7 replies to this topic

[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.

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.

Edit: Can't seem to get the upload to work, here's the question

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.

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.

Edit: File conversion

Update:
Alright, so. Did a little work on this but I'm not sure how to bring in the whole triangle aspect.

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 for the *delete me*

Attached Files

•  triangularflow.bmp   915.51KB   58 downloads

[Andree]

"...simplifying the navier stokes into dp/dz=u(d2vz/dy2)..."

(d2vz/dx2) is not zero and you cancel this...

[Guderian]

"...simplifying the navier stokes into dp/dz=u(d2vz/dy2)..."

(d2vz/dx2) is not zero and you cancel this...

Are you saying the Navier-Stokes simplifies to dp/dz=u((d2vz/dy2)+(d2vz/dx2)) or that it simplifies to dp/dz=u(d2vz/dx2)?

[Andree]

dp/dz=u((d2vz/dy2)+(d2vz/dx2))

you need to consider two coordinates, x and y, and solve 2nd order PDE with boundary conditions:
1. vz(x, H) = 0
2. vz(x, sqrt(3)*x) = 0

[Andree]

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

[Guderian]

dp/dz=u((d2vz/dy2)+(d2vz/dx2))

you need to consider two coordinates, x and y, and solve 2nd order PDE with boundary conditions:
1. vz(x, H) = 0
2. vz(x, sqrt(3)*x) = 0

Darn it, I was hoping that wasn't the case. The prof did say "You do not need to solve for Vz, use answer given and show it satisfies a certain EQN." Which probably means I can avoid solving the PDE.

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?

[Andree]

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?

almost there...
you have double integral, but for dx you integrate from -y/sqrt(3) to y/sqrt(3) and then you integrate for dy as you wrote from 0 to H

[Guderian]

You sir, are a gentleman and a scholar.