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- Electrical Power Transmission And Distribution - EPTD
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**Veer Surendra Sai University Of Technology VSSUT -**- Electrical and Electronics Engineering
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CLASS NOTES ON ELECTRICAL POWER TRANSMISSION AND DISTRIBUTION CHAPTER-1 TRANSMISSION LINE PARAMETERS-I INTRODUCTION The transmission line performance is based on its electrical parameters such as resistance, inductance and capacitance. As we know the transmission line are used for delivering electrical power from one end to other end or one node to other node, the path of power flow i.e. the transmission line can be represented as an electrical circuit having its parameters connected in a particular pattern. Since the transmission line consists of conductors carrying power, we need to calculate the resistance, inductance and capacitance of these conductors. RESISTANCE OF TRANSMISSION LINE The resistance of the conductor thus transmission line can be determined by (1.1). R l A (1.1) Where ' ' is the resistivity of the wire in -m, 'l ' is the length in meter and ' A' is the cross sectional area in m2. When alternating current flows through a conductor, the current density is not uniform over the entire cross section but is somewhat higher at the surface. This is called the skin effect and this makes the ac resistance a little more than the dc resistance. Moreover in a stranded conductor, the length of each strand is more that the length of the composite conductor thus increasing the value of the resistance from that calculated in (1.1). INDUCTANCE OF TRANSMISSION LINE In order to determine the inductance of transmission line, we shall first drive expression for the inductance of a solid conductor and it will be extended to a single phase transmission line. Then we shall derive expression for inductance of a group of conductors and then extend it to three phase transmission line. INDUCTANCE OF SOLID CONDUCTOR The inductance of solid conductor can be determined by calculating the flux linkage due to current flowing and using (1.2). 9

10 VEER SURENDRA SAI UNIVERSITY OF TECHNOLOGY, ODISHA, BURLA L (1.2) I Where L is the inductance in Henry, λ is the flux linkage in Weber-turns and I is the phasor current in Ampere. INDUCTANCE OF SOLID CONDUCTOR DUE TO INTERNAL FLUX Let us consider a solid conductor of radius 'r ' cm and the current flowing is ' I ' A as shown in Fig.-1.1. r o x dx FIG.-1.1 INTERNAL FLUX LINKAGE OF A ROUND CONDUCTOR. As we know Ampere's law states that the magnetomotive force (mmf) in ampere-turns around a closed path is equal to the net current in amperes enclosed by the path. Mathematically is written as (1.3). mmf H ds I (1.3) Where H is the magnetic field intensity in At/m, s is the distance along the path in meter. Let us consider a tubular element of thickness 'dx' of the conductor at a distance ' x' from the center of the conductor and the field intensity as ' H x ' at ' x' . It is constant at all points that are at a distance ' x' from the center of the conductor. Therefore ' H x ' is constant over the concentric circular path with a radius of ' x' and is tangent to it. Let the current enclosed by this path is ' I x ' . Hence by (1.3) we can write as follows. H x dx I x 2xH x I x Hx Ix 2x (1.4) (1.5) Assuming the current density to be uniform thorugh out the cross section of the conductor, the current at a radius of ' x' is given by (1.6)

CLASS NOTES ON ELECTRICAL POWER TRANSMISSION AND DISTRIBUTION 2x 2 I 2r 2 Ix (1.6) Substituting (1.6) in (1.5) we get Hx x I 2r 2 (1.7) The flux density at a distance of ' x' is given by Bx H x 0 r H x (1.8) Considering the unit length of the conductor i.e. one metre, the flux in the tubular element of thickness 'dx' of the conductor can be given by (1.9). d x Bx dx (1.9) Combining (1.7), (1.8) and (1.9) d x I xdx 2r 2 (1.9) The flux linkage at ' x' can be given by d x x 2 I d x x 3 dx 2 4 r 2r (1.10) The total internal flux linkage can be obtained by integrating (1.10) over the range of ' x' , i.e., from '0' to ' r ' as follows. r r d x 0 0 I x 3 dx 4 2r I 8 (1.11) (1.12) For relative permeability to be r 1 we have 0 4 X 10 7 , hence (1.12) can be written as follows, which is the flux linkage due to internal flux. int I X 10 7 Wb-T/m 2 (1.13) Hence the inductance of the conductor due to internal flux is obtained by using (1.2) and (1.13). Lint int I 1 X 10 7 H/m 2 (1.14) 11

12 VEER SURENDRA SAI UNIVERSITY OF TECHNOLOGY, ODISHA, BURLA INDUCTANCE OF SOLID CONDUCTOR DUE TO EXTERNAL FLUX Now we shall calculate the inductance of solid conductor due to flux linking with the conductor externally. Let us consider two points ' P1 '&' P2 ' at a distance of ' D1 '&' D2 ' from the center of the conductor and external to it. Let us consider a tubular element of thickness 'dx' of the conductor at a distance ' x' from the center of the conductor and the field intensity as ' H x ' at ' x' . as shown in Fig.-1.2. o r x dx FIG.-1.2 EXTERNAL FLUX LINKAGE OF A ROUND CONDUCTOR. The current enclosed by the tubular element is the total current i.e. ' I ' . Hence the fieled intensity is given by (1.15). Hx I (1.15) 2x The flux density at a distance of ' x' is given by Bx H x 0 r H x (1.16) Considering the unit length of the conductor i.e. one metre, the flux in the tubular element of thickness 'dx' of the conductor can be given by (1.17). d x Bx dx (1.17) Combining (1.15), (1.16) and (1.17) d x I 2x dx The flux linkage at ' x' can be given by (1.18)

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