To derive Euler's equation of inviscid fluid flow of a fluid element one needs to use the equation $$- \nabla P+ \rho \overrightarrow{g} = \overrightarrow{F} \tag{1} $$ where $ \nabla P $is the pressure gradient, $ \rho $ is the density of the fliud, $\overrightarrow{F}$ is the net force on the fluid, and then use the convective derivative $\dfrac{D}{Dt}$ to obtain the formula $$\rho \dfrac{D \overrightarrow{v}}{Dt}=- \nabla P+ \rho \overrightarrow{g} \tag{2} $$ where $\overrightarrow{v}$ is velocity vector field of the fluid.
I know how to arrive at (2) if I start from (1), which the book I am using now names Euler's equation of inviscid flow, but I have really forgotten how to derive (1).
**My attempt**
The scanty memory that I have tells me that I must assume that the fluid is at rest and hence consider that the only forces exerted on the fluid are pressure and gravitational pull and that the force due to pressure is given by $$ \nabla P dxdydz $$ and the force due to possible outward acceleration $\overrightarrow{a}$ is $$\overrightarrow{F} $$ so that the total outward force is $$ \nabla P dxdydz + \overrightarrow{F} $$
And by newton III, that is $F_{AB}=-F_{BA}$ , I think that this force is reacted back by the weight of the liquid $ g dm = g \rho dxdydz$ where $dm$ is elemental mass in the fluid and hence I can write $$ \nabla P dxdydz + \overrightarrow{F} = – g \rho dxdydz \overrightarrow{ k }$$
So my main confusion is about how the $dxdydz$ disappear for me to arrive at (1).
It is making me think that maybe I should put $\overrightarrow{F}=dm \overrightarrow{a} = \rho \overrightarrow{a} dxdydz= \overrightarrow{f} dxdydz$
where $\overrightarrow{f}=\rho \overrightarrow{a}$, so that i obtain
$$ \nabla P dxdydz + \overrightarrow{f} dxdydz = – g \rho dxdydz \overrightarrow{ k }$$
and then I can put $ \overrightarrow{g} = \langle 0,0,-g \rangle=- g \overrightarrow{ k } $
and hence obtain $$ – \nabla P+ \rho \overrightarrow{g} = \overrightarrow{f} \tag{3} $$
Guys please tell me if I am in the right direction.
Best Answer
The mass of a fluid within a volume $V$ is,
$$\int_V \rho \ dV$$
Where, $\rho$ is the density. Similarly, the momentum is,
$$\int_V \rho \cdot \vec v \ dV$$
Where $\vec v$ is the fluid velocity at a point inside the volume.
The mass within the volume changes with time if fluid flows in and out of the bounding surface $S$ of the volume.
$${{d} \over {dt}} \cdot \int_V \rho \ dV=-\int_S \rho \cdot \vec v \cdot \vec n \ dS$$
Where we take the dot product when appropriate and $\vec n$ is the surface normal. Similarly,
$${{d} \over {dt}} \cdot \int_V \rho \cdot \vec v \ dV=-\int_S \rho \cdot \vec v \ ( \vec v \cdot \vec n) \ dS$$
Using Newton's Second Law, the rate of change of momentum of a fixed portion of material is equal to the net force acting on that material. If we assume the force is due to pressure $p$ at the surface of $V$, we can say,
$$(1) \quad {{d} \over {dt}} \cdot \int_V \rho \cdot \vec v \ dV=-\int_S \rho \cdot \vec v \ ( \vec v \cdot \vec n) +p \cdot \vec n \ dS$$
Now, we can use the fact that,
$$\int_V \nabla f \ dV=\int_S \vec n \cdot f \ dS$$
along with the fact,
$$\int_V \nabla \cdot \vec B \ dV=\int_S \vec n \cdot \vec B \ dS$$
and rewrite $(1)$ as,
$$(2) \quad {{d} \over {dt}} \cdot \int_V \rho \cdot \vec v \ dV=-\int_V \nabla \cdot (\rho \cdot \vec v) \cdot \vec v +\nabla p\ dV$$
Moving terms, we get,
$$(3) \quad \int_V {{d} \over {dt}} \ (\rho \cdot \vec v)+\nabla \cdot (\rho \cdot \vec v) \cdot \vec v +\nabla p\ dV=0$$
Which can be simplified to,
$$(4) \quad {{d} \over {dt}} \ (\rho \cdot \vec v)+\nabla \cdot (\rho \cdot \vec v) \cdot \vec v +\nabla p=0$$
Which finally becomes,
$$(5) \quad {{D} \over {Dt}} \ (\rho \cdot \vec v)+\nabla p=0$$
Which is the Euler Equation for inviscid flow.
Your Method
I commented on your method
The scanty memory that I have tells me that I must assume that the fluid is at rest and hence consider that the only forces exerted on the fluid are pressure and gravitational pull and that the force due to pressure is given by $$ \nabla P dxdydz $$
This is completely correct
and the force due to possible outward acceleration $\overrightarrow{a}$ is $$\overrightarrow{F} $$
Ad-hoc, but correct
so that the total outward force is $$ \nabla P dx \cdot dy \cdot dz + \overrightarrow{F} $$
Yes, that is true
And by newton III, that is $F_{AB}=-F_{BA}$ , I think that this force is reacted back by the weight of the liquid $ g dm = g \rho dx \cdot dy \cdot dz$ where $dm$ is elemental mass in the fluid and hence I can write $$ \nabla P dx \cdot dy \cdot dz + \overrightarrow{F} = - g \rho dxdydz \overrightarrow{ k }$$
Use Newton's Second Law and I think this works