Quote:
Originally posted by Ewan
Yes. By dividing the flow into two you instead have two half flows with both flows at half speed. Given that pressure drop is a function of the square of flowrate, running a 1/2 flow will give you a 1/4 pressure drop. And the fact that you have two paths agains halfs pressure drop. So you will have 1/8th of the pressure drop of running the flow through 1 WB. This is assuming that flowrate is held constant.
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You've halved too many times.
Suppose you have two 'flow resistors' in series and the PQ relationship is defined by:
P = Q^2 * (R1 + R2)
Then if you put those two 'flow resistors' in parallel, (and assume that the necessary 'wyes' add no flow resistance) the equation becomes:
P = Q^2 / (1/R1 + 1/R2)
So assuming that:
R1 = R2 = 1
and therefore
Rs = R1 + R2 = 2
then:
Rp = 1/(1/R1 +1/R2) = 1/2
So:
Rp = 1/4 * Rs
Or, in words - the 'flow resistance' of the 'resistors' in parallel is a quarter of the flow resistance of the 'resistors' in series
IF the two resistors are equal and the added resistance of 'wyes' is neglected.
Edit: This is incorrect. See below.