Mathematical Analysis of an Ideal ZED

[MarkE]

MarkE, you introduced a rate into an Energy calculation.  Rate does not apply when calculating the Energy leaving the system as the piston rises in this case. 

The correct Energy = F*ds reduction for the case of the riser lifting the water on top of the piston is simply F_ave*S.

The Force starts at the maximum buoyancy Force that can be calculated from State 2.  It then declines linearly to exactly zero which is the case in State 3 since the water is flowing off of the piston as it is rising.  This is a straight forward calculation and involves no rates.

What are you putting into your coffee?  F*ds is not energy.  The integral of F*ds is energy.  The equation for up lift force as the riser goes up is just as I wrote it, and conforms to your description.  Similarly, the solution of the integral that I wrote is correct for these circumstances.  The second term in the integral is identically half the first term.

[MarkE]

Alrighty,, it is kind of funny looking because riser 2 is neutrally buoyant and the water is "hanging" on riser 3 and pulling the water in AR5 up, so I was wondering about what happens when the push from the water from the other side goes away, it looks like riser 2 would also have water hanging on aka a negative buoyant condition,, all of that of course since the "head" around the pod will need to drop down before the pod can lower allowing the system itself to lower,, well it is a case of the balancing act,, naturally the pod will drop down, or maybe be pushed down,,

It may seem non-intuitive to you, but the fluid in the pod chamber is drove the system to the State 3 condition.  That fluid remains under pressure.  The pressure is countered via the coupled force on the various sides of the risers.  If you open the pod chamber, the whole thing will collapse without hesitation. The pod and Riser 1 have net upward pressure on them, meaning that they are net pushing back down on everything in the pod chamber and AR2.  Open the pod chamber at the bottom and that positive pressure will drive fluid out.  That will unbalance the rest of the assembly and the risers will collapse down and the water volumes will equalize.  The reachable lower energy state is State 1:  32.5mm in AR[2-7], and no fluid in the pod chamber.

[mondrasek]

What are you putting into your coffee?  F*ds is not energy.  The integral of F*ds is energy.  The equation for up lift force as the riser goes up is just as I wrote it, and conforms to your description.  Similarly, the solution of the integral that I wrote is correct for these circumstances.  The second term in the integral is identically half the first term.

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?

[minnie]

   Hi,
      if you imagine a see-saw, equally weighted, try taking say one eighth of the weight from
one side. The now lighter side will stubbornly rise, that's what will happen if you try
this with your ZED.
     The pod can be discounted because it's just Archimedes, and the rest of the thing will
just behave like any ordinary telescopic ram, if you have massless, incompressible air.
    Anyone who seems to be getting more out than is put in has seriously got to hunt for
glitches in their work.
     Anyone proving that you really can get free energy out of this sort of device is going to
become a very famous person!
   I'm not saying that it can't be done, but it sure is one hell of a task
                            John.

[mondrasek]

Sure:

Take a volume where we are going to eject water replacing it with an incompressible fluid, where:

H is the height of the volume.
A is the cross-sectional area of the volume.
Ge is the acceleration due to gravity on earth.
pW is the density of water.
pX is the density of the incompressible fluid.

The pressure difference from bottom to top of the volume varies from 0 to H*Ge*(pW-pX).
The force required varies from 0 to H is A*Ge*(pW-Px)*H.
The work done is the integral of F*ds: = Integral( A*Ge*(pW-Px)*H dh)
The solution of the integral from 0 to H is of the form:  Kh*(H2^2 - H1^2) + F0*( 0.5*A*Ge*(pW-pX)*H^2

For pX = 0:  = 0.5*A*Ge*pW*H^2
And since the weight of water in the volume would be: Wdisplaced = A*Ge*pW*H, we get:  E = H/2*Wdisplaced.

From page 1 of this thread (Valentines day no less).  So the Energy integral resolves to H/2*Wdisplaced.  Or 0.5*H*Wdisplaced.  We have changed the variable that represent each value along the way, but I believe that Wdisplaced is the F_start value we calculate from the buoyancy Force.  And H is our lift height, or S as we have recently called it.  And so the Energy integral does resolve properly to 0.5*F_start*S.

With one problem.  "For pX = 0:  = 0.5*A*Ge*pW*H^2."  We are not pushing on the water in the piston directly with air (pX = 0).  We are pushing on it with the physical piston.  But is it okay to say that the piston is being moved by the air that is re-distributing in the ZED during the rise from State 2 to State 3 and therefore this equation applies?