Mathematical Analysis of an Ideal ZED

[mondrasek]

The first thing that you will want to be very careful with is the assumptions that you make with respect to the amount of work it takes to displace "air" or water in any of the chambers.  A mistake can easily throw energy off by the ratio of area of the cylinder area to the annular ring area of the chamber.  It is tempting to calculate force based on the smaller area in cases where it is actually a function of the much larger area.

Thanks for the input, MarkE.  Please let me know if you see any of those type of mistakes so far.

[mondrasek]

If we ASSUME that the ZED is acting exactly as a hydraulic cylinder, then it would have to follow Boyle's Law.  The Integral of PinVin must equal the Integral of PoutVout.  To find the lift that would result in this case requires that we find the Vout.  Again I must also ASSUME that the output Pressure that the outer riser can provide is a linear function, starting at an initial pressure found from the buoyant lift force applied to the cross sectional area of the outer riser, and ending at a pressure of zero.  The hydraulic lift force of ~82.590 grams results in an initial pressure value of ~795.980 Pa.  The average Pout becomes half of that, or ~397.990 Pa.  With Pin of ~872.257 Pa and Vin of ~2.1771 cc (from a previous post), we solve PinVin = PoutVout for Vout = ~4.771 cc.  The lift of the outer riser is then calculated to be ~4.688 mm.  This is drawn and analyzed (after redistributing the fluids properly) to see if it is neutrally buoyant or not.

The evaluation of this ZED shows it is definitely NOT neutrally buoyant.  In fact, it sucks, literally.  It is displaying a positive lift force from the pod of ~10.050 grams.  But the risers are both negatively buoyant.  The inner riser has a lift force of ~-7.238 grams and the outer riser has a lift force of ~-43.130 grams.  The total is ~-40.318 grams.  When we add the additional downward force of the ~8.168 grams the outer riser needed to weigh for the system to be neutrally buoyant in the setup position, the total lift in this analysis is now ~-48.486 grams, far below a neutral buoyant condition.

So this test failed.  The ZED could never rise to the height calculated by Boyle's Law.  If the ASSUMPTION of linear pressure transfers are correct (or close) the ZED could only rise a bit less than 2/3 of the required value necessary to satisfy Boyle's Law.  Therefor we are left with the possibilities that a) there is a mistake in the math, b) the ASSUMPTIONS are greatly skewing us away from expected results, or c) an Ideal ZED is NOT analogous to an Ideal Hydraulic Cylinder.

M.

[MarkE]

If we ASSUME that the ZED is acting exactly as a hydraulic cylinder, then it would have to follow Boyle's Law.  The Integral of PinVin must equal the Integral of PoutVout.  To find the lift that would result in this case requires that we find the Vout.  Again I must also ASSUME that the output Pressure that the outer riser can provide is a linear function, starting at an initial pressure found from the buoyant lift force applied to the cross sectional area of the outer riser, and ending at a pressure of zero.  The hydraulic lift force of ~82.590 grams results in an initial pressure value of ~795.980 Pa.  The average Pout becomes half of that, or ~397.990 Pa.  With Pin of ~872.257 Pa and Vin of ~2.1771 cc (from a previous post), we solve PinVin = PoutVout for Vout = ~4.771 cc.  The lift of the outer riser is then calculated to be ~4.688 mm.  This is drawn and analyzed (after redistributing the fluids properly) to see if it is neutrally buoyant or not.

The evaluation of this ZED shows it is definitely NOT neutrally buoyant.  In fact, it sucks, literally.  It is displaying a positive lift force from the pod of ~10.050 grams.  But the risers are both negatively buoyant.  The inner riser has a lift force of ~-7.238 grams and the outer riser has a lift force of ~-43.130 grams.  The total is ~-40.31771688 grams.  When we add the additional downward force of the ~8.168 grams the outer riser needed to weigh for the system to be neutrally buoyant in the setup position, the total lift in this analysis is now ~-48.486 grams, far below a neutral buoyant condition.

So this test failed.  The ZED could never rise to the height calculated by Boyle's Law.  If the ASSUMPTION of linear pressure transfers are correct (or close) the ZED could only rise a bit less than 2/3 of the required value necessary to satisfy Boyle's Law.  Therefor we are left with the possibilities that a) there is a mistake in the math, b) the ASSUMPTIONS are greatly skewing us away from expected results, or c) an Ideal ZED is NOT analogous to an Ideal Hydraulic Cylinder.

M.

Something else that one needs to be careful about is evaluating integrals.  Where the pressure which translates to force changes, we need to solve the integral.  If the force or pressure starts at zero and goes to some other value then the integral is trivial.  If the pressure / force starts and ends at non-zero values then we get both linear and quadratic terms.

[mondrasek]

Something else that one needs to be careful about is evaluating integrals.  Where the pressure which translates to force changes, we need to solve the integral.  If the force or pressure starts at zero and goes to some other value then the integral is trivial.  If the pressure / force starts and ends at non-zero values then we get both linear and quadratic terms.

Can you lend assistance with how to evaluate the PinVin integral properly since it starts from a non-zero condition in this setup?  Or would it be necessary for me to start over with a setup that initially has no water in the pod chamber?

Also, to be clear, is there any issue with evaluating the PoutVout where Pout starts at non-zero but should end at zero?

M.

[MarkE]

Can you lend assistance with how to evaluate the PinVin integral properly since it starts from a non-zero condition in this setup?  Or would it be necessary for me to start over with a setup that initially has no water in the pod chamber?

Also, to be clear, is there any issue with evaluating the PoutVout where Pout starts at non-zero but should end at zero?

M.

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. 

That should be intuitively satisfying because we "lift" only an infinitesimal amount of water by 0 height at the start, and another infinitesimal amount of water by all of the height at the end, so the average amount by which we "lift" all of the water that we displace is 0.5*H.  So far, so good.

But what happens when we already have displaced some water?  The math is still the same, just some terms are no longer zero.  Force needs to be defined as a function of what is being changed, displaced height for our problems, and then integrated.  The net work done is the difference between the integral at the start and end points.  Typically, this reduces to:

F = F0 + Kf*H  (Kf may be positive or negative depending on the circumstances)
Integral from H1 to H2 is:  F0*H2 + 0.5*Kf*H2^2 - F0*H1 - 0.5*Kf*H1^2.