Thats about 250w average, 300 max give or take.
Quote:
|
Lets say 50 gallons of coolant.
|
OK this is straight forward, but I don't see how the numbers are useful. :shrug: Eventually the water will still heat up enough that it crashes the comps if it isn't being cooled somehow. And if it is being cooled somehow, I really doubt you'll be able to accurately calculate the rate of heatloss.
Anyway, lets use metric because I can't remember how to do this in Imperial.
1gal = 3785.412 cm^3 H20
Specific heat h20 = 4.2 J/g
1g = 1cm^3
So 50*3785.4*4.2= 794,934J
Thats how many watts you'll need to raise the temp one degree. Then convert to hours assuming 250w load:
794,934/(3600)(250)= .88hrs/degree C (or 53min)
(Someone feel free to correct me if I forgot something)
Around room temp, this will work very well, but as the delta T between the air (or soil) and coolant widens, these numbers will deviate from reality. What really matters is not the water volume, but how well heat is conducted away from that water.