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Author Topic: Mathematical Analysis of an Ideal ZED  (Read 749472 times)

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #105 on: February 25, 2014, 11:48:07 PM »

Went ahead and added a riser gap that would give it the same SI as the pod gap. Wanted to see how big a difference it would make in the efficiency. Darn, it went from 153.94% to 153.56%.
Larry
Larry, that helps.  I will continue to go through the spreadsheet.  Please confirm that the drawing below is correct:

TinselKoala

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Re: Mathematical Analysis of an Ideal ZED
« Reply #106 on: February 26, 2014, 12:05:58 AM »
That would suck on so many levels.  But I think we could then resort to labels.  But it would be so much harder to follow.

My apologies to any colorblind individuals trying to follow along.

TK, I never said that the atmospheric air pressure makes any difference to any pressures inside the ZED.  Thankfully, it doesn't!  What having the outer annulus open to the atmosphere does do is allow for the Vin to NOT equal the Vout.  Because air also moves freely into and out of the system from the atmosphere and adds ANOTHER V (that is not costing us anything) that must be accounted for in an Energy Balance.

The air that freely crosses into and out of the system allows the water levels to redistribute to satisfy simple volumetric constraints.  And when allowed to do so it results in a lift force that is due to BUOYANCY, not the usual pressure * volume relationship found in a simple hydraulic cylinder.

Air freely crosses the boundary of your incorrectly drawn Red Box. The air does not enter the Zed system at all. The outer ringwall's only purpose is to keep the two outer incompressible fluid columns from running off down the drain. The level of the outermost liquid layer is partly determined by the pressure--- the _constant pressure_ -- exerted by the air, and this does not change, it does not move in and out of the system. The only thing the air pressure does is to press down on the outer fluid columns, and it presses down with the same pressure, a constant force, no matter the stage of the system. It does not add another "V" .... the volume of the outer air is essentially infinite, anyway, and so it does not change. Now, if you want to build a zed that is a hundred thousand feet tall, things will be (slightly) different wrt the outside air, because then the pressure it exerts on the top of the outer fluid layer will change, from zero altitude to the highest altitude of the water level. But for realistically sized Zeds, the volume and pressure of the outer air is constant, volume infinite and unchanging, and at the local atmospheric pressure. And don't forget, please, that the outer air pressure is also pressing down on the entire top surface of your apparatus, with that same pressure, but over a much larger area than the area of the outer fluid column.

The proper drawing of your red box would follow the outline of the surface of the outer liquid layer, I think, because that is the sealed surface. It rises and sinks, of course, but that's no problem as far as the boundary condition goes.

TinselKoala

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Re: Mathematical Analysis of an Ideal ZED
« Reply #107 on: February 26, 2014, 12:14:49 AM »
The interview went well.

Now that I am sure you know the concept of making ignorant claims against other people is wrong - I will apologize to you when you do.

Wayne

Save your apologies for the judge. I understand they take remorse into account when sentencing.


Now.... just what part of "Show me the sausages" don't you understand?
Quote
Show me the Sausages!    A philosopher designs a marvellous sausage machine. A scientist comes to marvel at this wonderful creation, and raises an eyebrow.
 The philosopher says, "Ah, behold the wonderful cogs and sprockets and temperature-controlled mixing chambers in my wonderful machine - surely you can see how it must produce the most fantastic sausages!"
 The scientist says "Yes, that is all very interesting. Show me the sausages."
 The philosopher says "How dare you, a mere scientist, question my wonderful philosophical reasoning?"
 Scientist: "I'm not questioning your reasoning - I want to know if your machine really produces sausages."
 Philosopher: "Can you point to any flaw in my argument that it produces sausages?"
 Sci: "I don't know - I just want to know if it produces sausages. Here is some meat. Why don't you feed it through and see if you get any sausages?"
 Phil: "And sully my wonderful machine with mere offal?"
 Sci: "You said it was a sausage machine. I want to see the sausages."
 Phil: "Are you questioning my ingredients?"
 Sci: "I'm just questioning whether it produces sausages or not. Show me the sausages."
 Phil: "Ah, so you cannot attack my premises and you cannot attack my argument. Therefore I'm right and you lose."
 Sci: "Don't be such a melodramatic prancing arse. Show me the sausages."
 Phil: "The sausages inevitably flow from the argument. You see my fine machine.  You can even inspect the meat & onions. The sausages necessarily flow."
 Sci: "Show me the sausages or I'm off to Tesco."
 Phil: "You are a mere scientist with no understanding of philosophical matters."
 Sci: "Bye."

http://answersingenes.blogspot.com/2011/06/show-me-sausages.html

You see, Travis? You aren't the only person with no sausages.
Don't be such a melodramatic prancing arse. Show me the sausages.

mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #108 on: February 26, 2014, 12:27:34 AM »
The proper drawing of your red box would follow the outline of the surface of the outer liquid layer, I think, because that is the sealed surface. It rises and sinks, of course, but that's no problem as far as the boundary condition goes.

But that would require that the red box changes shape (edit: but NOT change by an unpredicted volume) as the Energy Balance is performed.

Maybe so.  Could you explain further?

LarryC

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Re: Mathematical Analysis of an Ideal ZED
« Reply #109 on: February 26, 2014, 12:34:20 AM »
Larry, it is not my hang up.  It is that the math does not represent the model represented.  It's like estimating pi as 3.  Whether or not that is close enough depends on the circumstances.  In order to determine the magnitude of the error, one has to reverse engineer your spreadsheet, guess your intent, then substitute the correct relationships and evaluate the differences.  That is a big PITA and rather unreasonable.  It would be very helpful for you to state your assumptions, and for you to perform sanity tests on your own as to the validity of those assumptions.  Introducing ~16% error terms is a recipe for trouble.  I don't care if you fix the ring dimensions for constant area or keep them on a 0.5" grid as long as your calculations represent the model faithfully.

Let me make a suggestion that will make it easier to keep simple numbers on the spreadsheet:  Assign a constant to pi/4.  Then you can represent all your circular areas in integer units multiplied by the constant.  This should make it easier for you to audit your calculations.  The other thing that can be an immense help is to use named fields.  That saves a lot of chasing around.  Instead of a formula looking like: = $H$2*F19*E12 it would look like:  = riser_diameter*riser_length.  My last suggestion you may or may not like:  Using MKS units generally makes it easier to avoid mistakes between mass and force.  I can work in whatever units you are comfortable using.


MarkE,


Thanks for the suggestion, some will help.


But, don't understand your ~16% error, the .38% drop in efficiency was a ~.2% error.


PI/4 constant would help.


On the named fields, I do use them all the time for VBA modules. But, It is a good suggestion to use names on the constant parameter fields at the top for this example. I do like to use them at the multiple line level, in this case the 'Cycles', because when you copy one line to the next, the named fields do not increment its position. I use the 'Trace Precedent' to check formulas, it points to all the fields in the formula.


I agree with your points about MKS units. I don't use it because I've worked with many field engineers that use Imperial because the field workers that apply the specifications wouldn't understand and most times upper management wouldn't either. And I am more comfortable using Imperial, so thanks for working with my unit choice. 


Larry   




TinselKoala

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Re: Mathematical Analysis of an Ideal ZED
« Reply #110 on: February 26, 2014, 12:35:04 AM »
But that would require that the red box changes shape (edit: but NOT volume) as the Energy Balance is performed.

Maybe so.  Could you explain further?

Consider a simpler system for a moment.
A syringe full of air is connected by a short tube to a deflated balloon.

Where do you draw your red box now? Do you include the outside air that the balloon expands into when you depress the plunger? How much of it? Or does your red box follow the actual perimeter of the "hard parts" of the system? Here, the outer air is displaced by the expanding balloon, isn't it? But its only effect on the expansion is its pressure, which remains constant no matter how big the balloon is.
Since the balloon is increasing in total surface area, the total _force_ exerted by the outside air grows.. but that doesn't happen in your Zeds because the surface area of the fluid column is constant.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #111 on: February 26, 2014, 12:57:51 AM »

MarkE,


Thanks for the suggestion, some will help.


But, don't understand your ~16% error, the .38% drop in efficiency was a ~.2% error.


PI/4 constant would help.


On the named fields, I do use them all the time for VBA modules. But, It is a good suggestion to use names on the constant parameter fields at the top for this example. I do like to use them at the multiple line level, in this case the 'Cycles', because when you copy one line to the next, the named fields do not increment its position. I use the 'Trace Precedent' to check formulas, it points to all the fields in the formula.


I agree with your points about MKS units. I don't use it because I've worked with many field engineers that use Imperial because the field workers that apply the specifications wouldn't understand and most times upper management wouldn't either. And I am more comfortable using Imperial, so thanks for working with my unit choice. 


Larry   
Larry, 28^2/26^2.  Actually, I slipped and the area error is 8%, still that is 40X the 0.2% you think resulted, so that should raise suspicion right there.  The annular ring areas are the differences of squares, so by proportion using your original numbers:  28dia - 27dia = 55cir_area versus 26dia - 25dia = 51cir_area:  55/51 ~8%.

I am going through the spreadsheet now.  I have created a new worksheet for the 2 ZED where I am using named formulas and have substituted the exact geometry relations.  However, I see a fundamental error:  It looks like you failed to integrate when calculating your energy.  The force required to lift a column of water increases with the head.  In order to get the actual energy we have to perform the integration.  This should make intuitive sense if you consider punching a pin hole near the bottom of one of the columns.  When the column is very full, the stream is very strong, and as the column comes down to the pin hole the stream dribbles off to almost nothing.

mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #112 on: February 26, 2014, 01:19:17 AM »
Consider a simpler system for a moment.
A syringe full of air is connected by a short tube to a deflated balloon.

Where do you draw your red box now? Do you include the outside air that the balloon expands into when you depress the plunger? How much of it? Or does your red box follow the actual perimeter of the "hard parts" of the system? Here, the outer air is displaced by the expanding balloon, isn't it? But its only effect on the expansion is its pressure, which remains constant no matter how big the balloon is.
Since the balloon is increasing in total surface area, the total _force_ exerted by the outside air grows.. but that doesn't happen in your Zeds because the surface area of the fluid column is constant.

I really like the idea of changing the red box to follow the outline of the ZED system outer riser and only include the water in the outer annulus!  That does make sense.

Now as the input charge is entering the bottom of the pod chamber (Energy in), some Energy is also leaving the "red box" through that initial barrier in the outer annulus.

But wait, isn't that water rising up through the top surface of the red box and is therefore another PE source that is being generated?

I've definitely got to check out how it all balances when I can get back to it.  Unfortunately Wednesdays are a travel day for me so I don't know if I can get to it quickly.

Thanks so much for this correction!

M.

PS.  Is there any problem you see with calculating the Vout of the outer riser breaching the top of the red box as the volume of that cylinder that rises above (stroke distance * surface area of the outer riser)?

LarryC

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Re: Mathematical Analysis of an Ideal ZED
« Reply #113 on: February 26, 2014, 02:09:53 AM »
Larry, 28^2/26^2.  Actually, I slipped and the area error is 8%, still that is 40X the 0.2% you think resulted, so that should raise suspicion right there.  The annular ring areas are the differences of squares, so by proportion using your original numbers:  28dia - 27dia = 55cir_area versus 26dia - 25dia = 51cir_area:  55/51 ~8%.

I am going through the spreadsheet now.  I have created a new worksheet for the 2 ZED where I am using named formulas and have substituted the exact geometry relations.  However, I see a fundamental error:  It looks like you failed to integrate when calculating your energy.  The force required to lift a column of water increases with the head.  In order to get the actual energy we have to perform the integration.  This should make intuitive sense if you consider punching a pin hole near the bottom of one of the columns.  When the column is very full, the stream is very strong, and as the column comes down to the pin hole the stream dribbles off to almost nothing.


Hi MarkE,


Liked what you did with the drawing, but has a few minor issues, I'll respond later on that post.


You're too fast, so I'll also respond to your ~8% later.


But wanted to address your integration concerns as it is key to the process.
The Left avg. psi at E13 is the average pod psi from the end of equalization to the start of ready to stroke.
The Right avg. psi at H13 using the same technique. So left 3.233, right 3.992. Now when adding water to the right along you would multiply that average psi times the volume to get the input cost.


But, if at the same time we are allowing water to flow out the left at its average PSI, it is returning energy to the Right. This return of energy reduces the input energy applied to the transfer by the average pressure differential between the two units. So the Average PSI Differential shown at I13 is the average pressure required to transfer the fluid. I can show this process in a little different format if it would help.


In your pinhole example, if you had two columns of water with one having 25% less water and you had a small tube connected to the pinhole in each column. Would the water flow out as fast as your one column example or would the speed be reduced by 75%? Its all about the pressure differential.


Larry     

LarryC

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Re: Mathematical Analysis of an Ideal ZED
« Reply #114 on: February 26, 2014, 02:38:04 AM »
Larry, that helps.  I will continue to go through the spreadsheet.  Please confirm that the drawing below is correct:


MarkE,


Thanks for continuing, like your analysis technique.


Few minor issues in the drawing. The Riser Head bottom arrow should be even with the water height in the next column to the left. I have .48150 for D5 Riser Gap and the arrows are pointing at the pod water, should move to the right one column for the Riser gap.


Larry


MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #115 on: February 26, 2014, 02:46:17 AM »

Hi MarkE,


Liked what you did with the drawing, but has a few minor issues, I'll respond later on that post.


You're too fast, so I'll also respond to your ~8% later.


But wanted to address your integration concerns as it is key to the process.
The Left avg. psi at E13 is the average pod psi from the end of equalization to the start of ready to stroke.
The Right avg. psi at H13 using the same technique. So left 3.233, right 3.992. Now when adding water to the right along you would multiply that average psi times the volume to get the input cost.


But, if at the same time we are allowing water to flow out the left at its average PSI, it is returning energy to the Right. This return of energy reduces the input energy applied to the transfer by the average pressure differential between the two units. So the Average PSI Differential shown at I13 is the average pressure required to transfer the fluid. I can show this process in a little different format if it would help.


In your pinhole example, if you had two columns of water with one having 25% less water and you had a small tube connected to the pinhole in each column. Would the water flow out as fast as your one column example or would the speed be reduced by 75%? Its all about the pressure differential.


Larry   
Larry, if I have a capsule of water and lift that capsule from one height to another then I can just multiply the weight by the distance moved and I am done because the gravitational force on that capsule is constant.  But if what I do is change the height of a column by pumping water in or letting water out then I need to perform the integration. 

Let's suppose that we have two water towers each 1 sq meter in area, and 2m high.  First, let's fill each to 1m height:

Fstart = 0
Fend = 9789N
F/z = 9789N/m
Integral of:  (kf*z dz) = 0.5*kf*z^2
zstart = 0
zend = 1m

0.5*9789N/m*1m^2 - 0 = 4894.5J
We have that in each of two columns, so the total energy stored is:  2*4894.5J = 9789J

Now, let's pump the water back out of one column into the other, emptying the one and filling the other:

Fstart = 9789N
Fend = 19578N  Filled
Fend = 0 Emptied
Kf/z = 9789N/m

work added to column filled = 0.5*9789N/m*(Hend^2 - Hstart^2) = 0.5*9789N/m*(2^2-1^2) = 3*4894.5J
work removed from column emptied = 0.5*9789N/m*(Hend -Hstart) = 0.5*9789N/m*(0^2-1^2) = -4894.5J
net work performed: 9789J

So, we had 2*4894.5J = 9789J in the two columns combined at the start, we added 3*4894.5J to the column that we took up to 2m, and we removed 4894.5J from the column that we emptied.  We have the same total volume of water, but we doubled the stored energy, and we added every Joule of that extra energy from the outside. 

The converse of this is that when we equalize a column filled to a higher level with a column filled to a lower level, we lose energy to heat.  The worst case is where we take a column filled to 2H and equalize it with an empty column so that we now have 2 each of 1H.   This is the exact complement of the fill exercise that I just described and it loses half the stored energy:  That energy is lost forever: kaput, toodleoosky, goodbye Charlie. 

Your spreadsheet appears to be calculating energy by a series of linear adds and subtracts based on average pressure.  That yields incorrect results.  Integration is necessary.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #116 on: February 26, 2014, 03:04:41 AM »

MarkE,


Thanks for continuing, like your analysis technique.


Few minor issues in the drawing. The Riser Head bottom arrow should be even with the water height in the next column to the left. I have .48150 for D5 Riser Gap and the arrows are pointing at the pod water, should move to the right one column for the Riser gap.


Larry
Larry, OK I fixed that.  Because the stipulation is equal cross-section area in each column, I changed the formulas to derive the diameters with the areas fixed at 51circular inches, IE 51*pi/4.  Consequently, the formulas are:

Riser area = (26^2 + 51 )*pi/4  = 570.9844647900
Vessel area = (26^2 + 2*51)*pi/4  = 611.0397711233
Riser diameter = (26^2 + 51 )^0.5 = 26.96293752543
Annular clearance = (26.96293752543 - 26)/2 = .4814687627128
Riser diameter = (26^2 + 2*51 )^0.5 = 27.89265136196

The precision of these numbers are not particularly significant once we defined the annular cavities to all have the same 51 circular inch areas and perform our calculations based on that stipulated area rather than calculated area. 

MileHigh

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Re: Mathematical Analysis of an Ideal ZED
« Reply #117 on: February 26, 2014, 03:19:34 AM »
Quote from Wayne:

Quote
I am sorry MH - you should ask Monderask or Larry to explain.

Quote from MarkE to Larry:

Quote
Your spreadsheet appears to be calculating energy by a series of linear adds and subtracts based on average pressure.  That yields incorrect results.  Integration is necessary.

Webby, Mondrasek, Larry:

It's wake up time.  Can't you see what's happening?  You guys are trying your best but you aren't technical and you are simply wrong.  Wayne is using you and stroking you and trying to pretend that you guys know what you are doing but you clearly don't know what you are doing.  Wayne is using you to create the pretense that you are right for his own reasons.  The reason is to create the general impression that you guys 'get it' and his system is real.

You are lucky to have guys like MarkE and TK to give you guidance.  If they work out the whole sequence with you then you will be lucky to get that amount of hand-holding.

Be clear that I have nothing against you guys.  You are simply being used by Wayne for his own nefarious purposes.

To Wayne:

Quote
I am sorry MH - you should ask Monderask or Larry to explain.

Not so.  If MarkE and TK do a full cycle with your "explainers," holding their hands all the way then you will be exposed for the cynical manipulator you desperately try to be in order to keep your gravy train flowing.

The whole thing is reprehensible.  The Hammer of Justice, I can hear the knocking in the distance.

MileHigh

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #118 on: February 26, 2014, 03:54:38 AM »
MileHigh, it could be a long time before you see any satisfaction wrt to Wayne Travis.  It's the odd burned investor or investors who can become real nightmares for even experienced sharpies.  When a burned investor has the resources and the intent to get satisfaction, they have been known take sharpies down a long, dark, and painful road.  Such cases are rare, but they do happen.


MileHigh

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Re: Mathematical Analysis of an Ideal ZED
« Reply #119 on: February 26, 2014, 03:58:21 AM »
Mark:

Yes I know many times they get away with it.  It's unfortunate.  How many people may have sunk their life savings into Wayne?  I don't know if we will ever know.

MileHigh