Board 8 > Hey B8 math people, got a conceptual calculus thing that I can't remember.

(+) Yeah! (+)

So, I am doing a homework problem for my fluids class. I need to take the u relation in this first pic...



plug it into the bottom equation in this picture...



to get the equation in the blue box in this picture...



u-bar is shown in this picture...



This is where I have gotten to, but I have no idea how to get to the equation in the blue box from here. Need some nudging in the right direction.


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I don't like doing this, but @LordoftheMorons you are usually helpful with these things without outright telling me the answer, can you give any insight?
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I don't know anything about fluids (not in the physics curriculum). Is there some other equation you have relating p to u, v, w? It seems like you'd need one to get rid of dp'/dx (and presumably introduce the average values of u'^2, u'v', and u'w').
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oh hey is this the finnemore and franzini text

it looks like the fluids book I used in college
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this doesn't really look like conceptual calculus, more like mechanical grinding

which i love, but can't do at the moment. if you still need help in like 5 hours then tag me and i'll be happy to try to look through it <_<
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Oh, my original question was conceptual about the calculus involved, it changed as I was typing up the post.

LordoftheMorons posted...

Is there some other equation you have relating p to u, v, w?

Uh, basically it's vectors. V and P can be broken down into i, j, and k components with u, v, and w as the unit vectors.

so here are two other equations...

V(x, y, z, t) = iu(x, y, z, t) + jv(x, y, z, t) + kw(x, y, z, t)

a = dV/dt = pV/pt + (u*pV/px + v*pV/py + w*pV/pz)

Where p/p(x,y,z,t) is the partial derivative

That last one is also written as

a = dV/dt = pV/pt + (del * V)V

where the del operator is u*p/px + v*p/py + w*p/pz

(u*pV/px + v*pV/py + w*pV/pz) is called the convective acceleration.

Anyway, my original conceptual question had to do with taking the time derivative of a definite integral, u-bar. It just returns the original function, right?
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I was referring to the p that the gradient is acting on in the bottom of the second picture; I'm assuming it has some specific definition

If you take the time derivative of ubar it should be 0; in going from u to ubar you integrated out the time dependence.
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I was more thinking not using u-bar but the integral itself.

TBH, I'm not sure where that pressure gradient comes from. Let me look some more. The solution manual just mentions the convective acceleration, something about averaging them and then it lists the final equation.
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Well, you could take the derivative with respect to T and that might be nonzero. Once you're integrated over t though it's no longer a function of t.
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Oh right, that's the Navier-Stokes equation.



Which is a variation of Euler's equation

rho*g + del*p = rho*dV/dt

It has to do with the fact that a pressure difference will cause acceleration or deceleration.
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