Mathematical Analysis of an Ideal ZED

[MarkE]

Well MarkE,

There may be some factual basis to your statement, but the ZED is not a spring, it may have some attributes of a spring, but it is not "only" a spring.

Work does not only happen in one direction,, work can happen in any direction,, work really is not fixed to a direction,, a force over a distance, whether a straight line or a curve, is work.

I take it that you agree with my assessment in that you need to include the lever for the proper analogy with a spring to a ZED.

Webby1 we have been told to disregard all the set-up steps as one time affairs.  That's fine.  That leaves us with the behavior between State 2 and State 3, and then back from State 3 to State 2.  That behavior from a black box standpoint of the "ideal ZED" is indistinguishable from the behavior of a linear compression spring: a very complicated, and large, low energy compression spring, but a linear compression spring just the same. 

Unless you specify operation between states other than just State 2 and State 3, the "ideal ZED" can for purposes of external behavior, IE black box be completely replaced by a 0.48N/mm 2.492mm relaxed to maximally compressed linear compression spring. 

Where did I say that I agree?  You can add levers externally and define a new composite machine.  It will not fix the N*(X/N)² efficiency problem that plagues this compression spring emulating device. 

[MarkE]

Of course I am using the OD instead of the ID.  I am calculating the lift force due to BUOYANCY.  The correct water displacement value is achieved when you consider the water displaced by the air inside the riser AND the wall thickness.  That means you use the OD.

If you want to calculate the buoyancy Force on a fully submerged 1m³ cube that has a wall thickness of 100mm throughout, is it not displacing 1m³?  Or do you want to calculate the displacement using only the remaining internal volume?

Mondrasek, State 1 is in equilibrium is it not?  And why is that?  It is because the meniscus under the ID is at the same level as the water outside the riser.  The water displaced by the riser wall plays no role in the force balance.  What is true for State 1 is true for State 2, and State 3.  If you close off the bottom of the riser so that the meniscus cannot rise up underneath it, then you change the problem to where the OD determines the displacement.  Are you really going to argue that the riser wall that contributes no up force in State 1, suddenly changed its behavior because we introduce water into the pod chamber in State 2?

[mondrasek]

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

Untrue, MarkE.  Here is what you wrote in post 708:

So you get a movement S = F_START/K_RATE 0.30078N / 64.649N/m =  0.004653m and an output energy of
0.5*0.30078N*0.004653m   0.700mJ. 

The energy expended, is the loss of potential energy in the annular columns as they move from the State 2 condition to the State 3 condition with the riser at 2.492mm lift:  The internal energy goes from 2.560mJ to 1.536mJ which equals 1.024mJ internal energy loss.

The efficiency is:  0.700mJ/1.024mJ = 68.3%.

In the first paragraph you are discussing the no-pod, single layer.  See where you twice use the value for that lift of 0.004653m?

In the second paragraph you now reference the lift as only 2.492mm.  And include internal energy values from where exactly?

You then calculate an erroneous efficiency using these mismatched and false values.

[mondrasek]

Mondrasek, State 1 is in equilibrium is it not?  And why is that?  It is because the meniscus under the ID is at the same level as the water outside the riser.

The equal level of the two mensci results in a HEAD difference of zero.  Buoyancy Force = Weight of the Displace Water = (Density of Water)* HEAD * (Cross Section Area).  If HEAD is zero, buoyancy Force is zero.

The water displaced by the riser wall plays no role in the force balance.  What is true for State 1 is true for State 2, and State 3.  If you close off the bottom of the riser so that the meniscus cannot rise up underneath it, then you change the problem to where the OD determines the displacement.  Are you really going to argue that the riser wall that contributes no up force in State 1, suddenly changed its behavior because we introduce water into the pod chamber in State 2?

Complete hogwash.  And now I will clearly state that it is MarkE's intention to misdirect and provide false information in this thread.  He has done this several times in the past few days.

Buoyancy Force = Weight of the Displaced Water.  For the riser what is displacing the water is the Air inside, and the Material of the riser itself.

So yes, I will argue that the riser wall has no relevance at all on up Force in State 1 dues to no water HEAD.  But once a HEAD is introduced, yes, the wall thickness is included when calculating the Weight of the Displaced Water.  The riser wall did not change its behavior.  The water HEAD condition changed.

Why are you purposely trying to stall and now misdirect this Analysis?

[MarkE]

Your #745

Quote from: MarkE on Today at 01:54:01 AM

    The energy expended, is the loss of potential energy in the annular columns as they move from the State 2 condition to the State 3 condition with the riser at 2.492mm lift...

Okay, I am lost here.  The lift was 4.68536mm.  Did I miss something?

Or 4.653mm by your math.  Probably due to the use of different constants for water density and gravity?

Mondrasek, as you see in #745 you quoted where I talked about 2.492mm movement.  2.492mm movement applies to the 3 riser system just as I responded.  It looks like in #708 I wrote the 3 riser movement value of 2.492mm where I should have used the 4.653mm value.  However, the 4.653mm and only the 4.653mm value was used in the calculations.

#708

Re: Mathematical Analysis of an Ideal ZED
« Reply #706 on: Today at 01:54:01 AM »

    Quote

Quote from: mondrasek on March 06, 2014, 10:46:18 PM

    Another way to look at this is to calculate the Energy in the column of water that is on the piston.  That Energy is dissipated as the piston rises and the water spills out.  So that Energy must be equal to the Energy that is released when the ZED is allowed to rise.  That equation also resolves to 0.5*Fstart*S.

For the circumstance that you set-up that is correct:

So you get a movement S = FSTART/KRATE 0.30078N / 64.649N/m =  0.004653m and an output energy of
0.5*0.30078N*0.004653m   0.700mJ.

The energy expended, is the loss of potential energy in the annular columns as they move from the State 2 condition to the State 3 condition with the riser at 2.492mm lift:  The internal energy goes from 2.560mJ to 1.536mJ which equals 1.024mJ internal energy loss.

The efficiency is:  0.700mJ/1.024mJ = 68.3%.

This is the tyranny of N*(X/N)2.  Whenever you directly transfer energy between two potential energy stores you run into the N*(X/N)2problem.  In order to avoid the problem you have to retain N=1.0 and still transfer all of the energy.  You can only do that by converting the form of the energy from potential to kinetic.  That is how a pendulum works.  The insipidly stupid ZED design cannot do that.  It translates potential from one place to another.  There is even the "equalization phase" that HER/Zydro say is necessary.  Every time you take energy from one potential store and redistribute it to more stores than the original without first converting the form of the energy, you suffer losses.

Now let's talk about the quote above where you claim I am misrepresenting the physics.

Untrue, MarkE.  Here is what you wrote in post 708:

So you get a movement S = F_START/K_RATE 0.30078N / 64.649N/m =  0.004653m and an output energy of
0.5*0.30078N*0.004653m   0.700mJ. 

The energy expended, is the loss of potential energy in the annular columns as they move from the State 2 condition to the State 3 condition with the riser at 2.492mm lift:  The internal energy goes from 2.560mJ to 1.536mJ which equals 1.024mJ internal energy loss.

The efficiency is:  0.700mJ/1.024mJ = 68.3%.

In the first paragraph you are discussing the no-pod, single layer.  See where you twice use the value for that lift of 0.004653m?

Of course I did.  Apparently, you are still having difficulty understanding how to calculate energy even when I laid it out there for you to see.  The energy depends on the height squared.  You can get that by going through the detail work of writing out the integral, or given that form of integral has been solved many times, you can skip to the end and use the maximum pressure and volume values, that ... wait for it ... result in squaring the height.

In the second paragraph you now reference the lift as only 2.492mm.  And include internal energy values from where exactly?

You then calculate an erroneous efficiency using these mismatched and false values.

I calculated the values correctly, and showed you the work so that anyone could follow along.  Apparently, you were not able to follow and latched onto the typographical error without checking any of the work.  Are you able to follow along that the distance S is a function of the starting force and the rate at which that force changes per unit distance?  Do you suffer difficulty verifying that S = F_START/K_RATE 0.30078N / 64.649N/m =  0.004653m does in fact result in the correct displacement value?  pWater used is 998.2kg/m³, G0 used is 9.80665m/s².

Here again is the graphic for the single riser with the calculations as originally posted.