Question

The efficiency of the Otto cycle is given by which of these expressions?Where r = compression ratio.

• A. $1-\frac{1}{r^{\gamma -1}}$
• B. $1-\frac{1}{{\gamma }^{r-1}}$
• C. $\frac{1}{r}$
• D. $\frac{1}{\gamma }$

Answer 1

The correct option is A

$1-\frac{1}{r^{\gamma -1}}$

The amount of work done W is related to QH and QC by the first law of thermodynamics

$Q_H=W+Q_C$- - - - - - (1)

The efficiency then becomes

e =$\frac{W}{Q_H}$

$=\frac{Q_H-Q_C}{Q_H}=1-\frac{Q_C}{Q_H}$- - - - - - (2)

We can calculate the efficiency of this idealized cycle. Processes bc and da are constant volume, so the heats $Q_H$ and $Q_C$ are related simply to the temperatures:

$Q_H=n{C}_V(T_c-T_b)>0$- - - - - - (3)

$Q_C=n{C}_V(T_a-T_d)<0$ - - - - - - (4)

From (2), (3) and (4)

We get

$e=1-\frac{T_a-T_d}{T_c-T_b}$ - - - - - - - (5)

To simplify this further, we use the temperature-volume relationship for adiabatic processes for an ideal gas. For the two adiabatic processes ab and cd, $T_a(rV{)}^{\gamma -1}=T_b{V}^{\gamma -1}$- - - - - - (6)

and $T_d(rV{)}^{\gamma -1}=T_c{V}^{\gamma -1}$ - - - - - - (7)

$T_b=T_a{r}^{\gamma -1}$ - - - - - - (8)

$T_c=T_d{r}^{\gamma -1}$ - - - - - - (9)

Substituting for $T_c$ and $T_b$ in (5)

We get

$e=1-\frac{T_a-T_d}{T_a{r}^{\gamma -1}-T_d{r}^{\gamma -1}}$

$=1-\frac{(T_a-T_d)}{r^{\gamma -1}(T_a-T_d)}$

$=1-\frac{1}{r^{\gamma -1}}$