Hi Phuong, 

are your cells NER deficient? If not, no DNA DSB will be generated by UVC. 

Your bulb has an UV output of 4.9 W (http://www.ushio.com/files/specifications/germicidal-low-pressure-mercury-arc.pdf) . For the energy calculations, I'm not really sure how to do these, as I never performed such experiments. 

What is clear is that 5J.m-2 corresponds to a delivered quantity of energy (5J) per a unit of surface (m-2). If do you really want to control the dose, you must use a special UVC-dosimeter because the fluence rate (J.m-2.s-1) is depending on the distance between cells and your lamp.  5J.m-2 won't be enough to produce SSB nor DSB. You must considered an order of kJ.m-2 for producing SSB or DSB. UVA or UVB are more adapted for that.

@Olivier: Thanks for pointing out the UV output, I din't notice it. It's true if I need to do the calculation I need to use the output, not the input. And yes, I have NER-deficient cells to compare with wild type. I saw some papers reporting the DSB is also indirectly created by UVC, but it's somehow controversial (some paper said yes, some paper said no)..

@Daniel: I was asking around about the UVC-dosimeter but so far no labs around has one. The only UV source I have now is the UV lamp in the cell culture hood and the one for the chemidoc (to look at the gels), both of which I have no idea to measure or control the UV dose. Do you recommend anything else I can use or do to to measure the dose? Thanks a lot!

You can for example define the LD50 of your cells and do your experiments considering that. In mammalian cells, UVC DL50 is around few J.m-2.

Good luck,

Dear Phuong.

Just got aware of your question and would like to ask first, what kind of research you are doing? The reason for my ask is, that it is a decisive difference if you are simply doing a bachelor's work to learn scientific work, or if you are doing an extensive scientific research focusing on a specific topic. Do not get me wrong, no disparagement intended.

Background:

Your approach to get the required data is somewhat typical in the scientific literature and will lead to data that will only be comparable to itself. Meaning, the values that you calculate or measure won't be interchangeable to others. It will be just a value with some digits. With a simple bachelor's work it only is important to indicate, that you are aware of the challenges that I will explain in the following, fo a real research work it will be decisive to work with the information.

First of all it won't help you just look at the lamp type and suppliers address that you are using for your work. Off course you already took a look to the data sheet of this lamp, telling you it has a power consumption of 15W, 55V. Even though Olivier indicated on the UV output of 4.9W@254nm it is imperative for you to know, that this is a laboratory value... Meaning: The maximum the lamp theoretically is able to produce under optimum conditions.

What does this mean?

1. A lamps output will reduce by time due to an internal solarisation effect. That means, the older the lamp gets, the lower the maximum output will be!

2. The lamp's output decisively is dependent to it's working conditions. "Overheating" in terms of "covering the lamp by a box to protect yourself from getting irradiated" will cause a decisive output reduction, as UV lamps (low pressure mercury vapor discharge lamps) are very temperature sensitive. A ambient air temperature difference of just 1°C to 5 °C will lead to an UV output variation of up to 15%! See attached measurement illustration for your help. (Temperature varies from 19 to 24°C, no air movement)

3. Laboratory value also means, that you'll have to use an appropriate ballast. Not just any ballast that enlightens the lamp! Ballast intelligence (to control the lamp) is decisive for the live rate as well as for the output. A wrong ballast will e.g. overheat the lamp and it will be perfectly blue lightening ... but the wanted UV output is dramatically below expectations. Also the ballast's ignition behavior will deteriorate the lamp's live expectation, if not tuned to the lamp's filaments. So take care!

4. Take care of ignition behavior of the lamp. As it is no LED the lamps output also will vary. It is not like switch on = UV power on. It is a low pressure discharge and thus you'll need a specific internal working pressure. It takes some time to get it. Also see the same illustration as above.

5. Decisively take care of cooling effects! Do NEVER use a fan to cool the lamp or the chamber of your tests. Air flow on the discharge lamp will have an dramatic (negative) impact on the internal gas pressure and thus output goes down considerably!

6. If you'll find a UV meter anywhere the next problem will be, that you'll have to get sure if it is calibrated. Calibration should be done before your work, as you do not know if the UV meter was handled wrongly before your work. Calibration WILL COST MONEY.

7. If you'll got a calibrated and well working UV meter for your work... well, then the next problems will occur. :)

The closer you'll get to the lamp, the higher the error ratio will be. UV meters only have a very narrow field of vision. The closer you get, the smaller the lamp's segment will be that you focus. Furthermore... any mm in changing the position of the UV meter will cause a huge variety in measured data. Imperatively take care of the angle of sensor positioning towards the lamp. And further more... what is the REAL distance? Distance of top notch of the photo diode or and top point of the lamp in a rectangular distance?

8. If you are in the test design itself... do/did you take care of any material reflections? Do the walls of your petri dish reflect UV rays and thus influence the values? What about the nutrient medium itself? Big difference if you'll irradiate your cells within a culture medium or if you'll expose them e.g. on a metal plate without any nutrient solution.

9. Be ALWAYS aware, that UV rays are optical rays that can be reflected, bend, absorbed or transmitted, but they are NO LIGHT! The do reflect differently and get absorbed differently to light in terms of material values. E.g. UVC rays will get 100% absorbed by borosilicate. A conventional mirror will reflect visible light by 98% and more, but it won't reflect any UV rays.

10. well, there is much more to tell. But I advise you to do some reading before. There is plenty of work concerning this and I highly recommend the works of James R. Bolton concerning this topic and I also recommend reading of Kessler et al 2009 "Comparison of UV Power Measurement of Low Pressure UV-lamps by a worldwide Round Robin Test".

Hope I didn't confuse you to much, but keep in mind: There is much more to take care off but only knowing the way how to calculate the UV dose of a 15W bulb!

This will lead me to the beginning of writing: What is your intention?

I ve learned a lot from Martin Graupner... thanks a bunch..

I suggest you use an UV meter Because I think there is Not any exact way for this.The received UV to a surface( like your sample ) depends on source life ,distance,direction and so on .

you find it in

http://www.hagner.se/uv-ir-instruments-1/

NOTE: I've fixed some miss writings of my original post.

The original question was posted 3 years ago... and no one in fact gave a direct answer to the author's question.

.

I agree with all previous posts that a calibrated UV meter is desired due to several factors but...

If one wants a rough estimation:

Based on the datasheet information from USHIO UV bulbs [1]:

A G15T8 (15 W germicidal fluorescent bulb) emits 4.9 W of UV-C with a central peak at 253.7 nm

The UV dose is UV_Energy/Area

UV_Energy = UV_bulb_power * Exposure_time

Area = 4* pi * (UV_bulb distance)^2 (considering spheric diverence)

This leads to :

UV_dose= UV_bulb_power * Exposure_time / (4 * pi * UV_bulb_distance^2)

So the desired direct exposure time in seconds is:

Exposure_time =

( Desired_UV_dose * 4 * pi * (UV_bulb_distance)^2 )/ UV_bulb_power

Units:

UV_Energy: Joules

UV_bulb_power: Watts

Area: m^2

UV_bulb_distance: m

Exposure_time: s

pi: 3.141592

So for the desired 5 J/m^2 Energy we got

12.8 seconds of direct exposure for a 1 m distance

I do not recommend this simple formula for short distances because the lamp is cylindrical and the spherical divergence may not be accurate.

The UV power of lamp can vary with temperature and lamp life, which introduces more (difficult to predict) variables to the calculations.

But if you just want to kill SARS Coronavirus, which can be killed with a UV-C dose of 241 J/(m^2) according to [2], I believe that just doubling the computed exposure time should be sufficient to overcome these variations.

IMPORTANT:

DO NOT GET EXPOSED TO UV-C, IT IS HARMFUL TO YOUR EYES AND SKIN.

ALSO SOME UV-C BULBS CAN PRODUCE OZONE AND YOU MUST AVOID BE EXPOSED TO THAT TOO

DISCLAIMER: I'm not an expert in the area, so use this information at your own risk.

REFERENCES:

[1] https://www.ushio.com/files/specifications/germicidal-low-pressure-mercury-arc.pdf

[2] https://www.uv-light.co.uk/coronavirus/

Dear Ricardo

While I was looking for the calculation of UV exposure time, I found your answer. Thank you, for the calculation. I plan to use 5 germicidal fluorescent bulb and use a mirror for shielding the radiation in the opposite direction for surface disinfection. It looks like the attachment. How can I calculate the exposure time? I would be very grateful, if you give some ideas about the calculation of total UV energy and exposure time.

Thank you in advance.

Hi Kadir, do not use mirrors, use aluminum foil. To calculate the exposure time you must provide the UV power of your lamps and the distance to the target. Also you should give a good error marging because there are many factor thst can influence the exposure time.

Dear Ricardo, Thank you for your reply and the recommendation. I will consider to use aliminum foil. If I use 1 germicidal fluorescent bulb, I can use your calculation.

what If I use 5 same bulb like on the photo, how the calculation can be? Do I need to use "Cosine error in measurement" to calculate total UV energy on the surface?

Thank you

Kadir, the answer will depend on the power of your bulbs, you must search for it first. It also depends on the geometry of your setup. If the target is a piece of paper and all bulbs are facing it at the same distance, you just use my calculations and divide it by the number of bulbs, but if your target is a body and your bulbs are around it, then not all faces of the body will be illuminated by all bulbs. In that case you have to estimate how many bulbs are hitting the less illuminated face. Aways exagerate your estimation to avoid underexposure.

It's a good idea to read all previous posts because there are several considerations that other people made, specially from Martin Groupner.

Dear Ricardo,　Thank you so much for your comments and recommendations.

The power of bulb is 40W which has 19.8W of sterilization power.

My current target is kind of piece of paper and all bulbs are facing on it at the same distance. So The theoretical calculation would be

Exposure_time = ( Desired_UV_dose * 4 * pi * (UV_bulb_distance)^2 )/ (UV_bulb_power*number_of_bulbs).

Finally I will consider the loss factor in order to avoid underexposure.

Thank you

Dear Ricardo,

Through your post, finally I could get simple calculation to arrive dosage and timing. I need your help to arrive for the following situation.

Purpose:

UVC disinfection of travel bags when pass through a 4x3x3 (LxWxH) size rectangle tunnel. Planned to fix UV light in four side (Top/Down the mesh conveyor/right and left sides on the inner rectangle tunnel.

The speed of the conveyor is adjustable, based on the required time to achieve the dosage, we can adjust the speed. Lets Assume 3 inch/sec.(0.076m/sec)

UVC details:

1. Watts - 40W tube with 16W UVC output

2. Distance from UV source to the surfaces may vary from 0.2 meter to 0.5 meter.

3. Desired dosage is 70J/m2

4. No of bulbs - 4 (16W UVC output efficiency)

I need to calculate time required to achieve 70J/m2 .

Thanks in Advance.

Dear Swathy what is the size of your UV tunnel as no dimension has been mentioned

Dear Ricardo,

The formula states UV_Energy = UV_bulb_power * Exposure_time

However, as you mention, the product specs mentions as follows:

"A G15T8 (15 W germicidal fluorescent bulb) emits 4.9 W of UV-C with a central peak at 253.7 nm"

Which one is the UV_bulb_power (15W or 4,9W) to be used in the above formula?

Dear Ricardo,

As we know (Tseng and C. Li, 20070 ) D = I t , where D is dosage of UVc in J/m2, I is intensity, t is time in second . Again for calculating I ,we know (Arguelles 2020) , I= η4 . πr 2/P where , n is an attenuation factor where η ≤ 1, r is distance of object , P is power or wattage of UVc tube.

If we combine the upper two formula , we get t =D/I = D/ η4 . πr2/P , This formula is used for calculation of exposure time of UVC in given dose of UVC

Dear @Batu, The 15W germicidal tube consumes 15W from power grid and ouputs 4.9W of UV-C radiation. So you must use 4.9 in the formula.

Ricardo Braganca, Can you please tell me how the radiation intensity will vary from a UVC LED of 4mm diameter emitting radiation at 19mW power falling on a surface 5 mm away? You mentioned that the 1/4*pi*r^2 will not work for a small distance.