Mathematical Analysis of an Ideal ZED

[mondrasek]

MarkE, regarding your note inserted in the graphic that states:

3) No work extracted going from State 2 to State 3

We have already gone over this?  There is an ASSUMED non-physical device that MUST restrain the ZED from rising unimpeded (and wasting all that Energy rather than collecting it) that would account for the Work/Energy you keep throwing away.

This is an IDEAL Analysis and so a physical device should not need to be presented.  But if you insist that one does, I will oblige.  Please let me know if you need to see a physical manifestation of a "Worked on Device" or if you can agree that the Energy "lost from the system" due to the lift could have been collected.  Obviously we have all the correct ingredients:  A Force (from buoyancy) and a ds (distance that the riser lifts)?

[MarkE]

Yes it does.  And lets look at that relationship.

F = mA

m is mass.  Mass is a property of a physical material that does not change for the accepted IDEAL conditions of a constant temperature and obvious absence of a state of matter change.  Therefore m is a CONSTANT.

A is acceleration.  In this case it is the acceleration due to gravity.  It is also a CONSTANT.

Yes, in this case with your stipulations for the "ideal ZED" m of the riser and "air" is constantly zero.  Therefore the kinetic energy is constantly ... wait for it: zero.

So the Force (F) in F = mA is a mathematical fact which the calculation of cannot be disputed.  It is the product of two CONSTANTS (and yes, TK, one is a vector so the result is a vector).

So, what is Energy?  It is a resultant of the prior mathematical fact that is Force.  It is F*ds (where ds is distance).

F is Force which is the product of two constants.

No, F is whatever function defines it over the traversed distance S that it will be evaluated.  In the other thread I thought I read you saying that you work with CFD.  How could you work with CFD and misrepresent these fundamental concepts? Are you trolling?

ds is distance which another calculable (or measurable) physical fact and therefore a CONSTANT.

I find it hard to believe that you flunked calculus.  But, if you want to represent that you did, who am I to argue?

Ergo, you must solve for Force before you calculate the Energy.  And regardless of the outcome of that Energy value, it must be correct.

You must solve for the force at each incremental point over a path in order to solve for the energy applied.  Drag a heavy object with a real coefficient of friction for a distance and you apply real work.  That work all converts to heat.  You end up with zero kinetic or potential energy in the thing you dragged.

In your first attempt at this Analysis you solved first for Energy Balance.  This was erroneous and resulted in a physical State 3 that could not actually exist due to "unresolved" Forces in the system that did not sum to zero.

No, I calculated the energy loss for the change in internal energy.  You may rightly contest that I did not correctly solve the equilibrium height, because I did not. But the energy loss calculated was correct for the calculated movement, and only gets worse going to the higher true equilibrium height.

[MarkE]

MarkE, regarding your note inserted in the graphic that states:

3) No work extracted going from State 2 to State 3

We have already gone over this?  There is an ASSUMED non-physical device that MUST restrain the ZED from rising unimpeded (and wasting all that Energy rather than collecting it) that would account for the Work/Energy you keep throwing away.

Your assumption that some device can magically collect the lost energy is a fallacy.  Place any mechanism that you like in communication with the risers and show that you don't lose energy.  You cannot.  But go ahead and prove me wrong.  Every um of movement by the riser results in permanently lost energy.  I have shown the physical basis for this and the associated math.  If you want to argue differently, do more than exclaiming "No it isn't."

This is an IDEAL Analysis and so a physical device should not need to be presented.  But if you insist that one does, I will oblige.  Please let me know if you need to see a physical manifestation of a "Worked on Device" or if you can agree that the Energy "lost from the system" due to the lift could have been collected.  Obviously we have all the correct ingredients:  A Force (from buoyancy) and a ds (distance that the riser lifts)?

I do insist, because it is fundamental.  You cannot collect what you lose lifting, because the very act of lifting changes the N in N*X/N² to a value greater than 1.0. 

[mondrasek]

In the other thread I thought I read you saying that you work with CFD.

Sorry, but you must be remembering someone else.  I do no work with CFD, and have not since the days when grad students waited for their time slots on a NASA supercomputer in the early morning hours to run their sims.  But I was not directly involved with their work.  I was an undergrad at the time.

[mrwayne]

Sometimes simple things get missed.

It doesn't help if the missed things are counter intuitive.

If you assume a system can only be 100%, you must conclude that any loss means no Net Energy.

If the assumption holds true - no Net Energy.

................

Yet what happens when the system is 105% then 160% or 340%

Can you as a designer choose to use components that have some losses and still have NET.

and more over - if a standard car engine is 33% efficient - and powering a 330% efficient ZED -

Pretty unlikley to have enough losses that result in no Net gain.

...................

Just saying - you need to open the Box a bit.