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I think I'm missing something important here.

For a fluid with a velocity $u$ in a pipe with cross section area $S$, the volumetric flow rate is equal to:

$$Q_v = uS$$

If we now consider vectors, the volumetric flow rate is:

$$ Qv = \int \vec{u} ·\vec{n} dS$$

Where $\vec{n}$ is the normal vector to the cross sectional area. For example putting numbers, considering only one axis $x$:

$$ \vec{u} = 10 \vec{i} \qquad \vec{n} = \vec{i} \qquad S = 10 \\ Qv = 100$$

However from the divergence theorem we know that:

$$ \int \vec{u} · \vec{n} dS = \int \nabla · \vec{u} dV$$

And for incompressible fluids $\nabla · \vec{u} = 0$, so it would give $Q_v = 0$.

Obviously i'm doing something very wrong here, but I cannot see what. Am I considering something wrong in these equations? Is the normal vector in the divergence theorem refering to another surface in this case, or to both surfaces, these being, both $\vec{n} = \vec{i}$ and $\vec{n} = -\vec{i}$?

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The divergence theorem only applies to a closed surface that encloses the volume $V$.

If you construct a closed surface by "shrink wrapping" a section of pipe and closing off both ends, enclosing a volume $V$, the volumetric flow out the complete surface is zero, because the flow into one end is equal and opposite to the flow into of the other end - i.e. whatever goes into one end must come out of the other end.

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  • $\begingroup$ I see, so in the possible clossed surface, the normal vector would be both $\vec{i}$ and $-\vec{i}$, and so it would give 0 as well. Thank you :) $\endgroup$
    – Daniel V.
    Jan 5 '19 at 22:49

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