Fluids mechanics Helium balloon

Click For Summary

Discussion Overview

The discussion centers around the problem of determining the altitude at which a spherical helium balloon, with a diameter of 2.5 m and a mass of 6.7 kg, will settle when released into the U.S. Standard Atmosphere. The scope includes theoretical reasoning related to buoyancy and hydrostatic pressure in gases.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes using the balance of forces, stating that the weight of the balloon and the buoyant force must equal zero.
  • Another participant suggests using the hydrostatic pressure formula for gases to solve for altitude, indicating the need to find pressure at that altitude.
  • A third participant provides resources for air properties versus altitude, hinting at the use of standard atmospheric data.
  • One participant mentions calculating the density of the balloon and referencing a table, arriving at an altitude of 4000 m.

Areas of Agreement / Disagreement

Participants express differing approaches to solving the problem, with no consensus on the method or final altitude reached. Multiple competing views remain regarding the calculations and assumptions involved.

Contextual Notes

Limitations include the dependence on specific atmospheric models and the need for accurate pressure values at various altitudes, which are not fully resolved in the discussion.

teknodude
Messages
150
Reaction score
0
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 = buoyant 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
 
Physics news on Phys.org
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.
 
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
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
10K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 3 ·
Replies
3
Views
8K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K