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Darcy's Law

Darcy's Law

Darcy's Law

(OP)
I'm hoping that someone can clarify one of the variaibles in Darcy's law.  Does the cross sectional area (A) need to be constant over the length (l) in order for the standard equation to work or does a shape factor of sorts need to be applied?

My specific situation involves the design of a sand filter that is contained within a round pipe.  The direction of flow is from the center of the pipe to the edge so the cross sectional area parallel to flow is parabolic not rectangular.

RE: Darcy's Law

A is the cross-sectional area to flow through the length.

Not sure what you are trying to do. However, perhaps you can take the average of the cross-sectional area across the length.  

RE: Darcy's Law

I think you are trying to overcomplicate the evaluation.

You do not have to consider a radial flow plane.  You may consider a vertical flow plane from the center to the edge, having length of R2-R1 and a unit depth.

RE: Darcy's Law

(OP)
bimr,
Thanks for your reply.  What I'm trying to determine is whether the area (A) normal to the flow needs to be constant over the entire length.  For example, say you've got a funnel that has been filled with sand.  The area would change from the top to the bottom so would you need to modify the equation to suit, or would it be appropriate to use the average area?   

RE: Darcy's Law

don't change the equation, just integrate it. or for a good approximation, divide it into a number of finite slices and use average areas for each slice. If you feel that average area over the entire thing is a good enough estimate, than do that.

RE: Darcy's Law

One would think that the average would be good enough for an estimate.

However, note that on a sand filter, most of the filtration is in the first 2-Inches.

RE: Darcy's Law

For the funnel example, wouldn't you use the narrowest cross section?

Hydrology, Drainage Analysis, Flood Studies, and Complex Stormwater Litigation for Atlanta and the South East - http://www.campbellcivil.com

RE: Darcy's Law

An approximate area may be good enough, but an exact solution is just as simple:

Q = HKL2(pi)/LN(R2/R1)

Q = Volume of Flow
H = Head Loss
K = Conductivity
L = Length of pipe
(pi) = 3.14
LN = Natural Log
R2 = Radius of Filter Pipe
R1 = Radius of central inflow pipe

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