3-Phase Short-Circuit Current (Isc) at Any Point Within A LV Installation - Electrical Installation Guide
3-Phase Short-Circuit Current (Isc) at Any Point Within A LV Installation - Electrical Installation Guide
3-Phase Short-Circuit Current (Isc) at Any Point Within A LV Installation - Electrical Installation Guide
3phase shortcircuit current (Isc) at any point within a LV installation
From Electrical Installation Guide
Contents
1 Method of calculating ZT
2 Determination of the impedance of each component
2.1 Network upstream of the MV/LV transformer
2.2 Transformers
2.3 Busbars
2.4 Circuit conductors
2.5 Motors
2.6 Faultarc resistance
2.7 Recapitulation table
2.8 Example of shortcircuit calculations
3 References
In a 3phase installation Isc at any point is given by:
where
U20 = phasetophase voltage of the open circuited secondary windings of the power supply transformer(s).
ZT = total impedance per phase of the installation upstream of the fault location (in Ω)
Method of calculating ZT
Each component of an installation (MV network, transformer, cable, busbar, and so on...) is characterized by its impedance Z, comprising an element of resistance
(R) and an inductive reactance (X). It may be noted that capacitive reactances are not important in shortcircuit current calculations.
The parameters R, X and Z are expressed in ohms, and are related by the sides of a right angled triangle, as shown in the impedance diagram of Figure G33.
Fig. G33: Impedance diagram
The method consists in dividing the network into convenient sections, and to calculate the R and X values for each.
Where sections are connected in series in the network, all the resistive elements in the section are added arithmetically; likewise for the reactances, to give RT and
XT.
The impedance (ZT) for the combined sections concerned is then calculated from
Any two sections of the network which are connected in parallel, can, if predominantly both resistive (or both inductive) be combined to give a single equivalent
resistance (or reactance) as follows:
Let R1 and R2 be the two resistances connected in parallel, then the equivalent resistance R3 will be given by:
or for reactances
It should be noted that the calculation of X3 concerns only separated circuit without mutual inductance. If the circuits in parallel are close togother the value of X3
will be notably higher.
Determination of the impedance of each component
Network upstream of the MV/LV transformer
(see Fig. G34)
The 3phase shortcircuit fault level PSC, in kA or in MVA[1] is given by the power supply authority concerned, from which an equivalent impedance can be
deduced.
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Fig. G34: The impedance of the MV network referred to the LV side of the MV/LV transformer
A formula which makes this deduction and at the same time converts the impedance to an equivalent value at LV is given, as follows:
where
Zs = impedance of the MV voltage network, expressed in milliohms
Uo = phasetophase noload LV voltage, expressed in volts
Psc = MV 3phase shortcircuit fault level, expressed in kVA
The upstream (MV) resistance Ra is generally found to be negligible compared with the corresponding Xa, the latter then being taken as the ohmic value for Za. If
more accurate calculations are necessary, Xa may be taken to be equal to 0.995 Za and Ra equal to 0.1 Xa.
Figure G36 gives values for Ra and Xa corresponding to the most common MV[2] shortcircuit levels in utility powersupply networks, namely, 250 MVA and 500
MVA.
Transformers
(see Fig. G35)
The impedance Ztr of a transformer, viewed from the LV terminals, is given by the formula:
where:
U20 = opencircuit secondary phasetophase voltage expressed in volts
Sn = rating of the transformer (in VA)
Usc = the shortcircuit impedance voltage of the transformer expressed in %
The transformer windings resistance Rtr can be derived from the total loadlosses as follows:
so that in milliohms
where
Pcu = total loadlosses in watts
In = nominal fullload current in amps
Rtr = resistance of one phase of the transformer in milliohms (the LV and corresponding MV winding for one LV phase are included in this resistance value).
Note: for an approximate calculation, in the absence of more precise information on transformer characteristics, Cenelec 50480 suggests to use the following
guidelines:
if U20 is not known, it may be assumed to be 1.05 Un
in the absence of more precise information, the following values may be used: Rtr = 0.31 Ztr and Xtr = 0.95 Ztr
Example: for a transformer of 630kVA with Usc=4% / Un = 400V, approximate calculation gives:
U20 = 400 x 1.05 = 420V
Ztr = 420² / 630000 x 4% = 11 mΩ
Rtr = 0.31 x Ztr = 3.5 mΩ and Xtr = 0.95 x Ztr = 10.6 mΩ
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Fig. G35: Resistance, reactance and impedance values for typical distribution 400 V transformers (noload voltage = 420 V) with MV windings ≤ 20 kV
Busbars
The resistance of busbars is generally negligible, so that the impedance is practically all reactive, and amounts to approximately 0.15 mΩ/metre[3] length for LV
busbars (doubling the spacing between the bars increases the reactance by about 10% only).
In practice, it's almost never possible to estimate the busbar length concerned by a shortcircuit downstream a switchboard.
Circuit conductors
The resistance of a conductor is given by the formula:
where
ρ = the resistivity of the conductor material at the normal operating temperature
ρ has to be considered:
at cold state (20°C) to determine maximum shortcircuit current,
at steady state (normal operating temperature) to determine minimum shortcircuit current.
L = length of the conductor in m
S = c.s.a. of conductor in mm²
Fig. G35b: Values of ρ as a function of the temperature, cable insulation and cable core material, according to IEC609090 and Cenelec TR 50480 (in mΩ mm²/m).
Cable reactance values can be obtained from the manufacturers. For c.s.a. of less than 50 mm² reactance may be ignored. In the absence of other information, a
value of 0.08 mΩ/metre may be used (for 50 Hz systems) or 0.096 mΩ/metre (for 60 Hz systems). For busways (busbar trunking systems) and similar prewired
ducting systems, the manufacturer should be consulted.
Motors
At the instant of shortcircuit, a running motor will act (for a brief period) as a generator, and feed current into the fault.
In general, this faultcurrent contribution may be ignored. However, if the total power of motors running simultaneously is higher than 25% of the total power of
transformers, the influence of motors must be taken into account. Their total contribution can be estimated from the formula:
Iscm = 3.5 In from each motor i.e. 3.5m In for m similar motors operating concurrently.
The motors concerned will be the 3phase motors only; singlephasemotor contribution being insignificant.
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Faultarc resistance
Shortcircuit faults generally form an arc which has the properties of a resistance. The resistance is not stable and its average value is low, but at low voltage this
resistance is sufficient to reduce the faultcurrent to some extent. Experience has shown that a reduction of the order of 20% may be expected. This phenomenon
will effectively ease the currentbreaking duty of a CB, but affords no relief for its faultcurrent making duty.
Recapitulation table
(see Fig. G36)
Xa = 0.995 Za
Supply network
Figure G34
where with
Transformer
Figure G35
Rtr is often negligible compared to Xtr
for transformers > 100 kVA
Circuitbreaker Not considered in practice
Circuit conductors(2) Cables: Xc = 0.08 mΩ/m
See Subclause 4.2 Motors
Motors
(often negligible at LV)
Threephase maximum
circuit current in kA
U20: Phasetophase noload secondary voltage of MV/LV transformer (in volts).
Psc: 3phase shortcircuit power at MV terminals of the MV/LV transformers (in kVA).
Pcu: 3phase total losses of the MV/LV transformer (in watts).
Sn: Rating of the MV/LV transformer (in kVA).
Usc: Shortcircuit impedance voltage of the MV/LV transfomer (in %).
RT : Total resistance. XT: Total reactance
(1) ρ = resistivity at 20°C
(2) If there are several conductors in parallel per phase, then divide the resistance of one conductor by the number of conductors. The reactance remains practically
unchanged.
Fig. G36: Recapitulation table of impedances for different parts of a powersupply system
Example of shortcircuit calculations
(see Fig. G37)
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MV network
0.035 0.351
Psc = 500 MVA
Transformer 20 kV / 420
V
Pn = 1000 kVA 2.35 8.5
Usc = 5%
Pcu = 13.3 x 103 watts
Singlecore cables
5 m copper Xc = 0.08 x 5 = 0.40 2.48 9.25 Isc1 = 25 kA
4 x 240 mm²/phase
Main circuitbreaker Not considered in practice
Busbars 10 m Not considered in practice
Threecore cable
100 m Xc = 100 x 0.08 = 8 22 17.3 Isc3 = 8.7 kA
95 mm² copper
Threecore cable
20 m
Xc = 20 x 0.08 = 1.6 59 18.9 Isc4 = 3.9 kA
10 mm² copper
final circuits
RT : Total resistance. XT: Total reactance. Isc : 3phase maximum shortcircuit current
Calculations made as described in figure G36
Fig. G37: Example of maximum shortcircuit current calculations for a LV installation supplied at 400 V (nominal) from a 1000 kVA MV/LV transformer
References
1. ^ Shortcircuit MVA: EL Isc where:
EL = phasetophase nominal system voltage expressed in kV (r.m.s.)
Isc = 3phase shortcircuit current expressed in kA (r.m.s.)
2. ^ up to 36 kV
3. ^ For 50 Hz systems, but 0.18 mΩ/m length at 60 Hz
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Category: Chapter Sizing and protection of conductors
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