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# #Solar mooc problem set 1 alternate exercise 1 solution.

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Richard Stovall provides solution for Alternate Exercise #1 from Problem Set #1 of the SolPowerPeople #SolarMOOC

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### #Solar mooc problem set 1 alternate exercise 1 solution.

1. 1. Answer to Alternate Exercise #1 from Problem #1 SolPowerPeople, Inc. Austin, TX March 5, 2012
2. 2. Question 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?
3. 3. 1st Step Determine Height of back of module 39” ? 20°
4. 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. 5. 1st 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"
6. 6. Now solve for shadow length 39” 13.34” 20° ?° ?”
7. 7. 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
8. 8. Get angle from 30°N SunPath Chart 12°
9. 9. Now solve for shadow length 39” 13.34” 20° 12° ?”
10. 10. 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 tan function 39” 13.34” 20° 12° ?”
11. 11. 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° ?”
12. 12. Now solve for distance between rows ?” 66.76” (from prev slide) ?°
13. 13. Get azimuth angle from 30°N SunPath Chart 55°
14. 14. 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°
15. 15. 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)