CAMPOS_GUIA

Merlin Gerin technical guide Medium VoltageMV design guide We do more with electricity. Design Guide Goal This guide is a catalogue of technical know-how intended for medium voltage equipment designers. c Presenting and assisting in the selection of MV equipment in conformity with standards. c Providing design rules used to calculate the dimensions or ratings of an MV switchboard. How? c By proposing simple and clear calculation outlines to guide the designer step by step. c By showing actual calculation examples. c By providing information on units of measure and international standards. c By comparing international standards. In summary This guide helps you to carry out the calculations required to define and determine equipment dimensions and provides useful information enabling you to design your MV switchboard. Schneider Electric Merlin Gerin MV design guide 1 General contents MV design guide Presentation Metal-enclosed factory-built equipment Voltage Current Frequency Switchgear functions Different types of enclosures 5 5 6 8 9 9 10 Design rules Short-circuit power Short-circuit currents Transformer Synchronous generator Asynchronous motor Reminder Three phase calculation example 11 11 12 13 14 14 15 17 Busbar calculation Thermal withstand Electrodynamic withstand Intransic resonant frequency Busbar calculation example 21 24 27 29 31 Dielectric withstand Dielectric strength of the medium Shape of parts Distance between parts Protection index IP code IK code 38 38 39 39 41 41 41 Switchgear definition Medium voltage circuit breaker Current transformer Voltage transformer Derating 45 45 54 61 64 Units of measure Basic units Common magnitudes and units Correspondence between Imperial units and international system units (SI) 67 67 67 69 Standards Quoted standards IEC-ANSI comparison 71 71 72 References Schneider Electric documentation references 81 81 Index 83 Schneider Electric Merlin Gerin MV design guide 3 Presentation Metal-enclosed, factory-built equipment Introduction To start with, here is some key information on MV switchboards! reference is made to the International Electrotechnical Commission (IEC). In order to design a medium-voltage cubicle, you need to know the following basic magnitudes: c Voltage c Current c Frequency c Short-circuit power. The voltage, the rated current and the rated frequency are often known or can easily be defined, but how can we calculate the short-circuit power or current at a given point in an installation? Knowing the short-circuit power of the network allows us to choose the various parts of a switchboard which must withstand significant temperature rises and electrodynamic constraints. Knowing the voltage (kV) will allow us to define the dielectric withstand of the components. E.g.: circuit breakers, insulators, CT. Disconnection, control and protection of electrical networks is achieved by using switchgear. c Metal enclosed switchgear is sub-divided into three types: v metal-clad v compartmented v block. Schneider Electric Merlin Gerin MV design guide 5 : IEC 694. The rated voltage is always greater than the operating voltage and. c Impulse withstand voltage 1. is associated with an insulation level. (root mean square) value of the voltage that the equipment can withstand under normal operating conditions. Example: c Operating voltage: 20 kV c Rated voltage: 24 kV c Power frequency withstand voltage 50 Hz 1 mn: 50 kV rms. this is the maximum rms. in article 6.2/50 µs: 125 kV peak.B. Insulation level Ud (kV rms. factory-built equipment Voltage Operating voltage U (kV) This is applied across the equipment terminals. The voltage wave that results is simulated in a laboratory and is called the rated lightning impulse withstand voltage. accompany all changes in the circuit: opening or closing a circuit. Rated voltage Ur (kV) Previously known as nominal voltage. breakdown or shorting across an insulator. etc… It is simulated in a laboratory by the rated power-frequency withstand voltage for one minute. c Up: overvoltages of external origin or atmospheric origin occur when lightning falls on or near a line. N. c Ud: overvoltages of internal origin.Presentation Metal-enclosed. the dielectric testing conditions. article 4 sets the various voltage values together with. 1 mn) and Up (kV peak) This defines the dielectric withstand of equipment to switching operation overvoltages and lightning impulse. 6 Merlin Gerin MV design guide Schneider Electric . derating should be considered.2 12 17. factory-built equipment Standards Apart from special cases.8 to 15 20 to 22 25.2 28 12 38 50 70 Ud 60 75 95 125 170 Up 17.5 24 36 Insulation levels apply to metal-enclosed switchgear at altitudes of less than 1 000 metres. Above this.2 µs 50 µs Rated lightning withstand voltage 20 7. 7.2 12 17. 20 28 38 50 70 Normal operating voltage kV rms. 3.5 24 36 Ur Schneider Electric Merlin Gerin MV design guide 7 .2/50 µs 50 Hz kV peak list 1 40 60 75 95 145 list 2 60 75 95 125 170 Rated power-frequency withstand voltage 1 minute kV rms.5 Um t Rated voltage Rated power frequency withstand voltage 50 Hz 1 mm 0 1.8 to 36 kV rms.3 to 6.5 24 36 Rated impulse withstand voltage 1.Presentation Metal-enclosed. MERLIN GERIN equipment is in conformity with list 2 of the series 1 table in IEC 60 071 and 60 298. Each insulation level corresponds to a distance in air which guarantees equipment withstand without a test certificate. 11 g/m3 humidity and a pressure of 1 013 mbar.6 10 to 11 13. Rated voltage Rated lightning impulse withstand voltage 1.2/50 µs kV peak 60 75 95 125 170 Distance/earth in air cm 10 12 16 22 32 IEC standardised voltages U Um 0. 20°C. Rated voltage kV rms. 7. 5 • 1.: rated currents usually used by Merlin Gerin are: 400. Examples: c For a switchboard with a 630 kW motor feeder and a 1 250 kVA transformer feeder at 5. max.9 8 Merlin Gerin MV design guide Schneider Electric . If we do not have the information to calculate it.5 • 1.732 • 0.9 • 0. Rated normal current: Type of mechanism of material Max. It is the current that really passes through the equipment.9 η = motor efficiency = 0. the customer has to provide us with its value. Max. without exceeding the temperature rise allowed in standards. bare copper alloy or aluminium alloy 90 silver or nickel plated 115 tin-plated 105 N. value of current that equipment can withstand when closed. temperature of conductor (°C) contacts in air bare copper or copper alloy 75 silver or nickel plated 105 tin-plated 90 bolted connections or equivalent devices bare copper. 630.5 kV operating voltage. The operating current can be calculated when we know the power of the current consumers. v calculating the operating current of the transformer feeder: Apparent power: S = UIe I= 1 250 S = Ue 5. temp.Presentation Metal-enclosed. factory-built equipment Current Rated normal current: Ir (A) This is the rms. .40 °C 35 65 50 50 75 65 Operating current: I (A) This is calculated from the consumption of the devices connected to the circuit in question. 2 500 and 3 150 A.B. 1 250. values Max. The table below gives the temperature rises authorised by the IEC according to the type of contacts.732 = 130 A v calculating the operating current of the motor feeder: cosϕ = power factor = 0.9 I= P 630 = = 82 A Uecosϕη 5. rise = t°. factory-built equipment Minimal short-circuit current: Isc (kA rms.) Frequency fr (Hz) c Two frequencies are usually used throughout the world: v 50 Hz in Europe v 60 Hz in America. Several countries use both frequencies indiscriminately.) Peak value of maximal short-circuit: Idyn (kA peak) (value of the initial peak in the transient period) (see explanation in "Short-circuit currents" chapter.) Rms value of maximal short-circuit current: Ith (kA rms.) (see explanation in "Short-circuit currents" chapter. does not isolate Disconnecter switch ✔ switches isolates Fixed circuit breaker ✔ ✔ ✔ switches protects does not isolate Withdrawable circuit breaker switches protects isolates if withdrawn Fixed contactor ✔ ✔ switches does not isolate Withdrawable contactor ✔ switches isolates if withdrawn Fuse ✔ protects does not isolate ✔ = YES ✔ (once) Schneider Electric Merlin Gerin MV design guide 9 . 1 s or 3 s) (see explanation in "Short-circuit currents" chapter.Presentation Metal-enclosed. Switchgear functions Designation and symbol Disconnecter function Current switching operating fault isolates Earthing disconnecter isolates Switch (short-circuit closing capacity) switches. but always possible ✔ = YES ✔ ✔ ✔ 10 Merlin Gerin MV design guide Schneider Electric . factory-built equipment Different enclosure types Characteristics Cubicles Metal-clad Compartment Block-type External walls Number of MV compartments Internal partitions metal and always earthed ≥3 metal and always earthed 3 indifferent metal or not possible ✔ ≤2 indifferent metal or not Presence of bushings ✔ Shutters to prevent access to live compartments ✔ Ease of operations when live Arcing movement within the cubicle ✔ difficult.Presentation Metal-enclosed. cables. Determination of the short-circuit power requires analysis of the power flows feeding the short-circuit in the worst possible case.Design rules Short-circuit power Introduction Example 1: 25 kA at an operating voltage of 11 kV R E Zcc L Icc c The short-circuit power depends directly on the network configuration and the impedance of its components: lines. through which the short-circuit current passes.. motors. 63 kV T1 Isc1 A T2 Isc2 Isc3 Example 2: c Feedback via LV Isc5 is only possible if the transformer (T4) is powered by another source. U Isc : : operating voltage (kV) short-circuit current (kA rms.. Schneider Electric Merlin Gerin MV design guide 11 . or via MV/LV transformaters. c Three sources are flowing in the switchboard (T1-A-T2) v circuit breaker D1 (s/c at A) Isc1 + Isc2 + Isc3 + Isc4 + Isc5 v circuit breaker D2 (c/c at B) Isc1 + Isc2 + Isc3 + Isc4 + Isc5 v circuit breaker D3 (c/c at C) Isc1 + Isc2 + Isc3 + Isc4 + Isc5 A D1 B D2 10 kV C D3 D6 MT T3 Isc5 D4 D5 D7 M Isc4 BT T4 BT MT We have to calculate each of the Isc currents. etc).) Ref: following pages A U B Zs Ssc = e • U • Isc The short-circuit power can be assimilated to an apparent power. c Power feedback due to rotary sets (motors. Possible sources are: c Network incomer via power transformers. expressed in MVA or in kA rms for a given operating voltage. c Generator incomer. c The customer generally imposes the value of short-circuit power on us because we rarely have the information required to calculate it. transformers. c It is the maximum power that the network can provide to an installation during a fault. 5 kA peak ANSI ) . etc). UPS). 12 Merlin Gerin MV design guide Schneider Electric . v rms value of maximal short-circuit current: Ith = (kA rms.5 .5 • 25 kA = 63.6 • Isc at 60 Hz (IEC) or.12.75 kA peak IEC 60 056 or 2. It determines the breaking capacity and closing capacity of circuit breakers and switches. 2rIsc Time 2rIsc N.25 .7 • 25 kA = 67.1 when calculating maximal Isc.Design rules Short-circuit currents All electrical installations have to be protected against short-circuits.2 kA rms e • 10 The difference lies in the way in which we round up the value and in local habits. The value 19 kA rms is probably the most realistic.The IEC uses the following values: 8 . IEC 909 applies a coefficient of 1. Its value allows us to choose the setting of thresholds for overcurrent protection devices and fuses. It is defined in kA for 1 or 3 second(s) and is used to define the thermal withstand of the equipment. The short-circuit current must be calculated at each stage in the installation for the various configurations that are possible within the network. without exception.B.5 . 2. p 12 Introduction). this is in order to determine the characteristics that the equipment has to have withstand or break this fault current.5 • Isc at 50 Hz (IEC) or.1 takes account of a voltage drop of 10 % across the faulty installation (cables. c In order to choose the right switchgear (circuit breakers or fuses) and set the protection functions. 1 s or 3 s) (example: 25 kA rms.1)) and not just behind the breaking mechanism.: c A specification may give one value in kA rms and one value in MVA as below: Isc = 19 kA rms or 350 MVA at 10 kV v if we calculate the equivalent current at 350 MVA we find: 350 = 20.Idyn is equal to: 2.31.16 .1 • U = E e • Zcc Zcc (Cf: example 1. This coefficient of 1. direct component Current I peak= Idyn . especially when the length of cables is high and/or when the source is relatively impedant (generator. as well as the electrodynamic withstand of busbars and switchgear. Isc = Isc = 1. v peak value of the maximum short-circuit current: (value of the initial peak in the transient period) Ith Isc R X Idyn = (kA peak) MV cable figure 1 (example: 2. 2.20 . which more generally corresponds to a change in conductor cross-section.1). three short-circuit values must be known: v minimal short-circuit current: (example: 25 kA rms) Isc = (kA rms) This corresponds to a short-circuit at one end of the protected link (fault at the end of a feeder (see fig. These are generally used in the specifications.40 kA rms.7 • Isc (ANSI) times the short-circuit current calculated at a given point in the network. whenever there is an electrical discontinuity. v another explanation is possible: in medium and high voltage. 1 s) This corresponds to a short-circuit in the immediate vicinity of the upstream terminals of the switching device (see fig. etc). c Usc % is defined in the following way: The short-circuit current depends on the type of equipment installed on the network (transformers. is given by the following equation: Ir Isc = Usc Schneider Electric Merlin Gerin MV design guide 13 . generators. motors. we need to know the short-circuit voltage (Usc %). Example: c Transformer 20 MVA c Voltage 10 kV c Usc = 10 % c Upstream power: infinite 20 000 Sr Ir = = = 1 150 A e U no-load e•10 Isc = Ir = 1 150 = 11 500 A = 11.Design rules Short-circuit currents Transformer In order to determine the short-circuit current across the terminals of a transformer. lines. potentiometer U : 0 to Usc V primary secondary A I : 0 to Ir 1 the voltage transformer is not powered: U = 0 2 place the secondary in short-circuit 3 gradually increase voltage U at the primary up to the rated current Ir in the transformer secondary circuit. expressed in kA.5 kA U s c 10÷ 100 The value U read across the primary is then equal to Usc c The short-circuit current. c When the power gradually increases. 20/100 Ir fault appears Isc time healthy subtransient state state transient state permanent state short-circuit c The short-circuit current is given by the following equation: Isc = Xsc : Ir Xsc short-circuit reactance c/c c The most common values for a synchronous generator are: State Xsc Sub-transient X''d 10 . courant Example: Calculation method for an alternator or a synchronous motor c Alternator 15 MVA c Voltage U = 10 kV c X'd = 20 % Sr 15 Ir = = = 870 A e • U e • 10 000 Isc = 870 Ir = = 4 350 A = 4.35 kA Xcc trans.25 % Permanent Xd 200 .Design rules Short-circuit currents G Synchronous generators (alternators and motors) Calculating the short-circuit current across the terminals of a synchronous generator is very complicated because the internal impedance of the latter varies according to time. average duration 250 ms v permanent (this is the value of the short-circuit current in steady state). 14 Merlin Gerin MV design guide Schneider Electric . the current reduces passing through three characteristic periods: v sub-transient (enabling determination of the closing capacity of circuit breakers and electrodynamic contraints). c The short-circuit current is calculated in the same way as for transformers but the different states must be taken account of. takes account of motors when stopped and the impedance to go right through to the fault.20 % Transient X'd 15 . average duration.350 % Asynchronous motor M c For asynchronous motors v the short-circuit current across the terminals equals the start-up current Isc z 5 at 8 Ir v the contribution of the motors (current feedback) to the short-circuit current is equal to: I z 3 ∑ Ir The coefficient of 3. 10 ms v transient (sets the equipment's thermal contraints). 1 • U • Isc • e = U Zsc Isc = 1. Usc = 9 % 2 Usc(%) Z (Ω) = U • 100 Sr Sr (kVA) Usc (%) 100 to 3150 4 to 7.10 at 0.1 at 150 kV c Overhead lines R=ρ•L S X = 0.8.8.3.5 MV/LV 5000 to 5000 8 to 12 HV/MV c Cables X = 0.3 Ω/km ρ = 1.15 Ω/km Schneider Electric Merlin Gerin MV design guide 15 .1• U e • Zsc with Zsc = R2 + X 2 c Upstream network 2 Z= U Ssc R= X { 0.Design rules Short-circuit currents Reminder concerning the calculation of three-phase short-circuit currents c Three-phase short-circuit 2 Ssc = 1.10-6 Ω cm ρ = 3. Sr = 630 kVA.15 Ω/km three-phased or single-phased c Busbars X = 0.10-6 Ω cm HV MV/LV copper aluminium almélec c Synchronous generators 2 Z(Ω) = X(Ω) = U • Xsc (%) 100 Sr Xsc turbo exposed poles sub-transient 10 to 20 % 15 to 25 % transient 15 to 25 % 25 to 35 % permanent 200 to 350 % 70 to 120 % c Transformers (order of magnitude: for real values. refer to data given by manufacturer) E.: 20 kV/410 V.10-6 Ω cm ρ = 2.3 at 6 kV 0.4 Ω/km X = 0. Usc = 4 % 63 kV/11 V.g. Sr = 10 MVA.2 at 20 kV 0. in other words. Xa transformer RT. the contribution of an HV cable upstream of an HV/LV transformer will be: R2 = R1( U2 )2 et X2 = X1 (U2 )2 U1 U1 Z2 = Z1 (U2 )2 U1 ainsi This equation is valid for all voltage levels in the cable. XT impedance at primary n LV cable R2.Design rules Short-circuit currents c Synchronous motors and compensators Xsc high speed motors low speed motors compensators Sub-transient 15 % 35 % 25 % transient 25 % 50 % 40 % permanent 80 % 100 % 160 % c Asynchronous motors only sub-transient 2 Ir Z(Ω) = • U Sr Id Isc z 5 to 8 Ir Isc z 3∑ Ir. A HV cable R1.3 to 2 c Equivalent impedance of a component through a transformer v for example. X2 v Impedance seen from the fault location A: 1 a ∑ R = R2 + RT + R2 + R2 2 1 ∑ X = X2 + XT + X2 + Xa 2 n n n n n n2 n: transformation ratio c Triangle of impedances Z= (R2 + X2) Z X ϕ R 16 Merlin Gerin MV design guide Schneider Electric . contribution to Isc by current feedback (with I rated = Ir) c Fault arcing Id = Isc 1. for a low voltage fault. even through several series-mounted transformers. X1 Power source Ra. X. Impedance method All the components of a network (supply network.the equivalent value of R or X . short-circuit impedance (in ohms) Z = Zr + Zt1//Zt2 Z = Zr + Zt1 • Zt2 Zt1 + Zt2 (cf. Example 1: Network layout Tr1 Tr2 c The three-phase short-circuit current is: U e • Zsc A Equivalent layouts Zr Zt1 Za Zt2 Isc = Isc U Zsc : : : short-circuit current (in kA) phase to phase voltage at the point in question before the appearance of the fault. example 2 below) Za Zsc = Z//Za Zsc = Z • Za Z + Za Example 2: c Zsc = 0. transformer.Design rules Short-circuit currents Example of a three-phase calculation The complexity in calculating the three-phase short-circuit current basically lies in determining the impedance value in the network upstream of the fault location. cables. motors.the equivalent value of impedance .38 kA e • 0. alternator. in kV. R and Z are expressed in ohms.72 ohm c U = 10 kV 10 Isc = = 21. c The relation between these different values is given by: Z= (cf. bars. etc) are characterised by an impedance (Z) comprising a resistive component (R) and an inductive component (X) or so-called reactance.the short-circuit current.27 Schneider Electric Merlin Gerin MV design guide 17 . example 1 opposite) (R2 + X2) c The method involves: v breaking down the network into sections v calculating the values of R and X for each component v calculating for the network: . apparent power: 15 MVA . 1 to 20 MVA .X'd transient: 20 % .voltage: 10 kV .short-circuit voltage: Usc = 10 % v Alternator : .apparent power: 1 to 15 MVA. v the breaking and closing capacities of the circuit breakers D1 to D7.voltage 63 kV / 10 kV . c Equipment characteristics: v transformers: .Design rules Short-circuit currents Here is a problem to solve! Exercice data Supply at 63 kV Short-circuit power of the source: 2 000 MVA c Network configuration: Two parallel mounted transformers and an alternator.X"d sub-transient: 15 % c Question: v determine the value of short-circuit current at the busbars. Single line diagram Alternator 15 MVA X'd = 20 % X''d = 15 % T1 G1 63 kV Transformer 15 MVA Usc = 10 % T2 Transformer 20 MVA Usc = 10 % D3 D1 10 kV D2 Busbars D4 D5 D6 D7 18 Merlin Gerin MV design guide Schneider Electric . B. D5. In the case of a short-circuit upstream of a circuit breaker (D1. c Equivalent diagram Each component comprises a resistance and an inductance. so we can therefore deduce that the reactance is equal to the impedance (X = Z). we can deduce the value of Zt by applying the equation: Z= ( ∑R2 + ∑X2) N. reactance. We have to calculate the values for each component. D4. we have to calculate the various values of resistances and inductances. D6. we can say that Z = X. then separately calculate the arithmetic sum: Rt = R Xt = X c Knowing Rt and Xt.: Since R is negligible compared with X. T2 and G1. Schneider Electric Merlin Gerin MV design guide 19 . D6 and D7. The network can be shown as follows: Zr = network impedance Za = alternator impedance different according to state (transient or subtransient) Z20 = transformer impedance 20 MVA Z15 = transformer impedance 15 MVA busbars Experience shows that the resistance is generally low compared with. D3. D5.Design rules Short-circuit currents Here is the solution to the problem with the calculation method Solving the exercise c Determining the various short-circuit currents The three sources which could supply power to the short-circuit are the two transformers and the alternator. D7). D2. c To determine the short-circuit power. We are supposing that there can be no feedback of power through D4. this then has the short-circuit current flow through it supplied by T1. 33 0.5 Isc (in kA peak) N.29 Zer//Zat = Zer • Zat = 0. 2 Icc = U = 10 • 1 e •Zsc e Zsc Zet = 0.34 + 1.42 Zt = (Zr + Z15)//Za 20 Merlin Gerin MV design guide Schneider Electric .5 = 31 transient state Z = 0.47 sub-transient state Z = 0.34 + 1 Breaking capacity in kA rms.39 sub-transient state Z = 0.05 And now here are the results! Network Ssc = 2 000 MVA U op.34 • 1 Zer//Zat = Zer • Zat = Zer + Zat 0.4 Zr Za Z15 12. = 10 kV 15 MVA alternator U op.25 21.25 Circuit breaker Equivalent circuit Z (ohm) Closing capacity 2.67 0.15 Zt = [Zr + (Z15//Z20)]//Za D3 alternator Zr 17 Z = 0.5 Z15//Z20 = Z15 • Z20 = Z15 + Z20 0.5 Zat = 1. D4 to D7 Zr Za Z15 Z20 transient state Z = 0. = 10 kV Transient state (Xsc = 20 %) Sub-transient state (Xsc = 15 %) Busbars Parallel-mounted with the transformers Series-mounted with the network and the transformer impedance Parallel-mounting of the generator set Transient state Sub-transient state 2 2 Z15 = U •Usc = 10 • 10 15 100 Sr 2 2 Z20 = U •Usc = 10 • 10 20 100 Sr 2 Za = U • Xsc Sr 2 Zat = 10 • 20 15 100 2 Zas =10 • 15 15 100 0.40 • 2.5 = 37.29 Zer = 0.: a circuit breaker is defined for a certain breaking capacity of an rms value in a steady state.34 17 • 2.34 • 1.05 + 0.9 Zr Za Z20 14.4 • 2.25 transient state Z = 0. the calculations must be validated by laboratory tests.27 sub-transient state Z = 0. and as a percentage of the aperiodic component which depends on the circuit breaker's opening time and on R X of the network (about 30 %).9 • 2.5 = 53. For alternators the aperiodic component is very high.5 Zr + Zet = 0.67 • 0.B.27 z 0. = 10 kV 15 MVA transformer (Usc = 10 %) U op. = 10 kV 20 MVA transformer (Usc = 10 %) U op.67 + 0.5 Z15 Z20 Zt = Zr + (Z15//Z20) D1 15 MVA transformer 17.Design rules Short-circuit currents Component Calculation Zr = U2 Ssc = 102 2 000 Z = X (ohms) 0.40 21.5 = 42.35 Zt = (Zr + Z20)//Za D2 20 MVA transformer 12.33 Zas = 1 0.34 z 0.33 Zer + Zat 0. * N. c To carry out a busbar calculation. Physical busbar characteristics S d l : : : busbar cross section phase to phase distance distance between insulators for same phase ambient temperature (θn ≤ 40°C) permissible temperature rise* flat copper flat-mounted : cm2 cm cm θn (θ . We also have to check that the period of vibration intrinsic to the bars themselves is not resonant with the current period. a busbar calculation involves checking that it provides sufficient thermal and electrodynamic withstand and non-resonance. In summary: bar(s) of x cm per phase Schneider Electric Merlin Gerin MV design guide 21 . we have to use the following physical and electrical characteristics assumptions: Busbar electrical characteristics Ssc Ur U Ir : : : : network short-circuit power* rated voltage operating voltage rated current MVA kV kV A In reality.: It is is generally provided by the customer in this form or we can calculate it having the short-circuit current Isc and the operating voltage U: (Ssc = e • Isc • U.B.Design rules Busbar calculation Introduction c The dimensions of busbars are determined taking account of normal operating conditions. see chapter on "Shortcircuit currents"). The voltage (kV) that the installation operates at determines the phase to phase and phase to earth distance and also determines the height and shape of the supports. The rated current flowing through the busbars is used to determine the cross-section and type of conductors.: see table V in standard ICE 60 694 on the 2 following pages.θn) : : °C °C aluminium edge-mounted profile : material : arrangement : no.B. c We then ensure that the supports (insulators) resist the mechanical effects and that the bars resist the mechanical and thermal effects due to short-circuit currents. of bar(s) per phase * N. 7 When contact components are protected in different ways. the temperature and temperature rises that are allowed are those for the element for which table V authorises the highest values. 3 All the necessary precautions must be taken so that absolutely no damage is caused to surrounding materials. In this case. the limit values of temperature and temperature rise do not apply to vacuum devices. 2 and 3) Bolt connected or equivalent devices (Cf: 7) bare copper.Design rules Busbar calculation Temperature rise Taken from table V of standard IEC 60 694 Type of device.θn) with θn = 40°C 90 105 100 115 115 100 105 105 100 50 65 60 75 75 60 65 65 60 1 According to its function. Other devices must not exceed the values for temperature and temperature rise given in table V. of material and of dielectric (Cf: 1. the admissible values of temperature and temperature rise to take into consideration are the lowest for category concerned. 22 Merlin Gerin MV design guide Schneider Electric . 2 For vacuum switchgear. the same device may belong to several categories given in table V. bare copper alloy or aluminium alloy in air SF6 * oil silver or nickel plated in air SF6 oil tin-plated in air SF6 oil * SF6 (sulphur hexafluoride) Temperature θ (°C) (θ . 4 When the contact components are protected in different manners. In this case. 2 For vacuum switchgear. . the limit values of temperature and temperature rise do not apply to vacuum devices. Schneider Electric Merlin Gerin MV design guide 23 . the contacts must be considered as "bare". the admissible values of temperature and temperature rise to take into consideration are the lowest for category concerned. of material and of dielectric (Cf: 1.θn) with θn = 40°C 75 90 80 105 105 90 90 90 90 35 50 40 65 65 50 50 50 50 1 According to its function. . the temperature rise must be in conformity with publications concerning high voltage fuses. 3 All the necessary precautions must be taken so that absolutely no damage is caused to surrounding materials. according to specifications specific to each piece of equipment. 5 The quality of coating must be such that a protective layer remains in the contact zone: . the same device may belong to several categories given in table V. Should this not be true. 2 and 3) Contacts (Cf: 4) copper or bare copper alloy in air SF6 * oil silver or nickel plated (Cf: 5) in air SF6 oil tin-plated (Cf: 5 and 6) in air SF6 oil * SF6 (sulphur hexafluoride) Temperature θ (°C) (θ .after the making and breaking test (if it exists).Design rules Busbar calculation Temperature rise Extract from table V of standard IEC 60 694 Type of device.after the short time withstand current test.after the mechanical endurance test. 6 For fuse contacts. Other devices must not exceed the values for temperature and temperature rise given in table V. the temperatures and temperature rises that are allowed are those of the element for which table V authorises the lowest values. k3.14 1.12 k1 1.for a protection index greater than IP5 ambient temperature (θn ≤ 40°C) permissible temperature rise* busbar cross section busbar perimeter (opposite diagram) θn (θ . k5.40 2.89 2.θn)0.004 conditions coefficient product of 6 coefficients (k1. 5.61 • S0.10 0.90 µΩ cm α K temperature coefficient of the resistivity: 0. 4.5 • p0.08 1. see table below: 0.Design rules Busbar calculation Let's check if the cross-section that has been chosen: … bar(s) of … x … cm per phase satisfies the temperature rises produced by the rated current and by the short-circuit current passing through them for 1 to 3 second(s).83 2.20 1.68 0.70 a e In our case: e/a = the number of bars per phase = giving k1 = 24 Merlin Gerin MV design guide Schneider Electric . k2.39 ρ20 [1+ α (θ .63 0.05 0. of bars per phase 2 1.16 1.65 0. k4.91 2.18 1.50 e/a 0.85 2.80 1. 2.9 (θ . described below *(see table V of standard IEC 60 694 in the previous pages) Definition of coefficients k1.76 2.θn) S perimeter of a bar : : : : °C °C cm2 cm P p ρ20 : : : : : conductor resistivity at 20°C copper: aluminium: 1.20)] with: I : permissible current expressed in amperes (A) derating in terms of current should be considered: . 6: e c Coefficient k1 is a function of the number of bar strips per phase for: v 1 bar (k1 = 1) v 2 or 3 bars. 3.87 2.63 1. k6).83 µΩ cm 2.55 2.06 no.73 3 2.for an ambient temperature greater than 40°C .45 0. Thermal withstand… For the rated current (Ir) The MELSON & BOTH equation published in the "Copper Development Association" review allows us to define the permissible current in a conductor: I=K• 24.60 0. 9 (θ .98 In our case: n= giving k6 = In fact we have: k= • • • • • = I= • 24.9 ( - ) 0.15 c Coefficient k3 is a function of the position of the bars: v edge-mounted bars: k3 = 1 v 1 bar base-mounted: k3 = 0.95 v several base-mounted bars: k3 = 0.20)] I= A The chosen solution of • bar(s) cm per phase Is appropriate if Ir of the required busbars ≤ I Schneider Electric Merlin Gerin MV design guide 25 .75 c Coefficient k4 is a function of the place where the bars are installed: v calm indoor atmosphere : k4 = 1 v calm outdoor atmosphere: k4 = 1.61 • S0.θn)0.2 v bars in non-ventilated ducting: k4 = 0.80 c Coefficient k5 is a function of the artificial ventilation: v without artificial ventilation: k5 = 1 v ventilation should be dealt with on a case by case basis and then validated by testing.Design rules Busbar calculation c Coefficient k2 is a function of surface condition of the busbars: v bare: k2 = 1 v painted: k2 = 1. k6 is a function of the number of bars n per phase and of their spacing. The value of k6 for a spacing equal to the thickness of the bars: n k6 1 1 2 1 3 0.5 • 0.61 • [1+ 0.39 . c Coefficient k6 is a function of the type of current: v for a alternatif current of frequency ≤ 60 Hz.39 ρ20 [1+ α (θ .004 ( 0.5 • p0.20)] I=K• 24. Check that this temperature θt is compatible with the maximum temperature of the parts in contact with the busbars (especially the insulator). The equation below can be used to calculate the short-circuit temperature rise: ∆θcc = 0. rms value ) short-time withstand current duration (1 to 3 s) in s δ : density of the metal copper: aluminium: resistivity of the conductor at 20°C copper: aluminium: permissible temperature rise 0. it corresponds to 26. 26 Merlin Gerin MV design guide Schneider Electric .16 kA rms. 2 s 0. θt of the conductor after the short-circuit will be: θt = θn + (θ-θn) + ∆θsc θt = °C Check: θt ≤ maximum admissible temperature by the parts in contact with the busbars.24 • ( 10-6• ( )2 • )2 • • The temperature. 1 s v at 37 kA rms.83 µΩ cm 2. what does Ith1 correspond to for t = 1 s? (Ith2 )2 • t = constant (26.θn) : °C so Ith1 = ( constant ) = ( 137 • 10 ) t 1 7 ∆θsc = ∆θsc = °C Ith1 = 37 kA rms. for 1 s c In summary: v at 26.16 kA rms. it corresponds to 37 kA rms.16 • 103)2 •2 = 137 • 107 tk : 8.24 • ρ20 • Ith2 • tk (n • S)2 • c • δ with: ∆θsc c : : short-circuit temperature rise specific heat of the metal copper: aluminium: busbar cross section number of busbar(s) per phase is the short-time withstand current: (maximum short-circuit current.091 kcal/daN°C 0.Design rules Busbar calculation For the short-time withstand current (Ith) c We assume that for the whole duration (1 or 3 seconds): v all the heat that is given off is used to increase the temperature of the conductor v radiation effects are negligible.7 g/cm3 1.90 µΩ cm ρ20 : (θ .9 g/cm3 2.16 kA rms. 2 s.23 kcal/daN °C cm2 S n Ith : : : A rms Example: How can we find the value of Ith for a different duration? Knowing: (Ith)2 • t = constant c If Ith2 = 26. 1 s. 2 s. Design rules Busbar calculation Electrodynamic withstand We have to check if the bars chosen withstand the electrodynamic forces.5 for 50 Hz . Forces between parallel-mounted conductors The electrodynamic forces following a short-circuit current are given by the equation: F1 = 2 l • Idyn2 • 10-8 d with F1 Idyn : : force expressed in daN is the peak value of short-circuit expressed in A.25 4 1. to be calculated with the equation below: Idyn = k • Ssc = k • Ith Ue e F1 Idyn F1 Idyn Ssc Ith U l d k : : : : : : short-circuit power short-time withstand current operating voltage distance between insulators on the same phase phase to phase distance 2.6 for 60 Hz for IEC and 2. v number of supports =N v we know N. 2.5 3 1.7 according to ANSI kVA A rms kV cm cm d l Giving : Idyn = A and F1 = daN Forces at the head of supports or busducts Equation to calculate the forces on a support: F = F1 • d with F H h : : : H+h H h = e/2 F1 F H support force expressed insulator height distance from insulator head to busbar centre of gravity daN cm cm Calculation of forces if there are N supports c The force F absorbed by each support is at most equal to the calculated force F1 (see previous chapter) multiplied by a coefficient kn which varies according to the total number N of equidistant supports that are installed.10 ≥5 1. let us define kn with the help of the table below: giving F = (F1)• (kn) = N kn 2 0.14 daN c The force found after applying a coefficient k should be compared with the mechanical strength of the support to which we will apply a safety coefficient: v the supports used have a bending resistance daN F’ = check if F’ > F v we have a safety coefficient of F' = F Schneider Electric Merlin Gerin MV design guide 27 . they are subjected to a bending moment whose resultant strain is: η= with η : F1• l v • 12 I is the resultant strain.5 • h busbar cross section (in cm2) x' xx': perpendicular to the plane of vibration Check: η < η Bars Cu or Al (in daN/cm2) 28 Merlin Gerin MV design guide Schneider Electric . it must be less than the permissible strain for the bars this is: copper 1/4 hard: 1 200 daN/cm2 copper 1/2 hard: 2 300 daN/cm2 copper 4/4 hard: 3 000 daN/cm2 tin-plated alu: 1 200 daN/cm2 force between conductors distance between insulators in the same phase cm daN F1 l : : I/v : is the modulus of inertia between a bar or a set of bars (choose the value in the table on the following page) cm3 v : distance between the fibre that is neutral and the fibre with the highest strain (the furthest) phase 1 b v h x phase 2 c One bar per phase: 3 I= b•h 12 x' I b • h2 = v 6 c Two bars per phase: phase 1 v b phase 2 x 3 I = 2 ( b • h + S • d2) 12 I = v d h S : 2( b • h3 + S • d2) 12 1.Design rules Busbar calculation Mechanical busbar strength c By making the assumption that the ends of the bars are sealed. 25 9.41 4. modulus of inertia I/v.2 12.083 0.6 12.416 0.66 21.47 1.8 3.036 0.33 3. linear mass m.55 85.027 0.33 16.008 0.213 0.006 0.66 1. moment of inertia I for the bars defined below: Busbar dimensions (mm) S cm2 m Cu daN/cm A5/L I x' x Arrangement* x 100 x 10 10 0.41 4.013 0.12 12.4 1.036 0.83 7.75 7.6 6.8 19.35 1.2 80 x 5 4 0.018 0.021 0.33 2.66 14.83 10.33 66 26.Design rules Busbar calculation Choose your cross-section S.33 21.8 14.45 166.16 5. f E : : resonant frequency in Hz modulus of elasticity: for copper = 1.16 51.54 4.62 16.16 2.5 33 250 50 80 x 10 8 0.022 0.16 4.011 0.027 0.38 38.66 10.74 4.3 • 106 daN/cm2 for aluminium A5/L = 0.014 0.144 0.2 2.66 21.25 12.12 10.33 21.25 6.25 16.071 0.67 • 106 daN/cm2 linear mass of the busbar (choose the value on the table above) length between 2 supports or busducts daN/cm m : l : cm I moment of inertia of the busbar cross-section relative to the axis x'x.66 33.33 11.78 2.4 9.8 10.62 6.33 82.83 8.13 15.044 0.5 2.66 17.16 10.043 0.91 5.94 18.4 3.5 50 x 8 4 0.089 0.022 0.34 2.4 128 32 80 x 6 4.5 50 x 5 2.2 5.66 10.5 31.66 6.22 20.2 0. in other words between 42 and 58 and 80 and 115 Hz.5 76.66 83.48 25.33 5.83 1.6 64 16 80 x 3 2.09 0.88 42.3 6.007 0.33 41. Schneider Electric Merlin Gerin MV design guide 29 .011 0.25 cm4 cm3 cm4 cm3 cm4 cm3 cm4 cm3 cm4 cm3 cm4 cm3 I/v I x' x I/v I x' x I/v I x' x I/v I x' x I/v I x' I/v *arrangement: cross-section in a perpendicular plane to the busbars (2 phases are shown) Intrinsic resonant frequency The intrinsic frequencies to avoid for the busbars subjected to a 50 Hz current are frequencies of around 50 and 100 Hz.8 0.56 25 10 50 x 6 3 0.6 6.04 25.33 42.08 1.4 0.53 8.05 0.25 2.6 50 x 10 5 0.5 0.5 5 8.66 8.33 5. perpendicular to the vibrating plane cm4 (see formula previously explained or choose the value in the table above) : giving f= Hz We must check that this frequency is outside of the values that must be avoided. This intrinsic frequency is given by the equation: f = 112 E•I m•l4 Check that the chosen busbars will not resonate. for a time of tk = 3 seconds.500 A rms.500 A on a permanent basis and a short-time withstand current Ith = 31. Each cubicle has 3 insulators(1 per phase). Busbars comprising 2 bars per phase.θn) profile Top view Cubicle 1 Cubicle 2 Cubicle 3 Cubicle 4 Cubicle 5 : : : : : : : busbar cross-section (10 •1) phase to phase distance distance between insulators on the same phase ambient temperature permissible temperature rise (90-40=50) 10 18 70 40 50 cm2 cm cm °C °C flat busbars in copper 1/4 hard. with a permissible strain η = 1 200 daN/cm2 edge-mounted material arrangement: number of busbar(s) per phase: 2 d d c The busbars must be able to withstand a rated current Ir = 2. inter-connect the cubicles electrically. c Rated frequency fr = 50 Hz c Other characteristics: 1 cm 1 cm v parts in contact with the busbars can withstand a maximum temperature of θmax = 100°C v the supports used have a bending resistance of F' = 1 000 daN 10 cm 5 cm 12 cm d d 30 Merlin Gerin MV design guide Schneider Electric .Design rules Busbar calculation Busbar calculation example Here is a busbar calculation to check. Exercise data c Consider a switchboard comprised of at least 5 MV cubicles. Busbar characteristics to check: S d l θn (θ . 70 0. 5.89 2.80 Schneider Electric Merlin Gerin MV design guide 31 .40 2.10 number of bars per phase 2 1. 6: c Coefficient k1 is a function of the number of bar strips per phase for: v 1 bar (k1 = 1) v 2 or 3 bars.85 2.83 2.63 1.76 1.18 0.60 1.20)] with: I θn (θ . see table below: e/a 0. k2. 2.004 K e : condition coefficient product of 6 coefficients (k1.87 2.63 1. k4.50 2.68 1.45 2.08 0.20 In our case: e/a = number of bars per phase = giving k1 = 0.1 2 1. described below *(see table V in standard CEI 60 694 pages 22 and 23) Definition of coefficients k1.14 0.91 2.65 1. k3.05 0.80 3 2.16 0.Design rules Busbar calculation Let's check the thermal withstand of the busbars! For the rated current (Ir) The MELSON & BOTH equation allows us to define the permissible current in the conductor: I=K• 24.83 µΩ cm 40 50 10 22 °C °C cm2 cm ρ20 α a : temperature coefficient for the resistivity: 0.61 • S0. k6).39 ρ20 [1+ α (θ .73 1. 3.06 0.55 0.θn)0.θn) S p e : : : : : : permissible current expressed in amperes (A) ambient temperature permissible temperature rise* busbar cross-section busbar perimeter resistivity of the conductor at 20°C copper: 1.9 (θ .12 k1 1. 4. k5.5 • p0. 9 (θ .44 • 24.θn)0.39 ρ20 [1+ α (θ .2 v bars in non-ventilated ducting: k4 = 0. c Coefficient k6 is a function of the type of current: v for alternatif current at a frequency of 60 Hz. k6 is a function of the number of busbars n per phase and of their spacing.9 ( 90 . The value of k6 for a spacing equal to the thickness of the busbars: n k6 1 1 2 1 3 0.80 • 1 • 1 • 0.20)] I= 2 689 A The chosen solution: is appropriate: 2 busbars of 10 • 1 cm per phase 2 500 A < 2 689 A Ir < I either 32 Merlin Gerin MV design guide Schneider Electric .40 ) 0.8 • 1 • 1 = 1. we have: k = 1.98 In our case: n= 2 giving k6 = 1 In fact.39 .5 • 22 0.95 v several flat-mounted bars: k3 = 0.15 c Coefficient k3 is a function of the busbar position: v edge-mounted busbars: k3 = 1 v 1 bar flat-mounted: k3 = 0.5 • p0.004 ( 90 0.20)] I=K• 24.83 [1+ 0.75 c Coefficient k4 is a function of where the bars are installed: v calm indoor atmosphere: k4 = 1 v calm outdoor atmosphere: k4 = 1.Design rules Busbar calculation c Coefficient k2 is a function of the surface condition of the bars: v bare: k2 = 1 v painted: k2 = 1.61 • S0.61 • 10 1.80 c Coefficient k5 is a function of the artificial ventilation: v without artificial ventilation: k5 = 1 v cases with ventilation must be treated on a case by case basis and then validated by testing.44 I = 1. Design rules Busbar calculation For the short-time withstand current (Ith) c we assume that.9 g/cm3 in secs : 1. The equation below can be used to calculate the temperature rise due to short-circuit: ∆θcc = 0. 0.091 kcal / daN°C 10 2 31 500 A rms. cm2 S n Ith : : : tk δ ρ20 : : short-time withstand current duration (1 to 3 secs) density of the metal copper: resistivity of the conductor at 20°C copper: permissible temperature rise 3 8.θn): °C v The temperature rise due to the short circuit is: ∆θcc = Calculation of θt must be looked at in more detail because the required busbars have to withstand Ir = 2 500 A at most and not 2 689 A.24 • 1.9 ∆θcc = 4 °C The temperature θt of the conductor after short-circuit will be: θt = θn + (θ-θn) + ∆θcc = 40 + 50 + 4 = 94 °C for I = 2 689 A (see calculation in the previous pages) Schneider Electric Merlin Gerin MV design guide 33 . for the whole duration (3 seconds) : v all the heat given off is used to increase the temperature of the conductor v the effect of radiation is negligible.091 3 ( 2 •10 )2 • • 8. value of the maximum shortcircuit current) 0.83 µΩ cm 50 (θ .83 10-6• ( 31 500 )2 • 0.24 • ρ20 • Ith2 • tk (n • S)2 • c • δ with: c : specific heat of the metal copper: is the cross section expressed in cm2 number of bars per phase is the short-time withstand current (rms. 3 + 44.3 + 4 °C for Ir = 2 500 A The busbars chosen are suitable because: θt = 88.61 50 = 1.61 et Ir= constant • (∆θ)0.61 therefore 2 689 = 2 500 50 = ∆θ I = Ir ( (θ-θn))0.61 (∆ ) θ 50 ((∆ ) )0. for a rated current Ir = 2 500 A is: θt = θn + ∆θ + ∆θcc = 40 = 88. allows us to deduce the following: I = constant • (θ-θn)0. 34 Merlin Gerin MV design guide Schneider Electric .Design rules Busbar calculation c Let us fine tune the calculation for θt for Ir = 2 500 A (rated current for the busbars) v the MELSON & BOTH equation (cf: page 31).3 °C v temperature θt of the conductor after short-circuit.61 θ ( 2 689 2 500 ) 1 0.126 ∆θ ∆θ = 44.3 °C is less than θmax = 100 °C (θmax = maximum temperature that can be withstood by the parts in contact with the busbars). 5 3 1. The solution is OK Schneider Electric Merlin Gerin MV design guide 35 .10 ≥ 5 1.5 • 31 500 = 78 750 A F1 = 2 • (70/18) • 78 7502 • 10-8 = 482.Design rules Busbar calculation Let's check the electrodynamic withstand of the busbars. Forces between parallel-mounted conductors Electrodynamc forces due to the short-circuit current are given by the equation: F1 = 2 l • Idyn2 • 10-8 d (see drawing 1 at the start of the calculation example) l d : : distance between insulators in the same phase phase to phase distance 70 18 cm cm k : 2. 1 4 (kn) = 778 daN The supports used have a bending resistance F' = 1 000 daN calculated force F = 778 daN. v number of supports ≥ 5 = N v we know N.14 giving F = 683 (F1)• 1 .25 4 1.3 daN Forces at the head of the supports or busducts Equation to calculate forces on a support : F = F1 • H+h H with F H h : : : force expressed in daN insulator height distance from the head of the insulator to the busbar centre of gravity 12 5 cm cm Calculating a force if there are N supports c The force F absorbed by each support is at most equal to the force F1 that is calulated multiplied by a coefficient kn which varies according to the total number N of equi-distant supports that are installed.5 for 50 Hz according to IEC Idyn : peak value of short-circuit current = k • Ith = 2. let us define kn using the table below: N kn 2 0. 45 η = 195 daN / cm2 The calculated resultant strain (η = 195 daN / cm2) is less than the permissible strain for the copper busbars 1/4 hard (1200 daN / cm2) : The solution is OK Busbar dimensions (mm) S m daN/cm I x' x Arrangement x cm2 Cu A5/L cm4 cm3 cm4 cm3 cm4 cm3 cm4 cm3 cm4 cm3 cm4 cm3 100 x 10 10 0.66 14.5 33 250 50 I/v I x' x I/v I x' x I/v I x' x I/v I x' x I/v I x' I/v 36 Merlin Gerin MV design guide Schneider Electric . they are subjected to a bending moment whose resultant strain is: η= F1• l v • 12 I with η l : : is the resultant strain in daN/cm2 distance between insulators in the same phase is the modulus of inertia of a busbar or of a set of busbars (value chosen in the table below) 70 cm I/v : 14.83 1.33 16.66 83.3 • 70 12 • 1 14.45 cm3 η= 482.Design rules Busbar calculation Mechanical strength of the busbars Assuming that the ends of the bars are sealed.027 0.66 21.66 33.089 0.33 82.45 166. 089 daN/cm 70 cm I : moment of inertia of the busbar section relative to the axis x'x perpendicular to the vibrating plane 21.66 cm4 (choose m and I on the table on the previous page) f = 112 • 10 21. are suitable for an Ir = 2 500 A and Schneider Electric Merlin Gerin MV design guide 37 .089 •• 70 ) 4 6 f = 406 Hz f is outside of the values that have to be avoided. in other words 42 to 58 Hz and 80 to 115 Hz: The solution is OK In conclusion The busbars chosen.5 kA 3 sec.30.3 • 106 daN/cm2 m l : : 0.Design rules Busbar calculation Let us check that the chosen busbars do not resonate.66 ( 1.e. i. Inherent resonant frequency The inherent resonant frequencies to avoid for busbars subjected to a current at 50 Hz are frequencies of around 50 and 100 Hz. 2 Ith = 31. bars of 10 • 1 cm per phase. This inherent resonant frequency is given by the equation: E•I m•l4 f = 112 f E : : frequency of resonance in Hz modulus of elasticity for copper = linear mass of the bar length between 2 supports or busducts 1. insulating air interface between the live parts. the presence of humidity can cause a change in insulating performances. * We talk about "full gas" insulation.9 to 3 kV/mm Ionization limit (20°C. Its scope can be a constraint of the external medium (exposure to external elements). For ambient air this characteristic depends on atmospheric conditions and pollution. In the case of gases. 38 Merlin Gerin MV design guide Schneider Electric . Its effect is always the same: reducing the insulation performances by a factor of anything up to 10! c Condensation Phenomena involving the depositing of droplets of water on the surface of insulators which has the effect of locally reducing the insulating performance by a factor of 3. In the case of liquids.Design rules Dielectric withstand A few orders of magnitude Dielectric strength (20°C. 1 bar absolute): 2. in a liquid.ambient air between the live parts . altitude can cause a drop in insulating performance due to the drop in pressure. or so called "full gas performance"*. The dielectric strength of the medium This is a characteristic of the fluid (gas or liquid) making up the medium.) apart from air where a low concentration (humidity < 70%) gives a slight improvement in the overall performance level. liquid or solid insulation decrease as the temperature increases. is related to pressure. For solid insulators. For a device insulated in ambient air. possibly the breaking down of an internal surface. Great care must therefore be paid to expansion phenomena: a solid insulator expands by between 5 and 15 times more than a conductor. c Temperature The performance levels of gaseous. it always leads to a drop in performance. c Humidity In gases and liquids. The dielectric strength of air depends on the following ambient conditions c Pollution Conductive dust can be present in a gas. We are often obliged to derate the device. initial lack of cleanliness. c Pressure The performance level of gas insulation. it generally leads to a drop (SF6. thermal shocks can be the cause of micro-fissuration which can lead very quickly to insulator breakdown. 1 bar absolute): 2. N2 etc. pollution combined with humidity causes electrochemical conduction which will worsen discharge phenomena. Pollution level Pollution may originate: from the external gaseous medium (dust).6 kV/mm c The dielectric withstand depends on the following 3 main parameters: v the dielectric strength of the medium v the shape of the parts v the distance: . or be deposited on the surface of an insulator. related to pollution problems. operator safety. Schneider Electric Merlin Gerin MV design guide 39 . for various reasons. etc. *These indications are relative to a distance through a single air gap. the table in publication IEC 71-2 gives. without taking account of the breakdown voltage by tracking across the surfaces. we cannot test under impulse conditions.Design rules Dielectric withstand The shape of parts This plays a key role in switchgear dielectric withstand. It is essential to eliminate any "peak" effect which would have a disastrous effect on the impulse wave withstand in particular and on the surface ageing of insulators: Air ionization Ozone production Breakdown of moulded insulator surface skin Distance between parts Ambient air between live parts c For installations in which. they do not include any increase which could be required to take account of design tolerances. c Distances in air* between conductive parts that are live and structures which are earthed giving a specified impulse withstand voltage under dry conditions: V d 0 U Rated lightning impulse withstand voltage Up (kV) 40 60 75 95 125 Minimum distance in air phase to earth and phase to phase d (mm) 60 90 120 160 220 The values for distances in air given in the table above are minimum values determined by considering dielectric properties. according to the rated lightning impulse withstand voltage. wind effects. c These distances guarantee correct withstand for unfavourable configurations: altitude < 1 000 m. short circuit effects. the minimum distances to comply with in air either phase to earth or phase to phase. according to IEC 60 815*: Lf : tracking path Pollution level I-low Example of characteristic environments v industry free zone with very low density of housing equipped with heating installations v zones with low density of industry or housing but frequently subjected to wind and/or rain v agricultural regions 1 v mountain regions v all these zones can be located at distances of at least 10 km from the sea and must not be exposed to wind blowing in from the sea 2 v zones with industries producing particularly polluting smoke and/or with an average density of housing equipped with heating installations v zones with a high density of housing and/or industries but subjected frequently to winds and/or to rainfall v zones exposed to a sea wind. given in the table below. 40 Merlin Gerin MV design guide Schneider Electric . exposed to high winds carrying sand and salt and subjected to regular condensation.Design rules Dielectric withstand U Lf O Insulating air interface between live parts c There are 4 severity levels of pollution. but not too close to the coast (at a distance of at least several kilometres) 2 v zones with a high density of industries and suburbs of major cities with a high density of polluting heating installations v zones situated near to the sea. or at least exposed to quite high winds coming in from the sea 2 v generally fairly small areas. subjected to conductive dust and to industrial smoke producing conductive deposits that are particularly thick v generally fairly small areas. *IEC 60 815 guides you in choosing insulators for polluted environments 1 II-medium III-high IIII-very high The use of sprayed fertilisers or the burning of harvested land can lead to a higher level of pollution due to dispersion by the winds 2 The distances to the waters edge depends on the topography of the coast region and the extreme conditions of wind. very close to the coast and exposed to mist or to very high winds and to pollutants coming from the sea 2 v desert zones characterise by long periods without rain. It does not concern the circuit breaker on its own but the front panel must be adapted when the latter is installed within a cubicle (e.g. finer ventilation grills).5 kV. Definitions The protection index is the level of protection provided by an enclosure against access to hazardous parts. the penetration of solid foreign bodies and of water. Applicational scope It applies to enclosures for electrical equipment with a rated voltage of less than or equal to 72. The IP code Introduction Protection of people against direct contact and protection of equipment against certain external influences is required by international standards for electrical installations and products (IEC 60 529). The various IP codes and their meaning A brief description of items in the IP code is given in the table on the following page.Design rules Protection Index Temperature derating must be considered. The IP code is a coding system to indicate the protection index. Knowing the protection index is essential for the specification. installation. operation and quality control of equipment. Schneider Electric Merlin Gerin MV design guide 41 . 5 mm finger Ø 12.5mm X ~ 3 diameter ≥ 2.Design rules Protection index Item Code letter first characteristic figure Figures or letters IP Meaning for protection of equipment against penetration of solid foreign bodies (not protected) diameter ≥ 50 mm Representation of people against access to hazardous parts with (not protected) back of the hand 0 1 Ø 50mm 2 diameter ≥ 12.5mm 4 diameter ≥ 1 mm wire Ø 1mm 5 protected against dust wire 6 sealed against dust wire second characteristic figure 0 1 against penetration of water with detrimental effects (not protected) vertical water drops 2 water drops (15° inclination) 15° 3 rain 60° 4 water projection 5 spray projection 6 high power spray projection 7 temporary immersion 8 additional letter (optional) A B C D additional letter (optional) H M S W prolonged immersion against access to hazardous parts with: back of the hand finger tool wire additional information specific to: high voltage equipment movement during the water testing stationary during the water testing bad weather 42 Merlin Gerin MV design guide Schneider Electric .5 mm tool Ø 2. the IK indices have a different meaning to those of the previous third figures (cf. But since the adoption of IEC 60 529 as the European standard. each new index is given by a two figure number. no European country can have a different IP code. c Since the third figure in various countries could have different meanings and we had to introduce additional levels to cover the main requirements of product standards. spring-loaded hammer or vertical free-fall hammer (diagram below). Definitions c The protection indices correspond to impact energy levels expressed in joules v hammer blow applied directly to the equipment v impact transmitted by the supports. the only solution to maintain a classification in this field was to create a different code.Design rules Protection Index IK code Introduction c Certain countries felt the need also to code the protection provided by enclosures against mechanical impact. table below). To do this they added a third characteristic figure to the IP code (the case in Belgium. expressed in terms of vibrations therefore in terms of frequency and acceleration c The protection indices against mechanical impact can be checked by different types of hammer: pendulum hammer. c Since the IEC has up to now refused to add this third figure to the IP code. France and Portugal). This is a subject of a draft European standard EN 50102: code IK. striker latching mechanism relief cone pedulum pivot arming button support fall height attaching support specimen Schneider Electric Merlin Gerin MV design guide 43 . Previous 3rd figures of the IP code in NF C 20-010 (1986) IP XX1 IP XX3 IP XX5 IP XX7 IP XX9 IK code IK 02 IK 04 IK 07 IK 08 IK 10 NB: to limit confusion. Spain. 15 10 P IK 02 0.B.: 1 of the hammer head 2 Fe 490-2 according to ISO 1052. hardness 50 HR to 58 HR according to ISO 6508 3 hardness HR 100 according to ISO 2039-2 44 Merlin Gerin MV design guide Schneider Electric .35 10 P IK 04 0.Design rules Protection index The various IK codes and their meaning IK code energies in joules radius mm 1 material 1 steel = A 2 polyamide = P 3 hammer pendulum spring loaded 4 vertical IK 01 0.7 10 P IK 06 1 10 P IK 07 2 25 A IK 08 5 25 A IK 09 10 50 A IK 10 20 50 A ✔ ✔ ✔ ✔ ✔ = yes ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ N.2 10 P IK 03 0.5 10 P IK 05 0. c The circuit breaker is a device that ensures the control and protection on a network. c Since a circuit breaker is mostly in the "closed" position. It is always greater than the operating voltage. c Standardised values for Ur (kV) : 3. v rated cable-charging breaking current. § 4.24 .2 -12 . for a particular time constant (IEC) v the constant load current. v rated line-charging breaking current. the selection of controls and installation. the load current must pass through it without the temperature running away throughout the equipment's life. v rated back-to-back capacitor bank breaking current.7 • Isc (ANSI).Switchgear definition Medium voltage circuit breaker Introduction IEC 60 056 and ANSI C37-06 define on one hand the operating conditions. v rated small inductive breaking current.5 .5 • Isc for 50 Hz (IEC) 2. c The main circuit must be able to withstand without damage: v the thermal current = short-circuit current during 1 or 3 s v the electrodynamic current: 2.17.1 IEC 60 694) The rated voltage is the maximum rms. v rated capacitor bank inrush making current. the design and the manufacture.6 • Isc for 60 Hz (IEC) 2. v rated capacitor bank breaking current. Rated voltage (cf. the rated characteristics. It is capable of making.36 kV.6 . Special rated characteristics c These characteristics are not compulsory but can be requested for specific applications: v rated out-of-phase breaking current. and on the other hand the testing. value of the voltage that the equipment can withstand in normal service.7. Characteristics Compulsory rated characteristics c Rated voltage c Rated insulation level c Rated normal current c Rated short-time withstand current c Rated peak withstand current c Rated short-circuit duration c Rated supply voltage for opening and closing devices and auxiliary circuits c Rated frequency c Rated short-circuit breaking current c Rated transient recovery voltage c Rated short-circuit making current c Rated operating sequence c Rated time quantities. Schneider Electric Merlin Gerin MV design guide 45 . withstanding and interrupting operating currents as well as short-circuit currents. 5 .7 • Isc for special applications. § 4. It must be greater than or equal to the rated short-time withstand peak current.31.2 table 3 IEC 60 694). § 4.16 .6 • Isc for 60 Hz 2.3 .2 12 17.8 .7 IEC 60 694) The rated short-circuit is equal to 1 or 3 seconds.103 IEC 60 056) The making current is the maximum value that a circuit breaker is capable of making and maintaining on an installation in short-circuit. Rated short-time withstand current (cf.5 .2 µs Rated insulation level (cf. § 4.5 • Isc for 50 Hz 2. § 4.4.10 .Switchgear definition Medium voltage circuit breaker Upeak (%) 100 90 50 1.20 .4 IEC 60 694) With the circuit breaker always closed.40 .5 24 36 Impulse withstand voltage (Up in kV) 60 75 95 125 170 Power frequency withstand voltage (Ud in kV) 20 28 38 50 70 Rated normal current (cf.2/50 µs c The insulation level is characterised by two values: v the impulse wave withstand (1. the load current must pass through it in compliance with a maximum temperature value as a function of the materials and the type of connections.6 IEC 60 694) and making current (cf. Rated voltage (Ur in kV) 7.2/50 µs) v the power frequency withstand voltage for 1 minute. value of the maximum permissible short-circuit current on a network for 1 or 3 seconds. Rated peak withstand current (cf. IEC sets the maximum permissible temperature rise of various materials used for an ambient air temperature of no greater than 40°C (cf.25 . Rated short-circuit duration (cf.50 kA.5 IEC 60 694) Isc = Ssc U Isc : : : Ssc e •U (in MVA) (in kV) (in kA) short-circuit power operating voltage short-circuit current This is the standardised rms. c Values of rated breaking current under maximum short-circuit (kA): 6. § 4. § 4. Isc is the maximum value of the rated short-circuit current for the circuit breakers' rated voltage. The peak value of the short-time withstand current is equal to: 2. 46 Merlin Gerin MV design guide Schneider Electric . § 4.2 IEC 60 056 and 60 694) t (µs) 10 50 µs Standardised wave 1.12. 220 or 250 volts.15 s .230 .CO v quick 1: O .3 mn .240 volts. § 4.3 s . v for alternating current (ac): 120 .CO .0. open position displacement of contacts current flows opening-closing duration making-breaking duration contacts are touching in all poles and order O energising of closing circuit current starts to flow in first pole time final arc extinction in all poles separation of arcing contacts in all poles Schneider Electric Merlin Gerin MV design guide 47 .CO .48 .t' . The rated frequency is either 50 Hz or 60 Hz.CO v quick 2: O .104 IEC 60 056) c Rated switching sequence according to IEC.t .3 mn .CO .: other sequences can be requested.3 s .CO N. § 4. the release unit must enable the device to close) Rated frequency (cf. c The operating voltages must lie within the following ranges: v motor and closing release units: -15% to +10% of Ur in dc and ac v opening release units: -30% to +10% of Ur in dc -15% to +10% of Ur in ac v undervoltage opening release unit: the release unit gives the command and forbids closing the release unit must not have an action U 0% 35 % 70 % 100 % (at 85%. t Isc Ir t' Rated operating sequence (cf. (cf: opposite diagram) time O C O C O O CO : : represents opening operation represents closing operation followed immediately by an opening operation c Three rated operating sequences exist: v slow: 0 .CO.3 mn . a few countries use both frequencies. § 4.8 IEC 60 694) c Values of supply voltage for auxiliary circuits: v for direct current (dc): 24 .9 IEC 60 694) Two frequencies are currently used throughout the world: 50 Hz in Europe and 60 Hz in America.Switchgear definition Medium voltage circuit breaker Rated supply voltage for closing and opening devices and auxiliary circuits (cf.220 . c Opening/closing cycle Assumption: O order as soon as the circuit breaker is closed.110 or 125 . O .60 .CO .B.0. the nearest standard rating is 40 kA. What does Iasym equal? Iasym = Isym = 27 kA % DC 100 90 80 70 60 50 40 30 20 10 0 10 20 30 40 50 60 1 + 2( %DC )2 100 1 + 2 (0.7 kA c Using the equation [A]. the circuit breaker must break the rms. In this case use the τ1 graph. value of its periodic component. § 4. given by the term: "rated short-circuit breaking current" v the percentage of the aperiodic component corresponding to the circuit breaker's opening duration. value of the periodic component of the short-circuit (= its rated breaking current) with the percentage of asymmetry defined by the graphs below.8 kA.5 • Isc at 50 Hz or 2.8 kA for a %DC of 30%.Switchgear definition Medium voltage circuit breaker c Automatic reclosing cycle Assumption: C order as soon as the circuit breaker is open.086 τ1= 45 ms (standardised time constant) 70 80 90 τ (ms) t : circuit breaker opening duration (Top). 1.6 • Isc at 60 Hz. this being 10 ms at 50 Hz. 48 Merlin Gerin MV design guide Schneider Electric . c According to IEC.65) 2 } [A] τ4= 120 ms (alternating time constant) = 36. this is equivalent to a symmetric short-circuit current at a rating of: 36. Percentage of the aperiodic component (% DC) as a function of the time interval (τ) = 29.101 IEC 60 056) The rated short-circuit breaking current is the highest value of current that the circuit breaker must be capable of breaking at its rated voltage. The half-period corresponds to the minimum activation time of an overcurrent protection device. c As standard the IEC defines MV equipment for a %DC of 30%. According to the IEC.7 kA = 33. (with time delay to achieve 0. for a peak value of maximum current equal to 2. to which we add a half-period of the rated frequency. the graph gives a percentage of the aperiodic component of around 30 % for a time constant τ1 = 45 ms: %DC = e -(45 + 10) 45 Rated short-circuit breaking current (cf. closed position displacement of contacts open position current flows making-breaking duration opening-closing duration remaking duration reclosing duration final arc extinction in all poles separation of arc contacts in all poles and order C energising of opening release unit current flows time the contacts are touching in all poles the contacts touch in the first pole start of current flow in the first pole Example 1: c For a circuit breaker with a minimum opening duration of 45 ms (Top) to which we add 10 ms (Tr) due to relaying. increased by half a period at the power frequency (τr) c The circuit breaker rating is greater than 33.5 % Example 2: c Supposing that % DC of a MV circuit breaker is equal to 65% and that the symmetric short-circuit current that is calculated (Isym) is equal to 27 kA. c It is characterised by two values: v the rms.3 sec or 15 secs or 3 min). The recovery voltage wave form varies according to the real circuit configuration.100 kA. it is equal to 1.5 .7 • Isc.Switchgear definition Medium voltage circuit breaker c For low resistive circuits such as generator incomers. In this case use the τ4 graph. For all constants of between τ1 and τ4.50 . with a peak value of maximum current equal to 2. in other words the voltage across the terminals of the open pole. The ratio of this voltage to a simple voltage is called the first pole factor. 4 c Values of rated short-circuit breaking current: 6.16 20 .(Top + Tr) IDC IAC • 100 = 100 • e τ (1.5 kV. A circuit breaker must be able to break a given current for all recovery voltages whose value remains less than the rated TRV. Schneider Electric Merlin Gerin MV design guide 49 .12. 10 20 60 100 100 % aperiodic component %DC ≤ 20 ≤ 20 ≤ 20 ≤ 20 according to equation IAC IMC t (s) IDC * for circuit breakers opening in less than 80 ms IMC IAC Idc %DC : : : : making current periodic component peak value (Isc peak) aperiodic component value % asymmetry or aperiodic component: .5 for voltages up to 72. c First pole factor For three-phase circuits.31.5 . %DC can be higher.40 .10 .3 .102 IEC 60 056) This is the voltage that appears across the terminals of a circuit breaker pole after the current has been interrupted. I (A) c Short-circuit breaking tests must meet the five following test sequences: Sequence 1 2 3 4 5* % Isym. the TRV refers to the pole that breaks the circuit initially.8 . 4) c Symmetric short-circuit current (in kA): Isym = IAC r c Asymmetric short-circuit current (in kA): Iasym2 = I2AC + I2DC Iasym = Isym 1 + 2( %DC )2 100 Rated Transient Recovery Voltage (TRV) (cf. § 4. …. use the equation: -(Top + Tr) % DC = 100 • e τ1. ….25 . 57 0 td t3 t (µs) Uc = 1.6 30 41 62 Time (t3 in µs) 52 60 72 88 108 Delay (td in µs) 8 9 11 13 16 Increase rate (Uc/td in kV/µs) 0.43 UA . it is given for an asymmetry of 0%.18 0.2 12 17.5 24 36 TRV value (Uc in kV) 12.24 0.6 45 61 92 Time (t3 in µs) 104 120 144 176 216 50 Merlin Gerin MV design guide Schneider Electric . c If Ur is the rated circuit breaker voltage.715 Ur e td = 0.UB = 2U Uc = 1. § 4.3 20.4 • 1. at a voltage equal to twice the voltage relative to earth.47 0. standards require the circuit breaker to break a current equal to 25% of the fault current across the terminals. G U1 U2 G c In practice. the recovery voltage (TRV) at power frequency is equal to: v 2e Ur for networks with a neutral earthing arrangement e v 2. Td : t3 : Uc : TRV increase rate: time delay time defined to reach Uc peak TRV voltage in kV Uc/t3 in kV/µs X1 A B X2 Rated out-of-phase breaking current (cf.5 24 36 TRV value (Uc in kV) 18.5e Ur for other networks. U (kV) Uc Rated voltage (Ur in kV) 7.35 0.(-U2) = U1 + U2 si U1 = U2 so UA .25 • 2.2 12 17.5 • Rated voltage (Ur in kV) 7.4 30.31 0.15 t3 v a specified TRV is represented by a reference plot with two parameters and by a segment of straight line defining a time delay.5 • r • Ur = 1. the voltage across the terminals can increase up the sum of voltages in the conductors (phase opposition).34 0.26 0.Switchgear definition Medium voltage circuit breaker c Value of rated TRV v the TRV is a function of the asymmetry. e c Peak values for TRV for networks other than those with neutral earthing: e • Ur r Rate of increase (Uc/td in kV/µs) 0.42 0.UB = U1 .106 IEC 60 056) When a circuit breaker is open and the conductors are not synchronous. 2 12 17.109 IEC 60 056) The specification of a breaking current for a circuit breaker switch located upstream of capacitors is not compulsory.7 times the device's rated current. L A B Ic Rated single capacitor bank breaking current (cf.5 24 36 Rated breaking current for no-load cables (Ic in kA) 10 25 31. c Normal rated breaking current values for a circuit breaker located at the head of no-load cables: Rated voltage (Ur in kV) 7.108 IEC 60 056) The specification of a rated breaking current for a circuit breaker located at the head of no-load cables is not compulsory and is considered as not being necessary for voltages less than 24 kV. § 4.5 31. three-phased lines and to a rated voltage ≥ 72 kV.5 • Ur r e Rated back-to-back capacitor bank breaking current (cf. § 4.107 IEC 60 056) The specification of a rated breaking current for a circuit breaker switch situated at the head of no-load lines is limited to overhead.110 IEC 60 056) G U The specification of a breaking current for multi-stage capacitor banks is not compulsory. c If n is equal to the number of stages.Switchgear definition Medium voltage circuit breaker Rated cable-charging breaking current (cf. § 4 . Due to the presence of harmonics. Rated current (A) 400 630 1250 2500 3150 G U C Breaking current for capacitors (A) 280 440 875 1750 2200 By definition pu = Ur r e c The normal value of over-voltage obtained is equal to 2.5 pu. § 4. this being: X1 2. then the over-voltage is equal to: C1 C2 C3 2n 2n + 1 • pu with pu = Ur r e Schneider Electric Merlin Gerin MV design guide 51 .5 50 Rated line-charging breaking current (cf. the breaking current for capacitors is equal to 0. 2 µs Relative to earth (2 Ur + 5) kV 2Ur + 1 ⇒ 2(2Ur + 1) ⇒ 0 14 kV ⇒ 28 kV ⇒ 0 1 kV/s 0 1 mn t Normal operating conditions (cf. § 4.6 kV increase time 1. IEC 60 694) For all equipment functioning under other conditions than those described below. the frequency of the pick-up current is normally in the region of 2 . value Impulse test (4 Ur + 5) kV 4.9 pu + 5 = 31 kV at 6.112 IEC 60 056) The breaking of a low inductive current (several amperes to several tens of amperes) causes overvoltages. Equipment is designed for normal operation under the following conditions: c Temperature 0°C Instantaneous ambient minimal maximal average daily maximum value Installation Indoor -5°C +40°C 35°C Outdoor -25°C +40°C 35°C 52 Merlin Gerin MV design guide Schneider Electric . c The figure opposite shows the various voltages on the load side U Up Um Uc Uf Uif Ud t Uf Uc Um Uif Up Ud : : : : : : instantaneous network voltage value network voltage at the moment of breaking extinction point overvoltage relative to earth maximum overvoltage relative to earth maximum peak-to-peak amplitude of the overvoltage due to restrike. motors). c Insulation level of motors IEC 60 034 stipulates the insulation level of motors. The value of the circuit breaker's rated closing current must be greater than the making current for the capacitor bank. Insulation Between turns Test at 50 (60) Hz rms.6 kV (50% on the sample) increase time 0. Rated small inductive breaking current (cf. In service.9 pu + 5 = 31 kV at 6.5 µs (4 Ur + 5) kV 4. derating should be carried out (see derating chapter).111 IEC 60 056) The rated closing current for capacitor banks is the peak current value that the circuit breaker must be capable of making at the rated voltage. Power frequency and impulse withstand testing is given in the table below (rated insulation levels for rotary sets). § 4.5 kHz.Switchgear definition Medium voltage circuit breaker Rated capacitor bank inrush making current (cf. The type of circuit breaker will be chosen so that the overvoltages that appear do not damage the insulation of the current consumers (transformer. Switchgear definition Medium voltage circuit breaker c Humidity Average relative humidity for a period 24 hours 1 month Indoor equipment 95% 90% c Altitude The altitude must not exceed 1 000 metres. Electrical endurance The electrical endurance requested by the recommendation is three breaking operations at Isc. Merlin Gerin circuit breakers are capable of breaking Isc at least 15 times. Mechanical endurance The mechanical endurance requested by the recommendation is 2 000 switching operations. Merlin Gerin circuit breakers guarantee 10 000 switching operations. Co-ordination of rated values (cf. § IEC 60 056) Rated voltage Ur (kV) 3.6 Rated short-circuit breaking current Isc (kV) 10 16 25 40 8 12.5 16 25 40 8 12.5 16 25 40 50 8 12.5 16 25 40 8 12.5 16 25 40 8 12.5 16 25 40 Rated current in continuous service Ir (A) 400 630 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1600 1600 2500 2500 3150 7.2 400 400 630 630 630 1600 1600 1600 2500 2500 3150 12 400 400 630 630 630 1600 1600 1600 1600 2500 2500 2500 3150 3150 17.5 400 630 630 630 1600 2500 3150 24 400 630 630 630 1600 1600 2500 2500 3150 36 630 630 630 1600 1600 1600 2500 2500 3150 Schneider Electric Merlin Gerin MV design guide 53 Switchgear definition Current transformer Please note! Never leave a CT in an open circuit. This is intended to provide a secondary circuit with a current proportional to the primary current. Transformation ratio (Kn) Kn = Ipr = N2 Isr N1 N.B.: current transformers must be in conformity with standard IEC 185 but can also be defined by standards BS 3938 and ANSI. c It comprises one or several primary windings around one or several secondary windings each having their own magnetic circuit, and all being encapsulated in an insulating resin. c It is dangerous to leave a CT in an open circuit because dangerous voltages for both people and equipment may appear across its terminals. Primary circuit characteristics according to IEC standards Rated frequency (fr) A CT defined at 50 Hz can be installed on a 60 Hz network. Its precision is retained. The opposite is not true. Rated primary circuit voltage (Upr) c General case: Rated CT voltage ≥ rated installation voltage The rated voltage sets the equipment insulation level (see "Introduction" chapter of this guide). Generally, we would choose the rated CT voltage based on the installation operating voltage U, according to the chart: U Upr 3.3 5 5.5 6 6.6 10 11 13.8 15 20 22 30 33 7.2 kV 12 kV Core balance CT insulator air 17.5 kV insulator 24 kV 36 kV t cable or busduc (sheathed or not sheathed busduct) c Special case: If the CT is a core balance CT installed on a busduct or on a cable. The dielectric insulation is provided by the cable or busducting insulation and the air located between them. The core balance CT is itself insulated. 54 Merlin Gerin MV design guide Schneider Electric Switchgear definition Current transformer Primary operating current (Ips) An installation's primary operating current I (kA) (for a transformer feeder for example) is equal to the CT primary operating current (Ips) taking account of any possible derating. c If: S U P Q Ips : : : : : apparent power in kVA primary operating voltage in kV active power of the motor in kW reactive power of capacitors in kvars primary operating current in A c We will have: v incomer cubicle Ips = v generator set incomer Ips = v transformer feeder Ips = v motor feeder Ips = η : S e• U S e• U S e• U P e • U • cosϕ • η motor efficiency If you do not know the exact values of ϕ and η, you can take as an initial approximation: cos ϕ = 0.8 ; η = 0.8. Example: A thermal protection device for a motor has a setting range of between 0.6 and 1.2 • IrTC. In order to protect this motor, the required setting must correspond to the motor's rated current. c If we suppose that Ir for the motor = 45 A, the required setting is therefore 45 A; v if we use a 100/5 CT, the relay will never see 45 A because: 100 • 0.6 = 60 > 45 A. v if on the other hand, we choose a CT 75/5, we will have: 0.6 < 45 < 1.2 75 and therefore we will be able to set our relay. This CT is therefore suitable. v capacitor feeder 1.3 is a derating coefficient of 30% to take account of temperature rise due to capacitor harmonics. Ips = 1.3 • Q e •U v bus sectioning The current Ips of the CT is the greatest value of current that can flow in the bus sectioning on a permanent basis. Rated primary current (Ipr) The rated current (Ipr) will always be greater than or equal to the operating current (I) for the installation. c Standardised values: 10 -12.5 - 15 - 20 - 25 - 30 - 40 - 50 - 60 - 75 and their multiples and factors. c For metering and usual current-based protection devices, the rated primary current must not exceed 1.5 times the operating current. In the case of protection, we have to check that the chosen rated current enables the relay setting threshold to be reached in the case of a fault. N.B.: current transformers must be able to withstand 1.2 times the rated current on a constant basis and this as well must be in conformity with the standards. Schneider Electric Merlin Gerin MV design guide 55 Switchgear definition Current transformer In the case of an ambient temperature greater than 40°C for the CT, the CT's nominal current (Ipn) must be greater than Ips multiplied by the derating factor corresponding to the cubicle. As a general rule, the derating is of 1% Ipn per degree above 40°C. (See "Derating" chapter in this guide). Rated thermal short-circuit current (Ith) The rated thermal short-circuit current is generally the rms. value of the installation's maximum short-circuit current and the duration of this is generally taken to be equal to 1 s. c Each CT must be able to withstand the short-circuit current which can flow through its primary circuit both thermally and dynamically until the fault is effectively broken. Example: c If Ssc is the network short-circuit power expressed in MVA, then: c Ssc = 250 MVA c U = 15 kV 3 3 Ith 1 s = Ssc • 10 = 250 • 10 = 9 600 A 15 • e U•e Ith = Ssc U•e c When the CT is installed in a fuse protected cubicle, the Ith to use is equal to 80 Ir. c If 80 Ir > Ith 1 s for the disconnecting device, then Ith 1 s for the CT = Ith 1 s for the device. Overcurrent coefficient (Ksi) Knowing this allows us to know whether a CT will be easy to manufacture or otherwise. c It is equal to: Ksi = Ith 1 s Ipr c The lower Ksi is, the easier the CT will be to manufacture. A high Ksi leads to over-dimensioning of the primary winding's section. The number of primary turns will therefore be limited together with the induced electromotive force; the CT will be even more difficult to produce. Order of magnitude ksi Ksi < 100 100 < Ksi < 300 100 < Ksi < 400 400 < Ksi < 500 Ksi > 500 Manufacture standard sometimes difficult for certain secondary characteristics difficult limited to certain secondary characteristics very often impossible A CT's secondary circuit must be adapted to constraints related to its use, either in metering or in protection applications. 56 Merlin Gerin MV design guide Schneider Electric knowing that: P = R.5 mm2 c Consumption of copper cabling (line losses of the cabling).30 VA. Real power that the TC must provide in VA This is the sum of the consumption of the cabling and that of each device connected to the TC secondary circuit. Accuracy class (cl) c Metering: class 0. Rated output Take the standardised value immediately above the real power that the CT must provide.8 m 1 VA c Consumption of metering or protection devices.5 .Switchgear definition Current transformer Secondary circuit's characteristics according to IEC standards Rated secondary current (Isr) 5 or 1 A? c General case: v for local use Isr = 5 A v for remote use Isr = 1 A c Special case: v for local use Isr = 1 A N.0176 : L : S : if Isr = 5 A if Isr = 1 A length in metres of link conductors (feed/return) cabling section in mm2 5. Safety factor (SF) c Protection of metering devices in the case of a fault is defined by the safety factor SF. Consumption of various devices are given in the manufacturer's technical data sheet.I2 and R = ρ.5 c Switchboard metering: class 1 c Overcurrent protection: class 10P sometimes 5P c Differential protection: class X c Zero-sequence protection: class 5P. The value of SF will be chosen according to the current consumer's short-time withstand current: 5 ≤ SF ≤ 10.L/S then: (VA) = k • L S k = 0.44 : k = 0.5 . SF is the ratio between the limit of rated primary current (Ipl) and the rated primary current (Ipr). c The standardised values of rated output are: 2. SF = Ipl Ipr c Ipl is the value of primary current for which the error in secondary current = 10 %.B. Example: c Cable section: c Cable length (feed/return): c Consumed power by the cabling: 2. Schneider Electric Merlin Gerin MV design guide 57 .10 . (line loss: P = R I 2).: Using 5 A for a remote application is not forbidden but leads to an increase in transformer dimensions and cable section.15 . Schneider Electric CT's have a safety factor of 10. c The relay will function perfectly if: ALF real of CT > 2 • Ire Isr Ire Isr : : relay threshold setting rated secondary current of the CT c For a relay with two setting thresholds. we will generally have a high threshold set to 8 Ir max. we will generally have an instantaneous high threshold set at 14 Ir max. We will determine the required ALF in the following manner: Definite time overcurrent protection. Inverse definite time overcurrent protection c In all cases. giving the real ALF required > 28 v for a motor feeder. A safety factory of 5 is suitable.e. i. refer to the relay manufacturer's technical datasheet. (never true for feeders to other switchboards or for incomers): ALF real > 2 • Ir2 Isr Ir2 : instantaneous high setting threshold for the module 58 Merlin Gerin MV design guide Schneider Electric . However. ALF real > 20 • Ire c Special cases: v if the maximum short-circuit current is greater than or equal to 10 Ire: ALF real > 20 • Ire : Ire Isr relay setting threshold v if the maximum short-circuit current is less than 10 Ire: ALF real > 2 • Isc secondary Isr v if the protection device has an instantaneous high threshold that is used. the CT must guarantee accuracy across the whole trip curve for the relay up to 10 times the setting current. we have two constraints: having an accuracy limit factor and an accuracy class suited to the application. Accuracy limit factor (ALF) In protection applications. For these protection devices. the current transformer must be saturated before 10 Ir in the secondary. To be sure that this device will not be destoyed in the case of a primary fault. v For a transformer feeder.. according to the current consumer characteristic a lower safety factor can be requested.. giving a real ALF required > 16. 50 A for a 5 A device. c In accordance with the standards. we will use the highest threshold.Switchgear definition Current transformer c An ammeter is generally guaranteed to withstand a short-time current of 10 Ir. If (Rct + Rb + Rr) The exact equation is given by the relay manufacturer. c Class X is often requested in the form of: Vk ≤ a . Values characterising the CT Vk a Rct Rb Rr If : : : : : : Knee-point voltage in volts asymmetry coefficient max.Switchgear definition Current transformer Differential protection Many manufacturers of differential protection relays recommend class X CT's. c For a generator set differential: v if Isc is known: Isc short-circuit current for the generator set on its own If = Isc Kn relay v if the Ir gen is known: we will take If = 7 • Ir gen Kn v if the Ir gen is unknown: we will take If = 7 • Isr (CT) Isr(CT) = 1 or 5 A c For motor differential: v if the start-up current is known: we will take Isc = I start-up CT G CT relay If = Isc Kn v if the Ir motor is known: we will take CT M CT If = 7 • Ir Kn v if the Ir motor is not known: we will take If = 7 • Isr (CT) Isr(TC) = 1 or 5 A Reminder Ir : rated current Schneider Electric Merlin Gerin MV design guide 59 . resistance in the secondary winding in Ohms loop resistance (feed/return line) in Ohms resistance of relays not located in the differential part of the circuit in Ohms maximum fault current seen by the CT in the secondary circuit for a fault outside of the zone to be protected If = Isc Kn : : Isc Kn primary short-circuit current CT transformation ratio What values should If be given to determine Vk? c The short-circuit current is chosen as a function of the application: v generator set differential v motor differential v transformer differential v busbar differential. 60 Merlin Gerin MV design guide Schneider Electric .Switchgear definition Current transformer CT c For a transformer differential The Isc to take is that flowing through the CT's for a current consumer side fault. v if we do not know the exact value. therefore limited by the cable impedance. we will take: relay If = 20 Isr(CT) c For busbar differential v the Isc to take is the switchboard Ith If = Ith Kn CT c For a line differential The Isc to take is the Isc calculated at the other end of the line. we will take the switchboard Ith. If the impedance of the cable is not known. the fault current value If is less than 20 Isr(CT). In all cases. 5 1. Rated primary voltage (Upr) c According to their design. all of which is encapsulated in an insulating resin. or in a compensated network with an extinction coil without automatic elimination of the earthing fault 1.2 continuous. one or several secondary windings. N. The voltage transformer is intended to provide the secondary circuit with a secondary voltage that is proportional to that applied to the primary circuit.9 continuous 30 s continuous 30 s continuous 8h N. Generally.2 1.2 1. voltage transformers will be connected: v either phase to earth v or phase to phase 3000 V / 100 V e e Upr = U e 3000 V / 100 V Upr = U Schneider Electric Merlin Gerin MV design guide 61 .B. Characteristics The rated voltage factor (KT) The rated voltage factor is the factor by which the rated primary voltage has to be multiplied in order to determine the maximum voltage for which the transformer must comply with the specified temperature rise and accuracy recommendations.9 for 8 h and VT phase/phase 1. It comprises a primary winding.: lower rated durations are possible when agreed to by the manufacturer and the user. the voltage transformer must be able to withstand this maximum voltage for the time that is required to eliminate the fault. a magnetic core. voltage transformer manufacturers comply with the following values: VT phase/earth 1.: IEC standard 60 186 defines the conditions which voltage transformers must meet.2 Rated duration continuous Primary winding connection mode and network earthing arrangement phase to phase on any network neutral point to earth for star connected transformers in any network phase to earth in an earthed neutral network phase to earth in a network without an earthed neutral with automatic elimination of earthing faults phase to earth in an isolated neutral network without automatic elimination of earthing faults. According to the network's earthing arrangement. Normal values of the rated voltage factor Rated voltage factor 1.2 1.B.9 1.Switchgear definition Voltage transformer We can leave a voltage transformer in an open circuit without any danger but it must never be short-circuited. class 3 is very little used. the rated secondary voltage must be divided by e. c For single phase transformers intended to be connected in a phase to earth arrangement.25 .15 . Application not used industrially precise metering everyday metering statistical and/or instrument metering metering not requiring great accuracy Accuracy class 0. Measurement according to IEC 60 186 Classes 0.500 VA.400 .5 and 1 are suitable for most cases. E. Accuracy class This defines the limits of errors guaranteed in terms of transformation ratio and phase under the specified conditions of both power and voltage.30 . c The accuracy class is guaranteed for values: v of voltage of between 5% of the primary voltage and the maximum value of this voltage which is the product of the primary voltage and the rated voltage factor (kT x Upr) v for a secondary load of between 25% and 100% of the rated output with a power factor of 0.200 . (S = eUI in three-phase circuits) c Standardised values are: 10 .: 100 V e Rated output Expressed in VA.8 inductive.150 .g.2 0. Accuracy class Voltage error as ± % between 5% Upr and kT • Upr 3 6 between 2% and 5% Upr 6 12 Phase shift in minutes between 5% Upr and kT • Upr 120 24 between 2% and 5% Upr 240 480 3P 6P Upr = rated primary voltage kT = voltage factor phase shift = see explanation next page 62 Merlin Gerin MV design guide Schneider Electric .50 .300 .5 1 3 Protection according to IEC 60 186 Classes 3P and 6P exist but in practice only class 3P is used.100 .75 .1 0.Switchgear definition Voltage transformer Rated secondary voltage (Usr) c For phase to phase VT the rated secondary voltage is 100 or 110 V. It must not introduce any error exceeding the values guaranteed by the accuracy class. this is the apparent power that a voltage transformer can provide the secondary circuit when connected at its rated primary voltage and connected to the nominal load. voltage error (%) = (kn Usr .Switchgear definition Voltage transformer Transformation ratio (Kn) Kn = Upr = N1 Usr N2 for a TT Voltage ratio error This is the error that the transformer introduces into the voltage measurement. IT is expressed in minutes of angle. The thermal power limit or rated continuous power This is the apparent power that the transformer can supply in steady state at its rated secondary voltage without exceeding the temperature rise limits set by the standards. Schneider Electric Merlin Gerin MV design guide 63 .Upr)•100 Upr Kn = transformation ratio Phase error or phase-shift error This is the phase difference between the primary voltage Upr and the secondary voltage Usr. N. in other words to derate the device. for altitudes of over 1 000 metres v in terms of the rated current. the materials and the dielectric used. Derating.8 0. Exception of the Mexican market: derating starts from zero metres (cf.B. however. insulation is in air. we have to derate by 1. This applies for the lightning impulse withstand voltage and the power frequency withstand voltage 50 Hz .85 = 58. Normal conditions of use are described in the "Medium voltage circuit breaker" chapter.: if you do not want to supply 36 kV equipment. (see chapter on "Protection indices"). Insulation derating according to altitude Example of application: Can equipment with a rated voltage of 24 kV be installed at 2500 metres? The impulse withstand voltage required is 125 kV . Beyond these limits.7 0. refer to the cubicle selection guide (derating depends on the cubicle design). c Derating must be considered: v in terms of the insulation level. These different types of derating can be accumulated if necessary.1 mn.B.8 kV c No. c Merlin Gerin uses correction coefficients: v for circuit breakers outside of a cubicle.5 altitude in m 0 1000 2000 3000 4000 5000 64 Merlin Gerin MV design guide Schneider Electric .85 = 147.Switchgear definition Derating Introduction The various standards or recommendations impose validity limits on device characteristics. Altitude has no effect on the dielectric withstand of circuit breakers in SF6 or vacuum. the equipment that must be installed is: v rated voltage = 36 kV v impulse withstand = 170 kV v withstand at 50 Hz = 70 kV N. Standards give a derating for all equipment installed at an altitude greater than 1 000 metres. As a general rule. must be taken account of when the circuit breaker is installed in cubicles. because they are within a sealed enclosure.: there are no standards specifically dealing with derating. Correctilon coefficient k 1 0. dotted line on the graph below). table V § 442 of IEC 60 694 deals with temperature rises and gives limit temperature values not to be exceeded according to the type of device.6 0. we must have the appropriate test certificates proving that our equipment complies with the request. .85 v the impulse withstand must be 125/0. c For 2500 m: v k is equal to 0. use the graph below v for circuit breakers in a cubicle. However.05 kV v the power frequency withstand 50 Hz must be 50/0. The power frequency withstand 50 Hz is 50 kV 1 mn. it is necessary to reduce certain values. In this case.25 % U peak every 100 metres above 1 000 metres. when the ambient temperature exceeds 40°C and for a protection index of over IP3X.9 0. derating is of 1 % Ir per degree above 40°C. IEC standard 60 694 § 442 table 5 defines the maximum permissible temperature rise for each device. this temperature rise depends on three parameters: v the rated current v the ambient temperature v the cubicle type and its IP (protection index). Schneider Electric Merlin Gerin MV design guide 65 . c In fact.Switchgear definition Derating Derating of the rated current according to temperature As a general rule. Derating will be carried out according to the cubicle selection tables. material and dielectric with a reference ambient temperature of 40°C. because conductors outside of the circuit breakers act to radiate and dissipate calories. γ … L2 L3 N/A metre squared (m2) metre cubed (m3 ) radian (rad) solid angle time Ω.15 K Schneider Electric Merlin Gerin MV design Guide 67 .017 453 3 rad minute ('): 1' = 2π rad/21 600 = 2.848 137 • 10-6 rad minute (mn) hour (h) day (d) revolutions per second (rev/s): 1 tr/s = 2π Magnitude: space and time length l. β. f = 1/T use of the angström (10-10 m) is forbidden.908 882 • 10-4 rad second ("): 1" = 2π rad/1 296 000 = 4. β. (Iv) α.273.80665 m/s2 radian per second (rad/s) radian per second squared (rad/s 2) kilogramme (kg) gramme (g) : 1 g = 10-3 kg ton (t) : 1 t = 103 kg kilogramme per metre (kg/m) kilogramme per metre squared (kg/m2) kilogramme per metre cubed (kg/m3) metre cubed per kilogramme (m3/kg) kilogramme per metre cubed concentration by mass of component B (kg/m3 ) (according to NF X 02-208) N/A second (s) hertz (Hz) radian (rad) metre (m) Magnitude: periodic phenomena period T frequency f phase shift ϕ wavelength λ 1 Hz = 1s-1. γ … Ω. (S) V α. (L) m t I T n I. (L) L area volume plane angle A. (ω) Unit metre kilogramme second ampere kelvin mole candela radian steradian Symbol of the unit m kg s A K mol cd rad sr Dimension L M T I θ N J N/A N/A Common magnitudes and units Name Symbol Dimension SI Unit: name (symbol) metre (m) Comments and other units centimetre (cm): 1 cm = 10-2 m (microns must no monger be used.) gradian (gr): 1 gr = 2π rad/400 revolution (rev): 1 tr = 2π rad degree(°):1°= 2π rad/360 = 0.Units of measure Names and symbols of SI units of measure Basic units Magnitude Basic units length mass time electrical current thermodynamic temperature 2 quantity of material light intensity Additional units angle (plane angle) solid angle Symbol of the magnitude1 l. (ω) t N/A T steradian (sr) second (s) speed rad/s acceleration angular speed angular acceleration Magnitude: mass mass linear mass mass per surface area mass per volume volume per mass concentration density v a ω α m ρ1 ρA' (ρs) ρ v ρB d L T-1 L T-2 T-1 T -2 M L-1 M L-2 M L-3 M L3 M-1 M L-3 N/A T T-1 N/A L metre per second (m/s) metre per second squared acceleration due to gravity: (m/s2) g = 9. Use of a factor of nanometre (109 m) is recommended λ = c/f = cT (c = celerity of light) power level Lp e1 2 N/A decibel (dB) the symbol in brackets can also be used the temperature Celsius t is related to the themrodynamic temperature T by the relationship: t = T . instead the micrometre (µm)) are (a): 1 a = 102 m2 hectare (ha): 1 ha = 104 m2 (agricult. meas. N to avoid any confusion with the millinewton 1 N/m = 1 J/m2 1 J : 1 N. T γ.m = 1 W. θ E C S c λ Q Φ P hr 1 µΩ.m) Newton per metre (N/m) Joule (J) Joule (J) Comments and other units 1 N = 1 m.K)) Watt per metre-Kelvin (W/(m.m 1 W = 1 J/s 1 var = 1 W Kelvin and not degree Kelvin or °Kelvin t = T . CGS unit) 1 St = 10-4 m2/s (St = stokes.m and not m.cm2/cm = 10-8 Ω. CGS unit) p = mv F G.m) Siemens per metre (S/m) Farad per metre (F/m) Watt (W) Voltampere (VA) var (var) Kelvin (K) degree Celsius (°C) Joule (J) Joule per Kelvin (J/K) Joule per Kelvin (J/K) Watt per kilogramme-Kelvin (J/(kg. (P. µ ν p L2 M T-3 L-1 M T-2 Watt (W) Pascal (Pa) dynamic viscosity kinetic viscosity quantity of movement Magnitude: electricity current electrical charge electrical potential electrical field electrical resistance electrical conductivity electrical capacitance electrical inductance L-1 M T-1 L2 T-1 L M T-1 Pascal-second (Pa.15 K 1 W = 1 J/s 68 Merlin Gerin MV design guide Schneider Electric . Fm resistivity ρ conductivity γ permittivity ε active P apparent power S reactive power Q Magnitude: thermal thermodynamic temperature temperature Celsius energy heat capacity entropy specific heat capacity thermal conductivity quantity of heat thermal flux thermal power coefficient of thermal radiation T t.m/s) Ampere (A) Coulomb (C) Volt (V) Volt per metre (V/m) Ohm (Ω) Siemens (S) Farad (F) Henry (H) Tesla (T) Weber (Wb) Ampere per metre (A/m) Ampere per metre (A/m) Ampere (A) Ohm-metre (Ω.s 1 V = 1 W/A 1 Ω = 1 V/A 1 S = 1 A/V = 1Ω-1 1 F = 1 C/V 1 H = 1 Wb/A 1 T = 1 Wb/m2 1 Wb = 1 V. magnetism magnetic induction B magnetic induction flux Φ magnetisation Hi.Units of measure Names and symbols of SI units of measure Name Magnitude: mechanical force weight moment of the force surface tension work energy Symbol Dimension SI Unit: name (symbol) Newton Newton-metre (N.6 • 103 J (used in determining electrical consumption) 1 W = 1 J/s 1 Pa = 1 N/m2 (for the pressure in fluids we use bars (bar): 1 bar = 105 Pa) 1 P = 10-1 Pa.kg/s2 N.s) metre squared per second (m2/s) kilogramme-metre per second (kg.K)) Joule (J) Watt (W) Watt (W) Watt per metre squared-Kelvin (W/(m2.s Magnitude: electricity.273.s (P = poise. W) M. σ W E L M T-2 L2 M T-2 M T-2 L2 M T-2 L2 M T-2 power pressure P σ.s Watthour (Wh) : 1 Wh = 3.K)) I Q V E R G C L I TI L2M T-3 I-1 L M T-3 I-1 L2 M T-3 I-2 L-2 M-1 T3 I2 L-2 M-1 T4 I2 L2 M T-2 I-2 M T -2 I-1 L2 M T-2 I-1 L-1 I L-1 I I L3 M T-3 I-2 L-3 M-1 T3 I2 L-3 M-1 T4 I2 L2 M T-3 L2 M T-3 L2 M T-3 θ θ L2 M T-2 L2 M T-2 θ-1 L2 M T-2 θ-1 L2 T-2 θ-1 L M T-3 θ-1 L2 M T-2 L2 M T-3 L2 M T-3 M T-3 θ-1 1 C = 1 A. M magnetic field H magneto-motive force F. τ p η. m2 1 ft H2O = 2.ft.679 9 • 103 kg/m3 1 Ib.316 dm3 1 in3 = 1. in2 °F °R Ibf.°F Btu/Ib°F Oe Btu/ft2.h = 3.490 89 • 102 Pa 1 Ibf/ft2 = 47.609 344 km 1 852 m 1 yd = 0.ft cu. in3 fl oz (UK) fl oz (US) gal (UK) gal (US) Conversion 1 ft/s2 = 0. ft2 sq.ft2 = 42.s.546 09 dm3 1 gaz (US) = 3.880 26 Pa 1 Ibf/in2 = 6.678 26 W/(m2.355 818 J 1 Ibf.290 3 • 10-2 m2 1 sq.018 46 kg/m3 1 Ib/in3 = 27.°F Btu Ibf/ft Ibf.858 kg/m 1 Ib/ft2 = 4.304 8 m 1 in = 25.488 16 kg/m 1 Ib/in = 17.degree Fahrenheit British thermal unit pound force-foot pound force-inch British thermal unit per square foot.349 5 g (6) 1 Ib = 0.069 6 kg/m2 1 Ib/ft3 = 16.055 06 • 103 W 1 Ibf = 4. ' in.°C) 1 Btu = 1.882 43 kg/m2 1 Ib/in2 = 703.293 071 W 1 sq.°C 1 Btu/Ib.in.h.413 0 cm3 fl oz (US) = 29. for example) where the carat is used (1 carat = 3.s cu.°F = 67.°F = 5.785 41 dm3 mass linear mass mass per surface area mass per volume moment of inertia pressure pressure .ft = 9.in = 0.s = 1.s/ft2 Ib/ft.326 • 103 J/kg 1 Btu/ft3.hour British thermal unit per second pound-force foot inch (1) mile (UK) knot yard (2) once (ounce) pound (livre) pound per foot pound per inch pound per square foot pound per square inch pound per cubic foot pound per cubic inch pound square foot foot of water inch of water pound force per square foot pound force per square inch (3) British thermal unit per hour square foot square inch degree Fahrenheit (4) degree Rankine (5) pound force-second per square foot pound per foot-second cubic foot cubic inch fluid ounce (UK) fluid ounce (US) gallon (UK) gallon (US) (1) (2) (3) Symbol ft/s2 Btu/Ib Btu/ft3.638 71 • 10-5 m3 fl oz (UK) = 28.strain calorific power surface area temperature viscosity volume 12 in = 1 ft 1 yd = 36 in = 3 ft Or p.ft = 1 ft3 = 28. gold.s 1 cu.066 1 • 103 J/m3.488 164 Pa.880 26 Pa.186 8 • 103 J(Kg.894 76 • 103 Pa 1 Btu/h = 0.573 5 cm3 1 gaz (UK) = 4. " mile yd oz Ib Ib/ft Ib/in Ib/ft2 Ib/in2 Ib/ft3 Ib/in3 Ib.i.577 47 A/m 1 Btu/ft2.304 8 m/s2 1 Btu/Ib = 2.110 35 10-2 kg) Schneider Electric Merlin Gerin MV design guide 69 .ft = 1.hour.in = 6.in Btu/ft2.degree Fahrenheit British thermal unit per (pound.°C) 1 Oe = 79.s/ft2 = 47.32) (5) °R = 5/9 °K (6) Apart from mass of precious metals (silver.914 4 m 1 oz = 28.055 056 • 103 J 1 Ibf.140 g.in.67) TK = 5/9 q °R 1 Ibf.451 6 • 10-4 m2 TK = 5/9 (q °F + 459.4 mm 1 mile = 1.degree Fahrenheit) oersted British thermal unit per square foot.ft2 ft H2O in H2O Ibf/ft2 Ibf/in2 (psi) Btu/h sq.Units of measure Names and symbols of SI units of measure Correspondence between Imperial units and international system units (SI) Magnitude acceleration calory capacity heat capacity magnetic field thermal conductivity energy energy (couple) thermal flux force length Unit foot per second squared British thermal unit per pound British thermal unit per cubit foot.112 985 J 1 Btu/ft2.453 592 37 kg 1 Ib/ft = 1.: pound force per square inch (4) T K = temperature kelvin with q°C = 5/9 (q°F .h Btu/s Ibf ft.989 07 • 103 Pa 1 in H2O = 2.s 1 Ib/ft.h.°F = 4.154 6 W/m2 1 Btu/s = 1.448 222 N 1 ft = 0. c International Electrotechnical Vocabulary c High voltage alternating current circuit breakers c Current transformers c Voltage transformers c Alternating current disconnectors and earthing disconnectors c High voltage switches c Metal-enclosed switchgear for alternating current at rated voltage of over 1 kV and less than or equal to 72.Switzerland.5 kV c High-voltage alternating current combined fuse-switches and combined fuse-circuit breakers c High-voltage alternating current contactors c Specifications common to highvoltage switchgear standards c Calculation rules in industrial installations c Derating IEC 60 050 IEC 60 056 IEC 60 185 IEC 60 186 IEC 60 129 IEC 60 265 IEC 60 298 IEC 60 420 IEC 60 470 IEC 60 694 IEC 60 909 ANSI C37 04 Schneider Electric Merlin Gerin MV Design Guide 71 .Standards The standards mentioned in this document Where can you order IEC publications? Central Offices of the International Electrotechnical Commission 1. rue de Varembé Geneva . The documentation department (Factory A2) at Merlin Gerin can provide you with information on the standards. 76 8.S.7.17.8 38 15.2 µs Rated voltage (kV) 7.Isc 1 500 to 10 000 according to Ua and Isc no text 3 times Isc 2 000 standard test circuit ANSI peak voltage is 10% greater than the voltage defined by the IEC.25 1.5 23 K 1.ANSI comparison Overview of the main differences The following comparison is based on different circuit breaker characteristics.24 1.6 11. The two standards easily allow the SF6 to reconstitute itself.16 7.5 25 38 Umax (kV) 4.65 1 1 1 Outdoor equipment Rated installation level According to IEC Upeak (%) 100 90 50 1.5 24 36 Rated lightning withstand voltage (kV) 60 75 95 125 170 Rated power frequency withstand voltage 50 Hz 1 mm (kV) 20 28 38 50 70 10 50 µs Standardised wave 1.5 .2 12 17.24 .Standards IEC .25 15 38 Umin (kV) 3.3 1.6•Isc at 60 Hz 2.7•Isc for special cases around twice as severe 4 times K.5•Isc at 50 Hz 2.12 . the largest part of the graph is the initial part where the SF6 reconstitutes itself. Standardised values for Ur (kV) Indoor equipment class (kV) 4.6 .7 Isc IEC 30% without derating short-time withstand current peak value Transient Recovery voltage(1) electrical endurance mechanical endurance motor overvoltages (1) the 2. Rated voltages According to IEC c Standardised values for Ur (kV): 3.85 6.2/50 µs t (µs) 72 Merlin Gerin MV Design Guide Schneider Electric . The E2/t2 slope is 50% greater than the Uc/t3 slope. However.2 .2 13. Theme asymmetrical breaking capacity on faults across the terminals insulation level: impulse wave ANSI 50% with current derating imposes chopped waves for outdoor equipment 115% Uw/3 s 129% Uw/2 s 2.36 kV According to ANSI c The ANSI standard defines a class and a voltage range factor K which defines a range of rated voltages at constant power. 40 .16 7.Standards IEC .15 BIL for a duration tc = 3 µs Rated normal current According to IEC c Values of rated current: 400 .B.1600 .20 .20 .8 .8 38 N.3150 A According to ANSI c Values of rated current: 1200 .3000 A Short-time withstand current According to IEC c Values of short-circuit rated breaking capacity: 6.3 .31.29 BIL for a duration of tc = 2 µs 1.5 25.2000 .ANSI comparison According to ANSI Upeak (%) 100 90 70 50 10 tc Rated voltage (kV) t (µs) Rated lightning withstand voltage (kV) 60 95 95 150 110 125 150 150 200 Rated power frequency withstand voltage 50 Hz 1 mm (kV) 19 36 36 80 50 60 80 Onde coupée suivant ANSI pour le matériel d'extérieur Indoor equipment 4.2500 . c The impulse withstand is equal to: 1.25 .31.5 .63 kA According to ANSI c Values of short-circuit rated breaking capacity: v indoor equipment: 12.5 .2 13.50 .10 .25 .630 .1250 . c BIL: Basic Insulation Level The outdoor equipment is tested with chopped waves.12.5 .5 .16 .40 kA v outdoor equipment: Class (MVA) 250 350 500 750 1000 1500 2750 Breaking capacity (kA) I at Umax 29 41 18 28 37 21 40 KI at Umin 36 49 23 36 46 35 40 Schneider Electric Merlin Gerin MV Design Guide 73 .8 38 Outdoor equipment 15. 230 .7•Isc for special cases. v for alternating (ac): 120 . (K : voltage factor) Rated short-circuit duration According to IEC c The rated short-circuit duration is equal to 1 or 3 seconds.Standards IEC .110 or 125 . value.125 .5•Isc at 50 Hz v 2. According to ANSI c The rated short-circuit duration is equal to 3 seconds.48 .6•Isc at 60 Hz v 2.240 volts 74 Merlin Gerin MV Design Guide Schneider Electric .ANSI comparison Peak value of short-time current and closing capacity According to IEC c The peak value of short-time withstand current is equal to: v 2.60 . Rated supply voltage for closing and opening devices and auxiliary circuits According to IEC c Supply voltage values for auxiliary circuits: v for direct current (dc): 24 . c Operating voltages must fall within the following ranges: v Motor and closing release units: -15% to +10% of Ur in dc et ac v opening release units: -15% to +10% of Ur in ac. According to ANSI c The peak value of short-time withstand current is equal to: v 2.220 or 250 volts v for alternating current (ac): 120 .7 K Isc at peak value v 1.250 volts.220 . -30% to +10% of Ur in dc v undervoltage opening release units the release unit gives the command and forbids closing the release unit must not have an action U 0% 35 % 70 % 100 % (at 85%. the release unit must enable the device to close) According to ANSI c Supply voltage values for auxiliary circuits: v for direct current (dc): 24 .48 .240 volts.6 K Isc at rms. The difference is not important because without taking account of the asymmetry factor "S".Standards IEC . Short-circuit breaking capacity at the rated operating sequence c ANSI specifies 50% asymmetry and IEC 30%. When 30% is too low. there are specific cases (proximity of generators) for which the asymmetry may be greater than 50%. the designer has to check the circuit breaker breaking capacity.ANSI comparison c Operating voltage must fall within the following ranges: Voltage Motor and closing release units 48 Vsc 125 Vsc 250 Vsc 120 Vac 240 Vac Opening release units 24 Vsc 48 Vsc 125 Vsc 250 Vsc 120 Vac 240 Vac Voltage range (V) 36 to 56 90 to 140 180 to 280 104 to 127 208 to 254 14 to 28 28 to 56 70 to 140 140 to 220 104 to 127 208 to 254 Rated frequency According to IEC c Rated frequency: 50 Hz. it is equal to 10%. ANSI: Iasym = Isym IEC: Iasym = Isym (1 + 2 A2) = 1. According to ANSI c Rated frequency: 60 Hz. 30% is sufficient.08 Isym (A = 30%) According to IEC c Short-circuit breaking tests must meet the following 5 test sequences: Sequence n° % Isym % aperiodic component 1 2 3 4 5* 10 20 60 100 100 ≤ 20 ≤ 20 ≤ 20 ≤ 20 30 * for circuit breakers opening at least 80 ms Schneider Electric Merlin Gerin MV Design Guide 75 . c For both standard systems.22 Isym (A = 50%) (1 + 2 A2) = 1. In 95% of applications. 8 1. this being 44 kA of total current.6 1.017 2 0. 1.100 30 < 20 60 50 .100 100 < 20 KI to V/K < 20 SI to V 50 .050 4 0.067 cycles seconds c Rated short-circuit breaking capacity (kA) Sequence n° Example: c Isc = 40 kA c % asymmetry = 50% c Iasym = 1.1 for Merlin Gerin circuit breakers.2 s .3 1.5 1.58 V) Short-circuit breaking testing must comply with the 14 test sequences above.0 0. 1 2 3 4 5 6 7 8 9/10 11 12 13/14 current broken % aperiodic component 10 50 .033 3 0.O at KI rated Isc duration = KI for 3 s single phase testing at KI and KSI (0.100 electrical endurance reclosing cycle at ASI and AKSI C .100 KSI to V/K 50 .006 0.Standards IEC .22 1 + 2(50%)2 Sequence 6 will therefore be tested at 36 kA + 50% asymmetry.1 • 40 = 44 kA 44 44 c Isym = = = 36 kA 1.5 1 0.2 1.ANSI comparison According to ANSI c The circuit breaker must be able to break: v the rated short circuit current at the rated maximum voltage v K times the rated short-circuit current (maxi symmetrical interrupting capability with K: voltage range factor) at the operating voltage (maximum voltage/K) v between the two currents obtained by the equation: maxi symetrical current rated maxi voltage = =K rated short-circuit current rated voltage We therefore have a constant breaking power (in MVA) over a given voltage range. Moreover.1 1 0 0 ratio S Asymmetrical interrupting capability = S x symetrical interrupting capability. with: I R : : symmetrical breaking capacity at maximum voltage reclosing cycle coefficient (Reclosing factor) voltage range factor: K : K= Vmax Vmin S : asymmetrical factor: Iasym = 1. the asymmetrical current will be a function of the following table taking S = 1. Both at specified operating voltage Symetrical interrupting capability at specified operating voltage = 1.4 1.1 Isym for Merlin Gerin circuit breakers V : maximum rated voltage 76 Merlin Gerin MV Design Guide Schneider Electric .7 1. 5 16 25 40 50 8 12.5 16 25 40 Rated operating current Ir (A) 400 630 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1250 1600 1600 2500 2500 3150 7.6 4 11.5 16 25 40 8 12.5 400 630 630 630 1600 2500 3150 24 400 630 630 630 1600 1600 2500 2500 3150 36 630 630 630 1600 1600 1600 2500 2500 3150 According to ANSI Maximum rated voltage Umax (kV) 4.76 Rated short-circuit breaking current at Umax Isc (kA) 18 29 41 7 17 33 9.5 16 25 40 8 12.6 11.5 6.9 18 35 56 5.8 12 12 12 12 12 23 24 Rated short-circuit breaking current at Umin Isc (kA) 24 36 49 25 30 41 21 37 23 43 36 48 24 23 45 73 12 24 36 57 Merlin Gerin MV Design Guide Rated operating current Ir (A) 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 2000 600 1200 1200 1200 3000 3000 4000 2000 3000 2000 2000 8.25 600 15 2000 2000 2000 3000 15.ANSI comparison Coordination of rated values According to IEC Rated voltage Ur (kV) 3.5 16 25 40 8 12.4 11 22 36 Minimum rated voltage (kV) 3.5 5.8 38 Schneider Electric 77 .8 18 19 28 37 8.5 600 25.6 6.3 9.6 Rated short-circuit breaking current Isc (kA) 10 16 25 40 8 12.5 3.6 6.3 4.2 400 400 630 630 630 1600 1600 1600 2500 2500 3150 12 400 400 630 630 630 1600 1600 1600 1600 2500 2500 2500 3150 3150 17.5 11.85 4 2.Standards IEC . ANSI comparison Derating According to IEC c Refer to "Switchgear definition/Derating" chapter.00 0. The table of correction factors according to altitude (Altitude Corrections Factors: ACF).: "sealed system" type circuit breakers.96 N. According to ANSI c The ANSI standard C37 04 gives for altitudes greater than 1 000 metres: v a correction factor for the applicable voltage on the rated insulation level and on the rated maximum voltage. These values are not very high and should the customer request it. Altitude (ft) 3 300 5 000 10 000 (m) 1 000 1 500 3 000 ACF for: voltage 1. we must provide those for the device being considered.Isc.00 0. IEC and ANSI standards impose values well below this because they take account of oil breaking circuit breakers. According to ANSI c The electrical endurance is equal to 4 times K. According to ANSI c Mechanical endurance is of between 1 500 and 10 000 switching cycles according to the voltage and the breaking capacity.99 0. 78 Merlin Gerin MV Design Guide Schneider Electric . the manufacturer has responsibility of determining what is required in terms of materials (thicknesses.8 continous current 1. Isc S K : : : symmetrical breaking capacity at maximum voltage asymmetrical factor voltage range factor Mechanical endurance According to IEC c Mechanical endurance is of 2 000 switching cycles.B. Construction According to IEC c The IEC does not impose any particular constraints. v a correction factor for the rated operating current. however.95 0. it is not necessairy to apply the voltage ACF on the maximum rated voltage Electrical endurance Merlin Gerin circuit breakers can withstand Isc at least 15 times.S. etc) to meet performance requirements in terms of strength. According to ANSI c ANSI imposes a thickness of 3 mm for sheet metal. According to IEC c The electrical endurance is equal to 3 times Isc.Standards IEC. derating must be provided (see derating chapter). Altitude According to IEC c The altitude must not exceed 1 000 metres. Humidity According to IEC Average relative humidity value over a period 24 hours 1 month Indoor equipment 95 % 90 % According to ANSI c No specific constraints. Schneider Electric Merlin Gerin MV Design Guide 79 .Standards IEC .5°C + 40°C 35°C .: For all equipment operating under conditions other than those described above.25°C + 40°C 35°C ANSI N. According to ANSI c The altitude must not exceed 3 300 feet (1 000 metres). otherwise the equipment should be derated.ANSI comparison Normal operating conditions Equipment is designed to operate under the following normal conditions Temperature Standards IEC 0°C ambient instantaneous minimal maximal maximum average daily value minimal maximal Installation indoor outdoor .30°C + 40°C . otherwise the equipment should be derated.B. References Reference to Schneider Electric documentation c MV partner (Pierre GIVORD) c Protection of electrical networks (Christophe PREVE) c Protection of electrical networks (édition HERMES fax 01 53 10 15 21) (Christophe PREVE) c Medium voltage design (André DELACHANAL) c Cahiers techniques v n°158 calculating short-circuit currents v n°166 enclosures and protection indices (Jean PASTEAU) Schneider Electric Merlin Gerin MV Design Guide 81 . degree Fahrenheit 69 British thermal unit per hour 69 British thermal unit per pound 69 British thermal unit per second 69 British thermal unit per square foot.Index Alphabetical Index Denomination pages 67-69 57 62 58 57-62 68 53-79 67 67 67 48 68 67 29 14-16 48 A Acceleration Accuracy Accuracy class Accuracy limit factor Accuracy power Active power Altitude Angle Angular acceleration Angular speed Aperiodic component Apparent power Area Arrangement Asynchronous Automatic reclosing Discordance Distances Documentation 50 38-39 81 9 53-78 27 53-78 68-69 69 68 40 9 19 16 49-61 16 68 9 9 69 69 68 69 69 69 68-69 27 27 9-29-37-47-54-67 69 69 14-15 69 38-53-79 43 17 39 69 69 68 68 6 29 38 41 69 67-69 40 39-7 29-37-67-69 51 68 52 67 69 68 67 46 67-69 67-69 67-69 E Earthing disconnector Electrical endurance Electrodynamic withstand Endurance Energy Energy (torque) Entropy Environment Equipment Equivalent diagram Equivalent impedance F Factor Fault Arcs Field Fixed circuit breaker Fixed contactor Fluid ounce (UK) Fluid ounce (US) Flux Foot Foot of water Foot per second squared Force Forces Forces between conductors Frequency B Bending 28 Block 10 Breaking current 48-50-51-52-75 British thermal unit 69 British thermal unit per (pound.degree Fahrenheit) 69 British thermal unit per cubic foot.hour 69 Busbars 15-21-28 Busducting 27-29-37 C Cables Cable-charging Calculating a force Calculation Calorie capacity Calory power Capacitor bank Capacity Celsius Circuit breaker Closing Closing capacity Closing-opening Comparison Compartmented Concentration Condensation Conditions Conductance Conductivity Construction Coordination Cross section Cubic foot Cubic inch Cubicles Current Current transformer 15 51 27 15-17-21 69 69 51-52 68 68 45-48 52 74 47 72 10 67 38 52 68 68 78 53-77 21 69 69 10 8-67-68 54-55 69 69 67 64-65-78 38 38-39-40 59-60 60 9 9 Merlin Gerin MV design guide G Gallon (UK) Gallon (US) Generators H Heat capacity Humidity I IK code Impedance method Impulse testing Inch Inch of water Inductance Induction Insulation level Intrinsic resonance frequency Ionization threshold IP code K Knot L Length Level of pollution Lightning impulse Linear mass Line-charging Load Low inductive currents Luminous D Degree Fahrenheit Degree Rankine Density Derating Dielectric strength Dielectric withstand Differential Differential transformer Disconnector Disconnector switch Schneider Electric M… Magnetic field Magnetisation Magnitudes Making current Mass Mass per surface area Mass per volume 83 . Index Alphabetical Index Denomination pages 67 21 53-78 28 9 10 67 69 39 29-37 28-29-37 68 29-69 16 68 51-52 15 69 8 55 6-21 69 56 15 72 6 50 9-46-74 9 67 48 67 46-73 28 68 63 63-67 39 39-63 67 38-40 68 69 69 69 69 69 69 69 69 69 69 69 69 69 69 69 14-68 67 38-68-69 69 55 61 41-43 68 Merlin Gerin MV design guide R Radiation factor Rated current Rated frequency Rated insulation level Rated short circuit Rated values Rated voltage Ratio error Reactive power Resistance Resistivity Resonance Resultant strain 68 8-21-24-46-73 75 46-72 46-74 77 6-7-21-45-47-54-72-74 63 68 68 68 29-37 28 57 38-39 11-21 26 9-19 67 67 69 69 71 14 68 27-29 67-69 9 47-75 67 16 38-52-69-79 22-23 56-68 69 21 69 63 56 24 67-68 67-68 17 67 63 13-14-15 49 67 67 29-37 68-69 6-49-62-68 61 67-69 67 68 67 68 9 9 68 69 …M Materials Mechanical effects Mechanical endurance Mechanical withstand of busbars Metal enclosure Metal-clad Metre Mile (UK) Minimum distances Modulous of elasticity Modulous of inertia Moment of a force Moment of inertia Motors Movement Multi-stage S Safety factor Shape of parts Short circuit power Short time withstand current Short-circuit current Solid angle Speed Square foot Square inch Standards States Strain Supports Surface area Switch Switching sequence Symbols Synchronous compensators N Network O Oersted Operating current Operating current Operating voltage Ounce Over-current factor Overhead lines Overview Overvoltages P Peak Peak value Peak value of admissible current Period Periodic component Periodic phenomena Permissible short time withstand current Permissible strain Permittivity Phase error Phase shift Phase to earth Phase to phase Plane angle Pollution Potential Pound Pound force per square foot Pound force per square inch Pound force-foot Pound force-inch Pound force-second per square foot Pound per cubic foot Pound per cubic inch Pound per foot Pound per foot-second Pound per inch Pound per square foot Pound per square inch Pound square foot Pound-force Power Power level Pressure Pressure-strain Primary current Primary voltage Protection index T Temperature Temperature rise Thermal Thermal conductivity Thermal effects Thermal flux Thermal power Thermal short circuit current Thermal withstand Thermodynamic Thermodynamic temperature Three phase calculation example Time Transformation ratio Transformers Transient U Units Units of measurement V Vibration Viscosity Voltage Voltage transformer Volume Volume per mass Volume per mass W Wave lengths Weight Withdrawable circuit breaker Withdrawable contactor Work Y Yard Q Quantity 84 Schneider Electric . com The technical data given in this guide are given for information purposes only. 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