A small diesel engine uses a volume of 1.5 × 104 cm3 of fuel per hour to produce a useful power output of 40 kW.

Question 22

A small diesel engine uses a volume of 1.5 × 10⁴ cm³ of fuel per hour to produce a useful power output of 40 kW. It may be assumed that 34 kJ of energy is transferred to the engine when it uses 1.0 cm³ of fuel.

What is the rate of transfer from the engine of energy that is wasted?

A 102 kW                    B 142 kW                    C 182 kW                    D 470 kW

Reference: Past Exam Paper – March 2018 Paper 12 Q17

Solution:

Answer: A.

The ‘rate of transfer of energy’ is the power.

We want to find the power dissipated (wasted).

The engine uses a volume of 1.5×10⁴ cm³ of fuel per hour. This quantity is the volume per unit time (hour).

Volume / time = 1.5×10⁴ cm³ h^-1

From the question, when 1.0 cm³ of fuel is used, 34 kJ of energy is transferred.

Total energy transferred = 1.5×10⁴ × 34 = 510 000 kJ

But since ‘1.5×104 cm³’ is the volume per hour, ‘510 000 kJ’ is the energy transferred per hour.

But 1 hour = 3600 s

3600 s - - > 510 000 kJ transferred

1 s - - > 510 000 / 3600 = 141.67 kJ = 142 kJ

This is the total energy transferred per second, which is the power.

Total power transferred = 142 kJ

Useful power output = 40 kJ

Total power = Useful power + Power wasted

Power wasted = 142 – 40 = 102 kJ