The Efficiencies of Real Engines — Question 7

Question 7

Maximum power occurs below the reversible reservoir efficiency in a finite-contact model

A supplied engine model is internally reversible between adjustable hot working temperature $\theta$ and a 300 K cold reservoir. Heat arrives from a 600 K source through conductance $G=1.00$ W/K: $\dot Q_h=G(600-\theta)$. The cold contact is ideal. Thus $P=G(600-\theta)(1-300/\theta)$ for $300<\theta<600$ K.

1. Differentiate $P$ to find the working temperature and efficiency at maximum power.

2. Calculate maximum power and compare with the limit $\theta\to600$ K. Explain the tradeoff between efficiency and heat-transfer rate.

Question 7 - Solution

Maximum power occurs below the reversible reservoir efficiency in a finite-contact model

(a) Method and calculation

$$P/G=900-\theta-180000/\theta,\quad\frac{dP}{d\theta}=G[-1+180000/\theta^2]=0.$$$$\theta_*=\sqrt{180000}\approx424.264\,\mathrm K,\quad\frac{d^2P}{d\theta^2}=-\frac{360000G}{\theta^3}<0.$$

(b) Interpretation and further calculation

$$\eta_*=1-300/\theta_*=1-1/\sqrt2\approx.292893,\quad P_{\max}\approx51.4719\,\mathrm W.$$$$\theta\to600:\quad\eta\to.500,\quad\dot Q_h\to0,\quad P\to0.$$Maximum power and maximum limiting efficiency are different objectives within this stipulated contact model.

Physical check

This optimum is model-specific; it is not a universal efficiency formula for all real engines.