transmission-line-all

The HK Polytechnic University Transmission Lines & Cables 1 ' & $ % 1 Overhead Line Construction Technologies involved : • Electrical • Mechanical • Structural • Civil • Chemical Major components are : • Shield conductors for lightning protection (when necessary) • Tower (lattice or tubular) • Phase conductors • Insulators (V strings shown) • Foundation and grounding 2 ' & $ % 1.1 Phase Conductors 1.1.1 Conductor Material • In the early days, copper was used exclusively for conductor material, bus since the increase in the cost of copper, aluminium has largely displaced it. • Typical type of conductors are : AAC All-Aluminium Conductors AAAC All-Aluminium-Alloy Conductors ACSR Aluminium Conductor, Steel-Reinforced ACAR Aluminium Conductor, Alloy-Reinforced • The most common type is the ACSR conductor. It is made of composite stranding of aluminium conductor, steel reinforced. The steel provides mechanical strength and the aluminium conducts the current. • The reinforcement consists of a central core of galvanized steel wires that are sometimes greased for additional protection against corrosion. KWCn 1 The HK Polytechnic University Transmission Lines & Cables 3 ' & $ % • Typical 650 Ampere ACSR conductor – 54Al/7S – Aluminium strands ∗ 3 layers ∗ 54 conductors – Steel strands ∗ 7 conductors • Examples of ACSR conductors : 4 ' & $ % 1.1.2 Corona and Radio Interference • The high voltages used in EHV (400kV) and UHV (700kV) systems result in high voltage gradients at the conductor surface. • If the gradient exceeds 30 kV/cm, the breakdown level of air, a discharge takes place, which causes power loss called corona loss, and emission of electromagnetic waves, hence considerable radio interference and audio noise. • Corona loss, radio interference (interferes with radio receivers and TV sets) and audio noise (interferes with people) are at a minimum under fine weather conditions and increase by two orders of magnitude during severe weather. • For example, the corona loss on a 400 kV line is around 1 kW per mile in good weather and may reach 150 kW per mile during snow and fog. • In general, the average energy loss due to corona is only a small percentage of the conductor I2R loss. This is why corona loss is less a concern when compared with the problem caused by radio interference and audio noise. KWCn 2 The HK Polytechnic University Transmission Lines & Cables 5 ' & $ % 1.1.3 Bundle Conductors • Extra high voltage lines use bundle conductors – To reduce corona discharge – To increase current carrying capacity – To reduce the inductive reactance of the line • Bundles with two, three or four conductors are used. • The bundle acts, as far as the electric field lines are concerned, like a conductor of a diameter much larger than that of the component conductors. • This reduces the voltage gradient or conversely allows a higher voltage to be utilised for permissible levels of radio interference. • The distance between the conductors in the bundle is maintained by steel or aluminium bars (spacers). 6 ' & $ % 1.2 Shield Conductors – Earthwires • Shield conductors perform two important functions : 1. Prevention of direct lightning strikes to the phase conductors. 2. Provision of a low resistance path for the fault current in the event of flash-over from the line conductors to the structures. • Galvanized steel wires or line conductors are used for shield conductors. • The size is determined from the considerations of mechanical strength and fault current carrying capacity. • The number and location of the shield wires are selected so that almost all lightning strokes terminate on the shield wires rather than on the phase conductors. • Shield wires are grounded to the tower. KWCn 3 The HK Polytechnic University Transmission Lines & Cables 7 ' & $ % 1.3 Support Structures • Overhead lines are suspended from insulators which are themselves supported by towers or poles, i.e. the structure must support the line and fit within the available right-of-ways (ROW). • Some available structure configurations are shown below. 8 ' & $ % • Below shows (a) a typical 400 kV tower that forms the backbone of the UK supergrid and, for contrast, (b) the tower of the first North American 735 kV line. • A 750kV d.c. line is shown in (c) to illustrate the economic advantages, as far as transmission lines are concerned, of using direct rather than alternating current for bulk transfer of electric energy. KWCn 4 The HK Polytechnic University Transmission Lines & Cables 9 ' & $ % • The span between two towers is dependent upon the allowable sag in the line. • Before undertaking the actual construction of a line, it is important to calculate the permissible sag and the corresponding mechanical pull. • The summer and winter temperature range must be taken into account too because the length of the conductor varies with the temperature. • Typically, 400 kV towers are about 50m high and spaced about 400m apart. • Towers of three types may be distinguished: 1. the straight-run towers – support only the conductor and insulator weight. 2. the deviation towers – also resist some of the conductor tension. 3. the terminal towers – heavily constructed, found at the overhead line ends. 10 ' & $ % 1.4 Insulators • Insulators serve to support and anchor the conductors and to insulate them from ground. • Insulators are usually made of porcelain, but glass and other synthetic insulating materials are also used. • The most common type of insulator used is the cap and pin type as shown : • Insulators are connected in string, as shown above, by their cap and pin metallic parts to increase both the mechanical and electrical strength. KWCn 5 The HK Polytechnic University Transmission Lines & Cables 11 ' & $ % 1.5 Tower Grounding • Transmission-line towers are always solidly connected to ground. • Great care is taken to ensure that the ground resistance is low. • When lightning hits a tower, it creates a sudden voltage rise across the insulators as the lightning current discharges to ground. Such a voltage rise may produce a flash-over across the insulators and a consequent line outage. 12 ' & $ % 2 Transmission Line Parameters 2.1 Parameters • Series AC resistance • Series Inductance • Shunt Capacitance 2.2 AC Resistance • The DC resistance of a solid round conductor at a specified temperature is given by : Rdc = ρl A where ρ = conductor resistivity l = conductor length A = conductor cross-sectional area • The conductor resistance is affected by: spiraling, frequency and temperature. KWCn 6 The HK Polytechnic University Transmission Lines & Cables 13 ' & $ % • The stranded conductor resistance is larger the the solid conductor resistance because spiraling of the strands increases the actual length. • The AC resistance or effective resistance of a conductor is : Rac = Ploss |I|2 Ω where Ploss is the conductor real power loss in watts I is the rms conductor current • AC resistance is larger than DC resistance because the skin effect forces the current toward the conductor surface. • Resistance increases with the temperature and may be calculated from : R2 = R1 T + t2 T + t1 where R1 and R2 are conductor resistances at t1 and t2 Co and T is a temperature constant. For aluminum, T ≈ 228. • Because of the above effects, the conductor resistance is best determined from manufacturers’ data or the conductor tables. 14 ' & $ % • Typical conductor table KWCn 7 The HK Polytechnic University Transmission Lines & Cables 15 ' & $ % 2.3 Inductance of a Single Conductor • AC current in the conductor produces a magnetic field inside and around the conductor. The magnetic field lines are concentric circles. • The figure below shows the field distribution inside and outside of a conductor. Current flows normal to the plane. Right hand rule gives the field direction. Outside field Inside field I l Conductor r • When the current changes, the flux changes and a voltage is induced in the circuit. By definition, the inductance L is the ratio of its total magnetic flux linkage λ to the current I , given by : L = λ I 16 ' & $ % • The magnetic field intensity Hx, around a circle of radius x, is constant and tangent to circle. The Ampere’s law relatingHx to current Ix us given by :∮ Hxds = Ix or Hx = Ix 2pix A/m where Ix is the current enclosed at radius x. r Ix dx ds • From the relationship between the field intensity and flux density, the flux density can be calculated as : Bx = µ0Hx = µ0Ix 2pix Wb/m2 for non-magnetic conductor • The above equation is all that required for evaluating the flux linkage λ of a conductor. The inductance of the conductor can be defined as the sum of contributions from flux linkages internal and external to the conductor. KWCn 8 The HK Polytechnic University Transmission Lines & Cables 17 ' & $ % 2.3.1 Inductance Due to Internal Flux Linkage • A simple expression can be obtained for the internal flux linkage by neglecting the skin effect and assuming uniform current density throughout the conductor cross section, i.e. I pir2 = Ix pix2 • Substituting for Ix, the magnetic flux density Bx becomes : Bx = µ0I 2pir2 x Wb/m2 • The differential flux dφx for a small region of thickness dx and length l : dφx = Bx · dx · l = µ0Il 2pir2 xdx Wb • As the flux dφx links only the fraction of the conductor from the centre to radius x, only the fraction pix 2 pir2 of the total current is linked by the flux, 18 ' & $ % i.e. the differential flux linkage dλx is : dλx = pix2 pir2 dφx = µ0Il 2pir4 x3dx Wbt • The total internal flux linkage λint is found by integrating dλx from 0 to r : λint = µ0Il 2pir4 ∫ r 0 x3dx = µ0Il 8pi Wbt • Hence, the unit length inductance due to the internal flux linkage is : Lint = λint Il = µ0 8pi = 1 2 × 10−7 H/m where µ0 2pi = 2× 10−7 • Note: the internal flux linkage λint is proportional to the current I and the length l of the conductor but does not depend on the radius r of the conductor. KWCn 9 The HK Polytechnic University Transmission Lines & Cables 19 ' & $ % 2.3.2 Inductance Due to External Flux Linkage • Consider Hx external to the conductor at radius x > r as shown. • Since the circle at radius x encloses the entire current, Ix = I and the flux density at radius x becomes : Bx = µ0I 2pix Wb/m2 r I dxx 1D 2D • The differential flux dφx for a small region of thickness dx and length l : dφx = Bx · dx · l = µ0Il 2pix dx Wb 20 ' & $ % • Since the entire current I is linked by the flux outside the conductor, the differential flux linkage dλx is numerically equal to the differential flux dφx. • The external flux linkage λext between two points D1 and D2 is calculated by integrating the differential flux linkage dλx from D1 to D2 : λext = µ0Il 2pi ∫ D2 D1 1 x dx = µ0Il 2pi ln D2 D1 Wbt • Hence, the unit length inductance between two points external to a conductor is : Lext = λext Il = µ0 2pi ln D2 D1 where µ0 2pi = 2× 10−7 = 2× 10−7 ln D2 D1 H/m KWCn 10 The HK Polytechnic University Transmission Lines & Cables 21 ' & $ % 2.3.3 Total Inductance • The total flux passing through the plain placed between the conductor and a point with distance D from the conductor centre is the sum of the internal and external flux linkage : λtotal = µ0Il 8pi + µ0Il 2pi ln D r = µ0Il 2pi ( 1 4 + ln D r ) Wbt • Expressing 1 4 as ln 1 e−0.25 , the total unit length inductance becomes : Ltotal = λtotal Il = µ0 2pi ln D re−0.25 = µ0 2pi ln D GMRc H/m where GMRc = re−0.25 is known as the self Geometric Mean Radius. GMRc can be considered as the radius of a fictitous conductor which have no internal flux but with the same inductance as the actual conductor. 22 ' & $ % 2.4 Inductance of Single Phase Lines • The single phase line has two conductors. The radius of each conductor is r and the distance betweeen the conductors is D such that r << D. • The current flows from the source to the load in conductorA and returns in conductor B; therefore, IA = −IB . l I D A IB BA r D A B IA IB • Both currents generate magnetic fields. The field lines are concentric circles around the conductors. KWCn 11 The HK Polytechnic University Transmission Lines & Cables 23 ' & $ % • The flux linking with conductor A has two components. – λAA – the flux generated by conductor A current, IA – λAB – the flux generated by conductor B current, IB • The flux linking conductor A is determined by integration of the flux that paths through a plane between conductorA and an arbitrary selected distant point F . • These fluxes induced voltage in conductor A and can be described by self and mutual inductances. • The current in conductor A produces a flux of λAA betweenA and F . • The flux is calculated by the integration of the field generated by the current IA between A and F , i.e. λAA = µ0IAl 2pi ln DAF GMRc Wbt D A B AF AA F 24 ' & $ % D A B AF AB DBF DAB F • The current in conductor B produces a flux of λAB between A and F . • The flux is calculated by the integration of the field generated by the current IB between A and F , i.e. λAB = µ0IBl 2pi ln DBF DAB Wbt • The total flux λA linking conductorA is the sum of the flux generated by the current in conductorsA and B. The value is : λA = λAA + λAB = µ0l 2pi ( IA ln DAF GMRc + IB ln DBF DAB ) = µ0l 2pi ( IA ln 1 GMRc + IB ln 1 DAB ) + µ0l 2pi (IA lnDAF + IB lnDBF ) KWCn 12 The HK Polytechnic University Transmission Lines & Cables 25 ' & $ % • The second part of the above equation is zero for practical conditions : – F is a distant point such that : DAF ≈ DBF . – IA = −IB = I • Introducing DAB = D and the above approximation, λA = µ0l 2pi ( I ln 1 GMRc − I ln 1 D ) = µ0Il 2pi ln D GMRc • Hence, the unit length inductance of conductor A is : LA = λA IA = µ0 2pi ln D GMRc where GMRc = re−0.25 = 0.7788r is the self Geometrical Mean Radius. For stranded or ACSR conductor the GMRc has to be determined from the conductor tables. 26 ' & $ % 2.5 Inductance of Three Phase Lines • For a three phase line the current in phase conductor produces a magnetic field. The field lines are concentric circles around the conductors. • The flux linking with conductor A has three components. – λAA – the flux generated by phase A current, IA – λAB – the flux generated by phase B current, IB – λAC – the flux generated by phase C current, IC • The flux linking conductor A is determined by calculation of the flux that paths through a plane between conductor A and an arbitrary selected distant point F . • These fluxes induced voltage in conductor A and can be described by self and mutual inductances. D A B AF AA C F KWCn 13 The HK Polytechnic University Transmission Lines & Cables 27 ' & $ % • The current in conductor A produces a flux of λAA between A and F . • The flux is calculated by the integration of the field generated by the current IA between A and F , i.e. λAA = µ0IAl 2pi ln DAF GMRc Wbt • The current in conductor B produces a flux of λAB between A and F . • The flux is calculated by the integration of the field generated by the current IB between A and F , i.e. λAB = µ0IBl 2pi ln DBF DAB Wbt D A B AF AB DBF DAB F C • The current in conductor C produces a flux of λAC betweenA and F . 28 ' & $ % • The flux is calculated by the integration of the field generated by the current IC between A and F , i.e. λAC = µ0IC l 2pi ln DCF DAC Wbt • The total flux λA linking conductorA is the sum of the flux generated by the current in conductors A, B and C . The value is : λA = λAA + λAB + λAC = µ0l 2pi ( IA ln DAF GMRc + IB ln DBF DAB + IC ln DCF DAC ) = µ0l 2pi ( IA ln 1 GMRc + IB ln 1 DAB + IC ln 1 DAC ) + µ0l 2pi (IA lnDAF + IB lnDBF + IC lnDCF ) KWCn 14 The HK Polytechnic University Transmission Lines & Cables 29 ' & $ % • The second part of the above equation is zero for practical conditions : – F is a distant point such that : DAF ≈ DBF ≈ DCF . – For a balanced three phase system : IA + IB + IC = 0 • Therefore, the flux linking with conductor A is : λA = µ0l 2pi ( IA ln 1 GMRc + IB ln 1 DAB + IC ln 1 DAC ) • Similar equations can be derived for conductor B and C : λB = µ0l 2pi ( IA ln 1 DBA + IB ln 1 GMRc + IC ln 1 DBC ) λC = µ0l 2pi ( IA ln 1 DCA + IB ln 1 DCB + IC ln 1 GMRc ) • The above three equations show that a three phase line cannot be described by one inductance because the flux linkage depends on the phase currents. 30 ' & $ % 2.5.1 Symmetrical Spacing • If the line is arranged in a triangle with DAB = DBC = DCA = D λA = µ0l 2pi ( IA ln 1 GMRc + IB ln 1 D + IC ln 1 D ) DD D IA IBIC • Substituting IA = −(IB + IC) into the above equation : λA = µ0l 2pi IA ln D GMRc and the unit length inductance of the conductor A becomes : LA = λA IAl = µ0 2pi ln D GMRc Comparsion of the above with the equation for the inductance of single-phase lines shows that inductance per phase for a three-phase circuit with equilateral spacing is the same as for one conductor of a single-phase circuit. KWCn 15 The HK Polytechnic University Transmission Lines & Cables 31 ' & $ % 2.5.2 Transpose Line D IA IB IC CA DAB DBC • Practical transmission lines cannot maintain symmetrical spacing of conductors because of construction considerations. • With asymmetrical spacing, even with balanced currents, the voltage drop due to line inductance will be unbalanced. • One way to regain symmetry in good measure is to consider transposition. This consists of interchanging the phase configuration every one-third the length so that each conductor is moved to occupy the next physical position in a regular sequence as shown in the figure below. IA B C IA IB IC IC IB B BC C A A A 32 ' & $ % • Since in a transposed line each phase takes all three position, the inductance per phase can be obtained by finding the average value of all phases, i.e. L = LA + LB + LC 3 • Noting IB = IA/240o, IC = IA/120o and IA + IB + IC = 0 L= µ0l 2pi [ ln 1 GMRc − 1 3 ( ln 1 DAB + ln 1 DBC + ln 1 DCA )] = µ0l 2pi ln 3 √ DABDBCDCA GMRc = µ0l 2pi ln GMD GMRc where GMD = 3 √ DABDBCDCA • This again is of the same form as the expression for the inductance of one phase of a single-phase line. The Geometric Mean Distance, GMD, is the equivalent conductor spacing. KWCn 16 The HK Polytechnic University Transmission Lines & Cables 33 ' & $ % 2.5.3 Composite Conductors • In practical transmission lines, stranded conductors are used and most EHV lines are constructed with bundled conductors for reasons of economy. • It is useful in finding the equivalentGMR and GMD of parallel circuits to determine the inductance of composite conductors. • Consider a composite conductors consisted of two groups of conductors each having m and n number if strands respectively as shown in the figure below. n d c b a d’ b’ a’ m x y • The current I is assumed to be equally divided among the subconductors, i.e. the current per strand is I n in x and I m in y. 34 ' & $ % • Applying the previous results, the total flux linkage of subconductor a in x is λa = µ0l 2pi [ I n ( ln 1 GMRc + ln 1 Dab + ln 1 Dac + . . .+ ln 1 Dan ) − I m ( ln 1 Daa′ + ln 1 Dab′ + ln 1 Dac′ + . . .+ ln 1 Dam )] = µ0lI 2pi ln m √ Daa′Dab′Dac′ . . .Dam n √ GMRcDabDac . . .Dan The inductance of subconductor a is La = λa I/n = µ0l 2pi n ln m √ Daa′Dab′Dac′ . . . Dam n √ GMRcDabDac . . . Dan The inductance of other subconductors in x are similarly obtained, e.g. Ln = λn I/n = µ0l 2pi n ln m √ Dna′Dnb′Dnc′ . . .Dnm n √ DnaD