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Bit information about the problem We are dealing with the slide coating process - where basically a polymer is being put onto a slot, which is moving in the $x$-direction with velocity $v_0$. The volumetric flow rate $Q_w=Q/W$ where $Q$ is total volume-rate flow and W is the length in the $z$-direction. The polymer is solidifying the height is dropping from $h_0$ to $h_1$ for $x[0;L]$.

It is known: - constant density and viscosity - steady state - newtonian fluid - no gravity forces - lubrication approximation

Help required for

We are given the following expression for velocity (in $x$ direction),

$$ v_x=(3v_0 - 6\frac{Q_w}{h})(\frac{y}{h})^2+(6*\frac{Q_w}{h}-4v_0)\frac{y}{h}+v_0 $$

We want to introduce the following stream functions $$ v_x=-\frac{\delta \psi}{\delta y} $$ $$ v_y=-\frac{\delta \psi}{\delta x} $$

We are also given that $\psi(x,h(x))=0$. We wish to get an expression for $\psi(x,y)$ corresponding to $x[0;L]$ and $y[0;h]$.

My approach So my approach to this problem was something like this.

I integrated the given $v_x$ equation to get the following

$$ \psi(x,h)=\int v_xdy + \phi(x)=v_x=(3v_0 - 6\frac{Q_w}{h})(\frac{y^3}{3h^2})+(6*\frac{Q_w}{h}-4v_0)\frac{y^2}{2h}+v_0 y + \phi(x) $$

where $\phi(x)$ is a function.

Then I said that since we are not given an expression for $v_y$ therefore, I can differentiate the above w.r.t x, giving the following,

$$ \frac{\delta \psi}{\delta x} = 0+0+0+\frac{\phi(x)}{\delta x} = v_y $$

From there I just assumed that $v_y = 0$.

I am pretty sure the last part is definitely wrong. But is there any other way of doing it?

My thoughts: Should I determine an expression for $v_y$ from Navier-Stoke equation and only then use the stream function? Or I don't necessarily have to do that.

If additional information is required. Please write in the comment section, so I can provide it.

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  • $\begingroup$ It isn't clear to me what you are describing. Is the solidification taking place on the moving surface? Does your analysis take into account the volumetric throughput rate of the solid? $\endgroup$ – Chet Miller Oct 24 '18 at 0:36
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The Integration with respect to variable $y$ was correct. Now you have to Substitute $y=h(x)$ and use your given condition $\psi(x,y=h(x))=0$. From this you will know the x-dependent function $\phi(x)$. By differentiating this by $x$, you get $v_y$.

Remark: I know that the stream function $\psi$ is defined as

$v_x = \frac{\partial \psi}{\partial y}, v_y = - \frac{\partial \psi}{\partial x}$ (one positive sign and one negative sign). Is the Definition with both negative signs (both -) correct?

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  • $\begingroup$ Yes you are correct. I made a mistake. Thank you for your help and for pointing it out :) $\endgroup$ – MathCurious314 Oct 23 '18 at 21:28
  • $\begingroup$ Oh sorry but in the problem, we are told that h=h(x), so its already in the expression. We don't have an actual expression for h(x). $\endgroup$ – MathCurious314 Oct 23 '18 at 21:30

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