Electrical Installation Guide 2009 - the global specialist in energy ...

B - Connection to the MV publicdistribution network1 Supply of power at mediumvoltageThese currents are identified by subscripts 3, 2, 2E, 1, depending on the type ofshort-circuit, respectively three-phase, two-phase clear of earth, two-phase-to-earth,phase-to-earth.The method, based on the Thevenin superposition theorem and decomposition intosymmetrical components, consists in applying to the short-circuit point an equivalentsource of voltage in view of determining the current. The calculation takes place inthree steps.b Define the equivalent source of voltage applied to the fault point. It represents thevoltage existing just before the fault and is the rated voltage multiplied by a factortaking into account source variations, transformer on-load tap changers and thesubtransient behavior of the machines.b Calculate the impedances, as seen from the fault point, of each branch arriving atthis point. For positive and negative-sequence systems, the calculation does not takeinto account line capacitances and the admittances of parallel, non-rotating loads.b Once the voltage and impedance values are defined, calculate the characteristicminimum and maximum values of the short-circuit currents.The various current values at the fault point are calculated using:b The equations providedb A summing law for the currents flowing in the branches connected to the node:v I’’ k (see Fig. B6 for I’’ k calculation, where voltage factor c is defined by thestandard; geometric or algebraic summing)v I p = κ x 2 x I’’ k , where κ is less than 2, depending on the R/X ratio of the positivesequence impedance for the given branch; peak summingv I b = μ x q x I’’ k , where μ and q are less than 1, depending on the generators andmotors, and the minimum current interruption delay; algebraic summingv I k = I’’ k , when the fault is far from the generatorv I k = λ x I r , for a generator, where Ir is the rated generator current and λ is a factordepending on its saturation inductance; algebraic summing.BType of short-circuit3-phase2-phase2-phase-to-earthPhase-to-earth + 0 + 1 0I’’ kGeneral situationc Un3 Z 1c UnZ1 + Z2c Un 3 Z 2Z1 Z2 + Z2 Z0 + Z1 Z0c Un 3Z1 +Z 2 +Z 0Distant faultsc Un3 Z 1c Un2Z 1c Un 3Z 1 + 2Z 0c Un 32 Z1 + Z0Fig. B6 : Short-circuit currents as per IEC 60909CharacterizationThere are 2 types of system equipment, based on whether or not they react when afault occurs.Passive equipmentThis category comprises all equipment which, due to its function, must havethe capacity to transport both normal current and short-circuit current.This equipment includes cables, lines, busbars, disconnecting switches, switches,transformers, series reactances and capacitors, instrument transformers.For this equipment, the capacity to withstand a short-circuit without damageis defined in terms of:b Electrodynamic withstand (“peak withstand current”; value of the peak currentexpressed in kA), characterizing mechanical resistance to electrodynamic stressb Thermal withstand (“short time withstand current”; rms value expressed in kAfor duration between 0,5 and 3 seconds, with a preferred value of 1 second),characterizing maximum permissible heat dissipation.© Schneider Electric - all rights reservedSchneider Electric - Electrical installation guide 2009