Fluid Mechanics question,velocity potential

AI Thread Summary
The discussion revolves around understanding the theoretical aspects of velocity and force potential in fluid mechanics, particularly in the context of civil engineering. A user expresses confusion regarding the derivation of an expression for acceleration, specifically how to relate it to the velocity potential function. Key points include the correct application of partial derivatives and the relationship between velocity components and the potential function. Clarifications are provided on how to express acceleration in terms of velocity and potential derivatives. The conversation concludes with the user gaining clarity on the derivation process.
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I am new in this place, am studying civil engineering in Spain,Madrid, and is something I do not understand in a theoretical exposition of the velocity/force potential.

They suppose that the external force that it acts in each point of the fluid and the speed, derive from scalar , so they admit a potential :

\frac{\vec{F}}{m}= - \vec{\nabla} U

\vec{V}= \vec{\nabla} \Omega


If the acceleration depends on the coordinates of the point and the time \vec{a} ( u',v',w') = f(x,y,z,t) :

u'= \frac{du}{dt}= \frac{\partial u}{\partial x} \frac{\partial x}{\partial t} +\frac{\partial u}{\partial y } \frac{\partial y}{\partial t} + \frac{\partial u}{\partial z}\frac{\partial z}{\partial t} + \frac{\partial u}{\partial t} and thus with the other coordinates of the acceleration


And here my doubt comes, I do not understand as they obtain to this expression:

u'= \frac{\partial^2 \Omega}{\partial x^2} \frac{\partial \Omega}{\partial x} +\frac{\partial^2 \Omega}{\partial x \partial y} \frac{\partial \Omega}{\partial y} + \frac{\partial^2 \Omega}{\partial x \partial z}\frac{\partial \Omega}{\partial z} + \frac{\partial^2 \Omega}{\partial x \partial t}




if somebody can help to understand it me, would be thanked for. Thank you very much
 
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First of all, don't use partials where they don't belong!

Now, we have
\frac{dx}{dt}=u,\frac{dy}{dt}=v,\frac{dz}{dt}=w
Thus, we may write the expression for the acceleration in the x-direction, i.e, u' as:
u'=u\frac{\partial{u}}{\partial{x}}+v\frac{\partial{u}}{\partial{y}}+w\frac{\partial{u}}{\partial{z}}+\frac{\partial{u}}{\partial{t}}

Now, insert:
u=\frac{\partial\Omega}{\partial{x}},v=\frac{\partial\Omega}{\partial{y}},w=\frac{\partial\Omega}{\partial{z}}

See if you get it right now!
 
Compound functions always was my nightmare\frac{\partial}{\partial t} \left(\frac{\partial \Omega}{\partial x}\right)=\frac{\partial}{\partial x} \left(\frac{\partial \Omega}{\partial x}\right) \underbrace{\frac{dx}{dt}}_{u}+\frac{\partial}{\partial y} \left(\frac{\partial \Omega}{\partial x}\right) \underbrace{\frac{dy}{dt}}_{v} + \frac{\partial}{\partial z} \left(\frac{\partial \Omega}{\partial x}\right) \underbrace{ \frac{dz}{dt}}_{w}+ \frac{\partial}{\partial t} \left(\frac{\partial \Omega}{\partial x}\right) \frac{dt}{dt}


that is what it did not see, thank you very much to solve the doubt to me
 
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