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Topic: Simple Fluid Transport Problem, seemingly.  (Read 4646 times)

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Offline Guderian

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Simple Fluid Transport Problem, seemingly.
« on: April 30, 2008, 06:56:21 PM »
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.

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.

Again thanks for any help. Oh, and its nice to be part of a community.


Offline eugenedakin

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Re: Simple Fluid Transport Problem, seemingly.
« Reply #1 on: April 30, 2008, 08:04:24 PM »
Hello Guderian,

Sure, I will help you out. 

First, define each of the knowns:

for example:
H = Height of triangle = y

I will respond after you work this out, then we will proceed to the mathematics section  :)

I look forward to helping.

Sincerely,

Eugene
There are 10 kinds of people in this world: Those who understand binary, and those that do not.

Offline Guderian

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Re: Simple Fluid Transport Problem, seemingly.
« Reply #2 on: May 01, 2008, 01:41:40 AM »
Alright, so. Did a little work on this but I'm not sure how to bring in the whole triangle aspect.

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

Edit: Now that I look at this, I don't know if I can cancel out the d2vz/dx2, this would leave me with a 2nd order PDE which I did not believe my instructor would give us on this, although he did say "You do not need to solve for Vz, use answer given and show it satisfies a certain EQN."

Thanks again!

Update: Got to the PDE i've got he boundary conditions:

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

What am I missing...?

UPDATE:
Put everything together! Thanks guys!
« Last Edit: May 01, 2008, 07:35:40 AM by Guderian »

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