Question:You design an engine that takes in\({\bf{1}}{\bf{.5 \times 1}}{{\bf{0}}^{\bf{4}}}\;{\bf{J}}\)of heat at 650 K in each cycle and rejects heat at a temperature of 290 K. The engine completes 240 cycles in 1 minute. What is the theoretical maximum power output of your engine, in horsepower?

The temperature for heat input is\({T_1} = 650\;{\rm{K}}\)

The temperature for heat rejection is\({T_2} = 290\;{\rm{K}}\)

The het input for engine is\({Q_1} = 1.5 \times {10^4}\;{\rm{J}}\)

The rate of cycles per minute is \(r = 240\;{\rm{cycle}}/\min \)

The work done by engine is calculated by the difference between heat input and rejected heat, then this work is multiplied by rate of cycle to obtain theoretical maximum power output of engine.

The heat rejected by engine is given as:

\(\frac{{{Q_2}}}{{{Q_1}}} = \frac{{{T_2}}}{{{T_1}}}\)

Substitute all the values in the above equation.

\(\begin{array}{c}\frac{{{Q_2}}}{{\left( {1.5 \times {{10}^4}\;{\rm{J}}} \right)}} = \frac{{290\;{\rm{K}}}}{{650\;{\rm{K}}}}\\{Q_2} = 0.67 \times {10^4}\;{\rm{J}}\end{array}\)

The theoretical maximum power output of engine is given as:

\(P = \left( {{Q_1} - {Q_2}} \right)r\)

Substitute all the values in the above equation.

\(\begin{array}{l}P = \left( {1.5 \times {{10}^4}\;{\rm{J}} - 0.67 \times {{10}^4}\;{\rm{J}}} \right)\left( {240\;{\rm{cycle}}/\min } \right)\left( {\frac{{1\;\min }}{{60\;{\rm{s}}}}} \right)\\P = 33200\;{\rm{J}}/{\rm{s}}\\P = \left( {33200\;{\rm{J}}/{\rm{s}}} \right)\left( {\frac{{1\;{\rm{hp}}}}{{746\;{\rm{J}}/{\rm{s}}}}} \right)\\P = 44.5\;{\rm{hp}}\end{array}\)

Therefore, the theoretical maximum power output of engine is \(44.5\;{\rm{hp}}\).