# Fluid Statics Problem: Determining pressure change in adiabatic atmosphere

1. Jul 10, 2009

### kallistos

1. The problem statement, all variables and given/known data

At ground level in Colorado Springs, Colorado, the atmospheric pressure and temperature are 83.2 kPa and 25 C. Calculate the pressure on Pike’s Peak at an elevation of 2690 m above the city assuming an adiabatic atmosphere.

Known
Patm = 83200 Pa
T0 = 25° C
z-z0 = 2690 m
ρair at 25 C = 1.19 kg/m3
γ = 1.4
R = 286.9 J/(kg ⋅ K)

(Assume z0=0 for math work)
(γ is the ratio of specific heats)

Unknown
Pf = ?

2. Relevant equations

dP/dz = -ρg
P = ρRT
T=T0 - mz

Remember, temperature in the ideal gas equation must be absolute!

P/ρ^k = constant

3. The attempt at a solution

Calling the constant 'c':

P = c*ρ^k
ln (P) = ln (c*ρ^k)
ln (P) = ln (c) + k ln (ρ)

Substituting in ideal gas expression where ρ is:

ln (P) = ln (c) + k ln (P/RT)
ln (P) = ln (c) + k ln (P) - k (ln (R) + ln (T))

Consolidating all constants into some constant, C, and segregating temperature from pressure:

(1-k) ln (P) = -k * ln (T) + C

Now, differentiating with respect to P on the left and T on the right, we get:

(1-k) dP/P = -k * dT/T
dP/P = -k/(1-k)dT/T
dP/P = k/(k-1)dT/T

Integrating the left from P0 to P and the right from T0 to T:

ln (P) - ln (P0) = k/(k-1) (ln(T) - ln(T0))
ln (P/P0) = k/(k-1) ln (T/T0)

P = P0(T/T0)^(k/(k-1))

According to my professor, this equation and the adiabatic relationship, P/ρ^k = constant, are all I need to solve the problem. And I don't see how, given that we aren't given the final temperature at the new elevation. My instinct was to plug in T=T0 - mz, since we have the initial temperature, look up 'm' for dry adiabatic atmospheric conditions, and solve for 'p' that way. However, my professor says there is a way to solve for T without looking up the value for 'm'. He would prefer an answer this way as well. Any ideas?

---

NOTE 1: Logarithmic rules:

ln (ab) = ln (a) + ln (b)
ln (a/b) = ln (a) - ln (b)
ln (a^k) = k ln (a)

NOTE 2: Differentiation rule:

d/dx(ln(x)) = 1/x

2. Jul 10, 2009

### Coto

Hi kallistos,

Going from memory, I believe that k you are working with is actually $$\gamma$$ as you've defined it above is it not?

I would suggest using this:

$$\frac{P}{\rho ^{\gamma}} = c$$

and this:

$$P = \rho R T$$

to rewrite $$\frac{T}{T_0}$$

A quick check on my side yields $$\frac{T}{T_0} = \left (\frac{P}{P_0}\right ) ^{\frac{\gamma - 1}{\gamma}}$$.

Hope this helps.

3. Jul 10, 2009

### kallistos

Thanks for replying. See, I think this is more of math problem than it is a physical problem on my end. I'll play around with it, but just looking at your work, I don't know how to get from point a to point b.

4. Jul 10, 2009

### Coto

Well try writing this out:

$$P = \rho R T$$
$$P_0 = \rho_0 R T_0$$

and this out:
$$\frac{P}{\rho ^{\gamma}} = \frac{P_0}{\rho_0 ^{\gamma}}$$

Do you see why that last part is true?

5. Jul 10, 2009

### kallistos

The last part I understand... both expressions are equal to the constant, so both expressions are equal to each other.

6. Jul 10, 2009

### Coto

So try dividing the first two expressions, reorganize this to get $$\frac{T}{T_0}$$ as a function of the other stuff, and then use your last relation to get rid of the densities.

Does this help?

7. Jul 10, 2009

### kallistos

$$T = \frac{P}{\rho R}$$
$$T_0 = \frac{P_0}{\rho R}$$

$$\frac{T}{T_0} = \frac{\frac{P}{\rho}}{\frac{P_0}{\rho_0}}$$

I did the first few steps, but I'm not sure how to make the densities go away mathematically.

EDIT: It comes down to this; how do I rewrite the adiabatic-constant expression such that the density is on one side of the equation ALONE?

8. Jul 10, 2009

### Coto

So we can rewrite $$\frac{T}{T_0} = \frac{P}{P_0}\frac{\rho _0}{\rho}$$

and we can rewrite that last relation:
$$\left (\frac{\rho _0}{\rho}\right )^{\gamma} = \frac{P_0}{P} \implies \frac{\rho _0}{\rho} = \left (\frac{P_0}{P}\right )^{\frac{1}{\gamma}}$$

9. Jul 10, 2009

### kallistos

I think I got it now... it's one of those nights where I can do Calculus for sure, but basic algebra just goes... blank.

However, I don't see how this gets me to closer to my answer, since the final temperature is still an unknown.

10. Jul 10, 2009

### Coto

No worries. My day was filled with mind blanks.

Substituting back into your last equation
P = P0(T/T0)^(k/(k-1)) ($$k = \gamma$$)

for T/T0 means you don't need to know the final temperature at all. All you require is $$P_0$$ and $$\gamma$$.

11. Jul 10, 2009

### Coto

I should note that it seems a bit weird that the height didn't enter into the expression somewhere, however to be honest I never checked over your calculus part, I only looked over your very last equation and isolated how to get rid of that $$\frac{T}{T_0}$$.

12. Jul 10, 2009

### kallistos

Okay, let's reformat my equation so it looks neater:

$$P = P_0 \left (\frac{T}{T_0}\right )^{\frac{\gamma}{\gamma - 1}}$$

13. Jul 10, 2009

### kallistos

Now that I think about it, what I may have to do is plug in this:

$$\frac{dP}{P} = \frac{k}{(k-1)} \frac{dT}{T}$$

Into the hydrostatic equation:

$$\frac{dP}{dz} = -\rho g$$

Integrate with respect to T and Z, and rewrite in terms of P and Z algebraically, like you showed me.

14. Jul 10, 2009

### Coto

I think it may be possible that you missed something in the derivation of your equation. You will most likely require to use the expression we formulated, but it seems to me that the height should appear somewhere in the equation.

15. Jul 10, 2009

### kallistos

I'll try it out. :)

By the way, I was wondering if you would look over my derivation of:

$$\frac{dP}{P} = \frac{k}{(k-1)} \frac{dT}{T}$$

Not for veracity, since I actually KNOW this part is right, but for the sake of maximizing simplicity. Was there an easier way to get that above answer from:

$$\frac{P}{\rho ^k} = c$$

And the ideal gas law? Or did I actually have to employ all those messy logarithms?

16. Jul 11, 2009

### Coto

I think the trick would have been to realize that you have this:

$$\frac{P}{\rho ^{\gamma}} = \frac{P_0}{\rho_0 ^{\gamma}}$$

available to you.

You could then rewrite your differential equation for pressure purely in terms of pressure by using the above expression to rewrite $$\rho$$ in terms of $$P$$:

$$\frac{dP}{dz} = -f(P) g$$

where f(P) is just some function of pressure.

Integrating this expression would then yield your answer. But I haven't sketched this idea out, so maybe a problem would arise.

17. Jul 11, 2009

### kallistos

Well, we can rewrite the adiabatic expression as:

$$\rho^\gamma = \frac{P \rho_0^\gamma}{P_0}$$

Another algebra question. How do we get rid of gamma on the rho?

18. Jul 11, 2009

### Coto

Just take each side to the power of $$\frac{1}{\gamma}$$. As I'm sure you are aware:

$$(x^n)^{\frac{1}{n}} = x^{\frac{n}{n}} = x$$

19. Jul 11, 2009

### Coto

Hmm, so we have a couple of problems here.

First, make sure that you take the power of each side to $$^{\frac{1}{\gamma}}$$. In other words on the RHS of the differential equation you should have $$^{\frac{1}{\gamma}}$$ in place of the $$\gamma$$s.

Next, be careful on your integration step. Make sure to bring your pressure term on the RHS (in this case, $$P^{\frac{1}{\gamma}}$$) over to the LHS. In other words, on the same side as the differential that involves P.

After doing this, you can then integrate as you've done in your 2nd step. This will then yield the solution you're looking for.

Your new integral on the LHS will look like this:

$$\int^{P}_{P_0}P^{\frac{-1}{\gamma}}dP$$

20. Jul 11, 2009

### kallistos

Then here is what my differential looks like:

$$\frac{dP}{dz} = -\frac{P^{\frac{-1}{\gamma}} \rho_0}{P_0^{\frac{-1}{\gamma} g$$

Then we get:

$$\int^{P}_{P_0}P^{-\gamma} dP = -\int^{z}_{0}\frac{\rho_0}{P_0^\gamma} gdz$$

Then:

$$\frac{1}{1-\gamma} \left (P^{1-\gamma} - P_0^{1-\gamma} \right )= -\frac{\rho_0}{P_0^\gamma} gz$$

EDIT: This is what yours looks like:

$$\int^{P}_{P_0}P^{\frac{-1}{\gamma}}dP$$

EDIT 2: Oh christ, I see where I'm going wrong. Hold on.

Last edited: Jul 11, 2009