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Insolation question

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Posted by lemur on February 9, 2009, 11:00 pm
Is the insolation number the kwh that are potentially available from
the sun or the energy a typical PV panel will produce? What Im trying
to determine is if Gommit says the overall daily average for my area
will produce 4 kwh/sq m is that what I will get?  Or do I depend on a
15% pv efieciency (0.6 Kwh) per day?    If I  really needed to replace
400 Kwh per month, any experienced suggestions on how many sq m I

Thanks in advance

Posted by Ron Rosenfeld on February 10, 2009, 2:49 am

Solar panels are rated in Watts, so you don't have to do those messy
efficiency calculations.  What you need to know for an install is the panel
rating in watts, and how much space it will occupy.  The space occupied is
a function of the efficiency (more efficient panels occupy less space for
the same wattage rating), but the physical size will be in the

However, you did ask about how many sq m you would need.

These are very rough (and rounded-off) numbers, but give you an idea of how
to go about the calculations.

400 kWh/month = 13.2 kWh/day

I am assuming your power requirements are AC and not DC, so you will need
an inverter for the conversion.  If the inverter has an efficiency of 85%,
you will need to generate about 15.5 kWh/day.

Add on another 20% to account for the fact that nameplate ratings are NOT
done under real world conditions (see STC below), as well as any other
losses in your system, and your up to having your array generate about 18.6
kWh/day. (You might be able to cut this down somewhat, but the average
relationship between STC and PTC ratings -- see below -- seems to be about
11%, and you are bound to have other losses in your system in addition).

Since your insolation is 4hrs/day, you will need panels rated at 18.6/4 or
4.6 kW.  If your panels have a 15% efficiency, then they will occupy
approximately 4.6/15% or 30.7 sq M -- but you'll need to add a bit for
framing and so forth.

You still only need to buy 4.6 kW of panels.

In general, the insolation tables will specify an angle for the panels, so
you may need to do some calculations if your panels are not at the same
angle mentioned in the table.

As a rough initial estimate, a panel rated at 100W in an environment with
an average annual solar insolation of 4hrs/day will produce 400 wH/day.

However, this rough estimate will be high, due to a variety of factors.

This may be of value to you:


A brief discussion from

* PTC refers to PVUSA Test Conditions, which were developed to test and
compare PV systems as part of the PVUSA (Photovoltaics for Utility Scale
Applications) project. PTC are 1,000 Watts per square meter solar
irradiance, 20 degrees C air temperature, and wind speed of 1 meter per
second at 10 meters above ground level. PV manufacturers use Standard Test
Conditions, or STC, to rate their PV products. STC are 1,000 Watts per
square meter solar irradiance, 25 degrees C cell temperature, air mass
equal to 1.5, and ASTM G173-03 standard spectrum. The PTC rating, which is
lower than the STC rating, is generally recognized as a more realistic
measure of PV output because the test conditions better reflect
"real-world" solar and climatic conditions, compared to the STC rating. All
ratings in the list are DC (direct current) watts.

Neither PTC nor STC account for all "real-world" losses. Actual solar
systems will produce lower outputs due to soiling, shading, module
mismatch, wire losses, inverter and transformer losses, shortfalls in
actual nameplate ratings, panel degradation over time, and high-temperature
losses for arrays mounted close to or integrated within a roofline. These
loss factors can vary by season, geographic location, mounting technique,
azimuth, and array tilt. Examples of estimated losses from varying factors
can be found at:
http://rredc.nrel.gov/solar/codes_algs/PVWATTS/system.html .

Posted by Eeyore on February 10, 2009, 5:17 pm

lemur wrote:


Insolation is the averaged ACTUAL incident light energy at a given point
and hence takes into account clooud losses (averaged again) etc.

The energy a PV panel will produce depends on its orientation, whether it
tracks the sun or not, its efficiency (which varies with panel
temperature) and whether it it is run with a  maximum power point
controller. All of which gets very complicated.

And we've barely started yet.

Simple rule of thumb .... if you're not off-grid, just forget PV solar and
INSULATE your residence and employ low energy devices. PV solar may
actually be environmentally DAMAGING in some locations compared to all the
others above being positive.


Posted by stevey on February 11, 2009, 10:19 am
 The NREL --National Renewable Energy Labs had developed insolation map
and calculator called
PVWatts that show San Francisco area to get 5.25 KWh/sq.m-day (360
days a year).  This web
application can be accessed at http://www.pvwatts.org .

Your solar panel's rating already takes into account the efficiency
factor in watts per panel.
Thus, if your average monthly consumption of 1011 KWh will require an
array with 8.7KW
max. DC output rated panels to null out your electric bills(left with
monthly meter/account
charge).  I assumed a 15degree tilt and due-South  orientation.
Actually, within San Francisco there may be variations in insolation
due to fog levels in
micro-climates, say down to 4KWh.  In the SouthBay and East Bay, the
average insolation
I've used is  5.5 KWh daily.

Good Luck,
Steve  -solarMD

lemur wrote:

Posted by lemur on February 13, 2009, 1:36 am
 Thanks for help...I have a 45 watt pv panel that charges a battery to
power an LED night lite. I also have a small wind turbine. Neither of
these are cost effective with local power rates at $.10(ish) per

I also have a 180 sq ft (17 sq m) solar air heater. I KNOW how much
heat it produces and can convert btu's to kwh's....soooooo

If at my location, panels at lat + 15 degs,... ave daily insolation is

How may usable kwh could I expect from a typical 1 sq m pv panel? ..or
how many usable kwh from a 17 sq m?


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