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Consider a fluid element referenced in conservation-of-mass.

Linear deformation
The linear deformation of a fluid element is given three elangation components and six shearing components.

Elengation Components

$e_{xx} = \frac{\partial u}{\partial x}$, $e_{yy} = \frac{\partial v}{\partial y}$, and $e_{zz} = \frac{\partial w}{\partial z}$

Shearing components
$e_{xy} = e_{yx} = \frac{1}{2} (\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x})$,
$e_{yz} = e_{zy} = \frac{1}{2} (\frac{\partial v}{\partial z} + \frac{\partial w}{\partial y})$, and
$e_{xz} = e_{zx} = \frac{1}{2} (\frac{\partial u}{\partial z} + \frac{\partial w}{\partial x})$

Volumetric deformation

The volumetric deformation is given by

$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = div \space U$

Newton’s law of friction states that the frictional forces ($\tau$) exerted is directly proportional to the velocity ($du$) and inversly proportional to the distance($dy$). The proportionality constant is known as viscosity($\mu$).

$\tau = \mu \frac{du}{dy}$

For Newtonian fluids the rate of deformation is propotional to the viscuss stress and the Newton law of viscosity for compressible flow involves two proportionality constants.

Dynamic viscosity $\mu$ related to linear deformation

Second viscosity $\lambda$ related to volumetric deformation

The above declarations defines the nine viscous stresses as below

$\tau_{xx} = 2 \mu \frac{\partial u}{\partial x} + \lambda div \space U$

$\tau_{yy} = 2 \mu \frac{\partial v}{\partial y} + \lambda div \space U$

$\tau_{zz} = 2 \mu \frac{\partial w}{\partial z} + \lambda div \space U$

$\tau_{xy} = \tau_{yx} = \mu (\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x})$

$\tau_{yz} = \tau_{zy} = \mu (\frac{\partial v}{\partial z} + \frac{\partial w}{\partial y})$

$\tau_{xz} = \tau_{zx} = \mu (\frac{\partial w}{\partial x} + \frac{\partial u}{\partial z})$

Stoke’s hypothesis, states that

$\lambda = - \frac{2}{3} \mu$

So the linear elongation stresses can be re-written as below

$\tau_{xx} = 2 \mu \frac{\partial u}{\partial x} - \frac{2}{3} \mu \space div \space U$

$\tau_{yy} = 2 \mu \frac{\partial v}{\partial y} - \frac{2}{3} \mu \space div \space U$

$\tau_{zz} = 2 \mu \frac{\partial w}{\partial z} - \frac{2}{3} \mu \space div \space U$

The below x-momentum equation is taken from Momentum equation and forces.

$\begin{aligned} \rho \frac{Du}{Dt} = \frac{\partial (-p + \tau_{xx})}{\partial x} + \frac{\partial (\tau_{yx})}{\partial y} + \frac{\partial (\tau_{zx})}{\partial z} + S_{Mx} \end{aligned}$

When we substitute the stress values in the above equation

$\begin{aligned} \rho \frac{Du}{Dt} &= -\frac{\partial p}{\partial x} + \frac{\partial}{\partial x}[2 \mu \frac{\partial u}{\partial x} - \frac{2}{3} \mu \space div \space U] \\ &+ \frac{\partial}{\partial y} [\mu (\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x})] \\ &+ \frac{\partial}{\partial z} [\mu (\frac{\partial w}{\partial x} + \frac{\partial u}{\partial z})]\\ &+ S_{Mx}\\ &= -\frac{\partial p}{\partial x} - \frac{\partial}{\partial x}(\frac{2}{3} \mu \space div \space U)\\ &+ \frac{\partial}{\partial x} [\mu \frac{\partial u}{\partial x}] + \frac{\partial}{\partial y} [\mu \frac{\partial u}{\partial y}] + \frac{\partial}{\partial z} [\mu \frac{\partial u}{\partial z}]\\ &+ \frac{\partial}{\partial x} [\mu \frac{\partial u}{\partial x}] + \frac{\partial}{\partial y} [\mu \frac{\partial v}{\partial x}] + \frac{\partial}{\partial z} [\mu \frac{\partial w}{\partial x}]\\ &+ S_{Mx}\\ &= -\frac{\partial p}{\partial x} - \frac{\partial}{\partial x}(\frac{2}{3} \mu \space div \space U)\\ &+ div(\mu \space grad \space u)\\ &+ \frac{\partial}{\partial x} [\mu \frac{\partial u}{\partial x}] + \frac{\partial}{\partial y} [\mu \frac{\partial v}{\partial x}] + \frac{\partial}{\partial z} [\mu \frac{\partial w}{\partial x}]\\ &+ S_{Mx}\\ \end{aligned}$

For incompressible fluid the $div \space u = 0$
When the function is set to be continuous in an Open Set, then, $\frac{\partial}{\partial x} [\frac{\partial}{\partial y}] = \frac{\partial}{\partial y} [\frac{\partial}{\partial x}]$

When we apply the above two conditions, we get

$\begin{aligned} \rho \frac{Du}{Dt} &= -\frac{\partial p}{\partial x} + div(\mu \space grad \space u) + S_{Mx} \end{aligned}$

Similarly we can write the y-momentum and z-momentum equations as below

$\begin{aligned} \rho \frac{Dv}{Dt} &= -\frac{\partial p}{\partial y} + div(\mu \space grad \space v) + S_{My} \end{aligned}$

$\begin{aligned} \rho \frac{Dw}{Dt} &= -\frac{\partial p}{\partial z} + div(\mu \space grad \space w) + S_{Mz}\\ \end{aligned}$

And the Internal Energy Equation can be written as

$\begin{aligned} \rho \frac{Di}{Dt} &= -p \space div \space U + div(k \space grad \space T) + \Phi + S_{i}\\ \end{aligned}$

Where, $\Phi$ is the dissipation function and it is a non-negative function, since it consists of squared terms.

$\begin{aligned} \Phi &= \mu \lbrace \space 2 \lbrack (\frac{\partial u}{\partial x})^2 + (\frac{\partial v}{\partial y})^2 + (\frac{\partial w}{\partial z})^2 \rbrack \\ &+ (\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x})^2 + (\frac{\partial v}{\partial z} + \frac{\partial w}{\partial y})^2 + (\frac{\partial u}{\partial z} + \frac{\partial w}{\partial x})^2 \\ & - \frac{2}{3} \mu (div \space U)^2 \space \rbrace \end{aligned}$