91Ó°ÊÓ

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:

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

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

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

${\mathrm{η}}_{\text{diesel}}=\frac{\text{workdone}}{\text{heatsupplied}}$

substitute

simplify:

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

$=1-\frac{\left(T_4-T_1\right)}{\mathrm{γ}\left(T_3-T_2\right)}\left[âˆ\mu \frac{C_p}{C_v}=\mathrm{γ}\right]………………………\left(i\right)$

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

And cutoff ratio,

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

during adiabatic compression 1-2,

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

In constant pressure process 2-3

$$\begin{gathered} \frac{T_3}{T_2}=\frac{v_3}{v_2}=\mathrm{ÒÏ} \\ {T}_3=\mathrm{ÒÏ}{T}_2=\mathrm{ÒÏ}{T}_{1}r(\mathrm{γ}-1) \end{gathered}$$

During 3-4

$$\begin{gathered} \frac{T_3}{T_4}=\frac{v_4}{v_3}(\mathrm{γ}-1) \\ =\frac{r}{\mathrm{ÒÏ}}(\mathrm{γ}-1)\left(âˆ\mu \frac{v_4}{v_3}=\frac{v_1}{v_3}=\frac{v_1}{v_2}×\frac{v_2}{v_3}=\frac{r}{\mathrm{ÒÏ}}\right) \\ {T}_4=\frac{T_3}{\frac{r}{\mathrm{ÒÏ}}}(\mathrm{γ}-1)=\frac{\mathrm{ÒÏ}·T_{1}r(\mathrm{γ}-1)}{\frac{r}{\mathrm{ÒÏ}}(\mathrm{γ}-1)}=T_1{\mathrm{ÒÏ}}^{\mathrm{γ}} \end{gathered}$$

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

$$\begin{gathered} {\mathrm{η}}_{\text{diesel}}=\frac{1-\left(T_1{\mathrm{ÒÏ}}^{\mathrm{γ}}-T_1\right)}{r\left(\mathrm{ÒÏ}·T_1·r(\mathrm{γ}-1)-T_1·r(\mathrm{γ}-1)\right)}=\frac{1-\left({\mathrm{ÒÏ}}^{\mathrm{γ}}-1\right)}{\mathrm{γ}·r(\mathrm{γ}-1)(\mathrm{ÒÏ}-1)} \\ {\mathrm{η}}_{\text{diesel}}=1-\frac{1}{(\mathrm{γ}·r(\mathrm{γ}-1\left)\right)}\left[\frac{\left({\mathrm{ÒÏ}}^{\mathrm{γ}}-1\right)}{(\mathrm{ÒÏ}-1)}\right] \end{gathered}$$

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

$\mathrm{η}=1-\frac{1}{\mathrm{γ}}\frac{r_e^{-\mathrm{γ}}-r_c^{-\mathrm{γ}}}{r_e^{-1}-r_c^{-1}}$

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

$\mathrm{η}=1-\frac{1}{1.4}×\frac{2^{-1.4}-{18}^{-1.4}}{2^{-1}-{18}^{-1}}=0.419≈42\%$