Buoyancy and Balloons [SOLVED] (Topic)

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Buoyancy and Balloons [SOLVED]

Sep 16 2010, 3:49 am
By:

payne

Pages: < 1 2 3 >

Oct 7 2010, 8:03 pm

Vrael Post #21

I've been trying to solve that differential equation, and I'm almost positive it can't be solved explicitly. I asked around a bit, and have been fiddling with it from time to time over the past week or so, and I tried all the elementary functions and their derivatives, I tried the Laplace transform, and I finally went to a computer with Mathematica installed on it and the answer I got for the laplace transform was something like (e^-Ms/k)*(CoshIntegral(Ms/k) - ln(-k/M) - ln(M/k) - SinhIntegral(Ms/k)) / k, which isn't going to help find an explicit solution at all.

You can solve this numerically however, by choosing all the values in the equation and writing a program to iterate over a very small time interval, calculate the new acceleration/velocity/position based on the old a/v/Y, and on and on ect. The accuracy of the iterative program depends on how small your time interval is mostly. I can probably elaborate more on this if necessary.

None.

Oct 8 2010, 7:08 pm

CecilSunkure Post #22

Lol, so it sounds like my solution is the solution

I can try and help as well if you need.

None.

Oct 12 2010, 5:09 am

payne Post #23

:payne:

@Cecil: I am allowed to code a simulator.
Any help is welcome!

@Vrael: If my teacher sees some Laplace, he's going to be extremely happy, since aside from the project, this is the only theory we're learning, because he was saying it was applied everywhere... I guess he was right! :><:

Could you get me a picture of the equation, it'd help me visualize. Identifying which variables I'd need to set would be great too!

Anyways, I have an economy exam tomorrow, a chemistry exam after tomorrow, and some freaking important shit all the rest of the week, except on Saturday, which will basically be the only day I'll be able to study on for my math exam on Monday... I'm going to hate this week!

Insomnia ftw!

P.S. I have found that the top height the balloon will reach is something like 1150 (can't tell out of memory). But from there, it'll start oscillating... I'll have to calculate an average height based on the two first maximums of the oscillating stage. Shouldn't be hard at all: I have everything I need. But I just wanted to keep you guys informed.

None.

Oct 12 2010, 5:19 am

Vrael Post #24

The Laplace transform is a mathematical technique used to help solve differential equations. For a given function f(x),

L{ f(x) } = the integral from 0 to positive infinity of (e^-st)*f(x).

I don't think the Laplace transform is going to help you at all in your project, I couldn't solve the differential equation, and the Mathematica solver couldn't either. (Well technically it could, but the expression is ridiculously complex and not going to help). I can teach you more about it if you want, but beyond that it won't be useful to your project unless you just want to show your professor you tried it.

None.

Oct 12 2010, 5:23 am

payne Post #25

:payne:

Quote from

Vrael

Na, I'll try to stay concrete. After all, it's also for me: we're rated on the final oral we'll produce as well...

None.

Oct 13 2010, 8:31 pm

payne Post #26

:payne:

Weird... I've calculated that to reach its maximal altitude (1135 meters), it'd take the balloon 102 seconds...
Its speed, at some point, reached 17 m/s... this is ridiculous

I guess I should take in count the frictional force, but I can't find a formula that seems to apply to my situation. I remember seeing it somewhere once, but I just can't find it back. D:

EDIT: Well, since I couldn't find it, I decided to do my 2 pages of calculations on "Generalized formulas dependant of the altitude". I have established a(h), which contains among others B_F(h).
Once I find how to calculate the mass in function of temperature, which basically simply implies that I find p_int(h), I'll be able to move to the next part: finding the fuel costs to maintain 70�C.
The third and last part of the project is to calculate the time it'll take to the balloon to reach the ground once it cannot heat anymore (no more fuel). With this time, we'll be able to calculate the distance it has crossed during the step. Main concept that we'll use for this: heat trade (between balloon and air).

You now know exactly how the rest of the weeks should go. If you have any informations that could help me at some point, please feel free to share! :awesome:

Post has been edited 2 time(s), last time on Oct 13 2010, 10:22 pm by payne.

None.

Oct 28 2010, 1:50 am

payne Post #27

:payne:

Okay, I'm done with the first step, being "Reach the maximal height" (which is 1124 meters). However, I've calculated that it'll take the balloon less than 100 seconds to reach that height, which doesn't make sense.
I'm wondering if it could be because I still haven't taken in count the Friction Force yet. Could this force really big enough to make a significant difference in this time, knowing our balloon's volume is of 3000 m^3 ? One of my partners is trying to calculate this force, without much success, but for now, it looks like it isn't going to influence that much the sum of the forces applied to the balloon.

So, hum... anyone has an idea of how to calculate the friction?

P.S. I'm also done with the second step of the project, which is to completely use our fuel. All that's left to calculate now is the time it'll take to the balloon to "crash" with, as the only variable that affects its temperature (which is the only way to make its height vary), the exterior's temperature. Shouldn't be too hard since I'll basically reuse the same equations as the one I've used in the 2nd step.

Nearly done, gogogo!

(But it's kind of a bad news, because I'll still have to produce 2 pages of calculations per week until the end of the session... I guess I'll just polish some stuff, and take count of more variables.

)

None.

Oct 28 2010, 2:05 am

Vrael Post #28

Uhg. Essentially, surface area of the balloon times the drag coefficient for whatever material you're using times the velocity squared.
Ff = A*D*v^2

That might not be exactly right, but it should be close.

None.

Oct 28 2010, 2:08 am

payne Post #29

:payne:

If the balloon is a sphere, this means the surface area is half the total surface of the sphere, right?

None.

Oct 28 2010, 2:16 am

Vrael Post #30

Quote from

payne

If the balloon is a sphere, this means the surface area is half the total surface of the sphere, right?

That's a good first approximation, yeah.

None.

Nov 8 2010, 11:41 am

payne Post #31

:payne:

I'm currently trying to determine the time it'll take to the balloon to go from 1124 meters to 0.
The method I had in mind uses 2 variables that are defined by each other, thus creating a circular definition, which isn't going to help me in any way.
Any ideas?

Nota Bene: For this part, I consider that the only temperature exchange is done between the environment and the balloon. There aren't any source that will keep the interior temperature to 70 �C and it will thus decrease. As it decreases the altitude decreases, and this is how it is supposed to reach the ground after a certain amount of time which I need to calculate.

None.

Nov 8 2010, 1:47 pm

poison_us Post #32

Back* from the grave

Uhm, post the formulas. I'm in Calc I and I can solve derivatives with two variables, and I know people are in higher maths than me that could probably solve it much easier. I'm too lazy to sift through all the formulas to be able to know ;o

Nov 9 2010, 4:45 am

payne Post #33

:payne:

http://www.mediafire.com/?6dmd3qckks1o7ad

Everything's in there.
Again, let me remind you that the last part is incorrect: the calculations I'm using at the moment give a circular reference and thus doesn't offer me any solution.

None.

Nov 9 2010, 5:06 am

Vrael Post #34

well im pretty sure in reality people save fuel for the descent, think about it:

If the temperature in the balloon keeps decreasing and decreasing, the balloon will fall faster and faster, until it hits the ground SMACK.

If you dont mind a crash landing, you'll probably just be able to use the same setup as going upwards, but this time the buoyant force will be much smaller and the net acceleration will be downwards.

None.

Nov 11 2010, 4:16 am

payne Post #35

:payne:

Quote from

Vrael

My team and I did agree that we wanted our balloon to crash.
Anyways, problem is that using the same setup doesn't work: it creates a circular argument.

None.

Dec 2 2010, 12:56 am

payne Post #36

:payne:

I used Euler's method to approximate a value to the precedent value which permitted me to avoid the circular argument without much repercussions.

I'm done with my part of the project. The total time of the travel of the balloon is of exactly 20364.0536981009 seconds.
:wtfawesome:

Now I gotta write off this report of over 30 pages >_>

Post has been edited 1 time(s), last time on Dec 2 2010, 1:04 am by payne.

None.

Dec 6 2010, 7:50 am

payne Post #37

:payne:

http://www.mediafire.com/?r7mkbub7xa8i9s1

Have fun reading this 1.28 MB Microsoft Word file of 30 pages! :awesome:

At least, I suggest taking a look at the very first page's "Biography" section where we were authorized to put a picture of ourselves. :hurr:

None.

Dec 7 2010, 11:03 pm Dungeon-Master Post #38

Quote from

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 30 pages right there: http://www.mediafire.com/?r7mkbub7xa8i9s1

TL;DR but YAY a french word!!! (Mongolfi�re)!!!

None.

Dec 7 2010, 11:11 pm

payne Post #39

:payne:

Ya, the whole report is in French, unfortunately, and I don't feel like translating 30 pages into English.

In the end, this report isn't final. The professor troll'd us saying the final was for this monday, and when every teams (2) handed their project, he admitted it was a fake date: he wanted to be able to correct what we thought would be final in order to help us get a better note. In the end, everyone is winning and I'm pretty happy this was fake because we have some serious tweaks to bring to the organization of the report itself.

Final due for incoming Monday!

None.

Dec 13 2010, 9:40 pm

payne Post #40

:payne:

42 pages, final: http://www.mediafire.com/download.php?951z2mjirjd6rx5

None.

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