mathmari
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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 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)