Mathematical Analysis of an Ideal ZED

[mondrasek]

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*F_start*S.

[minnie]

   Yes, you can't build any pressure in this because it will spill,so you have to let the big riser rise.
   Buoyancy is really pressure,so the s.g. of the fluid providing the head is important and the
   pressure acting on the largest riser is also important, these two should be what goes in and
   what comes out. True the different densities of the internal fluid mix will have some effect
   due to their weight, but I can't see it making much difference to the output.
                                       John

[Magluvin]

Pentane is flammable, volatile, toxic and carcinogenic probably. Sodium polytungstate? Man, you are weird. How about using galinstan and kerosene?





Or just a big spring and a rock.

Would using the lightest workable liquid vs. air, cause us to use even more input due to the lighter liquid would have a lot more weight added to the system? More weight to move around, and more in before we get anything out. As compared to air. 

Was thinking also, being the air is compressible, initial input would be less loaded at the moment, rather than an abrupt requirement of 'full' input(surge) just to get things going. Just thinking. The total amount of possible excess input to begin a rise, as compared to an incompressible air substitute, might not be so much  in because of its compressibility. The incompressible air substitute definitely creates a much closer weight difference between it and the water, so more needs to be input than can be output because we are taking away the weight difference of the water, as compared to using air.

Fill a balloon with the oil and one with air and see which rises to the top of the pool from the bottom first.  So in the system being discussed, more oil would have to be moved than the amount of air mass, equals more loss to the system. Heck, might as well just use all water, if we dont care about the the object of changing the levels if weighted liquid in linked separate chambers as being discussed. In that case, then yes, buy a hand jack.

Also, once that air is up to pressure, that pressure contains the energy that put it there. So if we dont try to capture that after the riser is at bottom, then we just wasted it stupidly if it is just released foolishly without applying some use to it at release.  And beyond that, arrange the system to not release or use that compressed air that is at the pressure level of just before the riser is about to rise. Then we eliminated initial loss of compressing to the point of initial rise, again and again. We only have to go through 'that' loss 1 time. Pressure activated check valve?  Like a zener diode, wont let the current flow if below say 12v, for a 12v zener.


Mags

[MarkE]

MarkE.  I apologize for leaving out the word "integral" again.  It is a bad habit.

I think I saw this integral reduction shared by you.  It is the one we are using for the Energy in the water introduced during the change from State 1 to State 2.  That integral of F*ds is 0.5(P_start-P_end)*V for the special case where the start or end Pressure is zero.  Is that correct?

For the special case of zero to H_END, or H_START to zero, the work performed filling or emptying a column is:  E = 0.5*P_END*V_END filling, and 0.5*P_START*V_START emptying.  In all other cases the integral result is more complicated.

[MarkE]

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*F_start*S.

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

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%.

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.