Question

Assuming an engine operates as a reversible heat engine calculate the outlet temperature (temperature of rejected heat) that achieves a thermal efficiency of 63% if the engine takes in heat at 640°C .

Answer 1

Answer:

$64.8^{\circ}C$

Explanation:

The efficiency of a reversible heat engine is given by:

$\eta = 1 -\frac{T_C}{T_H}$

where

Tc is the cold temperature (outlet temperature)

Th is the hot temperature (temperature of the source)

In this problem, we know:

$\eta = 63\% = 0.63$

$T_H = 640^{\circ}C=913 K$

So, we can calculate the outlet temperature by re-arranging the formula:

$T_C = T_H ( 1 -\eta)=(913 K)(1- 0.63)=337.8 K=64.8^{\circ}C$