Problem — Diesel cycle efficiency

Problem

A Diesel engine has compression ratio $r = 18$ and cut-off ratio $\rho = 2{,}5$, with $\gamma = 1{,}4$. Calculate the ideal efficiency given by

$\eta_\text{Diesel} = 1 - \frac{1}{r^{\gamma-1}}\cdot\frac{\rho^\gamma - 1}{\gamma(\rho - 1)}$

Solution

Compression factor. With $\gamma - 1 = 0{,}4$:

$r^{\gamma-1} = 18^{0{,}4} \approx 3{,}18 \;\Rightarrow\; \frac{1}{r^{\gamma-1}} \approx 0{,}315$

Cut-off factor. With $\rho = 2{,}5$ and $\gamma = 1{,}4$, $\rho^\gamma = 2{,}5^{1{,}4} \approx 3{,}61$:

$\frac{\rho^\gamma - 1}{\gamma(\rho - 1)} = \frac{3{,}61 - 1}{1{,}4\,(2{,}5 - 1)} = \frac{2{,}61}{2{,}1} \approx 1{,}24$

Efficiency. Combining the two factors:

$\eta_\text{Diesel} = 1 - 0{,}315 \times 1{,}24 \approx 1 - 0{,}39 = \ev{0{,}61 \ (61\%)}$

The high compression ratio ($r = 18$, versus $\sim 9$ for petrol) pushes the ideal efficiency to $\sim 61\%$. The cut-off ratio $\rho$ penalises the Diesel engine slightly compared with an Otto engine of the same $r$ (the term $(\rho^\gamma-1)/[\gamma(\rho-1)] > 1$), but the much higher compression wins out. Real Diesel engines achieve $40$-$45\%$, still higher than petrol engines.

Note $\eta \approx 0{,}609$, i.e. $\approx 61\%$. Some sources round to $\approx 63\%$: the difference depends only on the number of digits kept in the intermediate steps ($18^{0{,}4}$ and $2{,}5^{1{,}4}$).

The exact calculation gives

Collegamenti

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