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# A Way To Generate A Minimum Of 10Kw, 24/7 !

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I was browsing on the net when I came across an advertisement for Vacuum elevators: . These vacuum elevators lift considerable weights to heights of 10 metres. This fascinated me because it showed that a vacuum could be created fast enough to use in a commercial elevator and strong enough to lift weights to a height of 10m. It occured to me that instead of lifting the cage and people, the same principle could be used to lift large amounts of water to heights of 10 m and then release the water to generate electricity. The idea was to use a falling counterweight to lift a large piston (n.b., the easiest way to form a vacuum is to increase volume. ) Once the vacuum was created water would flow into the overhead tank through a giant siphon ( to ensure that the tanks filled quickly) and then released through a turbine to generate electricity. Two tanks are used in tandem to generate electricity continuously. It is like a captive hydroelectricity generation system.

Edited by McQueen
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I was browsing on the net when I came across an advertisement for Vacuum elevators:

These are cool little elevators. Their main appeal is that they require little overhead, and nearly no underfoot, space, so can be installed in almost any house. After seeing one on a home-improvement TV show a couple of years ago, I flirted the prospect of in installing one in my 2-story house before giving it up as unnecessary, impractical, and simply gadget-crazy. :)

It occured to me that instead of lifting the cage and people, the same principle could be used to lift large amounts of water to heights of 10 m and then release the water to generate electricity.

The problem with such a scheme is that the energy needed to lift the water must be greater than the energy that the water could generate. Such a scheme producing more energy than it uses would be an example of perpetual motion, which is impossible for a classical machine.

Pumping water to elevated reservoir when energy that would otherwise be wasted is available, then generating hydroelectric energy with that water when needed, know as pumped-storage hydroelectricity, has been used since the 19th century, though they don’t use air-pumps like the small vacuum elevator McQueen links to, but more efficient water pumps.

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i don't know if this would be perpetual motion, wouldn't it need gravity as an imput of power?

but what would give enough weight to allow for the chamber to fall again, if the weight is enough to pull up the chamber and scyphon also

Edited by belovelife
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i don't know if this would be perpetual motion, wouldn't it need gravity as an imput of power?

but what would give enough weight to allow for the chamber to fall again, if the weight is enough to pull up the chamber and scyphon also

The overhead tank has one way valves in it that allow air to escape from the tank as the piston is raised but prevent air from re-entering the tank. Once the piston is raised a vacuum release valve allows air to re-enter the overhead tank in the area above the piston, and it is atmospheric pressure that pushes the piston down and raises the counterweight. Consider that if the piston is 1 m in diameter than the force required to raise it would be 3140 Kgs. The idea is to use a system of constant of potential as used in cable elevators. The counterweight weighs the same as the cage, if both counterweight and cage were identical in weight then the only force to be overcome would be the frictional force in the pulley, caused by the cable connecting the two weights. Since gravity plays a part with the counterweight K.E comes into play on the side of the counterweight and the counterweight can be correspondingly lighter. Since the two loads are approximately the same, a small electric motor can be used to assist in lowering of the piston and raising of the counterweight.

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i don't know if this would be perpetual motion, wouldn't it need gravity as an imput of power?

An important point is that gravity is not a source, or “input” of power. Gravity makes it possible to store work/energy by moving masses apart, and reclaim that work/energy by allowing them to be drawn together by their mutual attraction, but “gravity itself” – the law describing the energy stored by moving masses apart and converted to work when the masses move together – is not “used up” by such motion.

Consider that if the piston is 1 m in diameter than the force required to raise it would be 3140 Kgs.

I don’t know how you calculated this, McQueen – it would be good for you to show you work – but the units are wrong, as is your use of force as something required to raise a thing.

The kilogram, which is abbreviated kg, is the SI unit of mass, not force. Force is (that is, has dimension of) mass • acceleration, that is mass • distance • time-2. The SI unit for force is the Newton, which is abbreviated N. 1 N = 1 kg•m/s/s = 1 kg•m•s-2.

To raise something, a force must be applied to it over a distance. Force • distance is work. The SI unit for work is the Joule, abbreviated J. 1 J = 1 N • m = 1 kg•m•m/s/s = 1 kg•m2•s-2.

So, ignoring friction and other inefficiencies, raising a 1 m diameter, 1 m tall cylinder filled with water (which has a density of about 1000 kg/m3) a height of 10 m near the surface of the Earth (where the acceleration of gravity is about 9.8 m/s2) requires about

$\left( \frac12 \,\mbox{m} \right)^2 \pi \cdot 1 \,\mbox{m} \cdot 1000 \,\mbox{kg/m}^3 \cdot 9.8 \,\mbox{m/s}^2 \cdot 10 \,\mbox{m} = 76969 \,\mbox{J}$.

Power is work/time. It’s SI unit is the Watt, abbreviated W. So doing the lift described above in, say 10 seconds, requires

$76969 \,\mbox{J} / 10 \,\mbox{s} = 769.69 \,\mbox{W}$.

No matter how you build it or calculate its work/energy (energy has the same dimension as work, because it is the potential to perform work), a machine that puts out more power than is put into it is a perpetual motion machine, and ignoring the possibility of exploiting exotic phenomena like the Kasimir effect, is impossible. Although by calculating the work/energy of each part of a machine it’s possible to show this in detail, such complicated work isn’t necessary – if a machine makes more power than it uses, it’s impossible.

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An important point is that gravity is not a source, or “input” of power. Gravity makes it possible to store work/energy by moving masses apart, and reclaim that work/energy by allowing them to be drawn together by their mutual attraction, but “gravity itself” – the law describing the energy stored by moving masses apart and converted to work when the masses move together – is not “used up” by such motion.

I don’t know how you calculated this, McQueen – it would be good for you to show you work – but the units are wrong, as is your use of force as something required to raise a thing.

The kilogram, which is abbreviated kg, is the SI unit of mass, not force. Force is (that is, has dimension of) mass • acceleration, that is mass • distance • time-2. The SI unit for force is the Newton, which is abbreviated N. 1 N = 1 kg•m/s/s = 1 kg•m•s-2.

To raise something, a force must be applied to it over a distance. Force • distance is work. The SI unit for work is the Joule, abbreviated J. 1 J = 1 N • m = 1 kg•m•m/s/s = 1 kg•m2•s-2.

So, ignoring friction and other inefficiencies, raising a 1 m diameter, 1 m tall cylinder filled with water (which has a density of about 1000 kg/m3) a height of 10 m near the surface of the Earth (where the acceleration of gravity is about 9.8 m/s2) requires about

$\left( \frac12 \,\mbox{m} \right)^2 \pi \cdot 1 \,\mbox{m} \cdot 1000 \,\mbox{kg/m}^3 \cdot 9.8 \,\mbox{m/s}^2 \cdot 10 \,\mbox{m} = 76969 \,\mbox{J}$.

Power is work/time. It’s SI unit is the Watt, abbreviated W. So doing the lift described above in, say 10 seconds, requires

$76969 \,\mbox{J} / 10 \,\mbox{s} = 769.69 \,\mbox{W}$.

No matter how you build it or calculate its work/energy (energy has the same dimension as work, because it is the potential to perform work), a machine that puts out more power than is put into it is a perpetual motion machine, and ignoring the possibility of exploiting exotic phenomena like the Kasimir effect, is impossible. Although by calculating the work/energy of each part of a machine it’s possible to show this in detail, such complicated work isn’t necessary – if a machine makes more power than it uses, it’s impossible.

Hi CraigD,

Thanks for going into the whole thing in such depth, I agree with every word you have said, the whole post was a hoax right from the beginning (witness the title claiming of 10KW of power generation, whereas the maximum you can reasonably expect to get would be about 2.5 KW) BUT it was a hoax with a reason behind it. The salient fact is that you pointed out the flaw in the system and this is just what I was hoping would happen. I hope you don't take this amiss because I assure you that what follows is extremely interesting. Further, your initial premise on the working of the system is faulty. The idea is that once a vacuum is created in the tank, atmospheric pressure would push the water into the tank, no other work is involved. It is a little known fact that atmospheric pressure can support a column of water 10m high, this means that once the vacuum is created, if the height of the tank is about 8m, then atmospheric pressure could easily push the water from the lower tank into the upper tank. No that is not the problem ! Look at the figures for raising of a 1m dia piston against a vacuum of 1 Torr (i.e., against atmospheric pressure) they are absolutely staggering,7850Kgf (i.e., 76,930N)!! If a larger dia piston is used, as is likely, with a dia of 1.5m, then an astounding 17,662.5 Kgf (i.e., 173092.5N) is needed to lift the piston!!! Imagine that, the equivalent of 3 or 4 Mac trucks hanging from a height of 10m would be needed to lift the piston against atmospheric pressure! Surely, something can be done with these extraordinary figures.

The lift begins to rise a mere second after the man presses the button (i.e., the vacuum is created in such a short time using turbine fans.) So my new idea was to use the same system as is used in vacuum elevators to create a vacuum,namely a turbine fan. Now a cubic metre volume of air weighs just 1.25 Kgs, this, and the viscidness of air, is why it is so easy to push the air out of the container. So say you have 3 tubes each of 0.2m dia and 10m in height, inside each of these three tubes is a shuttle equipped with a small electrically operated turbine fan. The shuttle in one of the tubes is linked to a cable which (with suitable pulleys) winds round a spool at the top of the system and the cable then runs through another pulley that is attached to a counterweight situated at 10m above ground. When the counterweight is released the spool which is attached to a turbine fan turns creating a vacuum in the system ( just like in the vacuum elevator) the two other 0.2m dia tubes are attached to similar cable and pulley systems, however instead of running around a spool attached to a turbine they each wind around a generator, thus as the three weights drop, one turns the main turbine fan creating a vacuum in the system, and the other two generate electricity ! The small electrically operated turbine fans in the 0.2m dia tubes also push air out of the tubes, so in effect the counterweights meet no resistance as they are released it is almost as if they were in free fall. Just before the counterweights reach the bottom, a mechanical (or other type) catch seals the top of the shuttles, converting them into pistons. The situation now is as follows, the turbine fan has stopped spinning so atmospheric pressure is re-introduced in the system. Atmospheric pressure exerts a force of 314Kgf on each of the pistons (i.e., 3077.2N) if each of the counterweights weighs 30Kg. then the K.E of the each counterweight would be 2940J, remember while the counterweights are falling there is no resistance at all, apart from frictional forces. Therefore atmospheric pressure can easily lift the counterweights back to their original position. When the pistons reach the bottom of the shafts another catch pushes back the cover and starts the fans again converting them into shuttles and the whole process is repeated. The generators are fitted with a ratchet system of gears that allows the generators to continue spinning in the same direction even while being raised. This means that a conitnuous output of 2.94 x 2 = 5.88KW is possible. The point is the vacuum elevator has to clear a volume of (assuming the shaft is 1m dia) a minimum of 7.85 cubic metres of air and does this in one second, wuith the much smaller volume shafts we are using we would need to clear a maximum of about 2 cubic metres of air. Using larger diameter tubes would mean larger counterweights and more electricty. However, there is a limit to this. What do you think ?

Edited by McQueen

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