Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Fluid Mechanics question,velocity potential

  1. Aug 15, 2007 #1
    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 :

    [tex]\frac{\vec{F}}{m}= - \vec{\nabla} U[/tex]

    [tex]\vec{V}= \vec{\nabla} \Omega[/tex]

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

    [tex]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}[/tex] and thus with the other coordinates of the acceleration

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

    [tex]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}[/tex]

    if somebody can help to understand it me, would be thanked for. Thank you very much
  2. jcsd
  3. Aug 15, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    First of all, don't use partials where they don't belong!

    Now, we have
    Thus, we may write the expression for the acceleration in the x-direction, i.e, u' as:

    Now, insert:

    See if you get it right now!
  4. Aug 15, 2007 #3
    Compound functions always was my nightmare

    [tex]\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}[/tex]

    that is what it did not see, thank you very much to solve the doubt to me
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook