Is Exponential Form Better for Balloon with Inserted Load Calculation?

[Hak]

The method of separation of variables... All factors involving $\rho$ on the left, and everything involving $h$ on the right.

$$ \frac{d \rho}{\rho} = \left(T_0 \beta - \frac{g}{R} \right) \frac{dh}{T(h)} $$

Thanks. I get:

$$ ln |\rho| + c_1 = - \left(T_0 \beta - \frac{g}{R} \right) \frac{ln |1 - \beta h|}{T_0 \beta} + c_2$$, from which:

$$ \rho (h) = e^{- \left(T_0 \beta - \frac{g}{R} \right) \frac{ln |1 - \beta h|}{T_0 \beta} + c}$$.

This is the same result as before, with the difference that I had previously inserted an extra $h$ term, which was wrong.

 

[erobz]

Thanks. I get:

$$ ln |\rho| + c_1 = - \left(T_0 \beta - \frac{g}{R} \right) \frac{ln |1 - \beta h|}{T_0 \beta} + c_2$$, from which:

$$ \rho (h) = e^{- \left(T_0 \beta - \frac{g}{R} \right) \frac{ln |1 - \beta h|}{T_0 \beta} + c}$$.

This is the same result as before, with the difference that I had previously inserted an extra $h$ term, which was wrong.

It may be the same, but you are overcomplicating it and making it virtually unrecognizable. Use definite integrals on each side:

$$ \int_{\rho_o}^{\rho(h)} \frac{d \rho}{\rho} =\left(T_0 \beta - \frac{g}{R} \right)
\int_0^h \frac{dh}{T_o - T_o \beta h } $$

 

[Hak]

It may be the same, but you are overcomplicating it and making it virtually unrecognizable. Use definite integrals on each side:

$$ \int_{p_o}^{\rho(z)} \frac{d \rho}{\rho} =\left(T_0 \beta - \frac{g}{R} \right)
\int_0^h \frac{dh}{T_o - T_o \beta h } $$

OK. I get:
$$ln |\rho (z)| - ln|\rho_0| = \frac{ln |T_0| - ln |T_0 (1- \beta h)|}{T_0 \beta} \left(T_0 \beta - \frac{g}{R} \right) $$. Right?

Edit. I had forgotten part of the equation.

 

[erobz]

OK. I get:
$$ln |\rho (z)| - ln|\rho_0| = \frac{ln |T_0| - ln |T_0 (1- \beta h)|}{T_0 \beta}$$. Right?

Close, but you are forgetting the factor out front, and use log rules to turn differences of logs into a log of a quotient, and then rules about logs and factors, and exponents etc... once complete, exponentiate both sides.

 

[Hak]

Close, but you are forgetting the factor out front, and use log rules to turn differences into quotients, and logs and factors.

OK. I get:

$$ln |\frac{\rho(z)}{\rho_0}| = - ln |1 - T_0 \beta| \left(1 - \frac{g}{T_0 \beta R} \right) $$, from which:

$$\rho (z) = \rho_0 \cdot e ^ {- ln |1 - T_0 \beta| \left(1 - \frac{g}{T_0 \beta R} \right)}$$.
Right?

 

[erobz]

OK. I get:

$$ln |\frac{\rho(z)}{\rho_0}| = - ln |1 - T_0 \beta| \left(1 - \frac{g}{T_0 \beta R} \right) $$, from which:

Try again. The difference of logs on the RHS, where did it go?

Then...Rules about logs: specifically $a ln(x) = ln(x^a)$

Only then do you exponentiate both sides.

 

[Hak]

Try again. The difference of logs on the RHS, where did it go?

It is correct, or not? We have:

$$ln |T_0| - ln |T_0 (1 - \beta h)| \Rightarrow \ ln |\frac{T_0}{T_0 (1 - \beta h)}| = ln |\frac{1}{(1 - \beta h)}| = ln |(1 - \beta h)|^{-1} = - ln |(1 - \beta h)| $$.

Where is the mistake?

 

[erobz]

It is correct, or not? We have:

$$ln |T_0| - ln |T_0 (1 - \beta h)| \Rightarrow \ ln |\frac{T_0}{T_0 (1 - \beta h)}| = ln |\frac{1}{(1 - \beta h)}| = ln |(1 - \beta h)|^{-1} = - ln |(1 - \beta h)| $$.

Where is the mistake?

Sorry, Thats fine. I missed that you factored out $T_o$.

But also apply log power rule to the factor out front to bring it inside the logarithm as an exponent. You want to be ending with this form, before exponentiation:

$$\ln (a) = \ln(b^c) $$

 

[Hak]

But also apply log power rule to the factor out front to bring it inside the logarithm

OK, I get:

$$ln |(1- \beta h) ^ {(1 - \frac{g}{T_0 \beta R})}|$$.

Right?

 

[erobz]

OK, I get:

$$ln |(1- T_0 \beta) ^ {(1 - \frac{g}{T_0 \beta R})}|$$.

Right?

Yeah, for the right hand side. Then exponentiate both sides.

 

[Hak]

I get:
$$\rho (z) = \rho_0 |(1- \beta h) ^ {(1 - \frac{g}{T_0 \beta R})}|$$.

Right?

 

[erobz]

I get:
$$\rho (z) = \rho_0 |(1- \beta h) ^ {(1 - \frac{g}{T_0 \beta R})}|$$.

Right?

I get the negative of that exponent. $\left( \frac{g}{RT_o \beta} - 1 \right)$

 

[Hak]

I get the negative of that exponent. $\frac{g}{RT_o \beta - 1}$

Yes, you are right, I forgot to include it.

 

[Hak]

$$\rho (z) = \rho_0 |(1- \beta h) ^ {(\frac{g}{T_0 \beta R} - 1)}|$$ is the right expression.

 

[Hak]

Now, what should be done?

 

[erobz]

$$\rho (z) = \rho_0 |(1- \beta h) ^ {(\frac{g}{T_0 \beta R} - 1)}|$$ is the right expression.

you don't need the absolute values

 

[Hak]

you don't need the absolute values

OK, thanks. Now?

 

[erobz]

OK, thanks. Now?

Now, it should be clear that the function given for $\rho$ in the problem statement does not strictly adhere to the ideal gas law assumption that is used to solve the problem. Using ideal gas law, we see a power law. In the problem it was linear.

 

[Hak]

Now, it should be clear that the function given for $\rho$ in the problem statement does not strictly adhere to the ideal gas law assumption that is used to solve the problem. Using ideal gas law, we see a power law. In the problem it was linear.

Right, thank you. What conclusions can be drawn from this?

 

[erobz]

Right, thank you. What conclusions can be drawn from this?

I don't know, but it could be that the linear function approximates the power law reasonably well over these altitudes, or they simply chose a function for $\rho$ which makes the problem solvable algebraically. Yet another interpretation is that the linear function is empirical, and its the ideal gas law that was the issue. You have to plot the functions to see why they might have done what they have done.

 

[Hak]

I don't know, but it could be that the linear function approximates the power law reasonably well over these altitudes, or they simply chose a function for $\rho$ which makes the problem solvable algebraically. You have to plot the functions to see why they might have done what they have done.

I will try to trace them. However, I am noticing that in the expression of the problem a term $\alpha$ pops up, whereas here there is only one term in $\beta$. What does this mean?

 

[Hak]

The only real difference between the two functions seems to be that the function given by the text has an unlimited domain, whereas the one we have just calculated has an upper limit, so it has a limited domain. There does not seem to be any substantial difference.

 

[erobz]

I will try to trace them. However, I am noticing that in the expression of the problem a term $\alpha$ pops up, whereas here there is only one term in $\beta$. What does this mean?

$\alpha$ either characterizes the linear function they made up for $\rho ~\text{vs.}~ h$ or the linear function they observe for $\rho~ \text{vs.}~ h$, just as $\beta$ characterizes the approximately linear function they observe for $T$ vs. $h$ in the troposphere. You have to do some calculation to see what is what. I can't tell you anymore, because I'm not doing all that.

 

[Hak]

The only real difference between the two functions seems to be that the function given by the text has an unlimited domain, whereas the one we have just calculated has an upper limit, so it has a limited domain. There does not seem to be any substantial difference.

What do you think?

 

[erobz]

What do you think?

Then I think you have solved the mystery. The linear function approximates the power law ( strict ideal gas law treatment ) very well over the domain in question, and (this is the important part) ...it also makes the problem algebraically solvable. Or it too is an empirical result ( i.e. the atmosphere isn't really quite ideal ) like the temperature function, and it makes the problem solvable. It doesn't matter either way.

 

[Hak]

Then I think you have solved the mystery. The linear function approximates the power law ( strict ideal gas law treatment ) very well over the domain in question, and (this is the important part) ...it also makes the problem algebraically solvable. Or it too is an empirical result like the temperature function, and it makes the problem solvable. It doesn't matter either way.

OK, thank you, @erobz. You gave me a huge amount of help.

 

[hutchphd]

This is not rocket science.....it is balloon science. As the balloon ascends into cooler air that effect will increase the lift of the 100 deg (differential??) T this is not stated clearly at all. The air pressure is of course exponential with z so this prblem is sort of silly. Assume a uniform temperature and the simple exponential atmosphere and try for a good solution. This is a simple exercise, unless one chooses to make it otherwise. Yikes.

 

[Hak]

This is not rocket science.....it is balloon science. As the balloon ascends into cooler air that effect will increase the lift of the 100 deg (differential??) T this is not stated clearly at all.

I struggle to understand this statement. Could you explain it again? Thank you very much.

 

[Hak]

The air pressure is of course exponential with z so this prblem is sort of silly. Assume a uniform temperature and the simple exponential atmosphere and try for a good solution. This is a simple exercise, unless one chooses to make it otherwise. Yikes.

Excuse me, but I cannot understand what the crux of what you want to say is. Could you explain that too? Thank you again.

 

[hutchphd]

$$\rho(z,T)=\rho(z=0,T)e^{-\alpha z}$$and from the Ideal gas equation $$F_{lift}=\rho gV\frac{\Delta T} T$$the buoyancy is the mass of the air expelled because the ballon is heated. Also as mentioned N.B. $$\alpha\approx1/(20km)$$
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