Science Forums  # lewisw85

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1. 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 = impingemen
2. 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
3. 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 l
4. 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 h
5. 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 th
6. 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 part
7. 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
8. 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 ha
9. 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 curren
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