Calculate the rate at which heat must be rejected?

[jongood]

Homework Statement

An electric power plant produces 150 MW of power. If the coal releases 1350 × 106 kJ/h of
energy, determine the rate at which heat must be rejected from the plant.

Homework Equations

The Attempt at a Solution

So I am confused by what they are asking. So the power plant produces 150 MW of power, that is energy out. And the coal releases 1350 × 106 kJ/h so that is energy in. Am I right so far?

Now they are asking how fast the coal must release the energy in order to produce 150 MW of power?

( I didn't put any equations down because I can not find any equations in any of my notes or online. Also this class has no official textbook so I don't even know where to begin... thanks)

 

[cepheid]

You should be saying "power out" and "power in", but aside from that, you are correct.

Power plants are never 100% efficient, so I think you'll find that power out < power in. The shortfall is the energy per second that cannot be used to do useful work and is instead emitted as waste heat.

 

[jongood]

Thanks.

Is there an equation I should be using?

I used 1350*10^6 kJ/h * rate=150 MW

rate=150 MW/1350*10^6 kJ/h

and since 1350*10^6 kJ/h= 0.00375 MW


rate=150 MW/0.00375 MW
=40,000

Is this right? Seems too simple. Plus what would be the rate? 40,000 MW per hr?

 

[cepheid]

I meant that the *difference* between the input and output powers would be the power in waste heat (after all, that energy has to go somewhere).

First, convert both of the given quantities to the same units so that you can directly compare them.

 

[jongood]

power in= 1350,000,000 kJ/h
power out= 540,000,000 kJ/h

so the difference is 810,000,000 kJ/h.

I guess what I am confused about is the answer they are asking for .They are asking for the rate in which heat must be rejected? What is that exactly?

 

[cepheid]

Remember that power has dimensions of energy/time. So power IS a rate. In this case it is the rate at which energy is produced by the plant in joules/second aka watts, (or in kilojoules/hour if you prefer).

Take the difference between the input and output powers, and the result is the RATE at which the plant wastes energy in the form of heat (in J/s or kJ/h).