Question

A heat engine receives heat at 1000 K and rejects heat to environment at 300 K. Its efficiency is half that of a Carnot engine operating between these temperatures. The work output drives a refrigerator removing heat from a cold space at 260 K at 5.2 kW, rejecting heat at 300 K. The refrigerator COP is half of the Carnot refrigerator. Find the rate of heat supplied to the heat engine (kW). Round off to 2 decimals.

Hint:

When a heat engine drives a refrigerator, equate the engine work output to refrigerator work input. Use Carnot relations, then apply the given efficiency/COP fractions.

Answer 1

Correct Answer: 4.5

Carnot efficiency of heat engine:
\[ \eta_{CE} = 1 - \frac{T_L}{T_H} = 1 - \frac{300}{1000} = 0.7 \] Actual engine efficiency:
\[ \eta = \frac{0.7}{2} = 0.35 \] So:
\[ W = 0.35 Q_H \] Refrigerator Carnot COP:
\[ COP_{CR} = \frac{T_L}{T_H - T_L} = \frac{260}{300 - 260} = \frac{260}{40} = 6.5 \] Actual refrigerator COP = half of Carnot:
\[ COP = \frac{6.5}{2} = 3.25 \] Refrigerator removes:
\[ Q_L = 5.2\ \text{kW} \] Work required:
\[ W = \frac{Q_L}{COP} = \frac{5.2}{3.25} = 1.6\ \text{kW} \] This work is supplied entirely by the heat engine:
\[ 1.6 = 0.35 Q_H \] Thus:
\[ Q_H = \frac{1.6}{0.35} = 4.57\ \text{kW} \] Rounded to 2 decimals:
\[ Q_H = 4.58\ \text{kW} \]