Question

A heat engine uses fuel of energy content 43.1 MJ/kg and produces 17.4 kW of useful power. The heat rejection rate (through the exhaust and cooling systems) is 44.8 kW. Determine: (a) The efficiency of the engine (b) The usage rate of fuel in kg/h.

Answer 1

a)27.9%

b)
`\dot{m}=3.06 (Kg)/(h)`

Step-by-step explanation:

Given that

Fuel energy content = 73.1 MJ/kg

Useful power = 17.4 KW

Heat rejection rate = 44.8 KW

From first law of thermodynamics

Heat addition rate =Heat rejection rate + Power out put

Now by putting the values in the above formula

Heat addition rate = 44.8 + 17.4

Heat addition rate =62.2 KW

We know that efficiency is given as follows

`\eta =(power\ out\ put)/(Heat\ addition\ rate)`

So

`\eta =(17.4)/(62.2)`

`\eta =0.279`

So the efficiency is 27.9%.

Now to find usage rate of fuel

Lets take usage rate is
`\dot{m}`

Fuel energy content x usage rate of fuel = Heat addition rate

Now by putting the values

`73100* \frac{\dot{m}}{3600}=62.2`

`\dot{m}=3.06 (Kg)/(h)`