INGENIA

THM-23

Otto efficiency

Ideal spark-ignition cycle.

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GoverningOtto efficiency

Governing equation

η=1−r1−γ\eta=1-r^{1-\gamma}

r∈[4,16]r \in \left[4, 16\right]

where

r
r (—)
g
g (—)
eta
Otto efficiency (—)

The figure, part by part

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①

Governing law — red

η=1−r1−γ\eta=1-r^{1-\gamma}

r∈[4,16]r \in \left[4, 16\right]

r=4r = 4eta=0.426eta = 0.426

②

Domain of the variable — blue

rmin⁡=4,rmax⁡=16r_{\min} = 4,\quad r_{\max} = 16

r∈[4,16]r \in \left[4, 16\right]

r=4r = 4r=4r = 4

③

Response curve — green

η=1−r1−γ\eta=1-r^{1-\gamma}

r∈[4,16]r \in \left[4, 16\right]

r=4r = 4eta=0.426eta = 0.426

④

Rate of change — amber

detadr\frac{\mathrm{d}eta}{\mathrm{d}r}

r∈[4,16]r \in \left[4, 16\right]

r=4.075r = 4.075eta′=0.056eta^{\prime} = 0.056

Lecture brief

Historical brief

Carnot (1824), Clausius entropy, the first law and later van der Waals and Gibbs potentials turned heat into a state science. The sheets compute work, efficiency and vapour pressure. This sheet (THM-23 — Otto efficiency) is the form associated with Otto efficiency. Working symbols: rr, gg →\rightarrow etaeta. Ideal spark-ignition cycle. Pedagogical SI sheet with a live model and a swept parameter.

Purpose

Purpose: compute etaeta from rr, gg in Thermodynamics via η=1−r1−γ\eta=1-r^{1-\gamma} Ideal spark-ignition cycle. Use it when a real thermodynamics question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given r=8.000 —r = 8.000\,\mathrm{—}, g=1.400 —g = 1.400\,\mathrm{—}, the governing relation η=1−r1−γ\eta=1-r^{1-\gamma} yields eta=0.565 —eta = 0.565\,\mathrm{—}. One governing identity, SI units, a single sweep on the sheet. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Outputs

  • Otto efficiency eta0.565 —
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Narration of this film

One governing identity, SI units, a single sweep on the sheet.

Ideal spark-ignition cycle. Pedagogical SI sheet with a live model and a swept parameter.

Reading speed

Watch on YouTube