1. Answer to Alternate Exercise #1 from Problem #1 SolPowerPeople, Inc.Austin, TXMarch 5, 2012

2. Question You are designing a ground mount PV array at 30NLatitude with multiple rows that faces true south. The tiltof the modules are 20. The width of the modules are 39inches and they will be installed in landscape layout.What is the closest distance the rows of the modules canbe and not cause any shade during the hours of 8AM and4PM solar time? 3. 1st Step Determine Height of back of module 39 ? 20 4. 1st 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 5. 1st Step Determine Height of back of moduleopposite sin(q ) = hypotenuseopposite sin(20) =39" 39opposite 0.34202 = ?39" opposite = 39" 0.34202 20 opposite = 13.34" 6. Now solve for shadow length39 13.3420?? 7. Get angle from 30N SunPathChartNote: questiondidnt specifytime ofyear, so wemust assumeshortest day ofthe year Dec21st 8. Get angle from 30N SunPathChart12 9. Now solve for shadow length39 13.342012? 10. Now solve for shadow lengthWe know the opposite, and we have the angle, and wewant to solve for the adjacent, so we use the tan function39 13.3420 12? 11. Now solve for shadow length 13.34" tan(12) = adjacent 13.34" 0.21255 =adjacent 13.34" adjacent =390.21255 13.34adjacent = 62.76"2012? 12. Now solve for distance between rows ?66.76 (from prev slide)? 13. Get azimuth angle from 30N SunPath Chart 55 14. Now solve for distancebetween 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 15. Now solve for distancebetween rowsadjacentcos(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)