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.
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Homework Statement
A model AX-4 hot-air balloon consists of a rigid envelope of volume [tex]V=850m^3[/tex] with an opening at the lower end. The air inside is maintained at a temperature of [tex]100^\circ[/tex]. The balloon must lift a total load (envelope plus payload) of [tex]200kg[/tex].
Knowing that air density decreases with height according to the formula [tex]\rho=\rho_0(1-\alpha h)[/tex], with [tex]\alpha=0.049km^{-1}[/tex] and that the outside air temperature decreases as [tex]t=t_0(1-\beta h)[/tex] with [tex]\beta=0.026km^{-1}[/tex], calculate the maximum height the balloon can reach.
Relevant Equations
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My attempt is:

Archimedes' thrust on the balloon minus the weight of the air inside it is F=(\rho-\rho_{\rm in})Vg, where \rho_{in} is the density of the hot air inside.
Now we have, in general, \rho=\mu {n \over V}=\mu {P \over RT} from the perfect gas law (\mu would be the molar mass of the air, but we don't care about that, it just serves as a constant of proportionality).
But because inside and outside are communicating, the pressure is the same and we have
{\rho \over \rho_{in}}={t_{in} \over t} and thus.
F=(\rho-\rho_{in})Vg=\rho gV \left( 1-{t \over t_{in}}\right)= \rho_0gV(1-\alpha h)\left( 1-{t_0 (1-\beta h) \over t_{\rm in}}\right)
Let us write for convenience t_0/t_{in}=\gamma, and, equalizing this expression to the weight force of the load, we obtain:
mg=F= \rho_0gV(1-\alpha h)\left( 1- \gamma (1-\beta h) \right).
From here, I obtain a second-degree equation in h:

\rho_0 g V \alpha \beta \gamma h^2 - \rho_0 g V [\gamma (1- \beta) + \alpha] h - \rho g V (1-\beta) + mg = 0.

Now for a quick estimate of the quantities involved: the temperature on the ground is about t_0 \sim 300 K, from which \gamma \sim 300/373=0.80, while the density of air on the ground is \rho_0 \sim 1.2 kg m^{-3}

From here, I finally get h \approx 0.029 m, a value that seems exaggeratedly low to me. Where do I go wrong?
 
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Check your definition of the buoyant force.
 
erobz said:
Check your definition of the buoyant force.
How do you mean? I can't understand. Isn't it ##F = \rho V g##?
 
Hak said:
How do you mean? I can't understand. Isn't it ##F = \rho V g##?
Oh, I see...you took the force of weight of the air inside out as well.
 
erobz said:
Oh, I see...you took the force of weight of the air inside out as well.
So is that not correct?
 
Hak said:
So is that not correct?
No that's ok. You might be able to neglect it in comparison to the payload, but it's only the addition of a constant term, so it's not a big deal either way.
 
Okay. What might be wrong with my process? Do you have any ideas about that?
 
Hak said:
Okay. What might be wrong with my process? Do you have any ideas about that?
not yet.
 
erobz said:
not yet.
Okay, I will wait. Thank you.
 
  • #10
Hak said:
Okay, I will wait. Thank you.
Also, never mind about neglecting it, its actually a large weight in comparison to the payload.
 
  • #11
erobz said:
Also, never mind about neglecting it, its actually a large weight in comparison to the payload.
Yes, you are right, but such a low value of ##h## doesn't convince me, and I don't think it would change much by neglecting it...
 
  • #12
I think the ideal gas law relationship you are trying to force is over constraining it. You can write immediately the force ##F## as a function of altitude ##h##?
 
  • #13
erobz said:
I think the ideal gas law relationship you are trying to force is over constraining it. You can write immediately the force ##F## as a function of altitude ##h##?
I wouldn't know how to express it; I don't see any laws that can do it directly. Any suggestions?
 
  • #14
Hak said:
I wouldn't know how to express it; I don't see any laws that can do it directly. Any suggestions?
Its a given?

$$ \rho_{out} = \rho_o ( 1 - \alpha h ) $$
and
$$ \rho_{in} = \rm{const.}$$
 
  • #15
erobz said:
Its a given?

$$ \rho_{out} = \rho_o ( 1 - \alpha h ) $$
It is part of the known data of the problem.
 
  • #16
Hak said:
It is part of the known data of the problem.
So you can immediately write:

$$ F(h) = ( \rho_{out} - \rho_{in} ) V g = ( \rho_o ( 1 - \alpha h ) - \rho_{in} ) V g $$
 
  • #17
erobz said:
So you can immediately write:

$$ F(h) = ( \rho_{out} - \rho_{in} ) V g = ( \rho_o ( 1 - \alpha h ) - \rho_{in} ) V g $$
And then equalize it to the weight force? I don't know, I'm not really convinced because the text provides three other data. Why this data if it would not be used? Three seems like a high enough number to be simple distractors....
 
  • #18
Hak said:
And then equalize it to the weight force?
They are the only forces I see acting?
 
  • #19
erobz said:
They are the only forces I see acting?
What other forces are involved?
 
  • #20
Hak said:
What other forces are involved?
At equilibrium, I would say that is it. During ascent there is drag, but that is a non-issue for this question.
 
  • #21
erobz said:
At equilibrium, I would say that is it. During accent there is drag, but that is a non-issue for this question.
Right. So, how to get around the problem? As mentioned, there would be an overabundance of three data, way too many...
 
  • #22
Hak said:
Right. So, how to get around the problem? As mentioned, there would be an overabundance of three data, way too many...
Ignore it, its not necessary.
 
  • #23
erobz said:
Ignore it, its not necessary.
What should I ignore?
 
  • #24
Hak said:
What should I ignore?
Which piece of data (relationship) gets you directly to the result? Do you think you should ignore that one to instead make suppositions about atmospheric modeling with the ideal gas law?
 
  • #25
erobz said:
Which piece of data (relationship) gets you directly to the result?
I don't know. If you mean the process you recommended, although correct, it seems too simple. This problem comes from one of the most difficult exams in my country....
 
  • #26
Hak said:
I don't know. If you mean the process you recommended, although correct, it seems too simple. This problem comes from one of the most difficult exams in my country....
Then it must be over my head. I'll bow out.
 
  • #27
erobz said:
Then it must be over my head. I'll bow out.
Why do you want to bow out? It could be an interesting problem, and we could try to find the solution together....
 
  • #28
Hak said:
Why do you want to bow out? It could be an interesting problem, and we could try to find the solution together....
I see what I did...##\rho_{in}## is not constant.
 
  • #29
erobz said:
I see what I did...##\rho_{in}## is not constant.
Don't bow out, you could give me a precious help...
 
  • #30
Hak said:
Don't bow out, you could give me a precious help...
Its seems like what you did is correct.

I too would say:$$ \rho_o ( 1 - \alpha h ) \left( 1 - \gamma ( 1 - \beta h ) \right)Vg - mg = 0 $$
 
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  • #31
erobz said:
Its seems like what you did is correct.
Then I really don't know what to come up with. Let's see if someone else disproves my thesis, but this time my unfolding seems correct....
 
  • #32
Hak said:
a value that seems exaggeratedly low to me. Where do I go wrong?
When you substituted your numbers what values did you put in for ##\alpha## and ##\beta##? Show me the numbers.
 
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  • #33
kuruman said:
When you substituted your numbers what values did you put in for ##\alpha## and ##\beta##? Show me the numbers.
##\alpha = 49 \ m^{-1}## and ##\beta = 26 \ m^{-1}##.
 
  • #34
Hak said:
##\alpha = 49 \ m^{-1}## and ##\beta = 26 \ m^{-1}##.
I think you went the wrong way there...
 
  • #35
erobz said:
I think you went the wrong way there...
Why?
 
  • #36
Hak said:
Why?
## \alpha = 0.026 \frac{1}{km} = 0.026 \frac{1}{km}\frac{1 km}{1000 m} = 0.000026 \frac{1}{m} ##

etc...
 
  • #37
erobz said:
## \alpha = 0.026 \frac{1}{km} = 0.026 \frac{1}{km}\frac{1 km}{1000 m} = 0.000026 \frac{1}{m} ##

etc...
Damn! It's true!
 
  • #38
Hak said:
Damn! It's true!
@kuruman knocks it out of the park!
 
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  • #39
The balloon will stop rising when Its average density is equal to the density of the outside air. The average density is the total mass (air inside + envelope + load) divide by the external volume. I think we can safely ignore the buoyant force on the load. Find a relation between the density of the air inside and the density of the air outside when the pressure inside is equal to the pressure outside.
 
  • #40
kuruman said:
The balloon will stop rising when Its average density is equal to the density of the outside air. The average density is the total mass (air inside + envelope + load) divide by the external volume. I think we can safely ignore the buoyant force on the load. Find a relation between the density of the air inside and the density of the air outside when the pressure inside is equal to the pressure outside.
So, is my process incorrect?
 
  • #41
Hak said:
So, is my process incorrect?
I didn't say that. I told you what I would do to make the solution transparent. Put in the numbers and see what you get.
 
  • #42
kuruman said:
I didn't say that. I told you what I would do to make the solution transparent. Put in the numbers and see what you get.
I get, with my procedure, ##h \approx 784957 km## and ##h \approx 1.77 \times 10^{-3} m##. Where am I going wrong? Have you tried entering the numbers? If so, what do you get?

Edit. My actual value is ##h \approx 20.4 \ km##
 
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  • #43
erobz said:
Its seems like what you did is correct.

I too would say:$$ \rho_o ( 1 - \alpha h ) \left( 1 - \gamma ( 1 - \beta h ) \right) - mg = 0 $$
I must be missing something. How is ##\beta## relevant? The temperature inside is constant, and the density outside is a known function of height.
 
  • #44
haruspex said:
I must be missing something. How is ##\beta## relevant? The temperature inside is constant, and the density outside is a known function of height.
The outside air temperature decreases as t=t_0(1-\beta h)
 
  • #45
haruspex said:
I must be missing something. How is ##\beta## relevant? The temperature inside is constant, and the density outside is a known function of height.
I missed it at first too. The pressure inside the balloon is changing.

$$ 1 = \frac{P_{out}}{P_{in}} = \frac{\rho_{out} \cancel{R} T_{out}}{ \rho_{in} \cancel{R} T_{in}}$$
 
  • #46
Following my procedure, my value is ##h \approx 20.4 \ km##. This seems excessive.
 
  • #47
haruspex said:
I must be missing something. How is ##\beta## relevant? The temperature inside is constant, and the density outside is a known function of height.
Yes, there is an expression for ##\rho(h)##, but I think that this is the variation of density with height at constant temperature. We are given the variation of temperature with height in order to account for the effect of temperature on density.
 
  • #48
erobz said:
I missed it at first too. The pressure inside the balloon is changing.

$$ 1 = \frac{P_{out}}{P_{in}} = \frac{\rho_{out} \cancel{R} T_{out}}{ \rho_{in} \cancel{R} T_{in}}$$
Exactly, I looked at this very equation in my process.
 
  • #49
haruspex said:
I must be missing something. How is ##\beta## relevant? The temperature inside is constant, and the density outside is a known function of height.
The temperature inside is constant but the pressure is not. The density inside depends on the pressure outside and this pressure depends on both density and temperature outside.
 
  • #50
kuruman said:
The balloon will stop rising when Its average density is equal to the density of the outside air. The average density is the total mass (air inside + envelope + load) divide by the external volume. I think we can safely ignore the buoyant force on the load. Find a relation between the density of the air inside and the density of the air outside when the pressure inside is equal to the pressure outside.
Why my process is wrong, I cannot understand. I'll try to follow your way.

First, I would write down the given data and the unknown variable. Let h be the maximum height the balloon can reach, in kilometers. Then we have:
V=850m^3

m=200kg

T=100^\circ C=373K

\rho_0=1.2 kg/m^3

\alpha=0.049km^{-1}

t_0 \approx 300 K

\beta=0.026km^{-1}

Next, I would use the ideal gas law to relate the pressure, volume, temperature and number of moles of the air inside and outside the balloon. The ideal gas law is:
PV=nRT

where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature.

Since the balloon is open at the bottom, we can assume that the pressure inside is equal to the pressure outside at any height. Therefore, we can write:

P_{in}V_{in}=n_{in}RT_{in}

P_{out}V_{out}=n_{out}RT_{out}

where the subscripts in and out denote the inside and outside of the balloon, respectively.

Then, I would use the definition of density to express the number of moles in terms of mass and molar mass. The density is:
\rho=\frac{m}{V}

where \rho is the density, m is the mass, and V is the volume.

The molar mass is:

M=\frac{m}{n}

where M is the molar mass, m is the mass, and n is the number of moles.

Therefore, we can write:

n=\frac{m}{M}=\frac{\rho V}{M}

Substituting this into the ideal gas law equations, we get:

P_{in}V_{in}=\frac{\rho_{in} V_{in}}{M_{in}}RT_{in}

P_{out}V_{out}=\frac{\rho_{out} V_{out}}{M_{out}}RT_{out}

Next, I would simplify these equations by canceling out some terms. Since we are assuming that both the inside and outside air are composed of dry air with a constant molar mass of about 29 grams per mole, we can write:
M_{in}=M_{out}=M=0.029kg/mol

Also, since we are given that the volume of the envelope is constant, we can write:

V_{in}=V_{out}=V=850m^3

Therefore, we can simplify the equations as:

P_{in}=\frac{\rho_{in}}{M}RT_{in}

P_{out}=\frac{\rho_{out}}{M}RT_{out}

Then, I would equate these equations and solve for \rho_{in}/\rho_{out}. Since we are assuming that the pressure inside and outside are equal at any height, we can write:
\frac{\rho_{in}}{M}RT_{in}=\frac{\rho_{out}}{M}RT_{out}

Canceling out some terms and rearranging, we get:

\frac{\rho_{in}}{\rho_{out}}=\frac{T_{out}}{T_{in}}

Next, I would use the given formulas for \rho_{out} and T_{out} as functions of height to substitute them into this equation. We have:
\rho=\rho_0(1-\alpha h)

t=t_0(1-\beta h)

Therefore,

\rho_{out}=\rho_0(1-\alpha h)

T_{out}=t_0(1-\beta h)+273

where we have added 273 to convert the temperature from Celsius to Kelvin.

Substituting these into the equation for \rho_{in}/\rho_{out}, we get:

\frac{\rho_{in}}{\rho_0(1-\alpha h)}=\frac{t_0(1-\beta h)+273}{T_{in}}

Finally, I would use the definition of the average density of the balloon to express \rho_{in} in terms of the total mass and volume. We have:
\rho_{avg}=\frac{m_{total}}{V}

where \rho_{avg} is the average density, m_{total} is the total mass, and V is the volume.

The total mass is the sum of the mass of the air inside, the mass of the envelope, and the mass of the load. We can write:

m_{total}=m_{air}+m_{env}+m_{load}

The mass of the air inside is equal to the density of the air inside times the volume of the envelope. We can write:

m_{air}=\rho_{in}V

The mass of the envelope and the load are given as 200 kg. Therefore, we can write:

m_{env}+m_{load}=200kg

Substituting these into the equation for \rho_{avg}, we get:

\rho_{avg}=\frac{\rho_{in}V+200}{V}

Solving for \rho_{in}, we get:

\rho_{in}=\rho_{avg}-\frac{200}{V}

Substituting this into the equation for \rho_{in}/\rho_{out}, we get:

\frac{\rho_{avg}-\frac{200}{V}}{\rho_0(1-\alpha h)}=\frac{t_0(1-\beta h)+273}{T_{in}}

Here, I get totally stuck. Where do I go wrong?
 
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