Category Archives: solar pv

Solar PhotoVoltaic Primer

The cost of photovoltaic systems (panels and inverter) has dropped to about 2 to 3 dollars per watt. At this price systems have payback times in the 10 to 15 year range, regardless of size. This assumes a cost of about 10 cents a kilowatt hour (kW-hr) for electricity.

Here are a number of nuts and bolts issues for those interested in solar power. First and foremost you must have a location with southern exposure. Even a small amount of shade can seriously reduce energy production. For most this means a roof top location, but it needn’t be if you have the space to put the array on the ground. The simplest mounting puts the panels flat on the roof. The pitch of the roof is not all that important as long as it faces south.

sun's path

sun’s path

The amount of space needed for an array of course varies as to how much total power you want to produce. Different manufacturers make panels in different sizes (watts) but the total space needed is the same because all PV panels have the same efficiency, about 15 %. Five 100 watt panels will take up the same space as one 500 watt panel. One kW requires about 80 square feet of space.

A big decision is whether the array is isolated or connected to the electrical grid. Grid-tied systems here in Arkansas can take advantage of net metering. This means that the power produced by the panels can actually make a meter run backwards if they are producing more power than the home is consuming at any time. About the only disadvantage of a grid-tied system is that when the line goes down, so does the solar power production. This is necessary to protect power line workers.

PV Grid-tied system

PV Grid-tied system

The alternative to grid-tied is to go entirely off line by buffering production with batteries. This avoids the aforementioned problem, but greatly increases the cost and “hassle factor” of the system. This is only practical when connection to the grid is cost prohibitive, as in remote locations.

The total amount of energy produced by a system is obtained by the total wattage of a system. For example a 1 kilowatt system can produce a maximum of one kilowatt hour only when the sun angle is ideal. Averaged over a year, a simple rule of thumb is that you can get 4 hours of net production per day. Hence a 1 kW system can be expected to produce 4 kW-hrs per day, more some days, less others.

Let’s use an average consumption of 1000 kW-hrs per month (close to the average in Arkansas) to determined a system sized to replace 100 % of electric needs. 1000 kW-hrs per month means 33 kw-hrs per day. Divide that by 4 to get a a little over 8 kW system. To allow for some inefficiencies say we use a 9 kW system. At 2.5 dollars a watt, the total cost would be 22,500 $. The 30% federal tax rebate brings the final cost down to 15,750 $. Sales taxes and installation will add to the cost, but these numbers can be used to approximate a cost if you are interested in going solar.

large solar array

Solar Steel

People often think that solar photovoltaic panels are OK to put on a roof to cut ones electric bill a little but really doesn’t go far to fill the needs of the nation when it comes to electricity. Or that it’s OK for light weight usages like lighting in parking lots but can’t provide for heavy industries like steel mills. I would like to disabuse those folks of the idea that solar can’t keep us going.

First some fundamentals. Electrical energy is measured in Watt-hours (Wh) or multiples there of. If your monthly electric bill is about a hundred dollars, close to the Arkansas average, you are using a MegaWatt-hour (MWh), which is a million times a Watt-hour. This amount of electricity is available year around from a space about twenty seven feet on a side. It easily fits on a south facing roof. A system like this will not just lower your bill but eliminate it.

Let’s talk about power for heavy industry, and it doesn’t get much heavier than steel mills. Nucor Corporation operates twenty-three steel mills

electric arc furnace

electric arc furnace

across the United States producing twenty-two million tons of steel annually employing electric arc furnaces. If we can figure out how to do this with solar panels we can do anything.

It takes about one and a half Mwh electric to produce a ton of steel. On average each plant produces a million tons of steel a year, so we need one and a half TeraWatt-hours;

steel

steel

a TeraWatt is a million times a MegaWatt. How much land do we need per plant? It works out to one thousand five hundred acres. This is equal to the land use of less than four average farms in Arkansas. That’s it. The land occupied by four farms in Arkansas will provide enough sunlight to power a steel mill. Cool, huh?

When you look at total electric use in the United States over a year the numbers get really big. The national annual electric use is four PetaWatt-hours; a PetaWatt is a billion times a MegaWatt. So how much land would it take to generate all the electric power we use in the United States? A surprisingly small nine thousand square miles. This is an area smaller than Rhode Island.

The numbers I cite are good for the amount of sunlight in Arkansas using flat plate collectors. If the national power grid originated in Nevada using tracking panels, the area needed is less than five thousand square miles. There are counties in Nevada much larger than that. There is no question that sunlight alone can provide all the electric power we need in this country.

The obvious fly in the ointment is the need for storage when the sun doesn’t shine, or transmission to where the sun doesn’t shine, but both those limitations are under study and are an achievable goal in the near future. And that’s just solar Photovoltaics as an energy source. That amount of energy is available from wind turbines and the potential for geothermal is greater still.