# Vacuum pump and creation of vacuum

## Is it easier and faster to create a vacuum of 1 Torr in the 21st. Century than it was

• ### Yes, with a fraction of the energy

• Total voters
7
Sophiecentaur,
Thank you for the tutorial. I will attempt to continue and would appreciate your comments.
Perhaps a clearer discussion might ensue if I were to give a description of the pump itself. It is a sealed cylinder with a piston that is attached to a connecting rod to enable the piston to move up and down the cylinder. The pump has three non return valves a, b and c. The first non return valve (a) is situated on top of the cylinder this will allow air out of the cylinder but not into it, the second non return valve (b) is within the piston itself this will allow air from below the piston to be transferred above the piston but not vice versa, the last valve (c) is in the neck of the pump that attaches to the container to be evacuated, this allows air to enter from the container into the pump but not the other way around. Let us start with the piston at the bottom of its stroke. As it travels up the cylinder valve c opens and air from the container enters the pump. Valve (b) is closed and valve (a) remains closed until the piston reaches near the top of its stroke. As the piston returns back down the cylinder (valve (a) is closed) the air in the barrel below the piston enters the area above the piston through valve (b) on the next upward stroke that air is forced out through valve (a) and more air enters the cylinder through valve (c). If valve (a) has a diameter of 2mm then its total area would be 0.031 cm2. It has already been shown that the air at 1 Torr is denser (under greater pressure) than the air at AP and so can force its way out of the pump. The atmospheric pressure exerted on the area of valve (a) keeps the valve closed until the pressure in the cylinder is enough to overcome this resistance and push the air out. Since AP exerts a force of 1 Kg cm2 it follows that the force exerted on the 2mm diameter valve opening would be 3gms approx. (i.e., 0.12 x 3.14) So the energy needed to overcome this force would be 0.03 x 0.3 = 0.009 Newton metres (or Joules) and the power used would be 0.225 W. at an rpm of 1500. Is this correct or am I missing something. You can see why this problem has me worried.

sophiecentaur
Gold Member
2020 Award
It isn't usual to talk of forces in Kg although 'the weight of a Kilogram' may be what you mean. Could you, perhaps say 9.81N (or even 10N - as it's probably arbitrary). Then you'd be more like using SI units (not just being picky - it really does help you not to crash into Mars because someone thought that someone else meant something different from what they did actually mean)
"It has already been shown that the air at 1 Torr is denser (under greater pressure) than the air at AP and so can force its way out of the pump."??? It's not at 1Torr if it's compressed.

You can't say that the difference between two forces is energy. How do you know how much energy is use in forcing each dollop of air through the valve? It will depend upon many factors and you can't assume the pressure situation on either side of the valve. If the valve is light weight enough (or even resonating) you'd only be dealing with the energy needed to push a small volume of air against AP, which is PΔV, as I offered before.
I think you are getting bogged down in the minutiae of the mechanism and needing to make assumptions (guesses) about things. Why are you worried that it appears the pump only needs 1/4W? How long does it take to get the vacuum of the chamber to 1Torr?
Do the PV sum and see what energy that represents. Compare that value with 1/4W running for the time that your pump needs to operate.
I have to break off now for a day or two. Sea fishing is more refreshing than typing!

Thanks for the consideration. I recently watched the movie ‘Perfect Storm’ which has put me a bit off deep sea fishing, since I had already read ‘Old Man and the Sea’ as a boy and was quite affected by it.
I did realize that I had made a mistake by not converting Kilograms to Newtons . What I should have said was that air at 1 Torr, after it has been compressed by the action of the cylinder, is more dense than air at AP and so has no problem in forcing its way out of the pump. One Kg = 10 N approx. Area of 2 mm diameter valve aperture would be 0.031 Cm 2 , force exerted by AP = 1 Kg/cm2 . Force exerted by atmosphere on valve aperture = 10 N x 0.031 = 0.31 N. Work done in moving piston 30 cms. = 0.3 x 0.3 = 0.09 Nm (or Joules) and power at 1500 rpm = 2.25W. If the pump consists of four cylinders, then there are 4 power strokes at each rpm. That is at each stroke from Top Dead Centre to Bottom Dead Centre, 2 pistons are moving down and two pistons are moving up the cylinder (i.e., 2 intake and 2 exhaust strokes) after the pistons have reached their respective positions they have to reverse directions (i.e from BDC to TDC) to complete one full cycle. So in one rotation there are 4 power strokes. Since the capacity of each cylinder is 6 litres , this means that at 1500 rpm they are pumping out 600 litres of air per second and to empty out a 7000 cubic litre capacity would need:- 7000/600 = 11.6 seconds consuming 11.6 x 2.25 W = 26.25 W. This is the energy needed when dealing only with the atmospheric pressure but since the amount of air available in the container gets reduced at each stroke it should actually read P x Delta V and then friction forces and weight of the piston should be added in to get a more realistic figure.

The main problem with building a low power consumption roughing vacuum pump seems to be that of friction. I recently watched an episode of ‘Mad Scientists’ ( I know, I Know) where they showed a fire making device from Borneo, the laboratory version was a small diameter length of hollow glass pipe fitted with a piston and plunger, when the plunger was pushed down with enough force and speed, kindling placed at the bottom of the glass pipe was set alight. Exactly the same principle as used in the diesel engine. Using large bore cylinders and pistons to evacuate a vessel is going to result in a dramatic increase in temperature. Though how this rise in temperature will be determined is not obvious because on the down stroke, one way valves in the piston allow the air that is being compressed below the piston to escape into the top half of the cylinder above the piston. Again on the upstroke the one way valve at the top of the cylinder opens to release air as the pressure is built up to a sufficient level. Still, given that PTFE on PTFE has a co-efficient of friction less than that of ice on ice and can operate at temperatures up to 500 O C. this should not pose too much of a problem. Also new material such as the silica ceramic heat shielding tiles on the shuttle that are very light and heat resistant could be considered for pistons, connecting rod etc., Like most of you, I am just passing the time, thinking of non-obvious things.

sophiecentaur
Gold Member
2020 Award
The main problem with building a low power consumption roughing vacuum pump seems to be that of friction.

Absolutely. The work done on the gas is very little - certainly at the low pressure phase of operation. This is just what I have been saying all along. There is much less power involved when shifting small masses of air compared with shifting large masses of air. The best demonstration of this is when you put your hand over the nozzle of a vacuum cleaner and the motor speeds up dramatically. That's because it is no longer shifting any air and its load has reduced.
However, as the pressure difference increases, there will be higher forces on bearings, seals etc. and this will increase the amount of power lost. Working out the power expended on the gas is trivial compared with working out the friction losses. I should say that you couldn't hope to do that, in fact. You'd have to measure rather than calculate.

I don't think your point about adiabatic heating is relevant because the energy expended in compressing the small masses of gas in the cylinder should be 'returned' in the form of work done in pushing the gas out against AP. ALso, (when at lowest pressure) the remaining gas in the cylinder will be adiabatically cooled on the way down (lowering the pressure), reducing the energy needed to admit the gas from the chamber.
If you were to move the pump slowly then the change would be isothermal, in which case, there would be energy loss but during fast operation, there should be no energy loss. *

As a footnote, I think you have still not really appreciated my original argument about the effect with one huge cylinder. Thermodynamically speaking, the situation is similar if you have an initial volume of V, which changes to V', as if you take n strokes of a small cylinder of volume ΔV - where (V'-V) = nΔV. You could imagine a whole set of n small cylinders connected to the chamber, each one increasing the volume by ΔV. Would there be an overall difference between the situation of moving all the pistons at once or moving them one at a time? I don't think so.

* This is because the actual work done on making gas flow through the valves is very small. That component of work would be only ΔVΔP, on each stroke, where ΔP is the minute pressure drop across the valve whilst gas flows through it. The valves, of course, would need to be designed correctly to achieve this - not something you could actually calculate, either.

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I have got confirmation of sorts in the form of a pump that works at 1000 m3 h. (Which works out to 16.6 m3minute). with a power consumption of 1.5 Kw and pumps down to 10 -4 mbar which is much above 1 Torr (1.3mbar). To clarify, although the motor rating is about 5 Kw the power consumption is 1.5 Kw. Considering that the hypothetical problem set here was for a 7 cubic metre tank (i.e., about 2.5 times less than 16.6 cubic metre), and a final pressure of 1 Torr against 10-4 mbar. Which is about 10 -6 Torr. I think the original figure I had given of a power consumption of about 250 W is well within limits. Still nothing like taking practical measurements as you had stated. Thanks for the input.

sophiecentaur
Gold Member
2020 Award
And how much energy is required to evacuate your hypothetical 7m3 tank? (the PV value). What running time for your motor would this represent and at what power??

There still seems to be (units) confusion as to whether you are talking about Energy or Power. A given power of motor can supply as much energy as you want, by running it for the appropriate time. Are you, perhaps, referring to the Power needed to maintain this 1Torr, once it's been reached?

Have you actually measured the power consumption of the 5kW rated motor? What method did you use? A real Power or Watt-Hour Meter - or just V and I? Where do you get your "250W" from? I should have expected the power from the motor to start high and then to reduce to a steady value - is that what happened?
When people describe the observations they have made on these forums, you can never be sure whether they have the whole of the CERN facilities or a Multimeter from Maplin haha.

You can see why Peer - Reviewed papers are required before the establishment will accept evidence to support any theory. We could be talking at completely cross purposes (all too common, I find - and it's not always my fault ).