Help me, / Advanced Fluid dynamics Problems

In summary, the radial component of the inertia term in cylindrical coordinates can be evaluated using the Lagrangian derivative and the given identities, resulting in (u dot Del)uR = (Del Ur)Ur.
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Help me, please / Advanced Fluid dynamics Problems

Homework Statement



Evaluate the radial component of the inertia term (u dot Del)u in cylindrical
coordinates using the following identities:

x=RcosΘ
y=RsinΘ
uEx+vEy = UrEr+UΘEΘ

and any other vector identities from Appendix A as required. Here R
and y are cylindrical coordinates, Ur and UΘ are the corresponding
velocity components, and Er, EΘ are the unit base vectors.


Homework Equations


Lagrangian derivative:

The Attempt at a Solution

 
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The definition of the Lagrangian derivative is as follows: (u dot Del)u = (uEx + vEy) . (Del uE + Del vE) Using the identities given in the problem statement, this simplifies to: (UrEr + UΘEΘ).(Del UrEr + Del UΘEΘ) We can then use the following vector identities from Appendix A: Del UrEr = (Del Ur)Er + Ur(Del Er) and Del UΘEΘ = (Del UΘ)EΘ + UΘ(Del EΘ) Therefore, (u dot Del)u = (UrEr + UΘEΘ).((Del Ur)Er + Ur(Del Er) + (Del UΘ)EΘ + UΘ(Del EΘ)) Now, using the fact that (Del Er) and (Del EΘ) are both zero, we can simplify the equation to: (u dot Del)u = (Del Ur)Ur + (Del UΘ)UΘ The radial component of the inertia term is then: (u dot Del)uR = (Del Ur)Ur
 

FAQ: Help me, / Advanced Fluid dynamics Problems

What is fluid dynamics?

Fluid dynamics is a branch of physics that deals with the study of fluids in motion. It involves the analysis of how forces such as pressure, gravity, and viscosity affect the behavior of fluids such as liquids and gases.

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