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1. The Sun’s Energy Output
The Solar Constant
When we measure the midday intensity of
sunlight at the Earth’s surface, we find that
about 136.7 mW fall on every square
centimeter.
We call this number “The Solar Constant”
and designate it by the Greek letter sigma
().
At 1 A.U.:  = 136.7 mW/cm2.
Check yourself: Does everyone know what a
watt (W) is? A milliwatt (mW)?
A watt (W) is a unit of energy flow Joules per second.
A milliwatt (mW) is 10-3 W.
Does everyone know what an “A.U.” is?
An A.U. is the average Earth-Sun
separation, ~ 150,000,000 km.
Questions:
If the mean distance from the Earth to the
Sun is 1.5  108 km, and the solar radius is
1.4  106 km, then
1. What is the value of the solar
constant P at the photosphere, i.e., the
sun’s visible surface?
Answer to Question #1:
 = 136.7 mW/cm2 @ 1 A.U.
1 A.U. = 1.5 X 108 km
rSun = 1.4 X 106 km
p = ?
The same amount of energy per unit time
passes through the photosphere as through
a sphere with radius 1 A.U. 
Answer to Question #1 Continued:
a. Dimensionally:
Energy per Unit Time =   (Area)
b. Conservation of Energy:
P  (Photosphere Area) =   (1A.U. Sphere
Area of a Sphere = 4r2
4(rp2)(P ) = 4(r1A.U.2)()
 P =   (r1A.U./rP)2
c. Solving:
Area)
P = 136.7 mW/cm2  (1.5  108 km/1.4  106 km)2
~ 1.6  106 mW/cm2
.
Questions (continued):
If the mean distance from the Earth to the
Sun is 1.5  108 km, and the solar radius is
1.4  106 km, then
2. What is the total energy output per unit
time of the sun in W?
Answer to Question #2:
a. Dimensionally:
Total Energy per Unit Time = p  (Total
Surface Area of Sun)
b. Reminder: Area of a Sphere = 4r2
c. Solving:
With rSun = 1.4  106 km = 1.4  1011 cm,
(1.6  106 mW/cm2)  4(1.4  1011 cm)2
~ 3.9  1029 mW = 3.9  1026 W
390 Trillion-Trillion Watts
.
Question #2 Continued:
If an average American city has a peak
power consumption of 500 MW, estimate
how many average American cities this total
energy output (390 trillion-trillion watts) is
equivalent to.
1 MW = 106 W
Question #2 Continued:
3.9  1026 W
~ 7.8  1017 Avg.Cities
5  108 W/Avg.City
Question #2 Continued Again:
Estimate how much of this total energy
output is actually intercepted by the Earth.
Hint: rE = 6,400 km
Question #2 Continued Yet Again:
a. Dimensionally:
Energy Intercepted at Earth =
  Cross Section of Earth
=   r2
b. Solving:
136.7 mW/cm2   (6.4  108 cm)2
~ 1.8  1020 mW = 1.8  1017 W
.
180,000 trillion watts, enough to run almost
360 million average American cities!
#2 Continued:
1.) Dimensionally:
Energy Intercepted at Earth =
  Cross Section of Earth
=   r2
2.) Solving:
2   (6.4  108 cm)2
136.7
mW/cm
Question #2 is finished at last!!!
~ 1.8  1020 mW = 1.8  1017 W
.
180,000 trillion watts, enough to run almost
Question #2 is finished at last!!!
360 million average American cities!
Questions (Continued):
If the mean distance from the Earth to the
Sun is 1.5  108 km, and the solar radius is
1.4  106 km, then
3. In what form is this energy transmitted
into space?
Answer to Question #3:
The energy is transmitted as light (or, more
properly, electromagnetic radiation).
2. Harnessing the Sun’s Energy
Question:
How can we harness the energy from the
sun?
Some possibilities are:
Solar thermal collectors
Solar dynamic systems
Solar cells
What are Solar Cells?
+
A solar cell is a solid-state device that
directly converts sunlight into electricity.
-
What is the most common raw material from
which solar cells are made?
The most common raw material is white
sand, specially refined to remove unwanted
impurities.
“Refining” Sand: Can you fill in
the blanks?
O
Si
O
Silicon
Dioxide
Si
+
O
O
Words
Silicon
+
Oxygen
Chemical Symbols
SiO2
Si
+
O2
Incident sunlight
Reflected light
Absorbed light
SOLAR CELL
V oc
Energy absorbed from incident sunlight
electrically excites the solar cell to produce
a voltage. For silicon, V oc ~ 0.5 V
I
Load
+
V
-
When a load is placed across a solar cell,
electrical power is delivered to the load.
Power = Current  Voltage = I  V
Questions:
1. Is all of the sunlight falling on a solar
cell absorbed?
2. Is all of the energy absorbed by the
solar cell converted into electricity?
3. If the answer to Question #2 is, “No,”
then what other energies might be
involved?
Answer to Question #1:
No. Some of it is reflected back into space.
Answer to Question #2:
No. Silicon solar cells are nominally 20%
efficient.
Answer to Question #3:
The rest of the energy goes into heating
the solar cell.
Problem:
A given circular solar cell has a 1 cm radius.
It is 18% efficient. Because today is
cloudy, the solar constant is a mere 97
mW/cm2. What is the maximum power
output you can expect from the cell?
Answer:
The cell area (collecting area) is
r2 =  cm2
If the cell were 100% efficient, it would
produce
(97 mW/cm2 )  ( cm2)
~ 305 mW
But because it is only 18% efficient, it
produces
305 mW  0.18 = 55 mW
.
3. Using Solar Power
Question:
Now that you know something about
harnessing the sun’s energy with solar cells,
where do you suppose we can put that
energy to work?
Earth’s Surface
Solar System
Earth Orbit
Mars
Do you have any questions or topics you
would like to discuss?
For those interested in talking more,
contact me at:
joseph.c.kolecki@grc.nasa.gov