I need some math help

[beaverpants]

Ok id like to know how much heat my computer will put out per hour.. and how much that will warm up. Lets say 50 gallons of coolant.
And then how much will 4 computers heat up 50 gallons of coolant per hour? This is to figure out how much I need to cool the coolant.

The coolant will start off being 10c

planing water block on the CPU GPU on all 4 computers. and might WC the PSU HD RAM.

i have no idea how to do this math.

thanks guys.

[Brad]

by computer, do you mean athlon, P4, what speed, voltage, etc

we need lots more info

[beaverpants]

well im just wondering. b/c im still working out what im going to do. and what im going to cool.

i wanna cool 4 computers

1 MSI nForce 420 Athlon XP 1700+ at stock
would like to cool everything in this rig PSU RAM HD Chipset

2 Dual P3 650 stock speed.
at lest cool the CPU's

3 epox with AthlonXP 1600+ stock speed

4 athlon 1.2 ghz at stock

im still in the pre planing stage of my idea.

the plan is to cool all the computers from one pump in the basement. and cool the coolant with the cold (10c) water from my well..

the queson is how hot will the water get. as in how much cooling from the well water will be needed to keep it cold.

iv been looking but i cant find the math forumals needed.

if 50 gallons is not enuf for 4 computer i might go for 100 gallons of watter / coolent.

[bigben2k]

The cooling capacity is not a function of the volume of coolant, but of the heat dissipation of the rad.

If you want to rely on the volume of coolant, then you need to be aware that the coolant temperature will steadily increase, until it reaches a balance point, where the rad can dissipate all the heat. (because the rad is more efficient at a higher temperature).

If you want to run the numbers, then you're talking about Watts (and watts per hour) and BTUs, and the heat capacity of water (or coolant).

[beaverpants]

ok thanks.

my plan was not to use a rad.. to use a heat extanger from the cold house well water to the coolent. so i need to figure out how effective i need to make this.

[redleader]

Thats about 250w average, 300 max give or take.

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.

[myv65]

redleader is on the right path, but I don't think anyone here has yet interpreted your question quite right. Then again, maybe I've got it all wrong. LOL

As I understand it, you've got a well with practically unlimited water volume available. You wish to know how much of this water you would require in order to cool all your systems. That part is simple enough, but here's where the disconnect has occurred.

The amount of "system" volume you have is irrelevant. Without a place for the heat to go, no amount of water can absorb an indefinite input of heat. You state you do not wish to use a radiator. I take this to mean you intend to "get rid of" some of the circulating water and replace is with a quantity of "well water". Going further, I believe your real question is how much well water will I need to draw in a given hour/day/pick-your-time-units.

If my take is correct, then you plan to have some fixed volume that you'll circulate through the systems. To this fixed volume, you'll add fresh well water at either a constant, flow-controlled rate, or via discrete dumps representing a significant percentage of the total system volume.

What you'll probably find is there's more than one way to accomplish your task. In the "add a little well water continuously" case, you merely need a reservoir large enough that you can introduce the well water near the bottom (near the pump suction, too) along with an overflow weir that allows the least dense water (warmest) to gently spill over to a drain. The reservoir really doesn't need to be too big so long as the well water gets introduced to the pump suction. You will get best results if the reservoir is large enough to let the water essentially still itself whereby the warmer stuff will rise to the top. This would be pretty big (think hundreds of gallons).

In the "batch replace a large percentage of the fluid" method, you may either go with a really large volume or relatively small one. With a really large one, you may find the fluid reaches a steady-state temperature where the reservoir can dissipate the heat into the air. Just how big this would be depends on your total power input, room temp, reservoir construction, and tolerance for high chip temps. With a relatively small one, redleader's calcs show you how to find the temperature rise.

There is yet another option that would be the most space-efficient. Get yourself an oil-oil heat exchanger. You could then create a sealed circulation loop for the computers (using distilled water with a corrosion prevention additive) with the other side of the exchanger taking a little well water and dumping it to a drain. Again referencing redleader's numbers (which are correct for "typical" water data at a reasonable temperature), then you could use as little as 50 gph of well water with a temperature rise of less than 1°C to keep the sealed coolant loop cool. An exchanger of this sort tends to be pretty efficient, especially when compared to an air-fluid exchanger. You could probably get by with as little as 5 gph of well water while still keeping your sealed loop under 20°C. (Reasoning: 50 gallons would heat less than 1°C/hr, so 5 gallons/hour going through a rise of 10°C represents more energy transfer.)

Think hard about this last option. It offers you the smallest total system while allowing you to easily keep things nice and cool.

[SCompRacer]

There was a guy in the UK that buried a cylinder a few feet below the ground and recirculated his coolant. It was just one PC though. I'll have to find that link.

[beaverpants]

its the guy at http://www.zerofanzone.co.uk/ that did the tank

my idea right now is it conect a 50 gallon talk with cold water in to it. but it only flow's out when the house uses water. then make a heat excanger with lots of coppy pipe or put a rad underwater. and run the coolent for the computers threw the rad/ pipe. to let its heat off in to the cold water comeing from the well.

so lets say its 30 min betwhen when the water flows. so thare will be no added cold for that time... then what about at night? would 50 gallons of cold water and 50 gallons of coolent. not over heat over night? like have enuf heat mass to absorb the heat and not get to hot in lets say 8 houres. its my big consern.

so the more coolent you have the longer you can go befor it to hot.

thanks for the info guys

[Brad]

250w would be a little low for all that I think, but anyway, you could just run a big passive truck rad down in the basement to cool all the water, it'd take an age to warm up then

[Dix Dogfight]

redleader:
Your calculations are correct (although my teachers at school would cry if they saw your abuse of the units ;-) )

beaverpants:
Using the numbers redleader provides tell you that the temp of the water in the tank will increase with around one degree C per hour. 8h gives 8 degrees.

What you really need to think about is how much water the household uses per day. When you say 30min between "the water flows" what do you meen?
Do you change all, 50 gal, of water every 30 min or do you flush the toilet every 30 min which introduces only a little new water to the system.