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

Fri Dec 1 15:28:14 EST 2006

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
>>>
>>>
>>>
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>>>
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>
>   

-- 
Thomas E Adams
National Weather Service
Ohio River Forecast Center
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