how to convert radiation on horizontal plane to an inclined one

[J.P.M.]

Mr. J.P.M.

I an a newbie to solar here, just try to learn something here.
I admitted I might be out of league on this one.
Just feel free to tell me which phase I said is wrong and I would be happy to dig out more info.

I just try to answer the OP's question directly for what he wants to do.
I don't care and don't know whether he is supposed to do what he wants to do.
I know there are a lot of other factors to consider, not just the beam component.
But that's what the OP asked for:
I have data for the hourly radiation on a horizontal plane and weather conditions for a location for a few years. I need to convert the radiation on the horizontal plane to the radiation on an inclined plane.
And the answer to his question would be the formula I posted.

In spite of what you may think, the answer to the OP's question is not to be found in the formula you posted.

This subject has been addressed in a very thorough manner in other posts to this forum.

Read them, study them, and as a newbie, learn. I can point you in the direction of self help. I'm a poor teacher and a worse spoon feeder. Both waste my and most everyone else's time and good will.

[myfriendSUN]

My friendSUN:

No offense, but you're out of your league on this one and demonstrating that a little knowledge can be a dangerous thing.

I'd respectfully suggest you check out some of the references provided in this post for solar position, incident angle information and info on the various components of solar radiation that pass through the earth's atmosphere. Also, you seem to have gotten off track that the goal is solar radiation on a surface, as well as that idea that the total POA rad. is more than simply the beam component.

Mr. J.P.M.

I an a newbie to solar here, just try to learn something here.
I admitted I might be out of league on this one.
Just feel free to tell me which phase I said is wrong and I would be happy to dig out more info.

I just try to answer the OP's question directly for what he wants to do.
I don't care and don't know whether he is supposed to do what he wants to do.
I know there are a lot of other factors to consider, not just the beam component.
But that's what the OP asked for:
I have data for the hourly radiation on a horizontal plane and weather conditions for a location for a few years. I need to convert the radiation on the horizontal plane to the radiation on an inclined plane.
And the answer to his question would be the formula I posted.

[J.P.M.]

My friendSUN:

No offense, but you're out of your league on this one and demonstrating that a little knowledge can be a dangerous thing.

I'd respectfully suggest you check out some of the references provided in this post for solar position, incident angle information and info on the various components of solar radiation that pass through the earth's atmosphere. Also, you seem to have gotten off track that the goal is solar radiation on a surface, as well as that idea that the total POA rad. is more than simply the beam component.

[myfriendSUN]

sun angle alpha (at solar noon) = 90 - latitude + declination

declination depends on date of the year

Can we measure the sun angle at any incident or at each hour ???

http://pveducation.org/pvcdrom/prope...-high-accuracy

May be use the sun's position calculator to find the sun angle at any incident.

[myfriendSUN]

Those equations provide trig relationships that are only valid at solar noon each day, and assume all irradiation is direct normal, in other words, directly from the disc of the sun. They do very little to help with the more general solution that the OP appeared to want, using data that a typical weather station would provide.

For the OP to get what he want he needs to have the sun angle at each hour of the day and plug it in the formula.
The calculator from the link only works at solar noon each day.
The formula is true ALL the time.

[sensij]

I think this is the one you are looking for.
The formula is TRUE for any incident as far as you have the incident sun angle.

You are going to need to have the hourly sun angle "alpha" to do the calculation.

Those equations provide trig relationships that are only valid at solar noon each day, and assume all irradiation is direct normal, in other words, directly from the disc of the sun. They do very little to help with the more general solution that the OP appeared to want, using data that a typical weather station would provide.

[myfriendSUN]

Should I just use the formula given at http://pveducation.org/pvcdrom/prope...tilted-surface for Smodule? I am asking because it doesn't seem to take into account the time of the date and I thought that it would be necessary.

I think this is the one you are looking for.
The formula is TRUE for any incident as far as you have the incident sun angle.

You are going to need to have the hourly sun angle "alpha" to do the calculation.

[myfriendSUN]

My question is how I go from the direct radiation on a horizontal plane to the beam irradiation on a plane looking directly to the sun, in order to multiply after by cos(theta) to get to the inclined.
Or is there a -- trigonometric -- relation to transform directly the beam radiation falling on the horizontal surface to the beam radiation falling on a tilted surface?

Is this what you are looking for???

S(module) = S(horizontal)sin (alpha + beta) / sin alpha

where : alpha = incident angle of the sun, which change according to the time of the day
beta = tilt angle of panel

http://pveducation.org/pvcdrom/prope...tilted-surface

BUT you are going to need to have the hourly sun angle "alpha" to do the calculation

[J.P.M.]

JMHO - but you're likely taking the more complex route.
The simpler route is to use a tool like PVwatts.nrel.gov to get an estimate of the
kWh/m^2/day for each month (as well as kwh for the size PV system you plug in)

More JMHO: It seemed to me that he OP was looking for a way to calculate as much or more than just the bottom line result.

[J.P.M.]

OP:

Read/study what's been provided on this thread. All you are looking for is here. Honest. I believe I understand what you think you want. I've spent some parts of the better part of 40 or so years studying and using what you're dealing with.

Unfortunately, the subject is not quite as amenable to reduction to as simple a thing as you seem to want/need.

What Sensij provided is probably as good as anything for most applications. I'd respectfully suggest using it. For further explanation/backgroung see Duffie & Beckman, which also includes similar, but slightly different approaches to some aspects along with some discussion of the methods.

[foo1bar]

I want to transform it to the radiation falling on an inclined surface.

JMHO - but you're likely taking the more complex route.
The simpler route is to use a tool like PVwatts.nrel.gov to get an estimate of the
kWh/m^2/day for each month (as well as kwh for the size PV system you plug in)

[inetdog]

Good afternoon!
Thank you both for your answers! They were really helpful!

Sensij I don't want to calculate the horizontal POA irradiance. I already have it as a measurement from the meteorological station. I want to transform it to the radiation falling on an inclined surface. So what I want is to understand which trigonometric function I should use in order to transform the beam irradiation falling on the horizontal to the radiation falling on the inclined. So, I already have the function to calculate the cos(theta) but this only takes me half way, from -the direct radiation on a surface looking directly to the sun- to -the radiation falling on the inclined surface-. My question is how I go from the direct radiation on a horizontal plane to the beam irradiation on a plane looking directly to the sun, in order to multiply after by cos(theta) to get to the inclined.
Or is there a -- trigonometric -- relation to transform directly the beam radiation falling on the horizontal surface to the beam radiation falling on a tilted surface?

IF you have a device which takes incident solar radiation and converts some fraction of it to usable energy AND whose efficiency does not depend in any way on the angle at which the radiation is hitting the surface, then the calculation is very simple geometry:

If solar radiation gives you 1000 w/m₂ of energy where the area is the area perpendicular to the radiation direction (directly facing the sun), then all you need to do for an inclined panel is calculate how much area that panel covers in the line of the suns rays.
Look at it this way: The inclined panel will cast a shadow which shows how much of the sun's rays are being blocked. Consider a perpendicular panel that casts the exact same shadow. That will block (capture) the same amount of radiation.
The area of that panel is the area of the inclined panel time the cosine of the angle between the two panels.
If the inclined panel is fixed relative to the earth, that angle will be changing with time of day. There is no easy formula for that if all you have been given is the total incoming radiation over the course of the whole day.

The first complication is that a real solar collector (PV or thermal) may not work as well at absorbing the incident solar radiation when it comes in at other than a perpendicular. That will be a second correction factor which we have not taken into account yet.

The second complication, which is more troublesome, is that you have referred alternately to a horizontal plane and to a plane which is always perpendicular to the sun's rays.
Which are you actually talking about as the source of the measurements you are trying to convert??
If it is really a horizontal plane, then an inclined plane which is closer to perpendicular to the sun will actually intercept MORE sunlight that the horizontal plane. The amount by which it intercepts more energy will be cos(I)/cos(H) where I is the angle to the sun of the inclined plane and H is the angle of the sun to the horizontal.
For the special case of I = 0, the intercepted energy will be 1/cos(H) times the intercepted energy of the same area horizontal plane (e.g. flat roof).
But this factor will also change over the course of the day as the sun rises and sets.

[sensij]

What you want to do is not a simple trig relationship. You have measured GHI (probably). You will need an estimate of diffuse fraction to get the number you want. You will need a model to get from one to the other. Sorry. Read what J.P.M. wrote a couple times. If you aren't following, spend some more time with the textbooks until you do.

DISC model overview
DISC model spreadsheet

Once you have estimated the diffuse fraction you can use calculations for each component to get the sum of POA irradiance. You can just choose the beam component and focus on that, if you prefer.

POA calculation summary

Edit... if you are saying that your measurement is of DNI on a horizontal surface, then just skip to the POA calculation, and focus only on Eb. If calculating the angle of incidence is where you are stuck, try the link below. However, this would not be the typical approach. Most weather stations will measure and report only GHI, not DNI.

AOI calculation

[dimitrisbronowski]

If you really want to calculate horizontal POA irradiance, and then back to inclined array POA, the next section of the site I linked shows how to do it. Gory math that way.

Good afternoon!
Thank you both for your answers! They were really helpful!

Sensij I don't want to calculate the horizontal POA irradiance. I already have it as a measurement from the meteorological station. I want to transform it to the radiation falling on an inclined surface. So what I want is to understand which trigonometric function I should use in order to transform the beam irradiation falling on the horizontal to the radiation falling on the inclined. So, I already have the function to calculate the cos(theta) but this only takes me half way, from -the direct radiation on a surface looking directly to the sun- to -the radiation falling on the inclined surface-. My question is how I go from the direct radiation on a horizontal plane to the beam irradiation on a plane looking directly to the sun, in order to multiply after by cos(theta) to get to the inclined.
Or is there a -- trigonometric -- relation to transform directly the beam radiation falling on the horizontal surface to the beam radiation falling on a tilted surface?

[J.P.M.]

What you are looking for can be as simple or as complicated as your needs dictate. Most any decent solar energy text has info that can be summarized on one page. There are also many web sites that will do the calcs in a probably adequate way, but some info will need to be supplied to get meaningful #'s suitable for system output estimates. Know what is needed for what you want to get. GIGO.

Briefly, the solar radiation falling on a surface usually has 3 components: Direct beam, diffuse, and reflected from the surroundings or intentionally augmented with reflectors/mirrors.

The raw intensity of the total solar radiation at a location is a function of the altitude above sea level for that location, the path length of the radiation through the atmosphere (as f(zenith angle), the atmospheric conditions of dust, water vapor, and other junk, etc., the cloudiness at the time the intensity is sought and a few other things. It will vary from zero at night to some level of approx. 1,000 Watts/m^2 or maybe a bit more, but almost always less than that max. 1,000 Watts/m^2 measured perpendicular to the solar disk.

The direct beam component of the radiation received on a surface of arbitrary orientation is the direct normal component of the sunlight (what casts a shadow), projected onto the plane of the receiving surface, using the relation: Beam radiation on a surface of arbitrary orientation = Direct normal X Cos (angle of incidence). The angle of incidence is a bit of a complicated function of the relation between the orientation of the receiver (The "tilt" relative to the horizontal, and the azimuth of the receiver - the direction, usually relative to true north), and the solar position. The solar position is a function of the latitude and longitude of the location, the day of the year and time of day, usually by the hour, at the midpoint of that hour. There are many website that will give solar position using the described logic.

The diffuse component is the radiation that gets to the receiver after being broken up, waylaid and bounced around some (diffused) by the atmosphere. It may be all of the received radiation (as in under completely cloudy skies) or ~~ SOMEWHERE between 10 and 20% of the total reaching the ground under "clear" skies, or something in between for hazy, dirty, or partly cloudy skies. There are many, many different correlations for determining how much of the radiation on a receiver is diffuse, and/or/even its dominant direction - most all of them an estimate/ educated guess. Often, a good portion of the diffuse is considered, or calc'd, from those correlations (most all of them being quite empirical) as coming from the general direction of the solar disk. A less significant portion may be due to something called "horizon brightening".

The reflected component is composed of both direct and diffuse components and is very dependent on what's around the receiver, the reflective prop. of each reflecting surface, and that reflector's orientation, as well as the geometrical relationship to the receiver. Usually, but not always, unless part of the design as in booster mirrors, etc., the reflected component adds a few % to the total at most, more in snowy climates or very light colored surroundings, but most always less than 20 % under all but the most favorable conditions.

Perhaps more than you wanted to know, but, unfortunately or not, probably not as simple as most want to make it. The unfortunate part is, some complexity is necessary. The fortunate part is that it's not that sophisticated a concept, as much as it is involved, requiring some digging and time to understand what's needed for what is still going to be a dart throw given the uncertainties associated with the variables and the atmosphere.

Don't shot me, I'm only the piano player.

Take what you want of the above. Scrap the rest.