Question

A heat engine has thermal efficiency of 38%. It rejects heat to a cooling body at a rate of 550 kW. What is the rate at which heat is transferred into the engine (kW)?

Answer 1

By energy conservation;

$Q_S=W+Q_R$

$\Rightarrow W = Q_S-Q_R$

Now, thermall efficiency is given as;

$\eta = \frac{W}{Q_S}$

$\Rightarrow \eta = \frac{Q_S-Q_R}{Q_S} = 1-\frac{Q_R}{Q_S}$

$\Rightarrow Q_S = \frac{Q_R}{1-\eta}$

$\Rightarrow Q_S = \frac{550}{1-0.38}$

$\Rightarrow Q_S = 887.1\;\;kW\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...(Answer)$