lewisw85 1 Posted October 21 Report Share Posted October 21 What ion concentration does an ioniser produce in a room? I'm a bit stuck. Here's what I have. An ioniser can generate around 2.5 microamps per needle. Conservatively perhaps 1 microamp per needle. (See for example an experiment on youtube, or commercial documentation). The charge on an electron is 1.6E-19 coulombs. So 1 coulomb is 1/(1.6E-19) electrons equals 6.25E18 electrons. So 1 microamp is 6.25E12 electrons per second. Consider an idealised room as a cube 3m on each side. It's wall surface area is then 54 square meters. The ionised air molecules impinge on the walls (as in 'impingement rate'), the electrons go to ground, and the air molecules rebound. In a sense an electron gas, carried on air molecules, 'effuses' from the room leaving the air molecules to rebound. Making some assumptions, such as that the electron gas is evenly distributed, what is an estimate of the equilibrium or steady state electron/ion concentration in the air in the room? And if the ioniser is switched off, what is the decay constant or half life of the air ionisation? I think these are relatively simple calculations in the kinetic theory of gases, but I can't seem to manage them. Can anybody help? My thanks in advance for any input. (Note: A normal concentration for ions in city air might be around 1000 ions/cc. But I have a sense that an ioniser might produce much more than that.) Quote Link to post Share on other sites

lewisw85 1 Posted October 28 Author Report Share Posted October 28 J c o l l i s i o n = 1 4 n v ¯ = n 4 8 k B T π m . {\displaystyle J_{collision}={\frac {1}{4}}n{\bar {v}}={\frac {n}{4}}{\sqrt {\frac {8k_{BWell I've attempted an answer myself and here it is: In Wikipedia (thankyou) J c o l l i s i o n = 1 4 n v ¯ = n 4 8 k B T π m . {\displaystyle J_{collision}={\frac {1}{4}}n{\bar {v}}={\frac {n}{4}}{\sqrt {\frac {8k_{B}T}{\pi m}}}.} Using a VB style notation (apologies!), the impingement rate J is given by: J = n/4*sqrt(8*kb*T/pi/m) where: J = impingement rate n = air molecules/m3 = Loschmidt's Constant = 2.69E25 molecules/m3 (at STP) kb = Boltzman's constant = 1.38E-23 m2 kg / s2 / K T = 273 K approx m = average mass of 1 air molecule To find m, we can use mass of 1 mole of air = 29 g and Avogadro's constant A = 6.02E23 molecules/mole Hence m = 29/6.02E23 = 4.82E-23 g = 4.82E-26 kg (also found published) So J = 2.69E25/4 * sqrt(8*1.38E-23*273/pi/4.82E-26) J = 6.73E24*sqrt(199000) J = 3E27 collisions/m2/s (also found published) For R = total room air impingement rate (collisions/s) Ri = room ion impingement rate (ions or electrons/s) i = ioniser current = 1 uA = 6.24E12 electrons/s ni = ion concentration (ions/m3) And for the 3m cube room of wall area 56 m2 R = 56 * J = 56 * 3E27 = 1.68E29 collisions/s At equilibrium of ions in air, the ion impingement rate, Ri, is the ion current, is 6.24E12 ions/s, for 1 electron per ion. Now the ratio of ion concentration to air concentration (Loschmidt's constant) is equal to the ratio of the ion current to the wall molecular collision rate, an application of Dalton's Law I suppose That is ni/n = Ri/R So ion concentration ni = Ri * n / R = 6.24E12 * 2.69E25 / 1.68E29 ni = 1E9 ions/m3 = 1000 ions/cc This is a startling coincidence with my suggested figure, since we could have started with a different current, or size of room. The decay time constant T of ions in the room for an exponential decay is: T = total ions/ion current = 27 * ni / 6.24E12 = 27E9 / 6.24E12 s = 4.32 milliseconds So the half life of ions is 4.32 * ln(2) = 3 ms approx Loosely we could say that the ions travel 1.5m from the ioniser in the centre of the room to the walls in 3ms. This is 500m/s or 1800 kph or 1120 mph, which seems to tie up with the published speed of air molecules. A traditional ioniser has an 11 stage Cockroft-Walton voltage multiplier fitted with 10 nF capacitors and running directly off the AC mains. This has been measured to have a 6uA needle current. The above calculation suggests that's actually enough for 6 typical rooms, or perhaps a house, a hospital ward, or a bus. On the other hand one ioniser is then perhaps not enough for something like an attrium or a supermarket. A modern ioniser is often fully isolated and features an ion collector plate connected to the positive common rail of the circuit as the only ion collector. A traditional ioniser is effectively earthed by its neutral side so that the entire environment acts as the collector (hence the use of the wall area of the room in this calculation). This includes a person standing in the room and earthed through his feet. If the calculation is repeated for an isolated unit with a collector system, the ion concentration is much higher, but only relative to the collector. A person standing in the room experiences no ions. This could easily be a different situation medically, for example. On the other hand if the ion current is the same, the cleaning effect in terms of particles precipitated could be similar. Quote Link to post Share on other sites

montgomery 59 Posted October 28 Report Share Posted October 28 I don't know lewis and I haven't the slightest idea on the science you're talking about. But why would you ask that question on this forum? Seriously lewis, is there anyone here that you suppose could answer your question. That is, assuming that it's your question and not a cut and paste from some legitimate science publication or web posting? It's just that to me it seems to be somewhat suspicious that someone legitimately dealing with real science would be here? I hope my suspicions are wrong. Quote Link to post Share on other sites

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.