Hydro Differential pressure exchange over unity system.

[MT]

Hello TK,
thank you, I forgot to say in step 2 is riser locked but you figured it out already. I was just curious to know what pressure it can produce by using compartements as in ZED. It seems to me also that only 4. I'm trying to understand how the lift pressure builds up by adding second riser to a one riser model. Have lunch now, I'll be back...
Marcel

[parisd]

Marcel,

You cannot start with the assumption that air is not compressible. Air is compressible and if you want the air to be pressurised it need to be compressed, this is the equation of perfect gas P1V1=P2V2 valid at constant temperature, so to double air pressure in a closed system you need to reduce its volume by half.

Water is not compressible, if in a closed system (but with no air trapped in) it can easily be pressurized, in the Zed water has an atmosphere of air on top, so to pressurize that water you also need also to compress the air on top of it, this require a certain amount of water to be injected (all depending of the volume of air displaced by water). So the vertical force F on the horizontal surface doesnt come for free, it comes from the compression effect.

I have not yet figure how additional layers bring more efficiency.

Hi all,


I have one question see picture below.


Simplifications:
1. air is not compressible, always keeps volume
2. red riser have infinitely small weight

Step 1 is initial situation. Riser have no lift, there are 4 units of water in left and 4 in right side. Pressure inside of riser is the same as outside = 1atm.
Step 2. Adding 2 units of water. Since air is not compresible this action will push water in column 2 and 11 down by one unit and also raise columns 1 and 12 by one.

Question: what is the lifting force F after step 2? Equivalent weight of 4 units of water or 20units?

Marcel

[neptune]

Neptune .. MIdone is correct - in your example of the black box you would not have accounted in your energy budget for the PE of the spring & the work done joules to compress it [WD = f x d] - after system losses there would be a net loss of joules.
That was exactly my point. I had not accounted for the energy budget of the whole cycle, AND NEITHER HAD MIdone.

Parisd .. my 2 cents.

I believe this will probably turn out to be a measurement/interpretation error by Mr Wayne & his team - having said that this is the fun of doing experiments & garnering good data along the journey.

I will give you a clue where that measurement error might be occurring, IF there  is one.

It has to do with Pascal's principles & enclosed hydraulic systems - pressure is transmitted undiminished in an enclosed system & Wayne's system is a mixture of buoyancy & hydraulics, all under pressure - often pressure & force are confused which is unfortunate - so IMO a force is generated & magnified [it has to do with area ratio's] due to Pascal's principles - because it is basically a hydraulic system the force is multiplied in the system which is seen/interpreted as buoyancy potential increase to lift mass - but if, as some might know, when you take that magnified force output & add it to the weight force of the system fluids & compressed air you get a high internal force figure - when this is converted back to pressure [pgh] & then you recalculate the height potential [h] by rearranging the formula to find hypothetical 'h' i.e. Static Pressure = F / A = pgh therefore h = P / pg you may find the calculated 'h' doesn't match the heights of fluid & air in the actual system - this is an anomaly.

In short the Pascal force multiplication is added to the weight force of the system & by some math is converted to a system pressure which deduces a different 'h' than actual.

The upshot is that there appears a potential to lift virtual mass because of the force multiplier effect & the pressurized system.

The trouble is in 'normal world' that for both hydraulics & buoyancy to do work [f x d] then the Effort to Load [work done] ratio is zero sum energy wise i.e. f1 x d2 joules = f2 x d1 joules, less losses.

JMO's at this time - I watch with interest.


........................

See3d .. I don't know what formula's you are working thru but one comes to mind as a possibility.

That might be for a fluid drag formula to take account of viscosity - a simplified one I might build & use would be say ...

Coefficient of Drag 0.36 [change this factor as desired]

Viscosity Drag = Cd . 1/2 . density . velocity squared . area               i.e. Vd = Cd.0.5.1000.v^2.A

Where 1000 is density of water, v is velocity of riser etc, A is area of riser etc.

You would need to include an IF statement in the formula in your sim to account for changing directions & velocities of fluids i.e. viscosity dampening works up & down.

Here is a simple formula I quickly built & would consider using.

Input[Cd]*0.5*1000*if(Output[velocity].y1<0,Output[velocity].y1^2,Output[velocity].y1^2*-1)*0.01

Considerably simplified for a generic sim could be this stripped down one i.e. adding a viscosity dampening effect to buoyancy fairly similar to above in behaviour.

Input[Cd]*if(Output[velocity].y1<0,Output[velocity].y1^2,Output[velocity].y1^2*-1)*100

P.S. viscosity represented as a force is a small system energy loss but probably should be included for sim accuracy & to dampen oscillations.

The density figure has to be adjusted to say 1.225 kg/m^3 for air at 1 atmosphere etc - Vd = 0.36.1/2.p.v^2.A  where p = density

@Fletcher. You totally failed to understand my post. I was totally aware that my black box example did not allow for the spring recompression in its energy budget. My purpose was to show MIone that he had only analysed half a cycle of the ZED, and thus made the same mistake as I deliberately made in the black box example.

[fletcher]

MT .. Mr Wayne has categorically stated that the system would perform equally well using two incompressible fluids of different densities, so your example for analysis purposes of considering air as non compressible is valid.

He also said that a compressible gas is not the 'secret' ingredient for why it is OU in his view - as has been pointed out by Parisd gases follow the ideal gas laws [within reason] of a direct trade off of volume & pressure i.e. they are inverse in ideal conditions - this means that energy used to compress a gas is stored as PE in the gas, & after some internal losses where temperature/heat might enter or leave the system [minimal in a short time frame; search adiabatic & isothermal legs of a carnot cycle] - therefore the compressed gas will give back the PE as KE - this is the perfect spring analogy that many have talked about.

TK is correct IMO - the riser is locked & the fluid has to have more & more work done on it to enter the riser chamber as the pressure inside builds - but what is important is that the non compressible air will displace the fluid raising the head - to see the relationships clearly you have to work with volumes rather than 2d areas - the upshot is that any volume injected into the riser chamber has an energy cost - although the fluid head rises & does create more pressure the buoyancy/lift potential is 'short lived' & directly related to the volume injected which displaces an equal volume of fluid etc.

Some further thoughts to consider - pressure & force are two sides of the same coin but only one does Work - internally the thin wall of fluid is pushed upwards which raises the head - static pressure of a fluid = pgh in N/m^2 - if we assume the fluid has a density of 1000 kg/m^3 & g is 9.81 m/s^2 then it is easy to calculate the pressure in Newtons per meter squared [Pascals].

The buoyancy force is related to displacement volume - in your example you assume that the riser is virtually massless - this means that the buoyancy force = -mg N's of the volume displaced - in this case we have to use a 'virtual' volume [because the pressure is increased by the high thin head i.e pgh] - since the riser has next to no mass we can neglect weight force of the riser which would counter the buoyancy force.

So, we have a net upthrust force [buoyancy force less weight force] based on a virtual volume displacement due to high head [pgh = static pressure] - the penalty is that as soon as you release the riser to do Work [ f x d ] it has a very short stroke length - this is because everything is volume dependent [just like Archimedes].

N.B. Static Fluid Pressure = pgh   in N/m^2    or         Vpg / A          i.e   Ahpg / A = pgh       &     P = mg / A     

P = F / A  in N/m^2   ...   F = P x A in N

So now Pressures can be converted to Forces & Forces converted to Pressures.

Note: Forces x Distances do Work - Pressures do not.

Edited to add units.

   

[mondrasek]

Now this *is* Pod racing!  (Cheesy, eh?  But the best I could come up with.)

TK?  Please let me know what you think about the build.  TRUST ME!  I am not happy with it at all!  Maybe build 7.0 or so will not be so embarassing.

But WE do what we can, because we can, and we should (or must?)...

M.

PS.  @fletcher, Thank You for spending time on this thread.  You are a Gentleman.