A lawnmower tractor engine produces 18hp using 40kW of heat transfer from burning fuel. Find the thermal efficiency and the rate of heat transfer rejected to the ambient.

I converted the power to 13.24kW then found the efficiency by dividing 13.24kW by 40kW but the answer is incorrect?

Yes I am right there with you. I got 0.33 which is 33%. I think the answer in the textbook is wrong, check with your professor because that is what I am going to do.

To find the thermal efficiency and the rate of heat transfer rejected to the ambient, we need to use the given information correctly.

Firstly, let's correct the power calculation. The engine produces 18hp, which we can convert to kilowatts. To convert horsepower (hp) to kilowatts (kW), we need to use the conversion factor of 1 hp = 0.7457 kW. Therefore, 18 hp is equal to:

18 hp * 0.7457 kW/hp = 13.4234 kW (rounding to four decimal places)

Now, let's calculate the thermal efficiency. The thermal efficiency (η) is given by the formula:

η = (Useful output power / Heat input power) * 100%

In this case, the useful output power is 13.4234 kW (the power produced by the engine), and the heat input power is 40 kW. So:

η = (13.4234 kW / 40 kW) * 100% = 33.5585%

Therefore, the thermal efficiency is approximately 33.56% (rounding to two decimal places).

Now, let's calculate the rate of heat transfer rejected to the ambient. The rate of heat transfer rejected to the ambient is equal to the difference between the heat input power and the useful output power. So:

Heat transfer rejected = Heat input power - Useful output power

Heat transfer rejected = 40 kW - 13.4234 kW = 26.5766 kW (rounding to four decimal places)

Therefore, the rate of heat transfer rejected to the ambient is approximately 26.58 kW (rounding to two decimal places).

I hope this answers your question and guides you on how to solve similar problems in the future!