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What's the intuition or reasoning behind this?

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- Thread starter yosimba2000
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- #1

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What's the intuition or reasoning behind this?

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Chestermiller

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Chet

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I have found a similar explanation on this video which may give you a better understanding of what I'm asking. This is at time 1:50.

Sorry, the explanation still isn't working for me!

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Chestermiller

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[tex]\vec{V}=v(r)\vec{i}_z[/tex]

where ##\vec{i}_z## is the unit vector in the axial direction. This relationship satisfies the continuity equation exactly and, if you substitute it into the z component of the momentum equation in cylindrical coordinates, you end up with your professor's equation. The right hand side of this equation is independent of z. That means that dp/dz is independent of z. Right now, I don't see how to prove that dp/dz is not a function of r, but I'll think about it. I know that it is.

Chet

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Chestermiller

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Chet

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I do agree with all the equations in the video and what you've posted, but here's my real question:

I'm curious as to why, at time 1:50 in the video, a function of Z only being equal to a function of R only means that both of them have to be constants for this to be true. It doesn't haveWhat is the logic behind this?

- #7

Chestermiller

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If f(r) = g(z), give me a specific example of a case in which f and g are not constants.

I do agree with all the equations in the video and what you've posted, but here's my real question:

I'm curious as to why, at time 1:50 in the video, a function of Z only being equal to a function of R only means that both of them have to be constants for this to be true. It doesn't haveWhat is the logic behind this?

Chet

- #8

pasmith

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What's the intuition or reasoning behind this?

Start with: [tex]

\frac{\partial p}{\partial z} = f(r).

[/tex] Using the given information, [tex]

\frac{\partial f}{\partial r} = \frac{\partial}{\partial r}\left(\frac{\partial p}{\partial z}\right) = 0[/tex] and [tex]

\frac{\partial}{\partial z}\left(\frac{\partial p}{\partial z}\right) =\frac{\partial f}{\partial z} = 0.[/tex] Hence [itex]f(r) = \frac{\partial p}{\partial z}[/itex] is constant.

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