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

need?

Thanks in advance

Rob

*>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*

*>need?*

*>Thanks in advance*

*>Rob*

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

specifications.

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:

http://rredc.nrel.gov/solar/codes_algs/PVWATTS/version1/

A brief discussion from

http://www.gosolarcalifornia.ca.gov/equipment/pvmodule.php

* 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 .

--ron

lemur wrote:

*> Is the insolation number the kwh that are potentially available from*

*> the sun or the energy a typical PV panel will produce?*

Neither.

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.

Graham

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:

*> 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*

*> need?*

*> Thanks in advance*

*> Rob*

>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>need?>Thanks in advance>Rob