Question

(II) An inventor claims to have built an engine that produces 2.00 MW of usable work while taking in 3.00 MW of thermal energy at 425 K, and rejecting 1.00 MW of thermal energy at 215 K. Is there anything fishy about his claim? Explain.

Short answer

Yes, there is something fishy about the claim of the inventor since the engine’s real efficiency is more than its ideal efficiency.

Step-by-step solution

Understanding the efficiency of the engine

The ratio of the work done by the engine to the heat input at high temperature is termed the efficiency of the engine.

The expression for the efficiency is given as:

\(e = \frac{W}{{{Q_{\rm{H}}}}} = 1 - \frac{{{Q_{\rm{L}}}}}{{{Q_{\rm{H}}}}}\) … (i)

Here, W is the work done, \({Q_{\rm{H}}}\) is the heat input at high temperature, and \({Q_{\rm{L}}}\) is the heat exhausted at low temperature.

Given data

The work produced by the engine is \(W = 2\;{\rm{MW}}\).

The intake thermal energy is \({Q_1} = 3\;{\rm{MW}}\).

The initial temperature is \({T_1} = 425\;{\rm{K}}\).

The final temperature is \({T_2} = 215\;{\rm{K}}\).

The rejected thermal energy is \({Q_2} = 1\;{\rm{MW}}\).

Evaluation of the real efficiency of the engine

The real efficiency of the engine is given by:

\(e = \frac{W}{{{Q_1}}}\)

Substitute the values in the above expression.

\(\begin{array}{l}e = \left( {\frac{{2\;{\rm{MW}}}}{{3\;{\rm{MW}}}}} \right)\\e = 0.66 \times 100\% \\e = 66\% \end{array}\)

Evaluation of the ideal efficiency of the engine

The relation of ideal efficiency is given by:

\({e_{\rm{i}}} = 1 - \frac{{{T_1}}}{{{T_2}}}\)

Substitute the values in the above expression.

\(\begin{array}{l}{\eta _{\rm{i}}} = 1 - \left( {\frac{{215\;{\rm{K}}}}{{425\;{\rm{K}}}}} \right)\\{\eta _{\rm{i}}} = 0.49 \times 100\% \\{\eta _{\rm{i}}} = 49\% \end{array}\)

The engine made by the inventor is not possible because the engine’s real efficiency is more than its ideal efficiency. Thus, there is something fishy about his claim.