And your nested system is significantly different.... how?

The analysis I have shown, in effect, isolates the system being analyzed just as the red box did to yours.  It is similar to treating the enclosed system as a "black box."  Then we only need to concern ourselves with what crosses the barrier of that box.

The point of significant difference is this:  A correct analysis of your UTube system shows that the Work/Energy input=output (unity).  The results of the ZED system that I am analyzing show that input<>output (non unity).

1. You are confusing yourself with "Boyle's Law" pressure/volume calculations, because Travis and Red Sunset and Webby have all said that the air can be replaced with incompressible fluid. All chambers in your system should be filled with fluid that doesn't change in volume when it's under pressure.

All internal chambers have been replaced with an incompressible fluid.  There is the water (SG=1), and the air (SG=0) that is ASSUMED to be incompressible for this ideal case.  We can change those two fluids to real incompressible liquids of different specific gravity values if the assumption needs to be tested falsifiable.  But when you do the math you will see that the greater the difference in the SG of the two fluids used, the greater the output to input ratio once unity has been surpassed by proper ZED design.  Also, you can see that a compressible fluid such as air only causes a loss to the system.  So using two liquids will result in a higher performance, just so the difference in the SG of those fluids is > 1.

The outer chamber of the system is NOT filled with an incompressible fluid.  It is open to the atmosphere.  So an open system.  I find that interesting and a possible reason that Boyle's law may not apply?

1b. Air -- wet air -- is not an ideal gas and you will find that it deviates from strict Boyle-Charles law behaviour. How much? Enough to account for your numerical result? I don't know. Do you?

No air (or wet air) is used in the system under analysis.

2. You should be able to demonstrate some actual gain in something, somewhere, using just three layers. Travis has told us so!

The analysis of the 3-layer ZED does show that.  PinVin<PouVout.

3. What, exactly, is the real "output" of your system? Is it a lifted weight? If you are counting pressing against a stop as "output work", that's not kosher!

The output is being calculated as the Integral of Pout * Vout, ie. Work/Energy.  Pout is the average pressure of the ASSUMED linear declination of the maximum lift pressure (calculated from the buoyancy force at full charge applied to the surface area of the outer riser) as it falls to zero (where the system should be at neutral buoyancy after being allowed to rise due to the charge).

I have never calculated the real Vout.  Instead I have ASSUMED the system follows Boyle's law and would therefore have a Vout that can be calculated by the specified and calculated Integral of Pin, Vin, and the Integral of Pout.  Once that Vout is calculated I construct a model that has stroked to the height that would require and reevaluate.  If Boyle's law were true, then the system after applying this stroke value should be at neutral buoyancy.

It is not.

In the 2-layer model the system fails to satisfy Boyle's law by displaying PinVin>PoutVout (underunity).

In the 3-layer model the system fails to satisfy Boyle's law by displaying PinVin<PoutVout (overunity).

M. 

So what are you doing mentioning Boyle's Law at all? Boyle's Law is the relationship between Pressure and Volume of compressible (ideal) gases. You are assuming incompressibility at the top, then complaining that your system doesn't follow the law of compressible gases?

Maybe that's your problem then, since incompressible fluids do not follow Boyle's Law of the pressure-volume relationship of ideal, compressible gases.

So you are saying that Boyle's law does not apply to a fluid once it changes state from a vapor/gas to a fluid?  I thought that once the fluid is incompressible then the volume cannot change.  And then Boyle's law of P₁V₁=P₂V₂ would reduce to Integral of P₁ = Integral of P₂.  Are you saying this is incorrect?

And work is only performed when forces act over distances. Pressing against a stop with some pressure is not work. Does your butt do work on the chair you are sitting in?

I'm not sure how this applies?  Work is not being calculated while pressing against a stop.  A non unity quantity of Energy is.

Who is talking about phase changes here? Your device is filled completely with two incompressible fluids. No phase change happens. Boyle's Law does not apply to incompressible fluids!

Sorry if my previous explanation was not precise.  I did not mean to imply that any phase change occurs in the ZED system.  It does not.  I was only tring to explain that I thought Boyle's law would apply to liquids once they had changed phase from vapor to liquid.  If that is in error, I accept that and appreciate you pointing out my misunderstanding.

That's right: The pressure applied to the fluid is _unrelated_ to the volume of the fluid, it does not affect the volume, and no amount of applied pressure can change the volume of an _incompressible_ fluid, and no energy can be stored by "compressing" or applying pressure to an incompressible fluid.

I'm sorry if I don't understand your point here.  I would appreciate if you can try another way to help me grasp it.  Seriously.  But I see no Energy being store in "compressing" anything in the process I am analyzing.

How many pounds per gallon does a speed of 60 seconds per bushel represent? 
I am saying that no pressure change will result in a volume change of an incompressible fluid, therefore Boyle's Law is irrelevant and inapplicable! So Boyle's Law reduces to P₁V₁ = P₂V₁ since volume cannot change in an incompressible fluid, and this relation is contradicted by experiment (pressures can certainly be different for the same volume of water, a nearly incompressible fluid) because Boyle's Law is not applicable to _incompressible fluids_ !! It is an IDEAL GAS LAW and ideal gases are ideally compressible ... that is what Boyle's Law tells you!

I know!  But the analysis appears to show that the ZED system would need to stroke (under pressure) more than would be required by inputting the same volume of charge liquid under the same pressures that an equivalent hydraulic system would.  And so I don't understand!

Now if using Boyle's law to calculate the energy input due to the charge volume being introduced is wrong, then the analysis is wrong.  Completely wrong. 

Can you offer a correct method?

In a previous post you stated that the pressure against the stop caused by your calculated excess buoyancy represented a gain, I thought. And work and energy have the same units, the Joule in SI. If you are calculating Energy you are also calculating Work and vice versa.

I tried to state that by the method I was using to calculate it appears there must be more energy after the system "power stroke" than is predicted by the falsification test.  So either the test is erroneous or there must be excess energy left in the system.

TK, thank you for engaging and helping me to work this out.  It is much appreciated.

M.

Mondrasek, I think that the confusion here is in how does one measure the energy of a submerged buoyant object.  For an object of constant height, the amount of force that we must apply to the object to initially submerge it changes from the value at zero submersion to the fully submerged value.  From there on the buoyant force is constant and the amount of additional work required to submerge the object further is the difference between that constant buoyant force and the constant force of gravity on the mass of the object multiplied by the change in depth.

The attached graphic will hopefully help.

TK,

I would appreciate if you would stay on topic.  Can you please explain exactly how this post is relevant to the "Mathematical Analysis of an Ideal ZED" models so far presented?

M.

Do you have me confused with someone else?

No TK, I am just calling you out (regrettably).  Are you, or are you not, using the pseudonym of MarkE ?

But look again at Mark's diagram. Do you see what he is integrating? Do you see any need for an ideal gas law in determining the work required to submerge a _sealed_ buoyant object?

Finally!!!!!!!!!!!!!   We get to the Ideal gas law!  Which is PV=nRT.  Which for Isosthermic cases (ie. T₁=T₂) results in  Boyle's law:  PV_input=PV_output.

Now, if you were considering a Cartesian Diver (an _unsealed_ buoyant object)  and didn't stipulate beforehand that all your fluids were incompressible, that would be a different story.

Is the ZED system an "unsealed buoyant object" or not?

M.

Hi M.,


Your uncompressible air made it easy to do the flow spreadsheet for 2 zeds with 1 riser. I used the rule that 1X head change in the pod creates 2X head change in the riser to calculate the water heads.


Got an efficiency of ~154% with 2 zed units and ~74% with 1 Zed unit.


It makes it obvious that the average PSI differential between the Left and right zed units during water transfer is key to it's efficiency. I included the Archimedes spreadsheet again, compare the PSI differential from each, the Zed is superior.


I did have weight on the pod as it is critical that a pod sink point is specified for the water head calculations.


I recommend that anyone looking at these, understand the Archimedes calculations first, then the 2 Zed version will be easier.


M., please check for errors, thanks. This will be an interesting simple proof if we get it cleaned up.


Larry[size=78%]   [/size]

I finally got to spending some time with your model today and have a couple of questions.  The first is that when you admit water you want to fill up the pod chamber to 60mm height. 

MarkE, the volume of water admitted is only enough to fill the pod chamber to 37mm height.  That is shown in the second diagram here:

http://www.overunity.com/14299/mathematical-analysis-of-an-ideal-zed/msg387854/#msg387854

Do you (or anyone else) know how to calculate the amount of energy that crosses into a system when a specific volume of water is introduced, starting at a pressure of zero and building linearly to a final pressure of P_in?

Of course.

Here is what I would propose to try next:  Using the incompressible fluids in the model.  Is it correct to say that the Volume input (volume of water admitted into the bottom of AR1) should be equal to the Volume output as measured by the height change of the outer riser * the cross sectional area of the outer riser?

So it would seem, if I am understanding you correctly. However, consider the simple lever. If I "admit" a certain weight on the long end and it sinks, should that be equal to the Height Output as measured by the height change of the weight on the short end? I think nested hydraulic cylinders can act like a compound lever, and I think that the multiple layers might distribute an initial "volume input" over several outer risers, so the final "rise" might be small, but with increased force.

If so, a ZED that is drawn with that exact amount of rise and with the fluids re-distributed correctly should be neutrally buoyant, right?  If it is acting exactly as a simple ideal hydraulic cylinder?

Now you have lost me. Neutral buoyancy means that the mass of the displaced water is equal to the mass of the displacing object. Adding additional force pressing down on the "neutrally buoyant" object makes it sink, so is this extra force to be included in the figuring?  Since your risers are "massless" I think you are once again up against a place where your assumptions are non-physical and may be leading you off the correct track.

Try putting some Red Herrings in the water. That has helped Travis and Red Sunset along greatly.