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Trying to understand weather balloons better (ideal gas law)

by amwest
Tags: balloons, ideal, weather
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amwest
#1
May23-11, 11:30 PM
P: 7
Here's my understanding....
(P1V1)/T1 = (P2V2)/T2
P1 = starting pressure, typically sea level pressure which is 1atm.
V1 = Starting volume of the gas inside your balloon.
T1 = starting temperature, temperature at ground level.

P2 = final pressure, this should be the pressure at your flight altitude.
V2 = final volume, should be 100% of balloon capacity.
T2 = final temperature, temp at flight altitude.
Now i'm trying to understand this from a lighter than air gas standpoint, so...
Do i assume that:
P1, T1 are what i stated above, and do i assume that P2, T2 are proportional, if the gas isn't heated by a secondary source such as a hot air balloon?

Thanks in advance, will post further questions once these are validated or corrected!
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Andrew Mason
#2
May23-11, 11:54 PM
Sci Advisor
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P: 6,677
Quote Quote by amwest View Post
Here's my understanding....
(P1V1)/T1 = (P2V2)/T2
P1 = starting pressure, typically sea level pressure which is 1atm.
V1 = Starting volume of the gas inside your balloon.
T1 = starting temperature, temperature at ground level.

P2 = final pressure, this should be the pressure at your flight altitude.
V2 = final volume, should be 100% of balloon capacity.
T2 = final temperature, temp at flight altitude.
Now i'm trying to understand this from a lighter than air gas standpoint, so...
Do i assume that:
P1, T1 are what i stated above, and do i assume that P2, T2 are proportional, if the gas isn't heated by a secondary source such as a hot air balloon?

Thanks in advance, will post further questions once these are validated or corrected!
Temperature will be the ambient atmospheric temperature. Pressure will be the pressure that the balloon fabric can support and will be at least ambient atmospheric pressure. To determine whether the balloon will have positive buoyancy at a given T, P and V, what other parameter will you have to take into account (hint: apply ideal gas law)?

AM
amwest
#3
May24-11, 12:31 AM
P: 7
The weight of the gas?

amwest
#4
May24-11, 12:40 AM
P: 7
Trying to understand weather balloons better (ideal gas law)

Quote Quote by Andrew Mason View Post
Pressure will be the pressure that the balloon fabric can support and will be at least ambient atmospheric pressure.
AM
i understand that the pressure has to be low enough to not pop the balloon. What do you mean by at least ambient atm pressure?
rcgldr
#5
May24-11, 01:13 AM
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P: 7,119
Quote Quote by amwest View Post
What do you mean by at least ambient atm pressure?
I think the point here is that the pressure inside the balloon will be slightly higher or at least equal to the ambient pressure outside the balloon, depending on tension in the balloon's surface.

If there wasn't a limit on how much the balloon could expand (about 100 to 1 volume wise), a hydrogen filled weather balloon could reach the outer layer of atmosphere, above low orbiting objects.
amwest
#6
May24-11, 02:01 AM
P: 7
Quote Quote by rcgldr View Post
I think the point here is that the pressure inside the balloon will be slightly higher or at least equal to the ambient pressure outside the balloon, depending on tension in the balloon's surface.

If there wasn't a limit on how much the balloon could expand (about 100 to 1 volume wise), a hydrogen filled weather balloon could reach the outer layer of atmosphere, above low orbiting objects.
ok that makes sense.

So basicly a balloon is limited by:
V1 amount of gas needed to take off
then
V2 gas volume is maxed out at whatever altitude.
amwest
#7
May24-11, 02:13 AM
P: 7
So if i had V1 = 1m^3 of hydrogen weight of 0.0899kg/m^3
how do i determine what its V2 would be at different heights?
rcgldr
#8
May24-11, 02:59 AM
HW Helper
P: 7,119
Quote Quote by amwest View Post
how do i determine what its V2 would be at different heights?
The unknown is how much the pressure inside the balloon is affected by the tension in the balloon. You could assume zero tension and then assume pressure inside and outside the balloon is the same. You'll need to find pressure versus altitude forumulas / tables (wiki might have this).
amwest
#9
May24-11, 11:43 PM
P: 7
Quote Quote by rcgldr View Post
The unknown is how much the pressure inside the balloon is affected by the tension in the balloon. You could assume zero tension and then assume pressure inside and outside the balloon is the same. You'll need to find pressure versus altitude forumulas / tables (wiki might have this).
ok i've looked and found a few tables but not very complete and i've read that the tables are exponential but haven't found a fomula.
Andrew Mason
#10
May25-11, 10:43 AM
Sci Advisor
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P: 6,677
Quote Quote by amwest View Post
ok i've looked and found a few tables but not very complete and i've read that the tables are exponential but haven't found a fomula.
Try this link. If you want to factor temperature into it, you will have to find how temperature decreases with altitude.

AM


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