I was doing some reading about Navier-Stokes equations and stumbled upon this PDF from a course at Clarkson University:


In the last section (page 11/12), an equilateral triangle pipe is discussed in cartesian coordinates. If I understand correctly, the fluid velocity is assumed to be dependent on the triangular section.

Could someone explain to me what is going on in this (PDF) section? Why can we assume a solution of this form, and especially, how is the equation for the triangular section obtained in the above PDF derived?

Consider a pipe as shown in Figure 8 whose cross section is an equilateral triangle. The equation of the section is given as :

$$ f(x,y) = (x-a) \left( x - \sqrt{3}x + 2a \right) \left( x + \sqrt{3}x + 2a \right) = 0 $$

The solution is assumed to be $$v_z = w = A f(x,y) $$

Here is the image: Equilateral triangle in cartesian coordinates

I understand how to obtain the three line equations defining the triangle, and how to derive the solution given the equation of the section and the assumed form of the solution.

However, I have no idea why the equation of the section is given by multiplying the three equations (equated to zero) together. I am also unsure of the physical explanation behind the assumption that the speed is directly correlated to this defined section.


The equation they are trying to solve (from the Navier Stokes equation in the z direction) is$$0=-\frac{\partial p}{\partial z}+ \mu\left[\frac{\partial^2 v_z}{\partial x^2}+\frac{\partial^2 v_z}{\partial y^2}\right]$$ They guess a form for $v_z$ that automatically satisfies all three zero velocity boundary conditions at the three walls: $v_z=Af(x,y)$. This is the kind of thing they might try to do if they were trying to obtain an approximate solution for the velocity and the pressure. However, when they substitute the assumed form for $v_z$ into the differential equation, they find that they can satisfy it exactly. They use this to determine the value of the constant A. The rest is straightforward.

(There is a typo in the last equation for the pressure gradient, and there may be others).

  • $\begingroup$ Well, that's actually interesting, I find the "straightforward" part to be the exact opposite. I know what they are trying to do and how. Just not how they get the expression for f(x,y). It might be obvious, but it completely eludes me! $\endgroup$ Apr 4 '17 at 19:13
  • 1
    $\begingroup$ Hum, I see what you mean and that I missed in your answer, the solution is assumed not in relation to the triangular section, but to easily satisfies the boundary conditions. It makes sense. Thanks a lot! $\endgroup$ Apr 4 '17 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.