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myv65 · 06-01-2002, 12:32 AM

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.

06-01-2002, 12:32 AM	#7
myv65 Cooling Savant Join Date: May 2002 Location: home Posts: 365	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.