Question

The interior of a refrigerator has a surface area of 4.3 m2 . It is insulated by a 3.6 cm thick material that has a thermal conductivity of 0.0303 J/m · s ·◦ C. The ratio of the heat extracted from the interior to the work done by the motor is 6.3% of the theoretical maximum. The temperature of the room is 44.2◦C, and the temperature inside the refrigerator is 9.2 ◦C. Determine the power required to run the compressor.
Answer in units of W

Answer 1

Power is the ratio of the heat entered into the refrigerator to the coefficient of performance of the refrigerator. The power required to run the compressor in Watts is 249.149 W.

What is Power?

Power is the ratio of the heat entered into the refrigerator to the coefficient of performance of the refrigerator.

Given the surface area of the interior of the refrigerator is

A= 4.3 m ²

The refrigerator is insulated by a material of  thickness

t =3.6 cm =0.036 m

The thermal conductivity of insulating material is

k =0.0300 J/m-s

The change in temperature inside the refrigerator and room temperature is

dT= (273.15 + 44.2) -(273.15 + 9.2) =35 K

The rate of heat entered the refrigerator.

$Q=\frac{k\times A\times dT}{t}$

Substituting value into the equation, we get

Q =126.67 W

The coefficient of performance of the refrigerator is

$COP=\dfrac{(273.15+9.2)}{(273.15+44.2)-(273.15+9.2)}$

COP = 8.07

As, work done by the motor is 6.3% of the theoretical maximum,

$\dfrac{6.3\times8.07}{100} =0.50841$

The power required to run the compressor will be

$P=\dfrac{126.67}{0.50841} =249.149\:\rm W$

The power required to run the compressor is 249.149 W

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