Ewan, you gave up too easily. Although neither of us was really right. For constant flowrate, the pressure drop for two blocks in parallel would be 1/4 that of a single block, but 1/8th that of two blocks in series.
I did some equation handwaving and no one caught me.
The equation I listed earlier for calculating with flow resistances in parallel is wrong.
For resistance values relevant to the equation:
P = Q^2 * R
parallel resistances are calculated using the equation.
Rp = 1 / ( (1/R1^-2 + 1/R2^-2 + 1/R3^-2 ... ) ^2 )
A bit uglier than I'd suggested earlier, but
I believe this equation actually works.
Quote:
Originally posted by Gooserider
Excellent, but what happens in what I suspect might be a more common case of having a parallel setup where the two blocks are NOT equal? How would one figure out the amount of flow that would go through each block?
Example, with totally imaginary numbers...
Branch 1 CPU block, has a resistance of 2ft pressure
Branch 2 GPU block, with a resistance of 4ft pressure
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First of all; a waterblock has a dP vs Q
curve. A waterblock doesn't just have a pressure number. It is reasonable to say:
Branch 1 CPU block, has a resistance of 2ft pressure @ 2 gpm
but, just listing a pressure doesn't really say anything.
Anyway, assuming a flowrate of 2 gpm to go along with these hypothetical pressures, we can calculate 'flow resistances' for the two devices.
(Although this is only valid inasmuch as the PQ curve of the devices conforms to the equation P = Q^2 * R. Not everything does.)
2 = 2^2 * R1
4 = 2^2 * R2
R1 = 0.5 (ft-H2O/gpm^2)
R2 = 1 (ft-H2O/gpm^2)
Using my newly derived parallel resistance equation:
Rp = 1/(1/squareroot(R1) + 1/squareroot(R2))^2 = 0.172 (ft-H2O/gpm^2)
Now you can graph the curve:
P = Q^2 * 0.172
and the curve for a pump simultaneously, and find the point where the two curves intersect. This will give the dP and flowrate for the parallel system. Using the dP determined for the system you can then calculate the flowrate through each parallel leg using the known resistance value for each leg.
For example, if the pressure that the pump produces across the parallel system is 3 ft, then:
Q1 = squareroot( 3 / 0.5 ) = 2.45 gpm
Q2 = squareroot( 3 / 1 ) = 1.73 gpm
Qp = squareroot( 3 / 0.172 ) = 4.18 gpm
So block 1 gets about 59% of the flow,
and block 2 gets about 41% of the flow.
Of course for a real system you'd need to throw a radiator into the mix, as well as values for tubing, fittings, etc.
Now I've got to take a look at some previous work based on my incorrect assumption for calculating parallel resistance.
Edit: Corrected first paragraph.
2nd Edit: Corrected flow resistance units from ft-H2O/gpm to ft-H2O/gpm^2. Plus grammar.