# Question about vacuum chuck forces:

sonhouse
Science 19 Jul '16 17:56
1. sonhouse
Fast and Curious
19 Jul '16 17:56
I am designing a vacuum chuck, our substrates are about 210 mm X 120 mm alumina. We have a spinner/coater (photoresist) that we spead the liquid on the substrate and it spins at various rates, up to about 3000 RPM's. We use a poor design now and my chuck will use an o ring instead of the flat design we now use which has a number of problems.

My question is: Call it 8 inches by 4 inches for grins, and with full vacuum on the backside you would get around 350 pounds of force on the substrate.

The thing is, with the force we see now for the inferior chuck arrangment, we can pull the substrate off the chuck with fingers.

Now if it was a perfect seal and a perfect vacuum, say 1 E-5 torr or some such, wouldn't it take 300 pounds to pull off the substrate?

If so, and we pull upwards with say 5 pounds of force, would that equate to the inches of mercury used in the vacuum part of the system?

That is to say, the actual inches of mercury say with perfect vacuum causing a downwards clamping force of say 300 pounds but the vacuum was instead of 28 inches of mercury, was instead 2.8 inches of mercury, wouldn't that equate to a force needed to pull the substrate free, now being 30 pounds?

But we are clearly not anywhere near 30 pounds of force to free the substrate while under vacuum.

So if 3 pounds of upwards force is needed to pull the substrate loose while under vacuum, does that equate to around 0.28 inches of mercury? or 1% vacuum, or 7 odd torr? I don't have a gauge to measure the vacuum right now, could do that but have to modify present plumbing.

So if we have a vacuum pump capable of say 20 inches of mercury vacuum, wouldn't that require at the size of our substrates to need around 2/3 of perfect vacuum force or say around 200 pounds upwards force to pull off the substrate? Which would surely break the fragile alumina, its only 1 mm thick.

Sorry for the use of inches, just relating that to STP units, PSI and so forth.
2. 19 Jul '16 20:23
Its not very clear what you are asking, a diagram would help.
My first guess is that you are not taking area into account but should be.
3. sonhouse
Fast and Curious
19 Jul '16 23:11
Its not very clear what you are asking, a diagram would help.
My first guess is that you are not taking area into account but should be.
I thought I was making that clear, using inches, sorry, 8 X 4 inches on a perfect vacuum with atmosphere on top there would be 8 X 4 X 14.7 pounds of force holding the object in place, or 352.8 odd pounds of force.

So suppose it was a very strong metal, say 1/2 inch SS and you put a bolt partially threaded in the center and pull upwards with a force measurement tool.

I would presume you would find it took an upwards force of 352+ pounds to pull it loose from the vacuum chuck.

So what I feel on my own vacuum chuck is a force on the edges of maybe 3 pounds or so and that pulls away from the vacuum, like 1% of the force of pulling straight up from the center.

So how much leverage do I get pulling up on one corner Vs dead center? On our spin coater, the corners stick out a bit so they can be grabbed by fingers but not the center.

So a perfect vacuum, in these terms, not real high vacuum, just good roughing vacuum, might be something like 10 millitorr or so, or about 7 torr. That is a vacuum that a regular gauge that reads inches of mercury would show as total vacuum but of course in real life it is only roughing vacuum.

Anyway the difference between 10 millitorr and 1/1000ths of a millitorr would not change the force needed to pull it up away from the seal, it would still be around 350 pounds either way.

But if you go down from max reading of say 28 inches of mercury to 14 inches of mercury, about 380 torr, I would think the force needed to extract the part would also be cut in half or about 175 pounds.

Then cut it down to 1.4 inches of mercury with perfect seals, it would take about 18 pounds to pull apart I think.

So I am at say 3 pounds required to pull the part against the vacuum of the chuck but pulling on the edges.

If I was pulling on the center and found it took 3 pounds I would assume a vacuum of about 0.24 inches of mercury.

But how does that change going from pulling up from the center opposed to pulling from one corner?

I have used gauges on some other chucks and it shows say 20 inches on the chuck but going through 1/8th inch tubing which of course way reduces the flow but the parts there are 1/2 inch by 4 inches, 2 square inches and that would seem to require about 30 pounds of force to dislodge but it also takes only a very small force to release against that vacuum.

I am trying to figure out why the force to release against vacuum is so relatively low when the gauge numbers should be showing much more holding force but the actual holding force is much less than just the numbers suggests.

Of course leakage is a large part of it but I would think I would hear that much leakage. Any idea what is going on here?
4. DeepThought
20 Jul '16 01:21
Originally posted by sonhouse
I thought I was making that clear, using inches, sorry, 8 X 4 inches on a perfect vacuum with atmosphere on top there would be 8 X 4 X 14.7 pounds of force holding the object in place, or 352.8 odd pounds of force.

So suppose it was a very strong metal, say 1/2 inch SS and you put a bolt partially threaded in the center and pull upwards with a force mea ...[text shortened]... rge part of it but I would think I would hear that much leakage. Any idea what is going on here?
You're giving us lots of information, which seems to be the same thing over again in lots of different units (none of which are S.I.), but leaving out important details, such as what materials are in contact and not stating clearly what the actual question is:

1) What is the actual question, is it why the force required is so small when pulling on the screw (is that when 3lbs is required) or is it that the force when pulling on an edge is smaller (is that when only 3lbs is required) and if so what is the force needed when pulling on the screw.

2) You have a rectangle of aluminium resting on a thing with a hole in. How big is the hole (is it that that is 4"x8" ) and what material is the rectangle of aluminium in contact with? Is it metal or do you use rubber or some other material to ensure a good seal. If it's metal then how smooth are the surfaces - would you expect a good seal? If you don't know you should be able to get a handle on that by seeing how quickly the force needed to remove the rectangle (pulling on the screw you mentioned) falls to half it's value with the pump on when the pump has been turned off.

3) If you asking why the force required to remove the rectangle is so small, then I'd guess because the vacuum isn't as good as you think it is due to leakage. Can you use a sealant of some sort.

4) Is your estimate of the pressure internal to this thing based on a measurement or based on the rating of the vacuum pump?
5. sonhouse
Fast and Curious
20 Jul '16 03:37
Originally posted by DeepThought
You're giving us lots of information, which seems to be the same thing over again in lots of different units (none of which are S.I.), but leaving out important details, such as what materials are in contact and not stating clearly what the actual question is:

1) What is the actual question, is it why the force required is so small when pulling on the ...[text shortened]... ressure internal to this thing based on a measurement or based on the rating of the vacuum pump?
I was not pulling on a screw, just the corner which sticks out a bit. The substrate is alumina about 1 mm thick. I used pounds and such and inches because the gauges we use for the rough vacuum is in inches of mercury, not sure how to convert to SI.

I suppose 28 odd inches would be equivalent to a torr or so, not sure.

I don't have a gauge on this spinner/coater device. I am going to find one though. Like you say, the vacuum force must be very low if I can lift up the corner against vacuum with just a few pounds of upwards force.

I am making a chuck with an o ring which is the proper way to seal this kind of part. When I told our CEO about this project he asked why didn't we do that ten years ago?

I have only been there 5 years so the original design was before my time there.

What the made was an abortionđź™‚ a round chuck about 250 mm in diameter with narrow slots cut in the front surface fanning out where the alumina substrate completely covers the slots and those slots now go to a hole where vacuum hose goes to the pump through a control solenoid. One big problem is the solenoid gets stuck with dried up photo resist and we have to clean the thing ever other day. That was when I realized the problem was no o ring just slots to conduct vacuum.

Thinking about that, I guess the total area of the slots is what gives the downwards force and that is not a lot of area. My o ring solution would cover about 90 percent of the substrate so the force would be exponentially higher I think anyway. My biggest concern is that force bending and breaking the substrate which is very brittle.

It won't take a lot of bending force before breaking. So I have a solution for that but I am first making just an outer o ring seal. Using Parker 2-170 size o ring and protruding up about 1/2 mm so hope it doesn't break with that much rise, since the force would tend to bend the center of the substrate, 210 X 110 mm rectangular.

I think I just figured out why the force is so low, because of the very small surface area of the slots, maybe 1 mm wide and maybe 100 mm total, some straight lines, some circles and such but the surface area would be then about 100 mm ^2. Max.

So that would be about 0.15 inch square and that would be about 3 pounds of force to pull loose.

That had bugged me. I was thinking there would be more surface area under vacuum but the two surfaces facing are very smooth and so the only surface area under vacuum would be those slots. The rest of the area of the substrate is superfluous.

Glad you kept asking questions, I would have bugged myself nutty figuring that out.
6. 21 Jul '16 22:02
I am still not 100% certain what your set up is, but I think you are going about your calculations all wrong. If I understand you correctly, you are pulling a flat object off something else by hand. The key factor will be how good a seal the two objects make between them, how much they deform when pulled and where you pull. If you pull on the edge then air will get in from the side regardless. If you pull in the centre then you may get the sucker effect. A near vacuum somewhere in the setup should have little or no effect other than the fact that it will temporarily suck out the air that slips in when you pull. But the initial static forces should be no different from a situation where you had an object against a flat surface. In sum, the area doesn't really matter at all.
7. sonhouse
Fast and Curious
03 Aug '16 18:19
I am still not 100% certain what your set up is, but I think you are going about your calculations all wrong. If I understand you correctly, you are pulling a flat object off something else by hand. The key factor will be how good a seal the two objects make between them, how much they deform when pulled and where you pull. If you pull on the edge then ai ...[text shortened]... n where you had an object against a flat surface. In sum, the area doesn't really matter at all.
No, the total area doesn't matter except as a leak source. The actual vacuum pressure is from the little channels cut into the chuck metal. They are about 1 mm wide and roughly 50 mm long in various styles, straight lines and circles all connected to the central vacuum source. So the total force is just the area of the channels times the vacuum level. So if it was half vacuum, say 300 torr, and 50 mm^2 of channel surface area and two completely flat surfaces, the bottom metal chuck surface V the bottom of the substrate being worked, then with a very tight fit between the two surfaces there will be very little leakage to atmosphere and that seals fairly well like that, leaving only the surface area of the channels as the force vector. So using PSI, not sure what it is in metric talk, say 15 PSI is one atmosphere and we are positing half atmosphere or around 350 torr or say 7 PSI for grins, then, 50 mm^2 is about 1 kg of force, which is about what we feel when we try to pull up on the corner of the substrate, the corners of the rectangular substrate sticks out from the chuck a few mm so we can test the force of the vacuum in a crude way by just lifting up on the corners and feel what kind of force it takes to force the substrate away from the chuck.

So the total force is simply the level of vacuum actually reaching the channels times the area of the channels which like I said is roughly 50 mm^2. The actual force might be a bit higher because of non-perfect contact between the top surface of the chuck and the bottom surface of the substrate, any scratches or such, bits of dust, under the substrate would allow air to travel to the vacuum chuck in the center of the chuck which is a circular shaped chuck a bit smaller than the diagonal measurement of the substrate which is about 120 mm X 110 mm or so. Alumina.
8. 04 Aug '16 07:46
Originally posted by sonhouse
No, the total area doesn't matter except as a leak source. The actual vacuum pressure is from the little channels cut into the chuck metal.
Vacuums don't exert pressure.
I am still not certain about your setup, but I imagine you as having a flat surface with a hole in leading to a vacuum. On top of this hole you have placed a flat sheet of something. We are discussing the force needed to lift that flat sheet.
If we assume a perfect seal around the edge of the sheet (lets say we put wet rubber on one side) then the size of the hole or the strength of the vacuum really isn't all that important. What matters most is the area of the flat thing and atmospheric pressure because as far as it is concerned almost everything on the other side is vacuum.
That is how suction cups work. A good suction cup can be very difficult to remove without forcing a leak. Usually, the force needed to lift one is the force required to cause a leak. So ultimately atmospheric pressure and vacuums are irrelevant. What matters is how good the seal is and how easily it fails.
9. sonhouse
Fast and Curious
04 Aug '16 16:39
Vacuums don't exert pressure.
I am still not certain about your setup, but I imagine you as having a flat surface with a hole in leading to a vacuum. On top of this hole you have placed a flat sheet of something. We are discussing the force needed to lift that flat sheet.
If we assume a perfect seal around the edge of the sheet (lets say we put wet rubb ...[text shortened]... essure and vacuums are irrelevant. What matters is how good the seal is and how easily it fails.
The pressure I was referring to was the force needed to pull the two parts apart. The total area of the surface under the vacuum area is what tells how much pressure or pulling force you need to pull the pieces apart.

And I am assuming two very flat surfaces so there is little leakage around the channels that lead to the vacuum feed.

A photo or drawing would certainly clear things up. For instance, if the channels were 25 mm wide and just a square, 25 X 25 mm or 625 mm squared, or one square inch then the force needed to pull the two apart would be 14.7 times the percentage of vacuum present compared to full atmosphere, so if it was half atmosphere or say 300 torr instead of 760 torr which would clock in at around 15 inches out of 28 or so then the percentage would be around 50% or 0.5 times 14.7 psi or around 7 psi and so it would take 7 pounds (3kg) of force to separate the two pieces.

If if was 25 X 50 mm the force would then double to around 14 pounds (6 kg) to pull them apart and so forth.

We don't have anywhere near that surface area in the total area of channels leading to the vacuum supply so the force is therefore much less maybe 1 kg or so maximum.
10. 04 Aug '16 18:201 edit
Originally posted by sonhouse
The pressure I was referring to was the force needed to pull the two parts apart. The total area of the surface under the vacuum area is what tells how much pressure or pulling force you need to pull the pieces apart.
No, you are mistaken.
With a perfect seal, the whole area is as good as vacuum.
With an imperfect seal, what counts is how good the seal is.

In fact, I will go further and say that with a perfect seal, your imperfect vacuum actually exerts a force in the other direction thus the larger the area exposed to the vacuum the less force required to pull the two pieces apart.
11. sonhouse
Fast and Curious
04 Aug '16 18:35
No, you are mistaken.
With a perfect seal, the whole area is as good as vacuum.
With an imperfect seal, what counts is how good the seal is.
I deg to biffer with you. The area directly impinging on the vacuum is what determines the force needed to separate the two pieces.

If there was nothing but a hole say 4 mm in diameter, the force needed would be only that surface area times the vacuum level, which would be about 12 mm^2 and with 50% vacuum, about 62 grams to pull apart the pieces, even if the substrate was 1 meter square against a 1 meter square chuck and perfect sealing of the area under the substrate. That area would not participate in the vacuum force because there would be no air being sucked in by the vacuum except the 4 mm diameter hole under the substrate with say an o ring seal. In that case, with an o ring seal, the parts could be separated by the height of the o ring and there would be no difference in the pull of vacuum regardless of whether an o ring was there or not as long as the two surfaces were atomically smooth and no air flowed into the gap between the two substrates. Only the vacuum directly in the line of the 4 mm diameter hole would give the force needed to separate the two pieces. Of course there may be other forces at work with two extremely close gapped pieces but ignoring those forces and just dealing with the force produced by the vacuum alone, my numbers are right, only about 60 grams of force would hold the two parts together, regardless of the size of the total substrate, whether it would be a circular part 1 cm in diameter or a circular part 1 METER in diameter, the pulling force needed to separate the parts for that size, 4 mm hole, would still be 60 odd grams.
12. joe shmo
Strange Egg
04 Aug '16 19:165 edits
The vacuum pressure will always be between 1 atm and 0 atm. The pressure that can be supplied at any given flow ( in this case the total leakage of the chamber/connections or system ) is unique to the turbo-machine/or positive displacement pump that is doing the work. If there are no leaks; the pump will be able to pull a vacuum pressure corresponding to no flow conditions on its head vs discharge curve (a vertical system curve; giving you maximum force in this case). If the system is very poorly sealed or completely open the pump will be able to pull maximum flow, at no head ( horizontal system curve, giving no force in this case). In reality it will reside somewhere in between these two positions.

Pump curves are typically of the form: P(Q) = C - A*Q^2
System curves are typically of the form: S(Q) = B*Q^2

As a consequence they necessarily intersect ( the pump is pulling and the system resisting). The point of intersection is the operating point (pressure and flow) of the system.

The net force acting normal to the mating surfaces will be equal to: F_net = - Integral 0 to A (P_pump(A)*dA)

A1 is the Area of your slots in which a vacuum pressure ( less than 1 atm, greater than 0 atm ) acts.
13. sonhouse
Fast and Curious
04 Aug '16 20:14
Originally posted by joe shmo
The vacuum pressure will always be between 1 atm and 0 atm. The pressure that can be supplied at any given flow ( in this case the total leakage of the chamber/connections or system ) is unique to the turbo-machine/or positive displacement pump that is doing the work. If there are no leaks; the pump will be able to pull a vacuum pressure corresponding to n ...[text shortened]... the Area of your slots in which a vacuum pressure ( less than 1 atm, greater than 0 atm ) acts.
Which is exactly what I have been saying, it is the area of the slots times the vacuum that gives the total pressure. 14.7 PSI or 6.6 Kg/square inch or 6630 grams per square inch or 6630 grams per 645.16 square mm or about 10.2 grams/mm^2 with full vacuum and 5.1 grams/mm^2 with half vacuum. So 25 mm^2 of slots would give 128 grams of connecting force. So the total size of the substrate has no bearing on the issue, it could be just slightly larger than the area of the slots or 10 times bigger, the force on the substrate would be the same, 128 grams.
14. 04 Aug '16 20:24
Originally posted by sonhouse
That area would not participate in the vacuum force ....
And that is where you are going wrong. There is no such thing as a vacuum force. The vacuum does not suck.

The flat object has three different forces acting on it:
1. The atmosphere on one side - proportional to the area.
2. The solid support which gives a static non-moving force that keeps the flat object stationary but doesn't take part in our calculation.
3. The pressure from your non-perfect vacuum acting on the flat object. I say it again: pressure not suction. This increases in proportion to the amount of flat object exposed to the vacuum.

So, total force required to move the object is force 1. minus force 3.
15. sonhouse
Fast and Curious
04 Aug '16 20:42