Chapter 3: Solar Energy pps

Chapter 3 Solar Energy I 3.1 Introduction In this chapter, we will find out how to get the most solar energy ou t of the sun by considering the time, the altitude angle, the radiation and the photovoltaic (PV) materials. A simple PV model helps to understand the mechanism of energy conversion - solar to electricity. We will further use a simp le equivalent circuit to represent a PV cell to investigate its current-voltage characteristics. 3.2 Terminology Insolation - Incident Solar radiation Ecliptic Plane - The plane swept out by the earth in its orbit Equinox - One of the two days in a year when we have equal day and night in length Summer Solstice - The day of the year when we have the most hours of daylight Winter Solstice - The day of the year when we have the fewest hours of daylight Extraterrestrial (ET) - Something happening, existing, or coming from somewhere beyond the planet Earth. 51 3. Solar Energy I 52 Figure 3.1: A fixed earth and a sun that moves up and down. Source: Masters [1]. Figure 3.2: The altitude angle of the sun at solar noon. Source: Masters [1]. 3.3 Altitude Angle of the Sun Fig. 3.1 shows how the positions between the sun and the earth during a year. The earth is taken as the fixed reference point. The sun makes different solar declination δ to the earth, between ±23.45 ◦ . The declination can be expressed in terms of day in a 365-day year δ = 23.45 · sin[ 360 365 (n − 81)] (3.1) where the sine fun ction is measur ed in degrees and n = 81 is the approximated spring equinox (Mar 21). With the help of Fig. 3.1 and (3.1) we may gain some insight into what might be a good tilt angle for a solar collector. In general, to face the collector toward the equator and tile an angle equal to the local altitude is a good average position to receive sun ray. Altitude angle β N , as depicted in Fig. 3.2, is solar angle of the sun at solar noon. 3. Solar Energy I 53 ✻ N s Equator L L δ β N . . . . . . . . . . . . . . . . . . . . . . . . Zenith Figure 3.3: Tilt angle making sun’s r ays perpendicular to the PV module. The relationship between altitude angle, latitude and declination is given by β N = 90 ◦ − L + δ (3.2) Although (3.2) can be applied anywhere on the earth, it is more convenient to have another altitud e equation for places below the equator. It is given by β N = 90 ◦ + L − δ (3.3) Example: Find the optimum tilt angle for a north -facing photovoltaic module in Camperdown, NSW (latitude -33.54 ◦ [2]) at solar noon on November 1. Solution: November 1 is the 304th day of the year so the solar declination in (3.1) is δ = 23.45 · sin[ 360 365 (304 − 81)] = −15.1 ◦ (3.4) From (3.2), the altitude angle of the sun is equal to β N = 90 ◦ − 33.54 ◦ − (−15.1 ◦ ) = 71.56 ◦ The tilt angle should be facing north and at an angle equal to Tilt = 90 − β N = 90 − 71.56 = 18.44 ◦ 3. Solar Energy I 54 Figure 3.4: Altitude and its azimuth angle of the s un’s position. Source: Masters [1]. 3.4 Locate the Sun at Any Time of a Day The location of the su n can be found by the altitude angle β and its azimuth angle φ S , where the s ubscript of φ, S, indicates the angle of the sun. Fig. 3.4 shows the sun’s position described by the altitude angle and its azimuch angle. They are given by β = sin −1 (cosLcosδcosH + sinLsinδ) (3.5) φ S = sin −1 ( cosδsinH cosβ ) (3.6) Note that reference point of φ S is at noon. φ S is negative when th e sun goes towards west and positive when towards east. Suppose that th e sun rotates a 360 ◦ in 24 hours or (360/24=)15 ◦ per hour, therefore the hour angle of the sun can be expressed as Hour angle H = ( 15 ◦ hour ) · (hours before solar noon) (3.7) For example, at 9:00A.M. solar time H would be +45 ◦ . But during autumn and winter (in Australia) in the early morning and late afternoon, the magnitude of the sun’s azimuth is likely to be more than 90 ◦ away from the south. We may use the following test conditions to find out the azimuth angle if cos H ≥ tanδ tanL , then |φ S | ≤ 90 ◦ otherwize |φ S | ≥ 90 ◦ (3.8) Example: Find the altitude angle and azimuth angle for the sun at 2:00P.M. solar time in Camperdown, NSW (latitude -33.54 ◦ ) on the summer solstice. 3. Solar Energy I 55 Solution: The solar declination δ in summer solstice is -23.45 ◦ . From (3.7) we otbain the hour angle at H = 15 ◦ h · (−2h) = −30 ◦ Using (3.5), the altitude an gle is β = sin −1 [cos(−33.54)cos(−23.45)cos(−30) + sin(−33.54)sin(−23.45)] = sin −1 (0.8821) = 61.90 ◦ From (3.6), the azimuth angle is φ S = sin −1 ( cos(−23.45)sin(−30) cos61.90 ) = sin −1 (−0.9739) = −76.88 ◦ or = 180 − (−76.88) = 256.88 ◦ Applying (3.8) we have the following cosH = cos(−30) = 0.866 and tanβ tanL = tan(−23.45) tan(−33.54) = 0.654 Therefore, cosH > tanβ tanL We m ay conclude that the azimuth angle is φ S = −76.88 ◦ With the help of altitude and azimutch angles, we may able to draw a sun path diagram for a given latitude and locate th e sun for m aximum sun’s ray onto the collector of the PV panels. And we may able to estimate any shading at a site at particular time. 3.5 Insolation An extraterrestrial (ET) solar insolation, I 0 , describes the sun beam passing perpen- dicularly through an imaginary surface just outside of the earth’s atmosphere. It is expressed as I 0 = SC · [1 + 0.034cos( 360n 365 )] (W/m 2 ) (3.9) where SC is the solar constant and n is the day number. The predicted valu e of a clear sk y beam radiation at the earth’s surface can be modeled as I B = Ae −km (3.10) 3. Solar Energy I 56 Figure 3.5: O ptical depth , apparent extr aterrestrial flux and sky diffuse factor for the 21st Day of Each Month. Sour ces: Masters [1] and ASHRAE [3]. where I B is the beam portion of the radiation reaching the earth’s s urface (normal to the rays), A is an “apparent” extraterrestrial flux, and k is a dimensionless factor called the optical depth. The air mass ratio m is expressed as m = 1 sinβ (3.11) where β is the altitude angle of the sun. Fig. 3.5 shows a table with values of A and k used in the American Society of Heating, Refrigerating, and Air Conditioning Engineers Clear Day Solar Flux Model. We can write equations representing A and k in the table using curve fitting method A = 1160 + 75sin[ 360 365 (n − 275)] W/m 2 (3.12) k = 0.174 + 0.035sin[ 360 365 (n − 100)] (3.13) Example: Find the direct beam solar radiation normal to the sun’s rays at solar noon on a clear day in Atlanta (latitude 33.7 ◦ ) on Mar 21. Solution: Mar 21 is day number 81. From (3.12) the apparent extraterrestrial flux A is A = 1160 + 75sin[ 360 365 (81 − 275)] = 1175W/m 2 From (3.13), the optical depth is k = 0.174 + 0.035sin[ 360 365 (81 − 100)] = 0.178 On Mar 21, the solar declination from (3.1) is 0 ◦ , so from (3.2) the altitude angle of the sun at solar noon is β N = 90 ◦ − 33.7 + 0 = 56.3 ◦ 3. Solar Energy I 57 Figure 3.6: Illustrating the collector azimuth angle φ C and tilt angle Σ along with the solar azimuth angle φ S and altitude angle β. Source: Masters[1]. The air m ass ratio (3.11) is m = 1 sin 56.3 ◦ = 1.202 Using (3.10), the predicted value is I B = 1175e −0.178·1.202 = 949W/m 2 3.5.1 Direct-beam on Collector Surface Fig. 3.6 introduces the angles when a collector is tilted. T he incidence angle is given by cosθ = cosβcos(φ S − φ C )sinΣ + sinβcosΣ (3.14) Example: At solar noon in Atlanta (latitude 33.7 ◦ ) on Mar 21 the altitude angle of the sun was found to be 56.3 ◦ and th e clear-sky beam insolation was found to be 949 W/m 2 . 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. Solution: cosθ = cos56.3cos(0 − 20)sin52 + sin56.3cos52 = 0.923 3. Solar Energy I 58 From (3.10), I BC = I B cosθ = 949W/m 2 · 0.923 = 876W/m 2 3.5.2 Diffuse Radiation on Collector Surface Diffuse radiation, which is d ue to scattered radiation on clou ds, moisture, etc., on a collector is more difficult to calculate. We may relate the diffuse radiation on a horizontal surface I D H to the direct-beam radiation by a constant C, such that I D H = CI B (3.15) where C is a sky diffuse factor and it can be approximated from Fig. 3.5 as C = 0.095 + 0.04sin[ 360 365 (n − 100)] (3.16) The diffuse rad iation on the collector is related by I D C = I D H ( 1 + cosΣ 2 ) = CI B ( 1 + cosΣ 2 ) (3.17) Example: Recall the last example, find the diffuse radiation on the collector. Solution: From (3.16), we have C = 0.095 + 0.04sin[ 360 365 (81 − 100)] = 0.099 And from (3.17) I D C = CI B ( 1 + cosΣ 2 ) = 0.099 · 949W/m 2 ( 1 + cos52 2 ) = 75.90W/m 2 So the total beam on the collector including diffuse radiation is (949+75.9) W/m 2 = 1024.5 W/m 2 3.5.3 Reflected Radiation on Collector Surface The sun beam can be reflected on a bright surface like snow and strike onto the collector to boost its performance. The reflected radiation can be approximated as I RC = ρ(I BH + I D H )( 1 − cosΣ 2 ) = ρI B (sinβ + C)( 1 − cosΣ 2 ) (3.18) 3. Solar Energy I 59 Example: Recall the last example, find the reflected radiation on the collector if the reflectance of the surfaces in front of the panel is 0.15. Solution: I RC = ρI B (sinβ+C)( 1 − cosΣ 2 ) = 0.15·949(sin56.3+0.099)( 1 − cos52 2 ) = 25.47W/m 2 Therefore the total insolation on the collector is therefore I C = I BC + I D C + I RC = 949 + 75.9 + 25.47 = 1050W/m 2 3.6 Photovoltaic Materials and Electrical Characterisitics A material or device that is capable of absorbing and converting en ergy contained in the photons of light into electrical current and voltage is called photovoltaic (PV). The PV cells can be made up of mixture of silicon, boron and phosphorus, gallium and arsenic (GaAs), and cadmium and tellurium (CdTe). Electric current occurs when the electrons in a material is free to flow. Semicon- ductors like silicon is a perfect electrical ins ulator at 0K and its conductivity increases a little at room temperature. Energy of the electrons can be described using energy- band diagram. The top band region is called the conduction band where the electrons contributes to the electric current flow. The electrons, say in silicon, need to absorb sufficient energy to jump from lower energy band to higher energy band. Just beneath the conduction band is the forbidden band, and below is the filled band. In silicon, the band-gap energy, E g , is 1.12eV (electron-volts) to jump from filled band to cond u ction band. Where does the energy come fr om? In the PV cell, the energy comes from th e photons of the sun. When a solar cell absorbs a p hoton with high band-gap energy such that the electron can jump to the conduction band, the electr on leaves a hole in the nucleus of the cell. An adjacent electron fills this hole without change of energy level and creates another hole. This results in a flow of holes as well. The energy of photons E to create the electron-hole pair in an semiconductor is characterized by the following equations: c = λv (3.19) 3. Solar Energy I 60 and E = hv = h c λ (3.20) where c is the speed of light (3 × 10 8 m/s), v is the f requency (hertz), λ is the wavelength (m), E is the energy of a photon (J) and h is the Planck’s constant (6.626 × 10 −34 J-s). Example: What maximum wavelength can a photon have to create hole-electron pairs in GaAs (Band energy = 1.42eV)? What minimum frequency is that? Given 1 eV = 1.6 × 10 −19 J. Solution: From (3.20), the wavelength must be smaller than λ ≤ hc E = 6.626 × 10 −34 J· s × 3 × 10 8 m/s 1.42eV × 1.6 × 10 −34 J/eV = 0.875 × 10 −6 m = 0.875µm and from (3.19) the frequency must be greater than v ≥ c λ = 3 × 10 8 m/s 0.875 × 10 −6 m = 3.43 × 10 14 Hz From this example it is s hown that photon energy less than than 1.42eV is unable to excite an electron in GaAs cell. And photons with wavelen gth shorter than 0.875 µm has more than enough su fficient energy to excite an electron. In both cases, energy is wasted. Without an electric field drive, the leaving electron can possibly go back to its original nucleus and recombine the hole it left. Then the hole-electron pair disappears. To create this electric field, we make make the semiconductor electrically “unstable”. It is achieved by deliberately in sert different impurities to the pure silicon. One region is doped with very small amount of element from column V of the periodic table/chart; the other region is doped with atoms from column V of the table. Since silicon has 4 electrons at its outer orbit while, say phosphorus, has 5 electrons at its outer orbit, when they combined one electron is loosely bou nd and free to move around. So the column V element is called donor atom and the semiconductor doped with this element is called n-type material. Similarly, if we doped the silicon with column III element, a free hole will be created. Since it accepts electrons, it is called acceptor. This kind of semiconductor doped with colu mn III element is called p-type material. If we put th ese two dop ed materials together, a junction called depletion region is formed. It happens when the electrons in donors diffuse across the junction to recombine th e holes in the acceptors, leaving positively-charged, immobile carriers in [...]... open-circuit voltage is VOC = 0.0257ln( 3.0 + 1) = 0.62V 10−10 Since the short-circuit current is proportional to solar intensity, at half sun ISC = 1.5A and the open voltage is VOC = 0.0257ln( 1.5 + 1) = 0.602V 10−10 Plot of (3.23) we obtain the following V-I curves of the PV cell 3 Solar Energy I 63 Figure 3.8: Bibliography [1] G M Masters, Renewable and Efficient Electric Power Systems, John Wiley &... /kT − 1) (3.23) where I0 is the reverse saturation current of the diode, Id is the diode forward current, V is diode forward voltage drop, q is the electron charge (1.602 × 10−19 C), k is Boltz- 3 Solar Energy I 62 mann’s constant (1.381 × 10−23 J/K), and T is the junction temperature measured in Kelvin When the PV cell is open or RL = ∞, I = 0, we may solve (3.23) to get the open-circuit voltage VOC...3 Solar Energy I 61 I ? 6 ISC Id + RL Vo − Figure 3.7: A simple equivalent circuit for a PV cell the n-type region The same is for the p-type region The depletion region widens and the process stops until the... infoplease.com/atlas/latitude-longitude.html [3] ASHRAE, Handbook of Fundamentals, American Society of Heating, Refrigeration and Air Conditioning Engineers, Atlanta, 1993 [4] P Kruger, Alternative Energy Resources, John Wiley & Sons, 2004 [5] G Boyle, Renewable Energy: Power for a Sustainable Future, Oxford University Press, 2004 64 ... up a strong barrier of electric field to forbid holes and electrons to pass through it When the PV material is exposed to sunlight, photons are absorbed and holeelectron pairs may be formed The photon energy push electrons to n-side and holes to p-side If we connect a wire across the semicondutor terminals, electrons from nside will flow through the wires and recombine the holes in p-side A flow of electrons . Chapter 3 Solar Energy I 3.1 Introduction In this chapter, we will find out how to get the most solar energy ou t of the sun by considering the time,. depicted in Fig. 3.2, is solar angle of the sun at solar noon. 3. Solar Energy I 53 ✻ N s Equator L L δ β N . . . . . . . . . . . . . . . . . . . . . . . . Zenith Figure 3 .3: Tilt angle making sun’s. and azimuth angle for the sun at 2:00P.M. solar time in Camperdown, NSW (latitude -33.54 ◦ ) on the summer solstice. 3. Solar Energy I 55 Solution: The solar declination δ in summer solstice is