# Carnot heat engine at maximum power

I am trying to prove that the Carnot efficiency for a heat engine with maximum power is $$1-\sqrt{\frac{\tau_l}{\tau_h}}$$, where $$\tau_h$$ is the high temperature, $$\tau_2$$ the lower temperature, and we have the working temperature $$\tau_{hw},\tau_{lw}$$ such that $$\tau_h>\tau_{hw}>\tau_{lw}>\tau_l$$. During the hot isothermal stage, the rate of heat flow from hot reservoir to working fluid is proportional to the temperature difference $$x=\tau_h-\tau_{hw}$$, and for the cold it is proportional to $$y = \tau_{lw}-\tau_l$$. If we assume that both isothermal stages take time $$t$$, and that both isentropic stages take negligible time, then we can write $$Q_h=Ktx$$ for the hot isothermal stage, $$Q_l=Kty$$ for the cold stage.

I am aware of the derivation by Curzon and Ahlborn but I do not understand their derivation that well. So, I want to approach it a different way. I thought about expressing the power output as a function of $$K,x$$ and $$y$$ and then finding the optimal value for $$x,y$$. However, I am having trouble with this first step. Any help would be appreciated on finding this expression for the power output and also on how to write $$y$$ in terms of $$x,\tau_h,\tau_l$$. (We are also assuming the working fluid undergoes an ideal Carnot cycle between temperatures $$\tau_{hw}$$ and $$\tau_{lw}$$)

• There should be no square root in the Carnot efficiency formula. – Buzz Jun 11 at 1:03
• @Buzz This is for the efficiency at maximum power output – user267129 Jun 11 at 1:06
• Are you assuming that the isothermal steps are occurring simultaneously? – Chet Miller Jun 11 at 2:00
• Provide the reference for the "derivation by Curzon and Ahlborn" because, quite frankly, this does not make any sense to me. T – Bob D Jun 11 at 2:02
• In the Carnot engine, the two isothermal steps can't be simultaneous. One step occurs first and then the second isothermal step occurs. The working fluid can not be both expanding and compressing at the same time. On a different note, if the Carnot Cycle involving the inner temperatures is ideal, what can you say about the relationship between the two temperature differences that your are working with (x and y)? – Chet Miller Jun 11 at 12:17

Here is a very simple derivation from 1961,see Penfield ! Start with the picture of a Carnot engine sandwiched between two thermal resistances each being attached to a thermostat. According to Carnot's efficiency theorem the work $$W$$ delivered $$W \le \big(1-\frac{T_2}{T_1}\big)Q \tag{1}\label{1}$$ where $$Q$$ is the heat entering the engine at the higher temperature. The temperatures between which the Carnot engine operates satisfy the following dissipation equalities: $$T_1=T_c + Q_r R_c \\ = T_c +QR_c-WR_c \tag{2}\label{2}$$ $$T_2=T_h-QR_h \tag{3}\label{3}$$ Subsutitute $$\eqref{2}$$and $$\eqref{3}$$ into $$\eqref{1}$$ and solve for $$W$$: $$W\le \frac{T_h-T_c-QR}{T_h-QR}Q \tag{4}\label{4}$$ where$$R_c+R_h=R.$$ The RHS of $$\eqref{4}$$ has a maximum when $$Q_{opt}=\frac{\big(T_h-\sqrt{T_h T_c}\big)}{R}\\ =\frac{T_h-T_c}{R} \frac{\sqrt{T_h}} {\sqrt{T_h}+\sqrt{T_c}} \tag{5}\label{5}$$ Therefore the maximum work that can be extracted with an ideal reversible Carnot cycle is

$$W_{opt}=\frac{T_h-T_c}{R} \frac{\sqrt{T_h}-\sqrt{T_c}} {\sqrt{T_h}+\sqrt{T_c}} \tag{6}\label{6}$$

The ratio of the two is the efficiency $$\eta_{opt}= \frac{W_{opt}}{Q_{opt}}=1-\sqrt{\frac{T_c}{T_h}} \tag{7}\label{7}$$

Note that as long as the engine performs an ideal Carnot cycle $$\eqref{7}$$ holds for the ratio of the delivered power to the absorbed heating rate, as well

$$\dot{\eta}_{opt}= \frac{\dot{W}_{opt}}{\dot{Q}_{opt}}=1-\sqrt{\frac{T_c}{T_h}} \tag{8}\label{8}$$

: Penfield: "Available Power from a Nonideal Thermal Source", Journal Of Applied Physics, 1961, pp1793-1794