Ennaoui cours rabat part i

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

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

PVSEC-Part IFundamental and application of Photovoltaic solar cells and system

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

Ahmed EnnaouiHelmholtz-Zentrum Berlin für Materialien und Energie

[email protected]

Fundamental and application of Photovoltaic solar cells and system

@

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 radiationSome information: Helmholtz-Zentrum Berlin für Materialien und Energie

Highlight of Part I

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 skySolar altitude angle at solar noon and orientation of solar panelsg pSolar angles: the longitude, latitude, solar declination.Hour angle; azimuth, angular height and orientation of solar panelsSolar time (ST) and local standard time (LST) Solar time (ST) and local standard time (LST) Optimal orientation of fixed PV panelsThe sun as a blackbodyS l t t d l tSolar constant and solar spectrumDirect radiation, diffuse and albedoAir mass or air mass number

sunrise

sunset

Total radiation received by a surface

Helmholtz Zentrum Berlin

HZB & PVcomB in the Helmholtz Association

Helmholtz Zentrum Berlin für Materialien und Energie

FormerHahn-Meitner-Institute (HMI)

Employees: around 1,100 (full-time equivalency)

Budget: approx. 110 Mio. € (2009)Hahn Meitner Institute (HMI)

Number of employees in various scopes

• FOUNDED IN 01/01/2009

IntrastructureFormationSolar energy researchM t i l f t & l l• SYNERGETIC USE OF PHOTONS AND NEUTRONS

• FUNDAMENTAL RESEARCH

Materials for tomorow & large scale facilities

• DEVELOPMENT OF NEW MATERIALS

• RESEARCH FIELDS: SOLAR ENERGY MAGNETISM, MATERIALSBIOLOGY 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 prenewable energy system to carry a major load of the future energy supply

Six Helmholtz-Research topics

Energy

Earth and Environment

Key Technologies

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 BerlinPVcomB

PVcomB Baselines ProcessingNext conference 2012One Oral Presentation@E-MRS Spring Meeting May 14-18, 2012 Strasbourg, FranceOne oral Presentation@27th EU PVSEC24 - 28 September 2012 Frankfurt

Lab scale efficiency 16% already achievedCooperation with Bosch Solar via BMBF‐Projectp j

Objective: Scaling up Zn(S,O)/CIGS modules/Ennaoui/Man power: 1 Dipl. Ing. (Emi Suzuki) , 1 Dipl. (Umsür)

Quelle: PVComB/Rutger SchlatmannAhmed Ennaoui / head of a research group: Thin Film and nanostructured solar cells /Solar Energy Division / Helmholtz-Zentrum Berlin für Materialien und Energie

Work Programme 2012

Sustainable and controllable synthesis of nanomaterials for energy applications(Xianzhong Lin, PhD student , Umsür, Master student)

BMBF NanopV Projectg(Nanosciences, nanotechnologies, materials and new production technologies) (evaluated on the basis of two criteria: scientific quality and expected impact (economic, social, environmental)

p j

Printing solar cells

KesteriteInk

TEM HRTEM

5 nm

Electrophoresis

100 nm

Printing solar cells more economically similar to how news papers are printedInkjet printer integrated laser for annealing processingObjective Pilot lines for precision synthesis of nanomaterials

Ahmed Ennaoui / head of a research group: Thin Film and nanostructured solar cells /Solar Energy Division / Helmholtz-Zentrum Berlin für Materialien und Energie

1 Oral presentations + 1Poster @: E-MRS Spring Meeting May 14-18, Strasbourg 20121 Poster 27th EU PVSEC / 24 - 28 September Frankfurt 2012

Why this new concept of meetings ?Transfer of know-how between Moroccan academic.T t ib t i i th C i l D i d l i To contribute reengineering the Curriculum: Design and analysis 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 components (firstly: Morocco-Germany and later on with other EU components) to

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 To contribute the emergence of solar electricity from photovoltaics as a great contribution in the energy mix in Morocco within the nextyears, as an immediate response to the energy and climate concerns.Profound understanding of Silicon technology concentrator cell designProfound understanding of Silicon technology, concentrator cell 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 pattention 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

Prof. Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie

Optimal design of systems with insolation condition in Morocco and concept for residential.Energy storage and fuel cellsHelp Graduate student to create there own companies.

Introduction: Grid-connected PV systemsController, (charge regulator) regulates the voltage and currentTraditional System

Copyrighted Material, from internet

coming from the solar panels Determines whether this power isneeded for home use or whether it will charge a deep-cycle solarbattery to be drawn upon later on.

Traditional System

DC‐current from thecontroller can be used to runelectronic devices that don'trequire an AC‐current.

PhotovoltaicP > C

All other current must passthrough a DC to AC inverter,transforming it into electricity

all surplus electricity not beingd b h b

PhotovoltaicP < C

transforming it into electricityusable by general householdappliances.

drawn by your home can besent to your utility company'spower grid.

Introduction: Grid-connected PV systemsController, (charge regulator) regulates the voltage and currentTraditional System


coming from the solar panels Determines whether this power isneeded for home use or whether it will charge a deep-cycle solarbattery to be drawn upon later on.

Traditional System

DC‐current from thecontroller can be used to runelectronic devices that don'trequire an AC‐current.

PhotovoltaicP > C

All other current must passthrough a DC to AC inverter,transforming it into electricity

all surplus electricity not beingd b h b

PhotovoltaicP < C

transforming it into electricityusable by general householdappliances.

drawn by your home can besent to your utility company'spower 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 . ha = 0.14 and h is the location height above sea level in kilometers.

The intensity of the direct component of sunlight

AM0: in free space above the earth atmosphereAM1: at the equator (zenith angle 0)AM1.5: at zenith angle 48.2

AM1AM1.5

AM0

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

Objective of this course PVSEC-1

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

CAX

D

Furthermore: you will be able to answer to such questions: If it 9 p.m. at Position D, what time is it 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

EclipticCopyrighted Material, from internet

The Northern Hemisphere is tilted toward

the sun

The Northern Hemisphere is tilted awayfrom the sun

Eclipticplan

1 017 A U0.983 A.U. the sun from the sun 1.017 A.U 0.983 A.U.

1 AU = 1.496 x 108 km

Earth daytime and night time last 12 hours each

A line from the center of the Sun to the

North Pole: shorter day timessouth Pole : longer day times

South Pole closer to the Sun than the North Pole A line from the center of the Sun to the center of Earth passes right through the equator equinox.Sun’s rays normal to the Earth’s surface at equator (latitude 0 )

South Pole closer to the Sun than the North Pole.Winter solstice in the Northern HemisphereSummer solstice in the Southern HemisphereSun’s rays normal to the Earth’s surface at Tropic of Cancer (latitude +23 45 ) equator (latitude 0 ), of Cancer (latitude +23.45 )

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 LongitudeR b t N 34°0´ 47´´ W 06°49´ 57´´Rabat N 34°0´ 47´´ W 06°49´ 57´´KenitraCasablanca 33° 35´ 34´´ 7° 37´ 9´´Ifrane 31° 42´7´´ 6° 20´57´´

N

RabatLocal meridien P Ifrane 31 42 7 6 20 57

MeknesMohammadiaMarrakechEW

PLocal meridien, P

N 34 0 47W 06 49 57

Agadir 30° 25´ 12`` 9° 35´53´´OujdaFesHoceimaTangerGoulimine

S

Y t hhttp://www.geonames.org/search.html?q=rabathttp://maps.google.com/maps

Your smart phone

Longitude and inclinaison• The earth is divided into 360o longitudinal lines


The earth is divided into 360 longitudinal lines passing through poles.

• Zero longitudinal line passes through Greenwich

δ

• 1 day has 24 hours, and the earth spins 360º in thistime, so the earth rotates 15º every hour.

(1 hour = 15o of longitude) δ( g )e.g. point (A) on earth surface exactly 15o West of another point (B), will see the sun in exactly the same position after 1 hour = 15

AB

• The declination angle, δ varies seasonallyδ = 23.27 at summer and winter solstice δ = 0 at equinoxesδ takes all intermediate values

⎤⎡ )81(360i4523δ

n is the nth. day of the year since 1st. January

⎥⎦⎤

⎢⎣⎡ −= )81(365

sin45.23 nδ

Day Numbers for the First Day of Each Month

Optimal orientation of fixed PV panelsGeneral rule of thumb to be followed when installing fixed PV panels outdoors to maximize the

N PV panel tilted toward

annual energy production.Thumb

N PV panel tilted toward The equator (i.e. Toward south)

L1

June 21(summer solstice in

Northern Hemisphere)

Equator(l tit d 0°)

March 21 and September 21

+23,45°PV panel set in optimum position (i.e. horizontal )

L1

(latitude 0°) September 21(equinoxes

-23,45°

L2

EarthPV panel tilted toward

The equator (i.e. Toward North)

L2December 21

(winter solstice in south Hemisphere)

( )

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

Altit d l β LE iβNoon = 90 + L- δ

δL

P

Altitude angle, β LEquationLocal

horizontal

Example 1: Tilt Angle of a PV Module. Find the optimum tilt angle for a south-facing photovoltaic module in Rabat (latitude34° at solar noon on March 1st.Solution. March 1st. is the 60th. day of the year so the solar declination is:

⎤⎡⎤⎡ 360360

The tilt angle that would make the sun’s rays perpendicular to the module at noon would therefore be

°−=⎥⎦⎤

⎢⎣⎡ −=⎥⎦

⎤⎢⎣⎡ −= 3.8)8160(sin45.23)81(sin45.23

365360

365360 nδ

PV module Tilt 42 34°7.473.8349090 =−−=+−°= δβ Lnoon

°=−=−= 42.347.790β90Tilt noon

PV module Tilt 42.34°

°= 7.47noonβ

Solar Position and solar anglesSun’s position can be described by its altitude angle β (or h) and its azimuth angle AzConvention: 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 h or β Altitude angle in vertical plane between the sun s rays and projection of the sun s ray on a horizontal plane.Az (or ϕS) = Azimuth angle measured from south to the horizontal projection of the sun’s ray.

P

Hour angle HA (called also ω ) the number of degrees the earth must rotate before

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

solarnoon) before (hoursx h

15HA ⎟⎠⎞

⎜⎝⎛ °

=

°+=⎟⎠⎞

⎜⎝⎛ °

= 15(1h)x h

15 HA :timesolar AM11At

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

In the afternoon, the hour angle is negative.for example at 2:00 P M

°=⎟⎞

⎜⎛ °

= -30(-2h)x15HA

for example, at 2:00 P.M. solar time H would be −30°.

=⎟⎠

⎜⎝

= -30(-2h)x h

HA

Solar Angles

Angle Altitudeψ δ)cos(ω)cos(L)cos(δ)sin(L)sin(sin(h) +=

sinωcosδ AngleAzimuth

cos(h)sinωcosδsin(Az) =

Az < 0 West of S

P

Find altitude angle β and azimuth angle S at 3 PM solar time in Boulder, CO (L = 40˚) on the summer l ti At th l ti k th l d li ti δ 23 45°

Az < 0 West of S

solstice:.At the solstice, we know the solar declination δ = 23.45°

°=⎟⎠⎞

⎜⎝⎛ °

=⎟⎠⎞

⎜⎝⎛ °

= -45(-3h) xh

15 solarnoon) before (hours xh

15HA

0.7527(-45)(23.45)coscos(40)cos(23.45)sin(40)sinsin =+=β

°== 48.8(0.7527)sinβ -10.9848cos(48.8)

sin(-45) cos(23.45)sin(Az) −== °−== 80(-0.9848)sinφ -1S

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 orientationCopyrighted Material, from internet

SummerWinterS i /A t

sunrise

sunset

Spring/Autumn

sunset

32° angle 56° angle 80° angle

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

Solar Time vs. Clock TimeSolar time, ST: World Time Zones:

h // i h iWe are measuring relative to solar noon (sun is on our line of longitude)ST is depending on the exact longitude where solar time is calculated.

Local time, called civil time or clock time (CT)

http://wwp.greenwichmeantime.com/time-zone/http://wwp.greenwichmeantime.com/time-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)

ltit d (h β) altitude (h or β) measured in

degrees PM, afternoon

P

AM, before noon

World Map of Time Zones Copyrighted Material, from internet

West East

http://www.fgienr.net/time-zone/

Greenwich Civil Time: GCT time or universal timeTime along zero longitude line passing through Greenwich.Time starts from midnight at the Greenwich

Solar Time vs. Clock Time

W d t t l l l k ti (LCT) d l ti (ST)We need to connect local clock time (LCT) and solar time (ST)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 hich ca ses (2) Second adjustment resulting from the earth’s elliptical orbit which causes

the length of a solar day

Difference between a 24-h day and a solar day is given by: y y g yThe Equation of Time E

number day n degrees 81)(n364360B sinB 7.53B.1.5sin2B 9.87E =−=−=364

Combining longitude correction and the Equation of Time we get the relationship between local standard clock (CT) and solar time (ST)

[ ] E(min))( alLongitudin Local - )M( LT)(

4min (CT) Time Clock (ST) TimeSolar +°°°

+=

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

Solar Time vs. Clock TimeEquation of Time Eq

number day n degrees 81)(n364360B sinB 7.53B.1.5sin2B 9.87E =−=−=


*D i D li ht S i dd h t th l l ti

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

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

*During Daylight Savings, add one hour to the local time

Find Eastern Daylight Time for solar noon in Boston (longitude 71.1° W)

Example 1: Solar Time vs. Clock TimeFind Eastern Daylight Time for solar noon in Boston (longitude 71.1 W) on July 1st. Answer: July 1st. is day number n = 182. to adjust for local time,we obtain:

99.89 81)(182364360 81)(n

364360B °=−=−=

[ ] -3.589)1.5sin(99.-.89)7.53cos(99(99.89) 29.87sin2 1.5sinB-7.53cosB9.87sin2BE =−=−= [ ] )()(( )

For Boston at longitude 71.7° W in the Eastern Time Zone with local time meridian 75°

E(min) Longitude) Local - Meridian Time (Local4min (CT) TimeClock (ST) Time Solar +°+= E(min) Longitude) Local - Meridian Time (Localdegree

(CT) TimeClock (ST) Time Solar ++=

To adjust for Daylight Savings Time add 1 h, so solar noon will be at about 12:48 P.M.East 47.9A.M.:11 12.1min00:123.5)( 71.1) - 4(75 12 CT =−=−−°−=

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°

For the “Green community village” near Dubai (latitude angle = 25° N, local longitude angle = 55°12’ E, t d d ti UTC 4 d li ht i ti ) F b 3 t 14 00 D t i

Example 2: Solar Time vs. Clock Time

standard time zone = UTC +4, no daylight saving time) on February 3 at 14.00. Determine: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)Atlantic Standard Time (AST) is 4 hours behind of Coordinated Universal Time(UTC)

http://www.timeanddate.com/worldclock/search.htmlhttp://www.timeanddate.com/worldclock/results.html?query=Morocco

46,48- 81)(34364360 81)(n

364360B 34 n 3th February °=−=−==⇒a) The apparent solar time ,)(

364)(

364y

min -13.95)sin(-46.48 1.5 - 8)7.53(-46.48)sin2(-46.4 9.87sinB 1.5 - 7.53Bsin2B 9.87ET =−=−=

Local longitude angle = 55°12’ E = -55.2° (conversion in °, and negative since location is in EastStandard time zone UTC+4 Local Time meridian (LTM)= -60° (negative since location is in East)( ) ( g )No daylight saving time February LST = 14:00

b) H l d l d li ti δ

( ) [ ] 27:13degree4min.)55.2(60)13.95(00:14angle alLongitudin Local - M LT

degree4minE(min) LST AST =°−−°−+°−+=++=

16.97365

34)(284360sin23.45365

n)(284360sin 23.45δ −=+°

°=+°

°=b) Hour angle ω and solar declination δ.AST = 13:27 = 13.45 h (conversion of time in hours) ω =15° (hours from local solar noon) = 15° (ST-12)ω = 15°.( 13.45-12) = 21.75°

[ ] °°° 42 9816 97)5) () (21 7(2516 97)) i (i (25ihδ) ( )(L) (δ)i (L) i (i (h) -1[ ] °=−°+−°=⇒+= 42.9816.97)5).cos().cos(21.7cos(2516.97)).sin(sin(25sinhδ)cos(ω)cos(L)cos(δ)sin(L)sin(sin(h) -1

°=⎥⎦

⎤⎢⎣

⎡ −=⇒= − 28.98

cos(42.98)(21.75)16.97).sincos(sinAz

cos(h)sinω . cosδsin(Az) 1

The Sun’s path The Sun always rises in the eastIt rises higher and higher in the sky at noonAM, before noon: Line is some time in the morningAt solar noon: Sun reaches its maximum altitude

Ψ = zenith angle

Noon altitude: depending on your latitudePM, afternoon: Line is some time in the afternoonThe Sun starts to set (go down) in the West

β = altitude angle

Az = azimuth anglealtitude (h or β)

measured in degrees PM, afternoon

PP

http://solardat.uoregon.edu/PolarSunChartProgram.htmlAM, before noon

The Sun’s path Copyrighted Material, from internet

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

h

AAz


Th j ti f th th i h i d h d li h i t l lThe projection of the sun-path is shown in dashed line on horizontal palne

h

Az


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

h

Az

Objective of this course PVSEC-1

U d t di h th l ill i ti t l ti E th i thCopyrighted Material, from internet

Understanding how the solar illumination at any location on Earth varies over the courseof a year. You will know how to correctly set the orientation of fixed PV panels installedoutdoors to maximize annual energy production.

CAX

D

Furthermore: you will be able to answer to such questions: If it 9 p.m. at Position D, what time is it 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 AnalysisHow to locate the sun in the sky at any timeHow to locate the sun in the sky at any timeWhat sites will be in the shade at any time

Two interesting ways to represent sun course over the year:g y p yPolar diagrams and vertical diagrams.Determine the azimuth and altitude angles of trees, buildings, and other obstructionsEstimate 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 ( i hb ’ t i bl ki l l)(e.g., a neighbor’s tree is blocking a solar panel).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 sunshineperiods of a site solar masks due to neighboring buildings impact of the orientation periods of a site, solar masks due to neighboring buildings, impact of the orientation of the building, location of the windows, need and kind of solar protections….

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

Sun Path Diagrams for Shading AnalysisTool that helps you reading the movement of the sun throughout the day and during the seasons


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

β 46°β 58°

Equinox

Az 74°

Az 38°

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

Sun Path Diagrams for Shading AnalysisCopyrighted Material, from internet

Agadir

June 21 Sunrise/sunsetSunriseSunset

Today is March 25th. 2012Latitude: +30.42 (30°25'12"N)Longitude: -9.61 (9°36'36"W)Time zone: UTC+0 hoursLocal time: 12:39:40

Today (March 25th. 2012)Sun rises at 06:36 from North-East (Az = 90). Sun set happens Local time: 12:39:40

Country: MoroccoContinent: AfricaSub-region: Northern Africa

December 21

East (Az 90). Sun set happens at 18:53 when the sun is in North-West (Az = 270). On that day the elevation h = 50° at noon

December 21

Equinox March

September)September)

Variable J F M A M J J A S O N DInsolation,

kWh/m²/day 3.52 4.36 5.58 6.73 7.37 7.45 7.09 6.72 5.80 4.73 3.76 3.14

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

kWh/m /day



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 energiesCopyrighted Material, from internet

νλ c= )(

1239)(nm

eVEp λ=→==

λν cE p

hhShort Wavelength Long Wavelength

Max Planck1858 - 1947

Albert Einstein1879 - 1955

100 W

Q: What is Power [unit watts] ?

Example: 100 W light bulb is turning on for one hour Energy consumed is:100 W·h or 0 1 kW h

A: Rate at which energy is generated or consumed

Energy consumed is:100 W·h or 0.1 kW.h. Same amount would be generated from 40-watt light bulb for 2.5 hours

Absorption of Light by Atoms

The sun as a blackbodyCopyrighted Material, from internet

Absorption of Light by AtomsAbsorption occurs only when the energy of the light equals

the energy of transition of an electron

‐

Single electron transition in an isolated atom1 Electron 1 Photon (E = hν)1 Electron 1 Photon (E = hν)

Absorption of Light by Molecules

The sun as a blackbodyCopyrighted Material, from internet

Absorption of Light by Molecules

Smallest ΔE possible

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

Black Body RadiationPlanck law St f B lt L


⎤⎡⎟⎞

⎜⎛

=hc12hcE 5

2

λT),(λ A(m)λ

Planck lawRadiance of BB at fixed T (any λ)

Stefan-Boltzmann LawTotal amount of energyσ = 5.67 × 10-8 Wm-2K-4

Wien’s law λ at peak irradiance

42 T)F(W⎥⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛ 1Tk

hcexpBλ

λ( )KT

(mp ≈)λ 42 T)F(W.m σ=−

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 μmFT = 390 W m-2

ComputeCopyrighted Material, from internet

Consider the earth to be a blackbody with average surface temperature15°C and area equal to 5.1 x 1014 m2. Find the rate at which energy isradiated by the earth and the wavelength at which maximum power isradiated by the earth and the wavelength at which maximum power isradiated. Compare this peak wavelength with that for a 5800 Kblackbody (the sun).The earth radiates:

Watt10 x2.0273) x(15 )m10 x(5.1 )KWm10 x(5.64E 17421442-8 == −−

The wavelength at which the maximum power is emitted

0.5μ.58002898(sun) μm 10.1

2882898

T(K) 2898(earth)max ===== λλ

Radiation flux; Luminescence , EmittanceCopyrighted Material, from internet

dΦ

Intensity emitted by a source in a direction oxA surface element dS of a source S, and any direction Ox with respect to this element dS.

dΩdΦ

=][Watt.Str I Ox-1Ox

Source intensity (in Watts / Steradian)

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

2Rcos dS = d θΩ

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

Radiation flux; Luminescence , EmittanceCopyrighted Material, from internet

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

cosdS=d θΩ 2R = dΩ

Lox is the radiated power per unit of solid angle surrounding the Ox direction per unit j t d di l l t thi di ti i W tt/ 2 té diarea projected perpendicularly to this direction in Watt/m2.stéradian.

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.

L = IdS

= I

dS cosOxOx Ox

′ β

β pW/stWatt/m2.st

dS dS cos β

ΦΩΦ

d=d

d

=L Ox2

Ox

Ox

m2

βΩβ cosdSdcosdSOx

d = L dS cos d 2Ox OxΦ Ωβ

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 depends only on the temperature T of the surface. One can calculate the total flux:

∫∫∫=Φ⇒O2 dΩ cosβ L dS d dΩ cosβ L dS = Φd ∫∫∫=Φ⇒

2ππ.sOx dΩ cosβ L.dS d dΩ cosβ L.dS Φd

M =dd

= L cos dΦ

Ωβ∫∫∫dScos d

2 srβ

π∫∫∫

πLdβ 2β sin L π = dβ β sin β cos L π 2 = M 2π

2π

=∫∫ πLdβ 2β sin L π dβ β sin β cos L π 2 M00

=∫∫

L . π= M

Solar flux intercepted by the EarthCopyrighted Material, from internet

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)

dcosdSL=d 2 ΩθΦ

2dS'dS d'cos dS' = d θ

Ω →

d cosdSLd T ´dSdS´dSdS →→ ΩθΦ

2

0T

dS'dS2

d' cos dS' cos dS M = d θθ

→Φ

TT L.M π=

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:

2dπ

dS to an element dS of the earth s surface:

∫ ∫ θθπΦΦ →→

SS 2

0TdS'dS

2dS'S dS cos

d' cos dS' M = d = d

S

∫∫Σ

2

S

R π = Σ = dΣ = dS cosθ dS cos θ = projection of the element dS on the diametral plane of the sun

Solar flux intercepted by the EarthCopyrighted Material, from internet

dS' DR M = dΦ

20TdS'S ⎟

⎠⎞

⎜⎝⎛

→ D ⎠⎝Solar illumination of the earth is given by the equation

20T

dS'S

DR M =

dS'dΦ

= E ⎟⎠⎞

⎜⎝⎛→

24

DR T σ = E ⎟⎠⎞

⎜⎝⎛⇒ Inverse square law

of irradiance

σ = 5,67 . 10-8 W/(m2.K4), T = 5800 K

R = 6,96.108 m, D = 1,49.1011 mR 6,96.10 m, D 1,49.10 m

E = 1402 W/m2

The atmosphere will transmit a fraction (75%) of solar radiationp ( )

τ E = 0,75 E = 1052 W/m2

EarthEarth--Atmosphere Energy BalanceAtmosphere Energy BalanceCopyrighted Material, from internet

Solar radiation intersects Earth as a disk (πr2)

(Energy)in = Energy from sun (S) – Reflected Solar radiation= πr2 S - πr2 Sα πr S πr Sαr = radius of Earth (6360 km)

S = solar constant (1368 W/m2)α = albedo (earth’s reflectivity) (~30%)

Ein= πr2 S (1- α)Ein πr S (1 α)

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 EffectEarth’s natural greenhouse


PyrheliometerMeasures the direct solar beam(pointed at the sun)

Pyranometers

Measures temperature difference between an

PyranometersUsed to measure global solar radiation(both the direct solar beam and to diffuse sky radiation from the whole hemisphere)Measures temperature difference between anabsorbing (black) plate and a non-absorbing (white)plate. Thermopile converts temperature differenceof 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.

G(φ, λ, t1, t2) the total amount of solar irradation at latitude φ, longitude λ, betweenEnergy from the Sun at the Earth’s Surface

1 2time 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.

When the Sun is overhead, 100% of a beam of

Cosine Law of sunshine intensity

width, I0 strikes a piece of ground of width, I0.

As the Sun goes down and Zenith Angle, Z increases,progressively less of the sunbeam of width I strikes

I0progressively less of the sunbeam of width I0, strikesthe piece of ground. More of the sunbeam misses thatpiece of ground and is lost.

ZCOSINE LAW

ZII Z cos0 ×=

The fraction of the sunbeam that strikes the groundI /I hi h t i t h i l t (Z)

hh Z

I0Z

= IZ/I0, which trigonometry shows is equal to cos(Z)GROUND

Direct-beam radiation: that passes in a straight line through the atmosphere to the

Solar flux striking a collectorp g g p

receiver.Diffuse radiation : that has been scattered by molecules and aerosols in the

atmosphereatmosphere.Reflected radiation: that has bounced off the ground or other surface in front of the

collector.

Diffuse radiation Direct-beam

Diffuse radiation

collector, C

Reflected radiation

collector, C

Tilt angle

Extraterrestrial (ET) solar insolation, I0 (Watt/m2)

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

IEarth

I0

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 unitarea, perpendicular to the rays, outside the atmospheren is the day number.y

Base on NASA measurements SC = 1353 W/m2 (commonly accepted value 1377 W/m2)

Over a year’s time, less than half of the radiation that hits the top of the atmosphere

Attenuation and air massy , 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 fluxAttenuation 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 functiony p y

m) . kAexp(IB −=

IB = beam portion of the radiation that reaches the earth’s surface B pA = apparent extraterrestrial fluxk = optical depth m = air mass ratio:

sinβ1m =

β: altitudeφS solar azimuthφC Panel azimuth

where β is the altitude angle of the sun.sinβ

Panel Tilt

φC Panel azimuth

N

S Panel TiltS

Attenuation and air mass

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

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

Close fits to the values from the above table

m) . kAexp(IB −=

Close fits to the values from the above table )(W/m 275)(n

36536075sin1160A 2

⎥⎦⎤

⎢⎣⎡ −+= 100)(n

3653600.035sin0.175k ⎥⎦

⎤⎢⎣⎡ −+=

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


Direct Beam Radiation at the Surface of the Earth –

Attenuation and air mass

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

22 W/m1104)(W/m 275)(n365360 . 75sin1160A =⎥⎦

⎤⎢⎣⎡ −+=

360 ⎤⎡ 0.197 100)(n3653600.035sin0.175k =⎥⎦

⎤⎢⎣⎡ −+=

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

β: altitudeφS solar azimuthφC Panel azimuth

nPanel Tilt Panel Tilt

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

azimuth angles of the sun.

Special case of beam insolation on a horizontal surface

Insolation on a CollectorAt solar noon in Atlanta (latitude 33.7°C) on May 21 the altitude angle of the sun was found to be At solar noon in Atlanta (latitude 33.7 C) on May 21 the altitude angle of the sun was found to be 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 CollectorDiffuse Radiation on a Collector - find the diffuse radiation on the panel. Recall that it is solar 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.DH B 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 , pbright day with snow or water in front of the collector.

The amount reflected can be modeled as theproduct of the total horizontal radiationp(beam IBH , plus diffuse IDH) times the groundreflectance ρ. The fraction of that ground-reflected energy that will be intercepted bythe collector depends on the slope of thethe collector depends on the slope of thepanel , resulting in the following expressionfor reflected radiation striking the collectorIRC:RC

( ) radiation reflected no0reflector Horizontal ⇒=∑radiationreflectedtheof1sees""panelthethatpredictsit)(90panelVertical ⇒° radiationreflectedtheof

2 sees panelthethat predictsit )(90 panel Vertical ⇒

⇒

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