Inductance Of Transmission lins march.pptx

Engineering

1. Pr e s e n t e d B y : Er Kiranpreet Kaur Department of EE

2. Definition Of Inductance Flux Linkages of Conductors Flux linkages inside the conductor Flux linkages outside the conductor Flux li n k a g es o f on e c o n d u c tor in a grou p of conductors Inductance of a single phase two wire line Inductance of 3-phase overhead line Bundled conductors

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5. Consider a conductor of radius r carrying a current I. At a distance x from the center of this conductor, the magnetic field intensity H x The internal inductance per meter is i n t  H m  l I 8    int    H m   7  4   1  7 8  8  l int          If the relative permeability of the conductor is 1 (non-ferromagnetic materials, such as copper and aluminum), the inductance per meter reduces to

6. The total external flux linkages per meter can be found via integration The external inductance per meter is : To find the flux linkages external to a conductor, we will consider the portion of flux between two points P 1 and P 2 that lie at distances D 1 and D 2 from the center of the conductor.

7. Theoretically , the flux due to a conductor extends from the center of the conductor to right unto infinity. Assuming that the flux linkages will extend unto a point P very far from the group of the conductors , and the distances are as shown in fig.

8. Consider a group of conductors 1, 2, 3,…… n such that the sum of currents in all these conductors is zero. If the currents carried by respective strands are I1, I2, I3, …In we have I1+I2+I3+…+In = ψ1p1 = All flux linkages of conductor 1 due to tis own current I1 , internal and external , unto point P …. Ψ1p1 = 2 × 10 -7 I1 ln D1p/r1‘ Wb. T/Mt Ψ1p2 = Flux linkages with conductor 1 due to current in conductor-2 Ψ1p2 = 2 × 10 × I2 ln D2p/D12 -7

9. Similarly ψ1p3…….ψ1pn Ψ1p= 2 × 10 -7 { I1 ln D1p/r'1 + I2 ln D2p/D12 +...+ In ln Dnp/D1n} Flux linkages with conductor 1 due to I1 , I2…..In Net flux linkages ψ1p -7 Ψ1p = 2 × 10 { I1 ln 1/r1'+ I2 ln 1/D12 +…+ In ln 1/D1n } Wb-turns/m

10. The inductance of a single-phase line consisting of two conductors of radii r spaced by a distance D and both carrying currents of magnitude I flowing into the page in “A” conductor and out of the page in the “B“ conductor. Ñ  H x  d l  I x Since the path of radius x 2 encloses both conductors and the currents are equal and opposite, the net current enclosed is 0 and, therefore, there are no contributions to the total inductance from the magnetic fields at distances greater than D. A

11. The total inductance of a line per unit length in this transmission line is a sum of the internal inductance and the external inductance between the conductor surface ( r ) and the separation distance ( D ):  H m    1 l  l int l ext    ln  2  4 r  By symmetry, the total inductance of the other line is the same, therefore, the total inductance of a two-wire transmission line is  H m    4 r    W here r ’ = i s GMR (Geo m et r ic m ean rad ius) For a solid conductor G.M.R = 0.7788 times the radius of conductor. D is the distance between conductors ‘  l    1  l n D ‘  D ‘ 

12. In a 3-phase transmission line, the inductance of each conductor is considered instead of loop inductance. The conductor of a 3-phase overhead line may be placed symmetrically or unsymmetrically on the towers. With Symmetrical Spacing : A 3-phase line in which the space between any two conductor is the same as shown in fig. the line is called symmetrical line. Fig. shows the conductor of a 3-phase line conductor has a radius r meters and spacing between the conductors is D meters.

13. Under balanced three-phase phasor currents, the algebraic sum of the currents in the conductors is zero. Hence, Ia + Ib + Ic = The flux linkages of the conductor ‘a’ are ψa = 2 × 10 -7 [ Ia ln 1/Daa + Ib ln 1/Dab + Ic ln 1/Dac ] Wb-T/m Inductance of conductor a, La = ψa/Ia -7 = 2 × 10 ln D/r‘ H/m

14. A 3-phase line in which the space between the conductors is different as shown in fig., the line is called unsymmetrical line. Consider 3-phase line with conductors a, b and c each of radius r meters.

15. Let, the spacing between them be Dab, Dbc and Dca and the currents flowing through them be Ia, Ib and Ic respectively as shown in fig. From fig, the flux linkages of the conductor a Ψa = 2 × 10 -7 [ Ia ln 1/r' + Ib ln 1/D12 + Ic ln 1/D31 ] Wb-T/m

16. A bundle conductor is a conductor made up of two or more sub-conductors and is used as one phase conductor. Lines of 400kv and higher voltages invariably use bundled conductors. Sub-conductors of a bundled conductor are separated from each other by a constant distance varying from 0.2 m to 0.6 m depending upon designed voltage and surrounding conditions throughout the length of the line with the help of spacers.

17. It reduces corona loss. It reduces radio interference The bundled conductor lines transmit bulk power with reduced losses, thereby giving increased transmission efficiency Bundle conductor lines have a higher capacitance to neutral so they have higher charging current, which helps in improving power factor By bundling, the GMR is increased, the inductance per phase is reduced. As a result reactance per phase is reduced.