Fluids - Darcy's Law, permeability of a filter

[OldStudent0382]

Homework Statement

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

Homework 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=p_i-p_o

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

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!

 

[SteamKing]

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.

We can't help you resolve this problem unless you post your work.

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)

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.

 

[OldStudent0382]

We can't help you resolve this problem unless you post your work.

Okay -- I'll try to do my best with posting my work.

First, for the flow rate units:
Volumetric flow rate (Q) = k*μ*(1/A)*Δp*1/L
m³/s = [whatever the units are for k] * [kg/(ms)]*[1/m²]*[kg/(m s²)]*[1/m]

Which gives on the right hand side = k*[kg²/(m⁵*s³)]

I wasn't sure if the units of k were m/s or m² -- either way, the units do not add up to give m³/s for a volumetric flow rate. I don't know if my professor has been giving the equation wrong, or I'm just really bad at this (I'm not fairing to well in this class, so it's probably me). The only way I've been able to get the units to work, is if permeability are m² and I put area on top and viscosity on the bottom of the equation (A/μ instead of μ/A). This is only part of the reason I can't find and expression for permeability -- I don't know what the units are supposed to be AND I can't get the units to cancel out the way I'm expecting.

How I found the average velocity was taking the velocity profile
u_z=(1/4μ)*(∂P/∂z)*R²+C₁*ln(R)+C₂
Given the boundary conditions that velocity is 0 at the wall and at its maximum at the center, I found C₁=0 and C₂= r²/(4μ)*(Δp/L)

Then, the average velocity reducing to V_avg= R²/(8μ)*(dP/dz)=(0.0005^2)/(8*0.001) * (100000)/(0.1) = 31.25 m/s

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.

I'll try to explain my thought process on the cross section I chose as best as I can (I did evaluate both options, actually three, and chose this one as the easiest) -- let's say I took this cross section as 1 "unit" of the cylinder array - if you multiply out or copy and paste that cross section (through the center of the cylinders), the array will space itself correctly. If you took the outside of the 4 cylinders (a 2.8d * 2.8d square), and tried to build an array using that as a block, you'd need adjust for space between separate arrays. The third option I explored was to have the area past the 4 cylinders to extend 0.4d on each side (creating a 3.6d * 3.6d square) so it spaces out easily, which kind of was a more round about method of my first method. This is the only part of the problem that I'm fairly confident of -- everything else was purely a guess.

Here's a 2x2 of the filter cross section I chose:

And here's a 2x2 of the alternative:

Now -- my attempt at finding the permeability expression was to find the velocity of flow through 1 cylinder (since that is how much consumes my cross section) then find the ratio of the cylinder cross section to the filter cross section

k= V_avg*(π*d²/4)/(1.8²*d²) = V_avg*π/(4*1.8²)= 0.2424*V_avg which units are m/s

With V_avg=31.25, I find k to equal 7.575m/s -- plugging this number into the given equation for Q gives (what I think) is an extraordinary high number (I believe 7575 m³/s through a single 1mm cylinder).

Thanks for the help!

 

[Chestermiller]

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

 

[SteamKing]

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

According to this article on Darcy's Law:

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 = (P_o - P_i) in this article, and P_o < P_i.

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.

 

[OldStudent0382]

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

According to this article on Darcy's Law:

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 = (P_o - P_i) in this article, and P_o < P_i.

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.

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.

 

[Chestermiller]

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.

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

 

[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.

 

[OldStudent0382]

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

I found:
k=V_avg * μ * L/(ΔP)

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

 

[Chestermiller]

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

 

[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.

I appreciate your patience.

 

[Chestermiller]

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.

I appreciate your patience.

This answer is correct.

Chet

 

[OldStudent0382]

Thanks for your help!