Is Exponential Form Better for Balloon with Inserted Load Calculation?

[erobz]

OK, but what pressure value should I enter? I don't think I understand...

You found that the variation is temperature in the troposphere is approximately linear. That agrees with $T = ( T_o( 1- \alpha h )$ (at least as a realistic model). Whether or not the slope is approximately correct in this problem you must compare it with a reputable source that states its value. In my fluids text book they state the temperature in the troposphere decreases linearly with increasing altitude at a lapse rate of $5.87 \left[\frac{\text{K}}{ \text{km}}\right]$.

As for $P$ you don't enter a value, you enter the relationship( i.e. the model that relates the variables $P$, $\rho$ and $T$ )

 

[Hak]

You found that the variation is temperature in the troposphere is approximately linear. That agrees with $T = ( T_o( 1- \alpha h )$ (at least as a realistic model). Whether or not the slope is approximately correct in this problem you must compare it with a reputable source that states its value. In my fluids text book they state the temperature in the troposphere decreases linearly with increasing altitude at a lapse rate of $5.87 \left[\frac{\text{K}}{ \text{km}}\right]$.

As for $P$ you don't enter a value, you enter the relationship( i.e. the model that relates the variables $P$, $\rho$ and $T$ )

This relationship would be $P =\rho R T$?

 

[Hak]

This relationship would be $P =\rho R T$?

Also, may I know what your fluids text book is called?

 

[erobz]

Also, may I know what your fluids text book is called?

Engineering Fluid Mechanics: Crowe,Elger,Williams,Roberson, Ninth Edition.

 

[Hak]

Yes.

Then search "variation of temperature in troposphere vs altitude".

Then if you want to assume the ideal gas law model use that empirical result for the temperature to solve the hydrostatic differential equation for the density and compare with $\rho = \rho_o( 1 - \alpha h)$

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

From this differential equation, I obtain:

$$\rho(h) = k e^{- \frac{gh}{RT}}$$.

What should I get, comparing with the expression of $\rho$ given in the text?

 

[erobz]

From this differential equation, I obtain:

$$\rho(h) = k e^{- \frac{gh}{RT}}$$.

What should I get, comparing with the expression of $\rho$ given in the text?

How about you show some steps?

 

[Hak]

How about you show some steps?

Yes, I show the steps now, but is this result right?

 

[erobz]

Yes, I show the steps now, but is this result right?

I don't know, It's not an explicit result in my book. I would like to see the steps you have taken to solve the equation. If each main step was executed correctly, then the result would be too.

 

[Hak]

I don't know, It's not an explicit result in my book. I would like to see the steps you have taken to solve the equation. If each main step was executed correctly, then the result would be too.

OK, I will insert all the steps shortly. In case my result is right, what should I get from comparing the two expressions? In other words, what considerations should I make?

 

[erobz]

OK, I will insert all the steps shortly. In case my result is right, what should I get from comparing the two expressions? In other words, what considerations should I make?

One step at a time please. If you show some basic steps how you arrived at the solution ( verifying if it is correct or not- something you are very persistent in having others do in your problems), we can move forward with what it means.

 

[Hak]

Yes.

Then search "variation of temperature in troposphere vs altitude".

Then if you want to assume the ideal gas law model use that empirical result for the temperature to solve the hydrostatic differential equation for the density and compare with $\rho = \rho_o( 1 - \alpha h)$

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

Preface: I am very insistent that others check my work because this is the only way to understand whether my theories and procedures are reasonable or not. I hope I have not annoyed anyone.

Below are the steps I took to calculate the differential equation $\frac{dP}{dh} = -\rho g \Rightarrow \ RT \frac{d \rho}{dh} = -\rho g$.

This equation can be written as:

$$\rho' + \frac{g}{RT} \rho = 0$$.

It is a first-order linear ordinary differential equation of the type:

$$y'(h) + a_0 (h) y(h) = 0$$, which has solution $y(h) = k \cdot e^{-A(h)}$, with $A(h) = \int a_0(h) dh$.

Since, in this case, $A(h) = \frac{g}{RT} h$, we have:

$$\rho (h) = k \cdot e ^{- \frac{g}{RT} h}$$

What do you think?

 

[erobz]

Below are the steps I took to calculate the differential equation $\frac{dP}{dh} = -\rho g \Rightarrow \ RT \frac{d \rho}{dh} = -\rho g$.

$T$ isn't a constant, you can't take it out of the derivative.

$$ \frac{d}{dh} ( \rho T ) = - \frac{\rho g }{R} $$

Continue from here.

 

[Hak]

$T$ isn't a constant, you can't take it out of the derivative.

$$ \frac{d}{dh} ( \rho T ) = - \frac{\rho g }{R} $$

Continue from here.

You are right. I'm having difficulty solving this ODE, I can't figure out how to work with another dependent variable from $h$. Do you have any advice?

 

[erobz]

You are right. I'm having difficulty solving this ODE, I can't figure out how to work with another dependent variable from $h$. Do you have any advice?

$T$ is a function of $h$, use the product rule to complete the derivative on the LHS.

 

[Hak]

$T$ is a function of $h$, use the product rule to complete the derivative on the LHS.

Following your advice, I arrive at the equation:

$$\rho' T + T' \rho+ \frac{g}{R} \rho = 0 $$, from which:

$$\rho' T + T_0 \beta \rho + \frac{g}{R} \rho = 0 $$. Therefore:

$$\rho' T + \left(T_0 \beta + \frac{g}{R} \right) \rho = 0$$, whence:

$$\rho' + \frac{1}{T} \left(T_0 \beta + \frac{g}{R} \right) \rho = 0$$.

Following the procedure of the previous message, we have:

$$A(h) = \int \frac{1}{T(h)} \left(T_0 \beta + \frac{g}{R} \right) dh = \left(T_0 \beta + \frac{g}{R} \right) \int \frac{1}{T_0(1 + \beta h)} dh = \left(T_0 \beta + \frac{g}{R} \right) \frac{ln |1 + \beta h|}{T_0 \beta} h + c$$. Finally:

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

What do you think?

 

[erobz]

There is a sign error here, note that $T = T_o - T_o \beta h$.

$$\rho' T + T_0 \beta \rho + \frac{g}{R} \rho = 0 $$.

$$\rho' + \frac{1}{T} \left(T_0 \beta + \frac{g}{R} \right) \rho = 0$$.

Once you fix the sign error write

$$ \frac{d \rho}{dh} = \cdots $$

Then separate variables and integrate both sides.

 

[Hak]

Ok, I'm fixing the sign error.$$\rho' T + T' \rho + \frac{g}{R} \rho = 0 $$, from which:
$$\rho' T - T_0 \beta \rho + \frac{g}{R} \rho = 0 $$. Therefore:
$$\rho' T + \left(\frac{g}{R} - T_0 \beta \right) \rho = 0$$, whence:
$$\rho' + \frac{1}{T} \left(\frac{g}{R} - T_0 \beta \right) \rho = 0$$.
Following the procedure of the previous message, we have:
$$A(h) = \int \frac{1}{T(h)} \left(\frac{g}{R} - T_0 \beta \right) dh = \left(\frac{g}{R} - T_0 \beta \right) \int \frac{1}{T_0(1 - \beta h)} dh = - \left(\frac{g}{R} - T_0 \beta \right) \frac{ln |1 - \beta h|}{T_0 \beta} h + c$$. Finally:
$$\rho (h) = k \cdot e ^{\left(\frac{g}{R} - T_0 \beta \right) \frac{ln |1 - \beta h|}{T_0 \beta } h + c}$$.
Is it really necessary to trace the ODE back to a separable variable differential equation? Is my procedure wrong? I can't do it directly, however...

 

[erobz]

Is it really necessary to trace the ODE back to a separable variable differential equation? Is my procedure wrong?

Yes. You are not getting the proper result.

 

[Hak]

Yes. You are not getting the proper result.

Okay. I get to the equation:

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

I don't know how to handle this equation by integrating both members....

 

[erobz]

Okay. I get to the equation:

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

I don't know how to handle this equation by integrating both members....

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

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

 

[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. $$\alpha\approx1/(20km)$$
REVISED