91影视

Efficiency is a dimensionless number that estimates the performance of a device using thermal energy.

The output power is \(P = 135\;{\rm{hp}}\).

The actual efficiency is \({\eta _{{\rm{actual}}}} = 15\% = 0.15\).

The lower temperature is \({T_{\rm{L}}} = 85^\circ {\rm{C}}\).

The higher temperature is \({T_{\rm{H}}} = 495^\circ {\rm{C}}\).

The time is \(t = 1.0\;{\rm{h}}\).

(a)

The efficiency of the ideal cycle or Carnot cycle is calculated below:

\(\begin{aligned}{c}{\eta _{{\rm{ideal}}}} &= 1 - \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}}}}\\{\eta _{{\rm{ideal}}}} &= 1 - \frac{{\left( {\left( {85^\circ {\rm{C + 273}}} \right)\;{\rm{K}}} \right)}}{{\left( {\left( {495^\circ {\rm{C + 273}}} \right)\;{\rm{K}}} \right)}}\\{\eta _{{\rm{ideal}}}} &= 0.534\end{aligned}\)

The ratio of efficiency of the actual cycle and the ideal cycle is calculated below:

\(\begin{aligned}{l}R &= \frac{{{\eta _{{\rm{actual}}}}}}{{{\eta _{{\rm{ideal}}}}}}\\R &= \frac{{0.15}}{{0.534}}\\R &= 0.281\end{aligned}\)

Thus, the ratio of efficiency of the actual cycle and the ideal cycle is \(0.281\).

(b)

The work done by the engine is calculated below:

\(\begin{aligned}{l}W &= Pt\\W &= \left( {135\;{\rm{hp}}} \right)\left( {\frac{{746\;{\rm{W}}}}{{1\;{\rm{hp}}}}} \right)\left( {1\;{\rm{h}}} \right)\left( {\frac{{3600\;{\rm{s}}}}{{1\;{\rm{h}}}}} \right)\\W &= 3.63 \times {10^8}\;{\rm{J}}\end{aligned}\)

The expression for the heat given to the system is

\({Q_{\rm{H}}} = W + {Q_{\rm{L}}}\).

Here, \({Q_{\rm{L}}}\) is the heat rejected by the system.

The heat rejected by the engine can be calculated using the following expression.

\(\begin{aligned}{c}{\eta _{{\rm{actual}}}} &= \frac{W}{{{Q_{\rm{H}}}}}\\{\eta _{{\rm{actual}}}} &= \frac{W}{{W + {Q_{\rm{L}}}}}\\W + {Q_{\rm{L}}} &= \frac{W}{{{\eta _{{\rm{actual}}}}}}\\{Q_{\rm{L}}} &= \frac{W}{{{\eta _{{\rm{actual}}}}}} - W\\{Q_{\rm{L}}} &= W\left( {\frac{1}{{{\eta _{{\rm{actual}}}}}} - 1} \right)\end{aligned}\)

Substitute the value in the above equation.

\(\begin{aligned}{l}{Q_{\rm{L}}} &= \left( {3.63 \times {{10}^8}\;{\rm{J}}} \right)\left( {\frac{1}{{0.15}} - 1} \right)\\{Q_{\rm{L}}} &= \left( {2.1 \times {{10}^9}\;{\rm{J}}} \right)\left( {\frac{{2.38 \times {{10}^{ - 4}}\;{\rm{kcal}}}}{{1\;{\rm{J}}}}} \right)\\{Q_{\rm{L}}} &= 4.99 \times {10^5}\;{\rm{kcal}} \approx {\rm{5}}{\rm{.0}}\;{\rm{kcal}}\end{aligned}\)

Thus, the heat rejected by the engine is \(2.1 \times {10^9}\;{\rm{J}}\) or \({\rm{5}}{\rm{.0}}\;{\rm{kcal}}\).