Calculating Power and Efficiency of a Heat Engine with 6 kg/h Gasoline Input

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The discussion focuses on calculating the power output and efficiency of a heat engine using a gasoline input of 6 kg/h with a calorific value of 42000 kJ/kg. The initial calculations incorrectly assumed that multiplying the fuel rate by the calorific value directly yields the power output, neglecting the need to account for heat rejected (Ql). The correct approach involves determining the net power output by subtracting the heat rejected from the total heat input. Additionally, the efficiency formula was misquoted; the correct efficiency equation is 1 - (Ql/Qh), not the version provided. Accurate calculations are essential for determining the engine's performance metrics.
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A heat Engine receives 6 kg/h gasoline which has a calorific value of 42000 kj/kg. If heat rejected from the engine is 40 kw, calculate
a- the power done by the engine
b- the engine effiecincy

-this wt i got today in my exam and unfortuntely i didnt study it well but i did the following :
1- 6 kg/h -----> 1/600 kg/s
then i multiplied it by the 42000 kj/kg
so i can get kj/s ( which is the power )
so is that correct ??

2- i said effieciency = w/Qh = 1 - Qh/Ql

and Qh = W - Ql = 42000 - 40

so is that too correct ?

* i knw i didnt study well :cry: and sorry if i posted it in a wrong section
 
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Efficiency is 1 - Ql/Qh
 
LawrenceC said:
Efficiency is 1 - Ql/Qh

thanks for quick reply
sure i knw that
but my question is : wt i wrote is right or wrong ?? bcse i got diff units and no more details
 
The output of the engine is Qh-Ql. Multiplying fuel rate of use by caloric content only provides Qh. You must subtract Ql from it to arrive at power output. Ql is heat rejected and it is lost.

You have the equation for efficiency wrong. You wrote
"2- i said effieciency = w/Qh = 1 - Qh/Ql "
 

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