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There is a derivation of a formula in my textbook which I don't fully understand. Most of the formulas I encounter I understand without difficulty, but if someone can help me understand one step of the following derivation, I would be very grateful:

Consider a flow of fluid through an inclined core sample of length $\Delta l$, with a constant flow rate, $q$, maintained at a pressure differential $\Delta p$. The flow at an angle $\theta$ above the horizontal can be described by the following version of the Darcy equation:

$q = -A\frac{k}{\mu}\frac{d(p + \rho g z)}{dl}$

where $z$ is the elevation in the gravitational field. Since $z = l \sin \theta$, with $l$ as the direction of flow, the equation written for the pressure gradient, becomes:

$\frac{dp}{dl} = - \left(\frac{q \mu}{Ak} + \rho g \sin \theta\right)$

OK, so it is this last step here I don't quite understand. I see that from the given equation we have:

$q = -A\frac{k}{\mu}\frac{d(p + \rho g z)}{dl}$

$\frac{d(p + \rho g z)}{dl} = - \frac{q \mu}{Ak}$

$\frac{d(p + \rho g l \sin \theta)}{dl} = -\frac{q \mu}{Ak}$

But I don't see this last step from here on. How do you "extract" the $\sin \theta$ part from the differential expression on the left side of this equation?

If anyone can explain this to me, I would appreciate it a lot!

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    Thanks for the tip, Rahul!2012-08-23

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Are you sure about your definition of $x$? If $\theta$ is given as the angle from the horizontal, and $x$ is the horizontal component, then $x = l\cos \theta$, and $z = l \sin \theta$, which looks like what you need.

In your statement $\frac{d(p+\rho g l \sin \theta)}{dl}$, we simply apply differentiation rules. Since $\theta$ is independent of $l$, then we have

$\frac{d(p+\rho g l \sin \theta)}{dl} = \frac{dp}{dl}+\frac{d\rho g l \sin\theta}{dl} = \frac{dp}{dl} + \rho g \sin \theta \frac{dl}{dl}.$

This brings you to the form above -- if that's what you're asking.

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    Too many hours spent with fluid mechanics can do that to you; it certainly did it to me :D2012-08-23