Question

If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phase

Answer 1

Answer:

The heat input from the combustion phase is 2000 watts.

Explanation:

The energy efficiency of the heat engine ($\eta$), no unit, is defined by this formula:

$\eta = \frac{\dot W}{\dot Q}$ (1)

Where:

$\dot Q$ - Heat input, in watts.

$\dot W$ - Power output, in watts.

If we know that $\dot W = 600\,W$ and $\eta = 0.3$, then the heat input from the combustion phase is:

$\eta = \frac{\dot W}{\dot Q}$

$\dot Q = \frac{\dot W}{\eta}$

$\dot Q = \frac{600\,W}{0.3}$

$\dot Q = 2000\,W$

The heat input from the combustion phase is 2000 watts.