MHB How is the Integral Form of Momentum Conservation Derived in Fluid Dynamics?

[mathmari]

Hey!

I am looking at the conservation of momentum.

The force at $W$ from the tensions at the boundary $\partial{W}$ is $$\overrightarrow{S}_{\partial{W}}=-\int_{\partial{W}}p \cdot \overrightarrow{n}dA=-\int_{W}\nabla p dV$$ where $p(\overrightarrow{x}, t)$ the pressure and $\overrightarrow{n}$ the unit perpendicular vector.

The massive forces is $$\overrightarrow{B}_{W}=\int_{W}\rho \overrightarrow{b}dV$$ where $\overrightarrow{b}$ the density of massive forces.

So, the total force on the fluids in the volum $W$ is $$\overrightarrow{S}_{\partial{W}}+\overrightarrow{B}_{W}=\int_{W}( \rho \overrightarrow{b}-\nabla p)dV$$

From the second Newton's law we have that $\overrightarrow{F}=m\cdot \overrightarrow{a}$ and since $m=\int \rho dV$ and $\overrightarrow{a}=\frac{D\overrightarrow{u}}{Dt}$, where $\frac{D}{Dt}$ the material derivative, we have the following:

$$\int_{W}\rho \frac{D\overrightarrow{u}}{Dt}dV=\overrightarrow{S}_{\partial{W}}+\overrightarrow{B}_{W}=\int_{W}(\rho \overrightarrow{b}-\nabla p)dV$$

The differential form of the conservation of momentum is $$\rho \frac{D\overrightarrow{u}}{Dt}=-\nabla p+\rho\overrightarrow{b}$$

We are looking for the integral form of the conservation of momentum.

We have $$\rho \frac{\partial{\overrightarrow{u}}}{\partial{t}}=-\rho (\overrightarrow{u}\cdot \nabla )\overrightarrow{u}-\nabla p+\rho \overrightarrow{b}$$

From the differential form of the conservation of mass ($\frac{\partial{\rho}}{\partial{t}}+\nabla \cdot (\rho \overrightarrow{u})=0$) we get the following:

$$\frac{\partial}{\partial{t}}(\rho \overrightarrow{u})=-div(\rho \overrightarrow{u})\overrightarrow{u}-\rho(\overrightarrow{u}\nabla)\overrightarrow{u}-\nabla p+\rho\overrightarrow{b}$$

Could you explain to me how we get the last relation?? (Wondering)

 

[I like Serena]

We have $$\rho \frac{\partial{\overrightarrow{u}}}{\partial{t}}=-\rho (\overrightarrow{u}\cdot \nabla )\overrightarrow{u}-\nabla p+\rho \overrightarrow{b}$$

From the differential form of the conservation of mass ($\frac{\partial{\rho}}{\partial{t}}+\nabla \cdot (\rho \overrightarrow{u})=0$) we get the following:

$$\frac{\partial}{\partial{t}}(\rho \overrightarrow{u})=-div(\rho \overrightarrow{u})\overrightarrow{u}-\rho(\overrightarrow{u}\nabla)\overrightarrow{u}-\nabla p+\rho\overrightarrow{b}$$

Could you explain to me how we get the last relation?? (Wondering)

Hey! (Blush)

According to the product rule, we have:
$$\pd {} t(\rho \overrightarrow u) = \pd \rho t\overrightarrow u + \rho \pd {\overrightarrow u} t
$$
Now substitute the conservation of mass. (Wasntme)