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## Main Question or Discussion Point

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!

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

----------

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.

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!