Calculating radiation received at distance from light source?
December 16, 2014 5:50 PM Subscribe
How do I calculate the radiation energy that a surface / object receives at distance d from a light source of power W watts? I'm familiar, in theory, with the inverse square law but need help thinking about how to apply it.
The purpose is to set up a UV-C germicidal chamber and I want to make sure the light will be intense enough to actually sanitize the items within. Wikipedia notes that it takes 2-8 W*s/cm2 to kill various germs (where "kill" means "reduce the population by a power of 10," I think).
The light power will be fixed and I can calculate the surface area of the object to be sanitized, but how do I account for said surface's distance from the light, since the light power decreases as the square of that number? I think I'm looking to apply the formal physical concept of intensity, but I am having trouble with what to plug in where. Thanks!
The purpose is to set up a UV-C germicidal chamber and I want to make sure the light will be intense enough to actually sanitize the items within. Wikipedia notes that it takes 2-8 W*s/cm2 to kill various germs (where "kill" means "reduce the population by a power of 10," I think).
The light power will be fixed and I can calculate the surface area of the object to be sanitized, but how do I account for said surface's distance from the light, since the light power decreases as the square of that number? I think I'm looking to apply the formal physical concept of intensity, but I am having trouble with what to plug in where. Thanks!
Errr; sorry to use confusing terminology. I'm an astronomer, so my terminology is a bit different.
You can use Luminosity/Power and Brightness/Intensity interchangeably in this case and you'll be just fine.
posted by Betelgeuse at 6:37 PM on December 16, 2014
You can use Luminosity/Power and Brightness/Intensity interchangeably in this case and you'll be just fine.
posted by Betelgeuse at 6:37 PM on December 16, 2014
Best answer: Short answer, it depends on the geometry - the shape of your chamber and the shape and position of your lamp.
Long answer:
The basic procedure for calculating this is to find the power density (Watts per cm^2) at the location of your target, then multiply by the illuminated area of the target to find the power (Watts) delivered to the target.
In the absence of reflections, and assuming the lamp emits power uniformly over emission angle, the power emitted by the lamp is distributed evenly over any surface that is equidistant from the lamp. This is very simple for a spherical lamp, but more complicated for other lamp geometries.
For a spherical lamp in the middle of empty space, the power density as a function of distance from the center of the lamp is given by:
σ_P(r) = P / (4*pi*r^2)
Where P is the total power emitted by the lamp. Then given an illuminated surface area S of your object, the power received by your object would be
P_receive = S * σ_P(r) = S * P / (4 * pi * r^2)
It is, of course, unlikely that your germicidal chamber resembles a spherical lamp in empty space.
For other geometries, the power density may not go down with the square of the distance - for example if you have a very long cylindrical lamp in empty space, the received power would decrease with the inverse, rather than inverse square of the distance:
P_receive = S * σ_P(r) = S * P / (2*pi*r)
In a narrow chamber where one entire wall is fully covered by the lamp, the power density wouldn't change at all with distance (unless we allow for absorption by the chamber walls), so the received power would be
P_receive = S * σ_P(r) = S * P / A
where A is the area of the lamp's surface.
If you would like to clarify the geometry involved in your case, I can help you determine the correct model to use.
posted by Salvor Hardin at 7:00 PM on December 16, 2014
Long answer:
The basic procedure for calculating this is to find the power density (Watts per cm^2) at the location of your target, then multiply by the illuminated area of the target to find the power (Watts) delivered to the target.
In the absence of reflections, and assuming the lamp emits power uniformly over emission angle, the power emitted by the lamp is distributed evenly over any surface that is equidistant from the lamp. This is very simple for a spherical lamp, but more complicated for other lamp geometries.
For a spherical lamp in the middle of empty space, the power density as a function of distance from the center of the lamp is given by:
σ_P(r) = P / (4*pi*r^2)
Where P is the total power emitted by the lamp. Then given an illuminated surface area S of your object, the power received by your object would be
P_receive = S * σ_P(r) = S * P / (4 * pi * r^2)
It is, of course, unlikely that your germicidal chamber resembles a spherical lamp in empty space.
For other geometries, the power density may not go down with the square of the distance - for example if you have a very long cylindrical lamp in empty space, the received power would decrease with the inverse, rather than inverse square of the distance:
P_receive = S * σ_P(r) = S * P / (2*pi*r)
In a narrow chamber where one entire wall is fully covered by the lamp, the power density wouldn't change at all with distance (unless we allow for absorption by the chamber walls), so the received power would be
P_receive = S * σ_P(r) = S * P / A
where A is the area of the lamp's surface.
If you would like to clarify the geometry involved in your case, I can help you determine the correct model to use.
posted by Salvor Hardin at 7:00 PM on December 16, 2014
Response by poster: Thank you both!
There will be two use cases, both with a light that is in fact cylindrical (but obviously of finite length).
One of the use cases is itself a stout cylinder with the light suspended in the center, which on a practical level will be somewhere between (or maybe, better, a combination of) Salvor Hardin's first two models - like, a proper cylinder along the axis of the light with two semi-spheres emanating from each end.
The other is more like a lightproof square box, 2 ft on a side or so, with the light in a top corner or top/center. I think it's safest to count this as a point source with a sphere emanating out to the various walls where the objects to be sanitized will be placed.
posted by Joey Buttafoucault at 7:21 PM on December 16, 2014
There will be two use cases, both with a light that is in fact cylindrical (but obviously of finite length).
One of the use cases is itself a stout cylinder with the light suspended in the center, which on a practical level will be somewhere between (or maybe, better, a combination of) Salvor Hardin's first two models - like, a proper cylinder along the axis of the light with two semi-spheres emanating from each end.
The other is more like a lightproof square box, 2 ft on a side or so, with the light in a top corner or top/center. I think it's safest to count this as a point source with a sphere emanating out to the various walls where the objects to be sanitized will be placed.
posted by Joey Buttafoucault at 7:21 PM on December 16, 2014
Looking this up, I randomly found that Boston bans UV boxes in favor of autoclaves or dry heat sterilizers in nail salons:
Boston Public Health Commission
HOME > WHAT WE DO BY TOPIC > HEALTHY HOMES AND ENVIRONMENT > SAFE NAIL SALONS > New Sterilization Requirements
...
Contrary to popular belief, UV boxes do not actually sterilize tools. A UV box will only keep tools clean if they have already been properly sterilized prior to being stored inside the UV box. If a tool was not properly sterilized, bacteria and other organisms can still grow on the tools that were placed inside the UV box. This can cause the unintentional spread of infections from client to client. Because of this, the City of Boston has banned the use of UV boxes and instated the use of autoclaves and/or US FDA registered dry heat sterilizers; which actually sterilizes the multi-use tools.
unsurprisingly.
Are you confident UV will work in your application?
wiki: Many objects that are transparent to visible light absorb UV, glass for example completely absorbs all UV light. UV irradiation is routinely used to sterilize the interiors of biological safety cabinets between uses, but is ineffective in shaded areas, including areas under dirt (which may become polymerized after prolonged irradiation, so that it is very difficult to remove). It also damages some plastics, such as polystyrene foam if exposed for prolonged periods of time.
Assuming that you're not sterilizing nail tools and that UV will actually work for your application, (which is likely, as you're probably following current practices in your discipline, but a random reader may not be), you can apparently purchase UV test-strips that change color when they've passed a threshold of UV exposure. These would serve as a sanity check that your UV box is working or that it is time to replace your bulb.
The other is more like a lightproof square box, 2 ft on a side or so, with the light in a top corner or top/center. I think it's safest to count this as a point source with a sphere emanating out to the various walls where the objects to be sanitized will be placed.
Will you have reflective walls?
Increases in effectiveness and UV intensity can be achieved by using reflection. Aluminium has the highest reflectivity rate versus other metals and is recommended when using UV.
Including reflections, the problem may not have a tidy analytic solution; you may want to consult a biomed reference which looks perfect for this, engineering reference, and/or dump the problem into a raytracer like povray, blender, or something more formal, using the intensity at a spot in an absorbing box to normalize the intensity at the same spot in a reflective box. Also try google scholar, there are resources where other people have worked on this and describe their apparatus and results.
Or you can just use the same wattage (20 Watts, 200 Watts, 40 Watts, 400 Watts, dependantly) used in these randomly found PCR workstations, some of which are dead air and some of which are ventilated. Similarly, you can consult supply catalogs for applications similar to yours and use those values for wattages for lamps in similarly sized UV boxes.
posted by sebastienbailard at 2:40 AM on December 17, 2014 [1 favorite]
Boston Public Health Commission
HOME > WHAT WE DO BY TOPIC > HEALTHY HOMES AND ENVIRONMENT > SAFE NAIL SALONS > New Sterilization Requirements
...
Contrary to popular belief, UV boxes do not actually sterilize tools. A UV box will only keep tools clean if they have already been properly sterilized prior to being stored inside the UV box. If a tool was not properly sterilized, bacteria and other organisms can still grow on the tools that were placed inside the UV box. This can cause the unintentional spread of infections from client to client. Because of this, the City of Boston has banned the use of UV boxes and instated the use of autoclaves and/or US FDA registered dry heat sterilizers; which actually sterilizes the multi-use tools.
unsurprisingly.
Are you confident UV will work in your application?
wiki: Many objects that are transparent to visible light absorb UV, glass for example completely absorbs all UV light. UV irradiation is routinely used to sterilize the interiors of biological safety cabinets between uses, but is ineffective in shaded areas, including areas under dirt (which may become polymerized after prolonged irradiation, so that it is very difficult to remove). It also damages some plastics, such as polystyrene foam if exposed for prolonged periods of time.
Assuming that you're not sterilizing nail tools and that UV will actually work for your application, (which is likely, as you're probably following current practices in your discipline, but a random reader may not be), you can apparently purchase UV test-strips that change color when they've passed a threshold of UV exposure. These would serve as a sanity check that your UV box is working or that it is time to replace your bulb.
The other is more like a lightproof square box, 2 ft on a side or so, with the light in a top corner or top/center. I think it's safest to count this as a point source with a sphere emanating out to the various walls where the objects to be sanitized will be placed.
Will you have reflective walls?
Increases in effectiveness and UV intensity can be achieved by using reflection. Aluminium has the highest reflectivity rate versus other metals and is recommended when using UV.
Including reflections, the problem may not have a tidy analytic solution; you may want to consult a biomed reference which looks perfect for this, engineering reference, and/or dump the problem into a raytracer like povray, blender, or something more formal, using the intensity at a spot in an absorbing box to normalize the intensity at the same spot in a reflective box. Also try google scholar, there are resources where other people have worked on this and describe their apparatus and results.
Or you can just use the same wattage (20 Watts, 200 Watts, 40 Watts, 400 Watts, dependantly) used in these randomly found PCR workstations, some of which are dead air and some of which are ventilated. Similarly, you can consult supply catalogs for applications similar to yours and use those values for wattages for lamps in similarly sized UV boxes.
posted by sebastienbailard at 2:40 AM on December 17, 2014 [1 favorite]
Response by poster: Thanks for the very thoughtful response, sebastienbailard. To be clear, I am aware this won't result in sterilization of any of my equipment; I've been tying to think of it as a way to dramatically reduce the population of problematic organisms.
The primary use will actually be quite similar to cleaning "the interiors of biological safety cabinets"; essentially they are moderate-sized tanks open only on one end or only with a small aperture, which I am concerned are hard to sanitize by manual methods such as liquid solutions (I suppose I could get some spray bottles but even then it may be hard to aim properly). The thought is that I would dangle the bare bulb inside this chamber after mechanically cleaning off any visible grime and hope that's better than sloshing around some sanitizing solution which may or may not cover the entire tank for the recommended time.
The goal of this question was to get at least a vague idea of the amount of time that would be "enough" to realistically deliver the energy, then maybe leave the light on for 4-5 times that long anyway as a margin of safety. Thanks much for the biomed references, and those UV test strips also sound like an excellent mechanism for determining whether I'm in the right range.
posted by Joey Buttafoucault at 9:45 AM on December 18, 2014
The primary use will actually be quite similar to cleaning "the interiors of biological safety cabinets"; essentially they are moderate-sized tanks open only on one end or only with a small aperture, which I am concerned are hard to sanitize by manual methods such as liquid solutions (I suppose I could get some spray bottles but even then it may be hard to aim properly). The thought is that I would dangle the bare bulb inside this chamber after mechanically cleaning off any visible grime and hope that's better than sloshing around some sanitizing solution which may or may not cover the entire tank for the recommended time.
The goal of this question was to get at least a vague idea of the amount of time that would be "enough" to realistically deliver the energy, then maybe leave the light on for 4-5 times that long anyway as a margin of safety. Thanks much for the biomed references, and those UV test strips also sound like an excellent mechanism for determining whether I'm in the right range.
posted by Joey Buttafoucault at 9:45 AM on December 18, 2014
This thread is closed to new comments.
So, if you have a luminosity (or power - same thing) in Watts and a distance in centimeters, you can get a brightness in units of W/cm^2. Presumably, you then apply the light over some number of seconds to get the units right on the "2-8 W*s/cm^2."
An example:
Say you have a light with a power of 100 watts that is 10 centimeters away from the source you want to sterilize. Then it would have a brightness of:
100 W / 4 * pi * (10 cm)^2 = 0.08 W/cm^2
If you then wanted to get up to 2 W*s/cm^2, you would need to apply the light for:
2 / 0.08 = 25 seconds.
posted by Betelgeuse at 6:34 PM on December 16, 2014