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In my lecture notes, we went through a problem with a cylinder approaching a stationary wall with speed $v$. Since we are interested in the thin layer approximation, we utilised lubrication theory. Approximating the distance between the cylinder and wall as

$h(x) = d(1+x^2/2ad)$

where $d$ is the shortest distance between the cylinder and wall.

To find the flow velocity $u$, we used the parallel flow equations to get:

$u=1/2\mu (dp/dx)y(h-y)$

But what i don't understand is why we have imlemented the boundary conditions that the flow velocity is 0 and both $y=0$ and $y=h$, surely, the velocity should be $-v%$ at $y=h$?

MathematicianP
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  • This isn't really a math question. I think you'll have better luck at https://physics.stackexchange.com/ – user388557 Jan 19 '20 at 09:01
  • This is hard to answer without your notes (you should probably ask your professor) but using $-v$ at $y=h$ sounds reasonable to me, depending on the specifics of the problem. – David Jan 21 '20 at 04:27

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The cylinder is moving so it is also a function of time. The velocity is a vector usually denoted $\vec{v}$ or $\vec{u} = (u,v,w)$ where $u = \frac{\partial x}{\partial t}$ $v = \frac{\partial y}{\partial t}$. The boundary conditions would require, \begin{cases} u=0 & \text{ at } y=0, y=h(x,t) \\ v=0 & \text{ at } y=0\\ v=\frac{dh}{dt} & \text{ at } y=h(x,t)\\ \end{cases}

Alex
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