Pumukli - 18-6-2024 at 08:02
I have the following problem:
There are numerous ozone generators in the market that are advertised as "20 000 mg/h" or "30 000 mg/h" or an even higher number. Whatever.
The question: how could I calculate the achievable final ozone concentration in a given air space (empty room) if I know that the unit can produce say
30 000 mg ozone in an hour?
The problem seems non-trivial for me because of the half-life of ozone at room temp is around 15 minutes. If I start the generator at t(0) and stop it
t(0) + 1h the unit produced (if the advertisement was honest ;-) ) 30 000 mg ozone, but a nice amount of it already "decayed" by this time.
I'm wondering how could I calculate the final (peak) concentration of ozone in this case?
[Edited on 18-6-2024 by Pumukli]
bnull - 18-6-2024 at 12:10
It is non-trivial. Five bucks that you'll need to solve a differential equation. I'll take a look at it.
Edit: OK, my calculus is a little rusty. It's been years since I last used a Laplace transform. The result I'm getting is about five
times the number you gave (30000 mg/h). I wonder what I'm forgetting.
Edit2: Let's try it another way. The half-life is 15 minutes, so the decay function is$$f(t)=2^{\frac{-t}{15}},$$ with t in minutes.
Each minute the device produces 500 mg (=30000 mg/60 min) of ozone. At time t=0, we have 500*f(0)=500 mg. At time t=1, we have the decayed
500*f(1)=500*2-1/15 mg plus another fresh 500 mg. At time t=2, we have 500*f(2)=500*2-2/15 mg from the first batch plus
500*f(1)=500*2-1/15 mg from the second batch plus another fresh 500 mg, and so on and so on until t=60.
To make a long story short, we end up with the sum$$g(t)=\sum_{t=0}^{60}500*f(t)=\sum_{t=0}^{60}500*2^{\frac{-t}{15}}.$$Being too lazy to calculate
the sum above manually, I sent it to Wolfram and received the answer 10411 mg.
We can also treat (to the horror of any mathematically inclined person) the sum as an integral and
calculate$$\int_{0}^{60}500*2^{\frac{-t}{15}}dt.$$The answer is about 10144 mg being present at the end of one hour. At the end of ten hours, it's
about 10820 mg.
This is obviously the ideal case, with the room at 25 °C and no air drafts or oxidisable substances in the room. The real life values will probably
be well below that.
[Edited on 18-6-2024 by bnull]
B(a)P - 18-6-2024 at 15:33
A closed room with no external windows or doors will still exchange about 1/3 of a room volume per hour. So every hour you will also loose 1/3 of your
mass of ozone.
Twospoons - 18-6-2024 at 18:00
Or you could treat it as an equilibrium, setting the 1 minute production rate equal to the 1 minute destruction rate. 500mg/min = O3 conc*0.045, or
11072mg. This doesn't give conc vs time though, just an upper limit.
(strictly speaking this should be the limit of production per t = conc*destruction per t as t goes to zero, so the result above is a little high)
[Edited on 19-6-2024 by Twospoons]
Pumukli - 19-6-2024 at 08:28
Thank you guys! I can understand the logic behind both bnull's and Twospoons' calculations and it is reassuring that these different approaches yield
more or less the same result.
So for a layman the rule is: ONE-THIRD! :-)
One can expect at the end of the first hour one-third of the advertised amount of ozone produced at maximum in an hour present, under ideal conditions
(air tight walls, no reactive material in the "room", etc.), probably much less in real world situations. Or something like that, you understand what
I wanted to say. 