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