Fault Level Decay in Mining Distribution Networks: How Motor Contributions Distort Calculations

When a short circuit occurs on a power system, every rotating machine connected to the faulted bus contributes fault current. The physics is straightforward: at the instant of fault, each motor’s rotor continues spinning at near-synchronous speed due to its inertia. The rotating magnetic field in the motor’s airgap drives current into the short circuit, with the motor temporarily acting as a generator.

The magnitude of this contribution depends on the motor’s impedance, which changes over time as the fault persists:

For induction motors (which comprise the vast majority of mining loads), the fault contribution decays to zero within 3–10 cycles (60–200 ms at 50 Hz) because induction motors have no independent field excitation. Once the rotor flux decays, the motor stops contributing. Synchronous motors sustain fault current longer due to their DC field excitation, but also decay as the AVR reduces field current in response to the fault.

IEC 60909-0:2016, Clause 3.8 defines the motor contribution using the locked-rotor current ratio (I_LR/I_rated) and a correction factor that accounts for the motor’s subtransient reactance and the system voltage at the motor terminals.

IEC 60909-0:2016, Clause 11.5 provides the method for calculating the motor contribution to the initial symmetrical short-circuit current (I”k). The motor is modeled as a voltage source behind its subtransient reactance:

I”k_motor = (c × Un) / (√3 × Z_motor)

Where c is the voltage factor (1.0 for maximum fault current per Table 1), Un is the nominal voltage, and Z_motor is derived from the motor’s rated power and locked-rotor current ratio.

For the time-dependent decay, IEC 60909-0, Clause 11.5.3 provides the decrement function q(t) for induction motors:

q(t) = decay factor as a function of motor parameters and time

The decay is approximately exponential with a time constant determined by the motor’s subtransient and transient time constants (T”d and T’d).

Motor parameters for typical mining loads:

Large motors with low subtransient reactance (0.12–0.17 pu) produce the highest fault contributions per kW of rating. Mining plants are characterized by a concentration of such motors: SAG and ball mills, crushers, and large pumps. A single 2 MW SAG mill motor with X”d = 0.17 pu contributes approximately 5.9× its rated current of 210 A = 1,240 A (1.24 kA) at the instant of fault.

During a planned short-circuit test at a copper-gold mine’s 33 kV main distribution switchboard, fault current waveforms were captured using a high-speed digital recorder (sampling at 10 kHz, 16-bit resolution) on the bus CTs. The test involved a bolted three-phase fault applied through a making switch, with the fault cleared by the bus coupler breaker after a predetermined delay.

System parameters:

• Grid contribution (via 2×40 MVA, 132/33 kV transformers): 21.2 kA symmetrical
• Motor loads on the 33 kV bus: 6×2 MW SAG mill motors (direct online, across the line)
• Additional motor loads: 4×500 kW crusher motors, 8×250 kW pump motors
• Total connected motor load: 16 MW

Measured fault current decay envelope:

The motor contribution of 7.2 kA at fault initiation decayed with an effective time constant of approximately 35 ms, reaching negligible levels by 150–200 ms. The IEC 60909-0 calculated motor contribution was 6.8 kA — within 6% of the measured value, validating the standard’s methodology for this installation.

The total 28.4 kA initial fault current is 34% higher than the grid-only value of 21.2 kA. This additional 7.2 kA must be included in switchgear duty calculations: the circuit breakers must be rated to interrupt the initial fault current (which includes motor contribution), not just the steady-state value.

The decaying motor contribution creates a unique challenge for protection coordination. The fault current is not constant — it decreases over time as the motor contribution decays. This interacts with the relay’s time-current characteristic in ways that can cause unexpected behavior:

Scenario: Overcurrent relay set for grid-only fault current

An overcurrent relay on the 33 kV feeder was set with an instantaneous element pickup of 22 kA (105% of the 21.2 kA grid contribution). This setting was intended to provide fast tripping for bus faults while avoiding pickup on normal motor starting inrush.

When the fault occurred:

1. t = 0 ms: Fault current = 28.4 kA. Relay instantaneous element picks up (28.4 > 22).
2. t = 0–50 ms: Relay is timing. The relay requires a minimum of 30 ms sustained pickup to issue a trip signal (intentional short-time delay to ride through transients).
3. t = 50 ms: Fault current has decayed to approximately 25 kA. Still above pickup. Relay issues trip.
4. Breaker operating time: 60 ms (3 cycles at 50 Hz).
5. Total clearing time: 110 ms.

This sequence worked correctly. But consider a different relay setting: instantaneous pickup at 25 kA (a higher setting to improve selectivity with downstream relays).

1. t = 0 ms: Fault current = 28.4 kA. Relay picks up.
2. t = 30 ms: Fault current has decayed to 26.2 kA. Still above 25 kA. Relay would issue trip if timing allows.
3. t = 40 ms: Fault current = 25.3 kA. Marginally above pickup — relay may or may not confirm trip depending on CT accuracy and relay tolerance.
4. t = 60 ms: Fault current = 24.0 kA. Below 25 kA pickup. Relay instantaneous element drops out.
5. Relay falls back to the inverse-time element, which operates at a much longer time delay (e.g., 500 ms at 24 kA).
6. Total clearing time: 560 ms instead of 110 ms.

Switchgear and cables must be rated for the maximum prospective fault current, which includes motor contributions. The equipment duty calculations affected by motor contribution are:

The most critical rating is the making capacity. The peak asymmetrical fault current occurs within the first half-cycle and includes the full motor subtransient contribution plus the DC offset component. At this site, the peak making current was 72.4 kA — 34% higher than the 54.0 kA calculated without motor contribution. Switchgear rated for 63 kA making capacity would be inadequate; 80 kA rated switchgear was required.

For cable thermal withstand, the motor contribution is less significant because the I²t integral over the short motor contribution period (approximately 100 ms of elevated current) is small relative to the total fault duration (typically 200–500 ms). The cable withstand is governed by the sustained grid contribution over the full clearing time.

Not all installations are significantly affected by motor fault contribution. The impact depends on the ratio of connected motor load to grid fault level. The following rule of thumb identifies networks where motor contribution is significant:

Motor contribution ratio = Total motor MVA (at LRC) / Grid fault MVA

Typical motor contribution ratios by industry:

Mining and oil & gas installations consistently fall in the “significant to dominant” range. These are the industries where ignoring motor contribution produces the most dangerous errors in equipment rating and protection coordination.

1. Always include motor contribution in the initial symmetrical fault current calculation per IEC 60909-0:2016, Clause 11. Use the motor’s subtransient reactance and locked-rotor current ratio as input parameters. Do not default to grid-only values.
2. Calculate the motor contribution decay profile for relay coordination studies. The standard IEC 60909-0 method provides the initial and steady-state values, but protection coordination requires the time-dependent curve. Use the decrement functions from IEC 60909-0, Clause 11.5.3 or manufacturer-specific motor data.
3. Set instantaneous overcurrent elements with margin below the decayed fault current, not the initial value. The relay must remain picked up for its full operating time. If the motor contribution decays below pickup before the relay completes its timing cycle, the instantaneous trip will not occur. A 10–15% margin below the minimum expected fault current (grid contribution only) is recommended.
4. Verify switchgear making capacity against the peak asymmetrical fault current including motor contribution. The making capacity must exceed the peak current in the first half-cycle, which includes subtransient motor contribution and DC offset per IEC 62271-100, Clause 6.104.
5. For plants with more than 10 MW of direct-online motor load on a single bus, commission a detailed motor contribution study. Use actual motor parameters from manufacturer data sheets rather than the generic values in IEC 60909-0, Table 3.
6. Consider VFD-fed motors separately. Motors fed through variable frequency drives do not contribute fault current to the upstream bus because the VFD power electronics decouple the motor from the network. Only direct-online and soft-starter-fed motors contribute.

Standards referenced: IEC 60909-0:2016 (Clauses 3.8, 11.5, Table 3), IEC 62271-100:2021 (Clauses 6.102, 6.104), IEC 62271-1:2017 (Clause 6.5), AS/NZS 3008.1.1:2017 (Clause 4.5).