Problem — Efficiency of the Otto cycle

Problem

An Otto engine has compression ratio $r = 10$. For a diatomic gas ($\gamma = 7/5$), calculate the ideal efficiency $\eta = 1 - r^{-(\gamma-1)}$.

Solution

Otto cycle formula. The ideal efficiency depends only on the compression ratio $r$ and the adiabatic exponent $\gamma$:

$\eta = 1 - r^{-(\gamma - 1)}$

Exponent. With $\gamma = 7/5 = 1{,}4$:

$\gamma - 1 = 0{,}4$

Numerical substitution. With $r = 10$:

$\eta = 1 - 10^{-0{,}4}$

Compute $10^{-0{,}4}$: since $10^{0{,}4} \approx 2{,}512$, we get $10^{-0{,}4} \approx 0{,}398$. So:

$\eta = 1 - 0{,}398 = 0{,}602$

$\ev{\eta \approx 60{,}2\%}$

Increasing the compression ratio raises the efficiency, but in real engines $r$ is limited by detonation (self-ignition) of the petrol-air mixture.

Collegamenti

Argomenti: Heat engines Concetti: Efficiency · Thermodynamic transformations Competenze: Solving a thermodynamic cycle · Reading graphs Metodi: P-V and ln P - ln V diagram Oggetti: Ideal gas · Heat engine