3 ms·
One thing to note is that in your calculations, you are using 50cm distance, but in the study you linked, they are using 3 cm distance. Since this is inverse sq
by jennyyang 7y ago
One thing to note is that in your calculations, you are using 50cm distance, but in the study you linked, they are using 3 cm distance. Since this is inverse square law, that makes sense it would take longer to deactivate.
For another point of reference, this study[1] says it takes 1.8 J/cm^2 to deactivate the flu virus. I'm not sure which is more hearty but from my calculations, it seems as though it's similar (about 2 J/cm^2). If that's the case then if you brought the intensity of your light up (bring the bulb much closer and possibly wrap it in material that reflects UV-C if that's even possible) then it would only take seconds to deactivate the virus.
[1] https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4699414/ https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4699414/
- blackhaz 7y agoAs far as I understand, we are talking about flux - per area at specified distance, not luminosity. I did recalculate flux for 0.5 m.
- jennyyang 7y agoNo. The study I referenced used total energy per cm^2 required to deactivate the flu virus, which is J/cm^2. You get that by multiplying the flux by the exposure time. So whichever exposure time gets you the 1.8 J/cm^2 is enough to kill the flu virus and presumably SARS-CoV-2.
- blackhaz 7y agoWell, like I said - I recalculated for the flux of the 15 W lamp. An hour of zapping gets you to 1.72 J/cm^2 which is very close to your 1.8 J/cm^2 figure.
- jennyyang 7y agoWe are not disagreeing here, I think your 60 min calculation is correct. My point is that your calculations are at 50 cm and you get a much lower time for exposure if you bring your light bulb much closer and figure out a way to reflect the UV light so that more light from the bulb hits the intended target.