Calculate Surface Tension

One can look this up:

• Excess pressure in a bubble = 2 x surface tension / radius.

In general the surface may not be spherical and will have two principal radii of curvature, R1 and R2.

https://en.wikipedia.org/wiki/Laplace_pressure

An interesting example is where both sides of the surface are open to the atmosphere and the delta pressure = 0. The curvature could be zero (flat surface, R1 = R2 = infinity) or one radius of curvature could be negative and the other positive as illustrated in the picture below (catenoid of revolution).

Quite.

Surface tension requires a liquid. You are using the wrong term since your fluid, the air is not a liquid. It is a gas. Thus the surface tension is zero for no liquid exists or the stated pressure differential is the pressure required to restrain this gas.

I'd be more worried about what this sphere is made of and how one supports this vessel. Also, what is outside this sphere, more air?

PWSlack I do not understand your formula Sorry.

Yes air inside the sphere, and air outside the sphere

the inside has a pressure that is 0.048812 PSI HIGHER than the outside.

So the sphere is having a force against it that it is holding back this extra volume of air thus creating the positive pressure.

now the question is, what amount of force, is being applied to the sphere? I'm thinking surface tension. would be the measurement?

Example a sphere of 4 Meters in Diameter.

There is a pressure differential of 0.048812 PSI

Well if you're going to follow our latest trend of making up your own meanings for words, I'll answer in kind.

The force is seventy-two aglets. You must first convert your units into a consistent system and change meters into vamps. The rest is simple contour mapping.

2.5mmhg is equal to .04834 psi....basically soap bubble neighborhood...

https://physics.stackexchange.com/questions/38813/relationship-between-balloon-size-and-pressure

https://www.scientificsales.com/8246-Weather-Balloon-2000-Grams-Natural-p/8246.htm?gclid=CjwKCAiAtouOBhA6EiwA2nLKH4LiPxRDJZVQfBa3JnvwyuIHfg1PHV1vHlpQxZUWeUPorMH2UjW0OxoCxsoQAvD_BwE

https://www.topperlearning.com/answer/a-soap-bubble-of-radius-3-cm-is-charged-with-90nc-the-excess-pressure-inside-the-bubble-if-surface-tension-of-soap-solution-3-10-3-n-m-is-1-026n-m-2-2/4ofcs4gg

The pressure difference between the inside and outside of a bubble depends upon the surface tension and the radius of the bubble. The relationship can be obtained by visualizing the bubble as two hemispheres and noting that the internal pressure which tends to push the hemispheres apart is counteracted by the surface tension acting around the circumference of the circle.

For a bubble with two surfaces providing tension, the pressure relationship is:

http://hyperphysics.phy-astr.gsu.edu/hbase/surten2.html

https://www.eeeguide.com/surface-tension/

Yes, a soap bubble is a very good supposition that meets the liquid requirement and pressure differential. A four-meter diameter soap bubble seems dubious but knowing what tension is needed might be how one dissuades somebody from attempting the use of a soap bubble.

Still, this is a supposition. The OP has not clarified the material of this bubble.

Material is of no matter, it is something tht will not expand when the pressure inside is increased to make that pressure differential.

"PWSlack I do not understand your formula Sorry."

His formula is exactly the same as Rixter's

just replace R₁ and R₂ with R

Then differential pressure = 2 x Surface tension / R

The formula is not proprietary.

Assume the thickness of the wall is negligible compared with the diameter.

Carry out a theoretical section through the centre of the <...sphere...>, and analyse what is happening on this plane and perpendicular to it:

Area of section = pi x diameter²/4.

The force needed in the bubble walls to keep the <...sphere...> intact is therefore:

Force = Area of section x pressure difference.

The surface tension is therefore this force divided by the circumference of the sphere, which is pi x diameter.

One then allows for the fact that there are two liquid-air surfaces in a bubble.

Out drops the formula that can be looked up.

-0-0-0-

There is no need to repeat information from earlier postings in threads; doing so adds no value.

Surface tension requires a liquid.

The "surface" could be a rubber balloon...

Correct, it could be.

Were the wall to be formed from a solid, then the term "wall stress", or something similar, would be of interest, and therefore the term <...surface tension...> from the original posting in this thread would be both inappropriate and inapplicable.

• Surface tension has dimensions of force per unit length.
• Material stress has dimensions of force per unit area.

OK, Think a fabric ball, like below, how strong with the spec's aa described in first post. how strong would this fabric need to be to hold in that amount of pressure?

Is this a wind-up?

It's clear you don't mean surface tension as normally understood (a property of liquids), but just the tensile stress in the fabric. You need to know the thickness, then use the first formula in #12.

Why would you need to know the thickness?

I mean if it is a force with say a stretching force of "X" pounds, why would it matter of it was 0.001 thick or 1" thick isn't the amount of pulling force the same?

Because the fabric strength will be in psi (lb/in²). If the force is X lb/in, and thickness is 0.001", tensile stress = X/0.001 = 1000*X psi. If the thickness is 1", tensile stress = X psi.

<...need to know the thickness?...>

Simply, because without it one cannot calculate the tensile stress in a solid, which has dimensions of force per unit area.

<...surface tension...> is a property of a liquid, for which the negligible thickness is implicit. It has dimensions of force per unit length.

Please see post 10.

I don't know how I missed that Tornado! Thanks!

Now what confuses me still is Ok I have a number. How do I apply that number to the real world?

Like if you have some material. to make that envelope out of. Is there a way to somehow apply the number you came up with, to see if this material would be strong enough?

I know I am missing something, on how to use this nmber you got.

Joe

You could take a 1-inch-wide strip of the material and stretch it with a spring scale until it breaks. That gives its per-inch tensile strength.

That's what I thought, but then thought it can't be that simple.

Now say do you convert it back to per inch?

like say you have a strip 1" wide and 10" long, and it breaks at 100 pounds.

is it's strength then 10 pounds?

or how do you put it into practical use?

Joe

The length doesn't matter. The required strength of a bit less than 4 lb is already per inch of width. Ten pounds, for instance, would give a safety factor of a bit over 2.5.

Tornado's answers are good and correct, I just thought I'd try to explain a little more:-

Ignoring the weight of the ribbon itself, the tension in both these 1" ribbons is 10lb, and, the tension is 10lb throughout the length of the ribbons. So assuming that the ribbons don't break then this material is almost certainly strong enough to meet your 4lb per inch requirement.

Just a small point if you are testing a sample of the material, make sure that you test in at least three directions. Take the three one inch samples shown here:-

The material may not have an obvious "warp" and "weft", but, it was almost certainly made in a way which might affect its strength in different directions.

The total weigh on the supporting beam is 10 pounds.The weight on each ribbon is 5 pounds;consider a block and tackle arrangement.

That depends on how one reads the simple diagram.

How do you read it?

I see that the diagram might have a ribbon viewed on edge, making two support lines up. This view will have the ribbon reaching under the weight for support.

I also see the diagram might be viewed onto the flat, broad part of the ribbon, showing the two edges of the one ribbon going up to the support. Something hidden must attach the mass to the ribbon.

It is an ambiguous diagram but not bad for a quick sketch.

It's two completely different ribbons with two independent 10lb weights.

Both ribbons have the same tension (10lb), and, need to be the same strength.

I was trying to correct NSS's apparent misunderstanding from the previous post:-

"say you have a strip 1" wide and 10" long, and it breaks at 100 pounds.

is it's strength then 10 pounds?"

Strictly speaking, it can be compared to a pressure vessel. In case you have a thickness of the sphere material, a balloon in your case, and its strength you can use the pressure vessel calculations. That is what you seem to be after.

The analogy that a sphere can be considered two hemispheres is a good visualization as per Solar Eagle.

Here are the formula. The hemispheres have only half the stress as the cylindrical section of a pressure vessel.

https://web.mit.edu/course/3/3.11/www/modules/pv.pdf

Total circumferential tension F = PA = 0.048812 x [pi x (2 x 39.37)²] ≈ 951 lb.

Circumference = 2 x pi x r = 2 x pi x (2 x 39.37) ≈ 247.4 in.

Tension per inch of circumference ≈ 951/247.4 ≈ 3.844 lb.

If the differential pressure is 048812 PSI,then the strength of the sphere only has to be slightly higher than the differential pressure.

If you mean the skin has thickness t, uniformly stressed, by a simple force balance, skin tensile stress = X*Y/4/t psi.

If you mean surface tension as normally understood, this = X*Y/8 lbf/inch circumference. The extra factor of 2 because each surface contributes.

If it's a soap bubble, good luck making one 4m diameter!

https://www.ebay.com/itm/154064964593

Very nice, but it's not soap!

We can tell from the purple glow that this bubble is around 400nm thick....

https://www.thoughtco.com/make-giant-unpoppable-bubbles-603930

For comparison ...

• A sheet of paper is about 100,000 nanometers thick
• A strand of human DNA is 2.5 nanometers in diameter
• There are 25,400,000 nanometers in one inch
• A human hair is approximately 80,000- 100,000 nanometers wide
• A single gold atom is about a third of a nanometer in diameter
• On a comparative scale, if the diameter of a marble was one nanometer, then diameter of the Earth would be about one meter
• One nanometer is about as long as your fingernail grows in one second

How thick is a soap bubble?

https://askinglot.com/what-is-the-minimum-thickness-of-a-soap-bubble-film

I don't know, but is it relevant?

What is the surface tension of a bubble under water at any instant in time??How thick is the layer of the surface of the bubble?Sure it varies with depth,but at each point it is slightly higher than the ambient pressure.

Place a straw into a glass of soapy water and apply pressure till the liquidminiscus is balanced at the very end of the straw,but not yet made a bubble.At this point,the pressures are equal.Any pressure above that will create a bubble.

Not with you. What layer exists in or around a submerged bubble? If you know the surface tension it's easy to calculate the additional pressure (not forgetting that, unlike the bubble in this thread, there is only one surface).

Any pressure above ambient, however minute, will cause a bubble under a liquid.

I am sorry, I just do not understand this math equations stuff.

Can anyone simply answer me the original posting question?

Joe

42

{Sorry, but I just couldn't resist answering this way.}

https://www.omnicalculator.com/chemistry/surface-tension

But that applies to bubbles of liquid. That's not he's talking about. His photo in #30 makes it clear.

TESTS OF BALLOON MATERIALS

https://opensky.ucar.edu/islandora/object/technotes%3A464/datastream/PDF/view

All you folk are AWESOME!!! Now for a small additional question,

Ok we have this original sphere with this much pressure differential from inside to out.

Example a sphere of 4 Meters in Diameter.

There is a pressure differential of 0.048812 PSI

So OK starting with the sphere filled with "AIR", it is fully filled and has absolute zero in pressure differential.

How much in volume of outside air do we need to add to the inside to get that pressure differential?

Joe

OOPS! posted in error...

As nobody has answered this I'll do it. It's the volume of the sphere (which I'm sure you can work out) x pressure rise/original pressure. That is measured at the pressure and temperature of the atmosphere, say 14.5 psi and 20°C.

BTW 0.048812 PSI seems a tad over-precise

I would assume all units have to be the same?

Like vol of the sphere in cubic inches then?

Joe

That's right. As long as the units of actual pressure and the differential pressure are the same, they can be in any units, as they cancel. The volume input is in the same units as the volume of the sphere.

<...How much in volume of outside air...to add to the inside to get that pressure differential?...>

The answer comes from the Ideal Gas Equation, which is also non-proprietary.

In light of the OP's admission in #26, do you think that will help him much?