Fluid Mechanics of a weather balloon

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A discussion on the fluid mechanics of a weather balloon covers several problems involving buoyancy, pressure, and flow rates. For the weather balloon, the required volume to lift a 4600kg package is calculated to be approximately 4725 m^3, considering the densities of helium and air. The average density of a submarine accelerating upward at 0.325 m/s^2 needs further clarification, as initial guidance is requested. For pumping water to the Empire State Building, the calculated gauge pressure at the base is around 3.69x10^6 Pa. Lastly, assistance is sought for determining the speed of water from a faucet and the gauge pressure in a main pipe, with initial calculations provided but requiring further direction.
python023
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Here are a couple problems I would like to check and a couple I'm not sure how to do:

1.3)A 650kg weather balloon is designed to lift a 4600kg package. What volume should the balloon have after being inflated with helium at 0C and 1 atm pressure to lift the total load?

Density of helium = .179 pa.
.179 = (650 + 4600)/Volume.
Volume = 29000m^3?

1.4) A submerged submarine alters its buoyancy so that it initially accelerates upward at 0.325m/s^2. What is the submarines average density at this time?

The density of seawater is 1.025x10^3 kg/m^3. From here I have no idea where to go. Can someone give me a hint?

2.2) Water is to be pumped to the top of the Empire State Building, which is 366m tall. What gauge pressure is needed in the water line at the base of the building to raise the water to this height?

Based on a formula from my book, Absolute pressure = atmospherice pressure + (density x free-fall acceleration x depth). So,
1.01x10^5 pa + (1.00x10^3 x 9.81 x 366) = 3.69x10^6 pa?

3.3) The water supply of a building is fed through a main entrance pipe that is 6.0cm in diameter. A 2.0cm diameter faucet tap position 2.00m above the main pipe fills a 2.5x10^-2 m^3 container in 30.0s. What is the speed of the faucet and What is the gauge pressure in the main pipe?

The cross-sectional area of the main pipe is .0113 meters^2. The cross-sectional area of the faucet tap is .0013m^2. The flow rate from the faucet is .00083m^3/s. From here I do not know where to go, can someone guide me in the right direction?

Thank you.
 
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that's funny. i have the exact same problems in my textbook!
 
Holt Physics 2002. You wouldn't happen to have the answers too? :smile:
 
bump please
 
bump again
 
First, density can't be measured in pascals, which measures pressure.

You have to find ratios of the density of the fluid displaced vs the density of the fluid used, and when you multiply a density by volume you get either a force or mass, depending on whether you use mass or force density.
 
The .179 pa on the first problem was a typo. So for the first problem i don't simply use density = mass/volume? I don't believe I know what the displacement will be do I? I only know the mass of the objects, the density of helium and the density of air. So how should I set up my equation?
 
Draw a force diagram. You have the weight of the balloon and package directed down. You have the necessary Volume * (force density of air - force density of helium) as the bouyant force directed up.
 
So, 5250 kg = (1.29kg/m3-.179kg/m3)V
Volume = 4700m^3
?
 
  • #10
Looks close, 4725 m^3, assuming your densities are correct. That would be about an 18 meter diameter balloon.
 
  • #11
Yes that's what I had before rounding to 2 significant digits, on 1.1. Thank you.
Any insight on 1.4 or 3.3?
 
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