- #1
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
is 1.8Kw right?
e=W/q_h
.30 * 600
TY
dx
Is my equation wrong, what am i doing incorrectly tom?Originally posted by Tom
No, that would correspond to an efficiency of 0.33...
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
Is my equation wrong, what am i doing incorrectly tom?
Dx
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.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
No, Dx was quoting himself (not me) in his post above. The only "answer" I gave was...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.
Speaking of grains of salt,Originally posted by HallsofIvy
You might be better off to take Tom's answers with a grain of salt and check them carefully.