Is 13235 Calculation of the Effects of Short-Circuit Currents 1991 Ed

8/10/2019 Is 13235 Calculation of the Effects of Short-Circuit Currents 1991 Ed

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I ndian St andard

IS 13235 : 1991EC Pub 885 ( 1988 )( Superseding IS 6726 1

(Reaffkmed 1997)

CALCULATION OF THE EFFECTS OF

SHORT CIRCUIT URRENTS

( First Reprint FEBRUARY 1999 )

UDC 621-317-613

BUREAU OF INDIAN STANDARDS

MANAK BHAVAN 9 BAHADUR SHAH ZAFAR MAR0

Reaffirmed 2002


5/32

IS 13235 I 1991

IECPob865 1986)

I ndian St andard

CALCULATION OF THE EFFECTS OF

SHORT CIRCUIT URRENTS

NATIONAL FOREWORD

This Indian Standard which is identical with IEC Pub 865 ( 1986 ) Calculation of the effectsof short-circuit currents, issued by the International Electrotechnical Commission ( IEC ) wasadopted by the Bureau of Indian Standards on the recommendations of the ElectricalInstallations Sectional Committee ( ET 20 ) and approval of the Electrotechnical DivisionCouncil.

An important criterion for the proper selection of a circuit-breaker or any other fault protectivedevices for use at a point in an electrical circuit is the information on maximum fault current

likely at that point. The electromagnetic, mechanical and thermal stresses which a switchgearand the associated apparatus have to withstand depends on the fault current.of breaking and withstand capacities play a major role in

Proper selectione health of the electrical installa-

tions. Realizing this need and to provr ei uniform gPide fo. calculation of short-circuitcurrents, IS 5728 was brought out in 1970.# , .,,

Subsequent to the preparation of this standa d, considerable information has beencollated the world over on calculation of fault levels under different and specific circumstances.There was also a need to simplify calculation techniques in practical way commensurate withthe modern arithmetic tools available to engineers in the form of computers, digital transientnetwork analysers etc. With an objective to establish a general, practicable and conciseprocedure for short-circuit current calculations IS 5728 has been taken up for revision, aligningits contents with IEC 865 ( 1986 ).

This Indian Standard supplements the provisions of IS 13234 : 199l/IEC 909 (1988) by coveringstandardized procedures for the calculation of the electromagnetic and thermal effects of faultcurrents in electrical circuits.


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As in the Original Standard, this Page is Intentionally Left Blank


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IS 13235 1991IECPub865(19S)

V, Ratio between stress with and without unsuccessful three-phase auto-reclosure. see Figure 5,page 30

V, Ratio between dynamic and static conductor stress, see Figure 4, page 30Vcls Ratio between dynamic and static sub-conductor stress, see Figure 4Z Section modulusZ, Section modulus for sib-conductor

a*

Factor for force on support, see Table II

Factor for main conductor stress, see Table IIY Factor for natural frequency estimat; 7, see Table II

E p, lyFactor for peak short-circuit currentFactors for tensile force in flexible conductor, see Figure 6, page 3 I

um Bending stress in main conductor0, Bending stress in sub-conductora,,, Resulting conductor stress

2.2 Symbol s for Secti on Tw o - Thermal effect s

cm3cm3

N/mm2N/mm*N/mm2

3.

. 3.1

3.2

3.3

3.4

3.5

3.6

3.7

Ik Steady state short-circuit current (r.m.s.)I[ Initial symmetrical short-circuit curient (r.m.s.)II:Li

Thermal equivalent short-circuit current (r.m.s.)Individual thermal equivalent short-circuit current at repeated short-circuits (m.s.)

Ithr Rated short-time current (r.ma.)m Factor for the heat effect of t&d.c. component, # Figure 7aj, page 32

ih

Factor for the heat effect of the a.c. component, see Figure 761 page 32Thermal equivalent short-circ&t current density (r.m.s )

S thr Rated short-time current de&ity (r.m.s.) for I s& Short-circuit durationTki Individual short-circuit duration at repeated short-circuits% Rated short-time

2Conductor temperature at the beginning of a short-circuit

* e Conductor temperature at the end of a : nort-circuit

General definitions

kAkAkAkA

kA

A/mm2A/mm2

S

S

G

T

M ain conductor

A conductor or a composed arrangement of a number of conductors carrying the totalcurrent in one phase.

Sub-conductor

A conductor which carries a certain part of the total current in one phase and is a part of themain conductor.

support

An insulating device between a live conductor and the earthed structure or between mainconductors.Note. - Under action of electromagnetic forces, in the given position, the support might be exposed to bending,

tension and/or compression due to electromagnetic forces between the conductors.

Fixed support

A support of a conductor which does not permit the conductor to move angularly at thesupported point.

Simple support

A support which permits angular movement of the conductor.

Spacer E

A mechanical element between conductors which, at the point of installation, prevents sub-conductors be&g displaced towards each other.

Stiffening elemknt

A special spacer intended to prevent any movement between rigid sub-conductors at itsplace. I


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IS 13235 : 1991IECPab865(1986j

3.8 Rated shor t-ti me cur rent

The r.m.s. current which the electrical equipment is able to withstand during the ratedshort-time under specified conditions.Notes 1. - It is possible to state several pairs of values of rated short-time current and short-time; for thermal effect

I s is used in most I EC specifications.

2. - The rated short-time current as well as the correspouding short-time are stated by the manufacturer ofthe equipment.

3.9 Rated shor t-ti me cur rent density and rated shor t-ti mefor conductors

The r.m.s. value of the current density which a conductor is able to withstand for the ratedshort time.Note. - The rated short-time current density is determined according to Clause 10.

3.10 Thermal equivalent short-cir cuit curr ent

A constant r.m.s. value of current having the same thermal effect and the same duration asthe actual short-circuit current, which may comprise d.c. component and may subside in time.

Note. - If repeated short-circuits occur (i.e. rwated reclosures) YesuIting thermal equivalent short-circuit currentis evaluated (see Sub-clause 10.2).

3.11 Thermal equivalent short-ci rcuit cur rentdensity,,

The ratio of the thermal equivalent short-time current and the cross-section area of theconductor.

3.12 Dur ation of short-cir cuit cur rent

The time interval between the initiation of ti e short-circuit and the final breaking of thecurrent in all phases by the circuit-breakers or fuses.NOW. - The time interval in which the current does not flow need qot be considered.

SECTION ONE - THE ELECTROMAGNETIC EFFECT ON RIGID CONDUCTORSAND SLACK CONDUCTORS

4. General

With the calculation methods presented in this section, forces on insulators, stresses in rigidconductors and tensile forces in slack conductors can be estimated.

5. Mechadcalforces due to short-circuit currents

5. General

Currents in parallel conductors will induce electromagnetic forces between the conductors.When the parallel conductors are long compared to the distance between them, the forces willact evenly distributed along the conductors.

When the currents are in opposite directions the electromagnetic force is a repulsion whichtends to induce deformations that would increase inductance of the circuit.

The value of the force in a given direction can be calculated by considering the work donein the case of a virtual displacement in the actual direction. As the work is done by the elec-tromagnetic force, it must be equal to the change in the energy in the magnetic field caused bythis virtual displacement. This leads to the following equation for the force F, in direction x

(1)

where L s the self-inductance of the circuit and i he instantaneous value of the current. .


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IS 13235 1991IECPob845(1981)

For simple and idealized configurations, formulae for the calculation of inductance areavailable. However, in most cases approximations are necessary to give practical calculationmethods.

The force between two conductors is proportional to the square of the current, or to theproduct of the two currents. As the current is a function of time, the force will also be a

function of time. In the case of a short-circuit current without a d.c. component the force willvary with twice the frequency of the current. A d.c. component in the short-circuit current willgive rise to an increase of the peak value of the force and to a component of force varying withthe same frequency as the current. The peak value of the force is of particular interest in thecase of mechanically rigid structures.

The force will result in bending stress on rigid conductors, tension stress and deflection inflexible conductors and bending, compression or tension loads on the supports.

5.2 Calculation offorces

The calculation method in this section is based on the general equation (l), neglecting thevariations in distance during tkdshort-circuit. d

5,2.1 Forces between two parallel coh uctorsThe forces in newtons acting between two long. parallel conductors are given by the

formula:

F=0.2ilhA N (2)

where I is the distance, in metres, between supports, il and i; are the instantaneous values ofthe currents in the conductors in kiloamperes, a is the centre-line distance between theconductors in metres.

When the currents in the two conductors have the same direction, the forces are attractive.When the direction of the currents is opposite, the forces are repulsive.

5.2.2 Calculation of peak force between the main conductors during a three-phase short-circuit

In a three-phase system with the conductors arranged at the same intervals on the sameplane, the maximum force acts on the central conductor during a three-phase short-circuitand is:

Fm3 - 0.2 i& - - 0.87 N (3)aswhere.63 in kiloamperes is the peak value of the a.c. component of the short-circuit current inthe case of a balanced three-phase short-circuit.Nore. - Formula (3) can also be used for calculating the resulting peak force when the conductors arc in the comers

of an quilateral triangle and where 4 is the length of the side of the triangk.

52.3 Calculation of peak force between the main conductors during a two-phase short-circuit

The maximum force acting between the conductors carrying the short-circuit currentduring a two-phase short-circuit is:

LF *2 1m2 - 0.2 lP2 - N

aswhere ip2.in kiloamperes is the peak short-circuit current in the case of a line-to-line short-circuit. .


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IS I3235 : 1991TEClub865(1986),

5.2.4 Calcul ation ofpeak value off orces between r igid sub-conductors

The maximum force acts on the outer sub-conductors and is between two spacers:

when the short-circuit current is evenly distributed between the sub-conductors; n is thenumber of sub-conductors, I, is the distance between spacers or stiffening elements, in metres,and a, is the effective distance between sub-conductors, in metres.

52.5 Effective distance between conductors and sub-conductors

The forces between current-carrying conductors are dependent on the geometrical conligur-ation and the profile of the conductors. For this reason a, has been introduced in Sub-clauses 5.2.2, 5.2.3 and 5.2.4. However, if the dimensions of the cross-sections of theconductors are smaller than the distance between the conductor centres, in Sub-clauses 5.2.2and 5.2.3 a, may be replaced by a.

4 \Some values for a, are given in Table I. For other Astances and conductor dimensions the

formula I 1. c.

1 kt2 kt3 kt4 kt,-P -.+a,

-+- + . + -ai2 a13 aI4 ain

(6)

can be used

The values for k12, kin are to be taken from Figure 1, page 28.Note. - The forces due to bends and the influence of a large amount of magnetic material close to the conductors

are usually of low importance. For the calculations, see existing publications.

6. Calculation of stresses in rigid conductors and forces on supports6.1 General

The conductors may be supported in different manners, either fixed or simple or in 5:combination of both, and may have two, three, four or several supports. Depending on thekind of support and the number of supports, the stress in the conductors and the forces on thesupports will be different for the same short-circuit current. The formulas given also take theelasticity of the supports into account.

The stresses in theconductors and the forces on supports also depend on the ratio betweenthe natural frequency of the mechanical system and the frequency of the electromagneticforce. Especially in the case of resonance, or near to resonance, the stresses and forces in thesystem may be amplified.

6.2 Calculation of str esses in igid conductors

The assumption that the conductor,,is rigid means that the axial forces are disregarded.Under this assumption the forces acting are bending forces and the general formula for thebending stress in the supported main conductor is:

Frnlurn= &rV,B - N/mm2, 8Z

(7)

where for F , either the value Fm3 (three-phase short-circuit) or Fm2 (line-to-line short-circuit)is to be used. , I

The bendingstress ida sub-conductor is:

*sa Gs V, 45- N/mm216z,

, (8)


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IS 13235 1991IECPnb865(1986)

where Zand Z, are section moduli, V,, V,, and Vr are factors taking into account the dyna-mic phenomena and B is a correction factor depending on the type of supports, fixed orsimple, and the number of supports.

The actor pis to be taken from Table II and V,, V, and Vr rom Figures 4 and 5, page 30.The maximum possible values are V, - V,, - 1.In the case of three-phase auto-reclosure, the factor Vr - 1.8 is to be applied, otherwise

If,- 1.For further consideration see Sub-clause 6.4.

6.2.1 Permitted conductor stressA single conductor is assumed to withstand the short-circuit forces when:

cm L 4 40.2 (9)

where R,o.J is the stress corresponding to the yield point.The factor q s to be taken fr&r Table III. pWhen a main conductor coysists of two or more sub-conductors the total stress in the

conductor is:

otot = urn + G (10)

Nore. - or,, is the algebraic sum of a,,, and crS ndependent of the loading atrections (see Figure 2, page 28). but thelocation is to be considered.

The conductor is assumed to withstand the short-circuit forces when:

atot L 4 50.2 (11)

It is necessary to verify that the short-circuit does not affect the distance between sub-conductors too much, otherwise a value

os L 40.2 (12)is recommended.

In Table III the highest acceptable values for q for different cross-sections are given. Forq > 1.0 small permanent deformations may occur (approximately 1% of the distance betweensupports for q-values according to Table III) which do not jeopardize the safety of operation.

Note. - For the yield point of conductor materials, Q.2, the standards often state ranges with minimum andmaximum values. If only such limit values rather than actual readings are available, the minimum valueshould be used in Sub-clause 6.2.1 and the maximum value in Sub-clause 6.3.

6.2.2 Section modulus of main conductors composed of sub-conductors with rectangul ar cross-sectionThe bending stress and conseqnently the mechanical withstand of the conductor depend on

the section modulus.If the stress occurs in accordance with Figure 2a , page 28, the section modulus 2 is equal to

the sum of the section moduli Z, of the sub-conductors (Z, with respect to the axis x-x).

If the stress occurs in accordance with Figure 2b)and in the case there is only one or no stiff-ening element within a supported distance the section modulus 2 is equal to the sum of the

section modul Z, of the sub-conductors (Z, with respect to the axis y-y).When withiq a supported distance there are two or more stiffening elements, higher values

may be used. For main conductors composed of sub-conductors of rectangular cross-sectionwith a space between the bars equal to the bar thickness, 60% of the section modulus with


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IS 13235 : 1991IEC Pub 865 ( 1986

respect to the axis O-O may be used. For conductor groups having profile cross-section,approximately 50% of the section modulus with respect to the axis O-O may be used. SeeTable IV.

6.3 Calculation offorces on supports of rigid conductors

The dynamic force Fd is to be calculated from:

Fd = VF JraFrn

where for F, either the value Fm3 or F,.,Q s to be used.

The maximum possible values of I+ - V, can be estimated as follows:

v,* v,- 1 when o, + a, >0.8 Rpc.2

(13)

v,- VF- O-8 Rp0.2grn + 0s

when a,,, + o, d 0.8 R,,u.2

The value of VF. V, is not higher than 2.7 for a three-phase short-circuit, and 2.0 for a two-phase short-circuit.

For further consideration.see Sub- &use 6.4. p

The factor a is dependent on the kind and the number of supports and is to be taken fromTable II. See Note 2 to Table II. Regarding permitted load on insulator, see Clause 8.

6.4 Calculation with special regard to conductor oscillation

The formulae in Sub-clauses 6.2 and 6.3 give bending stresses and forces on supportsachieved from a calculation of a load which is as- umed static. In Sub-clauses 6.2 and 6.3 threefactors, V,, VF and V, are introduced to take into account the natural frequency of theconductor. Some figures are also given for V,, V, and V, which give the highest possiblestresses and forces.

6.4.1 Calculation_ojrelevant naturalfrequency

The relevant natural frequency of a single conductor can be calculated from:

(14)

The factor y is dependent on the kind and number of supports and is given in Table II.

For a main conductor composed of sub-conductors, J and m must apply to the mainconductor design. If the sub-conductors are of rectangular cross-section the relevant naturalfrequency of the main conductor can also be calculated as:

where:

The factor c is to betaken from graph a) or b) of Figure 3, page 29. In the case of no stiff-ening elements c = 1.

For the calculatid$ of sub-conductor stress, taking the relevant natural frequency intoaccount, the formula (14) is to be used. In this case replace Iby 4, m by mg and J by .I, and usey = 0.356.Note. - The second moments Jand J, are calculated according to Figures 2a or 2b . page28. *


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J S 13235 : 1991IEC Pub 865 ( 1986 )

6.4.2 Thefacto rs V,, V,, V,, and V,

The factors VF, V,, V,, and V, as functions of the ratio fC/ where f is the system frequencyin hertz, are a little different if a two-phase short-circuit is concerned and they are alsodependent on the mechanical damping of the conductor system. For practical calculationsthese factors can be taken from Figure 4, page 30, for a multi-phase system. Excepted are

structures with small relevant natural frequencies of conductor fc/f < 0.5, where K < 1.6; forthese structures VF = V, = V,, = 1 is to be applied.

For three-phase auto-reclosure the factor V, can be taken from Figure 5, page 30; in othercases V, = 1.

7. Calculation of tensile forces in slack conductors (line conductors)

A short-circuit current in a slack conductor will cause a tensile force in the conductor whichNilI affect insulators, support structures and apparatus. It is necessary to distinguish betweenthe tensile force Ft during the &rt-circuit (de&ribed in Sub-clause 7.1) and the tensile forceF after the short-circuit, when .the conductor falls, back to its initial position (described inSub-clause 7.2). To determine t& dynamic stress of the supports, the higher value Ft or Ff is tobe used.Note. - For span lengths exceeding 20 m the u;e of the following formulae is not recommended.

7.1 Tensil e force Ft duri ng t he short- circuit

The force 4 is calculated with:

Ft = F,(l + W) N (17)

where F,, is the static tension in the conductor

9-3[d-55-1] (18)

where g, M is the mass of the conductor per metre.

F;, is the force per metre in the outer phase conductor produced by the short-circuit currentand calculated according to

123F&-O.15 - N/m (1%

a

where Zk3 s the three-phase short-circuit current in kiloamperes (r.m.s.) and u, in metres, thecentre-line distance between conductors.

The factor vis a function of Cand cp and given in Figure 6, page 3 1.

(m I)2 400

Fit 1 100-+-

lOS1 EA

(20)

I - distance betw&m supports,tin metres

9 - resulting spring constant of two conductor supports in newtons per millimetre

E - Youngs mpQulus in newtons per square millim+eA - conductor s+are section in square millimetres

For insulators, damps, etc., of station components the value S - 100 N/mm may be used ifno more accurate value is known.


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IS 13235 : 1991IECPab865(1986)

7.2 Tensil efor ce Ff aft er t he t i ort -cir cuit ( a11 f span)

This force is only considered when:

F;, > 0.6g0

(21)

In this case is to be applied:

Ff = F& O 4-c if - - >2 (23)g0

where Fst6c is the static tension in the conductor at a conductor-temperature of 60C.Notes 1. - The resonance frequencies of these flexible conductors are low, and higher forces due to resonance are

not expected.2. - Due to conductor motion the stres may be lower than,given by the formulae. The use of lower stress is

acceptable when it can be demons&t .d etther by calct&rtion or by tests.

3. - In the case of twin-conductors whet&he single conductors touch each other effectively the tension forceFr s higher than in formula (17).

8. Per@tted load on post insulators and terminals

The dynamic force Fd for rigid conductors and Ft and Ff for flexible conductors, shall notbe higher than the rated withstand value given by the support and insulator manufacturer. Foran insulator stressed by a bending force the rated withstand value is given for a force acting on

the insulator head. For a force acting at a point higher than the insulator head, a value lowerthan the rated withstand value which gives a moment the insulator can withstand, is to beused.

Connectors of rigid conductors shall be rated on the basis of Fd, while connectors forflexible conductors &all be rated on the basis of force peaks of 1.5 Ft or 1 O Ff.

Note. - The factor I.5 takes into account the effect of oscillations absorbed by the mass of the insulators.

SECTION TWO - THE THERMAL EFFECT ON BARE CONDUCTORS

9. General

This section gives calculation methods for the thermal effect on bare conductors. For cablesand insulated conductors reference is given to the appropriate I EC Technical Committee andSub-Committee*.

The heating of conductors due to short-circuit currents involves several phenomena of anon-linear character and other factors that have to be either neglected or approximated inorder to make a mathematical approach possible.

For the purpose of this section the following assumptions have been made:- Proximity-effect (magnetic influence of nearby parallel conductors) has been disregarded.

--Technical Committee No. 20: Electric Cables. Sub-Committee 20A: High-voltage Cables.

.


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IS 13235 : 1991IECPob865( 1986)

- Resistance-temperature characteristic has been assumed linear.

- The specific heat of the conductor is considered constant,

- The heating is generally considered adiabatic.

10. Calculation of temperature rise10.1 General

The loss of heat from a conductor during the short-circuit is very low, and the heating cangenerally be considered adiabatic. The calculation in this section is based on adiabatic con-ditions.

When repeated short-circuits occur with a short-time interval between them (i.e. rapid auto-reclosure) the cooling down in the short dead-time is of relatively low importance, and theheating can still be considered adiabatic. In cases where the dead-time interval is of longerduration (i.e. delayed auto-reclosure) the heat loss may be taken into account.

The calculation does not taGinto account#he skin effect, i.e. the current is regarded asevenly distributed over the conductor cross-sectionarea. This approximation is not valid forlarge cross-sections, and there* for cross-sections above 600 mm2 the skin effect shall betaken into account. For calculation, refer to the literature.Note. - If the main conductor is composed of sub-conductors, uneven current distribution between the,sub-

conductors will influence the temperature rise of sub-conductors.

10.2 Calculation of thermal equivalent short-circuit current

The thermal equivalent short-circuit current is to be calculated using the short-circuitcurrent r.m.s. value and the factors m and n for the time-dependent heat effects of the d.c. and

a.c. components of the short-circuit current.The thermal equivalent short-circuit current can be expressed by:

Ith = Ii,/- (24)where m and n are numerical factors, I[the r.m.s. value of the initial symmetrical short-circuitcurrent; in a three-phase system, the balanced three-phase short-circuit is decisive. The valuesm and n are shown in Figure 7, page 32, as functions of the duration of the short-circuitcurrent. For a distribution network usually n = 1.Nore. - The relation I[/ /k is dependent on the impedance between the short-circuit and the source.

When a number of short-circuits occur with a short time interval in between, the resultingthermal equivalent short-circuit current is obtained from:

Ith==j/G (25)

where:n

Tk- c Tki (26)I- I

10.3 Calculation oftemperature rise and rated short-time current densityfor conductors

The temperature rise in a conductor caused by a short-circuit is a function of the duration ofthe short-circuit current, the thermal equivalent short-circuit current and the conductormaterial.

By use of the graphs in Figure 8, page 33, it is possible to calculate the temperature rise of aconductor when the short-time current density is known or vice versa. .


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Is 13235 : 1991IEcPob86 5(I X?6)

The highest recommended short-time temperature for different conductors is given inTable V.Note. - The maximum permitted temperature of the support has to be taken into account.

10.4 Calculation of the thermal short-circuit strength for different durations of the short-circuit

currentElectrical equipment has suffkient thermal short-circuit strength as long as the following

relations hold for the thermal equivalent short-circuit current Zth:

orIth < I thr for Tk < & CW

Ith < I thr for Tk > Tkr Vb)

where Ztht s the rated short-time current and Tk, the rated short-time.

The thermal short-circuit strength fo&are conductogl is sufficient when the thermal equiv-alent short-circuit current density Su, satisfies the follovkng relation:

1 ,

(28)

with Tkr - 1 sand for all Tk.The rated short-time current density &hr is shown in Figure 8, page 33.

i


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IS 13235 1991

IECPub i5(1986)

0.8

0.6

1 2 4 6 8 10

ad d -

017m

FIG. 1. - Factor kl, for effective conductor central distance 4.

4

Directionde la charge-ELoading

direction

. tI

i

b)

Directionde la charge

Loading

direction

FIG. 2. - Loading direction and bending axis with multiple conductor arrangements.


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0 0.04 0.0 0. 12

ml (n m ; I -

IS 13235 1991

IECpPbt365(1986)

b

0.0 0. 12

m l ( n m I ) -ow86

FIG. 3. - Factor c for the influenceof spacers or stiffening elements in formula (15).a) direction of oscillation vertical to the surface, with k number of stiffening elements;,b vibration in any direction with k number of spacers or direction of vibration along

the surface with k number of stiffening elements;c) arrangement of stiffening elements or spacers.

lb


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IS 13235 1991IECFab865(1986)

VF. VU. v i

2c

1s

Ct

,I-

,

vu. U

/c

L

L

se

i

1

fJf-OSlM6

FIG.~. - Factors VF, V, and V,, to be used with the two-phase and three-phase short-circuits;except arrangements with&< 0.5 in which K < 1.6.

FIG. 5. - iactor V, to be used with three-phase auto-reclosure.


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IS 13235 1991

IECPub865(1986)

FIG.~. -

I O- ' 2 46810 2 4 6 810' 2 4 68102 2 4 68103

Factor w for tensile force in flexible conductors. The connection between the factors I,V,and cp s given by:

C=(l+cpVG* VI.P

2+(P+(Pw 1-v


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I S 13235 : 1991IECPab865(1 M3fQ

6)

I

o

0.8

n

0. 6

0. 4

0. 2

1 8

m

1. 4

0. 6

0,i 0. 2

.$ rk-054/86

0. 01 0.02 0.05 0. 1 0. 2 0. 5 1 2 5 10 srk

OJ.r/86

FIG. 7. - a) Factor m, heat dissipation due to d.c. component.b Factor n, heat dissipation due to a.c. component.


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IS 13235 1991

IECPub865(1986)

TABLE

Effective conductor central distance a, in metres, or r ectangular section dimensions

c

I0.05 0.06ectangularsections 0.20

-0.080

0.16

-0.067

0.12.10.08

bIll-J A

0.005 0.020 0.024 0.027 0.033 0.0400.010 0.028 0.031 0.034 0.041 0.047

0.015 0.018 0.0220.020 0.023 0.027

'

- -0.016 0.018

-0.020

0.0130.019

-0.015

-0.054

-0.030

0.0050.010

-0.037

-0.043

-0.031

Q.017

.cs+

I h-0.014

fh 1-JLd

0.0050.010

-0.022

-0.026


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IS 13235 1991

IECPub865(19m)

TABLE II

Thefactors a. ? andy f or di ff erent support arr angements

Type of beam and supportsFactor Factor Factor

.a B Y

Single span beam

Continuous beams withequidistant simplesupports

AandBsimple supports

f

A: 0.5B: 0.5

1.0 0.157

A B

A: fixed supportB: simple support 0.73 0.245

AandB:fixed supports

Two spans0.73 0.245

Three ormore spans

0.73 0.356

Nom I. - Non-uniform span in continuous beams may be treated, with a sufftcient degree of accuracy, as though themaximum span applied throughout. This means that:- the end supports are not subjected to greater stress than the inner ones;

- spans should not differ in length from adjacent ones by more than 20%. If that does not prove to be possible,the conductors can be decoupled using flexible joints;

2. - The @alues are so calculated that a permissible plastic deformation occurs. They are so chosen that the sameq-value (see Sub-clause 6.2. I and Table 111) can be used for all conductor-support arrangements.


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4

2.00

1.70

d (;- - 0.05). 1.34

( m 0.075). 1.37

i- O.loO), 1.40- 0.12s). I.44

I- 0.160). 1.48- 0.200). 1.5 I

E

-L 1.83

1 d.+ I.50

Ifl&Z..

I.19

El

-L

lsl3235:1B1IEClob&S(l986)~

TABLE III

Acceptable q-values or diflerent cross-sections


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Is 13235 1991IECPnb i5(1986)

TABLE IV

Section moduli Z in cubic centimetres of main conductors with two and mor estiffening element:

mxm

0.05x0.01 0.06x0.01 0.08x0.01 0.10x0.01 0.12x0.01

I

9.9 11.9 15.8 19.8 23.8

Fm{ &, 6.9 8.7 A.4 13.9 d t7.3 20.8

I 1..

31.7 39.6

__

TABLE V

Recommended highest conductor temperatur es in the case of shor t ci rcu it

Type of conductors I Max. recommended conductor temperature in case ofshort-circuit IMechanical loaded bare conductors, solid or

stranded, Al or Cu

Aluminium alloy

Loaded steel

Unloaded steel

Cables and conductors, paper and plastics insulated

2WC

l7OC

250C

3OOYz

* Recommended temperatures are dealt with by 1 EC Technical Committee NO. 20: Electric Cables, and Sub-Com-mittee 20A: High-voltage Cables.


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IS 13235 : 1991IECPub865 1986)

APPENDIX

EXAMPLES OF CALCULATION OF THE EFFECTS OF SHORT-CIRCUIT CURRENTS

Exampl e 1. Mechanical effects on single rigid conductorsBasis for the calculations in this example is a three-phase system with one bus-bar per phase. The

bus-bars are mounted on edge. The cross-section is 60 x 10 mm2 and the material is an Al-Mg-Si-alloy.

The following data are given:

1 - l.Om E - 70 O oo N/mm2a - 0.2 m J = 0.5 cm4

3-

30kA z - l.Ocm3m - 1.62 kg/m & RPo.2 - 12t$N/mmz up to 180 N/mm2

Number of spans: 3 * g. .

The natural frequency of the bus-bars is:

fc

Theatiois 1.05.This gives the following values to the factors VF; V, and Vr:

-6 0.2 .VF -1.8 if&vr


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Is13235:1 mlECPkb86 J (l986)

Example 2. Mechanical efkts on multiple rigid conductors

Basis for the calculations m this example is the same three-phase system as in example 1, withthree bus-bars per phase. Between two supports t@ere &re two spacers, The spacers are0.06 m X 0.06 m. The cross-sections of the irAvidual bus-bars are & X 10 mm& The distancebetwken the indivtdual bus-bars is 10 mm. The other data are the same as in example 1.

The natural frequency of the main bus-bar is:fc - cfo -41% -52.3 - 506 Hz

When c is taken from Figure 3b), page 29, wkh k 2 as a functibn of the ratio:

m, 1.6 0.06. 2 _ oo4- - 301.6. 1.0mil *

This gives c - 0.96.

The natural frequency of the individual bus-bars is:

fc AThe ratio - and T gives the following values to the factors VF, V,. V, and V,:

v, - 1.8if yF* &< 0.8 (with the upper value of I+) 2)aiot

.

v, - 1.0

Km - 1.0

v, - 1.0

Maximum bending stress in the main bus-bar corresponding to one phase is:am- V

4DlV B- *

821.0 1.0 9 0.73

783 . 1.0-

8 . 3.023.8 N/mm2

When Zis found from Figure Zb), page 28.z-3cm3

Maximum force between individual bus-blirs is:

-()[email protected]

Maximum bending stress* theindividual bus-bars is:i

FIlr- - 1.0. 1.04 ,v, 16z, ;; pa - 15.6N/mm2.


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Is 13236 : 1991lECPub86S(lp86)

Total bending stress in the bus-bar material is:

atot - q,, + us - 23.8 N/mm2 + 15.6 N/mm2 - 39.4 N/mm2Maximum bending force on the supports is:

Fd- VF Y,o& - 1.8. 1.0. 1.1 783 - 1550NThe bus-bar is assumed to withstand the short-circuit force if:

qot - 39.4 N/mm2 < 1.5 - 120 N/mm2a, - 15.6 N/mm2 < 120 N/mm2 >

(with the lower value of Rpo.2)

Conclusions:The bus-bar will withstand the short-circuit force.The supports have to withstand a bending force of 1550 N.

Example 3. Mechanical effects on slack conductorsBasis for the calculations in this example is a three-phase flexible bus-bar connection with one

all-aluminium stranded conduct r of A - 240 mp2 per phase. The span length is I - 11.5 m. Theanchor points at each end of the span are pos&ulators on steel substructures with a resultantspring constant of S - 100 N/m&Therefore the static tensions of the conductors Fst 400 N maybe considered nearly invariable within a temperature range of -20C to 60C.

The following data are given additionally and necessarily:

Ikj- 15kA E = 60 000 N/mm* gn - 9.81 m/s2

a - 1.25m m = 0.67 kg/m

The force Fr, s:

F&-O.15 s3= 0.15 g -27N/ma .

The factor 9 is:

.-3[j/w -I]-3[j,/m-1] -9.68

The factor Cis:

c-(mI)Z 400 (0.67.11.5)* 400- I - 3.95

F: 1 100 4003 1 100

10s 1+EA+

10*100*11.5 60000*240

For c - 3.95 and (p - 9.68 Figure.6, page 3 1, shows y/ - 0.59.The tension in the conductor during the short-circuit is given by:

h-&(1 +Q,@-400(1 +9.68.0.59)-2684NBecause:

, 6 27-- -4.11 >2gnm 9.8 100.67

a tensile force F; ter the short-circuit can occur which is given by:

Ff-Fst m -400 &+8-3.95 -2284N


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Is 13236: 1991IltCRib865(~~

Conclusions:The supports (post insulators and steel substructure) have to withstand a dynamic bending force

of 2 684 N, given by the conductor tension during the short circuit.The clamping device for the conductor anchoring shall be rated on a basis of force peaks of

1.5.2684-4026N.

Exampl e 4. Thermal effects on bare conductorsBasis for the calculation is the same three-phase system as in example 1.

Following data are given:.Ig - 17kA zk - 17kA ip3 - 30kAK * 1.25 =k - 2.0s

Whent&-65Tand0, - 17OT, s;h, is found from Figure 86), page 33.

Stht - 80 A/mm*

This gives: 7 ?2I:h~gSthtA80A/mlp*060mm. lOmm,-48kA

I

m and n are found from Figures 7~) and 76), page 32. The bus-bars have sufitient thermal strengthwhen:

Conclusion:

The bus-bars have sufftcient thermal strength.

i


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Documents

Is 13235 Calculation of the Effects of Short-Circuit Currents 1991 Ed