Derive a formula for the efficiency of the Diesel cycle, in terms of the compression ratio V₁/ V₂and the cutoff ratio V₃/ V₂. Show that for a given compression ratio, the Diesel cycle is less efficient than the Otto cycle. Evaluate the theoretical efficiency of a Diesel engine with a compression ratio of 18 and a cutoff ratio of 2.

Diesel engine with a compression ratio of 18 and a cutoff ratio of 2.

The efficiency can be calculated from the temperature and specific heats as below:

$$\begin{gathered} Q_1=C_p\left(T_c-T_b\right) \\ {Q}_2=C_v\left(T_a-T_d\right) \\ \text{Efficiency,}\eta =\frac{Q_1+Q_2}{Q_1} \end{gathered}$$

Consider 1 kg of air.
Heat supplied at constant pressure $=C_p\left(T_3-T_2\right)$
Heat rejected at constant volume $=C_v\left(T_4-T_1\right)$
Work done = Heat supplied - Heat rejected

$=C_p\left(T_3-T_2\right)-C_v\left(T_4-T_1\right)$

${\eta }_{\text{diesel}}=\frac{\text{workdone}}{\text{heatsupplied}}$

substitute

simplify:

$=1-\frac{\left(C_v\left(T_4-T_1\right)\right)}{C_p\left(T_3-T_3\right)}$

$=1-\frac{\left(T_4-T_1\right)}{\gamma \left(T_3-T_2\right)}\left[\because \frac{C_p}{C_v}=\gamma \right]\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(i\right)$

Let compression ratio, $r=\frac{v_1}{v_2}$

And cutoff ratio,

$$\begin{gathered} \rho =\frac{v_3}{v_2} \\ \rho =\frac{\text{Volume at cut-off}}{\text{clearance volume}} \end{gathered}$$

during adiabatic compression 1-2,

$$\begin{gathered} \frac{T_2}{T_1}=\frac{v_1}{v_2}(\gamma -1)=r(\gamma -1) \\ {T}_2=T_{1}r(\gamma -1) \end{gathered}$$

In constant pressure process 2-3

$$\begin{gathered} \frac{T_3}{T_2}=\frac{v_3}{v_2}=\rho \\ {T}_3=\rho T_2=\rho T_{1}r(\gamma -1) \end{gathered}$$

During 3-4

$$\begin{gathered} \frac{T_3}{T_4}=\frac{v_4}{v_3}(\gamma -1) \\ =\frac{r}{\rho }(\gamma -1)\left(\because \frac{v_4}{v_3}=\frac{v_1}{v_3}=\frac{v_1}{v_2}\times \frac{v_2}{v_3}=\frac{r}{\rho }\right) \\ {T}_4=\frac{T_3}{\frac{r}{\rho }}(\gamma -1)=\frac{\rho ·T_{1}r(\gamma -1)}{\frac{r}{\rho }(\gamma -1)}=T_1{\rho }^{\gamma } \end{gathered}$$

Now substitute the values of T₂, T₃ and T₄ in equation (i), we get

$$\begin{gathered} {\eta }_{\text{diesel}}=\frac{1-\left(T_1{\rho }^{\gamma }-T_1\right)}{r\left(\rho ·T_1·r(\gamma -1)-T_1·r(\gamma -1)\right)}=\frac{1-\left({\rho }^{\gamma }-1\right)}{\gamma ·r(\gamma -1)(\rho -1)} \\ {\eta }_{\text{diesel}}=1-\frac{1}{(\gamma ·r(\gamma -1\left)\right)}\left[\frac{\left({\rho }^{\gamma }-1\right)}{(\rho -1)}\right] \end{gathered}$$

Now express efficiency of Diesel cycle in terms of compression ratio and the cut off ratio
Let, $r_c=V_1/V_2\&r_e=V_1/V_3$

$\eta =1-\frac{1}{\gamma }\frac{r_e^{-\gamma }-r_c^{-\gamma }}{r_e^{-1}-r_c^{-1}}$

Substitute the value compression ratio, r_c=18, and Cut-off ratio, r_e=2
Theoretical efficiency, $\eta =1-\frac{1}{\gamma }\frac{r_e^{-\gamma }-r_c^{-\gamma }}{r_e^{-1}-r_c^{-1}}$
Assume $\gamma =1.4$, we get

$\eta =1-\frac{1}{1.4}\times \frac{2^{-1.4}-{18}^{-1.4}}{2^{-1}-{18}^{-1}}=0.419\approx 42\%$