## How to convert from kJ/m^2 to uW/cm^2?

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• Hello, I have data I would like to replicte but it calls for UV-b irradiance of 13 kJ/m^2. I have tried a few ways I can think of and I don t think I have the
Message 1 of 6 , Mar 2, 2010
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Hello,

I have data I would like to replicte but it calls for UV-b irradiance of 13 kJ/m^2. I have tried a few ways I can think of and I don't think I have the correct answer yet.

Thank you
• kJ (or just J) is units for an amount of energy. uW (or just W) is an energy rate ie. J per unit time (in this case, 1 second) The 13kJ/m2 figure does not
Message 2 of 6 , Mar 3, 2010
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kJ (or just J) is units for an amount of energy.
uW (or just W) is an energy rate ie. J per unit time (in this case, 1 second)

The 13kJ/m2 figure does not include time. eg. it
could be 13kJ over 1 second... or 14 days.
If there is an implied one-second duration in
that then it is 13kW (since a joule is a Watt.second).

So, if that (one second) is the case then you are
looking at 13kW/m2 and you only have to correct
for the area (m2 to cm2). To do that just divide by 10,000 (ie 100x100cm).

So 13kW/m2 is 13000/10000 = 1.3W/cm2

That is a *huge* amount of UV-B. Maybe your time interval is not 1 second?
In industrial (ie. large scale) thermal energy
calculations it is common practice to consider
power in kW/hour. If that is the case here then
you would have to divide by a further factor of
3600 (ie. 60 * 60 to convert hours to seconds).
That would give you 3.61e-4 W/cm2 (ie. 361 microwatts/cm2).

That number sounds more likely/believable but we
are really doing this backwards here so it would
be best if you can look again at your original
data and confirm where the time interval comes into it. Then we can be sure.

Is that any help?
Cheers
- Andy

At 20:49 02/03/2010, you wrote:
>
>
>Hello,
>
>I have data I would like to replicte but it
>calls for UV-b irradiance of 13 kJ/m^2. I have
>tried a few ways I can think of and I don't
>think I have the correct answer yet.
>
>Thank you
>
>
>
>
>__________ Information from ESET NOD32
>Antivirus, version of virus signature database 4910 (20100302) __________
>
>The message was checked by ESET NOD32 Antivirus.
>
><http://www.eset.com>http://www.eset.com
• Hello, Thanks for the help. What you wrote and what I calculated are the same (pretty much). I forgot to write the figure of 13 kJ/m^2 was the total UV-b
Message 3 of 6 , Mar 3, 2010
• 0 Attachment
Hello,

Thanks for the help. What you wrote and what I calculated are the same (pretty much). I forgot to write the figure of 13 kJ/m^2 was the total UV-b irradiance over a 3 hour period each day (oops!). This is on the topic of growing plants, not herping. In a great study I am replicating the researchers used a totoal UV-b irradiance of 13.4 kJ/m^2 each day (in the time frame of 3 hours a day); thus I want to convert 13.4 kJ/m^2/3hr into uW/cm^2/sec so I can use the Uv-b meter SolarTech 6.2 (as it measures per second AFAIK).

Here is my math, would you mind checking it for me? thanks a ton!

1. ((13.4 kJ/m^/3hr) / 3) = 4.47 kJ/m^2

2. (4.47 kJ/m^2) x 1,000 = 4,470 joule/m^2

3. 4,470 joule/m^2 = 1.24 watt/m^2/hr

4. (1.24 watt/m^2/hr) x 1,000,000 = 1,240,000 uW/m^2/hr

5. (1,240,000 uW/m^2/hr) / 10,000 = 124 uW/cm^2/hr

6. (124 uW/cm^/hr) / 3,600 = 0.34 uW/cm^2/sec (as a single measurement from a UV-b meter like the SolarTech 6.2)

Thus:

((0.34 uW/cm^2/sec) x 3,600) x 3 = 367 uW/cm^2/3hr (the total irradiance of UV-b to reach per day equaling about 13.4 kJ/m^2/3hr)

So:

If my calculations are correct 13.4 kJ/m^2/3hr = 0.34 uW/cm^2/sec over a 3 hour period. Or, I could use about 0.1 uW/cm^2/sec for a 1 hour period to reach an equivalent of 13.4 kJ/m^2/3hr.

Thanks!

--- In UVB_Meter_Owners@yahoogroups.com, Andy B <sprocket@...> wrote:
>
> kJ (or just J) is units for an amount of energy.
> uW (or just W) is an energy rate ie. J per unit time (in this case, 1 second)
>
> The 13kJ/m2 figure does not include time. eg. it
> could be 13kJ over 1 second... or 14 days.
> If there is an implied one-second duration in
> that then it is 13kW (since a joule is a Watt.second).
>
> So, if that (one second) is the case then you are
> looking at 13kW/m2 and you only have to correct
> for the area (m2 to cm2). To do that just divide by 10,000 (ie 100x100cm).
>
> So 13kW/m2 is 13000/10000 = 1.3W/cm2
>
> That is a *huge* amount of UV-B. Maybe your time interval is not 1 second?
> In industrial (ie. large scale) thermal energy
> calculations it is common practice to consider
> power in kW/hour. If that is the case here then
> you would have to divide by a further factor of
> 3600 (ie. 60 * 60 to convert hours to seconds).
> That would give you 3.61e-4 W/cm2 (ie. 361 microwatts/cm2).
>
> That number sounds more likely/believable but we
> are really doing this backwards here so it would
> be best if you can look again at your original
> data and confirm where the time interval comes into it. Then we can be sure.
>
> Is that any help?
> Cheers
> - Andy
>
>
> At 20:49 02/03/2010, you wrote:
> >
> >
> >Hello,
> >
> >I have data I would like to replicte but it
> >calls for UV-b irradiance of 13 kJ/m^2. I have
> >tried a few ways I can think of and I don't
> >think I have the correct answer yet.
> >
> >Thank you
> >
> >
> >
> >
> >__________ Information from ESET NOD32
> >Antivirus, version of virus signature database 4910 (20100302) __________
> >
> >The message was checked by ESET NOD32 Antivirus.
> >
> ><http://www.eset.com>http://www.eset.com
>
• I think something got a bit twisted in there. That s not quite what I get. Watts already have time factored in them. If you divide by time you get Joules
Message 4 of 6 , Mar 3, 2010
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I think something got a bit twisted in there. That's not quite what I get.
Watts already have time factored in them. If you
divide by time you get Joules (total energy) and
then time is irrelevant (has been removed - so
you are left with just an amount of energy). So I
don't think you can combine W and time that way
around. Sure enough the number that pops out
there (0.34 uW/cm^2) is not really doable either
- it's a very tiny amount of UV-B that in
practice would be difficult to measure.

I see this:-

(13.4kJ/m2/3hr) / 3 => 4470 J/m2/hr

4470 J/m2/hr / (60*60) => 1.24 J/m2/sec (ie. 1.24 W/m2)

1.24 W/m2 / (100x100) => 1.24e-4 W/cm2 (ie. 124 uW/cm2)

So that would be 124 uW/cm2 (as a 6.2 meter
reading), still taking 3 hours. That's a
perfectly doable level of UV-B using a decent
lamp and adjusting distance; but you could apply
the same total energy as 248uW/cm2 for 1.5 hours; or 62 uW/cm2 for 6 hours.

One small point (and Steve may have more to say
if he is around as it has to do with how the
Solarmeter is calibrated for absolute uW -
whether that is with a particular irradiance
spectra), the Solarmeter, like any broadband
meter, has a response curve that is not flat so
you can only take that number as a rough guide.
If you are using sunlight and the experiment you
are replicating also used sunlight then it may be
safe enough (Steve?) but if the replication is
with a lamp that has a different spectral output
(eg. like the UV-B spike you get from a lamp)
then you only have a rough guide. How precise do you need to be?

- Andy

At 15:42 03/03/2010, you wrote:
>Hello,
>
>Thanks for the help. What you wrote and what I
>calculated are the same (pretty much). I forgot
>to write the figure of 13 kJ/m^2 was the total
>UV-b irradiance over a 3 hour period each day
>(oops!). This is on the topic of growing plants,
>not herping. In a great study I am replicating
>the researchers used a totoal UV-b irradiance of
>13.4 kJ/m^2 each day (in the time frame of 3
>hours a day); thus I want to convert 13.4
>kJ/m^2/3hr into uW/cm^2/sec so I can use the
>Uv-b meter SolarTech 6.2 (as it measures per second AFAIK).
>
>Here is my math, would you mind checking it for me? thanks a ton!
>
>1. ((13.4 kJ/m^/3hr) / 3) = 4.47 kJ/m^2
>
>2. (4.47 kJ/m^2) x 1,000 = 4,470 joule/m^2
>
>3. 4,470 joule/m^2 = 1.24 watt/m^2/hr
>
>4. (1.24 watt/m^2/hr) x 1,000,000 = 1,240,000 uW/m^2/hr
>
>5. (1,240,000 uW/m^2/hr) / 10,000 = 124 uW/cm^2/hr
>
>6. (124 uW/cm^/hr) / 3,600 = 0.34 uW/cm^2/sec
>(as a single measurement from a UV-b meter like the SolarTech 6.2)
>
>Thus:
>
>((0.34 uW/cm^2/sec) x 3,600) x 3 = 367
>uW/cm^2/3hr (the total irradiance of UV-b to
>reach per day equaling about 13.4 kJ/m^2/3hr)
>
>So:
>
>If my calculations are correct 13.4 kJ/m^2/3hr =
>0.34 uW/cm^2/sec over a 3 hour period. Or, I
>could use about 0.1 uW/cm^2/sec for a 1 hour
>period to reach an equivalent of 13.4 kJ/m^2/3hr.
>
>Thanks!
>
>--- In
><mailto:UVB_Meter_Owners%40yahoogroups.com>UVB_Meter_Owners@yahoogroups.com,
>Andy B <sprocket@...> wrote:
> >
> > kJ (or just J) is units for an amount of energy.
> > uW (or just W) is an energy rate ie. J per
> unit time (in this case, 1 second)
> >
> > The 13kJ/m2 figure does not include time. eg. it
> > could be 13kJ over 1 second... or 14 days.
> > If there is an implied one-second duration in
> > that then it is 13kW (since a joule is a Watt.second).
> >
> > So, if that (one second) is the case then you are
> > looking at 13kW/m2 and you only have to correct
> > for the area (m2 to cm2). To do that just divide by 10,000 (ie 100x100cm).
> >
> > So 13kW/m2 is 13000/10000 = 1.3W/cm2
> >
> > That is a *huge* amount of UV-B. Maybe your time interval is not 1 second?
> > In industrial (ie. large scale) thermal energy
> > calculations it is common practice to consider
> > power in kW/hour. If that is the case here then
> > you would have to divide by a further factor of
> > 3600 (ie. 60 * 60 to convert hours to seconds).
> > That would give you 3.61e-4 W/cm2 (ie. 361 microwatts/cm2).
> >
> > That number sounds more likely/believable but we
> > are really doing this backwards here so it would
> > be best if you can look again at your original
> > data and confirm where the time interval
> comes into it. Then we can be sure.
> >
> > Is that any help?
> > Cheers
> > - Andy
> >
> >
> > At 20:49 02/03/2010, you wrote:
> > >
> > >
> > >Hello,
> > >
> > >I have data I would like to replicte but it
> > >calls for UV-b irradiance of 13 kJ/m^2. I have
> > >tried a few ways I can think of and I don't
> > >think I have the correct answer yet.
> > >
> > >Thank you
> > >
> > >
> > >
> > >
> > >__________ Information from ESET NOD32
> > >Antivirus, version of virus signature database 4910 (20100302) __________
> > >
> > >The message was checked by ESET NOD32 Antivirus.
> > >
> > ><<http://www.eset.com>http://www.eset.com>http://www.eset.com
> >
>
>
>
>
>__________ Information from ESET NOD32
>Antivirus, version of virus signature database 4912 (20100303) __________
>
>The message was checked by ESET NOD32 Antivirus.
>
><http://www.eset.com>http://www.eset.com
• Hello, Thanks again, I see your points. The funny part is 124 uW/cm^2 is the figure I fist calculated too, but I thought that was too high so I re-worked my
Message 5 of 6 , Mar 3, 2010
• 0 Attachment
Hello,

Thanks again, I see your points. The funny part is 124 uW/cm^2 is the figure I fist calculated too, but I thought that was too high so I re-worked my math. I guess I should have left well enough alone.

Here is my original math:

1. ((13.4 kJ/m^2/3hr) / 3) = 4.47 kJ/m^2

2. (4.47 kJ/m^2) x 1,000 = 4,467 joule/m^2

3. 4,467 joule/m^2 = 1.24 watt/m^2

4. 1.24 watt/m^2 x 1,000,000 = 1,240,000 uW/m^2

5. (1,240,000 uW/m^2) / 10,000 = 124 uW/cm^2

RE: source of UV-b irradiance in the study:

UV-b irradiance came from a filtered Westinghouse FS-40 sunlamp (i.e., as described in the work of Mirecki and Teramura, 1984). Lamps were fitted with either presolarized 0.08 mm or 0.13 mm cellulose acetate (UV-b transmitting) or 0.13 mm Mylar Type A (as control filters).

RE: absolute uW:

The spectroradiometer used to quantitate the UV-b irradiance was an Optronic Laboratories, Inc. Model 742 spectroratiometer interfaced with a Hewlett Packard 85 printing calculator (it's an older study, circa 1986). The spectroradiometer was weighted with a generalized plant action spectrum (UV-bbe, Calwell, 1971), and normalized at 300 nm.

I would like the results to be as accurate as possible but I am happy with 'horse shoes and hand grenades'. If I am able to confidently provide 100-150 uW/cm^2 for 3 hours I would be happy, I don't need to be very accurate, a range is fine with me (and more attainable over a larger canopy than a single figure).

I thought you may find this interesting: in the study the reseradhers used a daily Uv-b irradiance (13.4 kJ/m^2/3hr) equivalent to the daily weighted UV-b irradiance under clear sky conditions and minium solar zenith in Columbia, South America at 0' latitude, 3 km elvention, June, 1980 (Green, et al., 1980)

Thanks so much for your time!

--- In UVB_Meter_Owners@yahoogroups.com, Andy B <sprocket@...> wrote:
>
> I think something got a bit twisted in there. That's not quite what I get.
> Watts already have time factored in them. If you
> divide by time you get Joules (total energy) and
> then time is irrelevant (has been removed - so
> you are left with just an amount of energy). So I
> don't think you can combine W and time that way
> around. Sure enough the number that pops out
> there (0.34 uW/cm^2) is not really doable either
> - it's a very tiny amount of UV-B that in
> practice would be difficult to measure.
>
> I see this:-
>
> (13.4kJ/m2/3hr) / 3 => 4470 J/m2/hr
>
> 4470 J/m2/hr / (60*60) => 1.24 J/m2/sec (ie. 1.24 W/m2)
>
> 1.24 W/m2 / (100x100) => 1.24e-4 W/cm2 (ie. 124 uW/cm2)
>
> So that would be 124 uW/cm2 (as a 6.2 meter
> reading), still taking 3 hours. That's a
> perfectly doable level of UV-B using a decent
> lamp and adjusting distance; but you could apply
> the same total energy as 248uW/cm2 for 1.5 hours; or 62 uW/cm2 for 6 hours.
>
> One small point (and Steve may have more to say
> if he is around as it has to do with how the
> Solarmeter is calibrated for absolute uW -
> whether that is with a particular irradiance
> spectra), the Solarmeter, like any broadband
> meter, has a response curve that is not flat so
> you can only take that number as a rough guide.
> If you are using sunlight and the experiment you
> are replicating also used sunlight then it may be
> safe enough (Steve?) but if the replication is
> with a lamp that has a different spectral output
> (eg. like the UV-B spike you get from a lamp)
> then you only have a rough guide. How precise do you need to be?
>
> - Andy
>
>
> At 15:42 03/03/2010, you wrote:
> >Hello,
> >
> >Thanks for the help. What you wrote and what I
> >calculated are the same (pretty much). I forgot
> >to write the figure of 13 kJ/m^2 was the total
> >UV-b irradiance over a 3 hour period each day
> >(oops!). This is on the topic of growing plants,
> >not herping. In a great study I am replicating
> >the researchers used a totoal UV-b irradiance of
> >13.4 kJ/m^2 each day (in the time frame of 3
> >hours a day); thus I want to convert 13.4
> >kJ/m^2/3hr into uW/cm^2/sec so I can use the
> >Uv-b meter SolarTech 6.2 (as it measures per second AFAIK).
> >
> >Here is my math, would you mind checking it for me? thanks a ton!
> >
> >1. ((13.4 kJ/m^/3hr) / 3) = 4.47 kJ/m^2
> >
> >2. (4.47 kJ/m^2) x 1,000 = 4,470 joule/m^2
> >
> >3. 4,470 joule/m^2 = 1.24 watt/m^2/hr
> >
> >4. (1.24 watt/m^2/hr) x 1,000,000 = 1,240,000 uW/m^2/hr
> >
> >5. (1,240,000 uW/m^2/hr) / 10,000 = 124 uW/cm^2/hr
> >
> >6. (124 uW/cm^/hr) / 3,600 = 0.34 uW/cm^2/sec
> >(as a single measurement from a UV-b meter like the SolarTech 6.2)
> >
> >Thus:
> >
> >((0.34 uW/cm^2/sec) x 3,600) x 3 = 367
> >uW/cm^2/3hr (the total irradiance of UV-b to
> >reach per day equaling about 13.4 kJ/m^2/3hr)
> >
> >So:
> >
> >If my calculations are correct 13.4 kJ/m^2/3hr =
> >0.34 uW/cm^2/sec over a 3 hour period. Or, I
> >could use about 0.1 uW/cm^2/sec for a 1 hour
> >period to reach an equivalent of 13.4 kJ/m^2/3hr.
> >
> >Thanks!
> >
> >--- In
> ><mailto:UVB_Meter_Owners%40yahoogroups.com>UVB_Meter_Owners@yahoogroups.com,
> >Andy B <sprocket@> wrote:
> > >
> > > kJ (or just J) is units for an amount of energy.
> > > uW (or just W) is an energy rate ie. J per
> > unit time (in this case, 1 second)
> > >
> > > The 13kJ/m2 figure does not include time. eg. it
> > > could be 13kJ over 1 second... or 14 days.
> > > If there is an implied one-second duration in
> > > that then it is 13kW (since a joule is a Watt.second).
> > >
> > > So, if that (one second) is the case then you are
> > > looking at 13kW/m2 and you only have to correct
> > > for the area (m2 to cm2). To do that just divide by 10,000 (ie 100x100cm).
> > >
> > > So 13kW/m2 is 13000/10000 = 1.3W/cm2
> > >
> > > That is a *huge* amount of UV-B. Maybe your time interval is not 1 second?
> > > In industrial (ie. large scale) thermal energy
> > > calculations it is common practice to consider
> > > power in kW/hour. If that is the case here then
> > > you would have to divide by a further factor of
> > > 3600 (ie. 60 * 60 to convert hours to seconds).
> > > That would give you 3.61e-4 W/cm2 (ie. 361 microwatts/cm2).
> > >
> > > That number sounds more likely/believable but we
> > > are really doing this backwards here so it would
> > > be best if you can look again at your original
> > > data and confirm where the time interval
> > comes into it. Then we can be sure.
> > >
> > > Is that any help?
> > > Cheers
> > > - Andy
> > >
> > >
> > > At 20:49 02/03/2010, you wrote:
> > > >
> > > >
> > > >Hello,
> > > >
> > > >I have data I would like to replicte but it
> > > >calls for UV-b irradiance of 13 kJ/m^2. I have
> > > >tried a few ways I can think of and I don't
> > > >think I have the correct answer yet.
> > > >
> > > >Thank you
> > > >
> > > >
> > > >
> > > >
> > > >__________ Information from ESET NOD32
> > > >Antivirus, version of virus signature database 4910 (20100302) __________
> > > >
> > > >The message was checked by ESET NOD32 Antivirus.
> > > >
> > > ><<http://www.eset.com>http://www.eset.com>http://www.eset.com
> > >
> >
> >
> >
> >
> >__________ Information from ESET NOD32
> >Antivirus, version of virus signature database 4912 (20100303) __________
> >
> >The message was checked by ESET NOD32 Antivirus.
> >
> ><http://www.eset.com>http://www.eset.com
>
• The 6.2 meter master NIST traceable calibration was done using a pair of typical tanning lamps peaking at 350nm UVA with ~7% UVB. Spectrum similar (but not
Message 6 of 6 , Mar 13, 2010
• 0 Attachment
The 6.2 meter master NIST traceable calibration was done using a pair of typical tanning lamps peaking at 350nm UVA with ~7% UVB. Spectrum similar (but not exact) to noon tropical sun. So you should be OK in that regard.

Without trying to redo all the below math... and assuming the 124 uW/cm2 intensity (dose rate) is correct... you can apply time factors accordingly to determine accumulated energy dose.

However... the 6.2 sensor sees "broadband" UVB... not weighted by any plant action spectrum. So I'm not sure what effect that might have.

Also... FS-40 lamps are "phototherapy" UVB lamps with lots of irradiance below solar 290nm. Maybe the filters they used block harmful rays below 290... the same rays bad for reptile eyes.

Anyway - have fun with your experimenting. We do get orders from Columbia for 6.2 meters from "plant" growers looking to get the biggest "bud" size. Gee - I wonder what they mean by that lol.

Steve

--- In UVB_Meter_Owners@yahoogroups.com, "jrdmck" <jrdmck@...> wrote:
>
>
>
>
>
>
> Hello,
>
> Thanks again, I see your points. The funny part is 124 uW/cm^2 is the figure I fist calculated too, but I thought that was too high so I re-worked my math. I guess I should have left well enough alone.
>
> Here is my original math:
>
> 1. ((13.4 kJ/m^2/3hr) / 3) = 4.47 kJ/m^2
>
> 2. (4.47 kJ/m^2) x 1,000 = 4,467 joule/m^2
>
> 3. 4,467 joule/m^2 = 1.24 watt/m^2
>
> 4. 1.24 watt/m^2 x 1,000,000 = 1,240,000 uW/m^2
>
> 5. (1,240,000 uW/m^2) / 10,000 = 124 uW/cm^2
>
>
>
> RE: source of UV-b irradiance in the study:
>
> UV-b irradiance came from a filtered Westinghouse FS-40 sunlamp (i.e., as described in the work of Mirecki and Teramura, 1984). Lamps were fitted with either presolarized 0.08 mm or 0.13 mm cellulose acetate (UV-b transmitting) or 0.13 mm Mylar Type A (as control filters).
>
>
>
> RE: absolute uW:
>
> The spectroradiometer used to quantitate the UV-b irradiance was an Optronic Laboratories, Inc. Model 742 spectroratiometer interfaced with a Hewlett Packard 85 printing calculator (it's an older study, circa 1986). The spectroradiometer was weighted with a generalized plant action spectrum (UV-bbe, Calwell, 1971), and normalized at 300 nm.
>
> I would like the results to be as accurate as possible but I am happy with 'horse shoes and hand grenades'. If I am able to confidently provide 100-150 uW/cm^2 for 3 hours I would be happy, I don't need to be very accurate, a range is fine with me (and more attainable over a larger canopy than a single figure).
>
>
>
> I thought you may find this interesting: in the study the reseradhers used a daily Uv-b irradiance (13.4 kJ/m^2/3hr) equivalent to the daily weighted UV-b irradiance under clear sky conditions and minium solar zenith in Columbia, South America at 0' latitude, 3 km elvention, June, 1980 (Green, et al., 1980)
>
>
> Thanks so much for your time!
>
>
> --- In UVB_Meter_Owners@yahoogroups.com, Andy B <sprocket@> wrote:
> >
> > I think something got a bit twisted in there. That's not quite what I get.
> > Watts already have time factored in them. If you
> > divide by time you get Joules (total energy) and
> > then time is irrelevant (has been removed - so
> > you are left with just an amount of energy). So I
> > don't think you can combine W and time that way
> > around. Sure enough the number that pops out
> > there (0.34 uW/cm^2) is not really doable either
> > - it's a very tiny amount of UV-B that in
> > practice would be difficult to measure.
> >
> > I see this:-
> >
> > (13.4kJ/m2/3hr) / 3 => 4470 J/m2/hr
> >
> > 4470 J/m2/hr / (60*60) => 1.24 J/m2/sec (ie. 1.24 W/m2)
> >
> > 1.24 W/m2 / (100x100) => 1.24e-4 W/cm2 (ie. 124 uW/cm2)
> >
> > So that would be 124 uW/cm2 (as a 6.2 meter
> > reading), still taking 3 hours. That's a
> > perfectly doable level of UV-B using a decent
> > lamp and adjusting distance; but you could apply
> > the same total energy as 248uW/cm2 for 1.5 hours; or 62 uW/cm2 for 6 hours.
> >
> > One small point (and Steve may have more to say
> > if he is around as it has to do with how the
> > Solarmeter is calibrated for absolute uW -
> > whether that is with a particular irradiance
> > spectra), the Solarmeter, like any broadband
> > meter, has a response curve that is not flat so
> > you can only take that number as a rough guide.
> > If you are using sunlight and the experiment you
> > are replicating also used sunlight then it may be
> > safe enough (Steve?) but if the replication is
> > with a lamp that has a different spectral output
> > (eg. like the UV-B spike you get from a lamp)
> > then you only have a rough guide. How precise do you need to be?
> >
> > - Andy
> >
> >
> > At 15:42 03/03/2010, you wrote:
> > >Hello,
> > >
> > >Thanks for the help. What you wrote and what I
> > >calculated are the same (pretty much). I forgot
> > >to write the figure of 13 kJ/m^2 was the total
> > >UV-b irradiance over a 3 hour period each day
> > >(oops!). This is on the topic of growing plants,
> > >not herping. In a great study I am replicating
> > >the researchers used a totoal UV-b irradiance of
> > >13.4 kJ/m^2 each day (in the time frame of 3
> > >hours a day); thus I want to convert 13.4
> > >kJ/m^2/3hr into uW/cm^2/sec so I can use the
> > >Uv-b meter SolarTech 6.2 (as it measures per second AFAIK).
> > >
> > >Here is my math, would you mind checking it for me? thanks a ton!
> > >
> > >1. ((13.4 kJ/m^/3hr) / 3) = 4.47 kJ/m^2
> > >
> > >2. (4.47 kJ/m^2) x 1,000 = 4,470 joule/m^2
> > >
> > >3. 4,470 joule/m^2 = 1.24 watt/m^2/hr
> > >
> > >4. (1.24 watt/m^2/hr) x 1,000,000 = 1,240,000 uW/m^2/hr
> > >
> > >5. (1,240,000 uW/m^2/hr) / 10,000 = 124 uW/cm^2/hr
> > >
> > >6. (124 uW/cm^/hr) / 3,600 = 0.34 uW/cm^2/sec
> > >(as a single measurement from a UV-b meter like the SolarTech 6.2)
> > >
> > >Thus:
> > >
> > >((0.34 uW/cm^2/sec) x 3,600) x 3 = 367
> > >uW/cm^2/3hr (the total irradiance of UV-b to
> > >reach per day equaling about 13.4 kJ/m^2/3hr)
> > >
> > >So:
> > >
> > >If my calculations are correct 13.4 kJ/m^2/3hr =
> > >0.34 uW/cm^2/sec over a 3 hour period. Or, I
> > >could use about 0.1 uW/cm^2/sec for a 1 hour
> > >period to reach an equivalent of 13.4 kJ/m^2/3hr.
> > >
> > >Thanks!
> > >
> > >--- In
> > ><mailto:UVB_Meter_Owners%40yahoogroups.com>UVB_Meter_Owners@yahoogroups.com,
> > >Andy B <sprocket@> wrote:
> > > >
> > > > kJ (or just J) is units for an amount of energy.
> > > > uW (or just W) is an energy rate ie. J per
> > > unit time (in this case, 1 second)
> > > >
> > > > The 13kJ/m2 figure does not include time. eg. it
> > > > could be 13kJ over 1 second... or 14 days.
> > > > If there is an implied one-second duration in
> > > > that then it is 13kW (since a joule is a Watt.second).
> > > >
> > > > So, if that (one second) is the case then you are
> > > > looking at 13kW/m2 and you only have to correct
> > > > for the area (m2 to cm2). To do that just divide by 10,000 (ie 100x100cm).
> > > >
> > > > So 13kW/m2 is 13000/10000 = 1.3W/cm2
> > > >
> > > > That is a *huge* amount of UV-B. Maybe your time interval is not 1 second?
> > > > In industrial (ie. large scale) thermal energy
> > > > calculations it is common practice to consider
> > > > power in kW/hour. If that is the case here then
> > > > you would have to divide by a further factor of
> > > > 3600 (ie. 60 * 60 to convert hours to seconds).
> > > > That would give you 3.61e-4 W/cm2 (ie. 361 microwatts/cm2).
> > > >
> > > > That number sounds more likely/believable but we
> > > > are really doing this backwards here so it would
> > > > be best if you can look again at your original
> > > > data and confirm where the time interval
> > > comes into it. Then we can be sure.
> > > >
> > > > Is that any help?
> > > > Cheers
> > > > - Andy
> > > >
> > > >
> > > > At 20:49 02/03/2010, you wrote:
> > > > >
> > > > >
> > > > >Hello,
> > > > >
> > > > >I have data I would like to replicte but it
> > > > >calls for UV-b irradiance of 13 kJ/m^2. I have
> > > > >tried a few ways I can think of and I don't
> > > > >think I have the correct answer yet.
> > > > >
> > > > >Thank you
> > > > >
> > > > >
> > > > >
> > > > >
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> > >
> > >
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