## Balloon Pressure Question

1. The problem statement, all variables and given/known data
Is it possible to find out the increase in pressure in a balloon if I know how much pressure I apply when I squeeze it? Or, alternatively, can I find out the change in volume of the balloon if I know how much pressure I'm applying? I know the initial volume, pressure, temperature of the balloon, and I know how much pressure I apply over a certain area when I squeeze the balloon.

3. The attempt at a solution
I understand how to solve this problem when I know the change in volume that occurs as a result of my squeezing the balloon. I would simply use the PV=nRT equation. I'm not sure however, how to approach the problem when all I know is the pressure I apply when squeezing the balloon. I would think I would need to know something about the pressure inside the balloon, and perhaps it's elasticity. After all, applying pressure on a metal box containing pressurized gas won't change it's pressure, so the elasticity of the enclosure must play some role.

Thanks in advance for any help! :)

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 Quote by hunnedlicks 1. The problem statement, all variables and given/known data Is it possible to find out the increase in pressure in a balloon if I know how much pressure I apply when I squeeze it? Or, alternatively, can I find out the change in volume of the balloon if I know how much pressure I'm applying? I know the initial volume, pressure, temperature of the balloon, and I know how much pressure I apply over a certain area when I squeeze the balloon. 3. The attempt at a solution I understand how to solve this problem when I know the change in volume that occurs as a result of my squeezing the balloon. I would simply use the PV=nRT equation. I'm not sure however, how to approach the problem when all I know is the pressure I apply when squeezing the balloon. I would think I would need to know something about the pressure inside the balloon, and perhaps it's elasticity. After all, applying pressure on a metal box containing pressurized gas won't change it's pressure, so the elasticity of the enclosure must play some role. Thanks in advance for any help! :)
One difficulty is that if you press on a balloon, it just stretches out in another place, perhaps negating, or at least reducing any change in pressure you may have created.

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 Quote by hunnedlicks 1. The problem statement, all variables and given/known data Is it possible to find out the increase in pressure in a balloon if I know how much pressure I apply when I squeeze it? Or, alternatively, can I find out the change in volume of the balloon if I know how much pressure I'm applying? I know the initial volume, pressure, temperature of the balloon, and I know how much pressure I apply over a certain area when I squeeze the balloon. 3. The attempt at a solution I understand how to solve this problem when I know the change in volume that occurs as a result of my squeezing the balloon. I would simply use the PV=nRT equation. I'm not sure however, how to approach the problem when all I know is the pressure I apply when squeezing the balloon. I would think I would need to know something about the pressure inside the balloon, and perhaps it's elasticity. After all, applying pressure on a metal box containing pressurized gas won't change it's pressure, so the elasticity of the enclosure must play some role. Thanks in advance for any help! :)
You're right. In order to solve the problem you are describing, you would have to analyze the deformational mechanics of the rubber sheet comprising the balloon (assuming plane strain). This would include the localized kinematics of the deformations (to determine the relationship between the displacements and the strains), and the relationship between the strains and the membrane stresses (Hooke's law, rubber elasticity relations, or something more accurate and complex). The displacements are likely to be large, and this would have to be taken into account in determining the strains. The membrane stresses would be included in the differential force balance, together with the pressure (which could be assumed to be uniform within the balloon). You would also have to specify the contact loading. This is similar to problems in determining the structural mechanical response of automobile tires under inflation loading, static contact loading with the ground, rolling contact loading with the ground, contact and cornering loading with the ground, etc.

## Balloon Pressure Question

You can squeeze a balloon without using your hands. What would happen to an inflated balloon if you took it up a mountain? What would happen to your metal box if you took it a long way underwater?

 Quote by Chestermiller You're right. In order to solve the problem you are describing, you would have to analyze the deformational mechanics of the rubber sheet comprising the balloon (assuming plane strain). This would include the localized kinematics of the deformations (to determine the relationship between the displacements and the strains), and the relationship between the strains and the membrane stresses (Hooke's law, rubber elasticity relations, or something more accurate and complex). The displacements are likely to be large, and this would have to be taken into account in determining the strains. The membrane stresses would be included in the differential force balance, together with the pressure (which could be assumed to be uniform within the balloon). You would also have to specify the contact loading. This is similar to problems in determining the structural mechanical response of automobile tires under inflation loading, static contact loading with the ground, rolling contact loading with the ground, contact and cornering loading with the ground, etc.
Thanks for the help, I'll look into the things you mentioned.

 place your inflated balloon inside a large cardboard box. Fill the box with smoke, use a rod with a flat paddle on the end to squash the balloon. To a first order approximation I would not expect the smoke to be displaced out of the box nor the level of smoke to go down. My expectation is that the pressure inside an inflated balloon is similar to atmospheric pressure and that it does not change a lot with additional inflation. Which I am going to assume will also apply to small amounts of deformation once inflated. Have you noticed how a balloon is hard to inflate at first and then gets easier? Can you explain why it gets easier given that the tension in the rubber sheet is increasing as the balloon inflates? The explanation for that is why I don't expect the pressure inside the balloon to change much once inflated. When you squeeze the balloon where is the energy going? Physics isn't always about numbers. It can start from a bit of hand-waving and simplification. ...then you have to go and check with an experiment.. The box and balloon should be easy the paddle I can make, its only the smoke part I need to figure out and then I can ask nature how she behaves.

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 Quote by NoPoke place your inflated balloon inside a large cardboard box. Fill the box with smoke, use a rod with a flat paddle on the end to squash the balloon. To a first order approximation I would not expect the smoke to be displaced out of the box nor the level of smoke to go down. My expectation is that the pressure inside an inflated balloon is similar to atmospheric pressure and that it does not change a lot with additional inflation. Which I am going to assume will also apply to small amounts of deformation once inflated. Have you noticed how a balloon is hard to inflate at first and then gets easier? Can you explain why it gets easier given that the tension in the rubber sheet is increasing as the balloon inflates? The explanation for that is why I don't expect the pressure inside the balloon to change much once inflated. When you squeeze the balloon where is the energy going? Physics isn't always about numbers. It can start from a bit of hand-waving and simplification. ...then you have to go and check with an experiment.. The box and balloon should be easy the paddle I can make, its only the smoke part I need to figure out and then I can ask nature how she behaves.
Can you explain why it gets easier given that the tension in the rubber sheet is increasing as the balloon inflates?

Answer: The stress-strain curve for rubber elasticity is non-linear, and the rubber exhibits a yield type region at higher stresses and strains. The stress in the rubber membrane is proportional to the balloon radius (for a spherical balloon).

When you squeeze the balloon where is the energy going?
Answer: It going into stored elastic deformational energy of the rubber (like a rubber band).

If you want to simplify the problem, start out by assuming that the balloon is spherical. Under these conditions, the rubber membrane experiences equal biaxial stretching, and you can use a strength of materials approach to get the membrane stresses, and to apply the force balance. Look up some rubber elasticity rheological properties for the kind of rubber in the balloon. This is a very solvable problem, and requires no "handwaving."

While true that the rubber in a balloon has non-linear properties that isn't why it gets easier to inflate as it inflates. The force that the stretched membrane exerts on the gas inside the balloon is a function of the membrane curvature (I don't know the math hence the hand waving) As the balloon inflates its curvature drops so even though the stress in the membrane is rising the reduced curvature dominates (more hand waving :( )

I agree that much of the energy from squashing the balloon is going into deforming the membrane. I don't know how much goes into increasing the internal pressure but I suspect very little. ( mad flapping of hands now taking place)

I need to find a safe smoke source so I can try it and see ....

 Quote by hunnedlicks 1. The problem statement, all variables and given/known data Is it possible to find out the increase in pressure in a balloon if I know how much pressure I apply when I squeeze it? Or, alternatively, can I find out the change in volume of the balloon if I know how much pressure I'm applying? I know the initial volume, pressure, temperature of the balloon, and I know how much pressure I apply over a certain area when I squeeze the balloon
I suspect that an answer of "no increase" is going to be close enough.

[typing isn't easy when your hands are flapping :) ]

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 Quote by NoPoke While true that the rubber in a balloon has non-linear properties that isn't why it gets easier to inflate as it inflates. The force that the stretched membrane exerts on the gas inside the balloon is a function of the membrane curvature (I don't know the math hence the hand waving) As the balloon inflates its curvature drops so even though the stress in the membrane is rising the reduced curvature dominates (more hand waving :( ) ]
As a person with much practical experience in modeling the deformational mechanics of automobile tire structures, I have no idea what you are alluding to. The curvature effect is already included in the determination of the stress. The non-linearity comes in primarily in the rheological equation for the rubber. Consider a perfectly spherical balloon, in which the pressure differential from inside to outside is $\Delta$p. The tensile stress in the rubber membrane is equal to

σ = r Δp / (2h)

where r is the current radius and h is the current thickness.

This is the only place where the curvature (as characterized by the balloon radius) comes in. Of course, r does also occur in the radial stretch ratio (strain), but that is involved with the rheological (stress-strain) response.

Chet

 Hi Chet, I can derive that formula (take a thin walled spherical vessel then consider each half) but I don't find it easy to see how it applies to inflating a balloon. ...I'll have a go and hope that someone can enlighten me... For a given amount of stress in the thin rubber sheet that makes up a balloon wall then the pressure inside the balloon will have an inverse relationship to the radius of the balloon. Since curvature and radius themselves have an inverse relationship then when inflating a balloon as the curvature goes down during inflation the pressure differential will fall too. Toy balloons really do get easier to inflate as they are inflated. This is different to what I experience when the rubber tube inside a bicycle tyre is inflated but is consistant with what I experience when I inflate the same inner tube without the restraining tyre and rim. My handwaving is because though I know the qualitative result from the experiment I don't know the equations. I can see the potential behaviour in that stress formula but I'm uncomfortable with drawing the conclusion that it is an explanation for what I observe. [edit in red to correct stupidity when typing]

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Gold Member
 Quote by NoPoke Hi Chet, I can derive that formula (take a thin walled spherical vessel then consider each half) but I don't find it easy to see how it applies to inflating a balloon. ...I'll have a go and hope that someone can enlighten me... For a given amount of stress in the thin rubber sheet that makes up a balloon wall then the pressure inside the balloon will have an inverse relationship to the pressure inside the balloon. Since curvature and radius themselves have an inverse relationship then when inflating a balloon as the curvature goes down during inflation the pressure differential will fall too. Toy balloons really do get easier to inflate as they inflate. This is different to what I experience when the rubber tube inside a bicycle tyre is inflated but is consistant with what I experience when I inflate the same inner tube without the restraining tyre and rim. My handwaving is because though I know the qualitative result from the experiment I don't know the equations. I can see the potential behaviour in that stress formula but I'm uncomfortable with drawing the conclusion that it is an explanation for what I observe.
Dear NoPoke,

I'm going to send you a private message outlining how to go about solving this problem. I'm reluctant to get everyone else in PF involved, because it is a little beyond introductory physics in scope.

Chet