Posted by sno on September 21, 2009, 3:04 am
Most of what I have heard/read from those who have really made a study
of it, is that until panels come down to around 1.00 per watt they are
not economical...of course if you live away from power lines and you
would have to pay for a line to be run to your house, at present prices
they may be....also rebates from government may be enough to bring the
It is my understanding that Germany has a law that requires the power
companies to pay for electricity from solar panels. at the same rate
they charge for electricity....for this reason they are installing
solar panels on a lot of roofs...so they are probably economical there
at todays prices.....also remember the price is not just the panels...
As others have mentioned a normal roof can produce enough electricity to
run a house....even in the northern states...especially if the house is
well insulated...and attention is paid to usage...
hope helps...have fun....sno
Correct Scientific Terminology:
Hypothesis - a guess as to why or how something occurs
Theory - a hypothesis that has been checked by enough experiments
to be generally assumed to be true.
Law - a hypothesis that has been checked by enough experiments
in enough different ways that it is assumed to be truer then a theory.
Note: nothing is proven in science, things are assumed to be true.
Posted by z on September 21, 2009, 4:11 am
Yeah around here folk pay between 9 and 11 cents a kiliwatt hour and the
power coop buys at 1/2 the rate, so its pretty hard to justify selling
back unless you somehow have uber cheap power like a free hydro or a wind
farm that you didn't have to buy. But for us in the serious boonies,
like you say.. can't afford the 300k+ to get a power hook up so its worth
Man i'd kill for $ a watt solar pannels .. but still its a pretty major
Posted by Martin Riddle on September 21, 2009, 1:59 am
The amount of wattage that strikes 1 square meter is about 800w.
Solar cell efficiency is about 15-20%.
Wiring and inverter losses (grid tie) is another 10%.
So worst case, 160w*0.70= 112w per square meter.
2000^2ft = 185^2m
185 * 112 = 20720watts
Now, you figure out the actual numbers with real parts.
Figuring spacing and mounting issues it is closer to 15000w
For your location, you need to look up the insolation, or rather the
number of average hours of sun per day.
now you have watt hours.
Posted by Morris Dovey on September 21, 2009, 8:02 am
Martin Riddle wrote:
This may be optimistic. The solar constant (includes all types of solar
radiation, not just the visible light) is the amount of incoming solar
electromagnetic radiation per unit area that would be incident on a
plane perpendicular to the rays, at a distance of one astronomical unit
from the sun. It has been measured by satellite to be roughly 1366 W/m^2
and fluctuates between 1412 W/m^2 in early January and 1321 W/m^2 in
early July due to the Earth's varying distance from the sun.
If we base calculations on just the cross sectional area, pi * r^2, and
then distribute that amount of energy over over the surface area of a
sphere, 4 * pi * r^2, we see that each m^2 of surface perpendicular to
the Sun's rays receives (on average) one-fourth of that 1366 W/m^2, or
approximately 342 W/m^2.
The portion of that 342 W/m^2 that actually reaches the panel will
depend on the latitude, the state of the atmosphere, and the altitude
(which affects the degree of atmospheric attenuation/filtering).
I'm not a PV expert, but I think that only a limited portion of the
total solar spectrum is useful. (Is this where the efficiency percentage
In any case, I think an optimistic upper limit for average solar input
will probably be closer to the 342 W/m^2 number.
Am I missing something?
DeSoto, Iowa USA
Posted by Ken Maltby on September 21, 2009, 9:57 am
"The maximum intensity of sunlight on earth is 0.089
w/cm2 at normal incidence on a bright day." -S.L.Soo
Ref; W.G.Pfann and W.VanRoosbroeck, "Radioactive
and Photoelectric p-n Junction Power Sources", J.of
Appl. Phys., 25. No.11 (1954),1422 [and other]
A Silicon solar cell, with an energy gap of 1.1ev, has
a practical efficiency of 14 to 15 per cent.
As to the use of the spectrum, a 35% theoretical
efficiency is the limit for silicon. It doesn't use 65%
of the solar wavelengths. [Not that this matters to
the energy installer, only to the designers of the
So far, you can't do better than the silicon cell on a
theoretical basis. All the differences in performance
seen over time have been developments that do
something to overcome the many obstacles that
keep the silicon cell from meeting its theoretical
The above relates to a single cell, it takes many cells
to make a panel and a number of panels to provide
any practical amount of energy. A panel's conversion
efficiency can only be a fraction of the cell efficiency.
But the efficiency of the PV panels isn't the main
problem with establishing what is possible, in the way
of energy production/extraction/conversion. The main
factor rapidly becomes; How much of that solar energy
actually arrives at the panels, with enough energy to
drive the output of the panels. There are some things
you can do to help here as well, like having the panels
pointed directly at the sun, all the time it is shining.
There are a number of other factors that you can't
control which effect "Solar Energy Per Square Foot In
Temperate Zone". Hopefully most of them are taken
into consideration in the official insolation tables.