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Homework Statement
An 800-MW electric power plant has an efficiency of 30%. It loses its waste heat in large cooling towers by evaporating a fine mist of water. If all waste heat went only into the evaporation of water, how many tons of water would be evaporated per day?
Homework Equations
W = |Qh|-|Qc|
e = 1 - (|Qc|/|Qh|)
Q = mL
Lwater = 2.26x10^6 J/kg
The Attempt at a Solution
I can't seem to get any of the answers listed and I can't quite figure out what to do. Here's what I tried.
W = 800MW
e = .30
Solving for |Qc| and |Qh| I got, |Qc| = 1866.69 MW, or 1866.69x10^6 J/s, |Qh| = 2666.67 MW
Since the wasted heat going to the cold reservoir is what is evaporating the water I plugged |Qc|= 1866.69x10^6 J/s into the Q for the equation Q = mL, so |Qc| = mL. L is the latent heat of water. Solving for m I got m = 825.97 kg/s. Since it's per day, I multiply m by 86400seconds and get 71,363.72 tons of water per day.
Would anybody like to take a crack at it and help me out? Thanks :)