Category Archives: Basic Fluids

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:

Slide2
Fig: Hagen-Poiseuille Flow Problem Definition

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:

Slide2
Fig 1: Illustration of Couette Flow

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!

cropped-logo12.png