Chapter 4.5 Inductance PDF

Module 4 Chapter 4.5: Inductance: Single-phase two-wire line and three-phase three-wire line with equal phase spacing All Rights Reserved 1 Inductance: Single-phase, two-wire • Flux linkage of conductor x Ix = I Iy = -I reference out of page return current 2 Inductance: Single-phase, two-wir...

Module 4 Chapter 4.5: Inductance: Single-phase two-wire line and three-phase three-wire line with equal phase spacing All Rights Reserved 1 Inductance: Single-phase, two-wire • Flux linkage of conductor x Ix = I Iy = -I reference out of page return current 2 Inductance: Single-phase, two-wire • Similarly: flux linkage of conductor y 3 Inductance: Single-phase, two-wire • The total inductance of the single-phase circuit, called loop inductance, is: 4 Three-phase, Three-wire Line • Below is a three-phase three-wire line consisting of three solid cylindrical conductors a, b, c, each with radius r, and with equal phase spacing D between any two conductors. • To determine inductance, assume balanced positive-sequence currents Ia, Ib, Ic that satisfy Ia + Ib + Ic = 0 5 Three-phase, Three-wire Line Then, flux linkage in phase a conductor is: Where: The inductance of phase a is: 6 Example 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 between conductors. Calculate in mH/km a) the inductance of each conductor due to internal flux linkages only, (Lint) 7 Example 1: Solution a) the inductance of each conductor due to internal flux linkages only, (Lint) From: Lint = ½ x 10-7 H/m = (1/2x10-7 H/m) (1000 m/ 1 Km) (1000 mH/1H) = 0.05 mH/Km per conductor 8 Example 2: 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 between conductors. Calculate in mH/km b) the inductance of each conductor due to both internal and external flux linkages 9 Example 2: Solution b) the inductance of each conductor due to both internal and external flux linkages From: H/m per conductor Lx = Ly = 2 x 10-7 ln D/r’x D = 0.5 m spacing r’ = e-1/4 r = 0.7788 r = 0.7788 (d/2) = 0.7788 (0.015/2) = 5.841 x 10-3 m Therefore: Lx = Ly = 2x10-7 ln (0.5/5.841x10-3)H/m (1000 m/1 Km) (1000 mH/H) = 0.8899 mH/Km per conductor 10 Example 3: 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 between conductors. Calculate in mH/km c) the total inductance of the line 11 Example 3: Solution c) the total inductance of the line L = Lx + Ly = 0.8899 + 0.8899 = 1.780 mH/Km per circuit 12