Mathematical Analysis of an Ideal ZED

[mrwayne]

Wayne--  Will you be serving up roasted crow in a couple days??

I guess that depends on the Vantage point....

I will let you know soon - or write me at home.

Wayne

[MarkE]

Mark D gives excellent advice.

And if it were not for the non Troll here on this we site - it would be a total waste of my time.

I am letting the Trolls do what they do - that's it.

They Lie constantly for self ego.......they ruin Stefans web site and try t suppress technology.

Mark D - give honest guidance, correction, and guidance.

Trolls..... Are Trolls....

When you help them - as you do.... you do not come across as nice and polite...

Take Care...

Wayne

"Troll" That is Wayne Travis speak for anyone who sees through his investment fraud.

[MarkE]

And you are incorrect again MarkE.  There is no need to specify any recovery mechanism for an IDEAL ANALYSIS. 

That is utter and total nonsense.  If one is going to analyze a machine or process then that machine or process must be defined.  Define it as dumping a bunch of the input energy each cycle, then you have defined in that loss.

You keep ignoring that fact, but doing so does not make your requirement correct.  We need only be concerned with calculating the ENERGY that enters or leaves the system.  Not how it is captured.  That is only necessary for a PRACTICLE DEVICE, not an IDEAL ANALYSIS.

Again that is total nonsense.  You are free to add what you like, but you must add it.

I could similarly ask you to define a mechanism that makes water incompressible, or friction losses zero.  They are the IDEAL conditions as are required by an Ideal Analysis.

Stipulating efficiency of something you are evaluating the efficiency of is circular reasoning.  If you want to go that route, we can stop before you begin by noting that all passive mechanisms are lossy.  Therefore any ideal machine you define is lossy.  Then we can just call it a day.

The spillway was never needed to calculate the Energy that is expended during the lift process.  And neither is any other physical recovery device needed to account for Energy leaving or entering the SUT.

Oh, but it was, because as shown, it cannot deliver all the energy expended.  That was a specific point you contended as I recall.

The Energy leaving the ZED during the State 3 to State 1 leg of the cycle is real and needs to be properly accounted for.  That Energy value is 0.2022mJ.  The complete cycle loss is correctly calculated as 0.2452mJ and the efficiency is 88.3% for the Ideal SUT.

No, you are wrong again.  I suggest you write out the energy balance because your cycle efficiency number does not correspond to either case:  Dumping the water without energy recovery in the ST3 => ST1 transition, or dumping the water with a high degree of energy recovery in that transition for a ST1 => ST2 => ST3 => ST1 cycle.

[MarkE]

Wayne--  Will you be serving up roasted crow in a couple days??

Persons who do not understand that gravity is conservative can always hope.

[mondrasek]

I suggest you write out the energy balance because your cycle efficiency number does not correspond to either case:  Dumping the water without energy recovery in the ST3 => ST1 transition, or dumping the water with a high degree of energy recovery in that transition for a ST1 => ST2 => ST3 => ST1 cycle.

ST1 => ST2 2.0984mJ of Energy enters the system.  Please note that the Energy is added by a non-specified input mechanism.

ST2 => ST3 1.6510mJ of Energy (maximum) exits the system during the lift.

ST3 => ST1 0.2022mJ of Energy exits the system during this reset phase.  Please note that this Energy could be recovered by a non-specified collection mechanism.

Energy Out/Energy In = (1.6510mJ + 0.2022mJ)/2.0984 = 88.3%.