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I'm trying to follow the book Turbulence by Davidson. Currently I'm having trouble in converting the differential NS equation to its integral form but I cannot see clearly how the Divergence theorem is applied to obtain it.

Given that the NS equation is:

$$\frac{Du}{Dt} =\frac{\partial u}{\partial t} + (u\cdot\nabla)u= -\nabla\left(\frac{P}{\rho}\right) + \nu\nabla^2u $$

And knowing that the divergence theorem gives

$$\oint F\cdot dS = \iiint \nabla\cdot F\,dV$$

Therefore the integral form will be:

$$ \frac{\partial }{\partial t} \iiint \rho u_i\,dV = -\oint u_i(\rho u\cdot dS) - \oint PdS +\text{viscous term}$$

I've been trying for hours to convert $ \iiint(u\cdot \nabla)u\,dV$ to $\oint u_i(\rho u\cdot dS) $ but I just can't seem to get it.

I have found that there is an identity:

$$\rho u\cdot\nabla u + \rho u \nabla\cdot u = \nabla\cdot(\rho u u ) $$

from this Physics Forums thread, but I have no idea where it comes from.

and since mass is conserved, it becomes

$$\rho u\cdot\nabla u = \nabla\cdot(\rho u u ) $$

so I've basically focused my attention on trying to convert $\nabla\cdot(\rho u u ) $ to $\oint u_i(\rho u\cdot dS)$ but my question is how come one of the $u$ can be broken off and placed at the front with the subscript $i$ and why is it $(pu\cdot dS)$ and not $(puu\cdot dS)$ instead?

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2 Answers 2

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Answer to question1: ui uj, the product is a tensor of 9 components, so Gauss Divergence theorem for a tensor has to be used rather than the one you used for a vector. In this case, since the tensor is symmetric, ui can come out and dangle outside the divergence operator. The image from Kundu, Fluid Mechanics. Gauss Divergence theorem!

Answer to question2: Look at page 35 of the turbulence book you mentioned, since ds is specified as a surface integral it is indeed integration over an area, and not over a line. Also the circular symbol on the integral specifies it is a closed integral, means the surface over which this integral is evaluated is closed, which is exactly what happens on a control volume. Rest of the derivation is same as you have derived. Turbulence book!

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  • $\begingroup$ Hello! It is preferable to type out screenshots or images of text; for formulae, one can use MathJax. Thanks! $\endgroup$
    – Jonas
    Dec 4, 2021 at 10:11
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Before proceeding, let us prove the identity $\nabla\cdot(\vec{a}\vec{b}) = \vec{b}\cdot\nabla\vec{a} + \vec{a}\nabla\cdot\vec{b}$. The derivation becomes easy when we use the cartesian tensor form, \begin{equation} \frac{\partial}{\partial x_j}(a_ib_j) = b_j\frac{\partial a_i}{\partial x_j} + a_i\frac{\partial b_j}{\partial x_j} \end{equation} If we now put $\vec{a} = \rho\vec{u}$, $\vec{b} = \vec{u}$ and assume that $\rho$ is a constant then we have $\nabla\cdot(\rho\vec{u}\vec{u}) = \rho\vec{u}\cdot\nabla\vec{u} + \rho\vec{u}\nabla\cdot\vec{u}$. If the fluid is incompressible, then $\nabla\cdot(\rho\vec{u}\vec{u}) = \rho\vec{u}\cdot\nabla\vec{u}$. Thus, \begin{equation} \iiint\nabla\cdot(\rho\vec{u}\vec{u})dV = \iiint \rho\vec{u}\cdot\nabla\vec{u}dV \end{equation} The integral on the left hand side can be transformed to surface integral and \begin{equation} \iint \rho\vec{u}\vec{u}\cdot\hat{n}dS = \iiint \rho\vec{u}\cdot\nabla\vec{u}dV. \end{equation}

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  • $\begingroup$ I'm confused as to why when substituting $ a = \rho u $ and $ b = u $ , how come the LHS of the identity involves a cross product instead of a dot product? Also, I get exactly the same final answer as you but our answers don't match that from the book? We both get $ \iint \rho uu . n dS $ instead of $ \oint u_i ( \rho u. dS)$ $\endgroup$
    – Flow
    Jul 2, 2019 at 6:23
  • $\begingroup$ Sorry, it was a typo. I am correcting it. $\endgroup$
    – Amey Joshi
    Jul 2, 2019 at 9:12

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