Derivation of Momentum Equation (Eulers Equation)

[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.

 

[fresh_42]

I have an example of this so-called material derivative (and a dozen other names):
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/#A-Material-Derivative

Here is an explanation:
http://www.fluiddynamics.it/capitoli/Eule.pdf

 

[pasmith]

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 $$-\frac{\partial p}{\partial z}\,\delta x\,\delta y\,\delta z.$$