Calculation of minimum levels of short-circuit current
Contents
If a protective device in a circuit is intended only to protect against short-circuit faults, it is essential that it will operate with certainty at the lowest possible level of short-circuit current that can occur on the circuit
In general, on LV circuits, a single protective device protects against all levels of current, from the overload threshold through the maximum rated short-circuit current breaking capability of the device. The protection device should be able to operate in a maximum time to ensure people and circuit safety, for all short-circuit current or fault current that may occur. To check that behavior, calculation of minimal short-circuit current or fault current is mandatory.
In addition, in certain cases overload protective devices and separate short-circuit protective devices are used.
Examples of such arrangements
Figure G43 to Figure G45 show some common arrangements where overload and short-circuit protections are achieved by separate devices.
As shown in Figure G43 and Figure G44, the most common circuits using separate devices control and protect motors.
Figure G45 constitutes a derogation to the basic protection rules, and is generally used on circuits of busways (busbar trunking systems), lighting rails, etc.
Variable speed drive
Figure G46 shows the functions provided by the variable speed drive, and if necessary some additional functions provided by devices such as circuit-breaker, thermal relay, RCD.
Protection to be provided | Protection generally provided by the variable speed drive | Additional protection if not provided by the variable speed drive |
---|---|---|
Cable overload | Yes | CB / Thermal relay |
Motor overload | Yes | CB / Thermal relay |
Downstream short-circuit | Yes | |
Variable speed drive overload | Yes | |
Overvoltage | Yes | |
Undervoltage | Yes | |
Loss of phase | Yes | |
Upstream short-circuit | Circuit-breaker
(short-circuit tripping) | |
Internal fault | Circuit-breaker
(short-circuit and overload tripping) | |
Downstream earth fault (indirect contact) | (self protection) | RCD ≥ 300 mA or CB in TN earthing system |
Direct contact fault | RCD ≤ 30 mA |
Conditions to be fulfilled
The protective device must fulfill:
- instantaneous trip setting Im < Isc_{min} for a circuit-breaker
- fusion current Ia < Isc_{min} for a fuse
The protective device must therefore satisfy the two following conditions:
- Its breaking capacity must be greater than Isc, the 3-phase short-circuit current at its point of installation
- Elimination of the minimum short-circuit current possible in the circuit, in a time tc compatible with the thermal constraints of the circuit conductors:
- (valid for tc < 5 seconds)
where S is the cross section area of the cable, k is a factor depending of the cable conductor material, the insulation material and initial temperature.
Example: for copper XLPE, initial temperature 90 °C, k = 143 (see IEC60364-4-43 §434.3.2 table 43A and Figure G52).
Comparison of the tripping or fusing performance curve of protective devices, with the limit curves of thermal constraint for a conductor shows that this condition is satisfied if:
- Isc (min) > Im (instantaneous or short time delay circuit-breaker trip setting current level), (see Fig. G47 )
- Isc (min) > Ia for protection by fuses. The value of the current Ia corresponds to the crossing point of the fuse curve and the cable thermal withstand curve (see Fig. G48 and Fig. G49)
Practical method of calculating Lmax
In practice this means that the length of circuit downstream of the protective device must not exceed a calculated maximum length:
The limiting effect of the impedance of long circuit conductors on the value of short-circuit currents must be checked and the length of a circuit must be restricted accordingly.
For protection of people (fault protection or indirect contacts), the methods to calculate the maximum circuit length are presented in chapter F, for TN system and IT system (second fault).
Two other cases are considered below, for phase-to-phase and phase-to-neutral short-circuits.
1 - Calculation of L_{max} for a 3-phase 3-wire circuit
The minimum short-circuit current will occur when two phase wires are short-circuited at the remote end of the circuit (see Fig. G50).
Using the “conventional method”, the voltage at the point of protection P is assumed to be 80% of the nominal voltage during a short-circuit fault, so that 0.8 U = Isc Zd, where:
Zd = impedance of the fault loop
Isc = short-circuit current (ph/ph)
U = phase-to-phase nominal voltage
For cables ≤ 120 mm^{2}, reactance may be neglected, so that ^{[1]}
where:
ρ = resistivity of conductor material at the average temperature during a short-circuit,
Sph = c.s.a. of a phase conductor in mm^{2}
L = length in metres
The condition for the cable protection is Im ≤ Isc with Im = trip current which guarantees instantaneous operation of the circuit breaker..
This leads to which gives
For conductors of similar nature, U and ρ are constants (U = 400 V for phase-to phase fault, ρ = 0.023 Ω.mm²/m^{[2]} for copper conductors), so the upper formula can be written as:
with Lmax = maximum circuit length in metres
For industrial circuit breakers (IEC 60947-2), the value of Im is given with ±20% tolerance, so Lmax should be calculated for Im+20% (worst case).
k factor values are provided in the following table, for copper cables, taking into account these 20%, and as a function of cross-section for Sph > 120 mm²^{[1]}
Cross-section (mm^{2}) | ≤ 120 | 150 | 185 | 240 | 300 |
---|---|---|---|---|---|
k (for 400 V) | 5800 | 5040 | 4830 | 4640 | 4460 |
2 - Calculation of L_{max} for a 3-phase 4-wire 230/400 V circuit
The minimum Isc will occur when the short-circuit is between a phase conductor and the neutral at the end of the circuit.
A calculation similar to that of example 1 above is required, but for a single-phase fault (230V).
- If Sn (neutral cross-section) = Sph
Lmax = k Sph / Im with k calculated for 230V, as shown in the table below
Cross-section (mm^{2}) | ≤ 120 | 150 | 185 | 240 | 300 |
---|---|---|---|---|---|
k (for 230 V) | 3333 | 2898 | 2777 | 2668 | 2565 |
- If Sn (neutral cross-section) < Sph, then (for cable cross-section ≤ 120mm^{2})
Tabulated values for Lmax
Based on the practical calculation method detailed in previous paragraph, pre-calculated tables can be prepared.
In practice, the tables Fig. F25 to Fig. F28 already used in chapter Protection against electric shocks and electrical fires for earth-fault calculation can also be used here, but applying the correction factors in Fig. G51 below, to obtain Lmax value related to phase-to-phase or phase-to-neutral short-circuits.
Note: for aluminium conductors, the lengths obtained must be multiplied again by 0.62.
Circuit detail | ||
---|---|---|
3-phase 3-wire 400 V circuit or 1-phase 2-wire 400 V circuit (no neutral) | 1.73 | |
1-phase 2-wire (phase and neutral) 230 V circuit | 1 | |
3-phase 4-wire 230/400 V circuit or 2-phase 3-wire 230/400 V circuit (i.e with neutral) | Sph / S neutral = 1 | 1 |
Sph / S neutral = 2 | 0.67 |
Examples
Example 1
In a 3-phase 3-wire 400 V installation the short-circuit protection of a 22 kW (50 A) motor is provided by a magnetic circuit breaker type GV4L, the instantaneous short-circuit current trip is set at 700 A (accuracy of ±20 %), i.e. in the worst case would require 700 x 1.2 = 840 A to trip.
The cable c.s.a. = 10 mm² and the conductor material is copper.
In Fig. F25 the column Im = 700 A crosses the row c.s.a. = 10 mm² at the value for Lmax of 48 m. Fig. G51 gives a factor 1.73 to apply to this value for 3-phase 3-wire circuit (no neutral). The circuit breaker protects the cable against short-circuit faults, therefore, provided that its length does not exceed 48 x 1.73 = 83 metres.
Example 2
In a 3L+N 400 V circuit, the protection is provided by a 220 A circuit breaker type NSX250N with micrologic 2 trip unit having instantaneous short-circuit protection set at 3000A (±20 %), i.e. a worst case of 3600 A to be certain of tripping.
The cable c.s.a. = 120 mm² and the conductor material is aluminium.
In Fig. F25 the column Im = 3200 A (first value > 3000A, as the table already includes the +20% on Im in its calculation) crosses the row c.s.a. = 120 mm² at the value for Lmax of 125 m. Being a 3-phase 4-wire 400 V circuit (with neutral), the correction factor from Fig. G51 to apply is 1. In complement, as the conductor is aluminium, a factor of 0.62 has to be applied.
The circuit breaker will therefore protect the cable against short-circuit current, provided that its length does not exceed 125 x 0.62 = 77 metres.
Notes
- ^ ^{1} ^{2} For c.s.a. > 120 mm^{2}, the resistance calculated for the conductors must be increased to account for the non-uniform current density in the conductor (due to "skin" and "proximity" effects). Suitable values are as follows:
- 150 mm^{2}: R + 15 %
- 185 mm^{2}: R + 20 %
- 240 mm^{2}: R + 25 %
- 300 mm^{2}: R + 30 %
- ^ Resistivity for copper EPR/XLPE cables when passing short-circuit current, eg for the max temperature they can withstand = 90°C (cf Figure G37).