Basics of Solar Energy

The Sun Always there; lots of Energy

How many photons (energy) reach the surface of the Earth on Average?

The energy balance in the atmosphere is qualitatively shown here:

The main components in this diagram are the following:

• Short wavelength (optical wavelengths) radiation from the Sun reaches the top of the atmosphere.

• Clouds reflect 17% back into space. If the earth gets more cloudy, as some climate models predict, more radiation will be reflected back and less will reach the surface

• 8% is backwards by air molecules

• 6% is actually directly reflected off the surface of the earth (on average) back into space

• So the total reflectivity of the earth is 31%. This is technically known as an Albedo . Note that during Ice Ages, the Albedo of the earth increases as more of its surface is reflective. This makes it difficult for the earth to warm back up - yet its a clever system.

What Happens to the 69% of the incoming radiation that doesn't get reflected back:

• 19% gets absorbed directly by dust, ozone and water vapor in the upper atmosphere. This region is called the stratosphere and its heated by this absorbed radiation. Loss of stratospheric ozone is causing the stratosphere to cool with time this has caused some of use stratospheric cooling as an argument against the occurence of global warming. The two are not connected at all.

• 4% gets absorbed by clouds located in the troposphere. This is the lower part of the earth's atmosphere where weather happens. This part of the equilibrium cycle is being changed as the troposphere, particularly at tropical latitudes, is getting cloudier.

• The remaining 47% of the sunlight that is incident on top of the earth's atmosphere reaches the surface. This is not a real significant energy loss Therefore it makes absolutely no sense to put solar panels in orbit and then "beam" the energy back to the surface.

How much energy from the sun reaches the surface of the Earth on Average?

Remember that we measure energy in units of Watt-hours. A watt is not a unit of energy, it is a measure of power.

ENERGY = POWER x TIME

1 Kilowatt Hour = 1KWH = 1000 watts used in one hour = 10 100 watt light bulbs left on for an hour

Incident Solar Energy on the ground:

Average over the entire earth = 164 Watts per square meter over a 24 hour day So the entire planet receives 84,000 Terrawatts of Power our current worldwide power use is about 20 TW. Since the Sun gives us 84,000 of these units and since the projected power needs of the world in the year 2050 is about 50 Terrawatts is this a solution?

Well its only a solution if we can build devices that can convert this incoming sunlight into outgoing electricity, heat, or fuel, as those are the three components of total energy use.

At the moment, we only have devices (PV cells) to convert sunlight into electricity and there is rather little grid connected PV power to the world portfolio. This is shown in the figure below where it can be seen that of the total global renewable energy portfolio (well dominated by Hydro) grid-connected solar power has currently making the least penetration, although it is starting to pick up in recent years as we will see.







Some example facility calculations:

There is a large amount of infrastructure (e.g. cost) required to convert from potential to deliverable energy.

• 8 hour summer day, 40 degree latitude 600 Watts per sq. meter

So over this 8 hour day one receives:

• 8 hours x 600 watts per sq. m = 4800 watt-hours per sq. m which equals 4.8 kilowatt hours per sq. m

• This is equivalent to 0.13 gallons of gasoline

• For 1000 square feet of horizontal area (typical roof area) this is equivalent to 12 gallons of gas or about 450 KWH

But to go from energy received to energy generated requires conversion of solar energy into other forms (heat, electricity) at some reduced level of efficiency.

We will talk more about PV cells in detail later. For now the only point to retain is that they are quite low in efficieny!

Collection of Solar Energy

Amount of captured solar energy depends critically on orientation of collector with respect to the angle of the Sun.

• Under optimum conditions, one can achieve fluxes as high as 2000 Watts per sq. meter

• In the Winter, for a location at 40 degrees latitude, the sun is lower in the sky and the average flux received is about 300 Watts per sq. meter

A typical household Winter energy use, with electrical heating, is around 2000-3000 KWHs per month or roughly 70-100 KWH per day.

Assume our roof top area is 100 square meters (about 1100 square feet).

In the winter on a sunny day at this latitude (40^o) the roof will receive about 6 hours of illumination.

So the incident solar energy over this 6 hour period is:

300 watts per square meter x 100 square meters x 6 hours

= 180 KWH (per day) more than you need.

But remember the efficiency problem. We have to build a device to convert incident solar power into deliverable power. For now, again we don't care what the device is, only that is efficiency lies somewhere in the range below:

• 5% efficiency 9 KWH per day
• 10% efficiency 18 KWH per day
• 20% efficiency 36 KWH per day

At best, this represents 1/3 of the typical daily Winter energy usage and it assumes the sun shines on the rooftop for 6 hours that day.

With sensible energy conservation and insulation and south facing windows, its possible to lower your daily use of energy by about a factor of 2. In this case, if solar shingles become 20% efficient, then they can provide 50-75 % of your energy needs

These considerations suggest that Solar PV rooftops (e.g. solar shingles) are likely a generically good idea.

What about the prospect of "solar farms" and remote power delivery?

The relative inefficiency can be compensated for with collecting area.

Examples:

A site in Eastern Oregon receives 1200 watts per square meter of solar radiation in July. Asuume that the solar panels are 10% efficient and that the are illuminated for 10 hours.

How many square meters would be required power Eugene at 300 Megawatts?

each square meter gives you 1200 x.1 = 120 watts

Remember, we want 300,000,000 Watts, not just 120.

There are one million square meters in one square km.

so, each square km gives your 120 x 1,000,000 = 120 MW.

Therefore 2.5 square kilometers would be enough to Power Eugene

Ten square kilometers would give you about 1200 MW keep this as a figure of merit reference under optimum conditions.

Now, of course, you wouldn't have continuous coverage, the individual collectors would have to be spaced out

And so, in practice, the actual land use is about twice as great per MW generated. In addition, 1200 Watts per square meter is July is about 2.5 times larger than the annual average. These considerations then produce the following rubric:

10,000 Megawatts of Average Annual Solar Power (when the sun is shining) requires 500 square kilometers of collecting area. (at 10% efficiency)

This is why any gain in efficiency is incredibly important.