(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The Comanche 3 power plant currently being constructed in Pueblo, CO, by Xcel Energy will

produce 750 MW of power. This plant uses supercritical boiler technology which will boost the

thermal efficiency to 42%. The plant uses Powder River Basin coal from Gillette, WY, that has a

heating value of 8,500 BTU/lbm. Find the amount of coal burned in this plant each year.

2. Relevant equations

[tex] n=\frac{what you get}{what you paid for}[/tex]

3. The attempt at a solution

So I made the conversion from 8500 BTU/lbm to 19.7710 MJ/kg (megajouls)

I did 750E6 times .42 which is 315E6 MW.

I did [tex] \frac{315E6 MW}{19.7710 MJ/kg}[/tex] which is 15.932 kg/s

I times that by 3600*24*365 to get g in a year and get 5.024E8 kg.

The answer is 2.86x109 kg, what did I do wrong?

**Physics Forums - The Fusion of Science and Community**

# Power plant

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Power plant

Loading...

**Physics Forums - The Fusion of Science and Community**