Thermodynamic cycle

A thermodynamic cycle returns a working substance to its initial state. Its net internal-energy change is zero, so the net work output equals net heat input.

The ideal Atkinson cycle considered here has reversible adiabatic compression, isochoric heat addition, a longer reversible adiabatic expansion and isobaric heat rejection. Its expansion ends at the initial pressure rather than at the initial volume. With constant ideal-gas heat capacities, $r=V_1/V_2$ and $r_p=p_3/p_2$, the temperatures are $T_2=T_1{r}^{\gamma -1}$, $T_3=r_p{T}_2$, and $T_{4^{\prime }}=T_1{r}_p^{1/\gamma }$. This ideal cycle is distinguished from the detailed valve timing and losses of a practical engine.

For the same initial state, compression ratio and constant-volume heat input, the ideal Atkinson cycle adds net work during the extended expansion and subsequent constant-pressure return. The extra work is the integral of $p_{\mathrm{expansion}}-p_1$ over the added volume, which is positive. Equivalently, concavity of $x^{1/\gamma }$ gives $\gamma (r_p^{1/\gamma }-1)<r_p-1$ when $r_p>1$ and $\gamma >1$, so its efficiency exceeds the corresponding Otto cycle efficiency. The comparison fixes these shared conditions rather than asserting universal superiority under arbitrary engine constraints.

The ideal Brayton cycle consists of adiabatic compression, constant-pressure heat addition, adiabatic expansion, and constant-pressure heat rejection. Its ideal working substance is an ideal gas with constant heat capacities, and the insulated stages are reversible thermodynamic processes.

For pressure ratio $r_p=p_h/p_l$ and heat capacity ratio $\gamma =C_p/C_V$, a reversible ideal-gas adiabat has $T\propto p^{(\gamma -1)/\gamma }$. Thus $T_2/T_1=T_3/T_4=k=r_p^{(\gamma -1)/\gamma }$, and the constant-pressure heat transfers have $Q_{\mathrm{out}}/Q_{\mathrm{in}}=(T_4-T_1)/(T_3-T_2)=1/k$. The thermal efficiency isThis formula requires constant heat capacities and internally reversible insulated stages. An arbitrary dissipative adiabatic process does not satisfy the same pressure-temperature relation.

For constant-pressure heat addition $A\to B$ and two reversible ideal-gas adiabats, set $r=(p_h/p_l{)}^{(\gamma -1)/\gamma }$. The Brayton cycle has ${\eta }_B=1-1/r$, maximum temperature $T_B$ and minimum $T_D=T_A/r$. A Carnot cycle using those extreme temperatures has the strictly greater efficiency displayed whenever $T_B>T_A$. Reversibility of the isobaric heating requires continuously matched reservoir temperatures, rather than heat transfer solely at the maximum temperature.

The thermal efficiency of a heat engine is the net work output divided by heat input. For a cycle absorbing $Q_{\mathrm{in}}$ and rejecting the positive amount $Q_{\mathrm{out}}$,

The ideal Otto cycle has adiabatic compression, constant-volume heat addition, adiabatic expansion, and constant-volume heat rejection. For an ideal gas with compression ratio $r=V_1/V_2$ and specific-heat ratio $\gamma$,

The ideal Diesel cycle consists of reversible adiabatic compression, constant-pressure heat addition, reversible adiabatic expansion, and constant-volume heat rejection.

For an ideal Diesel cycle with compression ratio $r=V_1/V_2>1$, cutoff ratio $\rho =V_3/V_2>1$ and constant heat-capacity ratio $\gamma$, its thermal efficiency is $\eta =1-({\rho }^{\gamma }-1)/[\gamma r^{\gamma -1}(\rho -1)]$. As $\rho \to 1^{+}$ the quotient tends to $\gamma$, giving $\eta \to 1-r^{1-\gamma }$. The maximum-to-minimum temperature ratio tends to $r^{\gamma -1}$, so the limiting efficiency equals that of a Carnot cycle between those temperatures. Both heat input and net work vanish in this limit: the result is a limit of their ratio, not a finite-work engine at exactly zero cutoff interval.