PROBLEM 53 & 54 June 2 2010

PROBLEM 53 & 54 June 2 2010 - from sunlight? Energy...

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Problems 53 and 54 Solutions Phys 002C June 2, 2010 Problem 53. (Like problem 30.31) Nuclear Energy 10 9 tons U = 10 12 kg 7 10 -3 U 235 = 7 10 9 kg U 235 available. Energy yield is at most E=208 MeV 1.6 10 13 J/MeV per atom = (208 MeV 1.6 10 13 J/MeV per atom) 6 10 23 atoms/mole /235 g/mole = 10 14 J/kg Total energy supply is at most 7 10 23 J. Current world consumption is approaching 10TW=10 13 W [Note a 2,000 calorie diet for 10 10 people is 1 TW!] Lifetime of U supply is therefore 7 10 23 J/10 13 W = 7 10 10 s = 2,000 y! Except for waste disposal and piracy, this is the solution. Problem 54. (Like ?) What area would be needed to supply the world’s 10 TW energy need
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Unformatted text preview: from sunlight? Energy from sunlight = ~1 kW/m 2 . Continuous available area is about R earth 2 =3.6 10 13 m 2 . Total solar power shining on Earth = 3.6 10 16 W = 3,000X world consumption. Trick is making a (100 km) 2 collector! Available solar cells say type SKU4557 $565 28 lbs 1 m 2 100W max (10% eff) x 5 h/day in the desert, 365 d/yx 5 years = 1000 kwh/$500. Cost is $1 per kwh including real estate cost (x2) compared to present $0.10/kwh which includes distribution costs....
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