Naked Science Forum

Non Life Sciences => Technology => Topic started by: SimpleEngineer on 30/09/2013 15:12:02

Title: Energy losses, Thermodynamics and Efficiency
Post by: SimpleEngineer on 30/09/2013 15:12:02
My colleagues and I constantly discuss ways to do things that seem ever so simple in our heads and on paper, that when these silly things called "The Laws of Thermodynamics" tell us to stop being silly and shows us the error of our ways.. (this never stops us trying to break them)

However we discussed the hydrolysis of water, how the energy you have to give (fair enough it can be free) would be more than the energy you would get back (sources tell us 85%-95% in comparison).. and this gives us wonder on what happens to that energy.. realizing that our typical foes of the three commandments tell us this is not possible!

We considered the forms of energy, and how realistically heat is usually the sources of loss.. BUT if we captured that heat (not hard in a water system) and used that energy to split the next lot of water.. Our thoughts began making all manner of perpetual motion designs involving hydrolysis cells, gas turbines and steam turbines..

BUT this cannot be the only source of loss, our wandering thoughts stressed over the balance of energy's through the hydrolysis cell,  the gains the losses and how to capture any slight implication of energy gain and we kept coming up with hard questions such as..

"If energy cant be created, then how does buoyancy work?" Free kinetic & potential energy?

"If energy cant be destroyed, then why are things not 100% efficient?" Where is the rest of the energy going?

Our perpetual motion machine was as such..

Sunlight turns water to oxygen and hydrogen, these are then burned across a gas turbine to give HP steam, through turbines and condensed within the reservoir of water that is then fed to the hydrolysis cell. Power from turbines used to pressurise the gasses and a self contained heat engine to take heat from the water to the gasses.

Considered within a single 'box' with only a section exposed to sunlight.

Mechanical losses, we call.. BUT.. if its all self contained.. Where does that energy go? if through heat it would be contained within the system. and anything this puts out could be turned to energy in one way or another.. (such as any extraneous kinetics from the turbines)

Can anyone help end our tussling with the LoT.. and set us back into the legal thought process.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: SimpleEngineer on 30/09/2013 15:22:06
Sorry my mistake.. not perpetual motion.. but a means to generate rotational energy (for electricity generation say) from sunlight.. and why wouldn't it be the same output from the same input

The perpetual motion was an idea dropped very early.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: alancalverd on 30/09/2013 22:43:05
Entropy.

As energy moves around a system, it becomes less potentially useful and entropy increases.

Imagine you have a million dollars. You can offer this to a group of people in exchange for their labour to make something you want. Say you employ a thousand people to build a posh yacht. Now each of them has a thousand dollars and you have a yacht. The amount of money hasn't changed but now nobody can afford a yacht.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: Kryptid on 01/10/2013 04:57:06
"If energy cant be created, then how does buoyancy work?" Free kinetic & potential energy?

I'm not exactly sure what you are asking here, as an object floating on the surface of the water is neither expending nor absorbing energy.

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"If energy cant be destroyed, then why are things not 100% efficient?" Where is the rest of the energy going?

You're right, it isn't being destroyed. It's being transformed into more than one form of energy. If you have an internal combustion engine that is 55% efficient (just a hypothetical number), then 55% of the chemical energy contained in the fuel can be converted into mechanical energy for pushing the car forward. The other 45% of the energy is either lost as unburned fuel in the exhaust, waste heat or vibrational/sound energy. Waste heat is one of the biggest problems when it comes to efficiency, as you cannot collect it all and change it into useful work. Some of it is conducted away by the air and some of it is lost as infrared radiation.

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Sunlight turns water to oxygen and hydrogen, these are then burned across a gas turbine to give HP steam, through turbines and condensed within the reservoir of water that is then fed to the hydrolysis cell. Power from turbines used to pressurise the gasses and a self contained heat engine to take heat from the water to the gasses.

Considered within a single 'box' with only a section exposed to sunlight.

Mechanical losses, we call.. BUT.. if its all self contained.. Where does that energy go? if through heat it would be contained within the system. and anything this puts out could be turned to energy in one way or another.. (such as any extraneous kinetics from the turbines)

It can't be completely contained, not even in a vacuum. If the transformation of energy heats or vibrates the box itself by even a tiny amount, then at least some part of the energy will be radiated away as infrared/microwave/other radiation.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: evan_au on 01/10/2013 10:46:29
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"If energy cant be created, then how does buoyancy work?" Free kinetic & potential energy?

To expand on Supercryptid's response:
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Waste heat is one of the biggest problems when it comes to efficiency
Back in the days of steam engines, Carnot worked out the maximum possible efficiency of an engine running on heat - it was one of the first applications of thermodynamics. The energy you can extract from a source of heat depends on the temperature difference between a hot body and a cold body. After you run the engine for a while, the hot body tends to cools down, and the cool body tends to warm up; when the temperature is equal, the efficiency drops to zero.

Practical heat engines can attain efficiencies of up to 50% by operating at very high temperatures, but petrol-driven automobile engines are much lower.

Trying to extract useful energy from "low-grade" heat sources such as the friction of lubricated bearings retrieves a very small fraction of the already-small amount of heat energy present.

See: http://en.wikipedia.org/wiki/Carnot%27s_theorem_%28thermodynamics%29

Cosmologists imagine the "heat death of the universe (http://en.wikipedia.org/wiki/Heat_death_of_the_universe)", where all matter and radiation converges to a common temperature. Although the amount of energy present is the same as today, it is evenly spread out through the universe, and heat engines cannot work.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen on 01/10/2013 13:24:00
Hi,
Interesting topic free energy versus perpetual energy, it has been on my mind a lot of late. I am not sure quoting Carnot is as relevant today as it was even 25 years ago, things have changed a lot, become more sophisticated, questions and answers are no longer as simple as they once were. I was in fact thinking about sailing ships. I was reading somewhere (and in fact a simple calculation based on the weight of a laden ship, number of days, distance etc., will probably bear me out) that the clipper ships could generate about 3000 hp (2.5 MW) continuously over a period of 3 months or so.  This in itself is fantastic considering that about 4 acres of solar PV panels would be needed to generate  a similar amount of power, during hours of peak sunlight or who knows how many tons of coal.  Even more peculiar is the fact that these ships were able to sail into  the wind almost as fast as they could sail with the wind behind them.  My question is how does this happen ? Take that same ship out of the water put it on castors that can turn 360 degrees in any direction, and create an artificial wind in front of it, what will happen, it will move backwards sideways, but never, ever under any circumstances forwards. What good is all the free energy in the world if you can only move when the wind is blowing from behind. The answer of course is that it is no good at all.  OK, now look at the question from another point of view. Take a 100 Kg weight up a height of 10m and when you drop it you get about 9.8KJ of energy. The point is, does the same question apply ? What good is this (.9.8 KJ if you only have it when the weight is falling), or does it fall into the category of ‘it is impossible and it is like trying to get perpetual motion or free energy', and so on.

Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: SimpleEngineer on 01/10/2013 14:26:59
Carnot Efficiency does not seem to apply to a closed system in which the hot and cold body are in fact the same process media in different states. The hot and cold body are in relative terms which we can manipulate with small applications of technology. All we can come up with is that the energy required to achieve the manipulations MUST increase.. however since there are no losses why is this so? the water might get too hot, in this case we could remove energy through power generation.. (free energy - impossible) the water might get too cold, but where does that energy go?

Buoyancy was questioned through the generation of bubbles underwater, and how the different pressures, density and even temperature seem to give rise to free energy as you can gain significant potential and kinetic energies just by having a lower density.

I personally dont understand entropy, I can do the calcs and pass an exam on it, but the practical applications do not make sense to me (I am not that bright ;) ) I dont have a tangible grip on how it behaves.. 
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen on 01/10/2013 15:03:03
My interest arises from the fact that after so many years of trying to switch to renewables, we are in no sense closer to a solution than when we first started with the result that things like fracking are being let loose on the public and YET there seemingly were free energy solutions, like sailing ships that worked very well, while generating  appreciable amounts of what must be considered free energy by any criteria.  Yes, we do have modern wind turbines, but they are  intermittent, sailing ships were so successful because they enjoyed more or less continuous uninterrupted power, since the oceans contained no significant obstructions to the wind. Surely, especially considering the circumstances we are in, with fossil fuels running out and the pollution caused by them reaching unacceptable levels,  we have to leave ourselves open to the possibility that   a  free energy solution similar to the free energy solution enjoyed by sailing ships does exist and can be found.  Or is that too daring a prospect to undertake?
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: SimpleEngineer on 01/10/2013 15:34:35
When the answer to how to extract this free energy is by strapping a dirty great gearbox and dynamo to a fan on top of a very large pole.. no I doubt we will get much further.. they are horrendously inefficient.. I KNOW how much energy it takes to build one.. and if you think you will be getting that back anytime soon...

The tidal and wave generators are seemingly much more efficient although not as practicable, and please let us not follow the chain of thought that led to someone suggesting we put wind turbines on moving platforms again..

My fear is that energy is not limitless..and the more we leech out of the earth's systems the less there will be? Is there any data about climate change and windmills? They MUST have some impact on the weather, however minor.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: Kryptid on 01/10/2013 16:24:44

Carnot Efficiency does not seem to apply to a closed system in which the hot and cold body are in fact the same process media in different states. The hot and cold body are in relative terms which we can manipulate with small applications of technology. All we can come up with is that the energy required to achieve the manipulations MUST increase.. however since there are no losses why is this so? the water might get too hot, in this case we could remove energy through power generation.. (free energy - impossible) the water might get too cold, but where does that energy go?

No physical device is a truly closed system, even if all of its components are in the same container. Waste heat (and therefore energy) can always escape in the form of radiation, even in a vacuum. Sure, you could put your device inside of a box that attempts to capture the waste heat and convert it into useful energy, but this new box itself will emit some degree of waste heat itself. You could put that box inside of a second box that does the same thing, but you'd need to have an infinite series of boxes-within-boxes to keep all of the waste heat from escaping.

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My fear is that energy is not limitless..and the more we leech out of the earth's systems the less there will be? Is there any data about climate change and windmills? They MUST have some impact on the weather, however minor.

You're right, it isn't limitless. All potential energy on Earth can be theoretically exhausted, but it will take a very long time (especially as a large amount of the energy available on Earth comes from the Sun). At the very least, we'd have to wait for the Sun to die before we run out of usable energy here (and the Earth itself may be consumed by the Sun before that happens).
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: alancalverd on 01/10/2013 20:28:10
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sailing ships were so successful because they enjoyed more or less continuous uninterrupted power, since the oceans contained no significant obstructions to the wind.


Apart from the Doldrums, and the fact that the prevailing wind is westerly everywhere, and often just stops completely for weeks at a time.... And however much power you can generate from a moving sail at sea, it isn't a lot of use to the rest of the population standing on land a thousand miles away.

Sailing ships were "successful" because there was buggerall else around until the advent of steam, whereupon they all disappeared in a few years.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: syhprum on 01/10/2013 21:31:50
I think we must distinguish between perpetual motion which we see around us all the time (the rotation of the earth for instance) and machines that produce power from nowhere which of course cannot exist.

PS supercriptid

55% thermal efficiency for an IC engine is not hypothetical, the largest ships Diesels can just about reach this figure
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: Kryptid on 01/10/2013 21:59:09
In regards to perpetual motion machines, I'd like to point out that there are three types. Two are forbidden by the laws of physics and one is not.

Perpetual Motion of the First Kind: A device or phenomenon that can increase the total energy content of a closed system. This violates the first law of thermodynamics (energy cannot be created nor destroyed). This kind is currently thought to be impossible.

Perpetual Motion of the Second Kind: A device or phenomenon that increases the available energy to do work in a closed system by decreasing the total entropy of that system. This violates the second law of thermodynamics (the total entropy in a closed system cannot decrease with time). This is actually only prevented by probability. In a technical sense, entropy can decrease in a closed system spontaneously. However, the chances of that occurring are so amazingly small that it is highly practical to consider it impossible (at least on a human time scale).

Perpetual Motion of the Third Kind: A device or phenomenon that can remain in motion indefinitely so long as it doesn't do any work. In practice, one could consider the motion of subatomic particles to be an example of this. A cloud of subatomic particles or atoms should always retain some motion forever due to both Heisenberg's Uncertainty Principle and the impossibility of reaching absolute zero. However, you can't extract useful work from a particle cloud which is already in its lowest energy state.

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55% thermal efficiency for an IC engine is not hypothetical, the largest ships Diesels can just about reach this figure.

I have to admit, the reason I said "hypothetical" is because I had no idea what a reasonable efficiency figure for internal combustion engines would be.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen on 02/10/2013 06:04:10
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In regards to perpetual motion machines, I'd like to point out that there are three types. Two are forbidden by the laws of physics and one is not.
What category would a sailing ship fall into, it certainly doesn't fit into any of the categories describing perpetual motion of the first, second or third kind? Would it be regarded as free energy, in a sense, although it does take a lot of work raising and lowering sails etc.,
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen on 02/10/2013 06:13:07
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Sailing ships were "successful" because there was buggerall else around until the advent of steam, whereupon they all disappeared in a few years.

The point, surely, is that even that 'buggerall' wouldn't have been available if not for certain circumstances. As I had pointed out a sailing ship that sails only before the wind is no good to anyone at all.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: evan_au on 02/10/2013 13:21:35
Question from McQueen:
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These ships were able to sail into  the wind almost as fast as they could sail with the wind behind them.  My question is how does this happen ? Take that same ship out of the water put it on castors that can turn 360 degrees in any direction, and create an artificial wind in front of it, what will happen, it will move backwards sideways, but never, ever under any circumstances forwards.

A sailing ship is not on castors; in a steady breeze, a vehicle on castors will be pushed roughly downwind, once the initial inertia is overcome.
A racing boat can sail within 30 degrees of directly upwind (http://en.wikipedia.org/wiki/Close_hauled#Close_hauled). When sailing into the wind, the sail acts like an aeroplane wing, and provides a sideways force. This sideways force is resisted by the force of the water on the hull, causing a small sideways motion to be accompanied by a larger motion into the wind.

If you saw the recent America's Cup races, most of the time these catamarans were up on hydrofoils; in this case the hydrofoils provide the sideways resistance.

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What category would a sailing ship fall into... Would it be regarded as free energy?
Energy is delivered from the Sun for "free", in that we did not pay to build & operate the Sun, or for transmission lines that cross millions of kilometers of vacuum.

The Sun delivers energy in various ways, including the direct photons, heating the atmosphere (producing winds), heating the sea (producing thermal gradients and currents plus rainfall for hydropower), photosynthesis (producing chemical energy, including fossil fuels); the rotation of the Earth interacting with the gravity of the Sun also produces tides. These could be considered "free", if it didn't cost anything to harness them.

However, photovoltaics, wind turbines & sails, ocean thermal power systems, planting/harvesting/processing crops, prospecting/mining/refining fossil fuels and tidal barrages cost considerable amounts of money to construct and maintain. So it comes down to cost-effectiveness.

Some modern cargo ships with engines also have sails that can be deployed to add some extra thrust (and save fossil fuels) when the wind is in a suitable direction. Some of the sail concepts are somewhat novel, such as rigid sails (http://www.gizmag.com/eco-marine-power-energysail-solar-sails/25005/), or kites (http://www.youtube.com/watch?v=qyLjISR6XQQ).
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen on 02/10/2013 14:58:53
Hi evan_au,

Nicely put, it figures, water is about a thousand times more dense than air, so it doesn’t really matter how hard the wind blows, water acts almost like a solid. Then, depending on the boats design, her sails etc., it will take the easiest way out which if correctly positioned will be forward on a tack. 

There are quite a few sources of energy that are termed renewable, some of these could be considered as sources of free energy. Here is a list that might be commonly found:-
1)   Solar, including PV and thermal and CSP and CPV.
2)   Wind
3)   Tidal
4)   Wave
5)   Biomass
6)   Bio-fuels ( including algae)
7)   Geothermal
8)   Hydroelectric
9)   Osmotic
10)   Ocean thermal
The amazing thing about this list is that it does not include either atmospheric pressure or gravity as a source of energy. Having said that look at this video:

It is quite astonishing the way in which the elevator cage with passenger is lifted using just a partial vacuum and atmospheric pressure, don’t you agree! When descending it is just gravity. If this had been described in words one in ten would have said it is impossible but since it isn’t ……!

Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: alancalverd on 02/10/2013 17:46:25
Big difference between square riggers, which really couldn't sail much into the wind, and aerofoil sails as on yachts and clippers. A modern sail is like a single-surface wing and generates "lift" by streamlined (nonturbulent) deflection of the wind. If you rotate the sail so that its chord (the line from the luff to the leech) points into the wind it produces a lift vector roughly perpendicular to the chord. If you now point the boat in that direction it will move forward, but in doing so it alters the apparent direction of the wind, which now seems to be coming from slightly ahead of the true wind. As the boat accelerates so the apparent wind moves further ahead until you reach the stall angle, with the apparent wind roughly 16 degrees either side of the true wind - interestingly, the same angle of attack at which a glider stalls. So by tacking +/- 16 degrees, you can progress upwind in a zigzag.

Windmills works in exactly the same way but in a rotating frame of reference, where the tips are travelling faster than the roots, so the wing is warped to maintain a constant angle of attack at each point. 

You can't use atmospheric pressure as a source of energy except by allowing air to flow in and out of a box, rather like the tide, as the weather changes - not terribly useful, though some clocks can be driven by their associated barometer. Atmospheric railways and lifts use a conventional engine to suck or blow: the air is a storage and transfer medium for the power of the engine   

     
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: evan_au on 02/10/2013 18:54:31
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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 (https://www.google.com.au/search?q=fish+tank+water+wheel&source=lnms&tbm=isch&sa=X&ei=xlpMUsLTBoSTiQfF5IG4CQ&ved=0CAkQ_AUoAQ&biw=1607&bih=836&dpr=1).

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.

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gravity as a source of energy

It is true that by rolling rocks down a hill, you can use their initial gravitational energy as a source of energy. But once all of the mountain rocks have filled all of the valley holes, there is no more energy to be gained*. This is a non-renewable form of energy on human timescales, and it is certainly not "free".

Rather than say "gravity is a source of energy", I would rather say that "gravity can temporarily store energy", such as is done in pumped-storage hydroelectricity (http://en.wikipedia.org/wiki/Hydroelectricity#Pumped-storage).

I liked the phrase in the YouTube video: "the cabin falls gently under gravity". If the cabin fell more than 2 floors under gravity, it would do some severe damage to the occupants. What is happening here is that air is slowly released back into the top of the cylinder through a valve, allowing the cabin to fall gently despite gravity.

Air leaking through the valve causes turbulent flow (http://en.wikipedia.org/wiki/Turbulent_flow) - this is a series of vortices which spawn smaller vortices, which in turn small even smaller ones. At sufficiently small scales (the Kolmogorov length scale (http://en.wikipedia.org/wiki/Kolmogorov_microscales)), this turns the initial source of potential energy (a pressure difference) into heat. As mentioned in the post on Carnot, heat is one of the least reusable forms of energy (even though much of our energy needs are met via this route).

Avoiding turbulence is why most proposals for efficient machines assume they are working in a vacuum, as turbulence of air or water irretrievably wastes energy.

If you are looking for efficiency, try electromagnetism, as electric motors can often achieve efficiencies of around 90%, twice that of very good heat engines. Better yet, if you want high efficiency, avoid motion of any kind, as that wastes energy through friction and turbulence.

*Dumping stars into a black hole is just rolling bigger rocks into a a bigger valley; it is still a non-renewable source of energy, and it is still not "free".
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: 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:
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You can't use atmospheric pressure as a source of energy except by allowing air to flow in and out of a box, rather like the tide, as the weather changes - not terribly useful, though some clocks can be driven by their associated barometer. Atmospheric railways and lifts use a conventional engine to suck or blow: the air is a storage and transfer medium for the power of the engine
Look at this link to the Grand Junction 90 inch, that was used to pump water out of the Thames, up to 7.5 million gallons a day! Considered one of the finest examples of the Newcomen atmospheric engine. The counterweight weighed 32.5 tons and was operated using only atmospheric pressure.
http://www.kbsm.org/engines/90-inch-engine
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If you are looking for efficiency, try electromagnetism, as electric motors can often achieve efficiencies of around 90%, twice that of very good heat engines. Better yet, if you want high efficiency, avoid motion of any kind, as that wastes energy through friction and turbulence
I read  somewhere that most coal fired power plants in Britain  operate nowhere near 90% but closer to 40%, I am not sure if this is true. In any case the idea is not to discuss efficiency but to see if it is possible to get something better than windmills or other present day renewable sources that work only intermittently and are therefore almost useless in practical terms. 
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: alancalverd on 03/10/2013 08:08:22
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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
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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. .
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: SimpleEngineer on 03/10/2013 09:12:05
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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 (https://www.google.com.au/search?q=fish+tank+water+wheel&source=lnms&tbm=isch&sa=X&ei=xlpMUsLTBoSTiQfF5IG4CQ&ved=0CAkQ_AUoAQ&biw=1607&bih=836&dpr=1).

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?



Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen 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.

Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: alancalverd 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.   
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen on 03/10/2013 11:32:38
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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 ?
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: Kryptid 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.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen on 04/10/2013 01:35:54
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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.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: alancalverd on 04/10/2013 08:20:30
What is the upper limit on the number of blue units per red one?
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen 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 (http://www.youtube.com/watch?v=XNOEP1XIFiM). He finally does succeed in climbing the building, quite an achievement.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: alancalverd 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.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen 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:
 [ 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 !
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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 (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. 
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: alancalverd 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.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen 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.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: alancalverd 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! 
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen 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. 
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen 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.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen 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 ?
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: evan_au 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?


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).
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen 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. 
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: Bored chemist 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?
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen 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 !
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: alancalverd 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.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: evan_au 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 (http://www.youtube.com/watch?v=xox9BVSu7Ok).
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen 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.
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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.
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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.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: SimpleEngineer 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.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen 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.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: SimpleEngineer 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.   
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen 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:
 [ Invalid Attachment ]
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!
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen 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.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: SimpleEngineer 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)
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen on 24/10/2013 13:05:41
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.

This is more like it, how positive and optimistic it is to actually be doing some calculations. What is not so nice though, is the armchair cynicism, can we really afford to be so offhand about things, I think if there is the slightest chance of something working we should get our noses to the grindstone until the idea has been completely, exhausted.

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)


Have a look at the SMB-C06 roots vacuum pump manufactured by Vacuum Pumps America, it has a throughput of 600 m^^3/hr and operates at 3500 rpm, the motor works on 2.2 KW and the final vacuum produced is on the order of 10^^-4  Torr.  The fantastic thing is that the ‘Home Grid’ application needs only one hundredth thousandth of that degree of vacuum !!! What do you think of that, one hundredth thousandth of that degree of vacuum, several orders of magnitude lower than what is available ! So yes, it lays the field wide open as to what is possible and what is not. I am pretty sure that it can be done.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: SimpleEngineer on 24/10/2013 14:15:25
Of course, I used the simpleset of simple calcs for my run through, however it may have shown why it is a very difficult system to calculate..

That pump you found is to be used as a booster pump, meaning that there is another beefier pump that will take the main power for the draw.. but a direct mechanical linkage to a piston type evacuater built correctly MAY provide enough vacuum but I am no pneumatics expert.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: SimpleEngineer on 24/10/2013 16:42:02
Actually.. I reran the calcs.. dropped a few clangers in there.. apologies. Will try to fix tomorrow.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen on 25/10/2013 06:40:34
Actually.. I reran the calcs.. dropped a few clangers in there.. apologies. Will try to fix tomorrow.

While trolling the net I have come across several vacuum pumps that will fulfill the criteria of 600^^3m/hr, dry running and using just 0.4 KW, that will reach 1 Torr without any backing or booster pump involved. This still leaves a margin of a factor of ten, since the vacuum that is required is more like 100 Torr than 1 Torr.  With the pump drawing such low power, it is possible that a separate vacuum pump can be used for each paired tube system, increasing the efficiency and  making the continuous generation of 15 KW possible.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: SimpleEngineer on 25/10/2013 09:08:19
That is pretty much what I am finding with the errors in my calcs.. I am about a factor of 10 off in a few areas.. *sigh* the eternal struggle with calculations using different units.. We noticed this when i passed my calcs in front of my colleague who has designed vacuum systems.. (He is resolute that the system will lose too much energy through friction yet cannot negate the principle)
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen on 25/10/2013 15:08:10
(He is resolute that the system will lose too much energy through friction yet cannot negate the principle)

Since you seem to be so obsessed with the idea that friction is going to clog up the works, I decided to investigate the problem. It seems that PTFE  ( polytetrafluoroethylene) has a sliding co-efficient of friction lower than ice, this applies to PTFE and steel and  almost any other material including itself.  The co-efficient of friction of PTFE (a type of teflon) is just 0.05, lower in fact than ice on ice, you can be sure that the pulley will also have a  PTFE component  in it, either as a coating for the bearings or as an integral material.  Could you kindly factor in the numbers and tell me how you think that friction will slow the whole system down and eventually bring it to a halt. Frankly the logic behind the statement continues to baffle me, given the large amount of K.E involved  both during descent and ascent of the counterweight. The descent being powered by gravity and the ascent by atmospheric pressure in the presence of a vacuum.
Here are a few more facts about PTFE
PTFE possesses the lowest friction coefficients of all solid materials; between 0.05 and 0.09:

    The static and dynamic friction coefficients are almost equal, so that there is no seizure or stick-slip action
    when increasing the load, the friction coefficient decreases until reaching a stable value
    The friction coefficient increases with the speed
    The friction coefficient remains constant at temperature variations.
Wear
The wear depends upon the condition of the other sliding surface and obviously depends upon the speed and loads. The wear is considerably reduced when adding suitable fillers to the PTFE (see filled PTFE).
 
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: SimpleEngineer on 25/10/2013 16:31:07
See work up..
Dimensions
Weight = 150kg
Density (steel) = 8000kg/m3
Volume = 0.018m3
h=1m
r= 0.075m
A= 0.018m2
movement = 10m
Volume air = 0.18m3
Forces
F=ma
Downwards force = 1470 N
Maximum upwards force
If pressure above is 0
Pressure below weight = 100kPa
Force acting on weight = 100000*0.018  = 1800 N
Maximum Pressure required for return = 18kPa
Acceleration and times
Downstroke
 a= 9.8m/s2
 t= 1.43s
Upstroke
amax = 2.2m/s2
tmin=3s
Vacuum Calcs
cycle time of 5s  (using vacuum pot to ensure vacuum is instantaneous)
so 0.18m3 of air in 5 secs
minimum increase in pressure.. say 1kPa, If vacuum pot is at 1 torr. (133pa) then 0.18m3 of atmospheric pressure air would occupy 555.5M3 to give a rise of 1kPa... This require a vacuum pump that can do 111.1M3/sec on outline calc (555.5/5)
Using S=V/t x ln(Po/Pi) = 238 M3/s (V= 555.5, t= 5secs , Po= 1133Pa, Pi=133Pa)
Using Power = S x dP/mech eff (typically 0.85)   
=   130kW or 280kW

Friction force with PTFE will still be 60N with the differential force on this example being 330N, this is fairly significant.. that does not take into account the friction of the air movements etc.

Work up may be wrong only had 15mins to write it (10 mins was drawing a picture that was too large to be posted :( )
Point out changes please.. As the power is large.

Something forgotten is the increase volume of air at lower pressures.. I have tried to take this into account
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: McQueen on 25/10/2013 16:49:34
Using Power = S x dP/mech eff (typically 0.85)   =   130kW or 280kW

All I can say is that using your calculations, the vacuum elevator would never have been built, would not have even been a pipe dream or a gleam in someone's eye. Don't forget that the vacuum pumps used in a vacuum elevator have to be pretty sophisticated to maintain a pressure differential that allows a constant velocity of 0.15m/sec, the final velocity for 'Home Grid' would be about 14m/sec! So I would surmise that the power needed for the vacuum pump would be far less than the 3 KW needed for the Vacuum elevator. I repeat a figure of about 0.5 KW, I could give links but that doesn't seem to bother you too much.
Title: Re: Energy losses, Thermodynamics and Efficiency
Post by: SimpleEngineer on 28/10/2013 10:40:02
oh it does bother me.. I have rerun the calcs you see above, and now get a solid solution.. This is for the system described.. a better designed system could obviously offer a massive reduction to power or other means of recouping enrgy.. but as for now I only have the calculation steps above to go with.