Today Electrical energy from sunshine

4. TodayElectrical energy from sunshine The most versatile, useful, simple form of energy, easily sent around. Power(Watts)= Voltage (Volts) x Current (Amperes) Heat from electricity= I2 R, with R=resistance in ohms Wires have resistance, so send electrical power at least I, highest V Transformers require AC current. All power point images are only for the exclusive use of Phys3070/Envs3070 Spring term 2014

5. Integral of yellow area = solar constant = 1365 W/m2 Red area– for full vertical sunlight.

6. 23.5 degrees The solar constant

7. 2. (10) If all of the sun’s power (1365 W/m2) to arrive at the earth (cross-sectional area pr2, with r=6370 km ) could be captured, how long would the sun have to shine to provide all the energy we use in a year? We consume power at 18 TW. Power = 1365 W/m2 * p (6.37*106 m)2 = 1.74*1017 W Energy used in one year = 18*1012 W * 3.16*107 sec = 56.88*1019 J Time = energy / power = 56.88*1019J / 1.74*1017J/sec = 32.7*102 sec = 0.91 hr

8. Sunlight on the ground One square meter pointing at the sun on top of the atmosphere collects 1.96 (=2) cal/min-cm2 (Solar Constant) x 4.184 J/cal x1/60 min/secx10,000 cm2/m2=1365 Watts/m2 Times ½ for day/night Times something for atmospheric absorption Times something for clouds Times about ½ for latitude Times something for the orientation of the collector. Our Colorado standard average mean daily insolation =5-5.5 kWh/m2-day (page 94 bottom), or 610-680 W/m2 averaged over an 8 hour day (not 24 hrs) and over a year on a horizontal collector. See Fig 4.3.

9. PV photsynthesis thermal

10. Photovoltaic = PV • There are materials that release an electron at some voltage when hit by a quantum of light, a single photon. Each such reaction gives about ½ eV of electrical energy. It takes 1.2 x1019 per second of these to give one Watt. • The efficiency of photon energy to electrical energy is about 11%, average of Table 4.3 • Each cell provides fraction of a volt, DC (power = volts times amps) • Needs a ‘converter’ to make AC for the grid.

11. Conversion/inversion A few DC volts and mA per cell must be collected and converted to 60 Hz Ac high voltages for transmission to the grid . Cost? Efficiency? Grid stability? (‘feed in tariffs’ Wednesday)

14. Data

16. Tilt the collector, perpendicular to the sun’s rays. Fixed, re-orient once in a while, tracking

18. Check it yourself www.sunviewer.net To see actual daily/weekly solar power graphs

19. Point----the 7000 Watts is the ‘capacity’, when the sun is just right.

22. Solar thermal 4.5 Concentrate the sun’s rays to raise something to a high temperature, using mirrors. • Heat to steam, to turbines, to the grid. • Parabolic steerable trough (Fig. 4.12) • Thermal tower (Fig. 4.10), hundreds of ‘heliostats’ • About >25% efficient (better than PV) • Heat can be stored for night time heating.

25. 1.The Valmont plant provides 229 MW of electricity to us. How many square meters of solar panels would be required in Boulder to replace all of this power? Assume some suitable storage system to smooth out the fluctuations, and use only the daily and seasonal averages for an annual average. A) (10) The solar constant at the top of the atmosphere is 1365 W/m2. How many square meters of suitably aligned panels above the atmosphere would you need to have up there? Assume 100% efficiency for turning solar power into electricity, for now. Area needed = 229*106W / 1365 W/m2 = 0.1678*106 m2 (0.41x0.41 km)

26. B) (15)Now, clearly state an assumed number for the fraction of that solar power that reaches Boulder, considering our latitude and the seasons, day/night, and effects of cloudy days. From Fig 4.3, Boulder receives 5-5.5 kW hr/m2 in one day, on the annual average. I will use 5.25 kW hr/m2. In one day at 1365 W/m2, Boulder (atop the atmosphere) would receive 1.365 kW/m2 * 24 hours = 32.76 kW hr / m2. So the actual fraction is 5.25 kW-hr/m2/ 32.76 kW hr/m2 = 0.1603 = 16% to a horizontal surface. Another way 47%(atmo. and weather, page 93) *0.50 (day/night) *0.50 (latitude, approximately) = 12%

27. C) (10)Your solar panels are too big to move to follow the sun. What is the best strategy to point your panels? Why? From Table 4.2, the best listed orientation is pointing south at 60 deg above the horizontal. For October 21, as an example, you gain a factor of (energy at 60 deg/ energy at horizontal ) = 2074 / 1348 = 1.54 (the units would cancel)

28. D) (15)The average efficiency of photovoltaic solar cells in Table 4.3 is 11%. Put all the numbers above and this efficiency into an estimate of how many Watts you could expect from one square meter in Boulder. All factors 1365 W/m2 *0.1603 (at a horizontal surface) *1.54 (tilted)*0.11 (efficiency to electricity) = 37.07 W/m2.

29. E) (10) Finally, how many square meters of your panels will you need to replace Valmont? Our football field is about 100 m by 40 m in area. How many football fields of your solar cells need we install? We need 229 MW, so area = 229*106W / (37.07 W/m2) = 6.18 *106 m2 Or 6.18*106m2 / (4000 m2 /field) =1.54 *103 =1540 football fields of tilted solar collector. (and some suitable storage system)

30. Extra credit opportunities If you go to some campus or other event related to the material of this course, write a half-page description of what you learned, and turn in (to me, not with HW). No more than 5! The CWA is coming. Then, if your final numerical core puts you near a break point in the letter grades, these may bump you up one notch.

31. Next CAPA Due 0900 Monday April 7. • Paper copies in the boxes in the 2B corridor • Wednesday: Status and future of solar energy • Friday: How to get good ideas into the market/Jeff Logan/NREL