Question

In an air-standard diesel cycle with compressior ratio 15, the explosion is happening for $\frac{1}{{20}^{th}}$ of the stroke. The air standard efficiency of the cycle is_____%. For air, $\gamma =1.41$

• A. 62.84

Answer 1

The correct option is A 62.84

For a diesel cycle
${\eta }_{airstd}=1-\frac{1}{r^{\gamma -1}}\times \frac{{\rho }^{\gamma }-1}{\gamma (\rho -1)}$
$r=15$

$\gamma =1.41$

$v_3-v_2=\frac{1}{2}(v_1--v_2)$

$\therefore 15=\frac{v_1}{v_2}$

$v_3-v_2=\frac{1}{20}(15v_2-v_2)$

$\therefore v_3=v_2$

$v_3-v_2=\frac{14}{20}{v}_2$

$\frac{v_3}{v_2}=\frac{34}{20}=1.7$

So,
${\eta }_{airstd}=1-\frac{1}{{15}^{1.41-1}}\times \frac{{1.7}^{1.41}-1}{1.41\times (1.7-1)}=0.6284or62.84$