A heat engine is operating between a refrigerator's outside fin coil unit and the kitchen air. Determine the maximum power generated by this heat engine.

Problem Description

(1) Determine the heat dissipated from the refrigerator

The coefficient of performance(COP) of a refrigerator is expressed as

      COP_R = Q_L /W_net,in = (Q_H - W_net,in)/W_net,in

Rearranging this equation gives

      Q_H = (1 + COP_R)W_net,in = (1 + 3.0)(1.0) = 4.0 kW

(2) Determine power generated by the heat engine

The thermal efficiency of a heat engine is expressed as

      η_th = W_net,out/Q_H

where
      W_net,out = power generated by the heat engine
      Q_H = heat transferred to the heat engine

Rearranging this equation yields

      W_net,out =  η_th Q_H

where Q_H is the heat transferred from the heat-exchange pipes to the heat engine, which is the heat dissipated from the refrigerator, hence

      Q_H = 4.0 kW

The thermal efficiency of a Carnot heat engine is

      η_th,rev = 1- (273 + 20)/(273 + 30) = 3.3%

The work output from this Carnot heat engine is

      W_net,out =  η_th,rev Q_H = (3.3%)(4.0) = 0.12 kW

The result shows the power generated from this Carnot heat engine is only enough to light a 100 W bulb. Hence, there is little benefit to use the heat dissipated by a refrigerator.