Buoyancy loop

[Harvey]

Very close - the only real difference in my idea was separate systems for generation and return - but I think your way is probably better. I don't think we need the side and bottom conveyors - those could be replaced with a slide for the shuttles to follow and there should be enough kinetic energy to slide it right into the airlock in one single drop. So when it goes over the edge, the airlock should be open and ready to receive it.

We could calculate the gravitational potential energy involved and see if a gain can be produced for a single bottle path.

When the airlock is flooded, the Gravitational Potential Energy (EGP) of the shuttle becomes negative according to the displacement weight of the water minus the EGP of the bottle in air at that height (using EGP = mgh where g = 9.81).

I just weighed a 2L bottle full of air @ 300 feet above sea level. Here it weighs 57g. So that is the mass of the bottle. The same bottle filled with tap water weighs 2182g. I am neglecting the thickness of the bottle wall which will undoubtedly displace some volume itself - so these calculations will be on the conservative side.

So our displacement weight will be 2182 - 57 = 2125g

The overall height of a 2L bottle is exactly 12 inches or 0.3048m.
The diameter is 4 inches or 0.1016m

Let us suppose we have a tank that is 2.5m deep. And let us suppose it stands on a support frame 0.5m tall. So all together, the overall height is 3m tall. When our shuttle (a 2L bottle as described) is at the top of the tank, floating on the water it will have a center height of 3.1016m minus that amount that is submerged which will equate to the 57g of water displaced. For ease of calculations and to be on the conservative side we can reduce the height to 3.1m

On the outside of the tank then, our shuttle would have a EGP of 0.057kg · 9.81m/s² · 3.1m = 1.73J. If the mechanical load of securing the shuttle in the airlock and closing the exterior gate is less than 1.73J, then we can use that energy to position the shuttle in the airlock and close the gate.

Next, we flood the airlock and any trapped air is allowed to escape. This can be done in a variety of ways - but it may be advantageous to do this with small ports, an inlet at the bottom and an outlet at the top until the airlock is fully flooded. This would ensure that maximum force is present on the shuttle when the interior gate is finally opened. If the submerged conveyor has cups built into to engage the shuttle, then the trapped air that is released could rise into those cups and aid in the transport of the conveyor.

After the airlock is fully flooded, we expect the potential energy now to be -2.125 · 9.81 · (0.6524 - 3.1) = 51.02J. This only a different way of saying that the EGP of the water that replaces the shuttle has that value. Where do we get the 0.6524 value from? That is the height off the ground, 0.5m plus the center of the bottle height in the airlock which is (0.3048 / 2). So that is the mass center of the shuttle at rest in the airlock. I used a negative in this case for the -2.125 to indicate that the force vector on the shuttle was away from the Earth. From these calculations, we would expect the shuttle to obtain a kinetic energy of 51.02J - 1.73J = 49.29J by the time it reaches the surface. If this were true, we would have plenty of energy left to open the interior gate and move the shuttle over the edge of the tank.

But there is a serious issue here - drag:Use of a Drag Coefficient to Calculate Drag Force due to Fluid Flow past an Immersed Solid

We would have to accurately model the drag on the shuttle as it moves through the water. This would no doubt lead us to apply methods of reducing that drag, which by far is greatly associated with waters adhesive characteristics. We would want our shuttle to be a slippery as possible in the fluid and the same with the submerged conveyor. Otherwise, the greatest amount of energy would be expended in friction with the water resulting in heating the water with nothing left for us to use.

This calculation used EGP as the means of evaluating the energy involved - but the more common approach would be to use fluid dynamics and pressures. It would be interesting to compare the two results and see if mistakes have been made in the above calculations and if so where.

[Harvey]

Hi Harvey

I am stuck with the time. How long will take the weight (500 Kg) to pressure water by gravity at height of 5m in tank with 0,2827 m2 of area?
In the Hyper Physics analysis Pressure Pressure I assume that the displacement from the hydraulic press is per one second, and takes 1 hour to reach the bottom.
To long, isn´t?
I am assuming that the work done in the curves, on top and bottom to connect to the lift side is for drag, other losses and to push the piston in place on the top.
I would like to hear your opinion on it?
Is it worth to apply it at GE contest?
Thank you
David

[ATTACH]6437[/ATTACH]

Hi David,

Gravity accelerates a mass at 9.81 meters per second squared regardless of the weight of the mass.

d = di + vi Δt + ½ a(Δt)²

If our initial distance (di) and our initial velocity (Vi) equal zero we can reduce this to distance equals one half acceleration multiplied by delta time squared.

d = ½ a(Δt)²

Transposed, we get (Δt)² = d / 0.5 * a

Moving the square to the right side we get:

t in seconds = √(5m / (0.5 * 9.8)) = 1.01 Seconds.

So, provided there are no negative forces working against gravity here, that is the time it will take for the 500kg weight to drop.

So what do we do if there are negative forces involved like fluid drag or resistance? In that case you must integrate that force differential over time to determine the fall rate. For example, if there is no where for the fluid to move at the initial distance, then the fluid is immediately pressurized and is in a hydraulic condition. In this state the negative force equals the positive force and the motion is zero. If now we immediately drop the negative force to zero, we can expect the full volume of 5m x 0.2827m² to be moved in 1.01 seconds. If instead there is a restriction on that flow rate, and the negative pressure does not drop to zero instantly, but instead transitions from the holding force to zero over some period of time, then that is your fall rate, that period of time. If you are attempting to determine that period of time, then you must evaluate the volumes involved and the flow rate of that liquid to determine the period.

For example, if you have 1.4135m³ of fluid and you wish to express that through a pipe that allows a flow rate of 0.70675 cubic meters per second, then you will expect the weight to fall in 2 seconds rather than 1.01 seconds because the flow rate impedes the gravitational acceleration. It is like having your foot on the brake when you press the accelerator pedal in a car. The force is there, but it is canceled by the negative force so the acceleration does not occur or a different acceleration occurs instead which is related to the force differentials.

[Matos de Matos]

If the submerged conveyor has cups built into to engage the shuttle, then the trapped air that is released could rise into those cups and aid in the transport of the conveyor.

After the airlock is fully flooded, we expect the potential energy now to be -2.125 · 9.81 · (0.6524 - 3.1) = 51.02J. This only a different way of saying that the [/SIZE][/SIZE]EGP[SIZE=1][SIZE=2] of the water that replaces the shuttle has that value. Where do we get the 0.6524 value from? That is the height off the ground, 0.5m plus the center of the bottle height in the airlock which is (0.3048 / 2). So that is the mass center of the shuttle at rest in the airlock. I used a negative in this case for the -2.125 to indicate that the force vector on the shuttle was away from the Earth. From these calculations, we would expect the shuttle to obtain a kinetic energy of 51.02J - 1.73J = 49.29J by the time it reaches the surface. If this were true, we would have plenty of energy left to open the interior gate and move the shuttle over the edge of the tank.

:

Hi Harvey:

Do you think that we can use the extra force of the air bubbling up in the cups to help pumping the water in the air lock ( > 0.002182m3) back to the tank?
We will need ≥ 53.459 J to pump it, and we have only 49.29 J.

David

[Harvey]

Hi Harvey:

Do you think that we can use the extra force of the air bubbling up in the cups to help pumping the water in the air lock ( > 0.002182m3) back to the tank?
We will need ≥ 53.459 J to pump it, and we have only 49.29 J.

David

You are right. I was imagining that only the water surrounding the shuttle would be drained when the airlock opened, but the airlock must completely fill to evacuate the shuttle and so that water will be dropped. Also the volume will be greater than the shuttle because it is the entire airlock volume . I just didn't think that through well enough

So there is no real advantage then.

Is there a way we could use wicks and capillary action to lift the water? Perhaps a series of trays, each with an inverted 'J' wick up to the next one


Yes the air would help move the conveyor, but I don't see how it can offset the total weight of the water even with the shuttles help. I'm afraid this concept too is flawed - if we are going to lose that much water, we may as well run it over a water wheel

Sorry for the diversion.

Perhaps there is a way to use city water pressure and a resonant water oscillation effect without actually flowing the water. If we keep the air at just the right temp it will compress and expand in the pipe and resonate. The temp can be ambient perhaps. I've read up on this, but have not studied it in full detail - but I have seen this happen first hand for long periods until the energy is fully dissipated. And in that case, the energy comes from the environment.

Just throwing out ideas.