How to convert this derivation of Poiseuille's law into the standard one? I am trying to derive Po

Laila Andrews 2022-05-19 Answered
How to convert this derivation of Poiseuille's law into the standard one?
I am trying to derive Poiseuille's law. I have reached a point in the derivation where I have:
V = ( p 1 p 2 ) ( R 2 ) 4 l u
Where l is length, u is viscosity, p is pressure, v is flow velocity and R is radius. What I am stuck on is shifting this to the volumetric flow rate:
V π R 2 = Q = ( p 1 p 2 ) ( R 4 ) 4 l u .
V π R 2 = Q = ( p 1 p 2 ) ( R 4 ) 4 l u .
However this is incorrect as Poiseuille's law is divided by 8. I know that I am probably missing something obvious, but I can't think of a reason as to why the whole equation needs to be halved. Any help understanding why (or whether my original derivation for V was inaccurate) would be appreciated.
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Answers (1)

Eliezer Olson
Answered 2022-05-20 Author has 16 answers
The expression for the flow speed as a function of distance from the pipe center is:
V ( r ) = ( p 1 p 2 ) ( r 2 R 2 ) 4 l μ
The average velocity of the flow can be calculated by performing an integration over the pipe cross section, and then dividing by the cross-sectional area:
V a v g = 0 R 0 2 π ( p 1 p 2 ) ( r 2 R 2 ) 4 l μ r d θ d r π R 2
The extra factor of r comes from the fact we are doing everything in polar coordinates. Since the function is independent of θ,
V a v g = 0 R 0 2 π ( p 1 p 2 ) ( r 2 R 2 ) 4 l μ 2 π r d r π R 2 = ( p 1 p 2 ) R 2 8 l μ
The factor of 8 comes from the combination of integrating the r 3 term, resulting in a 1 4 term, multiplied by the 2 from the angular integration, and multiplied again by the original 1 4 in the V ( r ) expression to yield a factor of 1 8
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