[GRASS-user] Clarification on units used in r.sun
Dylan Beaudette
dylan.beaudette at gmail.com
Mon Dec 11 15:53:10 EST 2006
Thanks for the clarification Jose!
Cheers,
Dylan
On Friday 01 December 2006 13:59, jararias at ujaen.es wrote:
> Hi again Dylan,
>
> you are using the same sensors than me!!. Just let me a little tricky
> suggestion: when you say that the LICOR is returning hourly integrated
> data that's not true at all because the sensor gives you the mean flux
> over the whole hour, i.e. it counts all the energy incoming on it during
> the hour. Then, it returns the equivalent average flux [W/m2]. AFTER THE
> MEASURE, you integrate in time the energy flux to obtain the hourly energy
> per squared meter. Remember that integration is a sum and, in this case,
> like multiply by the measurement period (1 hour).
>
> I really hope not to mislead you
>
> Cheers
>
> Jose
>
> > Dylan,
> >
> > And you might add for completeness:
> >
> > daily summed *energy* [Wh/m2] = *energy* [J/m2] (the total amount of
> > incoming radiation/energy for the 24-hour period).
> >
> > Tom
> >
> > Dylan Beaudette wrote:
> >> Hi Jose,
> >>
> >> This makes sense. Thanks again for the sage advice!
> >>
> >> >From what I have read about our sensor (it is a LiCor LI-200SZ
> >>
> >> pyranometer) It
> >> is returning hourly integrated data. Thus:
> >>
> >> irradiance [Wm^2] * hour_1 + irradiance [Wm^2] * hour_2 + ... irradiance
> >> [Wm^2] * hour_24 = daily summed *energy* [Wh/m^2] .
> >>
> >> This should now be comparible to the output of r.sun (mode 2) i think.
> >>
> >> Cheers,
> >>
> >> Dylan
> >>
> >> On Friday 01 December 2006 00:14, José Antonio Ruiz Arias wrote:
> >>> Hi Dylan,
> >>>
> >>>
> >>>
> >>> Probably you have cleared your doubt but I'll try to expound how I see
> >>> the
> >>> question. The most important thing is you firstly have to think if you
> >>> are
> >>> dealing with flux or energy.
> >>>
> >>>
> >>>
> >>> The flux is something (let say, particles or light photons) coming
> >>> through
> >>> a normal surface to the beam (for example, persons through a door).
> >>> Therefore the unit of flux has to be something like particles per
> >>> squared
> >>> meter and per second (how many particles have reached the normal
> >>> surface in
> >>> a second). In case of light (photons) we talk in terms of energy to
> >>> count
> >>> the particles, so we have J/m2*s = W/m2 (because W = J/s). Then, when
> >>> you
> >>> measure the solar radiation with a sensor, the measurement is usually
> >>> in
> >>> units of flux [W/m2]. What does it mean? Let suppose in a given moment
> >>> the
> >>> sensor reads 500 W/m2. That equals to 500 J/s/m2 = 500 J/m2/s, i.e. in
> >>> a
> >>> second you have received 500 Joules per squared meter so, in 2 seconds,
> >>> you
> >>> will have 1000 joules in a squared meter. In 3 seconds you will have
> >>> 1500
> >>> joules in a squared meter and so on. Now we have sum (or integrated)
> >>> the
> >>> flux throw the time and the magnitude can be seen as energy per unit of
> >>> surface in a certain period of time, let say hour, day, month,
> >>>
> >>>
> >>>
> >>> Watts per hour is a measurement of energy used in engineering not in
> >>> science. The only reason I think is because the magnitude of its value
> >>> is
> >>> more appropriate than J or MJ. The equivalence is 1 Wh = (1 J/s)*3600 s
> >>> =
> >>> 3600 J = 0.0036 MJ
> >>>
> >>>
> >>>
> >>> I work with radiometric sensors and sometimes I have used r.sun. The
> >>> approach I follow is to express both in J/m2/dia. With r.sun the output
> >>> is
> >>> in Wh/m2/dia so directly multiply by 0.0036 and you will have MJ/m2/dia
> >>> and
> >>> will know how much energy you have in a squared meter when the day
> >>> finish.
> >>> Now let suppose you are getting a measurement every 10 minutes, i.e.
> >>> every
> >>> 600 seconds. In that lapse of time you will have S[W/m2]*600seconds,
> >>> where
> >>> S is the measurement. In a day you will have Sday[J/m2/dia] = S1*600s +
> >>> S2*600s + S3*600s +
and so on for all measurements in the day.
> >>>
> >>>
> >>>
> >>> ________________________________
> >>>
> >>>
> >>>
> >>> José A. Ruiz Arias
> >>>
> >>> Departamento de Física
> >>>
> >>> Escuela Politécnica Superior
> >>>
> >>> Edificio A-3, Campus Lagunillas
> >>>
> >>> Universidad de Jaén
> >>>
> >>> 23071 Jaén Spain
> >>>
> >>> Tlf. +34 953212474
> >>>
> >>> Email: jararias at ujaen.es
> >>>
> >>> _____________________________________
> >>>
> >>>> -----Mensaje original-----
> >>>>
> >>>> De: grassuser-bounces at grass.itc.it
> >>>> [mailto:grassuser-bounces at grass.itc.it]
> >>>>
> >>>> En nombre de Dylan Beaudette
> >>>>
> >>>> Enviado el: jueves, 30 de noviembre de 2006 22:55
> >>>>
> >>>> Para: Glynn Clements
> >>>>
> >>>> CC: grassuser at grass.itc.it; GRASS devel
> >>>>
> >>>> Asunto: Re: [GRASS-user] Clarification on units used in r.sun
> >>>>
> >>>> On Wednesday 29 November 2006 12:31, Glynn Clements wrote:
> >>>>> Dylan Beaudette wrote:
> >>>>>> Quick question on the units used for the ouput of r.sun in mode 2
> >>>>
> >>>> (daily
> >>>>
> >>>>>> sums):
> >>>>>>
> >>>>>>
> >>>>>>
> >>>>>> In the manual pages for r.sun, the following 'unit' is included in
> >>>>>> the
> >>>>>>
> >>>>>> description:
> >>>>>>
> >>>>>> ----------------------------
> >>>>>>
> >>>>>> The solar radiation maps for given day are computed integrating the
> >>>>>>
> >>>>>> relevant irradiance between sunrise and sunset times for given day.
> >>>>
> >>>> The
> >>>>
> >>>>>> user can set finer or coarser time step step used for all-day
> >>>>
> >>>> radiation
> >>>>
> >>>>>> calculations. A default value of step is 0.5 hour. Larger steps
> >>>>>> (e.g.
> >>>>>>
> >>>>>> 1.0-2.0) can speed-up calculations but produce less reliable
> >>>>>> results.
> >>>>
> >>>> The
> >>>>
> >>>>>> output units are in Wh per squared meter per given day
> >>>>>> [Wh/(m*m)/day].
> >>>>>>
> >>>>>> -------------------------
> >>>>>>
> >>>>>>
> >>>>>>
> >>>>>> Is one to interpret this as "watt-hour per square meter per day" ?
> >>>>>>
> >>>>>> This would seem a little odd, as the the unit 'watt-hour' is not an
> >>>>>> SI
> >>>>>>
> >>>>>> unit.
> >>>>>
> >>>>> More odd (to me) is having (different) units of time in both the
> >>>>>
> >>>>> numerator and denominator. I would have thought it more logical to
> >>>>>
> >>>>> divide the result by 24 to give Watts per square metre [W/(m^2)].
> >>>>
> >>>> I need to check with some local experts, but judging from some recent
> >>>>
> >>>> tests -
> >>>>
> >>>> the output from r.sun is comparible to that from a weather station,
> >>>> with
> >>>>
> >>>> *hourly* averaged data in W/(m^2) which i think is analogous to the
> >>>>
> >>>> Wh/(m^2)
> >>>>
> >>>> units that r.sun uses.
> >>>>
> >>>>
> >>>>
> >>>> --
> >>>>
> >>>> Dylan Beaudette
> >>>>
> >>>> Soils and Biogeochemistry Graduate Group
> >>>>
> >>>> University of California at Davis
> >>>>
> >>>> 530.754.7341
> >>>>
> >>>>
> >>>>
> >>>> _______________________________________________
> >>>>
> >>>> grassuser mailing list
> >>>>
> >>>> grassuser at grass.itc.it
> >>>>
> >>>> http://grass.itc.it/mailman/listinfo/grassuser
> >>>>
> >>>>
> >>>>
> >>>> __________ Información de NOD32 1894 (20061130) __________
> >>>>
> >>>>
> >>>>
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> >>>>
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> >
> > --
> > Thomas E Adams
> > National Weather Service
> > Ohio River Forecast Center
> > 1901 South State Route 134
> > Wilmington, OH 45177
> >
> > EMAIL: thomas.adams at noaa.gov
> >
> > VOICE: 937-383-0528
> > FAX: 937-383-0033
> >
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--
Dylan Beaudette
Soils and Biogeochemistry Graduate Group
University of California at Davis
530.754.7341
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