Lecture 5B - Solar Resource - Part 2

Lecture 5B - Solar Resource - Part 2 - 1/28/11 ECE...

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Unformatted text preview: 1/28/11 ECE 4364/5374G: Alternate Energy Systems Lecture 5B: Solar Resource - 2 Professor Saifur Rahman Electrical & Computer Engineering Dept. Virginia Tech ECE 4364/5374G-5B (c) Saifur Rahman 1 Summary of Lecture 5A •  Extra-terrestrial radiation: 1353 w/sq m. •  Earth-Sun distance: shortest on Jan 3, longest of July 4. 147.1 to 152.1 million kilometers. Average: 149.6 million km, or 93 million miles. •  Declination Angle: +23.5 to –23.5 deg. •  Components of solar radiation. •  Atmospheric attenuation. ECE 4364/5374G-5B (c) Saifur Rahman 2 1 1/28/11 7.7 Measurement of terrestrial insola7on Available global radiation at any place and time. 7.7.1 The location factor, LF Assuming that the earth is a perfect sphere, one can use the spherical coordinate system (Lc,H,D) to locate a given point on the surface of the earth, both in time and space (see figure 7.4) Lc is the location on the earth (Lc= ! / 2 - latitude). H is the hour of the day (0 " H " 2! ). Hour 0 implies sunrise time on the equinox. D is the julian date (1 " D " 365 or ECE 4364/5374G-5B 366). (c) Saifur Rahman 3 Figure 7.4 Coordinate System ECE 4364/5374G-5B (c) Saifur Rahman 4 2 1/28/11 The equator is the trace of the X' -Y' plane that divides the earth into two equal parts. The latitude and the longitude are the two co-ordinates. The latitude is zero at the equator, and the longitude is zero at Greenwich, England. The latitude L goes from zero (at the equator) to 90° (at the poles). For the derivations used in this book, the latitude is considered positive in the northern hemisphere, and negative in the southern hemisphere. On the other hand, the longitude is considered positive to the west of Greenwich and negative to the east. The longitude decreases from 0 to -180° along the eastern hemisphere, while it increases from 0 to 180° along the western hemisphere. These two meet at the International Dateline between Japan and Hawaii. For the northern hemisphere, 0 ! Lc ! For the southern hemisphere, ECE 4364/5374G-5B ! 2 ! 2 .................................................... (7.4) ! Lc ! " (c) Saifur Rahman 5 Figure 7.5 Two-Coordinate System ECE 4364/5374G-5B (c) Saifur Rahman 6 3 1/28/11 ECE 4364/5374G-5B (c) Saifur Rahman 7 …………...……..7.5 ECE 4364/5374G-5B (c) Saifur Rahman 8 4 1/28/11 Assuming that the sun's rays (S) are perpendicular to the equator on the vernal and autumnal equinox's, the vector S will be parallel to the Y-axis. Then the components of S are ( Sx = 0.; Sy = S; Sz = 0). The normal energy flux, Sn is given by S n = S ny ........................................................................ 7.6 that is S n = S ny' cos (DAd)+ S nz' sin (DAd) ........................... 7.7 ny' = sin (Lc)sin (H) and nz' = cos (Lc) ..................................................................... 7.8 ECE 4364/5374G-5B (c) Saifur Rahman 9 LF = sin (Lc) sin (H) cos(DAd) + cos (Lc) sin (DAd) ................... 7.9 Sn = S sin(Lc) sin(H) cos (DAd)+ S cos (Lc) sin (DAd) ............. 7.10 Equation (7.10) incorporates information on location (Lc), time of day (H) and day of the year (DAd). P = SC AT [sin(Lc) sin(H)cos (DAd) + cos (Lc) sin (DAd)] ...... 7.11 d P = the power delivered to a location Lc on earth (kW/m2) SC = daily solar constant (kW/m2) which varies from one d day to another (outside the earth’s atmosphere) AT = the atmospheric transmittance ECE 4364/5374G-5B (c) Saifur Rahman 10 5 1/28/11 ECE 4364/5374G-5B (c) Saifur Rahman 11 7.7.2 Computa7on of Sunshine Intensity Based on equation (7.12) and (7.13) let us now compute the sunshine intensities at noon on the two solstices in the northern and southern hemispheres at 37° latitude. Nothern Hemisphere Summer solstice: ............ 21 June Solar noon: ...................... H = 90° Latitude: ........................... L = 37° Declination: ..................... DAd = 23.45° P = (SCd) (AT) [cos(37)sin(90° )cos(23.5)+ sin(37)sin (23.5)] = (SCd) (AT) [(0.7986) (1) (0.9171) + (0.6018) (0.3987)] = (SCd) (AT) [(0.7323 + (0.2399)] = (SCd) (AT) (0.9722) ECE 4364/5374G-5B (c) Saifur Rahman 12 6 1/28/11 Nothern Hemisphere (cont'd) Winter solstice: .............. 21 December Solar noon: ..................... H = 90° Latitude: .......................... L = 37° Declination: ..................... DAd = -23.45° P = (SCd) (AT) [0.7323-0.2399] = (SCd) (AT) (0.4924) ECE 4364/5374G-5B (c) Saifur Rahman 13 Southern Hemisphere Winter solstice: ........ 21 June Solar noon: ............... H = 90° Latitude: .................... L = -37° Declination: .............. DAd = 23.45° P = (SCd)(AT)[cos(-37)sin(90° )cos(23.45)+sin(-37)sin(23.45)] = (SCd) (AT) [0.7323-0.2399] = (SCd) (AT) (0.4924) Summer solstice: ..... 21 December Solar noon: ............... H = 90° Latitude: .................... L = -37° Declination: .............. DAd = -23.45° P = (SCd) (AT) [0.7323 + 0.2399] = (SCd) (AT) (0.9722) ECE 4364/5374G-5B (c) Saifur Rahman 14 7 1/28/11 Figure 7.9 Hour angles H and HA ECE 4364/5374G-5B (c) Saifur Rahman 15 7.7.3 The 7me ­of ­day factor ….…………...……………7.14 ECE 4364/5374G-5B (c) Saifur Rahman 16 8 1/28/11 The power available is: Northern hemisphere (L ! 0) Pn = SCd AT [cos (L) cos (HA) cos (DAd) + sin (L) sin (DAd)] ...................................................... 7.15 Southern hemisphere (L ! 0) P s = SCd AT [cos (L) cos (HA) cos (DAd) + sin (L) sin (DAd)] ....................................................... 7.16 ECE 4364/5374G-5B (c) Saifur Rahman 17 ECE 4364/5374G-5B (c) Saifur Rahman 18 9 1/28/11 ECE 4364/5374G-5B (c) Saifur Rahman 19 ECE 4364/5374G-5B (c) Saifur Rahman 20 10 1/28/11 ECE 4364/5374G-5B (c) Saifur Rahman 21 Compute the Impact of Atmospheric TransmiAance ECE 4364/5374G-5B (c) Saifur Rahman 22 11 1/28/11 ECE 4364/5374G-5B (c) Saifur Rahman 23 ECE 4364/5374G-5B (c) Saifur Rahman 24 12 1/28/11 Insola7on on a Tilted Plane ECE 4364/5374G-5B (c) Saifur Rahman 25 7.7.4 Energy on a 7lted plane At solar noon (HA = 0) the available power is given by (from 7.16): Pn = SCd.AT.[cos (L) cos (DAd) + sin (L). sin (DAd)] This can be simplified as, Pn = SCd.AT.[cos (L - DAd)] ........................................... 7.22 Equation (7.22) suggests that the available power at solar noon will be maximum when the latitude L equals the angle of declination Dad. We can introduce a new variable T(tilt angle, in degrees) in equation (7.22) to optimize the amount of available power. If T = 0, nothing changes. Pn = SCd.AT.[cos (L -T -DAd )] ....................................... 7.23 ECE 4364/5374G-5B (c) Saifur Rahman 26 13 1/28/11 Figure 7.7 Tilted Surface ECE 4364/5374G-5B (c) Saifur Rahman 27 Therefore for the maximum flux at solar noon the best tilt: L - T = DAd = for northern hemisphere ................................ 7.24 L + T = DAd = for southern hemisphere .............................. 7.25 Note that L is negative in the southern hemisphere. For horizontal (flat) surface (Pwr) when T = O; Pnh = SCd . AT . [cos (L) cos (DAd) + sin (L).sin (DAd)] ..... 7.26 and for tilted surface (Pnt ) when L - T = DAd Pnt = SCd . AT. [cos (0)] or .................................................... 7.27 Pnt = SCd . AT ......................................................................... 7.28 ECE 4364/5374G-5B (c) Saifur Rahman 28 14 1/28/11 End of Lecture 5 B Questions or comments? srahman@vt.edu ECE 4364/5374G-5B (c) Saifur Rahman 29 15 ...
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This note was uploaded on 03/16/2012 for the course ECE 5374G taught by Professor Srahman during the Spring '12 term at Virginia Tech.

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