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Why Reynolds Stress vanishes on boundary of closed volume?

  1. Jul 6, 2016 #1
    The rate of working of the Reynolds Stress can be written as:


    where ui is the fluctuating velocity and Ūi is the time-averaged velocity.

    It is stated in the textbook that, if we integrate the above equation over a closed volume V, the divergence term on the left integrates to zero since τRij (Reynolds Stress) vanishes on the boundary. What does this mean?

    The context is that, with this being zero, the author proves that globally, the integral over the closed volume of the two terms on the right must balance. Maybe if I understood the latter statement, I would understand this last sentence...?

  2. jcsd
  3. Jul 7, 2016 #2
    Do you agree that the Reynolds stresses vanish at the boundary?
    If yes, then what is the integral of the divergence of a quantity Q, when Q=0 at the boundary of the domain of integration? The divergence theorem will be helpful here.
    Do you agree that the integral of the first term in the equation is zero?
  4. Jul 8, 2016 #3
    Not really. If we are considering a small control volume within the flow domain, then I don't see why it should 'vanish'. We could specify a whole grid of these control volumes throughout the whole flow domain – the idea that the Reynolds Stress 'vanishes' is confusing to me.

    I understand that, if it does vanish on the boundary, then there can be no 'flux' of this quantity out of the control volume, and hence the surface integral would be zero, and from div. theorem, the volume integral would be zero too.

    If my above statement is correct, then it's the physical concept of this quantity (or indeed any other) 'vanishing' on the boundary that I don't understand.
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