- #1
ussername
- 60
- 2
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:
$$\rho \cdot v^{2}\left ( \vec{v}^{0}\cdot \left ( \frac{\partial \vec{v}^{0}}{\partial x} , \frac{\partial \vec{v}^{0}}{\partial y} , \frac{\partial \vec{v}^{0}}{\partial z}\right ) \right )=\rho \cdot v^{2}\cdot \left ( \vec{v}^{0}\cdot \left ( \mathrm{div} \vec{v}^{0} \right ) \right )$$
What is wrong about this derivation?
$$\rho \cdot \left ( v_{x}\cdot \frac{\partial \vec{v}}{\partial x} + v_{y}\cdot \frac{\partial \vec{v}}{\partial y} + v_{z}\cdot \frac{\partial \vec{v}}{\partial z}\right )$$
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:
$$\rho \cdot v^{2}\left ( \vec{v}^{0}\cdot \left ( \frac{\partial \vec{v}^{0}}{\partial x} , \frac{\partial \vec{v}^{0}}{\partial y} , \frac{\partial \vec{v}^{0}}{\partial z}\right ) \right )=\rho \cdot v^{2}\cdot \left ( \vec{v}^{0}\cdot \left ( \mathrm{div} \vec{v}^{0} \right ) \right )$$
What is wrong about this derivation?