# Derivation of Momentum Equation (Eulers Equation)

• aeroboyo
The second force in the x direction is due to the right face at x+ \delta x and is given by the pressure at (x+\delta x,y,z) times the area of this face. This force is in the negative x direction. Hence the contribution is -p(x+\delta x,y,z)\,\delta y\,\delta z. The contribution from the left face at x\,\delta x is the pressure at (x,y,z) times the area of this face and is directed in the positive x direction. Again, using a Taylor series we have -p(x,y,z) \approx -f

#### aeroboyo

hey, I'm having difficulty understanding how Eulers Equation is derived... its because my grasp of calculus is preety bad.

Eulers equation is:

dp = -qVdV, where p is pressure, q is density and V is velocity. And that's clearly a differential equation. But i'd like to know how its derived. I know i have to start with Newtons 2nd law:

F=ma

my textbook says that

The force in the x direction acting on a fluid particle is

F = p dydz - (p + dp/dx dx) dydz (1)

Hence F = dp/dx dxdydz = force on fluid element due to pressure. (d)

This is where my confusion begins... I don't understand why the Force would be pressure multiplied by dydz and then minus pressure plus dp/dx multiplied by dydz. In other words i really don't understand what's going on with equation one, or how it becomes equation (2).

Can any1 help? I think once i understand this example i'll be able to work out others. Thanks.

hey, I'm having difficulty understanding how Eulers Equation is derived... its because my grasp of calculus is preety bad.

Eulers equation is:

dp = -qVdV, where p is pressure, q is density and V is velocity. And that's clearly a differential equation. But i'd like to know how its derived. I know i have to start with Newtons 2nd law:

F=ma

my textbook says that

The force in the x direction acting on a fluid particle is

F = p dydz - (p + dp/dx dx) dydz (1)

Hence F = dp/dx dxdydz = force on fluid element due to pressure. (d)

This is where my confusion begins... I don't understand why the Force would be pressure multiplied by dydz and then minus pressure plus dp/dx multiplied by dydz. In other words i really don't understand what's going on with equation one, or how it becomes equation (2).

Can any1 help? I think once i understand this example i'll be able to work out others. Thanks.

Imagine a cuboid with sides $\delta x$, $\delta y$ and $\delta z$. There are two contributions to the force in the $x$ direction on the fluid in this cuboid. The first is due to the left face at $x$, and is given by the pressure at $(x,y,z)$ multiplied by the area of this face, which is $\delta y\,\delta z$. This force is in the positive $x$ direction. Hence the contribution is $p(x,y,z)\,\delta y\,\delta z$. The contribution from the right face at $x+ \delta x$ is the pressure at $(x + \delta x ,y,z)$ times the area of this face and is directed in the negative $x$ direction. Using a Taylor series we have to first order $$p(x + \delta x,y,x) \approx p(x,y,z) + \frac{\partial p}{\partial x}\,\delta x.$$ Thus to first order the net force in the $x$ direction is $$-\frac{\partial p}{\partial z}\,\delta x\,\delta y\,\delta z.$$

• fresh_42