Solar energy incident on Earth
Basics of solar technology
The Sun -> Always there; countless Energy
What Makes the sunlight Shine? Nuclear Fusion; something we may discover ways to do in the future the Earth and so solve our Energy Problem.
Just how many photons (power) get to the top of world typically?
The energy balance in the environment is shown here:
The main components inside diagram are the following:
- Brief wavelength (optical wavelengths) radiation from sunlight reaches the top of the atmosphere.
- Clouds mirror 17per cent back into room. In the event that earth gets even more cloudy, as some weather models predict, much more radiation would be reflected back and less will achieve the surface
- 6per cent is obviously straight shown off the area back to space
- And so the total reflectivity regarding the earth is 31percent. This is certainly technically known as an Albedo . Keep in mind that during Ice Ages, the Albedo of this planet increases as more of its area is reflective. This, naturally, exacerbates the problem.
- 19% gets consumed directly by dust, ozone and water vapour within the upper atmosphere. This area is called the stratosphere as well as its heated by this absorbed radiation. Reduced stratospheric ozone is causing the stratosphere to cool with time.
- 4% gets soaked up by clouds found in the troposphere. This is basically the reduced an element of the earth's environment where weather occurs.
- The rest of the 47% of this sunshine that's incident in addition to our planet's environment reaches the top. This is simply not a genuine significant power loss.
How much energy from the sun hits the surface of the Earth an average of?
Keep in mind that we measure power in units of Watt-hours. A watt is certainly not a unit of power, 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 in for one hour
Incident Solar Energy on the ground:
- Typical within the entire planet = 164 Watts per square meter over a twenty-four hour day
- 8 time summer time time, 40 level latitude 600 Watts per sq. meter
- 8 hours x 600 watts per sq. m = 4800 watt-hours per sq. m which equals 4.8 kilowatt hours per sq. m
- This might be comparable to 0.13 gallons of gasoline
- For 1000 sqft of horizontal location (typical roofing location) it is comparable to 12 gallons of fuel or about 450 KWH
Performance of PV Cells Demo
We'll talk more about PV cells in detail later on. For now really the only suggest retain is that they can be low in efficieny!
Collection of Solar Power
Amount of captured solar technology depends critically on direction of collector with regards to the angle of Sun.
- Under optimum conditions, one can attain fluxes as high as 1000 Watts per sq. meter
- In Winter, for a spot at 40 levels latitude, the sunlight is gloomier within the sky plus the typical flux received is approximately 300 Watts per sq. meter
Believe our roof top location is 100 square yards (about 1100 sqft).
When you look at the cold weather on a sunny day only at that latitude (40o) the roof will get about 6 hours of illumination.
So power generated over this 6 hour duration is:
300 watts per square meter x 100 square meters x 6 hours
= 180 KWH (per day) a lot more than you may need.
But remember the performance issue:
- 5% efficiency 9 KWH daily
- 10percent effectiveness 18 KWH per day
- 20percent efficiency 36 KWH each day
At best, this represents 1/3 of this typical daily Winter energy consumption and it assumes the sunlight shines on roof for 6 hours that day.
With sensible energy saving and insulation and south-facing house windows, its potential to lessen your everyday use of energy by about a factor of 2. In this case, if solar shingles come to be 20per cent efficient, they can offer 50-75 % of your energy requirements
Another instance calculation for solar technology which ultimately shows that relative inefficiency could be paid for with collecting location.
A website in Eastern Oregon receives 600 watts per square meter of solar power radiation in July. Asuume that the solar power panels are 10percent efficient and therefore the tend to be illuminated for 8 hours.
Exactly how many square yards would be required to generate 5000 KWH or electricity
each square meter provides you with 600 x.1 = 60 watts
in 8 hours you would gt 8x60 = 480 watt-hours or just around .5 KWH per square meter
you prefer 5000 KWH
you consequently require 5000/.5 = 10, 000 square meters of gathering location