Chapter 3: Solar Energy

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In this chapter, we will ﬁnd out how to get the most solar energy out 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 simple equivalent circuit to represent a PV cell to investigate its current-voltage characteristics.

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Nội dung Text: Chapter 3: Solar Energy

Chapter 3 Solar Energy I 3.1 Introduction In this chapter, we will ﬁnd out how to get the most solar energy out 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 simple 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 360 δ = 23.45 · sin[ (n − 81)] (3.1) 365 where the sine function is measured 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 βN , as depicted in Fig. 3.2, is solar angle of the sun at solar noon.
3. Solar Energy I 53 N 6 Equator L βN ..... ..... L ..... δ ..... .... s Zenith Figure 3.3: Tilt angle making sun’s rays perpendicular to the PV module. The relationship between altitude angle, latitude and declination is given by βN = 90◦ − L + δ (3.2) Although (3.2) can be applied anywhere on the earth, it is more convenient to have another altitude equation for places below the equator. It is given by βN = 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 360 (304 − 81)] = −15.1◦ δ = 23.45 · sin[ (3.4) 365 From (3.2), the altitude angle of the sun is equal to βN = 90◦ − 33.54◦ − (−15.1◦ ) = 71.56◦ The tilt angle should be facing north and at an angle equal to Tilt = 90 − βN = 90 − 71.56 = 18.44◦
3. Solar Energy I 54 Figure 3.4: Altitude and its azimuth angle of the sun’s position. Source: Masters [1]. 3.4 Locate the Sun at Any Time of a Day The location of the sun can be found by the altitude angle β and its azimuth angle φS , where the subscript 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) cosδsinH φS = sin−1 ( ) (3.6) cosβ Note that reference point of φS is at noon. φS is negative when the sun goes towards west and positive when towards east. Suppose that the sun rotates a 360◦ in 24 hours or (360/24=)15◦ per hour, therefore the hour angle of the sun can be expressed as 15◦ Hour angle H = ( ) · (hours before solar noon) (3.7) hour 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 tanδ (3.8) if cos H ≥ tanL , then |φS | ≤ 90◦ otherwize |φS | ≥ 90◦ 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 15◦ · (−2h) = −30◦ H= h Using (3.5), the altitude angle 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 φS = sin−1 ( cos(−cos61sin(−30) ) 23.45) .90 = sin−1 (−0.9739) = −76.88◦ or = 180 − (−76.88) = 256.88◦ Applying (3.8) we have the following tan(−23.45) tanβ cosH = cos(−30) = 0.866 and tanL = tan(−33.54) = 0.654 tanβ Therefore, cosH > tanL We may conclude that the azimuth angle is φS = −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 the sun for maximum 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, I0 , describes the sun beam passing perpen- dicularly through an imaginary surface just outside of the earth’s atmosphere. It is expressed as I0 = SC · [1 + 0.034cos( 360n )] (W/m2 ) (3.9) 365 where SC is the solar constant and n is the day number. The predicted value of a clear sky beam radiation at the earth’s surface can be modeled as IB = Ae−km (3.10)
3. Solar Energy I 56 Figure 3.5: Optical depth, apparent extraterrestrial ﬂux and sky diﬀuse factor for the 21st Day of Each Month. Sources: Masters [1] and ASHRAE [3]. where IB is the beam portion of the radiation reaching the earth’s surface (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 1 m= (3.11) sinβ 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 (n − 275)] W/m2 (3.12) 365 360 k = 0.174 + 0.035sin[ (n − 100)] (3.13) 365 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 360 (81 − 275)] = 1175W/m2 A = 1160 + 75sin[ 365 From (3.13), the optical depth is 360 k = 0.174 + 0.035sin[ (81 − 100)] = 0.178 365 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 βN = 90◦ − 33.7 + 0 = 56.3◦
3. Solar Energy I 57 Figure 3.6: Illustrating the collector azimuth angle φC and tilt angle Σ along with the solar azimuth angle φS and altitude angle β . Source: Masters[1]. The air mass ratio (3.11) is 1 m= = 1.202 sin 56.3◦ Using (3.10), the predicted value is IB = 1175e−0.178·1.202 = 949W/m2 3.5.1 Direct-beam on Collector Surface Fig. 3.6 introduces the angles when a collector is tilted. The incidence angle is given by cosθ = cosβ cos(φS − φC )sinΣ + sinβ cosΣ (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 the clear-sky beam insolation was found to be 949 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. Solution: cosθ = cos56.3cos(0 − 20)sin52 + sin56.3cos52 = 0.923
3. Solar Energy I 58 From (3.10), IBC = IB cosθ = 949W/m2 · 0.923 = 876W/m2 3.5.2 Diﬀuse Radiation on Collector Surface Diﬀuse radiation, which is due to scattered radiation on clouds, moisture, etc., on a collector is more diﬃcult to calculate. We may relate the diﬀuse radiation on a horizontal surface IDH to the direct-beam radiation by a constant C , such that IDH = CIB (3.15) where C is a sky diﬀuse factor and it can be approximated from Fig. 3.5 as 360 C = 0.095 + 0.04sin[ (n − 100)] (3.16) 365 The diﬀuse radiation on the collector is related by 1 + cosΣ 1 + cosΣ IDC = IDH ( ) = CIB ( ) (3.17) 2 2 Example: Recall the last example, ﬁnd the diﬀuse radiation on the collector. Solution: From (3.16), we have 360 C = 0.095 + 0.04sin[ (81 − 100)] = 0.099 365 And from (3.17) 1 + cosΣ 1 + cos52 ) = 0.099 · 949W/m2 ( ) = 75.90W/m2 IDC = CIB ( 2 2 So the total beam on the collector including diﬀuse radiation is (949+75.9) W/m2 = 1024.5 W/m2 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 1 − cosΣ 1 − cosΣ IRC = ρ(IBH + IDH )( ) = ρIB (sinβ + C )( ) (3.18) 2 2
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: 1 − cosΣ 1 − cos52 ) = 25.47W/m2 IRC = ρIB (sinβ + C )( ) = 0.15 · 949(sin56.3+0.099)( 2 2 Therefore the total insolation on the collector is therefore IC = IBC + IDC + IRC = 949 + 75.9 + 25.47 = 1050W/m2 3.6 Photovoltaic Materials and Electrical Characterisitics A material or device that is capable of absorbing and converting energy 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 insulator 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, Eg , is 1.12eV (electron-volts) to jump from ﬁlled band to conduction band. Where does the energy come from? In the PV cell, the energy comes from the photons of the sun. When a solar cell absorbs a photon with high band-gap energy such that the electron can jump to the conduction band, the electron 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 c E = hv = h (3.20) λ where c is the speed of light (3 × 108 m/s), v is the frequency (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 6.626 × 10−34 J· s × 3 × 108 m/s hc = 0.875 × 10−6 m = 0.875µm λ≤ = 1.42eV × 1.6 × 10−34 J/eV E and from (3.19) the frequency must be greater than 3 × 108 m/s c = 3.43 × 1014 Hz v≥ = 0.875 × 10−6 m λ From this example it is shown that photon energy less than than 1.42eV is unable to excite an electron in GaAs cell. And photons with wavelength shorter than 0.875 µm has more than enough suﬃcient 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 insert 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 bound 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 column III element is called p-type material. If we put these two doped materials together, a junction called depletion region is formed. It happens when the electrons in donors diﬀuse across the junction to recombine the holes in the acceptors, leaving positively-charged, immobile carriers in
3. Solar Energy I 61 - I ? + 6 Id ISC 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 recombined atoms build 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 hole- electron 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 n- side will ﬂow through the wires and recombine the holes in p-side. A ﬂow of electrons occurs. By convention, we deﬁne postive current ﬂows in the direction opposite to electron ﬂow. An simple equivalent circuit for a photovolatic cell is shown in Fig. 3.7. The PV cell consists of a current source driven by sunlight and a diode in parallel demonstrating the V-I characteristics. Two important parameters of a PV cell are short-circuit current and open-circuit voltage, which are deﬁned by I = ISC |Vo =0 (3.21) Vo = VOC |I =0 (3.22) Using the KCL, we may relate the currents in the equivalent circuit I = ISC − Id (3.23) = ISC − I0 (eqV /kT − 1) 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 kT ISC VOC = ln( + 1) (3.24) q I0 Example: Consider a 100cm2 photovoltaic cell with reverse saturation current I0 = 10−12 A/cm2 . In full sun, it produces a short-circuit current of 30mA/cm2 at 25◦ C. Find the open-circuit voltage at full sun and again for 50% sunlight. Plot the results. Solution: The reverse saturation current I0 is 10−12 A/cm2 × 100cm2 = 1 × 10−10 A. At full sun ISC is 0.030A/cm2 × 100cm2 = 3.0A. From (3.24) the open-circuit voltage is 3.0 VOC = 0.0257ln( + 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 1.5 VOC = 0.0257ln( + 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 & Sons, 2004. [2] Infoplease, Longitude. [Online]. Available: http://www. Latitude and infoplease.com/atlas/latitude-longitude.html [3] ASHRAE, Handbook of Fundamentals, American Society of Heating, Refrigera- tion 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