Preface to Problem H6 (not to be turned in) A couple (Jere and Jerrye Sue), residing in Daphne, AL, wants to "go green" and install a PV array which will replace

100% of their electrical usage with solar energy. To estimate the array size, J&JS compute their average daily electrical energy usage to be 48 kWh. From the web http://pwwatts.nrel.gov/ they determine that the average daily insolation in Daphne is 5.21 kWh/m². Assuming an array efficiency of 16%: PV Daily Electric Output = 0.16(5.21)=0.8336 kWh/m² Area = 48 0.8336 = 57.58m² 2 J&JS select a module similar to the Sharp NE-175U1. Panel Area = 1.6 x 0.925 = 1.48 m MPP... Voltage 35 V Current 5 A Power Efficiency 16% 175 W No. of Panels = 57.58/1.48 = 38.9 round up to 40. Form two sub-arrays, each with 20 panels, using a string of 5 panels connected in series, with 4 strings connected in parallel Ratings of each sub-array: 5 x 35 = 175 V; 4 x 5 = 20 A; 20 A x 175 V = 3.5 kW/n175 V; 20 A 4.625 x 6.4 m 175/175V 7 kW Array (formed by interconnecting two sub-array) Electrical Output: 175/175 V 7.0 kW 175 V; 20 A 4.625 x 6.4 m Array to be mounted on E-W axis, tilted at 30.6⁰ (or 0.5341 rad, the latitude of Daphne) from the horizontal towards the southern sky. 0.4102 9.25 x 6.4 m 30.35 x 21 ft The 7 kW output can be realized only if the solar radiation is perpendicular to the array surface. There are two issues that should be considered: Seasonal Variation: tilt of the earth's axis: 23.5° (0.4102 rad) Time of day Variation 1 2.0.4102, Correcting for Seasonal Variation, the sun varies from perpendicular to the panels at the equinoxes by ± 0.4102 rad at the solstices. PSV 7.cos(e) -de=6.805 kW -0.4102 Days are longer in the summer and shorter in the winter, but on average, days are 12 hours long. Let T = time of day; 6 s T ≤ 18. We define 0 = (T-12) /12. 0=-π/2 T = 6 T = 12 0=0 T = 18 0=+π/2 Averaging Psv over 6 ≤ T ≤ 18... P(0) = 6.805.cos(0) - 1/250s+*/2 PAV -(-) 6.805 - cos (0) de = 4.332 kW Since 52 kWh > 48 kWh, the array meets specifications. WAV PAV (12)=52 kWh =