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While learning fluid-dynamics, I stumbled across this problem that requires me to calculate the volumetric flow rate Q of a thin layer of oil leaking from a small hole in a barge, as depicted in the following figure:

enter image description here

If I understand the problem correctly, this is a Poiseuille-flow problem, as the oil is going up thanks to the difference in pressure caused by the hydrostatic pressure of the water.

I tried then to obtain the Navier-Stokes equation in the x direction for this problem: (my reference frame is such that the x axis is aligned with the direction of flow of the oil)

$$0=-\frac{p_{water}gcos\theta}{\rho_{oil}}-gcos\theta +\nu\frac{d^2u}{dy^2}$$

Where I used the fact that $\frac{\partial p}{\partial x}=\rho_{water}gcos\theta$ and that $u=u(y)$ only depends on y

Integrating using the boundary condition that u(0)=0 (non-slip condition) I get:

$$u=\frac{gcos\theta}{2\nu}(1-\frac{\rho_{water}}{\rho_{oil}})y^2+c_1 y$$

But I still have one constant left, how can I get rid of it? I tried equating the shear stresses in the oil-water interface but could not get anything helpful at all. Any help will be appreciated.

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    $\begingroup$ What boundary condition did you apply at the free surface? $\endgroup$ – Chet Miller Dec 5 '18 at 12:39
  • $\begingroup$ You mean in the oil-air interface at the top? Or the oil-water interface? $\endgroup$ – mobzopi Dec 5 '18 at 12:51
  • $\begingroup$ Why are you using $\rho_{water}$ in $Q$? Also where did the pressure gradient go? $\endgroup$ – Drew Dec 5 '18 at 12:58
  • $\begingroup$ Also your simplified Navier-Stokes equation is wrong... Look at the units of each term. $\endgroup$ – Drew Dec 5 '18 at 13:01
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    $\begingroup$ @ChesterMiller thanks chester, using the condition of zero shear stress I was able to solve it. $\endgroup$ – mobzopi Dec 5 '18 at 19:27
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If we take the positive x direction as up and to the left along the incline and perform a force balance on the oil layer between x and $x+\Delta x$, we obtain:

$$[p(x)-p(x+\Delta x)]h-\rho_{oil}g\cos{\theta}h\Delta x-\mu_{oil}\left(\frac{3\bar{v}}{h}\right)\Delta x=0$$where p is the pressure in the oil, and $\frac{3\bar{v}}{h}$ represents the shear rate at the wall for laminar flow in a channel of height h, with a zero shear stress at y=h and zero velocity at y=0; $\bar{v}$ is the average fluid velocity in the x direction. The first term is the result of pressure forces, the second term is the force of gravity, and the third term is the viscous shear stress at the wall. If we take the limit of this equation as $\Delta x$ approaches zero, we obtain: $$-\frac{dp}{dx}-\rho_{oil}g\cos{\theta}=\mu_{oil}\left(\frac{3\bar{v}}{h^2}\right)$$The pressure gradient within the film in the x direction is dictated by the hydrostatic pressure of the water immediately outside the layer: $$\frac{dp}{dx}=-\rho_{water} g \cos{\theta}$$So, combining these equations gives: $$g \cos{\theta}(\rho_{water}-\rho_{oil})=\mu_{oil}\left(\frac{3\bar{v}}{h^2}\right)$$The average velocity $\bar{v}$is related to the volumetric flow rate (per unit width into the page) by:$$\bar{v}=\frac{Q}{h}$$Therefore, $$Q=\frac{g \cos{\theta}(\rho_{water}-\rho_{oil})h^3}{3\mu_{oil}}$$This is very close to your original answer, except for the factor of 3 (vs 12) in the denominator.

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