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```					              Solar Radiation
• The sun is a gaseous body composed mostly of hydrogen

• Gravity causes intense pressure and heat at the core initiating
nuclear fusing reactions
• This means that atoms of lighter elements are combined into
atoms of heavier elements, which releases enormous quantities
of energy
• Even when planet Earth is 93 million miles away, we still
received an amazing quantity of usable energy from the sun.
• Considering 25% efficient PV modules, if we used 1% of the
surface of the earth we could meet 29 times our current total
energy demand –These some rough calculations I did, but I’ll be
different.
• Solar irradiance is the intensity of solar power, usually
expressed in Watts per square meter [W/m^2]
• PV modules output is rated based on Peak Sun (1000 W/m^2).
• Since the proportion of input/output holds pretty much linearly
for any given PV efficiency, we can very easily evaluate a system
performance check by measuring irradiance and the PV module
output.
of the square of the distance from the source –that is, twice the
distance ¼ of the energy, four times the distance 1/16 and so on
time. It is measured in Watt-hours per square meter [Wh/m^2]
• Solar Spectrum
most the energy
is electromagnetic
waves.

• Electromagnetic
Spectrum is the range
of all types of
electromagnetic
wavelength.
•    Atmospheric Effects: Solar radiation is absorbed,
scattered and reflected by components of the
atmosphere
•    The amount of radiation reaching the earth is less
than what entered the top of the atmosphere. We
classify it in two categories:
reaches the earth without scattering
atmosphere and clouds
• Air Mass represents how much atmosphere the solar
radiation has to pass through before reaching the Earth’s
surface
• Air Mass (AM) equals 1.0 when the sun is directly overhead at
sea level. AM = 1/ Cos Өz
• We are specifically concerned with terrestrial solar radiation –
that is, the solar radiation reaching the surface of the earth.
• At high altitudes or in a very clear days, Peak Sun may be
more than 1000 W/m^2 but it is a practical value for most
locations
• Peak Sun Hours is the number of hours required for a day’s
total radiation to accumulate at peak sun condition.
• Zenith is the point in the sky directly overhead a particular location –as
the Zenith angle Өz increases, the sun approaches the horizon.
AM = 1/ Cos Өz
•
• Example problem of Peak sun hours per day:
If during the day we have 4 hours at 500 Wh/m^2 and 6 hours at 250 Wh/m^2 we should
compute the peak sun hours per day as follow:
First, multiply 4hs x 500 W/m^2 and add to it 6hs x 250 W/m^2 – This will equal 3500
Wh/m^2
Second, we know that by definition Peak Sun is 1000 W/m^2, so if we divide the total
irradiation for the day by Peak Sun we will obtain Peak Sun hours. – That is,
Peak Sun Hours = Total Irradiation [Wh/m^2] / Peak Sun [W/m^2] = Peak Sun hours
In our specific problem:
Peak Sun Hours = 3500 Wh/m^2 / 1000 W/m^2 = 3.5 Peak Sun hours

• Note: most solar irradiation data is presented in Peak Sun Hours units
• Insolation; this is an equivalent term for solar irradiation and can be
expressed in KWh/m^2/day or peak sun hours

• Solar spectral distribution is important to
understanding how the PV modules that we’re going to
utilize respond to it
• Most Silicon based PV devices respond only to
visible and the near infrared portions of the spectrum
• Thin film modules generally have a narrower
response range
• Long-term solar irradiation measurements are the basis for
developing databases, which help us to calculate output.
• Being able to predict the output of our PV system, and this will
allow us to know whether it is working adequately or not
• Predicting output will help us to calculate the cost of the energy
generated over a given time period
• Pyranometers measure irradiance. Typically, you will use a
handheld pyranometer that uses a silicon cell or photodiodes and
you will set it adjacent to the array, in the same plane as the array –
not as precise but appropriate for construction
• Pyrheliometers measure direct solar radiation (and ignore
diffuse) and I’ve never ran into a situation where I had to use one
•   Two major motions of Earth affect the apparent
path of the sun across the sky:
1. Its yearly revolution around the sun
2. Its daily rotation about its axis
• These motions are the basis for solar timescale and
the reason why we have seasons, days and nights
• Ecliptic Plane is the plane of Earth’s orbit around
the sun
• Equatorial Plane is the plane containing Earth’s
equator and extending outward into space
• Solar Declination is the angle between the equatorial plane and the ecliptic plane
• The solar declination angle varies with the season of the year, and ranges
between –23.5º and +23.5º
Summer Solstice is at
maximum solar declination
(+23.5º) and occurs around
June 21st –Sun is at Zenith at
solar noon at locations 23.5º N
latitude
Winter Solstice is at minimum
solar declination (-23.5º) and
occurs around December 21st
At any location in the Northern
Hemisphere, the sun is 47º
lower in the sky at noon on
winter solstice than on the
summer solstice – Days are
significantly shorter than nights
Equinoxes occur when the
solar declination is zero.
Spring equinox is around
March 21st and the fall
equinox occurs around
September 21st –Sun is at
Zenith at solar noon on the
equator.
Around the equinoxes the
daily [rate of] change is at
maximum as oppose to
change of declination during
the solstices when it is at its
minimum
• Standard meridian is located at a multiple of 15º east
or west of zero longitude –located at Greenwich.
• 15º represents one hour of change, so we can infer
the 1º will be 4 minutes
• Both of these facts are interesting to know; however,
you will not apply either of these things to design or
install PV systems.
• You’re concerned with the average peak sun hours for
reading in solar time –but this irrelevant to you.

Solar Altitude Angle is the vertical angle between
the sun and the horizon –added to the Zenith angle is
equal to 90º
Azimuth Angle is the horizontal angle between a
reference direction.
In the solar industry we call south 180º and this
angle will range between 90º (east) and 270º (west)

• Solar Window is the area of sky between sun paths at summer solstice and winter
solstice for a particular location
Incidence Angle is the angle between the direction of
direct radiation and a line exactly perpendicular to the
array angle
• Array orientation is defined by two angles:

1. Tilt angle is the vertical angle between the horizontal and the
array surface
2. Array Azimuth Angle is the horizontal angle between a
reference direction –typically south- and the direction an
array surface faces
• Maximum energy gain will be achieved by orienting the array
surface at a tilt angle close to the value of the local latitude –In
high latitudes arrays should be very steep and vice versa
• For optimal performance the tilt angle should be adjusted from
the latitude angle by an amount equal to the average declination
during that time
• During the summer the average declination is +15º, so we
should have a tilt of latitude minus 15º to make the array
perpendicular to the average solar path –during the summer
• Array Azimuth angle will be optimal when that array is due south
• Sun trackers allow the PV array to change the tilt angle, the
azimuth angle, or both –generally is not considered cannot be
• Computer models and the average climate conditions are used to calculate
an optimal tilt angle factor –aka correction factor we have to subtract from
the latitude.
• In our area we use an optimal tilt angle factor of 15º
• By making our tilt angle equal to the latitude angle minus this angle factor, we
will improve the performance of our PV array
• Example for our area: If we look at dataset provided by NREL, we can see
that for a 0º tilt we would have 4.7 peak sun hours and for a latitude - 15º tilt
we would have 5.3 peak sun hours
In other words, we would have an output roughly 13% higher by using this
correction factor
You can calculate this percentage as follows:
(5.3 – 4.7) / 4.7 = .127 or 12.7%
With NREL dataset we can find the most convenient tilt for our system and use
the average peak sun hours of this tilt to calculate the annual production of our
system.
Annual energy production = Avg peak sun hours per year [hr/day] x 365 Days x
system size [Kw] = [KWh/year]
Example:
For a 4 KW system, located in San Francisco where we can expect 5.3 peak sun
hours per day the annual production of the system will be equal to 7738 KWh/yr
–(4KW x 5.3 peak sun hours per day x 365 days)
If we multiply this result by the number of years that we expect the system
to be producing energy and we divide the cost of the PV system by this
number, we will know how much it cost each KWh produced.
In our example: 7738 KWh/yr x 25 years = 193450 KWh . Let’s say that this
system cost \$26500 after rebate, then \$26500/ 193450 KWh =13.5 ¢/KWh
Typical Inclination/Orientation
Factors for Parallel Roof Mounted
PV Systems in California
ROOF PITCH
FACING    Flat   4:12    7:12 12:12   21:12   Vert.
South     0.89   0.97    1.00 0.97     0.89   0.58
SSE,SSW   0.89   0.97    0.99 0.96     0.88   0.59
SE, SW    0.89   0.95    0.96 0.93     0.85   0.60
ESE,WSW   0.89   0.92    0.91 0.87     0.79   0.57
E, W      0.89   0.88    0.84 0.78     0.70   0.52