Determine the time rate at which heat is supplied to the engine (1 Viewer)

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1. The problem statement, all variables and given/known data
a heat engine consists of an oil fired steam turbine driving an electric power generator with a power output of 120 megawatts. the efficiency of the heat engine is 40 percent. How do determine the time rate at which heat is supplied to the engine?


2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
 

dynamicsolo

Homework Helper
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To get help at the Homework Forum, you have to show your attempt to solve the problem first. So the first thing to start with would be:

We know the efficiency of the engine is 40% = 0.4. What is the definition of efficiency for a heat engine?
 
my teacher didnt teach us how to do this so im completly lost
 

dynamicsolo

Homework Helper
1,649
4
Have you been shown yet the equation

efficiency = (net work or energy output) / ( heat or energy input) ?

What would be the input and the output in your problem statement?
 
The efficiency is 40%, the output is 120 megawatts. Therefore, to fine the (heat or energy input) i can just divide (Efficiency)/(network or energy output) to find the anwswer? Also, do i have to change 120 megawatts or can i just leave it?
 

dynamicsolo

Homework Helper
1,649
4
The efficiency is 40%, the output is 120 megawatts. Therefore, to fine the (heat or energy input) i can just divide (Efficiency)/(network or energy output) to find the anwswer?
Check your algebra: if you're solving for the input (heat supply to the engine), what should your quotient be?

Also, do i have to change 120 megawatts or can i just leave it?
You can just leave it. The same formula also work for power, since it is the rate at which work or energy is transferred:

efficiency = (net power output) / (power input)

Don't forget that, while efficiency is often expressed as a percentage, it must be used in the equations as a fraction. So the efficiency here is 0.4 .
 

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