Is Exponential Form Better for Balloon with Inserted Load Calculation?

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The discussion focuses on calculating the buoyant force acting on a balloon with inserted load, using Archimedes' principle and the ideal gas law. The participants explore the relationship between the densities of the air inside and outside the balloon, emphasizing the importance of temperature and pressure variations with altitude. A key point raised is the need to correctly account for the buoyant force and the average density of the balloon system, which includes the air inside, the envelope, and the load. There is a consensus that simplifying assumptions can lead to significant errors in height calculations, and the correct interpretation of the variables involved is crucial. Ultimately, the conversation highlights the complexities of buoyancy calculations in variable atmospheric conditions.
  • #51
Hak said:
Here, I get totally stuck. Where do I go wrong?
A few notes, you don't need to convert by adding ##273##, just note that the temperatures used are to be absolute.

What is being suggested (as an alternative approach) is at equilibrium ##\rho_{avg} = \rho_{out} ##. You can write that immediately as:

$$\frac{\rho_{in} V + m}{V} = \rho_{out} $$

Work from there by subbing the relationships you found earlier for ##\rho_{in}## as a function of ##h## and the relationship you are given for ##\rho_{out}##.
 
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  • #52
erobz said:
A few notes, you don't need to convert by adding ##273##, just note that the temperatures used are to be absolute.

What is being suggested (as an alternative approach) is at equilibrium ##\rho_{avg} = \rho_{out} ##. You can write that immediately as:

$$\frac{\rho_{in} V + m}{V} = \rho_{out} $$

Work from there by subbing the relationships you found earlier for ##\rho_{in}## as a function of ##h## and the relationship you are given for ##\rho_{out}##.
What would be the relationship I previously found that would express ##\rho_{in}## as a function of ##h##. It currently escapes me. Could you clarify? Thank you.
 
  • #53
Hak said:
What would be the relationship I previously found that would express ##\rho_{in}## as a function of ##h##. It currently escapes me. Could you clarify? Thank you.

The relationship for the density of the air inside the balloon at a given altitude ##h##.
 
  • #54
erobz said:
The relationship for the density of the air inside the balloon at a given altitude ##h##.
Where did I find it?
 
  • #55
Hak said:
Where did I find it?
Don’t worry about where you found it. Can you find it now?
 
  • #56
erobz said:
Don’t worry about where you found it. Can you find it now?
All I can say about ##\rho_{in}## is: $$\frac{\rho_{in}}{\rho_{out}} = \frac{t_{out}}{t_{in}}$$. I cannot see the immediate dependence on ##h##.
 
  • #57
Hak said:
All I can say about ##\{rho_{in}## is: $$\frac{\rho_{in}}{\rho_{out}} = \frac{t_{out}}{t_{in}}$$. I cannot see the immediate dependence on ##h##.
Rearrange that expression...

##\rho_{in} = ? ##
 
  • #58
erobz said:
Rearrange that expression...

##\rho_{in} = ? ##
##\rho_{in} = \rho_{out} \frac{t_0}{t_{in}} (1-\beta h)##?
 
  • #59
Hak said:
##\rho_{in} = \rho_{out} \frac{T_{o}}{T_{in}} (1-\beta h)##?
We use capital ##T## for temperature, ##t## is time.

Keep going... There is another substitution to make.
 
  • #60
erobz said:
We use capital ##T## for temperature, ##t## is time.
OK, thanks. So, is ##\rho_{out} - \frac{m}{V} = \rho_{out} \frac{T_0}{T_{in}} (1-\beta h)##, with ##\rho_{out} = \rho_0 (1 - \alpha h)##, correct?
 
  • #61
Which value of ##T_0## should I use? With ##T_0 \approx 300 K##, I get two negative results. Therefore, this is not correct.
 
  • #62
Hak said:
Which value of ##T_0## should I use? With ##T_0 \approx 300 K##, I get two negative results. Therefore, this is not correct.
Maybe the relationship was with °C after all. So it is as you thought.

$$T = T_o [{}^{\circ} C] ( 1 - \beta h ) + 273 [K] $$
 
  • #63
erobz said:
Maybe the relationship was with °C after all. So it is as you thought.
So? What should I do?
 
  • #64
erobz said:
Maybe the relationship was with °C after all. So it is as you thought.

$$T = T_o [{}^{\circ} C] ( 1 - \beta h ) + 273 [K] $$
Which value of ##T_0## should I use? With ##T_0 \approx 27^\circ C = 300 K##, I get two negative values, again. Perhaps ##T_0## is wrong...
 
  • #65
Hak said:
Which value of ##T_0## should I use? With ##T_0 \approx 27^\circ C = 300 K##, I get two negative values, again. Perhaps ##T_0## is wrong...
I would say:

$$ T_{out}[\text{K}] = 27[{}^{\circ}C] ( 1 - \beta h ) + 273[\text{K}] $$

$$T_{in} [ \text{K}] = 100[{}^{\circ}C] + 273[ \text{K}] $$
 
  • #66
erobz said:
I would say:

$$ T_{out}[\text{K}] = 27[{}^{\circ}C] ( 1 - \beta h ) + 273[\text{K}] $$

$$T_{in} [ \text{K}] = 100[{}^{\circ}C] + 273[ \text{K}] $$
We have two negative results...
 
  • #67
Hak said:
We have two negative results...
You mean for the height ##h##?
 
  • #68
erobz said:
You mean for the height ##h##?
Yes.
 
  • #69
Hak said:
Yes.
Does this equation rearrange to the first equation we derived?
 
  • #70
erobz said:
Does this equation rearrange to the first equation we derived?
I don't think so, the results are different. Something is not right, the values are either too small or too large...
 
  • #71
Hak said:
I don't think so, the results are different. Something is not right, the values are either too small or too large...
Hak said:
OK, thanks. So, is ##\rho_{out} - \frac{m}{V} = \rho_{out} \frac{T_0}{T_{in}} (1-\beta h)##, with ##\rho_{out} = \rho_0 (1 - \alpha h)##, correct?
I didn't process this ##\uparrow##

The equation is.

$$ \frac{ \rho_{in}V + m }{V} = \rho_{out} $$

How did you get what you did above?
 
  • #72
erobz said:
I didn't process this ##\uparrow##

The equation is.

$$ \frac{ \rho_{in}V + m }{V} = \rho_{out} $$

How did you get what you did above?
I do not understand. Didn't we say we could match the average density ##\rho_{avg}## to the external density ##\rho_{out}##?
 
  • #73
Hak said:
I do not understand. Didn't we say we could match the average density ##\rho_{avg}## to the external density ##\rho_{out}##?
Yeah, that it what I think is meant by the equation you quote? it is saying at the equilibrium position the average density of the entire balloon system will equal the density of the surrounding atmosphere at ##h##.
 
  • #74
erobz said:
Yeah, that it what I think is meant by the equation you quote? it is saying at the equilibrium the average density of the entire balloon system will equal the density of the surrounding atmosphere at ##h##.
OK, then? I can't grasp the nettle...
 
  • #75
Hak said:
OK, then? I can't grasp the nettle...
Show your work on that equation please, if it doesn't work out then so be it, but we need to see how you rearranged the equation.
 
  • #76
Hak said:
All I can say about ##\rho_{in}## is: $$\frac{\rho_{in}}{\rho_{out}} = \frac{t_{out}}{t_{in}}$$. I cannot see the immediate dependence on ##h##.
I did not understand. Would you like me to show how I equated this equation above with the last one you quoted?
 
  • #77
Hak said:
I did not understand. Would you like me to show how I equated this equation above with the last one you quoted?
Start with the equation in post #71 and show the algebraic manipulations and substitutions you performed, because when I do those, I get the same result we got earlier.
 
  • #78
erobz said:
Start with the equation in post #71 and show the algebraic manipulations and substitutions you performed, because when I do those, I get the same result we got earlier.
I get:

$$\rho_{in} + \frac{m}{V} = \rho_{out} \Rightarrow \ \rho_{in} + \frac{m}{V} = \rho_{0} (1 - \alpha h)$$.

From ##\frac{\rho_{in}}{\rho_{out}} = \frac{T_{out}}{T_{in}}##, we get: ##\rho_{in} = \rho_{out} \frac{T_{out}}{T_{in}}##. Plugging in the previous one:

$$\rho_{out} \frac{T_{out}}{T_{in}} - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$. Finally:

$$\rho_{0} (1 - \alpha h)\frac{T_{0}}{T_{in}} (1 - \beta h) - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$.
 
  • #79
Hak said:
I get:

$$\rho_{in} + \frac{m}{V} = \rho_{out} \Rightarrow \ \rho_{in} + \frac{m}{V} = \rho_{0} (1 - \alpha h)$$.

From ##\frac{\rho_{in}}{\rho_{out}} = \frac{T_{out}}{T_{in}}##, we get: ##\rho_{in} = \rho_{out} \frac{T_{out}}{T_{in}}##. Plugging in the previous one:

$$\rho_{out} \frac{T_{out}}{T_{in}} - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$. Finally:

$$\rho_{0} (1 - \alpha h)\frac{T_{0}}{T_{in}} (1 - \beta h) - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$.
Which is the same result you initially found ( applying Newtons Second )?
 
  • #80
erobz said:
Which is the same result you initially found ( applying Newtons Second )?
Yes. I have discordant values, however. Now the two values come out to me as ##h \approx 32.6 \ m## and ##h \approx 11 \ km##. They are not convincing at all. Could you tell me what numerical result you get?
 
  • #81
Hak said:
$$\rho_{0} (1 - \alpha h)\frac{T_{0}}{T_{in}} (1 - \beta h) - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$.
As you did before:

$$\rho_{0} (1 - \alpha h)\frac{T_{0}}{T_{in}} (1 - \beta h) - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$.

## \gamma = \frac{T_{0}}{T_{in}} ##

$$\rho_{0} (1 - \alpha h) \gamma (1 - \beta h) - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$.

Multiply everything by ( or divide by) ##-1## :

$$-\rho_{0} (1 - \alpha h) \gamma (1 - \beta h) +\rho_{0} (1 - \alpha h) - \frac{m}{V} = 0$$
or (rearranging)
$$ \rho_{0} (1 - \alpha h) -\rho_{0} (1 - \alpha h) \gamma (1 - \beta h) - \frac{m}{V} = 0 $$

Factor:

$$\rho_{0} (1 - \alpha h) ( 1 - \gamma (1 - \beta h) ) - \frac{m}{V} = 0 $$

Multiply both sides by ##V##:

$$\rho_{0} (1 - \alpha h) ( 1 - \gamma (1 - \beta h) )V - m = 0 $$

That is the same relationship I got! Multiply everything by ##g## and we are back at the working that used Newtons Second Law directly.
 
  • #82
erobz said:
As you did before:

$$\rho_{0} (1 - \alpha h)\frac{T_{0}}{T_{in}} (1 - \beta h) - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$.

## \gamma = \frac{T_{0}}{T_{in}} ##

$$\rho_{0} (1 - \alpha h) \gamma (1 - \beta h) - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$.

Multiply everything by ( or divide by) ##-1## :

$$-\rho_{0} (1 - \alpha h) \gamma (1 - \beta h) +\rho_{0} (1 - \alpha h) - \frac{m}{V} = 0$$.
or
$$ \rho_{0} (1 - \alpha h) -\rho_{0} (1 - \alpha h) \gamma (1 - \beta h) - \frac{m}{V} = 0 $$

Factor:

$$\rho_{0} (1 - \alpha h) ( 1 - \gamma (1 - \beta h) ) - \frac{m}{V} = 0 $$

Multiply both sides by ##V##:

$$\rho_{0} (1 - \alpha h) ( 1 - \gamma (1 - \beta h) )V - m = 0 $$

That is the same relationship I got! Multiply everything by ##g## and we are back at the working that used Newtons Second Law directly.

Yes, correct. The only question is whether the method is correct, but it seems to me that it is.
 
  • #83
Hak said:
Yes, correct. The only question is whether the method is correct, but it seems to me that it is.
Where did all the stuff go about you were saying both signs are flopped ect... By deleting it you are making me look like a fool that is parroting everything you have already said for no reason...

Instead, simply say "I'm sorry, you are correct"
 
  • #84
erobz said:
Where did all the stuff go about you were saying both signs are flopped ect... By deleting it you are making me look like a fool that is parroting everything you have already said for no reason...
I did not understand this statement. Sorry, I'm not a native English speaker...
 
  • #85
Hak said:
I did not understand this statement. Sorry, I'm not a native English speaker...
You deleted all the stuff about disagreeing with the results being the same. Instead of just saying "yeah I goofed, sorry"

1696087560806.png
 
  • #86
erobz said:
You deleted all the stuff about disagreeing with the results being the same. Instead of just saying "yeah I goofed, sorry"
Yes, because I realised after not even five seconds that I had made a mistake. Of course, I didn't delete it to avoid saying 'I was wrong, sorry', but only because I realised that your statement was right. I'm not a cocky guy, next time I won't delete, sorry...
 
  • #87
May I ask what your final numerical result is?
 
  • #88
Let's wait for someone else to verify. I'm currently baking a cake for my youngest daughters birthday, Then I have a wedding to attend.
 
  • #89
erobz said:
Let's wait for someone else to verify. I'm currently baking a cake for my youngest daughters birthday, Then I have a wedding to attend.
Sorry, I couldn't have known. Here in Italy it's almost evening, I didn't notice the time difference. Happy birthday to your daughter, and have a wonderful evening at the wedding! Felicitations.
 
  • #90
Do you all confirm this procedure and its numerical solution? Or is there some other more subtle demonstration? Thank you very much.
 
  • #91
Hak said:
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?
 
  • #92
Hak said:
##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.
 
  • #93
erobz said:
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.
 
  • #94
Hak said:
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)?
 
  • #95
erobz said:
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?
 
  • #96
Hak said:
Does the second option mean that the one above is the correct symbolic method to solve the problem?
Yes.
Hak said:
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)##

$$ \frac{dP}{dz} = -\rho g $$
 
  • #97
erobz said:
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?
 
  • #99
Hak said:
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##.
 
  • #100
erobz said:
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...
 
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