Chapter One: Introduction

Chapter One: Introduction 1.1 Objectives The objectives of this project are: To understand the working and design of electrical power transmission lines. To study the parameters of transmission lines. To develop a software package to be used for the line parameter determination for standard operating voltage levels and line configuration in use by KPLC to be used as an input. 1.2 Transmission Lines Transmission lines are mainly used to convey electrical power between two points. They are usually a set of wires made of good electrical conductors such as copper or aluminium. In electrical power systems design, operation and expansion it is important to know the electrical and physical characteristics of conductors used in the construction of aerial distribution and transmission lines. Excellent insulation is very important between the conductors. With time, copper has been replaced by aluminium conductors because it is less expensive and has a lighter weight. 1.3 Types of Conductors The choice of a conductor greatly influences the performance of a transmission line. This depends mostly on i) The amount of power to be transmitted and ii) The frequencies involved In the transmission of electrical power different types of conductors are used depending on the operating voltage and line configuration. The most commonly used type of conductors are as follows: Stranded copper conductors Hollow copper conductors 1

ACSR aluminium conductor, steel-reinforced AAC all-aluminium conductor AAAC all-aluminium-alloy conductors ACAR aluminium conductor, alloy-reinforced Stranded conductors are more flexible and easier to handle than solid conductors. In the ACSR conductor, aluminium strands are wound about a core of stranded steel. ACAR has layers of electrical-conductor grade aluminium wound about a core of higher-strength aluminium. The number of strands depends on the number of layers and if all strands are of the same diameter. Table 1 in the appendix shows the different electrical characteristics of bare aluminium conductors steel-reinforced (ACSR). 1.4 Standard operating voltages Over the years, in the transmission of alternating-current power the operating voltages have greatly increased at a rapid rate. Starting from 1886 when a line was built at Cerchi, Italy, to transmit 150 hp 17 miles at 2000 volts. In the United States the voltages have increased from 3300V in 1890, 100kV in 1907, 150kV in 1913, 220kV in1923, 244kV in 1926, 287kV in 1936, 345kV in 1953, 500kV in 1965 and 765kV in 1969. The choice of the operating voltage also greatly determines the type of conductor to be used. Different countries use different operating voltages but which must lie within the range of the standard operating voltages. The most commonly used in Kenya are 11kV, 33kV, 66kV, 110kV, 132kV, 166kV and 220kV. This also applies for the frequency, whereby some use 60Hertz and others use 50 Hertz, which is used by KPLC. But in cases of very long transmission lines frequencies other than the two mentioned above can be used since they require a lower frequency. This is mainly because as the frequency is decreased, the capacitive reactance increases and the inductive reactance decreases. This is the only condition under which a different frequency can be used. 2

1.5 Line Parameters Electrical power can be transmitted through overhead lines or underground cables. Underground cables are mostly used in urban congested areas where there is more human interference.they have low maintenance costs, less voltage drops, are not affected by adverse weather conditions like the overhead lines. As compared to the overhead lines, they are very expensive to install over long distances and proper insulation is required at high voltages. This project mainly dwells on the parameters that greatly influence the standard operating voltage and line configuration to be used in the design and operation of a transmission line. These electrical parameters include resistance, inductance, capacitance and conductance. Conductance of overhead lines is assumed to be zero since it accounts for the leakage current, which is negligible. Hence, the resistance, inductance and capacitance of a transmission line have been discussed in this project. These parameters have been calculated taking into consideration the line configuration which includes, for instance, the length, cross sectional area, spacing between conductors of the line and so on. 3

Chapter Two: The Transmission Lines 2.1 Introduction A power source is placed on one end of the transmission line, while a load is placed on the other end. The line should then guide power to the load with minimum losses or reflections. The source end of the line is referred to as the sending-end, while the load-end the receiving end. The line usually has uniform electrical characteristics throughout and is normally not less than a quarter wavelength long. Examples of transmission lines are: two wire transmission lines, the shielded pair, parallel strip lines and coaxial lines. When a voltage Vsin(wt) is placed on the line, charges are set in motion and a current Isin(wt) flows towards the load. The effect on the sending end is not felt instantaneously at the load but tracks a finite speed along the line. At every point on the line, an electric field exists due to the presence of the charges while a magnetic field is created by the current. As the resistivity of any material making the transmission line will be non-zero, a continuous power loss occurs along the line. A power leakage is also experienced throughout if the line conductors are separated by a dielectric with some finite conductivity. These characteristics make it possible to predict the state at every point on the line at any given time. They are uniquely described by an electric network, comprising of resistance, inductance, capacitance and conductance that are distributed throughout the length of the line. The energy stored due to the magnetic field is represented by inductance per unit length and the power stored due to the electric field by a capacitance per unit length. The power lost as a result of finite resistance is modelled through a resistance per unit length and lastly the power leakage is represented by shunt conductance per unit length. Conductance between conductors or between conductors and the ground accounts for the leakage current at the insulators of overhead lines and through the insulation of cables. Since leakage at insulators of overhead lines is negligible, the conductance between conductors of an overhead line is assumed to be zero. The resistance and inductance uniformly distributed along the line form the series impedance. The conductance and capacitance existing between conductors of a single-phase line or from a conductor to neutral of a three phase line form the shunt admittance. Although the resistance, inductance, and capacitance are distributed, the equivalent circuit of a line is made up of lumped parameters. 4

2.1.1 Transmission Line Parameters 2.1.2 Resistance It is the opposition of line conductors to current flow. It is the main cause of power loss in a transmission line. Unless specifically qualified, it means effective resistance. Resistance due to d.c. current is given by R dc l = ρ A (2.0) where ρ is the resistivity at 20 o C l is the length of the conductor A is the cross sectional area of the conductor The international standard of conductivity is that of annealed copper. At 20 0 C for hard-drawn copper ρ is 1.77 10-8 Ω.m.For aluminium at 20 0 C ρ is 2.83 10-8 Ω.m. Because of skin effect, the d.c. resistance is different from ac resistance. The ac resistance is referred to as effective resistance, and is found from power loss in the conductor. The effective resistance is equal to the dc resistance of the conductor only if the distribution of current throughout the conductor is uniform. R = powerloss 2 I (2.1) where power is in watts I 2 is the rms current in the conductor in amperes The dc resistance of stranded conductors is greater than the value computed by equation (2.0) because spiralling of the strands makes them longer than the conductor itself. For each mile of conductor the current in all strands except the one in the centre flows in more than a mile of wire. The increased resistance due to spiralling is estimated as 1 % for three-strand conductors and 2 % for concentrically stranded conductors. 5

The variation of resistance of metallic with temperature is practically linear over the normal temperature range of operation. If temperature is plotted on the vertical axis and resistance on the horizontal axis as shown on Fig. 2.1, extension of the straight-line portion of the graph provides a convenient method of correcting resistance for changes in temperature. The point of intersection of the extended line with the temperature axis at zero resistance is a constant of the material. Figure 2.1 Graph of Resistance as a function of Temperature From the geometry of Fig. 2.1, Where R 1 and R 2 are resistances of the conductor at temperatures t 1 and t 2, respectively, in degrees Celsius and T is the constant determined from the graph. Values of the constant T are as follows: T = 234.5 T = 241 T = 228 for annealed copper of 100% conductivity for hard-drawn copper of 97.3% of conductivity for hard-drawn aluminium of 61% conductivity 6

2.1.3 Skin effect Uniform distribution of current throughout the cross section of a conductor exists only for direct current. As the frequency of alternating current increases, the nonuniformity of distribution becomes more pronounced. An increase in frequency causes nonuniform current density. This phenomenon is called skin effect. It is the tendency of alternating current to concentrate near the surface of a conductor. In a circular conductor the current density usually increases from the interior toward the surface. For conductors of sufficiently large radius, however, a current density oscillatory with respect to radial distance from the center may result. Due to this 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 cause of skin effect can be easily explained. A solid conductor may be thought to be consisting of a large number of strands each carrying a small part of the current. The inductance of each strand will vary according to its position.thus, the strands near the center are surrounded by a greater magnetic flux and hence have larger inductance than that near the surface. The high reactance of inner strands causes the alternating current to flow near the surface of conductor. This crowding of current near the surface is the skin effect. 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). 7

2.1.4 Inductance It is the flux linkages per unit current. When 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. Inductance, L = Henry (2.4) Where Ψ = flux linkages in Weber-turns I = current in amperes The inductance is also uniformly distributed along the length of the line as shown in Fig. 2.2. Figure 2.2 Circuit showing uniform distribution of resistance and inductance 2.1.5 Flux linkages due to a single current carrying conductor Consider a long straight cylindrical conductor of radius r meters and carrying I amperes (r.m.s). This current will set up magnetic field. The magnetic lines of force will exist inside the conductor as well as outside the conductor. Both these fluxes will contribute to the inductance of the conductor. 2.1.6 Flux linkages due to internal flux The magnetic field intensity at a point x meters from the center is given by; H x = (2.5) Assuming a uniform current density, 8

I x = I = I (2.6) Therefore, H x = I = I AT/m (2.7) If µ = (µ r µ 0 ) is the permeability of the conductor, then flux density at the considered point is given by; B x = µ r µ 0 H x wb/m 2 (2.8) = I = dx Weber (2.9) Now, flux dϕ through a cylindrical shell of radial thickness dx and axial length 1 m is given by; dϕ = B x 1 dx = dx Weber (2.10) This flux links with current I x only. Therefore, flux linkages per meter length of the conductor is dψ = dϕ = dx Weber-turns (2.11) Total flux linkages from center upto the conductor surface is Ψ int = dx (2.12) = Weber-turns per meter length (2.13) For a relative permeability of 1, µ = 4π 10-7 H/m, and Ψ int = 10-7 Wbt/m (2.14) L int = 10-7 H/m (2.15) 9

2.1.7 Flux linkages due to external flux The external flux extends from the surface of the conductor to infinity. Let us derive an expression for the flux linkages of an isolated conductor due only to that portion of the external flux which lies between two points distant D 1 and D 2 meters from the center of the conductor. Referring to Fig. 2.3, the field intensity at a distance x meters (from the center) outside the conductor is given by; Figure 2.3 A conductor and external points P 1 and P 2 H x = AT/ m (2.16) Flux density, B x = = wb/m 2 (2.17) Now, flux dϕ through a cylindrical shell of thickness dx and axial length 1 meter is dϕ = B x dx = dx webers (2.18) The flux dϕ links all the current in the conductor once and only once Therefore, flux linkages, dψ = dϕ = B x dx = dx Weber-turns (2.19) Total flux linkages of the conductor between P 1 and P 2 are obtained by integrating dψ from x = D 1 to x = D 2 to get, 10

Ψ ext = Weber-turns (2.20) Therefore, overall flux linkages, Ψ = Ψ int + Ψ ext = + (2.21) Therefore, Ψ = wb-turns/m length (2.22) Ψ ext = ln wb-turns/m length (2.23) The inductance due only to the flux included between P 1 and P 2 is L ext = 2 10-7 ln H/m (2.24) 2.1.8 Inductance of a single-phase two-wire line Figure 2.4 shows a circuit having two conductors of radii r1 and r2.one conductor is the return circuit for the other. Figure 2.4 Conductors of different radii The inductance of the circuit due to current in conductor 1 is determined by equation (2.24), with the distance D between conductors 1 and 2 substituted for D 2 and the radius r 1 of conductor 1 substituted for D 1. For external flux only L 1,ext = 2 10-7 ln H/m (2.25) For internal flux only 11

L 1,int = 10-7 H/m (2.26) The total inductance of the circuit due to the current in conductor 1 only is L 1 = H/m (2.27) The expression for inductance may be put in a more concise form by factoring equation (2.27) and by noting that ln = whence L 1 = 2 10-7 (2.28) Upon combining terms, we obtain L 1 = 2 10-7 ln (2.29) If we substitute for, Where = 0.7788 L 1 7 D = 2 10 ln r1 ' (2.30) Similarly, the inductance due to current in conductor 2 is L 2 7 D = 2 10 ln r ' 2 (2.31) The total inductance of the line is then L L If T T = L + L 1 = 4 10 2 7 = 2 10 7 2 D ln r1 ' r2 ' D D ln + ln = 2 10 r1 ' r2 ' 1/ 2 = 4 10 7 ln D r ' r ', the total inductance reduces to 1 7 2 2 D ln = 2 10 r ' r ' 1 2 7 2 1 D 2 ln 2 r ' r ' 1 2 (2.32) 12

L = 4 10-7 ln H/m (2.33) 2.1.9 Inductance of three-phase lines with equilateral spacing Figure 2.5 shows the conductors of a three phase line spaced at the corners of an equilateral triangle. If we assume balanced that there is no neutral wire, or if we assume balanced threephase phasor currents, I a + I b + I c = 0 we have, Ψ a = 2 10-7 Wbt/m (2.34) Since I a = - (I b + I c ), equation (2.34) becomes Ψ a = 2 10-7 Wbt/m (2.35) Ψ a = 2 10-7 Wbt/m (2.36) And L a = 2 10-7 H/m (2.37) Because of symmetry, the inductances of conductors b and c are the same as the inductance of conductor a. Since each phase consists of only one conductor, equation (2.37) gives the inductance per phase of the three-phase line. Figure 2.5 Cross-sectional view of the equilaterally spaced conductors of a three-phase line. 13

2.1.10 Inductance of three-phase lines with unsymmetrical spacing Figure 2.6 Transposition cycle A different inductance in each phase results in an unbalanced circuit. Balance of the three phases can be restored by exchanging 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. This is called transposition as shown in Fig. 2.6.The phasor expression for the flux linkages of a in position 1 when b is in position 2 and c in position 3 is as follows; Ψ a1 = 2 10-7 Wbt/m (2.38) With a in position 2 when b is in position 3 and c in position 1, Ψ a2 = 2 10-7 Wbt/m (2.39) With a in position 3 when b is in position 1 and c in position 2, Ψ a3 = 2 10-7 Wbt/m (2.40) The average value of the flux linkages of a is Ψ a = (2.41) 14

And with the restriction that I a = - (I b + I c ) the average inductance per phase is L a = 2 10-7 H/m (2.42) Where = which is the geometric mean of the three distances of the unsymmetrical line and is the GMR of the conductor. 2.1.11 Bundled conductors These are conductors which have one or more conductors per phase in close proximity compared with the spacing between phases. The bundle consists of two, three, or four conductors. This is done to reduce the effect of corona with its resultant power loss and particularly its interference with communications. It is excessive if the circuit has only one conductor per phase and is prevalent at extra-high voltage. The average inductance per phase is L a = 2 10-7 H/m (2.43) Where is the GMR of the individual conductors composing the bundle. For a two-strand bundle = = (2.44) For a three-strand bundle = = (2.45) For a four-strand bundle = = 1.09 (2.46) In computing D eq, the distance from the center of one bundle to the center of another bundle is sufficiently enough. 15

2.1.12 Parallel-circuit three-phase lines Figure 2.7 Typical arrangement of conductors of a parallel-circuit three-phase line If transposition is assumed, conductors a and a are in parallel to compose phase a. Similarly for phases b and c.we assume that a and a take the positions of b and b and then of c and c as those conductors are rotated similarly in the transposition cycle. To calculate D eq the GMD method requires that we use D p ab, D p bc, and D p ca, where the superscript indicates that these quantities are themselves GMD values and where D p ab means the GMD between the conductors of phase and those of phase b. The D s of equation (2.42) is replaced by D p s, which is the geometric mean of the GMR values of the two conductors occupying first the positions a and a, then the positions of b and b, and finally the positions of c and c. The inductance per phase is L = 2 10-7 H/m (2.47) 2.1.13 Capacitance This is as a result of the potential difference between the conductors; it causes them to be charged in the same manner as the plates of a capacitor when there is a potential difference between them. The capacitance between conductors is the charge per unit of potential difference. For short transmission lines, it is slight and is usually negligible but for longer lines of higher voltages it becomes increasingly important. Consider an isolated conductor carrying a uniformly distributed charge as shown in Fig. 2.8.The electric flux density is D = C/m 2 (2.48) 16

Where q is the charge on the conductor in coulombs per meter of length and x is the distance in meters from the conductor to the point where the electric flux density is computed. Figure 2.8 Lines of electric flux originating on the positive charges uniformly distributed over the surface of an isolated cylindrical conductor. The electric flux intensity is V/m (2.49) 2.1.14 The potential difference between two points due to a charge Consider a long straight wire carrying a positive charge of q C/m, as shown in Fig. 2.9.The instantaneous voltage drop between P 1 andp 2 is (2.50) v 12 = = = ln V where q is the instantaneous charge on the wire in coulombs per meter of length. 17

Figure 2.9 Path of integration between two points external to a cylindrical conductor having a uniformly distributed positive charge. 2.1.15 Capacitance of a two-wire line C = F/m (2.51) Where q is the charge on the line in coulombs per meter and v is the potential difference between the conductors in volts. The voltage v ab between the two conductors of the two-wire line shown in Fig. 2.9 is V ab = ln + ln V (2.52) V ab = V (2.53) Or by combining the logarithmic terms, V ab = ln V (2.54) The capacitance between conductors is 18

C ab = = F/m (2.55) If r a = r b = r, C ab = F/m (2.56) Figure 2.10 Cross section of a parallel-wire line Equation (2.56) gives the capacitance between the conductors of a two-wire line.for cases where the line is supplied by a transformer having a grounded center tap, the potential difference between the two conductors and the capacitance to ground, or capacitance to neutral, is the charge on a conductor per unit of potential difference between the conductor and the ground. Thus the capacitance to neutral for the two-wire line is twice the line-to-line capacitance. This is given by; C n = C an = C ab = F/m to neutral (2.57) The capacitive reactance and susceptance can also be calculated by using, X c = Ω.m to neutral (2.58) B c = Ʊ/m to neutral (2.59) 2.1.16 Capacitance of a three-phase line with equilateral spacing The three identical conductors of radius r of a three-phase line with equilateral spacing are as shown in Fig. (2.11) 19

Figure 2.11 Cross section of a three-phase line with equilateral spacing The voltage V ab of the three phase-line due only to the charges on conductors a and b is V ab = ln + ln V (2.60) And that due only to the charge q c is V ab = ln V (2.61) Which is zero since q c equidistant from a and b. Similarly, V ac = ln + ln V (2.62) V ab + V ac = ln V (2.63) Figure 2.12 Phasor diagram of the balanced voltages of a three-phase line From Fig. 2.12 we obtain the following relations between the line voltages V ab and V ac and the voltage V an from line a to the neutral of the three-phase circuit: 20

V ab = V an (0.866 + j0.5) (2.64) V ac = - V ca = V an (0.866 - j0.5) (2.65) V ab + V ac =3 V an (2.66) Substituting 3V an for V ab + V ac in equation (2.63), we obtain V an = ln V (2.67) Hence, C n = = F/m to neutral (2.68) 2.1.17 Capacitance of a three-phase line with unsymmetrical spacing Figure 2.13 Cross section of a three-phase line with unsymmetrical spacing For the line shown in Fig. 2.13 three equations are found for Vab for the three different parts of the transposition cycle. With phase a in position 1, b in position 2, and c in position 3, V ab = V (2.69) With a in position2, b in position3, and c in position 1, V ab = V (2.70) With a in position 3, b in position 1, and c in position 2, V ab = V (2.71) 21

Assuming that the charge per unit length on a conductor is the same in every part of the transposition cycle, the average voltage between conductors a and b, based on this assumption is, V ab = (2.72) = V (2.73) Where = (2.74) Similarly, the average voltage drop from conductor a to conductor c is V ac = V (2.75) Applying equation (2.66) to find the voltage to neutral, we have 3Van = Vab + Vac= V (2.76) Since q a + q b + q c = 0 in a balanced three-phase circuit, (2.77) 3V an = ln V And C n = = F/m to neutral (2.78) 22

2.1.18 Effect of earth on the capacitance of three-phase transmission lines Figure 2.14 Three-phase line and its image Earth affects the capacitance of a transmission line because its presence alters the electric field of the line. Consider a circuit consisting of a single overhead conductor with a return path through the earth as shown in Fig. 2.14. In charging the conductor, charges come from the earth to reside on the conductor, and a potential difference exists between the conductor and earth. The earth has a charge equal in magnitude to that on the conductor but of opposite sign. Applying the method of images to calculate the capacitance we get, V ab = (2.79) 23

And C n = F/m to neutral (2.80) 2.1.19 Bundled conductors Figure 2.15 shows a bundled-conductor line for which we can write an equation for the voltage as follows; Figure 2.15 Cross section of a bundled-conductor three-phase line V ab = (2.81) V ab = (2.82) C n = F/m to neutral (2.83) The is the same as for a two conductor bundle except that r has replaced.this leads us to the very important conclusion that a modified GMD method applies to the calculation of capacitance of a bundled-conductor three-phase line having two conductors per bundle. The modification is that we are using outside radius in place of the GMR of a single conductor. If we let stand for the modified GMR to be used in capacitance calculations, we have, C n = F/m to neutral (2.84) 24

For a two-strand bundle = = (2.85) For a three-strand bundle = = (2.86) For a four-strand bundle = = 1.09 (2.87) 2.1.20 Parallel-circuit three-phase lines The capacitance is, C n = F/m to neutral (2.88) Where is the outside radius of the conductor instead of the GMR. 25

Chapter Three: Representation of Lines 3.1 Introduction The transmission line parameters discussed in the first chapter greatly influence the voltage drop, line losses and efficiency of transmission. For example, the voltage drop in the line depends upon the line parameters. Similarly, the resistance of the transmission line conductors is the most important cause of power loss in the line and determines the efficiency. This chapter deals with the analytical expressions for the equivalent circuits for transmission lines, starting with the short lines, followed by the medium length lines and finally the long lines. These transmission line parameters are recognized as being uniformly distributed along the line. 3.2 Equivalent Circuits for Transmission Lines 3.2.1 Short Transmission Lines For all types of problems it is usually safe to apply the short transmission line analysis to lines up to 50 km in length or all lines of voltages less than about 20 kv. For short lines, shunt capacitance is so small that it can be omitted entirely with little loss of accuracy, and we need to consider only the series resistance R and the series inductance L for the total length of the line. The equivalent circuit of a short transmission line is shown in Fig. 3.1 where: I = load current R = loop resistance which is the resistance of both conductors X L = loop reactance V R = receiving end voltage Cos ϕ R = receiving end power factor (lagging) V S = sending end voltage Cos ϕ S = sending end power factor 26

Figure 3.1 Equivalent circuit of a short transmission line From the right angled triangle ODC, we get, (OC) 2 = (OD) 2 + (DC) 2 (3.1) 2 V S = (OE + ED) 2 + (DB + BC) 2 (3.2) = (V R cos ϕ R + IR) 2 + (V R sin ϕ R + IX L ) 2 (3.3) V S = (3.4) Percentage Voltage regulation = 100 (3.5) Sending end power factor, cos ϕ R = = (3.6) Power delivered = V R I R cos ϕ R (3.7) Line losses = I 2 R (3.8) Power sent out = V R I R cos ϕ R + I 2 R (3.9) Percentage Transmission efficiency = 100 = (3.10) 27

Taking V R as the reference phasor, the phasor diagram can be drawn.it is clear that V S is the phasor sum of V R and IZ. V S = V R + IZ (3.11) 3.2.2 Medium Transmission Lines This is an overhead transmission line whose length is about 50 to 150 km and the line voltage is moderately high, between 20kV and 100kV.Here the capacitance effects are taken into account due to the sufficient length and voltage. The capacitance is uniformly distributed over the entire length of the line. To make the calculations simple, the line capacitance is assumed to be lumped or concentrated in form of capacitors shunted across the line at one or more points and this gives more accurate results. The most commonly used methods for the solution of medium length lines are: i) End condenser method ii) Nominal T method iii) Nominal π method Only the last two methods will be considered for this case. 3.2.3 Nominal T method Here, the whole line capacitance is assumed to be concentrated at the middle point of the line and half the line resistance and reactance are lumped on as shown in Fig.3.4.Therefore,in this arrangement, full charging current flows over half the line. The phasor diagram for the circuit is shown in Fig. 3.5 taking the receiving end voltage V R as the reference phasor. Figure 3.2 Equivalent circuit of one phase of three phase transmission line 28

Let I R = load current per phase; R = resistance per phase; X L = inductive reactance per phase; V R = receiving end voltage per phase; Cos ϕ R = receiving end power factor (lagging); V S = sending end voltage per phase; V 1 = voltage across capacitor C; C = capacitance per phase; Receiving end voltage, V R = V R + j0 (3.12) Load current, I R = I R (cos ϕ R j sin ϕ R ) (3.13) Voltage across C, V 1 = V R + (3.14) = V R + I R (cos ϕ R j sin ϕ R ) ( ) (3.15) Capacitive current, I C = j w C V 1 = j 2 π f C V 1 (3.16) Sending end current, I S = I R + I C (3.17) Sending end voltage, V S = V 1 + I S = V 1 +I S (3.18) 3.2.4 Nominal method In this method, capacitance of each conductor, that is, line to neutral is divided into two halves; one half being lumped at the sending end and the other half at the receiving end as shown in Fig. 3.3.It is obvious that capacitance at the sending end gas no effect on the line drop but its charging current must be added into line current in order to obtain the total sending end current. 29

Figure 3.3 Equivalent circuit of the transmission line Let I R = load current per phase R = resistance per phase X L = inductive reactance per phase C = capacitance per phase Cos ϕ R = receiving end power factor (lagging) V S = sending end voltage per phase V R = V R + j0 (3.19) Load current, I R = I R (cos ϕ R j sin ϕ R ) (3.20) Charging current at load end is I C1 = j w V R = j π f C V R (3.21) Line current, I L = I R + I C1 (3.22) Sending end voltage, V S = V R + I L Z = V R + I L (R + j X L ) (3.23) Charging current at the sending end is I C2 = j w V S = j π f C V S (3.24) Sending end current, I S = I L + I C2 (3.25) 30

3.2.5 Long Transmission Line When the length of an overhead transmission line is about 150 km and line voltage is very high, greater than 100kV, it is considered as a long transmission line. The line parameters are considered uniformly distributed over the whole length of the line. A more convenient form of the equations for computing current and voltage of a long transmission line is found by introducing hyperbolic functions. The equivalent circuit diagram of a long transmission line is shown in Fig. 3.4. Figure 3.4 Equivalent circuit of one phase and neutral connection of a three phase line with impedance and shunt admittance of the line uniformly distributed. Consider a small element in the line of length dx situated at a distance x from the receiving end Let z = series impedance of the line per unit length y = shunt admittance of the line per unit length V = voltage at the end of element towards receiving end V + dv = voltage at the end of element towards sending end I + di = current entering the element dx I = current leaving the element dx Then for the small element dx, z dx = series impedance 31

y dx = shunt admittance Obviously, dv = I z dx (3.26) or = I z (3.27) Now, the current entering the element is I + di whereas the current leaving the element is I. The difference in the currents flows through shunt admittance of the element. That is, di = Current through shunt admittance of element = V y dx or = V y (3.28) Differentiating equation (3.27) with respect to x, we get, = z = z (V y) (3.29) or =y z V (3.30) The solution of this differential equation is V = cosh + sinh (3.31) Differentiating equation (3.31) with respect to x, we have = sinh + cosh (3.32) = I z (3.33) I z = sinh + cosh (3.34) I= { sinh + cosh } (3.35) 32

Equations (3.31) and (3.35) give the expression for V and I in the form of unknown constants k 1 and k 2. The values of k 1 and k 2 can be found by applying end conditions as under: At x = 0, V = V R and I = I R Putting these values in equation (3.31), we have, V R = k 1 cosh 0 + k 2 sinh 0 = k 1 + 0 (3.36) V R = k 1 (3.37) Similarly, putting x = 0, V = V R and I = I R in equation (3.35), we have, I R = { sinh + cosh } = {0 +k 2 } (3.38) k 2 = I R (3.39) Substituting for the values of k 1 and k 2 in equations (3.31) and (3.35), we get, V =V R cosh + I R sinh (3.40) and I= V R sinh + I R cosh (3.41) The sending end voltage (V S ) and sending end current (I S ) are obtained by putting x = l in the above equations as follows, V S = V R cosh + I R sinh (3.42) and I S = V R sinh + I R cosh (3.43) Now, l = = (3.44) and = = (3.45) 33

where Y = total shunt admittance of the line Z = total series impedance of the line Therefore, expressions for V S and I S become: V S = V R cosh + I R sinh (3.46) and I S = V R sinh + I R cosh (3.47) It is helpful to expand hyperbolic sine and cosine in terms of their power series. Cosh = (3.48) Sinh = (3.49) 3.3 Generalized Circuit Constants of a Transmission Line In any four terminal network, the input voltage and input current can be expressed in terms of output voltage and output current. Incidentally, a transmission line is a four-terminal network ; two input terminals where power enters the network and two output terminals where power leaves the network. Therefore, the input voltage (V S ) and input current (I S ) of a three phase transmission line can be expressed as: V S = AV R + BI R (3.50) I S = CV R +DI R (3.51) Where V S = sending end voltage per phase I S = sending end current V R = receiving end voltage per phase I R = receiving end current 34

and A, B, C and D (generally complex numbers) are the constants known as generalised circuit constants of the transmission line. The values of these constants depend upon the particular method adopted for solving a transmission line. Once the values of these constants are known, performance calculations of the line can be easily worked out. The following points may be kept in mind: (i) The constants A, B, C and D are generally complex numbers. (ii) The constants A and D are dimensionless whereas the dimensions of B and C are ohms and Siemens respectively (iii) For a given transmission line, A = D (3.52) (iv)for a given transmission line, AD BC = 1 (3.53) 3.4 Determination of Generalized Circuit Constants for different types of Transmission Line 3.4.1 Short lines Here, the effect of capacitance is neglected. Therefore, the line is considered to have series impedance. Fig. 3.1 shows the circuit of a three phase transmission line on a single phase basis. Here, I S = I R (3.54) And V S =V R +I R Z (3.55) Comparing these with equations (3.50) and (3.51), we have, A = 1 ; (3.56) B = Z ; (3.57) C = 0 ; (3.58) and D = 1 (3.59) 35

Incidentally; A = D (3.60) And AD BC = (1 * 1) (Z * 0) = 1 (3.61) 3.4.2 Medium lines Nominal T method In this method, the whole line to neutral capacitance is assumed to be concentrated at the middle point of the line and half the resistance and reactance are lumped on either side as shown in Fig.3.2. Here, V S = V1 + I S (3.62) And V 1 = V R + I R (3.63) Now I C = I S I R (3.64) = V 1 Y where Y = shunt admittance (3.65) = Y (3.66) I S = I R + Y (3.67) = YV R + I R (3.68) Substituting the value of V 1 in equation (3.62), we get, V S = V R + + (3.69) Substituting the value of I S, we get, V S = V R + I R (3.70) Comparing equations (3.70) and (3.68) with those of (3.50) and (3.51),we have, A = D = ; (3.71) 36

B = Z ; (3.72) C = Y; (3.73) Incidentally : AD BC = Y (3.74) = 1 + + Y Z Z Y - = 1 (3.75) 3.4.3 Medium lines Nominal method In this method, line-to-neutral capacitance is divided into two halve; one half being concentrated at the load end and the other half at the sending as Fig. 3.3. Here, Z = R + j X L = series impedance/phase (3.76) Y = j w C = shunt admittance (3.77) I S = I L +I C2 (3.78) Or I S = I L + V S (3.79) Also I L = I R + I C1 = I R + V R (3.80) Now V S = V R + I L Z = V R + (Putting the value of I L ) (3.81) Therefore, V S = V R + I R Z (3.82) Also I S = I L + V S = + V S (3.83) (Putting the value of I L ) Putting the value of V S from equation (3.82), we get, I S = I R + V R + (3.84) = I R + + + + (3.85) 37

= + Y (3.86) Comparing equations (3.82) and (3.86) with those of (3.50) and (3.51), we get, A = D = (3.87) B = Z; (3.88) C = Y ; (3.89) Also AD BC = - ZY (3.90) = 1 + + YZ ZY - = 1 (3.91) 3.4.4 Long lines Hyperbolic form of the equation Using this method, the sending end voltage and current of a long transmission line are given by: V S = V R cosh + I R sinh (3.92) I S = V R sinh + I R cosh (3.93) Comparing these equations with those of (3.50) and (3.51), we get, A = D = cosh ; (3.94) B = sinh ; (3.95) C = sinh ; (3.96) 38

Incidentally AD BC = cosh - sinh * sinh (3.97) = cosh 2 - sinh 2 = 1 (3.98) 39

Chapter Four: Software Design, Development and Testing 4.1 Definition A software package to be used for the line parameter determination for standard operating voltage levels and line configuration in use by KPLC to be used as an input can be visualised as one that can calculate the parameters themselves and the generalised constants. Then these generalised constants can be used as inputs in calculating the voltage profile of the short, medium and long transmission lines. The equations for calculating the parameters and the generalised constants have been derived in the previous chapters. For the different configurations as seen from the equations, the software will require the user to enter the resistivity, length, area, distance between conductors, GMR, GMD, radius of the conductors and distance above the ground in the case of evaluating the parameters. Some values of the GMR and GMD for standard conductors are usually provided by the manufacturers in form of tables. An example for bare aluminium conductor steel-reinforced is included in the appendices. For the determination of the generalised constants, the resistance, reactance and shunt admittance are entered by the user. As discussed earlier the conductance is negligible therefore the capacitance will be entered as the shunt admittance. 4.2 Resistance Here, the dc resistance is evaluated for two equations, (i) R = To get the value of the resistance, the software will require the user to enter the following values of the conductor Resistivity Length Area (ii) 40

The following values will be entered to get R 2 Resistance at temperature 1 Constant,T of the conductor Temperatures t 1 and t 2 4.3 Inductance For the inductance, different configurations are used like the bundled, three phase, single phase conductors and so on. In this cases mostly the user will be required to enter the following in order for the software to calculate the inductance. Distance between the conductors Radius of the conductors GMD of the conductors GMR of the conductors 4.4 Capacitance This is similar to the inductance hence for the evaluation of the capacitance, the user will be required to enter the values of, Distance between the conductors Radius of the conductors GMD of the conductors GMR of the conductors and Distance above the ground 4.5 The Short transmission line From previous discussions, the shunt capacitance is negligible for this type of line. The equations for determination of the constants of this line are given by: A = 1; 41

B = Z; C = 0; and D = 1 Therefore the user will enter, Total resistance per phase Total reactance per phase Length of the line 4.6 The medium lines These lines are between 50 and 150 km long. The equations for the medium lines are, A = D = ; B = Z ; C = Y; The following values are to be entered by the user for the software to calculate the parameters; Total resistance per phase Total reactance per phase Length of the line Total shunt admittance per phase 4.7 The Long lines Basically these are lines above 150km and the equation describing the constants are as follows; A = D = cosh ; 42

B = sinh ; C = sinh ; The software will require the following values for it to calculate the constants, Total resistance per phase Total reactance per phase Length of the line Total shunt admittance per phase 4.8 The software package The framework chosen for the development of this software was Visual studio.net using the C # programming language. This due to the fact that I was able to calculate the parameters and it provide a graphical user interface where the values were to be entered by the user. There are other high level programming languages that can also be used apart from this one. The code for the software is indicated in appendix A. 4.9 The program flowchart The flow chart on the next page illustrates how the program works. The software is started and the user chooses the category depending on what is to be calculated. The user enters the inputs and the calculated value is displayed. 43

Start Category Resistance Capacitance Inductance Constants No No No More than 150km User Inputs User Inputs User Inputs Less than 50km Length? From 50-150km All All All user user user Inputs read? Inputs read? Inputs read? Short line Medium line Long line Yes Yes Yes Calculate Resistance Calculate Capacitance Calculate Inductance Line Parameters Display END Calculate constants Display 44

4.10 Testing and Results The software was tested for the resistance, capacitance, inductance and the generalized constants using the following examples which were extracted from Elements of Power System Analysis and Principles of Power System books to illustrate the results of the software compared with the worked out theoretical examples. Example 1 Resistance Test Tables of electrical characteristics of all-aluminium Marigold stranded conductor list a dc resistance of 0.01558 Ω per 1000 ft at 20 0 C.Verify the resistance and get the dc resistance at 50 0 C.The conductor has 61 strands and its size is 1,113,000 cmil. Solution At 20 0 C with an increase of 2% for spiralling, the dc resistance is given by; R dc l = ρ A R = 1.02 = 0.05118 Ω per km Taking ρ = 2.83 10-11 Ω.km l = 1km A = 5.6396 10-10 km 2 The value given by the software is R = 0.05119 Ω per km This results into a % error = 100 = 0.01954 % Getting the dc resistance at 50 0 C we use, 45

R 0 = 0.05118 = 0.05737 Ω per km The value given by the software is R 0 = 0.05738 Ω per km This results into a % error = = 0.01743% Example 2 Inductance of a single phase two wire line test A single phase line has two parallel conductors 2 meters apart. The diameter of each conductor is 1.2cm.Calculate the inductance per km of the line. Solution Spacing of conductors, d =2m Radius of conductor r = 0.006m r = 0.006 x 0.7788 Inductance per km = 4 10-7 {ln(d/r )}x1000 =4 x 10-7 {ln (2/(0.006x0.7788)}) x1000= 2.42365x 10-3 H The value given by the software is L = 2.42365 x 10-3 H This results into a % error = = 0% Example 3 Inductance of a three phase line with equilateral spacing Find the inductance per km of a 3-phase transmission line using 1.24 cm diameter conducors when these are placed at the corners of an equilateral triangle of each side 2m. Solution Conductor spacing d = 2 m Conductor radius r= 0.0062m 46

Inductance /phase/km = 2 10-7 The value given by the software is =2 x 10-7 {ln (2/0.0062)} x 1000 = 1.15527 x 10-3 H L = 1.15527 x 10-3 H This results into a % error = = 0% Example 4 Inductance of a three phase line with unsymmetrical spacing The three conductors of a three 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. Solution The three conductors are placed at the corners of a triangle of sides D 12 = 2 m,d 23 = 2.5m and D 31 = 4.5 m.the conductor radius r = 1.24/2 = 0.62 cm = 0.0062 m. Equivalent equilateral spacing, = = = 2.82m Inductance/phase/m =2 10-7 = 2 10-7 = 12.24 x 10-7 H Inductance/phase/km =12.24 x 10-7 x 1000 = 1.224 x 10-3 H The value given by the software is Inductance/phase/km = 1.2242 x 10-3 H This results into a % error = = 0.01634% Example 5 Inductance of bundled conductors Each conductor of the bundled-conductor line shown in Fig. 4.1 is ACSR, 1,272,000-cmil Pheasant. Find the per phase inductance for d = 45 cm. 47

Figure 4.1 Bundled conductor Solution: a) The distances in ft are 0. 45 d = = 1476. 0. 3048 8 D = = 26. 25 0. 3048 ft ft For Pheasant conductors, GMR = 0.0466 ft given in Table 4.1.GMR b for a bundle of conductors is GMR b = GMR d = 0.0466 1.476 = 0. 2623 ft The geometric mean of the phase conductor spacing is D = 3 = eq 26. 25 26. 25 52. 49 33. 07 ft The inductance of the line is then L = 2 10 Deq ln GMR b = 2 10 33.07 ln = 9.674 10 0.2623 7 7 7 H / m L = 9.674 10 7 x 1000 = 9.674 x 10-4 H/km The value given by the software is L = 9.673 x 10-4 H/km This results into a % error = = 0.01033% 48

Example 6 Capacitance of a two wire line Example: Find the capacitance for a single phase line shown in figure 4.2 operating at 60 Hz. The conductor used for the line is Partridge, and the spacing is 20 ft. Figure 4.2 Single phase line Solution: The outside radius of the Partridge conductor computed from the outside diameter given in Table 4.1 is 0.642 r = in= 2 0.0268 =0.0268 x 0.3048/12 = 6.8072 x 10-4 m D = 20 x 0.3048 =6.096 m The capacitance is = = (2 x π x 8.85 x 10-12 )/ ln (6.096/0.00068072) = 6.11058 x 10-12 F/m = 6.11058 x 10-12 x 1000 = = 6.11058 x 10-9 F/km The value given by the software is C = 6.111 x 10-9 F/km This results into a % error = = 0.00687% 49

Example 7 Capacitance of a three phase line with unsymmetrical spacing A single-circuit three-phase line operated at 69 Hz is arranged as shown in figure 4.3.The conductors are ACSR Drake. Find the capacitance. Figure 4.3 Single-circuit three-phase line Solution: The outside radius for Drake conductor computed from the outside diameter given in Table 4.1 is 1108. r = in = 0. 0462 ft 2 The geometric mean distance for this line is D = 3 = eq 20 20 38 24. 8 ft C n = = F/m to neutral = (2π x 8.85 x 10-12 )/ (ln (24.8/0.0462) = 8.8466 x 10-12 F/m == 8.8466 x 10-12 x 1000 = 8.8466 x 10-9 F/km The value given by the software is C = 8.8487 x 10-9 F/km This results into a % error = = 0.0237% 50

Example 8 Capacitance of bundled conductors Each conductor of the bundled-conductor line shown in Fig. 4.1 is ACSR, 1,272,000-cmil Pheasant. Find the per phase inductance for d = 45 cm. Figure 4.4 Bundled conductor Solution Computed from the outside diameter given in Table 4.1 r = (1.382 x 0.3048)/ (2 x 12) = 0.01755m = = 0.0889 m = = 10.08 m Capacitance = (2 x π x 8.85 x 10-12 )/ ln (10.08/0.0889) = 11.754 x 10-12 F/m =11.754 x 10-12 x 1000 = 11.754 x 10-9 F/km The value given by the software is C = 11.7539 x 10-9 F/km This results into a % error = = 0.00085% Example 9 Determination of the generalized constants for long lines A 3-phase transmission line 200km long has the following constants: Resistance/phase/km = 0.16 ohm Reactance/phase/km = 0.25 ohm 51

Shunt admittance/phase/km = 1.5 x 10-6 S Calculate the A, B, C and D constants of the line. Solution Total Resistance/phase, Total Reactance/phase, R = 0.16 x 200 = 32Ω X L = 0.25 x 200 = 50 Ω Total Shunt admittance/phase, Y = j 1.5 x 10-6 x 200 = 0.0003 90 0 Series Impedance/phase, Z = R + j X L = 32 + j 50 = 59.4 < 58 0 For a long line the constants are given by A = D = cosh ; B = sinh ; C = sinh ; Now = = 0.133 < 74 0 ZY = 0.0178 < 148 0 Z 2 Y 2 = 0.0032 < 296 0 = = 445 < -16 0 = = 0.00224 < 16 0 Cosh = approximately = 1 + < 148 0 + < 296 0 52

= 1 + 0.0089 < 148 0 + 0.0000133 < 296 0 =1 + 0.0089 ( -0.848 + j 0.529 ) + 0.0000133 ( 0.438 j 0.9) A = D = 0.992 + j 0.00469 = 0.992 < 0.26 0 The value given by the software is A = D = 0.9925 < 0.276 0 This results into a % error = = 0.0504% Sinh = approximately = 0.133 < 74 0 + = 0.133 < 74 0 + 0.0004 < 222 0 =0.133 (0.275 + j 0.961 ) + 0.0004 (-0.743 j 0.67) = 0.0362 + j 0.1275 = 0.1325 < 74.15 0 B = sinh ; = 445 < -16 0 x 0.1325 < 74.15 0 = 58.9625 < 58.15 0 The value given by the software is B = 59.215 < 57.47 0 This results into a % error = = 0.426% C = sinh ; =0.00224 < 16 0 x 0.1325 < 74.15 0 = 0.0002968 < 90.15 0 The value given by the software is 53

C = 0.000299 < 90.09 0 This results into a % error = = 0.736% Example 10 Determination of the generalized constants for medium lines using nominal T method A balanced three phase load of 30 MW is supplied at 132 kv, 50Hz and 0.85 p.f. lagging by means of a transmission line. The series impedance of a single conductor is ( 20 + j52) ohms and the total phase-neutral admittance is 315 x 10-6 siemen.using nominal T method determine the A, B, C and constants of the line. Solution Series line impedance/phase, Z = ( 20 + j 52) Ω Shunt admittance/phase, Y = j 315 x 10-6 S For nominal T method, A = D = = 1 + x j 315 x 10-6 =0.992 + j0.00315 = 0.992 < 0.18 0 The value given by the software is A = D = 0.9918 < 0.18 0 This results into a % error = = 0.020% B = Z ; = ( 20 + j 52) = 19.84 + j 51.82 = 55.5 < 69 0 54

The value given by the software is B = 55.48 < 69.1 0 This results into a % error = = 0.036% C = Y = 0.000315 < 90 0 The value given by the software is C = Y = 0.000315 < 90 0 This results into a % error = = 0% Example 11 Determination of the generalized constants for short lines Determine the A, B, C and D constants of a three phase line 20km line with the following constants: Resistance/phase/km = 0.16 ohm Reactance/phase/km = 0.25 ohm Solution Total Resistance/phase, R = 0.16 x 20 = 3.2Ω Total Reactance/phase, X L = 0.25 x 20 = 5 Ω Series Impedance/phase, Z = R + j X L = 3.2 + j 5 = 5.94 < 58 0 A = 1 ; The value given by the software is A = 1 This results into a % error = = 0% B = Z = 5.94 < 58 0 ; The value given by the software is 55