Mathematical Analysis of an Ideal ZED

[mondrasek]

I understand what you are doing, and why you are doing it, and I approve. I am not trying to discourage you.

What I am doing is pointing out that you are leaving honest Wayne Travis a huge hole to wriggle out of.

Just like you tell me that the Cartesian Diver is irrelevant to the real Zed (it isn't) and that the spring-loaded automatic bollard doesn't do what you have been showing the single Zed does (it does)..... honest Wayne Travis will look at your completed analysis, and will respond in one of two ways. Or maybe both.

If your "OU" numbers hold up in the correctly-modelled spreadsheet (they won't) he will say "I told you so all along" and if your numbers do NOT hold up once you've modelled correctly and re-inserted reality in the form of masses, compressibility, viscosity and so on.... he will simply say, "Thanks fellas... but the real Zed, which is now horizontal and which we call the Rotary Taz.... doesn't work that way at all, you have left out some important steps and it's really too bad you aren't worth your salt. I'll pray for you anyway."

Cool.  I'm glad you approve of what we're doing.

I don't think I'm doing any of this with the expressed purpose of only helping Mr. Wayne.  Of course, if it is all true, and Mr. Wayne has been trying all along to help us all see the reality of his discovery, I would be honored to shake his hand.

I think I am doing it only to know the truth.  And therefore stop the open hostility in the forums on the subject.  The math either proves it or disproves it.  And I have seen no indication that the math has been rigorously performed so far.  So, why not do it?  I thank you and MarkE for your contributions to this effort.

And Minnie is right.  It is a lot of work, isn't it?

Finally, if anything extraordinary is found, we are not done, are we?  I think you have an excellent Feynman quote to present again?

Cheers,

M.

[mrwayne]

Hello Mondrasek,

I just caught up on the reading ..... I logged in first - that left three posts per page lol

I saw your last Post.

......................

Let me tell you what I will do when the Math is done: Nothing.

I have already discovered and met some pretty cool people.

I am very honored by you and the others work, and I look forward to meeting each of you in person.

.......

Sybil concerns me not.

[MarkE]

MarkE what you said in post 766 was this:

4.653 is the single riser.  I was answering your question with respect to the three riser.

This misdirected me to look at the 3-layer.  You did not say it was a transcription error or typo.  You said you were now talking about the 3-layer.

As for the rest of the Energies due to columns of water that you then refer to, I did not see them on your graphics or my 3-layer calcs that I thought you had directed me to move over to.  But I also did check my calcs for the no-pod, single layer and also did not find them.  I see that you are able to point out these values on your graphic, but again, we have a problem.  The Energies due to the columns of water do not match my own calcs for the no-pod, single riser Analysis.  My calcs and yours do match exactly for the 3-layer, so I am confident we are both calculating those the correct way.  But I do not get your values at all for the no-pod, single layer.  I have checked them several times now.  Could you please double check your calcs for the Energy in the system due to just the water columns for State 1?  The value I get is ~.879mJ.

ETA: Ran this through the spreadsheet and made some very slight corrections to the numbers and the drawing.
Here I check the work and make any needed corrections:

E = integral (F*ds).
K₁ = pi/4*pWater*G₀
The energy in the water columns at the end of State 1 is:
E_ST1 = 0.5*K₁*(116+100)*32.5² = 877uJ.  The 1.117mJ on the graphic neglected to multiply the diameters by pi/4.
it also neglected the component 0.5*K₁*(108)*1² but that amounts to a miniscule 0.4uJ, which changes the total to 878uJ.

State 2
The energy in the water columns at the end of State 2 is:
E_ST2 = 0.5*K₁*((116*59.293²) + (108*1²) + (100*30.157²) + (84*37²)
E_ST2 = K₁*261,559
E_ST2 = 2.011mJ
E_{ADDEDST1_ST2} = 1.133mJ

Lifting force end of State 2:
F_{up_st2} = Pressure * Area
P_{up_st2} = pWater*G₀*(59.293 - 1.420) = 566.5Pa
F_{up_st2} = 566.5Pa*pi/4*(0.028m)² = 0.3488N

Lifting rate State 2 to State 3:
H_{CHANGE_OUTER_per_mm_lift} = 28²/116 = 6.759mm/mm
H_{CHANGE_INNER_per_mm_lift} = (28²-108)/100 = 6.760mm/mm
Head_{CHANGE_per_mm_lift}= 13.519mm/mm
P_{CHANGE_per_mm_lift} = pWater*G₀*13.519mm/mm = -132.3Pa/mm
F_{CHANGE_per_mm_lift} = -132.3Pa/mm*pi/4*(0.028m)² = -81.485N/m
Lift distance = F_{up_st2}/F_{CHANGE_per_mm_lift} = 4.281mm

State 3
E_{ST3_INTERNAL} = 0.5*K₁*((116*30.353²) + (108*(4.282+1)²) + (100*30.366²) + (84*37²))
E_{ST3_INTERNAL} = K₁*158,545
E_{ST3_INTERNAL} = 1.219mJ
E_{ST2_to_ST3_INTERNAL} = 2.011mJ - 1.219mJ = 0.792mJ
E_{ST3_EXTERNAL} = F_{up_st2}*Lift distance + 0.5*F_{CHANGE_per_mm_lift}*Lift distance²
E_{ST3_EXTERNAL} = 0.3488N*4.282mm + 0.5*-81.485N/m*²*(4.282mm)²
E_{ADDEDST1_ST2} = 0.747mJ
E_{ST3_TOTAL} = 1.967mJ

Even with a perfectly idealized load, the process remains lossy:  0.747mJ theoretically recovered from 0.792mJ internal energy change.  So, why with a perfectly matched load did we fail to transfer 100% of the internal energy change?  The answer is still:  N*(X/N)² < X² for all N > 1.0.  The matched load reduce N close to 1.0 but cannot get there because of the fluid volume that increases under the riser wall.  If the riser wall is made very, very thin, then the efficiency can approach that of an ordinary spring.

[MarkE]

In the real ZED the production is the lift from the risers, in the version in the videos that lift is resisted by a hydraulic ram that charges up the external accumulator.

So the internal forces raise the risers, the risers moving up move the production ram and compress the fluid in the ram and that fluid passes into the accumulator and gets stored, from there it is distributed to the flow assist rams and they apply there force against the lever connecting the two bags together.

In the "ideal ZED" this part is not being considered.

First anything that is not external output does not constitute "production" in the ordinary use of the term.  Since the risers cycle internally within the ZED they do not constitute any energy delivered to the outside world.

Each of these subprocesses you mention is under unity.  The product of any set of numbers where all the number are less than one is less than the smallest of the numbers.  IOW, each of the processes imposes additional loss upon the system.

The "ideal ZED" has less loss than any 'real' ZED.  The "ideal ZED" is fundamentally an under unity device.  Therefore the less efficient 'real ZED' is also underunity.  QED.

[mondrasek]

Here I check the work and make any needed corrections:

E = integral (F*ds).
K₁ = pi/4*pWater*G₀
The energy in the water columns at the end of State 1 is:
E_ST1 = 0.5*K₁*(116+100)*32.5² = 877uJ.  The 1.117mJ on the graphic neglected to multiply the diameters by pi/4.
it also neglected the component 0.5*K₁*(108)*1² but that amounts to a miniscule 0.4uJ, which changes the total to 878uJ.

State 2
The energy in the water columns at the end of State 2 is:
E_ST2 = 0.5*K₁*((116*59.293²) + (108*1²) + (100*30.157²) + (84*37²)
E_ST2 = K₁*261,559
E_ST2 = 2.011mJ
E_{ADDEDST1_ST2} = 1.133mJ

Lifting force end of State 2:
F_{up_st2} = Pressure * Area
P_{up_st2} = pWater*G₀*(59.293 - 1.420) = 566.5Pa
F_{up_st2} = 566.5Pa*pi/4*(0.028m)² = 0.3488N

Lifting rate State 2 to State 3:
H_{CHANGE_OUTER_per_mm_lift} = 28²/116 = 6.759mm/mm
H_{CHANGE_INNER_per_mm_lift} = (28²-108)/100 = 6.760mm/mm
Head_{CHANGE_per_mm_lift}= 13.519mm/mm
P_{CHANGE_per_mm_lift} = pWater*G₀*13.519mm/mm = -132.3Pa/mm
F_{CHANGE_per_mm_lift} = -132.3Pa/mm*pi/4*(0.028m)² = -81.485N/m
Lift distance = F_{up_st2}/F_{CHANGE_per_mm_lift}4.282mm

State 3
E_{ST3_INTERNAL} = 0.5*K₁*((116*30.353²) + (108*(4.282+1)²) + (100*30.366²) + (84*37²))
E_{ST3_INTERNAL} = K₁*158,545
E_{ST3_INTERNAL} = 1.219mJ
E_{ST2_to_ST3_INTERNAL} = 2.011mJ - 1.219mJ = 0.792mJ
E_{ST3_EXTERNAL} = F_{up_st2}*Lift distance + 0.5*F_{CHANGE_per_mm_lift}*Lift distance²
E_{ST3_EXTERNAL} = 0.3488N*4.282mm 0.5*-81.485N/m*²*(4.282mm)²
E_{ADDEDST1_ST2} = 0.747mJ
E_{ST3_TOTAL} = 1.967mJ

The process remains lossy:  0.747mJ theoretically recovered from 0.792mJ internal energy change.  So, why with a perfectly matched load did we fail to transfer 100% of the internal energy?  The answer is still:  N*(X/N)² < X² for all N > 1.0.  The matched load reduce N close to 1.0 but cannot get there because of the fluid volume that increases under the riser wall.  If the riser wall is made very, very thin, then the efficiency can approach that of an ordinary spring.

Thank you MarkE.  And I hope you have a peaceful weekend!  Please do and let's reconvene next week?

BTW, about your ID vs. OD argument on how to properly calculate the buoyant Force on a riser.  It is, in fact, the OD that is needed in your calculation.  Since the riser is ASSUMED weightless by having a Specific Gravity = 0, the riser material is, in effect, just air.  So the riser, and the air inside, is just a big bubble.  Ergo the OD needs to be applied to your buoyant Force calculations.

Gonna sign off now.  Y'all have a wonderful evening.

Cheers,

M.