﻿ How Much Energy Wiill This System Produce?

# How Much Energy Wiill This System Produce?

## SimpleEnergyWorks.com

Brilliantquestion. And one which requires a careful answer… It all depends!

Some of the biggest confusion comes from Solar Installersoverpromising on the energy production that a system will be able toproduce. We want to be very careful to consider all of the factors sothat you will know exactly how much energy your system will produce. We will assume that you will be installing 5 X 200 watt solar modules for a total of 1000 watts. That is a 1 KiloWatt system. We normallysell 230 watt modules, but for ease of math we will be using a 1 kWsystem. The first (and most important) factor in how much energy your systemwill produce is the size of the system (ie. the number of watts total). This will always be the starting number from with we do the rest of our calculations.

The second factor is where are youinstalling the system? For example what is the exact location (latitude and longitude) of the installation. Each location has a recordedweather history (of approximately 40 years) which will be used toaverage how much sunlight you can expect per day.

The above map will give you a pretty good idea of how many hours ofgood quality sunlight (1000 watts per square meter) you will get at your specific location. If you want more specific data (for example monthly data, or exact location data) you can visit the NREL PVwatts calculator at:

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

Once you know how many watts and how many hours of good sunlight youget per day you need a few more bits of information including theazimuth, the roof slope, and how much shade you get at the locationwhere you will be installing the system.

The azimuth refers to whether the panels are facing due south (whichis best). Due south is 180 degrees. If your panels are facing 175degrees then there will be a minor (but measurable energy productionreduction). To figure out the azimuth you may need to take out anactual compass and find true north.

In our example we are assuming that we have a 4/12 roof pitch. Toconvert the pitch of a roof to an angle use the following table:

Roof pitch ratio/Roof pitch angle
1:12=4.5 degrees
2:12=9.5 degrees
3:12=14 degrees
4:12=18.5 degrees
5:12=22.5 degrees
6:12=26.5 degrees
7:12=30.5 degrees
8:12=33.5 degrees
9:12=37 degrees
10:12=40 degrees
11:12=42.5 degrees
12:12=45 degrees

An ideal angle is equal to your latitude. For instance, if you areat 35 degrees latitude then a 35 degree slope would be maximal formounting your solar modules. Any variation from that will decrease your annual energy production (a little). (The reason latitude works as ageneral guide is because at the equator the sun is shining directlyperpendicular, as you move up to latitude 35 degrees you will simplyneed to slope your modules up to that 35 degree angle to approximateperpendicular sun).

The final factor affecting solar pv energy production is how muchshade you get at the site. Pick a shadeless site if at all possible.Shade is the enemy of any PV system. For a site to be worthwhile youreally want good sunlight from 9am until 3pm. If you don’t have goodsunlight for those hours you may want to consider a different site(possibly even a ground mount solution).

Anyway, to get the exact amount of shade on a specific site you canjust use your eyes and approximate how much sun a site gets. You willneed to know which way is south and also be familiar with the path ofthe sun. If you need specific data you will need to get a hold of aSolar Pathfinder. This tool will give you a full year’s data on aspecific site shading. It allows you to analyze when shade will hit aspecific location for each month of the year. (It’s a very handy toolto have). Either order one (or have a solar installation company comeout and give an evaluation).

A chainsaw is a tool of last resort, but one which may come in handyas you are dealing with shade issues.

Once you have all of the above information (watts, location, azimuth, slope, shade %) then you can go to PVwatts and put the information into it’s “Derate” worksheet and it willautomatically give you an annual production estimate.

Lets run through an example using a 1000 watt (1kW) system withEnphase microinverters from SimpleEnergyWorks.    We will assume thatthe installation is in Knoxville, TN.  We put in some of our information on the first page:

We put a 1 into the DC rating (1 kW).

We use a “fixed tilt” (because we are not using a mechanical suntracker).

We put 30.5 degrees in as the “Array Tilt” or slope (because the roof has a 7:12 pitch).

We put in 180 for azimuth because the roof faces due south.

Now we push the “Derate Help” button to figure out how much we will“Derate” the system.  (Derating is simply taking all of the factors that will lower production and account for them indiviudally, then addingthem together so that you can see the real efficiency of the system)

Now you change the numbers underthe “component derate values” to those for your system:

The Yingli modules we use have a fluctuation rate of 3% (so insert.97 in the first field).

The Enphase Inverters have an efficiency rating of 95%, (so insert.95 in the second field).

Using Microinverters mismatch does not affect the system, so put in.995 (99.5%).

“Diodes and connections” also does not affect microinverter systemsso insert (.997)

Bump DC wiring up to 99% (so insert .99) because there is no DCwiring.

Bump “System Availability” up to 99.5% (.995) because microinvertersincrease it.

This is where you would put in a % of shading (if any).  (Leave at 1for now)

Push the recalculate button at the bottom and it will give you aDerate factor of .846.

So…. insert this .846 into the previous page where it asks for derate factor.

PVWatts will then calculate how much that system will produce overthe next 12 months.  It will look like the following:

So a 1 kW system in knoxville, TN with the before mentionedassumptions will produce 1333 kWh of energy over the next year.

If you are installing a 4.2 kW system simply multiply that 1 kWnumber by 4.2 and we find that the larger system will produce 5598 kWover the next 12 months.

I hope you enjoy running through these steps to figure out potentialsystem production.

This complex formula is the reason why it is hard to just give asimple answer of “how much will it produce?”  Estimates are easy, but to know exactly what it will produce you need to follow the steps!

Enjoy, and keep on learning.

Randy