Mathematical Analysis of an Ideal ZED

[MarkE]

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]

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]

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]

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 P_start and P_end for an incompressible fluid.

[MarkE]

The Integral of Pressure * Volume is Energy. 

The Integral of Pressure is equal to the average of P_start and P_end 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'.