I Viscosity Term in Navier-Stokes Equation

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I'm a bit confused about the viscosity term in the Navier-Stokes equation; my intuitive understanding of what it would is different from what it actually is.

I took the z component of the stress on an infinitesimal cube, but the same approach should apply in the x and y direction. I think my method might be a bit sloppy (maybe leading to the error), but I'm not sure how to be more precise. I end up missing the third term of the Laplacian, the second derivative of the z component of velocity with respect to z. What am I missing?


On the left side of the cube, the shear is:
##-\mu \frac {\partial v_z} {\partial y}##

and on the right side of the cube, the shear is:
##\mu (\frac {\partial v_z} {\partial y}+\Delta \frac {\partial v_z}{\partial y})##

Similarly, on the back, the shear is:
##-\mu \frac {\partial v_z} {\partial x}##

and on the front side, the shear is:

##\mu (\frac {\partial v_z} {\partial x}+\Delta \frac {\partial v_z}{\partial x})##

I then added up the shear stresses giving the following:
##\mu (\Delta \frac {\partial v_z}{\partial x}+\Delta \frac {\partial v_z}{\partial y})##
##\mu ( \frac {\partial^2 v_z}{\partial x^2}dx+\frac {\partial^2 v_z}{\partial y^2} dy)##

I also know that:
##a=\frac{F}{m}=\frac{\tau dA}{\rho dV}##

The differential lengths cancel, leaving me with:
\partial^2 v_z}{\partial x^2}+ \frac {\partial^2 v_z}{\partial y^2} )
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On the left side of the cube, the shear is:
##-\mu \frac {\partial v_z} {\partial y}##
This is not correct. The general equation for this shear stress according to the 3D version of the constitutive equation for a Newtonian fluid is $$\mu\left(\frac{\partial v_z}{\partial y}+\frac{\partial v_y}{\partial z}\right)$$
For the derivation of the tensorially correct form of the constitutive equation for a Newtonian fluid in 3D, see section 1.2 of Transport Phenomena by Bird, Stewart, and Lightfoot, Generalization of Newton's Law of Viscosity.
Based on the book, the correct equations should be:
$$\tau_{xz}=\mu(\frac{\partial v_z}{\partial x}+\frac{\partial v_x}{\partial z})$$
$$\tau_{yz}=\mu(\frac{\partial v_z}{\partial y}+\frac{\partial v_y}{\partial z})$$
$$\tau_{zz}=\mu(\frac{\partial v_z}{\partial z}+\frac{\partial v_z}{\partial z})$$

The book doesn't seem to explain how to relate this to acceleration, but I imagine I would use the same method I tried before, adding the shears on either side. I think the terms on the left side of the sum would give me the correct viscous term. This would mean the terms on the right would have to add up to 0, which I think it does if you assume it's incompressible:

$$\frac{\partial^2 v_x}{\partial x \partial z}+\frac{\partial^2 v_y}{\partial y \partial z}+\frac{\partial^2 v_z}{\partial z^2}=0$$
$$\frac{\partial}{\partial z}(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z})=0$$

The book explains that because under pure rotation there should be no shear, the terms must be symmetric, which I think helps me understand what's going on. The equations for stress are those for rate of change of strain times ##\mu##. In the rotation example, there's clearly no strain (as it's rotating as a rigid body), so no stress. So to find the shear stress, you need the rate of shear strain, and so you need both terms for how velocity varies. I think my confusion was from the fact I was using ##\tau=\mu\frac{du}{dy}## (from the classic intro example of a plate moving over a thin layer of fluid), which isn't general enough for this.

However, the equation for normal stress is still a bit confusing to me. The book uses an example of fluid moving through a contracting pipe; the velocity change leads to viscous stresses. I understand that there is deformation, and the changing deformation could lead to stresses due to viscous forces. However, it still seems to be off by a factor of two (why do you multiply ##\frac{\partial v_z}{\partial z}## by 2?).
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I came up with an explanation which I think makes sense but I'm not sure if it is correct. First I imagined a square within a square (as shown in the photo), and then I thought about the shear strain of the small square caused by the normal strain of the large square.


I called the length of a side of the large square ##s##. $$\frac{d\epsilon}{dt}=\frac{1}{s}\frac{dy}{dt}=\frac{\partial v_y}{\partial y}$$ Because the fluid is incompressible, there would be the following strain in the x direction (assuming there is no strain in the z direction):$$\frac{1}{s}\frac{dy}{dt}=-\frac{\partial v_x}{\partial x}$$

I knew that the normal strain would produce a change in angle in the smaller square, and I derived the connection in the following way: $$\tan (\theta)=\frac {y}{x}$$ $$\frac{d \tan(\theta)}{dt}=\frac{dy}{dt}x^{-1}+y(-1)x^{-2}\frac{dx}{dt}=\frac{\partial v_y}{\partial y}+\frac{\partial v_y}{\partial y}=2\frac{\partial v_y}{\partial y}$$

And because:
$$\frac{d \tan(\theta)}{dt}=\frac{\partial v_x}{\partial y} +\frac{\partial v_y}{\partial x}$$

It must be true that: $$\frac{\partial v_x}{\partial y}+\frac{\partial v_y}{\partial x}=2\frac{\partial v_y}{\partial y}$$

This would then relate the normal strain in the large square to the shear strain in the small square.

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