1
$\begingroup$

I've come across a problem which modifies the traditional problem of Couette flow by adding two layers of viscous fluid with different dynamic viscosities $\mu_{1}$ and $\mu_{2}$. Suppose we have two infinite plates, distance $h$ apart, with the plate at $h = 0$ stationary and the plate at $h$ moving at velocity $U$. Suppose we have a layer of fluid $A$ which occupies the region $0 \leq y \leq h_{1}$, and fluid $B$ is between $h_{1} \leq y \leq h$, each with dynamic viscosities $\mu_{1}$ and $\mu_{2}$ respectively.

I want to solve for the velocity profile. Now it is clear to me that we can assume the velocities $\mathbf{u_{1}}$ and $\mathbf{u_{2}}$ are directed only in the $x$ direction. So we assume that:

$$\mathbf{u_{i}} = (u_{i}(y), 0, 0). $$

Now, plugging into Navier Stokes we get that $u_{i}''(y) = 0$. The obvious boundary conditions are that

\begin{align} u_{1}(0) &= 0 \\ u_{2}(h_{1}) &= u_{1}(h_{1}) \\ u_{2}(h) &= U \end{align}

And, clearly we have that $u_{i}(y) = A_{i} + B_{i}y$ where the constants are determined by these boundary conditions. However, upon substitution I found that these conditions aren't enough to determine the flow completely. My question is, do we have another boundary condition at the interface of the fluid regarding the shear stresses in between the two fluids being equal? If so, I suspect it looks like

$$ \mu_{1} \frac{\partial u_{1}}{\partial y} = - \mu_{2}\frac{\partial u_{2}}{\partial y}$$

at $y = h_{1}$.

Having done this, I've determined the system in full, for example finding that $$ u_{1} = y\frac{U\mu _{2}}{h_{1}(\mu_{1} + \mu_{2}) - (h-h_{1})\mu_{1}}.$$ Is this the right track in proceeding the problem?

$\endgroup$
2
  • $\begingroup$ Your equation for the shear stress being continuous at the interface should have a + sign, not a - sign. In fact, the shear stress is constant over the entire gap. $\endgroup$ – Chet Miller Apr 3 '18 at 1:46
  • $\begingroup$ Perhaps you left a $y$ out of the right hand side of your proposed solution for $u_1$? $\endgroup$ – Mark Eichenlaub Apr 3 '18 at 2:14
1
$\begingroup$

You have the right idea, but

  • you have a sign error at the boundary, as mentioned by Chester in a comment
  • you left $y$ out of your equation for $u_1$

You should find

$$u_1 = Uy\frac{ \mu_2 }{h_1 \mu_2 + \mu_1 (h-h_1)}$$

It should be possible to detect these sorts of errors with a little care. For example, suppose $h = h_1$. Then the second fluid doesn't exist and $\mu_2$ has to disappear from your equation. Does it with your original answer? Does it with mine?

Or if $h_1 =0$, only fluid 2 exists; does your solution for $\mu_2$ do the right thing?

Also, you can detect things like forgetting the $y$ by doing a quick dimensional check.

$\endgroup$

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.