Laminar Flow through a Tringular Duct

[Guderian]

I'm currently a freshman chemical engineer at Penn State and I've gotten so frustrated at fluid transport that I have resorted to online forums for one last desperate attempt to try and figure out what exactly is going on. This question is part of a long agonizing 2 week homework assignment and I just want to finish it and be done with it. Anyway I'm attaching a file that has to do with a question discussing Laminar Flow in a triangular tube, I just need someone to walk me through this conundrum to try and find an answer.

Here's the jist of the question since I can't seem to get the upload to work:

Can't seem to get the upload to work, here's the question
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IN order to analyze the performance of such an apparatus, it is necessary to understand pressure driven flow in a duct whose cross section is an equilateral triangle.

a.) Verify that the velocity distribution for the laminar flow of a newtonian fluid in a duct of this type is given by : vz= ((P1-P2)/(4uLH))(y-H)(3x^2-y^2)

b.) From eqn 1, find the avg velocity, max velocity, and mass flow rate.
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For the picture, it's a triangular shaped duct, with an x-y coordinate system situated at the bottom of an upside down triangle. The top of the upside down triangle is y=H and the left and right side are -sqrt(3)x and sqrt(3)x respectively. Thanks again.

I'm assuming Navier-Stokes equations will work here since we've got a laminar newtonian fluid, and we'll be integrating these somehow to get this velocity profile in the z direction. It's just the whole triangular cross section that is throwing me off, what about shear stresses in particular how do those conceptually function versus lets say, a cylindrical tube? Any help in a timely fashion will be greatly appreciated.

What I've got:
Using boundary conditions y=H, v(z)=0 and y=0, v(z)=0 and simplifying the navier stokes into dp/dz=u(d2vz/dy2). Now, before I get all crazy and start integrating I need to know where the triangle aspect comes in and where the (3x^2-y^2) comes into play and how.

Also for part 2, when finding the area of this triangle, I'm assuming you put it in terms of H, so for the 1/A term you'd have 1/(H^2/(sqrt(3))) ? Then you would integrate in terms of dydx from 0 to H for y and -H/(sqrt(3)) to H/(sqrt(3)) for x correct?

Thanks again!

[siddharth]

For the picture, it's a triangular shaped duct, with an x-y coordinate system situated at the bottom of an upside down triangle. The top of the upside down triangle is y=H and the left and right side are -sqrt(3)x and sqrt(3)x respectively. Thanks again.

I'm assuming Navier-Stokes equations will work here since we've got a laminar newtonian fluid, and we'll be integrating these somehow to get this velocity profile in the z direction. It's just the whole triangular cross section that is throwing me off, what about shear stresses in particular how do those conceptually function versus lets say, a cylindrical tube? Any help in a timely fashion will be greatly appreciated.



Yeah, you have to solve the NS eqns for this geometry. First of all, if I understand your question correctly, note that v_z will be a function of *both* x and y.

Once you simplify the NS equations, you'd get

\mu \left(\frac{\partial^2 v_z}{\partial x^2} + \frac{\partial^2 v_z}{\partial y^2}\right) = \frac{\partial p}{\partial z}

You need to solve this equation with the appropriate boundary conditions.


What I've got:
Using boundary conditions y=H, v(z)=0 and y=0, v(z)=0 and simplifying the navier stokes into dp/dz=u(d2vz/dy2). Now, before I get all crazy and start integrating I need to know where the triangle aspect comes in and where the (3x^2-y^2) comes into play and how.


As I pointed out above, the NS equation you wrote is incorrect, because v_z depends on both x and y. Also, the physical boundary condition is the no-slip boundary condition. That is, the velocity of fluid at the walls are zero. You have written the boundary conditions at the base of the triangular duct, and the vertex. What about the walls of the triangular duct? This is where the triangle aspect comes in.

Can you sovle it from here?

[Guderian]

I'll need 4 boundary conditions for this right?

So:

1. vz(x, H) = 0 and
2. vz(x>=0, sqrt(3)*x) = 0 and vz(x<0, -sqrt(3)*x) = 0

What am I missing here?

[Guderian]

Figured it all out guys, thanks!

[siddharth]

I'll need 4 boundary conditions for this right?

So:

1. vz(x, H) = 0 and
2. vz(x>=0, sqrt(3)*x) = 0 and vz(x<0, -sqrt(3)*x) = 0

What am I missing here?

The second BC doesn't seem right. If the vertex of the triangular duct is at y=H, and the sides are at +/- sqrt(3), then the sides are y=-/+sqrt(3) + 3.

[foolosophy]

i wouldnt worry about it

just use the hydraulic diameter for the value of D in the circular pipe flow equations.