Problem — Realistic petrol engine (Otto vs Carnot)

Problem

A petrol engine operates with compression ratio $r = 9$ and effective combustion at $T_c \approx 1800\;\text{K}$, exhaust at $T_f \approx 700\;\text{K}$. Compare: (a) ideal Otto efficiency, (b) Carnot efficiency with these temperatures. Which of the two limits is “real” for the engine?

Solution

(a) Otto cycle efficiency. For the ideal Otto cycle (diatomic gas, $\gamma = 1{,}4$) the efficiency depends only on the compression ratio:

$\eta_\text{Otto} = 1 - \frac{1}{r^{\gamma-1}} = 1 - \frac{1}{9^{0{,}4}}$

With $9^{0{,}4} \approx 2{,}41$:

$\eta_\text{Otto} = 1 - \frac{1}{2{,}41} \approx 1 - 0{,}415 = \ev{0{,}58 \ (58\%)}$

(b) Carnot efficiency. With the cycle’s extreme temperatures:

$\eta_\text{Carnot} = 1 - \frac{T_f}{T_c} = 1 - \frac{700}{1800} \approx 1 - 0{,}389 = \ev{0{,}61 \ (61\%)}$

Which limit matters. The two values are close, but $\eta_\text{Otto} \approx 58\% < \eta_\text{Carnot} \approx 61\%$: the Otto limit is tighter. Carnot gives the absolute ceiling for any engine between those two temperatures, but the petrol engine isn’t a Carnot engine — it’s an Otto cycle, constrained by its geometric compression ratio. So $\eta_\text{Otto}$ is the limit that actually matters for the engine. Real-world efficiency is even lower ($\sim 25$-$30\%$) due to friction, real heat exchanges, and non-instantaneous combustion.

Collegamenti

Argomenti: Macchine termiche Concetti: Rendimento · Ciclo di Carnot · Trasformazioni termodinamiche Competenze: Risoluzione di un ciclo termodinamico Oggetti: Macchina termica · Gas ideale