalpy of 205 kJ/kg and 177 kJ/kg respectively. Enthalpy of the refrigerant leaving the condenser is 105 kJ/kg. If the mass flow rate of the refrigerant is 0.2 kg/s, the refrigeration effect will be
(a) 12.2 kW
(b) 14.4 kW
(c) 16.4 kW
(d) 20.2 kW

Q6. The expression for COP of reversed Brayton cycle is (r_P is the pressure ratio)
(a) 1/(〖r_p〗^((γ-1)/γ)-1)
(b) 1/(〖r_p〗^γ-1)
(c) 1/〖r_p〗^((γ-1)/γ)
(d) 1/(〖r_p〗^((γ-1)/γ)+1)


S1. Ans.(c)

As, we know that,
〖(COP)〗_HP= ( T_H )/(T_H – T_L )
7 = ( 1 )/(1 -T_L/T_H )
T_L/T_H =1-1/7=6/7
T_H/T_L =7/6

S2. Ans.(c)
As we know that, COP of refrigerator is
〖(COP)〗_R= ( T_L )/(T_H – T_L )

COP of the heat pump is
S3. Ans. (c)
Sol. Enthalpies at various points are
h₁ = 280kJ/kg
h₂ = 300 kJ/kg
h_3 = 180 kJ/kg
As, we know that COP of VCRS is
COP = (h_1 – h_3)/(h_2 – h_1 )=(280-180)/(300-280)=100/20 = 5

S4. Ans.(a)
The refrigerant ammonia is corrosive to copper and copper alloys but it is non corrosive to iron and steel.

S5. Ans.(b)
Refrigerant Effects = Enthalpy at outlet of evaporator – Enthalpy at outlet of condenser
RE = 177 – 105 = 72 kJ/kg
RE = 72 × 0.2 = 14.4 kW

S6. Ans (a)
(〖COP)〗_(Reversed Brayton)=1/(〖r_p〗^((γ-1)/γ)-1)
Here, r_P is the pressure ratio.

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Published by VINAYAK KUMAR

Graduated from BIT Sindri, Postgraduated from IIT BHU Varanasi

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