Assuming the Balloon Can Freely Expand, Calculate the Volume of the Balloon at This Altitude.
Summate how loftier balloon rises before stopping
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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 bounding main level (an altitude of 0.00 meters). Using the Standard Altitude/Density Tabular array 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 non pop when rising.
The mass density of helium at i.00 atm and 30°C is 0.180 kg/k3.
Homework Equations
I may exist incorrect almost 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 still just 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 elevation
I don't have a specific volume to use to solve for mass of helium and density of air changes at different altitudes and so I don't think this is the right form for this problem.
Using the nautical 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 similar .18 kg/g^3 match upwards with only over 1 atm in the nautical chart. I'm wondering if the temperature has any effect on the balloon. I hateful, temperature changes the higher y'all get and from my express knowledge, I don't think it has an effect.
Please don't solve information technology for me every bit I'd similar to figure this out on my own. I'm just having trouble finding relevant equations for this problem or something...
Answers and Replies
2. Information technology can exist increased by one if yous 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 feel whatsoever upward acceleration anymore).
3. So there is an old greek (name beginning with A) who has something to say about that; in that location is another relevant equation popping upwardly !
one. Your relevant equation is superseded past the table given. Your number of available relevant equations is now null, unfortunately.
2. It can be increased by ane if y'all enquire yourself the question "Coming to a stop" is the equivalent of what in terms of forces ? (There is no dynamics involved, it simply doesn't experience any upward acceleration anymore).
iii. 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 considering:
∑F = 0 = Fb - mg
then mg = Fb
Furthermore Fb = pvg (density of air * book of displaced fluid * grav) = mg (mass of fluid * grav)
pvg = mg
pv = one thousand
p = k/five = ph
Then it's not pressure, it'southward the buoyant force. However, I can't do much with the equation in that location. Yet, I know at present that if I have two gases of unlike density, the one with lower density volition rise. And then at a certain point the balloon will reach a betoken where density inside equals density outside and will cease to rise or autumn.
Density of helium for this problem is given. 0.18 kg/m^iii. So I just accept to notice 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 at present though. Thanks for your help!
Did yous find it's cerise cold upwards high in the atmosphere ? The helium might cool downwardly a fiddling on its way up ! There's a law connecting force per unit area, volume and temperature...
I'm bold that must exist the ideal gas law.
Pressure * Volume = n*R*Temp
I'm guessing calculus is needed at this bespeak to discover the rate of modify of pressure over temp. Volume, n, and R remain constant I assume considering the radius of the airship is 'very large.'
This volition take some reading every bit I never learned this yet. I'thousand assuming this means that equally temperature decreases, volume is compressed thus leading to greater pressure level. According to the chart, this means information technology would exist a type of elevate forcefulness.
Did you notice it'due south ruddy cold upwards loftier in the atmosphere ? The helium might absurd downwardly a little on its way upward ! There'southward a law connecting pressure, volume and temperature...
I'g assuming that must be the ideal gas police.Pressure * Volume = n*R*Temp
I'm guessing calculus is needed at this betoken to find the charge per unit of alter of pressure over temp. Volume, n, and R remain constant I assume considering the radius of the airship is 'very big.'
This will take some reading as I never learned this withal. I'm bold this means that as temperature decreases, volume is compressed thus leading to greater pressure. According to the nautical chart, this means it would be a type of drag force.
Y'all might desire to reconsider using the ideal gas law. The problem statement did specify that the mylar is "is very thin, very stiff, and completely inelastic." I have 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 airship?
[Edit: By the style, with my above proffer, I'm assuming that the airship was initially filled with helium to its full volume, per the problem argument saying, "A Mylar airship is filled with helium to a pressure level just greater than 1.00 atm."
This isn't typically washed with weather balloons. Ofttimes, the balloon is simply partially filled with helium (i.e. pressure of i atm when on the ground, but significantly less volume than the balloon's maximum volume). That allows the balloon'south volume to significantly expand equally it rises, assuasive the airship to reach much greater heights than it would otherwise, had information technology been initially filled to its full volume.
http://en.wikipedia.org/wiki/Weather_balloon
Simply in this problem, it sounds to me as if the airship was initially filled to its full book, since its initial pressure was "merely greater than 1.00 atm."]
I see. If volume is constant so so would its density since mass is also constant. Therefor it remains at .18 kg/m^iii. Thus information technology remains around xiv-16 grand meters. Not plenty info and as well restrictive of information to come upwards with anything more than than that. The way the answer function is setup looks like it isn't looking for an exact answer so I suppose that must exist good plenty.
Sounds expert to me.
This problem actually screams for a role (b) though. Information technology should and then ask you something like, "(b) Estimate the maximum altitude reached past the balloon if it was initially filled to only 1/iii its maximum book (filled at xxxo C, ane.00 atm pressure, at sea level, etc.).
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