How Do You Calculate the Lifting Capacity of a Helium Balloon?

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Discussion Overview

The discussion revolves around calculating the lifting capacity of a helium balloon, focusing on the principles of buoyancy and the forces involved in lifting a payload. Participants explore various equations and concepts related to buoyant force, density differences, and the implications of these factors in practical scenarios.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Thought experiments

Main Points Raised

  • One participant inquires about the calculation of lifting capacity for a gas-filled balloon and whether this relates to buoyant force.
  • Another participant suggests using Archimedes's equation to determine the buoyant force based on the volume of gas and the density difference between the gas and air.
  • There is a request for additional equations that might apply to the upward force exerted by a gas.
  • Participants discuss the relationship between buoyant force, the weight of the balloon, and the payload it can lift, emphasizing the need for the buoyant force to exceed the total weight of the balloon and its contents.
  • A thought experiment is presented regarding a model balloon in a sealed container, questioning the weight indicated by a scale when the balloon is filled with helium and hovers inside the container.
  • Another participant reiterates the thought experiment, asserting that the total weight of the system remains unchanged when helium is transferred to the balloon.
  • Further discussion arises about the implications of buoyancy in bubbling water and its potential effects on ships, suggesting a need for deeper exploration of buoyancy principles in non-uniform fluids.

Areas of Agreement / Disagreement

Participants generally agree on the application of Archimedes's principle and the basic calculations for buoyant force. However, there are nuances in understanding the implications of weight distribution and buoyancy in different scenarios, particularly in thought experiments, indicating that multiple views remain on these aspects.

Contextual Notes

Some participants express uncertainty regarding the completeness of the equations and principles discussed, particularly in relation to the buoyant force's dependence on various factors such as density and payload weight. The discussion also touches on the complexities of buoyancy in non-uniform fluids, which may not be fully resolved.

Who May Find This Useful

This discussion may be of interest to individuals studying physics, particularly those focused on fluid dynamics, buoyancy, and related experimental applications. It may also appeal to those curious about practical applications of theoretical principles in real-world scenarios.

Signifier
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Say I have some volume V of some gas with a pensity P less than the density of air. How can I calculate how much a balloon (of volume V) of this gas could lift?

Is this called the "buoyant" force, or is this something else? What I'd like to do is find the find the lifting "power" (...) of some volume of helium, or the upward force in the fluid of air that a volume of helium could exert. How much of a gas would I need to pack into a container in order to lift some mass up some distance, etc.

Is there a branch of physics these sorts of questions neatly fit under, BTW, that would help my search for information? I remember a strange attribute of my first three physics class, which were supposed to provide a solid introduction to all physics (minus quantum) was the complete lack of discussion on things like hydrostatics and buoyancy...

Thank you, sorry for the simplicity of these questions!
 
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use archimedes's equation: g * V * dp
where g - gravitational acceleration, V - volume, and dp the difference between the densities of the matter inside and outside the volume V.
 
Really, that's it? Are there any other equations that are used to find the amount of upward force some volume of some uniformly dense gas can exert?
 
That's really all there is to it. The buoyant force on an object (a balloon filled with gas, or whatever) is equal to the weight of the fluid it displaces. (That's Archimedes' principle.) The buoyant force is the same regardless of the contents of the balloon. Of course, to use the balloon to lift something else that buoyant force must be greater than the weight of the balloon (and its contents). That difference will tell you how much additional weight the balloon can lift.
 
So let me get this straight (sorry!): if I have a balloon filled with gas of density p, enclosed in a volume V, carrying a payload of mass m, then in order for the balloon to lift upward I must have

gV|dp| > gm

(where dp is the difference in densities between the gas inside and the atmosphere)? Yes, have I got it? Or is it:

gV|dp| > g(M+m)

where M is the mass of the balloon with all of its gas and m is the mass of the payload. This last expression would reduce to:

V(Patm - 2Pgas) > m

where Patm is the density of the atmosphere and Pgas is the density of the gas inside the balloon...

?
 
Signifier said:
So let me get this straight (sorry!): if I have a balloon filled with gas of density p, enclosed in a volume V, carrying a payload of mass m, then in order for the balloon to lift upward I must have

gV|dp| > gm

(where dp is the difference in densities between the gas inside and the atmosphere)? Yes, have I got it?
That's it. Realize that "m" also must include the mass of the balloon material as well as the load. Further realize that we are ignoring the (probably small) buoyant force of the payload.

Or is it:

gV|dp| > g(M+m)

where M is the mass of the balloon with all of its gas and m is the mass of the payload.
You've already included the mass of the gas within the balloon in the term V|dp|; don't count it twice.

This last expression would reduce to:

V(Patm - 2Pgas) > m
See? You're counting the mass of the gas twice. :wink:

Things with be much clearer if you work out the formula from the basic principles: The buoyant force must support the weight of everything.

Buoyant force = g P(air) V
Weight of everything = g(mass of balloon material + gas + payload)

Of course, mass of gas = P(gas) V.

So:
g (Pair - Pgas)V > g(mass of balloon material + payload)

Or,
(Pair - Pgas)V > (mass of balloon material + payload)
 
Here's a thought question, say you have a model balloon, complete with deflated balloon and helium tank inside a large sealed container filled with air (plus the model). The container (with the model in it) rests on a scale and the scale indicates a total weight of 100lbs. Then, via remote control, the helium in the model is transferred from the model's tank to the model's balloon filling it, so that the model balloon now hovers inside the container. What will the scale indicate for the weight of this closed (sealed) system with the model balloon hovering? Support your answer.
 
Jeff Reid said:
Here's a thought question, say you have a model balloon, complete with deflated balloon and helium tank inside a large sealed container filled with air (plus the model). The container (with the model in it) rests on a scale and the scale indicates a total weight of 100lbs. Then, via remote control, the helium in the model is transferred from the model's tank to the model's balloon filling it, so that the model balloon now hovers inside the container. What will the scale indicate for the weight of this closed (sealed) system with the model balloon hovering? Support your answer.

100lbs. The weight of the system, including the container, does not change when you redistribute the contents of the container. All you are doing when you blow up the balloon is move some helium out of the tank into the balloon and displace air molecules that were originally above the balloon to other parts of the container. Nothing has been added or taken away.

Now as long as we are changing the question from "simple" to more complex, what if the scale is the lower wall of the container, with the tank and the balloon both resting on the scale?

I've been pondering a somewhat similar buoyancy problem related to bubbling water. As seen on TV, some have suggested that bubbling water is less buoyant because of its lower density, speculating that ships have been lost in the Bermuda triangle because they sank in bubbling water. Archimedes principle is quoted as justification for this conclusion. No doubt the principle is well established for fluids of uniform density, but I doubt it applies in such a simple way to bubbling water. I'll start another thread sometime soon to expand on this.
 
Here's a thought question, say you have a model balloon, complete with deflated balloon and helium tank inside a large sealed container filled with air (plus the model). The container (with the model in it) rests on a scale and the scale indicates a total weight of 100lbs. Then, via remote control, the helium in the model is transferred from the model's tank to the model's balloon filling it, so that the model balloon now hovers inside the container. What will the scale indicate for the weight of this closed (sealed) system with the model balloon hovering? Support your answer.
100lbs. The weight of the system, including the container, does not change when you redistribute the contents of the container.

Say the container box weighs 50lbs, the air 49lbs, and the model 1lb. How does the air generate a 49lb downwards force on the box (model resting at bottom), and 50lbs downwards when the model is hovering? (I'm ignoring the small boyancy factor of the air against the model when it's deflated, although the answer below takes care of this as well).

Highlight below to see answer (hold down mouse button and pan pointer across the text):

The pressure of the air in the container decreases with altitude, less at the top, more at the bottom, and this pressure differential results in a net downwards force equal to the weight of the air. When the model is hovering, it displaces the air, increasing the pressure of the air, and also increasing the rate of pressure decrease versus altitude so the pressure differential generates a downforce equal to the weight of the air and the model. If the balloon were expanded further, it float to the top and would exert an upwards force on the container, and the pressure differential of the air would generate a downwards force equal to the sum of weight of the air, the model, and the upwards force on the top of the container.
 

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