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furshur
SFN Regular
USA
1536 Posts |
 Posted - 03/05/2007 : 13:11:20 [Permalink]
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On the subject of your flying saucer you write:
quote:
Therefore a perpetual mobile spinning in space would produce more or less electricity according to its spinning speed, and in space where there is nothing to stop it, generators could be designed that spin very fast, they would just need the first impulse, like satellites need, and then they would stay spinning on their own.
No. There is 'something' to stop the flying saucer. The generator will convert the rotational energy to electrical energy which will act as a brake on the flying saucer.
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If I knew then what I know now then I would know more now than I know. |
| Edited by - furshur on 03/05/2007 14:02:55 |
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BigPapaSmurf
SFN Die Hard
3192 Posts |
Posted - 03/05/2007 : 13:43:42 [Permalink]
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| Yep, really you would be doing yourself a favor if you tried making devices using a renewable resource like sunlight or geothermal energy. |
"...things I have neither seen nor experienced nor heard tell of from anybody else; things, what is more, that do not in fact exist and could not ever exist at all. So my readers must not believe a word I say." -Lucian on his book True History
"...They accept such things on faith alone, without any evidence. So if a fraudulent and cunning person who knows how to take advantage of a situation comes among them, he can make himself rich in a short time." -Lucian critical of early Christians c.166 AD From his book, De Morte Peregrini |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 03/05/2007 : 22:12:02 [Permalink]
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Well, it looks like I am carrying on a conversation with myself. On the off hand chance that Klaudio returns I'll make one more point.
I broke out my old fluid dynamics book and in about 5 minutes realized that it is not worth the effort to do any indepth calculations because of the major short coming of his process, which is that the contraption is in water.
When the bell is at the surface of the water the PE = g x H x m.
9.81 m/s/s x 100m x 8000kg = 7,848,000 joules
If this PE was completely converted into KE (say in a perfect vacuum with no other forces acting on the system except gravity) the velocity of the bell would be PE = KE = 1/2 x M x V x V so sqrt(V) = PE x 2 / M
= sqrt(7,848,000 [(m^2 kg)/s^2] x 2 / 8000 kg) = 44 m/s
= 158 km/hr
Of course the bell cannot travel at a velocity of 158 km/hr through water, it will reach terminal velocity and will not exceed that velocity no matter how deep the water is. Lets be generous and assume that the bell is hydrodynamically efficient and allow for the unrealistically high velocity of 50 km/hr. That means that 2/3 of the energy is lost due to drag. The journey to the surface does not fair any better due to the viscosity of water and the drag on the bell. This is before any useful energy is even tried to be extracted from the system.
Not only is this not perpetual motion this setup is actually VERY inefficient.
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If I knew then what I know now then I would know more now than I know. |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/06/2007 : 09:12:59 [Permalink]
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quote:
Originally posted by furshur
Well, it looks like I am carrying on a conversation with myself.
Well, I appreciate your efforts, 'cause I'm learning something.
As to your latest calculations, does the drag actually matter? Bianco's setup seems to generate power based upon the total up-and-down motion of the bell, and not its velocity. In fact, he thinks it's okay to ignore friction 'cause the whole thing is supposed to move very slowly.
Translating the up-and-down motion to a unidirectional output shaft rotation (to wind up a spring) is a no-brainer (I've built such a mechanism with LEGO). The wound-up spring is allegedly constantly "discharging" to drive the generators, compressor and whatnot. So all that's left is getting the bell to go up and down.
And I'd missed out on this drawing, before. That explains a lot. I was trying to envision how the bell's pressure would vary between 10 and 20 bar, but that's not at all what's supposed to happen.
But that is where everything falls apart. Falling through the water, the bell releases so-much gravitational potential energy that will get "caught" by Bianco's mechanisms and converted into useful power. Where his math is truly screwed up is in thinking that it will take less power than that to raise the bell back to the surface.
It seems to me that it's this claim, in particular, that's dead wrong:One aspect that he could see, that I did not consider in the drawing, was that the air that enters the bell will expand as the bell rises, producing an accelerating effect, and therefore more energy can be extracted in the ascent than in the descent... The problem being that because the distance is the same, and the resultant change in energy is the same, then the energy available to do work must be the same.
The total energy available per "cycle" will then be twice the energy generated by the descent of the bell. This amount will be dependent upon the efficiency of the converting mechanism housed atop the bell and the gravitational potential of the bell, and is independent of the medium through which the bell travels and also independent of the speed of the bell.
The amount of energy needed to raise the bell, on the other hand, is dependent upon the efficiency of the compressor, the gravitational potential of the bell, and the medium through which the bell travels, and the speed of the bell.
Am I correct so far? |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 03/06/2007 : 10:15:01 [Permalink]
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I think all of what you are saying is correct Dave.
This is a good point that illustrates klaudio's confusion:
quote:
It seems to me that it's this claim, in particular, that's dead wrong:
quote:
One aspect that he could see, that I did not consider in the drawing, was that the air that enters the bell will expand as the bell rises, producing an accelerating effect, and therefore more energy can be extracted in the ascent than in the descent...
The problem being that because the distance is the same, and the resultant change in energy is the same, then the energy available to do work must be the same.
It is accurate to say the air will expand and the bell will accelerate. That of course is not 'free' energy that is simply converting the PE of the compressed air into KE. If the air escaped out the bottom of the bell it would just be even less efficient.
As far as the speed of the bell, you of course are correct that a slower speed will have less losses than faster speeds, but the losses will still extreme.
Imagine taking a nice 2 mile walk through the park. Now imagine walking just waist deep in water for 2 miles. The energy lost to drag is huge.
At low velocities the flow past a body is laminar flow. At low flow conditions the drag is inversly proportional to the velocity and essentially linear due to the viscosity of the fluid being the dominate effect. At very low velocities there is still considerable drag. The liquid in effect acts like a series of sheets. The sheet nearest the moving body is traveling at the speed of the body. At some distance from the body the bulk liquid is stationary. Between these 2 points the sheets are shearing against each other causing drag and transfering energy from the body to the surrounding water.
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If I knew then what I know now then I would know more now than I know. |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 03/06/2007 : 10:32:00 [Permalink]
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I missed this at first Dave:
quote:
The total energy available per "cycle" will then be twice the energy generated by the descent of the bell. This amount will be dependent upon the efficiency of the converting mechanism housed atop the bell and the gravitational potential of the bell, and is independent of the medium through which the bell travels and also independent of the speed of the bell.
Dave, I may be missing your point here. If the converting mechanisum were 100% efficient, all of the PE would not be converted into usable energy. The ability to convert the PE is primarily dependent on the viscosity and density of the medium. Consider 2 extremes on the medium. A perfect vacuum would allow the 100% efficient converting mechanism to harness almost all of the energy. Now consider mercury as a medium the bell would not sink and you would have zero conversion.
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If I knew then what I know now then I would know more now than I know. |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/06/2007 : 11:08:44 [Permalink]
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furshur, Bianco's mechanism is needlessly complex. Let's replace his central shaft with it's nifty tracks into a giant gear rack over 100m high, and let's change his conversion mechanism into a 100%-efficient alternator with a gear that engages the rack. As the bell goes up and down, the alternator shaft rotates, and electrical power comes out.
The amount of power that comes out of the alternator will be the same, regardless of the speed of the bell. For example, we might get 10 watts for 10 seconds, or we might get 100 watts for one second, or a tenth of a watt over 1,000 seconds. Given the way alternators work, given a fixed number of rotations of the shaft (dependent on the distance the bell has to travel) will result in the motion of a specific number of electrons in the output wires. It doesn't matter how much time it takes for each rotation of the shaft, the total power, once all the rotations have been completed, should be the same.
So yes, I also understand that no motion equals no power. Bianco's contraption has power being generated by the effects of travel over a fixed difference. Let me put it another way: the total power available from the falling (or sinking) bell is its PE at a depth of 0m minus its PE at a depth of 100m. How long it takes to convert that PE difference into KE is immaterial, as long as the time isn't infinite (the bell has to move to convert any PE at all). The PE difference remains the same whether the bell is in free-fall or if the friction of the mechanism is so high that it takes a year.
Of course, we know alternators aren't 100% efficient. And we know that to really make use of the power will require regulating circuitry which simply won't function if the alternator doesn't turn quickly enough. But I'm just trying to show that even if we ignore all those real-life problems, Bianco can't get more power out of his system than he puts into it. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 03/06/2007 : 12:08:35 [Permalink]
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quote:
But I'm just trying to show that even if we ignore all those real-life problems, Bianco can't get more power out of his system than he puts into it.
That is the bottom line.
quote:
So yes, I also understand that no motion equals no power.
I did not mean that to sound condescending. I usually look at the obvious examples to help me to visualize complex problems.
quote:
The amount of power that comes out of the alternator will be the same, regardless of the speed of the bell. For example, we might get 10 watts for 10 seconds, or we might get 100 watts for one second, or a tenth of a watt over 1,000 seconds. Given the way alternators work, given a fixed number of rotations of the shaft (dependent on the distance the bell has to travel) will result in the motion of a specific number of electrons in the output wires. It doesn't matter how much time it takes for each rotation of the shaft, the total power, once all the rotations have been completed, should be the same.
Yes, I agree the velocity makes no difference on the power output. I probably confused the issue by discussing terminal velocity.
The point is that no matter how slow the bell moves there will be drag effects that will decrease the efficiency. Drag at low velocities is dependent on the density and viscosity of the medium so the medium does matter.
Here is some info on fluid dynamics:
Reynolds Number
Re = chararcteristic length * density * velocity / viscosity
The Reynolds Number describes aspects of a fluid. A Reynolds number of <10 is characteristic of laminar flow.
I am clearly doing a poor job of explaining myself these guys do a much better job.
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If I knew then what I know now then I would know more now than I know. |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/06/2007 : 13:46:43 [Permalink]
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I'm sorry, furshur, I didn't mean to say that you were sounding condescending. My fingers were upbeat and chipper when I typed that, honest.
But back to the discussion at hand. If we assume a medium through which the bell will sink (and rise, of course), then the power output per cycle will be constant. Where the density, viscosity and drag become important is calculating how much of that energy will be required to raise the bell out of the gravity well.
And that's the key. All Bianco has is a generator attached to a pole in a gravitional field such that the potential energy at one end of the pole is greater than the PE at the other end. We can all agree that moving the generator both up and down the pole will generate power (the same amount of power), and we can all agree that allowing the generator to fall from high PE to low PE will generate "free" power.
The question is this: regardless of how you "reset" the system for the next cycle (Bianco is filling a floatation device with compressed air to lift the generator, but you could winch it up or hire gnomes to hoist it up a ladder system), is the energy required to accomplish the "reset" more than the total power generated by one full cycle?
But wait... The power generated cannot be directly related to potential energy. PE is based in part on mass, but the generator generates power based on nothing but turns of its input shaft. Adding more mass to the generator will not, by itself, increase the total power output of the generator. Or consider it this way: turn the simplified system on its side and simply push the generator from one end of the pole to the other: the same amount of power will be generated as when the pole is vertical and you let the generator fall, but the PE difference from one end to the other will be zero.
So the amount of power generated must be dependent solely on the design and efficiency of the generator itself, along with the length of the pole.
But the amount of power required to raise the generator to begin the next cycle is dependent upon the mass of the generator, the drag imposed by the medium through which it moves, the length of the pole, the friction with the pole, the efficiency of the compressor, the size of the air pipes, etc.. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 03/06/2007 : 16:47:32 [Permalink]
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Ah, I see what you are saying. Yes the first drop is a kind of 'free' energy. Hooray! As you pointed out though we have to reset the system. Dang!
Lets add up the losses:
The power plant supplying the electricity for the compressor is not 100% efficient - wasted energy.
The generator is making a terrible racket - wasted energy.
The compressed air is heating up - wasted energy.
The friction of the moving parts - wasted energy.
And my favorite - the rising bell plowing through the water - wasted energy.
This is one of the worst perpetual motion ideas I've seen.
I think I will try to calculate how much energy is required to raise the bell. If I assume an efficiency for a compressor and assume that any energy extracted from the rising bell is 100% efficient I should be able to subtract it back out from the total energy balance and therefore I can ignore any power generation.
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If I knew then what I know now then I would know more now than I know. |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 03/07/2007 : 15:50:27 [Permalink]
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I think I understand what Klaudio was attempting in the following equation.
quote:
W = (D×G×H+Po)x Vf x ln[(DxGxH+Po)/Po]= 7848000 watt×s = 2.18 kWh
D density of water
G acceleration of gravity
H Hieght
Vf final volume
Po atmospheric pressure
This looks vaguely like the work done when compressing a gas using the ideal gas law at constant temperature.
If we neaten up Klaudio's math (and make an assummption) we get:
(DxGxH)+ Po = Pressure at 100 meters = Pf
Substituting Pf back into the equation we get-
Work = Pf x Vf X ln(Pf/Po)
I however do not think this is correct. I believe this is:
Ideal gas - PV = nRT
If we hold the temp constant then PV = constant.
So dW = PdV = nRT/V dv Work is force over distance so expanding gas does work.
Integrating this we get W = nRT ln(Vf/Vi)
nRT = PV soooo
W = Pf x Vf x ln(Vf/Vi)
This is actually the work done by the compressor to fill the bell (minus of course all of the fricion losses in the compressor).
I am out of time I will work on this some more later....
I sure wish klaudio would comment some day... |
If I knew then what I know now then I would know more now than I know. |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/08/2007 : 10:16:56 [Permalink]
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quote:
Originally posted by furshur
(DxGxH)+ Po = Pressure at 100 meters = Pf
But that's where the units don't match.
DxGxH is in kg/m3 times m/s2 times m, or kg/m/s2.
Po is in kg/m2.
How can one add those two together and end up with kg/m2 only?quote:
W = Pf x Vf x ln(Vf/Vi)
This is actually the work done by the compressor to fill the bell (minus of course all of the fricion losses in the compressor).
Well, if that's true, then klaudio has his equations listed backwards. He claims that W is the total work generated by ascent and descent of the bell. He claims that We is the work needed to "inflate the bell," and he defines that as nothing more than DxGxHxVf. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/08/2007 : 10:39:19 [Permalink]
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| Oh, I see where he screwed up. The pressure at the surface isn't 101,320 kg/m2, it is 101,325 Pa. And one Pascal is equal to one kg/m/s2. So the units would work out correctly for the addition, if klaudio had defined Po correctly (just like he didn't with G). So DxGxH+Po is, indeed, the pressure at 100m depth. In pure water. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 03/08/2007 : 10:46:55 [Permalink]
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quote:
Oh, I see where he screwed up. The pressure at the surface isn't 101,320 kg/m2, it is 101,325 Pa.
Yep, trying to follow his math with the multitude of mistakes is almost impossible.
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If I knew then what I know now then I would know more now than I know. |
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Dr. Mabuse
Septic Fiend
Sweden
9687 Posts |
Posted - 03/08/2007 : 14:57:02 [Permalink]
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quote:
Originally posted by Dave W.
The amount of power that comes out of the alternator will be the same, regardless of the speed of the bell. For example, we might get 10 watts for 10 seconds, or we might get 100 watts for one second, or a tenth of a watt over 1,000 seconds.
Bolding mine. Shouldn't that be Energy?
Power is measured in Watts, Energy is Joule (Watts*second) |
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A flying saucer design
