Question

4.1 A 60-Hz single-phase, two-wire overhead line has solid cylindrical copper conductors with 1.5 cm diameter. The conductors are arranged in a horizontal configuration with 0.5 m spacing. Calculate in mH/km (a) The inductance of each conductor due to internal flux linkages only; (b) The inductance of each conductor due to both internal and external flux linkages; (c) The total inductance of the line. (40 Points) F.A. (a) Lint = 0.05 mh per conductor km (b) Ly = Ly per conductor km (c) L = 1.780 per conductor 0.8899 mH mH km

Answer 1

- 1m f=60Hz , r=0.5 cm Solution givendata sola Internal quductance ) we know that flux in Kage current - is L = ds By the figure we can coute fuas - Lenclosed MME = Hol das = 1 22 Rets consider in Tegral length as Sareads I Ha - 2 2T22 we know that Mx B = M Hac 2 27 & calculating flux inds = Balads . do = Bada P.U length ds =) Ma da fuxper aut length-dex 2082 x3 da. I x² - fuxx Pux linkage El dir 2084 By Integrating eqh da = MI Hollra BTT OTC s : Hy = 1 a. Ho =40x107 ARX107 I le 2 de = Ixio anteenad Puductance = 8/2x107 Byegpo peux unege current tx I 0.5X10 t/m /¿ Per conductor 0.05 m. lo que

d = spacing ( AL) 20 015 d = 0.7788 & 2-sallius soln = we know the = Molo und Now By putting values L = axlotin 0:7788X105 X 102 = 2x107 In casi6o) 4.4497 x2x107 L = 8:899412 H/ 18 ) 0.8899 MH Ane km/ Soln ③ = we know that Total inductance Ly= MOME 내일 By puttiry values - 4= 4x107x 4x107x4.4497 = 1417980 X107 H/on 1,7798 mH km G=11780 KmH Kmle Aus ( 17 denotees per conductor
Proper stepwise solution is given above.

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