0
$\begingroup$

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!

  • 0
    FYI: You should use `\sin` to get a nice-looking symbol for the sine function, e.g. $\sin\theta$ vs. $sin\theta$.2012-08-23
  • 0
    Thanks for the tip, Rahul!2012-08-23

1 Answers 1

1

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.

  • 0
    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.2012-08-23
  • 0
    @Kristian I have edited my answer to answer your question a little better.2012-08-23
  • 0
    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!2012-08-23
  • 0
    I bet you've see them all the time without even thinking about it. Let $f(x) = ax^2+bx+c$. What's $df/dx$? $\frac{df}{dx} = \frac{d}{dx}\left(ax^2+bx+c\right) = \frac{d(ax^2+bx+c)}{dx}$.2012-08-23
  • 0
    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 :)2012-08-23
  • 0
    Too many hours spent with fluid mechanics can do that to you; it certainly did it to me :D2012-08-23