Velocity profile in a triangular duct (Derivation) How do I find a velocity profile of an incompressible fluid in a triangular duct. Can someone point me to a step-by-step solution so that I could understand the process of derivation of the final form of the velocity profile.
Edit #1: So, based on the parameters that I have. I have reduced the N-S equation to the following form: $$\frac{\partial p}{\partial x} = \mu (\frac{\partial^{2} u}{\partial y^{2}} + \frac{\partial^{2} u}{\partial z^{2}})$$
I assume boundary conditions are as follows: $$u_{x}(z=h)=0$$ and the triangle is defined by three half-planes: $$\bigg(\frac{\sqrt{3}}{2}\bigg)a \geq z$$ $$ z \geq \sqrt{3}y$$ $$z \geq -\sqrt{3}y$$.
And I'm stuck from here.
 A: For unidirectional, laminar flow the $x-$component of velocity $u$ depends only on coordinates $y$ and $z$. The components of velocity in the other directions are identically zero.  The $y-$ and $z-$momentum equations reduce to
$$\frac{\partial p}{ \partial y} = \frac{\partial p}{ \partial z}  = 0.$$
As you stated, the Navier-Stokes equations reduce to
$$\nabla^2u = \frac{\partial^2 u}{ \partial y^2} + \frac{\partial^2 u}{ \partial z^2} = \frac{1}{\mu} \frac{\partial p}{\partial x}.$$
Since, $u$ depends only on $y$ and $z$ and $p$ does not, it follows that the partial derivative of the pressure with respect to $x$ must be a constant
$$\frac{1}{\mu} \frac{\partial p}{\partial x} = G,$$
and we have a form of Poisson's equation 
$$\nabla^2 u = G.$$
This PDE is satisfied at all points in a region $\Omega \subset \mathbb{R}^2$ corresponding to a cross-section of the interior of the duct. Additionally, we have the no-slip boundary condition applied at the interior walls
$$u|_{\partial \Omega} = 0.$$
There is no closed-form solution to this problem on the specified triangular region in terms of elementary functions. 
Approximate numerical solutions can be obtained readily.  The most common approaches for problems of this type are finite element or boundary integral methods.
