Calculating Power Output and Heat Dissipation in a Coal-Powered Plant

[chrisama]

Homework Statement

A power plant consumes 20,000 tons of coal per day; the average heat value of the coal used is 26,000 Btu/kg and the overall efficiency is 28%.

What s the average electrical power output of the plant in kilowatts?

How much heat is dissipated per day, in Btus?

Homework Equations

I think that this equation may be helpful, but I'm not sure---P=E/t

The Attempt at a Solution

I realized that I must get the units into pounds, so I converted the tons into pounds and got 400,000. I also thought of dividing by 24 since 24 hours is the time frame, and power is granted by energy over time. However, I think that my attempt at this problem is totally wrong. I'd greatly appreciate your help.

 

[BishopUser]

Your relevant equation is very important. Power is energy per time. You are trying to find the power output (energy per time). The given info is 20,000 tons/day valued at 26,000 btu/kg. Looking at the given info i see both energy and time. It wants your answer in kilowatts (kJ/s). Try putting the given info together somehow converting where necessary to achieve the desired units. Also remember to apply the efficiency.

Your attempt is on the right track, but you got to take it further. Remember to write out the units along the way and hopefully you will see how to get from the given info to the output.

 

[chrisama]

Okay, I will now work it out on my own and display what I get soon.

 

[chrisama]

So, I worked out the equation and got 3,367,003.367 Kj/s for the first part.

This is how I did it:

I first got the time, which is 86,400 seconds.

Then I multiplied the 26,000btu/kg by 1.055kj/btu, which gave me 27,430 kj/kg.

Next, I converted the 20,000 tons into 40,000,000lbs. Then, I converted that into kg by dividing by 2.2. Once, I did that, I received 18,181,818 kg.

Then, I multiplied the 18,181,818kg by the 26,000 kJ/kg in order to get a unit of 2.9091E^11 KJ. I divided this number by 86,400 and received 3,367,003.367 kj/s.

I hope this is correct. I'm not trying to figure out the second part of the equation.

 

[chrisama]

I got it. thanx.