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Old Oct 8th 2014, 08:22 AM   #1
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Need basic help understanding liquid physics

Hello, I have never studied physics, but have always wondered why things work the way they do. A recent "thought experiment" I did left me confused, and I thought someone here could help explain the basics, so I know how my thoughts are messing me up. Here is the "experiment":
a)I have a 100 lb. weight floating on a raft, floating on a tidal creek;
b)Below the raft is a water filled 4Ē cylinder and piston, with a 1 sq. in. outlet at the bottom;
c)The outlet is attached to a hose at the surface;
d)When the tide drops, the raft lowers and compresses the piston against the bottom of the creek, which in turn forces the water out to the surface.
If on dry land, I would expect to find pressure at the outlet to be about 100psi, assuming the weight just dropped without buoyancy assistance. All the water would be forced out in seconds.
But in the creek with a 6 hour tide, the water should take 6 hours to clear. It still would have about 100psi, right? If the tide dropped and the piston settled on my foot, it would still weigh 100 lbs.
I canít wrap my head around this so that it makes sense. I am not looking for a complicated math formula, or precise pressure/volume/flowrate information. Just an idea of what would happen, and why. Thanks
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Old Oct 8th 2014, 11:39 AM   #2
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Needs a picture...

This needs a sketch of the scenario.
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Old Oct 8th 2014, 03:33 PM   #3
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Originally Posted by THERMO Spoken Here View Post
This needs a sketch of the scenario.
Hopefully this will help explain the scenario. I understand some of the effort/force/pressure on the output side will be lost going up the hose, but I suspect only a small percentage. I'm just trying to understand the forces at work.
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Old Oct 8th 2014, 09:55 PM   #4
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Please clarify - are you saying that as the tide drops and the piston presses against the bottom of the creek that the entire weight of the 100 pounds is supported by the piston (i.e., the raft flotation is high and dry up in the air)? Or does the raft continue to provide some bouyancy?
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Old Oct 9th 2014, 06:54 AM   #5
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I think the way you express your question demonstrates how I am confused. If the 100lbs dropped free-fall (through air) onto the piston, I would expect about 100psi. If the raft provides some buoyancy, wouldn't the weight still be the same, just for a longer period (tidal period)? If I put a scale under the piston (on the bottom), would it read 100 lbs?
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Old Oct 9th 2014, 09:35 AM   #6
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Raft - Piston and Cylinder

Hi Craig...

Thank you, the sketch was a help. I wrote up your PHF-Inquiry a little more succinctly that you had. I put it at my site.

http://www.thermospokenhere.com/wp/PHF_folder/raft_piston_cylinder.html

More information, specification is needed to make your inquiry quantitative.

You are in need (I suggest) of some reading in hydrostatics. Perhaps take a look at some examples:

http://www.thermospokenhere.com/wp/0...off_shore.html

Good Luck, TSH

Last edited by THERMO Spoken Here; Oct 9th 2014 at 10:45 AM. Reason: URL error...
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Old Oct 9th 2014, 02:32 PM   #7
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Your description seems very well stated.
To address Event #3: the pipe/hose is open-ended, and just spraying water into the air.
To address C1: I'm presuming the creek is shallow, perhaps 4 or 5 feet. I believe the weight of the water in the cylinder would be added to the mass, which would change the psi a little.
To address C2: Just remove the water line and the raft, and instead have the weight dropped from a crane or electro-magnet.
To address C3: THIS is probably what I don't understand most. The 100psi measured at the outlet is my expectation since I have 100 lbs pushing the water down to a 1 sq. in. outlet...100lbs per square inch. No?
To address C4: Again, no drawing really needed. Put a foot, or finger under the piston prior to the weight settling down on the piston. After it settles, I would have 100lbs on my body part. My assertion is that throughout it's travel path, the downward weight would always be 100 lbs. But C2 and C3 can't both be right. The freefall drop would last only seconds, the tidal drop 6 hours. I can't think that getting sprayed in the face by the first would feel the same as the second. This is what I'm trying to understand (I think!).
Again, thanks for your help.
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Old Oct 9th 2014, 03:11 PM   #8
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Craig,

I'm replacing a water heater right now.

If you want me to go on with this (a day later)... Please say so.

Otherwise, Good Luck...

TSH for PHF
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Old Oct 10th 2014, 01:48 AM   #9
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I suggest you draw a free-body diagram in order to understand what forces are at work here. The force acting down is gravity, at 100 pounds. Forces acting up are (a) the pressure acting on the piston, which is the area of the piston times the pressure, and pressure is determined by the depth of the piston in the water, and (b) the upward flotation force of the raft. If the vector sum of all three forces is zero, the mechanism is in equilibrium. But if the sum of forces is not zero,then the raft either sinks or rises according to F=ma. Note that your assumption that pressure on the piston is 100 psi assumes that the raft provides no flotation and the raft is suspended by the pressure - i.e. Sum of forces is zero and hence acceleration equals zero. But that's not what happens, because the pssure is dictated by the depth of the piston in the water: pressure on the piston head = density of water times depth of the piston head times acceleration of gravity g.
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Old Oct 10th 2014, 05:02 PM   #10
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Once again...

Craig,

I looked at all this again. I posted an updated response (Check it at the URL of Post #6 above).

You probably should (in fairness to me and ChipB) do a little "physics" as a self-start to "warp" your head around this.

Good luck, TSH for PHF
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