04- Flexible Cable and Load

8/12/2019 04- Flexible Cable and Load

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Flexible conductors & loads


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G. Trmouille Pages 2 / 44


TRAININGAREVA T&D

SYSTEMS - ERT


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Summary1 INTRODUCTION.......................................................................................................................4

2 FLEXIBLE CABLE BEHAVIOR ...............................................................................................5

3 CASE OF A SUBSTATION CONNECTION.............................................................................7

3.1 CASE OF FLEXIBLE CONNECTION BETWEEN EQUIPMENT .......................................................7 3.2 CONNECTION BETWEEN GANTRIES .....................................................................................7 3.3 FINITE ELEMENT SOFTWARE ...............................................................................................8

4 THE CATENARY CURVE FOR LEVEL SPANS FOR HANDS CALCULATION .. 11

4.1 CATENARY AND PARABOLIC SOLUTIONS FOR SAG IN LEVEL SPANS ..................................... 114.2 TOTAL CONDUCTOR LENGTH. ......................................................................................... 134.3 CONDUCTOR SLACK ....................................................................................................... 134.4 TOTAL AND HORIZONTAL TENSION .................................................................................. 144.5 LINEAR THERMAL AND ELASTIC ELONGATION................................................................... 144.6 LINEAR ELASTIC STRAIN -ALLALUMINIUM CONDUCTORS ................................................. 144.7 LINEAR THERMAL STRAINALLALUMINIUM CONDUCTOR ................................................ 154.8 SIMPLE CASE WITH A FLEXIBLE CASE CABLE SPAN (NO INSULATORS,NOT DROPPERS,NOSPACERS) ................................................................................................................................ 16

5 LOAD ON FLEXIBLE CONDUCTOR IN DAN/M.................................................................. 17

5.1 DEAD LOAD .................................................................................................................... 175.2 THE WIND ...................................................................................................................... 17

5.2.1 Typical approach: like IEC826 ..................................................175.2.2 Terrain roughness.....................................................................185.2.3 Assessment of meteorological measurements..........................185.2.4 Typical customer input ..............................................................195.2.5 Wind INPUT synthesis..............................................................19

5.2.6 The drag factor feeling..............................................................205.3 THE SHORT CIRCUIT ....................................................................................................... 20

5.3.1 origin formula .........................................................................205.3.2 In case of 2 conductors.............................................................205.3.3 The effect..................................................................................225.3.4 Application ................................................................................235.3.5 The simple approach ................................................................245.3.6 The IEC865 force evaluation ....................................................25

5.4 SEISMIC LOAD ................................................................................................................ 275.5 ICE LOAD ....................................................................................................................... 27

6 ALLOWABLE LOAD ON EQUIPMENT TERMINAL ............................................................ 28

6.1 DISCONNECTING SWITCH FROM

IEC62271-102DATED

2003

SHEET

43 ........................... 286.2 CIRCUIT BREAKER IEC62271-100DATED 2003 .............................................................. 316.3 CURRENT TRANSFORMERS IEC60044-1DATED 2003SHEET 45...................................... 336.4 INDUCTIVE VOLTAGE TRANSFORMERS IEC60044-2DATED 2003SHEET 45...................... 34

7 BALANCE SPAN / SAG / HEIGHT / ALLOWABLE STRESS / CLEARANCE ................... 35

7.1 STEP 1:CLEARANCE....................................................................................................... 357.2 STEP 2:COLLECT ALLOWABLE LOAD ............................................................................... 357.3 STEP 3INSTALLATION HYPOTESIS ................................................................................. 367.4 STEP 4:EVALUATE DE LOAD WITH PROPOSED CONFIGURATION ........................................ 36


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1 Introduction

The aim of this document is to explain the mechanical constraint on the short flexible bareconnection, located between equipments ( small span):

Allowable stress on equipment Sag tension, formula

CIGRE 324 : Sag-tension calculation methods for overhead lines., june 2007

CIGRE 214 The mechanical effects of short circuit currents in open air substations(part 2)

IEC 60826: Loading and strength of overhead transmission lines. IEC, Geneve

1991Are the reference standards


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Conductor tensions during the coldest periods of winter to allow for sufficient self-damping to prevent aeolian vibration-induced fatigue over the life of the line.

Conductor elongation diagramConductor elongation

As, relatively small changes in the length of suspended overhead conductors can causerather large changes in sag. A change of only 0.1 m in the length of a 300 m longconductor can change the sag by nearly 1 m. Small amounts of elastic, plastic, or thermalelongation yield the same result: a longer conductor, and relatively large changes in sag.

Linear Elastic (LE) Model Overhead conductors are modeled as linear springs with asingle elastic modulus and a single coefficient of thermal elongation. An effective elasticmodulus and effective coefficient of thermal elongation must be calculated for non-

homogeneous conductors (e.g. ACSR). Typical values of modulus and CTE are used.

Simplified Plastic Elongation (SPE) Model Overhead conductors are modeled aslinear springs. Plastic conductor elongation is calculated by adding a typical permanentchange in length (usually expressed as an equivalent temperature change). The amountof plastic elongation is based on experience rather than on laboratory tests or designloads.Normally, the conductor elastic modulus and coefficient of thermal elongation are singlevalued but for non-homogeneous conductors such as ACSR, the different tensions in thealuminum layers and steel core can be calculated for the typical plastic elongation but thevariation with design loading cannot. For high conductor temperatures, a typical knee-point temperature can be calculated but any dependence on conductor type, design load,and span length cannot.The excessively high initial loaded tensions that result from ignoring the initial non-linearbehavior of the aluminium layers are usually reduced by experienced engineers or usedas an additional safety margin in structure design.

The LE model ignores settlement & strand deformation as well as long-time or hightension plastic strain (though such plastic elongation may be allowed for by the use oflarge clearance buffers).

- The SPE modelignores the settlement and strand deformation that occur during initialtensioning of conductor but account for long-time or high tension creep plastic elongation


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by using a typical value of plastic conductor elongation (often expressed as a temperatureadder).

3 Case of a substation connection

3.1 Case of flexible connection between equipment

Due to small span (some meters), and small stress in the conductor (compare toconductor elastic limit), the elongation due to mechanical stress is neglect.

3.2 Connection between gantr ies

Nothing is neglect. Hand calculation is not possible due to:

Concentrated load due to spacer, anchor string Droppers .


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Finite element software are used to solve equations

3.3 Finite element software

HVS-SAMCEF

Aim of HVS-SAMCEF and calculation principleHVS-SAMCEF is a finite elements software which allows to analyse static or transient non-linearbehaviour of flexible conductors in different external conditions (climatic or electrical). It isdedicated to High Voltage Substations design.

The main hypothesis of the mathematic model is to neglect the bending and torsionnal stiffness ofthe cables.

HVS-SAMCEF is able to take into account :


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Great displacements (geometric non-linearities) Elasticity and inerty of the conductors Down droppers to the appartus terminals Concentrated mass

For the short-circuit cases : Short-circuit dynamic analysis Overheating of the conductors according to an adiabatic law Exact way of the short-circuit current flow (including droppers)

Industrial references of HVS-SAMCEF E.D.F. (Electricity of France) LABORELEC (Belgium laboratory of electrical industry) Janssen Engineering (Belgium) Fabrique Nationale / New Herstal (G.I.A.T. Industry - France) Tractebel (Belgium) Schneider (France) Alstom T&D (France)

HVS-SAMCEF has also been described in several international publications :

- The mechanical effect of short-circuit currents in open air substations , CIGRE, 1987- The mechanical effect of short-circuit currents in open air substations , CIGRE, 1996- and

INPUT DATA

Materials and GeometryHVS-SAMCEF calculates 2 adjacent phases in the same time.All points are given in a direct tridimensional orthogonal axis system (O,X,Y,Z).The origin 0 of the axis system is at the bottom of the tower.X axis is defined in the direction of the conductorY axis is perpendicular to the conductor from the first phase to the second

Z axis is in the vertical upside direction

The axis system will be a reference for the entire study.

Cases of study

Definition of the connections and cases availableThe connections length is defined by the imposed sag at the highest cables temperature inservice-state.This lead to the starting position of the connection, called maximum temperaturecase . For each other cases, the temperature of the conductors are then choosen in order toconsider the most unfavourable case.

The following loading cases can be calculated:

maximum temperature (equilibrium position) minimum temperature


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transversal wind (at a given temperature) frost (at a given temperature) frost + transversal wind (at a given temperature) short-circuit (at a given temperature)

short circuit + transversal wind (at a given temperature) short-circuit with reclosing (at a given temperature) short-circuit with reclosing + transversal wind (at a given temperature)

HVS-SAMCEF calculates for each loading case the actual cables tensions in Newton (N), and thecorresponding forces at the anchor points on the gantries at both sides. For each static case (i.e :without short-circuit), HVS-SAMCEF gives the displacement in meters (m) of the sag point. Foreach dynamic case, HVS-SAMCEF gives the minimum distance remaining between the 2 phases.The following tables extract the higher values between the 2 phases results.Each force is defined by its 3 componants in the (O,X,Y,Z) axis system.

Study at maximum temperature (equilibrium posit ion)The connection is defined in order to obtain the required sag in the following conditions oftemperature :

- The insulators strings are at maximum ambiant temperature of 50C- The temperature of the cables, supposed to be live, is their maximum service-state

temperature of 67C.- Wind pressure : None-Study at minimum temperatureThe conductors and insulator strings are considered to be at the minimum ambiant temperature20C(as per cls E9 1.04.00)- Wind pressure : None-Study with transversal windWind pressure : 750N/m

2.( as per IS 802)

Temperature of the conductors and insulator strings : : 20C( daily average temperature as perchapter E9).

Studies with short-circuitA bi-phase short-circuit current Icc is applied on the conductors and droppers. This short-circuitcurrent is supposed to have an assymetry of 100%. It is supposed to flow in each phase from thefirst to the last dropper, which is the most unfavourable case.

Wind pressure : 750N/m2.( as per IS 802)

Temperature of the conductors and insulator strings : 37C.Temperature coming from medium temperature ( between 7C and 50C) and medium cableheating ( 17C maxi)

Equivalent short circuit 40 kA 3 phased ( 34.6 kA biphased)Duration 1s

The gantry will be checked with 2 phases under short circuit ( lateral forces in same direction,which is a pessimistic assumption for grantry calcculations ), the third phases will be loaded

without short circuit.


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4 The Catenary Curve for Level Spans for hands

calculation

The shape of a catenary is a function of the conductor weight per unit length, w, and thehorizontal component of tension, H. The sag of the conductor, D, is a function of theseparameters, the span length, S, and the difference in elevation between the span supportpoints.

4.1 Catenary and parabolic solutions for sag in level spans

Conductor sag and span length are illustrated in Figure 6 for a level span. More detail isprovided in [3,4,5,6,7,8,9]

The Catenary Curve for Level Spans

The catenary equation for a conductor between supports at equal heights or at differentheights is the same. It is expressed in terms of the horizontal distance, x, from the vertex(or low point) of the catenary to a point on the catenary which is y(x) above the vertex.The catenary equation is given by


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The expression on the right side of equation (1) is an approximate parabolic equationbased upon the first term of the MacLaurin expansion of the hyperbolic cosine. Theapproximate parabolic equation is valid as long as x w/12H


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w 2.54 0.981daN

m:= S 10m:= D H( )

H

wcosh

w S

2 H

1

:=

500 1 103 1.5 103 2 103

0

0.5

1

1.5

2

D x( )

x

4.2 Total Conductor Length.

Application of calculus to the catenary equation allows the determination of conductorlength, L(x), measured along the conductor from the low point of the catenary in eitherdirection as a function of x.The equation for conductor length between supports is:

For a level span, the conductor length corresponding to half of the total conductor length

is at the spans midpoint (X = S/2), thus:

The parabolic equation for the total conductor length can also be expressed as a functionof sag, D, by substitution of the sag parabolic

4.3 Conductor SlackConductor slack is the difference between the total conductor length, L, and the chorddistance between supports. For a level span the distance between supports is the spanlength, S. Equating and rearranging the preceding exact equations for L and S, the slack(L-S) for the example problem is:


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4.4 Total and Horizontal Tension

As noted previously, at the low point of the catenary, the conductor tension is equal to thehorizontal component of tension. At every other location within the span, including the endsupports, the total conductor tension is greater than the horizontal component.

For a level span, at the supports, the vertical component of tension is equal to half theweight of the conductor:

At the end supports, the total tension is the vector sum of the horizontal and verticaltension components

4.5 Linear Thermal and Elastic Elongation

The original length of the conductor is assumed to be that produced by the manufacturer(manufactured length) at the time that the conductor was stranded and wound on thereel for delivery. On the reel as delivered, the conductor tension is nearly zero. If theconductor has a steel core, then the length of steel core and the aluminum layers wound

around it are nearly equal. Rawlins [19] has suggested that there may be residualstressesin the aluminium layers as a result of the stranding process but this is only important whenestimating high temperature sag.

4.6 Linear Elastic Strain - All Aluminium Conductors

As described in IEC technical brochure 1597, the elastic elongation of a strandedconductor can be represented by the equation:


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L = Length of conductor under horizontal tension H REF L =Reference length of conductor under horizontal tension for no ice orwind load at everyday temperature H REFE = Modulus of elasticity of the conductorA = Conductor cross-sectional area= Strain or relative elongation expressed in s or 110-6

H/A = Stress (tension divided by area)

IEC technical brochure 1597

100 200 300 4000

0.02

0.04

0.06

0.08

0.10.1

0

e_H x( )

400m0.1m x = Strain or relative elongation

4.7 Linear Thermal Strain All Aluminium Conductor

For all aluminium (alloy) conductors, the equation for linear thermal elongation is


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A=Aluminium coefficient of linear thermal elongation (2310-6 C-1)TC= change in conductor temperature in C

These cable length variations are important for the design of large span.

4.8 Simple case with a flexible case cable span ( no insulators, notdroppers, no spacers)

Initial condition:

Customer wants to have a 30m span connection. With a 3% sag at 50C.Cable: ASTER 581 mm

=> evaluate the Horizontal stress He=> evaluate the cable length Le

Evaluate the cable modification for a cable temperature of 85C and an overall load oncable 2.54 daN/m => 5 daN/m


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5 Load on flexible conductor in daN/m

5.1 Dead load

The dead load , in coming from the cable weight

Dead load ( N ) = Weight (kg) x gravity acceleration ( 9.81m/s, typically)

5.2 The wind

5.2.1 Typical approach: like IEC826

The characteristic value a of the unit-action, due to the wind blowing horizontally,perpendicularly to any element of the H.V. substation (conductors, insulators, H.V.apparatus, all or part of the support and gantry) is given by the following expression:

a = q0Cx G, (daN/m)

Where:q0= dynamic reference wind pressure;Cx = drag coefficient depending on the shape of the element being considered;G = Combined wind factor which takes into account the turbulence of the wind. It varies interms of the dynamic response of the element being considered (gust response). It alsodepends on the height of this element above the ground and, for conductors, on the span

length.

The dynamic reference pressure q0is given in terms of the reference wind velocity VRxatthe location of the line:


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= air mass per unit volume (equal to 1.225 kg/m at a temperature of 15 C and under a normalatmospheric pressure of 1 013 mbar);

= air density correction factor depending on air temperature and altitude typically: 1q0= dynamic wind pressure expressed in N/m and VRx in m/s.The wind velocity varies with height above ground. There are two possibilities to take care of thiseffect:- by selecting the combined wind factor G depending on the height as proposed in [1] and- by calculating the reference wind velocity VRx,z for various heights using the equation (5.3) or(5.6).

5.2.2 Terrain roughness

Terrain roughness has an effect on wind action. The greater this roughness, the more turbulentand slower is the wind. The terrain roughness has an influence both on the determination of thewind velocity for the design and on the determination of the gust factor.Four categories of terrain, of increasing roughness, are considered as indicated in Table

Open sea will have a lower terrain roughness than terrain type A.Lines crossing highly urbanized areas should be considered in a D terrain roughness. The terrainroughness is very difficult to assess for these areas. However, due to the described higherreliability of lines in these areas, design according to terrain category B or C are recommended.For a line which follows the ridge of a hill, a roughness which is one category less favourable thanthat chosen for the area should be adopted. For a line running along a valley, the C roughnessshould be chosen for all cases, whatever the characteristics may be

5.2.3 Assessment of meteorological measurements

Wind action is determined on the basis of the meteorological wind velocity Vx,10min defined as themean value of the wind during 10 min period recorded 10 m above ground in a terrain category x.

Usually, meteorological stations, except those along the coast or in urban areas, are placed inareas of roughness category B. Then the meteorological wind velocity will be VB,10min.Alternatively, the meteorological wind velocity may be recorded in a height of 10 m as a meanvalue over a period of time of t seconds. Let Vx,t be this velocity. If it is not measured at 10 m heightabove ground, the data should be first adjusted to this level, the conversion of recordings averagedon a period t to the 10 min reference period:

Vx,10min = Vx,t 1 / (Vx,t / Vx,10min)where Vx,t / Vx,10min can be obtained from Figure 9 as a function of the averaging period ofmeasurements for each terrain category x at the location of the meteorological station.


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From measurements in a terrain category x different from B the reference wind velocity VB,10min interrain category B is given by the following relation:

VB,10min = Vx,10 min kRwhere Vx,10min is the recorded wind velocity in terrain x. Values for the terrain factor KR are given in :

The variation of Vx,10min in terms of the height was not taken into account, as anemometers are,most of the time, placed at a height of about 10 m above surrounding ground. If this height z (m)differs from 10 m, the variation of wind velocity with height z can be derived from the followingequation (5.3) as a function of the wind velocity Vx,10min measured 10 m above ground.

5.2.4 Typical customer input

The customer give typically the unit action a for to contract, in this case we do not neet

5.2.5 Wind INPUT synthesis

In any case the Wind hypothesis to be consolidated in an internal document


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5.2.6 The drag factor feeling

He depend on the surface shape

5.3 The short circu it

5.3.1 origin formula

When a wire carrying an electrical current is placed in a magnetic field, each of themoving charges, which comprise the current, experiences the Lorentz force, and togetherthey can create a macroscopic force on the wire (sometimes called the Laplace force). Bycombining the Lorentz force law above with the definition of electrical current, thefollowing equation results, in the case of a straight, stationary wire:

wheredF = Force, measured in NewtonsI = current in wire, measured in amperesB = magnetic field vector, measured in tesladL = a vector, whose magnitude is the length of wire (measured inmetres), and whose direction is along the wire, aligned with thedirection of conventional current flow.

5.3.2 In case of 2 conductors


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The force of attraction or repulsion between two current-carrying wires (see Figure 1) isoften called Ampre's force law. The physical origin of this force is that each wire

generates a magnetic field (according to the Biot-Savart law), and the other wireexperiences a Lorentz force as a consequence.The best-known and simplest example of Ampre's force law, which underlies thedefinition of the ampere, the SI unit of current, is as follows: For two thin, straight,stationary, parallel wires, the force per unit length one wire exerts upon the other in thevacuum of free space is :


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Where kmis the magnetic force constant, r is the separation of the wires, and I1, I2 are theDC currents carried by the wires. The value of kmdepends upon the system of unitschosen, and the value of km decides how large the unit of current will be.

With 0 the magnetic constant, defined in SI units

Thus, for two parallel wires carrying a current of 1 A, and spaced apart by 1 m in vacuum,the force on each wire per unit length is exactly 2 10-7 N/m.

5.3.3 The effect

Movement of a flexible conductor during and after a line-to-line short-circuit


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Measured tensile force of a flexible bundle conductor during and after a line-to-line short-circuit (Fst , static force, Fpi, pinch force, Ft, tensile force, Ff drop force)

Some Videohttp://www.tdee.ulg.ac.be/doc-5.html

5.3.4 Application

I1

I2


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5.3.5 The simple approach

Ampirical simple approach

ELECTRA 12

F(short circuit) = (0.3 x 0.866 x 2.04 x 2.55 x Isc / d) x 10-8 result in daN/m, d, phase spacingin m, Isc, rated 3 phases short circuit value


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5.3.6 The IEC865 force evaluation


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5.4 Seismic load

The cable is flexible, typically no specific additional seismic load is considered

5.5 Ice load

It could be an important issue in the concerned countries.

Heavy radial ice buildup on a relatively small bare overhead conductorThe snow is considered as an increasement of the conductor weight


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6 Allowable load on equipment terminal

6.1 Disconnecting switch from IEC62271-102 dated 2003 Sheet 43

The rated mechanical terminal load shall be stated by the manufacturer.Disconnectors and earthing switches shall be able to close and open while subjected totheir rated static mechanical terminal loads.The maximum static mechanical terminal load to which the terminal of a disconnector orearthing switch is allowed to be subjected under the most disadvantageous conditions isthe rated static mechanical terminal load of this disconnector.Recommended rated static mechanical terminal loads are given in table.The maximum external dynamic mechanical load to which the terminal of a disconnectoror earthing switch is allowed to be subjected is the rated dynamic mechanical load of this

disconnector.Disconnectors and earthing switches shall be able to withstand their rated dynamicmechanical terminal load under short-circuit.The rating of the disconnector or earthing switch for terminal loads depends not only onthe design, but also on the strength of the insulators used.The required cantilever strength of an insulator shall be calculated taking intoconsideration the height of the terminal above the top of the insulator as well as additionalforces acting on the insulator .Should be obtain from equipment manufacturer:

mechanical terminal load

external load acting on each terminal

NOTE 1 The external load is the result of the combined mechanical forces to which thedisconnector or earthing switch may be subjected. Wind forces acting on the equipmentitself are not included as they do not contribute to the external load.NOTE 2 A disconnector or earthing switch may be subjected to several mechanical forcesdifferent in value, direction and point of action.NOTE 3 The terminal loads as defined here do not usually apply to enclosed switchgear.

static mechanical terminal load

Static mechanical terminal load at each terminal equivalent to the mechanical force towhich this terminal of the disconnector or earthing switch is subjected by the flexible orrigid conductor connected to this terminal


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Dynamic mechanical terminal load

Combination of the static mechanical load and the electromagnetic forces under short-circuit conditions

The most disadvantageous conditions should be considered by the user when specifyingthe rated terminal loads.NOTE It is recommended to calculate the required static terminal load on the basis of : minimum specified ambient air temperature, and 10 C plus ice load plus wind load, or 5 C plus wind load (tropical countries).To calculate the required static and dynamic terminal loads as well as the requiredstrength of insulators, the forces resulting from the conductors connected to thedisconnector or earthing switch, including the forces of wind and ice (if applicable) onthese conductors should be considered.

Recommended static mechanical terminal loads


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6.2 Circuit breaker IEC62271-100 dated 2003

The static terminal load test is performed to demonstrate that the circuit-breaker operatescorrectly when loaded by stresses resulting from ice, wind and connected conductors.The static terminal load test is applicable only to outdoor circuit-breakers having ratedvoltages of 52 kV and above.If the manufacturer using calculations, can prove that the circuit-breaker can withstand thestresses, tests need not be performed.Ice coating and wind pressure on the circuit-breaker shall be in accordance with 2.1.2 ofIEC 60694.Some examples of forces due to flexible and tubular connected conductors (not includingwind or ice load or the dynamic loads on the circuit-breaker itself) are given as a guide intable.The tensile force due to the connected conductors is assumed to act at the outermost endof the circuit-breaker terminal.For simultaneous action of ice, wind and connected conductors, the resultant terminalforces, FshA, FshB and Fsv are defined as rated static terminal loads.

Examples of static horizontal and vertical forces fo r static t erminal load test

Horizontal force Vertical force


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STATIC TERMINAL LOAD FORCE


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6.3 Current transformers IEC 60044-1 dated 2003 sheet 45

These requirements apply only to current transformers having a highest voltage for equipmentof 72,5 kV and above.In table guidance is given on the static loads that current transformers shall be capablewithstanding. The figures include loads due to wind and ice.The specified test loads are intended to be applied in any direction to the primary terminals.

Static withstand test loads

Modalities of application of test loads to be appliedto the primary terminals

Horizontal to each terminal

Vertical to each terminal

NOTE: The test load shall be applied to the centre of the terminal.


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6.4 Induct ive voltage transformers IEC 60044-2 dated 2003 sheet 45

These requirements apply only to inductive voltage transformers having a highest voltagefor equipment of 72,5 kV and above.In table, guidance is given on the static loads that inductive voltage transformers shall becapable of withstanding. The figures include loads due to wind and ice.The specified test loads are intended to be applied in any direction to the primaryterminals.

Static withstand test loads

NOTE 1: The sum of the loads acting in routinely operating conditions should not exceed

50 % of the specified withstand test load.

NOTE 2: In some applications voltage transformers with through current terminals shouldwithstand rarely occurring extreme dynamic loads (e.g. short circuits) not exceeding 1,4times the static test load.

NOTE 3: For some applications it may be necessary to establish the resistance to rotationof the primary terminals. The moment to be applied during the test shall be agreedbetween manufacturer and purchaser.


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7 Balance Span / Sag / Height / allowable stress / clearance

Allowable stress on VT and CT

Example Um=420 kV, BIL 1425 kV, SIL 1050 kV, twin ASTER 851mmm, Isc 40 kA/1s

7.1 Step 1: clearance

IEC 61936 &

71-2

Minimum clearance P-G, rod structure if any N

Minimum clearance P-G, conductor structure n

Minimum clearance P-P, rod conductor P

Minimum clearance (P-P conductor, // conductor) p

Minimun clearance ground level - insulator base H-N

Minimum height of live parts over access area H

Danger zone see drawings DLWorking clearance Dv horizontal Dv

see drawings Dv Dv vertical

7.2 Step 2 : col lect allowable load

VT allowable load ( daN) CT allowable load (daN)

Normal operation (dead weight only)

75 daN 200 daN

Static load 150 daN 400 daN


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Dynamic 1.4 x 150 daN 1.4 x 400 dAN

7.3 Step 3 installation hypotesis

Collect data from proposed design (tender AREVA proposal, customer drawings)equipment drawing (with height of equipment) , selected phase phase distance

7.4 Step 4 : Evaluate de load with proposed configuration

The load evaluation refer to a calculation method, the contract typically, require to follow adedicated method

In this case we assume:Resulting load, in case of combination

Wind Wind + Short circuit


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Dead load = weightIn some contract, the wind

speed to consider in case ofcombination with the shortcircuit, is reducted compared tothe wind speed without shortcircuit

In order to calculate the load of terminal with the typical formula

H wS2

8 D:=

Calculate win case of static load (Wind + dead load)=> calculate the resulting traction on the terminal, and compare to the limit

Calculate win case of dynamic load (Wind + dead load)=> calculate the resulting traction on the terminal, and compare to the limit

Data this numerical application :

Wind load evaluation:The characteristic value aof the unit-action, due to the windblowing horizontally, perpendicularly to any element of the H.V. substation (conductors,insulators, H.V. apparatus, all or part of the support and gantry) is given by the followingexpression: a = q C

Where the dynamic pressure q = 1/2 V q = V/16.3 ( V in m/s)

:volumic air mass, equal to 1.225 kg/m3 at 15C under normal atmospheric pressure of1013mbar ( IEC 826-2)

The calculus of the dynamic pressure didnt depend on calculation standard, but the dragfactor C depends on the standard., Could be assumed as 1.2, without any otherinformation


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Wind pressure

q1

16.3

155

3.6

2

1daN

m2

114daN

mm=:=

cable data

d 37.95mm:= span 5m 3.125m+ 0.3m( ):= sag 1.2m:= typicaly 10% is a stating point

dead_load 2.3540.981 2daN

m:= conductor cross section A 851mm

2:=

Wind transversal load

w_wind 1.2 q d 2 10.358daN

m=:=

Dead Load

w_weight dead_load:= H_DL span( ) w_weightspan

2

8 sag:=

H_DL span( ) 29 daN=

resulting static load

w_static w_wind2

w_weight2

+ 11.31

mdaN=:=

evaluation : Static terminal load

H_stat span( ) w_staticspan

2

8 sag:=

H_stat span( ) 72 daN=

L_stat span( )2H_stat span( )

w_wind

shw_wind span

2 H_stat span( )

w_wind 0.5:=

L_stat span( ) 43 daN=

Tstat H_stat span( )( )2

L_stat span( )( )2

+ 83.991daN=:=


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evaluation : Dynamic terminal load

Isc 40000:=

w_sc 0.3 0 .867 2.04 2.552 Isc2

6.5 1 10 8 1 daN

m:=

w_sc 8.493daN

m=

resulting dynamic load

w_dyn w_wind w_sc+( )2

w_weight2

+ 19.41

mdaN=:=

H_dyn span( ) w_dynspan

2

8 sag:=

H_dyn span( ) 124 daN=

L_dyn span( ) 2H_dyn span( )w_sc

sh w_sc span2 H_dyn span( )

w_sc 0.5:=

L_dyn span( ) 34 daN=

Tdyn H_stat span( )( )2

L_stat span( )( )2

+ 83.991daN=:=


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initial reference contitions

qe 50C:=

we 2.354daN

m:= span 30m:= sag_e 3%:= is the sag at the initial reference temperature

He wespan

2

8 span sag_e 2943N=:=

cable length Le span 8span sag_e( )

2

3 span+ 30.072m=:=

Le2He

wesh

we span

2 He

30.072m=:=


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modification of the load, modification of the temperature qf 85C:=

wf 5daN

m:= Lf Hf ( )

2Hf

wfsh

wf span

2 Hf

:= DL Hf( ) Lf Hf ( ) Le:=

?L is due to the temperature variation, and the electic deformation

aluminium coefficient of linear thermal expension a 2.3 105

1

C:=

DL_thermal Le a qf qe( ):=

due to mechanical stress

modulus of electicity of the conductor E 58.3 109Pa:=

DL_load Hf( ) LeHf He( )

E A:=


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f Hf( ) DL Hf( ) DL_thermal DL_load Hf( )+( ):=

should equal to zero, to obtain the solution

2 103

4 103

6 103

8 103

1 104

1

0.5

0

0.5

1

f Hf( )

Hf

f 5360N( ) 2.663 104

m=

5.3 103

5.32 103

5.34 103

5.36 103

5.38 103

5.4 103

0.01

5 103

0

5 103

0.01

f Hf( )

Hf

Lf 5360N( ) 30.098m=

Hf 5360N:=

the final vertical sag is Sag_f% wf span

2

8 Hf span 3.498%=:=


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Hyperbolic cosine

Approximation

Clearances


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Documents

04- Flexible Cable and Load