If you take a water balloon filled normally at the surface of the ocean and take it down, like 100 meters would it deflate due to the higher water pressure? What if you popped it?
Everything is compressible with enough force. I would say that threshold is when that compressibility is relevant to the discussion. So in this case, a water ballon under water. There is no depth in the ocean where the pressure will be high enough to make a notable difference.
I would imagine in many applications when dealing with water and fluid dynamics, unless measurements have to be extremely tight and calculated, water would be treated as incompressible. I'm guessing a lot of the time, the force applied to water is not high enough for it to matter.
Depends on how much air is in there with the water.
Water itself is incompressible so the balloon won’t shrink that much, but there’s some air in there so it would shrink a bit.
The balloon’s popping action comes from being filled and expanded. It’s an elastic, so if something breaks it, it snaps back.
The only way that wouldn’t happen is if there was mostly air in there, and you went to the appropriate depth where that air was compressed down to its original volume.
10 Comments
Zwuzelmaus@feddit.org · 10 pts · 5h
Liquid Water does not change it's density much with pressure, but yes.
plant@lemmy.dbzer0.com · 7 pts · 5h
Water is incompressible.
M137@lemmy.today · 6 pts · 4h
Not it's not, common myth/misunderstanding.
From Wikipedia (couldn't find a way to link directly to that part of the page):
https://en.wikipedia.org/wiki/Properties_of_water
Go to the "Physical properties of water" and scroll down for the part in the image.
BassTurd@lemmy.world · 9 pts · 4h
Ok, so for all intents and purposes, water is incompressible.
dohpaz42@lemmy.world · 5 pts · 3h
So what’s the threshold to say when something that compresses is compressible?
BassTurd@lemmy.world · 9 pts · 3h
Everything is compressible with enough force. I would say that threshold is when that compressibility is relevant to the discussion. So in this case, a water ballon under water. There is no depth in the ocean where the pressure will be high enough to make a notable difference.
I would imagine in many applications when dealing with water and fluid dynamics, unless measurements have to be extremely tight and calculated, water would be treated as incompressible. I'm guessing a lot of the time, the force applied to water is not high enough for it to matter.
FuglyDuck@lemmy.world · 4 pts · 4h
Depends on how much air is in there with the water.
Water itself is incompressible so the balloon won’t shrink that much, but there’s some air in there so it would shrink a bit.
The balloon’s popping action comes from being filled and expanded. It’s an elastic, so if something breaks it, it snaps back.
The only way that wouldn’t happen is if there was mostly air in there, and you went to the appropriate depth where that air was compressed down to its original volume.
nooneescapesthelaw@mander.xyz · 4 pts · 1h
Let's first state our assumptions:
Very thin membrane, the balloon skin cannot be compressed more than it already is
Zero trapped air
Isothermal conditions, water temperature at the bottom of the ocean is the same as the top, that way we don't need to deal with thermal expansion
Balloon is a perfect sphere
Calculations:
Pressure on the balloon is densitygravitydepth ~= 100010100 = 1 Mpa or 10 atmospheres
The bulk modulus of water is 2.2 Gpa so (change in volume)/(initial volume) = (change in pressure)/(bulk modulus) = 0.0046% decrease in volume
0.0046% decrease in volume is like a 0.015% decrease in radius
So yes the balloon does get a little bit smaller
someguy3@lemmy.world · 3 pts · 4h
At 100 m the compression would be nil.
If you take it deep enough for there to be measurable compression and then pop it, nothing will happen because it's already compressed.
Steve@startrek.website · 1 pts · 1h
The latex would still snap back to its original shape, which would at least look similar to a normal pop.