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Fluid Mechanics Question: erivation of Momentum EQ

  1. Nov 1, 2009 #1
    Fluid Mechanics Question::Derivation of Momentum EQ

    I have been doing some serious review of my FM. I am working my text from cover to cover in an attempt to solidify what I 'learned' in my 6-week summer FM class. What a load of crap that was.

    I am attempting to follow the text's derivation of the Differential form of the Conservation of linear momentum. I will start here from the beginning for those who would like to help, but might not quite remember the derivation off the top of their head :wink:

    The figure below shows a differential volume element. It was the only one available, so it shows the mass flux instead of the momentum flux on the faces of the element.

    Picture8-2.png

    In words, we have the momentum equation:

    [sum of forces] = [rate of change of momentum in CV] + [momentum flux out of CV] - [momentum flux into CV]

    Let (u, v, w) be the scalar (x, y, z) components of velocity of the fluid passing through the differential Control Volume (CV).

    [tex]
    \sum \mathbf{F} =
    \frac{\partial}{\partial{t}}(\rho\mathbf{V})dx dy dz +
    \sum(\dot{m}\mathbf{V})_{out} -
    \sum(\dot{m}\mathbf{V})_{in}
    [/tex]

    OR

    [tex]
    \sum \mathbf{F} =
    [\frac{\partial}{\partial{t}}(\rho\mathbf{V}) +
    \frac{\partial}{\partial{x}}(\rho u \mathbf{V}) +
    \frac{\partial}{\partial{y}}(\rho v \mathbf{V}) +
    \frac{\partial}{\partial{z}}(\rho w \mathbf{V})]
    dx dy dz
    \,\,\,\,\,\,(1)[/tex]

    looking at only the term in brackets from (1) we have ... (this is where I lose them)

    [tex]
    \frac{\partial}{\partial{t}}(\rho\mathbf{V}) +
    \frac{\partial}{\partial{x}}(\rho u \mathbf{V}) +
    \frac{\partial}{\partial{y}}(\rho v \mathbf{V}) +
    \frac{\partial}{\partial{z}}(\rho w \mathbf{V})
    =
    [\frac{\partial{\rho}}{\partial{t}} + \nabla\cdot(\rho\mathbf{V}] +
    \rho(\frac{\partial{\mathbf{V}}}{\partial{t}} +
    u\frac{\partial{\mathbf{V}}}{\partial{x}} +
    v\frac{\partial{\mathbf{V}}}{\partial{y}} +
    w\frac{\partial{\mathbf{V}}}{\partial{z}} )
    \,\,\,\,\,\,\,(2)[/tex]

    I have no idea how they got to the last line.

    Can we just look at one term here at a time? Let's look at just the very first term on the left hand side (LHS) of (2) [itex]\partial({\rho\mathbf{V}})/\partial{t}[/itex]

    It appears that this is the last term in the RHS of (2) that is:

    [tex]
    \frac{\partial}{\partial{t}}(\rho\mathbf{V})=
    u\frac{\partial{\mathbf{V}}}{\partial{x}} +
    v\frac{\partial{\mathbf{V}}}{\partial{y}} +
    w\frac{\partial{\mathbf{V}}}{\partial{z}}
    [/tex]

    I don't understand. It appears that they have assumed that [itex]\rho[/tex] is constant across the differential Control Volume. I don't like this. Here is why:

    a) They did not state that this was an assumption

    b) If we are assuming that [itex]\rho[/tex] is constant across the CV, then why don't we
    just assume that V is also constant across the CV and call it a day?

    Am I missing something else? Is there some other way that they could have arrived at

    [tex]
    \frac{\partial}{\partial{t}}(\rho\mathbf{V})=
    \rho(u\frac{\partial{\mathbf{V}}}{\partial{x}} +
    v\frac{\partial{\mathbf{V}}}{\partial{y}} +
    w\frac{\partial{\mathbf{V}}}{\partial{z}})
    [/tex]

    without assuming constant density?

    Let's talk about this.

    Thanks,
    Casey

    EDIT:
    I see that I have made an error. The 1st term on the LHS of (2) is the partial with respect to time and therefore does not
    equal the last term on the RHS of (2) which is clearly the total derivative with respect to time.

    BUT it still looks like they are factoring out the density which doesn't make sense to me. Shouldn't we have to use the
    product rule on the term

    [tex]\frac{\partial}{\partial{t}}(\rho\mathbf{V})[/tex]

    since both rho and V are functions of (x, y, z, t) ?
     
    Last edited: Nov 1, 2009
  2. jcsd
  3. Nov 1, 2009 #2
    Re: Fluid Mechanics Question::Derivation of Momentum EQ

    Yet another one for the archives ...
     
  4. Nov 2, 2009 #3

    minger

    User Avatar
    Science Advisor

    Re: Fluid Mechanics Question::Derivation of Momentum EQ

    Here's what I have from my notes. Looking at the control volume, the net surface forces in the x-direction are:
    [tex]
    \begin{equation}
    \begin{split}
    \left[ p - \left( p + \frac{\partial p}{\partial x}dx\right)\right]\,dy\,dz + \left[\left(\tau_{xx} + \frac{\partial \tau_{xx}}{\partial x}dx\right) - \tau_{xx}\right]\,dy\,dz + \cdots \\
    \left[\left(tau_{yx} + \frac{\partial \tau_{yx}}{\partial y}dy\right)-\tau_{yx}\right]\,dx\,dz + \left[\left(\tau_{zx} + \frac{\partial \tau_{zx}}{\partial z}dz\right)-\tau_{zx}\right]\,dx\,dy
    \end{split}
    \end{equation}
    [/tex]
    So the total of surface and body forces are:
    [tex]
    \begin{equation}
    F_x = \underbrace{\left[-\frac{\partial p}{\partial x} + \frac{\partial \tau_{xx}}{\partial x} + \frac{\partial \tau_{yx}}{\partial y} + \frac{\partial \tau_{zx}}{\partial z}\right]\,dx\,dy\,dz}_{\mbox{Surface Forces}} + \underbrace{\rho f_x\,dx\,dy\,dz}_{\mbox{Body Forces}}
    \end{equation}
    [/tex]
    We have found the force, now from Newton's second law, it gets equated to mass and acceleration. Mass is:
    [tex]
    m = \iiint_S \rho\,dS = \rho\,dx\,dy\,dz
    [/tex]
    While acceleration is:
    [tex]
    a = \frac{dV}{dt}
    [/tex]
    The acceleration of a fluid element is its time rate of change of velocity. Since we are in the Lagrangian frame of reference, we'll use the total derivative
    [tex]
    a_x = \frac{Du}{Dt}
    [/tex]
    After we note that dx dy dz on the left cancel out with the mass term on the right, we get:
    [tex]
    \rho\frac{Du}{Dt} -\frac{\partial p}{\partial x} + \frac{\partial \tau_{xx}}{\partial x} + \frac{\partial \tau_{yx}}{\partial y} +\frac{\partial \tau_{zx}}{\partial z} + \rho f_x
    [/tex]
    The conservative form is (I can show proof if needed) then:
    [tex]\rho\frac{Du}{Dt} = \frac{d(\rho u)}{dt} + \vec{\nabla}\cdot(\rho u\vec{V})
    [/tex]
     
  5. Nov 3, 2009 #4
    Re: Fluid Mechanics Question::Derivation of Momentum EQ

    I saw a similar approach in my book by Anderson. I know that I just need to move on, but I had hoped to work out the math in my OP. That is, I would really lime to understand how we went from the LHS of (2) to the RHS of (2).

    The part that is really getting me is this: how did they get around using the product rule on this

    [tex]
    \frac{\partial}{\partial{t}}(\rho\mathbf{V})
    [/tex]

    ?

    Thanks for looking at this minger.

    ~Casey
     
  6. Nov 4, 2009 #5
    Re: Fluid Mechanics Question::Derivation of Momentum EQ

    There appears to be an error on the RHS of equation (2). The first quantity in brackets should be multipled by the vector V. I just worked it out and wrote down the intermediate steps. See the attached pdf file.
     

    Attached Files:

  7. Nov 4, 2009 #6
    Re: Fluid Mechanics Question::Derivation of Momentum EQ

    So you did use the chain rule. For some reason I started doing exactly what you did, but then I stopped. I think I was intimidated by the fact that all of the variables are functions of (x,y,z,t).

    But now looking at it, it is not that bad since we are holding x,y,z constant. It's derivations like these that make me wish I had just a little time off to do some serious review of my calculus. Perhaps things would come more 2nd nature.

    Thanks for your time guys :smile:
     
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