It's also going to depend on ballon material and thickness.
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I doubt it's a simple answer. A balloon thins dramatically as it expands. If you knew stress-strain relationship in detail, it should be simple to work it out.
If you make the basic assumption that the balloon is like a "thin walled" pressure vessel at constant temperature; I would think that you should be able to use the ideal gas law
PV=nRT
thus P=nRT/V V=volume of the balloon. As you fill the balloon you can determine the mass of air used and the change in volume can be measured.
Realizing that the wall thickness of the balloon changes as a function of pressure, adds more complexity in this approach.
You should curve-fit an equation based on internal gauge pressure and diameter, taking several data points. You could then obtain an equation for the particular balloon that you investigated.
This becomes a function of external atmospheric pressure and elasticity of the balloon itself.
I have often applied thermodynamic tables, state 1 is the outside ambient air conditions, state 2 becomes the internal balloon pressure. It then becomes a very simple isentropic throttle problem, state 1 compresses air to state 2. Using nitrogen tables if you don't have air tables gives a guy decent answers. The inherent assumption is that the compression process is reversable.
Compute the specific volume based on the isentropic process from your estimate of fluid quality. This would completely descripe state 1 and 2, from which mass is obtained. You need to assume that the balloon initally assumes a purely spherical shape, this give you an estimate of volume. With sphere volume understood, diameter can be easily computed.
Interestingly enough, the process becomes an iterative solution. I have treated various balloon shapes as segments of parabola, etc, but found the most challenging mathematical arc is the ellipse. Ultimately you need to obtain volume of the enclosure which involves various integrating strategies. In practical terms, you don't need to go this far, infact using thermodynamics is a fairly uncommon approach to begin with. Without significant error, spherical profile is a good balloon estimate.
Hope this helps.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
The elasticity of the balloon is not a constant, and probably not linear.
The process will not be reversable. Significant hysterisis exists in the latex. Just blow up a balloon and deflate it, then compare it to a new (never inflated) balloon.
To my way of thinking, the two simultaneous modes of energy storage (gas compression and wall stretch) make any appeal to gas laws too difficult.
I wonder if the way into this one is to start with the balloon, rather than the gas. I guess this is what JStephen was hinting at.
First port of call might be to determine experimentally the force-strain curve for the first stretch of a new sample of the latex. Now transform the axes of that plot into diameter (from initial volume and strain) and pressure (I think a hoop stress equation will take you from force to gauge pressure here).
I can already see one possible flaw here - tensile tests generally work in one direction, so if you're not careful, the sample will neck in two dimensions (in the balloon, it's constrained in all dimensions except thickness)and your curve won't be representative.
MintJulep, that is because you had stretched the plastic past the elastic limit! This would not enter into the thermodynamic model, only the mathematical physic one. In this regard, it would be an issue....I guess.
Distracted, throw down some numbers. State 1 is STP (101.3 kPa, T=18C) and State 2 is T=18C (assuming adiabatic) with internal balloon pressure of......?
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
This might make an interesting Science-Fair type project.
You can measure the volume in the balloon fairly accurately by pouring a measured amount of water into it. You can measure pressure fairly accurately by using the manometer principle.
So the way you would work it- hook the balloon to a hose, and immerse in water (to support the balloon). Put a funnel on the hose. Pour, say, one cup of water in the funnel, raise or lower the funnel to whatever pressure you want, and measure the height. Add another cup, measure the height, and so on.
Of course, the whole problem changes with balloon shape. Consider the long skinny balloons, where the inflated part starts at one point, and gets longer as you add air.
I agree, it might proove an interesting science problem.
But the big issue with this computation, theoretically speaking, is the heat transfer or thermal absorption of heat from the surrounding environment. You can easily see this by simply blowing up a balloon and sticking it in the basement away from sunlight. The balloon will deflate as the internal gas pressure decreases.
Alternatively, take the same balloon and put it near the stove. Absorption of heat will increase internal pressure and expand the balloon. Most analytical types screw this part up in exams, the volume of the balloon is constant so the diameter is directly proportional to internal pressure.
You can of course, blow up a balloon and measure pressure, temperature, diameter. Using Boyles/Charles Law, the gas volume can be computed at an alternate state from which the diameter can be extrapolated. This gets around the elasticity issue of the medium as brought up an earlier discussion. For my money though, the thermodynamic approach solves most of these issues since air or nitrogen tables give you all the properties required to form the closed solution set. There are still other issues, what is the "shape" of the confining volume, that are required to link diameter to volume, but ultimately you must arrive at a volumetric quantity at a pressure and temperature.
This problem can be much more complex than originally thought. But a very good discussion to all participants, very interesting none-the-less.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
"Most analytical types screw this part up in exams, the volume of the balloon is constant so the diameter is directly proportional to internal pressure."
I must be the analytical type that would screw this up in an exam. Volume in a balloon is not hardly constant. Diameter is not proportional to internal pressure...with most balloons, maximum pressure is required right at the start, before it starts "ballooning". I forsee a highly nonlinear effect here, with maximum pressure at the start and at the popping point.
You bring up temperature, but balloons are temperature sensivite, so I think you'd find big changes in the elastic properties as you changed temperatures in your example. Balloon properties are time-sensitive, and if you leave a balloon blown up for a day, it's quite a bit different than one that you just now blow up.
Balloon properties are time-sensitive, and if you leave a balloon blown up for a day, it's quite a bit different than one that you just now blow up.
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I suppose that the rubber does exhibit some relaxation or creep over time, but the main reason that balloons get smaller over time is that they leak.
You're right about volume being invariant. I meant to say MASS, mass is constant assuming no losses. Of course there is alway a loss so mass is invariant over a short period of time.
Volume as correctly mentioned, is dependent on pressure. I will post a computation as an example.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
I think you get an empirical relationship between the diameter of the balloon and the membrane force in the balloon from an experiment by stretching a flat circular membrane in 2 dimensions and recording the ratio of the change in area to the corresponding linear membrane force,t around the periphery. Call this relationship
t=f(A/Ai), where Ai is the initial area of the test piece.
For the balloon this is then
1) t=f{(D/Di)^2}. Now we also know from the pressurized spherical system that:
PiDt=PiD^2/4(P-Po) or 2)t=D(P-Po)/4 (Po is the pressure outside the balloon) after eliminating one D and Pi.
Now you have two equations in the 2 variables D and t. By making D the independent variable one can get corresponding t in eq 1 and P can be obtained by substitution in eq 2.
According to the above no thermodynamics are involved in this analysis, since the gas will accomodate the structural restraints.
