In some artictle I've found Navier-Stokes Momentum Equation written in a following form
$$\frac{\partial u}{\partial t}+\text{div}(u\otimes u)+\nabla p – \nu\Delta u =f$$
Since usually it appears with $u\cdot \nabla u $ term instead of $\text{div}(u\otimes u)$. It all came up from using following form of the Newton Law $$\underbrace{\frac{d}{dt}\int_B(\rho u)(t,\cdot)dx}_{\text{change of the linear momentum}}=\underbrace{-\int_{\partial B} (\rho u\otimes u )(t,\cdot)n(\cdot)dS}_{\text{flux of the momentum through boundary}}+\underbrace{F_B}_{\text{applied force}}\\
F_B=\underbrace{\int_B(\rho f)(t,\cdot)dx}_{\text{volume forces}}+\underbrace{\int_{\partial B}\text{T}(t,\cdot)n(\cdot)dS}_{\text{surface forces (the tension)}}$$
Then we end up with integral form
$$\int_B\Big(\frac{\partial}{\partial t}(\rho u)(t,\cdot)+\text{div}(\rho u\otimes u)(t,\cdot)-(\rho f)(t,\cdot)-\text{divT}(t,\cdot)\Big) dx=0.$$
Defining stress tensor $T$ and assuming constant density we end up with the first equation, but since I'm not really fluent with tensors I could use some help understanding the difference between this formulation and the original one, and how to make calculations to get one form another.
Thank you!
Best Answer
The most generic form of the Navier-Stokes equation is $$ \frac{\partial (\rho \mathbf{u})}{\partial t} + \nabla \cdot(\rho \mathbf{u} \otimes \mathbf{u}) = \nabla p - \mathbf{f} + \nabla \cdot \mathbf{S}, \tag{1} $$ in which $\mathbf{S}$ is the shear stress tensor ($\mathbf{S}=\mu\nabla \mathbf{u}$ in your case). The continuity equation is $$ \frac{\partial \rho}{\partial t}+\nabla \cdot (\rho \mathbf{u})=0. \tag{2} $$ Using the product rule for derivatives in equation $(1)$ leads to $$ \rho \frac{\partial \mathbf{u}}{\partial t} + \mathbf{u}\frac{\partial \rho}{\partial t} + \rho \mathbf{u} \cdot \nabla \mathbf{u} + \mathbf{u} \nabla \cdot(\rho \mathbf{u}) = \nabla p - \mathbf{f} + \nabla \cdot \mathbf{S}, \tag{3} $$ and using the continuity equation we see that the second and fourth terms in LHS cancel each other, leading to $$ \rho\left( \frac{\partial \mathbf{u}}{\partial t} + \mathbf{u} \cdot \nabla \mathbf{u} \right) = \nabla p - \mathbf{f} + \nabla \cdot \mathbf{S}. \tag{4} $$ Since it's usually assumed that the continuity equation holds, equation $(4)$ is completly equivalent to equation $(1)$. In practice, however, there are some "differences":
Equation $(1)$ is called the conservative form of Navier-Stokes equation, while equation $(4)$ is called the non-conservative form. These names are a bit misleading: both equations are the momentum conservation equations. However, when solving numerically the governing equations of fluid dynamics, it's sometimes more useful to use equation $(1)$. It's basically due to the fact that accross a shock wave the velocity $\mathbf{u}$ is discontinuous (and, therefore, equation $(4)$ has the gradient of a discontinuous function), while $\rho \mathbf{u} \otimes \mathbf{u}$ is continuous even accross the shock. See that equation $(1)$ is called conservative form because it has derivatives of a conserved quantity (the momentum, i.e., $\rho \mathbf{u}$) while equation $(4)$ has derivatives of a non-conserved quantity (the velocity).
Equation $(4)$ explicitly shows the transport of momentum in the term $\mathbf{u} \cdot \nabla \mathbf{u}$. Notice that the conservation equation of the property $\phi$ (which can be enthalpy, vorticity, chemical species, etc.) will have a term $\mathbf{u} \cdot \nabla \phi$. Therefore, equation $(4)$ "looks like" every other conservation equation. It's usually defined the material derivative of property $\phi$ as $$ \frac{D \phi}{Dt} = \frac{\partial\phi}{\partial t} + \mathbf{u} \cdot \nabla\phi, $$ which can be interpreted as the rate of change of the property $\phi$ along time in a particle of fluid while this particle is transported by the flow. Then, the conservation equation of any property $\phi$ can be written generically as $$ \frac{D \phi}{Dt} = \text{source terms}, $$ in which the source terms for the case $\phi=\mathbf{u}$ (i.e., the Navier-Stokes equation) are $(\nabla p - \mathbf{f} + \nabla \cdot \mathbf{S})/\rho$.
Summarizing: both equations are the same. When you need to solve them numerically, equation $(1)$ can be more suitable. If you want to interpret the physical meaning of the terms of Navier-Stokes equation, equation $(4)$ is more suitable.