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hw2-key

# hw2-key - EE 295 Homework 2 Solutions March 1 2010 1...

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Unformatted text preview: EE 295 - Homework 2 Solutions March 1, 2010 1 Readings 1. Chapters 3-4 of Bergen & Vittal 2 From the textbook (Bergen/Vittal) 2.1 Problem 3.8 Aluminum 52,620 ccmil conductor w/7 strands, each with d = 0 . 0867 in. Outside diamter is: D = 0 . 2601 in. Find GMR: Solution : The GMR will result from the distances among the individual conductors. The GMR for each individual conductor is: r ii = 0 . 7788 d 2 The center conductor is distance d from every conductor. Thus the product of the distances from the center conductor is: R c = r ii d 6 The outer conductors have the following distance product: R o = r ii ( d 3 )(2 d )( √ 3 d ) 2 Thus the GMR is: R s = GMR = ( R c R 6 o ) 1 / 7 = 1 . 088 d = 0 . 0943 in = 0 . 00786 ft This is pretty close to the book value for this conductor ( . 00787 in). Problem 3.9 Find the per phase inductance per meter for a 765-kV line with d ab = d bc = 45 and d ac = 90 Solution (in feet): D m = (45 2 90) 1 / 3 = 56 . 7 ft R b = ( r ii · 1 . 5 2 (1 . 5 √ 2)) 1 / 4 = (0 . 0479 · 1 . 5 2 (1 . 5 √ 2)) 1 / 4 = 0 . 6915 ft The line inductance is therefore found from equation 3.33: l = 2 × 10- 7 ln D m R b = 8 . 81 × 10- 7 H/m 1 2.2 Problem 3.10 To do this problem, you really need to do problem 3.9 as well (probably an unfair request on myTo do this problem, you really need to do problem 3....
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hw2-key - EE 295 Homework 2 Solutions March 1 2010 1...

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