Inter row shading 4-19-12
- 1. Inter-row Shading Advanced Site Survey Concepts S
- 2. Inter-row Shading S Altitude and Azimuth Angles S Solar Sun Path Chart S Trigonometry- SOH CAH TOA S Application of concepts to problem sets S Arc-tan function S Designing on a sloped roof
- 3. Solar Azimuth and Altitude angles There are two primary numbers provided by the solar sun path chart that are required to do inter row shading calculations: solar altitude angle and solar azimuth angle. Solar altitude is how high in the sky the sun is in relation to the horizon. Solar azimuth is where the sun is in the sky in relation to a reference direction, usually south for solar applications.
- 4. Solar altitude and azimuth Inter-row shading: First consideration is solar altitude angle Second consideration is azimuth angle
- 5. A Simple real world graphic
- 6. Solar Sun Path Chart for 30° N 84° at noon on June 21st- Summer 11am on Solstice April 21 and Aug 21- 66° 3pm on April 36° at 21 and Aug noon 21- 44° Dec 21st- Also- 1:20pm Winter on Feb and Solstice Oct- 44°
- 7. Solar Sun Path Chart for 45° North 68.5° at noon on Summer Solstice
- 8. How do I get my very own Solar Sun Path Chart? S http://solardat.uoregon.edu/SunChartProgram.php S Enter negative for Southern hemisphere (or to cheat on the Azimuth degree line)
- 9. What else do we need to know to design an array that avoids inter-row shading? Trigonometry
- 10. Basic Trig
- 11. SOH CAH TOA which one? H Θ H Adjacent Opposite 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. SOH CAH TOA
- 13. SOH CAH TOA S=O/H C=A/H T=O/A H= ? To solve for the Hypotenuse use Cosine: C(30 )= 12/H O=? To solve for the Opposite use 30° Tangent: A= 12 T(30) = O/12
- 14. SOH CAH TOA SOH CAH TOA H = 30 S = O/H O= ? 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. 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. 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. Next- application to inter-row shading formula
- 18. Inter-row Shading ? What is the ideal distance between rows?
- 19. Question S 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. 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. 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. Now solve for shadow length 39” 13.34” 20° ?° ?”
- 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. 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. 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. Now solve for distance between rows ?” 66.76” (from prev slide) ?°
- 27. Get azimuth angle from 30°N SunPath Chart 55°
- 28. Now solve for distance between rows So now we know the angle, and the hypotenuse, and we need to solve for the adjacent. We must use cosine function. ?” 66.76” (from prev slide) 55°
- 29. Now solve for distance between rows adjacent cos(q )° = hypotenuse adjacent cos(55)° = 66.76" ?” 66.76” (from prev slide) adjacent 0.5735 = 55° 66.76" adjacent = 66.76"´ 0.5735 adjacent = 38.28" ANSWER: ~38.28” (depending on how much rounding you did)
- 30. First we must understand what 3/12 means
- 31. It refers to the rise over the run 3” 12” For every 3” of rise, there is 12” of run
- 32. So how do we solve for the angle? 3” ?° 12”
- 33. SOH CAH TOA 3” ?° 12” SOH or CAH or TOA We have the opposite and the adjacent We know that the tangent of the angle is equal to the opposite over the adjacent So: 3/12= TanΘ 3/12 = .25
- 34. Therefore, 3 divided by 12 is the tangent 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. to convert from the tangent of an angle to the actual angle value use the arctan function.
- 38. There it is! 3” 14.04° 12”