It is written in this article: http://www.maths.ed.ac.uk/~yktsang/4520/basic_fluid.pdf "In the continuum model of fluids, physical quantities are considered to be varying continuously in space, for example, we may speak of a velocity field ~u(~x, t) or a temperature field T(~x, t). The “local” values of such quantities at a single point P in space should be understood as average values over a small region of size Lp about P."

Then in the example he took a small volume Dx and Dy: " We now consider the various forces acting on a fluid element of sides ∆x, ∆y and ∆z in our channel flow problem"

My question is, this fluid element is it equal to Lp? If no, does the pressure in his study is an average value over a small region of size Lp ?


Yes, $\Delta x$, $\Delta y$, and $\Delta z$ all have size $L_p$. And yes, his pressure is an average over a small region of size $L_p$.

  • $\begingroup$ but how for example P(x) is an average value over a small region of Lp and we are studying over a small region? $\endgroup$
    – Tonylb1
    Oct 20 '15 at 20:23
  • $\begingroup$ For one thing, in this case he wants to find the force on the side of the cube, which is related to the average pressure, so an average of $p$ over a length scale $L_p$ makes sense here. Also, he says there should be a big separation in scale between $L_p$ and the molecular scale, so the local average of the pressure shouldn't change much if you average $p$ over a much smaller window like $L_p/100$, so you can imagine you are doing that to compute the pressure on the box. $\endgroup$ Oct 20 '15 at 20:37
  • $\begingroup$ We do the same study for hydrostatic pressure and we get p(z+dz) = p(z) + mgdz and after this equation is applied between 2 points of fluid. These 2 points also represents the average of 2 small volume of fluids but we made the study only on one small volume. $\endgroup$
    – Tonylb1
    Oct 20 '15 at 20:43
  • $\begingroup$ Can we say that dz is small such we can consider the point at position z in the fluid is the average between p(z+dz) and p(z) is p(z)? $\endgroup$
    – Tonylb1
    Oct 20 '15 at 20:46
  • $\begingroup$ You are supposed to think of $p(z)$ being the average of a box centered on $z$ and $p(z+dz)$ being the average of a box centered on $z+dz$. It is important that these will be different so you can get a net force on your box. $\endgroup$ Oct 20 '15 at 20:47

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