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I'm trying to understand the Navier Stokes Equation by solving a fluid dynamics problem. Not too sure if this is an appropriate place for my question.

I have a thin layer of paint with constant thickness $h$ spread evenly on a very long and wide wall. There is also a uniform flow of air flowing up directly opposite the flow of the paint.

I'm trying to find the velocity on the film, but I'm having trouble defining the boundary condition that this updraft of air creates.

I first started of with the y-momentum of the NSE since we are primarily concerned with the vertical direction.

$$\frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y}+ w\frac{\partial v}{\partial z} = \frac{-1}{\rho} \frac{\partial P}{\partial y}+ \nu(\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} + \frac{\partial^2 v}{\partial z^2}) + g_y$$

I got rid of the unsteady term above because the problem mentioned a constant thickness. Got rid of pressure since pressure is atmospheric and the same inside and outside of the boundary layer. Got rid of the terms that are not in the vertical direction since we're concerned with 1D flow. And, by continuity $\frac{\partial v}{\partial y} = 0$. Which leaves me with

$$0 = \nu\frac{\partial^2 v}{\partial y^2} + g_y$$

And if I integrate twice I get:

$$V(y) = \frac{1}{2\nu}(-g_y)y^2 + c_1y + c_2$$

Since I'm trying to find the velocity field within the fluid, I know I need to integrate twice. But, my question is how do I incorporate the shear stress by the air $\tau_{air}$? Do I just add it on as a term? Or do I integrate twice and use $\tau_{air}$ as a boundary condition? How do I define such a boundary condition?

I will note that I was able to define these boundary conditions because of the no-slip condition.

$$v(x=0,y)=v_b$$ and $$u(x=0, y)=0$$

I was reading Kundu's textbook and also found this, but I am not sure if it is right. At $x = h$ we have $ \nu_e \frac{\partial v}{\partial x} = \frac{\tau_0}{\rho_0}$

mathnoob
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1 Answers1

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I think the following is close to the correct answer. In my question above, I incorrectly cancelled out $\frac{\partial^2 v}{\partial x^2}$. It should have been $\frac{\partial^2 v}{\partial y^2}$ that gets approximated to $0$ since we are dealing with an infinitely long wall.

Starting off with $$0 = \mu \frac{\partial^2 v}{\partial x^2} + \rho g_y$$

Integrating once we get: $$\mu \frac{\partial v}{\partial x} = - \rho g x + C_1 x $$

Using the boundary condition from the air-fluid interface, we get that shear stress from air $\tau_{air} \approx -\mu \frac{\partial v}{\partial x}$ at $x=h$. Note that $\nu = \frac{\rho}{\mu}$.

Therefore, $$C_1 = -\tau_{air} + \rho g h$$

Integrating once more we get: $$v(x) = - \frac{g x^2}{2 \nu} + \frac{C_1 x}{\mu} + C_2$$

Using our B.C. associated with the no-slip condition At $x=0$ (at the wall) $v(x=0)=0$ (fluid parcel is rotational).

Substituting and rearranging gives us the final answer for our velocity field.

$$v(x) = ( \frac{x^2}{2} + h - \tau_{air} ) \frac{g x}{\nu} $$

mathnoob
  • 135