Mathematical Analysis of an Ideal ZED

[MarkE]

Yes, I contend that there is an error in the verbiage in the graphic.  The error is in the statement on line #3 which reads:  As water levels equalize, energy is lost.

You have failed to correctly describe the Energy that is "lost" by this Ideal 3-layer ZED system when transitioning from State 2 to State 3.  The Energy does not need to be simply "lost."  That Energy can be collected.

Only if you calculate the Energy that is (described incorrectly as "lost") necessary to cause the calculated "lift" of the outer ZED riser will the Energy balance calculations be correct.

As you described the device the energy absolutely is lost.  Look at your own drawings.  You later added a clever yet absurd mechanism to reclaim almost all of that lost energy.  The mechanism is for all intent and purpose an unfolded mirror of the original device.  I evaluated your proposed device as you presented it.  You might recall that I asked you many times to present your complete analysis and you refused each time, stating that you felt you had provided enough information and would not be bothered. 

[MarkE]

MarkE, what amount of Energy is necessary for the SUT to rise 2.590mm?

As you set it up: 

>5.51mJ to get to State 2, as 2.10mJ added going from State 1 to State 2 and an initial stored energy of 3.41J in the water columns in State 1.  It is >5.51mJ as the energy consumed forcing out the "air" that must be vented to establish State 1 is not accounted.

Here is your post #27, including your erroneous evaluation, and erroneous application of Boyle's Law to incompressible fluids, along with the accompanying graphic you supplied.  As we eventually learned, you erroneously treated State 1 as a neutral condition and based on that faulty assumption reasoned that any energy added going from State 1 to State 2 would show up as a difference in stored energy at State 3.  Even if State 1 had been neutral, that logic would too have been erroneous.  That erroneous logic had built into it an assumption of over unity.

Re: Mathematical Analysis of an Ideal ZED
« Reply #27 on: February 21, 2014, 09:58:25 PM »

    Quote

I was hoping to have someone check my math and process of analyzing the 2-layer system before ever posting the 3-layer.  That has not happened and so here is the next model if anyone is interested.  It utilizes the exact same 2-layer model and adds an additional third riser on the outside.  That way the same calculations for the 2-layer portion to find the water levels after introducing the Vin volume "charge" could be re-used.

The PinVin I calculate now rises to ~2.103 mJ.  If PoutVout is to be equal to that per Boyle's law, then the system should stroke ~1.9094 mm.  That is drawn on the right hand side and analyzed to see if it is neutrally buoyant.  It is not, and actually is still pushing upward with ~31.8276 grams of force.  So the ZED would stroke further than shown until it could come to rest again with neutral buoyancy.  And that would require that PouVout would be greater than the PinVin of ~2.103 mJ.

[minnie]

    I assume the compression spring would behave in the same way if furnished
     with the "absurd mechanism".
    It's a bit of a "you can't have the penny and the bun" scenario.
                    John.

[MarkE]

    I assume the compression spring would behave in the same way if furnished
     with the "absurd mechanism".
    It's a bit of a "you can't have the penny and the bun" scenario.
                    John.

The absurd pizza pan spillway mechanism works the same with a compression spring as it does with the Nested Russian Dolls of Ignorance.  In both cases one has to be very careful not to reintroduce the N*(X/N)² problem. 

In an idealized case we suppose that we try to make this glorified spring into some sort of water pumping device.  We alternate between State 2 and State 3.  Let's start in State 3.  We push down on the whole assembly exerting a force that goes from zero to 1.471N over 2.59mm, expending 1.905mJ.  Then we lock down the assembly as we add our charge of water.  Now, we flood the 2.59mm high pizza pan ever so carefully so as not to spill.  If we fill from an infinitely large reservoir that is filled to the bottom of the pan's State 2 position, then we have to add 0.5*pWater*G₀*pi/4*271.49mm²*2.59mm² energy in order to fill the pan.  That is equal to: 1.901mJ.  Now, we let go, and the whole affair rises, dispensing the added water onto the spillway, delivering 1.901mJ.

We've completed a cycle.  Let's see how the energy stacks up:

Stored Energy in State 3 at start = Stored Energy in State 3 at end.
Energy added:  1.905mJ going from State 3 to State 2, plus 1.901mJ to fill the pan in State 2
Energy removed:  2*0.5*pWater*G₀*pi/4*271.49mm²*2.59mm² = 3.802mJ

Neglecting all real losses we net: 3.806mJ in and 3.802mJ out for letting our water take a ride on the spring loaded pizza pan on its way from the top of the reservoir to the bottom of the spillway using this Rube Goldberg inspired device.  What we use and how we use it to push the assembly down between State 3 and State 2, how we lock during State 2, and what we use for a pump in State 2 are all exercises for the interested reader.

[MarkE]

So,, I am confused here MarkE,, you are saying that 1.905mJ down  on the risers will reset them to state 2, but when they lift up they deliver 3.802mJ,, so why can we not take the water from the raised pizza pan and deliver that back into a lowered pan,, that would then be all that water falling down 2.59mm refilling the lower pan and providing more output.  Of course as it lifted that water would need to be transferred at loss,, but then that loss also removes the need to pump that water back into the pizza pan in the first place.

I don't know what is confusing you:

External Energy Added:
1.905mJ applied externally to push the risers down from State 3 to State 2.
1.901mJ added externally to get water out of the reservoir to fill the pan.

External Energy Removed:
3.802mJ total extracted in the difference of water originating at the infinite reservoir level with the pan bottom position lifted to the top of the spillway 2.59mm higher.

Go ahead and let the water run from the spillway back to the reservoir, and you lose all of the work that you added.