Easy Gravity wheel

[quantumtangles]

@ 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 â€" Vrunner)
Delta Mom = 1020kg/s x (28 m/s â€" 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]

@ 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:

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.

[quantumtangles]

@ 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.

[fletcher]

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]

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