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I am trying to make some derivations of open channel flow equations. And the problem is, I quite don't get some of the operations that are given in books on the following subject. For example:

$Q=Q(x)$

$A=A(x)$

$U(x)=Q/A$

$g=9.81$

$\frac{1}{gA} \frac{d}{dx}(\frac{Q^2}{A})=\frac{1}{g} \frac{Q}{A} \frac{d}{dx}(\frac{Q}{A})=\frac{d}{dx}(\frac{U^2}{2g})$

In the example above: If the Q and A are dependent on x, can I simply move Q out of the d/dx?? Just like that?

Regards

1 Answers 1

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Not in general:

$\frac d {dx} \left (\frac {Q^2} A\right) = \frac{2AQQ'-A'Q^2}{A^2}$

$Q \frac{d}{dx}\left(\frac Q A\right)=Q\frac{AQ'-A'Q}{A^2}$ So the only way these two things are equal is if $QAQ'=0$. In that case, you could move the $Q$ out.

What book do you see this equation in?

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    Well, I am quite familiar with them. But what I am trying to do is to understand the derivation from non-conservative to conservative form. But I just cand understand it. No matter how I do derive it, I don't get the form that is in the books. So I thought, that maybe it is some similar thing like with this here.2012-09-14