Mathematical Analysis of an Ideal ZED

[MarkE]

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]

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]

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]

O U Kids.

http://www.youtube.com/watch?v=c8WYI7QCj0k

[LarryC]

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