Calculating required UVC exposure time for a specific dosage

Thread Starter

redgear

Joined Oct 17, 2019
131
Hi all,

I am planning to build a disinfection cabinet using UVC radiation. Many of my friends and family work in the health sector and I plan to gift it to them. I am a beginner, i'm sorry if my questions are too easy or stupid. I have been collecting information online about the same and I am pretty confused. The cabinet will have a surface area of 7300 cm^2. I am planning to use 6 X Philips TUV T5 16W tubes with UVC wattage of 4W and irradiance of 45 uW/cm^2 at 1m. The maximum distance between the object and the tube will be 30cm in my case, as irradiance is inversely proportional to distance, I am simply multiplying the irradiance value by 3, so I get 135 uW/cm^2. The total UVC wattage will be 24W(6*4) and irradiance will be 810 uW/cm^2(6*135)(correct me if I'm wrong). I will be covering the interiors of the box with polished aluminium sheets to increase reflectivity.

Different resources suggest different dosages and different formulas to calculate the dosage. Studies based on SARS-CoV-1 suggest a dosage of 7-2410 J/m^2. The IUVA recommends 1-3 J/cm^2 for real world conditions but only 10-20 mJ/cm^2 under controlled lab conditions. Since the cabinet will be closed and effectively almost all UVC radiation will reach the object can I achieve 99.9% deactivation using UVC dosage in order of few hundred mJ?

The next problem is properly calculating the exposure time. I find different formulas that give completely different results.
  • According to this study, the exposure time formula is given by : \[ Exposure Time(s) = \frac{UV Dose(J/cm^2)}{Irradiance(W/cm^2)} \] substituting values I get around 62 minutes.
  • According to this Research gate answer, the exposure time is given by : \[ Exposure Time(s) = \frac{Desired UV dose(J/m^2) * 4 * pi * UV bulb distance^2(m)} {UV bulb power(W)} \] substituting values, I get 23.55 minutes.
  • A Youtube video uses the formula, \[ Exposure Time(s) = \frac{Surface Area(cm^2) * Dosage(J/cm^2)}{Power(W)} \] substituting values I get around 15 minutes.

As you can see, the differences are huge, idk which is the correct one.

Do I still need a Ballast for the tube if I step down the voltage to the tube's operating voltage?

I'm also planning to include a timer, will any 555 timer or timer relay be enough?

Since, UVC is harmful to humans, I will have to add some mechanism to stop the equipment from being operator if the doors of the cabinet are opened, I planning to use a contact switch.

Thanks
 

KeithWalker

Joined Jul 10, 2017
1,038
As you can see, the differences are huge, idk which is the correct one.

Do I still need a Ballast for the tube if I step down the voltage to the tube's operating voltage?

I'm also planning to include a timer, will any 555 timer or timer relay be enough?

Since, UVC is harmful to humans, I will have to add some mechanism to stop the equipment from being operator if the doors of the cabinet are opened, I planning to use a contact switch.

Thanks
I don't know which formula is the most accurate. It important that the tubes stay on long enough to sterilize properly so I would use the one that gives the longest time. The extra time will not cause any harm.
The fluorescent tube will not work without the ballast and probably will not start if you lower it's operating voltage.
With a device as critical as this, I would use a manufactured timer. There are many on the market that operate a relay and that you can manually adjust the time setting.
You are correct; you must include an interlock on the door of the cabinet to disconnect the power when it is opened. Use the simplest fail-safe method you can Complexity will increase the possibility of a failure.
Regards,
Keith
 

Thread Starter

redgear

Joined Oct 17, 2019
131
I don't know which formula is the most accurate. It important that the tubes stay on long enough to sterilize properly so I would use the one that gives the longest time. The extra time will not cause any harm.
The fluorescent tube will not work without the ballast and probably will not start if you lower it's operating voltage.
With a device as critical as this, I would use a manufactured timer. There are many on the market that operate a relay and that you can manually adjust the time setting.
You are correct; you must include an interlock on the door of the cabinet to disconnect the power when it is opened. Use the simplest fail-safe method you can Complexity will increase the possibility of a failure.
Regards,
Keith
Thanks for the reply.
I agree, leaving the tubes on for the longest time is the simplest solution. But it would be nice to give a minimum amount of time they need to wait before can take it out.
Alright, so I will be needing 6 ballasts assuming one each for 6 tubes.
I was thinking on the lines of using a microprocessor with a FET driver for the timer. What do you think of it?
What do you think of a contact switch that is on only if the cabinet door is closed? I guess it would be the most reliable. In addition I am planning to add one more magnetic switch(just as a fail safe)
 

KeithWalker

Joined Jul 10, 2017
1,038
A microprocessor and FET would do the job but simpler is usually more reliable. I found this 555 timer circuit in the internet that would do the job. It uses the c-mos version of the 555 to get such a long delay.
It's a good idea to have dual diversity on the door interlock. With both switches in series, it will be safe if either one fails.90minTimer.jpg
 
Last edited:

MrAl

Joined Jun 17, 2014
7,665
Hi all,

I am planning to build a disinfection cabinet using UVC radiation. Many of my friends and family work in the health sector and I plan to gift it to them. I am a beginner, i'm sorry if my questions are too easy or stupid. I have been collecting information online about the same and I am pretty confused. The cabinet will have a surface area of 7300 cm^2. I am planning to use 6 X Philips TUV T5 16W tubes with UVC wattage of 4W and irradiance of 45 uW/cm^2 at 1m. The maximum distance between the object and the tube will be 30cm in my case, as irradiance is inversely proportional to distance, I am simply multiplying the irradiance value by 3, so I get 135 uW/cm^2. The total UVC wattage will be 24W(6*4) and irradiance will be 810 uW/cm^2(6*135)(correct me if I'm wrong). I will be covering the interiors of the box with polished aluminium sheets to increase reflectivity.

Different resources suggest different dosages and different formulas to calculate the dosage. Studies based on SARS-CoV-1 suggest a dosage of 7-2410 J/m^2. The IUVA recommends 1-3 J/cm^2 for real world conditions but only 10-20 mJ/cm^2 under controlled lab conditions. Since the cabinet will be closed and effectively almost all UVC radiation will reach the object can I achieve 99.9% deactivation using UVC dosage in order of few hundred mJ?

The next problem is properly calculating the exposure time. I find different formulas that give completely different results.
  • According to this study, the exposure time formula is given by : \[ Exposure Time(s) = \frac{UV Dose(J/cm^2)}{Irradiance(W/cm^2)} \] substituting values I get around 62 minutes.
  • According to this Research gate answer, the exposure time is given by : \[ Exposure Time(s) = \frac{Desired UV dose(J/m^2) * 4 * pi * UV bulb distance^2(m)} {UV bulb power(W)} \] substituting values, I get 23.55 minutes.
  • A Youtube video uses the formula, \[ Exposure Time(s) = \frac{Surface Area(cm^2) * Dosage(J/cm^2)}{Power(W)} \] substituting values I get around 15 minutes.

As you can see, the differences are huge, idk which is the correct one.

Do I still need a Ballast for the tube if I step down the voltage to the tube's operating voltage?

I'm also planning to include a timer, will any 555 timer or timer relay be enough?

Since, UVC is harmful to humans, I will have to add some mechanism to stop the equipment from being operator if the doors of the cabinet are opened, I planning to use a contact switch.

Thanks
Hi,

When you are dealing with lighting you often have to work with a solid angle that sweeps out an area and that gives you the exposure.
What happened with my bulb was that i calculated just about 1.2mw per square inch and i thought that was way too low, but then i read that as little as 5uw per square inch for 3 seconds is enough to disable the viruses assuming all of them are exposed for the whole time (would vary with surface type).

The 4pi in one of your formulas is probably using a solid angle calculation which can tell you the exposure energy.
I did it a different way though, i used degrees instead of radians so i divide by 180 twice because the light spreads out as an approximate lower hemisphere, and to sweep out a given area at a given distance it is convenient for me to know the exposure at a surface that is swept out by a 1 degree solid angle. It came out larger than 1 square inch because of the distance i used in the calculation, so i divided and came up with 0.0012 watt seconds per square inch. From what i have read, that is more than enough for a 3 second exposure but then again i have a carpet so i let it go longer (the carpet is not really completely flat but has fibers that make up the surface).

So let's see, if you have 24 watts then that is close to 0.0013 watts per square over any surface that cuts perpendicular to the solid angle volume approximately (assuming that 1 degree solid angle). Now if you position the light 57 inches away from the source and assuming a point source, that gives you 0.0013 watts over a 1 inch square of that surface. If you leave the light on for 10 seconds that would be an exposure of 0.013 watt seconds over that 1 inch square.

In the above i assumed a transformation of a spherical surface to a flat surface is 1 to 1 for small solid angles.
What this means is that the distance of 57 inches forms a sphere around the light. If any part is closer than 57 inches of course it gets a square proportionally higher exposure, and if farther it gets proportionally less exposure.

A neighbor of mine has two 10 watt bulbs and leaves them on for a couple hours.

You can not use a power supply to run a tube. It must have a ballast. The ballast can deal with the negative resistance of the tube when it starts up. Without that, you could easily burn out the tube after only one use because you wont be able to react fast enough to lower the voltage. Most lights that use tubes though come with the ballast i think because they plug into the wall or screw into a standard lamp base.

I ran into a problem too though, i could not get any reasonably priced UVc LEDs.
Oh yeah, that's another issue, it has to be a UVc lamp/bulb/LED not UVa or UVb.






negative res
 

jpanhalt

Joined Jan 18, 2008
9,710
@MrAl

Have you considered that the common envelope for many LED's is polycarbonate ?

1592942795700.png

I don't know what is used for the UVC LED's you mention, as you have not given part numbers, but I suspect finding a plastic that transmits 254 nm is difficult, and even more difficult at sub-220 nm.
 

jpanhalt

Joined Jan 18, 2008
9,710
The UVC region below 200 nm was largely unexplored when I was in school, except for vacuum UV. Maybe you want to review this about "Burdick-Jackson" solvents that are considered the best in the industry: https://macro.lsu.edu/HowTo/solvents/UV Cutoff.htm

It is a tough region of the spectrum in which to work. I view those companies hawking sanitizers with UVC <200 nm with skepticism in ordinary situations, e.g., surfaces in the home.
 

MrAl

Joined Jun 17, 2014
7,665
@MrAl

Have you considered that the common envelope for many LED's is polycarbonate ?

View attachment 210548

I don't know what is used for the UVC LED's you mention, as you have not given part numbers, but I suspect finding a plastic that transmits 254 nm is difficult, and even more difficult at sub-220 nm.
Hi,

As a matter of fact yes i did. They use a special material i think is special glass not plastic.
However, i did not consider the effects of air but i think that is minimal at short distances like 8 feet or so.
But maybe you are thinking about the wattage as compared to a regular visible light bulb wattage. If we calculate the way i did for a visible light bulb wattage then maybe we have to apply some derating factor to account for a lesser power output due to the different type of light output. I am not sure about that really, but i am apt to think in terms of equivalent power output and i can tell you that the 60 watt UVc LED 'bulb' i have lights up a whole room much more than a 60 watt curly bulb or 60 watt equivalent daylight white LED 'bulb'. It is so bright it actually seems better than a regular bulb so i was asking the question of just how much power concentration does it take to cause human eye damage or human skin damage. From what i read, skin damage can be 'felt' by the human so they would know not to allow exposure again, but i dont know how to tell how much it hurts the human eye at a distance of say 8 feet. I was hoping to find that out at some point also.
 

MrAl

Joined Jun 17, 2014
7,665
The UVC region below 200 nm was largely unexplored when I was in school, except for vacuum UV. Maybe you want to review this about "Burdick-Jackson" solvents that are considered the best in the industry: https://macro.lsu.edu/HowTo/solvents/UV Cutoff.htm

It is a tough region of the spectrum in which to work. I view those companies hawking sanitizers with UVC <200 nm with skepticism in ordinary situations, e.g., surfaces in the home.
Hi,

I didnt see your second reply until now.

When you talk about ordinary situations you mean surfaces that are not flat so the entire surface is exposed to the light?
Yes the example i was thinking about was carpet in a home. The carpet has spaces between the fibers where the light might not reach effectively enough.

However as i am sure you know, UVc has been proven to neuter viruses and so any population dies off without being able to reproduce.
This brings up a possibly interesting question...
If they are already in a cell and they die off, does that mean we have dead viruses in our cells after that? I guess so.
 

MrAl

Joined Jun 17, 2014
7,665
Hello again,

Here is a simpler way to calculate the exposure energy.

Starting with the surface area of a sphere which is 4*pi*r^2, for a point source of light spreading in every possible direction, the watts per square inch would be:
Watts/(4*pi*d^2)
where d is the distance of the source.
Now because we assume all the light radiates in only the lower hemisphere, that changes to:
watts/(2*pi*d^2)
so with d in inches the irradiance reaching a point at a distance 'd' is:
watts/(2*pi*d^2) watts per square inch.
A true 24 watt source at a distance of 10 inches has flux density of:
24/(2*pi*100)=0.038197 watts per square inch.
If you run it for 100 seconds the exposure is:
3.8197 watt seconds per square inch.
 

Thread Starter

redgear

Joined Oct 17, 2019
131
Hello again,

Here is a simpler way to calculate the exposure energy.

Starting with the surface area of a sphere which is 4*pi*r^2, for a point source of light spreading in every possible direction, the watts per square inch would be:
Watts/(4*pi*d^2)
where d is the distance of the source.
Now because we assume all the light radiates in only the lower hemisphere, that changes to:
watts/(2*pi*d^2)
so with d in inches the irradiance reaching a point at a distance 'd' is:
watts/(2*pi*d^2) watts per square inch.
A true 24 watt source at a distance of 10 inches has flux density of:
24/(2*pi*100)=0.038197 watts per square inch.
If you run it for 100 seconds the exposure is:
3.8197 watt seconds per square inch.
Hey,
Thanks for the much simpler calculation.
 

jpanhalt

Joined Jan 18, 2008
9,710
However as i am sure you know, UVc has been proven to neuter viruses and so any population dies off without being able to reproduce.
This brings up a possibly interesting question...
If they are already in a cell and they die off, does that mean we have dead viruses in our cells after that? I guess so.
Can you explain a bit more what you mean by "neutering?" Viruses depend upon the cells they infect to reproduce. There is no sexual reproduction. Most people do not consider them to be living. There is also production of defective particles that are not infections. Modern techniques can detect both (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4139191/ ).

Yes, it can take awhile for an infected individual to clear all viral particles, defective viruses, and fragments of viruses. That is, particles that may be detected by tests for antigens but are not infectious, which was one of the problems encountered early on with some of the tests for coronavirus.
 

MrAl

Joined Jun 17, 2014
7,665
Can you explain a bit more what you mean by "neutering?" Viruses depend upon the cells they infect to reproduce. There is no sexual reproduction. Most people do not consider them to be living. There is also production of defective particles that are not infections. Modern techniques can detect both (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4139191/ ).

Yes, it can take awhile for an infected individual to clear all viral particles, defective viruses, and fragments of viruses. That is, particles that may be detected by tests for antigens but are not infectious, which was one of the problems encountered early on with some of the tests for coronavirus.
Yes that was a case of choosing a bad word for that.
What it does is simply renders them unable to "reproduce" but i dont know the actual mechanism by which this works. Is it before they enter the cell or after, but in any case they cant multiply any longer so the ones that are left over eventually die and hopefully that is all of them if the exposure did the trick.
 

MrAl

Joined Jun 17, 2014
7,665
Hey,
Thanks for the much simpler calculation.
Oh you are welcome, and besides, the original method was not really the right way to do it anyway so it is better to use this "spherical surface" method. If the other way is done right they will come out the same though, but the way i had shown it is not really the right way so use this new method for best results.
One thing i have not found out just yet is how to measure the efficacy of the light bulb. I dont think my exposure meter can handle UVc or even any UV for that matter, but i'll have to check.
However, i think there might be a special test paper you can use to check the exposure and it think it is geared to the application of sanitizing too so ti would be good to get some. I'll have to check for that too but you can do a search if you like.
 

jpanhalt

Joined Jan 18, 2008
9,710
Yes that was a case of choosing a bad word for that.
What it does is simply renders them unable to "reproduce" but i dont know the actual mechanism by which this works. Is it before they enter the cell or after, but in any case they cant multiply any longer so the ones that are left over eventually die and hopefully that is all of them if the exposure did the trick.
The major mechanism for 254 nm light has been known for 50 years. Not hard to look up. Maybe it would be more fruitful if you provided some mechanistic research on how light <200 nm works without harming directly or PRODUCING REACTIVE products that are HARMFUL to humans.
 

MrAl

Joined Jun 17, 2014
7,665
The major mechanism for 254 nm light has been known for 50 years. Not hard to look up. Maybe it would be more fruitful if you provided some mechanistic research on how light <200 nm works without harming directly or PRODUCING REACTIVE products that are HARMFUL to humans.
Well if you look up what has been known for 50 years then you can look up that too.
 

MrAl

Joined Jun 17, 2014
7,665
Glad to be of help.
Well i was making a point that you seemed to be complaining but you did not provide any info either. I know it alters their genetic structure but i dont know what else you would have wanted me to say. I already said it stops them from being able to reproduce and that seems ok by itself, but i add now that it is because their genetic structure is altered that's why they cant reproduce any longer.
Because they can not reproduce any longer the colony eventually dies off and that's the end of the viral load. Of course that means that the light has to be able to reach each particle or some of them might still survive. Then it would become a question of the remaining dosage.

Sorry if i did not supply enough information but i am not sure what you are looking for, and i am not an expert in this area anyway.
 
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