Potential theory for water waves

[Nikitin]

hi :)

http://ocw.mit.edu/courses/mechanic...s-fall-2009/course-text/MIT2_017JF09_ch06.pdf

In page 37 they use Newton's 2nd law for a fluid element (while ignoring viscous forces) to derive the bernoulli equation for unsteady flow.

Well, what I am confused about is the last step: They integrate through the x direction, y direction and then z direction to get 3 separate bernoulli equations,,, but then they just add it all up to one? I don't understand the mathematics behind this as I'm completely new to potential theory.

 

[Nikitin]

Furthermore, on page 40 they equate the dynamical pressure {for waterwaves} to be equal to the change in the potential function?? To see what I mean:

From this (Bernoulli) $p + 0.5 \rho V^2 (\approx 0) + \rho g z +\rho \frac{\partial(\phi)}{\partial(z)} = C$, they go to this: $p_d = p+\rho g z = -\rho \frac{\partial(\phi)}{\partial(z)}$

How did they do that?

 

[Hypersphere]

hi :)

http://ocw.mit.edu/courses/mechanic...s-fall-2009/course-text/MIT2_017JF09_ch06.pdf

In page 37 they use Newton's 2nd law for a fluid element (while ignoring viscous forces) to derive the bernoulli equation for unsteady flow.

Well, what I am confused about is the last step: They integrate through the x direction, y direction and then z direction to get 3 separate bernoulli equations,,, but then they just add it all up to one? I don't understand the mathematics behind this as I'm completely new to potential theory.

This is nothing mysterious, it is just that z=0 for the equations in $\hat{x}$ and $\hat{y}$, so that equation works for all three cases. If that doesn't make sense, do the three integrals and compare the results.

 

[AlephZero]

It works through "The first force balance (x direction)" and gets to the result

"Bermouilli forrmula" = $C_1$ when y and z are constant

The same process in the y direction gives
"Bermouilli forrmula" = $C_2$ when z and x are constant

and in the z direction
"Bermouilli forrmula" + $\rho g z$ = $C_3$ when x and y are constant

But $\rho g z$ is a constant when z is constant, so you can merge it into $C_1$ and $C_2$.

In other words the final equation is actually valid in the x, y, and z directions.

This seems like a tortuous way to get to the result, but I guess that depends what you already know and what the teaching objectives for this part of the course are.

 

[Nikitin]

Oh how stupid of me. This is just the same as integrating partial derivatives of a function and then comparing the results to find the actual function. OK, thanks!