Presentation is loading. Please wait.

Presentation is loading. Please wait.

(Contents are solely for educational purpose)

Similar presentations


Presentation on theme: "(Contents are solely for educational purpose)"— Presentation transcript:

1 (Contents are solely for educational purpose)
聚光型太陽電池發展趨勢 (OE_10290) 陳怡嘉 國立東華大學光電工程學系 (Friday) 14:10-17:00 (Contents are solely for educational purpose)

2 能源需求 太陽光 太陽能 單位換算 太陽光譜 空氣質量 太陽能電池 聚光型太陽能電池

3 Global Primary Energy Consumption
Individual Accumulated

4 Global Primary Energy Consumption
In 2008, total worldwide energy consumption was 474 exajoules (132,000 TWh). This is equivalent to an average power use of 15 terawatts (2.0×1010 hp).[1] Based upon some attempted estimates, making strong assumptions, the annual potential for renewable energy are of the order of: solar energy 1,575 EJ (438,000 TWh), wind power 640 EJ (180,000 TWh), geothermal energy 5,000 EJ (1,400,000 TWh), biomass 276 EJ (77,000 TWh), hydropower 50 EJ (14,000 TWh) and ocean energy 1 EJ (280 TWh).[2][3][4] [1] "Consumption by fuel, 1965–2008" (XLS). Statistical Review of World Energy BP. 8 June Archived from the original on 26 July Retrieved 24 October 2009. [2] World Energy Assessment (WEA). UNDP, New York [3] Johansson, T. B., McCormick, K., Neij, L., & Turkenburg, W. (2004). The Potentials of Renewable Energy: Thematic Background Paper. Thematic Paper prepared for the International Conference on Renewable Energies, Bonn. Retrieved 6 July 2008, from [4] de Vries BJM, van Vuuren DP, Hoogwijk MM (2007). "Renewable energy sources: Their global potential for the first-half of the 21st century at a global level: An integrated approach". Energy Policy 35: 2590–2610. doi: /j.enpol

5 Conversion between energy and power
Unit Conversion Energy How? 3600 Value of energy in TWh can be obtained by multiplying value of energy in EJ by Conversion between energy and power How? Value of equivalent power in TW can be obtained by multiplying the corresponding value of energy in EJ by Why? Explained later.

6 Electric Energy Consumption
Consumption of electric energy is measured in watt-hours (written W·h, equal to Watt x Hour) 1 W·h = 3600 joule = calorie. Electric and electronic devices consume electric energy to generate desired output (i.e. light, heat, motion, etc.). During operation, some part of the energy is consumed in unintended output, such as waste heat. See Electrical Efficiency . In 2008, the world total of electricity production and consumption was 20,279 TWh (terawatt-hours). This number corresponds to an average consumption rate of around 2.3 terawatts continuously during the year. The total energy needed to produce this power is roughly a factor 2 to 3 higher because the efficiency of power plants is roughly 30-50%, see Electricity generation. The generated power is thus in the order of 5 TW. This is approximately a third of the total energy consumption of 15 TW, see World energy consumption. 16816TWh (83%) of electric energy was consumed by final users. The difference of 3464TWh (17%) was consumed in the process of generating power and consumed as transmission loss and all most consumed at misuse.

7 Global and Electric Energy Consumptions
Total worldwide energy consumption World total of electricity consumption Electric energy consumed by final users (The difference of 3464 TWh (17%) was consumed in the process of generating power and consumed as transmission loss and all most consumed at misuse.) Energy needed to produce the world total of electricity consumption (Efficiency of power plants is roughly 30-50%, say, 44%)

8 能源需求 太陽光 太陽能 單位換算 太陽光譜 空氣質量 太陽能電池 聚光型太陽能電池

9 Global Solar Energy Flows
petawatts (PW) (1 PW = 1,000,000,000,000,000 Watts) PW = 1015 W

10 Standard SI Prefixes for Watt
SI multiples for watt (W) Submultiples Multiples Value Symbol Name 10−1 W dW deciwatt 101 W daW decawatt 10−2 W cW centiwatt 102 W hW hectowatt 10−3 W mW milliwatt 103 W kW kilowatt 10−6 W µW microwatt 106 W MW megawatt 10−9 W nW nanowatt 109 W GW gigawatt 10−12 W pW picowatt 1012 W TW terawatt 10−15 W fW femtowatt 1015 W PW petawatt 10−18 W aW attowatt 1018 W EW exawatt 10−21 W zW zeptowatt 1021 W ZW zettawatt 10−24 W yW yoctowatt 1024 W YW yottawatt Common multiples are in bold face

11 Outstanding solar potential compared to all other energy sources
Compared with 174 PW (174,000 TW) of total incoming solar radiation

12 Yearly Solar Fluxes & Human Energy Consumption
Equivalent to non-reflected solar power of 122 PW Yearly Solar fluxes & Human Energy Consumption Solar 3,850,000 EJ Wind 2,250 EJ Biomass potential 100–300 EJ Primary energy use (2010) 539 EJ Electricity (2010) 66.5 EJ EJ = 1018 J

13 Equivalence of Solar Fluxes
Power Energy Flux Total 174 PW 5,487,264 EJ/yr 1367 W/m2 Non-reflected 122 PW 3,850,000 EJ/yr 957 W/m2 Total incoming solar radiation 174 PW Reflected radiation 52 PW Non-reflected radiation 174 – 52 = 122 PW

14 Equivalence of Solar Fluxes
Power Energy Flux Total 174 PW 5,487,264 EJ/yr 1367 W/m2 Non-reflected 122 PW 3,850,000 EJ/yr 957 W/m2 Irradiance (power flux) normal to the sun radiation Circumference of earth sphere  km ( = D) Area of cross section of earth sphere passing through sphere center = r2 =   (40000 / 2 / )2 =  108 km2

15 Earth Dimensions Radius 6378.1 km (equatorial) 6356.8 km (polar)
Circumference km (equatorial)  km (meridional)  Surface area km2

16 400 km 377 km 23 km

17 Equivalence of Solar Fluxes
Power Energy (EJ) Energy (TWh) Solar Energy 174,000 TW 5,487,264 EJ/yr 1,530,000,000 TWh/yr Global Energy consumption 15 TW 474 EJ/yr 132,000 TWh/yr

18 能源需求 太陽光 太陽能 單位換算 太陽光譜 空氣質量 太陽能電池 聚光型太陽能電池

19 1KLOE=4032度 Joule, 1 J = 1 N × 1 m joule = volt × coulomb

20 (一)1卡(Cal)=4.1868焦耳(Joule) (二)1BTU=1,055焦耳(Joule)=252卡(Cal) (三)1千瓦小時(kWh)=3.6×106 焦耳 (四)1千公秉油當量(KLOE,相當於1千公秉原油所含之熱量) =9.0百萬千卡=9.2×106 Kcal=9.2×109 Cal (五)1千公噸標準煤當量(KTCE,相當於1千公噸標準煤所含的熱量)=6.4百萬千卡=6.4×106 Kcal=6.4×109 Cal (六)1公噸油當量(TOE,相當於1公噸原油所含之熱量)=1千萬千卡=1.0×107 Kcal=1.0×1010 Cal=4.1868×1010 焦耳(Joule)

21 英國熱能單位(Btu): 1個英國熱能單位等於將1磅的水溫度升高1華氏度所需要的熱量。它是用來計量燃料能量和產熱設備輸出功率的標準度量單位。
常用的能量單位包括: 英國熱能單位(Btu): 1個英國熱能單位等於將1磅的水溫度升高1華氏度所需要的熱量。它是用來計量燃料能量和產熱設備輸出功率的標準度量單位。 卡路里(Cal): 1卡路里等於將1克水溫度升高1攝氏度所需要的能量。 焦耳(J): 能量計量的基本單位。1焦耳定義為使用1牛頓力使物體沿力的方向移動1米所消耗的能量。 千瓦小時(kWh): 千瓦小時一般用來度量一段時間內電能的消耗量。用這個單位可以有效地計量家用電器(比如電冰箱)的耗電量。每月的電費帳單也是採用千瓦小時來計量的。1千瓦等於1000瓦,1千瓦小時就是一個功率為1000瓦的用電設備1小時的耗電量。 噸油當量: 1噸油當量表示燃燒一噸(1000公斤)或者7.4桶石油所獲得的能量。1噸油當量等價於1270立方米天然氣或者1.4噸煤所蘊藏的能量,即41.87千兆焦耳(GJ)或者11.63兆瓦時能量。

22 初級能源的能量值 煤 - 約250萬英國熱能單位/噸 原油 - 約560萬英國熱能單位/桶 油 - 約578萬英國熱能單位/桶 天然氣
- 約250萬英國熱能單位/噸 原油 - 約560萬英國熱能單位/桶 - 約578萬英國熱能單位/桶 天然氣 - 約1,030英國熱能單位/立方尺 液態天然氣 - 約250萬英國熱能單位/桶

23 功率是指能量消耗的速度。功率的度量單位有馬力和瓦特。和能量的單位一樣,功率的單位也可以相互轉換。
馬力: 馬力是功的單位。這個單位最初用來度量將煤從煤礦中提升上來所需要的能量。為了更好的理解馬力這個單位,可以做如下的換算,即1馬力等於將33000磅的物體在1分鐘內提升1英尺所做的功。 瓦特: 瓦特是功率的單位,表示在特定的時間內,能源消耗的速度。1瓦特等於每秒1焦耳。

24 能量單位換算 1 卡路里 = 4.1868 焦耳 1 英國熱能單位 = 1,055 焦耳 = 252 卡路里 1 千瓦小時
1 卡路里 = 焦耳 1 英國熱能單位 = 1,055 焦耳 = 252 卡路里 1 千瓦小時 = 3.6 x 106 焦耳 1 兆瓦時 = 3.6 x 109 焦耳 1 千兆瓦時 = 3.6 x 1012 焦耳 1 噸油當量 = x 1010 焦耳 1 百萬噸油當量 = x 1016 焦耳

25 能源需求 太陽光 太陽能 單位換算 太陽光譜 空氣質量 太陽能電池 聚光型太陽能電池

26 Solar Spectra

27 Blackbody Radiation Spectrum Overlay
The irradiance of the sun on the outer atmosphere when the sun and earth are spaced at 1 AU - the mean earth/sun distance of 149,597,890 km - is called the solar constant. Currently accepted values are about 1360 W m-2 (the NASA value given in ASTM E a is 1353 ±21 W m-2). The World Metrological Organization (WMO) promotes a value of 1367 W m-2. The solar constant is the total integrated irradiance over the entire spectrum (the area under the curve in Fig. 1 plus the 3.7% at shorter and longer wavelengths). Excellent Explanation

28 Black body Radiation Black-body radiation is the type of electromagnetic radiation within or surrounding a body in thermodynamic equilibrium with its environment, or emitted by a black body (an opaque and non-reflective body) held at constant, uniform temperature. The radiation has a specific spectrum and intensity that depends only on the temperature of the body. Peter Theodore Landsberg (1990). "Chapter 13: Bosons: black-body radiation". Thermodynamics and statistical mechanics (Reprint of Oxford University Press 1978 ed.). Courier Dover Publications. pp. 208 ff. ISBN Planck's law describes the spectral radiance of electromagnetic radiation at all wavelengths from a black body at temperature T. As a function of frequency ν, Planck's law is written as or It can be converted to an expression for I'(λ,T) in wavelength units by substituting ν by c / λ and evaluating or    and from , we have so or The above equation is energy per unit wavelength per unit solid angle.

29 Black body Radiation Curves

30 Black body Radiation Curve at 5800 K
The spectrum of the Sun's solar radiation is close to that of a black body with a temperature of about 5,800 K peaking at a wavelength of 500 nm.

31 Interactive tool for Black Body Radiation
There is an online, interactive tool from the University of Colorado for investigating the spectrum of various blackbodies. Here is the link to run it online: PhET Interactive Simulation of the Blackbody Spectrum.

32 Simple Solar Spectral Model for Direct and Diffuse Irradiance
Black body radiation spectrum as overlay curve Terrestrial absorption RAYLEIGH SCATTERING AEROSOL SCATTERING AND ABSORPTION WATER VAPOR ABSORPTION OZONE AND UNIFORMLY MIXED GAS ABSORPTION Fraunhofer lines

33 Rayleigh Scattering and Mie Scattering
Selective scattering (or Rayleigh scattering) occurs when certain particles are more effective at scattering a particular wavelength of light. Molecules much smaller than the wavelength of the radiation, like oxygen and nitrogen for example, are more effective at scattering shorter wavelengths of light (blue and violet). The selective scattering by air molecules is responsible for producing our blue skies on a clear sunny day. Another type of scattering (called Mie Scattering) is responsible for the white appearance of clouds. Cloud droplets with a diameter of 20 micrometers or so are large enough to scatter all visible wavelengths more or less equally. This means that almost all of the light which enters clouds will be scattered. Because all wavelengths are scattered, clouds appear to be white.

34 Rayleigh Scattering and Mie Scattering

35 Scattering

36 Absorption by Terrestrial Molecules (Telluric lines)

37 AM1.5 Global Solar Spectrum

38 AM1.5 Global Solar Spectrum

39 Transparent Window of Atmosphere
Black-body temperature K (Earth) The Earth’s atmosphere is not completely opaque to longwave. There is a transparent transmission band extending from 8 to 13 microns, in the center of the terrestrial thermal black body curve. This allows up to 30% of the longwave to escape.

40 Fraunhofer Lines In the Sun, Fraunhofer lines are seen from gas in the outer regions of the Sun, which are too cold to directly produce emission lines of the elements they represent. (seen as absorption lines) "Spectrum of blue sky" by Spectrum of blue sky.png : Remember the dotDerivative work : Eric Bajart - Spectrum of blue sky.png. Licensed under GFDL via Wikimedia Commons -

41 Labeling of Fraunhofer Lines
"Fraunhofer lines" by Fraunhofer_lines.jpg: nl:Gebruiker:MaureenVSpectrum-sRGB.svg: PhroodFraunhofer_lines_DE.svg: *Fraunhofer_lines.jpg: Saperaud 19:26, 5. Jul. 2005derivative work: Cepheiden (talk)derivative work: Cepheiden (talk) - Fraunhofer_lines.jpgSpectrum-sRGB.svgFraunhofer_lines_DE.svg. Licensed under Public Domain via Wikimedia Commons - Dark lines in the solar spectrum are caused by absorption by chemical elements in the Solar atmosphere. Some of the observed features were identified as telluric lines originating from absorption in oxygen molecules in the Earth's atmosphere.

42 Telluric Spectrum The solar spectrum from 2958 to 5400 Å is essentially clear of any terrestrial lines. Red of 5400 Å weak H2O lines begin to appear, at first two weak H2O bands (Camy-Peyret et al. 1985). By 5790 Å H2O is joined by O2. Thus for solar flux spectra obtained from ground based facilities, a correction for the telluric spectrum should be undertaken. The discrete terrestrial absorbers are illustrated in the figure. It is apparent from the figure that telluric lines have a substantial effect on a large part of the visible and near-infrared spectrum. The Astrophysical Journal Supplement Series, 195:6 (8pp), 2011 July

43 Illustration of Telluric Contamination
The near infrared night sky of Castanet-Tolosan observatory. The sky line emission background is here detected during an observation of SS433 object with a LISA spectrograph (IR version - R = 800). HPS street lamp view from my balcony observatory (Castanet-Tolosan observatory) ! Emission lines identification in the sky background of SS433 spectrum. "blue" labeled line are artificial (HPS lamps). Note the telluric airglow emission at 6300 A and the rich atmospheric OH spectrum.

44 Standard Solar Spectra
The AM1.5 Global spectrum is designed for flat plate modules and has an integrated power of 1000 W/m2 (100 mW/cm2). The AM1.5 Direct (+circumsolar) spectrum is defined for solar concentrator work. It includes the direct beam from the sun plus the circumsolar component in a disk 2.5 degrees around the sun. The direct plus circumsolar spectrum has an integrated power density of 900 W/m2. The AM0 spectrum is the standard spectrum for space applications and has an integrated power of W/m2.

45 Global, Direct, and Diffuse Insolation
Direct, diffuse, and total insolation for a standard atmosphere, with relative air mass of 1.5.

46 Global, Direct, and Diffuse Insolation
Global Horizontal Radiation - also called Global Horizontal Irradiance; total solar radiation; the sum of Direct Normal Irradiance (DNI), Diffuse Horizontal Irradiance (DHI), and ground-reflected radiation; however, because ground reflected radiation is usually insignificant compared to direct and diffuse, for all practical purposes global radiation is said to be the sum of direct and diffuse radiation only GHI = DHI + DNI * cos (Z) where Z is the solar zenith angle.

47 能源需求 太陽光 太陽能 單位換算 太陽光譜 空氣質量 太陽能電池 聚光型太陽能電池

48 Solar Spectra at Various Air Masses
"Simulated direct irradiance spectra for air mass=0 to 10 with SMARTS 2.9.5" by Solar Gate - My own calculations and graphing. Licensed under CC BY-SA 3.0 via Wikimedia Commons -

49 Air Mass (AM) The air mass coefficient defines the direct optical path length through the Earth's atmosphere, expressed as a ratio relative to the path length vertically upwards, i.e. at the zenith. AM0 The solar spectrum outside the atmosphere, approximated by the 5,800 K black body, is referred to as "AM0", meaning "zero atmospheres". AM1 The spectrum after travelling through the atmosphere to sea level with the sun directly overhead is referred to, by definition, as "AM1". This means "one atmosphere". AM1.5 The spectrum after travelling through “1.5 atmosphere” thickness to sea level, corresponding to a solar zenith angle of =48.2°. 48.2o

50 Air Mass Coefficient (Approximation)
The air mass coefficient defines the direct optical path length through the Earth's atmosphere, expressed as a ratio relative to the path length vertically upwards, i.e. at the zenith.(reasonably accurate for values of up to around 75o)

51 Purpose of Introducing Air Mass
The original purpose of introducing the idea of air mass is to specify the zenith angle of solar irradiance so that the final irradiance power flux can be estimated by consideration of both (1) the attenuation of the irradiance passing through the atmosphere and (2) the attenuation by increased angle of incident. angle of incident However, the estimation is complicated and the precise values still have to resort to experiment measurements. (especially true at high zenith angles)

52 Air Mass Coefficient (Spherical Model)

53 Air Mass Coefficient (Modified Spherical)
Schoenberg, E Theoretische Photometrie, Über die Extinktion des Lichtes in der Erdatmosphäre. In Handbuch der Astrophysik. Band II, erste Hälfte. Berlin: Springer. Spherical model of the Earth’s atmosphere: Geometry for computing a diagonal path through the Earth’s atmosphere. Atmospheric effects on optical transmission, at an angle z to the normal, can be modelled as a path of length s, as if the atmosphere is concentrated uniformly in approximately the lower 9 km. This path is known as the Airmass. The Air mass coefficient is defined as the ratio s/yatm. This model does not apply at non-optical wavelengths where higher layers of the atmosphere affect transmission, for example: the ozone layer at higher ultraviolet frequencies and the ionosphere at lower radio frequencies. Essentially all the atmospheric effects are due to the atmospheric mass in the lower half of the Troposphere. Courtesy of Neil Clarke

54 Spherical Plane Parallel

55 Kasten & Young Spherical Kasten and Young (1989) Plane Parallel

56 Air Mass Coefficient (Schoenberg)
Hardly any difference for z < 70o. Kasten F, Young AT. Revised optical air mass tables and approximation formula. Applied Optics [Internet] ;28:4735–4738. Available from:

57 Standard Solar Irradiance Power
Formula where solar intensity external to the Earth's atmosphere Io = kW/m2, and the factor of 1.1 is derived assuming that the diffuse component is 10% of the direct component. Z AM formula ASTM G-173 degree W/m2 - 1353 1347.9 1 1040 23° 1.09 1020 30° 1.15 1010 45° 1.41 950 48.2° 1.5 930 1000.4 60° 2 840 70° 2.9 710 75° 3.8 620 80° 5.6 470 85° 10 270 90° 38 20 Meinel, A. B. and Meinel, M. P. (1976). Applied Solar Energy Addison Wesley Publishing Co.

58 Equivalence of Solar Fluxes
Power Energy Flux Total 174 PW 5,487,264 EJ/yr 1367 W/m2 Non-reflected 122 PW 3,850,000 EJ/yr 957 W/m2 Compared with formula value of 1040 W/m2 at AM1. Non-reflected irradiance (power flux) Circumference of earth sphere  km ( = D) Area of cross section of earth sphere passing through sphere center = r2 =   (40000 / 2 / )2 =  108 km2

59 Peak Sun Hours The average daily solar insolation in units of kWh/m2 per day is sometimes referred to as "peak sun hours". The term "peak sun hours" refers to the solar insolation which a particular location would receive if the sun were shining at its maximum value for a certain number of hours. Since the peak solar radiation is 1 kW/m2, the number of peak sun hours is numerically identical to the average daily solar insolation. For example, a location that receives 8 kWh/m2 per day can be said to have received 8 hours of sun per day at 1 kW/m2. Being able to calculate the peak sun hours is useful because PV modules are often rated at an input rating of 1kW/m2.

60 Exercise Please estimate the area of solar panel required to provide annual global electricity consumption of 80 EJ, assuming the solar panel is located at a location that receives 5 kWh/m2 per day, and the energy conversion efficiency of solar panel is 44%.

61 Solar Energy is Abundant
The above map shows the area of PV cells of current efficiency (10%) required to supply ALL the WORLDS current energy requirements (electrical/transport/heating!). The RED area shows the area required if the efficiency approached 100%.

62 Solar Irradiance http://en.wikipedia.org/wiki/Insolation
"SolarGIS-Solar-map-South-And-South-East-Asia-en" by SolarGIS © 2012 GeoModel Solar. Licensed under CC BY-SA 3.0 via Wikimedia Commons -

63 Search for Alternative Energy Source
(工研院能資所2000年 統計)


Download ppt "(Contents are solely for educational purpose)"

Similar presentations


Ads by Google