My tec liquid chiller project

[Uncle Jimbo]

He's using 2 server PSUs, sounds like he knows what he's doing in that area.



Did you notice he said he is using 2 PA120.3 radiators? That brings him down to around 0.045-0.050 C/W for that part of your estimation surely?

I did the calculation for each 'side' and then combined to get total cooling, my assumption is that there is one PA120 on each side. His total sandwich has a number of interfaces so this is all guesswork until we see some real numbers.

With the big surface area he has, Rt between the cold chamber and the TEC cold side should be insignificant. TEC hot side to the hot chamber is critical but again, with his care about pressure, I assume that is not significant. So we basically have Rt for the hot chamber plus Rt for the PA120 on the water loop. We can say pretty accurately that with big fans, the PA120 will deliver .02 water to ambient, so the big question is the hot side chamber.

If that was a pin grid block, I would expect something under .015, but with the home-made channel, that's probably optimistic. Only some testing will show what he actually has there. My guess of .035 is based on surface area transfer at 2GPM of .4C/W per inch, but it could be half of that and it would not surprise me. So my number is probably worst case, and his total per side WC Rt could be .03 or .025.

The air cooling is a SWAG too - but those look like standard sinks with 80mm fans, which are usually around .3C/W. Rt per side is a straight up parallel so Rt(tot) is the Rt per sink / 4 or .075.

Calculating unequal parallel equivalent resistance is 1/Req = 1/R1 + 1/R2 + ...
so in this case, for my SWAG, 1/Rtot = 1/.055 + 1/.075. For LCD, we can use 1/Rtot = 1/(11*.005) + 1/(15*.005) or .005/Rtot=1/11+1/15 => 15/165 + 11/165 => 26/165 = .1576. Solving for Rtot yields .031 so I SWAGed .035 to account for everything I don't know.

Total Rt to ambient is half of that, or .0175