1. Renewable Energy Solar Hydroelectric Wind Wave All originate with the sun

2. Sun emits EM E at av rate (P)= 3.9 x 10 26 J/s R E is radiated in all directions in a spherical shape from the sun. Insolation = incoming solar radiation. Intensity -power/m 2 hitting Earth. I = power/Area. A = 4  r 2. r orbit ~ 1.5 x 10 11 m. on Earth ~ 1380 W/m 2.

3. What happens to Insolation

4. Amount of Energy at surface Amount hitting surface & absorbed varies: Angle of rays (max I when 90 o to surface) season, cloud cover, type color of surface, reflection

5. Solar Energy Solar Panels Photovoltaic PV Cells heating only.Produce current

6. Solar PV 1:30 http://www.youtube.com/watch?v=Sur89b7afso Clip 1:30 Solar Themal 1:30 Clip 1:30 http://www.youtube.com/watch?v=64mtITOuXiA

7. 4. A solar panel is installed to heat 1.4 m 3 of water from 20 – 50 o C. If the average power from the sun hits it with 0.9 kW/m 2, a. Determine the number of Joules needed to heat the water. b. Estimate the area of the solar panel needed to heat the water in 2 hours. The density of water is 1000 kg/m 3,assume an ideal panel.

8. 1.8 x 10 8 J. A = 28 m 2.

9. Solar Advantage Renewable Clean Disadvantage Night and Clouds Requires large Area.

10. Hydroelectric - Dams Falling water spins turbine PE >>KE >> Electric Can calculate energy mgh is E of water Mass is also density x volume. At home look at example Hamper pg 192 – 193.

11. Hydro elec 3 min http://www.youtube.com/watch?v=wvxUZF4l vGw http://www.youtube.com/watch?v=wvxUZF4l vGw Clip 2:15

12. Pumper Storage German Version 2.5 min http://www.youtube.com/watch?v=GJ7ltJlMY9E

13. Hydro Sample Prb The flow rate of water over a dam of height 40-m is 500 L/s. Find the power generated. density of water is 1 kg/L or 1000 kg/ m 3.

14. Power = rate E transformed. GPE = mgh P = E/t (each second 500 L of water fall = V/t) P = mgh/t Need mass water D = mass/voldV = masssub dV for mass. P = (dV) gh t (1 kg/L) (500 L/s) (10 m/s 2 ) (40 m) P = 200,000 W

15. IB Prob Pumped Storage

16. Wind Turbine Cylindrical volume of air with velocity v, spins blades. KE air > KE blades> elec E. Power P, = ½ A  v 3. –  = air density – A = area of blade sweeps out. Blade = radius. – v = wind velocity

17. Can calculate Area circle. KE/s = ½  (  r 2 ) v 3. Power P, = ½ A  v 3. Mass = (  ) vol

18. Wind 1.21 min http://www.youtube.com/watch?v=0Kx3qj_o RCchttp://www.youtube.com/watch?v=sLXZk n2W-lk&feature=player_detailpage http://www.youtube.com/watch?v=0Kx3qj_o RCchttp://www.youtube.com/watch?v=sLXZk n2W-lk&feature=player_detailpage

19. 2. Wind Wind with density of 1.2 kg/m 3.goes through a windmill with 1.5m blades with v i = 8m/s. The wind slows to 3 m/s, and density changes to 1.8 kg/m 3 after leaving the blades. Find the max power of the windmill.

20. The  E from the wind is the work done on the blades. P i – P f A =  r 2. P = ½ A  v 3. Solve for each speed & density. (2171 – 171)W = 2000 W.

21. Wave Power uses up & down motion of waves (KE>PE) Pelamis Oscillating Hot dogWater column

22. Power in waves comes from Oscillation Wave alternate from KE (falling ) to PE rising Power per lengthof wavefront P/L = ½ A 2  g v. A = amplitude (half height) g ~ 10 m/s 2. v = wave velocity

23. 3. Waves. A wave with v = 4.8 m/s and height = 10 m is 2 m long. Find the power. A = ampl = 5 m.  = 1000 kg/m 3. g = 10 m/s 2. P/l = ½ A 2  gv. = ½ (5 m) 2 (1000)(10)(4.8 m/s) = 6 x 10 5 W/m For 2 m wave = 1.2 MW

24. Oscillating Water Column 2:45 http://www.youtube.com/watch?v=gcStpg3i5V8 http://www.youtube.com/watch?v=mcTNkoyvLFs Pelamis no sound 2 min

25. Hints. Often use density to get a mass. Must approximate shapes for volume calculation. E p = mgh for both hydroelectric & derivation of wave energy. Rates L/s or kg/s often can be used somehow to get at power which is a rate. Ex P = mgh=  Vgh. V is a flow rate. ttt

26. Hwk Hamper Read 189 – 198 Do pg 194-198 #13,15,16

27. Derivation of Equation A wave on the surface of water is assumed to be a square-wave of height 2A, as shown. The wave has wavelength λ, speed v and has a wavefront of length L. For this wave, (i)show that the gravitational potential energy EP stored in one wavelength of the wave is given by EP = ½ A 2 gρL. where ρ is the density of the water and g is the acceleration of free fall.

28. Mass water =  V =  1/2 A L). Height fall = A mgh =  1/2 A L) A = 1/2A 2  g