Coefficient of Performance

  • Apply thermodynamic principles to design and performance analysis of heat engines, heat pumps, combustion systems, fuel cells, chemical reactors, building environments including heating, ventilation and air-conditioning systems, electronics, and biomedical systems.

  • Introduce the concepts of refrigerators and heat pumps and the measure of their performance.

  • Evaluate the maximum possible coefficient of performance for refrigerators and heat pumps based on the reversed Carnot cycle.

Coefficient of Performance is a ratio of heating or cooling provided to electrical energy consumed.
Coefficient of Performance for heat pumps \[COP = \frac{Q_{H}}{W}\] Coefficient of Performance for refrigerators and air conditioners \[COP = \frac{Q_{L}}{W}\] The main purpose of the refrigerator is to produce the cooling effect and it consumes power for this. The main purpose of the heat pump is to cool the room and it consumes power for it. From the above formula of the COP you will find that:
COP of Heat Pump = 1 + COP of the Refrigerator
The COP of the refrigerator and air-conditioner can be less than one or greater than one. The above formula also shows that the COP of the heat pump can never be less than one; it is always more than one.
It is worth mentioning here that efficiency of the engine can never be more than one, but the COP of the heat pump is always more than one. You should know that the terms efficiency and the COP are entirely different terms. The efficiency describes the capacity of the engine to produce power from the fuel, while the COP indicates how effectively the energy is used to produce the desired effect which is producing heat in case of the heat pump and coolness in case of the refrigerator.
The coefficient of performance or COP (sometimes CP) of a heat pump is the ratio of the heating or cooling provided over the electrical energy consumed. The COP was created to compare heat pump systems according to their energy efficiency.
A level of performance that describes a process that uses the lowest amount of inputs to create the greatest amount of outputs. Efficiency relates to the use of all inputs in producing any given output, including personal time and energy.
The COP can be more that one since the heat transferred can be more than power required by the refrigerator but the efficiency can never be more than one since the power generated by engine cannot be more than the total heat content of the fuel.

Solved Example:

73-1-01

A reversible heat engine receives 2 KJ of heat from reservoir at 1000 K and certain amount of heat from another reservoir at 800 K. It rejects 1 kJ heat to reservoir at 400 K. The net work output is: (GATE ME 2014- Shift I)

Solution:
Since, the heat engine is given as reversible, no entropy change takes place. \[ - \dfrac{Q_1}{T_1} - \dfrac{Q_2}{T_2} + \dfrac{Q_3}{T_3} = 0\] \[ - \dfrac{2}{1000} - \dfrac{Q_2}{800} + \dfrac{1}{400} = 0\] \[Q_2 = 0.4\ kJ\] Work done, \[W = Q_1 + Q_2 - Q_3 = 2 + 0.4 - 1 = 1.4\ kJ\]

Correct Answer: B

Solved Example:

73-1-02

Which of the following statement is wrong?

Correct Answer: C

Solved Example:

73-1-03

The relative coefficient of performance is:

Correct Answer: A

Solved Example:

73-1-04

The COP of a domestic refrigerator:

Correct Answer: B

Solved Example:

73-1-05

The domestic refrigerator uses following type of compressor:

Correct Answer: D

Solved Example:

73-1-06

Presence of moisture in a refrigerant affects the working of: (SSC JE EE Paper 7- March 2017 Evening)

Correct Answer: D

Solved Example:

73-1-07

For better COP of refrigerator, the pressure range corresponding to temperature in evaporator and condenser must be:

Correct Answer: A

Solved Example:

73-1-08

The bank of tubes at the back of domestic refrigerator are: (NPCIL SA/ST ME GJ Nov 2019- Shift I)

Correct Answer: A

Solved Example:

73-1-09

The value of COP in vapor compression cycle is usually:

Correct Answer: B

Solved Example:

73-1-10

In a refrigeration system, heat absorbed in comparison to heat rejected is:

Correct Answer: B

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73-1-11

Condensing temperature in a refrigerator is the temperature:

Correct Answer: D

Solved Example:

73-1-12

Formation of frost on evaporator in refrigerator:

Correct Answer: A

Solved Example:

73-1-13

At lower temperatures and pressures, the latent heat of vaporisation of a refrigerant:

Correct Answer: B

Solved Example:

73-1-14

If a heat pump cycle operates between the condenser temperature of +27$^\circ$C and evaporator temperature of -23$^\circ$C, then the Carnot COP will be: (UKPSC JE Mech 2013- Paper II)

Correct Answer: D

Solved Example:

73-1-15

Flow of refrigerant is controlled by: (SJVNL JE Mech 2018)

Correct Answer: D

Solved Example:

73-1-16

Freon group of refrigerants are: (RSMSSB LDC Paper I- Aug 2018)

Correct Answer: D

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73-1-17

Ammonia is:

Correct Answer: D

Solved Example:

73-1-18

Choose the wrong statement.

Correct Answer: A

Solved Example:

73-1-19

For obtaining high COP, the pressure range of compressor should be:

Correct Answer: B

Solved Example:

73-1-20

The coefficient of performance is the ratio of the refrigerant effect to the:

Correct Answer: D

Solved Example:

73-1-21

The C.O.P of a refrigeration cycle with increase in evaporator temperature, keeping condenser temperature constant, will:

Correct Answer: A

Solved Example:

73-1-22

The C.O.P. of a refrigeration cycle with lowering of condenser temperature, keeping the evaporator temperature constant, will:

Correct Answer: A

Solved Example:

73-1-23

Which of the following refrigerants has lowest freezing point: (MPSC Asst. Engg. EE Mains 2019- Paper 2)

Correct Answer: D

Solved Example:

73-1-24

The COP of a vapor compression plant in comparison to vapor absorption plant is:

Correct Answer: A

Solved Example:

73-1-25

The C.O.P. of a domestic refrigerator in comparison to domestic air conditioner will be:

Correct Answer: C

Solved Example:

73-1-26

The change in evaporator temperature in a refrigeration cycle, as compared to change in condenser temperature, influences the value of C.O.P.:

Correct Answer: A

Solved Example:

73-1-27

A reversible heat engine rejects 80% of the heat supplied during a cycle of operation. If the engine is reversed and operates as a refrigerator, then the coefficient of performance shall be: (ESE Mech 2017)

Solution:
\[Q_2 = 80\%\ \mathrm{of}\ Q_1 = 0.8\ Q_1\] \[W = Q_1 - Q_2 = Q_1 - 0.8 Q_1 = 0.2\ Q_1\] \[(COP)_R = \dfrac{Q_2}{W} = \dfrac{0.8\ Q_1}{0.2\ Q_1} =4\]

Correct Answer: C

Solved Example:

73-1-28

A solar collector receiving solar radiation at the rate of 0.6 kW/m\(^2\) transforms it to the internal energy of a fluid at an overall efficiency of 50%. The fluid heated to 350K is used to run a heat engine which rejects heat at 313K. If the heat engine is to deliver 2.5 kW power, the minimum area of the solar collector required would be, nearly: (VIZAG ME Mech 2013)

Solution:
\[\eta = 1 - \dfrac{T_2}{T_1} = 1 - \dfrac{313}{350} =0.106\] \[Q = \dfrac{W}{0.106} = 23.6\ kW\] \begin{align*} \eta_{overall} &= \dfrac{Q}{\mathrm{Solar\ Radiation}}\\ 0.5 &= \dfrac{23.6}{0.6 \times A}\\ A &= 79\ m^2 \end{align*}

Correct Answer: D