[GRASS-user] Clarification on units used in r.sun

[Dylan Beaudette]

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) __________
> >
> >
> >
> > Este mensaje ha sido analizado con  NOD32 antivirus system
> >
> > http://www.nod32.com

-- 
Dylan Beaudette
Soils and Biogeochemistry Graduate Group
University of California at Davis
530.754.7341