Question

The thermal efficiency of a general heat engine is 35 percent, and it produces 60 hp. At what rate is heat transferred to this engine, in \(\mathrm{kJ} / \mathrm{s} ?\)

Short answer

Answer: The heat transfer rate to the engine is 127.834 kJ/s.

Step-by-step solution

Convert the output from horsepower to Watts

We are given the power output of the engine in horsepower (hp). In order to work with metric units, we will convert it to Watts (W) using the conversion factor 1 hp = 745.7 W. Therefore, the power output in Watts is given by: Power (W) = 60 hp * 745.7 W/hp

Calculate the input power using the thermal efficiency

The thermal efficiency is defined as the ratio of the useful output power to the total input power supplied. In this case, we have the thermal efficiency as 35% (or 0.35 in decimal form) and the output power in Watts. We can use this information to calculate the input power. Let Q_in be the input power. 0.35 = (Output Power) / Q_in Now we can solve for Q_in: Q_in = Output Power / 0.35

Convert the input power from Watts to \(\mathrm{kJ} / \mathrm{s}\)

Since 1 W = 1 J/s, we have the input power in Joules per second. To convert this to \(\mathrm{kJ} / \mathrm{s}\), we need to divide the input power by 1000. Hence, Heat transfer rate = Input Power in \(\mathrm{kJ} / \mathrm{s}\) = Q_in / 1000 Now we can put everything together and calculate the heat transfer rate to the engine.

Calculate the heat transfer rate to the engine

First, calculate the power output in Watts: Power (W) = 60 hp * 745.7 W/hp = 44,742 W Next, calculate the input power: Q_in = Output Power / 0.35 = 44,742 W / 0.35 = 127,834 W Finally, convert the input power to \(\mathrm{kJ} / \mathrm{s}\): Heat transfer rate = Q_in / 1000 = 127,834 W / 1000 = 127.834 \(\mathrm{kJ} / \mathrm{s}\) So, the heat transfer rate to the engine is 127.834 \(\mathrm{kJ} / \mathrm{s}\).