LESSON NO. 2.pptx

LESSON NO. 2 INDUCTANCE AND CAPACITANCE OF TRANSMISSION LINES LESSON OBJECTIVES: AT THE END OF THE LESSON, THE STUDENT WOULD BE ABLE TO 1. Discuss the components of inductive and capacitive transmission lines 2. Calculate the geometric mean radius and distance of single and stranded transmission conductors A typical transmission lines Principal Elements of a High Voltage Transmission Line 1. Conductor 2. Transformer 3. Line Insulator 4. Support: a. wood poles b. concrete poles c. steel tower/steel poles 5. Protective Device: a. circuit breaker Device b. fuse cut-off c. relays d. Ground wire e. lightning Arrester 6. Voltage Control Device 7. Other Equipment for voltage control System Types of Conductors:- Mainly we are having four types of conductors. 1. Solid conductors. 2. Homogeneous Stranded conductors. 3. Composite Stranded conductors 4. Bundled conductors Solid conductor: 1. Solid conductors is single piece of conductor 2. These conductors may be of copper (or) Aluminum 3. It is having high mechanical strength and tensile strength 4. Circular solid conductor will be preferred to make electrical field same throughout the surface of the conductor. 5. There is no application of solid conductors as transmission lines 6. Flat (or) solid conductors are used for power transformer winding Homogeneous stranded conductors A number of strands are taken and are twisted together to increase the current carrying capacity by maintaining the same operating voltage. 1. All strands are of same material 2. Compared to stranded conductor it is having high mechanical strength and low tensile strength. 3. Stringing is easy in stranded conductor so transportation is easy Composite stranded conductor Strands of different materials are twisted together to form the stranded conductor. Its purpose is to improve the tensile strength. 1. As the layer size is increased, the number of strands increase in the following manner : 1+6+12+24+30+… 2. If d is the diameter of each conductor and n is the number of layers then the diameter of the stranded conductor is D = (2n – 1) d 3. Generally ACSR conductor is represented by x/y x = number of aluminum Or steel strand y = number of aluminum or steel strand In power conductor the aluminum strand is greater than the steel strand Bundled conductor When voltage is above 230kV, corona loss and interference with the communication lines is more. Corona occurs when the surface potential gradient of a conductor exceeds the dielectric strength of the surrounding air. This causes ionization of the area near the conductor. The high voltage surface gradient is reduced by using two or more conductors per phase in close proximity. This is called conductor bundling. The conductors of a bundle are separated at regular intervals with spacer dampers that prevent clashing of the conductors and prevent them from swaying in the wind. Types of conductors used in overhead Transmission Lines: 1. AAC – All Aluminum Conductor - have higher tensile strength than the ordinary electrical conductor grade of Aluminum 2. AAAC - All Aluminum Alloy Conductor 3. ACSR – Aluminum Conductor Steel Reinforced - consist of a central core of steel strands surrounded by layers of Aluminum strands(commonly used in high voltage) 4. ACAR - Aluminum Conductor Alloy Reinforced - has a central core of high strength Aluminum surrounded by layers of electrical conductor grade Aluminum 46KV up high voltage Types of Conductors 1. Copper. Copper is an ideal material for overhead lines owing to its high electrical conductivity and greater tensile strength. It is always used in the hard drawn form as stranded conductor. Although hard drawing decreases the electrical conductivity slightly yet it increases the tensile strength considerably. 2. Aluminum. Aluminum is cheap and light as compared to copper but it has much smaller conductivity and tensile strength. The relative comparison of the two materials is briefed below: (i) The conductivity of aluminum is 60% that of copper. The smaller conductivity of aluminum means that for any particular transmission efficiency, the X-sectional area of conductor must be larger in aluminum than in copper. (ii) The specific gravity of aluminum (2·71 gm/cc) is lower than that of copper (8·9 gm/cc).Therefore, an aluminum conductor has almost one-half the weight of equivalent copper conductor. For this reason, the supporting structures for aluminum need not be made so strong as that of copper conductor. (iii) Aluminum conductor being light, is liable to greater swings and hence larger cross-arms are required. (iv) Due to lower tensile strength and higher co-efficient of linear expansion of aluminum, the sag is greater in aluminum conductors. Transmission line model A transmission line is used for the transmission of electrical power from generating substation to the various distribution units. Characteristics parameters (per unit length) 1. Series resistance (R) 2. Shunt conductance (G) 3. Series inductance (L) 4. Shunt capacitance (C) Series resistance The DC resistance of a conductor at a temperature T is given by: 𝑹 = 𝝆𝒍/𝑨 where ρ= conductor resistivity at temperature T, Ω-m l = the length of the conductor A = the current-carry cross-sectional area of the conductor The AC resistance of a conductor is given by: 𝑹𝒂𝒄 =𝑷𝒍𝒐𝒔𝒔 / 𝑰2 where Ploss – real power dissipated in the conductor in watts I – rms conductor current Ipk = √2 Irms Transmission line conductor resistance depends on Spiraling: The purpose of introducing a steel core inside the stranded aluminum conductors is to obtain a high strength-to-weight ratio. A stranded conductor offers more flexibility and easier to manufacture than a solid large conductor. However, the total resistance is increased because the outside strands are larger than the inside strands on account of the spiraling. The layer resistance-per-length of each spirally wound conductor depends on its total length as follows: Skin Effect An increase in frequency causes non uniform current density. This result is called skin effect. - also defined as the tendency of current to move outward the surface of conductor. AC resistance is greater than dc resistance of the line because of skin effect. Skin effect increases with the increase of frequency, area, permeability, temperature and size of conductor. Rt = Ro [1 + ά(T2 – To)] When an AC current is applied to the conductor, the current concentrates near the surface of the conductor and its strength decreases as you go towards the center of the conductor. The depth till which current flows in a conductor is called as Skin Depth. Shunt conductance 1. Conductance is associated with power losses between the conductors or between the conductors and ground. 2. Such power losses occur through leakage currents on insulators and via a corona 3. Leakage currents are affected by contaminants such dirt and accumulated salt accumulated on insulators. Meteorological factors such as moisture 4. Corona loss occurs when the electric field at the surface of a conductor causes the air to ionize and thereby conduct. 5. Corona loss depends on: a. Conductor surface irregularities b. Meteorological conditions such as humidity, fog, and rain 6. Losses due to leakage currents and corona loss are often small compared to direct I2 R losses on TLs and are typically neglected in power flow studies. Corona discharge When corona occurs it causes power loss into the transmission line. Which reduced the efficiency of the transmission line. Can be calculated using an equation known as Peek's Formula. EXAMPLE NO. 1 1. What is the DC resistance in Ω/mile of a 1.25 inch diameter solid aluminum wire with resistivity 2.65x10-8Ω-m. 2. The resistance of a copper wire at 0oC is 30Ω. Find the resistance of the nickel wire at 00C if its resistance at 400C is equal to the resistance of copper at the same temperature. The alpha of copper = -234.5 The alpha of Nickel = -147 3. The power loss of a 50Km, 0.6 inch copper conductor is 150 watts at a current of 1.27A peak current. Calculate the frequency of the current flowing in the conductor. Resistivity of copper is 1.68x10-8 Ω-m 4. A 3 phase transmission line is designed to deliver 200MVA at 220kv over a distance of 63Km. The total transmission line loss should not exceed 2.5 percent of the rated line if the resistivity of the conductor line is 2.8x10-8Ω-m, determine the conductor diameter. Line inductance When a current flows through a conductor, a magnetic flux is set up which links the conductor. The current also establishes a magnetic field proportional to the current in the wire. Due to the distributed nature of a TL we are interested in the inductance per unit length (H/m). The inductance of a transmission line is calculated as flux linkages per ampere. If permeability μ is constant, sinusoidal current produces sinusoidally varying flux in phase with the current. The resulting flux linkages can then be expressed as a phasor λ and If i, the instantaneous value of current, is substituted for the phasor I, then λ should be the value of the instantaneous flux linkages produced by i. Flux linkages are measured in weber-turns, Wbt. i = 5 SIN (377t + 30)A I = 5∟30 A Flux linkages between two points external to an isolated conductor As a step in computing inductance due to flux external to a conductor, 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 at DI and D2 meters from the center of the conductor. In the Figure P I and P2 are two such points. The conductor carries a current of I A Inductance of a single phase two wire line Flux linkages of one conductor in a group A more general problem than that of the two-wire line is presented by one conductor in a group of conductors where the sum of the currents in all the conductors is zero. Such a group of conductors is shown in the figure. Conductors 1 , 2, 3, …, n carry the phasor currents I1, I2, I3 …, In " The distances of these conductors from a remote point P are indicated on the figure as D1P , D2P, D3P …DnP Let us determine λ1P1 the flux linkages of conductor 1 due to I1 internal flux linkages but excluding all the flux beyond the point P. Inductance of composite conductor lines EXAMPLE NO. 2 1. Calculate the flux linkage and the inductance of an isolated conductor carrying a current of 10A at points 2 meters and 3 meters 2. The total inductance of a conductor at 2m is 9.264x10- 7 H/m. what is the radius of the conductor? 3. Two conductor in the figure is separated 2 meters apart if each of the conductor has a radius of 2.5cm. Calculate the total inductance(loop inductance) of the system. 4. A three phase conductor are symmetrically spaced with 2 meters each and 1.2cm of equal radius. Calculate the total inductance of the system. 5. Calculate the GMR of the given set of conductors if r = 1.5cm 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. The figure shows the transposed line. The phase conductors are designated as A, B and C and the positions occupied are numbered 1, 2 and 3. The effect of transposition is that each conductor has the same average inductance. 6. Determine the inductance and reactive inductance of a 3 phase line operating at 50 Hz and conductors arranged as given in figure. The conductor diameter is 0.8 cm. find the inductance and reactive inductance at 20Km. 7. A 3 phase 60Hz TL has its conductor arranged in triangular formation so that the two sides is 25 ft and the third side is 42 ft, the conductor is an ACSR 24/7 strand 556.5MCM. Determine the inductance and inductive reactance of the line per mile. If the line is 12 miles. Find the inductance and inductive reactance. 8. A three phase line is designed with equilateral spacing of 16 ft. it is decided to build the same line with horizontal spacing (D13 = 2D12 = 2D23). The conductors are transposed, what would be the spacing between adjacent conductor in order to obtain the same inductance in the original design. 9. Two conductors of a single phase TL are 0.5cm and 0.75cm in diameter respectively. The space between the conductors are 1.5 meters. Calculate a. The size of the conductor in CMIL b. The inductance of each line c. Total inductance of the single phase line. 10. Determine the inductance of a single phase transmission line consisting of three conductors of 2.5 mm radii in the ‘go’ conductor and 5 mm radii in the return conductor. The configuration of line is as shown in figure 11. A single phase line consist of two circuits in parallel. Conductors a and a’ in parallel form the lead while conductor b and b’ in parallel form of return circuit. Calculate the total inductance of the line per km assuming that the current is equally shared by the two parallel conductors. The diameter of each conductor is 30mm. Bundled conductors At extra high voltages (EHV), that is, voltages above 230KV, corona with its resultant power loss and particularly its interference with communications is excessive if the circuit has only one conductor per phase. The high voltage gradient at the conductor in the EHV range is reduced considerably by having two or more conductor per phase in close proximity compared with the spacing between phases. Such a line is said to be composed of bundled conductors. The bundle consist of two, three or four conductors called subconductor. The current will not divide exactly between the conductors of the bundle unless there is a transposition of the conductors within the bundle, but the difference is of no practical importance and the GMD method is accurate for the calculations. Advantage of bundled conductor over single conductors 1. The bundle conductors transmit bulk power with reduced losses, thereby giving increase transmission efficiency 2. Since the bundle conductor lines have a higher capacitance to neutral in comparison with the single conductor lines, therefore, they have higher charging current, which helps improving the power factor. 3. Since by bundling, the self GMD or GMR is increased, the inductance per phase, in comparison with single conductor lines, is reduced, as a result reactance per phase is reduced. 4. Bundled conductors reduced corona loss and radio interference. 12. Find the inductive reactance of a 3 phase bundled conductor line with 2 conductors per phase with spacing 40cm. Phase to phase separation is 7m in horizontal configuration. All conductors are ACSR with diameter 3.5cm. ASSIGNMENT NO. 1 1. Find the GMR of the given conductor 2. A part of transposition cycle of a 3 phase double circuit line is shown. Radius of each conductor is 0.9cm. The conductors are solid copper. Find the inductance per phase per km of the line. 3. Calculate the inductance per km or a 3 phase overhead TL using 1.24cm diameter conductors when these are placed at the corners of a equilateral triangle of each side 2 meters Capacitance of transmission lines If a long, straight cylindrical conductor lies in a uniform medium such as air and is isolated from other charges so that the charge is uniformly distributed around its periphery, the flux is radial. All points equidistant from such a conductor are points of equipotential and have the same electric flux density Potential difference between two points Capacitance of single phase two wire line If the line is supplied by z transformer having ground center tap Line to line Line to neutral Capacitance of 3 phase line with equilateral spacing For three phase unsymmetrical spacing Effects of ground (three phase) Bundled conductors EXAMPLE NO. 3 1. Find the capacitive susceptance per mile of a single-phase line operating at 60 Hz. The conductor is DRAKE , and spacing is 20 ft between centers. 2. A three phase TL has flat horizontal spacing with 2 meters between adjacent conductors. At a certain instant the charge on one of the outside conductors is 60uC/Km and the charge on the center conductor and on the other outside conductor is -30uC/Km. the radius of each conductor is 0.9 cm. neglect the effect of the ground and find the voltage drop between the two identically charged conductor at the instant specified. 3. Calculate the capacitance to neutral per meter of a single phase line composed of two single conductor having a diameter of 0.229 inch. The conductors are 10 ft apart and 25 feet above the ground. 4. A three phase 60hz line has flat horizontal spacing. The conductors have an outside diameter of 3.28 cm with 12m between conductors. Determine the capacitive reactance to neutral in ohm-meter and the capacitive reactance of the line in ohms if its length is 130 miles. a. Assume negligible ground effect b. If ground effect is considered if the conductor is placed 20m from the ground 5. determine the capacitance and charging current per Km of the 460Kv line using two bundle conductor per phase. The diameter of each conductor is 5cm 6. Six conductors of double circuit transmission line are arranged as shown. the conductor used is OSTRICH. Find the capacitive reactance to neutral and the charging current per km per phase at 132KV and 60Hz, assuming that the line is regularly transposed. Neglect the effect of earth. 7. A 60hz 3 phase line composed of one ACSR “bluejay” conductor per phase has flat horizontal spacing of 11m conductor. Compare the capacitive reactance in ohms per km per phase of this ine to that of two conductor bundle of ACSR 26/7 conductor having the same cross sectional area and 11m spacing. The spacing between bundle is 40cm ASSIGNMENT NO. 2 1. A three phase 60Hz DOVE TL has its conductors arranged in a triangular formation so that two of its conductors are 30 feet and the third is 50 feet. Determine the capacitance to neutral and capacitive to neutral in miles. 2. Determine the capacitance of the arrangement shown a. if the effect of the earth is neglected b. If the effect of the earth is considered. The height of the conductors is 10 meters from the ground and each has a radius of 5cm Voltage and Current relations on Transmission Lines The transmission lines are categorized as three types 1) Short transmission line– the line length is up to 80 km and the operating voltage is < 20 kV. 2) Medium transmission line– the line length is between 80 km to 160 km and the operating voltage is > 20 kV and < 100kV 3) Long transmission line – the line length is more than 160 km and the operating voltage is > 100 kV Whatever may be the category of transmission line, the main aim is to transmit power from one end to another. Like other electrical system, the transmission network also will have some power loss and voltage drop during transmitting power from sending end to receiving end. Due to smaller distance and lower line voltage, the capacitance effect are extremely small and can be neglected. Its performance depends on the resistance and inductance of the line.. Through in an actual line, the resistance and inductance are distributed over the whole length but in case of short lines, the total resistance and inductance are assumed to be lumped at one place. For short length, the shunt capacitance of this type of line is neglected and other parameters like electrical resistance and inductor of these short lines are lumped, hence the equivalent circuit is represented as given in the next slide. For convenience, it is considered that the parameters of the conductors are lumped into one conductor, and the return conductor is assumed to have no resistance and inductive reactance. The sending end voltage Vs = VR + IsZL Sending power factor pfs = cos (θIS - θVS) Sending power Ps = VsIspfs = PR + PLOSS PLOSS = I2sZLINE Voltage regulation = (Vs – VR) / VR The lower the voltage regulation, the better it is because little variation in receiving end voltage due to the variation in load current. Regulation is defined as the change in voltage at the receiving end (load) when the full load is thrown off, the sending end (supply) frequency remaining unchanged. Eff = PR / PR + PLOSS (line losses) Where PR = VRIRcos θR Using the phasor diagram with IR as the reference point EXAMPLE NO. 4 1. A 10 mile, 60hz single phase TL using DOVE conductor equilaterally spaced with 5 feet spacing between centers. It delivers 2500KW at 13.8KV to a balance load. a. Determine the per phase impedance of the line b. What must be the sending end voltage when the power factor is 0.866 lagging, Unity power factor and 0.9 leading c. Determine the percent regulation of the line at different power factor d. Transmission efficiency e. Draw the phasor diagram depicting the operation of the line in each case. Assume wire temperature to be 50oC 2. A single phase line is transmitting 1100KW power to a factory at 11KV and at 0.8 lagging. It has a total resistance of 2Ω and a reactance of 3Ω. Determine the voltage at the sending end, percentage regulation and efficiency. Three phase short TL: The assumption on short 3 phase TL are 1. System is Y connected 2. Transmission is balance 3. Per phase basis Sending power, Ps = √3 Vs,LLIs,LLpfs (note that IΦ = ILL in Y ) Or Ps = 3 VSN Is pfs Receiving power = √3 VR,LL IR,LL pfR Or PR = 3 VRN IR pfR For Y connected, Is = IR VSN = Vs/ √3 VSN = sending end to neutral VRN = receiving end to neutral PLOSS = 3IS2 R 3. A 20 mile, three phase transmission line is composed of 336.4MCM, 26/7 ACSR strand. The conductor are spaced horizontally with 3 ft between adjacent conductor. It is supplying a balanced load of 4000KW at 13.8KV with 0.8 lagging power factor at 60HZ a. Calculate the sending end voltage and the power factor b. Voltage regulation of the line c. Efficiency of the TL d. If a capacitor bank is connected in parallel with the load that draws a line current of 120A, calculate the sending end voltage and the sending end power factor. 4. Calculate the equilateral spacing between three phase line with sending end voltage of 8100V, 0.78 lagging and receiving end voltage of 7620V, 0.8 lagging per phase. If load voltage is used as reference and the sending end current is 60A. The radius of the conductor is 0.035 ft at a distance of 19.5Km Medium length transmission lines Using the circuit constant V S = A VR + B I R IS = C V R + D I R 5. A 120km three phase TL is supplying a load that draws 180A at 100KV with 0.866 lagging. The conductors are spaced 6m horizontally between adjacent phases that uses 336.4MCM 30/7 strand. Assuming that the capacitance are lumped at the middle of the line. Determine a. The line constants b. Sending end voltage c. Sending end current d. Percent regulation ASSIGNMENT NO. 3 A three phase TL has the ffg data: Length = 96KM The conductors are spaced horizontally with 8m with adjacent conductors. A substation at the receiving end draws 200A at 0.8 power factor lagging at 225KV. Using nominal T network. The conductor is 556,500MCM ACSR 18/1 strand at 500C Calculate a. The line constant b. Sending end current c. Voltage regulation

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