Guest Aemilius Posted July 10, 2013 Report Posted July 10, 2013 (edited) Hey CraigD.... CraigD "I think the best way to understand this thought experiment is to purposefully avoid the details of how its critical 'water dimple' is made.... Details of how the depression is made? I think the only thing to consider here is the effect the magnet has on the surface of the fluid, *maintaining the depression in the resevoir that forms as a direct result of the diamagnetic property of the fluid. There's no reason to avoid that, especially since it's clearly a defining feature of the experiment.... How the depression is actually made is beyond the scope of the thought experiment so I'm not sure why you're bringing that up.... For the experiment we only need to consider the (enabling) effect the depression may have on the system. *The word "maintain" means to "Keep (something) at the same level....". All three methods depicted in the diagrams, magnetism, air pressure and water pressure, can maintain a fluid depression, making it a suitable adjective. CraigD "....simply assuming it can be done, and consider the essential mechanics of the system. We can do this in a simplified, but precise and correct, way." Right, I think I see where this is going now.... It will all be much easier to understand after thoroughly gutting the experiment by removing from consideration a (if not the) defining feature of it. Then we can comfortably move on to consider the essential mechanics of the system in a simplified, but precise and correct, way.... confident that the result will conform to the foregone conclusion we knew all along we must arrive at. Avoiding mention of the magnetic/diamagnetic repulsive effect and the role it plays in maintaining the depression, by just assuming it can be done in any number of ways, as I said before, conveniently lumps together the magnetic method that doesn't require any input of energy (gas, electricity, etc.) to maintain a depression.... ....with all the other methods of maintaining a depression that do, like a Powered Air Pump that requires input of energy (gas, electricity, etc.) to maintain a depression.... ....or a Powered Water Pump that requires input of energy (gas, electricity, etc.) to maintain a depression. CraigD "Let’s begin with the system in the state sketched. Let’s assume the size and difference in height of the surface on the right side and the bottom of the tube (which determine the force acting on the water in the tube) and the frictional forces in the tube are exactly sufficient for its rate of flow to be 1 m3/s. So, after 1 s of operation, the dimple will contain 1 m3/s. of water.Now for the dimple to retain its shape, whatever is producing it must exert sufficient force over sufficient distance to displace some or all of the water below that 1 m3/s. of water of water to restore the dimple to its previous shape. This force, times the distance over which it acts, is the amount of work the system must performs in 1 second – that is, it is its power. For the same of being able to use specific quantities in this example, let’s assume that the cross sectional area of the tube is 1 m2, the difference in height of the surface on the right side and the bottom of the tube is 1 m, and, to keep our numbers short and simple, that the acceleration of gravity is exactly 9.8 m/s/s, , and the density of the fluid (let’s just assume it’s pure water) 1000 kg/m3. The system, then, has moved a 1000 kg volume of water a height of 1 m in 1 s, so its power, , plus the power to overcome its friction (which is too hard to calculate, and only distracts from this discussion)." If less than this power is put into the system, the dimple will fill with water until the tube’s height difference is zero, and the siphon will stop." And there it is! By avoiding consideration of the qualitatively different effect the magnetic field has on the fluid compared to the other methods, you've successfully arrived at your foregone conclusion. Let's go ahead and run your version of the experiment then.... Before the experiment starts, the fluid in the resevoir and the fluid at the bottom of the depression on the surface maintained by the magnetic field are at rest in equilibrium and the siphon is full (shown in the diagram below at the moment the left end of the tube is exposed to atmospheric pressure). As the experiment starts, fluid will begin to flow. The next thing that happens in your version of events (below) is that after the fluid begins to flow, the depression will quickly fill with fluid, reducing the difference in height between the columns to zero. At that point the operation of the siphon will come to a stop, at rest in equilibrium.... No more flow. If the fluid in the region just beneath the magnet was at rest in equilibrium just before the experiment started in the form of a depression maintained by the magnetic field, how could the siphon simply fill the depression like a bucket, stop flowing, and then be at rest in equilibrium at the same level as the rest of the fluid in the resevoir as if the magnet wasn't even there? If your version of the sequence of events was correct, at the end of the experiment, the magnet will end up in close proximity to the fluid without repelling it, the fluid will end up in close proximity to the magnet without being repelled by it, and the whole mechanism of the (well known) interaction between a diamagnetic fluid and a magnetic field will have been mysteriously defeated by the momentary flow of fluid in the siphon. Looking at the obvious impossibility of your result, I think my version of the sequence of events is actually much more likely.... that as the experiment starts, fluid will begin to flow and start filling the depression, causing the magnetic field to begin acting to restore the depression, leading to more flow filling the depression, causing the magnetic field to continue restoring the depression, and so on. I think if the fluid was allowed to flow too quickly it could overwhelm the capability of the magnetic field to restore the depression, but if the flow rate was carefully controlled to match the rate at which the magnetic field is capable of restoring the depression, the fluid level in the depression would remain constant while the siphon was in operation. Thanks again for responding and discussing it with me.... Your move. Edited July 11, 2013 by Aemilius Quote
CraigD Posted July 11, 2013 Report Posted July 11, 2013 I think we’re settling into a usual mode of discussion for a perpetual motion machine of the 1st kind (not a bad thing, if it’s a productive discussion), which starts with (1) a simplified description showing its impossibility, then (2) the insistence that a more detailed description would conclude it is possible, then (3) a more detailed description. So, on to phase 3! Avoiding mention of the magnetic/diamagnetic repulsive effect and the role it plays in maintaining the depression, by just assuming it can be done in any number of ways, as I said before, conveniently lumps together the magnetic method that doesn't require any input of energy (gas, electricity, etc.) to maintain a depression....It’s true that permanent magnets don’t require energy to exert force on charged or magnetic materials. We must broaden our perspective on work/energy for this system beyond drawing using gas, electricity, etc. This, I think, guides us in the right direction:How, after flowing for only a moment and filling the depression, has the diamagnetic property of the fluid just vanished? How has the magnet just stopped repelling the fluid which now rests in equilibrium just beneath it?It hasn’t, and it hasn’t. Consider, however, what’s happened when the water “flowed to fill the depression”: When you move something that a magnet repels toward the magnet, you must apply a force over the distance you move it. Multiplying this force by this distance gives the work of this action, and the energy stored it stores in the thing+magnet system. If you stop applying the force and allow the thing to return to rest at its starting position, the same amount of energy is extracted from the system as was put into it. So, in this part of the Cole syphon thought experiment, the energy that must be put into the system is to move (flow) the water toward the magnet (into the depression). The force for this work-energy comes from the net force of the columns of water in the tube. The work done by the magnet “restoring the dimple” is equal to the work done moving the water into it, plus frictional losses. So the magnet is being powered not by an external electric current or similar, but by the syphon. It’s a bit tricky to intuitively grasp this with the current sketches, which show the water approaching the magnet from above and going away from it below, so let’s re-imagine (or better, re-sketch) it like this: Ignoring changes in friction, we’re allowed to do this with a syphon without changing its key mechanics – a syphon flows no matter where its tube goes, because it depends only on the difference in the height of its two effective ends. Hopefully, in this re-drawn sketch, it’s intuitively obvious that the water follows effectively the same path going toward the magnet, which stores energy, as going away from it, which removes energy. The energy calculations would work out the same no matter what path the water followed approaching and going away from the magnet, but would be more complicated, and harder to intuit. The mechanical accounting gets neat and simple here, if we keep the language high-level and non-mathematical. The pressure on the water in the left end of the tube is higher on the than on the right where it enters the water by exactly the amount needed to depress a column of water the left end’s distance below the un-dimpled surface of the water. To the water in the tube, then, the system “looks” (or, perhaps better said, feels) as if the height of the left end of the tube the same as the height of the right end. With the difference in tube end heights being zero, the syphon won’t flow. JMJones0424 1 Quote
Guest Aemilius Posted July 11, 2013 Report Posted July 11, 2013 (edited) Hey CraigD.... I have removed this post. I'd thought I might be able to end this thread quickly with a few questions that would lead to consensus, but there're too many issues with your last post that need to be addressed. I'll post a full response to your last post as soon as it's completed. Thanks again for agreeing to continue discussing it with me. Edited July 15, 2013 by Aemilius Quote
Moontanman Posted July 11, 2013 Report Posted July 11, 2013 (edited) Ok, this is not a perpetual motion machine, that is obvious and this is why. Let's assume the magnet works as planned and creates a dimple in the water and your siphon flows to fill the dimple, it shouldn't take long before the water in each basin goes back to equilibrium because the magnets ability to create a dimple will degrade as the water level rises in the basin with the magnet. So even if the magnet creates a dimple the rising water will stop this as soon as the water level reaches a level that keeps the magnet from depressing the water below the level of the siphon... So even in perfect circumstances the siphon would reach equilibrium as soon as the dimple ceased to exist due to rising water levels... edit: Ok, I assumed two basins, my bad, I missed the point... Edited July 11, 2013 by Moontanman Quote
Guest Aemilius Posted July 13, 2013 Report Posted July 13, 2013 (edited) Hey CraigD.... I'm just going to throw this out there and see where it goes.... After the experiment starts (exposing the left end of the siphon full of fluid to atmospheric pressure), runs and then ends, what condition will the system be in? If it's not one of these, would you mark up one of the diagrams and post it? ... Edited July 14, 2013 by Aemilius Quote
Guest Aemilius Posted July 14, 2013 Report Posted July 14, 2013 (edited) CraigD "It’s a bit tricky to intuitively grasp this with the current sketches, which show the water approaching the magnet from above and going away from it below, so let’s re-imagine (or better, re-sketch) it like this: The CraigD Siphon Ignoring changes in friction, we’re allowed to do this with a syphon without changing its key mechanics – a syphon flows no matter where its tube goes, because it depends only on the difference in the height of its two effective ends. Hopefully, in this re-drawn sketch, it’s intuitively obvious that the water follows effectively the same path going toward the magnet, which stores energy, as going away from it, which removes energy. The energy calculations would work out the same no matter what path the water followed approaching and going away from the magnet, but would be more complicated, and harder to intuit. The mechanical accounting gets neat and simple here, if we keep the language high-level and non-mathematical. The pressure on the water in the left end of the tube is higher on the than on the right where it enters the water by exactly the amount needed to depress a column of water the left end’s distance below the un-dimpled surface of the water. To the water in the tube, then, the system “looks” (or, perhaps better said, feels) as if the height of the left end of the tube the same as the height of the right end. With the difference in tube end heights being zero, the syphon won’t flow." You missed the significance of the air gap earlier and now you've missed the significance of the siphon passing through the magnet instead of around it, as shown in The CraigD Siphon above. In The CraigD Siphon the fluid in the siphon is only exposed to one pole of the magnet which is facing the end of the siphon. This will result in a net gain in downward force (red) exerted by the magnet on the fluid in the siphon. Contrast this with The Cole Siphon where the fluid in the siphon is simultaneously exposed to both poles of the magnet. The upward repulsive force exerted on the fluid in the siphon immediately above by the magnet is precisely matched by the equal and opposite downward repulsive force exerted on the fluid in the siphon immediately below by the magnet. This will result in a net gain of zero in upward or downward force exerted on the fluid in the siphon by the magnet. Conclusions.... The CraigD Siphon - A net gain in downward force is exerted by the magnet on both the fluid in the resevoir and the column of fluid in the siphon on the left. The magnet will prevent the column of fluid in the siphon on the left (exposed to only one pole) from falling under the force of gravity alone. The Cole Siphon - A net gain in downward force is exerted by the magnet on the fluid in the resevoir, but a net gain of zero in upward or downward force is exerted by the magnet on the column of fluid in the siphon on the left. The magnet will not prevent the column of fluid in the siphon on the left (exposed to both poles) from falling under the force of gravity alone. You've said the two arrangements (The CraigD Siphon and The Cole Siphon) are essentially the same because the fluid follows effectively the same path going toward the magnet, and that the energy calculations work out the same no matter what path the fluid follows approaching and going away from the magnet. As shown above, they clearly do not follow the same effective path toward the magnet, and the energy calculations clearly do not work out the same no matter what path the fluid follows approaching and going away from the magnet, evidenced by the dramatically different conditions each produces. I just wanted to point that out. Something else we can clear up has to do with your remark earlier that.... "When you move something that a magnet repels toward the magnet, you must apply a force over the distance you move it." In light of the above, it's obvious that this line of reasoning applies to the The CraigD Siphon, but does not apply to The Cole Siphon. Conclusions.... The CraigD Siphon - Some additional form of energy must be put into the system to advance the fluid toward the magnet because the force of gravity alone is insufficient. The Cole Siphon - No additional form of energy must be put into the system to advance the fluid toward the magnet because the force of gravity alone is sufficient. The Cole Siphon Anyway, back to the question from my last post.... After the experiment starts (exposing the left end of the siphon full of fluid to atmospheric pressure), runs and then ends.... What condition do you think the system will be in? If it's not one of these, could you mark up one of the diagrams and post it? ... Edited July 16, 2013 by Aemilius Quote
CraigD Posted July 16, 2013 Report Posted July 16, 2013 (edited) I’ll respond to you later post, Aemilius, but want to make an important point in response to a fragment of this oneAfter the experiment starts (exposing the left end of the siphon full of fluid to atmospheric pressure) The performance of a siphon doesn’t depend on atmospheric pressure, despite their resemblance to tube barometer, which do. It depends only on the difference in height of their tube ends, the acceleration of gravity, and the complicated effects friction on the fluid in its tube. A siphon doesn’t need air pressure acting on the surface of the fluid in its reservoirs – assuming it doesn’t do untoward things to the fluid, like allow it to boil, a siphon works in vacuum, so long as it has gravity and an opposing force (eg: isn’t in free fall). Without gravity – we can also say “without weight” – a siphon won’t work, regardless of the pressure of the air around it. Awareness of the presence of air around a siphon actually complicates describing its mechanics, as air is also a fluid (we are taught in school to consider air a gas rather than a fluid, but this distinction applies to the connectedness of its molecules, not it’s dynamic behavior – aerodynamics, the physics of the flow of air, and hydrodynamics, of the flow of water, are both special cases of the more general fluid dynamics, the physics of both), like what for simplicity’s sake I’ll call the water that flows through its tube. The column of air between the ends of its tube and outside it has a net height equal to the column of water inside it, which opposes the force of the siphon. The siphon works the way it does because the density of the air is lower than that of the water – were we to contrive (perhaps with thin membranes to allow H2O to float above N2) an arrangement where the “air” part of the apparatus has higher density than the “water” part, it would flow backwards, from low to high. -Edit: the text within the spoiler box has a major mistake - see the PS in this post for details. That said, let’s ignore the air part of these sketches, and be concerned only with the water part, the major force-at-a-distance effect, gravity, and the unusual one, a permanent magnet. You've said the two arrangements (The CraigD Siphon and The Cole Siphon) are essentially the same because the fluid follows effectively the same path going toward the magnet, and that the energy calculations work out the same no matter what path the fluid follows approaching and going away from the magnet. As shown above, they clearly do not follow the same effective path toward the magnet, and the energy calculations clearly do not work out the same no matter what path the fluid follows approaching and going away from the magnet, evidenced by the dramatically different conditions each produces. I’m sticking with the claim that the CraigD and Cole siphons are, ignoring friction, effectively identical, without attempting to formally calculate or model in detail the magnetic forces involved, because such work would be difficult. I think we’re getting close to an informal, intuitive understanding of why neither the Cole nor the CraigD siphon can flow. After some groping, I’ve come up with two explanation approaches: #1This one makes an intuitively appealing replacement, then draws from fairly obvious potential energy laws. Magnetic field can be considered to emerge in a complicated way from electrostatic fields resulting from the interaction of charged particles. So let’s replace the magnet in this thread’s sketches with a compact object with a large net charge, and the magnetized water with water with a net charge – that is, ionized water, or water with ions dissolved in it. The forces on the water from this “electrostatic pusher” now follow a simple law, Coulumb’s law, [math]F = k_1 \frac{Q_1 Q_2}{r^2}[/math] Which is similar in form to the universal gravity equation, [math]F = k_2 \frac{m_1 m_2}{r^2}[/math] So we can equate “approaching and going away from the magnet-analogous pusher” with changing [imath]r[/imath]. We know from simple classical mechanics that the energy calculations around change in [imath]r[/imath] depend only on its starting and ending value, not the intermediate values of its “path”, given by the GPE formula, [math]E = k_3 \frac{m_1 m_2}{r}[/math] So it doesn’t matter how the siphon tubes are arranged, so ignoring friction, the CraigD and Cole siphons are effectively identical. #2This explanation approach is essentially an appeal to symmetry. In the CraigD siphon, it’s obvious why (not surprisingly as I imagined this thought experiment variation for just this purpose): the force of the magnet on the water in the tube is exactly equal and opposite to the force due the net height of the column of water in the tube, which is also equal to the difference in height of its two ends. The water inside the left side of the tube is subject to the same downward magnetic force as the water outside it – it’s “part of the dimple”, so can’t “spill” above it. Nor, if the left side of the tube is lowered to below the surface of the dimple, can the siphon flow, because then the pressure (force divided by area) of the column of water in the siphon is equal and opposite that of the magnet on the surface of the dimple. In the Cole siphon, it’s less obvious. The upward force of the magnet (it doesn’t matter if the N or the S pole is up) on the water in the tube above it opposite its downward force on the water in the tube below. Though opposite, however, the upward and downward force aren’t equal, so there is a net force by the magnet on the water in the tube, equal (if the tube reached the surface of the dimple) or greater (if it doesn’t – that is, if there’s a air gap) to the force of the siphon due to the difference in height of the tube ends (and gravity). Here’s why: The pressure necessary to depress a column of water the depth of the dimple is given by the same formula mentioned in post #22, and quoted in (edited) post #1:[math] P = h p g [/math]In the case of the dimple-making pressure of the magnet, [imath]h[/imath] is the depth of the dimple. In the case of the siphon, it’s the difference in the height of the left end of the tube and the undimpled surface of the water. The left side of the tube can’t be deeper than the dimple, so the pressure due to the siphon can’t be greater than the pressure of the magnet on the water in the tube below it. Now, consider the water in the tube above the magnet. The force on a given small cylinder of it from the magnet is equal and opposite for a given distance above or below. But the tube below the magnet ends before the tube above it. Though water below the bottom of the tube continues to experience the force of the magnet, the lack of force from the tube walls prevents it from transferring this force to the column inside the tube – though complicated, this is why siphons must have tubes, rather than it being possible to create one with just an arcing column of water between a high and a low reservoir. Now we need to show that the net opposite force of the magnet on the water in the tube equals or exceeds that of siphon. Fortunately, this comes simply and directly from the hydrostatic pressure equation, because the force on a column of water (for ease of visualization, we can think of it as having the same cylindrical dimensions as the tube, but for reasons beyond the scope of this explanation, that doesn’t matter – any column will do) below the surface of the dimple is the same as on the column of water the same distance above the magnet, which is the same as the force of the siphon. The distance of the bottom of the tube from the magnet is less than or equal to the distance of the dimple from the magnet, so the net force of the magnet is greater than of the siphon. We might be tempted to try rearranging the tube to reduce the net opposite force of the magnet on the water, but I’m hoping we can visually imagine this as a gradual transition from the “straight through the magnet” Cole siphon arrangement to the “as far away from the magnet as possible” CraigD siphon arrangement, and intuitively accept that it remains impossible to have the siphon discharge within the magnet-created dimple without the same forces that create the dimple opposing and exceeding those of the siphon. Edited July 19, 2013 by CraigD Noted mistake C1ay 1 Quote
CraigD Posted July 16, 2013 Report Posted July 16, 2013 As a related aside, this Livescience.com article, this, and this are science news articles about Yuanming Liu and colleagues’ semi-famous (it ranked #39 in time.com’s “50 best inventions of 2009”, between a foldable speaker and an edible race car ;)) baby mouse levitating experiment. Though I’ve tried in this thread to avoid confusion from delving into the details of using magnets on water, it’s an interesting subject, and worth a bit of thought into how it works. A mouse is essentially a complicated, bulky bag of H2O water, about 50% water by mass. H2O, like all molecules, consists of particles of opposite charge – protons and electrons, which are attracted or repelled by particles of opposite or same charge. Normally, the position of these particles is so mixed up that no large net force is exerted or experienced by water, but a sufficiently strong magnetic field – which in the case of nearly all strong magnets, including Liu’s 17 T electromagnet, is produced largely by a material with its protons and electrons un-mixed up – can un-mix these charged particles, resulting in the same strong magnetic field that un-mixed its particles exerting a force larger than that of gravity on the water, resulting in – in this case – a levitating 0.01 kg baby mouse. Liu used a super-cold super-conducting solenoid-type electromagnet, which to the best of my understanding is technically a permanent magnet (being super-conducting, has no resistance, so needs no energy input), unlike stronger magnets such as the 100 T LANL’s Pulsed Field Facility magnet, which draws 320 MW of electricity, or the 2800 T field created with (highly baby-mouse-unfriendly) explosives in 1998 at the VNIEF. The record for a solid magnet is, I think, 1.25 T, for a NdFeB permanent magnet. From what I know of Aemilius, I suspect that no matter how much “it’s impossible” physics we talk, he won’t be satisfied ‘til he’s built something involving magnets and water. Calculating how much of a dimple can be made with an affordable permanent magnet (good-size NdFeB ones are actually not very expensive) poses a challenge for a physics amateur of my ilk, and I suspect, even a pro, as there’s “phase transition-y” complexity involved, involving the above mentioned H2 un-mixed up-ing (of more conventionally put, magnetizing). It might be easier just to assemble some NdFeB magnets and as still pool of water as can be managed, and see what can be done. As best I’ve been able to figure looking at the few photos of it I’ve been able to find, Liu’s apparatus puts the mouse inside it’s coil-wrapped tube, not over or under it, so using something like it to make the water dimple sketched in this thread poses some redesign challenges. I think you’d have to keep have an inside-the-solenoid reservoir for the dimple and the left side of the tube, an outside one for the right side of the tube, and a separate syphon to make them effectively a single reservoir, snuck into the open top of the solenoid in a way that avoids the water in it getting magnetized. Or possibly one or more simple tubes passing through solenoid’s coils. The various articles mention a “levitating zone” that the mouse could fall out of, so I’ve hope one of these approaches would be possible. Turtle and C1ay 2 Quote
Guest Aemilius Posted July 16, 2013 Report Posted July 16, 2013 (edited) CraigD "I’ll respond to you later post, Aemilius, but want to make an important point in response to a fragment of this one." Aemilius "After the experiment starts (exposing the left end of the siphon full of fluid to atmospheric pressure)...." CraigD "The performance of a siphon doesn’t depend on atmospheric pressure, despite their resemblance to tube barometer, which do. It depends only on the difference in height of their tube ends, the acceleration of gravity, and the complicated effects friction on the fluid in its tube. A siphon doesn’t need air pressure acting on the surface of the fluid in its reservoirs – assuming it doesn’t do untoward things to the fluid, like allow it to boil, a siphon works in vacuum, so long as it has gravity and an opposing force (eg: isn’t in free fall). Without gravity – we can also say 'without weight' – a siphon won’t work, regardless of the pressure of the air around it." Well, you can take all that up with Pascal. "The moment the left end of the siphon is exposed to atmospheric pressure...." signifies the beginning of the experiment so, if you like, for the moment, while you settle your dispute with Pascal, let's just say "The moment the left end of the siphon is uncapped....". I find it incredibly odd that out of my entire post you chose that to focus on for the opening of your last post, but it is a thought experiment after all and you're entitled to focus on whatever you choose. You've certainly given me a great deal to think about with your "all over the map" response. Kind of reminds me of that joke you told me about milk production, and how when it was the physicist's turn to speak, he walked to the board, drew a circle, and began.... “assume the cow is a sphere”. In this case though, the "physicist" seems to begin with.... "assume the sphere is a cow"! For some reason you seem to be intent on changing every single element and condition of this simple thought experiment.... You want to ignore the atmospheric pressure part of the diagrams.You want to replace the permanent magnet in the diagrams with a compact object with a large net charge.You want to replace the diamagnetically enhanced fluid with water that has a net charge on it.You even want to change the “straight through the magnet” Cole Siphon arrangement to the “as far away from the magnet as possible” siphon arrangement! I have to say you bring up some good points too, though that will take me a while to digest (ran to Safeway and bought a case of Pepto Bismol). CraigD "The record for a solid magnet is, I think, 1.25 T, for a NdFeB permanent magnet." Actually, the current record is 5 Teslas for a room temperature permanent magnet, which is why I chose that field strength for the thought experiment. CraigD "From what I know of Aemilius, I suspect that no matter how much “it’s impossible” physics we talk, he won’t be satisfied ‘til he’s built something involving magnets and water." You may not know as much of me as you think. Anyway, you mentioned earlier in post 50 that without putting a certain threshold level of energy into the system that "....the dimple will fill with water until the tube’s height difference is zero, and the siphon will stop." I was seeking clarification of your position when I asked you what conditon the system would be in after the experiment started, ran and then ended, but you didn't respond. So, I ask again.... After the experiment starts (uncapping the left end of the siphon full of fluid), runs and then ends.... What condition do you think the system will be in? If it's not one of these, could you mark up one of the diagrams and post it? ... Edited July 17, 2013 by Aemilius CraigD 1 Quote
Guest Aemilius Posted July 17, 2013 Report Posted July 17, 2013 (edited) Hah! How you doing C1ay? When I saw you were viewing the thread I bet my brother (he's following all this) $25.00 you wouldn't write a word and that you'd give my post a "negative reputation" mark. As soon as we shook hands, I refreshed the page and there it was! I'll buy a nice big steak and a six pack tonight.... Thanks man! Edited July 17, 2013 by Aemilius Quote
Guest Aemilius Posted July 17, 2013 Report Posted July 17, 2013 (edited) Hey CraigD.... I know there are ways a siphon can operate under abnormal conditions, but I won't argue the role atmospheric pressure plays in the operation of a practical siphon. No matter what you say I'll side with Pascal. The logic of this part of your post though is inescapable.... CraigD "Now, consider the water in the tube above the magnet. The force on a given small cylinder of it from the magnet is equal and opposite for a given distance above or below. But the tube below the magnet ends before the tube above it. Though water below the bottom of the tube continues to experience the force of the magnet, the lack of force from the tube walls prevents it from transferring this force to the column inside the tube...." Sharp eye, that's a real problem. I didn't see it before, but I definitely see the logic of it now just as you describe it. So for now, since I can't show the forces exerted by the two poles of the magnet on the fluid in the siphon to be truly equal and opposite, I have to withdraw the question.... "After the experiment starts (uncapping the left end of the siphon full of fluid), runs and then ends, what condition do you think the system will be in?" After I illustrate a procedure for correcting the problem in a modified diagram though, I'll be asking you that question again (probably a day or two). Take care. Edited July 17, 2013 by Aemilius Quote
C1ay Posted July 17, 2013 Report Posted July 17, 2013 Hah! How you doing C1ay? When I saw you were viewing the thread I bet my brother (he's following all this) $25.00 you wouldn't write a word and that you'd give my post a "negative reputation" mark. As soon as we shook hands, I refreshed the page and there it was! I'll buy a nice big steak and a six pack tonight.... Thanks man!Sounds great. I'd rather see you waste the money on a 5 tesla magnet though so you can learn from experience. DFINITLYDISTRUBD and Moontanman 2 Quote
Guest Aemilius Posted July 18, 2013 Report Posted July 18, 2013 (edited) Hey CraigD.... CraigD "Now, consider the water in the tube above the magnet. The force on a given small cylinder of it from the magnet is equal and opposite for a given distance above or below. But the tube below the magnet ends before the tube above it. Though water below the bottom of the tube continues to experience the force of the magnet, the lack of force from the tube walls prevents it from transferring this force to the column inside the tube...." I still say that was a great observation. I'm fairly certain you're familiar enough with the system not to need any accompanying text or explanation to evaluate this sequence of diagrams, but if you do just say so. It has some interesting implications, but I'll wait for you to share your opinion before getting into that. Tell me what you think when you have a moment.... Edited July 18, 2013 by Aemilius Quote
Moontanman Posted July 18, 2013 Report Posted July 18, 2013 Hey CraigD.... I'm fairly certain you're familiar enough with the system not to need any accompanying text or explanation to evaluate this sequence of diagrams, but if you do just say so. It has some interesting implications, but I'll wait for you to share your opinion before getting into that. Tell me what you think when you have a moment.... Has the dimple effect been observed or is this all hypothetical? Quote
Guest Aemilius Posted July 18, 2013 Report Posted July 18, 2013 (edited) Hey Moontanman.... Moontanman "Has the dimple effect been observed or is this all hypothetical?" No, it's not just hypothetical. The air gap in the record breaking 5 Tesla permanent magnet I chose to base the experiment on is only a few millimeters accross. Maybe it could operate a small siphon with a diameter a little smaller than a drinking straw if the details could be worked out, I'm not really sure though. The emphasis in the scientific community right now seems to be all about decreasing the effect of gravity rather than increasing it the way I've illustrated it here (downward force on the surface of the fluid in the resevoir). Here's something on how a dimple can be created.... "Place a neodymium magnet in a shallow dish. Fill the dish with water so the magnet is completely covered (about .5cm water above the magnet). Bounce light from a distant small source off the water surface and onto a wall or screen. (sunlight works well.) You will see a uniform oval projected spot of sunlight reflection from the water surface. In the projected oval of light, right at the location of the submerged magnet, you'll find a small bright splotch. The bright splotch is caused by a concave dimple in the water surface. The magnet repels the water slightly, which creates the concavity. Try using less water so there is just 1mm between the magnet face and the surface. This gives a bigger effect, but some people might then suspect that surface tension plays a role. Try looking down into the bowl so you see the reflection of the ceiling. If you move your head back and forth, you will detect a small distortion at the location of the magnet. C. Brown suggests placing the bowl in front of a screened window, then looking at the reflection of the wire mesh within the bowl. The distorted water surface will cause Moire' patterns to be seen." And here there's the levitation of a two inch wide water drop (obviously using a much stronger magnetic field).... "The researchers also levitated water drops up to 2 inches wide (5 cm). This suggests the variable gravity simulator could be used to study how liquids behave under reduced gravity, such as how heat is transferred or how bubbles behave." It seems logical to assume that the upward force needed to levitate a water drop that size directed downward could displace an equal volume of water on the surface of a container full of water. I tried an online calculator that gave a volume of about 65 cubic centimeters (65 grams, sounds a bit high to me). Maybe someone could double check that and post it. If the water drop was flattened though the volume would be considerably less, the article didn't say. Anyway, I thought it was an interesting idea. The 5 Tesla permanent magnet would probably run around $100,000 or more, but CraigD and C1ay still seem convinced I'm actually going to try and build one no matter how often I say it's just a thought experiment. I'll build it.... Just as soon as I win the lottery. Barring some kind of "Eureka!" moment, I think I'm pretty much done with it for now (depending on CraigD's next post), plus I think I've got C1ay following me around "neg repping" my posts, he must be all riled up about something I said.... At least I got a steak and a six pack out of it! Later man. Edited July 18, 2013 by Aemilius Quote
Guest Aemilius Posted July 18, 2013 Report Posted July 18, 2013 (edited) I said in the last post that "Barring some kind of "Eureka!" moment, I think I'm pretty much done with it...." I take that back. No "Eureka!" moment (well, maybe a little one, we'll see), but as an artist, when you've been concentrating on an idea for a while images have a tendency to just "pop up", and a real whopper of an idea just did. I don't want to get out ahead of CraigD though so I'll bring them up and post them as time goes on. One of them may even be possible right now using cheap commercially available magnets. Edited July 18, 2013 by Aemilius Quote
CraigD Posted July 19, 2013 Report Posted July 19, 2013 As I and wikipedia have noted before (I used the phrase “terribly difficult”, wikipedia the more understated “non-trivial”), modeling magnetic fields for a give arrangement of permanent magnets and material materials such as your latest magnet + “magnetic shield” is … err … terribly difficult, so I won’t try to do that - or if I do, I’ll try it privately, ‘til I’ve got something presentable. I’ve found “gravity simulators”, in the form of simple human-readable computer programs, useful and convincing for explaining the impossibility of various gravity-based perpetual motion machines, so would love to have a similar “magnet simulator”. Instead, let’s try a RAA proof. Let’s assume (falsely, to the best of my understanding – which is that, despite requiring more complicated equations, magnets in magnetic fields obey the same energy laws as charged bodies in electrostatic fields) that there is a way to just “turn off” the magnetic field in a desired volume of space, without affecting it outside of that volume, and that it looks like your sketch, Aemilius. We could then make a perpetual motion machine of the first kind simply by putting this magnet-and-shield device on a closed loop of water with a hydroelectric generator, like this:Because the net force on the water on one side of the device is greater than on the other, water should flow, allowing energy to be extracted by the generator. Unlike the latest sketch of the Cole siphon, where the magnet’s PPM-creating effect is indirect, it’s obvious in this PPM that the magnet device is adding energy to the system, which, being just a permanent magnet and a “shield”, it shouldn’t be able to do. So the device is impossible. The Cole siphon depends on the same impossible “more net force on one side than the other” characteristic of the device. So it, too, is impossible. PS: I need to correct a mistake in a previous post:The performance of a siphon doesn’t depend on atmospheric pressure, despite their resemblance to tube barometer, which do. It depends only on the difference in height of their tube ends, the acceleration of gravity, and the complicated effects friction on the fluid in its tube. A siphon doesn’t need air pressure acting on the surface of the fluid in its reservoirs – assuming it doesn’t do untoward things to the fluid, like allow it to boil, a siphon works in vacuum, so long as it has gravity and an opposing force (eg: isn’t in free fall). Without gravity – we can also say “without weight” – a siphon won’t work, regardless of the pressure of the air around it.Well, you can take all that up with Pascal. "The moment the left end of the siphon is exposed to atmospheric pressure...." signifies the beginning of the experiment so, if you like, for the moment, while you settle your dispute with Pascal, let's just say "The moment the left end of the siphon is uncapped....". I find it incredibly odd that out of my entire post you chose that to focus on for the opening of your last post, but it is a thought experiment after all and you're entitled to focus on whatever you choose.I jumped on the atmospheric pressure mention because I though it a good, quick, easy “leaning opportunity”. It was, but not in the way I expected – I was practically completely wrong in my claim about a siphon not depending on atmospheric pressure! As the wikipedia article you link as “Pascal” rightly explains, the difference in height of the inlet and the highest point in siphon cannot be greater than [math]h =\frac{P}{\rho g}[/math]where [imath]P[/imath] is the atmospheric pressure, [imath]\rho[/imath] the density of the fluid, and [imath]g[/imath] the acceleration of gravity. For standard values of [imath]P = 101325 \,\mbox{N/m}^2[/imath], [imath]\rho = 1000 \,\mbox{kg/m}^3 [/imath] and [imath]g = 9.80665 \,\mbox{m/s/s}[/imath], this gives [imath]h \dot= 10.33 \,\mbox{m}[/imath] – a good thing to know before trying to build a big siphon! It’s useful to think of a siphon as being analogous to a rope or chain over a pulley, which is what I was doing when I incorrectly stated that air pressure isn’t important. I wasn’t completely wrong in my claim, as because water and other fluids do have a small but non-zero tensile (pulling) strength, it is possible to for a siphon to work in a vacuum. Because the tensile strength of fluids are small, though, such a siphon must be finely built and weak, so practically speaking, siphons must have air pressure to work. As long as the air pressure is the same on both ends of a siphon, and its high point doesn’t exceed the maximum, however, air pressure can be ignored, simplifying the description back to one that depends only on gravity, fluid density, and the inlet-outlet height difference. JMJones0424 1 Quote
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