Hydro Differential pressure exchange over unity system.

[neptune]

Referring to post 1376 on page 92 by LarryC. People have expressed doubts as to whether we can get the same head pressure from a number of layers in the diagram on the right as we would from a single tall column of water, and there has been talk of using submersible pressure gauges etc. Here is a very simple experiment to prove the point. Imagine we have the apparatus set up as in the right hand diagram. Now with the water levels as shown we extract the piston, and throw it away. The tube in which the piston fitted is now extended upwards to a hight of several feet. Now instead of depressing a piston with a weight, we pour water from a jug into the extended piston tube. We keep pouring until the water levels in the various U tubes move to the same position as happened when the piston was pushed down by the weight. At this stage, we find that the hight of the water column in the piston tube, is equal to the sum of the heights of the vertical columns in the U tubes. To me, this proves that the pressure of the column in the piston tube is equal to the pressure of the columns in the U tubes.

[mondrasek]

Referring to post 1376 on page 92 by LarryC. People have expressed doubts as to whether we can get the same head pressure from a number of layers in the diagram on the right as we would from a single tall column of water, and there has been talk of using submersible pressure gauges etc. Here is a very simple experiment to prove the point. Imagine we have the apparatus set up as in the right hand diagram. Now with the water levels as shown we extract the piston, and throw it away. The tube in which the piston fitted is now extended upwards to a hight of several feet. Now instead of depressing a piston with a weight, we pour water from a jug into the extended piston tube. We keep pouring until the water levels in the various U tubes move to the same position as happened when the piston was pushed down by the weight. At this stage, we find that the hight of the water column in the piston tube, is equal to the sum of the heights of the vertical columns in the U tubes. To me, this proves that the pressure of the column in the piston tube is equal to the pressure of the columns in the U tubes.

Simple and brilliant, @neptune!  Now, can anyone send TK one more polycarbonate fluorescent light tube protector tube?

M.

[see3d]

When you revise the sim, could you _please_ also account for the internal work, so that the total work "output" includes this internal work and its storage in the compression of the air and the lifted mass of the water? I can live with water density in pounds per cubic inch and hidden counterbalances and an unrealistically low-mass riser  ... but not accounting for _all_ the work seems ... well, it seems like an omission.

At the top of the cycle when the lifted weight is as high as it is going to get, there is energy (work) stored in the system still, and this work isn't fully accounted for, and it is necessary to reset the system. If you released the internal pressure at this point... the weight would fall... and then if you resealed the system... it would have less air/water in it than you started the cycle with, showing that this stored internal work is indeed necessary to reset the system to the starting point of the cycle, and if it is bled off in any way... it will have to be replaced by additional outside work to bring the system back to the starting state.

TK,

Thank you for your response.

I will try to make it more clear what the Riser counterbalance is about in a picture for the next release.  Just imagine a rope up from the Riser up to and over a pulley with a counterbalance weight suspended from the other end.  It can also be visualized as a lever in reverse of the water counterbalance weight that I did show.  However, there is a buoyant force at the bottom of the Riser wall at the depth of the initial H1 and H2.  If in the design, this is large enough, it could offset the weight of the Riser, but I doubt it would in a 1 layer design.  Counterbalancing is is a practical problem with a simple solution.

The stored potential energy is shown in the graph when the WorkOut/WorkIn < 1.  I also have another graph that I did not think would be interesting that shows the stored potential energy added to the WorkOut/WorkIn. 

I considered it a practical waste to overdrive the input beyond lifting the weight to the limit stop.  Just as it is a waste to have the Riser weight counterbalanced to more than zero.  In a physical machine, I would try to keep these down to about 1% at each end of the stroke.  I will will try to better show the balance of all the unused potential energy.

In my model, there is no provision for using any excess potential energy beyond lifting the load weight.  It is all returned on the down stroke.  Perhaps that is easier to understand if I explain how I envision a closed loop:

On a ZED in the down initialized state, add a 1 pound output load weight and a 1 pound input weight.  After the output load weight is just raised to the top stop, remove it and the input load weight.  The ZED will return to the initial down state.  If the input weight has dropped less than the output load weight has risen, then the cycle can repeat forever.

Please keep the good questions coming.

Now, back to work fixing my calculations.

~Dennis

[neptune]

That part's clear enough.
The part I don't understand is what he does with all the extra hydraulic fluid he's generating. You inject 15 units to one Zed and get 30 back. He's said that many times, and that is supposed to be a full cycle for that Zed, right?So you take 15 volume units of that hydraulic fluid and inject it to the other Zed and it returns 30 back. Now you have 30 cu in extra hydraulic fluid in your accumulator. Pretty soon you are going to have to start selling that off, or the barrels are going to be all over the back yard.

Or perhaps the ratio of 30/15 is for the two Zeds together. But if each full cycle takes 15 out of the accumulator and puts 30 back in... you still have the same problem: too much hydraulic fluid.

Yes, you have too much hydraulic fluid. So the excess fluid is then used to drive a hydraulic motor, which , in turn drives an alternator, feeding a load. That is where the "free energy" comes from.

[LarryC]

Attached is the third picture from my previous post with additional Psi pointer information.

This will be much harder to understand where the force and Psi are coming from than the 3 U water levels. If anybody has questions and they don't want to post here, please PM me.

Regards, Larry