Vacuum pump and creation of vacuum

[McQueen]

Air weighs about 800 times less than water, 1 litre of water weighs 1000 gms approx. while 1 litre of air weighs 1.25 gms approx. So theoretically it should need less energy ( in fact a lot less) to create a rough vacuum in a given volume than it would need to remove a liquid (water) from the same volume. If you remember Otto Von Guericke in the 17th. Century, (more than 400 years ago) performed his famous sphere experiment in Magdeburg by removing the air from two hemispheres of bronze that were joined together creating a vacuum in the resulting sphere by using a simple piston vacuum pump. 18 horses could not pull the spheres apart, (when he was in Berlin he repeated the experiment with 24 horses.). The vacuum that Guericke created probably had a value of about 1 Torr. (1/760 atmospheric pressure) . Looking once more at Guericke’s piston pump, one wonders if today one could make a reciprocating vacuum piston pump out of graphite and Teflon, which would be almost friction free, lightweight and heat resistant and could be run at high rpm. The pump could be a multi-cylinder (say 4 cylinder pump) with a box fitted to lengthen the stroke of the piston giving each cylinder a capacity of about 6 litres. Using such a pump it should be possible to empty a 7 cubic metre container to a pressure of 1 Torr in about 38 seconds. Thinking about this air at atmospheric pressure contains about 2.69 x 10²⁵ molecules per cubic meter while air at 1 Torr contains about 3.5 x 10 ²² molecules per cubic meter. Again the distance (mean free path) between air molecules at atmospheric pressure is about 6.7 x 10 ^{- 6} cms and at 1 Torr it is 5 x 10^-3 cms . Therefore there will be enough molecules to pump out even when the pressure reaches 1 Torr. , since air expands to fill available space and will exert an equal pressure on all points within the container. Consider that at each stroke of the pump it has to shift a maximum of only 7.5 gms of weight (i.e. weight of 6 litres of air) atmospheric pressure is sealed off so it doesn’t count. Such a pump should need to lift only the weight of the air and its own weight (weight of piston connecting rod etc.) My question is how much power will it take to run. A minimum of about 250 Watts I am guessing. Any comments ?

 

[f95toli]

There are tiny, lightweight diaphragm pumps that can reach pressures of about 1-2 mBar (typically used as backing pumps); their power consumption is of the order of 50-100W.

They are sold by Edwards and many other companies.

 

[russ_watters]

The power can be literally anything, with the variable being how long it takes to pump out the vessel.

The statement about water vs air in the beginning is uselessly vague and not necessarily true. What was the point of that? And I'm not sure I see the relevance of the historical anecdote or the use of the word "lift" toward the end...this isn't an attempt at free energy, is it...?

 

[McQueen]

Russ Waters,

The power can be literally anything, with the variable being how long it takes to pump out the vessel.
The statement about water vs air in the beginning is uselessly vague and not necessarily true. What was the point of that? And I'm not sure I see the relevance of the historical anecdote or the use of the word "lift" toward the end...this isn't an attempt at free energy, is it...?

I am unhappy to note you have not read the post properly as I had made a definite mention of the time estimate of 38 seconds. I am ( I hope not without some effect ) shocked that you should think that the difference between the mass of air and water is pointless or that a reference to Von Guericke is frivolous. I am tempted to ask what you don’t find frivolous ! A good scientist should try, at all costs to avoid pedantry and keep his/her mind open and fresh! But anyway both of these facts are important in context of the OP because (a) if there is a such a large difference in the mass of air and water, obviously it will take more energy and work to shift a given volume of water than to shift a given volume of air and yet if you look at the power consumed by contemporary Vacuum pumps ( even roughing pumps as mentioned in the OP ) there seems to be a phenomenal difference between the work to be done and power consumed. A large volume pump as that mentioned in the OP that would remove the air to a pressure of 1 Torr from a container with a volume of 7 cubic metres in a very short time, might consume anything between 26 hp to 15 hp (i.e., 19.3 Kw and 11.1 Kw ) which seems a bit high given the weight (mass) of air to be moved. WHY ? (b) surely technology has improved during the intervening 4.? centuries and consequently the performance should have improved also, at least it should have. Again returning to the example of the time taken to empty the 7 cubic metre capacity tank to a pressure of 1 Torr should, when I re-think it, take only 10 seconds with a 4 cylinder piston pump with a capacity of 6 litres per cylinder, running at 1500 rpm. Again both graphite ang Teflon (PTFE) have a frictional co-efficient lower than ice, so why isn’t the power being consumed reduced drastically. Is this of interest or not ?

 

[only1universe]

Russ Waters,


I am unhappy to note you have not read the post properly as I had made a definite mention of the time estimate of 38 seconds. I am ( I hope not without some effect ) shocked that you should think that the difference between the mass of air and water is pointless or that a reference to Von Guericke is frivolous. I am tempted to ask what you don’t find frivolous ! A good scientist should try, at all costs to avoid pedantry and keep his/her mind open and fresh! But anyway both of these facts are important in context of the OP because (a) if there is a such a large difference in the mass of air and water, obviously it will take more energy and work to shift a given volume of water than to shift a given volume of air and yet if you look at the power consumed by contemporary Vacuum pumps ( even roughing pumps as mentioned in the OP ) there seems to be a phenomenal difference between the work to be done and power consumed. A large volume pump as that mentioned in the OP that would remove the air to a pressure of 1 Torr from a container with a volume of 7 cubic metres in a very short time, might consume anything between 26 hp to 15 hp (i.e., 19.3 Kw and 11.1 Kw ) which seems a bit high given the weight (mass) of air to be moved. WHY ? (b) surely technology has improved during the intervening 4.? centuries and consequently the performance should have improved also, at least it should have. Again returning to the example of the time taken to empty the 7 cubic metre capacity tank to a pressure of 1 Torr should, when I re-think it, take only 10 seconds with a 4 cylinder piston pump with a capacity of 6 litres per cylinder, running at 1500 rpm. Again both graphite ang Teflon (PTFE) have a frictional co-efficient lower than ice, so why isn’t the power being consumed reduced drastically. Is this of interest or not ?

I'm struggling to see the application of basically lubricating an old experiment.

 

[McQueen]

Let me explain, for those who are interested but do not have the time to look it up or the necesssary information at their fingertips, a little about vacuums, and why Otto Von Guericke’s work with them is so relevant. Von Guericke was able to create a vacuum of 1 Torr (Torricellian vacuum) using only the most basic equipment. Which is quite an achievement, he also did a lot of work with static electricity ( the familiar revolving ball of sulphur in the British museum, which makes your hair stand on end was one of his experiments.). Secondly the use of a vacuum is indispensable in modern industry. For instance vacuums are used in numerous applications from thin film production to packaging and from electron microscopes to certain processes in air craft and spacecraft production. However the degree of vacuum needed for these applications varies greatly. The vast majority of industrial uses require relatively low vacuums, anything from about 10 Torr to 1 Torr, which is why I consider the OP to be both important and relevant. More sophisticated applications require higher order of vacuums that might range to about 10 ^{– 9} mm of mercury . At 10 ^{– 9} mm of mercury, the distance between individual molecules of air is about 50 Km! On the other hand at 1Torr the mean free path between molecules is only on the order of 5 x 10 ^–3 centimeters. It follows that to make vacuums of such an high order, requires considerable energy and the use of sophisticated equipment. This does not mean that the performance and energy consumption of machines we use to create lower order vacuums cannot be improved. In fact looking at the facts, it is surprising that they have not been improved. Is there an explanation for this lacuna or is there really some important physical consideration that I have overlooked ?

 

[f95toli]

It follows that to make vacuums of such an high order, requires considerable energy and the use of sophisticated equipment.

One torr is definitely not a good vacuum; that kind of pressure can be achieved with just about any pump. Even a very basic rotary pump -which is about as sophisticated as the pumps used to increase the water pressures in homes- can reach a pressure of about 0.01torr or so.
A "proper" vacuum pump (say an ordinary turbopump) will get you down to 10^-6 torr.
It is only when you start talking about pressures lower than that, that you need sophisticated equipment (not that a turbo pump is cheap, but that has more to do with the fact that the market is relatively small).
And again, a standard diaphragm pump (which consumes very little power) can easily reach a pressure of 1 torr.

 

[Lsos]

I don't think it's so much the mass of what you're evacuating that is important, but the fact that you're moving it "uphill" against pressure. Pressure doesn't care if the substance you're moving weighs 1 gram or 1 kilogram, it's going to try to stop you just as hard. I suppose a more massive substance would be harder to move due to inertia, but I doubt that's a significant effect.

Someone correct me if I'm off, and/or if I'm completely missing the point of what OP is asking.

 

[McQueen]

f95toli

Thanks for clarifying that point, yes 1 Torr is definitely NOT a hard vacuum, the point I was trying to make is that the vast majority of Industrial vacuums are from about 10 / 15 Torr to 1 Torr, yet a (comparatively ) huge amount of power is used to prepare these rough vacuums. Again I agree with you that a diaphragm pump could easily reach a pressure of 1 Torr but look at its capacity and rpm and ask yourself whether this (tiny) pump could evacuate a volume of 7 cubic metres in under a minute ? To my mind the point that Lsos, is trying to make might hold water when dealing with pressure below 1 Torr for instance at a pressure of 10 ^–3 Torr the distance between air molecules would be about 5 centimetres, so the pump that we have described would be able to collect very few air molecules (if any) during each stroke BUT for a vacuum of 1 Torr none of this applies. The pump is sealed off against atmospheric pressure and the pressure of the molecules that the pump is evacuating will be more than the atmospheric pressure weighing against them. If this had not been the case I doubt if Von Guericke’s crude hand operated pump could have managed. In any case a simple mathematical calculation should prove the truth of what I am saying.

 

[McQueen]

Going into the problem in a little more detail. When the pump has pumped out half the air in the 7 cubic metre capacity tank, there will be half as many molecules available to pump out. So it will take double the time to pump out those remaining molecules of air, when the tank is three quarters empty, there will be only a quarter of the molecules remaining so it should take four times the time to pump out those remaining molecules of air in order to reach a final pressure of 1 Torr in the tank. On the other hand at a pressure of 1 Torr there are still 3.5 x 10 ¹⁶ molecules available per cubic centimetre remaining, so that a pump with a cylinder capacity of 6 litres should contain about 3.5 x 10 ¹⁹ molecules of air, sufficient for the pump to pump out. Further since air exerts pressure equally in all directions it means that pump does not have to ‘hunt’ for them as would be the case if the vacuum were on the order of 10 ^-3 Torr or greater. As I see it the pump will continue to work at the same rate (dependent on rpm) while the mass of air moved will decrease. According to the industry the pump will lose about 10% efficiency when the tank is three quarters empty, but this should not drastically effect the time or the power needed to achieve a vacuum of 1 Torr. So the problem basically resolves down to a question of whether it is possible to build a vacuum pump using advanced materials such as graphite and PTFE that would use just 150W to 250W to evacuate a container with a 7 cubic metre volume, to a pressure of 1 Torr, in well under a minute.

 

[Lsos]

Going into the problem in a little more detail. When the pump has pumped out half the air in the 7 cubic metre capacity tank, there will be half as many molecules available to pump out. So it will take double the time to pump out those remaining molecules of air...

There will be half as many molecules to pump out AND you will have 2x as much pressure to fight against when pumping them out. The pump will definitely have to work harder and likely longer.

When the pressure is equal inside and out, then the pump does no work other than, well...pumping losses. If the pressure inside was greater than outside, then the pump could actually do negative work...that is, it could receive energy from the air that would be flowing past it on its own accord. But you can clearly see that the more air you remove, the more this situation deteriorates.

 

[xts]

Guys! A little bit of common sense!

7 cubic metres of vacuum vs normal pressure (residual 1 Torr makes no difference) in 38 seconds means 7m³ * 100kPa / 38s = 18kW - that's an absolute minimum.

I also doubt if von Guericke really reached 1 Torr - for his experiment with two feet in diameter hemispheres, 1/2 atm would be sufficient to stop 9 (12) horses pulling on each side. Even strongest Saxonian percherons...

 

[xts]

Air weighs about 800 times less than water, 1 litre of water weighs 1000 gms approx. while 1 litre of air weighs 1.25 gms approx. So theoretically it should need less energy ( in fact a lot less) to create a rough vacuum in a given volume than it would need to remove a liquid (water) from the same volume.

Why do you think so? Have you seen Torricelli's barometer? You don't need any force nor power to remove mercury from a quarter of a metre long pipe. But 'easier to remove' air do not leave it on its own will...

 

[russ_watters]

Yeah, I missed the 38 seconds buried in a long post...but xts is right. What it looks like you are missing is that you aren't just lifting air out against its mass, but working against the pressure of the atmosphere.

There isn't much new to be done with vacuum pumps because they are relatively simple and the laws of physics governing them hasn't changed.

 

[xts]

18kW - that's an absolute minimum.

I must admit - I lied That'd be an adiabatic minimum.
In isothermic conditions you may make it twice cheaper. But theoretical limit of 9kW is still much beyond your 250W.

 

[sophiecentaur]

If you started with a piston right at the top of a cylinder with no inlet, the work done in pulling the piston down to leave a volume of V against an atmospheric pressure P could be as little as PV if the seal were perfect and the initial volume were exactly zero. Any energy n top of this would be due to the imperfections of the system, mainly the 'compression ratio' of the pump plus the chamber to be evacuated and any leaks, of course.
For the first few strokes, the pressure difference would not be near P but, pretty soon, the pressure difference would be more or less P, so the total energy would be approximately PV times the number of strokes of the pump that were needed. The lowest achievable vacuum would be limited by the compression ratio of the pump, before the non-return valve (and how well the valve works, too)
All these comments only apply, of course, to moderate levels of vacuum but it's a start, if all you want is some basic figures to start with.

 

[McQueen]

The speed at which the 7 cubic metre volume container is emptied to a pressure off 1 Torr ( I can’t think how xts can ignore this final pressure) is solely dependent (when dealing with a piston pump) on the capacity of the cylinder, number of cylinders and the rpm. If normal materials are used, rpm will be restricted to a minimum by frictional heat, or alternatively the pump would have to have a water cooled jacket etc., raising its complexity , weight and power consumption. Friction plays a large role in the amount of power consumed, in order to obtain an air tight sealing an air tight fit is needed, this increases friction reduces rpm and rate of work and increases power consumption. By using lightweight materials with a very low co-efficient of friction and good sealing qualities, many of these problems are eliminated. As for the work done it comes down to the differential pressure, (inlet pressure versus outlet pressure), if the inlet pressure is greater than the outlet pressure (as in the case of 1 Torr compressed into a smaller area increasing its density) then the work done is minimal. To take full atmospheric pressure at 100 Kpa does not seem to be correct.

 

[sophiecentaur]

Compressing to 1torr is not what counts (the positive energy for that is negligible. What counts, surely, is the work in shifting the majority of the air out against the atmosphere.
The points you make are valid but basically practical. My approach at least gives a meaningful baseline. Compressing the portion of gas which is extracted to AP wastes energy but the amount would depend on the details, I think.

 

[sophiecentaur]

I have thought some more on the practicalities of this. I still say that the baseline energy needed to achieve a low vacuum is PV. Low, very low or very very low will make very little difference to this figure. However. What limits the final pressure will be the compression ratio of the pump (including all the volume on the pump side of the non return valve) and the effectiveness of the valve - i.e. how well it opens and whether it leaks back. With a reciprocating or rotary pump, the same thing applies. As far as I can see, the steady state power requirement to maintain its 'best' vacuum will be equal to the volume (at AP) expelled each stroke times the number of strokes per second times P. For a 'perfect' valve, this would end up as zero (because no gas would leak back inside) and the final pressure inside would be given just by the compression ratio of the intgernal workings of the pump.
@McQueen
I think that you will appreciate that there will be very little difference between the total pressure change for a 'good' pump and a 'poor' pump. What's the effective difference between 99.9% and 99.9999% of PV? That's why I am ignoring the actual final pressure for the baseline energy calculation.
If you have a poor pump, I agree that it will require more power to maintain a given vacuum than a good pump because the poor pump will be leaking more gas back in, which constantly needs to be removed (more work each stroke). A poor pump would need to be run faster.

 

[McQueen]

@sophiecentaur,
While I agree with your comments with regard to the valve, cylinder capacity etc., I still think that the order of vacuum to be achieved plays a significant part. As I had mentioned before in a vacuum of 10 ^-3 Torr the mean free path of the molecules (i.e., distance between air molecules) would be 5 cms and at 10^-6Torr the distance between molecules would be 50 metres ! So the chances of a molecule entering the pump cylinder are slight. On the other hand at 1 Torr the mean free path between air molecules is 10 ^-3 cms. (10 microns). So the flow is viscous, there are plenty of molecules available in the pump cylinder to be pumped out. If, as you had pointed out, the one way valve to the atmosphere works properly and leakage is at a minimum, very little energy should be required to work the pump and everything would depend on the the cylinder capacity, the number of cylinders and the rpm. If this is true why is so much energy needed to operate such a pump. The main reason seems to be the heat generated by friction and the need to operate the pump more slowly or alternatively to make the whole machine heavier by using a water jacket for cooling. If on the other hand a high speed rotary pump is used, its capacity would be smaller, moving less air in a given time. How does pressure x volume fit into all this ? I am curious. What I understood from your post is that initially the energy needed to take the piston from the top of the cylinder down would be negligible but as the container being pumped out empties, energy needed to move the piston will be close to atmospheric pressure. Is this right.

 

[sophiecentaur]

IF you just consider my original scenario of a closed cylinder with a piston right at the top. I think you would agree that the work done in moving the piston so that it leaves a volume of V (a vacuum) would be PV. There isn't any other 'magic' sink of energy / work involved. If there is a small volume of air in the top at the start then the only difference in energy needed will be that fraction - negligible, if you are talking of a high vacuum.
Can we go that far, for a start? Or can you say where I am wrong?

btw, your last paragraph mixes up its units and I don't see where it is going.

 

[McQueen]

For the first few strokes, the pressure difference would not be near P but, pretty soon, the pressure difference would be more or less P, so the total energy would be approximately PV times the number of strokes of the pump that were needed.

Sorry, Sophiecentauri, I have been giving a little more thought to what you had written in your post and have come to the following conclusion. At the final pressure attained of 1 Torr the number of molecules occupying a 6 litre volume (6000 cc cylinder capacity) would be 2.1 x 10 ²⁰ approx. As this space is compressed during the upward stroke of the piston the volume would be reduced from 6000 c.c approx to 11.25 cc (if there is a 2mm clearance between the piston at TDC and the top of the cylinder) so the number of molecules available would be approx 2.1 x 10²² approx. Therefore, higher than Atmospheric pressure, ( Air at atmospheric pressure contains approx 2.69 x 10 ¹⁹ molecules) cm² . In actual fact to calculate more exactly it is necessary to use calculus and calculate with diminishing pressure over time. Taking only AP the maximum load on the pump would be 15Nm for a pump at 1500 rpm this would work out to 125 W/sec. (i.e AP of 0.5 Kg time length of stroke of 30cms. ) Add on the weight of the piston, frictional forces etc., and the answer should approach power needed to run the pump, which is still much less than the 9Kw suggested.

 

[sophiecentaur]

The only difference between successive strokes of a small pump with a large vacuum vessel and a large cylinder with no added vacuum vessel is that the energy is expended in an exponential fashion for the small pump but it's linear for the single cylinder.
I agree that you would need to use the analytical approach, rather than just the numbers - the shutter comes down for me at a page of numbers but an equation makes things clearer.
It just strikes me that it is much easier to consider the work done in pushing air out into the atmosphere than worrying about how you actually achieve this push. The work done on each stroke will be PΔV, where ΔV is the volume of air actually pumped out. That volume will actually be the volume of the pump divided by the compression ratio. This gives an exponential relationship between stroke number and the inside pressure.

Once again, your units are all to hell. You can't have W/sec or a load of 15Nm or a pressure of 0.5Kg ! We can't continue unless you 'behave' yourself with the units. It can just turn out nonsense.

 

[McQueen]

I was under the impression that watt per second was the rate at which power was being used, should I have used Joules instead of Newton metres, and instead of pressure, force ?

 

[sophiecentaur]

Basics:
Watts (W) is the unit of Power.
A Watt is One Joule (J) per second.
Force is Newtons (N)
Weight is also Newtons
Mass is Kilograms
OK?
Let's continue

 

[McQueen]

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]

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!

 

[McQueen]

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 into get a more realistic figure.

 

[McQueen]

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]

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.

 

[McQueen]

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]

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 ).