Every hour the Earth intersects 1.4 Kilowatt hours of solar energy per square meter... AND since the radius of the Earth is 6,000 kilometers the surface area intersected with solar radiation is about 3.14 x 6,000,0002 or 113,000,000,000,000 m2. This means that about 160,000,000,000,000 KWH of radiant energy intersect the path of our earth every hour. Of course not all of this energy makes it through our atmosphere. HOWEVER where sunlight is direct and not blocked with clouds about 1KW of power is available per square meter. The question is how can we measure sunlight to make a sound investment calculation. Weather you're planning on installing solar panels to generate photo-voltaic power or solar collectors to generate solar thermal heat you'll need a tool to measure sunlight to predict the payback period of your investment.
Professionals use something known as a Pyranometer to measure solar insolation or solar flux that heats up a blackened thermopile incased inside a set of clear hemispheres. The thermopile generates a current proportional both direct and indirect radiant energy. A data logger is used to log flux variations throughout the day. From this data a graph may be plotted that gives a clear picture of the sunlight availability over a period of time. More affordable Pyranometers may be made with silicone sensors incased in white plastic capsules. Alternatives to the Huckse Pyranometer and the Apogee Pyranometer are the Sky Eye and the photodiode assemblies below.
Over the course of 12 hours on a clear sunny day on Long Island in the month of November a solar insulation curve generated with a Sky Eye and a photodiode looks like this...
Tree shadows in the early morning and late afternoon as well as a cloud shadows interfere with ideal readings for unobstructed sunlight but with a little imagination and the help of a few broken red lines the curve may be extended to reach the 600 points. At these points the sunlight is 600 from the direct radiation of solar noon and the radiant energy levels at this time of day, under ideal conditions, would be half the radiant energy level of solar noon. If you examine the graph closely you'll notice that the sunlight energy available at solar noon is 680 Watts/m2. At 4;15PM the angle of the sun is at an angle of incidence of 600 from the direct radiation at solar noon and the energy available at this time of the day should be half 680 Watts/m2 and as far as I can see from the graph this is true. At sunset the angle of incidence on the solar application is 900 and the level of radiant energy drops to 0. This finding may be inferred from both graphs, but it could be better demonstrated without interference from clouds and shadows. The data from the photodiode curve below was taken simultaneously with the data taken by the Sky Eye. It has been magnified to enhanced the illustration below. Can you see the similarity between the Sky Eye readings and the Photodiode readings?
At first glance
the plot of the graph made with the photodiode seems a bit crude BUT it still
accurately represents the radiant energy available within a limited range. The
600 point is off the chart but If this point were
extrapolated I believe it would read a light intensity half the peak light
intensity at solar noon. Solar Noon peak voltage is 3.3 mV. and the
600 point is about 1.6mV.. I'm sure you could plot
the rest of this curve more precisely if you had the sun for the job....
Remember the most important points on the curve are the:
PEAK ENERGY SOLAR NOON DIRECT RADIATION angle of incidence 00
HALF ENERGY (.66 x (time between sunrise and solar noon) HALF RADIATION angle of incidence 600
ZERO ENERGY ( at sunrise and sunset sun horizontal to sensor) ZERO RADIATION angle of incidence 900
The following videos may not be 100% accurate but they will give you the basic idea of what's involved with measuring sunlight.
The photodiode assembly is a nice tool that may be used to estimate the intensity of solar flux. It's simply pressed into DVM probes and the meter is set to the mV scale. To calculate the intensity of light in terms of watts/m2 the user multiplies the output voltage by a factor of 80 or so.... Unfortunately I can not remember where I purchased these PV cells.