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Author Topic: Easy Gravity wheel  (Read 58933 times)

quantumtangles

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Re: Easy Gravity wheel
« Reply #75 on: August 03, 2011, 05:02:29 PM »
@ Vidar

That is most interesting and helpful.

In other words there is no benefit to using less dense working fluids.

I suspected this and said so in earlier posts about the invention.

I only used the example of less dense working fluid to answer critics who claimed working fluid had to be "lifted" to the top of tank A.

I always preferred the idea of using seawater as working fluid (due to its greater density).

However, what I have had difficulty calculating is the amount of energy required to force one cubic metre of seawater per second (density 1020kg/m3) into the base of a 30m high cylinder full of seawater.

one cubic metre per second is drawn from the top of the vessel at the same time as one cubic metre per second is crammed into the base of it.

You calculated required energy in joules, which makes sense.

But somewhere I recollect warnings about trying to convert joules to watts (I need to know the value in watts).

How would you go about calculating the precise energy in watts required to force one cubic metre of water per second into the base of a 30m high cylinder full of seawater? (mass out at the top equals mass in at the base).

Would the removal of 1m3/s from the top of the tank create a suction effect that would minimise the energy required to draw water into the base of the tank? Would adherence caused by covalent bonds in the water molecules 'glue' them together en bloc?

The pressure at the base of the 30m high cylinder is 300,186 Pascals gauge pressure (1020kg/m3 x 30m x 9.81 m/s/s) + Patmos of 101325 Pascals = 401,511 Pascals = 401.511kPa

I use compressed air in vessel B to force seawater back into the base of vessel A. But the amount of energy required is critical.

I took the direct approach of ensuring that the pressure in vessel B always exceeded the pressure at the base of vessel A (in other words I did the maths to ensure vessel B always maintained pressure of 500kPa).

So it is not a buoyancy question (as the working fluid is seawater just as the rest of the fluid in tank A is also seawater).

Rather it is a question about the energy required to cram one cubic metre per second of seawater into the base of a 30m high cylinder, a cylinder which is already full of seawater (note that one cubic metre per second of seawater is simultaneously being drawn out of the top of tank A by a pump assisted siphon so that mass in equals mass out).

Many thanks Vidar. I am getting closer to knowing the truth of the matter. if you have any thoughts whatever on how to calculate the energy in watts to cram this cubic metre of water per second into the base of cylinder A I would much appreciate them.

Low-Q

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Re: Easy Gravity wheel
« Reply #76 on: August 03, 2011, 06:12:08 PM »
1 Joule is the same as 1W in a period of one second. In one hour you get 3600 Joule of energy at 1W/s. So the term Watt is not energy alone. Watt is energy only over a given period of time. So Joule should be the term to use in any context. Hope that clears it up:). To lift 9810N 30 meters require approx 30kJ of energy. Or 30kW/s. You need 30kW to lift 1m^3 of water 30 meters up pr second.
« Last Edit: August 03, 2011, 06:36:04 PM by Low-Q »

quantumtangles

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Re: Easy Gravity wheel
« Reply #77 on: August 03, 2011, 09:03:50 PM »
@ LowQ

Many thanks.

If it really only takes 30kW to move one cubic metre of water up 30m, then my machine should work.

The same cubic metre of water falling 25m onto an impulse turbine generates about 200kW of electricity.

Here is the maths for electrical output.

Pwatts = h(25m) x g (9.81 m/s/s) x rho (1020kg/m3) x 0.9 (eff) x 1m3/s (flow)
= 225kW

So it seems we spend 30kW and get output of 225kW.

Interesting.

Pirate88179

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Re: Easy Gravity wheel
« Reply #78 on: August 03, 2011, 09:07:57 PM »
@ LowQ

Many thanks.

If it really only takes 30kW to move one cubic metre of water up 30m, then my machine should work.

The same cubic metre of water falling 25m onto an impulse turbine generates about 200kW of electricity.

Here is the maths for electrical output.

Pwatts = h(25m) x g (9.81 m/s/s) x rho (1020kg/m3) x 0.9 (eff) x 1m3/s (flow)
= 225kW

So it seems we spend 30kW and get output of 225kW.

Interesting.

I believe the input would be 30 kw per second.  I think the time has to be factored in.

Bill

Low-Q

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Re: Easy Gravity wheel
« Reply #79 on: August 03, 2011, 09:22:50 PM »
Just have in mind that Watt isn't energy unless you apply time. My guess is that the water falling 25 meters will for sure generate 225kW, but for how long? You talk about impulse turbine! That is exactly where the flaw is. You have a short burst of 225kW.

That is why I said that mesuring everything in Joules is better. Watt alone does not show how much energy there is because you miss time in the equation. Sorry. It will most probably not work. Do a recalculation in Joules. That water falling 25 meters creates less energy than it takes to push it up 30m.

This is btw a very good example why gravitywheels works inside the inventors mind, but does not work in real world. What do the inventor miss? Time!

Vidar

quantumtangles

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Re: Easy Gravity wheel
« Reply #80 on: August 03, 2011, 11:05:52 PM »
@ LowQ & Pirate

Many thanks. Of course you are correct about time.

However the equation for calculating power output in watts uses SI units.

Acceleration due to gravity (9.81 m/s/s) is time based (per second per second).

The mass flow rate of one cubic meter per second is precisely 1020 kg/s.

The only units that are not time related are height (or head) in metres (25m) and the unitless fraction representing turbine efficiency (always a figure between zero and one for any system, and always between 0.69 and 0.95 for Pelton impulse turbines). I somewhat generously allowed 0.9 for my turbine because larger turbines have better efficiency and a 0.9m diameter turbine is enormous by Pelton turbine standards).

So I completely agree that time is the critical factor. However I respectfully suggest that time is already 'factored in' if one uses SI units.

There is a way of cross checking the electrical output result obtained from the standard hydroelectric flow/head equation, and that is by using the Pmech equation. In order to deploy the Pmech equation, one must first calculate RPM, which in turn requires one to calculate teh force in Newtons that is applied to the buckets of the turbine.

Beginning with the force equation, F= m*a

F = 1020kg/s x 9.81 m/s/s
F = 10006.2 Newtons (enormous force).

But velocity will be greater than 9.81 m/s/s.

Bernoulli's equation gives us the velocity of the water jet applied to the turbine.

P = ½ r . V2

P = Pressure (401,536 Pa)
rho = density (1020kg/m3)
V = velocity (m/s) mystery value

401536 = ½ 1020 . V2
401536 = 510 . V2
V2 = 401,536 / 510
V = 28 m/s

Force of water Jet (Fjet)

1 m3/s of seawater (1020kg/m3) represents a flow rate of 1020kg/s.

F = 1020kg/s x 28 m/s/s
= 28560 Newtons

Fjet Momentum Change Calculation

Turbine speed may not exceed 50% of water jet speed

Vjet = 28 m/s
Vrunner = 14 m/s

Delta Mom = mass flow rate x Delta V
Delta Mom = mass flow rate x (Vjet – Vrunner)
Delta Mom = 1020kg/s x (28 m/s – 14 m/s)
Delta Mom = 14280 N

Fjet = 14280N.

Turbine RPM

The RPM figure for the turbine is based on runner velocity
of 14 m/s (half of water jet velocity*).

First we need to know the circumference of the turbine.

Diameter = 0.9m
radius = 0.45m
2.pi.r = 2.827433388m circumference

Vrunner = 14 m/s
RPS = 4.951487 revolutions per second x 60
= 297 RPM

Power Output

Applying (297 RPM) and also (Fjet = 14280 Newtons) to the Pmech equation:

Pmech = Fjet x Njet x pi x flowrate x RPM x 0.9 x 0.87m / 60
= 14280N x 1(jet) x pi x 1m3/s x 297RPM x 0.9(eff) x 0.87m / 60
= 173Kw

So far as I am aware, this is 173kW per second (it is not 173kW per year, or per hour or per month).

All of the units which lead to the result of the calculation in the Pmech equation are in seconds or are based on seconds (apart from revolutions per minute but you will note that I divided by 60 to obtain the result, thus converting the RPM into revolutions per second. The SI unit for angular velocity is radians per second but the equation I used amounts to precisely the same thing again because of the division by 60.

So I think the output is somewhere between 173 and 225 kW per second.

Please let me know your views on this.


* Note on turbine velocity being half of water jet velocity

x = vb / vj
 
 
x = ratio
vb = Cup velocity at pitch circle diameter of turbine
vj = Jet velocity
 
 
F = mb. vj . (1-x) (1+ z.cos g)
 
h = mb . (vj . vj) . x . (1-x) . (1+z.cosg) / ½ . mb (vj . vj)
 
P = F . vb = mb . vj . (1-x) . (1+z.cos g) . vb  = mb . vj . x . (1-x) . (1+z.cos g)
 
dh / dx = 2(1-2x). (1+z.cos g) = 0
 
x = 0.5
 
 
h = system efficiency as a unit-less fraction between zero and one
F = force of water on cups (N)
mb = mass flow rate into cup (kg/s)
vj = Jet velocity (m/s)
vb = runner tangential velocity at pitch circle diameter (m/s)
z = efficiency factor for flow in buckets (unit-less fraction between zero and 1)
g = angle of sides of cups
x = speed ratio of vj to vb

fletcher

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Re: Easy Gravity wheel
« Reply #81 on: August 03, 2011, 11:08:03 PM »
@ LowQ

Many thanks.

If it really only takes 30kW to move one cubic metre of water up 30m, then my machine should work.

The same cubic metre of water falling 25m onto an impulse turbine generates about 200kW of electricity.

Here is the maths for electrical output.

Pwatts = h(25m) x g (9.81 m/s/s) x rho (1020kg/m3) x 0.9 (eff) x 1m3/s (flow)
= 225kW

So it seems we spend 30kW and get output of 225kW.

Interesting.

You might like to try the math again ?

Vidar's logic always makes good sense to me - in this case I think he made a slight typo - understandable as it's just hit his radar & he is actually explaining thoughtfully why gravity is conservative for another example [applies here as well].

Keeping it simple:

Problem A: Energy required to raise 1 m^3 of water [say 1000kg] 30 meters N.B. we can think of this two ways [they are equivalent].

1. what is the volume [diameter of cylinder] of the vessel & therefore how much displacement volume occurs injecting 1 m^3 - this will see a rise in water level h - so we need to know the entire tanks mass x h to find the increase in Potential Energy in Joules.

2. we don't need to know area & volume & tank contents mass - all we need to know is that 1 m^3 will be raised 30 meters - this will see a rise in Potential Energy in Joules.

i.e. Pe = mgh => 1000kg x 9.80665N x 30 m = 294,200 Joules or 294 KJ's [per sec].

This is how much energy in Joules [Work Done] must be supplied to force the water to gain Potential Energy.

Problem B: What energy [Work Done] can we get out of the descending water mass.

Ans: Potential Energy Joules at 100% efficiency will convert to an equivalent Kinetic Energy in Joules i.e. energy of position is converted into an equivalent energy of motion.

Therefore Pe at start is mgh => 1000kg x 9.80665N x 25m = 245,200 Joules or 245 KJ's [per sec]

However, we all know that there are frictional losses in water ducting & flow systems, head loss etc - added to that it that engines & pumps are generally inefficient [electric engine perhaps 80% then deduct further efficiency for pump losses, lets say engine & pump 70% efficient].

Therefore Kinetic Energy from gravity is 245KJ's per sec x 70% = 171.5 KJ's per sec useable Output.

But we needed an Input of 294KJ's per sec.

Clearly this simple analysis shows IMO that there is a shortfall in Joules of energy per sec to make this theory viable & the reality self sustaining.

EDIT:

Quote from: quantumtangles
So I think the output is somewhere between 173 and 225 kW per second.

This seems to agree closely with the 171.5 to 245 KJ's per sec from the simple math approach.
« Last Edit: August 03, 2011, 11:39:36 PM by fletcher »

quantumtangles

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Re: Easy Gravity wheel
« Reply #82 on: August 04, 2011, 01:13:41 AM »
@ Fletcher

That is extremely helpful. I have been hoping for months for some feedback on the maths.

The problem I have tried to solve is that of making hydroelectricity viable using recirculating water (without the need for naturally occurring falling water).

I have known from the outset that lifting water was not viable. My initial sums indicated that 4.5 times more energy would be required to lift water up than could be derived from it falling onto a turbine.

But yes, 300kW is about right if the system were supernaturally efficient (for output of 173kW).

It makes perfect sense that input of at very least 294kW would be required to obtain electrical output of 173kw to 225kW (if water has to be lifted 30m upwards).

But what if air pressure were used to force recirculated water sideways back into the base of tank A? I thought in April 2011 that that might work and I am still unsure about it.

Put bluntly, the pressure at the base of tank A (arising from the height of the column of seawater and its density) is only about 4 atmospheres (about 401,000 Pascals).

Generally speaking, fluid always moves from areas of high pressure to areas of low pressure.

If all I have to do is ensure tailgate water in tank B (after striking the turbine) is pressurised to in excess of this, then the tailgate water will force its way through a one way valve sideways and back into tank A.

All that is needed is an air compressor powerful enough to pressurise the standing tailgate water in tank B to over 400kPa. Surely (I said to myself) such an air compressor would not need to consume 300kW or more?

This is what has haunted me since April 2011. The thought that an air compressor can indeed side step the need to lift water 30m upwards (by pushing water 'sideways' into the base of tank A at the same time as the pump assisted siphon draws an equal volume of water up and away from the surface of tank A).

Very interesting and helpful analysis. Many thanks but please let me know what you make of the sideways pressure idea.
« Last Edit: August 04, 2011, 01:41:17 AM by quantumtangles »

fletcher

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Re: Easy Gravity wheel
« Reply #83 on: August 04, 2011, 10:56:36 AM »
I think I remember coming across your thread about this idea - you were looking for a very select bunch of rather exclusively qualified people to comment IIRC - perhaps you could point me back to it & I could take a look at any drawings you have done, time willing - might save some time & also not hijack this thread.

Pushing water sideways [horizontally] thru a one-way valve takes the same amount of energy to enter the tank as lifting the water thru the height of the tank - this may not be what you meant & I don't want to go off on too many tangents - but ... hydraulic pressure is force x area - water pressure increases with depth in a linear manner [that'll be where you get your 4 atmosphere's at 30 meters etc] - liquids are also virtually incompressible so this linear relationship is constant & the density doesn't effectively change with depth - the next thing is that water pressure always acts normally [i.e. at right angles] to surfaces - IOW's it has no 'sheering moment' - that means that the water pressure at 30 meters will be the same on any surface at that depth - that is, whether it be the bottom of the tank, the side walls, or any other face or direction water pressure can apply itself to.

This may not be relevant to what you are proposing but it probably does illustrate that 'pushing water sideways' is no easier [Work Done in Joules] than entering the tank from above etc.


Low-Q

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Re: Easy Gravity wheel
« Reply #84 on: August 04, 2011, 11:31:39 AM »
Mass that is moving sideways, where gravity is the key force, you will in best case not gain or spend energy on moving a mass horizontally. In worst case, which will be the practical outcome, is to spend energy and loose energy output by any movement of mass in any direction.

The energy output of your system will regardless of complexity be at most 25/30 of the energy input. Even if there was a 30m fall, your output would not be higher than the input. If that was the case you shouldn't have to rise the water at all. Just take energy out of it while it stands still in the same spot all the time. How plausible is that?

Vidar

quantumtangles

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Re: Easy Gravity wheel
« Reply #85 on: August 04, 2011, 02:09:19 PM »
Thanks all. I hope this was on topic as the turbine is a wheel and gravity is involved.

The initial idea evolved with a series of posts under Recirculating Fluid Turbine Invention. Been driving me mad for months. Just want certainty.


fletcher

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Re: Easy Gravity wheel
« Reply #86 on: August 05, 2011, 12:02:43 AM »
I'm about to go off & start reading his thread/topic Vidar.

Before having read & understood it [so I don't really know yet] I think he is suggesting that compressing air & using that to drive a piston which pushes a volume of water into the bottom of a tank possibly takes less energy in Joules [Work Done] than other mechanical means - it would have to be way less as we've all pointed out - I think he is also suggesting an advantage may be gained from a siphon effect of some sort ?

Perhaps there is some advantage [small though it might be if it exists] in using compressed air in terms of the Carnot Cycle & Adiabatic heating & Isothermal Cooling of the tank of compressed air - I'll expand these thoughts on the other thread perhaps - any advantage might be able to be engineered into the situation quantum is proposing - who knows if it has a place, will need to read his logic first.

quantumtangles

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Re: Easy Gravity wheel
« Reply #87 on: August 05, 2011, 12:39:02 AM »
@ Vidar & Fletcher et al

I really appreciate your input. Most of all I really need your help.

Low-Q

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Re: Easy Gravity wheel
« Reply #88 on: August 05, 2011, 02:11:04 AM »
I'm about to go off & start reading his thread/topic Vidar.

Before having read & understood it [so I don't really know yet] I think he is suggesting that compressing air & using that to drive a piston which pushes a volume of water into the bottom of a tank possibly takes less energy in Joules [Work Done] than other mechanical means - it would have to be way less as we've all pointed out - I think he is also suggesting an advantage may be gained from a siphon effect of some sort ?

Perhaps there is some advantage [small though it might be if it exists] in using compressed air in terms of the Carnot Cycle & Adiabatic heating & Isothermal Cooling of the tank of compressed air - I'll expand these thoughts on the other thread perhaps - any advantage might be able to be engineered into the situation quantum is proposing - who knows if it has a place, will need to read his logic first.
Ofcourse, input of energy will make this work, but that energy is going through a process with loss, so one can only hope for an energy output which is less than the energy you put in. A siphon will definitely not provide the extra energy. Try to emty the ocean with a siphon. That is impossible because the ocean is the bottom level on this planet (except a few areas which is held dry with water pumps).

The gravity wheel, or other types of gravity "powered" devices is a dead end, but gravity can be handy in combination with other factors such as waterfalls, as waterfalls is created by the sun, ocean, mountains and gravity.

Vidar

quantumtangles

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Re: Easy Gravity wheel
« Reply #89 on: August 05, 2011, 02:42:22 AM »
@ Vidar

Your example of trying to empty the ocean with a siphon is somewhat different. Functional siphons require tubes (leading up to the siphon crest from the water source) that are shorter than the tubes leading down from the siphon crest to discharge the water.

Any attempt to place a siphon in the ocean would result in the 'long end' of the siphon being under water and therefore it could not possibly work.

By contrast (in the system under review) the long end of the siphon tube is below the level of the water intake without actually being 'under water'. Accordingly the siphon in question should work.

One of the key questions (re the proposed system as a whole) is whether it can work if 'unlimited' energy is available to circulate the water. I say this because initially I feared P1V1=P2V2 equalisation would prevent circulation regardless how much energy was expended.

If recirculation is possible with unlimited energy (which seems to be the position) the second key question is: how much energy is required.

Calculating power output is trivial to mathematicians and physicists. But calculating required power input is not so straightforward.

Certainly any idea of lifting water 30m is doomed.

Even though 'water lifting' is path dependent (methods of lifting may vary from the oestentatiously inefficient to the slightly more efficient) it certainly cannot work. Power expenditure will be in the order of 4.5 times power output.

But by pressurising tank B to more than the base of tank A, water should flow from B into the base of tank A.

But then the problem moves further round the circuit.

If tank B is under high pressure, how can the upper siphon work? How can low pressure water from A enter the high pressure environment of Tank B?

At first it seems tank B has to be both a high pressure area (to evacuate tailgate water after it has struck the turbine) and also a low pressure area (to allow the siphon to keep flowing).

The solution is to have two distinct pressure areas inside tank B.

The main body of tank B is always kept at higher pressure than the base of tank A. Fine. But there is also a second much smaller very high pressure area inside the exit nozzle of the siphon, that keeps it flowing because the air compressor drives siphon water out.

It all boils down to expenditure versus output (bearing in mind the turbine will have to fight against high air pressure (air of greater density) which will reduce RPM and therefore output.

I find the issues fascinating and much appreciate your views on the way to debunking or establishing the concept.

Kind regards and thanks,