Tut 1 solutions - SOLA2540-9001 Applied PV Page 1 of 9...

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SOLA2540-­‐9001 Applied PV Page 1 of 9 SOLA2540 & SOLA9001 Applied Photovoltaics Tutorial 1 Solutions Question 1 (b) The latitude angle is positive/north (g) See the graph, note that δ and ϕ are both > 0. (h) Note that α ! ! !"# = 180 ! α ! ! !"# (i) One need to know that in the southern hemisphere, ϕ becomes negative.
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SOLA2540-­‐9001 Applied PV Page 2 of 9 Question 2 (a) see lecture notes (b) June 21 st , φ = -34°, δ = 23.45° α F_Nth = 90° + ( φ δ ) = 90° - 34° - 23.45° = 32.55° I HZ = 400Wm -2 I β = 400sin(32.55 + β )/sin(32.55) Wm -2 = 743sin(32.55 + β ) Wm -2 (c) Dec 21 st , φ = -17°, δ = -23.45° α F_Nth = 90° + ( φ δ ) = 90° - 17° + 23.45° = 96.55° (greater than 90° – sun is behind the panel) I HZ = 950Wm -2 I β = 950sin(96.55 + β )/sin(96.55) Wm -2 = 956sin(96.55 + β ) Wm -2 (d) The direct component on the tilted surface is greater than the direct component on the horizontal surface where: sin( α + β ) sin α > 1 sin( α + β ) > sin α (e) As we know that sin ࠵? + ࠵? 1 , the peak of the function sin ࠵? + ࠵? occurs at ࠵? + ࠵? = ! ! + ࠵?࠵? , where it is safe to only consider the condition when ࠵? + ࠵? = ! ! . Place whose latitude is intensity (W/m 2 ) angle ( o ) Noon solar radiation intensity onto the panel for a given tilt angle Sydney Cairns
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