Heat Engine Efficiency - Heat & Thermodynamics Step by Step

A heat engine converts only part of its heat input into useful work. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Read the engine budget

The engine takes in 30 joules of heat and produces 12 joules of useful work.

Q_{\text{in}} = 30\ \text{J}\qquad W_{\text{out}} = 12\ \text{J}

Efficiency is useful work over heat input

Efficiency counts the fraction of heat input that becomes useful work.

\eta = \frac{W_{\text{out}}}{Q_{\text{in}}}

Compute the efficiency

Divide 12 joules by 30 joules. The efficiency is two fifths.

\eta = \frac{12\ \text{J}}{30\ \text{J}} = \tfrac{2}{5}

More useful work raises efficiency at fixed input

Hold the heat input at 30 joules. The diagram shows the middle row; more useful work means a larger efficiency fraction.

\begin{array}{c|c|c}W_{\text{out}}&Q_{\text{in}}&\eta\\6\ \text{J}&30\ \text{J}&\tfrac{1}{5}\\12\ \text{J}&30\ \text{J}&\tfrac{2}{5}\\18\ \text{J}&30\ \text{J}&\tfrac{3}{5}\\\end{array}

More input lowers the fraction if work is fixed

Now hold useful work at 12 joules. Spreading the same work over more input gives a smaller efficiency fraction.

\begin{array}{c|c|c}W_{\text{out}}&Q_{\text{in}}&\eta\\12\ \text{J}&24\ \text{J}&\tfrac{1}{2}\\12\ \text{J}&30\ \text{J}&\tfrac{2}{5}\\12\ \text{J}&40\ \text{J}&\tfrac{3}{10}\\\end{array}