# Derivation of formula for pressure gradient (fluid mechanics)

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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FYI: You should use \sin to get a nice-looking symbol for the sine function, e.g. $\sin\theta$ vs. $sin\theta$. – Rahul Aug 23 '12 at 19:15
Thanks for the tip, Rahul! – Kristian Aug 23 '12 at 19:58

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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Yes, since $z$ is the elevation in the gravitational field (vertical field), this should be a $sin$-expression. I mistakenly wrote $x$ when I posted the question. It has been changed to $z$ now, which is correct. – Kristian Aug 23 '12 at 19:10
@Kristian I have edited my answer to answer your question a little better. – Emily Aug 23 '12 at 19:15
Thanks a lot. I can't even remember the last time I encountered a differential expression such as this (where there is a sum of two terms inside the $d$-numerator), and that's why I became uncertain. Really appreciate your help! – Kristian Aug 23 '12 at 19:21
I bet you've see them all the time without even thinking about it. Let $f(x) = ax^2+bx+c$. What's $df/dx$? $\frac{df}{dx} = \frac{d}{dx}\left(ax^2+bx+c\right) = \frac{d(ax^2+bx+c)}{dx}$. – Emily Aug 23 '12 at 19:24
Ed - you're absolutely right! I think I'm just a bit out of it tonight! Believe it or not, I did actually get an A in Calculus 1, 2 and 3 :) – Kristian Aug 23 '12 at 19:58