Refrigeration And Air Conditioning MCQ Questions & Answers

71

The wet bulb temperature during sensible heating of air

A) Remains constant B) Increases C) Decreases D) None of these

Discuss

Answer & Solution

Answer: Option B

Solution:

The wet bulb temperature during sensible heating of air Increases

72

The C.O.P. of a refrigerator working on a reversed Carnot cycle is (where T₁ = Lowest absolute temperature and T₂ = Highest absolute temperature)

A) $$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2} - {{\text{T}}_1}}}$$ B) $$\frac{{{{\text{T}}_2} - {{\text{T}}_1}}}{{{{\text{T}}_1}}}$$ C) $$\frac{{{{\text{T}}_1} - {{\text{T}}_2}}}{{{{\text{T}}_1}}}$$ D) $$\frac{{{{\text{T}}_2}}}{{{{\text{T}}_2} - {{\text{T}}_1}}}$$

Discuss

Answer & Solution

Answer: Option A

Solution:

The C.O.P. of a refrigerator working on a reversed Carnot cycle is $$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_2} - {{\text{T}}_1}}}$$
Where
T₁ = Lowest absolute temperature,
T₂ = Highest absolute temperature

73

The relative coefficient of performance is equal to

A) $$\frac{{{\text{Theoretical C}}{\text{.O}}{\text{.P}}{\text{.}}}}{{{\text{Actual C}}{\text{.O}}{\text{.P}}{\text{.}}}}$$ B) $$\frac{{{\text{Actual C}}{\text{.O}}{\text{.P}}{\text{.}}}}{{{\text{Theoretical C}}{\text{.O}}{\text{.P}}{\text{.}}}}$$ C) Actual C.O.P. × Theoretical C.O.P. D) None of these

Discuss

Answer & Solution

Answer: Option B

Solution:

The ratio of the actual C.O.P. to the theoretical C.O.P. is known as relative coefficient of performance.
The relative coefficient of performance is equal to $$\frac{{{\text{Actual C}}{\text{.O}}{\text{.P}}{\text{.}}}}{{{\text{Theoretical C}}{\text{.O}}{\text{.P}}{\text{.}}}}$$

74

If a gas is to be liquefied, its temperature must be

A) Increased to a value above its critical temperature B) Reduced to a value below its critical temperature C) Equal to critical temperature D) None of the above

Discuss

Answer & Solution

Answer: Option C

Solution:

If a gas is to be liquefied, its temperature must be Equal to critical temperature

75

The advantage of dry compression is that

A) It permits higher speeds to be used B) It permits complete evaporation in the evaporator C) It results in high volumetric and mechanical efficiency D) All of the above

Discuss

Answer & Solution

Answer: Option D

Solution:

The advantage of dry compression is that
- It permits higher speeds to be used
- It permits complete evaporation in the evaporator
- It results in high volumetric and mechanical efficiency

76

The leaks in a refrigeration system using Freon are detected by

A) Halide torch which on detection produces greenish flame lighting B) Sulphur sticks which on detection gives white smoke C) Using reagents D) Smelling

Discuss

Answer & Solution

Answer: Option A

Solution:

The leaks in a refrigeration system using Freon are detected by Halide torch which on detection produces greenish flame lighting

77

The higher temperature in vapour compression cycle occurs at

A) Receiver B) Expansion valve C) Evaporator D) Compressor discharge

Discuss

Answer & Solution

Answer: Option D

Solution:

The higher temperature in vapour compression cycle occurs at Compressor discharge

78

Highest temperature encountered in refrigeration cycle should be

A) Near critical temperature of refrigerant B) Above critical temperature C) At critical Temperature D) Much below critical temperature

Discuss

Answer & Solution

Answer: Option D

Solution:

Highest temperature encountered in refrigeration cycle should be Much below critical temperature

79

In actual air-conditioning applications for R-12 and R-22, and operating at a condenser temperature of 40°C and an evaporator temperature of 5°C, the heat rejection factor is about

A) 1 B) 1.12 C) 2.15 D) 5.12

Discuss

Answer & Solution

Answer: Option B

Solution:

T1 = 5°C or 273 + 5 = 278K
T2 = 40°C or 273 + 40 = 313K
$$\eqalign{ & {\text{C}}{\text{.O}}{\text{.P}}{\text{.}} = \frac{{{\text{T1}}}}{{{\text{T2}} - {\text{T1}}}} \cr & = \frac{{278}}{{313 - 278}} \cr & = \frac{{278}}{{35}} \cr & {\text{C}}{\text{.O}}{\text{.P}}{\text{.}} = 7.942 \cr & {\text{Heat rejection factor (HRF)}} \cr & = 1 + \frac{1}{{{\text{C}}{\text{.O}}{\text{.P}}{\text{.}}}} \cr & = 1 + \frac{1}{{7.942}} \cr & = \frac{{8.942}}{{7.942}} \cr & = 1.12 \cr} $$

80

The capillary tube is not used in large capacity refrigeration systems because

A) Cost is too high B) Capacity control is not possible C) It is made of copper D) Required pressure drop cannot be achieved

Discuss

Answer & Solution

Answer: Option B

Solution:

The capillary tube is not used in large capacity refrigeration systems because Capacity control is not possible

Refrigeration And Air Conditioning MCQ Questions & Answers | Mechanical Engineering