Hydro Posted January 26 Author Report Share Posted January 26 (edited) On 1/9/2024 at 10:29 AM, Hydro said: I found several more articles on this topic. Some are not in English. In my system this language is defined as Serbian. The automatic translation seems to be clear. For now I’m posting only the most interesting from my point of view Advanced mathematics from a professor at the University of Michigan. Edited January 26 by Hydro Link to comment Share on other sites More sharing options...

Hydro Posted February 2 Author Report Share Posted February 2 (edited) On 04.01.2024 at 05:45, Hydro said: Yes, it seems that the task is not easy. Nobody has anything to say? I myself, when I saw it for the first time, simply fell into a stupor. I was especially shocked by the phrase: "Unlike traditional free-flow turbines, a machine working on a similar principle does not slow down the outward flow by extracting its kinetic energy, but speeds it up by extracting its potential energy" However, no matter how hard I try, I cannot find any errors. And then, while searching for an answer, I came across an article by professor from Michigan. And there all this is only confirmed. Edited February 2 by Hydro Link to comment Share on other sites More sharing options...

OceanBreeze Posted February 4 Report Share Posted February 4 On 2/2/2024 at 8:17 PM, Hydro said: I myself, when I saw it for the first time, simply fell into a stupor. I was especially shocked by the phrase: "Unlike traditional free-flow turbines, a machine working on a similar principle does not slow down the outward flow by extracting its kinetic energy, but speeds it up by extracting its potential energy" However, no matter how hard I try, I cannot find any errors. And then, while searching for an answer, I came across an article by professor from Michigan. And there all this is only confirmed. Hydro, I’m sorry to say that my time is being taken up by other matters and I just do not have enough time to spend on this forum and thread, to give you a detailed answer. However, I will say that a detailed mathematical analysis isn’t too difficult, and the work done by the anonymous professor at the University of Michigan hardly qualifies as “higher mathematics”. It is just standard hydrodynamic theory and integration of partial differential equations. It is first year calculus. I only have time to reply to your one comment “I was especially shocked by the phrase: "Unlike traditional free-flow turbines, a machine working on a similar principle does not slow down the outward flow by extracting its kinetic energy, but speeds it up by extracting its potential energy" This makes perfect sense, and the source you quoted explains it very well: “e.g. in a reservoir with a high level and slow flow velocity upstream of the dam, the acceleration of the flow is accompanied by a significant decrease in the level downstream of the dam.” Where is the mystery in that? It is just water going over a dam! It is dropping from a height and so giving up a great deal of gravitational potential energy while speeding up and its kinetic energy is increasing. However, it does lose some of that increased kinetic energy due to the braking of the hydrodynamic resistance of the turbine. Any hydrodynamic dam does require to be well-designed to balance out all of these effects to obtain maximum efficiency. I hope that clears up some of your questions. This reminds me of the common misconception that a wind turbine extracts energy from the wind AT the turbine blades by slowing down the wind. I emphasized the AT for a reason: If the turbine slowed the wind when it reaches the blades, what happens to the wind following behind? If the entire wind stream were to be slowed, the turbine would spin very slowly and extract low energy! The ideal turbine would only slow the wind passing through just enough to keep the blades spinning at a high speed. I once taught this in a class at the Coast Guard Academy, true story (with some embellishments) ME: “Energy is extracted AT the rotor due to the pressure drop at that point. There is a high pressure at the front of the rotor and a low pressure at the back.” [ uhoh, that annoyingly bright kidet in the back of the class has his hand up] KIDET: “Why do we calculate the power of the wind turbine using the difference between the upstream wind velocity and the downstream wind velocity, if all the energy comes from pressure?” ME: “Wonderful question! Power output of the turbine can be calculated this way: P = p A/4 [(V1 + V2) (V1^2-V2^2)] where p is air density, A is the area of the turbine blades, V1 is upstream wind velocity (free stream) and V2 is the far down stream wind after slowed by the turbine blades.” [Right now that annoyingly bright kidet in the back of the room is raising his hand again!] KIDET asks: “If that is how the turbine extracts wind power, how does the turbine KNOW what the far downstream wind velocity is?” ME: (losing patience) “Pay attention you! I said we can calculate the turbine Power using the upstream and downstream wind velocities, I did not say that is how the WT extracts the wind power! Look, coastie: Dimensionally, Force = Kg x m/s^2. Pressure is Force/Area =( Kg x m/s^2 ) / m^2 = Kg/ (m x s^2) Energy is Pressure x Volume =[ Kg/ (m x s^2)] x m^3 = (Kg x m^2) / s^2 As you can see, pressure is just energy per volume. To repeat, The WT extracts wind energy right at the rotor due to the pressure drop at that point. There is a high pressure at the front of the rotor and a low pressure at the back and pressure is just energy per volume. [KIDET raises his hand again] Recess time, everybody! Whew! Link to comment Share on other sites More sharing options...

Hydro Posted February 6 Author Report Share Posted February 6 I completely agree with you that mathematics is very simple and this is not higher mathematics. But this in no way detracts from its value. This is exactly the point! After all, the goal here is not higher mathematics, but namely the calculation of a gravity flow (I repeat - free flow!) hydroelectric power station without a dam (sic!). No matter how simple this calculation may be. Judging by the calculations made by "anonymous" (**) professor at the University of Michigan, the formula #6 at the end of paragraph of the article (which was derived independently in the previous article) is quite important. And not only in theoretical terms (it is not yet in any textbook on hydraulics/hydrodynamics), but also in practical meaning - it can be used to determine the power of any hydraulic turbines based on the condition of flow optimization. Moreover, all formulas for calculating power of hydraulic turbines are derived from it. This is, strictly speaking, the basic formula. There, in the comments, there is even an example of how the basic formula for hydropower is simply derived from this formula. In addition, plotting diagrams based on it shows the presence of an extremum that determines the optimum efficiency of such turbines. And this graph is very indicative. In our case, water does not fall from a height; it is a free flow, without head. And the essence of this phrase lies precisely in “traditional free-flow turbines” And, if traditional free-flow turbines have a theoretical efficiency in terms of kinetic energy, limited by the Betz limit (namely by slowing down the flow), then judging by the diagram from the article they can provide energy equivalent to 59% of the dynamic pressure V1^2/2g (in the ideal case), whereas according to this principle, deltaE' is obtained (in the ideal case). That is, a “traditional free-flow turbine” (Darier turbine or Savonius rotor) can theoretically produce only 0.5 kW * 0.59 = 0.295 kW in a flow with a depth of 1 m, a width of 1 m and a speed of 1 m/s. The same non-traditional turbine under the same conditions, judging by the formulas, can theoretically produce 3 kW. And no one seems to have any complaints about the formulas? Or is there? https://youtu.be/uOPkCS5Q2Wg?si=27qzAQB7aP3-vlKh (**) by the way, the professor is not anonymous at all, I simply translated the text from Serbian into English and removed the Cyrillic alphabet, names and addresses in order to ensure the most unbiased approach and not to create unnecessary advertising for the authors on the site. In fact, these are real people with names. Here, for example, is an original article posted in the electronic scientific library https://cyberleninka.ru/article/n/a-highly-efficient-method-for-deriving-energy-from-a-free-flow-liquid-on-the-basis-of-the-specific-hydrodynamic-effect Link to comment Share on other sites More sharing options...

OceanBreeze Posted February 6 Report Share Posted February 6 3 hours ago, Hydro said: I completely agree with you that mathematics is very simple and this is not higher mathematics. But this in no way detracts from its value. This is exactly the point! After all, the goal here is not higher mathematics, but namely the calculation of a gravity flow (I repeat - free flow!) hydroelectric power station without a dam (sic!). No matter how simple this calculation may be. Judging by the calculations made by "anonymous" (**) professor at the University of Michigan, the formula #6 at the end of paragraph of the article (which was derived independently in the previous article) is quite important. And not only in theoretical terms (it is not yet in any textbook on hydraulics/hydrodynamics), but also in practical meaning - it can be used to determine the power of any hydraulic turbines based on the condition of flow optimization. Moreover, all formulas for calculating power of hydraulic turbines are derived from it. This is, strictly speaking, the basic formula. There, in the comments, there is even an example of how the basic formula for hydropower is simply derived from this formula. In addition, plotting diagrams based on it shows the presence of an extremum that determines the optimum efficiency of such turbines. And this graph is very indicative. In our case, water does not fall from a height; it is a free flow, without head. And the essence of this phrase lies precisely in “traditional free-flow turbines” And, if traditional free-flow turbines have a theoretical efficiency in terms of kinetic energy, limited by the Betz limit (namely by slowing down the flow), then judging by the diagram from the article they can provide energy equivalent to 59% of the dynamic pressure V1^2/2g (in the ideal case), whereas according to this principle, deltaE' is obtained (in the ideal case). That is, a “traditional free-flow turbine” (Darier turbine or Savonius rotor) can theoretically produce only 0.5 kW * 0.59 = 0.295 kW in a flow with a depth of 1 m, a width of 1 m and a speed of 1 m/s. The same non-traditional turbine under the same conditions, judging by the formulas, can theoretically produce 3 kW. And no one seems to have any complaints about the formulas? Or is there? https://youtu.be/uOPkCS5Q2Wg?si=27qzAQB7aP3-vlKh (**) by the way, the professor is not anonymous at all, I simply translated the text from Serbian into English and removed the Cyrillic alphabet, names and addresses in order to ensure the most unbiased approach and not to create unnecessary advertising for the authors on the site. In fact, these are real people with names. Here, for example, is an original article posted in the electronic scientific library https://cyberleninka.ru/article/n/a-highly-efficient-method-for-deriving-energy-from-a-free-flow-liquid-on-the-basis-of-the-specific-hydrodynamic-effect For the case I discussed earlier, where H1 > H2, such as water going over a dam, the mathematics work fine. For your drawing above, presumably the initial condition is H1 = H2 as the machine is free-floating in a body of water, such as a channel. Now, the math does not work because there is nothing I can see that accounts for the work that has to be done to move a large amount of water out of the way, to create a space where the level H2 is lower than H1. The water isn't going to "fall" to this lower level by the force of gravity! Unless you can show me where the work necessary to remove the water from H1 to H2 is accounted for, I have to label this as just another crackpot idea that will not work. In short, when H1 = H2 there is no energy extraction. To create an H2 < H1 requires work. Once that work is taken into consideration, the entire concept fails as it will require more energy to do that work that any amount of energy extracted. Surely, you must be able to see that! I will allow you the courtesy of posting a reply, then I am going to close this thread. Link to comment Share on other sites More sharing options...

Hydro Posted February 6 Author Report Share Posted February 6 Yes, many people miss this. I didn’t understand it the first time either. The article shows this, but I had to read it several times to understand it. The author draws attention to the following. Here are excerpts from the article with the chart: “Now let us imagine that we are extracting part of kinetic energy from a cubic meter of water, which is flowing within a current, and use it to “move aside” the cubic meter of water that follows it (downstream). That is we will speed up the downstream cubic meter of water by slowing down the upstream volume of water. As a result, a level difference arises between them and potential energy emerges in the difference between these levels, which can be extracted from the current. The following question arises: will the amount of the extracted potential energy be more, less or equal to the energy used to speed up the second cubic meter of water – or, in other words, the energy expanded to increase its kinetic energy?” "Because the amount of water entering the device is equal to the amount of outflowing water, and the speed of the outflowing stream is higher than that of the inflowing stream, then the sectional area of the outflowing stream will be less than that of the inflowing stream. Therefore, its depth H2 will be less than the depth of the inflowing stream H1 by the value h. As a result of this, potential energy appears between the different levels of the inflowing and outflowing streams" "Then the most interesting aspect occurs. It can be seen that the left side of the equation, which is in brackets, will increase in a linear fashion when it depends on h or in a hyperbola when it depends on V2, whereas the right part will decrease, and in a parabola at that. Which side will gain the upper hand? Let us plot a graph showing energy’s dependence on the drop between the levels h (Fig. 2). The graph will be plotted to show the various levels of the inflowing stream’s velocity V1 after designating it as a constant. It is remarkable that the graph showing the energy’s dependence on the drop between the levels h has an extremum. On the rising branch of the graph, the energy balance will be positive (the power factor > 1), i.e. the extracted potential energy will be mostly expended as kinetic energy on speeding up the outflowing stream, and the device will self-accelerate until it reaches the maximum." That is, the work expended on accelerating the outgoing flow is slightly less than the energy released due to the difference in water levels before and after the turbine The author calls this a "positive feedback loop". According to the mechanism of occurrence, I compare this with a nuclear chain reaction, when the energy spent on knocking out the next neutron from the nucleus is less than the energy released during the decay of the nucleus. A chain reaction with a "positive feedback loop" occurs. But it does not go to infinity, but only to the so-called “saturation point.” In the diagrams this is indicated as an extreme point on the graph. The author called this the "critical point", corresponding to the critical speed and depth. Further acceleration does not occur and the turbine operates stably in this mode. Maybe my analogy is inaccurate, but probably many analogies can be given to this from physics. Link to comment Share on other sites More sharing options...

OceanBreeze Posted February 6 Report Share Posted February 6 “Now let us imagine that we are extracting part of kinetic energy from a cubic meter of water, which is flowing within a current, and use it to “move aside” the cubic meter of water that follows it (downstream). That is we will speed up the downstream cubic meter of water by slowing down the upstream volume of water." Let us imagine no such thing! Water always seeks its own level. In order to "move aside" a cubic meter of water requires work to be done on the water, requiring quite a lot of energy. The above is claiming that energy can be obtained by slowing down the incoming flow and extracting some of the kinetic energy and using that to speed up the outgoing flow sufficiently to create a difference in height. In other words, the incoming flow is doing the work on the outgoing flow; the so-called turbine is transparent to this process as it has no power source other than the water flow. That should be enough to convince you this is describing an over-unity device! If that isn't enough, then consider the water must also do mechanical work on the "turbine", another loss of energy. This is just one more crackpot, unworkable over-unity device and now I am glad I didn't read through all of the copy-and-paste articles you posted. Thread Closed and moved to Silly Claims. Link to comment Share on other sites More sharing options...

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