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Power plant

by Winzer
Tags: plant, power
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Winzer
#1
Jan17-08, 07:20 PM
P: 605
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?
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mgb_phys
#2
Jan17-08, 07:39 PM
Sci Advisor
HW Helper
P: 8,953
You need to divide by 0.42 not multiply.
You want to know how many MJ of coal your are paying for, not what you get out of it!


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