Energy. on this world and elsewhere. Instructor: Gordon D. Cates Office: Physics 106a, Phone: (434)

Size: px
Start display at page:

Download "Energy. on this world and elsewhere. Instructor: Gordon D. Cates Office: Physics 106a, Phone: (434)"

Transcription

1 Energy on this world and elsewhere Instructor: Gordon D. Cates Office: Physics 106a, Phone: (434) Course web site available at click on classes and find Physics or at Lecture #19 October 31, 2017

2 Wind

3 Capacity additions as a function of time

4 Computing the power carried by wind A Power = 1/2 A ρ v 3 Here the area A is the area swept out by the blades of the windmill.

5 Computing the power carried by wind Consider the energy per unit time passing the indicated plane. Power associated with the wind = kinetic energy contained in box of air time it takes box to pass the plane

6 Computing the power carried by wind Consider the energy per unit time passing the indicated plane. Power associated with the wind = kinetic energy contained in box of air time it takes box to pass the plane

7 Computing the power carried by wind What is the kinetic energy carried by the air? KE = ½ (mass of air) (velocity of air) 2 L v mass of air = L 3 density = L 3 ρ KE = ½ (L 3 ρ) v 2

8 Computing the power carried by wind Assume the length of the side of the box is L How much time does it take the box to pass the plane? L We know that (velocity x time) equals distance: v t = d, Here the distance is just L, thus: t = d/v = L/v

9 Computing the power carried by wind Power = KE time it takes box to pass ½ (L 3 ρ) v = 2 = ½ L (L/v) 2 ρ v 3 Power = 1/2 A ρ v 3

10 Computing the power carried by wind Power = KE time it takes box to pass ½ (L 3 ρ) v = 2 = ½ L (L/v) 2 ρ v 3 Power = 1/2 A ρ v 3

11 Computing the power carried by wind A Power = 1/2 A ρ v 3 Here the area A is the area swept out by the blades of the windmill.

12 What is the power hitting a real windmill? r = 40 meters

13 Generated power versus wind speed How much power is carried by the wind for a wind speed of 15 meters per second? efficiency at 15 m/s = 2/10.94 = 18.3%

14 Generated power versus wind speed At 10 m/s, the graph indicates power output of roughly 1.4 MW. The power of the wind at 10/ms = 3.42 MW efficiency = 1.4 MW / 3.42 MW = 40.9%!!!

15 Windmill with broken regulating mechanism

16 Windmill with broken regulating mechanism

17

18 Solar Energy

19 The price of solar cells

20 The price of solar cells

21 Two approaches to generation of electricity from solar energy Solar thermal or Concentrated Solar Power(CSP) Photovoltaics (PV)

22 Two approaches to generation of electricity from solar energy Solar thermal or Concentrated Solar Power (CSP) Photovoltaic (PV)

23 Solar-thermal electricity and climate change in the popular media

24 Solar-thermal electricity and climate change in the popular media

25 Solar-thermal electricity and climate change in the popular media Is the above concept of generating electricity just part of an imagined dystopian future?

26 California s Ivanpah Solar Thermal Plant

27 Solar Thermal or Concentrated Solar Power (CSP) The Ivanpah Solar Plant: 377 MW facility nearing completion in the Mojave desert in California. Supposed to deliver 970 GW-hours to various power companies This would represent an average power of GW, indicating 29% of nameplate capacity

28 Photovoltaic electrical generating plant in China: the Tengger Desert Solar Park, 1500 MW PV based solar plants are now the fastest growing technology worldwide.

29 Solar thermal generation of electricity

30 Types of Solar Thermal Parabolic Trough - Most existing facilities, and currently approved facilities in California are of this variety. Solar-tracking dish-based collectors with Sterling Engines This is the SES design, and their are currently 1 or 2 proposed large-scale facilities in California. Power-tower design. - Several have been built, but in the tens of Megawatt range.

31 The Solar Power Tower Design Multiple mirrors called heliostats track the sun and reflect light onto a central collector. Temperatures are so high that molten salt is used as the liquid that gets heated. The molten salt, in turn, is used to produce steam that drives turbines to make electricity. The molten salt stores heat so well that the plant can produce electricity for several days without sunlight.

32 Solar Two - 10 MW demonstration project This system is no longer in operation for the production of electricity.

33 20 Megawatt System near Seville Spain

34 Solar Two - 10 MW demonstration project

35 The concept of intensity In the context of sunlight: Intensity of sunlight = Incident power of sunlight area over which it falls

36 Intensity of sunlight hitting the earth Above the atmosphere: about 1.4 kw/m2 - This is for a surface FACING the sun. On the Earth s surface: about 1.0 kw/m2 Averaged over day and night, all the seasons, for the 48 contiguous states in the U.S.: around 200 W/m 2

37 Area needed to generate electricity using solar power Power = efficiency x Intensity x Area Area = Power efficiency x Intensity

38 Area needed to generate electricity using solar power 83.4 mi. x 83.4 mi., area needed to produce (on average) all electricity currently used (assuming 200 W/m 2 ).

39 Area needed to generate electricity using solar power 83.4 mi. x 83.4 mi., area needed to produce (on average) all electricity currently used (assuming 200 W/m 2 ). 118 mi. x 118 mi. (twice the above mentioned land area).

40 Area needed to generate electricity using solar power 441 million acres devoted to cropland (for comparison) mi. x 83.4 mi., area needed to produce (on average) all electricity currently used (assuming 200 W/m 2 ). 118 mi. x 118 mi. (twice the above mentioned land area).

41 So what efficiency am I assuming? 441 million acres devoted to cropland (for comparison) mi. x 83.4 mi., area needed to produce (on average) all electricity currently used (assuming 200 W/m 2). 118 mi. x 118 mi. (twice the above mentioned land area). Area = Power efficiency x Intensity Efficiency = Power area x Intensity I ve suggested we use 200 W/m 2 for intensity, and the area I give on my slide is (83.4 mi) 2, so what should we use for power?

42 What is the average power? To get this, let s find the total energy used over a year, and divide by the time over which it is used

43

44

45 Area needed to generate electricity using solar power assuming 12.9% efficiency and 200 W/m million acres devoted to cropland (for comparison) mi. x 83.4 mi., area needed to produce (on average) all electricity currently used (assuming 200 W/m 2 ). 118 mi. x 118 mi. (twice the above mentioned land area).

46 Area needed to generate electricity using solar power assuming 12.9% efficiency and 200 W/m million acres devoted to cropland (for comparison) mi. x 83.4 mi., area needed to produce (on average) all electricity currently used (assuming 200 W/m 2 ). 118 mi. x 118 mi. (twice the above mentioned land area).

47