Quote from: dogie on October 16, 2014, 09:24:21 PM

Quote from: MrTeal on October 16, 2014, 08:57:06 PM

Depends on the restriction of the blocks, but I'm guessing they aren't terribly restrictive. Flow rates wouldn't have to be excessively high to get reasonable performance in a BTC mining application. The specific heat capacity of water is ~4200J/(l*°K), or 4200(W*s)/(l*°K). To find out what flow rate you need to move a certain amount of heat with a specific temperature rise, it's just P/(C*ΔT)
So, if we want the outlet temp of the 4th unit (and inlet of the 5th) to be 30°K above the inlet of the first unit, with the first 4 units dumping 4kW of heat into the water (which is way more than you'd actually see) you'd get the following numbers.

Flow = 4000W / (4200(Ws)/(l°K) * 30°K)
Flow = 0.03175l/s
Flow = 1.9l/min

Not sure how many of these dogie has, but it'd be a cool test to do anyway. You wouldn't even need to use that rad, you could just run three of the 360mm ones he sells in series, and maybe a couple of the pumps to generate sufficient head.

*Edit: That's a minimum flow rate to keep the inlet water temperature of the last unit at a reasonable (say 65C with 35C inlet water) temperature. Depending on the block design you might need a higher rate to actually get 1kW per device from the block into the water.

Lets try some maths:

flowrate = Q / ((heat capacity)* density *(Tout - Tin))

Q = 5KW
Heat capacity of water = 4.18 kJ/kg/K
Density = ~1000kg/m^3
Tout max = 65C to keep chips < 75C [pretty optimistic heat exchanger]
Tin = 30C [depends on ambient, I wouldn't be getting mine below 40C]
flowrate = ? m^3/s

flowrate = 5 / (4.18 * 1000 * 35)
flowrate = 0.0000342 m/3^s = 0.0341 L/s

With a 6mm internal diameter [forced to this size by the heatsink channels]...

A= 0.0047 m^2
flowrate = A*v
v = velocity m/s
0.0000342 = 0.0047 * v
Therefore v would be 0.0084 m^s.

Check your math, you're comparing W and kJ directly.

Edit: I see you realized your mistake and corrected it.