# Thread: Adding energy to a closed system

1. ## Adding energy to a closed system

If u had a indistructable pipe of 10 ft vertical ,with an object in it that is not compressable. Fill the pipe with water. Having the object weigh as much as water displaced. Now compress the water in the pipe to increse its density. The object is now lighter and will move to the top of the pipe.
My question is:
Is the pressure needed to compress water vs pressure left after object moved to the top of the pipe equal.
Guessing energy is lost somewhere or transformed to make the object move .  Reply With Quote

2. If you are compressing both the water and the object in the water the density of both will increase. Nothing is non-compressible.

I would consider whether or not the water and the object change density at the same rate will under compression. If the water changes density at an increased rate relative to the object it is possible to create buoyancy with the object in the water.  Reply With Quote

3. You can change the density by heating the pipe vessel up as well. This is how a Galileo Thermometer works.

"A Galileo thermometer (or Galilean thermometer) is a thermometer made of a sealed glass cylinder containing a clear liquid and several glass vessels of varying mass. As the temperature changes, the individual floats rise or fall in proportion to their respective weights and the density of the surrounding liquid. It is named after Galileo Galilei because he discovered the principle on which this thermometer is basedthat the density of a liquid changes in proportion to its temperature."   Reply With Quote

4. ## Geavity energy generator

If u compress a fluid it creates a higher boyent force to any object in it that compress at a higher pressure.
When decompressing the fluid you will get back the energy u put in to compress it(not taken in consideration friction), regardless of any movment by the object. Moving the object higher when the fluid is compressed and let it drop when fluid is decompressed allows for force gain from gravity. The system will be using more then gaining but if u dont look at friction you can see that this is a way to show you could harness energy from gravity. Gravity energy generator?  Reply With Quote

5. Density = Mass / Volume

or

p = m / V (a)

Density change
E = - dp / (dV / V0)
= - (p1 - p0) / ((V1 - V0) / V0) Equation A

Where:
E = bulk modulus - liquid elasticity (N/m2)

The minus sign corresponds to the fact that an increase in the pressure leads to a decrease in volume.

With (A) - the final volume after pressure change can be expressed as
V1 = V0 (1 - (p1 - p0) / E) (Ab)
Combining (Ab) with (a) - the final density can be expressed as:
ρ1 = m / (V0 (1 - (p1 - p0) / E))

These are the equations you need to calculate the amount of pressure you need to float an object..  Reply With Quote

6. sorry but i'm wandering why you think i need this math to get the knowledge. my question was not how much force ill get, but merely that gravitational force will be greater force acting on the object because of the fluid density. at the same time again sorry im probably that guy that skipped school where your the one that went to it so i understand when u get frustrated at people like me that just try to skip learning and try to get to understanding but cant. after a bit i changed the question to a more basic one. (not taking in consideration friction)does it take the same energy to compress a fluid as it makes when it decompresses?  Reply With Quote

7. Originally Posted by vipera2006 sorry but i'm wandering why you think i need this math to get the knowledge.
Math is the language of science and physics. Math will provide you the answers you are looking for..  Reply With Quote

8. I like your answer , also it seems like you understand fluids. question what viscosity will the water have if you pressurize it as close to frozen as you can with out it freezing?  Reply With Quote

9. Originally Posted by Cragyon Density = Mass / Volume

or

p = m / V (a)

Density change
E = - dp / (dV / V0)
= - (p1 - p0) / ((V1 - V0) / V0) Equation A

Where:
E = bulk modulus - liquid elasticity (N/m2)

The minus sign corresponds to the fact that an increase in the pressure leads to a decrease in volume.

With (A) - the final volume after pressure change can be expressed as
V1 = V0 (1 - (p1 - p0) / E) (Ab)
Combining (Ab) with (a) - the final density can be expressed as:
ρ1 = m / (V0 (1 - (p1 - p0) / E))

These are the equations you need to calculate the amount of pressure you need to float an object..
Excellent Calculation.  Reply With Quote

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