Putting vacuums in series (pneumatics)

[Capernicus]

So if you put two air pressure vessels in series,  its still the same psi.  (pressure/resistance= air flow.)

But if you put two vacuums in series (I mean just vessels,  not pumps/motors.) - is it the same psi - or do they add?

If you were pulling on a vacuum, the more you pull it the more force it has against you pulling?  this is what I'm trying to figure out,  but I dont have anything to test it with,  but I do have a 3d printer,  but I was wondering does someone have the answer for me already?

[TimFox]

When you pull on a piston, where there is a partial vacuum on the other side, the force you are pulling against is the difference in air pressure from your side to the lower pressure on the vacuum side of the cylinder (i.e., the "gauge pressure" of the vacuum chamber).  However, if the vacuum seal is good, and no air leaks around it, all you are doing is pulling against the atmospheric pressure.  If there is a gas sealed inside, it acts as a spring with an approximately linear Hookes' law force where the extension is proportional to the net force.
If you connect that cylinder to the piston of a second identical cylinder, I assume the force you exert splits evenly on the two pistons.
Can you draw exactly what you mean?

[Capernicus]

Thanks so much for the help,  I just need something simple cleared up.

My idea is,  having 2 evacuated vessels in series just counts as a longer cylinder,  so it would be more force?

But 2 vessels with the same high pressure, only count as 1!  (no force increase.)

[TimFox]

In your drawing, with only one “vacuum”, assume for discussion that the absolute pressure at the right side of the cylinder is 0.5 atmospheres when the volume is 10 cm³.  When you pull the piston to the left until that volume increases to 100 cm³, then the absolute pressure in the vacuum side decreases to 0.05 atmospheres.  The force resisting you increases from (0.5 atm) x (piston area) to (0.95 atm) x (piston area).  This assumes a sealed piston and cylinder.  If the chamber is connected to a pump that maintains a constant (negative gauge) pressure, then the force doesn’t change.

[Capernicus]

So if i double my position towards left of the piston in the cylinder,  given all other parametra the same -  its not double the force?

[langwadt]

So if i double my position towards left of the piston in the cylinder,  given all other parametra the same -  its not double the force?

force = pressure * area

no magic involved

[Capernicus]

So if i double my position towards left of the piston in the cylinder,  given all other parametra the same -  its not double the force?

force = pressure * area

no magic involved

Oh Ok thanx for telling me,   I was thinking it was going to be harder to pull the piston the more evacuated space was behind it...  im getting confused...

[TimFox]

The relevant pressure in the force equation is the pressure difference between the two sides of the piston, where the outside is constant at 1 atm (approximately 100 kPa) and the inside varies inversely with the volume.  If the initial vacuum is extremely low absolute pressure, then the change in force will be small.

[Capernicus]

So you can pretty much pull a blocked syringe for a mile and it never gets harder to pull it.

[TimFox]

That depends on the initial vacuum pressure, as I stated.  This is a quantitative problem, and you will have trouble maintaining the sliding seal over absurd dimensions.  There is an old saying about giving an inch and taking a mile.

[Capernicus]

I mean pressure 0.    So then it makes no distance whatsoever.  (outside 1atm)

Long vacuums have no force difference over short ones, thanks for explaining.

[TimFox]

I don’t understand your interpretation of my answer.  I think you mean that the force on the piston depends only on the pressure difference, not on the volume of the vacuum chamber.  However, the internal pressure follows the gas law when the volume changes.
“Vacuum” does not necessarily mean 0 pressure, which is good since I never achieved lower than about 10^-13 atm in laboratory chambers.  It means pressure lower than local atmospheric pressure in most applications for vacuum activated devices.

[Capernicus]

Yes I understand what you said.

But how can u even change the pressure of ~0 pressure?  it just stays ~same...  ~0...  no matter what the pull length was,  or how much evacuated space there was,  its just the same back force regardless.

[TimFox]

How do you obtain zero pressure?  Even in my extreme example in a pumped laboratory chamber with no sliding seals, there is a small pressure.  You can’t say it is negligible if you insist on mile-long extension.  Some estimates of the pressure in interplanetary space are better than my lab, roughly 10^-16 atm, but the distances are also very large.

[Capernicus]

Just push the piston to the end, and put the plug in.  (mechanical)  Or close a bellow,   or squish an eyedropper.

[TimFox]

No.  Have you ever pumped a vacuum?

[Capernicus]

I never have, hence my simple and confused questions.
I havent even pumped a car tyre before in my whole life.

[TimFox]

Your mechanical methods work to produce a “partial vacuum” which can be adequate for a vacuum actuator, but still have measurable absolute pressure.
Practical vacuum actuators in automobiles use a negative pressure that is only a small fraction of atmospheric pressure to get a usable force on a sealed diaphragm, avoiding sliding seals.

[langwadt]

How do you obtain zero pressure?  Even in my extreme example in a pumped laboratory chamber with no sliding seals, there is a small pressure.  You can’t say it is negligible if you insist on mile-long extension.  Some estimates of the pressure in interplanetary space are better than my lab, roughly 10^-16 atm, but the distances are also very large.

still the change is very small, 1atm - 0 vs. 1atm - ~0

[Capernicus]

What the hell I'm going to do an experiment.



I'm going to resin print it and see if when I pull the left piston out to just inside,  the right one is harder to pull out or not.

Then I can finally put it to rest - and move on.

[TimFox]

With the vacuum you can achieve, there should be a measurable absolute pressure inside, and your sliding seals will leak.  It will be hard to interpret the experimental results.

[Capernicus]

I bet it'll be air tight,  I trust my anycubic photon,  but I spose, I haven't actually done it before, I'm just guessing itll be air tight.

The vacuum will be near 0% if its air tight, solid against solid,  except that little bit at the end, where the noddies meet, otherwise it wouldnt be air-tight, I think.

I could add a softbody membrane to it to be sure... let me think, maybe...

[Capernicus]

Ive got an easier one, made out of odds and ends around the house. 




Itll only evacuate the kebab sticks diametre tho,  so maybe u stuff 3 or so in, depending on the radius of the straw?

Then once uve stuck it in,  you put some tacky on the end, and then u should have the vacuum prepared.

[langwadt]

I bet it'll be air tight,  I trust my anycubic photon,  but I spose, I haven't actually done it before, I'm just guessing itll be air tight.

The vacuum will be near 0% if its air tight, solid against solid,  except that little bit at the end, where the noddies meet, otherwise it wouldnt be air-tight, I think.

I could add a softbody membrane to it to be sure... let me think, maybe...

just do the math, force = pressure * area,  pressure * volume = constant (for constant temperature)

[T3sl4co1l]

Pulling on a vacuum costs zero work.  Pushing against the atmosphere however requires work.

The force, of which, is defined by the rate of volume change.  For a piston in linear motion, the cross section is constant, and volume proportional to displacement.

So the force is constant.

This can be a good way to make constant force springs; though it depends on atmospheric pressure (so, varies with altitude).  It's more common to do it with compressed gas ("gas spring"), the pressure of which will increase some with compression, which itself can be a valuable feature (if the minimum volume is zero, then force increases hyperbolically with displacement, making a softer stop than a hard piece of metal, or a rubber bumper).

I don't know what you mean by vacuum vessels "in series".  You can connect them in parallel, joined with a pipe; this simply increases the total volume and equalizes the pressure between them.  Surrounding a rigid vessel in any other pressure (greater or lesser than what's inside) has very little effect, because of its rigidness (small change in volume for large change in pressure).  The situation is different for, say, a balloon (volume varies with pressure differential, in a nonlinear manner), or a bellows or other very flexible structure.  (High altitude balloons are usually little more than loose polyethylene bags; the gas expands at high altitude (lower pressure), eventually inflating the bag to full.)

Tim

[EPAIII]

You are going to 3D print a vacuum chamber, cylinder, and piston? No, wait, two cylinders and two pistons. And get a good vacuum seal? GOOD LUCK!

As others have said above, the pressure against the pulling force on the pistons is strictly due to the difference in the pressure on the two sides of those pistons AND to the effective area of the pistons. In your drawing BOTH of the pistons have one side at atmospheric pressure and the other side at the same, lower vacuum pressure. Your drawing has no dimensions, but both pistons seem to have the same dimensions and therefore the same effective area. So, at all times, both of those pistons will feel the same force. The math for this is:

Force = Pressure X Area

Total simplicity. Pressure is measured in units like pounds per square inch (PSI) or Newtons per square cm or some other combination of force and area and of course, area is measured as area.

Now, for how that force changes with the length of the pull, that will be complicated by the amount of air (or other gas) that was in the volume of your apparatus before you start pulling the pistons. As you pull on either of the pistons you will increase the volume that the initial amount of gas occupies. But, since that initial amount of gas is not removed by your simple apparatus, there will always be some remaining pressure inside that apparatus. Professional and scientific systems for creating "high" vacuums DO remove some percentage of the remaining gas at each stage of creating such "high" vacuums. So they can approach a "perfect" vacuum. I say "approach" because no one has ever created a perfect vacuum. Even the vast areas between the galaxies of the universe still has some small amount of gas.

If your 3D printed apparatus does reach a small remaining amount of pressure inside it, you will quickly reach a point where you will not be able to measure the remaining differences in force needed to further separate the pistons unless you spend a lot of money on the force scales needed. And even then, after selling your house, boat, and car, you still will quickly run out of the needed equipment to measure further differences.

One more thing about creating "high" vacuums: the very materials that you construct your apparatus with will have amounts of gas absorbed into their very structure. That gas will bleed out of the walls and pistons and everything else. This is a real problem which continued for a long time - not hours or days, but months and years. Things like vacuum tubes had devices built into them that were designed to continue to remove gas every time they were used.

3D print a vacuum chamber? Yea, sure. Good luck!