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Reynolds Transport Theorem and Conservation Laws.

  1. Aug 12, 2007 #1
    The Reynolds Transport theorem (RTT) is usually applied to derive the conservation of mass, momentum and energy in a fluid. But when I try to apply the RTT to other physical quantities, I get weird results. Can anyone see where I'm going wrong?

    As a simple example, take the physical quantity to be position, [itex]x[/itex]. Assume that the fluid density [itex]\rho[/itex] and velocity [itex]u[/itex] are independent of time, so that the time derivative of the volume integral [itex]\int \rho\,x\,dV[/itex] is zero. Applying the RTT to [itex]\rho\,x[/itex] gives,

    \frac{\partial\,\rho\,x}{\partial\,t} + \nabla\cdot(\rho\,x\,u) = 0

    which can be rearranged as,

    \left[\frac{\partial\,\rho}{\partial\,t} + \nabla\cdot(\rho\,u)\right]\,x
    + \rho\left\{\frac{\partial\,x}{\partial\,t} + u\cdot\nabla\,x\right\}

    This can be simplified by applying the mass equation,

    \frac{\partial\,\rho}{\partial\,t} + \nabla\cdot(\rho\,u) = 0

    to yield the result,

    \rho\,D_t(x) = 0

    where [itex]D_t[/itex] is the material derivative,

    D_t = \frac{\partial}{\partial\,t} + u\cdot\nabla

    I think that [itex]\partial\,x/\partial\,t = 0[/itex] and [itex]\nabla\,x = I[/itex], where [itex]I[/itex] is the identity matrix, in which case the result simplifies to,

    \rho\,u = 0

    This is incorrect:confused: [itex]u[/itex] does not need to be zero in order for the time derivative of [itex]\int \rho\,x\,dV[/itex] to be zero.
  2. jcsd
  3. Aug 12, 2007 #2


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    Saying that [itex]\int \rho\,x\,dV[/itex] is zero, makes no sense, especially if one expands dV into dx dy dz.

    One cannot transport position.

    See - http://en.wikipedia.org/wiki/Reynolds_transport_theorem


    http://www.math.gatech.edu/~ho/Transphy.pdf - in the first paragraph [tex]\phi(x,t)[/tex] should read [tex]F(x,t)[/tex] which is the integrand in the equation below.

    I think one is misapplying RTT.

    The idea is that the rate of change in mass or other physical quantity like energy depends on the flow (transport) of mass or other physical quantity across the boundary of that volume. Position is not transported.
  4. Aug 13, 2007 #3
    Astronuc, thanks for your comments. I was actually claiming that the time-derivative of [itex]\int \rho\,x\,dV[/itex] is zero, rather than the integral itself. I have subsequently realised that the time-derivative is not zero; perhaps that's what you meant? I'll give a longer reply in the next post.
  5. Aug 13, 2007 #4
    Frame of reference for the Reynolds Transport Theorem

    I think I've more or less got to grips with this problem. The probem arose because I misunderstood the frame of reference for the Reynolds Transport Theorem (RTT).

    The RTT is simply obtained by differentiating a volume integral with respect to time,

    [tex]\frac{d\,\int F(x,t)\,dV}{dt} = \int\,\frac{\partial\,F}{\partial\,t} + \nabla\cdot(F\,u)\,dV[/tex]

    where [itex]u[/itex] is the velocity. It may be applied to any function [itex]F(x,t)[/itex], including [itex]F = x[/itex].

    In order to convert the integral expression into a differential equation, one must make a physical postulate for the value of [itex]d\,\int F(x,t)\,dV\,/dt [/itex] . For the mass equation without source terms, it is set equal to zero. I think that source terms are equivalent to a non-zero value for the integral, though I haven't figured out a quantitative relation between the source and the integral.

    A key point which is not made entirely clear in many discussions of the RTT is whether the control volume V is fixed in space, or whether it is transported by the fluid (so that it changes shape and volume) ?

    Some texts, (eg Flanders, Differential Forms with applications to the physical sciences, Dover edition, pg 188), claim that V is fixed in space, and introduce the boundary-flux term [itex]\int \nabla\cdot(F\,u)\,dV[/itex] in a heuristic manner. Other texts claim that V is transported by the fluid, and obtain the boundary-flux term automatically when evaluating [itex]dV(t)/dt[/itex].

    The correct answer is that the volume is transported, ie [itex]V = V(t)[/itex]. For a crisp derivation, try a text with a differential-geometry slant, eg Frankel, The Geometry of Physics, 1st edition, Sec. 4.3 and Eq. 3.17.

    At the start of this thread, I set [itex] F = \rho\,x[/itex], and postulated that [itex]d\,\int \rho\,x\,dV\,/dt = 0 [/itex]. This postulate was based on the incorrect fixed-volume frame. If one adopts the correct transported-volume frame, one sees that as the volume is transported from one location to another, the average value of [itex]x[/itex] within the volume will change, ie [itex]x = x(t)[/itex], and hence the integral is time-dependent and its time-derivative is non-zero.
    Last edited: Aug 13, 2007
  6. Aug 13, 2007 #5


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    I was careless. I focussed on the integral, rather than the time derivative thereof. The use of x (location) still does not make sense though.

    Please refer to this discussion - http://en.wikipedia.org/wiki/Physical_quantity

    Location is not a physical attribute, but size (length) is.

    RTT is about the change of a physical property within a control volume.

    I'll comment on the other post later.
  7. Aug 14, 2007 #6
    I agree with Astronuc. You are transporting something. Velocity is not a physical something. I would suggest you read through a proof of RTT because you are going to take the dot product of b and the local velocity vector, i.e. the flux. It would make no sense to have the b = velocity and dot that with itself, would it? (b = an extensive parameter) [Velocity is not an extensive parameter].
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