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

AI Thread Summary
The discussion focuses on deriving the integral form of momentum conservation in fluid dynamics. It begins with the expression for the total force on a fluid volume, combining pressure and body forces. The relationship between mass, acceleration, and forces is established using Newton's second law, leading to the differential form of momentum conservation. The conversation then shifts to how the integral form can be derived from the conservation of mass, specifically using the product rule for differentiation. The final expression connects the time derivative of momentum density to the forces acting on the fluid.
mathmari
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Hey! :o

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)
 
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mathmari said:
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)
 
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