Case Design for Liquid Cooling

[Blauhung]

Late vote, but if it happens i will buy the bigger one that can fit the EVGA mobo


As for the whole parallel water cooling discussion.....
At the flow rates that just about everyone will be using in water cooling, you can very safely assume that the temperature of water entering the radiator and leaving the radiator are approximately equal as all of the coolant has hit a steady state.

here's the energy ballance on a series water cooling flow with 2 loads and 1 radiator

Qcpu1 + Qcpu2 = Qrad (the heat dissipated by the radiator at steady state is = to the heat generated by the 2 cpu's)
Qcpu = Tdelta * Fh2o / Rblock (since the input and output temp can be considered constant, the heat absorbed by the water in the block is going to be a function of the temp difference between the water and the CPU, the flow rate of the water, and resistance to heat transfer of the block)
Qrad = Tdelta *Fh2o / Rrad

so you end up with
((T1-Tw)*F1)/R1 + ((T2-Tw)*F2)/R2 = ((Tw-Ta)*Fr)/Rr (i substituted in the temp delta's here Tw being water temp, T1/2 being cpu temp, and Ta being air temp)

if you are in series, then the flow rates will all be equal so they all cancel out, you can also assume the temp and resistance to heat transfer on each block to be equal so you end up with
2*(Tc-Tw)/Rb = ((Tw-Ta)/Rr (simplified T1/T2 into Tc or core temp, and R1/2 into Rb or block resistance)
(Tc-Tw)/(Tw-Ta)=Rb/(2*Rr) for series

For paralell, you can assume that the flow rates at each block are 1/2 the flow at the rad
((Tc-Tw)*F/2)/Rb + ((Tc-Tw)*F/2)/Rb = ((Tw-Ta)*F)/Rr (F cancels again)
(Tc-Tw)/(Tw-Ta)=Rb/Rr for parallel

Now to solve these you need to list what you know
-lets assume that the Rb:Rr ratio is ~1 just to save on some math
-Qcpu = 100w (another easy #)

so for series here are our sets of equations
Q=(Tc-Tw)*F/R
(Tc-Tw)/(Tw-Ta)=1/2
after a bit of substitution you can get that
Tcser = QR/F +2QR/F+Ta = 3QR/F + Ta


and for parallel
Q=(Tc-Tw)*(1/2*F)/R
(Tc-Tw)/(Tw-Ta)=1
subs
Tcpar = 2QR/F + 2QR/F + Ta =4QR/F + Ta

so assuming that all the Q's R's and F's are equal between the 2

Tcseries/Tcpar = (3QR/F +Ta)/(4QR/F +Ta)

and this kinda shows that your core temp will be lower in series then in parallel.


In summary, your water temp is approximately constant, and running stuff in parallel really only results in lowering your flow rates in each block which drastically lowers your energy transfer efficiency and makes the thing you are trying to cool hotter.

[Hunter 101]

There we go

Other than that, are there any updates Mick, on the manufacturing front?

[Kaldskryke]

...
Qcpu = Tdelta * Fh2o / Rblock (since the input and output temp can be considered constant, the heat absorbed by the water in the block is going to be a function of the temp difference between the water and the CPU, the flow rate of the water, and resistance to heat transfer of the block)
...

I know I'm new here and all (note my postcount) but how do we know the heat flux is linearly proportional to flow rate?

AFAIK most correlations for heat transfer coefficients for tubes are something more like Nu = 0.023*Re^(0.8)*Pr^(1/3), such as the seider-tate correllation which certainly implies a non-linear relationship between local velocity and heat transfer

Furthermore, when you correctly assume halved volumetric flow through your parallel portions that doesn't mean the local fluid velocity within those portions will be halved, and it's the local fluid velocity that matters, right?

I'd love to be corrected on this because it certainly makes the math simpler

[Blauhung]

I know I'm new here and all (note my postcount) but how do we know the heat flux is linearly proportional to flow rate?

AFAIK most correlations for heat transfer coefficients for tubes are something more like Nu = 0.023*Re^(0.8)*Pr^(1/3), such as the seider-tate correllation which certainly implies a non-linear relationship between flow and heat transfer

you are right, but once you start getting into Nusselt and Reynolds #'s and stuff people tend to gloss over even more then what most probably did at my math there. You can assume its somewhat linear for the flows we are running at. If i pulled out my books you can show that reducing the flow rate by half kinda kills your heat transfer efficiency by about that much too provided you aren't doing anything weird by changing into a different flow regime within the block.

And lol, i have no clue how i came on and looked at this thread right as you were writing this. As for local fluid velocity, that more a function of the flow path of the block and pretty much all gets rolled into the efficiency correlations for the heat exchanger you are looking at.

[Kaldskryke]

you are right, but once you start getting into Nusselt and Reynolds #'s and stuff people tend to gloss over even more then what most probably did at my math there. You can assume its somewhat linear for the flows we are running at. If i pulled out my books you can show that reducing the flow rate by half kinda kills your heat transfer efficiency by about that much too provided you aren't doing anything weird by changing into a different flow regime within the block.

Thanks .

EDIT: wouldn't running blocks in parallel decrease the total loop restriction and thus increase total flowrate?

[Blauhung]

Thanks .

EDIT: wouldn't running blocks in parallel decrease the total loop restriction and thus increase total flowrate?

yup, but not by much, the overall loop resistance in serries would be
Ro = Rb1 + Rb2 + Rrad

and in Parrallel
Ro = 1/(1/Rb1 + 1/Rb2) + Rrad

The overall resistance to flow would be lower, so your flow through your radiator would be slightly higher, and yes the flow through the blocks would be slightly higher then half, and without going through the math you can't say for sure, but again its not going to be enough to have it work out to be better then serries