Undergrad Analytical Open Channel Rectangular Fluid Flow

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An analytical solution for open channel rectangular fluid flow involves a steady-state, incompressible flow profile bounded by three walls with the top open to the atmosphere. The geometry is infinite in the x-direction, with y ranging from -w to w and z from -h to h. When the width and height are equal, the flow exhibits a parabolic profile along the y-axis, diminishing near the walls. As the width significantly exceeds the height, the velocity profile becomes nearly uniform in the center, with parabolic flow occurring in the z-direction close to the walls. The solution for this flow mirrors that of a closed channel, adjusted for the pressure gradient due to the channel's inclination.
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Hi All,

I'm looking for an analytical solution to the open channel rectangular fluid flow profile. The flow is bounded by three walls but the top is open to atmosphere. Assume steady state flow that is parallel and incompressible.I've already found information involving a rectangular flow channel bounded by 4 rectangular plates(top closed):

Geometry: Infinite along x direction. y ranges from -w to w and z ranges from -h to h. z denotes depth of our channel.

-If w=h, we have parabolic flow along y axis(along any given plane within depth z). Flow drops off as you approach rectangular walls at y= -w or +w.

-As w/h becomes very large(e.g w/h ≥ 10) , velocity profile flattens along centre of y-axis and falls off very close to the walls(within distance h from either wall). Flow along the y direction is almost uniform until we are very close to the side walls. We also get parabolic flow in z direction within this limit of w>>h.

Any help would be much appreciated.
 
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The problem of open-channel rectangular flow must involve an inclined channel for the flow to be steady. The pressure gradient is zero, since it is open to the air, and the shear stress at the free surface is zero.

In closed rectangular channel flow, the boundary condition at the centerline is also zero shear stress. So the solution for open-channel rectangular flow must be the same as that for half the channel in closed channel flow, but with the pressure gradient replaced by ##\rho g \sin{\theta}##, where ##\theta## is the angle of inclination.
 
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