- #1

joshmccraney

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Assume we have a rectangular cavity (2D) filled with a liquid of dimensions ##L \times H## and that the top plate of the cavity moves with some velocity ##V_0##. Also assume ##L \gg H##. I'll also assume ##L \gg H## implies flow is roughly 1-dimensional, and thus a pressure gradient drives the flow near the still wall in the opposite direction rather than the ##y## component that would "bounce" off the vertical cavity walls. When computing the velocity profile, from the Navier-Stokes equation I get $$v_x \frac{\partial v_x}{\partial x} = -\frac{\partial P}{\partial x} + \mu \frac{\partial^2 v_x}{\partial y^2} + \mu \frac{\partial^2 v_x}{\partial x^2}$$ subject to ##v_x(0) = v_0## and ##v_x(H) = 0##, where I assume the flow is fully developed and steady so ##v_x = f(y)##. Notice now the equation reduces to $$\frac{d P}{d x} = \mu \frac{d^2 v_x}{d y^2}$$

Clearly all we now must do is solve for the pressure gradient since the boundary conditions determine the two integration constants. To find this, I imagine the net flow rate must be zero since no flow enters or leaves. Thus $$\frac{1}{HL} \int_0^L \int_0^H \, v_x dy \, dx = 0$$. Then we should be able to solve for ##\frac{d P}{d x}##.

Does this look right to you? Thanks for your time!