Inter row shading 4-19-12

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An example of how to determine the ideal distance between rows for a solar PV array with multiple rows for a given location.

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Inter row shading 4-19-12

  1. 1. Inter-row ShadingAdvanced Site Survey Concepts S
  2. 2. Inter-row ShadingS Altitude and Azimuth AnglesS Solar Sun Path ChartS Trigonometry- SOH CAH TOAS Application of concepts to problem sets S Arc-tan function S Designing on a sloped roof
  3. 3. Solar Azimuth and Altitude anglesThere are two primary numbers provided by the solar sunpath chart that are required to do inter row shadingcalculations: solar altitude angle and solar azimuth angle.Solar altitude is how high in the sky the sun is in relation tothe horizon.Solar azimuth is where the sun is in the sky in relation to areference direction, usually south for solar applications.
  4. 4. Solar altitude and azimuth Inter-row shading: First consideration is solar altitude angle Second consideration is azimuth angle
  5. 5. A Simple real world graphic
  6. 6. Solar Sun Path Chart for 30° N84° atnoon onJune21st-Summer 11am onSolstice April 21 and Aug 21- 66° 3pm on April36° at 21 and Augnoon 21- 44°Dec21st- Also- 1:20pmWinter on Feb andSolstice Oct- 44°
  7. 7. Solar Sun Path Chart for 45° North68.5° atnoon onSummerSolstice
  8. 8. How do I get my very own Solar Sun Path Chart?S http://solardat.uoregon.edu/SunChartProgram.phpS Enter negative for Southern hemisphere (or to cheat on the Azimuth degree line)
  9. 9. What else do we need to know todesign an array that avoids inter-row shading? Trigonometry
  10. 10. Basic Trig
  11. 11. SOH CAH TOA which one? H Θ H AdjacentOpposite Θ Adjacent Opposite The location of the angle theta (Θ) determines which side is the opposite or adjacent. The hypotenuse is always the longest side.
  12. 12. SOH CAH TOA
  13. 13. SOH CAH TOA S=O/H C=A/H T=O/A H= ? To solve for the Hypotenuse use Cosine: C(30 )= 12/HO=? To solve for the Opposite use 30° Tangent: A= 12 T(30) = O/12
  14. 14. SOH CAH TOA SOH CAH TOA H = 30 S = O/HO=? C= A/H θ T= O/A A= ? Which formula is used to solve for the Opposite length? Which formula is used to solve for the Adjacent length?
  15. 15. Which formula is used to solve for the Opposite length? SOH CAH TOA Angle θ = 30° S = O/H S(30) = O/30 .5 = O/30 (.5) x 30 = (O/30) x 30 15 = O
  16. 16. Which formula would be used to solve for the Adjacent length? SOH CAH TOA Angle θ = 30° C(30) = A/30 .867= A/30 (.867) x 30 = (A/30) x 30 26.01 = A
  17. 17. Next- application to inter-row shading formula
  18. 18. Inter-row Shading ? What is the ideal distance between rows?
  19. 19. QuestionS You are designing a ground mount PV array at 30°N Latitude with multiple rows that faces true south. The tilt of the modules are 20°. The width of the modules are 39 inches and they will be installed in landscape layout. What is the closest distance the rows of the modules can be and not cause any shade during the hours of 8AM and 4PM solar time?
  20. 20. 1 st Step Determine Height of back of module We have the angle theta, and we have the 39” hypotenuse, and ? we need to know the opposite, so we should use 20° the sine function…
  21. 21. 1 st Step Determine Height of back of module opposite sin(q )° = hypotenuse opposite sin(20)° = 39" 39” opposite 0.34202 = ? 39" opposite = 39"´ 0.34202 20° opposite = 13.34"
  22. 22. Now solve for shadow length 39” 13.34” 20° ?° ?”
  23. 23. Get angle from 30°N SunPath Chart Note: question didn’t specify time of year, so we must assume shortest day of the year Dec 21st 12 altitude angle
  24. 24. Now solve for shadow length We know the opposite, and we have the angle, and we want to solve for the adjacent, so we use the tangent function 39” 13.34” 20° 12° ?”
  25. 25. Now solve for shadow length 13.34" tan(12)° = adjacent 13.34" 0.21255 = adjacent 13.34" adjacent = 39” 0.21255 13.34” adjacent = 62.76" 20° 12° ?”
  26. 26. Now solve for distance between rows ?” 66.76” (from prev slide) ?°
  27. 27. Get azimuth angle from 30°N SunPath Chart 55°
  28. 28. Now solve for distance between rowsSo now we know the angle, and thehypotenuse, and we need to solve for theadjacent. We must use cosine function. ?” 66.76” (from prev slide) 55°
  29. 29. Now solve for distance between rows adjacentcos(q )° = hypotenuse adjacentcos(55)° = 66.76" ?” 66.76” (from prev slide) adjacent0.5735 = 55° 66.76"adjacent = 66.76"´ 0.5735adjacent = 38.28" ANSWER: ~38.28” (depending on how much rounding you did)
  30. 30. First we must understand what 3/12 means
  31. 31. It refers to the rise over the run 3” 12” For every 3” of rise, there is 12” of run
  32. 32. So how do we solve for the angle? 3” ?° 12”
  33. 33. SOH CAH TOA 3” ?° 12”SOH or CAH or TOAWe have the opposite and the adjacentWe know that the tangent of the angle is equal tothe opposite over the adjacentSo: 3/12= TanΘ 3/12 = .25
  34. 34. Therefore, 3 divided by 12 is thetangent of the angle which is 0.25 3/12= TanΘ 3/12 = .25 3” ? 12” How do we solve for the angle? In order to convert from the tangent of an angle to the actual angle value use the arc- tan function.
  35. 35. to convert from the tangent of an angle to the actual angle value use the arctan function.
  36. 36. There it is!3” 14.04° 12”

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