The Nuclear Power Thread

[jrmichler]

I agree with the post above, but get 70,000 GPM.

100 MW = 100E6 watts X 3.4 BTU/hr/W = 3.4E8 BTUH.
3.4E8 BTUH / 10 deg F X 1.0 lb deg F / BTU = 3.4E7 lbs/hr of water.
3.4E7 lbs/hr / 8.34 lbs/gallon / 60 min/hr = 70,000 GPM.

If you get rid of the heat by evaporating water in a cooling tower or spray pond, that much heat would evaporate 700 GPM of water. There would be a need to use more water than that in order to control solids buildup.

 

[gmax137]

I agree with the post above, but get 70,000 GPM.

100 MW = 100E6 watts X 3.4 BTU/hr/W = 3.4E8 BTUH.
3.4E8 BTUH / 10 deg F X 1.0 lb deg F / BTU = 3.4E7 lbs/hr of water.
3.4E7 lbs/hr / 8.34 lbs/gallon / 60 min/hr = 70,000 GPM.

If you get rid of the heat by evaporating water in a cooling tower or spray pond, that much heat would evaporate 700 GPM of water. There would be a need to use more water than that in order to control solids buildup.

Thanks for the check, @jrmichler . You're quite right, I'm not sure what I fat fingered to get 90K.

 

[berkeman]

If you get rid of the heat by evaporating water in a cooling tower or spray pond, that much heat would evaporate 700 GPM of water.

Sounds like a good strategy, but I wonder if they will spray the water fountains upwind of the barracks (heating up the troops) or upwind of the neighboring town (heating up the neighbors)...

 

[jrmichler]

Sounds like a good strategy, but I wonder if they will spray the water fountains upwind of the barracks (heating up the troops) or upwind of the neighboring town (heating up the neighbors)

To help put the evaporation into perspective, I live on an 840 acre lake. Normal summer evaporation from the lake in summer is about 2300 GPM. That's average over 24 hours, peak evaporation during the day is much higher.