### 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.

### Solved Example:

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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, $W = Q_1 + Q_2 - Q_3 = 2 + 0.4 - 1 = 1.4\ kJ$

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Which of the following statement is wrong?

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The relative coefficient of performance is:

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The COP of a domestic refrigerator:

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The domestic refrigerator uses following type of compressor:

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

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

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

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The value of COP in vapor compression cycle is usually:

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

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Condensing temperature in a refrigerator is the temperature:

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Formation of frost on evaporator in refrigerator:

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

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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)

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

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

Ammonia is:

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Choose the wrong statement.

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

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

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

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

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

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

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

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

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

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