Mixed up between Hydraulic and Atmospheric Pressure on Fluids?

Model:
A vertical thin Tube_a of water (or any fluid) linked (at the bottom only) to a larger vertical Tube_b of water

One equation for hydraulic fluids for the above model is:
(For visual of 1st model below see "Fluid Pressure Calculations" at : http://hyperphysics.phy-astr.gsu.edu/hbase/pman.html#meac )
P_a = density * g * h + P_b
(Where P_a & P_b are the pressures applied downward at the surface of thin Tube_a & wider Tube_b)
Let say A_a (Area of Tube_a) = 1 and A_b = 3 and F_a = 2 and F_b = 3
P_a = F_a / A_a = 2 /1 = 2
P_b = 3 / 3 = 1
Thus P_a = density * g * h + P_b
2 = density * g * h + 1
************** therefore h = 1 / density * g (PARTAA)
-----------------------------------------------------------------
But then another equation for Hydraulic Press (to lift up a car) says Work_a = Work_b therefore F_a * D_a = F_b * D_b
(For visual of 2nd model below see: http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html#hpress and http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html#car )
(F is the force and D is the distance of vertical travel ....similar to h in the previous equation) (in fact I believe h = D_b - D_a)
Thus F_a * D_a = F_b * D_b
(replacing Force with P * A in the previous equation)
(2 * 1) * D_a = (1 * 3) * D_b
****************** Therefore D_a = 3/2 * D_b (PARTBB)
Since h in the previous equation is a vertical distance that is in some way related to D_a - D_b but then PARTAA does not match with PARTBB????
since h is related to the density of the fluid (h = 1/ density * g)?
So where did I make a mistake??
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Where you not thinking. Try to do your homwork on paper, please. Der alte Hexenmeister. "Many people would sooner die than think... and they do." --- Bertrand Russell.
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Der alte Hexenmeister wrote:

I dont like when people critisize instead of answer a simple question. Ok you can critisize on relativity instead of answer because it can get complex but it doesn't help.
Likewise i did it on paper with no success. Other's mistake the one with the density equation only to deal with atmospheric pressure applied at both ends which is wrong if you look again (unless I missed something).
And therefore if you answered based on that assumption, you can see how your simple reply can be useless and that you might be wrong.
If you really know the difference then please give it, for my brain juice cannot figure it out.
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snipped-for-privacy@hotmail.com wrote:

You don't have to worry about relativiters, they're Earth's 1905 moron grunt elevator operators from Africa.
The problem is the modelling. The tubes equation is a stantion equation. so it only work's normal to the gravity field. Put the hydraulic press is an invertible press so it works on conservation of energy, not pressure.

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snipped-for-privacy@netscape.net wrote:

I doubt it, both models are almost identical, both have pressure applied on both ends of each tube and both determine the final height of the fluid after the applied pressure.
In #1 h = h2 - h1 in #2 h = d1 + d2 (#1 uses the base for height where as #2 uses the intial height of the pistons (or fluid) when both pressures are the same on the fluid.

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Oh... If it is your likes, dislikes amd emotions you wish to discuss, I'll leave you to it, I don't like discussing people's whims. Der alte Hexenmeister. "Many people would sooner die than think... and they do." --- Bertrand Russell.

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Der alte Hexenmeister smarted off & wrote:

So why bother replying then?
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Actually I did on paper, what do you think i just wrote here.
It's a bunch of numbers and calculations....so where is the mistake?
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http://erptoday.com/ET/Manufacturing-Software-Vendor-Directory.aspx?ref=aw&gclid=CKffxZPPpYICFQdwEAod73qFXw http://www.denisonhydraulics.com / http://home.wxs.nl/~brink494/frm_e.htm http://www.ehow.com/how_13315_have-hydraulics-added.html http://www.poclain-hydraulics.com /
Der alte Hexenmeister.
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In sci.physics, snipped-for-privacy@hotmail.com
wrote on 30 Dec 2005 14:09:48 -0800

The mistake is in taking Androcles seriously... :-)
--
#191, snipped-for-privacy@earthlink.net
It's still legal to go .sigless.
  Click to see the full signature.
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The Ghost In The Machine wrote:

thanks, thats the 2nd time you're correct in a similar discussion.
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snipped-for-privacy@hotmail.com wrote:

The hydraulic equations assumes that both cylinders are at the same height. The other one does not.
A more comprehensive one is Bernoulli's equation, which says that
F1/A1 + (1/2)p*v1^2 + p*g*h1 = F2/A2 + (1/2)p*v2^2 + p*g*h2
where F1, F2 are the forces delivered at points 1 and 2 A1, A2 are pipe cross-sections at points 1 and 2 v1, v2 are fluid velocities at points 1 and 2 (for static cases, these are zero) h1, h2 are the heights at points 1 and 2 p is the fluid density.
PD
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PD wrote:

Any assumption would be impossible....the pistons rest a same fluid height with zero or even pressure on both sides.
The fluid's height then readjust itself, if the pressures applied are not the same on both sides.

I resolved the Bernouilli since there's no velocity thus v =0 and it's the exact same as the 1st equation:
P1 + pgh1 = P2 + pgh2 P1 = P2 +pg (h2-h1) P1 = P2 +pg h (h = h2-h1) which is the exact as the 1st equation (1st web link)
So I'm back with the same problem and even the 2nd web link for the hydraulic press says P1 = P2 (thus it's probably neglecting the weight(pressure) of the fluid (P1 = P2 + 0)
we have h = h2 -h1 (1st web link) and h = d1 + d2 (2nd web link)
h2 & h1 are measure from the base where as d1 & d2 are measured from where both piston's would have the same height....
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I found the answer which is that they are both the very same...they simply omitted minor forces from the equation (meaning they considered the weight of the liquid as minimal for the press equation .... so they omited it).
Manometer model at the very bottom of:
http://www.cartage.org.lb/en/themes/Sciences/Physics/Mechanics/FluidMechanics/Statics/Measurement/Measurement.htm
(they use z1&z2 instead of d1&d2 and specify that bernouilli alone is insufficient to determine their values bernouilli determines the value of z1+z2 (=h), you then need z1/z2 = Area2/Area1)
snipped-for-privacy@hotmail.com wrote:

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On 30 Dec 2005 10:44:00 -0800, snipped-for-privacy@hotmail.com wrote:

Not a mistake exactly, just an over extended assumption.
It is well known that the head or difference in liquid levels in a manometer is proportional to the pressure difference into the two legs and inversely proportional to the liquid density and value of g.
You chose the diameter of the second leg of the manometer to be root3 the diameter of the first leg, so their surface areas are in the proportion 1:3 and you specified the pressures on each leg to be in the proportion 2:1
THEN, you started doing the numbers for work on each leg's surface, supposing that your pressures and tube diameters were compatible.
They are not.
It makes it easier to see, probably, if I make a see-saw/teeter-totter analogy of it. If one surface falls three times as much as the other surface rises, this is like teeter-totter with odd legs - one three times as long as the other. The long leg has a force of two pressing down on it. The short leg has a force of three pressing down on it.
For stability, the moments must balance i.e length a X force a = length b X force b or 3 X 2 = 1 X 3 Well well!!!! The moments don't balance.
What this means, without laboring the analogy too much, is that the liquid in the thin pipe disappears altogether and bubbles appear in the thick pipe at any force greater than 1 on the thin leg's area of 1 in other words, pressure = 1 maximum on this leg.!
This could serve as an example (particularly to physics students) that mind experiments are all very well if you are Einstein, but if you are not, it is better to touch base with reality pretty often.
Brian Whatcott Altus OK
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