Past exam of the physics B course of the University of Cambridge / 2017 / Paper 2 / Section C / C10

A Carnot cycle consists of four reversible stages: isothermal expansion at the hot-reservoir temperature $T_h$, adiabatic expansion to $T_c$, isothermal compression at $T_c$, and adiabatic compression back to the starting state. The adiabatic stages exchange no heat. For one mole of an ideal gas, the two isothermal stages have heats $Q_h=R{T}_h\ln (V_B/V_A)$ and $Q_c=R{T}_c\ln (V_C/V_D)$ in magnitude. The reversible adiabatic relations ensure $V_B/V_A=V_C/V_D$.

More generally, the total entropy change over any reversible cycle is zero. The two heat exchanges therefore satisfyWith work output $W=Q_h-Q_c$, its thermal efficiency isThe temperatures must be absolute. The second law makes this the maximum efficiency for an engine exchanging heat only with those two reservoirs; irreversible entropy production lowers the efficiency.

Label the initial state 1, the end of compression 2, the end of constant-pressure heating 3, and the end of adiabatic expansion 4. The volume sequence is $V_1\to V_2\to V_3\to V_1\to V_1$, with $V_2<V_3<V_1$. The Diesel cycle has a compression adiabat $1\to 2$, a horizontal constant-pressure heat-input segment $2\to 3$, an expansion adiabat $3\to 4$, and a vertical constant-volume cooling segment $4\to 1$.

For an ideal gas, a reversible adiabat obeys $p{V}^{\gamma }=\mathrm{constant}$, so it falls more steeply than an isotherm. The expansion adiabat lies above the compression adiabat at equal volume between $V_3$ and $V_1$, giving a clockwise cycle with positive work output.

For one mole, $pV=RT$, and constant heat capacities satisfy $C_p-C_V=R$. The specified diatomic-gas model hasA reversible adiabatic process satisfies $T{V}^{\gamma -1}=\mathrm{constant}$. Constant-pressure expansion satisfies $T\propto V$. Applying these in order givesEvery volume ratio is dimensionless, so the expressions have units of temperature. For the intended volume ordering, the minimum temperature is $T_1$ and the maximum is $T_3$.

Use heat into the gas as positive. The two adiabatic stages have $Q_{12}=Q_{34}=0$. During constant-pressure heating and constant-volume cooling,The first law over a complete cycle gives $W_{\mathrm{out}}=Q_{23}+Q_{41}$, since the internal energy returns to its initial value. ThereforeSubstitute the temperatures:HenceEquivalently, define compression ratio $r=V_1/V_2$ and cutoff ratio $\rho =V_3/V_2$:These formulas use the ideal constant-heat-capacity model and reversible adiabatic strokes. The heat-addition and heat-rejection strokes occur over changing gas temperatures, unlike the isothermal heat exchanges of a Carnot engine.

For $\rho \to 1^{+}$, Taylor expansion gives ${\rho }^{\gamma }-1=\gamma (\rho -1)+O((\rho -1{)}^2)$. The vanishing-cutoff limit of the Diesel cycle isThe maximum temperature tends to $T_2=T_1{r}^{\gamma -1}$, while the minimum stays at $T_1$. A Carnot engine between those limiting extrema hasso the limiting efficiencies coincide. At exactly $V_3=V_2$, however, both heat input and net work vanish, and the cycle retraces its compression adiabat. The efficiency there is an undefined $0/0$; the boxed value is the continuous limit for neighbouring nonzero cycles, not finite work at zero heat input.

For a finite cutoff ratio $\rho >1$, the Carnot comparison uses $T_{\max }=T_3=T_1{r}^{\gamma -1}\rho$, giving ${\eta }_C=1-1/(r^{\gamma -1}\rho )$. The Diesel cycle has smaller efficiency. For example,which makes its rejected/input heat ratio larger than the Carnot ratio.