Is Exponential Form Better for Balloon with Inserted Load Calculation?

[Hak]

Do you all confirm this procedure and its numerical solution? Or is there some other more subtle demonstration? Thank you very much.

$h \approx 11 \ km$ seems an exaggerated value, don't you think?

 

[erobz]

$h \approx 11 \ km$ seems an exaggerated value, don't you think?

Not according to an internet search of maximum height reached by hot air balloon.

 

[Hak]

Not according to an internet search of maximum height reached by hot air balloon.

All right, thank you. Then I await confirmation, tell me if anyone is not convinced by this process and result. Thank you.

 

[erobz]

All right, thank you. Then I await confirmation, tell me if anyone is not convinced by this process and result. Thank you.

What is it that you want to investigate? Whether or the claims $T = T_o(1 - \beta h )$ and $\rho = \rho_o( 1-\alpha h )$ are realistic atmospheric models in the troposphere, or whether or not you've solved it correctly for the models given in the problem (because the latter has obviously been verified - at least for the process)?

 

[Hak]

What is it that you want to investigate? Whether or the claims $T = T_o(1 - \beta h )$ and $\rho = \rho_o( 1-\alpha h )$ are realistic atmospheric models in the troposphere, or whether or not you've solved it correctly for the models given in the problem (because the latter has obviously been verified - at least for the process)?

Both. Does the second option mean that the one above is the correct symbolic method to solve the problem?

 

[erobz]

Does the second option mean that the one above is the correct symbolic method to solve the problem?

Yes.

Both.

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 $\rho = \rho_o( 1 - \alpha h)$

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

 

[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 $\rho = \rho_o( 1 - \alpha h)$

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

OK, thank you. So, I have to substitute the expression of $\rho$ in the differential equation you reported below, and then what? What would be the empirical values to substitute?

 

[Hak]

OK, thank you. So, I have to substitute the expression of $\rho$ in the differential equation you reported below, and then what? What would be the empirical values to substitute?

This is what I found out.
https://seattlecentral.edu/qelp/set...ude because,above by incoming solar radiation.

 

[erobz]

OK, thank you. So, I have to substitute the expression of $\rho$ in the differential equation you reported below, and then what? What would be the empirical values to substitute?

I would substitute in for the Pressure $P$, since the objective is to solve for $\rho$.

 

[Hak]

I would substitute in for the Pressure $P$, since the objective is to solve for $\rho$.

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

 

[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.

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