............................................... Topics in Physics:

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............................................... Topics in Physics:

1. The Sun’s Energy Output

The Solar Constant Check yourself: Does everyone know what a watt (W) is? A milliwatt (mW)? We call this number “The Solar Constant” and designate it by the Greek letter sigma (  ). 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. At 1 A.U.:  = 136.7 mW/cm 2.

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. 1 A.U.

Questions: If the mean distance from the Earth to the Sun is 1.5  10 8 km, and the solar radius is 1.4  10 6 km, then 1. What is the value of the solar constant  P at the photosphere, i.e., the sun’s visible surface?

1 A.U. 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:  = 136.7 mW/cm 2 @ 1 A.U. r Sun = 1.4 X 10 6 km 1 A.U. = 1.5 X 10 8 km  p = ?

Answer to Question #1 Continued: a. Dimensionally: Energy per Unit Time =   (Area) ~ 1.6  10 6 mW/cm 2 b. Conservation of Energy:  P  ( Photosphere Area ) =   ( 1A.U. Sphere Area ) c. Solving:  P =  136.7 mW/cm 2  (1.5  10 8 km/1.4  10 6 km) 2 4  (r p 2 )(  P ) = 4  (r 1A.U. 2 )(  )   P =   (r 1A.U. /r P ) 2 Area of a Sphere = 4  r 2.

Questions (continued): If the mean distance from the Earth to the Sun is 1.5  10 8 km, and the solar radius is 1.4  10 6 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) 390 Trillion-Trillion Watts ~ 3.9  10 29 mW = 3.9  10 26 W b. Reminder: Area of a Sphere = 4  r 2 c. Solving: With r Sun = 1.4  10 6 km = 1.4  10 11 cm, (1.6  10 6 mW/cm 2 )  4  (1.4  10 11 cm) 2.

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 = 10 6 W

About 780,000 trillion average American cities! 3.9  10 26 W 5  10 8 W/Avg.City ~ 7.8  10 17 Avg.Cities Question #2 Continued:

Question #2 Continued Again: Estimate how much of this total energy output is actually intercepted by the Earth. Hint: r E = 6,400 km

Question #2 Continued Yet Again: a. Dimensionally: Energy Intercepted at Earth =   Cross Section of Earth =    r 2 180,000 trillion watts, enough to run almost 360 million average American cities! ~ 1.8  10 20 mW = 1.8  10 17 W b. Solving: 136.7 mW/cm 2   (6.4  10 8 cm) 2.

Question #2 is finished at last!!! #2 Continued: 1.) Dimensionally: Energy Intercepted at Earth =   Cross Section of Earth =    r 2 180,000 trillion watts, enough to run almost 360 million average American cities! ~ 1.8  10 20 mW = 1.8  10 17 W 2.) Solving: 136.7 mW/cm 2   (6.4  10 8 cm) 2 Question #2 is finished at last!!!.

Questions (Continued): If the mean distance from the Earth to the Sun is 1.5  10 8 km, and the solar radius is 1.4  10 6 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? 136.7 mW/cm 2 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? Words + Chemical Symbols + Silicon Dioxide SiO 2 Silicon Si Oxygen O2O2 Si O O O O

Reflected 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 Absorbed light Incident sunlight

Load I When a load is placed across a solar cell, electrical power is delivered to the load. +V-+V- Power = Current  Voltage = I  V +_+_

Questions: 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? 1. Is all of the sunlight falling on a solar cell absorbed?

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/cm 2. What is the maximum power output you can expect from the cell?

Answer: The cell area (collecting area) is  r 2 =  cm 2 If the cell were 100% efficient, it would produce (97 mW/cm 2 )  (  cm 2 ) But because it is only 18% efficient, it produces ~ 305 mW 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 Earth Orbit Solar System 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