1. dP/dz = -kg

Hi,

this is physics, but it's the math that is a problem to me. Problem is added in pdf-file. Can anyone quide me from EQ1 to EQ2 and further to EQ3?

If I use a computer program it gives me arctan...

Thanks!

- Trond

2. To get from equation 1 to equation 2, you must differentiate the mean pressure function with respect to z:

$\displaystyle \frac{\partial}{\partial{z}} p(z) = \frac{0\left(1 + \left(\frac{z}{h}\right)^2\right) - p_0\left(0 + \left(\frac{2zh^2 - 0z^2}{h^4}\right)\right)}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2}$

$\displaystyle \frac{\partial}{\partial{z}} p(z) = \frac{-p_0\frac{2z}{h^2}}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2}$

$\displaystyle \frac{\partial}{\partial{z}} p(z) = -2\frac{p_0z}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2h^2}$

That's to EQ2.

All they do for EQ3 is just substitute p_0 for P, and since I'm not too familiar with this topic, I'd just go with it.

3. Originally Posted by Aryth
To get from equation 1 to equation 2, you must differentiate the mean pressure function with respect to z:

$\displaystyle \frac{\partial}{\partial{z}} p(z) = \frac{0\left(1 + \left(\frac{z}{h}\right)^2\right) - p_0\left(0 + \left(\frac{2zh^2 - 0z^2}{h^4}\right)\right)}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2}$

$\displaystyle \frac{\partial}{\partial{z}} p(z) = \frac{-p_0\frac{2z}{h^2}}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2}$

$\displaystyle \frac{\partial}{\partial{z}} p(z) = -2\frac{p_0z}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2h^2}$

That's to EQ2.

All they do for EQ3 is just substitute p_0 for P, and since I'm not too familiar with this topic, I'd just go with it.
I also find strange that g appears in EQ 1 and not in EQ 2. But it wouldn't be constant anyway... I don't really know what happened to g in the derivation.

4. I believe you are right. I think they are 2 separate equations derived independently and not one from the other.

5. EQ 2 and 3 are only the left side of EQ 1. Further I set the solution of the diff like right side of EQ 1. So g will be back in EQ 4.

Thanks, this helped me further on. (By the way, P is pressure. And the next steps will end up with a function of temperature with respect to z (after substitude with a gas law). z is height from sea level).