I am working on the effect of UV-C radiation on the removal of pesticide residue of strawberries and I have a problem with the calculations of Kj/m2 of my lamps, meaning I don't know to convert the watt of my lamps to Kj/m2 according to time exposure? can you help me?
The following questions and answers are relevant
https://www.researchgate.net/post/The-basic-way-to-calculate-UV-dose-from-a-UVC-15Watt-bulb
https://www.researchgate.net/post/How-to-calculate-UV-Irradiance-W-m2-over-a-range-of-wavelengths-provided-power-in-watts-is-known
https://www.researchgate.net/post/How-can-I-calculate-the-UV-dose-in-J-m2
https://www.researchgate.net/post/How-can-we-calculate-UV-light-intensity
Dear Jila,
The applied dose per area is the applied power per area multiplied by the application period: UV-C dose (J/m2) = UV-C radiation density (W/m2) * time (s)
If you apply a typical UV-C lamp with 10 W integral UV-C power, you just have to take into account the irradiated area by such a lamp.
By the way, most Hg discharge lamps (low-pressure and almagam) mainly emit at 253.7 nm, meaning that not all pesticides can be destroyed. Better are Xe2* excimer lamps, which emit at 172 nm, i.e. a photon energy which is able to destroy all kind of chemical bounds. As an alternative you can combine 253.7 nm radiation with ozone or H2O2, which is known as advanced oxidation processes. You can find many ppts and some paper on my RG homepage.
All the best for your research,
Thomas
- Badges
- Science topic
- Similar topics
- Phytochemistry
- Extracts