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Where Does The Energy Come From ?


deschoe

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Actually, now that I have read Craig's description again, I think I can see what deschoe is trying to do and I can offer an explanation in mechanical terms as to why it won't work:- 

 

Surface tension acts equally on both points where the hoop emerges from the liquid so contributes no net force to it. So we can ignore that. 

 

But Deschoe thinks that the buoyancy on the side where the liquid level is higher (due to capillary rise) will be greater than on the other side.

 

But buoyancy works by displaced liquid falling, i.e. reducing in gravitational potential energy as the buoyant object rises. The liquid can change places with the object and because the weight of the displaced liquid exceeds that of the rising object, the liquid will fall and the object will rise. 

 

But, in this case, rotation of the hoop or ring does not provide a space into which liquid can fall, so as to change places with it. So no motion arises.  

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you need no hoop, you can do it with cubes as shown.you have to understand, why a floater in a capillary rises more than an outside one. otherwise you poke around like a blind man in the fog 

Well of course it does! The surface tension reduces the pressure on the water surface so it rises higher. We've already been over that. And so the water can fall, by displacing the "floater" upwards until it reaches the top of this - slightly higher - surface inside the capillary.

 

There is no mystery here. The water inside the capillary is just as capable of losing gravitational potential, by being displaced downwards below the rising floater, as anywhere else in the liquid. It is under slightly lower pressure, sure, but its density, which is what determines its ability to create buoyancy, remains unchanged.

 

In energy terms, the water just below the meniscus has a bit more gravitational potential energy than that at the surface of the bulk of the water, because it a bit farther from the centre of the Earth: it has farther to fall. So it can do a bit more work on the rising floater and cause it to gain more potential energy. So it rises a bit further.

 

Where does this extra gravitational potential energy come from? It comes from the chemical energy in the attraction (mainly hydrogen bonding) between water molecules and glass, which I referred to at the beginning of this thread.

 

Rather as with a magnet: you can get it to attract an object, to raise it and thus do work against gravity. But only once. You cannot get continuous motion out of that: once the magnetic energy has been minimised by attracting the body, that's it. Done. Finished. Of course, you CAN do it repeatedly with an electromagnet, by turning it on and off. But every time you do that you consume more electrical energy in energising the electromagnet. 

 

Face it: Conservation of Energy is one of the most powerful and universal concepts in physics for analysing a problem and it  never fails. You cannot obscure its operation even with elaborate constructions involving surface tension. :)

Edited by exchemist
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First :

 

Capillary action (sometimes capillaritycapillary motion, or wicking) is the ability of a liquid to flow in narrow spaces without the assistance of, or even in opposition to, external forces like gravity. The effect can be seen in the drawing up of liquids between the hairs of a paint-brush, in a thin tube, in porous materials such as paper and plaster, in some non-porous materials such as sand and liquefied carbon fiber, or in a cell. It occurs because of intermolecular forces between the liquid and surrounding solid surfaces. If the diameter of the tube is sufficiently small, then the combination of surface tension (which is caused by cohesion within the liquid) and adhesive forces between the liquid and container wall act to propel the liquid.

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First :

 

Capillary action (sometimes capillaritycapillary motion, or wicking) is the ability of a liquid to flow in narrow spaces without the assistance of, or even in opposition to, external forces like gravity. The effect can be seen in the drawing up of liquids between the hairs of a paint-brush, in a thin tube, in porous materials such as paper and plaster, in some non-porous materials such as sand and liquefied carbon fiber, or in a cell. It occurs because of intermolecular forces between the liquid and surrounding solid surfaces. If the diameter of the tube is sufficiently small, then the combination of surface tension (which is caused by cohesion within the liquid) and adhesive forces between the liquid and container wall act to propel the liquid.

Very good. Just as I have been saying. Intermolecular forces. Often especially strong when water is involved, due to its propensity for hydrogen bonding to suitable materials like glass -  as well as to itself, of course. This is a form of chemical energy (energy is released when such bonds form and is able to do work, such as lifting a liquid a limited distance). 

 

Next.........?

Edited by exchemist
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Capillary pressure is important in reservoir engineering because it is a major factor controlling the fluid distributions in a reservoir rock. Capillary pressure is only observable in the presence of two immiscible fluids in contact with each other in capillary-like tubes. The small pores in a reservoir rock are similar to capillary tubes and they usually containing two immiscible fluid phases in contact with each other. Thus capillary pressure becomes an important factor to be considered when dealing with reservoir rocks. When two immiscible fluids are in contact with each other in a capillary tube, a clear interface exists between them. This interface arises from interfacial tension effects. The interface is a curved surface and the pressure on the concave side exceeds that in the convex side. This pressure difference is known as capillary pressure. In the presence of two immiscible fluids, one of them preferentially wets the tube surface and it is called the “wetting” fluid, the other fluid is the “non-wetting” fluid. Capillary pressure can be mathematically expressed as: Pc = Pnw − Pw ...........................................................................................................(7-2) Where Pnw and Pw are the pressures of the non-wetting and wetting fluids across the interface, respectively. Figure 7-1 shows a capillary tube in contact with two immiscible fluids where the contact interface can be seen as well as a diagram of pressures throughout the system. 7-2 Figure 7-1. Pressure Relations in Capillary Tubes for Air-Water and Oil-Water Systems. (After Amyx et al – 1960) An important expression relating the capillary pressure with the radius of the capillary tube and the interfacial tension can be derived by balancing the pressures in the system including the hydrostatic and interfacial tension (considering a system with oil and water, where water is the wetting fluid): r cos P wo wo c 2σ θ = ...................................................................................................(7-2) Where σ wo is the interfacial tension between the fluids, θ wo is the contact angle, and r is the capillary tube radius. Considering the same system presented in Figure 7-1 and Equation 7-1, it can be shown that the capillary pressure is also related to the height at which the fluid rises inside the capillary tube. This relation is given by: Pc = Po − Pw = (ρ w − ρ o )gh ....................................................................................(7-3) Where Po and Pw are the oil (non-wetting fluid) and water (wetting fluid) pressures across the interface, respectively; ρ o and ρ w are the oil and water densities, respectively; g is the acceleration due to gravity, and h is the height of the column of water in the capillary tube with respect to a reference point (in the case of Figure 7-2, the level of water in the large vessel). 7-3 Capillary pressure may also be described using a more complex equation called the Laplace equation:         = + 1 2 1 1 r r Pc σ ........................................................................................................(7-4) Where Pc is capillary pressure, σ is interfacial tension, and 1 r and 2r are the principal radii of curvature. Figure 7-2 shows the distribution and measurement of the principal radii

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Capillary pressure is important in reservoir engineering because it is a major factor controlling the fluid distributions in a reservoir rock. Capillary pressure is only observable in the presence of two immiscible fluids in contact with each other in capillary-like tubes. The small pores in a reservoir rock are similar to capillary tubes and they usually containing two immiscible fluid phases in contact with each other. Thus capillary pressure becomes an important factor to be considered when dealing with reservoir rocks. When two immiscible fluids are in contact with each other in a capillary tube, a clear interface exists between them. This interface arises from interfacial tension effects. The interface is a curved surface and the pressure on the concave side exceeds that in the convex side. This pressure difference is known as capillary pressure. In the presence of two immiscible fluids, one of them preferentially wets the tube surface and it is called the “wetting” fluid, the other fluid is the “non-wetting” fluid. Capillary pressure can be mathematically expressed as: Pc = Pnw − Pw ...........................................................................................................(7-2) Where Pnw and Pw are the pressures of the non-wetting and wetting fluids across the interface, respectively. Figure 7-1 shows a capillary tube in contact with two immiscible fluids where the contact interface can be seen as well as a diagram of pressures throughout the system. 7-2 Figure 7-1. Pressure Relations in Capillary Tubes for Air-Water and Oil-Water Systems. (After Amyx et al – 1960) An important expression relating the capillary pressure with the radius of the capillary tube and the interfacial tension can be derived by balancing the pressures in the system including the hydrostatic and interfacial tension (considering a system with oil and water, where water is the wetting fluid): r cos P wo wo c 2σ θ = ...................................................................................................(7-2) Where σ wo is the interfacial tension between the fluids, θ wo is the contact angle, and r is the capillary tube radius. Considering the same system presented in Figure 7-1 and Equation 7-1, it can be shown that the capillary pressure is also related to the height at which the fluid rises inside the capillary tube. This relation is given by: Pc = Po − Pw = (ρ w − ρ o )gh ....................................................................................(7-3) Where Po and Pw are the oil (non-wetting fluid) and water (wetting fluid) pressures across the interface, respectively; ρ o and ρ w are the oil and water densities, respectively; g is the acceleration due to gravity, and h is the height of the column of water in the capillary tube with respect to a reference point (in the case of Figure 7-2, the level of water in the large vessel). 7-3 Capillary pressure may also be described using a more complex equation called the Laplace equation:         = + 1 2 1 1 r r Pc σ ........................................................................................................(7-4) Where Pc is capillary pressure, σ is interfacial tension, and 1 r and 2r are the principal radii of curvature. Figure 7-2 shows the distribution and measurement of the principal radii

So you can look stuff up on the internet. Jolly good. But what is the relevance of this screed to the subject at hand? 

Edited by exchemist
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so you learned there is negative pressure in a capillary. do you agree ?

No, I told you, many posts ago, that there will be a slightly reduced pressure in the capillary due to the upward force of the capillary rise, which very slightly offsets atmospheric pressure. Go back and re-read, please.  

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ok, then please tell me, when you arrive in reality and you understand, that in a capillary prevails negative pressure, of course

I repeat, the pressure is very slightly less than atmospheric. If you want to call that "negative" pressure, with respect to the ambient atmospheric pressure, that is fine by me, though in absolute terms is still very much positive of course.

 

Let's get a few numbers, so we can agree what we are talking about:-  

 

One atmosphere of pressure will support a column of water ~10m high.

 

If we have capillary rise, in a given tube, of say 5mm, we can see that the pressure reduction due to the attraction of the top surface meniscus of the water towards the glass, is 5 x 10⁻4 of an atmosphere. In other words, the pressure immediately below the meniscus has been reduced by 0.05% below atmospheric. So if we say atmospheric pressure is approx 1bar, the pressure at this point inside the capillary would be 0.9995 bar.

 

(The pressure inside the capillary at the level of the water surface outside it is of course exactly one atmosphere.)  

 

Agreed?   

Edited by exchemist
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ok I agree. lets come to the next stepp. do you agree, that a vertikal swimming floater has a little water hill, which climbs up the floater because of adhesion and this water hill is an additional weight that acts upon the floater ( outside a capillary ) ?

Depends on the material the floater is made of. But if it is glass, then sure. The capillary attraction will exert a downward force and pull the floater down slightly until it gains enough extra buoyancy to support the raised water around the line of contact with the water surface.

Edited by exchemist
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ok, and do you also agree, that if you put the floater into a capillary, then this little water hill is sucked into the water inside the capillary, because of the pressure relations, you described, so less weight acts upon the floater inside the capillary than upon the outside floater ?

Edited by deschoe
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ok, and do you also agree, that if you put the floater into a capillary, then this little water hill is sucked into the water inside the capillary, because of the pressure relations, you described, so less weight acts upon the floater inside the capillary than upon the outside floater ?

No. As I've just said, the downward pull on the floater will be the same whether the floater is inside the capillary or outside it. 

 

The degree to which floater floats higher inside the capillary will be due purely to the higher water surface inside it. 

 

Nothing is "sucked" in by the pressure reduction, because the raised water inside the capillary has already balanced that.   

Edited by exchemist
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so you think the water pressure at the lower end of the capillary floater is higher than the water pressure at the lower end of the outside floater even the lower end of the capillary floater is at a higher level than the lower end of the outside floater. this would mean pressure jumps inside the water. ???

Edited by deschoe
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Actually, now that I have read Craig's description again, I think I can see what deschoe is trying to do and I can offer an explanation in mechanical terms as to why it won't work:- 

 

Surface tension acts equally on both points where the hoop emerges from the liquid so contributes no net force to it. So we can ignore that. 

 

But Deschoe thinks that the buoyancy on the side where the liquid level is higher (due to capillary rise) will be greater than on the other side.

 

But buoyancy works by displaced liquid falling, i.e. reducing in gravitational potential energy as the buoyant object rises. The liquid can change places with the object and because the weight of the displaced liquid exceeds that of the rising object, the liquid will fall and the object will rise. 

 

But, in this case, rotation of the hoop or ring does not provide a space into which liquid can fall, so as to change places with it. So no motion arises.  

 

 

The forever rotating hoop is clearly a perpetual motion fantasy. What I was musing about earlier would be a capillary tube that is bent over on itself like a siphon with the inlet submerged in the water and the outlet just a tiny distance from the water. My naïve thought was maybe a drop would form on the outlet end, making contact with the water surface and then being pulled out of the tube, then another drop and so on. Then I realized that the capillary force that allows the water to be drawn into the tube in the first place, would not allow a drop to emerge out of the tube. So even this “perpetual drops of water” crazy idea could not possibly work. And, if something that basic cannot work, certainly a more complex scheme involving a hoop or two floaters cannot possibly work. But, as you say, it may be interesting to work out the physics and math to show exactly why it can’t. And I see you are doing that now. Interesting! I won't interrupt.

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