# Is Exponential Form Better for Balloon with Inserted Load Calculation?

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

Last edited:
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.

Hak said:
Okay, I will wait. Thank you.
Also, never mind about neglecting it, its actually a large weight in comparison to the payload.

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...

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##?

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?

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
$$\rho_{in} = \rm{const.}$$

erobz said:
Its a given?

$$\rho_{out} = \rho_o ( 1 - \alpha h )$$
It is part of the known data of the problem.

Hak said:
It is part of the known data of the problem.
So you can immediately write:

$$F(h) = ( \rho_{out} - \rho_{in} ) V g = ( \rho_o ( 1 - \alpha h ) - \rho_{in} ) V g$$

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....

Hak said:
And then equalize it to the weight force?
They are the only forces I see acting?

erobz said:
They are the only forces I see acting?
What other forces are involved?

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.

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...

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.

erobz said:
Ignore it, its not necessary.
What should I ignore?

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?

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....

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.

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....

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.

erobz said:
I see what I did...##\rho_{in}## is not constant.
Don't bow out, you could give me a precious help...

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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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....

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.

DeBangis21 and erobz
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}##.

Hak said:
##\alpha = 49 \ m^{-1}## and ##\beta = 26 \ m^{-1}##.
I think you went the wrong way there...

hutchphd
erobz said:
I think you went the wrong way there...
Why?

hutchphd

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