# Pascal's law and pistons

1. Aug 16, 2014

### gloo68

I have some questions in regards to a diagram that has a twist to the regular set up that is usually used with demonstrating Pascal's law and pistons.

In the diagram I have attached, if you added a column of water on top of the large piston which is sealed at the top from the atmosphere, there are three questions (assume the red walls of the large piston and the column on it have the same density as water):

1. I assume that even though the smaller piston on the left has an additional column of water to add pressure greater than level 2, the column of water on the large piston equalizes the hydrostatic difference and thus the large piston cannot be moved up?

2. I also assume the column size (diameter) on each of the columns doesn't matter as long as the column heights are the same?

3. If i add an incremental force down on the large piston, I can push the column of water on the small piston upward?

2. Aug 17, 2014

### jim hardy

To my simplistic thinking:

1. The sealed column of water atop the large piston is not free to move down into the lower volume where it could "equalize" something.

2. It might as well be just a weight of whatever is the mass of its water content lying above L1. So its area does matter.
Take it to the extremes - what if its area were zero? would your system be balanced?
What if it occupied the whole piston area? Would system be balanced?

3. Sure. Sum vertical forces.

Force on area of large piston excluding sealed column = (area of large piston not including column) X pressure at L2

Force on rest of area which is sealed column's top piece: = (area of sealed column) X pressure at inside surface of column's top ;
as drawn that pressure is zero. If piston moves down it will become positive, if piston moves up up it will become negative.

As drawn piston sees a net force up. If it's free to move your system is not in equilibrium.

Plausible ?

3. Aug 17, 2014

### gloo68

@Jim - thank you. I was more curious about # 3 because I was thinking that both columns were adding a hydrostatic pressure upward on the bottom of the large piston. I was then thinking that it wouldn't be just some small incremental force, but some magnitude large enough to first overcome the upward hydrostatic force.

So if # 3 is true, then if we filled the rest of the large piston level to Level 1, the water level in the small piston column would (or could if enough space), theoretically be pushed up by the difference of L1 -L2?

4. Aug 17, 2014

### jim hardy

i think i got your question... and i think you've missed a small point...

Remember that your system as drawn is not in equilibrium.
If you let it go the large piston would rise until force up on it from pressure beneath equals force down on top of sealed cylinder. Level in your open column at left would fall.

If you filled the volume above large piston with water to level of L1, you'd have equilibrium.
You could let it go and it'd stay put.
That's because the vertical forces on both areas, large piston and top of sealed column, would be zero. Level in your column at left would stay put, not rise or fall.

Once again look at the pressures on opposite sides of each surface:
If you filled above piston to L1
Large piston would see same pressure on both sides, (L1-L2) inches of water.
Column's sealed top would see atmospheric pressure on both sides
so there'd be no vertical force.

The key to these problems is to first get the pressure P on both sides of the horizontal areas. Then compute the vertical forces by F = P X A

Have you any electrical background? Kirchoff's Voltage Law ?
What helped me in freshman physics to visualize this was to remember how my ears hurt when i'd dive to bottom of a swimming pool. Clearly pressure is a function of depth below surface, ie height of water column. You can express pressure in feet or inches of fluid.
When you go down pressure increases, when you go up it decreases. (Duh. Not belaboring the obvious here, just trying to explain how i learned to "keep it simple".)

So i think in terms of pressure changes.
Using Kirchoff's Voltage Law in electrical we walk around the circuit writing down each voltage change. They'll add to zero.
I do the same thing in hydraulic circuits and call it Kirchoff's Pressure Law - 'the sum of pressure changes around any closed loop equals zero.'

Now back to your original sketch:
We'll swim the circuit
Start at top of left column where it's open to atmosphere.
Swim down to bottom of tank. Pressure has increased from atmospheric to L1 inches (or feet) of water. At about ten feet our ears would hurt noticeably.
Now swim horizontally to just below small sealed column. That's horizontal, so we're at same pressure, L1.
Now swim up to bottom surface of large piston. Pressure decreased by amount L2, to L1-L2.
That's the pressure on bottom surface of large piston. Force up is that pressure X area.
If you've not filled the volume above with water, there's no pressure on the top surface so force on piston is all up.
If you DID fill above the piston with water there's pressure on top surface too so net force on that part of piston is area X pressure difference.
Now continue swimming up to top of sealed column , a distance of L1-L2. Pressure decreased by L1-L2, which takes us back to atmospheric .
So that surface sees same pressure, atmosphereic, on both sides. No vertical force .
Observe sum of our pressure changes L1 - L2 - (L1 - L2) = 0.

You can express pressures either gage or absolute, won't matter.

That stupid mental exercise is how i work such problems. I actually imagine the sensation in my ears. But i don't admit that to just everybody.

It's very useful. Try it yourself and see if you agree that filling above your piston to L1 balances things out perfectly.

(Shhhh ! Your secret's safe with me)

old jim

Last edited: Aug 17, 2014
5. Aug 17, 2014

### gloo68

Jim, you say that the large piston would rise now and the water level in the left (open piston) goes down now. But I am assuming there is no way the water column on the left (which is closed), could fall and create a vacuum, and also then, if the large piston rises, the water level in the sealed column would then rise higher than the smaller side piston which would mean an imbalance in hydrostatic pressure btw the 2 sides no? The large side piston column would be higher than the small side?

6. Aug 17, 2014

### jim hardy

Whoa, you show it open at the top.

If this is a sealed system nothing can move.

If left column is sealed at L1 then nothing can move.

If left columns is NOT sealed, then:
when you let go your piston will rise
level in open column on left will decrease to some new height L3.
When vacuum at top of sealed column on right creates a down force that's equal to the up force against piston, it'll stop.
Since up force on piston requires pressure on bottom of piston, that means when things settle L3 will still be higher than the piston's new position.

If you assign diameters and heights it'll be a nice algebra exercise for you to figure out where they'd settle.

I think your concept of "hydrostatic pressure" is not well developed.
It's forces that balance.

7. Aug 17, 2014

### gloo68

ok sorry Jim, my last post I meant to say the column on the right on top of the large piston. If that one is sealed, then how can the small diameter piston move down? Wouldn't that move the column on the large piston up and then the hydrostatic pressure on the right side be much higher?

8. Aug 17, 2014

### jim hardy

That's the small one that's sealed at top , as shown on sketch?
The invisible piston at top of left cylinder at Level1, and the large piston are both free to move, but the small sealed cylinder on right is fixed to large piston? Is that our system ?

Easy - it pushes the large piston up, just as in a hydraulic jack. Hollow cylinder is shown attached at its bottom so it must go along with large piston.

What would make pressure there higher?
Your pressure reference point is where the system is open to atmosphere, which is at top of left cylinder.
Any place in the system that's higher than your reference point has hydrostatic pressure less than your reference point.
Any place in the system that's lower than your reference point has hydrostatic pressure greater than your reference point .
(assuming no pumps)
Any two places in the system that have the same elevation have same hydrostatic pressure. Doesn't matter which side they're on.

it's that simple.

9. Aug 17, 2014

### gloo68

The pressure i assume is higher on the side of the large piston because the sealed cylinder level is now higher. Remember that this sealed piston top was at same water level before as that on small diameter piston. If the small diameter water level can push down and raise the large piston, wouldn't that take the sealed column level higher up??

10. Aug 17, 2014

### jim hardy

No. Pressure depends only on elevation. At any given elevation it is the same on both sides of your sketch. That's what Pascal's principle says- going sideways there's no pressure change, only up and down.

In the sketch you just drew, the sealed cylinder is now higher than the surface of the fluid in left column, where fluid contacts atmosphere. So there's a vacuum at top of sealed cylinder equal to "new high level" minus level of surface in left cylinder.

When those levels were the same the pressures there were also the same.
Now the levels are not the same so neither are the pressures.
That "small piston" is invisible.
Is there an invisible piston on left small cylinder or is it open to atmosphere?
If there is a piston there are you pushing down on it ? Or pulling up on it, as you'd have had to do in your initial sketch ?

Is the water level pushing down, as you just said, or are you pushing on that invisible piston that may or may not be there? I assumed you're not pushing or pulling on it.

In your initial sketch with no invisible piston, the "water level" will push down and raise the large piston. You'll arrive at something like your most recent sketch.

Of course - the sealed column is mechanically attached to the large piston. It cannot do otherwise.
What has that to do with pressure?

You seem to think that the top of the sealed column is some sort of starting point for pressure.
It's not. It is sealed therefore the pressure in there can be anything, pressure or vacuum.

Your reference point for pressure can only be someplace where you know what the pressure is -
and in your system that's only one place, the surface in the left column where it's atmospheric, which is zero gage. (or F/A if you're using that invisible piston)

Pascal tells us that pressure depends only on how far above or below that reference you are , and NOT how far sideways from it you are.

Label the level of the surface in left (unsealed) column L3. Pressure there is atmospheric, zero .
At elevation L3 inside the sealed cylinder atop piston, the pressure is atmospheric, zero. Above that level it's vacuum.
At level 2 the pressures inside your left unsealed column and inside your right sealed column are the same.

Your concept of Pascal's principle seems flawed.
It states that a difference in pressure results only from a difference in elevation.
http://en.wikipedia.org/wiki/Pascal's_law
at any given level there's no Delta h .

I suspect this is a classroom exercise that's been given you to drive home the point.

It's time to apply some algebra and prove it to yourself . It might help you to draw some pressure gages on your system and figure what they'd read.

http://www.grc.nasa.gov/WWW/k-12/WindTunnel/Activities/Pascals_principle.html

Gages on left side of tank would read same.

11. Aug 17, 2014

### gloo68

Thank you so much for your help Jim...really appreciate your time and expert input. I was suspecting that because it was sealed that there was something not correct. You explained my suspicion about the vacuum in the sealed cylinder. I understand now why you say the large piston will go higher.

No, I am not a student (too old to even remember most of my elementary physics). I was working on an idea for a prototype and when i get stumped...I really on this forum to explain away my ignorance on topics.

Thanks again Jim....you may see some more silly questions from me....hope you will continue to enlighten me.

G

12. Aug 17, 2014

### jim hardy

Wow now i feel bad for getting frustrated... I'm sorry....

That is what we all want - to see "the light go on".

I too am rusty .

Glad you asked, and seriously thanks for telling us you succeeded in getting past that "stump".. Makes this old guy feel a little less useless..

and we all vicariously enjoy your victory.

Come back often

and sorry if i got grumpy on you

i was trained to holler and arm wave to get points across. No insult intended.

Interesting isn't it - how hard it is to put ideas precisely into words ?

old jim

13. Aug 17, 2014

### gloo68

oh no no Jim....I didn't feel you get curt on grumpy at all !!!!! I thought you were really patient and helpful bud.

I was just an average student...and probably average intelligence. I don't think i would have made it past engineering school. But now that I am approaching 50, I am fascinated with the sciences and all the important things it has given mankind. I find myself searching for answers and being enlightened with things that i felt was rammed down my throat back 30 years ago.

Again, thank you and looking forward to your tutelage.

G