We shall derive Euler’s equation in 3-Dimension in Cartesian coordinate system. Euler’s Equation can also be derived from the Navier Stokes Equation with Steady and Non-Viscous flow.

Consider the three dimensional fluid element,

Now the fluid element is of a size $δx, δy$ and $δz$ along the X, Y and Z direction respectively.

This fluid element behaves as an infinitesimal control volume (CV) as $δx→0, δy→0$ and $δz→0.$

Let fluid enter the left hand face of the CV with a velocity u and leaves the right hand face with a velocity $\bigg(u+\dfrac{∂u}{∂x}δx \bigg)$.

By, Newton’s second law in the X direction,

$$∑F_x = ma_x$$

$\bigg(P-\dfrac{∂P}{∂x} \dfrac{δx}{2}\bigg)δyδz-\bigg(P+\dfrac{∂P}{∂x} \dfrac{δx}{2}\bigg)δyδz = ρ\bigg\{u \dfrac{∂u}{∂x}+v \dfrac{∂u}{∂y}+w \dfrac{∂u}{∂z}\bigg\} δxδyδz \\ ∴-\dfrac 1ρ \dfrac{∂P}{∂x}=\bigg[ u \dfrac{∂u}{∂x}+v \dfrac{∂u}{∂y}+w \dfrac{∂u}{∂z}\bigg]$

Similarly in Y direction,

$\bigg(P-\dfrac{∂P}{∂y} \dfrac{δy}{2}\bigg)δxδz-(ρgδxδyδz)-\bigg(P+\dfrac{∂P}{∂y} \dfrac{δy}{2}\bigg)δxδz = ρ\bigg\{u \dfrac{∂v}{∂x}+v \dfrac{∂v}{∂y}+w \dfrac{∂v}{∂z}\bigg\} δxδyδz\\ ∴-\dfrac 1ρ \dfrac{∂P}{∂y}-g=\bigg[ u \dfrac{∂v}{∂x}+v \dfrac{∂v}{∂y}+w \dfrac{∂v}{∂z}\bigg]$

In Z direction,

$\bigg(P-\dfrac{∂P}{∂z} \dfrac{δz}{2}\bigg)δxδz-\bigg(P+\dfrac{∂P}{∂z} \dfrac{δz}{2}\bigg)δxδz = ρ\bigg\{u \dfrac{∂w}{∂x}+v \dfrac{∂w}{∂y}+w \dfrac{∂w}{∂z}\bigg\} δxδyδz\\ \therefore-\dfrac 1ρ \dfrac{∂P}{∂x}=\bigg[ u \dfrac{∂w}{∂x}+v \dfrac{∂w}{∂y}+w \dfrac{∂w}{∂z}\bigg]$

These are the Euler’s equation along respective axis.

Adding the three equations we get,

$-\dfrac 1ρ \dfrac{∂P}{∂x}-\dfrac 1ρ \dfrac{∂P}{∂y}-g-\dfrac 1ρ \dfrac{∂P}{∂z}=\bigg[ u \dfrac{∂u}{∂x}+v \dfrac{∂u}{∂y}+w \dfrac{∂u}{∂z}\bigg]+\bigg[ u \dfrac{∂v}{∂x}+v \dfrac{∂v}{∂y}+w \dfrac{∂v}{∂z}\bigg]+\bigg[ u \dfrac{∂w}{∂x}+v \dfrac{∂w}{∂y}+w \dfrac{∂w}{∂z}\bigg] \\ ∴-\dfrac 1ρ \bigg(\dfrac{∂P}{∂x}+\dfrac{∂P}{∂y}+\dfrac{∂P}{∂z}\bigg)-g=\bigg[ u \dfrac{∂u}{∂x}+v \dfrac{∂u}{∂y}+w \dfrac{∂u}{∂z}\bigg]+\bigg[ u \dfrac{∂v}{∂x}+v \dfrac{∂v}{∂y}+w \dfrac{∂v}{∂z}\bigg]+\bigg[ u \dfrac{∂w}{∂x}+v \dfrac{∂w}{∂y}+w \dfrac{∂w}{∂z}\bigg]$

In mathematical terms,

$∴-\dfrac 1ρ (∇P)-g=(\bar{U} ∙∇) \bar{U} \\ ∴ρ(\bar{U} ∙∇) \bar{U} =-∇P-ρg$

…this is the Euler’s Momentum Equation.