Question

Carnot heat engine receives heat at 900 K and rejects the waste heat to the environment at 300 K. The entire work output of the heat engine is used to drive a Carnot refrigerator that removes heat from the cooled space at 215 °C at a rate of 250 kJ/min and rejects it to the same environment at 300 K. Determine (a) the rate of heat supplied to the heat engine and (b) the total rate of heat rejection to the environment.

Answer 1

A heat engine is a device that turns heat into mechanical energy that may subsequently be utilized to do mechanical labour. The total rate of heat rejection to the environment is 298.4458 kJ/min.

What is a Heat Engine?

A heat engine is a device that turns heat into mechanical energy that may subsequently be utilized to do mechanical labour. It accomplishes this by transferring a functioning ingredient from a higher to a lower state temperature.

For the following condition, the diagram can be made as shown below. Also, the Heat Engine Sink temperature, $T_H$ = 900 K

Heat Engine surrounding temperature, $T_L$ = 300 K

Refrigerator surrounding temperature, $T'_C$ = 215 C + 273 =  488 K

Refrigerator sink temperature, $T'_H$ = 300 K

The heat removed by the refrigerator, $Q'_{in}$ = 250 kJ/min

A.) For a Carnot refrigerator, the COP or the coefficient of performance is given by,

$COP = \dfrac{T'_C}{T'_H-T'C}$

COP = (488)/(488-300)

COP = 2.596

The COP is also the ratio of the Heat input and the work output.

$COP = \dfrac{Q'_{in}}{W}$

2.596 = 250/W

W = 96.302 kJ/min

The Carnot engine efficiency is expressed as,

$\eta = 1- \dfrac{T_C}{T_H}$

η = 1 - (300/900)

η = 0.6667

Also, the Carnot engine efficiency is defined as,

$\eta = \dfrac{W}{Q_{in}}$

$0.6667 = \dfrac{96.302}{Q_{in}}\\\\Q_{in} = \rm 144.4458\ kJ/min$

Hence, the rate of heat supplied to the heat engine is 144.4458 kJ/min.

B.) The total work done by the engine is 96.302 kJ/min, while the heat input is 144.4458 kJ/min. Therefore, the heat output of the system is,

$W = Q_{in} - Q_{out}\\\\Q_{out} = 48.1438\rm\ kJ/min$

Thus, the total rate of heat rejection to the environment,

$\text{Total Heat rejection} = Q_{out}+Q'_{in}$

                               = 48.4458 kJ/min + 250 kJ/min

                                = 298.4458 kJ/min

Hence, the total rate of heat rejection to the environment is 298.4458 kJ/min.

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