The Diesel cycle is the thermodynamic cycle which approximates the pressure and volume of the combustion chamber in a Diesel engine. In the Diesel cycle, the working medium in the combustion chamber is

1. compressed adiabatically from $$V_1$$, $$p_1$$ to $$V_2$$ < $$V_1$$, $$p_2$$ > $$p_1$$,
2. expanded isobarically at $$p_2$$ from $$V_2$$ to $$V_3$$ > $$V_2$$,
3. expanded adiabatically from $$V_3$$, $$p_2$$ to $$V_1$$ and $$p_4$$ > $$p_1$$,
4. cooled isochorically at $$V_1$$ from $$p_4$$ to the initial pressure $$p_1$$.

The heat absorbed by the medium is given by $$Q=\int_{2 \rightarrow 3} C_p dT$$. Assuming the working medium is an ideal monatomic gas, show that the efficiency of the Diesel cycle can be written in the form

$\eta = 1 - \gamma^{-1} {R_{31}^{\gamma} - R_{21}^{\gamma} \over R_{31} - R_{21}}$

Efficiency of the Diesel Cycle

Because the process $$1 \rightarrow 2$$ and $$3 \rightarrow 4$$ are adiabatic, heat is not lost during those processes, so the efficiency can therefore be written in terms of the heat gained and lost, $$Q_H$$ and $$Q_L$$, as indicated on the diagram.

From the equation given above we can say that the heat gained $$Q_H$$ is given in terms of specific heat and temperature by:

\begin{equation} Q_H = C_p (T_3 – T_2) \label{eq:q1} \end{equation}

We can likewise deduce the heat lost as $$Q=\int_{4 \rightarrow 1} C_V dT$$ such that

\begin{equation} Q_L = C_V (T_1 – T_4) \label{eq:q2} \end{equation}

The thermal efficiency is defined as1 $$\eta={Q_H+Q_L \over Q_H}$$, which we can express using \eqref{eq:q1} and \eqref{eq:q2} in the form of specific heats and temperatures \begin{equation} \eta=1+{C_V(T_1-T_4) \over C_p(T_3-T_2)} \label{eq:etaq1q2} \end{equation}

Using the ideal gas law $$PV=nRT$$ where the $$n$$s and $$R$$s cancel, and the adiabatic index $$\gamma = {C_p \over C_V}$$ we can further write

$\eta = 1 + \gamma^{-1} {P_1 V_1 - P_4 V_4 \over P_3 V_3 - P_2 V_2}$

From the diagram it can be seen that the volume V is the same in state 4 as in 1, i.e. $$V_4=V_1$$. Likewise for pressure, $$P_3=P_2$$. Therefore we can reduce our number of variables so that

$\eta = 1 + \gamma^{-1} {V_1(P_1 - P_4) \over P_3(V_3-V_2)}$

Dividing the fraction through by $$P_3V_1$$:

$\eta = 1+\gamma^{-1} {\frac{P_1}{P_3} - \frac{P_4}{P_3} \over R_{31} - R_{21}}$

where $$R_{31} := \frac{V_3}{V_1}$$ and $$R_{21} := \frac{V_2}{V_1}$$.

For an adiabatic expansion, $$PV^{\gamma}$$ is constant. Exploiting this fact, we can see from the diagram that the following relations hold:

\begin{eqnarray} \frac{P_1}{P_3}&=&\left( {V_2 \over V_1} \right)^{\gamma} &=& R_{21}^{\gamma} \nonumber \\ \frac{P_4}{P_3}&=&\left( {V_3 \over V_1} \right)^{\gamma} &=& R_{31}^{\gamma} \nonumber \end{eqnarray}

Finally, putting these back in and playing with minus signs:

$\eta = 1 - \gamma^{-1} {R_{31}^{\gamma} - R_{21}^{\gamma} \over R_{31} - R_{21}} ; \blacksquare$

1. See MIT’s brief page about the Diesel cycle. It’s also whence I nicked my diagram. ↩︎