Problem — Why engines run at high temperature

Problem

Industrial turbines and jet engines push to extreme temperatures ($1500\,^\circ\text{C}$ and beyond) despite the strain on materials. Why is it so worthwhile to raise $T_c$ rather than lower $T_f$?

Solution

The two levers. The maximum efficiency is

$\eta = 1 - \frac{T_f}{T_c}$

It can be increased either by raising $T_c$ or by lowering $T_f$. But the two levers aren’t symmetric in practice.

$T_f$ is nearly locked. The cold reservoir is the environment: air or cooling water, i.e. $T_f \approx 300\;\text{K}$. Lowering it would require another refrigerator, which consumes work — wiping out the gain. In most cases $T_f$ is constrained by the outside world and can’t be reduced for free.

$T_c$ is the real lever. That leaves $T_c$. And the gain is substantial because $\eta$ depends on the ratio $T_f/T_c$: starting from $T_f \approx 300\;\text{K}$,

$\eta(T_c = 800\;\text{K}) = 1 - \tfrac{300}{800} = 0{,}63 \qquad \eta(T_c = 1800\;\text{K}) = 1 - \tfrac{300}{1800} = 0{,}83$

Every extra kelvin on $T_c$ pushes the ceiling higher without additional energy cost (only material costs).

Conclusion.

$\ev{T_f \approx 300\;\text{K}\ \text{constrained} \Rightarrow \text{raising } T_c\ \text{is the only effective lever}}$

This is why we invest in alloys and blade cooling to withstand ever-higher temperatures: that’s where efficiency is bought, not on the cold side.

Collegamenti

Argomenti: Macchine termiche Concetti: Rendimento · Ciclo di Carnot Competenze: Impostazione simbolica Oggetti: Macchina termica