# Hagen-Poiseuille flow

Here, I am going to go over the solution to fully developed laminar pipe flow. This is a canonical problem and provides an exact solution to the Navier-Stokes equations. It is often referred to as the Hagen-Poiseuille flow problem.

The problem is depicted in the figure below:

In this problem, we are examining laminar flow through a pipe. The problem states that the flow in the pipe is being driven by a constant pressure gradient in the axial direction (dP/dz=constant). We assume that the flow is purely axial ($v_r = v_\theta =0$), steady state ($\partial / \partial t =0$), incompressible (ρ=constant), axisymmetric ($\partial / \partial \theta =0$). We also neglect gravity.

First we start with the axial Navier-Stokes momentum equation in cylindrical coordinates:

$\rho v_z \frac{\partial v_z}{\partial z} = -\frac{dp}{dz}+\frac{\mu}{r} \frac{d}{dr}\left(r \frac{dv_z}{dr}\right)$

By using our assumptions we can reduce this equation to give us:

$\frac{dp}{dz}=\frac{\mu}{r} \frac{d}{dr}\left(r \frac{dv_z}{dr}\right)$

Because we know that dP/dz is a constant, this function is easily integrated twice. The first integration:

$r\frac{dv_z}{dr}=\frac{r^2}{2\mu}\frac{dP}{dz}+A$

And the second integration:

$v_z=\frac{r^2}{4\mu}\frac{dP}{dz}+A \ln(r)+B$

Now we apply our boundary conditions!

No slip condition at r=R : $v_z=0=\frac{dP}{dz}\frac{R^2}{4\mu}+A\ln(r)+B$

Finite axial velocity at r=0: $v_z=0+A \ln(0)+B$

Since ln(0) is a discontinuity, we know that in order for these equations to be satisfied, A must be equal to zero (ie A=0). Then combining these two equations, we get that:

$B=\left(-\frac{dP}{dz}\right)\left(\frac{R^2}{4\mu}\right)$

Thus our final solution is:

$v_z=\left(\frac{dP}{dz}\right)\left(\frac{1}{4\mu}\right)\left(r^2-R^2\right)$

And there we have our answer!

# Couette Flow (Moving Wall)

In this post I am going to go through the solution to the moving wall Couette flow problem.

An illustration of the problem is given below:

In this problem, the fluid between two parallel plates is being driven by the motion of the top plate. Here, we assume that the flow is axial (v=w=0), incompressible (ρ=constant), fully developed (u only a function of y). We also assume that gravity can be neglected and that no pressure gradient is present (dp/dx=0).

First we start with the Navier-Stokes momentum equation in the x-direction:

$\rho\left(u\frac{\partial u}{\partial x}+ v \frac{\partial u}{\partial y}\right) = \frac{\partial p}{\partial x} + \mu\left( \frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial y^2}\right)$

By applying our assumptions listed above, you should be able to see that the equation simply becomes:

$\frac{d^2 u}{dy^2}=0$

To solve the problem we must integrate this equation and solve using the boundary conditions defined by the problem. The integration results in:

$u=Ay+B$

Our boundary conditions come from the problem and our super smart knowledge of the no-slip condition ;). Ie.

@ $y=h$, $u=V$

@ $y=0$, $u=0$

After subbing in we get two equations, with two unknowns (the integration constants)

$V=Ah+B$

$0=A(0)+B$

Therefore we can see that B=0 and A=V/h. This leads to the final solution of this plane Couette flow problem:

$u=\frac{Vy}{h}$

Now we have shown that the velocity profile in this case is the linear profile above. We can also calculate the shear stress:

$\tau=\mu \frac{du}{dy} = \mu \frac{V}{h}$

Any thoughts or questions please respond in the comments!