Mathematical Analysis of an Ideal ZED

[MarkE]

The Mondrasek three layer ZED.

Yes, Virginia using incompressible fluids it behaves just like the serpentine hydraulic piston that it is.
One must take into account a couple of key points:

1) The increasing area of the annular rings means that there is force gain from the innermost annular ring to all other annular rings.  So, when pumping water into the inner most ring, the so called pod chamber, while we displace a like weight in each of the other chambers due to using incompressible fluids, the weight that reflects back to the inner most ring decreases as we move out.  The total force that opposes the input energy source at the end of filling 37mm is the weight of the 37mm added to the innermost ring plus the loss of the same weight as a counterbalance in AR2 times the area ratio of AR1/AR2, plus the same weight times the area ratio of AR1/AR3, etc out to AR7.

2) The correct energy values are always obtained by integrating F*ds.  When we do this, we get 3.412mJ total stored energy in the various water columns at the end of the first state where we fill annular rings 2-7 up to 32.5mm high.

3) The added energy required to pump 37mm of head into AR1 works out to 2.099mJ.  This is identically the difference between the 3.412mJ stored at the end of the first state and the energy that one obtains by calculating and summing the stored energy in each of the annular rings at the end of State 2: 5.5111mJ.  IOW, ignoring things like friction loss, the device is completely conservative pumping water in.  Gravity has not been cheated. 

4) Releasing the risers and allowing them to rise causes the the water levels in the various annular rings to move towards equalized heights.  As I have already shown, anytime we take two columns one filled higher than the other and allow them to move towards equalization, we lose energy to heat.

So the bottom line here is that the three layer device is conservative until one lets the risers move, and then it is lossy.  If we change out the incompressible air with a compressible gas then we will add pumping losses on top of the other losses.  There is nothing in the ZED that cheats gravity.  There is nothing in the ZED that increases efficiency lifting and dropping weights over an electric winch.  The best ZED is no ZED.  QED.

What we have seen from the HER/Zydro team is an unwillingness to evaluate energy properly as any ME would learn in college: by performing the integral of F*ds.  HER/Zydro is a case of garbage in and garbage out.  HER/Zydro's representatives who post here claim that they are fully competent.  So, we are either witness to Dunning-Kruger in action, or the refusal to analyze properly as any competent ME would is a deliberate choice.

[mondrasek]

The Mondrasek three layer ZED.

Yes, Virginia using incompressible fluids it behaves just like the serpentine hydraulic piston that it is.
One must take into account a couple of key points:

1) The increasing area of the annular rings means that there is force gain from the innermost annular ring to all other annular rings.  So, when pumping water into the inner most ring, the so called pod chamber, while we displace a like weight in each of the other chambers due to using incompressible fluids, the weight that reflects back to the inner most ring decreases as we move out.  The total force that opposes the input energy source at the end of filling 37mm is the weight of the 37mm added to the innermost ring plus the loss of the same weight as a counterbalance in AR2 times the area ratio of AR1/AR2, plus the same weight times the area ratio of AR1/AR3, etc out to AR7.

2) The correct energy values are always obtained by integrating F*ds.  When we do this, we get 3.412mJ total stored energy in the various water columns at the end of the first state where we fill annular rings 2-7 up to 32.5mm high.

3) The added energy required to pump 37mm of head into AR1 works out to 2.099mJ.  This is identically the difference between the 3.412mJ stored at the end of the first state and the energy that one obtains by calculating and summing the stored energy in each of the annular rings at the end of State 2: 5.5111mJ.  IOW, ignoring things like friction loss, the device is completely conservative pumping water in.  Gravity has not been cheated.

MarkE, thank you for double checking everything up to this point.  I'm glad that we agree so far.

4) Releasing the risers and allowing them to rise causes the the water levels in the various annular rings to move towards equalized heights.  As I have already shown, anytime we take two columns one filled higher than the other and allow them to move towards equalization, we lose energy to heat.

MarkE, you did not finish the test.  And I must insist that you do.  Because it is only when again measuring the Energies AFTER the lift that I am finding things do not add up.  And, unfortunately, the rise (stroke) is the hardest part (for me at least) to calculate.  I cannot simply calculate the final resting position that the ZED will rise to if allowed to do so where all the buoyant forces induced by the water charge sum to zero.  I would have to do this iteratively and it would take forever.  You and LarryC would probably write a VBA program to do that.  I lack that ability.

So I took a different approach:  I ASSUMED first that all the added Energy from the input charge would convert to motion of the outer riser by F*ds.  I then re-drew the ZED model with that exact amount of rise and re-distributed the water.  If all the added Energy had been converted to motion of the outer riser then the sum of the buoyant forces in the system should be zero at that state.  When I did that analysis the sum of the buoyant forces was NOT zero.  It was a positive value that meant the ZED would need to rise even further.

[MarkE]

MarkE, thank you for double checking everything up to this point.  I'm glad that we agree so far.

MarkE, you did not finish the test.  And I must insist that you do.  Because it is only when again measuring the Energies AFTER the lift that I am finding things do not add up.  And, unfortunately, the rise (stroke) is the hardest part (for me at least) to calculate.  I cannot simply calculate the final resting position that the ZED will rise to if allowed to do so where all the buoyant forces induced by the water charge sum to zero.  I would have to do this iteratively and it would take forever.  You and LarryC would probably write a VBA program to do that.  I lack that ability.

So I took a different approach:  I ASSUMED first that all the added Energy from the input charge would convert to motion of the outer riser by F*ds.  I then re-drew the ZED model with that exact amount of rise and re-distributed the water.  If all the added Energy had been converted to motion of the outer riser then the sum of the buoyant forces in the system should be zero at that state.  When I did that analysis the sum of the buoyant forces was NOT zero.  It was a positive value that meant the ZED would need to rise even further.

Are you saying that you agree with the analysis through State 2?  It is OK if you don't, but I will then want to know specifically what you object against.

You have stipulated that the pods and risers are massless.  Unless you specified some sort of payload someplace that I missed, going from State 2 to State 3 therefore does no work, but we know that it is lossy, because any increase in the internal volume requires the water columns to move towards equalization.  The cylinder volume from the Riser 3 OD inward increases only at the expense of a reduced water column in AR7.  Water volume from AR7 and AR6 go towards equalization, as do AR5 and AR4, and AR3 and AR2.  The internal volume can increase no more than the ratio of the area of riser3 to the entire area including AR7 multiplied by the water volume added in State 2.  And we know that equalizing columns is a lossy process.  So before I go off to show the specific changes going to a State 3, I need more information from you about what useful work  you intend this thing to do going from State 2 to State 3.  As long as it can be shown that work is less than the energy loss going between those two states, then the machine is lossy.

[LarryC]

I checked the results before I posted.  They agree with the spreadsheet to five digits.  The 51 is the constant annular ring area expressed in circular inches that you agreed to use:  IE the area of the annular gap between the pod and the innermost ring wall.  40.06 is what you get when you convert from circular inches to square inches, which the constant K1 rolled-up along with the density of water. 

Maybe you are not familiar with the concept of circular area units.  They get used in power electronics quite a bit.  A circular area unit is the area a square would take that has the width of a given circle's diameter.  The relationship between circular area and absolute area is:  absolute area = circular area * pi/4.  With a pod of 25" diameter, the circular area is 25² = 625.  The ring wall at 26" diameter is 26² = 676.  The area difference is of course the sum of the two diameters = 51 circular inches.  We can work in these more convenient units throughout the problem before applying the common constants pi/4 and the density of water, and our conversion from cubic inches volume and inches height to cubic feet and feet height.

My reduction simply reproduced the net total of the spreadsheet formulas in algebraic form.

MarkE,


Okay, now we have 66.14% Zed Efficiency for both approaches.

[mondrasek]

Are you saying that you agree with the analysis through State 2?  It is OK if you don't, but I will then want to know specifically what you object against.

I agree with your analysis through State 2.  Everything adds up exactly as I also found.

You have stipulated that the pods and risers are massless.  Unless you specified some sort of payload someplace that I missed, going from State 2 to State 3 therefore does no work, but we know that it is lossy, because any increase in the internal volume requires the water columns to move towards equalization.  The cylinder volume from the Riser 3 OD inward increases only at the expense of a reduced water column in AR7.  Water volume from AR7 and AR6 go towards equalization, as do AR5 and AR4, and AR3 and AR2.  The internal volume can increase no more than the ratio of the area of riser3 to the entire area including AR7 multiplied by the water volume added in State 2.  And we know that equalizing columns is a lossy process.  So before I go off to show the specific changes going to a State 3, I need more information from you about what useful work  you intend this thing to do going from State 2 to State 3.  As long as it can be shown that work is less than the energy loss going between those two states, then the machine is lossy.

I utilized the same analysis method for the output rise as was used for the input of the water charge:  F*ds as expressed for the case of a Volume of a Fluid that is being moved by a change in Pressure that either starts or ends at zero:  P_average*V.  The riser initially will want to move with a Pressure that can be calculated from the buoyant force sum of the pod and risers.  That Pressure should drop linearly to zero as the ZED reaches equilibrium at the end of the rise.  The physical device that would restrain the initial Pressure and allow it to drop to zero while performing the rise is not important for the analysis I think.  Please let me know if you think otherwise.