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How Much Air Can An Ioniser Clean?

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How much air can an ioniser clean? Here's my suggestion.


An ioniser can generate around 2.5 microamps per needle. Conservatively perhaps1 microamp per needle. (See for example an experiment on youtube, or commercial documentation).


The charge on an electron is 1.6 x 10-19coulombs.


So 1 coulomb is 1/(1.6 x 10-19) electrons = 6.25 x 1018 electrons.


So 1 microamp is 6.25 x 1012 electrons per second.


Next the ISO9 clean room standard for city air is 35 x 106 particles per cubic meter.


So 1 particle occupies 1/(35 x 106) cubic meters, (at least arithmetically)


Assuming that the current is distributed by Brownian motion and self repulsion, as 1 electron on 1 particle (as in Millikan's Oil Drop experiment for example), we get that the charge at least theoretically could be distributed into:


(6.25 x 1012)/(35 x 106) cubic meters = 176,000 cubic meters of air


By cube root this corresponds to a cube of air 56 meters on each side.


In addition this volume is charged every second!


The assumption then is that if the charged particle comes near to an earthed surface, it is drawn to the surface and sticks, so cleaning the air.


Does this make sense? Can anyone improve this estimate? This seems to imply that ionisers could be put on lamposts to clean the air in a city, never mind in a room.


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Just as a crosscheck of your numbers, I find that a typical 10 ampere, 220-Watt unit can produce 200,000,000 ions/cubic centimeter at a distance of 1 foot from the unit. That is, in a small volume of say .03 cubic meter, or 30,000 cc there are some 6E12 ions.


Now, consider what volume could be ionized with that many ions to achieve your designated density of 35E6 particles per cubic meter. That works out to about 172,000 cubic meters, in close enough agreement with what you came up with.


But there are several other factors that you did not take into consideration:


1). Will one ion per particle actually be able to clean the air, or does the ratio of ions/particle need to be much higher in order to be effective?


2) 172,000 cubic meters is the volume that could be ionized if the ions can be evenly distributed throughout. To do that, that volume of air would need to pass through the small local volume where the ions are produced. In other words, the unit needs a fan and the fan speed will be the dominant factor for the air delivery rate to the ionizer.

Taking the second consideration first, a typical ionizer fan produces a flow of 300 cfm, or 8.5 cubic meters/minute. If there was perfect mixing within the 172,000 cubic meters, that volume could be ionized in about 20,000 minutes or 14 days. (Not every second, as you wrote) In reality, the mixing would be very imperfect; most of the ions would be located near the machine and then spread out with continuously decreasing density.


As for the first consideration, there is no reason at all to think one ion per particle would work. In fact, this wasn’t even the case in Millikan's Oil Drop experiment. In that experiment the charge on each particle was measured to be some integer multiple of the basic electron charge, so more than one ion per particle. What the ratio should be for effective air cleaning is another matter altogether, and I don’t know the answer with any certainty.


What I can say is the typical unit I mentioned earlier, which produced 6E12 ions, is intended for a standard-size room of about 170 cubic meters. That works out to 35E9 ions per cubic meter or about 1000 ions per particle to meet ISO9 standards. Even this is an over simplification as this actually is a very complex non-linear process that depends on the physics of the ion-particle interactions and the duration of the particles and the ions in the space and would need to be modeled using differential equations. Here is a link to an interesting analysis that has the theory and math.





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First of all, thankyou so much for your very interesting reply.


Now the article seems to suggest 1 to 10 ions/electrons per particle as appropriate, and then chooses 5, if I understand it.


At 5 electrons/ions per particle, in my very rough basic calculation, the charge would distribute into 176,000/5=35,200 cubic meters instead.


And then the cube of air charged is 32.7 meters on each side, still very large.


And then again, in the oil drop experiment context, 'some integer multiple' does not seem to imply a value of electrons/ions per particle as high as 1000, and smaller values must have a measurable effect. So it might be that the typical unit mentioned is significantly over-rated.


Intuitively, the larger the particle, the greater the number of charges it can carry. This implies something like a similar (equal?) field intensity at its surface, according to its radius, and that in its turn related (equal?) to the field intensity of, say, an oxygen molecule carrying 1 electron, and that in its turn to the field intensity at the ioniser needle. That seems to be a different idea though.


As to the fan, the simplest assumption is that the 'electrostatic wind', or self repulsion, generated by the ions eventually distributes them evenly in a volume, as if fan assisted, helped by natural currents. But that might take so long that a steady state even distribution is unwarranted. So I tried my ioniser, its wind caused a lighter flame to sputter, but did not blow it out. A fan would! So the electrostatic effect is significant, but not quite as large! But anyway, this calculation is not really about how long, but how much. At a truly wild guess, lets say even distribution after 10 minutes with a fan, and 60 minutes without.


My basic contention is that it appears that a 'room ioniser' might have a significant effect in a much larger space. Am I right I wonder? I don't think it can do anything in the open air. Or could we make a 'big' ioniser? Ha!

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Thank you for bringing up an interesting subject for discussion. Lately we have been getting a preponderance of crackpot posts and it is a pleasure to have something worth replying to.


Unfortunately, it appears you did misunderstand the paper. The range of 1 to 10 that you mentioned refers to the number of ion charges that are acquired by the particles, not the ratio of ions to particles in the room.


Here is the relevant quote from the paper:


“This indicates that at typical ion concentrations achievable with an ioniser in a real room, the number of unit charges acquired by sub-micron particles are likely to be of the order 1 to 10 during their residence in the room. For the purposes of simplifying the model the characteristic number of unit charges carried by these small particles is approximated as qc = 5.”


As for the ratio of ions to particles in the room you can see this by referring to the Graphs and Tables but here are a few relevant quotes:


“Figure 7 shows the predicted ion concentrations from a second simulation where sub-micron particles are included in the model with a generation rate two orders of magnitude greater than the larger particles of 2.17 x 10^5 particles/m3 s and an initial concentration of 1.8 x 10^6 particles/m3.”


“Although the charging of particles is not considered to any great extent in this study, it is clear from Figure 9 that for an ioniser to result in any significant electrodyamic effects the ion generation rate must be higher that ~10^8 ions/s, with an ion concentration in the room space of 10^10 ions/m3 or greater.”


As you can see, for effective cleaning the number of ions in the room must be several orders of magnitude greater than the number of particles. A ratio of 1000/1 is actually a bit on the conservative side.






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Ok. I see my mistake. So why this factor of 1000, or whatever? My guess might be that there is a problem with the rate of collisions between the particles and the ions. The ions have a low density or partial pressure, and at 1 ion per particle the particles have an equal low density. The ions and the uncharged particles rarely collide, and so they co-exist for a long time. Then a large excess of ions is needed to ensure that enough collisions, presumably 5, occur to transfer 5 electrons of charge to any given particle in some reasonable period of time, on average of course. This must relate to some quantity like mean free path. So we have a low density gas of highly mobile ions, a related low density of rather static particles, and a high density of unionised air, as a carrier if you like, of equal mobility, or so, to the ions. Goodness! I think I just went out of my depth. Is there a simpler approach to that situation, to make an estimate? Can we ignore the bulk of the air for example? I am beginning to think that ionisers are designed by guesswork and precedence, and so no two ionisers are in fact much alike.

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I don’t think there is a simpler approach to estimate the right ion/particle ratio for effective air cleaning. The paper I referenced details the experiments that were conducted as well as the mathematical models, which involve differential equations. The differential equations themselves are not too difficult but the coefficients they use were most likely obtained from experiment. Apparently, some experiments came first, then the mathematical models were formulated and refined using coefficients obtained from those experiments and some statistical analysis.


If you want a simpler way of thinking about this, you might try thinking of blindly trying to shoot down a stealth airplane. You know the plane is somewhere in your airspace but you don’t know exactly where it is. The best you can do is put up as much flak as you can, in the hope of hitting it and even one hit may not be enough to bring it down. After running this experiment several times, you find that the flak is only effective when there are at least 1000 shells fired for each aircraft you want to hit and it takes an average of 5 hits to bring a plane down. Obviously, the more flak you can put up, the more likely you will be successful at downing an aircraft. That is a very crude analogy but way easier to think about than a statistical analysis!


As a side note: If those particles ever develop stealth technology, the ionisers will be useless against them. In the allied air campaign over Iraq, not a single F-117A Nighthawk stealth fighter was shot down, despite extremely heavy Iraqi anti-aircraft fire. In fact, there has been only one incidence of an F-117A Nighthawk stealth fighter being lost to ground fire. That was on 27 March 1999, by the 3rd Battalion of the 250th Air Defense Missile Brigade of the Army of Yugoslavia (now Serbia, Montenegro, Slovenia, Croatia, Macedonia and Kosovo)


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Ha! My mental analogy was a solar system, with asteroids rarely hitting planets, and electrostatic forces replacing gravity! I think yours is rather better, if we assume that forces do not encourage collisions.

We can still get some sort of result. At 1000 ions per particle, we get that the charge at least theoretically could be distributed into:

(6.25 x 10E12)/(35 x 10E9) cubic meters = 176 cubic meters of air

By cube root this corresponds to a cube of air 5.6 meters on each side.

And this volume is charged every second as before.

That still seems to suggest that ioniser is removing particles at a rate of around 1 room per second, so to speak.

I also note that I seem to have assumed an ioniser standard, with a single needle carrying 1uA of current, which is rather arbitrary, but there don't seem to be any ioniser standards that I can find. That also means that the experiments in the paper just used an example ioniser, which could have been poorly designed, or overated.

Another approach might be to consider that the ionised air continuously discharges to ground via the walls, and with an ioniser this results in an equilibrium ion concentration. We measure the ion concentration of, let's say, fresh sea air, take that as 'international standard ionised air', and size the ioniser to maintain that ionic concentration in a discharging environment.

At least that suggests a surface area square law. A room twice the 'usual standard size on a side' requires four times the 'usual standard ioniser current', or equivalently, 4 standard room ionisers.

I also note that a device called an electrostatic precipitator, used to clean power station flue smoke, seems to be rather like an ioniser with a collection plate and a forced flow. Electrostatic precipitators seem to have some rigorous engineering design standards. However they do seem to work in part, or mostly, by subjecting the smoke to an unionised electric field, so they are not quite like a free air ioniser. On the other hand, if you add a fan and a filter or collection plate to a free air ioniser, you seem to have arrived at an electrostatic precipitator design via a different route.


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Another approach might be to draw up an analog equivalent circuit for an ioniser in a room something like this:


Image not accepted. It's a current generator feeding a resistor and capacitor in parallel.


This avoids differential equations, at least to begin with. The current is in ions or electrons per second, not amps, the potential in ions/cubic meter, not volts, the capacitance in cubic meters, and the resistance is proportional to the reciprocal of the wall or grounded area in square meters, I think. There might well be some other constants of proportionality as well.


In this model the particles in the air have been neglected and the ion concentration in the air treated as an end in itself. Hmm! However a fan might be equivalent to an increase in wall area, and a filter might be the same (if earthed?). An open window becomes perhaps a second current generator (or sink) in parallel. Particles might be an additional volume since they accept charge, but a flow of particles from a source (a fire? josticks?) might be an increase in area as well, since they can be carried to the walls, along with their charge. A fully ducted air conditioning system with inlets and separate exhausts might possibly break the analogy.


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A re-formulation of the original question might be:


How many room ionisers are needed in a much larger space?


Using the results above, and assuming that the ionisers are correctly sized for one room, we can measure a large volume in room units, cube root that, and square it to the number of ionisers required. This is very much a rough engineering estimate. For example:


A hotel foyer two rooms, wide, deep and high might require 4 room ionisers.


A hospital reception two rooms wide and deep but only of standard height might need 2 or 3 room ionisers.


An attrium 3 rooms wide, deep, and high might need 9 room ionisers.


A bus of length two rooms might need two ionisers to allow for the increased ventilation.


A station concourse as big as a 125 room hotel might need to be fitted with 25 room ionisers.


This rule maintains the ion concentration as it would be in a room, and not the removal of particles by them sticking to surfaces. Obviously in some cases a pro rata reduced number of double or triple current rated ioniser units would make practical sense.

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I think it's important to say that the analysis above applies to a negative ioniser, which consists of a high or static voltage generator (circa say 20,000 volts), some series resistors (circa 10 megohms) converting it to a current generator, and needles, a brush or a wire loop emitting negative ions at a much lower voltage (circa a few thousand volts). The point then is that this generator is earthed at its positive end, and is not fully isolated or floating.


This means that the earthed walls, or an earthed person standing in the room, experience negative ions or particles.


Because of a lack of standards other proprietary arrangements seem to be possibly occurring. For example the positive end could be connected to an arrangement of plates (which might be described as a filter or collection plate), assisted by a fan, and the arrangement could be ungrounded or floating. In this case a person in the room and the walls experience no negative charges and the above analysis does not apply. The device itself is then perhaps described as an ioniser air filter or cleaner or electrostatic precipitator. It's self contained. But the plates or filter could also be fully or partially grounded (by say 100 megohms), in which case it has two modes of action. It's also possible to reverse the polarity and have positive needles and ions. It's also possible it seems to have needles at both polarities. The device then generates both positive and negative ions separately and simultaneously. That arrangement seems to be used or described as an anti-static generator. Whew!

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