To get more than an infinitesimal amount of work out of a Carnot engine, we would have to keep the temperature of its working substance below that of the hot reservoir and above that of the cold reservoir by non-infinitesimal amounts. Consider, then, a Carnot cycle in which the working substance is at temperature$T_{hw}$as it absorbs heat from the hot reservoir, and at temperature$T_{cw}$as it expels heat to the cold reservoir. Under most circumstances the rates of heat transfer will be directly proportional to the temperature differences:

$\frac{Q_h}{\Delta t}=K\left(T_h-T_{hw}\right)\text{and}\frac{Q_c}{\Delta t}=K\left(T_{cw}-T_c\right)$

I've assumed here for simplicity that the constants of proportionality $\left(K\right)$are the same for both of these processes. Let us also assume that both processes take the

same amount of time, so the$\Delta t^{\text{'}}$'s are the same in both of these equations.*

$\left(a\right)$Assuming that no new entropy is created during the cycle except during the two heat transfer processes, derive an equation that relates the four temperatures$T_h,T_c,T_{hw}$and $T_{cw}$

$\left(b\right)$Assuming that the time required for the two adiabatic steps is negligible, write down an expression for the power (work per unit time) output of this engine. Use the first and second laws to write the power entirely in terms of the four temperatures (and the constant $K$), then eliminate$T_{cw}$using the result of part $\left(a\right)$.

$\left(c\right)$When the cost of building an engine is much greater than the cost of fuel (as is often the case), it is desirable to optimize the engine for maximum power output, not maximum efficiency. Show that, for fixed $T_h$and $T_c$, the expression you found in part $\left(b\right)$has a maximum value at $T_{hw}=\frac{1}{2}\left(T_h+\sqrt{T_h{T}_c}\right)$. (Hint: You'll have to solve a quadratic equation.) Find the corresponding$T_{cw}.$

$\left(d\right)$Show that the efficiency of this engine is $1-\sqrt{T_c/T_h}$Evaluate this efficiency numerically for a typical coal-fired steam turbine with $T_h=600°\mathrm{C}$and$T_c=25°\mathrm{C}$, and compare to the ideal Carnot efficiency for this temperature range. Which value is closer to the actual efficiency, about $40\%$, of a real coal-burning power plant?

The fundamental issue with a Carnot cycle engine is that the two isothermal stages of the cycle go very slowly since we are transferring heat between two systems that are almost at the same temperature. Making the temperature of the working material considerably different from that of the reservoir where it absorbs and subsequently expels heat is one technique to speed up the cycle. That is, if the system absorbs heat $Q_h$from a hot reservoir at $T_h$, the temperature of the working material (usually a gas) when it absorbs heat is $T_{hw}<T_h$, and the gas is $T_{cw}>T_c$.

To keep things simple, we'll assume that the rate of heat transfer is proportional to the temperature differential between the gas and the reservoir for both hot and cold reservoirs. That is to say:

$\begin{array}{r}\frac{Q_h}{\mathrm{\Delta }t}=K\times \left(T_h-T_{hw}\right)\\ \frac{Q_c}{\mathrm{\Delta }t}=K\times \left(T_{cw}-T_c\right).\end{array}$

$K$is a constant in both circumstances, and $∆t$is assumed to be the same (that is, the durations of both isothermal stages in the cycle are the same).

We may deduce the following relationship:

$\frac{Q_h}{K\times \left(T_h-T_{hw}\right)}=\frac{Q_c}{K\times \left(T_{cw}-T_c\right)}$

Since the only entropy generated in the cycle is along the two isothermal stages (no entropy is generated along the adiabatic stages), then the gas must have expended exactly the same amount of entropy when expelling heat to the cold reservoir as it absorbed when absorbing heat from the hot reservoir, because the state of the engine is the same at the end of the cycle as it was at the start. That is to say:

$\begin{array}{l}\frac{Q_h}{T_{hw}}=\frac{Q_c}{T_{cw}}\\ Q_c=\frac{T_{cw}}{T_{hw}}\times Q_h\\ \begin{array}{l}\frac{Q_h}{K\times \left(T_h-T_{hw}\right)}=\frac{T_{cw}\times Q_h}{K\times T_{hw}\times \left(T_{cw}-T_c\right)}\\ T_{hw}\times \left(T_{cw}-T_c\right)=T_{cw}\times \left(T_h-T_{hw}\right)\\ -T_{hw}\times T_c=T_{cw}\times \left(T_h-T_{hw}-T_{hw}\right)\\ T_{cw}=\frac{T_{hw}\times T_c}{2T_{hw}-T_h}.\end{array}\end{array}$

We can calculate the engine's power output if the time necessary for the two adiabatic phases is much smaller than the time required for the two isothermal steps.

The work is created over a $2∆t$time span and is as follows:

$\begin{array}{l}P=\frac{W}{2\mathrm{\Delta }t}\\ W=Q_h-Q_c\\ P=\frac{Q_h-Q_c}{2\mathrm{\Delta }t}\\ \begin{array}{l}P=\frac{K\mathrm{\Delta }t\times \left(T_h-T_{hw}\right)-K\mathrm{\Delta }t\times \left(T_{cw}-T_c\right)}{2\mathrm{\Delta }t}\\ P=\frac{K}{2}\times \left(T_h-T_{hw}-T_{cw}+T_c\right)\\ P=\frac{K}{2}\times \left(T_h+T_c-T_{hw}-\frac{T_{hw}\times T_c}{2T_{hw}-T_h}\right).\end{array}\end{array}$

The output power by

$$\begin{gathered} \begin{array}{l}\frac{dP}{d{T}_{hw}}=\frac{K}{2}\times \frac{d}{d{T}_{hw}}\times \left(T_h+T_c-T_{hw}-\frac{T_{hw}\times T_c}{2T_{hw}-T_h}\right)\\ \frac{dP}{d{T}_{hw}}=\frac{K}{2}\times \left(-1-\frac{\left(2T_{hw}-T_h\right)\times T_c-2T_{hw}\times T_c}{{\left(2T_{hw}-T_h\right)}^2}\right)\\ \frac{dP}{d{T}_{hw}}=\frac{K}{2}\times \left(-1+\frac{T_h\times T_c}{{\left(2T_{hw}-T_h\right)}^2}\right)\end{array} \\ \begin{array}{l}\frac{K}{2}\times \left(-1+\frac{T_h\times T_c}{{\left(2T_{hw}-T_h\right)}^2}\right)=0\\ \frac{T_h\times T_c}{{\left(2T_{hw}-T_h\right)}^2}=1\\ T_h\times T_c={\left(2T_{hw}-T_h\right)}^2\\ \sqrt{T_h\times T_c}=\left(2T_{hw}-T_h\right)\\ T_{hw}=\frac{1}{2}\times \left(T_h+\sqrt{T_h{T}_c}\right)\\ \begin{array}{l}{T}_{cw}=\frac{1}{2}\times \frac{T_c\times T_h+T_c\times \sqrt{T_h\times T_c}}{T_h+\sqrt{T_h\times T_c}-T_h}\\ T_{cw}=\frac{1}{2}\times \left(\frac{T_c\times T_h}{\sqrt{T_h\times T_c}}+\frac{T_c\times \sqrt{T_h\times T_c}}{\sqrt{T_h\times T_c}}\right)\\ T_{cw}=\frac{1}{2}\times \left(\sqrt{T_h\times T_c}+T_c\right)\\ T_{cw}=\frac{1}{2}\times \left(T_c+\sqrt{T_h\times T_c}\right).\end{array}\end{array} \end{gathered}$$

The engine efficiency is

$\begin{array}{l}e=1-\frac{Q_c}{Q_h}\\ e=1-\frac{\frac{T_{cw}}{T_{hw}}\times Q_h}{Q_h}\\ e=1-\frac{T_{cw}}{T_{hw}}\\ e=1-\frac{T_c+\sqrt{T_h\times T_c}}{T_h+\sqrt{T_h\times T_c}}\\ \begin{array}{l}e=1-\frac{T_c+\sqrt{T_h\times T_c}}{T_h+\sqrt{T_h\times T_c}}\times \frac{T_h-\sqrt{T_h\times T_c}}{T_h-\sqrt{T_h\times T_c}}\\ e=1-\frac{\left(T_c+\sqrt{\left.T_h\times T_c\right)}\left(T_h-\sqrt{T_h\times T_c}\right)\right.}{T_h^2-\left(T_h\times T_c\right)}\\ e=1-\frac{\left(T_h\times T_c\right)-\left(T_h\times T_c\right)-\left(T_c\sqrt{T_h\times T_c}\right)+\left(T_h\sqrt{T_h\times T_c}\right)}{T_h^2-\left(T_h\times T_c\right)}\\ e=1-\frac{\left(T_h-T_c\right)\times \sqrt{T_h\times T_c}}{T_h\times \left(T_h-T_c\right)}\\ e=1-\frac{\sqrt{T_c}}{\sqrt{T_h}}.\end{array}\end{array}$

For coal-fire steam turbine,the efficiency of

$T_h={600}^{\circ }\mathrm{C}=873\mathrm{K}$and $T_c={25}^{\circ }\mathrm{C}=298K$is

$$\begin{gathered} e=1-\frac{\sqrt{298}}{\sqrt{873}} \\ e=0.4157. \end{gathered}$$

It's extremely near to the actual efficiency of a coal-fired power plant, which is around $40\%$.For these temperatures, the "optimal" Carnot efficiency is:

$$\begin{gathered} e_{max}=1-\frac{T_c}{T_h} \\ {e}_{max}=1-\frac{298}{873} \\ {e}_{max}=0.6586. \end{gathered}$$