Understanding UV lamp specifications

[741]

I'm trying to understand UV lamp specifications. Here is an example from Helios, https://www.heliosquartz.com/wp-content/uploads/2016/01/Helios-Quartz_UV-LAMPS_eng.pdf
Measurements "represent average values at 1 meter".

For example type HCL5W/G23: Is the final column the total UV output from the lamp i.e. the total UV flux (therefore regardless of distance)?

The next-to-last column gives uW/cm2, so this will be measured at the given 1 metre distance I assume.

So as I see it, the 11W is distributed over about 0.5m2 which is about 22W/m2 or 22/(100*100) = 22mW/cm2. <--- My mistake, not 11W

Here is a copy & paste of the top row from the data sheet
HCL5W/G23   12,5mm   83mm   5W   180mA   34V   9µW/cm2   1W

At 1m radius, diameter is 2, circumference is 3.14*2m = 6.28m, and length of area illuminated is then 83mm* this value, i.e.
   6.28*0.083 =  0.52 m2.

The 1W is distributed over about 0.5m2 which is about 2W/m2 or 2/(100*100) = 0.2mW/cm2 = 200uW/cm2. <---

But the value shown is just 9 uW/cm2.

[RoGeorge]

Only a fraction of the electrical energy that is put into the lamp is transformed into the desired UV light.  The rest is wasted on other side effects, e.g. heat.

[Nusa]

That applies to incandescent bulbs in general, including the typical household bulbs that most of us are still using in at least some fixtures. Even the best of them still lost 90% of the energy as heat. Which is why it was so easy for alternative technologies to save energy.

To the OP, it doesn't change the response, but you seem to be jumping rows when reading the chart. You started with 11W, quoted the output for the 9W lamp, then mentioned the output value for the 5W lamp. Very hard to follow your argument when you do that. In any case, notice the output wattage is actually listed on the chart in the very last column. Subtract that from input wattage and you have your non-UV energy consumption for that bulb.

[RoGeorge]

Also, the lamp model given as example, the power and the calculations doesn't make sense to me.

I'm reading that table like this (for the first line):
5W electrical power in, 1W optical UV power out.

For an ideal 1W light inside of a 1 meter sphere (not circle), that 1W will generate about 7.96\$\mu\$W/cm² at the sphere surface.

However, the datasheet says 9\$\mu\$W/cm² instead of only 7.96\$\mu\$W/cm², maybe they considered some other shape than a sphere, or maybe the 1W is the guaranteed total power, but the measured lamp generates slightly more, IDK.

[Zero999]

That applies to incandescent bulbs in general, including the typical household bulbs that most of us are still using in at least some fixtures. Even the best of them still lost 90% of the energy as heat. Which is why it was so easy for alternative technologies to save energy.

UVC lamps aren't incandescent though. Mercury UVC lamps have efficiencies of 30% of more. A good data sheet will provide both the power output as well as power input. The over all assembly will be a bit less efficient as some energy will be lost in the ballast and reflector.

Page 12 gives the efficiencies of their germicidal lamps.
https://www.heliosquartz.com/wp-content/uploads/2016/01/Helios-Quartz_UV-LAMPS_eng.pdf

[741]

"Only a fraction of the electrical energy.." - Sure, I realise that. I (thought I) used the 1W value which is UV emission rather than the 5W overall 'power' value.
(In fact - as another respondent pointed out I used 11W). I goofed with the 11W, which came from another example I was trying earlier, so thank you for telling me about that.

I meant to use the top row; here is a copy & paste of the top row from the data sheet
HCL5W/G23   12,5mm   83mm   5W   180mA   34V   9µW/cm2   1W   

I have now ammended the original post.

Regarding the sphere and presumably spherically symmetrical point source: That is the kind of idealised model I have seen in some text books. I would have thought this is not a useful way to present data for a tubular source.
I am really surprised that this is the way the data is shown, but your answer is very close to theirs so that is the most likely explanation.

Below I have repeated the 1m sphere calculation for all examples. The fairly consistent 'error' might be related to emission at other frequencies.


To judge whether a UV tube is suitable for a given use, I'd want to know the results for examples like that I tried to do at the start - "flux at some radial distance".
Q: Are my calculations & reasoning correct is this respect?

Many thanks for all the help! 
I found what looks like good information here: https://docs-emea.rs-online.com/webdocs/1506/0900766b81506d14.pdf

For the case of water sterilisation, an equation of the same form I had at the top (circumferance at X metres * length) does appear - but that is only to find the 'required' dose. Also, they use several lamps, around the circumference of the water tube. So it's not the same situation as my example.

To find what dose is delivered, they use one of two charts. The 1st is for distance under 0.5m. The second chart, where I assume things are sort of like a 1-D line of point sources applies for over 1m.

For a point source, I can see the square law applies. For a linear source we have a continuum of point sources. The flux would presumably be 'the integral of' the point source case, but I've not worked this out yet.

[redgear]

For an ideal 1W light inside of a 1 meter sphere (not circle), that 1W will generate about 7.96\$\mu\$W/cm² at the sphere surface.

Can you provide the formula you used for calculation?

[RoGeorge]

An ideal light source is supposed to be a point equally radiating light in all direction, like a sphere.

The surface aria of a sphere is A = 4πr² = 4*3.14*1meter*1meter = 4*3.14*100centimeters*100centimeters = 125663.71 square centimeters.

1W spread over all that surface area will be 1W/125663.71cm² = 1000000µW/125663.71cm² = 7.96µW/cm²