Mathematical Analysis of an Ideal ZED

[minnie]

    Webby,
             Every aspect of the ideal ZED is factual. I can't see any way for there to be an
    anomaly if the math is done correctly.
         I'm sure Mark will see the double n's and double ,,.
     As for Travis, look for yourself. I was quite shocked with what I found.
                   John.

[MarkE]

I had an interesting thought this morning.  Referring back to the State 1X diagram that MarkE presented to be in a true state of rest, unlike the State 1 diagram.  State 1X would occur if the Ideal ZED in State 1 is unrestrained, since there are buoyant forces from the displaced water due to the riser walls extending below the water level.

In order to come to rest at State 1X the risers all lift and cause each to have a slight negative buoyancy due to a negative water head.  The sum of the positive buoyancies due to water displaced by the riser walls and the negative buoyancies due to the negative water displaced by the negative water head is, of course, zero.  But it also creates a negative pressure (vacuum) in the pod chamber.  This means that the water charge needed to go from State 1X to State 2 would begin at a negative pressure, and not zero as I had intended when designing this model for Analysis.  But is this significant at all?

This effects the initial force and the amount of energy that it takes to get to State 2.  The stored internal energy at State 2 is unaffected.  If one wanted to cycle between State 1 and State 2 or State 1X and State 2, then it would matter to those cycles.

[MarkE]

I posted an answer to the part that was needed, Mark.  That answer was the OD of the main body plus the displaced volume of the riser skirts that went below the water line.  That is a 9mm distance.

I would like to know when you understood that the system works.

LOL, you are a fun guy Tom.  There were five questions.  Did you answer one?  Who knows, because you threw out a number without saying which question you were trying to answer.

If you are now saying that you were calculating based on the OD of the cylinder, then you still have your physics wrong.

[minnie]

     Yes Webby,
                  the lift is LOSEY,  I think you get some left over because you restrict the upward
   movement.
           Carefully explain why you think there should be an anomaly in the math.
    The ZED is very lossy as you know, therefore if you end up with 5% excess any machine
   would have to be the size of a mountain to get anything useful when you consider the
   losses involved in converting your 5% into something useful.
        If you lift a weight with a bottle jack and take the weight off you haven't got much left!
     You rattled me years ago when you reckoned I could run my house on a 5kva genset.
    My electric shower alone is 9 kW. Yeah the genset might keep the fridges running but
     that would be about it!
                                John.

[MarkE]

ETA, the "π" here is pi, the equation

.25π×44^2×(32.5−10)+  (.25π×(44^2−42^2)×(10−1)) = 35427.740354532

is correct.

35427.740354532×.000001×9.789 = 0.34680215N

You calculated the correct displaced volume.

The dimensions that matter:

44mm OD
42mm ID
32.5mm outside meniscus
10.0mm inside meniscus
1mm height bottom of the cylinder wall

One can either divide this into two cylinders one inside the other:  The submerged portion of the riser wall and the displaced cylinder under the cylinder ID.  Or, one can divide it vertically as you have done, the riser OD down to the inner meniscus, and then the riser ring below the meniscus:

Concentrically:
V_CIR = (32.5-1)*(OD² - ID²)  + (32.5-10)*ID² = 31.5*OD² - 9*ID² 

Or vertically:
V_CIR  = (32.5-10)*OD²  + (10-1)*(OD² - ID²) = 31.5*OD² - 9*ID² 

W = V_CIR*pi/4*pWater*G₀

What all this does is blows your BS spreadsheet manipulations completely out of the water.  Because if you follow the R4 spreadsheet, you will see that it correctly accounts for the cylinder volumes working them concentrically.