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Chapter 4: Solar Resource SOLUTIONS Gmasters Chapter 4: Solar Resource SOLUTIONS Chapter 4: Problems ________________________________________________________ 4.1 Consider the design of a “light shelf” for the south side of an office building locateD at a site with latitude 30o. The idea is that the light shelf should help Keep Direct sunlight from entering the office. It also bounces light up onto the ceiling to Distribute natural Daylight more uniformly into the office. o Upper LATITUDE 21.5 window 2 ft Light shelf 4 ft X1 X2 Lower window Not allowed S Figure P4.X As shown, the winDow Directly above the light shelf is 2-feet high and the window directly below is 4-ft. What shoulD the Dimensions X1 and X2 be to be sure that no direct sunlight ever enters the space at solar noon? SOLN: The worst day will be at the winter solstice when the sun is lowest in the sky, at which point the declination is -23.45o. From (4.7) the noon altituDe angle of the sun is o βΝ = 90 – L + δ = 90 - 30 - 23.45 = 36.55 36.55o o! 2 ft 45 2 X1 = o = 2.7 ft X2! tan 36.55 X21 !ft 4 ft 4 ft 4 X = = 5.4 ft 2 tan 36.55o S 4.2 Rows of buildings with photovoltaics covering vertical south-facing walls need to have adequate spacing to assure one building doesn't shade the collectors on another. gmasters Pg.4. 1 2/7/13 Chapter 4: Solar Resource SOLUTIONS !"#$%"$&'()*%+$,-./--%$01$(2$(%)$3$425$For$no$shading$between$8$am$and$4$pm,$d/H$≥$?$)67$8$9$ PVs$ H S d a. Using the shadow diagram for 30oN in AppenDix F, roughly what ratio of separation distance (d) to building height (H) would assure no shading anytime between 8 am anD 4 pm? b. If the spacing is such that d = H, During what months will the rear builDing receive full solar exposure. a. SOLN: 8 am anD 4 pm are just about 3 squares back, so with the peg being H, that means the separation D is about 3 H… D/H ≥ 3. (or, more accurately, d/H ≥ 2.8) b. SOLN: Looks like full sun for all of spring anD summer, but shaDing begins in October and runs into February . 4.3 Consider the challenge of designing an overhang to help shade a south-facing, sliding-glass patio Door. You woulD liKe to shaDe the glass in the summer to help control air-conditioning loads, and you woulD also like the glass to get full sun in the winter to help proviDe passive solar heating of the home. Suppose the slider has a height of 6.5 ft, the interior ceiling height is 8 ft, and the local latitude is 40o. a. What should be the overhang projection P to shade the entire window at solar noon during the solstice in June? gmasters Pg.4. 2 2/7/13 Chapter 4: Solar Resource SOLUTIONS b. With that overhang, where would the shade line, Y, be at solar noon on the winter solstice? P" βN" Y" 8)*"""6.5"*" South" Figure P4.X a. SOLN: To shade the window in June o Summer solstice βN = 90 − L + δ = 90 − 40 + 23.45 = 73.45 Y 8 P = = o = 2.377 ft tanβN tan 73.45 b. SOLN: In winter, when you want the sun, the shade line will be at Winter solstice β = 90 − 40 − 23.45 = 26.55o N o Y = P tanβN = 2.377 tan26.55 = 1.188 ft So, the shade line is 8 ft - 1.188 ft = 6.8 ft above the floor, so the entire window is in the sun. c. The shadow distance Y for a south-facing window when it is not solar noon is given by P tanβ Y = cosφS Will the bottom of the shadow line designed for solar noon still shade the window at 10 am? SOLN: 10:00 am is 2 h before solar noon, so H = 2h x 15o/h = 30o. From (4.8) sinβ = cos L cosδ cos H + sin L sinδ = cos 40 ⋅cos23.45 ⋅cos 30 + sin 40 ⋅sin23.45 β = 59.82o From (4.9) cosδ sin H cos23.45 ⋅sin 30 sinφ = = = 0.912 S cosβ cos59.82 o φS = 65.846 From the above equation gmasters Pg.4. 3 2/7/13 Chapter 4: Solar Resource SOLUTIONS P tanβ 2.377 tan59.82 Y = = = 9.99 ft cosφS cos65.846 So, the shadow still covers the window (ignoring edge effects). In fact, it will be shadeD all day long so designing for solar noon is a gooD strategy. 4.4 Suppose you are concerneD about how much shaDing a tree will cause for a proposeD photovoltaic system. StanDing at the site with your compass and plumb bob, you estimate the altituDe angle of the top of the tree to be about 30o anD the wiDth of the tree to have azimuth angles that range from about 30o to 45o west of south. Your site is at latitude 32o. Using a sun path diagram (AppenDix C), describe the shaDing problem the tree will pose (approximate shaDeD times each month). SOLN: The tree shaDes the site from roughly 2 pm to 3 pm from roughly miD-November through miD-January. 4.5 Suppose you are concerned about a tall thin tree located 100 ft from a proposeD PV site. You Don’t have a compass or protractor anD plumb bob, but you do notice that an hour before solar noon on June 21 it casts a 30 ft shadow directly toward your site. Your latitude is 32o N. a. How tall is the tree? b. What is its azimuth angle with respect to your site? c. Using an appropriate sun path Diagram from AppenDix C, roughly what are the first and last days in the year when the shadow will land on the site? SOLN: a. It is the summer solstice so δ = 23.45o and it is an hour before solar noon so the hour angle is 15o. From (4.8) the altitude angle of the sun is gmasters Pg.4. 4 2/7/13 Chapter 4: Solar Resource SOLUTIONS sinβ = cos L cosδ cos H + sin L sinδ = cos 32cos23.45cos15 + sin 32sin23.45 = 0.962 β = 74.23o From h/30 = tan 74.23o the tree height is h = 30 tan 74.23o = 106 ft b. From (4.9) the azimuth angle of the tree is ⎛ cosδ sin H ⎞ ⎛ cos23.45sin15⎞ o φS = Asin = Asin⎜ ⎟ = 60.9 ⎝⎜ cosβ ⎠⎟ ⎝ cos 74.23 ⎠ c. The altitude angle of the tree relative to the site is β = Atan(106/100) = 46.7o From the sun path Diagram for 32o and using the altitude of the tree as 46.7o anD the azimuth 60.6o it looKs liKe the site is shaDeD briefly each Day from early September to early April. S! 61o! PV! 106’! PV! o 74.2o! 46.7 ! !0!!!!!!!!!!!!30’!!!!!!!!!!!!!!!!!100’! 4.6 Using Figure 4.18, what is the greatest Difference between local stanDarD time and solar time for the following locations? At approximately what Date woulD that occur? a. San Francisco, CA (longituDe 122o, Pacific Time Zone) b. Boston, MA (longituDe 71.1o, Eastern Time Zone) c. Boulder, CO (longitude 105.3o, Mountain Time Zone) d. Greenwich, England (longitude 0o, Local time meriDian 0o) SOLN: Using (4.14) ST – CT = 4(Local Meridian – Local Long) + E The extremes of the equation of time Emax ≈ +16.5 min arounD n = 303, or about October 30, anD Emin ≈ - 14.6 min arounD n = 44, or about February 13. Using time zones from Table 4.3: a. San Francisco: MAX = 4(120 – 122) + E = -8 –14.6 = -22.6 minutes arounD February 13 b. Boston: MAX = 4(75 – 71.1) + E = 15.6 + 16.5 = 32.1 minutes arounD October 30 c. Boulder: 4(105 – 105.3) + E = - 1.2 - 14.6 = -15.8 minutes arounD February 13 gmasters Pg.4. 5 2/7/13 Chapter 4: Solar Resource SOLUTIONS (check -1.2 + 16.5 = 15.3 arounD October 30… not quite as baD) d. Greenwich: MAX = 4(0 – 0) + E = 16.5 minutes arounD October 30 4.7 Calculate the following for (geometric) sunrise in Seattle, latitude 47.63o, longitude 122.33o W (in the Pacific Time Zone), on the summer solstice (June 21st). a. Find the azimuth angle of sunrise relative to due south. b. Find the time of sunrise expressed in solar time. c. Find the local time of sunrise. Compare it to the website http://aa.usno.navy.mil/data/docs/RS_OneDay.html. Why do they differ? SOLN: a. From (4.16) and (4.9) realizing β = 0 at sunrise: (4.16) cos H = − tan L tanδ = − tan 47.63tan23.45 = −0.4755 so H = 118.39o cosδ sin H cos23.45sin118.39 (4.9) sinφ = = = 0.807 S cosβ cos0 o o so φS could be either 53.81 or 180-53.81 = 126.19 Test this using (4.11) tanδ tan23.45 cos H = cos118.39 = −0.475 and = = 0.396 tan L tan 47.63 tanδ since cos H < then φ > 90 tan L S So the azimuth angle at sunrise is 126.19o toward the north-east (which is obvious from any sun path diagram). b. Sunrise can be found from the hour angle H 118.39o Sunrise = o = o = 7.893h = 7h 53.56min before solar noon 15 / h 15 / h That is, in solar time, Sunrise = 4:06.44 am.
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