MEC-27

Heat-engine efficiency

η = Wnet / Qin = 1 − Qout/Qin. First-law bound of a cyclic engine.

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Heat enginesCarnot 1824

Governing equation

\eta=\dfrac{W_{\mathrm{net}}}{Q_{\mathrm{in}}}=1-\dfrac{Q_{\mathrm{out}}}{Q_{\mathrm{in}}}

$Q_{\mathrm{out}} \in \left[20, 7000\right]$

where

Q_{\mathrm{in}}: Heat in (kJ)
Q_{\mathrm{out}}: Heat out (kJ)
\eta: Efficiency (—)
W_{\mathrm{net}}: Net work (kJ)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

\eta=\dfrac{W_{\mathrm{net}}}{Q_{\mathrm{in}}}=1-\dfrac{Q_{\mathrm{out}}}{Q_{\mathrm{in}}}

$Q_{\mathrm{out}} \in \left[20, 7000\right]$

$Q_{\mathrm{out}} = 20$ $\eta = 0.98$

Along the curve

②

Domain of the variable — blue

Q_{\mathrm{out}}_{\min} = 20,\quad Q_{\mathrm{out}}_{\max} = 7000

$Q_{\mathrm{out}} \in \left[20, 7000\right]$

$Q_{\mathrm{out}} = 20$ $Q_{\mathrm{out}} = 20$

Along the curve

③

Response curve — green

\eta=\dfrac{W_{\mathrm{net}}}{Q_{\mathrm{in}}}=1-\dfrac{Q_{\mathrm{out}}}{Q_{\mathrm{in}}}

$Q_{\mathrm{out}} \in \left[20, 7000\right]$

$Q_{\mathrm{out}} = 20$ $\eta = 0.98$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}\eta}{\mathrm{d}Q_{\mathrm{out}}}

$Q_{\mathrm{out}} \in \left[20, 7000\right]$

$Q_{\mathrm{out}} = 63.9$ $\eta^{\prime} = -0.001$

Along the curve

Lecture brief

Historical brief

Machine design grew from Coulomb torsion and Hertz contact (1881) through Soderberg fatigue and the heat-engine cycle. The lab writes shaft, bearing, contact and thermodynamic limits in SI. This sheet (MEC-27 — Heat-engine efficiency) is the form associated with Carnot 1824. Working symbols: $Q_{\mathrm{in}}$ , $Q_{\mathrm{out}}$ $\rightarrow$ $\eta$ , $W_{\mathrm{net}}$ . A cyclic device returns to the same state, so ΔU = 0 and Wnet = Qin − Qout.

Purpose

Purpose: compute

\eta

W_{\mathrm{net}}

from

Q_{\mathrm{in}}

Q_{\mathrm{out}}

in Mechanical via

\eta=\dfrac{W_{\mathrm{net}}}{Q_{\mathrm{in}}}=1-\dfrac{Q_{\mathrm{out}}}{Q_{\mathrm{in}}}

η = Wnet / Qin = 1 − Qout/Qin. First-law bound of a cyclic engine. Use it when a real mechanical question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

Q_{\mathrm{in}} = 1000.000\,\mathrm{kJ}

Q_{\mathrm{out}} = 650.000\,\mathrm{kJ}

, the governing relation

\eta=\dfrac{W_{\mathrm{net}}}{Q_{\mathrm{in}}}=1-\dfrac{Q_{\mathrm{out}}}{Q_{\mathrm{in}}}

yields

\eta = 0.3500\,\mathrm{—}

W_{\mathrm{net}} = 350.0\,\mathrm{kJ}

. Two reservoirs, a piston, a work arrow. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Heat in Q_{\mathrm{in}}1000 kJHeat out Q_{\mathrm{out}}650 kJ

Outputs

Efficiency \eta0.3500 —
Net work W_{\mathrm{net}}350.0 kJ

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Narration of this film

Two reservoirs, a piston, a work arrow.

A cyclic device returns to the same state, so ΔU = 0 and Wnet = Qin − Qout.

Reading speed

Watch on YouTube