2.  Sun makes biomass through photosynthesis.  Biomass gathers and grows over time.  Biomass buried under great pressure and heat.  Biomass becomes coal, oil and natural gas over eons.  Coal, oil and natural gas are extracted.  Coal, oil and natural gas are used as fuels. Primary energy sources PRACTICE: Describe the origin of fossil fuels.

3. Primary energy sources PRACTICE: Proved reserves are resources that we are sure we can obtain. Production means actual reserves that have been obtained and placed on the market. Use the table to estimate how long oil reserves might last. SOLUTION: Expectancy = Reserves / Production = 1200.7  10 9 / 29.6  10 9 = 40.6 years.  This is an estimate because both numbers are subject to yearly change.

4. Specific energy and energy density of fuel sources  Specific energy E SP is how much energy (J) you can get per unit mass (kg) from a fuel. Its units are J kg -1.  Energy density E D is how much energy (J) you can get per unit volume (m 3 ). Its units are J m -3. EXAMPLE: Fission of each uranium-235 produces 3.5  10 -11 J of energy. The density of U-235 is 1.8  10 4 kg m -3. Calculate the E SP and the E D of U-235. SOLUTION: (a) m = (235 u)(1.661  10 -27 kg u -1 ) = 3.90  10 -25 kg. E SP = 3.5  10 -11 J / 3.90  10 -25 kg = 9.0  10 13 J kg -1. (b) E D = (9.0  10 13 J kg -1 )(1.8  10 4 kg m -3 ) = 1.6  10 18 J m -3.

5. Solving specific energy and energy density problems PRACTICE: If coal is transported in rail cars having a capacity of 1.5 metric tons, how many cars per day must supply the power plant of the previous example? SOLUTION:  From the previous example we calculated that we need 320000 kg of coal per day.  Since a metric ton is 1000 kg, we have (320000 kg ) / (1.5  1000 kg / car) or 213 cars d -1 !

6. Solving specific energy and energy density problems PRACTICE: If a nuclear power plant powered by uranium-235 has the same output and the same efficiency as the coal-fired plant of the previous example, how many kilograms of nuclear fuel will it burn per day? Per year? SOLUTION:  From the previous example we calculated that we need 10368000 MJ of energy per day.  Thus (10368000 MJ) / (1 kg / 90000000 MJ) or 0.1152 kg d -1 ! This is 42 kg y -1.

7. Solving specific energy and energy density problems PRACTICE: Explain why it is advantageous to have a submarine which is nuclear powered, as opposed to diesel. SOLUTION:  There are two main reasons: (1)Nuclear reactors don’t use oxygen, so the sub can stay under water for months at a time. (2) Nuclear fuel is extremely compact for the amount of energy it contains. Thus the sub can cruise far before refill.

8. In the balloon-water electricity engine name all of the energy conversions that occur. SOLUTION: 1 - Coal (or wood) becomes heat. 2 - Heat expands air in balloon to decrease its density. 3 - Buoyant force causes balloon to rise, increasing potential energy of the water. 4 - Falling water converts potential to kinetic energy. 5 - Moving water collides with fan blades and imparts kinetic energy to them. 6 - Moving fan blades generate electricity. Sankey diagrams In the balloon-water electricity engine name all of the energy losses that occur. SOLUTION: 1 - Coal loses heat to air outside balloon. 2 - Hot balloon loses heat to outside air. 3 - Excess hot air escapes bottom of balloon. 4 - Falling water heats up air as it falls. 5 - Falling water continues past blades colliding with the ground. 6 - Internal friction of fan shaft impedes rotation. 7 - Internal resistance of wiring generates I 2 R heat loss.

9. Solving problems relevant to energy transformations  This is a fission reaction.  Heat and some kinetic energy of the product neutrons.  Looking at the reaction we see that one neutron initiates one fission… but each reaction produces two neutrons.  Thus the fission process can produce a self-sustaining chain reaction.

10.  Product neutrons are too fast to cause fissioning.  The moderator slows them down to the right speed.  The control rods absorb neutrons produced by fission.  This allows a reactor to just maintain its reaction rate at the self- sustaining level, rather than becoming a dangerous chain reaction.

11.  Reactor coolant from the pile is circulated through a heat exchanger.  The heat exchanger heats up water to boiling to run a steam turbine.  The steam turbine turns a generator to produce electricity.  The reactants: U-235 and 1n: 218950 + 940 = 219890 MeV.  The products: Xe-144, Sr 90 and 2n: 134080 + 83749 + 2(940) = 219709 MeV.  The mass defect is 219890 – 219709 = 181 MeV. (This is the energy released through mass loss according to E = mc 2 ). mass loss

12. Solving problems relevant to energy transformations PRACTICE: Since the air is still moving after passing through the rotor area, obviously not all of the kinetic energy can be used. The maximum theoretical efficiency of a wind turbine is about 60%. Given a turbine having a blade length of 12 m, a wind speed of 15 ms -1 and an efficiency of 45%, find the power output. The density of air is  = 1.2 kg m -3. SOLUTION: A =  r 2 =  (12 2 ) = 452 m 2. Power= (0.45)(1/2)A  v 3 = (0.45)(1/2)  452  1.2  15 3 = 411885 W = 410 kW.

13. Solving problems relevant to energy transformations  A =  r 2 =  (7.5 2 ) = 177.  Power in = (1/2)A  v 3 = (1/2)(177)(1.2)9 3 = 77420 W  Power thru = (1/2)A  v 3 = (1/2)(177)(2.2)5 3 = 24338 W  Power ext = power in - power thru  Power ext = 77420 – 24338 = 53082 W = 53 kW.  P out = (efficiency)P extracted  P out = (0.72)(53 kW) = 38 kW

14. Solving problems relevant to energy transformations  Power is proportional to v 3.  The wind speed had doubled (from 6 to 12 m s -1 ).  Thus the power should increase by 2 3 = 8 times.  Output will now be 8(5 kW) = 40 kW.

15.  Wind power doesn’t produce greenhouse gas.  Wind power is a renewable resource.  Wind depends on the weather.  2 GW / 0.8 MW = 2500 turbines required! Solving problems relevant to energy transformations

16.  The total volume of water is V = 1700(2500)(50) = 2.125  10 8 m 3.  The mass of the water is m =  V = (1000 kg m -3 )(2.125  10 8 m 3 ) = 2.125  10 11 kg.  The average water height is h = 75 + 50 / 2 = 100 m.  The potential energy yield will then be E P = mgh = (2.125  10 11 )(10)(100) = 2.125  10 14 J. PRACTICE: A reservoir for a hydroelectric dam is shown. (a) Calculate the potential energy yield. (b) If the water flow rate is 25 m 3 per second, what is the power provided by the moving water?  Each cubic meter has a mass of 1000 kg.  Thus each second m = 25(1000) = 25000 kg falls.  Thus each second the reservoir relinquishes E P = mgh = (25000)(10)(100) = 2.5  10 7 J.  Since power is measured in W (or J s -1 ) the power provided is 2.5  10 7 W = 25 MW. (c) If the water is not replenished, how long can this reservoir produce power at this rate?  The total volume of water is V = 1700(2500)(50) = 2.125  10 8 m 3.  The volume flow rate is 25 m 3 s -1.  Thus the time is given by t = (2.125  10 8 m 3 ) / (25 m 3 s -1 ) = 8.5  10 6 s = 2361 h = 100 d.

17. Describing solar energy systems PRACTICE: The sun radiates energy at a rate of 3.90  10 26 W. What is the rate at which energy from the sun reaches earth if our orbital radius is 1.5  10 11 m?  The surface area of a sphere is A = 4  r 2.  Recall that intensity is the rate at which energy is being gained per unit area. I = P/ [4  r 2 ] = 3.90  10 26 / [4  (1.5  10 11 ) 2 ] I = 1380 W m -2.  This is 1380 J/s per m 2.

18. PRACTICE: Explain why the solar intensity is different for different parts of the earth. SOLUTION:  The following diagram shows how the same intensity is spread out over more area the higher the latitude. 1380 W/m 2 Describing solar energy systems

19. Solving problems relevant to energy transformations

20. EXAMPLE: A photovoltaic cell of 1.00 cm 2 and of 10.5%. (a)If the cell is placed in a position where the sun's intensity is I = 1250 W m -2, what is the power output of the cell? A = (1 cm 2 )(1 m / 100 cm) 2 = 0.0001 m 2. P IN / A = I so P IN = IA = 1250(0.0001) = 0.125 W.  The cell is only 10.5% efficient so that P OUT = 0.105P IN = 0.105(.125) = 0.0131 W. (b) If the cell is rated at 0.500 V, what is its current output? P = IV so that I = P/ V = 0.0131/.5 = 0.0262 A. (c) If ten of these cells are placed in series, what will the voltage and the current be? In series the voltage increases. In series the current stays the same. Thus V = 10(0.500) = 5.00 V and I = 0.0262 A. (d) If ten of these cells are placed in parallel, what will the voltage and the current be? In parallel the voltage stays the same. In parallel the current increases. Thus V = 0.500 V and I = (10)(0.0262) = 0.262 A.

21. Solving problems relevant to energy transformations EXAMPLE: A photovoltaic cell has an area of 1.00 cm 2 and an efficiency of 10.5%. (e) How many cells would you need to operate a 100 W circuit? P out = 0.0131 W/ cell. Thus (100 W) / (0.0131 W/ cell) = 7630 cells!  Obviously, a solar-powered calculator doesn’t need as much power to operate!

22.  Efficiency = P out / P in so that  P in = P out / Efficiency = 47  10 -3 / 0.08 = 0.5875 W.  Thus I = P in / A = 0.5875 W / 6.5  10 -4 m 2 I = 900 W m -2 = 0.90 kW m -2. Solving problems relevant to energy transformations

23.  Do NOT use latitude as a reason for this question.  The question specifically states that it is concerned with only a particular region having a variation, so its latitude does not change.  Cloud cover variation is one reason.  Season is another reason. Solving problems relevant to energy transformations

24. PRACTICE: Draw a Sankey diagram for a photovoltaic cell having an efficiency of 18%.  Note the approximate proportions.  Perhaps, like the nuclear reactor, we could also show the energy needed to process the materials that go into making a photovoltaic cell. ENERGY IN INCIDENT SUNLIGHT USEABLE ELECTRICITY WASTED HEAT Solving problems relevant to energy transformations

25.  Because of the tilt of Earth’s axis southern exposures get more sun in the northern hemisphere, and northern exposures get more sun in the southern hemisphere.

26.  From Topic 3 the energy needed for ∆T = 25 K is Q = mc∆T = 140(4200)(25) = 1.47  10 7 J.  But P OUT = Q / t = 1.47  10 7 J / (6 h)(3600 s / h) so that P out = 681 W.  Since Efficiency = P OUT / P IN = 0.35 then P IN = P OUT / Eff. = 681/0.35 = 1944 W.  From I = P IN / A we get A = P IN / I so that A = 1944 / 840 = 2.3 m 2. Solving problems relevant to energy transformations

27. Conduction, convection and thermal radiation Topic 8: Energy production 8.2 – Thermal energy transfer EXAMPLE: A brick has a thermal conductivity of k = 0.710 J / m·s·C°, and the dimensions shown.  Two heat sources are placed in contact with the ends of the brick, as shown. Find the rate at which heat energy is conducted through the brick. SOLUTION: A = 0.750  0.500 = 0.375 m 2. Why?  Then  Q /  t = kA  T / d = 0.710  0.375  (100 – 20) / 2.50 = 8.52 J s -1. 20°C 100°C 2.50 m0.500 m 0.750 m

28. Topic 8: Energy production 8.2 – Thermal energy transfer  Higher temperature = higher intensity so Y is above X.  Higher temperature = smaller wave- length at peak Solving problems involving Wien’s displacement law

29. FYI  Since no body is at absolute zero (K = 0) it follows from the Stefan-Boltzmann law that all bodies radiate.  Be sure that T is in Kelvin, not Celsius. Topic 8: Energy production 8.2 – Thermal energy transfer EXAMPLE: Mercury has a radius of 2.50  10 6 m. Its sunny side has a temperature of 400°C (673 K) and its shady side -200°C (73 K). Treating it like a black-body, find its power. SOLUTION:  A sphere = 4  (2.50  10 6 ) 2 = 7.85  10 13 m 2.  For T use T AVG = (673 + 73) / 2 = 373 K P =  AT 4 = (5.67  10 -8 )(7.85  10 13 )373 4 = 8.62  10 16 W. Solving problems involving the Stefan-Boltzmann law

30. PRACTICE: The sun radiates energy at a rate of 3.90  10 26 W. What is the rate at which energy from the sun reaches Earth if our orbital radius is 1.5  10 11 m? SOLUTION: A SPHERE = 4  r 2.  Recall that intensity is the rate at which energy is being gained per unit area.  Then I = P / [ 4  r 2 ] = 3.90  10 26 / [ 4  (1.5  10 11 ) 2 ] = 1380 W m -2.  This value is called the solar constant. Topic 8: Energy production 8.2 – Thermal energy transfer The solar constant intensity = power / A intensity P sun = 1380 W m -2 the solar constant

31. PRACTICE: Explain why the solar intensity is different for different latitudes. SOLUTION:  The following diagram shows how the same intensity is spread out over more area the higher the latitude. Topic 8: Energy production 8.2 – Thermal energy transfer 1380 W/m 2 The solar constant

32. Topic 8: Energy production 8.2 – Thermal energy transfer PRACTICE: The sun radiates energy at a rate of 3.90  10 26 W. What is the rate at which energy from the sun reaches Jupiter if its orbital radius is 7.8  10 11 m? (This is Jupiter’s solar constant.) SOLUTION: Use I = P / [ 4  r 2 ] I = P / [ 4  r 2 ] = 3.90  10 26 / [ 4  (7.8  10 11 ) 2 ] I = 51 W m -2.  This is 51 J / s per m 2. The solar constant

33. Topic 8: Energy production 8.2 – Thermal energy transfer EXAMPLE: Assuming an albedo of 0.30, find, for Earth: (a) the power of the sunlight received. SOLUTION: Use I = 1380 W m -2 and I = P / A.  The radius of Earth is r = 6.37  10 6 m so its cross-sectional area is A =  r 2 =  (6.37  10 6 ) 2 = 1.27  10 14 m 2.  An albedo of 0.30 means that 70% of the sunlight is absorbed (because 30% is scattered). Thus P =(0.70)IA =(0.70)(1380)(1.27  10 14 ) = 1.23  10 17 W.  Thus Earth intercepts energy from the sun at a rate of 1.23  10 17 W. Solving problems involving albedo

34. EXAMPLE: Assuming an albedo of 0.30, find, for Earth: (b) the predicted temperature due to the sunlight reaching it. SOLUTION:  From the previous example P = 1.23  10 17 W.  But this power is distributed over the whole rotating planet, which has an area A sphere = 4  r 2.  From Stefan-Boltzmann we have P =  AT 4 1.23  10 17 = (5.67  10 -8 )4  (6.37  10 6 ) 2 T 4 T = 255 K (-18°C). Topic 8: Energy production 8.2 – Thermal energy transfer Solving problems involving albedo

35. EXAMPLE: If the average temperature of Earth is 289 K, find its emissivity. SOLUTION: We determined that Earth absorbs energy from the sun at a rate of P = 1.23  10 17 W.  Since the temperature of Earth is relatively constant, we can assume it is radiating power at the same rate of P BODY = 1.23  10 17 W.  At 289 K the power radiated by a black-body of the same size as Earth is: P B-B = (5.67  10 -8 )4  (6.37  10 6 ) 2 289 4 = 2.02  10 17 W.  Thus e = P BODY / P BLACK-BODY = 1.23 / 2.02 = 0.61. Solving problems involving emissivity and Earth’s average temperature Topic 8: Energy production 8.2 – Thermal energy transfer

36.  albedo = P SCATTERED / P INCIDENT  P INCIDENT = 340 W m -2  P SCATTERED = 100 W m -2  albedo = 100 / 340 Solving problems involving albedo

37. Absorption of infrared radiation by greenhouse gases Topic 8: Energy production 8.2 – Thermal energy transfer PRACTICE: Consider the absorption graphs below: (a) Which portion of the electromagnetic spectrum is represented? (b) Which greenhouse gas contributes the most to the greenhouse effect? (c) Which gas is the least significant contributor? SOLUTION: (a) Infrared (heat). (b) Water vapor! (c) Oxygen and ozone.

38. Energy balance in Earth surface / atmosphere system Topic 8: Energy production 8.2 – Thermal energy transfer  P / A = (1)(5.67  10 -8 )288 4 = 390 W m -2 (amount radiated)  P / A = 140 W m -2 (amount absorbed from atmosphere)  390 – 140 = 250 W m -2 (amount absorbed from sun).

39. Energy balance in Earth surface / atmosphere system Topic 8: Energy production 8.2 – Thermal energy transfer  P / A = (0.72)(5.67  10 -8 )(242 + 6) 4 = 154 W m -2.

40. Energy balance in Earth surface / atmosphere system Topic 8: Energy production 8.2 – Thermal energy transfer  Amount absorbed from sun 250 W m -2 is still same.  P / A = 154 W m -2 + 250 W m -2 = 404 W m -2.

41. Energy balance in Earth surface / atmosphere system Topic 8: Energy production 8.2 – Thermal energy transfer  From Stefan-Boltzmann we have P =  AT 4.  Thus P / A =  T 4 or 404 = 5.67  10 -8 T 4, and  T = 291 K, or an increase of 291 – 288 = 3 K.