How Does Air Pressure Change with Altitude?

[stripedcat]

I have the solution so to speak, but I'd like some explanation for it. You do not have to type out the explanation yourself if you have a link to something that might shed insight. I don't know all of the format commands for the math yet, so forgive me for the formatting issues that are bound to crop up.

Okay, here we go!

Under ideal conditions, air pressure decreases continuously with the height above sea level at a rate proportional to the pressure at that height. The barometer reads 30 inches at sea level and 15 inches at 18,000 feet. Find the barometric pressure at 35,000 feet.

dp/dh = kp, P(0)=30

$$P(h) = 30e^{kh}$$

First question is, how are we getting that second equation there? I see where the number and variables come from but I'm not quite sure how they were arranged that way.

Our first h value is 18,000

$$P(18,000) = 30e^{18,000k}$$

As stated in the problem, this will be = 15, right? Not sure if that's important, just wanted to verify.

k = ln(1/2)/18,000

I don't know where they got that ln (1/2) from though I do know it's going to become -ln(2)/18,000

I could continue the problem, but that last bit there is what I really need answered. I have knowledge gaps in my math, so I'm sure I'm missing something.

 

[chisigma]

I have the solution so to speak, but I'd like some explanation for it. You do not have to type out the explanation yourself if you have a link to something that might shed insight. I don't know all of the format commands for the math yet, so forgive me for the formatting issues that are bound to crop up.

Okay, here we go!

Under ideal conditions, air pressure decreases continuously with the height above sea level at a rate proportional to the pressure at that height. The barometer reads 30 inches at sea level and 15 inches at 18,000 feet. Find the barometric pressure at 35,000 feet.

dp/dh = kp, P(0)=30

$$P(h) = 30e^{kh}$$

First question is, how are we getting that second equation there? I see where the number and variables come from but I'm not quite sure how they were arranged that way.

Our first h value is 18,000

$$P(18,000) = 30e^{18,000k}$$

As stated in the problem, this will be = 15, right? Not sure if that's important, just wanted to verify.

k = ln(1/2)/18,000

I don't know where they got that ln (1/2) from though I do know it's going to become -ln(2)/18,000

I could continue the problem, but that last bit there is what I really need answered. I have knowledge gaps in my math, so I'm sure I'm missing something.

Wellcome on MHB stripedcat!...

Your first question is how to solve the 'initial value problem'...

$\displaystyle \frac{d p}{d h} = k\ p,\ p(0)= 30\ (1)$

The equation in (1) is the most simple example of ODE, in which the variables are separable so that You can write...

$\displaystyle \frac{d p}{p} = k\ d h\ (2)$

... and integrating both terms of (2) You obtain...

$\displaystyle \int \frac{d p}{p} = k\ \int d h \implies \ln p = k\ h + \ln c \implies p = c\ e^{k\ h}\ (3)$

The next step is to find the constant c and k in (3). Are You able to do that?...

Kind regards

$\chi$ $\sigma$

 

[stripedcat]

Thank you for the welcome!

$\displaystyle \frac{d p}{d h} = k\ p,\ p(0)= 30\ $

$\displaystyle \frac{d p}{p} = k\ d h\ $

Alright so we arrange it as you have done, multiply both sides by dh and divide both sides by p. No problem. I know how dp/p became ln |p|, p^-1 dp = ln p

Now for the right side: k dh, that would become k(h+c) wouldn't it? Thus making it ln p = k(h+c)

Both sides by e maing it p = e^k(h+C), I know to bring down that C to make it p = Ce^kh

Okay, got that now! Thanks.

Constant? 30 e? I'm not sure what you're asking but to press on with what I think occurs next.. I did figure out the ln ½ bit

ln(30e^18000k) =18000k + ln(30)
ln(15) = 18000k + ln(30)

Play the number flipping game to isolate k, ln(15) – ln(30) = 18000k, which becomes (ln(15) – ln(30))/18000 = k

ln(15)-ln(30) = 1/2ln = -ln(2) so -ln(2)/18000 = k

Now that we have k...

[math]P(h) = 30e^{(-ln(2)/18000)h}[/math]

Is that right? Plugging in the 35000 for h gives 7.8~

 

[SuperSonic4]

Now for the right side: k dh, that would become k(h+c) wouldn't it? Thus making it ln p = k(h+c)

It's better to say $$kh + C$$ as you've written [math]kh+kc[/math]. It's ok this time because you get to say [math]C' = kc[/math]

Both sides by e maing it p = e^k(h+C), I know to bring down that C to make it p = Ce^kh

You have $$p = e^{k(h+c)} = e^{kh+kc} = e^{kc}e^{kh}$$. However, [math]p = Ce^{kh}[/math] is right.

Constant? 30 e? I'm not sure what you're asking but to press on with what I think occurs next.. I did figure out the ln ½ bit

[math]30e[/math] is a constant but chisigma is asking you to find the values of $$c$$ and $$k$$ using the information given

ln(30e^18000k) =18000k + ln(30)
ln(15) = 18000k + ln(30)

Play the number flipping game to isolate k, ln(15) – ln(30) = 18000k, which becomes (ln(15) – ln(30))/18000 = k

ln(15)-ln(30) = 1/2ln = -ln(2) so -ln(2)/18000 = k

I'm not sure what you're doing here but that first line is incorrect - logarithms do not work that way. Because you have an exponent in the logarithm you need to use the power law: [math]\ln(a^b) = b\ln(a)[/math]. In your case you end up with [math]\ln(30e^{18000k}) = 18000k\ln(30)[/math].

edit: I see what you did now, that works

Spoiler

$$15 = 30e^{18000k}$$

Divide both sides by 30 (this is where you'll come across the 1/2 as in the OP)

$$\frac{1}{2} = e^{18000k}$$

$$\ln \left(\dfrac{1}{2}\right) = 18000k$$

$$k = \dfrac{\ln \left(\dfrac{1}{2}\right)}{18000}$$