Energy from the Sun Part I

1. . . . . . . . . . . . . . • Topics in Physics: . . . . . . . . Energy from the Sun Part I . . . . . . . . . . . . . . . . . . Mr. K., NASA/GRC/LTP . . . . . . Edited: Ruth Petersen . .

2. 1. The Sun’s Energy Output

3. 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)?

4. 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?

5. 1 A.U. An A.U. is the average Earth-Sun separation, ~ 150,000,000 km.

6. 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 Pat the photosphere, i.e., the sun’s visible surface?

7. 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 = ? 1 A.U. The same amount of energy per unit time passes through the photosphere as through a sphere with radius 1 A.U. 

8. Answer to Question #1 Continued: • a. Dimensionally: • Energy per Unit Time =   (Area) • b. Conservation of Energy: • P  (Photosphere Area) =   (1A.U. Sphere Area) • Area of a Sphere = 4r2 4(rp2)(P ) = 4(r1A.U.2)()  P =   (r1A.U./rP)2 • c. Solving: • P = 136.7 mW/cm2  (1.5  108 km/1.4  106 km)2 ~ 1.6  106 mW/cm2 .

9. 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?

10. 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

11. 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

12. Question #2 Continued: 3.9  1026W ~ 7.8  1017 Avg.Cities 5  108W/Avg.City About 780,000 trillion average American cities!

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

14. 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!

15. Question #2 is finished at last!!! Question #2 is finished at last!!! Question #2 is finished at last!!! • #2 Continued: • 1.) Dimensionally: • Energy Intercepted at Earth = •   Cross Section of Earth • =  r2 • 2.) 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!

16. 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?

17. Answer to Question #3: • The energy is transmitted as light (or, more properly, electromagnetic radiation).

18. 2. Harnessing the Sun’s Energy

19. Question: • How can we harness the energy from the sun?

20. Some possibilities are: Solar thermal collectors Solar dynamic systems Solar cells

21. 136.7 mW/cm2 What are Solar Cells? + A solar cell is a solid-state device that directly converts sunlight into electricity. -

22. 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.

23. + Silicon Dioxide Silicon Oxygen Si SiO2 O2 “Refining” Sand: Can you fill in the blanks? O O Si Si O O Words + Chemical Symbols +

24. Incident sunlight Reflected light Absorbed light V oc SOLAR CELL Energy absorbed from incident sunlight electrically excites the solar cell to produce a voltage. For silicon, V oc ~ 0.5 V

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

26. 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?

27. 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.

28. 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?

29. 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 .

30. 3. Using Solar Power

31. 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?

32. Solar System Earth’s Surface Mars Earth Orbit

33. Do you have any questions or topics you would like to discuss?

34. For those interested in talking more, contact me at: joseph.c.kolecki@grc.nasa.gov