How to Calculate Heat Input for a 30% Efficient Heat Engine

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Homework Help Overview

The discussion revolves around calculating the rate of heat input for a heat engine with an efficiency of 30% and a power output of 600W. Participants are exploring the relationship between efficiency, work output, and heat input.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to apply the efficiency formula, questioning the correctness of their calculations and the meanings of the symbols used. There is confusion regarding the interpretation of efficiency and its application in the context of the problem.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's reasoning and calculations. Some guidance has been offered regarding the correct interpretation of the efficiency formula, but there is no explicit consensus on the correct approach yet.

Contextual Notes

There is a noted lack of clarity regarding the definitions of symbols used in the equations, and some participants express uncertainty about the correct formula to use for calculating heat input.

Dx
A heat engine has efficiency of 30% and its power is 600W. what is the rate of heat input?

is 1.8Kw right?

e=W/q_h
.30 * 600

TY
dx:wink:
 
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No, that would correspond to an efficiency of 0.33...
 
Originally posted by Tom
No, that would correspond to an efficiency of 0.33...

Is my equation wrong, what am i doing incorrectly tom?
Dx
 
Originally posted by Dx
Is my equation wrong, what am i doing incorrectly tom?
Dx

I can't tell, since I don't know the meanings of the symobls you used. Efficiency is power output divided by power input.
 
Originally posted by Dx
A heat engine has efficiency of 30% and its power is 600W. what is the rate of heat input?

is 1.8Kw right?

e=W/q_h
.30 * 600

yourt the man Tom! Thats what my symbols mean. e= efficiency, W=work ouot/Q_h=work in. so why did i get this wrong, tom.
Im lost in the math and don't know where? please help?
Dx
 
You might be better off to take Tom's answers with a grain of salt and check them carefully. He seems to be doing the same thing you are: taking the numbers given and putting them together in the simplest way without regard for the correct formula.


In this case, the formula you give is: e=W/q_h and the values are efficiency= e= 0.30, Work out= W= 600.

In other words, 0.30= 600/q_h. Solving that, 0.30*q_h= 600 so
q_h= 600/0.30= 2000 Joules.

It should occur to you that, since efficiency is always less than 1, there must be more energy put in that you get out.

.30 * 600
 
Originally posted by HallsofIvy
You might be better off to take Tom's answers with a grain of salt and check them carefully. He seems to be doing the same thing you are: taking the numbers given and putting them together in the simplest way without regard for the correct formula.

No, Dx was quoting himself (not me) in his post above. The only "answer" I gave was...

"I can't tell, since I don't know the meanings of the symobls you used. Efficiency is power output divided by power input."

...which is no different from what you posted.
 
Originally posted by HallsofIvy
You might be better off to take Tom's answers with a grain of salt and check them carefully.
Speaking of grains of salt,
600 watts/30% = 2,000 watts, not joules. :smile:
 
Dx said:
A heat engine has efficiency of 30% and its power is 600W. what is the rate of heat input?

is 1.8Kw right?

e=W/q_h
.30 * 600

TY
dx:wink:
can anybody please tell me the formula for heat rate for fuels?
 

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