Ennaoui cours rabat part i

Photovoltaic Solar Energy Conversion (PVSEC)
‫إﻧﺘﺎج اﻟﻜﻬﺮﺑﺎء ﻣﻦ اﻟﻄﺎﻗﺔ اﻟﺸﻤﺴﻴﺔ‬
‫ﻴ‬ ‫إ ج ﻬﺮﺑ ﻦ‬

Courses on photovoltaic for Moroccan academic staff; 23-27 April, ENIM / Rabat
23 27

PVSEC-Part I
Fundamental and application of Photovoltaic solar cells and system
Ahmed Ennaoui
Helmholtz-Zentrum Berlin für Materialien und Energie
ennaoui@helmholtz-berlin.de
@

This material is intended for use in lectures, presentations and as handouts to students, it can be provided in Powerpoint format to allow
customization for the individual needs of course instructors. Permission of the author and publisher is required for any other usage.

PVSEC-1: all about solar radiation
Some information: Helmholtz-Zentrum Berlin für Materialien und Energie
Short on my ongoing research activities at HZB and PVComB
Why this new concept of meetings ?
Highlight of Part I
Earth's motion around the sun and tracking the sun in the sky
Solar altitude angle at solar noon and orientation of solar panels
g p
Solar angles: the longitude, latitude, solar declination.
Hour angle; azimuth, angular height and orientation of solar panels
Solar time (ST) and local standard time (LST)
Optimal orientation of fixed PV panels
The sun as a blackbody
Solar
S l constant and solar spectrum
t t d l t
Direct radiation, diffuse and albedo sunrise

Air mass or air mass number sunset

Total radiation received by a surface

HZB & PVcomB in the Helmholtz Association

Helmholtz Zentrum Berlin
für Materialien und Energie Employees: around 1,100
(full-time equivalency)
Former Budget: approx. 110 Mio. € (2009)
Hahn Meitner Institute
Hahn-Meitner-Institute (HMI)
Number of employees in various scopes

Intrastructure
Formation
• FOUNDED IN 01/01/2009 Solar energy research

• SYNERGETIC USE OF PHOTONS AND NEUTRONS Materials for t
M t i l f tomorow & large scale
l l
facilities
• FUNDAMENTAL RESEARCH

• DEVELOPMENT OF NEW MATERIALS

• RESEARCH FIELDS: SOLAR ENERGY
MAGNETISM, MATERIALS
BIOLOGY MATERIALS

HZB & PVcomB in the Helmholtz Association

Quelle: PVComB/Rutger Schlatmann

About Helmholtz Association

Strategic Goal: Create the scientific and technological base for competitive
g g p
renewable energy system to carry a major load of the future energy supply

Six Helmholtz-Research topics

Energy Key Technologies

Earth and Environment Structure of Matter

Health Transport and Space

Financing of activities (programmes) instead of financing single institutes (centres)
Programme Oriented Funding (POF)

Solar Energy Division in HZB


Goal Strategy of PVcomB

PVcomB
Kompetenzzentrum Dünnschicht- und Nanotechnologie für Photovoltaik Berlin

PVcomB Baselines Processing
Next conference 2012
One Oral Presentation@E-MRS Spring Meeting May 14-18, 2012 Strasbourg, France
One oral Presentation@27th EU PVSEC
24 - 28 September 2012 Frankfurt

Lab scale efficiency 16% already achieved
p
Cooperation with Bosch Solar via BMBF‐Project j
Objective: Scaling up Zn(S,O)/CIGS modules/Ennaoui/
Man power: 1 Dipl. Ing. (Emi Suzuki) , 1 Dipl. (Umsür)
Ahmed Ennaoui / head of a research group: Thin Film and nanostructured solar cells /Solar Energy Division / Helmholtz-Zentrum Berlin für Materialien und Energie

Sustainable and controllable synthesis of nanomaterials for energy applications
(Xianzhong Lin, PhD student , Umsür, Master student)
Work Programme 2012
g BMBF NanopV Project
p j
(Nanosciences, nanotechnologies, materials and new production technologies)
(evaluated on the basis of two criteria: scientific quality and expected impact (economic, social, environmental)
Printing solar cells
TEM HRTEM
Kesterite
Ink

5 nm

Electrophoresis

100 nm

Printing solar cells more economically similar to how news papers are printed
Inkjet printer integrated laser for annealing processing
Objective Pilot lines for precision synthesis of nanomaterials

1 Oral presentations + 1Poster @: E-MRS Spring Meeting May 14-18, Strasbourg 2012
1 Poster 27th EU PVSEC / 24 - 28 September Frankfurt 2012
Ahmed Ennaoui / head of a research group: Thin Film and nanostructured solar cells /Solar Energy Division / Helmholtz-Zentrum Berlin für Materialien und Energie

Why this new concept of meetings ?
Transfer of know-how between Moroccan academic.
To
T contribute reengineering the Curriculum: Design and analysis
t ib t i i th C i l D i d l i
of a new Graduate Degree at Moroccan Universities
To create synergies between Moroccan Academic and Industrial
components (firstly: Morocco Germany and later on with other EU components) to
Morocco-Germany
promote innovative R&D in the field of photovoltaics,
from fundamental breakthroughs to proof of concept devices.
To contribute the emergence of solar electricity from photovoltaics
as a great contribution in the energy mix in Morocco within the next
years, as an immediate response to the energy and climate concerns.
Profound understanding of Silicon technology, concentrator cell design
technology design.
Thin film technologies (2nd generation PV).
High efficiency concepts (3rd generation PV) .
Going beyond the existing bulk crystalline silicon technologies, with a specific.
g y g y g p
attention to new class of PV materials like e.g. chalcogenides.
(copper indium gallium diselenide (CIGS) family of compounds)
Photovoltaic system and components and application.
Optimal design of systems with insolation condition in Morocco and concept for residential
residential.
Energy storage and fuel cells
Help Graduate student to create there own companies.
Prof. Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie

Introduction: Grid-connected PV systems
Copyrighted Material, from internet

Controller, (charge regulator) regulates the voltage and current
Traditional System
coming from the solar panels Determines whether this power is
needed for home use or whether it will charge a deep-cycle solar
battery to be drawn upon later on.

DC‐current from the
controller can be used to run
Photovoltaic electronic devices that don't
P>C require an AC‐current.

Photovoltaic
P<C All other current must pass
through a DC to AC inverter,
all surplus electricity not being
transforming it into electricity
drawn b your h
d by home can b be
usable by general household
sent to your utility company's
appliances.
power grid.

Intensity of sun light on ground

The intensity of the direct component of sunlight
ID = 1353 kW/m2 . [1 - a.h] . 0.7(AM0.678) + a . h
a = 0.14 and h is the location height above sea level in kilometers.

AM0: in free space above the earth atmosphere
AM1: at the equator (zenith angle 0)
AM1.5: at zenith angle 48.2

AM1
AM1.5
AM0

http://pvcdrom.pveducation.org/SUNLIGHT/AIRMASS.HTM

Objective of this course PVSEC-1

Understanding how the l illumination t
U d t di h th solar ill i ti at any l ti on E th varies over th course
location Earth i the
of a year. You will know how to correctly set the orientation of fixed PV panels installed
outdoors to maximize annual energy production.

X A
C

D

Furthermore: you will be able to answer to such questions: If it 9 p.m. at Position D, what time is it
at position C? Position B? If it is 1 p.m. at Position X, at which location is the time 5 p.m.

Earth's motion around the sun and tracking the sun in the sky

Ecliptic
The Northern plan The Northern
Hemisphere Hemisphere is
is tilted away tilted toward
from the sun 0.983 A.U. 1.017 A.U
1 017 A U the sun
1 AU = 1.496 x 108 km

North Pole: shorter day times Earth daytime and night time
south Pole : longer day times last 12 hours each

South Pole closer to the Sun than the North Pole
Pole. A line from the center of the Sun to the
Winter solstice in the Northern Hemisphere center of Earth passes right through the
Summer solstice in the Southern Hemisphere equator equinox.
Sun’s rays normal to the Earth’s surface at Tropic Sun’s rays normal to the Earth’s surface at
+23.45
of Cancer (latitude +23 45 ) equator (latitude 0 )
),

Where are we?
Places located east of the Prime Meridian have an east longitude (E) address.
Places located west of the Prime Meridian have a west longitude (W) address.
Morocco : Northern hemisphere located within the latitude of 32 N and longitude of 05º W.
Locations Latitude Longitude
Rabat
R b t N 34°0´ 47´´ W 06°49´ 57´´
N
Kenitra
Rabat Casablanca 33° 35´ 34´´ 7° 37´ 9´´
Local meridien, P
meridien Ifrane 31 42 7
31° 42´7´´ 6 20 57
6° 20´57´´
N 34 0 47
P Meknes
W 06 49 57 Mohammadia
W E Marrakech
Agadir 30° 25´ 12`` 9° 35´53´´
Oujda
Fes
Hoceima
S Tanger
Goulimine
Your
Y smart phone
t h
http://www.geonames.org/search.html?q=rabat
http://maps.google.com/maps

Longitude and inclinaison
• The earth is divided into 360o longitudinal lines
passing through poles. δ
• Zero longitudinal line passes through Greenwich
• 1 day has 24 hours, and the earth spins 360º in this
time, so the earth rotates 15º every hour.
(
(1 hour = 15o of longitude)
g ) δ
e.g. point (A) on earth surface exactly 15o West A
of another point (B), will see the sun in exactly
B
the same position after 1 hour = 15

• The declination angle, δ varies seasonally
δ = 23.27 at summer and winter solstice
δ = 0 at equinoxes
δ takes all intermediate values
⎡ 360 ⎤
δ = 23.45 sin ⎢
i (n − 81)⎥
⎣ 365 ⎦
Day Numbers for the First Day of Each Month
n is the nth. day of the year since 1st. January

Optimal orientation of fixed PV panels
General rule of thumb to be followed when installing fixed PV panels outdoors to maximize the
annual energy production.
Thumb

N PV panel tilted toward
The equator (i.e. Toward south) June 21
(summer solstice in
L1 Northern Hemisphere)

PV panel set in optimum position (i.e. horizontal )
+23,45°
L1
Equator March 21 and
(latitude
(l tit d 0°) L2 September 21
(equinoxes
-23,45°

December 21
Earth L2 (winter solstice in
south Hemisphere)
PV panel tilted toward
The equator (i.e. Toward North)
( )

The PV must be tilted toward the equator at an angle with respect to the ground that
is equal to the latitude L at which the PV panel is located,

Altitude angle at solar noon

L
P δ
βNoon = 90 + L- δ
Altit d angle, β
Altitude l Equation
E i L
Local
horizontal

Example 1: Tilt Angle of a PV Module. Find the optimum tilt angle for a south-facing photovoltaic module in Rabat (latitude
34° at solar noon on March 1st.
Solution. March 1st. is the 60th. day of the year so the solar declination is:
⎡ 360 ⎤ ⎡360 ⎤
δ = 23.45sin⎢ (n − 81)⎥ = 23.45sin⎢ (60 − 81)⎥ = −8.3°
⎣ 365 ⎦ ⎣365 ⎦
The tilt angle that would make the sun’s rays perpendicular to the module at noon would therefore be
PV module Tilt 42.34°
42 34°
βnoon = 90° − L + δ = 90− 34− 8.3 = 47.7

Tilt = 90− βnoon = 90− 47.7= 42.3
°
β noon = 47.7°

Solar Position and solar angles
Sun’s position can be described by its altitude angle β (or h) and its azimuth angle Az
Convention: the azimuth angle is considered to be positive before solar noon.
Every hour that passes is an increase of the hour angle of 15°.
ψ = Zenith angle between sun's ray and a line perpendicular to the horizontal plane.
h or β = Altitude angle in vertical plane between the sun's rays and projection of the sun's ray
sun s sun s
on a horizontal plane.
Az (or ϕS) = Azimuth angle measured from south to the horizontal projection of the sun’s ray.

P

Solar Angles
Hour angle HA (called also ω ) the number of degrees the earth must rotate before
sun will be over your line of longitude.⎞
⎛ 15 °
HA = ⎜ ⎟ x (hours before solarnoon)
⎝ h ⎠
⎛ 15 ° ⎞
At 11 AM solar time : HA = ⎜ ⎟ x (1h) = +15 °
⎝ h ⎠

The earth needs to rotate another 15°, or 1 hour, before it is solar noon
15 hour

In the afternoon, the hour angle is negative.
for example, at 2:00 P M
example P.M.
solar time H would be −30°.
⎛ 15 ° ⎞
HA = ⎜ ⎟ x (-2h) = -30 °
⎝ h ⎠

Solar Angles

Altitude Angle
ψ
sin(h) = sin(L)sin( δ) + cos(L)cos( δ)cos(ω)

Azimuth Angle
cosδ sinω
sin(Az) =
cos(h)
P
Az < 0 West of S

Find altitude angle β and azimuth angle S at 3 PM solar time in Boulder, CO (L = 40˚) on the summer
solstice:.At th solstice, we k
l ti At the l ti know th solar d li ti δ = 23 45°
the l declination 23.45°
⎛ 15 ° ⎞ ⎛ 15 ° ⎞
HA = ⎜ ⎟ x (hours before solarnoon) = ⎜ ⎟ x (-3h) = -45 °
⎝ h ⎠ ⎝ h ⎠
sin β = sin(40)sin (23.45) + cos(40)cos (23.45)cos (-45) = 0.7527
cos(23.45) sin(-45)
sin(Az) = = −0.9848 β = sin -1 (0.7527) = 48.8 ° φ S = sin -1 (-0.9848) = −80 °
cos(48.8)

Solar angles

Sunrise and sunset can be found from a simple use of:

Find the time at which sunrise (geometric and conventional) will occur in Boston (latitude 42.3°)
on July 1 (n = 182). Also find conventional sunset.

Solar Position vs solar panel orientation

Winter Summer
Spring/Autumn
S i /A t
sunrise

sunset
56° angle
32° angle 80° angle

http://solarelectricityhandbook.com/solar-angle-calculator.html

Solar Time vs. Clock Time
Solar time, ST: World Time Zones:
http://wwp.greenwichmeantime.
h // i h i
We are measuring relative to solar noon (sun is on our line of longitude) com/time-zone/
ST is depending on the exact longitude where solar time is calculated. http://wwp.greenwichmeantime.
com/time-
Local time, called civil time or clock time (CT) zone/africa/morocco/index.htm

Each time zone is defined by a Local Time Meridian located (LTM)
The origin of this time system passing through Greenwich, (0° longitude)
Clock time can be shifted to provide Daylight Savings Time (summer time)
altitude (h or β)
ltit d
measured in
degrees PM, afternoon

P

AM, before noon

World Map of Time Zones Copyrighted Material, from internet

West East

Greenwich Civil Time: GCT time or universal time
Time along zero longitude line passing through Greenwich.
http://www.fgienr.net/time-zone/
Time starts from midnight at the Greenwich


We
W need to connect l l clock ti (LCT) and solar ti (ST)
dt t local l k time d l time
We have to take into consideration:
(1) Longitudinal adjustment related to time zones
(2) Second adj stment res lting from the earth’s elliptical orbit which ca ses
adjustment resulting hich causes
the length of a solar day

Difference between a 24-h day and a solar day is g
y y given by:
y
The Equation of Time E
360
E = 9.87 sin2B − 7.53B.1.5 sinB B= (n − 81) degrees n = day number
364

Combining longitude correction and the Equation of Time we get
the relationship between local standard clock (CT) and solar time (ST)
4min
Solar Time (ST) = Clock Time (CT) + [LT M(°) - Local Longitudinal (°)] + E(min)
( °)

World Time Zones:
http://wwp.greenwichmeantime.com/time-zone/
http://wwp.greenwichmeantime.com/time-zone/africa/morocco/index.htm

Equation of Time E
q
360
E = 9.87 sin2B − 7.53B.1.5 sinB B= (n − 81) degrees n = day number
364


The Equation of Time adjusts for
the earth’s tilt angle

*During D li ht S i
*D i Daylight Savings, add one hour to the local time
dd h t th l l ti
World Time Zones:
http://wwp.greenwichmeantime.com/time-zone/
http://wwp.greenwichmeantime.com/time-zone/africa/morocco/index.htm

Example 1: Solar Time vs. Clock Time
Find Eastern Daylight Time for solar noon in Boston (longitude 71.1 W)
71.1°
on July 1st.
Answer: July 1st. is day number n = 182. to adjust for local time,
we obtain:
360 360
B= (n − 81) = (182 − 81) = 99.89°
364 364
E = 9.87sin2B − 7.53cosB - 1.5sinB = 9.87sin2[2 (99.89)] − 7.53cos(99.89) - 1.5sin(99.89) = -3.5
( ) ( ) ( )

For Boston at longitude 71.7° W in the Eastern Time Zone with local time meridian 75°
4min
Solar Time (ST) = Clock Time (CT) + (Local Time Meridian - Local Longitude)° + E(min)
degree

To adjust for Daylight Savings Time add 1 h, so solar noon will be at about 12:48 P.M.
CT = 12 − 4(75 - 71.1)° − ( −3.5) = 12 : 00 − 12.1min = 11 : 47.9A.M. East

Time Belts of the U.S : Eastern Standard Time - E.S.T. is calculated to the 75th meridian west longitude. Central Standard Time - C.S.T. is calculated to the 90th
meridian west. Mountain Standard Time - M.S.T. is calculated to the 105th meridian west. Pacific Standard Time - P.S.T. is calculated to the 120th meridian
west. Alaska was standardized in 1918 on 150th Meridian west, but in actual practice, other zones are and have been in use: 120° 150°, 165°

Example 2: Solar Time vs. Clock Time
For the “Green community village” near Dubai (latitude angle = 25° N, local longitude angle = 55°12’ E,
standard time zone = UTC +4, no daylight saving ti ) on February 3 at 14.00. Determine:
t d d ti 4 d li ht i time) F b t 14 00 D t i
a. the apparent solar time.
b. solar declination and hour angle , solar altitude and solar azimuth angles.
UTC = Universal Time and GMT = Greenwich Mean Time.
Atlantic Standard Time (AST) is 4 hours behind of Coordinated Universal Time(UTC)
http://www.timeanddate.com/worldclock/search.html
http://www.timeanddate.com/worldclock/results.html?query=Morocco
a) The apparent solar time 360 360
February 3th ⇒ n = 34 B =
y (n − 81) =
( ) (34 − 81) = - 46,48°
( ) ,
364 364
ET = 9.87 sin2B − 7.53B - 1.5 sinB = 9.87 sin2(-46.48) − 7.53(-46.48) - 1.5 sin(-46.48) = -13.95 min
Local longitude angle = 55°12’ E = -55.2° (conversion in °, and negative since location is in East
Standard time zone UTC+4 Local Time meridian (LTM)= -60° (negative since location is in East)
( ) ( g )
No daylight saving time February LST = 14:00
4min
AST = LST + E(min) + (LT M - Local Longitudinal angle ) = 14 : 00 + (−13.95°) + [− 60° − (−55.2°)]. 4min = 13 : 27
degree degree
Hour angle ω and solar d li ti δ
b) H l d l declination δ.
360 °(284 + n) 360 °(284 + 34)
AST = 13:27 = 13.45 h (conversion of time in hours) δ = 23.45 ° sin = 23.45 °sin = − 16.97
365 365
ω =15° (hours from local solar noon) = 15° (ST-12)
ω = 15°.( 13.45-12) = 21.75°
sin(h) = sin(L)sin(δ) + cos(L)cos(δ) ( ) ⇒ h = sin-1 [sin(25°) i (−16 97) + cos(25°)
i (h) i (L) i ( (L) (δ)cos(ω) i i (25 ).sin( 16.97) (21 75).cos(−16 97)] = 42 98°
(25 ).cos(21.75) ( 16.97) 42.98
cosδ . sinω ⎡ cos(−16.97).sin(21.75) ⎤
sin(Az) = ⇒ Az = sin −1 ⎢ ⎥ = 28.98°
cos(h) ⎣ cos(42.98) ⎦

The Sun’s path
The Sun always rises in the east
It rises higher and higher in the sky at noon Ψ = zenith angle
AM, before noon: Line is some time in the morning
At solar noon: Sun reaches its maximum altitude
Noon altitude: depending on your latitude
β = altitude angle
PM, afternoon: Line is some time in the afternoon
The Sun starts to set (go down) in the West
Az = azimuth angle
altitude (h or β)
measured in
degrees PM, afternoon

P

AM, before noon
http://solardat.uoregon.edu/PolarSunChartProgram.html

The Sun’s path Copyrighted Material, from internet

The projection of the sun-path is shown in dashed line on horizontal palne

h

Az

The Sun’s path

The j ti
Th projection of th sun-path iis shown iin d h d li on h i t l palne
f the th h dashed line horizontal l

h

Az

The Sun’s path

The projection of the sun-path is shown in dashed line on horizontal palne

h

Az

Objective of this course PVSEC-1

Understanding how the l illumination t
U d t di h th solar ill i ti at any l ti on E th varies over th course
location Earth i the
of a year. You will know how to correctly set the orientation of fixed PV panels installed
outdoors to maximize annual energy production.

X A
C

D

Furthermore: you will be able to answer to such questions: If it 9 p.m. at Position D, what time is it
at position C? Position B? If it is 1 p.m. at Position X, at which location is the time 5 p.m.

Sun Path Diagrams for Shading Analysis
How to locate the sun in the sky at any time
What sites will be in the shade at any time

Two interesting ways to represent sun course over the y
g y p year:
Polar diagrams and vertical diagrams.
Determine the azimuth and altitude angles of trees, buildings, and other obstructions
Estimate the amount of energy lost to shading using the sun path diagram
Present information in solar time (sun at its zenith at noon) or in standard time
(time on the clock).
The shading of solar collectors is an area of legal and legislative concern
(e.g., a neighbor’s t is blocking a solar panel).
( i hb ’ tree i bl ki l l)
Architect can locate the site of a project (latitude, longitude) and can study positions
of the sun (azimuth, altitude) and its movement in the sky to determine sunshine
periods of a site solar masks due to neighboring buildings impact of the orientation
site, buildings,
of the building, location of the windows, need and kind of solar protections….

http://solardat.uoregon.edu/PolarSunChartProgram.html
http://www.jaloxa.eu/resources/daylighting/sunpath.shtml
http://www.youtube.com/v/IjOhtmmq7aM&hl=en_US&fs=1&rel=0


Tool that helps you reading the movement of the sun throughout the day and during the seasons
seasons.

β 46°
β 58°

Equinox

Az 74°

Az 38°

0°
http://learn.greenlux.org/packages/clear/thermal/climate/sun/sunpath_diagrams.html


Sunrise/sunset
June 21
Sunrise
Sunset
Agadir
Today is March 25th. 2012
Latitude: +30.42 (30°25'12"N)
Longitude: -9.61 (9°36'36"W) Today (March 25th. 2012)
Time zone: UTC+0 hours Sun rises at 06:36 from North-
Local time: 12:39:40 East (Az = 90). Sun set happens
Country: Morocco at 18:53 when the sun is in North-
Continent: Africa West (Az = 270). On that day the
Sub-region: Northern Africa elevation h = 50° at noon

December 21

Equinox
March
September)
Variable J F M A M J J A S O N D
Insolation,
3.52 4.36 5.58 6.73 7.37 7.45 7.09 6.72 5.80 4.73 3.76 3.14
kWh/m /day
kWh/m²/day

http://www.gaisma.com/en/location/agadir.html


Figure below

Solution
Workshop
Friday

The sun path diagram with superimposed obstructions makes it easy to
estimate periods of shading at a site.

Colors of light have different wavelengths and different energies

Short Wavelength Long Wavelength
c hc 1239
λ= E p = hν = → E p (eV ) =
ν λ λ (nm)

Max Planck
1858 - 1947

Albert Einstein
100 W
1879 - 1955
Q: What is Power [unit watts] ?
A: Rate at which energy is generated or consumed

Example: 100 W light bulb is turning on for one hour
Energy consumed is:100 W·h or 0 1 kW h
0.1 kW.h.
Same amount would be generated from 40-watt light bulb
for 2.5 hours


Absorption of Light by Atoms
Absorption occurs only when the energy of the light equals
the energy of transition of an electron

‐

Single electron transition in an isolated atom
1 Electron 1 Photon (E = hν)


Absorption of Light by Molecules

Smallest ΔE possible

Molecules have multiple atoms bonded together
More energy states in molecules than atoms
More electron are excited light with a range of frequencies are absorbed

Black Body Radiation

Planck law Wien’s law Stefan-Boltzmann Law
St f B lt L
Radiance of BB at fixed T (any λ) Total amount of energy
λ at peak irradiance
2hc2
E(λ, T) = 5
1 σ = 5.67 × 10-8 Wm-2K-4
λ ⎡ ⎛ hc ⎞ ⎤ A
λp (m) ≈
(m) F(W −2 ) = σT 4
F(W.m
⎢exp⎜
⎜ λk T ⎟ − 1⎥
⎟ T(K )
⎣ ⎝ B ⎠ ⎦
c = 3.0 × 108 ms-1 ; h = 6.63 × 10-34 Js ; k =1.38 × 10-23 JK-1 ; A = 0.002897 [m.K] ; σ = 5.67 × 10-8 [Wm-2K-4]

Sun (visible)
λMAX = 0.5 μm
FT = 64 million W m-2
Earth (infrared)
λMAX = 10 μm
FT = 390 W m-2

Compute

Consider the earth to be a blackbody with average surface temperature
15°C and area equal to 5.1 x 1014 m2. Find the rate at which energy is
radiated by the earth and the wavelength at which maximum power is
radiated. Compare this peak wavelength with that for a 5800 K
blackbody (the sun).
The earth radiates:
E = (5.64 x 10 -8 Wm −2 K −4 ) (5.1 x 1014 m 2 ) (15 x 273) 4 = 2.0 x 1017 Watt

The wavelength at which the maximum power is emitted
2898 2898 2898
λmax (earth) = = = 10.1μm λ (sun) = = 0.5μ.
T(K) 288 5800

Radiation flux; Luminescence , Emittance

Intensity emitted by a source in a direction ox
A surface element dS of a source S, and any direction Ox
with respect to this element dS.
dΦ Ox Source intensity
I Ox [Watt.Str -1 ] =
dΩ (in Watts / Steradian)

S radiates throughout the space
g p
Φ the radiation flux and dΦox portion of Φ radiated into a
solid angle dΩ.
Iox Source intensity in the direction Ox

dΩ = dS cosθ
R2

For a hemispherical space, the solid angle = 2π Steradian
Solid angle for all the space = 4π Steradian

Radiation flux; Luminescence , Emittance

Lox: Luminescence of a source area dS
Flux from the projected area dS '= dS cosβ

dΩ = dS cosθ
dΩ
R2
Lox is the radiated power per unit of solid angle surrounding the Ox direction per unit
area projected perpendicularly t thi di ti i W tt/ 2.stéradian.
j t d di l l to this direction in Watt/m té di
The flux emitted by a surface element dS in a solid angle dΩ surrounding a direction
Ox, tilted β with respect to the normal to this surface.
p
Watt/m2.st W/st
I Ox I Ox
L Ox = =
dS′ dS cos β
m2
dΦ Ox
dΩ = d Φ Ox
2
L Ox =
dS cos β dΩ dS cos β

d 2 ΦOx = L Ox dS cosβ dΩ

The emittance, M of a diffuse source

Diffuse sources are governed by LAMBERT RULE: regardless of the
direction of observation Lox = L . This is the case where the luminance L
depends only on the temperature T of the surface. One can calculate
surface
the total flux:
d 2 Φ O = L dS cosβ dΩ ⇒ dΦ = L dS ∫∫∫
Ox L.dS L.dS cosβ dΩ
2ππ.s

dΦ
M =
dS
d
= L ∫∫∫ π
2 sr
cos β dΩ

π π
M = 2 π L ∫ cos β sin β dβ = π L ∫ 2 sin 2β dβ = πL
2
0 0

M = π.L

Solar flux intercepted by the Earth

A surface element dS on the surface of the sun (Sun: R = 696,000 km )
A a surface element dS´ on earth (earth-sun distance: D = 149,637,000 km)

d 2 Φ dS→dS´ = L T dS cosθ dΩ dS→dS´
dS' cosθ'
dΩ dS →dS' =
d2
M T = π.L T
M 0 dS cosθ dS' cos θ '
d Φ dS→dS' = T
2

π d2

d2ΦdS dS´ = flux emitted by the element dS in a solid angle dΩ surrounding the direction
dS to an element dS´ of the earth's surface:
dS earth s

M0 dS' cosθ '
dΦS→dS' = ∫d2ΦdS→dS' = T ∫ cosθ dS
π S d2
S
dS cos θ = projection of the element dS
on the diametral plane of the sun ∫
S
cosθ dS = ∫ dΣ = Σ = π R 2
Σ

Solar flux intercepted by the Earth

2
⎛R ⎞ 0
dΦ S→dS' = M ⎜ ⎟ dS'
T
⎝D⎠
Solar illumination of the earth is given by the equation
2 2
dΦ S→dS' ⎛R⎞ 4 ⎛R ⎞ Inverse square law
E= = M0
T ⎜ ⎟ ⇒E= σ T ⎜ ⎟
dS' ⎝D⎠ ⎝D⎠ of irradiance

σ = 5,67 . 10-8 W/(m2.K4), T = 5800 K
R = 6,96.108 m, D = 1,49.1011 m
E = 1402 W/m2
The atmosphere will transmit a fraction (75%) of solar radiation
p ( )
τ E = 0,75 E = 1052 W/m2

Earth-
Earth-Atmosphere Energy Balance

Solar radiation intersects Earth as a disk (πr2)

(Energy)in = Energy from sun (S) – Reflected Solar radiation
= πr2 S - πr2 Sα
r = radius of Earth (6360 km)
S = solar constant (1368 W/m2)
α = albedo (earth’s reflectivity) (~30%)
Ein= πr2 S (1- α)
(1

Earth radiates as a sphere with area 4πr2 (m2)
Stephan-Boltzmann equations defines outgoing energy based on radiating temperature

(Energy)out = 4πr2 σT4 units (m2)(Wm-2K-4)(K4) Eout= Total energy emitted by the Earth

Black body the in = out incoming = outgoing πr2 S (1- α) = 4πr2 σT4
Te= 255K (-18 C)

Earth’s actual surface temperature Ts = 288K (15 C) λmax (µm) = 2877/288 = 10 µm (Infra Rot)
Ts - Te = 288 – 255 = 33

Interactions within atmosphere alter radiation budget (Earth is not a black body) Greenhouse Effect
Earth’s natural greenhouse


Pyrheliometer
Measures the direct solar beam
(pointed at the sun)

Pyranometers
Used to measure global solar radiation
(both the direct solar beam and to diffuse sky
radiation from the whole hemisphere)
Measures temperature difference between an
absorbing (black) plate and a non-absorbing (white)
plate. Thermopile converts temperature difference
of plates to a voltage difference

http://en.wikipedia.org/wiki/File:Solar_Spectrum.png

Average irradiance in watts per square meter (W/m2) available can be measured
I-V characteristic of a solar cell is a sensor for solar radiation.

Energy from the Sun at the Earth’s Surface
G(φ, λ, t1, t2) the total amount of solar irradation at latitude φ, longitude λ, between
time t1 and time t2 on surfaces of any orientation.
The relative proportion of beam irradiation and diffuse irradiation.
The spectral breakdown of the radiation at the surface.

Cosine Law of sunshine intensity
When the Sun is overhead, 100% of a beam of
width, I0 strikes a piece of ground of width, I0.

As the Sun goes down and Zenith Angle, Z increases, I0
progressively less of the sunbeam of width I0, strikes
the piece of ground. More of the sunbeam misses that
piece of ground and is lost.

COSINE LAW
Z
I Z = I 0 × cos Z

h Z I0
The fraction of the sunbeam that strikes the ground h Z
= IZ/I0, which t i
hi h trigonometry shows i equal t cos(Z)
t h is l to (Z)
GROUND

Solar flux striking a collector
Direct-beam radiation: that passes in a straight line through the atmosphere to the
p g g p
receiver.
Diffuse radiation : that has been scattered by molecules and aerosols in the
atmosphere.
atmosphere
Reflected radiation: that has bounced off the ground or other surface in front of the
collector.

Diffuse radiation
Direct-beam

collector, C

Tilt angle
Reflected radiation

Extraterrestrial (ET) solar insolation, I0 (Watt/m2)
Estimate of the extraterrestrial (ET) solar insolation I0, that passes perpendicularly
insolation,
through an imaginary surface just outside of the earth’s atmosphere

I0
Earth

Day-to-Day extraterrestrial solar
insolation, Ignoring sunspots

SC is the solar constant which is the average power of the sun's radiation that reaches a unit
area, perpendicular to the rays, outside the atmosphere
n is the day number.
y
Base on NASA measurements SC = 1353 W/m2 (commonly accepted value 1377 W/m2)

Attenuation and air mass
Over a year’s time, less than half of the radiation that hits the top of the atmosphere
y , p p
reaches the earth’s surface as direct beam I0.
On a clear day, and sun high in the sky, beam radiation at the surface can exceed 70%
of the extraterrestrial flux
Attenuation of incoming radiation is a function of the distance that the beam has to
travel through the atmosphere, which is easily calculable
A commonly used model: attenuation as an exponential decay function
y p y
I B = Aexp(−k . m)
IB = beam portion of the radiation that reaches the earth’s surface
p
A = apparent extraterrestrial flux
k = optical depth β: altitude
1 φS solar azimuth
m = air mass ratio: m =
sinβ φC Panel azimuth
where β is the altitude angle of the sun.
N
Panel Tilt
S

Optical Depth k and the apparent Extraterrestrial Flux A.
E traterrestrial Fl A
The Sky Diffuse Factor C can be used later for diffuse radiation

Measurements for th 21 t D of E h M th after ; S
M t f the 21st Day f Each Month ft Source: ASHRAE (1993).
(1993)

I B = Aexp(−k . m)
Close fits to the values from the above table
⎡ 360 ⎤ ⎡ 360 ⎤
A = 1160 + 75sin⎢ (n − 275)⎥ (W/m 2 ) k = 0.175 + 0.035sin ⎢ (n − 100)⎥
⎣ 365 ⎦ ⎣ 365 ⎦


ASHRAE, 1993, Handbook of Fundamentals, American Society of Heating, Refrigeration and Air Conditioning Engineers, Atlanta.

Direct Beam Radiation at the Surface of the Earth –
Find the direct beam solar radiation normal to the sun’s rays at solar noon on a clear day in
Atlanta (latitude 33.7°C) on May 21.
May 21 is day number 141

⎡ 360 ⎤
A = 1160 + 75sin . ⎢ (n − 275) ⎥ (W/m 2 ) = 1104 W/m 2
⎣ 365 ⎦
⎡ 360 ⎤
k = 0.175 + 0.035sin ⎢ (n − 100)⎥ = 0.197
⎣ 365 ⎦

Altitude angle:

The air mass ratio:

The value of clear sky beam radiation at the earth’s surface:

Direct-Beam Radiation, IBC
The translation of direct-beam radiation IB (normal to the rays) into beam
direct beam
insolation striking a collector face IBC is a simple function of the angle of incidence

β: altitude
φS solar azimuth
φC Panel azimuth

Panel Tilt Panel Tilt
n

θ = incidence angle between a normal to the collector face and the incoming beam.
At any p
y particular time θ will be a function of the collector orientation, the altitude and
azimuth angles of the sun.

Special case of beam insolation on a horizontal surface

Insolation on a Collector
At solar noon in Atlanta (latitude 33.7°C) on May 21 the altitude angle of the sun was found to be
33.7 C)
76.4° and the clear-sky beam insolation was found to be 902 W/m2. Find the beam insolation at
that time on a collector that faces 20° toward the southeast if it is tipped up at a 52° angle.

The beam radiation on the collector

Insolation on a Collector
Diffuse Radiation on a Collector - find the diffuse radiation on the panel. Recall that it is solar
noon in Atlanta on May 21 (n = 141), and the collector faces 20° toward the southeast and is
tipped up at a 52° angle. The clear-sky beam insolation was found to be 902 W/m2.
Diffuse insolation on a horizontal surface: IDH = C x IB where C is a sky diffuse factor.
y

The diffuse sky factor, C

The diffuse energy striking the collector

Total beam insolation (697 W/m2) plus diffuse on the collector (88W/m2) 785 W/m2.

Reflected Radiation, IRC
Reflection can provide a considerable boost in performance, as for example on a
p p , p
bright day with snow or water in front of the collector.
The amount reflected can be modeled as the
p
product of the total horizontal radiation
(beam IBH , plus diffuse IDH) times the ground
reflectance ρ. The fraction of that ground-
reflected energy that will be intercepted by
the collector depends on the slope of the
panel , resulting in the following expression
for reflected radiation striking the collector
IRC:

Horizontal reflector (∑ = 0) ⇒ no reflected radiation
1
Vertical panel (90°) ⇒ it predicts that the panel " sees"
sees of the reflected radiation
2

⇒

More detail in workshop and problem solving activities

NEXT
PVSEC-2
Fundamental and application of
pp
Photovoltaic solar cells and system

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