Mathematical Analysis of an Ideal ZED

[mondrasek]

What air is being used that is 'incompressible??

Mags

Mags, we are working through an "Ideal case" Analysis.  So we are neglecting the normal losses dues to friction, compressibility, and surface tension, etc., for now. 

If the Analysis fails to show any possible gains under "Ideal" conditions, there is no way it can ever show gains once those losses are included.

If the Analysis DOES show possible gains under "Ideal" conditions, then we can move on to see if Engineering can minimize the "real world" (less than Ideal) losses to the extent that the gains are not completely lost.

M.

[MarkE]

No movement no energy.

That's a relief.  So what energy were you referring to when you claimed that I did not account for energy of the restraints?

Do you contend that energy causes movement?

When did I say anything like that?

Where are your buoyant lift forces in your spreadsheet, where is the buoyant lift from the pod in your spreadsheet?

The water masses are all in the spreadsheet.

Do you contend that a partially sunk float will not move up?

Mondrasek stipulated as such.  See State 1.

It is looking to me that what you did in your analysis is a uniform spread of the water volume countered by the weight of the water columns.

Water countered itself???

Archimedes is dead in the water I think.

Then you aren't thinking very clearly.  Archimedes' Principle dictates the behavior of the always less efficient than a common brick ZED.

[MarkE]

It was not, that was in response to you saying there is no energy to be had from the risers and pod lifting, but there can be if you apply the restraint at a 99.9% value required.  That is just common sense MarkE,, why just waste what you have when you do not need to.

A restraint prevents motion.  It does not create energy.

[MarkE]

The incompressible air has an ASSUMED Specific Gravity = 0.

In fact, using a compressible fluid such as real air causes more losses due to the relationships described in the Ideal Gas Law, PV=nRT.  The change in V due to P does change the T, by creating heat when compressed.  It is a loss mechanism only and cannot be completely recovered.  So using two incompressible liquids leads to a better performance. 

That's right, and that's part of the reason that your ZED is an "ideal ZED".  Using air as in a real ZED is less efficient.

And the bigger the difference between those two fluid's Specific Gravity values, the better.

Nope, the efficiency improves as the SG's converge.  When the SG's are the same, IE the "air" is replaced with water, the losses improve a lot.

I agree that the outside air pressure/volume does not need to be considered.  Except that the system is open to the outside air and so air can enter and exit the system freely as Pressure and Volume changes inside the system require to satisfy the Volume constraint of each internal fluid to remain constant.

Yes, the "air" is just a fluid that moves back and forth to fill out the volume changes.

[MarkE]

Of course.  There is no head difference between the ID and OD surface on any riser.  The pod has no water in contact with it at all.  So all risers and the pod are being acted on by zero buoyant Forces.  Also, the sum of the buoyant Forces on all the risers and the pod is exactly zero.  There are zero Forces acting on the system and therefore zero motion would occur.

BTW, this is not a "stipulation."  This is a physical fact derived from the geometry and the assumption of incompressible fluids.

I withdraw my objection.  Yes, State 1 is in equilibrium naturally.  I will have to rework the problem on that basis.  It will not however materially change the outcome:  Preparing State 1 is lossy.  State 1 to State 2 is ideally conservative, and State 2 to State 3 is always lossy.