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

[mado134]

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 and sorry if i posted it in a wrong section

 

[LawrenceC]

Efficiency is 1 - Ql/Qh

 

[mado134]

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

 

[LawrenceC]

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 "

 

[asteriatic]

Dear student,

First of all, it is okay to make mistakes in exams. What's important is that you learn from them and try to understand the concepts better in the future. Let's go through your calculations and see if they are correct.

1) Yes, your first step is correct. You converted the input of 6 kg/h to kg/s by dividing it by 3600 seconds in an hour. This gives us a value of 0.00167 kg/s.

Next, you multiplied it by the calorific value of gasoline (42000 kj/kg) to get the power in kj/s. This gives us a value of 70.2 kj/s or 70.2 kW (since 1 kW = 1000 W).

So, the power done by the engine is 70.2 kW.

2) Your calculation for efficiency is also correct. However, there are some points to clarify.

The efficiency of a heat engine is defined as the ratio of the work done (W) to the heat input (Qh). So, your equation should be: efficiency = W/Qh.

In this case, the work done by the engine is equal to the power (since work = power x time). So, W = 70.2 kW.

The heat input (Qh) is equal to the energy released by burning 1 kg of gasoline, which is 42000 kJ.

So, the efficiency of the engine can be calculated as: efficiency = 70.2/42000 = 0.00167 or 0.167%.

I hope this helps you understand the concept better. Keep studying and practicing, and you will do better in the future. Best of luck!

Sincerely,