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

[José Antonio Ruiz Arias]

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

> 

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