# Fluids - Darcy's Law, permeability of a filter

1. Mar 19, 2015

### OldStudent0382

1. The problem statement, all variables and given/known data
Derive an analytical expression for the permeability (k) of the structured porous medium (which is a size selective filter) constituted by a periodic array of cylinders of diamter d separated by a distance h = 1.8d

2. Relevant equations
Q=k*(μ/A)*(Δp/L)
Q = volumetric flow rate
k= media permeability
μ=fluid viscosity
A=cross sectional area
L=length between inlet and outlet
Δp=pi-po

Please note that h should be 1.8d and not 0.8d

3. The attempt at a solution
I'm not sure where to start with this, since I can't even seem to get the units to work out given the equation for the volumetric flow rate.

My first attempt was to take the cross sectional area of 1 cylinder (since 1 array consists of 1 quarter of 4 cylinders and divide by the cross section of the array (1.8^2 * d^2), then multiply by the average velocity of the flow through 1 of the cylinders (which I found to be 31.25 m/s)

I am at a complete loss as to how to approach this problem. I know I'm overlooking something very basic.

THANK YOU!

2. Mar 19, 2015

### SteamKing

Staff Emeritus
It seems like if you take a sample area of the filter such that A = b2, you'll have 4 cylindrical holes instead of what you assumed.

It appears that b = 1.8 d + d = 2.8 d, given the layout in the cross-section detail.

3. Mar 19, 2015

### OldStudent0382

4. Mar 19, 2015

### Staff: Mentor

This looks like a pretty straightforward problem. What is the pressure drop vs flow rate relationship for viscous flow through a circular tube of length L and diameter d, if the fluid viscosity is μ and the volumetric flow rate is Q? If you have trouble determining this relationship using the equations that they have provided, look up Hagen Poiseuille on Google.

Chet

5. Mar 19, 2015

### SteamKing

Staff Emeritus
The statement of Darcy's Law in the OP is what's screwed up.

http://en.wikipedia.org/wiki/Darcy's_law

Q = -k(A / μ)*ΔP / L

which is why your units for Q were not working out. k has units of m2, BTW. The negative sign is used because ΔP = (Po - Pi) in this article, and Po < Pi.

Other than this, who knows if your text has other glaring mistakes in it?

As far as the calculations are concerned, it's obvious that the pitch of the holes in your alternative 2x2 filter section is incorrect. The layout of the filter section you chose still has the same area of holes as the cross section given in the problem statement. I don't see any advantage to working this problem using your filter section over that given in the problem statement.

Rework your calculations using the correct expression of Darcy's Law and see what drops out.

6. Mar 20, 2015

### OldStudent0382

Thanks -- the equation given in the problem is the same one in my professor's handouts and what he gave in the lectures (and of course, this was rushed in the last 5 minutes, so he couldn't go over a problem).

I know this should be a really easy problem and I'm just getting hung up on something stupid.

7. Mar 20, 2015

### Staff: Mentor

So answer my question in post #4. Until you can provide this answer, you are not going to be able to solve this problem.

Chet

Last edited: Mar 20, 2015
8. Mar 20, 2015

### OldStudent0382

taking the integral of the velocity profile * area I get:

Q=∂P/∂z * π*R4/(8*μ) -> (|ΔP|/L)*π*r4/(8*μ)

Thanks for all the help -- this already has been far more than I expected to receive.

9. Mar 20, 2015

### OldStudent0382

I can't edit my last post, so I'll add a new reply:

I found:
k=Vavg * μ * L/(ΔP)

Am I on the right track or am I a lost cause?

10. Mar 20, 2015

### Staff: Mentor

In terms of the pore diameter, this equation reads

$$Q=\frac{πd^4}{128μ}\frac{Δp}{L}$$

For flow through a porous medium, Darcy's law is based on the superficial flow velocity, which is equal to the total volumetric flow rate divided by the combined cross sectional area of both rock and pores. In this problem, the superficial velocity is equal to Q, the volumetric flow rate per pore, divided by A, the cross sectional area of rock and pore (per pore). The cross sectional area per pore is equal to $A=(1.8d)^2$. So if you divide the left hand side of the above equation by the symbol A, and the right hand side of the equation by $(1.8d)^2$, what do you get? From this, you should be able to back out an equation for the permeability exclusively in terms of d.

Chet

Last edited: Mar 20, 2015
11. Mar 20, 2015

### OldStudent0382

Dividing by (1.8d)^2 gives

$$\frac {Q}{A} = \frac {\pi*d^4}{128u} * \frac {Δ P}{L} * \frac {1}{(1.8d)^2}$$

which reduces to
$$\frac {Q}{A} = \frac {1}{32*1.8^2} \frac {\pi*d^2}{4} \frac {1}{μ} * \frac {Δ P}{L}$$

Would k just be this part?:
$$\frac {\pi*d^2}{414.72}$$

I also tried taking the ratios of the cross sections and multiplying it by what dropped out when I determined the flow rate from the velocity profile (k came out to 0.2424/8* r^2 after removing the viscosity/pressure drop/length terms) and it looks like they're giving me the same answer.

12. Mar 20, 2015