{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

PROBLEM 53 &amp; 54 June 2 2010

# PROBLEM 53 &amp; 54 June 2 2010 - from sunlight Energy...

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online