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teknodude
Oct21-06, 05:22 PM
A spherical helium balloon is 2.5 m in diameter and has a total mass of 6.7 kg.
When released into the U. S. Standard Atmosphere, at what altitude will it settle?

My first thought was that diagram has the weight of the balloon acting downward, buoyant force upwards.


-weight + Fb = 0

Fb = bouyant force

-mg + (specific weight of air)(h2-h1)A =0

My initial equation was this:

-mg - PA(pressure) + Fb = 0 but i believe this one is wrong.

The answer is 4000 m

teknodude
Oct21-06, 09:30 PM
alright i think i have to solve for the altitude using the Hydrostatic pressure in gases formula.

P = Pa ( 1- B*z/T0)^(g/R&B)

Pa= atm pressure
B= lapse rate 0.00650 K/m
R = 287 (m^2)/(s^2*K)
T0 = 288.16K
z = altitude
Only problem is that i need to find the pressure at that to find the altitude z.

FredGarvin
Oct22-06, 06:56 AM
The hint is Standard US Atmosphere. If you have a textbook, there should be a table for the air properties vs. altitude. If you don't have one, look here:

http://www.pdas.com/e2.htm
http://scipp.ucsc.edu/outreach/balloon/atmos/1976%20Standard%20Atmosphere.htm

teknodude
Oct22-06, 02:42 PM
what a weird problem to assign especially for a section on bouyancy. I calculated the density of the balloon and looked it up on the table in my book and got 4000m.

Thanks