Derive an expression for efficiency of diesel cycle.
1 Answer

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Diesel cycle comprises of the following operations:

(i) 1-2......Adiabatic compression.

(ii) 2-3......Addition of heat at constant pressure.

(iii) 3-4......Adiabatic expansion.

(iv) 4-1......Rejection of heat at constant volume.

Consider 1 kg of air.

Heat supplied at constant pressure =$C_p (T_3-T_2 )$

Heat rejected at constant volume =$C_v (T_4-T_1 )$

Work done = Heat supplied – Heat rejected

=$C_p (T_3-T_2 )-C_v (T_4-T_1 )$

$η_diesel=\frac{work done}{heat supplied}$

=$\frac{C_p (T_3-T_2 )-C_v (T_4-T_1 )}{C_p (T_3-T_2 )}$

=$\frac{1-(C_v (T_4-T_1 )}{C_p (T_3-T_2 )}$

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

Let compression ratio,r=$\frac{v_1}{v_2}$ , and cut-off ratio,ρ=$\frac{v_3}{v_2}$ i.e.$\frac{Volume at cut-off}{clearance volume}$

Now, during adiabatic compression 1-2,

$\frac{T_2}{T_1}$ =$\frac{v_1}{v_2}^(γ-1)$=$r^(γ-1)$ or $T_2=T_1 r^(γ-1)$

During constant pressure process 2-3,

$\frac{T_3}{T_2}$ =$\frac{v_3}{v_2}$ =ρ or $T_3=ρT_2=ρT_1 r^(γ-1)$

During adiabatic expansion 3-4

$\frac{T_3}{T_4}$ =$\frac{v_4}{v_3}^(γ-1)$=$\frac{r}{ρ}^(γ-1)$ (since $\frac{v_4}{v_3} =\frac{v_1}{v_3} =\frac{v_1}{v_2} ×\frac{v_2}{v_3} =\frac{r}{ρ}$)

$T_4=\frac{T_3}{\frac{r}{ρ}}^(γ-1)$ =$\frac{ρ∙T_1 r^(γ-1)}{\frac{r}{ρ}^(γ-1)}$ =$T_1 ρ^γ$

By inserting values of $T_2,T_3$ and $T_4$ in eqn. i, we get

$η_diesel=\frac{1-(T_1 ρ^γ-T_1 )}{γ(ρ.T_1.r^(γ-1)-T_1∙r^(γ-1))} =\frac{1-(ρ^γ-1)}{γ∙r^(γ-1) (ρ-1)}$

$η_diesel=1-\frac{1}{(γ.r^(γ-1)} [\frac{(ρ^γ-1)}{(ρ-1)}]$

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