Tariq, there is information missing - what is the size of your beam spot?

I'll answer later in the week as I'm at a conference.

sohan

Hi Omairi, 

you need to give more details as the total energy or dose of uv will depend upon wavelength, irradiance and area. All of the above will vary as the distance between lamp and the your target surface (cells for example) changes. The final dose will be expressed in millijoules/ joules per cm2 /m2 for example.

I think a 7 days exposure from a 5cm distance using all of the wavelengths you mentioned above will be too high dose, more for the smaller wavelengths.

Here is a link for more details. 

http://www.eit.com/instruments/UV-EB%20West%202009.pdf

Unless you have a perfect point source there is no accurate way to "calculate" a radiation exposure.  You need to actually measure the incident radiation with an appropriate device.   If the lamps are commercially available you probably don't need to take a spectral output measurement but you certainly should have that information available for reviewers when it comes time to write up.

UV Dose is the product of UV light intensity and time Dose is sometimes referred to as fluence

DOSE = INTENSITY x TIME= millijoules/(sec)(cm2) x time= mJ/cm2

NOTE: 10 J/m2 = 1,000 microWsec/cm2 = 1 mJ/cm2

E= hc/λ

Where h= Planck Constant (6.626 x 10-34 m2 kg/s); c = speed of light in vacuum (3 x 108 m/s); λ (Lambda) = wavelength of the radiation, in this 365nm, which is equals to 3.65 x 10-7 m.

So, E = 6.626 x 10-34 x 3 x 108)/ 3.65 x 10-7 = 6.626 x 10-34 x 3 x 108)/3.65 x 107 = 5.446 x 10-19

Units = [(m2 kg/s) (m/s)]/(m) = (m2 kg/s x m/s) x (1/m) = m2 kg/s2 = J

E = 5.446 x 10-19 J

Irradiation duration: 7 x 24 x 60 x 60 = 604800 s

E = 5.446 x 10-19 x 6.048 x 105 = 3.294 x 10-13 J (this much radiation is falling on metre square, but your source is 5 cm is away)

Consideration of squared law as your source is.

5 cm = 0.05 m, which means radiation only covers (0.05)2 = 0.0025 m2

This means you have to divide the above equation by 1/0.0025 = 400 so as to make it in 1 m2 , as below

E = 3.294 x 10-13/400 = 8.235 x 10-16 J

ANSWER = 8.235 x 10-16 J at 365nm

Others are

9.953 x 10-16 J at 302nm

1.183 10-15 J at 254nm

The makes sense because as the wave length (λ) gets smaller (365 à 254) the radiation dosage increases.

Can someone check please.

Thank you Dr Jheeta, Premi, Shinkle and Tafesse for your illuminating help, those resources that Dr. Premi redirected me to also made things clearer for me. 

Dr. Jheeta: thank you again for taking some of your time and  for your help with the calculations, this is extremely useful for the rest of the data I am currently working on..

I enjoyed talking with you on our last conference and I look forward for meeting you again in future ones.

Is the 8W power you specify the electrical power? If so, you can't determine the UV power from it - you will need a radiometer because lamp efficiency varies with wavelength. You should probably use an irradiance detector, which would include a cosine diffuser (which decreases sensitivity with the angle from the Normal). Ocean optics makes some nice spectral detector systems, including this one: http://oceanoptics.com/product/usb2000rad/

You likely need Fluence rate, not irradiance, so you will need to correct from measured irradiance to fluence rate based on geometrical considerations. If you are using "stick" lamps (it appears that you are using lines from a mercury vapor lamp), then you can use the Keitz formula to get a good estimate of fluence rate from irradiance.

https://www.researchgate.net/publication/254217959_Ultraviolet_Lamp_Output_Measurement_A_Concise_Derivation_of_the_Keitz_Equation

Once you have fluence rate, the Dose is just Fluence rate x time.

Article Ultraviolet Lamp Output Measurement: A Concise Derivation of...

Hi Tareq,

how is your work proceed. Some notes due to the lack of reference especially for UV-C really comprehensive good and easy procedure hardly exists. If you still want to read some suggestions.

Article About the possibility of calibrating optical detectors by so...

I forgot to mention that you most likely deals with different wavelength distributions. The 256 is a shark Hg line whereas the outhers are broader emissions induced also by a low pressure Hg lamp.

@ Dr. Sohan Jheeta Sir

Thank you sir for giving elaborated calculation here. It would be of great help. Although, I just wanted to ask one question here. What is the role of 15W here?? Don't we need to incorporate this value in anytime during calculation?? Because UV lamp of different wavelength 's are available. So if somebody have 8W UV lamp..that will definitely going to change the UV dose. So how will we calculate the fluence dose of UV on cells????

Thank you.

Hi Priya

First, please note I think you meant to use the word 'detailed' rather than 'elaborate'. Anyhow, 15W is ok for everyday use, like comparing things and figures (15w and 8W); it's an indication as to how much power is rated or being consumed by that apparatus. It is not accurate.

Second, for accuracy you need to know the optimum wavelength (lambda) of the lamp being used. From this, using the elaborate calculation as previously, you can work out the dosage in J/m2.

Third, power by definition is a rate of energy being used and so 15W is actually 15 J/s. You can then work out the dosage by multiplying 15 by length of time of irradiation. BUT this will not accurate as 15 is a factory setting.

So, what is the operating wavelength of your lamp?

Please note currently I'm in transit to Africa, where the internet is virtually nil or at best patchy. I try to answer any further questions.

I hope this helps.

Kind regards

Dr. Jheeta..

Sir, It is same as mentioned by Dr. Tareq.

So as per your calculation, the fluence dose of UV on cells will be 8.235 x 10-16 J/m2 ??

Thanks Sir for clearing some of my doubts.

So I assume your lamp is operating 365nm,

You also need to know how to workout dosage (J/M2) , is as follows:

in 1s the energy delivered is 8.235 x 10-16 J/m2 (ie dosage)

in 2s the energy delive. is 2 x 8.235 x 10-16 = 1,647 x 10-15 J/m2 (dosage)

in 3s 3 x 8.235 x 10-16 = 2.471 x 10-15 J/m2 (dosage)

ect (dosage increases with time of exposure)

You plot dosage on the x-axis (independent values) and your other obtained results should on the y-axis (dependent values), because y dependents on x

let me if you know what i mean

Dr. Jheeta

Thanks for your great response.

If the wavelength is in a certain range (e.g., 400 nm - 600 nm), how can i calculate the energy does? Energy does is in a certain range? (In that case, can we use average term?

Also, when you calculate E, is it necessary to calculate the number of photons?

Sorry i have missed your post somehow. Generally your operational manual should tell you the operational peak level that you should use. To adivse you further, what is the figure in the manual?

Also, what are you trying to achieve? Give me some details of your experiment please?

Generally, doesage can be either in J/m2 or photons/m2 - the latter is generally called Fluence. it doesn't really matter very much so as long as you keep the same units throughout your experimentation.

Hi, I need to perform the UV crosslinking of printed protein in microarray at 254nm and energy dose 1.25J/cm2. We do not have an UV crosslinker. Can we do the UV crosslinker with a transilluminator? How?

Thanks, Carla

Thanks Dr. Jheeta,

Each second the exposure will be different.

So we can plot the curve. we will get the equation

so if my exposure time from a particular uv source is 5 min for example then should i take average.

or from equation I can directly calculate.

Please can anybody explain with formulation the efficacy of UV disinfection in water step wise so that we can follow , iam looking for effective UV of 4 log reduction at least . As per my knowledge minimum contact time required for UV to work is 10 seconds with 6 inches distance to completely disinfect . Better way is we can sample water before UV treatment and after UV treatment and calculate log reduction of bacteria . But there should be simple formula to calculate the dosage required .

The only effective UV light that which kills everything in sight is a lamp that delivers 121.6 nm wavelength radiation. In order to work out dosage I need to know the specification (wattage, current, voltage etc) of the you have or you are going to purchase. I can't write down any formulation because it takes too long.

Kind regards

Sohan

Sohan Jheeta , Sir, how can we calculate j/cm2 of uvgi (254 nm) light when we have 15watts and 34 volts operating voltage.

Hi Tariq

I need to know the answers to the following questions:

1) Why do you want to work out the J/cm2? (you could use number of photons impinging on cm2. ie photons/cm2. this makes much more sense. Photons/cm2 is called fluence which is essentially same as J/cm2).

2) from your technical manual I need to know either "spot" radius/diameter of the beam on the target/surface and/or how far is the source/lamp from the target/surface?

Kind regards

Sohan

Hello Sohan Jheeta,

I am trying to figure out the UV dose in mJ/cm2 for a UV/H2o2 AOP system called the TrojanUVPhox.

lamp wattage = 300 W

lamp wavelength = 254 nm

the wavelength energy = 7.82598 x 10^-19 J

arc length of lamp = 150 cm

lamp quarts diameter = 1.9 cm

there are 144 lamps total in the reactor.

length of TrojanUVPhox dual reactor = 3.9 m

diameter of TrojanUVPhox dual reactor = 1.3 m

volume of TrojanUVPhox dual reactor = 5.1766 m^3

the HRT for the TrojanUVPhox is 118.2 seconds given 10 MGD flow.

Is this enough information to calculate how much of a UV dose that the TrojanUVPhox can provide?

Hi Jasminn

I need the following information:

1) what is the operating voltage of one of the lamps?

2) How far is the lamp from the target you are trying to irradiate?

3) what is the area of the target you are trying to irradiate?

4) what does "lamp quarts diameter = 1.9 cm" mean"

Kind regards

Sohan

Sohan Jheeta mi pregunta es teniendo yo una lampara de 15 W con una longitud de onda de 254nm para irradiar a un hongo fitopatogeno Botrytis cinerea con un tiempo de irradiación de 8 min manejando distancias de 30 cm ¿como puedo hallar la intensidad y dosis de radiación? Tareq L. S. Omairi

Hi Johanna

I need to know the following:

1) what is the operational voltage of the 15 w and/or 8 W lamp? check the operating manual

2) what is the operational current of these lamps? check the operating manual

3) when you turn on the light lamp, what is the area of the light spread on the target? this should also be in your operating manual. If the manual doesn't have the information, then you will have to measure the lit area and let m know. Thanks

Kind regards

Sohan

Hi Sohan Jheeta

I am assembling a UV Disinfection Unit and was wondering how to calculate the dosage at a given time and distance from surface. The unit has 10 bulbs each with a length of 84.3cm and 87 watts of power. If exposure time is 900 seconds and distance from surface is 447cm, do you know how I might be able to calculate dosage (J/cm2)?

If you knew the formula, that would be very helpful.

Thank you,

Danny

Hi Elisa

First, why do you want to use Ws/cm2? Thanks

Kind regards

Sohan

One more correction to be added. UV tubes do emit light in different wavelengths. So 30W tube does not mean that all the 30W is UV wavelength of your requirement. Generally UV tubes are 30% efficient (UVC, I am speaking of). So the power available in that wavelength range is less than 30%.

Equation E= hc/λ is per photon. So we need to know how many photons ...?

eg: 30W can be divided by say 3.. you get 10W of UVC. consider the surface of area of tube light and prepare another area of cylinder away from your tube. X power per unit area from tube comes out of a tube and now the power is fixed. As you go away the cylinder area is proportional to R and hence the power will be inversly proportional to R.

Inverse square relation is for the point source, for linear source, we must use 1/R rule.

Any comments on this?

calculate intensity by lux meter

P. Y. Nabhiraj: One more correction to be added. UV tubes do emit light in different wavelengths. So 30W tube does not mean that all the 30W is UV wavelength of your requirement. Generally, UV tubes are 30% efficient (UVC, I am speaking of). So the power available in that wavelength range is less than 30%.

Sohan Jheeta: this is incorrect. The efficiency figure is generally quoted as 100 or 1000 hours 30%. This means your lamp will work efficiently as it should for 100 or 1000 hours, then its wavelength will drop to 30%, which effectively means you have to replace the lamp—nothing lasts forever.

P. Y. Nabhiraj: Equation E= hc/λ is per photon. So we need to know how many photons ...?

Sohan Jheeta: you can measure it either as J/s by using E= hc/λ; or you can use the following formula to convert wavelength into photons/s

P(w) = IV --> I = P/V (this information is available from a manual)

I = NQ

where “I” is current in Amp; and N is the number of photons arriving in a beam with coulomb (C) charge (Q) onto a given surface area. C being a constant number and is equal to 1.602*10-19 C.

P. Y. Nabhiraj: eg: 30W can be divided by say 3.. you get 10W of UVC.

Sohan Jheeta: as above

P. Y. Nabhiraj: consider the surface of area of tube light and prepare another area of cylinder away from your tube. X power per unit area from tube comes out of a tube and now the power is fixed. As you go away the cylinder area is proportional to R and hence the power will be inversely proportional to R.

Sohan Jheeta: made un-necessarily too complex. Surface area of tube is not necessary. what is necessary is the physical area lit-up by the tube

P. Y. Nabhiraj: Inverse square relation is for the point source, for linear source, we must use 1/R rule. Any comments on this?

Sohan Jheeta: all you need the area lit-up by the tube and then convert this to J/m^2 or photons/m^2

Thanks for the input.

Kind regards

Sohan

Sohan Jheeta : Tube efficiency I meant what percent of electrical power is converted into UVC. If you see the datasheet of Eg: Philips.

on page 11: https://www.assets.signify.com/is/content/PhilipsLighting/Assets/philips-lighting/20200402-philips-uv-purification-lamps-and-gear-catalog.pdf

at 100 hours say for example for a tube of 11W the datasheet says the UV-C output is 3.6W. Rest is spent as light of different wavelengths and filament heating. So I said UVC production is about about 30% of total power.

Can you please comment on this? datasheet says 3.6W at 100 h. I think it means, after 100 hours of usage UVC would be 3.6W.

Hi Nabhiraj

This morning I have checked with my colleagues. We agree with you that “UVC at 100h (W) being 3.6 W” does mean after continuous use of the lamp, the power use will drop to 3.6 W. This means the efficiency (ie J/s or Photons/s or J/m^2 … etc) will also drop by 30% - ie E= (hc/λ)/3 J/s. What is really important is (1) it is better to run thorough “microbiological response to UV light” tests so as to standardise your apparatus and to optimise it as a germicidal machine. Such practice would also become your salient point about your apparatus. Further, such microbiological test should be run every so often. (2) you should also install UV light sensor as was suggested the previous person, Joy Mukherjee; (3) install motion detector.

Intensity: you can increase the intensity (1) by moving the lamp closer to the target; (2) use two or pumps with the identical specification to cover the same area; and (3) increase the time of irradiation.

All the while please keep in mind health and safety—perhaps invest in motion detector.

Kind regards

Sohan

I have a uv chamber 8 inches ×10 inches.. and has a 8w output bulb at 254 nm .. how long should the sterilization cycle be... if for sars cov-2 we need 250j per mter sq.

Hi Rishikesh

very important: Are you using blubs or tube?

what is the voltage and current?

In the manufacturer's specification/manual there should be two figures on wattage-one high, the other low, do you have it?

Thanks

Kind regards

Sohan

Hello Sohan Jheeta,

I am using a UV crosslinker machine (UVC 254nm) for my gel but I need to use the time option instead of energy specification (e.g. 15 or 30 mins). Do you know how I can calculate the total UV dose in J/cm2 in this setting?

Thanks in advance,

There are a few points in this discussion I would like to clarify.

A low-pressure mercury arc lamp operating at appropriate temperature (~40C cold spot) will have an electrical to 254nm efficiency of about 33% when new. That is a good rule of thumb.

Mercury lamps often have fluctuating performance over the first hours, so manufacturers quote and measure output after a 100-hour burn-in. So that is considered "new".

Lamp output will drop with age, as mercury embeds onto the quartz tube and emission paste is sputtered off the filaments onto the tube. High quality lamps utilize an internal coating of zirconia or alumina to reduce this embedding. Without the coating the output will drop to about 50% of the new value in a year (50% x 33% = 16.5% efficiency at 8760 hours). Lamps with good coatings can maintain 80% or more of their initial output after a year (26% efficiency).

The calculation methods described by other posters, using a 1/R relation, are not correct because they assume that the emission from a lamp is radial, which it is not; it is roughly Lambertian (cosine) due to self-absorption in the plasma.

To calculate the flux from a mercury lamp reaching a point in space, you should use the Keitz equation. This includes the finite lamp length and the approximately Lambertian emission from the plasma.

Article Ultraviolet Lamp Output Measurement: A Concise Derivation of...

Hi Michael

You are absolutely right with the efficiency of UV lamps and about the inverse square law. it is not even possible to obtain near perfect answer which is what people are looking for. Rightly so, within short space of time the UV light output decreases markedly. People need UV light meter to keep tabs on the lamp efficiency in the long run... Even the top quality lamp would only operate at around 35% capacity.

On the plus side, it is possible to calculate J/m2 at 35% efficiency.

Kind regards

Sohan

Hello, I recommend you to measure with a radiometer equipment the real radiation emitted by your lamps, since the nominal power descends as time passes. Alternatively, you may calculate this value by actinometry. After that, you will be able to calculate the real radiant power (W/cm2) that reaches the surface of the material irradiated.

Sohan Jheeta

Dear Dr Jheeta

Thanks for your helpful calculations. I have one more question here. What if I have more than one lamp for the irradiation process (same wavelength, same power), Don't we need to multiply the result of this equation E= hc/λ by the number of the lamps

Dear Liela,

Yes, you need to multiply the result with number of lamps.

Dear Dr. P. Y. Nabhiraj,

When the bulbs work together, there will be interference between the rays (overlapping). So is that correct to multiply by the number of bulbs?

Dear Dr Liela Ali,

if you are using the reflectors(assuming all light reflects back and forth). The light intensity inside the box has to increase by the number of bulbs...

However, practically it will be little less.

If you consider placing the object between the light and then one half of the light goes on opposite direction of the object, then the overlapping area gets light equivalent to that of one bulb.

The energy emitted by the bulb has to add up to the energy of other bulb in the vicinity. I hope now it is clear. Please use good reflectors. then you can simply add.

Dear Dr. P. Y. Nabhiraj,

Yes, sir, everything is obvious now. I am grateful for your valuable assistance.

Dr,

Download Paper with title:

Pollyallyl diglycol carbohydrates (PADC) polymer as a UVC dosimeter: A new technique combining thermal and UVC treatments.

The dose converted to kJ/cm2

If there are missing in understanding, Don't hesitate to contact with me,

[email protected]

;Thanks alot!

Dear dr. Sohan Jheeta ,

I'm a PhD student and struggling at the moment with the same calculus, and i got a question. In the question of dr. Tareq, he said that the distance of the irradiated surface is 5 cm, is it okay to convert it into m2, as you've done (0.05)2 = 0.0025 m2. I am asking this because in my situation the distance of the irradiated sample is 0.035 m. Thank you in advance ! Have a great day !

Download Paper with title:

Pollyallyl diglycol carbohydrates (PADC) polymer as a UVC dosimeter: A new technique combining thermal and UVC treatments.

The dose converted to kJ/cm2

@Sohan Jheeta

Thank you dr. Ebrahim Sedqy and dr. Sohan Jheeta. Have a wonderful day !