Timeline for Pressure inside a submerged rigid box, and how to change it
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jan 25, 2017 at 0:54 | history | edited | sammy gerbil | CC BY-SA 3.0 |
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| Jan 25, 2017 at 0:40 | comment | added | sammy gerbil | Incompressible fluids are idealisations. A pump moves fluid in or out. | |
| Jan 25, 2017 at 0:16 | comment | added | Kerrek SB | Oh, I think I get it: The pump could have an effect on the pressure even if it didn't move any amount of fluid, right? It would just be transferring force from the interior pressure onto the box walls. The stress itself does contain energy density, and you can spend work on transferring it. (I guess that's why it's measured by a "stress-energy" tensor.) | |
| Jan 25, 2017 at 0:12 | comment | added | Kerrek SB | Does that require a real, compressible fluid to work then? For an idealized, incompressible fluid there'd be no room to put more in or to take out (rememeber that the desired state of box 1 has no gaps). | |
| Jan 25, 2017 at 0:09 | comment | added | sammy gerbil | You can reduce (or increase) fluid pressure inside a sealed vessel using a pump. This works by taking fluid out or forcing more fluid in. | |
| Jan 25, 2017 at 0:01 | comment | added | Kerrek SB | I guess what I'm always having trouble with is imaging a dynamically changing pressure field where no energy is transferred - and why such processes wouldn't be reversible. Clearly poking a hole in the box is easy, but getting the pressure back out is not, even though no energy was transferred. | |
| Jan 24, 2017 at 23:55 | comment | added | Kerrek SB | And is there any mechanism to go the other way -- i.e. take the sealed box 2 and reduce its internal pressure? Without changing the contained (idealized) volume, so that it attains a state like that of box 1? | |
| Jan 24, 2017 at 23:54 | comment | added | sammy gerbil | If you open the box, or make a hole in it, the internal pressure will equalise with the pressure in the water outside. The water inside will be compressed very sightly, and a very small amount of water will flow in from outside until the internal and external pressures are equal. | |
| Jan 24, 2017 at 23:49 | comment | added | Kerrek SB | Yes, right, but now I want to open the lid of the low-pressure box. If there's air inside, this will obviously cause movement. But if there's already water inside, then when happens? | |
| Jan 24, 2017 at 23:49 | history | edited | sammy gerbil | CC BY-SA 3.0 |
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| Jan 24, 2017 at 23:42 | comment | added | sammy gerbil | Whether the boxes are full of air or water makes no difference. Any change in the internal fluid pressure in a sealed box depends on the box being compressed. If there is no change in volume of the box, there is no change in internal fluid pressure. | |
| Jan 24, 2017 at 23:34 | comment | added | Kerrek SB | To be clear, both boxes are meant to be full of water. Box one is filled and shut at sea level, Box 2 is only shut deep underwater. So it's the water pressure inside the boxes I am thinking about. | |
| Jan 24, 2017 at 23:26 | history | answered | sammy gerbil | CC BY-SA 3.0 |