Radiation definitions

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JavierOL
JavierOL Solar Expert Posts: 27
Hi!

I have a doubt about solar irradiance: I am measuring the global and diffuse irradiance in a horizontal surface with a pyranometer. To know the value of the direct irradiance is enough to do this:

Gbeam,h = Gglobal,h - Gdiff,h ??

I have seen in some articles that the direct irradiance has to be multiplied by the cosinus os the zenith angle... like this: Gglobal,h = Gbeam,h * cos(Oz) + Gdiff,h

And another question: the direct irradiance in the horizontal is the same that the direct normal Solar irradince??

Thank you in advance!

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  • BB.
    BB. Super Moderators, Administrators Posts: 33,497 admin
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    Re: Radiation definitions

    You have lots of "issues" with irradiance numbers...

    There are different units--But you can usually figure out the unit conversions to what makes sense for you conventions...

    MJ/Day/Year, kWH per square meter per day/year...

    And what we use here "hours of noon time equivalent sun" per day... Where 1 hour of sun = 1 kWH per Square Meter of sunlight (at ~noon) for an "average" standard temperature/pressure/etc. conditions.

    Next comes the cosine issues... I use PV Watts a lot--It will use ~20 years of data (for many locations) and give you a "typical" "real day" (picked from 20 years of samples on that "day") and can convert between fixed, 1 axis or 2 axis tracking. And the Flat plate can be at any angle/orientation with respect to the earth. So, the "plate" does matter how it is oriented and if fixed or not. You can play with PV Watts and see what happens for different regions. You will also find that some regions with clear morning and afternoon thunderstorms generate more power with southeast facing arrays (and cooler morning vs hot afternoons such as Florida). Others may have morning marine layers that burn off late in the mornings so that south west facing arrays produce more energy (my region).

    I have not looked, but I believe that the PV Watts algorithms are "open sourced" (for the most part) and you can search through the NREL.gov site and find them (I looked many years ago and found them at that time).

    Then lastly, there is the weather--PV Watts is based on actual measurements on the ground. So weather, visibility, humidity, etc. are all taken into account for the "final number". PV Watts can also output hourly data so you can see the details if you wish.

    And, while the PV Watts data may be over 50 years old in some cases (and probably not "recent" in any case), there are now satellite surveys available. At least that gives us more data vs the fixed locations per US State we have with PV Watts.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • JavierOL
    JavierOL Solar Expert Posts: 27
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    Re: Radiation definitions

    Hi Bill, I think you've answered the wrong question...
  • NorthGuy
    NorthGuy Solar Expert Posts: 1,913 ✭✭
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    Re: Radiation definitions
    JavierOL wrote: »
    Hi Bill, I think you've answered the wrong question...

    Your question is about words. There's a substance behind words. When you understand the substance, you can easily put your words in place. That's what I would do if I were you.

    Like Bill said, find the algotithm that PVWatts uses (or similar) and try to understand it.
  • BB.
    BB. Super Moderators, Administrators Posts: 33,497 admin
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    Re: Radiation definitions

    Do you want to ask the question again? "Normal" to plate usually means 90 degrees to the plate (normal is a right angle in math/engineering/science)... So A fixed plate would have a different kWH per square meter vs a two axis tracking plate...

    If you are measuring "global" in the sense of all radiation from the sky (that is measurable as heat vs UV and other radiation bands) vs that that is direct from the sun?

    There is reflection/refraction issues that will affect absolute accuracy from diffuse radiation measurements. A "black" plate pyrometer will have different reflective behavior vs a glass topped solar panel (if that is what you are comparing against).

    And "sky shine" can be quite dramatically different if there are clouds/dust/etc. in the air (obviously).

    Are you measuring the energy for a specific purpose? Here is a nice website that talks about some of the issues:

    http://pveducation.org/pvcdrom/properties-of-sunlight/measurement-of-solar-radiation

    I am sure you know more about the process than I--For me, I am trying to understand what the data is to be used for (biological/plant, solar panels tracking vs fixed, solar thermal, aging of paints, etc.). Knowing what the data is being used for, generally, helps (for me anyways) to define how to measure/collect the data in the first place.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • JavierOL
    JavierOL Solar Expert Posts: 27
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    Re: Radiation definitions
    BB. wrote: »
    Do you want to ask the question again? "Normal" to plate usually means 90 degrees to the plate (normal is a right angle in math/engineering/science)... So A fixed plate would have a different kWH per square meter vs a two axis tracking plate...

    If you are measuring "global" in the sense of all radiation from the sky (that is measurable as heat vs UV and other radiation bands) vs that that is direct from the sun?

    There is reflection/refraction issues that will affect absolute accuracy from diffuse radiation measurements. A "black" plate pyrometer will have different reflective behavior vs a glass topped solar panel (if that is what you are comparing against).

    And "sky shine" can be quite dramatically different if there are clouds/dust/etc. in the air (obviously).

    Are you measuring the energy for a specific purpose? Here is a nice website that talks about some of the issues:

    http://pveducation.org/pvcdrom/properties-of-sunlight/measurement-of-solar-radiation

    I am sure you know more about the process than I--For me, I am trying to understand what the data is to be used for (biological/plant, solar panels tracking vs fixed, solar thermal, aging of paints, etc.). Knowing what the data is being used for, generally, helps (for me anyways) to define how to measure/collect the data in the first place.

    -Bill


    Well, I think we have not understood...my question is simpler. I am measuring the global and the difuse irradiance that reach to the horizontal surface (in W/m2). To do this, I am using two pyranometers, one with a shadow ring to measure the diffuse.

    If I'm right, the global irradiance could be descomposed like: Gglobal=Gdirect+Gdiffuse.

    So, I need to know the value for the direct irradiance. What I do is substract the diffuse irradiance to the global, and the direct irradiance is as follows:

    Gdirect,h = Gglobal,h - Gdiffuse,h; *the h indicates the horizontal surface

    My doubt is if this way to calculate the direct irradiance is the right one, assuming that I know the values for the global and diffuse irradiance.

    Thank you
  • BB.
    BB. Super Moderators, Administrators Posts: 33,497 admin
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    Re: Radiation definitions

    You probably are off to a good start--But it would also depend on the acceptance angles of the diffuse sensor... It is capable of efficiently recording the 180 degree (horizon to horizon)--And is that what you are even interested in?

    Some (a few, many, all?) high end sensors are designed to track the sun--So that would seem to indicate that many sensors have a directional sensitivity component.

    Which gets back to the tool measuring the solar irradiation vs the "something" you will processing the collected data to simulate (or whatever).

    Shading the sun from the one sensor would need tracking (either moving the shade, or slewing the whole sensor+shade assembly).

    You could do a sanity check by using a black tube to only allow direct sunlight on the "sun sensor" and see if blocking the diffuse light gives you reasonably sane results when compared to Total=Diffuse+Direct.

    If you are doing this accurately--The sensor optics will be critical to the understanding of the measurements.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • NorthGuy
    NorthGuy Solar Expert Posts: 1,913 ✭✭
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    Re: Radiation definitions
    JavierOL wrote: »
    I am measuring the global and the difuse irradiance that reach to the horizontal surface (in W/m2). To do this, I am using two pyranometers.

    I don't think pyranometer is a right tool for that. Pyranometers are usually spherical and the results of measurements do not generally depend on the tilt. On the other hand, the surface is flat and the result will change if you tilt it.

    However, for the entities that you measure with the pyranometer, the formula is correct.
  • JavierOL
    JavierOL Solar Expert Posts: 27
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    Re: Radiation definitions

    Well, the point is that first I measure the irradiance in the horizontal, and then I use the mathematical algorithms to change from a horizontal to a tilted surface.

    Thanks for the clarification.
  • NorthGuy
    NorthGuy Solar Expert Posts: 1,913 ✭✭
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    Re: Radiation definitions
    JavierOL wrote: »
    Well, the point is that first I measure the irradiance in the horizontal, and then I use the mathematical algorithms to change from a horizontal to a tilted surface.

    Thanks for the clarification.

    No. You do not measure horizontal. You measure on a sphere. To measure horizontal, you would need a sensor with flat horizontal surface, such as a solar panel. There's no way to translate spherical measurements into horizontal with a formula. That is why pyronomer is not an adequate tool for this.
  • JavierOL
    JavierOL Solar Expert Posts: 27
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    Re: Radiation definitions

    Ok...so, Im using the pyranometer KIPP&ZONEN CMP11, and in the datasheet says: "is a high quality radiometer designed for measuring short-wave irradiance on a plane surface which results from the sum of direct and diffuse radiation".

    I thought that plane surface means horizontal...if not, how can I calculate the irradiance on the horizontal?
  • NorthGuy
    NorthGuy Solar Expert Posts: 1,913 ✭✭
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    Re: Radiation definitions
    JavierOL wrote: »
    Ok...so, Im using the pyranometer KIPP&ZONEN CMP11, and in the datasheet says: "is a high quality radiometer designed for measuring short-wave irradiance on a plane surface which results from the sum of direct and diffuse radiation".

    I thought that plane surface means horizontal...if not, how can I calculate the irradiance on the horizontal?

    May be it does measure horizontal. It's very easy to tell. If the sensor is flat it measures horizontal. Sorry if I misled you.

    From horizontal, you'll need quite a complex calculations to re-calculate it to some angled surface. For calculations, you need to know the exact position of the sun, which is changing all the time.
  • JavierOL
    JavierOL Solar Expert Posts: 27
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    Re: Radiation definitions

    Ok! I am already using some radiation models to calculate irradiance in tilted surface. But you know something about this way to descompose global irradiance:

    Gglobal,h = Gdirect,h * cos(Oz) + Gdiff,h ?? (Oz is the zenith angle)


    Thank you
  • NorthGuy
    NorthGuy Solar Expert Posts: 1,913 ✭✭
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    Re: Radiation definitions
    JavierOL wrote: »
    Ok! I am already using some radiation models to calculate irradiance in tilted surface. But you know something about this way to descompose global irradiance:

    Gglobal,h = Gdirect,h * cos(Oz) + Gdiff,h ?? (Oz is the zenith angle)

    That's what you would do if Gdirect was measured on a surphace that is perperndicular to the sun. If your pyranometer does indeed measures on horizontal surface then what you get is already "Gdirect * cos(Oz)".
  • JavierOL
    JavierOL Solar Expert Posts: 27
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    Re: Radiation definitions
    NorthGuy wrote: »
    That's what you would do if Gdirect was measured on a surphace that is perperndicular to the sun. If your pyranometer does indeed measures on horizontal surface then what you get is already "Gdirect * cos(Oz)".

    Ok! I understand thanks!!