Mathematical Analysis of an Ideal ZED

[mondrasek]

No.  You are mixing circumstances of compressible and non-compressible substances.  Work is always the integral of F*ds.  If we take a capsule of fluid and subject it to 1psi or a million psi we have not done any work on that fluid.  If we apply pressure against a cross section of fluid through a distance, then we do work.  When we lift columns of fluid we can obtain the work performed and stored by solving the F*ds integral which will work out for a single column to:  E=0.5*total_weight*height = pave*volume = 0.5*density*volume*height = 0.5*density*area*height². 

The energy is not stored in compression of the fluid for the simple reason that the fluid is incompressible.  The energy is stored in the gravitational potential of the raised mass.  Larry  asserted that raising some cross-section by 1' to end up with the 1+2+4 configuration "cost only 1/3" of some other configuration.  But it doesn't.  The force went from 0 to 3X what it would have raising an isolated column by 1'.  Identically, the amount of work performed was 3X that required to raise an isolated column by 1'.  The force and the energy both scaled by 3X versus the isolated column.  Had we done the exercise totally emptying the middle column, then the force would have gone from zero to 9X over a 3X stroke.  Kf would still be 3*pWater*area, and the integral would be:  0.5*3*pWater*area*(3²-0) = 27*0.5*pWater*area, IE 27X the energy of raising an isolated column by 1' and 3X the energy of raising an isolated column by 3'.

Have you met Fletcher?  (sorry, but he is AFK) for a bit.  But I think he would like this exchange immensely!

And so do I.  Thank you for all your input and hard work.  The introduction of the maths you have been presenting to this topic has been a breath of fresh air to say the least!

M.

PS.  I did not actually read through and "digest" your post, and I apologize.  I'm otherwise preoccupied.  But I do plan to check it out shortly (hopefully in the morning).  You have taught me alot so far, and again, I thank you.  I do appreciate your considerate contributions to this forum!

[LarryC]

Darn, Gif display problem.

[minnie]

   Hi,
      I've got a feeling if you used liquids with different sg's instead of air the thing would
   work just like a hydraulic jack. Wouldn't be much use because pressure would be so
   small and then it would vent.
      mrwayne  seems to have gone quite quiet all of a sudden.
   In industrial applications hydraulic pumps can deliver pressures of up to 20,000 psi
   which seems very far removed from what we're messing with here!
                        John.

[mondrasek]

      I've got a feeling if you used liquids with different sg's instead of air the thing would
   work just like a hydraulic jack.

Minnie, that is exactly the gist of the analysis I performed.  We are assuming the air is incompressible, so we are, in effect, using two liquids with different SG's.  I then tried to see if the ZED performed exactly like a hydraulic jack by inputting a known amount of energy in the form of a specific volume of fluid over a specific pressure range.  Then I checked to see if the ZED (jack) would rise by the appropriate amount to provide an equal energy output in the form of the outer riser (jack piston) rising a balancing calculated volume for the specific pressure range that was developed due to the buoyant forces the input created.  It did not act just like a jack (a simple hydraulic cylinder).  The 2-layer ZED (jack) would fail to stroke far enough, so under unity.  The 3-layer ZED (jack) would stroke further than predicted, so over unity.

[LarryC]

Larry, I've looked at the latest spreadsheet.  There is still a good deal of work to do here.  Please refer to the drawing below:

We need to either insure that the starting and ending energy states are identical, or else account for the stored energy in each state.    No matter what, we do need to calculate the work added and going from each state to the next.  Knowing the stored energy at all states provides a good sanity check.  Please be aware that each time water equalizes from a taller single column to two or more lower columns that we lose stored energy.  The drawing includes formulas for calculating stored energy under the assumptions previously stated:

20C
G0=9.80665m/s/s
zero thickness walls
25" diameter pod
26" diameter pod chamber
51 circular inch riser gap and riser head areas

MarkE,


Attached spreadsheet with your Stored energy Ft Lbs request.