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Author Topic: Mathematical Analysis of an Ideal ZED  (Read 754003 times)

MarkE

• Hero Member
• Posts: 6830
Re: Mathematical Analysis of an Ideal ZED
« Reply #165 on: February 27, 2014, 09:38:18 PM »

MarkE,

Our Zed spreadsheet output formulas are F*ds as Force * Stroke. The input Ft Lbs formulas use P average * Volume of the fluid moving into the Pod retainer. That also increases the head in the risers.

The pressure rises faster than would account for just the input volume height, due to the riser head change. Are you now saying we have to account for the riser head change? That would be double dipping as its already factored in P average.

Larry
Larry, under conditions where force changes as a function of distance (height) as it does in your example, then the integral of F*ds becomes integral(f(z)dz).  The incompressible fluid in the three columns transmits pressure between each.  At the end state: The right most column has a head of 4' that presses down trying to push the entire fluid volume towards the left.  The middle column has a head of 2' the presses down trying to push the entire volume towards the right.  The net pressure is: (4-2)ft*0.65psi/ft.  The leftmost column adds one more foot of head that the input energy source has to push against.  Now, the total pressure that the input source has to work against at the end is:  (4-2+1)ft*0.65psi/ft = 3*0.65psi/ft.  The total force at that point is:  3*0.65psi/ft*area.  Therefore:  the force that the input source has to work against changes from 0 to 3*0.65psi/ft*area as the input head changes from 0 to 1ft.  The input energy is therefore the integral of:  3*0.65psi/ft*area/ft*z dz, evaluated from 0 to 1ft input z.  That evaluates to:  0.5*3*0.65psi*area*ft in ft. lbs. work as the drawing shows. That input energy identically matches the change in stored energy from 18*0.5*0.65*area to 21*0.5*0.65*area.  That is the inescapable physical reality.

If you plug non-physical formulas into a calculator, spreadsheet, or computer codes, you simply create bogus results.

If you are having difficulty with these concepts, then I suggest looking at the situation in two places:

The incremental work that you have to do to pump in the first microinch, and the work that you have to do pumping in the last microinch.  When you start the two 3 inch heads cancel the net force seen by the pump at the left most tube:  F~= 0, and the work to pump in 1uinch head of fluid is also ~0. When you end, the input source is working against (virtually) the right head plus the left head less the middle head and the work is: E ~= F*s ~=3*0.65psi/ft*area*1uinch.

TinselKoala

• Hero Member
• Posts: 13958
Re: Mathematical Analysis of an Ideal ZED
« Reply #166 on: February 27, 2014, 09:58:03 PM »
Quote
If you plug non-physical formulas into a calculator, spreadsheet, or computer codes, you simply create bogus results.

Heh... IGI-IGO:

Incompressible garbage in, incompressible garbage out.

Sounds like the plumbing in this old house.

mondrasek

• Hero Member
• Posts: 1301
Re: Mathematical Analysis of an Ideal ZED
« Reply #167 on: February 27, 2014, 10:04:44 PM »
MarkE and LarryC,

I took some liberties with your current analysis to illustrate something that I think might be helpful to some who are following along.  Feel free to ignore if it does not raise any talking points for you two.

M.

TinselKoala

• Hero Member
• Posts: 13958
Re: Mathematical Analysis of an Ideal ZED
« Reply #168 on: February 27, 2014, 10:47:41 PM »

LarryC

• Hero Member
• Posts: 911
Re: Mathematical Analysis of an Ideal ZED
« Reply #169 on: February 27, 2014, 10:54:17 PM »
Larry, under conditions where force changes as a function of distance (height) as it does in your example, then the integral of F*ds becomes integral(f(z)dz).  The incompressible fluid in the three columns transmits pressure between each.  At the end state: The right most column has a head of 4' that presses down trying to push the entire fluid volume towards the left.  The middle column has a head of 2' the presses down trying to push the entire volume towards the right.  The net pressure is: (4-2)ft*0.65psi/ft.  The leftmost column adds one more foot of head that the input energy source has to push against.  Now, the total pressure that the input source has to work against at the end is:  (4-2+1)ft*0.65psi/ft = 3*0.65psi/ft.  The total force at that point is:  3*0.65psi/ft*area.  Therefore:  the force that the input source has to work against changes from 0 to 3*0.65psi/ft*area as the input head changes from 0 to 1ft.  The input energy is therefore the integral of:  3*0.65psi/ft*area/ft*z dz, evaluated from 0 to 1ft input z.  That evaluates to:  0.5*3*0.65psi*area*ft in ft. lbs. work as the drawing shows. That input energy identically matches the change in stored energy from 18*0.5*0.65*area to 21*0.5*0.65*area.  That is the inescapable physical reality.

If you plug non-physical formulas into a calculator, spreadsheet, or computer codes, you simply create bogus results.

If you are having difficulty with these concepts, then I suggest looking at the situation in two places:

The incremental work that you have to do to pump in the first microinch, and the work that you have to do pumping in the last microinch.  When you start the two 3 inch heads cancel the net force seen by the pump at the left most tube:  F~= 0, and the work to pump in 1uinch head of fluid is also ~0. When you end, the input source is working against (virtually) the right head plus the left head less the middle head and the work is: E ~= F*s ~=3*0.65psi/ft*area*1uinch.

MarkE,

I think you picked up my P average from the example and using it as .65psi/ft. Water is .43psi/ft. That's part of the confusion. So going from 0 to 3*.43 makes sense. But When the pressure change is linear, as in the example and simulations, the Integral resolves to Pin average * Vin, this was also stated by M. earlier. Now if you don't agree, I can write a program with your micro inch height change and calculating the force, but it should be obvious.

Larry

MarkE

• Hero Member
• Posts: 6830
Re: Mathematical Analysis of an Ideal ZED
« Reply #170 on: February 27, 2014, 11:24:12 PM »

MarkE,

I think you picked up my P average from the example and using it as .65psi/ft. Water is .43psi/ft. That's part of the confusion. So going from 0 to 3*.43 makes sense. But When the pressure change is linear, as in the example and simulations, the Integral resolves to Pin average * Vin, this was also stated by M. earlier. Now if you don't agree, I can write a program with your micro inch height change and calculating the force, but it should be obvious.

Larry
Larry, yes I picked up the wrong constant for water density.  However, it drops out of the equations in terms of relative work.  The initial stored energy is 18X 0.5*pWater*area/height, the final stored energy is 21X the same quantity, and the energy input is 3X the same quantity which is the exact difference between stored energy at the end versus the beginning.  Energy is conserved.  There is no gain.

Here is the drawing updated removing the specific density coefficient for water.  Substitute whatever fluid for the water that you like, and plug in the corresponding density in the units of your preference.  The equations still work the same way.  The relative quantities do not change.

You are welcome to code whatever you like as long as you can show that it if physically reasonable.  We know that at the end of the pumping cycle that we are supporting net 3ft of fluid.  Therefore we can write a simple computer program:

NumberOfSteps = 1000 ;
FluidHeightEnding = 1#;
FluidDensity = 1# ;
FluidArea = 1# ;
FluidHeightIncrement = FluidHeightEnding/NumberOfSteps ;
Kf = 3#*FluidDensity*FluidArea*FluidHeightIncrement ;
EnergyIn = 0 ;

for(StepCount = 0;StepCount < NumberOfSteps;StepCount++)
{
ForceAverage = 0.5*(StepCount+StepCount+1)*Kf ; //Average force between the start and end of the step
EnergyIncrement = ForceAverage*FluidHeightIncrement ; //Average force * incremental height added.
EnergyIn += EnergyIncrement ;
}

Whether you use 100 steps, 1000 steps, or 1,000,000 steps, for 3 units height you will get exactly 1.5*FluidDensity*FluidArea each time, which is identically the 3*0.5*FluidDensity*FluidArea of the 0+3+3 versus the 1+2+4 configuration.

LarryC

• Hero Member
• Posts: 911
Re: Mathematical Analysis of an Ideal ZED
« Reply #171 on: February 28, 2014, 12:33:47 AM »
Larry, yes I picked up the wrong constant for water density.  However, it drops out of the equations in terms of relative work.  The initial stored energy is 18X 0.5*pWater*area/height, the final stored energy is 21X the same quantity, and the energy input is 3X the same quantity which is the exact difference between stored energy at the end versus the beginning.  Energy is conserved.  There is no gain.

Here is the drawing updated removing the specific density coefficient for water.  Substitute whatever fluid for the water that you like, and plug in the corresponding density in the units of your preference.  The equations still work the same way.  The relative quantities do not change.

It is conservative. Never said there was a gain with the multiple connected column setup. Said that you can get the same PSI with 1/3 the input Ft Lbs (using P average * Volume). If you released, it would have the same 1/3 output Ft Lbs.

Smaller fluid volumes in and out for the same PSI reduces cycle time. Cycle time reduction increases HP. To take advantage of its properties you need a setup that can use the rise in water in the first column (Archimedes) and also use the PSI generated by the second and third columns (Riser dome).

Larry

MarkE

• Hero Member
• Posts: 6830
Re: Mathematical Analysis of an Ideal ZED
« Reply #172 on: February 28, 2014, 12:40:45 AM »

It is conservative. Never said there was a gain with the multiple connected column setup. Said that you can get the same PSI with 1/3 the input Ft Lbs (using P average * Volume). If you released, it would have the same 1/3 output Ft Lbs.

Smaller fluid volumes in and out for the same PSI reduces cycle time. Cycle time reduction increases HP. To take advantage of its properties you need a setup that can use the rise in water in the first column (Archimedes) and also use the PSI generated by the second and third columns (Riser dome).

Larry

Larry comparing force or pressure with energy is a pointless exercise.  They are not comparable quantities.  I can get lots and lots of force and / or pressure with zero work or lots of work.

Changing time scales without holding energy constant does not lead to power.  The system you have presented is both ordinary and conservative.

You need to fix your spreadsheets so that they reflect the actual energy values.

mondrasek

• Hero Member
• Posts: 1301
Re: Mathematical Analysis of an Ideal ZED
« Reply #173 on: February 28, 2014, 12:59:34 AM »
Larry comparing force or pressure with energy is a pointless exercise.  They are not comparable quantities.

The Integral of Pressure * Volume is Energy.

The Integral of Pressure is equal to the average of Pstart and Pend for an incompressible fluid.

MarkE

• Hero Member
• Posts: 6830
Re: Mathematical Analysis of an Ideal ZED
« Reply #174 on: February 28, 2014, 01:42:32 AM »
The Integral of Pressure * Volume is Energy.

The Integral of Pressure is equal to the average of Pstart and Pend for an incompressible fluid.
No.  You are mixing circumstances of compressible and non-compressible substances.  Work is always the integral of F*ds.  If we take a capsule of fluid and subject it to 1psi or a million psi we have not done any work on that fluid.  If we apply pressure against a cross section of fluid through a distance, then we do work.  When we lift columns of fluid we can obtain the work performed and stored by solving the F*ds integral which will work out for a single column to:  E=0.5*total_weight*height = pave*volume = 0.5*density*volume*height = 0.5*density*area*height2

The energy is not stored in compression of the fluid for the simple reason that the fluid is incompressible.  The energy is stored in the gravitational potential of the raised mass.  Larry  asserted that raising some cross-section by 1' to end up with the 1+2+4 configuration "cost only 1/3" of some other configuration.  But it doesn't.  The force went from 0 to 3X what it would have raising an isolated column by 1'.  Identically, the amount of work performed was 3X that required to raise an isolated column by 1'.  The force and the energy both scaled by 3X versus the isolated column.  Had we done the exercise totally emptying the middle column, then the force would have gone from zero to 9X over a 3X stroke.  Kf would still be 3*pWater*area, and the integral would be:  0.5*3*pWater*area*(32-0) = 27*0.5*pWater*area, IE 27X the energy of raising an isolated column by 1' and 3X the energy of raising an isolated column by 3'.

mondrasek

• Hero Member
• Posts: 1301
Re: Mathematical Analysis of an Ideal ZED
« Reply #175 on: February 28, 2014, 02:18:58 AM »
No.  You are mixing circumstances of compressible and non-compressible substances.  Work is always the integral of F*ds.  If we take a capsule of fluid and subject it to 1psi or a million psi we have not done any work on that fluid.  If we apply pressure against a cross section of fluid through a distance, then we do work.  When we lift columns of fluid we can obtain the work performed and stored by solving the F*ds integral which will work out for a single column to:  E=0.5*total_weight*height = pave*volume = 0.5*density*volume*height = 0.5*density*area*height2

The energy is not stored in compression of the fluid for the simple reason that the fluid is incompressible.  The energy is stored in the gravitational potential of the raised mass.  Larry  asserted that raising some cross-section by 1' to end up with the 1+2+4 configuration "cost only 1/3" of some other configuration.  But it doesn't.  The force went from 0 to 3X what it would have raising an isolated column by 1'.  Identically, the amount of work performed was 3X that required to raise an isolated column by 1'.  The force and the energy both scaled by 3X versus the isolated column.  Had we done the exercise totally emptying the middle column, then the force would have gone from zero to 9X over a 3X stroke.  Kf would still be 3*pWater*area, and the integral would be:  0.5*3*pWater*area*(32-0) = 27*0.5*pWater*area, IE 27X the energy of raising an isolated column by 1' and 3X the energy of raising an isolated column by 3'.

Have you met Fletcher?  (sorry, but he is AFK) for a bit.  But I think he would like this exchange immensely!

And so do I.  Thank you for all your input and hard work.  The introduction of the maths you have been presenting to this topic has been a breath of fresh air to say the least!

M.

PS.  I did not actually read through and "digest" your post, and I apologize.  I'm otherwise preoccupied.  But I do plan to check it out shortly (hopefully in the morning).  You have taught me alot so far, and again, I thank you.  I do appreciate your considerate contributions to this forum!

LarryC

• Hero Member
• Posts: 911
Re: Mathematical Analysis of an Ideal ZED
« Reply #176 on: February 28, 2014, 07:15:27 AM »
Darn, Gif display problem.

minnie

• Hero Member
• Posts: 1244
Re: Mathematical Analysis of an Ideal ZED
« Reply #177 on: February 28, 2014, 03:27:52 PM »

Hi,
I've got a feeling if you used liquids with different sg's instead of air the thing would
work just like a hydraulic jack. Wouldn't be much use because pressure would be so
small and then it would vent.
mrwayne  seems to have gone quite quiet all of a sudden.
In industrial applications hydraulic pumps can deliver pressures of up to 20,000 psi
which seems very far removed from what we're messing with here!
John.

mondrasek

• Hero Member
• Posts: 1301
Re: Mathematical Analysis of an Ideal ZED
« Reply #178 on: February 28, 2014, 03:49:12 PM »
I've got a feeling if you used liquids with different sg's instead of air the thing would
work just like a hydraulic jack.

Minnie, that is exactly the gist of the analysis I performed.  We are assuming the air is incompressible, so we are, in effect, using two liquids with different SG's.  I then tried to see if the ZED performed exactly like a hydraulic jack by inputting a known amount of energy in the form of a specific volume of fluid over a specific pressure range.  Then I checked to see if the ZED (jack) would rise by the appropriate amount to provide an equal energy output in the form of the outer riser (jack piston) rising a balancing calculated volume for the specific pressure range that was developed due to the buoyant forces the input created.  It did not act just like a jack (a simple hydraulic cylinder).  The 2-layer ZED (jack) would fail to stroke far enough, so under unity.  The 3-layer ZED (jack) would stroke further than predicted, so over unity.

LarryC

• Hero Member
• Posts: 911
Re: Mathematical Analysis of an Ideal ZED
« Reply #179 on: February 28, 2014, 04:28:15 PM »
Larry, I've looked at the latest spreadsheet.  There is still a good deal of work to do here.  Please refer to the drawing below:

We need to either insure that the starting and ending energy states are identical, or else account for the stored energy in each state.    No matter what, we do need to calculate the work added and going from each state to the next.  Knowing the stored energy at all states provides a good sanity check.  Please be aware that each time water equalizes from a taller single column to two or more lower columns that we lose stored energy.  The drawing includes formulas for calculating stored energy under the assumptions previously stated:

20C
G0=9.80665m/s/s
zero thickness walls
25" diameter pod
26" diameter pod chamber
51 circular inch riser gap and riser head areas

MarkE,