Solar Energy I
Is solar energy is a viable means to supply power to
homes and automobiles? Solar cells, also commonly
known as photovoltaic cells, use energy from the Sun
(especially photons in the visible range) to excite
electrons in a semiconductor P-N junction. This
excitation generates an electrical current. This is quite
similar to how the light meter in your camera works.
The P-N junction is sensitive to light.
Figure 1:
Equipment for Solar Energy I
alligator clamps
heat lamp (halogen light) with stand
light meter
multimeter
ruler
small motor
solar cells
solar panel
Objectives [At the end of this lesson students will be able to...]
•
•
describe the concept of solar energy (light to electricity) conversion.
explain the limitations of solar power and energy.
Start-up questions
1. What are the advantages of solar energy?
2. What are the disadvantages and problems of solar energy?
3. Is the solar vehicle a reality? Can we really use sun to power our cars?
Solar Energy
Solar powered cars use sun energy to run an electric motor. The excess energy from bright
sunlight also charges up storage batteries. In the off-sun time (either at night or cloudy periods),
the batteries will be used to power the motor or other devices.
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Let's look at the efficiency of using a solar cell to convert light into electricity directly. In this
case, we will demonstrate the idea of converting light energy into electricity using an electrical
light bulb (100 W clear GE). This is a two-step conversion process (a really stupid idea if you
think about it). The first step is to apply 120 V AC to the light bulb and convert electrical energy
into light energy. Then, we will convert the light energy into electrical energy using a
photovoltaic cell (commonly called a solar cell). [Why is this stupid? But this is a good
demonstration of stupidity nevertheless.]
First step: We will measure the electrical power into the light bulb by monitoring the voltage and
current from the line to the light bulb. Then we will use a light meter to measure the amount of
light (photo) energy it produces.
Voltage of light bulb =
Current of light bulb =
Electrical power consumption of light bulb =
Light meter reading =
(1 ft·candle=1 lumen/ft², 1 lumen=1/680 lightwatt)
Distance of bulb to light meter (also where the solar cell will be) =
Area of sphere at where the meter is = 4πr² =
Total light power from the light bulb =
Efficiency of light bulb for visible light production =
Second step: We will use the light from the light bulb to produce electricity using the solar cell.
This will be done by measuring the power absorbed by the solar cell and the power generated by
the solar cell.
Total light power from the light bulb =
Area of solar cell =
Area of entire sphere =
Power absorbed by the solar cell =
Load resistor =
Load voltage =
Power to load produced by the solar cell =
Efficiency of solar cell =
Overall efficiency =
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Most solar devices are experimental at this time.
The experimental automobile at Auburn is called the
Sol of Auburn. It has the following specifications:
Figure 2: Sol of Auburn
Dimensions: 6m (20') long, 2m (6.6') wide and 1.25m (4') high.
Motor: 6kW(8HP) DC, 600rpm, 96V, 92% efficient at operating level, 94% peak, 12kg
Solar array: Peak power at 1400W at a solar power density of 1kW/m², 96V.
Batteries: 136kg (300lb), 96V, capacity at 52 A-hr
Chassis: Tubular aluminum frame, 1.5" diameter and 0.065" wall, carbon fiber, kevlar
composites and carbon body.
Brakes: motorcycle hydraulic disk brakes.
Wheels: Two 36 spokes and Two 48 spokes, 51 cm wheels.
Tires: 51×4cm (20"×1.75") tires at 85psi.
Maximum speed: 57 mph.
Weight of car (without driver): 800 lbs.
Driver: Under 176 lbs, under 6'.
Assessment questions
1. What is the overall efficiency of the Sol of Auburn's conversion of solar energy into
electricity?
2. Can you relate the problems of solar cars discussed above with the Sol of Auburn?
Example problems
1. I have invented a new improved solar cell which has very high efficiencies (which makes
me a millionaire or possibly a billionaire, but I still teach at Auburn just for fun). In my
experiments, I found that when the solar power input was 1.00 kW, my solar cell output
was 0.30 kW. What is the efficiency?
Answer: 30.%
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2. I also obtained the following data:
Input Power (kW)
2
3
4
5
6
7
8
9
10
Output Power (kW)
0.60
0.90
1.15
1.30
1.55
1.75
1.95
2.10
2.20
Plot the conversion graph and discuss what is happening.
3. I am a farmer who lives in a very remote area where I have no city water and no
electricity from Alabama Power Company. I have a well for all of my water supply. I use
a motor to operate a well pump and I use a solar photovoltaic panel to generate the
electricity for the motor. The pump is rated at 2.00 HP (needs 2.00 HP of electricity from
the solar panel to operate) and it is 90.0% efficient in pumping water. My solar cells are
10.0% efficient and provide sufficient power to operate the pump when the power density
of the sun is exactly at 500. W/m².
a. What is the size of my solar panel?
Solution: I need 2.00HP (or 1,490 watts) to operate the pump from the solar
panel. My panel is 10.0% efficient and the input power density is 500. W/m². So
my panel power output density is 50.0 W/m². To provide 1,490 W to the pump, I
need 1,490 W ÷ 50 W/m² = 29.8 m² of panel.
b. If my pump runs at 120. V, how much current does it need?
Solution: Power = 1,490 Watts = Volt × current = 120 V × current
Current = 12.4 Amp.
c. As I mentioned, my pump is 90.0% efficient in pumping water. If my well is 100.
m deep, how many kilograms of water can it pump up in exactly 1 hour? (You
may like to recall that potential energy of a substance is related to mgh, mass
times gravitational constant times height.)
Solution: Input power to the pump is 1,490 W. Efficiency is 90.0%. Output
mechanical power is 1,340 W. In one hour at 1,340 W, the energy from the pump
(energy is power times time, 1340 W × 3600 s) is 4,820 Joules. Mechanical lifting
energy = mgh = 4,820 J = mass × 9.81 m/s² × 100 m, or mass = 4.92 kg.
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Homework
I want to use solar cells to produce all the electricity I need for my home. I want to be completely
independent of the Alabama Power Company. I find that my average power demand during the
day time hours of exactly 9 am to 6 pm is 5,000. W and at night from exactly 6 pm to 9 am the
following morning is 2,000. W. My plan is to use sufficient solar cells in the daytime to provide
the power needs and also to charge up a set of batteries for use after the sun has set. I am using
12. V car batteries. Each battery is capable of storing 50. Amp-hour of charge when fully
charged (that is at each full charge, each battery has enough stored electrical energy to produce
50. Amp for 1.0 hour or 1.0 Amp for 50. hour or any product of current-time that gives 50. Amphour). It just so happens that I live in an area where the sun always shines from exactly 9 am to 6
pm with a power density of 1.0 kW/m² and the solar energy is not available from exactly 6 pm to
9 am. My solar cell efficiency is 9.0%.
a. What is the minimum area of solar cells that I need to satisfy my daytime needs and to
charge up my batteries?
b. How many batteries do I need?
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