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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:

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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!

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