Staredit Network > Forums > Lite Discussion > Topic: Buoyancy and Balloons [SOLVED]
Buoyancy and Balloons [SOLVED]
Sep 16 2010, 3:49 am
By: payne
Pages: 1 2 3 >
 

Sep 16 2010, 3:49 am payne Post #1

:payne:

To all the Math/Physic nerds of SEN...
I'm in a program named "Math's Integration". Over all, we choose a subject at start and work on it all through the year, handing about 2-3 pages of calculations per week (obligatory!).

My subject: What will be the distance travelled by a balloon having 'x' litres of 'x' fuel (taking in count the initial elevation from the ground and the arrival's fuel used).
'x' being a value we just haven't decided yet, so if you want to calculate stuff, just set a constant as 'x' and you'll be fine.

The thing is... I can't get to figure out how to apply the Buoyancy equations on this situation...
I'm trying to start with an extremely simplified situation, though I still fail to apply the equation: I'm not counting any friction, wind, balloon deformation due to pressure changes, non-uniform air density inside the balloon due to the heat not being spread equally and instantly, etc.

B = -p*V*g
If I set heights, find approximated values of the air density, and apply the equation, I'm fine, but this really won't get me anywhere.

I do understand the balloon will eventually reach an equilibrium state... but how to calculate its height?
Am I supposed to find what is the density that makes the Buoyancy equal to the negative force that represents the gravity and then find a table that permits me to evaluate an approximate height based on this air density?
I'm really, really lost!
Please, please, intelligent people, help me! :(

EDIT: PARTIALLY SOLVED: http://www.staredit.net/261831/

EDIT: COMPLETELY SOLVED: http://www.staredit.net/271126/

Get my final report of 42 pages right there: http://www.mediafire.com/download.php?951z2mjirjd6rx5

Post has been edited 9 time(s), last time on Dec 13 2010, 9:40 pm by payne.



None.

Sep 16 2010, 5:05 am CecilSunkure Post #2



You can consider the fuel to be directly proportional to acceleration. You then can break up the direction vector into it's three velocity components x, y and z. You then apply acceleration (gravity z, fuel z, wind x and wind y) to each of the components, and move the balloon each timestep by the newly calculated component velocities. I'm unsure if you're allowed to code a simulation to solve this, but you can just store the current height each timestep in a variable, unless the new height is lower than the previous. At the end of the simulation, you'll have the final position of the balloon, and the maximum height.

If you know the equation of the parabola the balloon will travel in, you would just solve for the root(s) of the equation.

Try to get Mini Moose or Vrael in here, and they can give you a better answer. I also have jabok on MSN, I think he went or is in college for math.



None.

Sep 16 2010, 5:32 am Vrael Post #3



This really depends on how many factors you're taking into account. Essentially though, the basic idea here (I think) is that the buoyancy force minus the gravitational force is equal to the mass of the balloon times the acceleration of the balloon. That is where I would start, but that does not take into account mass change due to burning of fuel, drag on the balloon due to its velocity (though I'd imagine this would not affect the distance much since balloons typically travel at low velocities), air density changes, ect.

Up until about 36,000 feet the density of air remains approximately the same, if your balloon will be breaching this barrier then past that point you'll need to take into account changes in air density (unless you're being super-precise and need to do this anyway). The buoyancy force is caused by a pressure differential. As you descend in a fluid like air, the pressure increases by p*g*h, where p = density, g = gravity, and h = height descended. On the bottom of the balloon the pressure is then greater than at the top of the balloon, causing a net upwards force. If this net upwards force is greater than the force of gravity downward, the balloon moves up. Eventually as the balloon gets high enough, it will reach a point where the buoyancy force upwards is equal to the gravitational force downwards, in which case it will probably just continue upwards for a bit dependant on its velocity at the time it reaches that particular equilibrium point, and then descend a bit and re-ascend as it passes equilibrium, oscillating for a bit, then level off at the equilibrium point, where F(buoyancy) = F(gravity), so the acceleration is 0 (in the vertical direction). I think this will be your height, but calculating this height may be hard. I would start by finding the density of the heated air in your balloon, and finding a height where the density of the outside air is close to the density of the hot air.

If you want to add in factors like mass loss due to fuel burning, and drag forces, ect, you'll start getting into differential equations, and I'm not 100% sure how that will affect the maximum height without putting more thought into it. There are some obvious things, like less weight due to fuel will allow it to fly higher by decreasing the downwards force, but I'd need to almost do the entire problem for you to get you a good answer with these factors taken into account.



None.

Sep 16 2010, 11:16 am payne Post #4

:payne:

Quote from Vrael
This really depends on how many factors you're taking into account. Essentially though, the basic idea here (I think) is that the buoyancy force minus the gravitational force is equal to the mass of the balloon times the acceleration of the balloon. That is where I would start, but that does not take into account mass change due to burning of fuel, drag on the balloon due to its velocity (though I'd imagine this would not affect the distance much since balloons typically travel at low velocities), air density changes, ect.

Up until about 36,000 feet the density of air remains approximately the same, if your balloon will be breaching this barrier then past that point you'll need to take into account changes in air density (unless you're being super-precise and need to do this anyway). The buoyancy force is caused by a pressure differential. As you descend in a fluid like air, the pressure increases by p*g*h, where p = density, g = gravity, and h = height descended. On the bottom of the balloon the pressure is then greater than at the top of the balloon, causing a net upwards force. If this net upwards force is greater than the force of gravity downward, the balloon moves up. Eventually as the balloon gets high enough, it will reach a point where the buoyancy force upwards is equal to the gravitational force downwards, in which case it will probably just continue upwards for a bit dependant on its velocity at the time it reaches that particular equilibrium point, and then descend a bit and re-ascend as it passes equilibrium, oscillating for a bit, then level off at the equilibrium point, where F(buoyancy) = F(gravity), so the acceleration is 0 (in the vertical direction). I think this will be your height, but calculating this height may be hard. I would start by finding the density of the heated air in your balloon, and finding a height where the density of the outside air is close to the density of the hot air.

If you want to add in factors like mass loss due to fuel burning, and drag forces, ect, you'll start getting into differential equations, and I'm not 100% sure how that will affect the maximum height without putting more thought into it. There are some obvious things, like less weight due to fuel will allow it to fly higher by decreasing the downwards force, but I'd need to almost do the entire problem for you to get you a good answer with these factors taken into account.

This answer is very great. It helped a lot!
To find the equilibrium point, I had made (g = B), which, in the end, pushed me to calculate "9,8 = 100 * x * 9,81" ('x' being the the unknown density of the air inside the balloon at the equilibrium height and '100' being the random volume value I decided to assign to my balloon). From there, I obtained a density of 1/100, which is extremely low. I couldn't find this value in any table, so it wasn't very useful.
And based on my research, a minimal volume for balloons is equal to 250, so it means the density would be even lower.
What am I doing wrong? :/

EDIT: I found a table the evaluates a 0,1 density somewhere between 11 Km and 20 Km of altitude... this is ridiculous! O.o



None.

Sep 16 2010, 11:24 pm Vrael Post #5



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

1.2041 kg/(m^3) is the approximate density of air at 20 degrees C (or 293 Kelvin) and 101kPa (kilo-pascals).

Your force equation is wrong actually. The sum of forces on the balloon must be 0 at the equilibrium point, so we have

0 = Fb - Fg
0 = Fb - 9.8*(mass of the balloon)
Fb = weight of fluid displaced, and since we know density = mass/volume, we can calculate the weight of air displaced based on the densities.
If say, the air balloon volume is 250 m^3, 250*1.2041 = 301.025kg of air at normal temperature.
The hot air has a lower density, but the same volume, lets say the density is p. 250*p (kg) = our new mass of air.
Then (301.025 - 250p) = mass of air displaced, and then 9.8 times this value is the weight of air displaced, also known as the buoyancy force.
so Fb = (301.025 - 250p)*9.8, for a balloon with volume 250 m^3.

going back to the first equation,
0 = (301.025 - 250p)*9.8 - 9.8(mass of the balloon)
If you know the mass of the balloon, you can solve for the new density value.



None.

Sep 16 2010, 11:55 pm payne Post #6

:payne:

Oh shi-! I thought it was ''weight of the fluid that displaces the environment-fluid''. :O
That for sure will my calculations out! :P

Okay, now let's say I have found my new density value, how do I determine the height of the equilibrium point? All I can think of is to try to use PV=nRT to find the pression, and then use the Barometric formula1, 2.
My teacher also told me the next step I should try to achieve after this is to try to find the speed at which the balloon elevates itselft. Should I use ''Forces Sum = m * a'' combined with some Motion Equation ?



1. http://fr.wikipedia.org/wiki/Formule_du_nivellement_barom%C3%A9trique#Formule_internationale_du_nivellement_barom.C3.A9trique
2. http://en.wikipedia.org/wiki/Barometric_formula
It looks like the French version of the article contains a formula which gives a direct relation between Height and Pression.



None.

Sep 17 2010, 8:29 pm Vrael Post #7



Hrm. I'm not sure how to numerically calculate the height of the equilibrium point, that's probably something you'll need to find in a textbook. What I would do is solve for the density of the outside air when Fb = Fg, then try to find a textbook/webpage that lists air density values at different elevations.

As for the speed of the balloon, velocity is the integral of acceleration with respect to time. a = dv/dt. Since you have Fb - Fg = ma,

(Fb - Fg)/m = a.
v = t*(Fb-Fg)/m

This is the basic equation, but this does not take into account the fact that the "m" of the balloon will decrease as the balloon uses up its fuel. I'm also not 100% sure if the mass includes the mass of the air inside the balloon. I think it does include it, but I'm not 100% sure.

Lets say that the fuel is burnt off at a rate of b kg/sec, and the starting mass of the whole balloon (including fuel) = M
then Mass at time t = M(t) = M - bt

Then, since mass is changing, Fg changes, and m changes. We get
Fg = 9.8*(M - bt)
and m = (M - bt)
so our equation for a changes
a = dv/dt = (Fb - 9.8*(M - bt))/(M - bt)
dv = (Fb - 9.8*(M - bt))/(M - bt) * dt
you can integrate it yourself, I don't have pen/paper with me, though I think there's a natural log in the answer



None.

Sep 20 2010, 9:51 pm payne Post #8

:payne:

Okay, back to work!
I've tried to apply your interpretation of the Buoyancy Force and found out something that didn't make sense at all: I required an interior density smaller than ''-1''... :/
So I talked to my professor, and while he didn't know much about the B.Force, we figured out this equation -should- work:

m * a = 0 = Fb - Fg = gravity * (volume of the balloon * densityext - mass of air inside of the balloon)

You think it makes sense? I believe the equation you've suggested me was taking in count the density of the air inside of the balloon when muliplying with the volume.



None.

Sep 21 2010, 5:15 am Vrael Post #9



Fb is dependant on the weight of fluid displaced, which we can determine from the densities.
lets say:
P1 = density of air inside balloon
P2 = density of air outside balloon
g = 9.8 m/s^2
V = volume inside balloon
M1 = mass of air inside balloon
M2 = mass of air inside equivelent volume as the balloon

Density times volume gives us mass, and mass times the acceleration due to gravity gives us weight.
So we have to determine the equivelent mass of outside air for a volume of equal size.
M2 = (P2)*V
M1 = (P1)*V
Then the mass of fluid displaced, is M2 - M1, and g times that will give us the weight of fluid displaced.
So,
Fb = g*(M2 - M1)
substituting in for M2 and M1,
Fb = g*(P2*V - P1*V)
Fb = g * V * (P2 - P1)
So your equation was wrong.
Fg = Total Mass of the Balloon * 9.8, lets call it M3 *g

So the total equation becomes g*V*(P2 - P1) - g*M3 = M3*a



None.

Sep 23 2010, 5:46 am payne Post #10

:payne:

If I use your formula (g*V*(P2 - P1) - g*M3 = M3*a), and set Forces Sum to 0, and try to determine the maximal density of the air inside the balloon for it to elevate, I still get non-sense (the variable disapears...).
I used the formula my professor deducted (though he says he isn't sure at 100% about the veracity of his formula): g*V*(P2) - g*M3 = M3*a.
Applying this one works pretty well and gives me reasonable results.

... Just wanted you to know. :/

Anyways, I found that the maximal density of the air inside the balloon should be 1,09666...
I am now at trying to find out the minimal temperature this implies. Using tables found on the Internet (I use 1% Relative Humidity) , I find a value of approximatively 49 Celsius degrees... which makes sense.
Time to try to calculate it "by hand". :P (Shouldn't be too hard.)

I'll keep you guys up to date. ;)

EDIT: Nevermind. How the hell am I supposed to find the temperature based on the density? :ermm:

EDIT: And once that's done, I have to calculate the speed at which the balloon will elevate itself with an interior temperature of 70C (arbitrary number). I'll also need to determine the maximal altitude it'll reach.
... no idea how I am going to finish this f*cking project, seriously. It's just a total mindfuck. :-(
I've already had a 0 because I hadn't produced at least 2 pages of calculations for the last class, and now I am supposed to write 2+2 pages of calculations for this morning... I'm only at 2 pages and I have no more inspiration. I wish I could still drop that course. :flamer:

Post has been edited 2 time(s), last time on Sep 23 2010, 6:56 am by payne.



None.

Sep 23 2010, 7:50 pm Vrael Post #11



Well, your proffessor's equation is wrong, since it doesn't take into account the density differential. If the buoyancy force was only dependent on the outside density of air, then there would be no point in heating the air inside to lower the density, the force would exist anyway.

So, the maximum inside density to get lets say 1kg of mass off the ground with a 100 m^3 balloon, would be like this:

9.8*100*(1.2041 - P1) - 9.8*1 = 0 (0 since any buoyancy larger would accelerate the balloon upwards)

980(1.2041 - P1) = 9.8
1.2041 - P1 = .01
1.1941 = P1

Of course, this is a very large balloon lifting a very small weight, so the density of the balloon doesnt need to be much lower than the outside air. What values are you using for the volume of the balloon and the mass of the balloon (V and M3)?

If you just need pages of calculations, what I suggest you do is just make giant tables for different values of maximum air density required to lift the balloon, like
V = 100, M = 1, P1 = ?
V = 100, M = 2, P1 = ?
...
V = 100, M = 50, P1 = ?

next table
V = 200, M = 1, P1 = ?
V = 200, M = 2, P1 = ?
...
V = 200, M = 50, P1 = ?

ect.

If you know any programming, you can easily take like 20 minutes to make a program to calculate about a bazillion values like this then copy paste it into a word document and print it.


Also remember, the max height will be dependent on both the minimum density of heated air inside, and the minimum density of the outside air
g*V(Po.min - Pin.min) - g*M3 = 0
At some point the density difference wont be large enough to lift the balloon any further because the density of outside air will decrease as the balloon rises.

Post has been edited 3 time(s), last time on Sep 23 2010, 8:11 pm by Vrael.



None.

Sep 24 2010, 3:42 am payne Post #12

:payne:

V = 3000 m^3
M3 = 450 + (pint*V)

450 Kg being the mass of the things attached to the balloon.

As for your equation, as I've said, I can't get to calculate anything with it since the unknown variables is being suppressed at some step.
... while my professor's actually giving me a reasonable answer. :/
Though I do understand why it could be a false equation.

Mindfuck situation ftw!

P.S. I can't try to bullshit my professor with a huge table... he won't count it as calculation. I really need to come up with brand new formulas each time, do at most 2 examples of applications of it applied to my situation, and repeat with an other formula until I reach 2 pages. :(



None.

Sep 24 2010, 3:59 am Vrael Post #13



Err, don't include the mass of heated air as part of M3. I realize in an earlier post I said that it might have been included as part of the mass of the balloon, sorry for confusing you about that, it does not count. Instead, what you could do is
M3 = mass of fuel + mass of balloon fabric + mass of balloon bucket
Then you can change the acceleration equation to the differential equation I outlined above, since M3 = K - fuel burn rate*time and acceleration will be time dependant, you can integrate to get velocity and height as a function of time.



None.

Sep 24 2010, 4:05 am payne Post #14

:payne:

Quote from Vrael
Err, don't include the mass of heated air as part of M3. I realize in an earlier post I said that it might have been included as part of the mass of the balloon, sorry for confusing you about that, it does not count. Instead, what you could do is
M3 = mass of fuel + mass of balloon fabric + mass of balloon bucket
Then you can change the acceleration equation to the differential equation I outlined above, since M3 = K - fuel burn rate*time and acceleration will be time dependant, you can integrate to get velocity and height as a function of time.
Why shouldn't I include balloon's air mass? Because it's an "open system" ?
And do I use this M3 mass that doesn't include the air's mass for the "m * a" part? :O



None.

Sep 24 2010, 8:11 pm Vrael Post #15



I believe its because the air inside the balloon is part of the fluid of the atmosphere.

And yes, use M3 without the mass of the air inside the balloon in the sum of forces equation.

I'll work on a more thorough physical explanation for you as for why the air inside the balloon doesn't count as part of the mass in the force equation, but you're right, numerically it doesn't make sense. I checked my equations with some data on wikipedia, and if you don't include the mass of air inside the balloon they match up exactly.

Edit:
I think the reason the air does not count as part of the accelerating mass is because its actually part of the ambient fluid, meaning it still exerts atmospheric pressure on the balloon. Since the air inside is not isolated from the air outside the balloon, the atmospheric pressure from the outside will translate inside and up into the hotter air inside the balloon, which then exerts that as upwards pressure on the balloon fabric. So in a sense, the air inside the balloon doesn't count towards the total mass because it is doing the pushing, and its supported by the atmosphere below it.

Post has been edited 1 time(s), last time on Sep 27 2010, 8:16 pm by Vrael.



None.

Sep 30 2010, 12:38 am payne Post #16

:payne:

Thank you very much for all your help, Vrael!
I've just found the perfect website for my problem.
Unfortunately for you, it is in French, but you could still use Google Translate if you're curious.

http://tpemontgolfieres.e-monsite.com/rubrique,2-4-conjecture-sur-l-altitude,561802.html
It basically explains everything I need to take in count in my calculations.

This can be locked.



None.

Sep 30 2010, 5:17 am payne Post #17

:payne:

In fact, I do have a little question: what happens to the air inside the balloon when it elevates?
Does the variation of pression affect it, considering pint is constant.

EDIT: Here is the document that will support my calculations pages: http://www.mediafire.com/?c4cto20ah0a57d7
I've found out my balloon would reach approx. 5586 meters of altitude! :awesome:
I'm now at trying to find the time it'll take it to reach this altitude with an interior temperature of 70 Celsius.
... no idea how I'll do it yet! >.<

Post has been edited 1 time(s), last time on Sep 30 2010, 9:20 am by payne.



None.

Sep 30 2010, 8:55 pm Vrael Post #18



Can you explain your question a bit better? The hot air rises to the top of the balloon, leaving colder air towards the bottom.

As for the time, its probably the equation I set up before.

V = integral of acceleration with respect to time.
Y = integral of velocity with respect to time
Y' = V, Y'' = a, where the ' = derivative
a = F/m
a = (Fb - Fg)/m
a = (g*V*(P2 - P1) - g*M3) / M3
If you use a constant temperature, P1 will probably be constant. g is constant, V is constant, but P2 and M3 will not be constant, since the higher you go the lower P2 is, and the longer you fly the more fuel you burn off. So if we say
M3 = M - kt (where k is the burn rate)
and
P2 = P0 - f(Y)
where f(Y) is some function of the height, X, of the balloon. Lets assume f(Y) = cY, a constant times Y for simplicity
we end up with a second order linear differential equation
Y'' = (g*V*(P0 - cY - P1) - g(M - kt))/(M - kt)
Y'' = g*V*(P0 - P1)/(M - kt) - g - g*V*c*Y/(M - kt)
lets call g*V*(P0-P1) = a since its constant, and g*V*c = b
Y'' = a/(M-kt) - g - bY/(M-kt)
Y'' + bY/(M-kt) = a/(M-kt) - g
(M - kt)Y'' + bY = a - g(M - kt)

Uhg. This equation is not homogeneous, so I'll need to find a textbook or something to solve it probably. I don't remember how off the top of my head.

Edit: If you can find the solution, then you can just plug in Y = 5586meters or whatever your height was and solve for t = time.



None.

Oct 2 2010, 6:37 am payne Post #19

:payne:

Hum... I just saw something's weird: what I've done in that Excel document... well, I use a ton of constants, but I've never used the temperature of the air inside the balloon... That shouldn't be possible since under a temperature of 49C, I've calculated the balloon wouldn't raise!
Mind-fuck...



None.

Oct 2 2010, 9:43 pm Vrael Post #20



Well that could simply be because the buoyant force is dependant on the density of the air, not the temperature. Density is dependant on temperature, but if you had already calculated the density from the temperature that might be why temperature isn't appearing in your equation.



None.

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