# I Derivation of non-dimensional Navier Stoke equation

1. Jun 17, 2017

### ussername

Take the first three terms of Navier Stoke equation:
$$\rho \cdot \left ( v_{x}\cdot \frac{\partial \vec{v}}{\partial x} + v_{y}\cdot \frac{\partial \vec{v}}{\partial y} + v_{z}\cdot \frac{\partial \vec{v}}{\partial z}\right )$$

Define the length $v$ of the velocity vector field:
$$\vec{v}=v\cdot \vec{v}^{0}$$
where $\vec{v}^{0}$ is the unit vector field with same direction as $\vec{v}$. Thus it is:$$v_{x}=v\cdot v_{x}^{0}$$$$v_{y}=v\cdot v_{y}^{0}$$$$v_{z}=v\cdot v_{z}^{0}$$

Now both terms $v\cdot \vec{v}^{0}$ can change with $x$ coordinate, so their derivation is:
$$\frac{\partial (v\cdot \vec{v}^{0})}{\partial x}=\vec{v}^{0}\cdot \frac{\partial v}{\partial x}+v\cdot \frac{\partial \vec{v}^{0}}{\partial x}$$

When I substitute these derivations into the NS equation (1. equation), I get something like this:
$$\rho \cdot\vec{v}^{0}\cdot v\cdot \left ( \vec{v}^{0}\cdot \left ( \frac{\partial v}{\partial x} , \frac{\partial v}{\partial y} , \frac{\partial v}{\partial z}\right ) \right )+\rho \cdot v^{2}\left ( \vec{v}^{0}\cdot \left ( \frac{\partial \vec{v}^{0}}{\partial x} , \frac{\partial \vec{v}^{0}}{\partial y} , \frac{\partial \vec{v}^{0}}{\partial z}\right ) \right )$$

But according to the literature it should be just the second term:
$$\rho \cdot v^{2}\left ( \vec{v}^{0}\cdot \left ( \frac{\partial \vec{v}^{0}}{\partial x} , \frac{\partial \vec{v}^{0}}{\partial y} , \frac{\partial \vec{v}^{0}}{\partial z}\right ) \right )=\rho \cdot v^{2}\cdot \left ( \vec{v}^{0}\cdot \left ( \mathrm{div} \vec{v}^{0} \right ) \right )$$

2. Jun 17, 2017

### Orodruin

Staff Emeritus
I don't think you are understanding it right. If the point is to get rid of the dimensions then $v$ should be some constant reference speed, not the speed of the fluid at each point.

Edit: I also suggest you stop using $\cdot$ when you don't mean scalar product, it makes your post difficult to read.

3. Jun 17, 2017

### ussername

So what is the definition of the unit velocity vector?

Edit: If $v$ is the average constant speed, than the length of $\vec{v}^0$ is not generally unity.

Last edited: Jun 17, 2017
4. Jun 17, 2017

### Orodruin

Staff Emeritus
Indeed. It is just a dimensionless velocity. Note that $v$ can be any fixed non-zero speed. You would typically fix it to some speed that appears in your problem, such as the speed of the flow at infinity.

5. Jun 21, 2017

### ussername

The non-dimensional Navier Stokes equation is:
$$\frac{L\cdot f}{v}\frac{\partial \vec{v}^*}{\partial \vec{\tau}^*}+ v_x^*\frac{\partial \vec{v}^*}{\partial x^*}+ v_y^*\frac{\partial \vec{v}^*}{\partial y^*}+v_z^*\frac{\partial \vec{v}^*}{\partial z^*}=\frac{g\cdot L}{v^2}\vec{g}^*+\frac{p^0}{\rho \cdot v^2} \bigtriangledown^*p^*+\frac{\eta }{L\cdot v\cdot \rho } \bigtriangledown^{*2}\vec{v}^*$$
I understand that by setting physical parameters ($v,\rho,L...$) I can achieve in my scaled flow e.g. identical ratio of friction force to net force as in the original flow.

What I don't understand is how can I determine which terms in this equation are negligible. I know the physical dimensional parameters ($v,\rho,L...$) but I generally don't know the non-dimensional vectors ($v_x^*\frac{\partial \vec{v}^*}{\partial x^*}+ v_y^*\frac{\partial \vec{v}^*}{\partial y^*}+v_z^*\frac{\partial \vec{v}^*}{\partial z^*}$, $\bigtriangledown^{*2}\vec{v}^*$...).
So how can I know which values are standing behind every term in the equation?