Question

A heat engine operates between a source at 477°C and a sink at 27°C. If heat is supplied to the heat engine at a steady rate of 65,000 kJ/min, determine the maximum power output of this heat engine.

Answer 1

Answer:

$T_C = 27+273.15 = 300.15 K$

$T_H = 477+273.15 = 750.15 K$

And replacing in the Carnot efficiency we got:

$e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%$

$W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}$

Explanation:

For this case we can use the fact that the maximum thermal efficiency for a heat engine between two temperatures are given by the Carnot efficiency:

$e = 1 -frac{T_C}{T_H}$

We have on this case after convert the temperatures in kelvin this:

$T_C = 27+273.15 = 300.15 K$

$T_H = 477+273.15 = 750.15 K$

And replacing in the Carnot efficiency we got:

$e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%$

And the maximum power output on this case would be defined as:

$W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}$

Where $Q_H$ represent the heat associated to the deposit with higher temperature.