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Photovoltaic Solar Energy Conversion

Advanced course 1

ENIM Rabat Morocco

إنتاج الكهرباء من الطاقة الشمسية

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- 1. 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 allowcustomization for the individual needs of course instructors. Permission of the author and publisher is required for any other usage.
- 2. PVSEC-1: all about solar radiationSome information: Helmholtz-Zentrum Berlin für Materialien und Energie Short on my ongoing research activities at HZB and PVComBWhy this new concept of meetings ?Highlight of Part IEarths motion around the sun and tracking the sun in the skySolar altitude angle at solar noon and orientation of solar panels g pSolar angles: the longitude, latitude, solar declination.Hour angle; azimuth, angular height and orientation of solar panelsSolar time (ST) and local standard time (LST)Optimal orientation of fixed PV panelsThe sun as a blackbodySolarS l constant and solar spectrum t t d l tDirect radiation, diffuse and albedo sunriseAir mass or air mass number sunsetTotal radiation received by a surface
- 3. 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
- 4. HZB & PVcomB in the Helmholtz AssociationQuelle: PVComB/Rutger Schlatmann
- 5. 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 SpaceFinancing of activities (programmes) instead of financing single institutes (centres)Programme Oriented Funding (POF)
- 6. Solar Energy Division in HZBQuelle: PVComB/Rutger Schlatmann
- 7. Goal Strategy of PVcomB PVcomBKompetenzzentrum Dünnschicht- und Nanotechnologie für Photovoltaik BerlinQuelle: PVComB/Rutger Schlatmann
- 8. 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 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) 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
- 9. 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
- 10. 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-Germanypromote 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
- 11. Introduction: Grid-connected PV systems Copyrighted Material, from internet Controller, (charge regulator) regulates the voltage and currentTraditional 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 dont 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 companys appliances. power grid.
- 12. Introduction: Grid-connected PV systems Copyrighted Material, from internet Controller, (charge regulator) regulates the voltage and currentTraditional 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 dont 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 companys appliances. power grid.
- 13. Intensity of sun light on ground Copyrighted Material, from internet 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 AM0http://pvcdrom.pveducation.org/SUNLIGHT/AIRMASS.HTM
- 14. Objective of this course PVSEC-1Understanding how the l illumination tU d t di h th solar ill i ti at any l ti on E th varies over th course location Earth i theof a year. You will know how to correctly set the orientation of fixed PV panels installedoutdoors to maximize annual energy production. X A C DFurthermore: you will be able to answer to such questions: If it 9 p.m. at Position D, what time is itat position C? Position B? If it is 1 p.m. at Position X, at which location is the time 5 p.m.
- 15. Earths motion around the sun and tracking the sun in the sky Copyrighted Material, from internet 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 eachSouth Pole closer to the Sun than the North Pole Pole. A line from the center of the Sun to theWinter solstice in the Northern Hemisphere center of Earth passes right through theSummer 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 +23.45of Cancer (latitude +23 45 ) equator (latitude 0 ) ),
- 16. 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 hhttp://www.geonames.org/search.html?q=rabathttp://maps.google.com/maps
- 17. Longitude and inclinaison Copyrighted Material, from internet• 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 ) δe.g. point (A) on earth surface exactly 15o West Aof another point (B), will see the sun in exactly Bthe 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 Monthn is the nth. day of the year since 1st. January
- 18. Optimal orientation of fixed PV panelsGeneral rule of thumb to be followed when installing fixed PV panels outdoors to maximize theannual 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,
- 19. Altitude angle at solar noon L P δ βNoon = 90 + L- δ Altit d angle, β Altitude l Equation E i L Local horizontalExample 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: ⎡ 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 = 47.7Tilt = 90− βnoon = 90− 47.7= 42.3 ° β noon = 47.7°
- 20. 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 suns ray and a line perpendicular to the horizontal plane.h or β = Altitude angle in vertical plane between the suns rays and projection of the suns ray sun s sun son a horizontal plane.Az (or ϕS) = Azimuth angle measured from south to the horizontal projection of the sun’s ray. P
- 21. Solar Angles Hour angle HA (called also ω ) the number of degrees the earth must rotate beforesun 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 ⎠
- 22. 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 SFind altitude angle β and azimuth angle S at 3 PM solar time in Boulder, CO (L = 40˚) on the summersolstice:.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)
- 23. Solar anglesSunrise 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.
- 24. Solar Position vs solar panel orientation Copyrighted Material, from internet Winter Summer Spring/Autumn S i /A t sunrise sunset 56° angle 32° angle 80° anglehttp://solarelectricityhandbook.com/solar-angle-calculator.html
- 25. Solar Time vs. Clock Time Solar time, ST: World Time Zones: http://wwp.greenwichmeantime. h // i h iWe 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.htmEach 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
- 26. 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
- 27. Solar Time vs. Clock Time 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
- 28. Solar Time vs. Clock Time Equation of Time E q 360 E = 9.87 sin2B − 7.53B.1.5 sinB B= (n − 81) degrees n = day number 364 Day Numbers for the First Day of Each Month 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 tiWorld Time Zones:http://wwp.greenwichmeantime.com/time-zone/http://wwp.greenwichmeantime.com/time-zone/africa/morocco/index.htm
- 29. 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. EastTime 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 90thmeridian 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 meridianwest. Alaska was standardized in 1918 on 150th Meridian west, but in actual practice, other zones are and have been in use: 120° 150°, 165°
- 30. Example 2: Solar Time vs. Clock TimeFor 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 ia. 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 δb) 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 365ω =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) ⎦
- 31. The Sun’s pathThe Sun always rises in the eastIt rises higher and higher in the sky at noon Ψ = zenith angleAM, before noon: Line is some time in the morningAt solar noon: Sun reaches its maximum altitudeNoon altitude: depending on your latitude β = altitude anglePM, afternoon: Line is some time in the afternoonThe Sun starts to set (go down) in the West Az = azimuth anglealtitude (h or β) measured in degrees PM, afternoon P AM, before noon http://solardat.uoregon.edu/PolarSunChartProgram.html
- 32. The Sun’s path Copyrighted Material, from internet The projection of the sun-path is shown in dashed line on horizontal palne h Az
- 33. The Sun’s path Copyrighted Material, from internet 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
- 34. The Sun’s path Copyrighted Material, from internet The projection of the sun-path is shown in dashed line on horizontal palne h Az
- 35. Objective of this course PVSEC-1 Copyrighted Material, from internetUnderstanding how the l illumination tU d t di h th solar ill i ti at any l ti on E th varies over th course location Earth i theof a year. You will know how to correctly set the orientation of fixed PV panels installedoutdoors to maximize annual energy production. X A C DFurthermore: you will be able to answer to such questions: If it 9 p.m. at Position D, what time is itat position C? Position B? If it is 1 p.m. at Position X, at which location is the time 5 p.m.
- 36. 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.htmlhttp://www.jaloxa.eu/resources/daylighting/sunpath.shtmlhttp://www.youtube.com/v/IjOhtmmq7aM&hl=en_US&fs=1&rel=0
- 37. Sun Path Diagrams for Shading Analysis Copyrighted Material, from internet 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
- 38. Sun Path Diagrams for Shading Analysis Copyrighted Material, from internet Sunrise/sunset June 21 Sunrise SunsetAgadirToday is March 25th. 2012Latitude: +30.42 (30°2512"N)Longitude: -9.61 (9°3636"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 happensCountry: Morocco at 18:53 when the sun is in North-Continent: Africa West (Az = 270). On that day theSub-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/dayhttp://www.gaisma.com/en/location/agadir.html
- 39. Sun Path Diagrams for Shading Analysis Copyrighted Material, from internet
- 40. Sun Path Diagrams for Shading Analysis Copyrighted Material, from internet Figure belowSolutionWorkshop Friday The sun path diagram with superimposed obstructions makes it easy to estimate periods of shading at a site.
- 41. Colors of light have different wavelengths and different energies Copyrighted Material, from internet Short Wavelength Long Wavelength c hc 1239 λ= E p = hν = → E p (eV ) = ν λ λ (nm) Max Planck 1858 - 1947Albert 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
- 42. The sun as a blackbody Copyrighted 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 atom 1 Electron 1 Photon (E = hν)
- 43. The sun as a blackbody Copyrighted Material, from internetAbsorption 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
- 44. Black Body Radiation Copyrighted Material, from internet Planck law Wien’s law Stefan-Boltzmann Law St f B lt LRadiance 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
- 45. Compute Copyrighted Material, from internetConsider 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. Compare this peak wavelength with that for a 5800 Kblackbody (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 WattThe wavelength at which the maximum power is emitted 2898 2898 2898λmax (earth) = = = 10.1μm λ (sun) = = 0.5μ. T(K) 288 5800
- 46. Radiation flux; Luminescence , Emittance Copyrighted Material, from internetIntensity emitted by a source in a direction oxA surface element dS of a source S, and any direction Oxwith respect to this element dS. dΦ Ox Source intensity I Ox [Watt.Str -1 ] = dΩ (in Watts / Steradian)S radiates throughout the space g pΦ the radiation flux and dΦox portion of Φ radiated into asolid angle dΩ.Iox Source intensity in the direction Ox dΩ = dS cosθ R2For a hemispherical space, the solid angle = 2π SteradianSolid angle for all the space = 4π Steradian
- 47. Radiation flux; Luminescence , Emittance Copyrighted Material, from internetLox: Luminescence of a source area dSFlux from the projected area dS = dS cosβ dΩ = dS cosθ dΩ R2Lox is the radiated power per unit of solid angle surrounding the Ox direction per unitarea projected perpendicularly t thi di ti i W tt/ 2.stéradian. j t d di l l to this direction in Watt/m té diThe flux emitted by a surface element dS in a solid angle dΩ surrounding a directionOx, 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
- 48. The emittance, M of a diffuse sourceDiffuse sources are governed by LAMBERT RULE: regardless of thedirection of observation Lox = L . This is the case where the luminance Ldepends only on the temperature T of the surface. One can calculate surfacethe 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
- 49. Solar flux intercepted by the Earth Copyrighted 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) 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 d2d2ΦdS dS´ = flux emitted by the element dS in a solid angle dΩ surrounding the directiondS to an element dS´ of the earths surface: dS earth s M0 dS cosθ dΦS→dS = ∫d2ΦdS→dS = T ∫ cosθ dS π S d2 SdS cos θ = projection of the element dS on the diametral plane of the sun ∫ S cosθ dS = ∫ dΣ = Σ = π R 2 Σ
- 50. Solar flux intercepted by the Earth Copyrighted Material, from internet 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 irradianceσ = 5,67 . 10-8 W/(m2.K4), T = 5800 KR = 6,96.108 m, D = 1,49.1011 mE = 1402 W/m2The atmosphere will transmit a fraction (75%) of solar radiation p ( ) τ E = 0,75 E = 1052 W/m2
- 51. Earth- Earth-Atmosphere Energy Balance Copyrighted Material, from internet 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) (1Earth 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 EarthBlack 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 = 33Interactions within atmosphere alter radiation budget (Earth is not a black body) Greenhouse EffectEarth’s natural greenhouse
- 52. Copyrighted Material, from internet 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.pngAverage irradiance in watts per square meter (W/m2) available can be measuredI-V characteristic of a solar cell is a sensor for solar radiation.
- 53. Energy from the Sun at the Earth’s SurfaceG(φ, λ, t1, t2) the total amount of solar irradation at latitude φ, longitude λ, betweentime 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.
- 54. Cosine Law of sunshine intensityWhen the Sun is overhead, 100% of a beam ofwidth, I0 strikes a piece of ground of width, I0.As the Sun goes down and Zenith Angle, Z increases, I0progressively less of the sunbeam of width I0, strikesthe piece of ground. More of the sunbeam misses thatpiece of ground and is lost. COSINE LAW Z I Z = I 0 × cos Z h Z I0The 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
- 55. Solar flux striking a collector Direct-beam radiation: that passes in a straight line through the atmosphere to the p g g preceiver. Diffuse radiation : that has been scattered by molecules and aerosols in theatmosphere.atmosphere Reflected radiation: that has bounced off the ground or other surface in front of thecollector. Diffuse radiation Direct-beam collector, C Tilt angle Reflected radiation
- 56. 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 sunspotsSC is the solar constant which is the average power of the suns radiation that reaches a unitarea, perpendicular to the rays, outside the atmospheren is the day number. yBase on NASA measurements SC = 1353 W/m2 (commonly accepted value 1377 W/m2)
- 57. 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 calculableA 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 pA = apparent extraterrestrial fluxk = optical depth β: altitude 1 φS solar azimuthm = air mass ratio: m = sinβ φC Panel azimuthwhere β is the altitude angle of the sun. N Panel Tilt S
- 58. Attenuation and air mass 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 ⎦ Day Numbers for the First Day of Each MonthASHRAE, 1993, Handbook of Fundamentals, American Society of Heating, Refrigeration and Air Conditioning Engineers, Atlanta.
- 59. Attenuation and air massDirect 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 inAtlanta (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:
- 60. 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 andazimuth angles of the sun. Special case of beam insolation on a horizontal surface
- 61. 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 33.7 C)76.4° and the clear-sky beam insolation was found to be 902 W/m2. Find the beam insolation atthat 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
- 62. 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. yThe diffuse sky factor, CThe diffuse energy striking the collector Total beam insolation (697 W/m2) plus diffuse on the collector (88W/m2) 785 W/m2.
- 63. 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 1Vertical panel (90°) ⇒ it predicts that the panel " sees" sees of the reflected radiation 2 ⇒More detail in workshop and problem solving activities
- 64. NEXT PVSEC-2Fundamental and application of ppPhotovoltaic solar cells and system

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