Shear Flow Against a Wall (Fluid Mechanics)

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The discussion revolves around deriving the velocity component V(x; y) for a shear flow against a wall, defined by the equation U = Uo(2y/ax - y^2/(ax)^2), under the assumption of incompressibility. Participants emphasize the importance of setting the divergence of the velocity vector to zero, leading to the relationship ∂V/∂y = -∂U/∂x. A suggested approach involves defining a new parameter, η = y/ax, to simplify calculations. This transformation aids in managing the derivatives more effectively. Overall, the thread focuses on utilizing fluid mechanics principles to solve the problem at hand.
Isobel2
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Homework Statement



Assume a shear flow against a wall, given by U= Uo (2y/ax - y^2/((ax)^2) where a is a constant. Derive the velocity component V (x; y) assuming incompressibility.



Homework Equations


Haven't been able to find any in my course notes.


The Attempt at a Solution


Some googling has taught me that shear flow is the flow induced by a shear stress force gradient. But I really need some sort of equation to solve this I think.
 
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Hi Isobel2. Welcome to Physics Forums.

You need to make use of MaxManus' suggestion, and set the divergence of the velocity vector equal to zero:

\frac{\partial V}{\partial y}=-\frac{\partial U}{\partial x}

But, before you start trying to do this by brute force, first define the following parameter:

\eta=\frac{y}{ax}

so that U = U_0(2\eta - \eta ^2)

Also note that \frac{\partial U}{\partial x}=\frac{\partial U}{\partial \eta}\frac{\partial \eta}{\partial x}=-\frac{\partial U}{\partial \eta}\frac{\eta}{x}
\frac{\partial V}{\partial y}=\frac{\partial V}{\partial \eta}\frac{\partial \eta}{\partial y}=\frac{\partial V}{\partial \eta}\frac{1}{ax}

Working with the parameter η in this way will make the "arithmetic" much simpler and less prone to error.
 
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