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Author Topic: Energy losses, Thermodynamics and Efficiency  (Read 12903 times)

Offline Supercryptid

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #25 on: 03/10/2013 22:36:23 »
What if the bubble forms underwater.. as in a submerged PV cell.. the cell splits the water by hydrolysis causing bubbles to form on the surface, when the bubble gets to a certain size it will float to the surface.. does it get to the surface at a lower temperature than the water? does it cool the water by gaining energy? I cant see how the PV cell gives it any energy rather than that which was needed to split the molecule. So where does this energy come from?

I don't know this with any certainty (as I cannot seem to find a reference to it elsewhere online), but I suspect that electrolysis of water that is at high pressure requires more energy than low pressure. Imagine that in order to do electrolysis, you must pull the hydrogen atoms and oxygen atom far away enough from each other that they are essentially no longer bound. However, if there are many other water molecules around it pushing back against the hydrogen atoms, then more energy must be expended in order to counteract this external force that is trying to push them back together. The higher the pressure, the more energy that is require to overcome it. If that is true, then the "extra" energy must be supplied by the PV cell itself.

If I'm wrong, someone please correct me.
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #26 on: 04/10/2013 01:35:54 »
Quote
I don't know this with any certainty (as I cannot seem to find a reference to it elsewhere online), but I suspect that electrolysis of water that is at high pressure requires more energy than low pressure. Imagine that in order to do electrolysis, you must pull the hydrogen atoms and oxygen atom far away enough from each other that they are essentially no longer bound.
I am not too sure about this point either. I do know that at present the electrolysis of water to produce hydrogen is prohibitively expensive and it is only because the prospect of ‘free energy’ has been brought up that I introduced the idea at all.
Returning to the ‘Home Grid’ design I have put forward for generating electricity for individual homes. The concept, apart from the  fact that the kinetic energy of a falling weight is being used to run a generator to run a vacuum device,  seems to be a bit vague in its present form. So here is another diagram.

This represents the Home Grid system seen from above.  All four units, each unit consists of a counterweight and shuttle/piston connected together by a cable and running over a generator spool to generate electricity.  When the counterweight descends the shuttle rises and vice-versa, all of the units are identical in their composition and working, except for the fact that the unit represented in red uses all the electricity it generates to create a partial vacuum  throughout the system i.e., in the red and blue units. The blue units then use that vacuum to produce usable electricity by raising and lowering a counterweight,  that is output from the system. The system can have one or more blue units, in the present depiction with 3 blue units, around 15 KW can be generated, if only one blue unit were present then only 5 KW could be generated and so on. The manner in which the vacuum elevator operates and the volume over which the vacuum is created in the vacuum elevator seems to indicate that such a configuration is possible.  Then again the amount of electricity that could be generated can be decided upon by choosing an appropriate weight for the counterweight.
The generators will need to be cooled, fortunately when the air is being evacuated from the tubes, the temperature drops rapidly, this can be used either directly or indirectly to cool the generators.
« Last Edit: 04/10/2013 01:49:25 by McQueen »
 

Offline alancalverd

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #27 on: 04/10/2013 08:20:30 »
What is the upper limit on the number of blue units per red one?
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #28 on: 04/10/2013 08:48:37 »
I would say that 3 : 1 is pretty close, because the volume evacuated by the vacuum device should not increase over a certain limit. By the way here is an interesting vid from the BBC Bang Goes the Theory with Jeremy: http://www.youtube.com/watch?v=XNOEP1XIFiM. He finally does succeed in climbing the building, quite an achievement.
 

Offline alancalverd

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #29 on: 04/10/2013 14:30:09 »
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the volume evacuated by the vacuum device should not increase over a certain limit.

No. If there are no leaks, any pump will evacuate any volume, given enough time.

You need to review the difference between power and energy in order to understand the vacuum elevator and the flaws in your thinking. Think of jacking up a car. You expend a few watts for several minutes and end up with a one ton vehicle a meter off the ground - 10 kJ of potential energy. Now if you tie the car to a dynamo and let it drop under gravity it could deliver 10 kW for 1 second, or 1W for almost 3 hours.
« Last Edit: 04/10/2013 14:53:39 by alancalverd »
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #30 on: 04/10/2013 16:07:33 »
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You need to review the difference between power and energy in order to understand the vacuum elevator and the flaws in your thinking.
Surely that is one of the most basic considerations to take into account.  It seems to me that there is a slight misconception in your thinking. Think of the piston/shuttle and the counterweight as two weights suspended by a pulley, now surely if there is a discrepancy in weight, the rate at the which the heavier load descends would be dictated solely by the acceleration due to gravity and the height from which it is descending from and not how much the lesser weight weighs. (n.b: This would hold true if frictional losses are taken into account). Look at following image:

If m1>m2, the body ‘A’ will move downward with acceleration ‘a’ and the body ‘B’ will move up with same   acceleration.  So as you can see provided there is a discrepancy between the two weights, ( or the force acting on the two loads)  larger than the frictional losses, the counterweight should move upwards at the same speed at which it descends !
Quote
Think of jacking up a car. You expend a few watts for several minutes and end up with a one ton vehicle a meter off the ground - 10 kJ of potential energy. Now if you tie the car to a dynamo and let it drop under gravity it could deliver 10 kW for 1 second, or 1W for almost 3 hours.
Also, the whole point of the vacuum elevator is that the control over the vacuum is so absolute that the speed at which the lift cage rises can be carefully controlled. Hence the figures you arrive at i.e., velocity of elevator = 30 ft/minute which is about 15 cms/sec  (0.54 Kmh.)and  power would be 3KW/60 = 500 W.  But believe me it takes quite a lot of manipulation with the turbines to achieve that speed. If the lift cage were allowed to rise at the speed induced by atmospheric pressure it would attain a velocity of 14m/sec or 50 Kmh, which means that when it came to a stop there would be a good chance that anyone riding in the lift cage would continue on through the roof !
Lastly if you look at a video of a vacuum elevator, it will be apparent that the vacuum is established in the system almost immediately. http://www.youtube.com/watch?v=7e2OPcWIBXQ
Since the volume over which a vacuum has to be established  in the ‘Home Grid’ system is about one third of the volume in which a vacuum has to be established in the vacuum elevator, I am calculating that the vacuum will be established even faster. Remember that the power being used is exactly the same 3 KW but the volume to be cleared is about one third of the original volume. 
« Last Edit: 04/10/2013 16:22:24 by McQueen »
 

Offline alancalverd

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #31 on: 04/10/2013 16:27:07 »
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Think of the piston and the counterweight as two weights suspended by a pulley, now surely if there is a discrepancy in weight, the rate at the which the heavier load descends would be dictated solely by the acceleration due to gravity and the height from which it is descending from and not how much the lesser weight weighs.

No.

The accelerating force is gm1 - gm2. The total mass is (m1 + m2) so the acceleration is a = F/m = g(m1-m2)/(m1 +m2). 

The video does not show you how long it took to establish the vacuum, nor is the pressure differential mentioned. I could show you a video of a rifle being fired, but that gives you no idea of how much energy was expended making the cartridge, only how much energy was available when it was complete. 

If you don't completely evacuate the chamber, you will get a natural braking effect as the residual air is compressed - much simpler than trying to restrict the flow.
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #32 on: 04/10/2013 16:52:53 »
To illustrate approximately how powerful  a turbine fan can be, the engine on a jet plane moves about a 1000 cubic metres of air per second, so there is always the possibility that the system will work.
 

Offline alancalverd

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #33 on: 04/10/2013 17:46:14 »
And each 6-inch blade delivers more power than a Formula 1 engine. So what?

A word of friendly advice: if you do invent or discover a machine that produces more energy than it consumes, don't publicise it here or anywhere. You won't be able to patent it, but just go into production - using your money, not mine! 
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #34 on: 04/10/2013 22:06:28 »
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A word of friendly advice: if you do invent or discover a machine that produces more energy than it consumes, don't publicise it here or anywhere. You won't be able to patent it, but just go into production - using your money, not mine! 
I don’t see why there is so much cause for pessimism, after all if sunlight and wind work, why shouldn’t gravity and atmospheric pressure ?   The point  is that you can’t expect a high partial vacuum in such a short time, but a rough vacuum can be created and it is all that is needed. True the video does not state how long it takes for the lift cage to move after the button has been pressed, but it is more or less common sense that it must be less than a few  seconds because if it were more people would get restless and maybe try to open the door,  if that was the case, there would have been some mention of it. The second point is that even if the counterweight produces slightly less power on the ascent than on the descent, it is still a big step forward.  A 28 cm dia piston would have a force of 615.4  Kg acting on it if the differential pressure was 1 Kg/cm sq.  Suppose the final pressure was half that  or  200 Kgf, or even 175 Kgf it would still be enough to lift the counterweight back to its original position. Even in the worst case scenario with a final pressure of 175 Kgf it would still take about 3 seconds for the counterweight to ascend.  Basically as long as the pressure differential can be maintained throughout the ascent of the counterweight, there is nothing to stop it working, and it is almost a given that that can be done.  Surely that is what should count and  is  the advantage that the design has. 
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #35 on: 05/10/2013 01:32:31 »
I have been going through my calculations again, and find that the design will work exactly as described with a continuous output of 10KW, if a partial vacuum of 100mb (76 torr) can be created. Also an error in the earlier calculation: for the elevator to travel at 30 ft/min the turbine would have to operate at 50W and not 500W as was quoted earlier, which somehow does not sound quite right. One can't help thinking what 3000W would do.
« Last Edit: 05/10/2013 01:34:43 by McQueen »
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #36 on: 05/10/2013 07:39:12 »
Hi Alancalverd,
Here is an update. I wrote to one of the vacuum elevator manufacturers asking how much time in seconds, it takes for the lift cage to start moving after the button is pressed, she replied that it was considerably less than a second. Does this change anything ?
 

Offline evan_au

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #37 on: 06/10/2013 06:28:10 »
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if sunlight and wind work [to produce energy], why shouldn’t gravity and atmospheric pressure?

  • Sunlight works to generate energy because (i) the photons have inherent energy, or the (ii) the temperature of the body exposed to concentrated sunlight is greater than the temperature of a cold body, and you can use a heat engine. Generation continues as long as the Sun shines.
  • Wind works to generate energy because there is a difference in pressure between a high pressure weather pattern and a low pressure weather pattern (typically a difference of around 2% of atmospheric pressure, from regions which may be 1000km or more apart). An irregular progression of high and low pressure regions is created by the rotation of the Earth and the temperature differences between night and day. Somewhat erratic generation continues as long as the Sun shines.
  • Gravity works to generate energy if there is an object from a high location to a low location. As soon as it reaches the low location, it takes more energy to return it to the high location than you extracted from its original fall. So gravity is a "once-only" energy source, not a repeatable source.
  • Atmospheric pressure works to generate energy if there is a high-pressure region and a low-pressure region. As soon as the excess air from the high-pressure region reaches the low-pressure region, the pressures equalise. It takes more energy to restore the high-pressure and low-pressure regions than you extracted from their original equalisation. So atmospheric pressure is a "once-only" energy source, not a repeatable source.
  • There is another sense in which atmospheric pressure is a repeatable source of energy: If you are able to capture air pressure when a high-pressure weather pattern is overhead, and release it when a low-pressure weather pattern is overhead (which may be a day to a week later). However, this pressure difference is only about 2% (barring tornadoes). The energy you can extract from this is much lower than you get from a steam engine, which may work from a low-pressure region as low as 0.1 atmospheres, or as high as 10 atmospheres, ie the sort of pressures you need to move a vacuum elevator. It is a bit inconvenient to get into a vacuum elevator, press the button, only to wait a few days until the weather changes - and then it only moves you 2% of the distance to the next floor! Generation of this fairly weak and erratic source of energy continues as long as the Sun shines.

So I don't see a way that gravity and/or atmospheric pressure could produce a repeatable 10kW in a device that would fit in an average house. On the other hand, there are rooftop solar panels and backyard windmills that can produce 10kW of power (when conditions are right).
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #38 on: 06/10/2013 14:59:51 »
Hi evan_au,
Have just been trolling the web and was shocked to see the number of sites dealing with alternate energy, free energy sites and so on, the numbers of postings  would surely dwarf naked scientist. I was at a bit of a loss wondering why my post was so controversial,  and although I must have known about these alternate energy forums, it is easy to forget.  Having said that, I can understand why your posts seem to be on the  desultory  side it can be daunting trying to claw through someone’s ideas and designs, especially if you suspect they are fixated.  To understand the way in which I visualise the use of  atmospheric pressure in my design all that needs be done is to understand how the Newcomen engine works. Here is an animation: http://www.animatedengines.com/newcomen.html
As can be seen the heavy counterweight is lifted when a vacuum is created in the cylinder by the condensation of steam and when atmospheric pressure pushes the piston down the cylinder raising the counterweight. When the vacuum is evacuated and atmospheric pressure is neutralised, gravity pulls the piston back up.  I am planning to use the kinetic energy generated by the falling counterweight ( about 14.7 KJ) to turn a generator ( about 5 KW) to supply a multi stage turbine fan ( about 3 KW ) that will provide the  vacuum, instead of condensation of steam. By all accounts it takes less than a second to generate enough vacuum to lift the lift cage weighing about 350 Kgs. If the system which I call ‘Home Grid’ is correctly designed it should develop the same amount of power when the counterweight is descending and ascending, hence continuous generation of power. I think 10 KW of continuous power would be well within the limits,  windmills can do it, why not other machines.
I would also like to add that when a particularly sticky engineering problem comes along, the only way to deal with it is to collect as much data and information about the project as possible and then launch into the project, and with all due respect to all the die- hard believers in the second law for the conservation of energy, I think right now, with the situation we are in, would be a good time to exercise the same kind of logic that we use in building a bridge, to evaluate   projects like mine.
In order to clarify once more my design not only provides for the counterweight descending, it also provides for its ascending once again, which after all is the hall mark of a machine that works. 
« Last Edit: 23/10/2013 16:39:41 by McQueen »
 

Offline Bored chemist

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #39 on: 06/10/2013 15:13:17 »
Reading such beautifully written posts brings on nostalgia ! Still you are wrong when you say that atmospheric pressure cannot be considered a source of energy:

If atmospheric pressure was the source of the energy, why would they need coal?
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #40 on: 06/10/2013 15:29:24 »
Why burn the coal at all, why not just place it on the piston and see it do its work !
 

Offline alancalverd

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #41 on: 06/10/2013 16:55:43 »
Hi Alancalverd,
Here is an update. I wrote to one of the vacuum elevator manufacturers asking how much time in seconds, it takes for the lift cage to start moving after the button is pressed, she replied that it was considerably less than a second. Does this change anything ?

No. The air brakes on a truck or the vacuum brakes on a train work instantaneously (we hope) when required to do so, but it takes time to charge (or evacuate) the reservoir between applications of the brakes.

Quote
Why burn the coal at all, why not just place it on the piston and see it do its work !

Because when the piston reaches the bottom of its stroke, you have to schlep it all up the hill again.
 

Offline evan_au

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #42 on: 06/10/2013 19:12:58 »
That's a nice animation of the Newcomen engine. It's mode of operation is as I remember it; what the animation does not convey is how slowly it operated.

I might be a die-hard believer in the second law for the conservation of energy, but I am an even stronger believer in the rapid dissipation of energy by vortices in fluids. The Home Grid will need a continuous input of energy (provided in the form of burning coal, for the Newcomen engine) to overcome losses due to friction, viscosity and turbulence.

I am happy to promote sound engineering principles, such as used by bridge builders. These include applying the laws of physics and calculating forces. But bridge builders who ignore the complex behaviour of moving air sometimes come unstuck.
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #43 on: 07/10/2013 00:18:55 »
Quote
That's a nice animation of the Newcomen engine. It's mode of operation is as I remember it; what the animation does not convey is how slowly it operated.
That's a funny thing to say because as it happens the Wartsila Sulzer diesel engine works at an rpm of about 22 - 100 rpm and has a piston stroke that is about 2.5 m in length, (10 ft.) yet it is quoted everywhere as being the most efficient engine in the world.
Quote
I am an even stronger believer in the rapid dissipation of energy by vortices in fluids. The Home Grid will need a continuous input of energy (provided in the form of burning coal, for the Newcomen engine) to overcome losses due to friction, viscosity and turbulence.
The kinetic energy provided by the descending counterweight is greater than what is needed to run the vacuum generating device used in the vacuum elevator to evacuate an area two to three times the volume of the Home Grid system.  It should not be a problem, as for dissipation by vortices, if modern vacuum pumps did not have adequate measures to prevent blow back, at least while generating , what is in modern terms, such a slight vacuum, then technology has been in vain.
Quote
No. The air brakes on a truck or the vacuum brakes on a train work instantaneously (we hope) when required to do so, but it takes time to charge (or evacuate) the reservoir between applications of the brakes.
Well modern sewage vacuum devices which work over several miles, work in real time, no spooling up.
« Last Edit: 07/10/2013 00:24:22 by McQueen »
 

Offline SimpleEngineer

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #44 on: 07/10/2013 11:01:58 »
Ahhh the good ol two stroke diesel engine..

As for the vacuum drive... I understand the principle of removing air, (not as hard as many would wish to believe), find it hard to see how this could be achieved as efficiently as you describe, or sustainably.. the losses in your system (friction, resistance, etc) would suggest that, yes it would start and carry on for a bit, but the strokes would get shorter and shorter until they stop and the energy required to get it started would more than likely be more than you could possibly get out of it..

I would LOVE to see this tried in practice though.. but would HATE to have to specify the tolerances and talk about them to contractors.. lol..

Reminds me a little of a particular hydroelectric plant I used to visit.. two lakes at different heights.. drop the water from the top one when the power is required for peak loading, pump it back up when there is excess energy to be used.. obviously your system might be able to provide a longer (but lower) continuous output, but operated like this would spread household use over a longer period, evening out those nasty spikes.
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #45 on: 07/10/2013 12:08:07 »
I feel that it will either work or it won't, I can't see it slowly running down, b'cos basically it's just like the Newcomen Engine, those worked for a long time. In the early days the Newcomen engines created laughable vacuums, with the piston being sealed by rags etc., but everything gradually improved, in any case creating a vacuum with steam can't be compared to trying to suck the air out.
 

Offline SimpleEngineer

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #46 on: 07/10/2013 14:53:21 »
It should slowly run down, each stoke will take out energy due to friction, giving you less energy to generate the vacuum from, any leaks would reduce your vacuum, increasing energy losses meaning less energy returned from the counterweight, less vacuum etc.

I can see how your thinking goes, and in an ideal system it should go forever, but realities tell us something different. I dot know how your turbines would draw through air at a fixed rate regardless of energy input, but I could say a rotation of say, a few variable vane pumps (offset rubber impeller) would deliver a fixed value evacuation per movement, and this will slow down as the energy available to move the air will decrease due to friction and resistances.   
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #47 on: 23/10/2013 16:32:06 »
It should slowly run down, each stoke will take out energy due to friction, giving you less energy to generate the vacuum from, any leaks would reduce your vacuum, increasing energy losses meaning less energy returned from the counterweight, less vacuum etc.
I have a feeling that you have paid only the most cursory glance at how the 'Home Grid' system works. So although I had decided not to, i will try once again to explain:
The scientific  definition of a perpetuum mobile machine is as follows:- "hypothetical machine which, once activated, would continue to function and produce work"  indefinitely with no input of energy.”
A more accurate description would be “perpetual motion machine - a machine that can continue to do work indefinitely without drawing energy from some external source;”
   Machines which extract energy from seemingly perpetual sources—such as ocean currents—are capable of moving "perpetually" (for as long as that energy source itself endures), but they are not considered to be perpetual motion machines because they are consuming energy from an external source and are not isolated systems. Similarly, machines which comply with both laws of thermodynamics but access energy from obscure sources are sometimes referred to as perpetual motion machines, although they also do not meet the criteria for the name.
   The starting point is to decide which of these definitions best describes the working of the ‘Home Grid’ system.  As you can see I have settled on the third definition which I have underlined. The source of external energy to begin with is therefore gravity, which means that at the start of the operation the counterweight has to be hoisted into place.  Once that is done however, we find that  the height at which the counterweight has been placed, determines if there is enough energy  available to run a vacuum machine for long enough to create a vacuum within the system. Fortunately the ‘vacuum elevator’ which works on a similar principle operates to a height of 10m.  All things being equal the total volume of the ‘Home Grid’ system to be evacuated  is smaller than the volume to be evacuated in the ‘Vacuum elevator’ system.  Let us take as an example a counterweight of 150 Kg suspended at a height of 10m, then as explained, since it is connected by cable to an open ended shuttle, the only forces opposing the descent of the counterweight is the friction present at the sides of the shuttle and at the point where the cable passes over the generator sheave. To all purposes the counterweight is almost in free fall and its K.E can be calculated using the equation: mgh.  This gives a kinetic energy of 14.7 KJ, since the time taken for the counterweight to descend is 1.4 seconds, more than 7 KJ is available to turn the 5KW generator. Again since ( we questioned the manufacturers) it takes less than 1 second after the button is pressed for the vacuum elevator cage to start ascending, we assume that the 1.4 seconds available to use, is more than adequate time to create the vacuum, (a look at commercially available vacuum lifters should bear this out). The shuttle is now sealed, it has a vacuum below it and atmospheric pressure above it. Even if a vacuum of only 500mbar , or less, has been achieved it is still enough energy  to  lift the counterweight back to its original height. This can be calculated using the equation:   a= (m1 – m2)g/(m1 + m2) to find the acceleration of the piston as it moves down. So enough energy is available to both lift the counterweight back to its original position AND produce about 5KW of electrical output !
Look at the picture below:

This is a genuine perpetual motion machine design that is perpetually hopeful. Here only a single source of energy, namely gravity is used, in opposition to itself. Therefore it will never work, the forces equalise. In the ‘Home Grid’ system two forces are used atmospheric pressure in the presence of a vacuum and gravity. Remember that 24 horses, straining with all their might could not separate the spheres at Magdeburg, designed by Otto Von Guericke.  Atmospheric pressure in the presence of a vacuum is a powerful force my friend, easily the equal of gravity.  Infact to lift a 1 metre square board against a vacuum of 1 torr would require 10 tons , 22,000 lbs !
P.S. I will try to post a clearer explanation of how the 'Home Grid' system works when I have more time. But here is a tip, the counterweight,cable and pulley  system that power the vacuum do not contribute useful electrical output!
« Last Edit: 23/10/2013 16:42:44 by McQueen »
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #48 on: 24/10/2013 06:49:50 »
I would LOVE to see this tried in practice though.. but would HATE to have to specify the tolerances and talk about them to contractors.. lol..
This sounds a bit more like it !  The only problem is that I don’t see why it can’t be done. Money continues to be thrown at wind and solar pv and solar thermal and to a lesser extent at tidal and wave power, why not give new ideas like ‘Home Grid’ a chance, surely it would not cost too much ?
As for the vacuum drive... I understand the principle of removing air, (not as hard as many would wish to believe), find it hard to see how this could be achieved as efficiently as you describe, or sustainably.. the losses in your system (friction, resistance, etc) would suggest that, yes it would start and carry on for a bit, but the strokes would get shorter and shorter until they stop and the energy required to get it started would more than likely be more than you could possibly get out of it..
The first cylinders used by Thomas Savery, were just suitably caulked wooden casks, these soon gave way to iron cylinders that were so badly cast it was almost impossible to get an air tight fit, many of these first cylinders were more ovoid than circular, but they still worked. The reason that they worked in the Newcomen engine was because the surface area of the piston was calculated in such a manner that the amount of force acting on the  piston ensured that it descended almost as rapidly as air leaked into the cylinder and under the piston, even taking into account the lifting of the counterweight.  This of course is the million dollar question, can modern technology build up a sufficient vacuum in the time given, about 1.4 seconds, to ensure that the counterweight rises again. This is the only question that remains to be answered and even here the outlook is bright because not one but  two manufacturers of  ‘vacuum elevators’ informed us that the lift cage starts to rise within a second of the button being pressed.  The other concerns, such as the system running down because of friction and inertia, seem far fetched, especially considering that the Newcomen engines, performed pretty well for almost two hundred years. (i.e., given the existence of a sufficient vacuum there is no reason for the system not to work).  In any case a true engineer does not sit there imagining what might happen, he takes out his slide rule or electronic calculator and gets down to calculating data and then , if so indicated, puts the design to the test by building something.  Even a simple engineer should know that.
 

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #49 on: 24/10/2013 11:14:06 »
There has not been enough information given to actually do ANY calculations. But I will give my ideas a go.

the system is 150kg cylindrical block (probably benefit from a different shape but for calc means a cylinder) at 8000kg/m3 this give a cylinder with a height of 1m (arbitrary) and a radius of 0.077m.

The energy generated would be (from mgh) 14,700J from the drop, (vented top) the air pressure would supply 5.4kN at max vacuum.. we only need 2.94kN for return journey (barring resistances) meaning we would only need to draw 0.54 of a vacuum (so reduction to 46kPa) (assuming the remainding air is free to remove due to evacuation by the shuttle)

The volume of air would be 0.18m3 we would need to remove 54% of this.. 0.0972m3 per movement, 1 a second will give a vacuum requirement of 175m3/hr which requires 9kW (Power=flow.pressure difference). That's a very basic calc.. another works on pump down time (S=v/t . ln(Po/Pi) ) which gives 468M3/hr.. this would need a 25kW vacuum pump..

So depending on which you look at its still not possible. you will gain 7.35kW in total from each unit and need minimum of 9kW to operate it. (based on this system of course)

Calculations can go towards, slower running, different shapes and volumes, mechanical linkages etc. I based this off electrical power (and even if you got the difference down to a minimum there are efficiency to worry about)
 

The Naked Scientists Forum

Re: Energy losses, Thermodynamics and Efficiency
« Reply #49 on: 24/10/2013 11:14:06 »

 

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