Buoyancy-Based Power Generation - Full Disclosure

[quantumtangles]

I have chosen parameters for your invention to get some feel for the power output during the downward cycle of the B units.

Primary tank height = 30m
Diameter = 1m
Contains fresh water of density 1000 kg/m3
Volume = 23.56 m3

B units x 200 (= 100 B units on each side of the system connected by a loop of buoyant lightweight material).

Let us also assume the diameter of the circle represented by all the lightweight buoyant material holding the B units together in a chain has a circumference of 314.159m.

If the circumference of the chain holding the B units has a circumference of 314.159m, then:
C= 314.159m
C = 2 x pi x r
314.159 = 2 x pi x r
314.159 = 2 x  pi x r
314.159 = 6.28319 x r
r = 314.159 / 6.28319
radius = 50m
diameter = 100m

Downward cycle of the B units

We must assign a weight to each B unit if we are to calculate the force that will be applied by the B units in the all important downward part of the cycle.

Too much weight and we will slow the machine down on the upward buoyant cycle. Too little weight and the machine will not generate much energy on the downward part of the cycle.

If each B unit weighs 1 kg, then having 100 B units pulling downwards due to gravity on the right hand side of the primary tank at any one time will amount to a weight of 100kg moving downwards due to gravity.

Let us assume for the moment the system can move at enormous speed, and that all 100 of the B units can descend downwards to the base of the system within one second.

This would mean that one revolution of the entire chain of B units (including those moving upwards due to buoyancy) would take place every two seconds. I know this may be difficult to achieve in practice, but lets go for gold here.

If one complete revolution of the B units takes 2 seconds = 30 revolutions per minute (30 RPM)

On the downward part of the cycle, the following energy would be generated due to gravity:

Force = mass (kg/s) x acceleration m/s/s
F = m*a
F = 100kg/s x 9.81 m/s/s
F = 981 Newtons

Applying this force in Newtons and converting it to mechanical power output in watts for the downward part of the cycle, using an efficiency co-efficient of 0.85 to take B unit friction (and only B unit chain friction) into account:

P mech (watts) = Force (981N) x pi x 0.85 (eff) x 30rpm x 100m (diameter B chain)  / 60
Pmech watts = 130.98 kW

So this would be the power output at 30RPM if the system could rotate without having to fight through the primary tank (if it could maintain this mass flow circulation without needing buoyancy to lift the B units).

It should be noted that the assumptions made include an assumption of high speed in the B unit chain (30RPM is a high speed) and that in practice the overall angular velocity of the B units (as they ascend through the water in the primary tank) must be lower.

How much lower remains to be calculated, but that figure will affect our RPM and therefore our cyclical power output.

Now we need to do the maths relating to the upward cycle of the B units.

What shape are the B units? They should be shaped to reduce water resistance on the upward cycle (air resistance will be negligible in the downward cycle but we can calculate all of these things).

How can the B units enter the base of the primary tank without causing water loss? The base of the primary tank will also be under pressure.

Pressure at the base of the primary cylinder is as follows:
P = h x g x d
P = height(m) x gravity(9.81 m/s/s) x density (kg/m3)

P = 30m x 9.81 m/s/s x 1000 kg/m3
P = 294,300 Pascals Gauge

Pressure due to the height of the column of fresh water in the primary tank (294,300 Pa Gauge) + atmospheric pressure (101,325 Pa) = 395625 Pa Absolute, so it will rush outwards forcefully unless prevented from doing so. This is just under 4 atmospheres of pressure.

[quantumtangles]

A candidate solution (to the problem of how the B units can enter the base of the primary tank without water loss) is this.

An airlock at the base of the primary tank could be pressurised using a float activated air compressor to maintain a pressure of 500,000 Pascals inside the air lock. The compressor would only activate if water entered the air lock.

There would be two flaps in the airlock. An outer and inner gate.

The B units would enter the airlock from outside the primary tank. The outer flap (like a cat flap) would close when the B unit entered to keep air pressure as high as possible.

Then the inner flap of the airlock would open allowing the B unit to enter the high pressure water at the base of the primary cylinder.

The water would be unable to flow out into the airlock chamber because the pressure in the airlock would be maintained to 500,000 Pa by the air compressor.

The Abac Genesis air compressor consumes 11kW, but produces 800,000 Pa at a rate of about 0.02 m3/s.

0.02 m3/s is equivalent to 20 cubic litres per second, so the right configuration of B units just might work (though I have not done any buoyancy maths yet).

So the B units should be able to enter the primary tank through an airlock without water escaping.

The water at pressure of just under 400,000 Pa would not be able to escape from the primary tank into an airlock pressurised to 500,000 Pa. Simply not possible.

The output nozzle of the air compressor might also be directed to inject air into the B units if needed to give them extra buoyancy. Again I have not looked at buoyancy maths yet, but this is a candidate solution to the water loss problem.

[sm0ky2]

theres no real need to complicate the reload procedure....

instead of an airlock, we can just pressurize the smaller tank, prior to opening the inner door.

we can compare the energy it takes to pressurize the smaller tank,
with the energy it takes to pump the small ammount of water back to the top of the larger tank. Whichever is smaller, would be the most efficient method of dealing with water loss.

another tactic would be to allow for an air-gap at the top of the larger tank. Since both tanks are at a pressure-equibrillium, the smaller tank would be more resistant to further compression of the water, and the water displaced by the B-units would compress the air at the top of the large tank. this will further decrease the water lost in the process.

[quantumtangles]

I agree. It makes no difference whether you use an air lock or pressurise the smaller tank. There should be no water loss if this is done correctly. Both ideas amount to the same thing with different labels.

However, you probably have to leave the primary tank open to atmospheric pressure. If you seal it, the first problem is that the chain keeping the B units together will have to leave the tank through some sort of seal (I think this would be difficult if not impossible).

But my main objection to sealing the primary tank is that there is no reason to seal it, and pumping pressure into sealed tanks is a bad idea if there is no reason for it.

Leaving the primary tank open to atmospheric pressure would be fine. Lower pressure in this tank could not be a bad thing.

You mention pumping water up to the top of the primary tank from the base. This will consume franchise sized amounts of energy. The general rule with pumping water upwards is that you will spend 5 times as much energy pumping it uphill than you will ever get out of it at the bottom of the hill. So anything and I mean anything is better than pumping water uphill in an energy generator.

But this is all detail. My main concern is the speed in m/s at which the buoyant B units will ascend the primary tank.

The speed, when we do the maths, could be as low as 0.5 m/s, in which event the RPM will be low, and therefore the power output will be low.

How would you propose to try and increase the upward velocity of the B units to prevent low RPM?

The circuit of B units is a circuit, and thus what might have been high downward velocity due to gravity will be retarded by low upward velocity due to buoyancy and resistance. It will be upward velocity due to buoyancy and low velocity due to resistance.

This is the key to whether or not the system is viable. Upward speed of the B units against the resistance of the water.

It really is all about buoyancy velocity in terms of power output.

[quantumtangles]

You can remove the problem of slow upward velocity.

Timing how long it takes for the B units to rise in the primary tank, (let us say it takes the B units 5 times as long to rise as it does for them to fall when unconnected to the rising units) you can use a greater number of chains of B units, and connect only the downward moving belts of units to the driveshaft.

In other words, all the myriad chains of B units are constantly moving up and down, but only downward moving units are connected to the drive-shaft.

Upward moving units still float slowly up, but are disconnected from the drive-shaft when they are doing so. This prevents the floaters slowing down the fast movers (the gravity units).

By calculating the time difference between rising and falling B units (before they are connected together) simply add extra chains of B units and configure the system so that at least one set of B units is continuously falling (when connected to the drive shaft) at the same time as x number of slowly ascending B units are in various stages of 'getting to the top' of the cylinder and ready to fall.

If the time lag is x 5, then five times as many are needed. If the relative time lag is 7 units of time, then 7 times as many unit chains are needed etc.

There would be a continuous downward avalanche of B units.

Then you only get downward gravity action on the drive-shaft, and high RPM (provided a mechanism to disengage all ascending units 'from' and engage all descending units 'to' the drive-shaft can be engineered).

It would be like a gear box that engages only downward moving belts of B units.

Sometimes I amaze even myself 

PS I multiplied by gravity twice in my earlier Pmech equation having already used gravity to calculate force using Newton's equation. I don't think that's right. I must stop improvising equations. To solve for the error, divide 130kw by 9.81 m/s/s = 13.25kW Not sure any more. Too tired.