When the supply valve A is slightly opened, a very viscous Newtonian liquid in the rectangular tank overflows, Fig. 7–39a. Determine the velocity profile of the liquid as it slowly spills over the sides.

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Fluid Description. We will assume the liquid is an incompressible Newtonian fluid that has steady laminar flow. Also, after falling a short distance from the top, the liquid along the sides will continue to maintain a constant thickness a, Fig. 7–39b.

Analysis. With the coordinate axes established as shown, there is only a velocity u in the x direction. Furthermore, due to symmetry, u changes only in the y direction, not in the x or z direction, Fig. 7–39c.
Because the flow is steady and the liquid is incompressible, the continuity equation becomes

[latex] \begin{aligned} & \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0 \ & \frac{\partial u}{\partial x}+0+0=0 \end{aligned} [/latex]

Integrating yields

u = u(y)

Using this result, the Navier–Stokes equations in the x and y directions reduce to

[latex] ho\left(\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}+w \frac{\partial u}{\partial z} ight)= ho g_x-\frac{\partial p}{\partial x}+\mu\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2} ight) [/latex]

[latex] 0+0+0+0= ho g-\frac{\partial p}{\partial x}+0+\mu \frac{\partial^2 u}{\partial y^2}+0 [/latex] (1)

[latex] \begin{gathered} ho\left(\frac{\partial v}{\partial t}+u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}+w \frac{\partial v}{\partial z} ight)= ho g_y-\frac{\partial p}{\partial y}+\mu\left(\frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}+\frac{\partial^2 v}{\partial z^2} ight) \ 0+0+0+0=0-\frac{\partial p}{\partial y}+0+0+0 \end{gathered} [/latex]

This last equation shows that the pressure does not change in the y direction; and since p is atmospheric on the surface of the liquid, it remains so within the liquid, that is, p = 0 (gage). As a result, Eq. 1 now becomes

[latex] \frac{\partial^2 u}{\partial y^2}=-\frac{ ho g}{\mu} [/latex]

Integrating twice, we obtain

[latex] \frac{\partial u}{\partial y}=-\frac{ ho g}{\mu} y+C_1 [/latex] (2)

[latex] u=-\frac{ ho g}{2 \mu} y^2+C_1 y+C_2 [/latex] (3)

To evaluate the constant [latex] C_2 [/latex], we can use the no-slip condition; that is, the velocity of the liquid at y = 0 must be u = 0, Fig. 7–39c. Therefore, [latex] C_2 [/latex] = 0. To obtain [latex] C_1 [/latex], we realize that there is no shear stress [latex] au_{x y} [/latex] on the free surface of the liquid. Applying this condition to the fourth of Eqs. 7–73 yields

[latex] au_{x y}=\mu\left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x} ight) [/latex] (7-73)

[latex] 0=\mu\left(\frac{\partial u}{\partial y}+0 ight) [/latex]

Therefore, du/dy = 0 at y = a. Substituting this into Eq. 2 we get [latex] C_1=( ho g / \mu) a [/latex], and so Eq. 3 now becomes

[latex] u=\frac{ ho g}{2 \mu}\left(2 y a-y^2 ight) [/latex]

Once it is fully developed, this parabolic velocity profile, shown in Fig. 7–39c, is maintained as the liquid flows down the edge of the tank. It is the result of a balance of downward gravity and upward viscous forces.

Question 7.12: When the supply valve A is slightly opened, a very viscous N...

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