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Today-Ch. 5

Today-Ch. 5. Hydropower Geothermal power Wave power Tidal power Ocean thermal power Biomass (more with Ch. 8) All power point images are only for the exclusive use of Phys3070/Envs3070 Spring term 2014. reminder.

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Today-Ch. 5

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  1. Today-Ch. 5 Hydropower Geothermal power Wave power Tidal power Ocean thermal power Biomass (more with Ch. 8) All power point images are only for the exclusive use of Phys3070/Envs3070 Spring term 2014

  2. reminder • The Betz factor of 0.59 for the maximum fraction of wind power that can be extracted is NOT included in the text equation with the 6.1x10-4.

  3. mgh

  4. GeothermalDeep heat from radioactive decay. Steam, geysers ‘Hot dry rock’, inject water

  5. Cooling towers Use waste heat, often as steam, to heat up water until it evaporates. One kg of evaporated hot water carries away 2.26 MJ of energy. For 1 GWe3 GWt you need to get rid of 2 GW = 2 GJ/sec, or 2 GJ/sec / 2.26 MJ/kg =885 kg / sec

  6. Ocean power 1. Tidal energy Dams, build a pressure head Turbines in the flow 2. Wave 3. Ocean thermal

  7. Power in wind= ½ r* Area *v3, with v in meters per second, r=density of air (1.2 kg/m3) P/area=6.1 x10-4 v(m/s)3 kW/m2 But- at most we can use 59% of this (Betz limit) Power(kW)/m2 (from the windmill) = 0.59 x0.61 v3 in W/m2 Typically 60% of the optimum for a good machine (Fig. 5.6)

  8. Wave power

  9. Ocean thermal power Water is most dense at 4 deg C, and thus this is the temperature at the bottom of oceans. Build a heat engine, with surface warm water temps near 30 deg C. Too much plumbing, pumping, very poor Carnot efficiency. Instead– pump up cold water for AC.

  10. Biofuels Wood/pulp/woodwaste, cheaper, cleaner than coal, 7700 MWe Municipal waste Manure/farm waste Ethanol/Biodiesel--

  11. Where does waste go?

  12. Example Wood byproducts provide 7700 MWe per year. How much coal is thus not dug, hauled, and burned?

  13. numbers Assume efficiency =0.30. Coal holds 2.81x1010 J/ton. In one year, wood provides 7.7x103x106 J/sec x 3.16x107 sec = 24.3 x 1016 J of electrical energy At 30%, this needed 24.3x1016/0.30 =81.1x1016 J of heat. Tons of coal = J / (J/ton) =81.1x1016 J / 2.81x1010 J/ton=29 x106 tons (almost 3% of the billion tons per year we dig and burn)

  14. (10) You are deciding what kind of heating system to put into your new house in Boulder. Your choices are electrical heat (using power from burning coal at the Valmont plant at 32% • efficiency), burning coal in a home furnace at 80% efficiency, or installing a heat pump (driven by Valmont electricity) with a C. O. P. of three. • Compare and contrast the tons of coal needed to • provide 4 x 107 BTU per year.

  15. A. electricity from coal 4*107Btu/year= 0.32*2.66*107Btu/ton * X tons/year X=4.699 tons of coal/year

  16. B. A coal furnace 4*107 Btu/year = 0.80 * 2.66*107 Btu/ton * Y tons/year Y=1.88 tons/year

  17. C. An electric heat pump COP=Coefficient Of Performance =Qh/W (pp 79-82) Qh = heat energy to the hot place W = work paid for to do the pumping Qh = 4*107 Btu ( in one year) W=Qh/3 = 4*107 Btu/3 = 0.32*2.66 *107 Btu/ton *Z tons Z=4*107 /3 * 1/0.32 * 2.66*107 =1.57 tons of coal in one year

  18. CAPA 1A The city of Boulder produces GW of electrical power from hydro plants in Boulder Canyon. Actually, 11.7 GW-hr per year = 11.7*109 W-hour/year / 8760 hour/year =1.33 *106 W = 1.33 MW

  19. CAPA 2E A useful rule-of-thumb for a flat plate collector is about 1 or 2 Btu per square foot per day of collected energy. In Boulder, Fig. 5.3, we get 5.25 kW-hr/m2 per day, during the 8 hour day. This is 5.25 (kW-hr/m2)/day*3413 Btu/kW-hr *1/(3.28 ft/m)2 = 1666 Btu/square foot per day.

  20. Friday Nuclear fission Chapter 6 Dr. Sandy Bracken/ Leadership

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