Mathematical Analysis of an Ideal ZED

[TinselKoala]

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]

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]

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 V_out 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]

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]

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