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INDUCTANCE & CAPACITANCE CALCULATIONS

The document discusses various types of conductors used in transmission lines, including solid, stranded, hollow, and bundled conductors, along with their properties. It also covers key parameters like resistance, inductance, and capacitance, explaining the skin and proximity effects that influence these properties in alternating current (AC). Additionally, the text details the mathematical treatment of inductance associated with both single and parallel conductors, illustrating the impact of design on conductivity and efficiency in power transmission.

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By D.Rajesh Asst. Prof. EEE Department 15 October 2020 D.Rajesh Asst.Prof. EEE Department 1
 Types of conductors, line parameters,  Calculation of inductance and capacitance of single and double circuit transmission lines, Three phase lines with bundle conductors.  Skin effect and  proximity effect. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 2
 Solid conductors  Stranded conductors  Hallow conductors  Bundled conductors 15 October 2020 D.Rajesh Asst.Prof. EEE Department 3
 Solid conductors 1. Very costly & high tensile strength 2. Difficult to string these conductors between poles 3. For A.C. these exhibit more skin effect 4. Less flexibility, so rarely used 15 October 2020 D.Rajesh Asst.Prof. EEE Department 4
 Stranded conductors Three or more strands of conductor material is twisted to form a single conductor. They can carry high currents and are more flexible than solid conductors. Ex: ACSR,AAC,AAAC,ACAR 15 October 2020 D.Rajesh Asst.Prof. EEE Department 5
 Hallow conductors Hollow conductors are used in the transmission lines to: (1) Improve heat dissipation (2) Reduce corona loss (3) Reduce skin effect (4) Reduce the line inductance 15 October 2020 D.Rajesh Asst.Prof. EEE Department 6
 Bundled Conductors  Bundled Conductors are used in transmission lines where the voltage exceeds 230 kV.  At such high voltages, ordinary conductors will result in excessive corona and noise which may affect communication lines.  The increased corona will result in significant power loss. Bundle conductors consist of three or four conductors for each phase.  The conductors are separated from each other by means of spacers at regular intervals. Thus, they do not touch each other.  A bundled conductor reduces the reactance of the electric transmission line. It also reduces voltage gradient, corona loss, radio interference, surge impedance of the transmission lines. By making bundle conductor, the geometric mean radius (GMR) of the conductor increased. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 7
 (i) Resistance.  (ii) Inductance.  (iii) Capacitance. A transmission line has resistance, inductance and capacitance uniformly distributed along the whole length of the line. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 8
The resistance of transmission line conductors is the most important cause of power loss in a transmission line. The resistance R of a line conductor having resistivity ρ, length l and area of crossection a is given by ; R = ρ l/a It is the opposition of line conductors to current flow. The resistance is distributed uniformly along the whole length of the line as shown in Fig. (i). However, the performance of a transmission line can be analysed conveniently if distributed resistance is considered as lumped as shown Fig. (ii). 15 October 2020 D.Rajesh Asst.Prof. EEE Department 9
 When an alternating current flows through a conductor, a changing flux is set up which links the conductor. Due to these flux linkages, the conductor possesses inductance.  Mathematically, inductance is defined as the flux linkages per ampere i.e., Inductance, L = ψ I henry where ψ = flux linkages in weber-turns I = current in amperes 15 October 2020 D.Rajesh Asst.Prof. EEE Department 10
 Any two conductors of an overhead transmission line are separated by air which acts as an insulation, therefore, capacitance exists between any two overhead line conductors. The capacitance between the conductors is the charge per unit potential difference i.e.,  Capacitance, C = 𝑞 𝑣 farad where q = charge on the line in coulomb v = p.d. between the conductors in volts 15 October 2020 D.Rajesh Asst.Prof. EEE Department 11
When a conductor is carrying steady direct current (d.c.), this current is uniformly distributed over the whole X-section of the conductor. However, an alternating current flowing through the conductor does not distribute uniformly, rather it has the tendency to concentrate near the surface of the conductor as shown in Fig. This is known as skin effect. The tendency of alternating current to concentrate near the surface of a conductor is known as skin effect. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 12
 Due to skin effect, the effective area of cross-section of the conductor through which current flows is reduced. Consequently, the resistance of the conductor is slightly increased when carrying an alternating current.  The skin effect depends upon the following factors : (i) Nature of material (ii) Diameter of wire − increases with the diameter of wire. (iii) Frequency − increases with the increase in frequency. (iv) Shape of wire − less for stranded conductor than the solid conductor.  It may be noted that skin effect is negligible when the supply frequency is low (< 50 Hz) and conductor diameter is small (< 1cm). 15 October 2020 D.Rajesh Asst.Prof. EEE Department 13
 The inductance of a circuit is defined as the flux linkages per unit current. 1. Flux linkages due to a single current carrying conductor 2. Flux linkages in parallel current carrying conductors. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 14
1. Flux linkages due to a single current carrying conductor Consider a long straight cylindrical conductor of radirus r metres and carrying a current I amperes (r.m.s.) as shown in Fig. This current will set up magnetic field. The magnetic lines of force will exist inside the conductor aswell as outside the conductor. Both these fluxes will contribute to the inductance of the conductor. (i) Flux linkages due to internal flux. (ii) Flux linkages due to external flux. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 15
(i) Flux linkages due to internal flux. The magnetic field intensity at a point x metres from the centre is given by; 𝐻𝑥 = 𝐼𝑥 2π𝑥 [According to Ampere’s law, m.m.f. (ampere-turns) around any closed path equals the current enclosed by the path. The current enclosed by the path is 𝐼𝑥 and m.m.f. = 𝐻𝑥 × 2π x. ∴ 𝐻𝑥 × 2π x = 𝐼𝑥.] Assuming a uniform current density, 𝐼𝑥 π𝑥2 = 𝐼 π𝑟2 𝐼𝑥= π𝑥2 π𝑟2 𝐼 = 𝑥2 𝑟2 𝐼 put 𝐼𝑥 in 𝐻𝑥 𝐻𝑥= 𝑥2 𝑟2 𝐼* 1 2π𝑥 = 𝑥 2π𝑟2 𝐼 𝐴𝑇/𝑚 15 October 2020 D.Rajesh Asst.Prof. EEE Department 16
 If μ (= 𝜇0𝜇𝑟 ) is the permeability of the conductor, then flux density at the considered point is given by;  𝐵𝑥 = 𝜇0𝜇𝑟𝐻𝑥 wb/𝑚2 = 𝜇0𝜇𝑟 𝑥 2π𝑟2 𝐼 = 𝜇0𝑥𝐼 2π𝑟2 𝑤𝑏/𝑚2 [∵ 𝜇𝑟 = 1 for non-magnetic material] Now, flux dφ through a cylindrical shell of radial thickness dx and axial length 1 m is given by; dφ= 𝐵𝑥 × 1 × dx= 𝜇0𝑥𝐼 2π𝑟2 dx 𝑤𝑒𝑏𝑒𝑟 This flux links with current 𝐼𝑥 (= π𝑥2 π𝑟2 𝐼) only. Therefore, flux linkages per metre length of the conductor is 𝑑ψ= π𝑥2 π𝑟2 dφ = 𝜇0𝑥3𝐼 2π𝑟4dx 𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠 Total flux linkages from centre upto the conductor surface is ψ𝑖𝑛𝑡 = න 0 𝑟 𝜇0𝑥3𝐼 2π𝑟4 dx = 𝜇0𝐼 2π𝑟4 𝑥4 4 𝑟 0 = 𝜇0𝐼 2π𝑟4 𝑟4 4 = 𝜇0𝐼 8π 𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠 per meter length 15 October 2020 D.Rajesh Asst.Prof. EEE Department 17 Internal inductance is independent of conductor radius.
 (ii) Flux linkages due to external flux. Now let us calculate the flux linkages of the conductor due to external flux. The external flux extends from the surface of the conductor to infinity. The field intensity at a distance x metres (from centre) outside the conductor is given by ; 𝐻𝑥 = 𝐼 2π𝑥 AT / m Flux density, 𝐵𝑥 = 𝜇0𝜇𝑟𝐻𝑥 = 𝜇0𝐼 2π𝑥 𝑤𝑏/𝑚2 [∵ 𝜇𝑟 = 1 for non-magnetic material] Now, flux dφ through a cylindrical shell of thickness dx and axial length 1 metre is dφ= 𝐵𝑥 × 1 × dx= 𝜇0𝐼 2π𝑥 𝑤𝑒𝑏𝑒𝑟 The flux dφ links all the current in the conductor once and only once. ∴ Flux linkages, dψ = dφ = 𝜇0𝐼 2π𝑥 dx 𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠 Total flux linkages of the conductor from surface to infinity, ψ𝑒𝑥𝑡 = න 𝑟 ∞ 𝜇0𝐼 2π𝑥 dx ∴ Overall flux linkages, ψ =ψ𝑖𝑛𝑡 + ψ𝑒𝑥𝑡 = 𝜇0𝐼 8π + ‫׬‬𝑟 ∞ 𝜇0𝐼 2π𝑥 dx = 𝜇0𝐼 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx 𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠 per meter length 15 October 2020 D.Rajesh Asst.Prof. EEE Department 18
2. Flux linkages in parallel current carrying conductors. The conductors A,B, C etc. carrying currents 𝐼𝐴, 𝐼𝐵, 𝐼𝐶 etc. Let us consider the flux linkages with one conductor, say conductor A. There will be flux linkages with conductor A due to its own current and there will be flux linkages with this conductor due to the mutual inductance effects of, 𝐼𝐵, 𝐼𝐶 , 𝐼𝐷, 𝐼𝐸 etc. The total flux linkages with conductor A. Flux linkages with conductor A due to its own current = 𝜇0𝐼𝐴 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx …….(i) Flux linkages with conductor A due to current 𝐼𝐵 = 𝜇0𝐼𝐵 2π ‫׬‬𝑑1 ∞ 1 𝑥 dx …….(ii) Flux linkages with conductor A due to current 𝐼𝐶 = 𝜇0𝐼𝐶 2π ‫׬‬ 𝑑2 ∞ 1 𝑥 dx …….. (iii) ∴ Total flux linkages with conductor A = (i) + (ii) + (iii) + ...... = 𝜇0𝐼𝐴 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx + 𝜇0𝐼𝐵 2π ‫׬‬ 𝑑1 ∞ 1 𝑥 dx + 𝜇0𝐼𝐶 2π ‫׬‬ 𝑑2 ∞ 1 𝑥 dx + ⋯ 15 October 2020 D.Rajesh Asst.Prof. EEE Department 19
 Consider a single phase overhead line consisting of two parallel conductors A and B spaced d metres apart.  Conductors A and B carry the same amount of current (i.e. 𝐼𝐴 = 𝐼𝐵), but in the opposite direction because one forms the return circuit of the other. ∴ 𝐼𝐴 + 𝐼𝐵 = 0  In order to find the inductance of conductor A (or conductor B), we shall have to consider the flux linkages with it.  There will be flux linkages with conductor A due to its own current 𝐼𝐴 and also due to the mutual inductance effect of current 𝐼𝐵 in the conductor B.  Flux linkages with conductor A due to its own current = 𝜇0𝐼𝐴 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx …….(i) Flux linkages with conductor A due to current 𝐼𝐵 = 𝜇0𝐼𝐵 2π ‫׬‬ 𝑑 ∞ 1 𝑥 dx …….(ii) 15 October 2020 D.Rajesh Asst.Prof. EEE Department 20
 Total flux linkages with conductor A is  ψ𝐴 = 𝜇0𝐼𝐴 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx + 𝜇0𝐼𝐵 2π ‫׬‬ 𝑑 ∞ 1 𝑥 dx = 𝜇0 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx 𝐼𝐴 + 𝐼𝐵 ‫׬‬ 𝑑 ∞ 1 𝑥 dx = 𝜇0 2π 1 4 + log𝑒 ∞ − log𝑒 𝑟 𝐼𝐴 + 𝐼𝐵 log𝑒 ∞ − log𝑒 𝑑 = 𝜇0 2π 𝐼𝐴 4 + log𝑒 ∞ 𝐼𝐴 + 𝐼𝐵 − 𝐼𝐴 log𝑒 𝑟 − 𝐼𝐵 log𝑒 𝑑 = 𝜇0 2π 𝐼𝐴 4 − 𝐼𝐴 log𝑒 𝑟 − 𝐼𝐵 log𝑒 𝑑 ( ∴ 𝐼𝐴 + 𝐼𝐵 = 0) = 𝜇0 2π 𝐼𝐴 4 − 𝐼𝐴 log𝑒 𝑟 + 𝐼𝐴 log𝑒 𝑑 (∴ 𝐼𝐴 = −𝐼𝐵) = 𝜇0 2π 𝐼𝐴 4 + 𝐼𝐴 log𝑒 𝑑 𝑟 = 𝜇0𝐼𝐴 2π 1 4 + log𝑒 𝑑 𝑟 𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠 per meter length 15 October 2020 D.Rajesh Asst.Prof. EEE Department 21
 Inductance of conductor A, 𝐿𝐴 = ψ𝐴 𝐼𝐴 = 𝜇0 2π 1 4 + log𝑒 𝑑 𝑟 H/m = 4Π∗10−7 2π 1 4 + log𝑒 𝑑 𝑟 H/m 𝐿𝐴 =10−7 1 2 + 2log𝑒 𝑑 𝑟 H/m ……..(i) Loop inductance = 2 𝐿𝐴 H/m =10−7 1 + 4log𝑒 𝑑 𝑟 H/m ………(ii) Note that eq. (ii) is the inductance of the two-wire line and is sometimes called loop inductance. However, inductance given by eq. (i) is the inductance per conductor and is equal to half the loop inductance. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 22
 Expression in alternate form. The expression for the inductance of a conductor can be put in a concise form. 𝐿𝐴 = ψ𝐴 𝐼𝐴 = 𝜇0 2π 1 4 + log𝑒 𝑑 𝑟 H/m = 4Π∗10−7 2π 1 4 + log𝑒 𝑑 𝑟 H/m =2* 10−7 1 4 + log𝑒 𝑑 𝑟 H/ = 2* 10−7 log𝑒 𝑒 1 4 + log𝑒 𝑑 𝑟 H/m =2* 10−7 log𝑒 𝑑 𝑟𝑒 − 1 4 H/m ………(iii) If we put 𝑟𝑒− 1 4=r′ The radius r′ is that of a fictitious conductor assumed to have no internal flux but with the same inductance as the actual conductor of radius r. The quantity 𝑒− 1 4 = 0·7788 so that r′ = r 𝑒− 1 4 = 0·7788 r The term r′ (= r 𝑒− 1 4) is called geometric mean radius (GMR) of the wire. Note that eq. (iii) gives the same value of inductance 𝐿𝐴 as eq. (i) Loop inductance =2* 2* 10−7 log𝑒 𝑑 𝑟𝑒 − 1 4 H/m Note that r′ = 0·7788 r is applicable to only solid round conductor. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 23
Three conductors A, B and C of a 3-phase line carrying currents 𝐼𝐴, 𝐼𝐵, and 𝐼𝐶 respectively. Let 𝑑1, 𝑑2 and 𝑑3 be the spacings between the conductors as shown. Let us further assume that the loads are balanced i.e. 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶 = 0. Consider the flux linkages with conductor A. There will be flux linkages with conductor A due to its own current and also due to the mutual inductance effects of 𝐼𝐵 and 𝐼𝐶 . Flux linkages with conductor A due to its own current = 𝜇0𝐼𝐴 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx …….(i) Flux linkages with conductor A due to current 𝐼𝐵 = 𝜇0𝐼𝐵 2π ‫׬‬ 𝑑3 ∞ 1 𝑥 dx …….(ii) Flux linkages with conductor A due to current 𝐼𝑐 = 𝜇0𝐼𝑐 2π ‫׬‬ 𝑑2 ∞ 1 𝑥 dx …….(ii) 15 October 2020 D.Rajesh Asst.Prof. EEE Department 24
 Total flux linkages with conductor A is ψ𝐴 = (i) + (ii) + (iii) = 𝜇0𝐼𝐴 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx + 𝜇0𝐼𝐵 2π ‫׬‬ 𝑑3 ∞ 1 𝑥 dx+ 𝜇0𝐼𝑐 2π ‫׬‬ 𝑑2 ∞ 1 𝑥 dx = 𝜇0 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx 𝐼𝐴 + 𝐼𝐵 ‫׬‬ 𝑑3 ∞ 1 𝑥 dx + 𝐼𝑐 ‫׬‬ 𝑑2 ∞ 1 𝑥 dx =𝜇0 2π 1 4 + log𝑒 ∞ − log𝑒 𝑟 𝐼𝐴 + 𝐼𝐵 log𝑒 ∞ − log𝑒 𝑑3 + 𝐼𝐵 log𝑒 ∞ − log𝑒 𝑑2 = 𝜇0 2π 𝐼𝐴 4 + log𝑒 ∞ 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶 − 𝐼𝐴 log𝑒 𝑟 − 𝐼𝐵 log𝑒 𝑑3 − 𝐼𝐶 log𝑒 𝑑2 = 𝜇0 2π 𝐼𝐴 4 − 𝐼𝐴 log𝑒 𝑟 − 𝐼𝐵 log𝑒 𝑑3 − 𝐼𝐶 log𝑒 𝑑2 ( ∴ 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶= 0) ψ𝐴= 𝜇0 2π 1 4 − log𝑒 𝑟 𝐼𝐴 − 𝐼𝐵 log𝑒 𝑑3 − 𝐼𝐶 log𝑒 𝑑2 15 October 2020 D.Rajesh Asst.Prof. EEE Department 25
 (i) Symmetrical spacing. If the three conductors A, B and C are placed symmetrically at the corners of an equilateral triangle of side d, then, 𝑑1=𝑑2=𝑑3 = 𝑑 ψ𝐴= 𝜇0 2π 1 4 − log𝑒 𝑟 𝐼𝐴 − 𝐼𝐵 log𝑒 𝑑 − 𝐼𝐶 log𝑒 𝑑 = 𝜇0 2π 1 4 − log𝑒 𝑟 𝐼𝐴 − (𝐼𝐵 + 𝐼𝐶) log𝑒 𝑑 = 𝜇0 2π 1 4 − log𝑒 𝑟 𝐼𝐴 − (−𝐼𝐴) log𝑒 𝑑 ( ∴ 𝐼𝐵 + 𝐼𝐶= −𝐼𝐴 ) = 𝜇0𝐼𝐴 2π 1 4 + log𝑒 𝑑 𝑟 weber−turns/m Inductance of conductor A, 𝐿𝐴 = ψ𝐴 𝐼𝐴 H /m = 𝜇0 2π 1 4 + log𝑒 𝑑 𝑟 H /m = 4Π∗10−7 2π 1 4 + log𝑒 𝑑 𝑟 H/m 𝐿𝐴 =10−7 1 2 + 2log𝑒 𝑑 𝑟 H/m Derived in a similar way, the expressions for inductance are the same for conductors B and C. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 26
 (ii) Unsymmetrical spacing. When 3-phase line conductors are not equidistant from each other, the conductor spacing is said to be unsymmetrical. Under such conditions, the flux linkages and inductance of each phase are not the same. A different inductance in each phase results in unequal voltage drops in the three phases even if the currents in the conductors are balanced. Therefore, the voltage at the receiving end will not be the same for all phases. In order that voltage drops are equal in all conductors, we generally interchange the positions of the conductors at regular intervals along the line so that each conductor occupies the original position of every other conductor over an equal distance. Such an exchange of positions is known as transposition. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 27
 A3-phase transposed line having unsymmetrical spacing. Let us assume that each of the three sections is 1 m in length.  Let us further assume balanced conditions i.e., 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶 = 0. . Let the line currents be :  𝐼𝐴 =I∠00 = I (1+ j 0)  𝐼𝐵 = I∠−1200 = I (− 0·5 − j 0·866)  𝐼𝐶 = I∠−2400= I (− 0·5 + j 0·866)  The total flux linkages per metre length of conductor A is ψ𝐴= 𝜇0 2π 1 4 − log𝑒 𝑟 𝐼𝐴 − 𝐼𝐵 log𝑒 𝑑3 − 𝐼𝐶 log𝑒 𝑑2 Putting the values of 𝐼𝐴, 𝐼𝐵 and 𝐼𝐶 , we get, ψ𝐴= 𝜇0 2π 1 4 − log𝑒 𝑟 𝐼 − I (− 0·5 − j 0·866) log𝑒 𝑑3 − I (− 0·5 + j 0·866) log𝑒 𝑑2 = 𝜇0 2π 𝐼 4 − 𝐼 log𝑒 𝑟 + 0·5 Ilog𝑒 𝑑3 + j 0·866 I log𝑒 𝑑3 + 0·5I log𝑒 𝑑2 −j 0·866I log𝑒 𝑑2 = 𝜇0 2π 𝐼 4 − 𝐼 log𝑒 𝑟 + 0·5 I(log𝑒 𝑑3 + log𝑒 𝑑2 )+ j 0·866 I( log𝑒 𝑑3 − log𝑒 𝑑2) = 𝜇0 2π 𝐼 4 − 𝐼 log𝑒 𝑟 + 𝐼 log𝑒 𝑑2𝑑3 + j 0·866 I log𝑒 𝑑3 𝑑2 = 𝜇0 2π 𝐼 4 + 𝐼 log𝑒 𝑑2𝑑3 𝑟 + j 0·866 I log𝑒 𝑑3 𝑑2 = 𝜇0𝐼 2π 1 4 + log𝑒 𝑑2𝑑3 𝑟 + j 0·866 log𝑒 𝑑3 𝑑2 15 October 2020 D.Rajesh Asst.Prof. EEE Department 28
 ∴ Inductance of conductor A is 𝐿𝐴 = ψ𝐴 𝐼𝐴 = ψ𝐴 𝐼 H /m = 𝜇0 2π 1 4 + log𝑒 𝑑2𝑑3 𝑟 + j 0·866 log𝑒 𝑑3 𝑑2 H /m = 𝜇0 2π 1 4 + log𝑒 𝑑2𝑑3 𝑟 + j 0·866 log𝑒 𝑑3 𝑑2 H /m = 4Π∗10−7 2π 1 4 + log𝑒 𝑑2𝑑3 𝑟 + j 0·866 log𝑒 𝑑3 𝑑2 H /m = 10−7 1 2 + 2 log𝑒 𝑑2𝑑3 𝑟 + j 1.732 log𝑒 𝑑3 𝑑2 H /m Similarly inductance of conductors B and C will be : 𝐿𝑩= 10−7 1 2 + 2 log𝑒 𝑑3𝑑1 𝑟 + j 1.732 log𝑒 𝑑1 𝑑3 H /m 𝐿𝑪=10−7 1 2 + 2 log𝑒 𝑑1𝑑2 𝑟 + j 1.732 log𝑒 𝑑2 𝑑1 H /m Inductance of each line conductor = 1 3 (𝐿𝐴+ 𝐿𝑩+ 𝐿𝑪) = 10−7 1 2 + 2 log𝑒 3 𝑑1𝑑2𝑑3 𝑟 H /m 15 October 2020 D.Rajesh Asst.Prof. EEE Department 29 If we compare the formula of inductance of an unsymmetrically spaced transposed line with that of symmetrically spaced line, we find that inductance of each line conductor in the two cases will be equal if d = 3 𝑑1𝑑2𝑑3 The distance d is known as equivalent equilateral spacing for unsymmetrically transposed line.
The use of self geometrical mean distance (abbreviated as self-GMD) and mutual geometrical mean distance (mutual-GMD) simplifies the inductance calculations, particularly relating to multiconductor arrangements. The symbols used for these are respectively 𝐷𝑠 and 𝐷𝑚. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 30
 Inductance/conductor/m = 10−7 1 2 + 2log𝑒 𝑑 𝑟 H/m = 1 2 ∗ 10−7 + 2 ∗ 10−7log𝑒 𝑑 𝑟 H/m If we replace the original solid conductor by an equivalent hollow cylinder with extremely thin walls, the current is confined to the conductor surface and internal conductor flux linkage would be almost zero. Consequently, inductance due to internal flux would be zero and the term 1 2 ∗ 10−7 shall be eliminated. The radius of this equivalent hollow cylinder must be sufficiently smaller than the physical radius of the conductor to allow room for enough additional flux to compensate for the absence of internal flux linkage. It can be proved mathematically that for a solid round conductor of radius r, the self-GMD or GMR = 0·7788 r. Inductance/conductor/m = 2 ∗ 10−7log𝑒 𝑑 𝐷𝑠 where 𝐷𝑠 = GMR or self-GMD = 0·7788 r It may be noted that self-GMD of a conductor depends upon the size and shape of the conductor and is independent of the spacing between the conductors. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 31
 The mutual-GMD is the geometrical mean of the distances from one conductor to the other and, therefore, must be between the largest and smallest such distance.  (a) The mutual-GMD between two conductors (assuming that spacing between conductors is large compared to the diameter of each conductor) is equal to the distance between their centres i.e. 𝐷𝑚 = spacing between conductors = d  (b) For a single circuit 3-φ line, the mutual-GMD is equal to the equivalent equilateral spacing i.e., 𝑑1𝑑2𝑑3 1 3 𝐷𝑚 = 𝑑1𝑑2𝑑3 1 3 15 October 2020 D.Rajesh Asst.Prof. EEE Department 32
 (c) The principle of geometrical mean distances can be most profitably employed to 3-φ double circuit lines. Consider the conductor arrangement of the double circuit shown in Fig. Suppose the radius of each conductor is r. Self-GMD of conductor = 0·7788 r Self-GMD of combination aa´ is 𝐷𝑠1 = ( 𝐷𝑎𝑎 × 𝐷𝑎 ư 𝑎 × 𝐷 ư 𝑎 ư 𝑎 × 𝐷 ư 𝑎𝑎 ) 1 4 Self-GMD of combination bb´ is 𝐷𝑠2 = ( 𝐷𝑏𝑏 × 𝐷𝑏 ሗ 𝑏 × 𝐷 ሗ 𝑏 ሗ 𝑏 × 𝐷 ሗ 𝑏𝑏 ) 1 4 Self-GMD of combination cc′ is 𝐷𝑠3 = ( 𝐷𝑐𝑐 × 𝐷𝑐 ư 𝑐 × 𝐷 ư 𝑐 ư 𝑐 × 𝐷 ư 𝑐𝑐 ) 1 4 Equivalent self-GMD of one phase 𝐷𝑠 = 𝐷𝑠1 ×𝐷𝑠2×𝐷𝑠3 1 3 The value of 𝐷𝑠 is the same for all the phases as each conductor has the same radius. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 33
 Mutual-GMD between phases A and B is 𝐷𝐴𝐵 = ( 𝐷𝑎𝑏 × 𝐷𝑎 ሗ 𝑏 × 𝐷 ư 𝑎𝑏 × 𝐷 ư 𝑎 ሗ 𝑏 ) 1 4 Mutual-GMD between phases B and C is 𝐷𝐵𝐶 = ( 𝐷𝑏𝑐 × 𝐷𝑏 ư 𝑐 × 𝐷 ሗ 𝑏𝑐 × 𝐷 ሗ 𝑏 ư 𝑐 ) 1 4 Mutual-GMD between phases C and A is 𝐷𝐶𝐴 = ( 𝐷𝑐𝑎 × 𝐷𝑐 ư 𝑎 × 𝐷 ư 𝑐𝑎 × 𝐷 ư 𝑐 ư 𝑎 ) 1 4 Equivalent mutual-GMD, 𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3 Mutual GMD depends only upon the spacing and is substantially independent of the exact size, shape and orientation of the conductor. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 34
(i) Single phase line Inductance/conductor/m =2 ∗ 10−7log𝑒 𝐷𝑚 𝐷𝑠 where 𝐷𝑠 = GMR or self-GMD = 0·7788 r 𝐷𝑚 = Spacing between conductors = d (ii) Single circuit 3-φ line Inductance/phase/m = 2 ∗ 10−7log𝑒 𝐷𝑚 𝐷𝑠 where 𝐷𝑠 = GMR or self-GMD = 0·7788 r 𝐷𝑚 = 𝑑1𝑑2𝑑3 1 3 (iii) Double circuit 3-φ line Inductance/phase/m = 2 ∗ 10−7 log𝑒 𝐷𝑚 𝐷𝑠 where 𝐷𝑠 = 𝐷𝑠1 ×𝐷𝑠2×𝐷𝑠3 1 3 𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3 15 October 2020 D.Rajesh Asst.Prof. EEE Department 35
 A single phase line has two parallel conductors 2 metres apart. The diameter of each conductor is 1·2 cm. Calculate the loop inductance per km of the line.  Loop inductance per metre length of the line =10−7 1 + 4log𝑒 𝑑 𝑟 H = 10−7 1 + 4log𝑒 200 0.6 H =24.23* 10−7 Loop inductance per km of the line = 24.23* 10−7 *1000 =2.423mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 36
A single phase transmission line has two parallel conductors 3 m apart, the radius of each conductor being 1 cm. Calculate the loop inductance per km length of the line if the material of the conductor is (i) copper (ii) steel with relative permeability of 100.  Loop inductance per metre length of the line =10−7 𝜇𝑟 + 4log𝑒 𝑑 𝑟 H/m (i) With copper conductors, 𝜇𝑟 = 1 ∴ Loop inductance/m=10−7 1 + 4log𝑒 300 1 H/m =23.8 * 10−7 H Loop inductance per km of the line = 23.8* 10−7*1000=2.38mH (ii) With steel conductors, 𝜇𝑟 = 100 ∴ Loop inductance/m=10−7 100 + 4log𝑒 300 1 H/m =122.8 * 10−7H Loop inductance per km of the line = 122.8 * 10−7*1000=12.28mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 37
Find the inductance per km of a 3-phase transmission line using 1·24 cm diameter conductors when these are placed at the corners of an equilateral triangle of each side 2 m. conductor spacing d = 2 m =200cm conductor radius r = 1·24/2 = 0·62 cm. Inductance/phase/m=10−7 1 2 + 2log𝑒 𝑑 𝑟 H/m =10−7 1 2 + 2log𝑒 200 0.62 H/m =12* 10−7 H Inductance/phase/km = 12* 10−7 *1000H =1.2mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 38
The three conductors of a 3-phase line are arranged at the corners of a triangle of sides 2 m, 2·5 m and 4·5 m. Calculate the inductance per km of the line when the conductors are regularly transposed. The diameter of each conductor is 1·24 cm. Three conductors of a 3-phase line placed at the corners of a triangle of sides 𝐷12 = 2 m, 𝐷23 = 2·5 m and 𝐷31 = 4·5 m. The conductor radius r = 1·24/2 = 0·62 cm. Equivalent equilateral spacing, 𝐷𝑒𝑞=3 𝐷12 ∗ 𝐷23 ∗ 𝐷31 = 3 2 ∗ 2.5 ∗ 4.5=2.82m=282cm Inductance/phase/m =10−7 1 2 + 2log𝑒 𝐷𝑒𝑞 𝑟 H/m =10−7 1 2 + 2log𝑒 282 0.62 H/m =12.74* 10−7 Inductance/phase/km = 12.74* 10−7 *1000H =1.274mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 39
Calculate the inductance of each conductor in a 3-phase, 3-wire system when the conductors are arranged in a horizontal plane with spacing such that 𝐷31 = 4 m ; 𝐷12 = 𝐷23 = 2m. The conductors are transposed and have a diameter of 2·5 cm. The conductor radius r = 2·5/2 = 1·25 cm. Equivalent equilateral spacing, 𝐷𝑒𝑞=3 𝐷12 ∗ 𝐷23 ∗ 𝐷31 𝐷𝑒𝑞= 3 2 ∗ 2 ∗ 4=2.52m=252cm Inductance/phase/m =10−7 1 2 + 2log𝑒 𝐷𝑒𝑞 𝑟 H/m =10−7 1 2 + 2log𝑒 252 1.25 H/m =11.1* 10−7 Inductance/phase/km = 11.1* 10−7 *1000H =1.11mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 40
Two conductors of a single phase line, each of 1 cm diameter, are arranged in a vertical plane with one conductor mounted 1 m above the other. A second identical line is mounted at the same height as the first and spaced horizontally 0·25 m apart from it. The two upper and the two lower conductors are connected in parallel. Determine the inductance per km of the resulting double circuit line. Conductors a, a′ form one connection and conductors b, b′ form the return connection. The conductor radius, r = 1/2 = 0·5 cm. Inductance per conductor per metre= 2 ∗ 10−7 log𝑒 𝐷𝑚 𝐷𝑠 G.M.R. of conductor = 0·7788 r = 0·7788 × 0·5 = 0·389 cm Self G.M.D. of aa′ combination is 𝐷𝑠 = ( 𝐷𝑎𝑎 × 𝐷𝑎 ư 𝑎 × 𝐷 ư 𝑎 ư 𝑎 × 𝐷 ư 𝑎𝑎 ) 1 4 𝐷𝑠 = ( 0.389 × 100 ×0.389 × 100) 1 4 =6.23cm Mutual G.M.D. between a and b is 𝐷𝑚 = ( 𝐷𝑎𝑏 × 𝐷𝑎 ሗ 𝑏 × 𝐷 ư 𝑎𝑏 × 𝐷 ư 𝑎 ሗ 𝑏 ) 1 4 𝐷𝑎 ሗ 𝑏= 𝐷 ư 𝑎𝑏= 252 + 1002 = 103 𝐷𝑚 = ( 25 × 103 × 103 × 25 ) 1 4 =50.47cm Inductance per conductor per metre= 2 ∗ 10−7log𝑒 𝐷𝑚 𝐷𝑠 = = 2 ∗ 10−7 log𝑒 50.47 6.23 =0.42 × 10−6H ∴ Loop inductance per km of the line=2* 0.42 × 10−6 *1000=0.84mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 41
Find the inductance per phase per km of double circuit 3-phase line shown in Fig. The conductors are transposed and are of radius 0·75 cm each. The phase sequence is ABC. Inductance/phase/m = 2 ∗ 10−7 log𝑒 𝐷𝑚 𝐷𝑠 where 𝐷𝑠 = 𝐷𝑠1 ×𝐷𝑠2×𝐷𝑠3 1 3 𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3 Self-GMD of conductor = 0·7788 r=0.7788*0.75=0.584cm Self-GMD of combination aa´ is 𝐷𝑠1 = ( 𝐷𝑎𝑎 × 𝐷𝑎 ư 𝑎 × 𝐷 ư 𝑎 ư 𝑎 × 𝐷 ư 𝑎𝑎 ) 1 4 𝐷𝑎 ư 𝑎= 𝐷 ư 𝑎𝑎= 62 + 42=7.21m 𝐷𝑠1 = ( (0.584×10−2 )×7.21×(0.584×10−2 )×7.21) 1 4=0.205m Self-GMD of combination bb´ is 𝐷𝑠2 = ( 𝐷𝑏𝑏 × 𝐷𝑏 ሗ 𝑏 × 𝐷 ሗ 𝑏 ሗ 𝑏 × 𝐷 ሗ 𝑏𝑏 ) 1 4 = ((0.584×10−2 )× 5.5×(0.584×10−2 )×5.5) 1 4=0.18m Self-GMD of combination cc′ is 𝐷𝑠3 = ( 𝐷𝑐𝑐 × 𝐷𝑐 ư 𝑐 × 𝐷 ư 𝑐 ư 𝑐 × 𝐷 ư 𝑐𝑐 ) 1 4 = 𝐷𝑠1= 0.205m 𝐷𝑠 = 𝐷𝑠1 ×𝐷𝑠2×𝐷𝑠3 1 3 𝐷𝑠 = 0.205×0.18×0.205 1 3 =0.195m 15 October 2020 D.Rajesh Asst.Prof. EEE Department 42
𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3 Mutual-GMD between phases A and B is 𝐷𝐴𝐵 = ( 𝐷𝑎𝑏 × 𝐷𝑎 ሗ 𝑏 × 𝐷 ư 𝑎𝑏 × 𝐷 ư 𝑎 ሗ 𝑏 ) 1 4 𝐷𝑎 ሗ 𝑏= 𝐷 ư 𝑎𝑏= 32 + (4 + 0.75)2=5.62m 𝐷𝑎𝑏= 32 + (0.75)2=3.1m 𝐷𝐴𝐵 = ( 3.1 × 5.62 × 5.62× 3.1 ) 1 4=4.17m Mutual-GMD between phases B and C is 𝐷𝐵𝐶 = ( 𝐷𝑏𝑐 × 𝐷𝑏 ư 𝑐 × 𝐷 ሗ 𝑏𝑐 × 𝐷 ሗ 𝑏 ư 𝑐 ) 1 4 = 𝐷𝐴𝐵 =4.17m Mutual-GMD between phases C and A is 𝐷𝐶𝐴 = ( 𝐷𝑐𝑎 × 𝐷𝑐 ư 𝑎 × 𝐷 ư 𝑐𝑎 × 𝐷 ư 𝑐 ư 𝑎 ) 1 4 = ( 6 × 4 × 4 × 6 ) 1 4=4.9m Equivalent mutual-GMD, 𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3= 4.71×4.17×4.9 1 3=4.4m Inductance/phase/m = 2 ∗ 10−7 log𝑒 𝐷𝑚 𝐷𝑠 = 2 ∗ 10−7 log𝑒 4.4 0.195 =0.623 ∗ 10−3mH Inductance/phase/km= 0.623 ∗ 10−3 *1000=0.623mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 43
Calculate the inductance per phase per metre for a three-phase double-circuit line whose phase conductors have a radius of 5·3 cm with the horizontal conductor arrangement as shown in Fig. Inductance per conductor per metre= 2 ∗ 10−7log𝑒 𝐷𝑚 𝐷𝑠 where 𝐷𝑠 = 𝐷𝑠1 ×𝐷𝑠2×𝐷𝑠3 1 3 𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3 Self-GMD of combination aa´ is0·7788 r = 0·7788 × 5·3 × 10−2 = 0·0413 m 𝐷𝑠1 = ( 𝐷𝑎𝑎 × 𝐷𝑎 ư 𝑎 × 𝐷 ư 𝑎 ư 𝑎 × 𝐷 ư 𝑎𝑎 ) 1 4 = (0·041× 24 ×0·041× 24 ) 1 4=0.995m Self-GMD of combination bb´ is 𝐷𝑠2 = ( 𝐷𝑏𝑏 × 𝐷𝑏 ሗ 𝑏 × 𝐷 ሗ 𝑏 ሗ 𝑏 × 𝐷 ሗ 𝑏𝑏 ) 1 4 = (0·041× 24 ×0·041× 24 ) 1 4=0.995m Self-GMD of combination cc′ is 𝐷𝑠3 = ( 𝐷𝑐𝑐 × 𝐷𝑐 ư 𝑐 × 𝐷 ư 𝑐 ư 𝑐 × 𝐷 ư 𝑐𝑐 ) 1 4 = (0·041× 24 ×0·041× 24 ) 1 4=0.995m Equivalent self-GMD of one phase 𝐷𝑠 = 𝐷𝑠1 ×𝐷𝑠2×𝐷𝑠3 1 3=0.995m 15 October 2020 D.Rajesh Asst.Prof. EEE Department 44
𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3 Mutual-GMD between phases A and B is 𝐷𝐴𝐵 = ( 𝐷𝑎𝑏 × 𝐷𝑎 ሗ 𝑏 × 𝐷 ư 𝑎𝑏 × 𝐷 ư 𝑎 ሗ 𝑏 ) 1 4 𝐷𝐴𝐵 = ( 8 × 32 × 16× 8 ) 1 4=13.45m Mutual-GMD between phases B and C is 𝐷𝐵𝐶 = ( 𝐷𝑏𝑐 × 𝐷𝑏 ư 𝑐 × 𝐷 ሗ 𝑏𝑐 × 𝐷 ሗ 𝑏 ư 𝑐 ) 1 4 = ( 8 × 32 × 16× 8 ) 1 4=13.45m Mutual-GMD between phases C and A is 𝐷𝐶𝐴 = ( 𝐷𝑐𝑎 × 𝐷𝑐 ư 𝑎 × 𝐷 ư 𝑐𝑎 × 𝐷 ư 𝑐 ư 𝑎 ) 1 4 = ( 16 × 8 × 40 × 16 ) 1 4=16.917m Equivalent mutual-GMD, 𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3= 13.45×13.45×16.917 1 3=14.518m Inductance/phase/m = 2 ∗ 10−7 log𝑒 𝐷𝑚 𝐷𝑠 = 2 ∗ 10−7log𝑒 14.518 0.995 =5.36 ∗ 10−7 H 15 October 2020 D.Rajesh Asst.Prof. EEE Department 45
In a single phase line (See. Fig.)conductors a and a′ in parallel form one circuit while conductors b and b′ in parallel form the return path. Calculate the total inductance of the line per km assuming that current is equally shared by the two parallel conductors. Conductor diameter in 2·0 cm. Inductance/phase/m = 2 ∗ 10−7 log𝑒 𝐷𝑚 𝐷𝑠 Mutual G.M.D., 𝐷𝑚 = ( 𝐷𝑎𝑏 × 𝐷𝑎 ሗ 𝑏 × 𝐷 ư 𝑎𝑏 × 𝐷 ư 𝑎 ሗ 𝑏 ) 1 4 = ( 120 × 140 × 100× 120 ) 1 4=119cm Self G.M.D., 𝐷𝑠 = ( 𝐷𝑎𝑎 × 𝐷𝑎 ư 𝑎 × 𝐷 ư 𝑎 ư 𝑎 × 𝐷 ư 𝑎𝑎 ) 1 4 𝐷𝑎𝑎= 𝐷 ư 𝑎 ư 𝑎=0.7788*r=0.7788* 2 2 =0.7788cm 𝐷𝑠 = (0.7788 × 20 × 0.7788× 20 ) 1 4=3.94cm Inductance/phase/m = 2 ∗ 10−7 log𝑒 𝐷𝑚 𝐷𝑠 = 2 ∗ 10−7 log𝑒 119 3.94 =6.81* 10−7 H Loop Inductance/phase/km=2*6.81* 10−7 *1000H =1.363mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 46
Any two conductors of an overhead transmission line are separated by air which acts as an insulation, therefore, capacitance exists between any two overhead line conductors. The capacitance between the conductors is the charge per unit potential difference i.e., Capacitance, C = 𝑞 𝑣 farad where q = charge on the line in coulomb v = p.d. between the conductors in volts 15 October 2020 D.Rajesh Asst.Prof. EEE Department 47
 Electric Potential The electric potential at a point due to a charge is the work done in bringing a unit positive charge from infinity to that point. (i) Potential at a charged single conductor (ii) Potential at a conductor in a group of charged conductors. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 48
(i) Potential at a charged single conductor. Consider a long straight cylindrical conductor A of radius r metres. Let the conductor operate at such a potential (𝑉𝐴) that charge 𝑄𝐴 coulombs per metre exists on the conductor. The electric intensity E at a distance x from the centre of the conductor in air is given by: 𝐸 = 𝑄𝐴 2𝜋𝑥𝜀0 volts/m where 𝑄𝐴 = charge per metre length 𝜀0 = permittivity of free space As x approaches infinity, the value of E approaches zero. Therefore, the potential difference between conductor A and infinity distant neutral plane is given by : Work is done in bringing a unit positive charge against E from infinity to conductor surface. 𝑉𝐴 = න 𝑟 ∞ 𝐸 𝑑𝑥 𝑉𝐴 = න 𝑟 ∞ 𝑄𝐴 2𝜋𝑥𝜀0 𝑑𝑥 = 𝑄𝐴 2𝜋𝜀0 න 𝑟 ∞ 1 𝑥 𝑑𝑥 15 October 2020 D.Rajesh Asst.Prof. EEE Department 49
(ii)Potential at a conductor in a group of charged conductors. Consider a group of long straight conductors A, B, C etc. operating at potentials such that charges 𝑄𝐴, 𝑄𝐵, 𝑄𝐶 etc. coulomb per metre length exist on the respective conductors The potential at A Potential at A due to its own charge (i.e. 𝑄𝐴) = න 𝑟 ∞ 𝑄𝐴 2𝜋𝑥𝜀0 𝑑𝑥 Potential at conductor A due to charge 𝑄𝐵 = න 𝑑1 ∞ 𝑄𝐵 2𝜋𝑥𝜀0 𝑑𝑥 Potential at conductor A due to charge 𝑄𝐶 = න 𝑑2 ∞ 𝑄𝐶 2𝜋𝑥𝜀0 𝑑𝑥 15 October 2020 D.Rajesh Asst.Prof. EEE Department 50
 Overall potential difference between conductor A and infinite neutral plane is 𝑉𝐴 = (i) + (ii) + (iii) + ....... = ‫׬‬ 𝑟 ∞ 𝑄𝐴 2𝜋𝑥𝜀0 𝑑𝑥+ ‫׬‬ 𝑑1 ∞ 𝑄𝐵 2𝜋𝑥𝜀0 𝑑𝑥 +‫׬‬ 𝑑2 ∞ 𝑄𝐶 2𝜋𝑥𝜀0 𝑑𝑥+……… = 1 2𝜋𝜀0 ‫׬‬ 𝑟 ∞ 𝑄𝐴 𝑥 dx + ‫׬‬ 𝑑1 ∞ 𝑄𝐵 𝑥 𝑑𝑥 + ‫׬‬ 𝑑2 ∞ 𝑄𝐶 𝑥 𝑑𝑥 + ⋯ … . . = 1 2𝜋𝜀0 𝑄𝐴 log𝑒 ∞ − log𝑒 𝑟 + 𝑄𝐵 log𝑒 ∞ − log𝑒 𝑑1 + 𝑄𝐶 log𝑒 ∞ − log𝑒 𝑑2 + ⋯ . = 1 2𝜋𝜀0 log𝑒 ∞ 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶 + 𝑄𝐴 log𝑒 1 𝑟 + 𝑄𝐵 log𝑒 1 𝑑1 + 𝑄𝐶 log𝑒 1 𝑑2 + ⋯ . Assuming balanced conditions i.e., 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶=0 𝑉𝐴= 1 2𝜋𝜀0 𝑄𝐴 log𝑒 1 𝑟 + 𝑄𝐵 log𝑒 1 𝑑1 + 𝑄𝐶 log𝑒 1 𝑑2 + ⋯ . 15 October 2020 D.Rajesh Asst.Prof. EEE Department 51
Consider a single phase overhead transmission line consisting of two parallel conductors A and B spaced d metres apart in air. Suppose that radius of each conductor is r metres. Let their respective charge be + Q and − Q coulombs per metre length. The total p.d. between conductor A and neutral “infinite” plane is 𝑉𝐴 = ‫׬‬ 𝑟 ∞ Q 2𝜋𝑥𝜀0 𝑑𝑥+ ‫׬‬ 𝑑 ∞ −𝑄 2𝜋𝑥𝜀0 𝑑𝑥 = 1 2𝜋𝜀0 Q log𝑒 ∞ − log𝑒 𝑟 − Q log𝑒 ∞ − log𝑒 𝑑 = Q 2𝜋𝜀0 log𝑒 𝑑 − log𝑒 𝑟 = Q 2𝜋𝜀0 log𝑒 𝑑 𝑟 volts 15 October 2020 D.Rajesh Asst.Prof. EEE Department 52
The total p.d. between conductor B and neutral “infinite” plane is 𝑉𝐵 = ‫׬‬ 𝑟 ∞ −Q 2𝜋𝑥𝜀0 𝑑𝑥+ ‫׬‬ 𝑑 ∞ 𝑄 2𝜋𝑥𝜀0 𝑑𝑥 = 1 2𝜋𝜀0 −Q log𝑒 ∞ − log𝑒 𝑟 +Q log𝑒 ∞ − log𝑒 𝑑 = −Q 2𝜋𝜀0 log𝑒 𝑑 − log𝑒 𝑟 = −Q 2𝜋𝜀0 log𝑒 𝑑 𝑟 volts Both these potentials are w.r.t. the same neutral plane. Since the unlike charges attract each other, the potential difference between the conductors is 𝑉𝐴𝐵 = 2𝑉𝐴= 2Q 2𝜋𝜀0 log𝑒 𝑑 𝑟 ∴ Capacitance, 𝐶𝐴𝐵 = Q 𝑉𝐴𝐵 = Q 2Q 2𝜋𝜀0 log𝑒 𝑑 𝑟 𝐶𝐴𝐵 = 𝜋𝜀0 log𝑒 𝑑 𝑟 F/m 15 October 2020 D.Rajesh Asst.Prof. EEE Department 53
 Capacitance to neutral. Since potential of the mid-point between the conductors is zero, the potential difference between each conductor and the ground or neutral is half the potential difference between the conductors. Thus the capacitance to ground or capacitance to neutral for the twowire line is twice the line-to-line capacitance ∴ Capacitance to neutral 𝐶𝑁 = 𝐶𝐴𝑁 = 𝐶𝐵𝑁 = 2𝐶𝐴𝐵 𝐶𝑁= 2𝜋𝜀0 log𝑒 𝑑 𝑟 F/m 15 October 2020 D.Rajesh Asst.Prof. EEE Department 54
 In a 3-phase transmission line, the capacitance of each conductor is considered instead of capacitance from conductor to conductor.  Here, again two cases arise viz., symmetrical spacing and unsymmetrical spacing.  (i) Symmetrical Spacing. The three conductors A, B and C of the 3-phase overhead transmission line having charges 𝑄𝐴, 𝑄𝐵, and 𝑄𝐶 per metre length respectively. Let the conductors be equidistant (d metres) from each other. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 55
 overall potential difference between conductor A and infinite neutral plane is  𝑉𝐴 = ‫׬‬𝑟 ∞ 𝑄𝐴 2𝜋𝑥𝜀0 𝑑𝑥+ ‫׬‬𝑑 ∞ 𝑄𝐵 2𝜋𝑥𝜀0 𝑑𝑥 +‫׬‬𝑑 ∞ 𝑄𝐶 2𝜋𝑥𝜀0 𝑑𝑥 = 1 2𝜋𝜀0 ‫׬‬ 𝑟 ∞ 𝑄𝐴 𝑥 dx + ‫׬‬ 𝑑 ∞ 𝑄𝐵 𝑥 𝑑𝑥 + ‫׬‬ 𝑑 ∞ 𝑄𝐶 𝑥 𝑑𝑥 = 1 2𝜋𝜀0 𝑄𝐴 log𝑒 ∞ − log𝑒 𝑟 + 𝑄𝐵 log𝑒 ∞ − log𝑒 𝑑 + 𝑄𝐶 log𝑒 ∞ − log𝑒 𝑑 = 1 2𝜋𝜀0 log𝑒 ∞ 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶 + 𝑄𝐴 log𝑒 1 𝑟 + (𝑄𝐵 + 𝑄𝐶 ) log𝑒 1 𝑑 Assuming balanced conditions i.e., 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶=0 𝑄𝐵 + 𝑄𝐶= −𝑄𝐴 ∴ 𝑉𝐴= 1 2𝜋𝜀0 𝑄𝐴 log𝑒 1 𝑟 + (−𝑄𝐴 ) log𝑒 1 𝑑 𝑉𝐴 = 𝑄𝐴 2𝜋𝜀0 log𝑒 𝑑 𝑟 F/m ∴ Capacitance of conductor A w.r.t neutral, 𝐶𝐴 = 𝑄𝐴 𝑉𝐴 = 𝑄𝐴 𝑄𝐴 2𝜋𝜀0 log𝑒 𝑑 𝑟 𝐶𝐴 = 2𝜋𝜀0 log𝑒 𝑑 𝑟 F/m Note that this equation is identical to capacitance to neutral for two-wire line. Derived in a similar manner, the expressions for capacitance are the same for conductors B and C. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 56
 (ii) Unsymmetrical spacing.  Assuming balanced conditions i.e., 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶=0  Potential of 1st position, 𝑉1= 1 2𝜋𝜀0 𝑄𝐴 log𝑒 1 𝑟 + 𝑄𝐵 log𝑒 1 𝑑3 + 𝑄𝐶 log𝑒 1 𝑑2  Potential of 2nd position, 𝑉2= 1 2𝜋𝜀0 𝑄𝐴 log𝑒 1 𝑟 + 𝑄𝐵 log𝑒 1 𝑑1 + 𝑄𝐶 log𝑒 1 𝑑3  Potential of 3rd position, 𝑉3= 1 2𝜋𝜀0 𝑄𝐴 log𝑒 1 𝑟 + 𝑄𝐵 log𝑒 1 𝑑2 + 𝑄𝐶 log𝑒 1 𝑑1  Average voltage on condutor A is 𝑉𝐴 = 1 3 (𝑉1+ 𝑉2+ 𝑉3) = 1 3∗2𝜋𝜀0 𝑄𝐴 log𝑒 1 𝑟3 + (𝑄𝐵+ 𝑄𝐶) log𝑒 1 𝑑1𝑑2𝑑3 15 October 2020 D.Rajesh Asst.Prof. EEE Department 57
 As 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶=0, therefore 𝑄𝐵 + 𝑄𝐶= −𝑄𝐴 = 1 2∗3𝜋𝜀0 𝑄𝐴 log𝑒 1 𝑟3 + (−𝑄𝐴) log𝑒 1 𝑑1𝑑2𝑑3 = 1 3 ∗ 𝑄𝐴 2𝜋𝜀0 log𝑒 𝑑1𝑑2𝑑3 𝑟3 = 𝑄𝐴 2𝜋𝜀0 log𝑒 𝑑1𝑑2𝑑3 𝑟3 1 3 = 𝑄𝐴 2𝜋𝜀0 log𝑒 𝑑1𝑑2𝑑3 1 3 𝑟 ∴ Capacitance from conductor to neutral is 𝐶𝐴 = 𝑄𝐴 𝑉𝐴 = 𝑄𝐴 𝑄𝐴 2𝜋𝜀0 log𝑒 3 𝑑1𝑑2𝑑3 𝑟 = 2𝜋𝜀0 log𝑒 3 𝑑1𝑑2𝑑3 𝑟 15 October 2020 D.Rajesh Asst.Prof. EEE Department 58
A single-phase transmission line has two parallel conductors 3 metres apart, radius of each conductor being 1 cm. Calculate the capacitance of the line per km. Given that 𝜀0 =8·854 × 10−12 F/m. Conductor radius, r = 1 cm Spacing of conductors, d = 3 m = 300 cm capacitance of the line = 𝜋𝜀0 log𝑒 𝑑 𝑟 F/m = 𝜋∗8·854 × 10−12 log𝑒 300 1 =0.485* 10−11 F/m capacitance of the line per km= 0.485* 10−11 *1000 = 0.485* 10−8 F/km = 0.485* 10−2 μF/km 15 October 2020 D.Rajesh Asst.Prof. EEE Department 59
A 3-phase overhead transmission line has its conductors arranged at the corners of an equilateral triangle of 2 m side. Calculate the capacitance of each line conductor per km Given that diameter of each conductor is 1·25 cm. 𝜀0 =8·854 × 10−12 F/m. Conductor radius, r = 1·25/2 = 0·625 cm Spacing of conductors, d = 2 m = 200 cm Capacitance of each line conductor= 2𝜋𝜀0 log𝑒 𝑑 𝑟 F/m == 2𝜋8·854 × 10−12 log𝑒 200 0.625 F/m=0.0096* 10−9 F/m capacitance of the line per km 0.0096* 10−9 *1000 0.0096* 10−6 F/km = 0.0096 μF/km 15 October 2020 D.Rajesh Asst.Prof. EEE Department 60
A 3-phase, 50 Hz, 66 kV overhead line conductors are placed in a horizontal plane as shown in Fig. The conductor diameter is 1·25 cm. If the line length is 100 km, calculate (i) capacitance per phase, (ii) charging current per phase, assuming complete transposition of the line. 𝜀0 =8·854 × 10−12 F/m. Conductor radius, r = 1·25/2 = 0·625 cm The equivalent equilateral spacing is 𝑑 = 3 𝑑1𝑑2𝑑3 = 3 2 × 2 ⋅ 5 × 4 ⋅5 = = 2·82 m Conductor spacing , d = 2·82 m = 282 cm (i) Line to neutral capacitance== 2𝜋𝜀0 log𝑒 𝑑 𝑟 F/m == 2𝜋8·854 × 10−12 log𝑒 282 0.625 F/m=0.0091* 10−9 F/m capacitance of the line per km 0.0091* 10−9 *1000 0.0091* 10−6 F/km = 0.0091 μF/km ∴ Line to neutral capacitance for 100 km line is C = 0·0091 × 100 = 0·91 μF 15 October 2020 D.Rajesh Asst.Prof. EEE Department 61
 (ii) Charging current per phase is 𝐼𝑐 = 𝑉𝑝ℎ 𝑋𝑐 = 𝑉𝑝ℎ 1 2𝜋𝑓𝑐 = 66000 3 ∗ 2𝜋*50*0·91 μ =10.9A 15 October 2020 D.Rajesh Asst.Prof. EEE Department 62
A 3-phase, 50 Hz, 132 kV overhead line has conductors placed in a horizontal plane 4 m apart. Conductor diameter is 2 cm. If the line length is 100 km, calculate the charging current per phase assuming complete transposition. 𝜀0 =8·854 × 10−12 F/m. The diameter of each conductor is 2 cm so that conductor radius r =2/2 = 1 cm = 1× 10−2 m. Now 𝑑1 = AB = 4m; 𝑑2 = BC = 4 m; 𝑑3 = AC = 8 m The equivalent equilateral spacing is 𝑑 = 3 𝑑1𝑑2𝑑3 = 3 4×4×8 = 5.04 m (i) Line to neutral capacitance== 2𝜋𝜀0 log𝑒 𝑑 𝑟 F/m == 2𝜋8·854 × 10−12 log𝑒 5.04 1× 10−2 F/m=0.0085* 10−9 F/m capacitance of the line per km= 0.0085* 10−9 *1000 = 0.0085* 10−6 F/km = 0.0085 μF/km ∴ Line to neutral capacitance for 100 km line is C = 0.0085 × 100 = 0·885 μF 15 October 2020 D.Rajesh Asst.Prof. EEE Department 63
 Charging current per phase is 𝐼𝑐 = 𝑉𝑝ℎ 𝑋𝑐 = 𝑉𝑝ℎ 1 2𝜋𝑓𝑐 = 132000 3 ∗ 2𝜋*50*0·885 μ =21.18A 15 October 2020 D.Rajesh Asst.Prof. EEE Department 64
 Definition: Proximity Effect is the phenomena of non-uniform current distribution on the surface of adjacent current carrying conductor due to the effect of another current carrying conductor in its proximity. Since in cables, the conductors are very near to each other, this effect is dominant. Whereas in overhead lines as lines are far apart when compared with cable, this effect can be neglected. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 65
 Cause of Proximity Effect:  The main reason for proximity effect is production of magnetic field in the surrounding of a current carrying conductor.  We know that any current carrying conductor produces magnetic field in its surrounding as per Biot Savart Law.  When this magnetic field links with the adjacent conductor, it gives rise to a circulating or eddy current in it. Because of this eddy current, the current distribution on the nearby becomes non-uniform.  This leads to additional resistance to the flow of current in the nearby conductor. Hence due to proximity effect, the net ac resistance of conductor increases. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 66
 Two conductors carrying current in the same direction. The magnetic field created by conductor A will link with conductor B. Similarly, the magnetic field created by conductor B will link with conductor A.  The direction of magnetic field at any point P inside the conductor B will in going inside the plane of device.  Since current is flowing in upward direction, this means that free electrons are moving in downward direction. The direction of force on the free electrons due to the magnetic field of conductor A will be given by the cross vector product of velocity vector of electron and magnetic field vector at point P. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 67
 From the cross vector product, the free electrons of conductor B will be drifted toward the inner surface i.e. toward conductor A and hence the current will flow mostly in the farthest surface of conductor B.  The same reasoning when applied to conductor A, it can be seen that current in the conductor A will mostly flow in the farthest surface. This effect of re- distribution of current on the conductor surface is called the proximity effect. This is so called proximity effect as this is due to the effect of proximity (nearby) conductor. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 68
 when two current conductors carry current in the same direction, the current gets distributed to the farthest half of both the conductor.   Similarly when two conductors carry current in opposite direction, the current in conductors gets concentrated toward the inner facing surface of both the conductors 15 October 2020 D.Rajesh Asst.Prof. EEE Department 69
 Factors affecting Proximity Effect:  The proximity effect mainly depends on the factors like conductor material, conductor diameter, frequency and conductor structure. The factors are explained below in details  Frequency – The proximity increases with the increases in the frequency. This effect is not observed in case of DC. In case of DC, as the current is uniformly distributed over the entire cross-section of conductor, the magnetic force on the free electrons inside conductor is less prominent. Hence this effect is not observed.  Diameter – The proximity effect increases with the increase in the conductor.  Structure – This effect is more on the solid conductor as compared to the stranded conductor (i.e., ASCR) because the surface area of the stranded conductor is smaller than the solid conductor.  Material – If the material is made up of high ferromagnetic material then the proximity effect is more on their surface. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 70
 How to reduce Proximity Effect?  The proximity effect can be reduced by using the ACSR (Aluminum Core Steel Reinforced) conductor. In ACSR conductor the steel is placed at the centre of the conductor and the aluminium conductor is positioned around steel wire.  The steel increased the strength of the conductor but reduced the surface area of the conductor. Thus, the current flow mostly in the outer layer of the conductor and no current is carried in the centre of the conductor. Thus, reduced the proximity effect on the conductor. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 71

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INDUCTANCE & CAPACITANCE CALCULATIONS

  • 1.
    By D.Rajesh Asst. Prof. EEEDepartment 15 October 2020 D.Rajesh Asst.Prof. EEE Department 1
  • 2.
     Types ofconductors, line parameters,  Calculation of inductance and capacitance of single and double circuit transmission lines, Three phase lines with bundle conductors.  Skin effect and  proximity effect. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 2
  • 3.
     Solid conductors Stranded conductors  Hallow conductors  Bundled conductors 15 October 2020 D.Rajesh Asst.Prof. EEE Department 3
  • 4.
     Solid conductors 1.Very costly & high tensile strength 2. Difficult to string these conductors between poles 3. For A.C. these exhibit more skin effect 4. Less flexibility, so rarely used 15 October 2020 D.Rajesh Asst.Prof. EEE Department 4
  • 5.
     Stranded conductors Threeor more strands of conductor material is twisted to form a single conductor. They can carry high currents and are more flexible than solid conductors. Ex: ACSR,AAC,AAAC,ACAR 15 October 2020 D.Rajesh Asst.Prof. EEE Department 5
  • 6.
     Hallow conductors Hollowconductors are used in the transmission lines to: (1) Improve heat dissipation (2) Reduce corona loss (3) Reduce skin effect (4) Reduce the line inductance 15 October 2020 D.Rajesh Asst.Prof. EEE Department 6
  • 7.
     Bundled Conductors Bundled Conductors are used in transmission lines where the voltage exceeds 230 kV.  At such high voltages, ordinary conductors will result in excessive corona and noise which may affect communication lines.  The increased corona will result in significant power loss. Bundle conductors consist of three or four conductors for each phase.  The conductors are separated from each other by means of spacers at regular intervals. Thus, they do not touch each other.  A bundled conductor reduces the reactance of the electric transmission line. It also reduces voltage gradient, corona loss, radio interference, surge impedance of the transmission lines. By making bundle conductor, the geometric mean radius (GMR) of the conductor increased. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 7
  • 8.
     (i) Resistance. (ii) Inductance.  (iii) Capacitance. A transmission line has resistance, inductance and capacitance uniformly distributed along the whole length of the line. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 8
  • 9.
    The resistance oftransmission line conductors is the most important cause of power loss in a transmission line. The resistance R of a line conductor having resistivity ρ, length l and area of crossection a is given by ; R = ρ l/a It is the opposition of line conductors to current flow. The resistance is distributed uniformly along the whole length of the line as shown in Fig. (i). However, the performance of a transmission line can be analysed conveniently if distributed resistance is considered as lumped as shown Fig. (ii). 15 October 2020 D.Rajesh Asst.Prof. EEE Department 9
  • 10.
     When analternating current flows through a conductor, a changing flux is set up which links the conductor. Due to these flux linkages, the conductor possesses inductance.  Mathematically, inductance is defined as the flux linkages per ampere i.e., Inductance, L = ψ I henry where ψ = flux linkages in weber-turns I = current in amperes 15 October 2020 D.Rajesh Asst.Prof. EEE Department 10
  • 11.
     Any twoconductors of an overhead transmission line are separated by air which acts as an insulation, therefore, capacitance exists between any two overhead line conductors. The capacitance between the conductors is the charge per unit potential difference i.e.,  Capacitance, C = 𝑞 𝑣 farad where q = charge on the line in coulomb v = p.d. between the conductors in volts 15 October 2020 D.Rajesh Asst.Prof. EEE Department 11
  • 12.
    When a conductoris carrying steady direct current (d.c.), this current is uniformly distributed over the whole X-section of the conductor. However, an alternating current flowing through the conductor does not distribute uniformly, rather it has the tendency to concentrate near the surface of the conductor as shown in Fig. This is known as skin effect. The tendency of alternating current to concentrate near the surface of a conductor is known as skin effect. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 12
  • 13.
     Due toskin effect, the effective area of cross-section of the conductor through which current flows is reduced. Consequently, the resistance of the conductor is slightly increased when carrying an alternating current.  The skin effect depends upon the following factors : (i) Nature of material (ii) Diameter of wire − increases with the diameter of wire. (iii) Frequency − increases with the increase in frequency. (iv) Shape of wire − less for stranded conductor than the solid conductor.  It may be noted that skin effect is negligible when the supply frequency is low (< 50 Hz) and conductor diameter is small (< 1cm). 15 October 2020 D.Rajesh Asst.Prof. EEE Department 13
  • 14.
     The inductanceof a circuit is defined as the flux linkages per unit current. 1. Flux linkages due to a single current carrying conductor 2. Flux linkages in parallel current carrying conductors. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 14
  • 15.
    1. Flux linkagesdue to a single current carrying conductor Consider a long straight cylindrical conductor of radirus r metres and carrying a current I amperes (r.m.s.) as shown in Fig. This current will set up magnetic field. The magnetic lines of force will exist inside the conductor aswell as outside the conductor. Both these fluxes will contribute to the inductance of the conductor. (i) Flux linkages due to internal flux. (ii) Flux linkages due to external flux. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 15
  • 16.
    (i) Flux linkagesdue to internal flux. The magnetic field intensity at a point x metres from the centre is given by; 𝐻𝑥 = 𝐼𝑥 2π𝑥 [According to Ampere’s law, m.m.f. (ampere-turns) around any closed path equals the current enclosed by the path. The current enclosed by the path is 𝐼𝑥 and m.m.f. = 𝐻𝑥 × 2π x. ∴ 𝐻𝑥 × 2π x = 𝐼𝑥.] Assuming a uniform current density, 𝐼𝑥 π𝑥2 = 𝐼 π𝑟2 𝐼𝑥= π𝑥2 π𝑟2 𝐼 = 𝑥2 𝑟2 𝐼 put 𝐼𝑥 in 𝐻𝑥 𝐻𝑥= 𝑥2 𝑟2 𝐼* 1 2π𝑥 = 𝑥 2π𝑟2 𝐼 𝐴𝑇/𝑚 15 October 2020 D.Rajesh Asst.Prof. EEE Department 16
  • 17.
     If μ(= 𝜇0𝜇𝑟 ) is the permeability of the conductor, then flux density at the considered point is given by;  𝐵𝑥 = 𝜇0𝜇𝑟𝐻𝑥 wb/𝑚2 = 𝜇0𝜇𝑟 𝑥 2π𝑟2 𝐼 = 𝜇0𝑥𝐼 2π𝑟2 𝑤𝑏/𝑚2 [∵ 𝜇𝑟 = 1 for non-magnetic material] Now, flux dφ through a cylindrical shell of radial thickness dx and axial length 1 m is given by; dφ= 𝐵𝑥 × 1 × dx= 𝜇0𝑥𝐼 2π𝑟2 dx 𝑤𝑒𝑏𝑒𝑟 This flux links with current 𝐼𝑥 (= π𝑥2 π𝑟2 𝐼) only. Therefore, flux linkages per metre length of the conductor is 𝑑ψ= π𝑥2 π𝑟2 dφ = 𝜇0𝑥3𝐼 2π𝑟4dx 𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠 Total flux linkages from centre upto the conductor surface is ψ𝑖𝑛𝑡 = න 0 𝑟 𝜇0𝑥3𝐼 2π𝑟4 dx = 𝜇0𝐼 2π𝑟4 𝑥4 4 𝑟 0 = 𝜇0𝐼 2π𝑟4 𝑟4 4 = 𝜇0𝐼 8π 𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠 per meter length 15 October 2020 D.Rajesh Asst.Prof. EEE Department 17 Internal inductance is independent of conductor radius.
  • 18.
     (ii) Fluxlinkages due to external flux. Now let us calculate the flux linkages of the conductor due to external flux. The external flux extends from the surface of the conductor to infinity. The field intensity at a distance x metres (from centre) outside the conductor is given by ; 𝐻𝑥 = 𝐼 2π𝑥 AT / m Flux density, 𝐵𝑥 = 𝜇0𝜇𝑟𝐻𝑥 = 𝜇0𝐼 2π𝑥 𝑤𝑏/𝑚2 [∵ 𝜇𝑟 = 1 for non-magnetic material] Now, flux dφ through a cylindrical shell of thickness dx and axial length 1 metre is dφ= 𝐵𝑥 × 1 × dx= 𝜇0𝐼 2π𝑥 𝑤𝑒𝑏𝑒𝑟 The flux dφ links all the current in the conductor once and only once. ∴ Flux linkages, dψ = dφ = 𝜇0𝐼 2π𝑥 dx 𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠 Total flux linkages of the conductor from surface to infinity, ψ𝑒𝑥𝑡 = න 𝑟 ∞ 𝜇0𝐼 2π𝑥 dx ∴ Overall flux linkages, ψ =ψ𝑖𝑛𝑡 + ψ𝑒𝑥𝑡 = 𝜇0𝐼 8π + ‫׬‬𝑟 ∞ 𝜇0𝐼 2π𝑥 dx = 𝜇0𝐼 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx 𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠 per meter length 15 October 2020 D.Rajesh Asst.Prof. EEE Department 18
  • 19.
    2. Flux linkagesin parallel current carrying conductors. The conductors A,B, C etc. carrying currents 𝐼𝐴, 𝐼𝐵, 𝐼𝐶 etc. Let us consider the flux linkages with one conductor, say conductor A. There will be flux linkages with conductor A due to its own current and there will be flux linkages with this conductor due to the mutual inductance effects of, 𝐼𝐵, 𝐼𝐶 , 𝐼𝐷, 𝐼𝐸 etc. The total flux linkages with conductor A. Flux linkages with conductor A due to its own current = 𝜇0𝐼𝐴 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx …….(i) Flux linkages with conductor A due to current 𝐼𝐵 = 𝜇0𝐼𝐵 2π ‫׬‬𝑑1 ∞ 1 𝑥 dx …….(ii) Flux linkages with conductor A due to current 𝐼𝐶 = 𝜇0𝐼𝐶 2π ‫׬‬ 𝑑2 ∞ 1 𝑥 dx …….. (iii) ∴ Total flux linkages with conductor A = (i) + (ii) + (iii) + ...... = 𝜇0𝐼𝐴 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx + 𝜇0𝐼𝐵 2π ‫׬‬ 𝑑1 ∞ 1 𝑥 dx + 𝜇0𝐼𝐶 2π ‫׬‬ 𝑑2 ∞ 1 𝑥 dx + ⋯ 15 October 2020 D.Rajesh Asst.Prof. EEE Department 19
  • 20.
     Consider asingle phase overhead line consisting of two parallel conductors A and B spaced d metres apart.  Conductors A and B carry the same amount of current (i.e. 𝐼𝐴 = 𝐼𝐵), but in the opposite direction because one forms the return circuit of the other. ∴ 𝐼𝐴 + 𝐼𝐵 = 0  In order to find the inductance of conductor A (or conductor B), we shall have to consider the flux linkages with it.  There will be flux linkages with conductor A due to its own current 𝐼𝐴 and also due to the mutual inductance effect of current 𝐼𝐵 in the conductor B.  Flux linkages with conductor A due to its own current = 𝜇0𝐼𝐴 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx …….(i) Flux linkages with conductor A due to current 𝐼𝐵 = 𝜇0𝐼𝐵 2π ‫׬‬ 𝑑 ∞ 1 𝑥 dx …….(ii) 15 October 2020 D.Rajesh Asst.Prof. EEE Department 20
  • 21.
     Total fluxlinkages with conductor A is  ψ𝐴 = 𝜇0𝐼𝐴 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx + 𝜇0𝐼𝐵 2π ‫׬‬ 𝑑 ∞ 1 𝑥 dx = 𝜇0 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx 𝐼𝐴 + 𝐼𝐵 ‫׬‬ 𝑑 ∞ 1 𝑥 dx = 𝜇0 2π 1 4 + log𝑒 ∞ − log𝑒 𝑟 𝐼𝐴 + 𝐼𝐵 log𝑒 ∞ − log𝑒 𝑑 = 𝜇0 2π 𝐼𝐴 4 + log𝑒 ∞ 𝐼𝐴 + 𝐼𝐵 − 𝐼𝐴 log𝑒 𝑟 − 𝐼𝐵 log𝑒 𝑑 = 𝜇0 2π 𝐼𝐴 4 − 𝐼𝐴 log𝑒 𝑟 − 𝐼𝐵 log𝑒 𝑑 ( ∴ 𝐼𝐴 + 𝐼𝐵 = 0) = 𝜇0 2π 𝐼𝐴 4 − 𝐼𝐴 log𝑒 𝑟 + 𝐼𝐴 log𝑒 𝑑 (∴ 𝐼𝐴 = −𝐼𝐵) = 𝜇0 2π 𝐼𝐴 4 + 𝐼𝐴 log𝑒 𝑑 𝑟 = 𝜇0𝐼𝐴 2π 1 4 + log𝑒 𝑑 𝑟 𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠 per meter length 15 October 2020 D.Rajesh Asst.Prof. EEE Department 21
  • 22.
     Inductance ofconductor A, 𝐿𝐴 = ψ𝐴 𝐼𝐴 = 𝜇0 2π 1 4 + log𝑒 𝑑 𝑟 H/m = 4Π∗10−7 2π 1 4 + log𝑒 𝑑 𝑟 H/m 𝐿𝐴 =10−7 1 2 + 2log𝑒 𝑑 𝑟 H/m ……..(i) Loop inductance = 2 𝐿𝐴 H/m =10−7 1 + 4log𝑒 𝑑 𝑟 H/m ………(ii) Note that eq. (ii) is the inductance of the two-wire line and is sometimes called loop inductance. However, inductance given by eq. (i) is the inductance per conductor and is equal to half the loop inductance. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 22
  • 23.
     Expression inalternate form. The expression for the inductance of a conductor can be put in a concise form. 𝐿𝐴 = ψ𝐴 𝐼𝐴 = 𝜇0 2π 1 4 + log𝑒 𝑑 𝑟 H/m = 4Π∗10−7 2π 1 4 + log𝑒 𝑑 𝑟 H/m =2* 10−7 1 4 + log𝑒 𝑑 𝑟 H/ = 2* 10−7 log𝑒 𝑒 1 4 + log𝑒 𝑑 𝑟 H/m =2* 10−7 log𝑒 𝑑 𝑟𝑒 − 1 4 H/m ………(iii) If we put 𝑟𝑒− 1 4=r′ The radius r′ is that of a fictitious conductor assumed to have no internal flux but with the same inductance as the actual conductor of radius r. The quantity 𝑒− 1 4 = 0·7788 so that r′ = r 𝑒− 1 4 = 0·7788 r The term r′ (= r 𝑒− 1 4) is called geometric mean radius (GMR) of the wire. Note that eq. (iii) gives the same value of inductance 𝐿𝐴 as eq. (i) Loop inductance =2* 2* 10−7 log𝑒 𝑑 𝑟𝑒 − 1 4 H/m Note that r′ = 0·7788 r is applicable to only solid round conductor. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 23
  • 24.
    Three conductors A,B and C of a 3-phase line carrying currents 𝐼𝐴, 𝐼𝐵, and 𝐼𝐶 respectively. Let 𝑑1, 𝑑2 and 𝑑3 be the spacings between the conductors as shown. Let us further assume that the loads are balanced i.e. 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶 = 0. Consider the flux linkages with conductor A. There will be flux linkages with conductor A due to its own current and also due to the mutual inductance effects of 𝐼𝐵 and 𝐼𝐶 . Flux linkages with conductor A due to its own current = 𝜇0𝐼𝐴 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx …….(i) Flux linkages with conductor A due to current 𝐼𝐵 = 𝜇0𝐼𝐵 2π ‫׬‬ 𝑑3 ∞ 1 𝑥 dx …….(ii) Flux linkages with conductor A due to current 𝐼𝑐 = 𝜇0𝐼𝑐 2π ‫׬‬ 𝑑2 ∞ 1 𝑥 dx …….(ii) 15 October 2020 D.Rajesh Asst.Prof. EEE Department 24
  • 25.
     Total fluxlinkages with conductor A is ψ𝐴 = (i) + (ii) + (iii) = 𝜇0𝐼𝐴 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx + 𝜇0𝐼𝐵 2π ‫׬‬ 𝑑3 ∞ 1 𝑥 dx+ 𝜇0𝐼𝑐 2π ‫׬‬ 𝑑2 ∞ 1 𝑥 dx = 𝜇0 2π 1 4 + ‫׬‬ 𝑟 ∞ 1 𝑥 dx 𝐼𝐴 + 𝐼𝐵 ‫׬‬ 𝑑3 ∞ 1 𝑥 dx + 𝐼𝑐 ‫׬‬ 𝑑2 ∞ 1 𝑥 dx =𝜇0 2π 1 4 + log𝑒 ∞ − log𝑒 𝑟 𝐼𝐴 + 𝐼𝐵 log𝑒 ∞ − log𝑒 𝑑3 + 𝐼𝐵 log𝑒 ∞ − log𝑒 𝑑2 = 𝜇0 2π 𝐼𝐴 4 + log𝑒 ∞ 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶 − 𝐼𝐴 log𝑒 𝑟 − 𝐼𝐵 log𝑒 𝑑3 − 𝐼𝐶 log𝑒 𝑑2 = 𝜇0 2π 𝐼𝐴 4 − 𝐼𝐴 log𝑒 𝑟 − 𝐼𝐵 log𝑒 𝑑3 − 𝐼𝐶 log𝑒 𝑑2 ( ∴ 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶= 0) ψ𝐴= 𝜇0 2π 1 4 − log𝑒 𝑟 𝐼𝐴 − 𝐼𝐵 log𝑒 𝑑3 − 𝐼𝐶 log𝑒 𝑑2 15 October 2020 D.Rajesh Asst.Prof. EEE Department 25
  • 26.
     (i) Symmetricalspacing. If the three conductors A, B and C are placed symmetrically at the corners of an equilateral triangle of side d, then, 𝑑1=𝑑2=𝑑3 = 𝑑 ψ𝐴= 𝜇0 2π 1 4 − log𝑒 𝑟 𝐼𝐴 − 𝐼𝐵 log𝑒 𝑑 − 𝐼𝐶 log𝑒 𝑑 = 𝜇0 2π 1 4 − log𝑒 𝑟 𝐼𝐴 − (𝐼𝐵 + 𝐼𝐶) log𝑒 𝑑 = 𝜇0 2π 1 4 − log𝑒 𝑟 𝐼𝐴 − (−𝐼𝐴) log𝑒 𝑑 ( ∴ 𝐼𝐵 + 𝐼𝐶= −𝐼𝐴 ) = 𝜇0𝐼𝐴 2π 1 4 + log𝑒 𝑑 𝑟 weber−turns/m Inductance of conductor A, 𝐿𝐴 = ψ𝐴 𝐼𝐴 H /m = 𝜇0 2π 1 4 + log𝑒 𝑑 𝑟 H /m = 4Π∗10−7 2π 1 4 + log𝑒 𝑑 𝑟 H/m 𝐿𝐴 =10−7 1 2 + 2log𝑒 𝑑 𝑟 H/m Derived in a similar way, the expressions for inductance are the same for conductors B and C. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 26
  • 27.
     (ii) Unsymmetricalspacing. When 3-phase line conductors are not equidistant from each other, the conductor spacing is said to be unsymmetrical. Under such conditions, the flux linkages and inductance of each phase are not the same. A different inductance in each phase results in unequal voltage drops in the three phases even if the currents in the conductors are balanced. Therefore, the voltage at the receiving end will not be the same for all phases. In order that voltage drops are equal in all conductors, we generally interchange the positions of the conductors at regular intervals along the line so that each conductor occupies the original position of every other conductor over an equal distance. Such an exchange of positions is known as transposition. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 27
  • 28.
     A3-phase transposedline having unsymmetrical spacing. Let us assume that each of the three sections is 1 m in length.  Let us further assume balanced conditions i.e., 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶 = 0. . Let the line currents be :  𝐼𝐴 =I∠00 = I (1+ j 0)  𝐼𝐵 = I∠−1200 = I (− 0·5 − j 0·866)  𝐼𝐶 = I∠−2400= I (− 0·5 + j 0·866)  The total flux linkages per metre length of conductor A is ψ𝐴= 𝜇0 2π 1 4 − log𝑒 𝑟 𝐼𝐴 − 𝐼𝐵 log𝑒 𝑑3 − 𝐼𝐶 log𝑒 𝑑2 Putting the values of 𝐼𝐴, 𝐼𝐵 and 𝐼𝐶 , we get, ψ𝐴= 𝜇0 2π 1 4 − log𝑒 𝑟 𝐼 − I (− 0·5 − j 0·866) log𝑒 𝑑3 − I (− 0·5 + j 0·866) log𝑒 𝑑2 = 𝜇0 2π 𝐼 4 − 𝐼 log𝑒 𝑟 + 0·5 Ilog𝑒 𝑑3 + j 0·866 I log𝑒 𝑑3 + 0·5I log𝑒 𝑑2 −j 0·866I log𝑒 𝑑2 = 𝜇0 2π 𝐼 4 − 𝐼 log𝑒 𝑟 + 0·5 I(log𝑒 𝑑3 + log𝑒 𝑑2 )+ j 0·866 I( log𝑒 𝑑3 − log𝑒 𝑑2) = 𝜇0 2π 𝐼 4 − 𝐼 log𝑒 𝑟 + 𝐼 log𝑒 𝑑2𝑑3 + j 0·866 I log𝑒 𝑑3 𝑑2 = 𝜇0 2π 𝐼 4 + 𝐼 log𝑒 𝑑2𝑑3 𝑟 + j 0·866 I log𝑒 𝑑3 𝑑2 = 𝜇0𝐼 2π 1 4 + log𝑒 𝑑2𝑑3 𝑟 + j 0·866 log𝑒 𝑑3 𝑑2 15 October 2020 D.Rajesh Asst.Prof. EEE Department 28
  • 29.
     ∴ Inductanceof conductor A is 𝐿𝐴 = ψ𝐴 𝐼𝐴 = ψ𝐴 𝐼 H /m = 𝜇0 2π 1 4 + log𝑒 𝑑2𝑑3 𝑟 + j 0·866 log𝑒 𝑑3 𝑑2 H /m = 𝜇0 2π 1 4 + log𝑒 𝑑2𝑑3 𝑟 + j 0·866 log𝑒 𝑑3 𝑑2 H /m = 4Π∗10−7 2π 1 4 + log𝑒 𝑑2𝑑3 𝑟 + j 0·866 log𝑒 𝑑3 𝑑2 H /m = 10−7 1 2 + 2 log𝑒 𝑑2𝑑3 𝑟 + j 1.732 log𝑒 𝑑3 𝑑2 H /m Similarly inductance of conductors B and C will be : 𝐿𝑩= 10−7 1 2 + 2 log𝑒 𝑑3𝑑1 𝑟 + j 1.732 log𝑒 𝑑1 𝑑3 H /m 𝐿𝑪=10−7 1 2 + 2 log𝑒 𝑑1𝑑2 𝑟 + j 1.732 log𝑒 𝑑2 𝑑1 H /m Inductance of each line conductor = 1 3 (𝐿𝐴+ 𝐿𝑩+ 𝐿𝑪) = 10−7 1 2 + 2 log𝑒 3 𝑑1𝑑2𝑑3 𝑟 H /m 15 October 2020 D.Rajesh Asst.Prof. EEE Department 29 If we compare the formula of inductance of an unsymmetrically spaced transposed line with that of symmetrically spaced line, we find that inductance of each line conductor in the two cases will be equal if d = 3 𝑑1𝑑2𝑑3 The distance d is known as equivalent equilateral spacing for unsymmetrically transposed line.
  • 30.
    The use ofself geometrical mean distance (abbreviated as self-GMD) and mutual geometrical mean distance (mutual-GMD) simplifies the inductance calculations, particularly relating to multiconductor arrangements. The symbols used for these are respectively 𝐷𝑠 and 𝐷𝑚. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 30
  • 31.
     Inductance/conductor/m =10−7 1 2 + 2log𝑒 𝑑 𝑟 H/m = 1 2 ∗ 10−7 + 2 ∗ 10−7log𝑒 𝑑 𝑟 H/m If we replace the original solid conductor by an equivalent hollow cylinder with extremely thin walls, the current is confined to the conductor surface and internal conductor flux linkage would be almost zero. Consequently, inductance due to internal flux would be zero and the term 1 2 ∗ 10−7 shall be eliminated. The radius of this equivalent hollow cylinder must be sufficiently smaller than the physical radius of the conductor to allow room for enough additional flux to compensate for the absence of internal flux linkage. It can be proved mathematically that for a solid round conductor of radius r, the self-GMD or GMR = 0·7788 r. Inductance/conductor/m = 2 ∗ 10−7log𝑒 𝑑 𝐷𝑠 where 𝐷𝑠 = GMR or self-GMD = 0·7788 r It may be noted that self-GMD of a conductor depends upon the size and shape of the conductor and is independent of the spacing between the conductors. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 31
  • 32.
     The mutual-GMDis the geometrical mean of the distances from one conductor to the other and, therefore, must be between the largest and smallest such distance.  (a) The mutual-GMD between two conductors (assuming that spacing between conductors is large compared to the diameter of each conductor) is equal to the distance between their centres i.e. 𝐷𝑚 = spacing between conductors = d  (b) For a single circuit 3-φ line, the mutual-GMD is equal to the equivalent equilateral spacing i.e., 𝑑1𝑑2𝑑3 1 3 𝐷𝑚 = 𝑑1𝑑2𝑑3 1 3 15 October 2020 D.Rajesh Asst.Prof. EEE Department 32
  • 33.
     (c) Theprinciple of geometrical mean distances can be most profitably employed to 3-φ double circuit lines. Consider the conductor arrangement of the double circuit shown in Fig. Suppose the radius of each conductor is r. Self-GMD of conductor = 0·7788 r Self-GMD of combination aa´ is 𝐷𝑠1 = ( 𝐷𝑎𝑎 × 𝐷𝑎 ư 𝑎 × 𝐷 ư 𝑎 ư 𝑎 × 𝐷 ư 𝑎𝑎 ) 1 4 Self-GMD of combination bb´ is 𝐷𝑠2 = ( 𝐷𝑏𝑏 × 𝐷𝑏 ሗ 𝑏 × 𝐷 ሗ 𝑏 ሗ 𝑏 × 𝐷 ሗ 𝑏𝑏 ) 1 4 Self-GMD of combination cc′ is 𝐷𝑠3 = ( 𝐷𝑐𝑐 × 𝐷𝑐 ư 𝑐 × 𝐷 ư 𝑐 ư 𝑐 × 𝐷 ư 𝑐𝑐 ) 1 4 Equivalent self-GMD of one phase 𝐷𝑠 = 𝐷𝑠1 ×𝐷𝑠2×𝐷𝑠3 1 3 The value of 𝐷𝑠 is the same for all the phases as each conductor has the same radius. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 33
  • 34.
     Mutual-GMD betweenphases A and B is 𝐷𝐴𝐵 = ( 𝐷𝑎𝑏 × 𝐷𝑎 ሗ 𝑏 × 𝐷 ư 𝑎𝑏 × 𝐷 ư 𝑎 ሗ 𝑏 ) 1 4 Mutual-GMD between phases B and C is 𝐷𝐵𝐶 = ( 𝐷𝑏𝑐 × 𝐷𝑏 ư 𝑐 × 𝐷 ሗ 𝑏𝑐 × 𝐷 ሗ 𝑏 ư 𝑐 ) 1 4 Mutual-GMD between phases C and A is 𝐷𝐶𝐴 = ( 𝐷𝑐𝑎 × 𝐷𝑐 ư 𝑎 × 𝐷 ư 𝑐𝑎 × 𝐷 ư 𝑐 ư 𝑎 ) 1 4 Equivalent mutual-GMD, 𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3 Mutual GMD depends only upon the spacing and is substantially independent of the exact size, shape and orientation of the conductor. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 34
  • 35.
    (i) Single phaseline Inductance/conductor/m =2 ∗ 10−7log𝑒 𝐷𝑚 𝐷𝑠 where 𝐷𝑠 = GMR or self-GMD = 0·7788 r 𝐷𝑚 = Spacing between conductors = d (ii) Single circuit 3-φ line Inductance/phase/m = 2 ∗ 10−7log𝑒 𝐷𝑚 𝐷𝑠 where 𝐷𝑠 = GMR or self-GMD = 0·7788 r 𝐷𝑚 = 𝑑1𝑑2𝑑3 1 3 (iii) Double circuit 3-φ line Inductance/phase/m = 2 ∗ 10−7 log𝑒 𝐷𝑚 𝐷𝑠 where 𝐷𝑠 = 𝐷𝑠1 ×𝐷𝑠2×𝐷𝑠3 1 3 𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3 15 October 2020 D.Rajesh Asst.Prof. EEE Department 35
  • 36.
     A singlephase line has two parallel conductors 2 metres apart. The diameter of each conductor is 1·2 cm. Calculate the loop inductance per km of the line.  Loop inductance per metre length of the line =10−7 1 + 4log𝑒 𝑑 𝑟 H = 10−7 1 + 4log𝑒 200 0.6 H =24.23* 10−7 Loop inductance per km of the line = 24.23* 10−7 *1000 =2.423mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 36
  • 37.
    A single phasetransmission line has two parallel conductors 3 m apart, the radius of each conductor being 1 cm. Calculate the loop inductance per km length of the line if the material of the conductor is (i) copper (ii) steel with relative permeability of 100.  Loop inductance per metre length of the line =10−7 𝜇𝑟 + 4log𝑒 𝑑 𝑟 H/m (i) With copper conductors, 𝜇𝑟 = 1 ∴ Loop inductance/m=10−7 1 + 4log𝑒 300 1 H/m =23.8 * 10−7 H Loop inductance per km of the line = 23.8* 10−7*1000=2.38mH (ii) With steel conductors, 𝜇𝑟 = 100 ∴ Loop inductance/m=10−7 100 + 4log𝑒 300 1 H/m =122.8 * 10−7H Loop inductance per km of the line = 122.8 * 10−7*1000=12.28mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 37
  • 38.
    Find the inductanceper km of a 3-phase transmission line using 1·24 cm diameter conductors when these are placed at the corners of an equilateral triangle of each side 2 m. conductor spacing d = 2 m =200cm conductor radius r = 1·24/2 = 0·62 cm. Inductance/phase/m=10−7 1 2 + 2log𝑒 𝑑 𝑟 H/m =10−7 1 2 + 2log𝑒 200 0.62 H/m =12* 10−7 H Inductance/phase/km = 12* 10−7 *1000H =1.2mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 38
  • 39.
    The three conductorsof a 3-phase line are arranged at the corners of a triangle of sides 2 m, 2·5 m and 4·5 m. Calculate the inductance per km of the line when the conductors are regularly transposed. The diameter of each conductor is 1·24 cm. Three conductors of a 3-phase line placed at the corners of a triangle of sides 𝐷12 = 2 m, 𝐷23 = 2·5 m and 𝐷31 = 4·5 m. The conductor radius r = 1·24/2 = 0·62 cm. Equivalent equilateral spacing, 𝐷𝑒𝑞=3 𝐷12 ∗ 𝐷23 ∗ 𝐷31 = 3 2 ∗ 2.5 ∗ 4.5=2.82m=282cm Inductance/phase/m =10−7 1 2 + 2log𝑒 𝐷𝑒𝑞 𝑟 H/m =10−7 1 2 + 2log𝑒 282 0.62 H/m =12.74* 10−7 Inductance/phase/km = 12.74* 10−7 *1000H =1.274mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 39
  • 40.
    Calculate the inductanceof each conductor in a 3-phase, 3-wire system when the conductors are arranged in a horizontal plane with spacing such that 𝐷31 = 4 m ; 𝐷12 = 𝐷23 = 2m. The conductors are transposed and have a diameter of 2·5 cm. The conductor radius r = 2·5/2 = 1·25 cm. Equivalent equilateral spacing, 𝐷𝑒𝑞=3 𝐷12 ∗ 𝐷23 ∗ 𝐷31 𝐷𝑒𝑞= 3 2 ∗ 2 ∗ 4=2.52m=252cm Inductance/phase/m =10−7 1 2 + 2log𝑒 𝐷𝑒𝑞 𝑟 H/m =10−7 1 2 + 2log𝑒 252 1.25 H/m =11.1* 10−7 Inductance/phase/km = 11.1* 10−7 *1000H =1.11mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 40
  • 41.
    Two conductors ofa single phase line, each of 1 cm diameter, are arranged in a vertical plane with one conductor mounted 1 m above the other. A second identical line is mounted at the same height as the first and spaced horizontally 0·25 m apart from it. The two upper and the two lower conductors are connected in parallel. Determine the inductance per km of the resulting double circuit line. Conductors a, a′ form one connection and conductors b, b′ form the return connection. The conductor radius, r = 1/2 = 0·5 cm. Inductance per conductor per metre= 2 ∗ 10−7 log𝑒 𝐷𝑚 𝐷𝑠 G.M.R. of conductor = 0·7788 r = 0·7788 × 0·5 = 0·389 cm Self G.M.D. of aa′ combination is 𝐷𝑠 = ( 𝐷𝑎𝑎 × 𝐷𝑎 ư 𝑎 × 𝐷 ư 𝑎 ư 𝑎 × 𝐷 ư 𝑎𝑎 ) 1 4 𝐷𝑠 = ( 0.389 × 100 ×0.389 × 100) 1 4 =6.23cm Mutual G.M.D. between a and b is 𝐷𝑚 = ( 𝐷𝑎𝑏 × 𝐷𝑎 ሗ 𝑏 × 𝐷 ư 𝑎𝑏 × 𝐷 ư 𝑎 ሗ 𝑏 ) 1 4 𝐷𝑎 ሗ 𝑏= 𝐷 ư 𝑎𝑏= 252 + 1002 = 103 𝐷𝑚 = ( 25 × 103 × 103 × 25 ) 1 4 =50.47cm Inductance per conductor per metre= 2 ∗ 10−7log𝑒 𝐷𝑚 𝐷𝑠 = = 2 ∗ 10−7 log𝑒 50.47 6.23 =0.42 × 10−6H ∴ Loop inductance per km of the line=2* 0.42 × 10−6 *1000=0.84mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 41
  • 42.
    Find the inductanceper phase per km of double circuit 3-phase line shown in Fig. The conductors are transposed and are of radius 0·75 cm each. The phase sequence is ABC. Inductance/phase/m = 2 ∗ 10−7 log𝑒 𝐷𝑚 𝐷𝑠 where 𝐷𝑠 = 𝐷𝑠1 ×𝐷𝑠2×𝐷𝑠3 1 3 𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3 Self-GMD of conductor = 0·7788 r=0.7788*0.75=0.584cm Self-GMD of combination aa´ is 𝐷𝑠1 = ( 𝐷𝑎𝑎 × 𝐷𝑎 ư 𝑎 × 𝐷 ư 𝑎 ư 𝑎 × 𝐷 ư 𝑎𝑎 ) 1 4 𝐷𝑎 ư 𝑎= 𝐷 ư 𝑎𝑎= 62 + 42=7.21m 𝐷𝑠1 = ( (0.584×10−2 )×7.21×(0.584×10−2 )×7.21) 1 4=0.205m Self-GMD of combination bb´ is 𝐷𝑠2 = ( 𝐷𝑏𝑏 × 𝐷𝑏 ሗ 𝑏 × 𝐷 ሗ 𝑏 ሗ 𝑏 × 𝐷 ሗ 𝑏𝑏 ) 1 4 = ((0.584×10−2 )× 5.5×(0.584×10−2 )×5.5) 1 4=0.18m Self-GMD of combination cc′ is 𝐷𝑠3 = ( 𝐷𝑐𝑐 × 𝐷𝑐 ư 𝑐 × 𝐷 ư 𝑐 ư 𝑐 × 𝐷 ư 𝑐𝑐 ) 1 4 = 𝐷𝑠1= 0.205m 𝐷𝑠 = 𝐷𝑠1 ×𝐷𝑠2×𝐷𝑠3 1 3 𝐷𝑠 = 0.205×0.18×0.205 1 3 =0.195m 15 October 2020 D.Rajesh Asst.Prof. EEE Department 42
  • 43.
    𝐷𝑚 = 𝐷𝐴𝐵×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3 Mutual-GMD between phases A and B is 𝐷𝐴𝐵 = ( 𝐷𝑎𝑏 × 𝐷𝑎 ሗ 𝑏 × 𝐷 ư 𝑎𝑏 × 𝐷 ư 𝑎 ሗ 𝑏 ) 1 4 𝐷𝑎 ሗ 𝑏= 𝐷 ư 𝑎𝑏= 32 + (4 + 0.75)2=5.62m 𝐷𝑎𝑏= 32 + (0.75)2=3.1m 𝐷𝐴𝐵 = ( 3.1 × 5.62 × 5.62× 3.1 ) 1 4=4.17m Mutual-GMD between phases B and C is 𝐷𝐵𝐶 = ( 𝐷𝑏𝑐 × 𝐷𝑏 ư 𝑐 × 𝐷 ሗ 𝑏𝑐 × 𝐷 ሗ 𝑏 ư 𝑐 ) 1 4 = 𝐷𝐴𝐵 =4.17m Mutual-GMD between phases C and A is 𝐷𝐶𝐴 = ( 𝐷𝑐𝑎 × 𝐷𝑐 ư 𝑎 × 𝐷 ư 𝑐𝑎 × 𝐷 ư 𝑐 ư 𝑎 ) 1 4 = ( 6 × 4 × 4 × 6 ) 1 4=4.9m Equivalent mutual-GMD, 𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3= 4.71×4.17×4.9 1 3=4.4m Inductance/phase/m = 2 ∗ 10−7 log𝑒 𝐷𝑚 𝐷𝑠 = 2 ∗ 10−7 log𝑒 4.4 0.195 =0.623 ∗ 10−3mH Inductance/phase/km= 0.623 ∗ 10−3 *1000=0.623mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 43
  • 44.
    Calculate the inductanceper phase per metre for a three-phase double-circuit line whose phase conductors have a radius of 5·3 cm with the horizontal conductor arrangement as shown in Fig. Inductance per conductor per metre= 2 ∗ 10−7log𝑒 𝐷𝑚 𝐷𝑠 where 𝐷𝑠 = 𝐷𝑠1 ×𝐷𝑠2×𝐷𝑠3 1 3 𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3 Self-GMD of combination aa´ is0·7788 r = 0·7788 × 5·3 × 10−2 = 0·0413 m 𝐷𝑠1 = ( 𝐷𝑎𝑎 × 𝐷𝑎 ư 𝑎 × 𝐷 ư 𝑎 ư 𝑎 × 𝐷 ư 𝑎𝑎 ) 1 4 = (0·041× 24 ×0·041× 24 ) 1 4=0.995m Self-GMD of combination bb´ is 𝐷𝑠2 = ( 𝐷𝑏𝑏 × 𝐷𝑏 ሗ 𝑏 × 𝐷 ሗ 𝑏 ሗ 𝑏 × 𝐷 ሗ 𝑏𝑏 ) 1 4 = (0·041× 24 ×0·041× 24 ) 1 4=0.995m Self-GMD of combination cc′ is 𝐷𝑠3 = ( 𝐷𝑐𝑐 × 𝐷𝑐 ư 𝑐 × 𝐷 ư 𝑐 ư 𝑐 × 𝐷 ư 𝑐𝑐 ) 1 4 = (0·041× 24 ×0·041× 24 ) 1 4=0.995m Equivalent self-GMD of one phase 𝐷𝑠 = 𝐷𝑠1 ×𝐷𝑠2×𝐷𝑠3 1 3=0.995m 15 October 2020 D.Rajesh Asst.Prof. EEE Department 44
  • 45.
    𝐷𝑚 = 𝐷𝐴𝐵×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3 Mutual-GMD between phases A and B is 𝐷𝐴𝐵 = ( 𝐷𝑎𝑏 × 𝐷𝑎 ሗ 𝑏 × 𝐷 ư 𝑎𝑏 × 𝐷 ư 𝑎 ሗ 𝑏 ) 1 4 𝐷𝐴𝐵 = ( 8 × 32 × 16× 8 ) 1 4=13.45m Mutual-GMD between phases B and C is 𝐷𝐵𝐶 = ( 𝐷𝑏𝑐 × 𝐷𝑏 ư 𝑐 × 𝐷 ሗ 𝑏𝑐 × 𝐷 ሗ 𝑏 ư 𝑐 ) 1 4 = ( 8 × 32 × 16× 8 ) 1 4=13.45m Mutual-GMD between phases C and A is 𝐷𝐶𝐴 = ( 𝐷𝑐𝑎 × 𝐷𝑐 ư 𝑎 × 𝐷 ư 𝑐𝑎 × 𝐷 ư 𝑐 ư 𝑎 ) 1 4 = ( 16 × 8 × 40 × 16 ) 1 4=16.917m Equivalent mutual-GMD, 𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴 1 3= 13.45×13.45×16.917 1 3=14.518m Inductance/phase/m = 2 ∗ 10−7 log𝑒 𝐷𝑚 𝐷𝑠 = 2 ∗ 10−7log𝑒 14.518 0.995 =5.36 ∗ 10−7 H 15 October 2020 D.Rajesh Asst.Prof. EEE Department 45
  • 46.
    In a singlephase line (See. Fig.)conductors a and a′ in parallel form one circuit while conductors b and b′ in parallel form the return path. Calculate the total inductance of the line per km assuming that current is equally shared by the two parallel conductors. Conductor diameter in 2·0 cm. Inductance/phase/m = 2 ∗ 10−7 log𝑒 𝐷𝑚 𝐷𝑠 Mutual G.M.D., 𝐷𝑚 = ( 𝐷𝑎𝑏 × 𝐷𝑎 ሗ 𝑏 × 𝐷 ư 𝑎𝑏 × 𝐷 ư 𝑎 ሗ 𝑏 ) 1 4 = ( 120 × 140 × 100× 120 ) 1 4=119cm Self G.M.D., 𝐷𝑠 = ( 𝐷𝑎𝑎 × 𝐷𝑎 ư 𝑎 × 𝐷 ư 𝑎 ư 𝑎 × 𝐷 ư 𝑎𝑎 ) 1 4 𝐷𝑎𝑎= 𝐷 ư 𝑎 ư 𝑎=0.7788*r=0.7788* 2 2 =0.7788cm 𝐷𝑠 = (0.7788 × 20 × 0.7788× 20 ) 1 4=3.94cm Inductance/phase/m = 2 ∗ 10−7 log𝑒 𝐷𝑚 𝐷𝑠 = 2 ∗ 10−7 log𝑒 119 3.94 =6.81* 10−7 H Loop Inductance/phase/km=2*6.81* 10−7 *1000H =1.363mH 15 October 2020 D.Rajesh Asst.Prof. EEE Department 46
  • 47.
    Any two conductorsof an overhead transmission line are separated by air which acts as an insulation, therefore, capacitance exists between any two overhead line conductors. The capacitance between the conductors is the charge per unit potential difference i.e., Capacitance, C = 𝑞 𝑣 farad where q = charge on the line in coulomb v = p.d. between the conductors in volts 15 October 2020 D.Rajesh Asst.Prof. EEE Department 47
  • 48.
     Electric Potential Theelectric potential at a point due to a charge is the work done in bringing a unit positive charge from infinity to that point. (i) Potential at a charged single conductor (ii) Potential at a conductor in a group of charged conductors. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 48
  • 49.
    (i) Potential ata charged single conductor. Consider a long straight cylindrical conductor A of radius r metres. Let the conductor operate at such a potential (𝑉𝐴) that charge 𝑄𝐴 coulombs per metre exists on the conductor. The electric intensity E at a distance x from the centre of the conductor in air is given by: 𝐸 = 𝑄𝐴 2𝜋𝑥𝜀0 volts/m where 𝑄𝐴 = charge per metre length 𝜀0 = permittivity of free space As x approaches infinity, the value of E approaches zero. Therefore, the potential difference between conductor A and infinity distant neutral plane is given by : Work is done in bringing a unit positive charge against E from infinity to conductor surface. 𝑉𝐴 = න 𝑟 ∞ 𝐸 𝑑𝑥 𝑉𝐴 = න 𝑟 ∞ 𝑄𝐴 2𝜋𝑥𝜀0 𝑑𝑥 = 𝑄𝐴 2𝜋𝜀0 න 𝑟 ∞ 1 𝑥 𝑑𝑥 15 October 2020 D.Rajesh Asst.Prof. EEE Department 49
  • 50.
    (ii)Potential at aconductor in a group of charged conductors. Consider a group of long straight conductors A, B, C etc. operating at potentials such that charges 𝑄𝐴, 𝑄𝐵, 𝑄𝐶 etc. coulomb per metre length exist on the respective conductors The potential at A Potential at A due to its own charge (i.e. 𝑄𝐴) = න 𝑟 ∞ 𝑄𝐴 2𝜋𝑥𝜀0 𝑑𝑥 Potential at conductor A due to charge 𝑄𝐵 = න 𝑑1 ∞ 𝑄𝐵 2𝜋𝑥𝜀0 𝑑𝑥 Potential at conductor A due to charge 𝑄𝐶 = න 𝑑2 ∞ 𝑄𝐶 2𝜋𝑥𝜀0 𝑑𝑥 15 October 2020 D.Rajesh Asst.Prof. EEE Department 50
  • 51.
     Overall potentialdifference between conductor A and infinite neutral plane is 𝑉𝐴 = (i) + (ii) + (iii) + ....... = ‫׬‬ 𝑟 ∞ 𝑄𝐴 2𝜋𝑥𝜀0 𝑑𝑥+ ‫׬‬ 𝑑1 ∞ 𝑄𝐵 2𝜋𝑥𝜀0 𝑑𝑥 +‫׬‬ 𝑑2 ∞ 𝑄𝐶 2𝜋𝑥𝜀0 𝑑𝑥+……… = 1 2𝜋𝜀0 ‫׬‬ 𝑟 ∞ 𝑄𝐴 𝑥 dx + ‫׬‬ 𝑑1 ∞ 𝑄𝐵 𝑥 𝑑𝑥 + ‫׬‬ 𝑑2 ∞ 𝑄𝐶 𝑥 𝑑𝑥 + ⋯ … . . = 1 2𝜋𝜀0 𝑄𝐴 log𝑒 ∞ − log𝑒 𝑟 + 𝑄𝐵 log𝑒 ∞ − log𝑒 𝑑1 + 𝑄𝐶 log𝑒 ∞ − log𝑒 𝑑2 + ⋯ . = 1 2𝜋𝜀0 log𝑒 ∞ 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶 + 𝑄𝐴 log𝑒 1 𝑟 + 𝑄𝐵 log𝑒 1 𝑑1 + 𝑄𝐶 log𝑒 1 𝑑2 + ⋯ . Assuming balanced conditions i.e., 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶=0 𝑉𝐴= 1 2𝜋𝜀0 𝑄𝐴 log𝑒 1 𝑟 + 𝑄𝐵 log𝑒 1 𝑑1 + 𝑄𝐶 log𝑒 1 𝑑2 + ⋯ . 15 October 2020 D.Rajesh Asst.Prof. EEE Department 51
  • 52.
    Consider a singlephase overhead transmission line consisting of two parallel conductors A and B spaced d metres apart in air. Suppose that radius of each conductor is r metres. Let their respective charge be + Q and − Q coulombs per metre length. The total p.d. between conductor A and neutral “infinite” plane is 𝑉𝐴 = ‫׬‬ 𝑟 ∞ Q 2𝜋𝑥𝜀0 𝑑𝑥+ ‫׬‬ 𝑑 ∞ −𝑄 2𝜋𝑥𝜀0 𝑑𝑥 = 1 2𝜋𝜀0 Q log𝑒 ∞ − log𝑒 𝑟 − Q log𝑒 ∞ − log𝑒 𝑑 = Q 2𝜋𝜀0 log𝑒 𝑑 − log𝑒 𝑟 = Q 2𝜋𝜀0 log𝑒 𝑑 𝑟 volts 15 October 2020 D.Rajesh Asst.Prof. EEE Department 52
  • 53.
    The total p.d.between conductor B and neutral “infinite” plane is 𝑉𝐵 = ‫׬‬ 𝑟 ∞ −Q 2𝜋𝑥𝜀0 𝑑𝑥+ ‫׬‬ 𝑑 ∞ 𝑄 2𝜋𝑥𝜀0 𝑑𝑥 = 1 2𝜋𝜀0 −Q log𝑒 ∞ − log𝑒 𝑟 +Q log𝑒 ∞ − log𝑒 𝑑 = −Q 2𝜋𝜀0 log𝑒 𝑑 − log𝑒 𝑟 = −Q 2𝜋𝜀0 log𝑒 𝑑 𝑟 volts Both these potentials are w.r.t. the same neutral plane. Since the unlike charges attract each other, the potential difference between the conductors is 𝑉𝐴𝐵 = 2𝑉𝐴= 2Q 2𝜋𝜀0 log𝑒 𝑑 𝑟 ∴ Capacitance, 𝐶𝐴𝐵 = Q 𝑉𝐴𝐵 = Q 2Q 2𝜋𝜀0 log𝑒 𝑑 𝑟 𝐶𝐴𝐵 = 𝜋𝜀0 log𝑒 𝑑 𝑟 F/m 15 October 2020 D.Rajesh Asst.Prof. EEE Department 53
  • 54.
     Capacitance toneutral. Since potential of the mid-point between the conductors is zero, the potential difference between each conductor and the ground or neutral is half the potential difference between the conductors. Thus the capacitance to ground or capacitance to neutral for the twowire line is twice the line-to-line capacitance ∴ Capacitance to neutral 𝐶𝑁 = 𝐶𝐴𝑁 = 𝐶𝐵𝑁 = 2𝐶𝐴𝐵 𝐶𝑁= 2𝜋𝜀0 log𝑒 𝑑 𝑟 F/m 15 October 2020 D.Rajesh Asst.Prof. EEE Department 54
  • 55.
     In a3-phase transmission line, the capacitance of each conductor is considered instead of capacitance from conductor to conductor.  Here, again two cases arise viz., symmetrical spacing and unsymmetrical spacing.  (i) Symmetrical Spacing. The three conductors A, B and C of the 3-phase overhead transmission line having charges 𝑄𝐴, 𝑄𝐵, and 𝑄𝐶 per metre length respectively. Let the conductors be equidistant (d metres) from each other. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 55
  • 56.
     overall potentialdifference between conductor A and infinite neutral plane is  𝑉𝐴 = ‫׬‬𝑟 ∞ 𝑄𝐴 2𝜋𝑥𝜀0 𝑑𝑥+ ‫׬‬𝑑 ∞ 𝑄𝐵 2𝜋𝑥𝜀0 𝑑𝑥 +‫׬‬𝑑 ∞ 𝑄𝐶 2𝜋𝑥𝜀0 𝑑𝑥 = 1 2𝜋𝜀0 ‫׬‬ 𝑟 ∞ 𝑄𝐴 𝑥 dx + ‫׬‬ 𝑑 ∞ 𝑄𝐵 𝑥 𝑑𝑥 + ‫׬‬ 𝑑 ∞ 𝑄𝐶 𝑥 𝑑𝑥 = 1 2𝜋𝜀0 𝑄𝐴 log𝑒 ∞ − log𝑒 𝑟 + 𝑄𝐵 log𝑒 ∞ − log𝑒 𝑑 + 𝑄𝐶 log𝑒 ∞ − log𝑒 𝑑 = 1 2𝜋𝜀0 log𝑒 ∞ 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶 + 𝑄𝐴 log𝑒 1 𝑟 + (𝑄𝐵 + 𝑄𝐶 ) log𝑒 1 𝑑 Assuming balanced conditions i.e., 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶=0 𝑄𝐵 + 𝑄𝐶= −𝑄𝐴 ∴ 𝑉𝐴= 1 2𝜋𝜀0 𝑄𝐴 log𝑒 1 𝑟 + (−𝑄𝐴 ) log𝑒 1 𝑑 𝑉𝐴 = 𝑄𝐴 2𝜋𝜀0 log𝑒 𝑑 𝑟 F/m ∴ Capacitance of conductor A w.r.t neutral, 𝐶𝐴 = 𝑄𝐴 𝑉𝐴 = 𝑄𝐴 𝑄𝐴 2𝜋𝜀0 log𝑒 𝑑 𝑟 𝐶𝐴 = 2𝜋𝜀0 log𝑒 𝑑 𝑟 F/m Note that this equation is identical to capacitance to neutral for two-wire line. Derived in a similar manner, the expressions for capacitance are the same for conductors B and C. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 56
  • 57.
     (ii) Unsymmetricalspacing.  Assuming balanced conditions i.e., 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶=0  Potential of 1st position, 𝑉1= 1 2𝜋𝜀0 𝑄𝐴 log𝑒 1 𝑟 + 𝑄𝐵 log𝑒 1 𝑑3 + 𝑄𝐶 log𝑒 1 𝑑2  Potential of 2nd position, 𝑉2= 1 2𝜋𝜀0 𝑄𝐴 log𝑒 1 𝑟 + 𝑄𝐵 log𝑒 1 𝑑1 + 𝑄𝐶 log𝑒 1 𝑑3  Potential of 3rd position, 𝑉3= 1 2𝜋𝜀0 𝑄𝐴 log𝑒 1 𝑟 + 𝑄𝐵 log𝑒 1 𝑑2 + 𝑄𝐶 log𝑒 1 𝑑1  Average voltage on condutor A is 𝑉𝐴 = 1 3 (𝑉1+ 𝑉2+ 𝑉3) = 1 3∗2𝜋𝜀0 𝑄𝐴 log𝑒 1 𝑟3 + (𝑄𝐵+ 𝑄𝐶) log𝑒 1 𝑑1𝑑2𝑑3 15 October 2020 D.Rajesh Asst.Prof. EEE Department 57
  • 58.
     As 𝑄𝐴+ 𝑄𝐵 + 𝑄𝐶=0, therefore 𝑄𝐵 + 𝑄𝐶= −𝑄𝐴 = 1 2∗3𝜋𝜀0 𝑄𝐴 log𝑒 1 𝑟3 + (−𝑄𝐴) log𝑒 1 𝑑1𝑑2𝑑3 = 1 3 ∗ 𝑄𝐴 2𝜋𝜀0 log𝑒 𝑑1𝑑2𝑑3 𝑟3 = 𝑄𝐴 2𝜋𝜀0 log𝑒 𝑑1𝑑2𝑑3 𝑟3 1 3 = 𝑄𝐴 2𝜋𝜀0 log𝑒 𝑑1𝑑2𝑑3 1 3 𝑟 ∴ Capacitance from conductor to neutral is 𝐶𝐴 = 𝑄𝐴 𝑉𝐴 = 𝑄𝐴 𝑄𝐴 2𝜋𝜀0 log𝑒 3 𝑑1𝑑2𝑑3 𝑟 = 2𝜋𝜀0 log𝑒 3 𝑑1𝑑2𝑑3 𝑟 15 October 2020 D.Rajesh Asst.Prof. EEE Department 58
  • 59.
    A single-phase transmissionline has two parallel conductors 3 metres apart, radius of each conductor being 1 cm. Calculate the capacitance of the line per km. Given that 𝜀0 =8·854 × 10−12 F/m. Conductor radius, r = 1 cm Spacing of conductors, d = 3 m = 300 cm capacitance of the line = 𝜋𝜀0 log𝑒 𝑑 𝑟 F/m = 𝜋∗8·854 × 10−12 log𝑒 300 1 =0.485* 10−11 F/m capacitance of the line per km= 0.485* 10−11 *1000 = 0.485* 10−8 F/km = 0.485* 10−2 μF/km 15 October 2020 D.Rajesh Asst.Prof. EEE Department 59
  • 60.
    A 3-phase overheadtransmission line has its conductors arranged at the corners of an equilateral triangle of 2 m side. Calculate the capacitance of each line conductor per km Given that diameter of each conductor is 1·25 cm. 𝜀0 =8·854 × 10−12 F/m. Conductor radius, r = 1·25/2 = 0·625 cm Spacing of conductors, d = 2 m = 200 cm Capacitance of each line conductor= 2𝜋𝜀0 log𝑒 𝑑 𝑟 F/m == 2𝜋8·854 × 10−12 log𝑒 200 0.625 F/m=0.0096* 10−9 F/m capacitance of the line per km 0.0096* 10−9 *1000 0.0096* 10−6 F/km = 0.0096 μF/km 15 October 2020 D.Rajesh Asst.Prof. EEE Department 60
  • 61.
    A 3-phase, 50Hz, 66 kV overhead line conductors are placed in a horizontal plane as shown in Fig. The conductor diameter is 1·25 cm. If the line length is 100 km, calculate (i) capacitance per phase, (ii) charging current per phase, assuming complete transposition of the line. 𝜀0 =8·854 × 10−12 F/m. Conductor radius, r = 1·25/2 = 0·625 cm The equivalent equilateral spacing is 𝑑 = 3 𝑑1𝑑2𝑑3 = 3 2 × 2 ⋅ 5 × 4 ⋅5 = = 2·82 m Conductor spacing , d = 2·82 m = 282 cm (i) Line to neutral capacitance== 2𝜋𝜀0 log𝑒 𝑑 𝑟 F/m == 2𝜋8·854 × 10−12 log𝑒 282 0.625 F/m=0.0091* 10−9 F/m capacitance of the line per km 0.0091* 10−9 *1000 0.0091* 10−6 F/km = 0.0091 μF/km ∴ Line to neutral capacitance for 100 km line is C = 0·0091 × 100 = 0·91 μF 15 October 2020 D.Rajesh Asst.Prof. EEE Department 61
  • 62.
     (ii) Chargingcurrent per phase is 𝐼𝑐 = 𝑉𝑝ℎ 𝑋𝑐 = 𝑉𝑝ℎ 1 2𝜋𝑓𝑐 = 66000 3 ∗ 2𝜋*50*0·91 μ =10.9A 15 October 2020 D.Rajesh Asst.Prof. EEE Department 62
  • 63.
    A 3-phase, 50Hz, 132 kV overhead line has conductors placed in a horizontal plane 4 m apart. Conductor diameter is 2 cm. If the line length is 100 km, calculate the charging current per phase assuming complete transposition. 𝜀0 =8·854 × 10−12 F/m. The diameter of each conductor is 2 cm so that conductor radius r =2/2 = 1 cm = 1× 10−2 m. Now 𝑑1 = AB = 4m; 𝑑2 = BC = 4 m; 𝑑3 = AC = 8 m The equivalent equilateral spacing is 𝑑 = 3 𝑑1𝑑2𝑑3 = 3 4×4×8 = 5.04 m (i) Line to neutral capacitance== 2𝜋𝜀0 log𝑒 𝑑 𝑟 F/m == 2𝜋8·854 × 10−12 log𝑒 5.04 1× 10−2 F/m=0.0085* 10−9 F/m capacitance of the line per km= 0.0085* 10−9 *1000 = 0.0085* 10−6 F/km = 0.0085 μF/km ∴ Line to neutral capacitance for 100 km line is C = 0.0085 × 100 = 0·885 μF 15 October 2020 D.Rajesh Asst.Prof. EEE Department 63
  • 64.
     Charging currentper phase is 𝐼𝑐 = 𝑉𝑝ℎ 𝑋𝑐 = 𝑉𝑝ℎ 1 2𝜋𝑓𝑐 = 132000 3 ∗ 2𝜋*50*0·885 μ =21.18A 15 October 2020 D.Rajesh Asst.Prof. EEE Department 64
  • 65.
     Definition: Proximity Effectis the phenomena of non-uniform current distribution on the surface of adjacent current carrying conductor due to the effect of another current carrying conductor in its proximity. Since in cables, the conductors are very near to each other, this effect is dominant. Whereas in overhead lines as lines are far apart when compared with cable, this effect can be neglected. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 65
  • 66.
     Cause ofProximity Effect:  The main reason for proximity effect is production of magnetic field in the surrounding of a current carrying conductor.  We know that any current carrying conductor produces magnetic field in its surrounding as per Biot Savart Law.  When this magnetic field links with the adjacent conductor, it gives rise to a circulating or eddy current in it. Because of this eddy current, the current distribution on the nearby becomes non-uniform.  This leads to additional resistance to the flow of current in the nearby conductor. Hence due to proximity effect, the net ac resistance of conductor increases. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 66
  • 67.
     Two conductorscarrying current in the same direction. The magnetic field created by conductor A will link with conductor B. Similarly, the magnetic field created by conductor B will link with conductor A.  The direction of magnetic field at any point P inside the conductor B will in going inside the plane of device.  Since current is flowing in upward direction, this means that free electrons are moving in downward direction. The direction of force on the free electrons due to the magnetic field of conductor A will be given by the cross vector product of velocity vector of electron and magnetic field vector at point P. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 67
  • 68.
     From thecross vector product, the free electrons of conductor B will be drifted toward the inner surface i.e. toward conductor A and hence the current will flow mostly in the farthest surface of conductor B.  The same reasoning when applied to conductor A, it can be seen that current in the conductor A will mostly flow in the farthest surface. This effect of re- distribution of current on the conductor surface is called the proximity effect. This is so called proximity effect as this is due to the effect of proximity (nearby) conductor. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 68
  • 69.
     when twocurrent conductors carry current in the same direction, the current gets distributed to the farthest half of both the conductor.   Similarly when two conductors carry current in opposite direction, the current in conductors gets concentrated toward the inner facing surface of both the conductors 15 October 2020 D.Rajesh Asst.Prof. EEE Department 69
  • 70.
     Factors affectingProximity Effect:  The proximity effect mainly depends on the factors like conductor material, conductor diameter, frequency and conductor structure. The factors are explained below in details  Frequency – The proximity increases with the increases in the frequency. This effect is not observed in case of DC. In case of DC, as the current is uniformly distributed over the entire cross-section of conductor, the magnetic force on the free electrons inside conductor is less prominent. Hence this effect is not observed.  Diameter – The proximity effect increases with the increase in the conductor.  Structure – This effect is more on the solid conductor as compared to the stranded conductor (i.e., ASCR) because the surface area of the stranded conductor is smaller than the solid conductor.  Material – If the material is made up of high ferromagnetic material then the proximity effect is more on their surface. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 70
  • 71.
     How toreduce Proximity Effect?  The proximity effect can be reduced by using the ACSR (Aluminum Core Steel Reinforced) conductor. In ACSR conductor the steel is placed at the centre of the conductor and the aluminium conductor is positioned around steel wire.  The steel increased the strength of the conductor but reduced the surface area of the conductor. Thus, the current flow mostly in the outer layer of the conductor and no current is carried in the centre of the conductor. Thus, reduced the proximity effect on the conductor. 15 October 2020 D.Rajesh Asst.Prof. EEE Department 71