Thermodynamic issue w/ power plant
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Thermodynamic issue w/ power plant

[From: ] [author: ] [Date: 11-05-19] [Hit: ]
Help me out cuz Im stuck on this issue.-First find the amount of the dumped heat.If the efficiency is 25 % then 75% is dumped.so 3.The specific heat of water is approx 4.2 J /deg/cm^3.......
A power plant produces electrical energy at the rate of 1300 MW with an efficiency of 0.25. The excess heat is dumped into a river that carries 1500 cubed meters per second of water...how much does the river's temperature increase? Help me out cuz I'm stuck on this issue.

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First find the amount of the dumped heat. If the efficiency is 25 % then 75% is dumped.
This is three times the amount of the delivered power so the plant must dump 3900Mw
so 3.9 * 10^9 J/s

The specific heat of water is approx 4.2 J /deg/cm^3.
= 4.2 * 10^6 J/deg/m^3

So the temperature rise is the amount of heat divided by the amount of water divided by the specific heat.
=3.9 * 10 ^9 /(1500 * 4.2 * 10^6)
=0.62 deg.
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