Refrigeration and Air-conditioning Miscellaneous Easy Questions and Answers | Page - 5

Refrigeration and Air-conditioning Miscellaneous

A vapour absorption refrigeration system is a heat pump with three thermal reservoirs as shown in the figure. A refrigeration effect of 100 W 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

1. 167
1. 100
1. 80
1. 20

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From question, since refrigeration effect of 100 W is required
So, 100 = 250
Work obtained 300 – 250

Work obtained = = 100 × 50 = 20 W
250

Now for this amount of work, heat is absorbed from reservoir 3 and rejected to sink 2.
∴ η = W = 1 - 300 = 1
Q₃ 400 4

⇒ Q₃ = 4W = 4 × 20 W = 80 W

Correct Option: C

From question, since refrigeration effect of 100 W is required
So, 100 = 250
Work obtained 300 – 250

Work obtained = = 100 × 50 = 20 W
250

Now for this amount of work, heat is absorbed from reservoir 3 and rejected to sink 2.
∴ η = W = 1 - 300 = 1
Q₃ 400 4

⇒ Q₃ = 4W = 4 × 20 W = 80 W

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 1000 W, respectively. The COP for the heat pump is

1. 7.5
1. 6.5
1. 4.0
1. 3.0

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Heat addition, Q₂ = 750 W
Heat rejected, Q₁ = 1000 W
[C.O.P]_H.P = Q₁ = Q₁
W Q₁ - Q₂

= 1000 = 1000 = 4
1000 - 750 250

T₁ = 273 + 27º = 300ºK

T₂ = 273 – 13 = 260K
[C.O.P]_H.P = Q₁
Q₁ - Q₂

for actual heat pump
= 1000 = 1000 = 4
1000 - 750 250

Correct Option: C

Heat addition, Q₂ = 750 W
Heat rejected, Q₁ = 1000 W
[C.O.P]_H.P = Q₁ = Q₁
W Q₁ - Q₂

= 1000 = 1000 = 4
1000 - 750 250

T₁ = 273 + 27º = 300ºK

T₂ = 273 – 13 = 260K
[C.O.P]_H.P = Q₁
Q₁ - Q₂

for actual heat pump
= 1000 = 1000 = 4
1000 - 750 250

List-I
A. Liquid to suction heat exchange
B. Constant volume heat addition
C. Normal shock
D. Ammonia-water
List-II
1. Vapour absorption refrigeration
2. Vapour compression refrigeration
3. Diesel cycle
4. Otto cycle
5. Converging nozzle
6. Converging diverging nozzle

1. A – 2, B – 4, C – 6, D – 1
1. A – 5, B – 4, C – 6, D – 1
1. A – 2, B – 4, C – 5, D – 1
1. A – 2, B – 4, C – 6, D – 5

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A – 2, B – 4, C – 6, D – 1

Correct Option: A

A – 2, B – 4, C – 6, D – 1

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

1. 1/11
1. 1/10
1. 1/9
1. 1/1.1

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Clearance ratio = Clearance volume
Swept volume

= 100 = 1
1000 - 100 9

Correct Option: C

Clearance ratio = Clearance volume
Swept volume

= 100 = 1
1000 - 100 9

A refrigerator uses R134a as its refrigerant and operates on a ideal vapour-compression 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
At P = 0.8 MPa, h = 272.05 kJ/kg (superheated vapour)
At P = 0.8 MPa, h = 93.42 kJ/kg (saturated liquid)

1. 8.931 kW
1. 89.31 kW
1. 9.831 kW
1. 9.813 kw

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h₁ = 236.04 kJ/kg
h₂ = 272.05 kJ/kg
h₃ = 93.42 kJ/kg
s₁ = 0.9322 kJ/kg
ṁ = 0.05 kg/s
Heat Rejection = ṁ(h₂ – h₃) = 0.05(272.05 – 93.42)
Heat Rejection = 8.931 kW

Correct Option: A

h₁ = 236.04 kJ/kg
h₂ = 272.05 kJ/kg
h₃ = 93.42 kJ/kg
s₁ = 0.9322 kJ/kg
ṁ = 0.05 kg/s
Heat Rejection = ṁ(h₂ – h₃) = 0.05(272.05 – 93.42)
Heat Rejection = 8.931 kW