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  4. Energy losses, Thermodynamics and Efficiency
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Energy losses, Thermodynamics and Efficiency

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Offline alancalverd

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #20 on: 03/10/2013 08:08:22 »
Quote
Still you are wrong when you say that atmospheric pressure cannot be considered a source of energy:.....the Newcomen atmospheric engine.


Sadly, I have to say I was right. Newcomen engines use the condensation of steam to evacuate a chamber to produce a pressure differential with the ambient atmosphere. A brilliant idea because you can generate the steam with a low pressure boiler - nothing more than a big kettle - which simpllifies the engineering, as long as you don't need anything to move quickly. Perfect for pumping water out of tin mines, which is what he originally designed it for: you can use the recovered water to chill the vacuum cylinder and top up the boiler.  But read the last line of the Kew Museum website
Quote
Returned to steam 1976
As I said above, air is the working fluid but the prime energy source is burning wood or coal. Hence the expression "nineteen to the dozen" - a good pump would shift 19 tons of water per dozen bushels of Welsh steam coal.

The conversion of mechanical to electrical energy in a big power station is well over 90% efficient - hence the reason why "micropower" is not a good idea for sustainability as small dynamos (less than 100 kW) rarely exceed 80%. But the Carnot efficiency of the prime mover (converting coal energy to mechanical energy in high temperature steam) just approaches 60% (35% is good for a small boiler) hence 54% overall thermal efficiency is considered extremely successful and only achievable in the largest and most modern stations - the ones the EU insists that we shut down. .
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Offline SimpleEngineer (OP)

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #21 on: 03/10/2013 09:12:05 »
Quote from: evan_au on 02/10/2013 18:54:31
Quote
generation of bubbles underwater...seem to give rise to free energy
It is true that bubbles rising through a water column could turn a water-wheel and generate power, such as is done in the common fish-tank ornament.

However, pumping the air down under the water consumes all of the energy you could capture as the bubbles rise to the surface.

Actually most of the energy from the bubbles rising to the surface is lost in friction and turbulence.


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?



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Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #22 on: 03/10/2013 09:44:03 »
Agreed, Newcomen engines were horribly inefficient as James Watt proved with his improved steam engine. The thing to remember is that the working principle was different and that soon after Watt improved it the design was abandoned in favour of steam pressure, as far as can be seen no-one ever looked at the technology again. Until, that is the vacuum elevator came along, using roughly the same principle with an electric motor instead of steam as the prime mover. It was low pressure steam that was used in the Newcomen engine, the piston was lifted because the rocker beam was unevenly balanced and as soon as equilibrium was reached in the air pressure above and below the piston, gravity pulled the beam down and the piston up.
Here are a few more facts and figures about the vacuum elevator, that you might ( or might not) be interested in.  The lift cage is 32 ins (81cms approx) across, it can work up to a height of 10m. and no more. It uses 3 KW of energy to turn three turbines. It can lift a weight of 204.5 Kg, but taking into account safety regulations can probably lift about 30% to 40% more than that.  The turbines seem to take a few milliseconds (judging by the video) to evacuate air from the cylinder above the lift cage. This raises an interesting question look at the diagram below:-
 [ Invalid Attachment ]
As can be seen the counterweights are connected to an open ended shuttle, so that when the counterweights are released there is practically no resistance to their descent except for the friction at the generator spool and the friction where the open ended shuttle is touching the tube. Suppose the height at which the counterweight is situated is 10m and that the generators which they are turning by means of the cable (just as in a car alternator) are capable of generating 5 KW.  http://www.aurasystems.com/pages/prod_exploded.html
Lastly the vacuum motor seen (attached by ducting to the tubes containing the shuttles) at the bottom right of the picture needs 3 KW.  Then the sequence is as follows the counterweight weighing 150 Kg  is released, generating 14.7 KJ of energy and taking about 1.4 seconds. This means that 7.25 KW approx. are available to turn the generator which needs only 5KW ! The generator which is generating 5 KW of electricity supplies the vacuum motor which needs only 3KW of power.  All this time the counterweight has been descending and the open ended shuttle has been rising. When the shuttle is almost at the top of its tube it is sealed by means of an actuator or similar converting the open ended shuttle into a closed piston.
 [ Invalid Attachment ]
Atmospheric pressure is now introduced above the piston with the result that there exists a partial vacuum below the piston and atmospheric pressure above. If a vacuum of 1 torr is achieved. The piston (30 cms dia) would have 706.5 Kgf acting on it while it needs to lift only 150 Kg ( i.e., the weight of the counterweight) . As the shuttle travels downwards under the force of atmospheric pressure the counterweight travels upwards till it reaches its original position. The piston is once more unsealed turning it into an open ended shuttle and the whole cycle keeps repeating.  Looking at the diagram above one of the tubes and counterweights is used exclusively to create a vacuum in the system while the other tubes and counterweights, use this vacuum in a parasitical manner to produce usable electricity by spinning their generators, the generators are geared to produce electricity both while the counterweights are descending and ascending.
Another even better design but slightly more complicated, is if the energy of the descending counterweight was stored in ultracapacitors or flywheels, when the counterweight has reached the bottom the energy in the ultracapacitors is used to activate the vacuum machine so that a partial vacuum exists all during the descent of the sealed shuttle (piston).

The great thing about this design is that in excess of 10 KW can be generated continuously, in a manner similar to the way wind turbine motors generate energy.


* hg.jpg (38.92 kB, 570x799 - viewed 2176 times.)

* hg1.jpg (40.33 kB, 570x799 - viewed 2072 times.)
« Last Edit: 04/10/2013 01:48:39 by McQueen »
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Offline alancalverd

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #23 on: 03/10/2013 10:56:47 »
Vacuum systems are still in widespread use, the only difference being that the prime mover nowadays tends to be a mechanical or electrical pump rather than steam. You will find three vacuum-driven gyros in the cockpit of most small aircraft, and atmospheric office document transport systems are still on sale.

But the "continuous 10 kW" in your example still comes from the prime mover, and it isn't continuous.   
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Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #24 on: 03/10/2013 11:32:38 »
Quote
But the "continuous 10 kW" in your example still comes from the prime mover, and it isn't continuous.   

Surely the significant fact here is that the 'prime mover' is being powered by the kinetic energy of the descending and ascending counterweight and not from an electrical outlet, if the generators are continuously putting out power both while the counterweights are ascending and descending, why is the output not continuous ?

What I really want to know is , is our technology up to it ? I think it is, things like valves, vacuum pumps ( turbines, blowers etc,.) friction reducing materials, pulleys etc., have changed beyond comprehension, in fact it would even be possible to monitor the whole process, through strategic placing of pressure sensors.  If that is indeed the case and it can be done, the system has a small footprint, needs little maintenance and what maintenance there is will be basic. Put one in every house (or garden) it would generate enough electricity, to cook, bathe, wash clothes, run all of the household appliances, heat the house AND still have enough left over to (a) either charge your cars lithium ion battery OR make and compress enough hydrogen through electrolysis to use in a hydrogen fuel cell. So what's all the fuss about ?
« Last Edit: 03/10/2013 12:12:34 by McQueen »
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Offline Kryptid

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #25 on: 03/10/2013 22:36:23 »
Quote from: SimpleEngineer on 03/10/2013 09:12:05
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.
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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.
 [ Invalid Attachment ]
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.

* PLAN.jpg (17.05 kB, 406x357 - viewed 1155 times.)
« Last Edit: 04/10/2013 01:49:25 by McQueen »
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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?
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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.
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Offline alancalverd

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #29 on: 04/10/2013 14:30:09 »
Quote
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.
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Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #30 on: 04/10/2013 16:07:33 »
Quote
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:
 [ Invalid Attachment ]
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. 

* uneven wieght.jpg (7.06 kB, 150x320 - viewed 1096 times.)
« Last Edit: 04/10/2013 16:22:24 by McQueen »
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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.
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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.
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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! 
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Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #34 on: 04/10/2013 22:06:28 »
Quote
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. 
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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 »
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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 ?
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Offline evan_au

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #37 on: 06/10/2013 06:28:10 »
Quote
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).
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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 »
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Re: Energy losses, Thermodynamics and Efficiency
« Reply #39 on: 06/10/2013 15:13:17 »
Quote from: McQueen on 02/10/2013 22:40:20
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?
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