# Example - =-0.6015/0.9050 =-0.6646 Φ = 131.7°(if in a.m or Φ = 228.3°(if in p.m Ψ = 180° when the surface facing south γ = | Φ-ψ | =

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Example A house owner in Champaign, IL (at 88 deg W longitude and 40 deg N latitude) installed a solar PV panel (10 m 2 ) for his home. The panel is set on roof facing south and titled at 40 deg. What is the power of the PV panel at 3:00 pm CST on Feb. 21? Assuming the efficiency of the PV panel is 15%. Solution: Champaign, IL (40 deg N, 88 deg W) From Table 7-2 : On Feb. 21, EOT = -13.9 min, δ = -10.8°, A = 1187 W/m 2 , B = 0.142 LCT = local standard time - (L L -L S )*(4min/deg) = 3:00 p.m. -(88-90)*4 = 3:08 p.m. LST = LCT + EOT = 3:08 p.m. + (-13.9min) = 2:54 p.m. H = 2*15° + 54/4 = 43.5° Sin β = cos L cos H cos δ + sin L sin δ = cos(40°)cos(43.5°)cos(-10.8°)+sin(40°)sin(-10.8°)=0.5458-0.1204 =0.4254 β = 25.2° cos Φ = ( sin δ cosL-cos δ sinLcosH)/cos β = (sin(-10.8°)cos(40°)-cos(-10.8°) sin(40°) cos(43.5°))/cos(25.2°)

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Unformatted text preview: = -0.6015/0.9050 = -0.6646 Φ = 131.7° (if in a.m.) or Φ = 228.3° (if in p.m.) Ψ = 180° when the surface facing south γ = | Φ-ψ | = |228.3 -180| = 48.3° Titled angle α = 40° cos θ = cos β cos γ sin α + sin β cos α = cos(25.2°)cos(48.3°)sin(40°) + sin(25.2°)cos(40°) = 0.3869 + 3262 = 0.7131 θ = 44.5° From Table 7-2: On Feb. 21, A = 1187 W/m 2 , B = 0.142 C N = 0.98 (Figure 7-7) G D = G ND *cos θ = 833.46W/m 2 *0.7131 = 594.3W/m 2 Power = G D *area*efficiency = 594.3*10*15% = 891.5 W 4 4 ° ° 2 / 4 . 833 98 . 2 . 25 sin 142 . exp 1187 sin exp m W C B A G N ND = • ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = β...
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## This note was uploaded on 01/30/2011 for the course TSM 436 taught by Professor Wang,x during the Spring '08 term at University of Illinois, Urbana Champaign.

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Example - =-0.6015/0.9050 =-0.6646 Φ = 131.7°(if in a.m or Φ = 228.3°(if in p.m Ψ = 180° when the surface facing south γ = | Φ-ψ | =

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