Design, Install &Commission
grid connected photovoltaic
power systems
SESSION 2
SOLAR RADIATION
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Note that this is a
composite image, for
comparison of sizes.
The Earth is actually
150 million kilometers
(93 million miles) from
the Sun - which is why
the Sun looks rather
smaller to us (further
away) than it appears in
this picture.
J.P. Teasel version 06/12/10
Photo courtesy SOHO, the Solar and Helioscopic
Observatory
Design, Install &Commission grid connected photovoltaic power systems
Courtesy Johnson window film
Sunlight
•The intensity of sunlight is called irradiance, ( also known as insolation).&
for PV systems the units are watts per square meter (W/m2).
•A square meter is about 11 square feet. A typical, clear sky,
solar-noon value of irradiance falling on the surface of the
earth at sea level is 1,000 W/m2.
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
solar energy.
• 99% of Solar Energy is between 200 and
4000 nanometers in wavelength.
• almost all of solar energy is light and heat
• Depending on what wavelengths are agreed
upon as “visible light” (between 380 and 750
nanometers) about 44% of the sun’s energy
is in our visible light range.
• About 6-7% is in the ultra-violet (ultra =
beyond; "beyond"-violet) range. Most of this is
shielded by the ozone layer (which appears to
be slowly recovering).
• The rest (just over 50%) we call near infrared (infra = below; "below"-red). We perceive
infra-red radiation as heat. It is the trapping of
infra-red radiation going back out into space
which is the main cause of global warming
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Design, Install &Commission grid connected photovoltaic power systems
• about 1370 W/m2 of sun
power reaches the Earth’
atmosphere
• about 30% is reflected back to
space, mostly by clouds.
• about 70% is absorbed by
Earth's surface and
atmosphere and is then reradiated in the form of thermal
radiation.
• this interaction keeps the
Earth in radiative balance, as
required by basic physical
laws, and maintains the Earth's
average surface temperature at
about 16°C.
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Solar constant
• at Earth's distance from the sun, an average power of about 1370 W/m2 (1.366
kilowatts per square metre kW/m²) reaches the top of the Earth/atmosphere system.
• actual direct solar irradiance (measured by satellite) fluctuates by about 6.9%
during a year (from 1.412 kW/m² in early January to 1.321 kW/m² in early July)
due to the Earth's varying distance from the Sun
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Design, Install &Commission grid connected photovoltaic power systems
• in many areas, the time period that the irradiance exceeds 1,000 W/m2 can be
three hours or more. (This has an impact on the electrical design of the system)
peak value at
sea level
nominal
value
courtesy J Wiles http://www.nmsu.edu/~tdi/pdf-resources/cc101.pdf
• one of the standard test conditions (STC) value of irradiance (1,000
W/m2)is factors used to rate PV module and PV array output.
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
This image shows how different types of
solar radiation (x-rays to infrared radiation)
penetrate into the Earth's atmosphere. It is
this solar radiation that ionizes the upper
atmosphere, creating the ionosphere.
National Earth Science Teachers
Association (NESTA)
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
pyranometers (from the Greek "pyr" (fire) and "ano"
(sky).
The "pyranometer" is basically
a flat plate (covered with a
transparent dome) that is
coated with an extremely
absorptive surface. As the sun
strikes it, the surface gets hot.
The temperature of the surface
is measured with a thermopile,
giving an output voltage related
to the amount of solar radiation
striking the surface.
J.P. Teasel version 06/12/10
E Instruments International
http://www.e-inst.com/images/hvac/mainSolarRadiation
Design, Install &Commission grid connected photovoltaic power systems
parameter
Symbol
Quantity & unit
Irradiance
G
kW/m²
W/m²
Measuring device
Pyranometer (thermopile
or reference solar PV cell
Solar constant
Gsc
1367 W/m²
Peak value at sea level
Go
1.0 kW/m²
1000 W/m²
Nominal value
J.P. Teasel version 06/12/10
0.8 kW/m²
800 W/m²
Design, Install &Commission grid connected photovoltaic power systems
Geometric Effects
Latitude, time of day
and season all affect
the amount of energy
impinging on a surface
laid flat on the earth's
surface.
• imagine a flat plate collector having an area, A, of 1 m2 tilted at an angle, β, from the
horizontal and faces the sun perpendicularly.
• at a time, of say 0900 hours, there are 12 rays of the sun's beam coming from an
altitude angle, γ, (or incidence) that strikes the collector at this position.
• however, if the collector is laid horizontally on the Earth's surface, i.e. at angle β = 0
from the horizontal, the collector only captures 9 rays.
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Design, Install &Commission grid connected photovoltaic power systems
Max capture of solar radiation at solar noon
Collector A
receiving all 12
rays
• it then becomes apparent that if the sun is overhead and the
collector is laid flat on the horizontal, it will capture all of the sun's
rays as shown
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power
systems
Placement of solar module in relation to Sun’s
altitude ( at Solar noon)
Graphic courtesy GSES
• to obtain maximum irradiance solar module must always face the sun
• ideally, the solar module should be tilted (β deg) to the horizontal facing
north
• such that that there is a 90° angle between the sun ( at solar noon) and
the module
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
 Question:
 If the altitude of the sun was 67°, and one wanted
to tilt a solar module directly at it, what would the
tilt angle be?
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
 answer:
 If the altitude of the sun was 67°, and one wanted
to tilt a solar module directly at it, what would the
tilt angle be?
 180° - 90° - 67° = 23° from the horizontal (ground)
 In a straight line there is 180°, to get the most sunlight on the module it
needs to be perpendicular to the sun; i.e. we need to have 90° between
the sun and module
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
• Solar noon is the time of day when the sun is at its highest altitude
in Australia between 11 .00 to 13:00 Hrs
• In the northern hemisphere modules best positioned facing due South,
Southern hemisphere facing North
• Optimum tilt angle is
generally latitude plus
5° to 15 ° but is
dependent on the
exact location
• The amount of solar
radiation falling on a
tilted plane for Sydney
is shown in the
following table
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Altitude and azimuth angles
the location of the Sun is specified by two angles
• solar altitude (γ) angle is the angle between the sun’s rays and the horizontal plane
• solar azimuth (ά) is the angle between the projections of the sun’s rays on a horizontal
plane and the north direction.
Azimuth is measured clockwise going from 0 (true north) to 359°.
East is 90°;South 180°and West 27°
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
The sun path diagram used to determine the position of the sun in the sky
any time of day, any day of the year
• Azimuth angles, shown on
the circumference of the
diagram
• Altitude angles presented,
by concentric circulars
• Sun path lines from east
to west for different dates of
the year
• Time of lines cross the
sun path lines
• Location information that
refers to latitude
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Sun path diagram for latitude 32°south (sydney)
http:www.squ1.com/archive/index.php? http:www.squ1.com/archive/solar/solar-position.html
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
http:www.learn.londonmet.au.uk/packages/clear/index.html)
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Magnetic North & True North
A compass needle points to magnetic north, which is defined by the direction of
the horizontal component of the geomagnetic field (H). The geomagnetic
declination (D), sometimes called variation, is the angle between true north and
magnetic north
• Solar modules must be inline with
true north to optimise maximum
performance
• e.g. Sydney , the magnetic deviation
is approximately 13° East.
• Simply this means True North is
approximately 13 ° west of magnetic
north
J.P. Teasel version 06/12/10
True north
Magnetic
deviation
Magnetic
north
West
East
Design, Install &Commission grid connected photovoltaic power systems
Atmospheric effects
Large faction of solar radiation reaching top of earth’s atmosphere is
reflected i.e. known as ‘albedo’
The sun
Solar radiation at top
atmosphere Go =
1367W/m²
Diffuse solar radiation
Reflected solar
radiation(albedo) from
atmosphere
Reflected solar radiation
Solar radiation at sea
level G = 1000 W/m²
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Atmospheric effects continued
•Radiation reaching the earth's atmosphere is composed of direct and diffused
radiation
•Diffused though not as intense as direct radiation can still heat for solar collectors
and energy for solar cell
•Radiation reaching earth’s surface has a different spectral content compared to
radiation out the earth’s atmosphere
•Water and Co2 in the atmosphere absorb some wavelength bands
• relative distance that radiation must travel through the atmosphere to reach a
given location is the AIR mass
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Air Mass
AM = 1/cos θ
Θ is the angle between sun and the line to a point
directly overhead
•Outside earth’s
atmosphere AM = O
(AMO)
•AMI corresponds to
the sun being directly
overhead
•The standard condition
for solar modules is
AM1.5@1 kW/m²
irraiance at a cell
• Cell temperature is normally greater than the
temperature 25°
ambient temperature
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Irradiation & Peak Sun Hours
Irradiation is the total quantity of radiant solar energy per unit
over given period i.e. daily, monthly annually
SI unit for energy is the joule
courtesy :apricus.com
The joule is a very small unit solar radiation is often expressed in mega
joule (MJ)
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Design, Install &Commission grid connected photovoltaic power systems
Irradiation & Peak Sun Hours cont…..
• Irradiance is a measure of the rate of energy received per unit area, and has
units of Watts per square metre (W/m2), where 1 Watt (W) is equal to 1 Joule (J)
per second.
e.g. If the solar radiation on a horizontal surface for a whole day is 25 MJ/m² and
we want to work out the number of hours for which radiation at the rate 1000
W/m² is gained for a day, we need to convert MJ to kWh
The joule is a watt second
1 kWh= 60 seconds x 60 minutes =3600 joules
25 x 106 ÷ 3600 x103 = 6.94kWh/m²
• Daily radiation commonly referred to as peak sun hours (PSH)
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
The number of PSH for a day is the number of hours for which energy at the
rate of 1kW/m² is received on the installation
To convert MJ/m² to peak sun hours first divide the number of MJ/m² by 3.6
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
question
Peak Sun Hours
 The total daily energy received on the roof of a
house for a grid-connect installation is 26.3MJ/m2.
What is the number of peak sun hours (PSH) at the
site? Round to one decimal place.
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
answer
Peak Sun Hours
 The total daily energy received on the roof of a house for
a grid-connect installation is 26.3MJ/m2. What is the
number of peak sun hours (PSH) at the site? Round to
one decimal place.

The correct answer is 7.37 peak sun hours
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Solar Radiation for
the area where you
are designing may
be obtained from :
‘Australian Solar
Radiation
Handbook’
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Solar irradiation map of Australia
Source : courtesy of Bureau of Meteorology website:
http://born.gov.au/cgi-bin/climare/cgi_bin_scripts/solar-radiation.cgi
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Solar irradiation
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Solar Altitude
•As a result of earth’s rotation around the sun
•The sun moves between the Tropic of
Cancer(23.45°N) Northern Hemisphere &
•Tropic of Capricorn (23.45°S) Southern
Hemisphere
•When the sun is over either two tropics is
known as the solstice
•When the sun is over the equator equinox
•Sun reaches Tropic of Cancer at northern
solstice (June 22nd
•Tropic of Capricorn at the southern solstice
December 22nd
•The sun crosses the equator at the
equinoxes in March 21st and September 23rd
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Solar altitude
The formula for calculating the altitude (γe) of the
sun when it is over the equator
γ e = 90°- latitude (in degrees)γ
The formula for calculation the altitude γt
Of the sun when it is over the tropics (Cancer &
Capricorn)
γ t = 90°- latitude (in degrees) ±23.45°γ
Whether you use ( + or –) depends on the
hemisphere (Nor S) and over which tropic you
want to determine the sun’s altitude
NB:It is assuming you facing the equator i.e.
facing north in the southern hemisphere
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Solar altitude for Darwin at Equinox &
Solstices





Example Darwin is 12.46°S.
Therefore the altitude of the sun when it is
over the two tropics and the equator are
as follows:
Equator (March 21st and September 23rd)
γ e = 90°- latitude (in degrees) =
90°-12.46°=77.54°
Tropic of Cancer (June 22nd)
 γ t = 90°- latitude (in degrees) -23.45°

= 90°-12.46°- 23.45°= 54.09°
Tropic of Capricorn( December 22nd)
 γ t = 90°- latitude (in degrees) + 23.45°

= 90°-12.46° +23.45°= 109.99°

NB: The altitude of the sun (respect to facing
south) could also be expressed as
 79.01 °i.e. (180 °- 100.99 °)
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
For Sydney
the sun will
never be
directly
overhead
From
March29th
to September
14th the sun
will be in a
northern
direction
J.P. Teasel version 06/12/10
You will need
to look for
potential
obstacles that
could cause
shading
Design, Install &Commission grid connected photovoltaic power systems
This table provides altitude at the two equinoxes and also indicates when
the sun is in the northern sky with respect to the latitude. This may assist
locating possible obstacles that will shade the installation PV module.
J.P. Teasel version 06/12/10
Session 2 Question 1 formative in class
Tilted Surface
If the sun's rays are impinging vertically on a plane flat on the earth's surface
at 1kW/m2, what the effect of tilting that surface to an angle of 60° to the
horizontal?
The output from the panel would increase The output from the panel would
decrease The output of the panel would stay the same None of the above
You answered The output from the panel would decrease
When the panel is tilted, the effect surface area that the rays are hitting is less.
Therefore the output from the panel will decrease, as it is receiving less solar
radiation.
You can see this effect by using a book and a lamp. Set the lamp up so that the
light is pointing directly down onto a flat surface (i.e. table or desk). The light source
needs to be perpendicular to the surface (at 90° to each other).
Hold the book flat directly under the lamp so that you can see the shadow it
creates.
Tilt the book at different angles to the lamp and note the effect of tilting the book
has on the shadow size. The smaller the shadow, the less solar radiation that is
hitting your book.
The largest shadow is created when the book is at 90° to the light source
J.P. Teasel version 06/12/10
Session 2 Question 2 formative in class
Altitude and Azimuth
What do the terms altitude and azimuth mean?
altitude: angle between projection of the sun's rays and the horizontal
plane azimuth: is the angle between the projection of the sun's rays on a
horizontal plane and the north direction. A 360 degree bearing, e.g.
where East is 90 degree, West is 270 degree, South is 180 degree and
North 0 degree.
Briefly :Altitude is the angle between the suns' rays and the horizontal.
Azimuth is the angle between the sun and north.
J.P. Teasel version 06/12/10
Session 2 Question 3 formative in class
Irradiance and Irradiation Units
What are the units for irradiance and irradiation?
kWh/m2 and kW/m2 respectively kWh/m2 and MJ/m2 respectively
MJ/m2 and kWh/m2 respectively MJ/m2 and kW/m2 respectively
kW/m2 and kWh/m2 respectively
kW/m2 and kWh/m2 respectively
Irradiance measures power per unit area and irradiation measures
energy per unit area, so therefore
Irradiance is measured in W/m2 or kW/m2
Irradiation is measured in MJ/m2 or kWh/m2
So the correct answer is kW/m2 and kWh/m2 respectively
J.P. Teasel version 06/12/10
Session 2 Question 4 formative in class
Irradiance and Irradiation Units
What are the units for irradiance and irradiation?
kWh/m2 and kW/m2 respectively kWh/m2 and MJ/m2 respectively
MJ/m2 and kWh/m2 respectively MJ/m2 and kW/m2 respectively
kW/m2 and kWh/m2 respectively
kW/m2 and kWh/m2 respectively
Irradiance measures power per unit area and irradiation measures
energy per unit area, so therefore
Irradiance is measured in W/m2 or kW/m2
Irradiation is measured in MJ/m2 or kWh/m2
So the correct answer is kW/m2 and kWh/m2 respectively.
J.P. Teasel version 06/12/10
Session 2 Question 5 formative in class
Irradiance and Irradiation Units
What are the units for irradiance and irradiation?
kWh/m2 and kW/m2 respectively
kWh/m2 and MJ/m2 respectively
MJ/m2 and kWh/m2 respectively
MJ/m2 and kW/m2 respectively
kW/m2 and kWh/m2 respectively
kW/m2 and kWh/m2 respectively
Irradiance measures power per unit area and irradiation measures
energy per unit area, so therefore
Irradiance is measured in W/m2 or kW/m2
Irradiation is measured in MJ/m2 or kWh/m2
So the correct answer is kW/m2 and kWh/m2 respectively
J.P. Teasel version 06/12/10
Session 2 Question 6 formative in class
Solar Radiation Categories
What are the categories of solar radiation that reach the Earth's
surface?
Direct, diffuse and albedo
Direct and albedo
Direct and diffuse
Diffuse and albedo
Direct and diffuse
Direct radiation is the radiation that reaches the earth's surface directly
from the sun; i.e. it is a direct "beam" of light, it has not been reflected off
anything.
Diffuse radiation has been scattered via clouds, objects or particles in the
atmosphere, but still manages to reach the earth's surface. It is generally
not as intense as direct radiation.
Albedo is the radiation that is reflected off the top of the earth's
atmosphere and does not reach the earth's surface.
The correct answer is direct and diffuse (radiation).
J.P. Teasel version 06/12/10
Session question 7 formative in class
J.P. Teasel version 06/12/10
Session 2 question formative in class
J.P. Teasel version 06/12/10
Design, Install &Commission grid connected photovoltaic power systems
Take home Questions Session 2
1. Approximately, what is the peak value for irradiance on the earths
surface?
2.
If the sun’s rays was falling vertically on a plane flat on the surface of
the earth, what would the effect if the plane was tilted 60° to the
horizontal?
3.
Define the two angles that are used to specify the sun’s position in the
sky
4.
If a plane was laying flat on the earth’s surface, at what angle would
the sun’s rays impinge on that plane at 15:00 Hrs on 22nd December
(22ndJune) at Darwin
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Design, Install &Commission grid connected photovoltaic power systems
5.
Define the following terms: solar constant; irradiance; irradiation; air
mass peak sun hour
6.
What factors effect the amount of solar radiation reaching the earth’s
surface
7.
Name the categories of solar radiation reaching the surface of the
earth.
8.
explain why it is important to have the surface of a flat plate of the
Earth at sea level
9.
What is the reference condition used for air mass when rating solar
cells
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Design, Install &Commission grid connected photovoltaic power systems
10. What would be the number of peak sun hour days if the profile of the
power from the sun was as follows?
time
Irradiance
W/m²
07:00 - 08:00
200
08:00 - 09:00
250
09:00 - 10:00
350
10:oo - 11:00
450
11:00 - 12:00
600
12:00 - 1300
700
13:00 - 14:00
650
14:00 - 15:00
500
15:00 - 16:00
400
16:00 - 17:00
250
J.P. Teasel version 06/12/10
Hint Work out
the total
energy for the
day in
kWh/m²