quote:

Originally posted by AKKAMAAN:
Wasn't there one little piece of information missing on the input data....those 100psi, is that absolute pressure or pressure over atmospheric pressure....on such a low system pressure (100psi), that could make some relativ differens huh???
Just some thoughts.....

I think the main point is that the piston doesn't move very far.
Assuming both pressures are absolute then if the bulk modulus os oil, β=200,000psi and ΔP=100-14.7=85.3 then
ΔP*V/β=0.0004265*V which when divided by (Ap-Ar) is going to be small.

Using e the answer is 0.0004265909640566168*V which is a little bigger. It is more accurate theoretically to use e IF you all the other details to 5-6 digits of precision which you won't. So why bother? The simple linear estimate I use is good enough for most cases because you will not know the bulk modulus of oil to 5 or 6 decimal places.

The reason why e is in compression/expansions is because they are not linear. As oil is compressed it gets to be harder to compress the next increment of oil.

http://en.wikipedia.org/wiki/E_(mathematical_constant)

I have another under pressure problem that shows why e and natural logs show up in hydraulics and it is good to know but trust me, as long as the change in pressure is small relative to the bulk modulus of oil just use the linear approximation.

Lets restate the problem. Lets say the pressure in the tank is 10,000 psi now the linear answer is (10,000-14.7)/200,000=0.0499265 and the perfect answer is
exp((10000-14.7)/200000)-1=.05119383078998552
You can see that error is now about 2.48%. I didn't bother with the symbolic part of multiplying by V/(Ap-Ar)

Most of us deal with pressures below 3000 psi so our errors will not be as big and linear approximations can be used. Our controllers get used in hydro-forming where the pressures can be very high 30000 to 90000 psi, the bulk modulus of water is very high, 320000 psi. but 90000 psi is a big percentage of 320000 psi so the exact formula should be used.

QUOTE]Originally posted by Peter Nachtwey:
[QUOTE]
Thank you Peter!
Very good and filling answer....I got it!! ...yes you are right about that little detail...why bother?....I just saw that Woodygb assumed both pressures absolute...


I kinda figured it was e log, so that make perfect sense now....
Thx again

kram,
The piston will move to left only! why?
acc. to pascal's law F = P*A
acc. to that F1>F2
but for F1 = F2+F3
100*A1 = 100*A2+14*A3 eq. (k)
here p2<p hence system can't even stay balanced, why?
from the above eqn. k we can say that A3 = 7.14(A1-A2),
while in our problem A3 is exactly equal to the (A1-A2) hence piston will move towards left only! and yes very nice question!

i agree with woody, the piston will move to the left until the pressure inside decreases to 14 psi (atm), same as outside. This decrease in pressure is caused by the area differential (different volume as well)in either side of the piston.
Look at that piston as the rod and piston of a double acting cylinder the force created from the left (rod) is canceled in the right side (same area) and the remaining area (annular)is the same area exposed to the exterior, so the only pressure available to get equilibrium is 14 psi.

Regards
kamilRhu

Hydraulic Calculations

Ok, as I look at this the piston moves to the left. Simply because area on the right side of the piston is bigger than the area on the left side. The piston should move all the way to the left.

I think the another question to ask about this is: What is the meduim in the tank? (liquid or gas?) Liquid is practically non-compressable where is very compressable.

Although the medium isn't clearly stated it is implied.

Originally posted by Kram.

quote:

I figure after more thought that if that was a sealed fluid medium that hydraulic lock would occur

And the solution ...

quote:

Piston will move to the left ...( pressure in the system will drop as the volume increases) ..until all pressures are equal..which is at Atmospheric pressure.

( With the one proviso that there is enough free movement of the piston and rod for the pressures to balance.)

Applies for any medium.

Regards Woody

quote:

Originally posted by MCSlaven:
The piston should move all the way to the left.

NO!!! at least not for a liquid.
Obviously you didn't understand Woodygb's solution or my quick linear estimation.
Woodygb's formula is correct. The piston will move very little as a percentage of the tank volume if you use the scale in the original picture.

quote:

I think the another question to ask about this is: What is the meduim in the tank? (liquid or gas?) Liquid is practically non-compressable where is very compressable.

Woodygb used β, the bulk modulus of oil, to take into account how compressible the fluid is.

Ok, I assumed it moved all the way to the left, and I realize you should never assume anything (lol), because to have a liquid medium maintaining 100psi, means there is a pump somewhere providing flow into the tank.

Without the pump, and considering liquid is almost non-compressable, the cylinder still moves to the left, but just a small distance, since there is no pump.

I guess the next thing is to put some real numbers to this. Liquid is sligltly compressable. If there were 10000 gallons in the tank then, the compressability of the liquid may move the pistion more. Although at 100psi, you will not get much compression.

However if the medium is gas, then this is highly compressable. Because of this, we may see more movement in the pistion depending on the size difference of the piston areas.

Woodygb's solution is correct.

quote:

Originally posted by MCSlaven:
I guess the next thing is to put some real numbers to this. Liquid is sligltly compressable. If there were 10000 gallons in the tank then, the compressability of the liquid may move the pistion more. Although at 100psi, you will not get much compression.

I picked some numbers roughly to scale.
If I assign:
Vol: 1 gal or 231 in^3
CylDia: 231 in^3 to the initial volume and 3 in dia for the piston

Pt: 100$ /* Tank pressure */
Pa: 14.7$ /* Atmospheric pressure */
%beta: 200000$ /* Bulk modulus of oil *.
CylDia: 3$ /* inches */
RodDia: 1$ /* inches *
Vol: 231$ /* Initial tank volume before releases */
Ap: (%pi/4)*CylDia^2$ /* Piston area */
Ar: (%pi/4)*RodDia^2$ /* Rod area */
dx: (exp((Pt-Pa)/%beta)-1)*Vol/(Ap-Ar), numer; /* The distance moved in inches */
0.0156835280004359 /* inches */

That isn't far but you could multiply that 0.0156x10000 to get 156 inches if the area of the rod and piston are kept to the same size but a tank that can hold 10000 gal would be a cube almost 5000 inches on a side and 156 inches is still relatively small. If the tank size and the areas both get scaled up by 10000 then the distance moved is still just 0.01568 inches.

quote:

However if the medium is gas, then this is highly compressable. Because of this, we may see more movement in the pistion depending on the size difference of the piston areas.

[/quote]
Yes, but I bet that the piston would slam all the way over to the left as Nahum first showed.
Can anybody work out how far the piston will move to the left if the tank is filled with gas?

Woodygb should give some one else a chance.

Thanks Peter.

Im new to the forums and fluid power also, still in school studying it. I love it.

Im famliar with most of the equations so far, with the exception of (exp((Pt-Pa)/%beta)-1)*Vol/(Ap-Ar)

This has been a great little "what if" simulation to work with.

quote:

Originally posted by MCSlaven:
Im new to the forums and fluid power also, still in school studying it. I love it.

College? If so, what year?

quote:

Originally posted by MCSlaven:
Im famliar with most of the equations so far, with the exception of (exp((Pt-Pa)/%beta)-1)*Vol/(Ap-Ar)

(exp((Pt-Pa)/%beta)-1) this calculates how much the volume of oil will expand or contract based on the change in pressure and the bulk modulus of oil. In the example the oil volume increases by 0.0426% if I remeber right. Then Woodygb multiplied this by the volume to get the change in volume and then divided by the difference in the areas to get the distance moved. He solved this problem symbolically since he didn't have actual dimensions. You will not find the whole equation in a book because it is a combination of formulas to solve this particular problem.
Ask your instructor about (exp((DeltaP)/%beta)-1. If this is for college the professor should know where it came from. This is a test for your instructor.

quote:

Originally posted by MCSlaven:
This has been a great little "what if" simulation to work with.

More home work. How far does the piston move if the fluid in the tank is air instead of liquid? I gave my answer above.

Have fun. This is more useful than solving Sudoku puzzles.

Im at Washtenaw Community College. Im actually in the Automation Technology course, but they have several specializations for that track, and I've taken the Fluid Power track.

Just finishing up a 2nd year.

Good, then you have had your first year of calculus.

Woodygb used information from this and applied it to the problem. The work sheet was for a problem posed by someone else a few weeks back.
http://www.deltamotion.com/pet...ima/EnergyInOil.html
The original program is
http://www.deltamotion.com/pet...xima/EnergyInOil.wxm
The program I used is called wxMaxima
http://wxmaxima.sourceforge.ne.../index.php/Main_Page
wxMaxima is a free and very power math tool. Free is a good price for students. If you have a student version of Matlab, good, but wxMaxima can solve problems that Matlab can't.

My daughter asked me if I could help her with an algebra math problem. I said sure no problem. I looked at the math problem for about 5 minutes then looked at her and said, "when the hell did we start using the alphabet in math" and then quickly walked away.