Shear Flow Against a Wall (Fluid Mechanics)

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SUMMARY

The discussion focuses on 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), where 'a' is a constant. Participants emphasize the importance of incompressibility, leading to the condition that the divergence of the velocity vector must equal zero. The transformation using the parameter η = y/ax simplifies the calculations, allowing for a clearer path to solving the problem. Key equations from fluid mechanics, including the Navier-Stokes equations and Bernoulli's principle, are referenced to aid in the derivation.

PREREQUISITES
  • Understanding of shear flow dynamics
  • Familiarity with the Navier-Stokes equations
  • Knowledge of incompressible fluid properties
  • Basic calculus for partial derivatives
NEXT STEPS
  • Study the derivation of the Navier-Stokes equations in fluid mechanics
  • Learn about the implications of incompressibility in fluid flow
  • Explore the application of the Bernoulli Equation in shear flow scenarios
  • Investigate the use of non-dimensional parameters in fluid dynamics
USEFUL FOR

Students and professionals in fluid mechanics, particularly those studying shear flow and its mathematical modeling, will benefit from this discussion.

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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