Calculating the Energy from Sunlight over a 12-Hour Period
(Written
in response to an inquiry recently received)
Incident sunlight
is usually thought of in terms of power per unit area. The typical units
are mW/cm2. At the earth's surface, the nominal value of
the solar constant is 137 mW/cm2. This value corresponds
to high noon with the sun directly overhead (as would occur at the equator
or in the tropics).
The energy from
sunlight may be obtained from this number and a little geometry. If
we take energy in mJ (millijoules), then from the units alone we obtain
mJ
= (mW/cm2) x (Area in cm2) x (Time in sec)
(since mW = mJ/sec).
If the sun were always directly overhead, the amount of energy incident
upon a 1 cm2 solar collector oriented perpendicular to the
sun's rays in 12 hours would be
(137
mW/cm2) x (1 cm2) x (4.3 x 104 sec.) = 5.9 x 106
mJ.
But we know that
the sun is usually not directly overhead. It moves East-West throughout
the course of a day and North-South throughout the course of a year.
The yearly variation is easily taken care of. We know that the sun moves
± 23.5° above and below the equator over the course of a
year. Typically, it is on the equator every March 21 and September 21.
We may estimate ,
the sun's position north of the equator,
by calculating
= 23.5° sin (2{T/365.25})
where dT is the
number of days counted from the vernal equinox (April 21). (Notice that
for northern winter months,
is negative). We may them estimate the solar constant for any particular
day by calculating
D
= (137mW/cm2) cos (L - )
where L is our latitude.
The daily variation
is a little more involved. Let the sun be at altitude A (angle above
the eastern horizon) at a given time t during the day. Then at time
t + dt, the altitude will have increased to A + dA. The differential
time is related to the differential altitude by
dt
= (12/) dA hr
since the sun travels
through radians (180°)
in 12 hours.
If we again try to calculate the amount of energy incident upon a 1
cm2 solar collector oriented perpendicular to the sun's noon-time
rays in 12 hours, we would have to use for the equivalent collecting
area
Area = (sin A) cm2
since we want only the amount of area projected perpendicular to the
sun's rays at time t. The differential energy dE incident upon our solar
collector in time dt is
dE
= D
(sin A) dt
= D
(sin A) (12/) (3,600
sec/hr) dA
= D
(4.3 x 104 sec/) [(sin
A) dA].
The expression in
square brackets may be integrated from A = 0 to A =
to yield
E
= D
(4.3 x 104 sec/)
If, for comparison,
we set D
= 137 mW/cm2 (i.e., we were to take our collector to the
tropics), we find
E
= 3.7 x 106 mJ.
The ratio between
this value and the one derived above is 2/.
At my latitude, 42° N, the midsummer value is
E
= 1.8 x 106 mJ
and the midwinter
value is
E
= 7.2 x 105 mJ.
The ratio of the
midwinter to the midsummer values is
(7.2
x 105 mJ)/( 1.8 x 106 mJ) = 0.4,
which means that
over a 12-hour period my solar collector collects a mere 40% of the
energy in midwinter that it would in midsummer. And this particular
calculation still does not take into account the longer days in summer
and the shorter days in winter! The next article will tell how that
calculation is to be done...