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Calculate how high balloon rises before stopping

  • Thread starter ichivictus
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Homework Statement



A Mylar balloon is filled with helium to a pressure just greater than 1.00 atm on a day when the temperature is 30°C. It is released near sea level (an altitude of 0.00 meters). Using the Standard Altitude/Density Table below, estimate the maximum altitude reached by the balloon.

Assume that the Mylar is very thin, very strong, and completely inelastic. Assume further that the radius of the balloon is very large. The balloon does not pop when rising.

The mass density of helium at 1.00 atm and 30°C is 0.180 kg/m3.

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Homework Equations



I may be wrong about these but this is what I believe is relevant so far:

P = Force/Area = pgh


The Attempt at a Solution



We haven't reached this in lecture yet but I am just studying ahead.

When the balloon comes to a stop, its weight will equal the atmospheric pressure.

Patm = mg
P = pgh = mg
p is the atmospheric density
ph = m
h = mass of helium / density of air at that height

I don't have a specific volume to use to solve for mass of helium and density of air changes at different altitudes so I don't think this is the right course for this problem.

Using the chart I match the standard density with helium density. The altitude is inbetween 14000 and 16000 and pressure inbetween 140 and 100 (x100Pa).

It looks like .18 kg/m^3 match up with just over 1 atm in the chart. I'm wondering if the temperature has any effect on the balloon. I mean, temperature changes the higher you go and from my limited knowledge, I don't think it has an effect.

Please don't solve it for me as I'd like to figure this out on my own. I'm just having trouble finding relevant equations for this problem or something...
 

Answers and Replies

  • #2
SteamKing
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The statement, "When the balloon comes to a stop, its weight will equal the atmospheric pressure." is incorrect. Is pressure the same as weight? The units of the two are different: weight is force and pressure is force per unit area.
 
  • #3
BvU
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1. Your relevant equation is superseded by the table given. Your number of available relevant equations is now zero, unfortunately.
2. It can be increased by one if you ask yourself the question "Coming to a stop" is the equivalent of what in terms of forces ? (There is no dynamics involved, it just doesn't experience any upward acceleration anymore).
3. Then there is an old greek (name beginning with A) who has something to say about that; there is another relevant equation popping up !
 
  • #4
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1. Your relevant equation is superseded by the table given. Your number of available relevant equations is now zero, unfortunately.
2. It can be increased by one if you ask yourself the question "Coming to a stop" is the equivalent of what in terms of forces ? (There is no dynamics involved, it just doesn't experience any upward acceleration anymore).
3. Then there is an old greek (name beginning with A) who has something to say about that; there is another relevant equation popping up !
Ah yes Archimedes! At static equilibrium the buoyant force will equal its weight because:

∑F = 0 = Fb - mg
so mg = Fb
Furthermore Fb = pvg (density of air * volume of displaced fluid * grav) = mg (mass of fluid * grav)
pvg = mg
pv = m
p = m/v = ph


So it's not pressure, it's the buoyant force. However, I can't do much with the equation there. However, I know now that if I have 2 gases of different density, the one with lower density will rise. So at a certain point the balloon will reach a point where density inside equals density outside and will cease to rise or fall.

Density of helium for this problem is given. 0.18 kg/m^3. So I just have to find what altitude has the same density for air. I really don't think the 30C matters.

So I'm sort of back to where I started... I think I get it now though. Thanks for your help!
 
  • #5
BvU
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Did you notice it's ruddy cold up high in the atmosphere ? The helium might cool down a little on its way up ! There's a law connecting pressure, volume and temperature...
 
  • #6
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Did you notice it's ruddy cold up high in the atmosphere ? The helium might cool down a little on its way up ! There's a law connecting pressure, volume and temperature...
I'm assuming that must be the ideal gas law.

Pressure * Volume = n*R*Temp

I'm guessing calculus is needed at this point to find the rate of change of pressure over temp. Volume, n, and R remain constant I assume considering the radius of the balloon is 'very large.'

This will take some reading as I never learned this yet. I'm assuming this means that as temperature decreases, volume is compressed thus leading to greater pressure. According to the chart, this means it would be a type of drag force.
 
  • #7
collinsmark
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Did you notice it's ruddy cold up high in the atmosphere ? The helium might cool down a little on its way up ! There's a law connecting pressure, volume and temperature...
I'm assuming that must be the ideal gas law.

Pressure * Volume = n*R*Temp

I'm guessing calculus is needed at this point to find the rate of change of pressure over temp. Volume, n, and R remain constant I assume considering the radius of the balloon is 'very large.'

This will take some reading as I never learned this yet. I'm assuming this means that as temperature decreases, volume is compressed thus leading to greater pressure. According to the chart, this means it would be a type of drag force.
You might want to reconsider using the ideal gas law. The problem statement did specify that the mylar is "is very thin, very strong, and completely inelastic." I take that as an instruction to assume that the balloon's volume should be considered constant. With that in mind, once filled, would a change in temperature affect the overall density of the balloon?

[Edit: By the way, with my above suggestion, I'm assuming that the balloon was initially filled with helium to its full volume, per the problem statement saying, "A Mylar balloon is filled with helium to a pressure just greater than 1.00 atm."

This isn't typically done with weather balloons. Often, the balloon is only partially filled with helium (i.e. pressure of 1 atm when on the ground, but significantly less volume than the balloon's maximum volume). That allows the balloon's volume to significantly expand as it rises, allowing the balloon to reach much greater heights than it would otherwise, had it been initially filled to its full volume.

http://en.wikipedia.org/wiki/Weather_balloon

But in this problem, it sounds to me as if the balloon was initially filled to its full volume, since its initial pressure was "just greater than 1.00 atm."]
 
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  • #8
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I see. If volume is constant then so would its density since mass is also constant. Therefor it remains at .18 kg/m^3. Thus it remains around 14-16 thousand meters. Not enough info and too restrictive of information to come up with anything more than that. The way the answer part is setup looks like it isn't looking for an exact answer so I suppose that must be good enough.
 
  • #9
collinsmark
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I see. If volume is constant then so would its density since mass is also constant. Therefor it remains at .18 kg/m^3. Thus it remains around 14-16 thousand meters. Not enough info and too restrictive of information to come up with anything more than that. The way the answer part is setup looks like it isn't looking for an exact answer so I suppose that must be good enough.
Sounds good to me. :smile:

This problem really screams for a part (b) though. It should then ask you something like, "(b) Estimate the maximum altitude reached by the balloon if it was initially filled to only 1/3 its maximum volume (filled at 30o C, 1.00 atm pressure, at sea level, etc.).
 

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