Q#22 (Past Exam Paper – March 2018 Paper 12 Q17)
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
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
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