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Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Mechanical Engineering MCQ


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30 Questions MCQ Test - Test: Heat Pump & Refrigeration Cycles & Systems - 2

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Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 1

A heat pump working on a reversed Carnot cycle has a C.O.P. of 5. lf it works as a refrigerator taking 1 kW of work input, the refrigerating effect will be:

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 1

or heat rejected = 5 × work done 

And heat rejected = refrigeration effect + work input  

or, 5 × work input – work input = refrigeration effect  

or, 4 × work input = refrigeration effect

or refrigeration effect = 4 × 1 kW = 4 kW 

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 2

If an engine of 40 percent thermal efficiency drives a refrigerator having a coefficient of performance of 5, then the heat input to the engine for each kJ of heat removed from the cold body of the refrigerator is:  

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 2

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Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 3

A refrigerator works on reversed Carnot cycle producing a temperature of – 40°C. Work done per TR is 700 kJ per ten minutes. What is the value of its COP? 

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 3

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 4

In a refrigeration plant, if the condenser temperature increases, the power input to the compressor will  

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 5

The COP of a Carnot heat pump used for heating a room at 20°C by exchanging heat with river water at 10°C is:

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 5

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 6

A refrigerating machine working on reversed Carnot cycle consumes 6kW to produce a refrigerating effect of 1000kJ/min for maintaining a region at – 40oC.The higher temperature (in degree centigrade) of the cycle will be: 

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 6

or, T − 233= 83.88 or,T = 316.88 K = 43.88°C

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 7

A Carnot refrigerator works between the temperatures of 200 K and 300 K. If the refrigerator receives 1 kW of heat the work requirement will be: 

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 7

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 8

Operating temperature of a cold storage is –2°C From the surrounding at ambient temperature of 40°C heat leaked into the cold storage is 30 kW. If the actual COP of the plant is 1/10th of the maximum possible COP, then what will be the power required to pump out the heat to maintain the cold storage temperature at –2°C?

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 8

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 9

One ton refrigeration is equivalent to:

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 10

When the lower temperature is fixed, COP of a refrigerating machine can be improved by:

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 10

In heat engines higher efficiency can be achieved when (T1 – T2) is higher. In refrigerating machines it is the reverse, i.e. (T1 – T2) should be lower. 

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 11

A refrigerating machine working on reversed Carnot cycle takes out 2 kW of heat from the system at 200 K while working between temperature limits of 300 K and 200 K. COP and power consumed by the cycle will, respectively, be:

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 11

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 12

In the system given above, the temperature T = 300 K. When is the thermodynamic efficiency σE of engine E equal to the reciprocal of the COP of R? 

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 12

∴ Rmust act as a Heat pump

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 13

The COP of a Carnot heat pump used for heating a room at 20°C by exchanging heat with river water at 10°C is:

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 13

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 14

A refrigeration system operates on the reversed Carnot cycle. The temperature for the system is: Higher temperature = 40°C and Lower temperature = 20°C.

The capacity of the refrigeration system is 10 TR. What is the heat rejected from the system per hour if all the losses are neglected? 

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 14

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 15

Assertion (A ): T he COP of an air-conditioning plant is lower than that of an ice plant. 

Reason (R): The  tem peratures required in the ice plant are lower than those required for an air-conditioning plant. 

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 15

The COP of an air-conditioning plant is higher than that of an ice plant. 

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 16

For a given value of TH (Source temperature) for a reversed Carnot cycle, the variation of TL (Sink temperature) for different values of COP is represented by which one of the following graphs?

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 16

COP is on y-axis and TL on x-axis

⇒ Curve (C) is the corerct representation of above equations since it passes through the origin

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 17

Efficiency of a Carnot engine is 75%. If the cycle direction is reversed, COP of the reversed Carnot cycle is  

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 17

If the cycle direction is reversed it will be a heat pump not refrigerator. Students make a common mistake here and calculated (COP)R. We know that the definition of refrigeration is producing a temperature below atmospheric temperature. In heat engine lower temperature is atmospheric temperature. When we reverse this cycle then lower temperature will be atmospheric and higher temperature will be more than atmospheric means it will be a Heat Pump not a refrigerator.

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 18

A Carnot heat pump works between temperature limits of 277º C and 27º C. Its COP is

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 18

[We may put T1 and T2 in ºC or in K but T1 – T2 will be same] 

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 19

A refrigerator works on reversed Carnot cycle producing a temperature of – 40°C. Work done per TR is 700 kJ per ten minutes. What is the value of its COP? 

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 19

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 20

In the above figure, E is a heat engine with efficiency of 0.4 and R is a refrigerator. Given that Q2 + Q4 = 3Q1 the COP of the refrigerator is:

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 20

For heat engine, efficiency = 

And for refrigerator, 

W + Q = Q4    or (Q1 – Q2) + Q3 = Q4 or Q1  + Q3 = Q2 + Q4 = 3Q1

Therefore  2 Q1 = Q3

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 21

The coefficient of performance (COP) of a refrigerator working as a heat pump is given by:

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 21

The COP of refrigerator is one less than COP of heat pump, if same refrigerator starts working as heat pump i.e. (COP)heat pump = (COP)refrigerator + 1 

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 22

A refrigerator operates between the temperatures of –23ºC and 27ºC. If one TR = 3.5 kW, the minimum power required per TR to operate the refrigerator is 

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 22

= 0.7 kW 

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 23

Which one of the following statements is correct?

In a domestic refrigerator periodic defrosting is required because frosting 

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 24

Consider the following statements:  In the thermoelectric refrigeration, the coefficient of performance is a function of: 

1. Electrical conductivity of materials  

2. Pelti er coefficie nt  

3. Seeb eck coefficie nt  

4. Temperature at cold and hot junctions  

5. Thermal conductivity of materials. 

Of these statements: 

[IES-1997]

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 25

In a reversible cycle, the source temperature is 227°C and the sink temperature is 27°C. The maximum available work for a heat input of 100 kJ will be:

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 25

Maximum efficiency for 227° and 27°C sources = 

Maximum work available for a heat input of 100 kJ  = 0.4 × 100 = 40 kJ. 

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 26

In a 0.5 TR capacity water cooler, water enters at 30°C and leaves at 15°C.What is the actual water flow rate?

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 26

 or 0.5 12660 4.2 (30 15 ) or 100 kg/hr

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 27

It is proposed to build refrigeration plant for a cold storage to be maintained at – 3°C. The ambient temperature is 27°C. If 5 × 106 kJ/h of energy is to be continuously removed from the cold storage, the MINIMUM power required to run the refrigerator will be:

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 27

 

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 28

Assertion (A ): When solid CO2 (dry ice) is exposed to the atmosphere, it gets transformed directly into vapour absorbing the latent heat of sublimation from the surroundings

Reason (R): The triple  point of CO2 is at about 5 atmospheric pressure and at 216 K. 

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 29

In a one ton capacity water cooler, water enters at 30°C at the rate of 200 litres per hour. The outlet temperature of water will be (sp. heat of water = 4.18 kJ/kg K)

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 29

3.516 × 36 00 = 4.18× 200 × (300 − x )

or x   = 14 .98°C ≈ 15 °C

Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 30

A heat pump is shown schematically as

Detailed Solution for Test: Heat Pump & Refrigeration Cycles & Systems - 2 - Question 30

In heat pump, heat is rejected to source, work done on compressor, and heat absorbed from sink. 

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