Fluid mechanics: Euler's equation derivation

[happyparticle]

Homework Statement

Derivation of the Euler's equation using $\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$

Relevant Equations

$\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$

Euleur's equation:
$\frac{D\vec{u}}{Dt} = \frac{-\nabla P + \vec{g}}{\rho}$

For a volume fixed in space with the positive component of the z axis pointing up, I have the force of gravity pointing down.

Using the following equation I have to derive the Euler's equation.
$\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$

Where $\phi$ is the flux, $F_v$ and $F_s$ are the volume and surface forces.

I think that the flux moving outside of the volume is:
$- \oint_S \rho \vec{u} \otimes \vec{u} dS$

Also, since the force of gravity is a volume force
$F_v = \rho V \vec{g}$
and
$F_s = \oint_S (-P \hat{n}) dS$ is the pressure on the surfaces, pointing outside of the volume.

So far I have
$\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = - \oint_S \rho \vec{u} \otimes \vec{u} dS + \rho V \vec{g} + \oint_S (-P \hat{n}) dS$

I'm wondering if I'm right. Also, I'm not sure which component of the pressure should I keep.
For instance, for a cylinder where the surface S is the top, where the flux is measured. I think I could consider only the z components of both the pressure and gravity.
In this case, I would have $F_s = \oint_S (-P \hat{n}) \cdot \hat{z} dS$ and $F_v = \rho V g \hat{z}$.

 

[Chestermiller]

Homework Statement: Derivation of the Euler's equation using $\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$
Relevant Equations: $\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$

Euleur's equation:
$\frac{D\vec{u}}{Dt} = \frac{-\nabla P + \vec{g}}{\rho}$

For a volume fixed in space with the positive component of the z axis pointing up, I have the force of gravity pointing down.

Using the following equation I have to derive the Euler's equation.
$\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$

Where $\phi$ is the flux, $F_v$ and $F_s$ are the volume and surface forces.

I think that the flux moving outside of the volume is:
$- \oint_S \rho \vec{u} \otimes \vec{u} dS$

Also, since the force of gravity is a volume force
$F_v = \rho V \vec{g}$
and
$F_s = \oint_S (-P \hat{n}) dS$ is the pressure on the surfaces, pointing outside of the volume.

So far I have
$\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = - \oint_S \rho \vec{u} \otimes \vec{u} dS + \rho V \vec{g} + \oint_S (-P \hat{n}) dS$

I'm wondering if I'm right. Also, I'm not sure which component of the pressure should I keep.
For instance, for a cylinder where the surface S is the top, where the flux is measured. I think I could consider only the z components of both the pressure and gravity.
In this case, I would have $F_s = \oint_S (-P \hat{n}) \cdot \hat{z} dS$ and $F_v = \rho V g \hat{z}$.

That is not what I get. Here is my analysis of momentum balance on the control volume: $$\int{\frac{\partial(\rho \mathbf{u})}{\partial t}dV}=-\int{\rho \mathbf{u}(\mathbf{u}\centerdot}\mathbf{n})dS+\int{\rho \mathbf{g}dV}-\int{P\mathbf{n}dS}$$The left hand side is the rate of accumulation of momentum within the control volume. The first term on the right hand side is the net rate of momentum flowing into the control volume. The second term on the right hand side is the gravitational force on the fluid within the control volume. The third term on the right hand side is the surface force acting on the fluid within the control volume. If we apply the divergence theorem and Gauss' theorem to the surface integrals in this equation, we obtain:
$$\int{\frac{\partial (\rho \mathbf{u})}{\partial t}dV}=-\int{\nabla \centerdot (\rho \mathbf{u} \otimes \mathbf{u})}dV+\int{\rho \mathbf{g}dV}-\int{\nabla P dV}$$