Entropy in the Real World: Engines — Question 10

Question 10

A required engine output sets the share of heat needed from a hotter source

A cyclic engine receives total heat 1000 J per cycle: fraction $x$ from 600 K and fraction $1-x$ from 400 K. It rejects heat only to 300 K. The desired output is 350 J per cycle.

1. Find the minimum $x$ compatible with reversible operation.

2. If total entropy generation is .200 J/K per cycle, find the required $x$ and rejected heat. State whether the result lies in the physical fraction range.

Question 10 - Solution

A required engine output sets the share of heat needed from a hotter source

(a) Method and calculation

$$Q_{c,\min}=300\left(\frac{1000x}{600}+\frac{1000(1-x)}{400}\right)=750-250x.$$$$W_{\max}=1000-Q_{c,\min}=250+250x\ \Rightarrow\ x\ge.400.$$

(b) Interpretation and further calculation

$$Q_c=750-250x+300(.200),\quad W=190+250x.$$$$350=190+250x\ \Rightarrow\ x=.640,\quad Q_c=650\,\mathrm J.$$The fraction is admissible, $0\le x\le1$. Irreversibility requires a greater contribution from the hotter source for the same work output.

Physical check

The required high-temperature heat share is a design constraint derived jointly from energy and entropy.