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Arc flash: IEEE 1584

IEEE 1584-2018 arcing current and reduced arcing current (for protection timing)

Iarc,V=10k1+k2lgIbf+k3lgG(k4Ibf6+k5Ibf5+k6Ibf4+k7Ibf3+k8Ibf2+k9Ibf+k10)Iarc=f(Iarc,600,Iarc,2700,Iarc,14300;Voc)(≤ 600 V: equation 25)VarCf=k1Voc6+k2Voc5+k3Voc4+k4Voc3+k5Voc2+k6Voc+k7Imin=Iarc(1−0.5VarCf)
where
Iarc,V
intermediate arcing current at 600, 2700 or 14 300 V (kA)
k1…k10
Table 1 coefficients (Table 2 for VarCf, k1…k7)
Ibf
bolted fault current (kA)
G
conductor gap (mm)
Iarc
arcing current at V_oc (kA)
Voc
system voltage (kV)
VarCf
arcing current variation factor
Imin
reduced arcing current (kA)
Explain this equation

Gives the arcing current and reduced arcing current from IEEE 1584-2018, the currents used to read clearing times from fuse and relay curves.

An arc has its own voltage drop, acting like a resistance in series with the fault. At HV it is small compared with the system voltage, so the arcing current is close to Ibf; at 400 V it is a large fraction, so the arcing current can be half to three-quarters of Ibf. A longer gap G means a higher arc voltage.

The 10^(…) term is a log-log fit of arcing current against bolted current and gap; the polynomial in Ibf trims the fit at high currents. VarCf is a published polynomial in Voc describing the spread of measured arcing currents; Imin = Iarc × (1 − 0.5 VarCf) is the lower bound.

Sense check: at 11 kV, 12.5 kA, VCB, 152 mm, Iarc is about 11.4 kA (91 %); Annex D.2 at 480 V and 45 kA gives 28.8 kA (64 %).

Watch out: a few per cent less current can move a time-current curve from instantaneous to inverse-time.

Works out the arcing current and the reduced (minimum) arcing current from IEEE 1584-2018, so you can read both clearing times from the protection curve.

code

Enter your values to see the result.

Good to know

At HV the arcing current is close to the bolted current (typically 85–95 %); at 400–480 V it is much lower (often 50–75 %) because the arc voltage is a large fraction of the supply voltage. VarCf is the published variation of the test arcing current; the reduced value Iarc × (1 − 0.5 VarCf) covers the lower end. Where several sources feed the fault, each protective device sees only its share: scale the arcing current by (device bolted current ÷ total bolted current) before reading its curve. Clearing time must be read from the upstream protection characteristic at the arcing current that actually flows through that device (scale the arcing current by the ratio of device current to total fault current where several sources feed). Run the study twice: full arcing current with its clearing time, and reduced arcing current with its (often much longer) clearing time; take the higher energy. IEEE 1584 Annex B suggests 2 s may be used as a reasonable maximum where clearing is slower and the person can move away; do not rely on it where escape is restricted (confined space, ladder, inside a cubicle, elevated platform).

Common mistake

Reading the trip time at the bolted fault current. An instantaneous element set just below the bolted current may not operate at the arcing current, which can turn a 60 ms clearance into a time-delayed one of 0.5 s or more.

Limits

IEEE 1584-2018 model range: three-phase a.c., 50 or 60 Hz; Voc 208 V–15 kV; Ibf 500 A–106 kA at 208–600 V and 200 A–65 kA at 601 V–15 kV; gap 6.35–76.2 mm (≤600 V) and 19.05–254 mm (>600 V); working distance at least 305 mm; enclosure height or width up to 1244.6 mm and opening area up to 1.549 m², width greater than four times the gap; no limit on arcing time. Single-phase, d.c. and systems above 15 kV are outside the model. Coefficients from IEEE 1584-2018 Tables 1 and 2; reproduces Annex D.1 and D.2.

Sources

For competent high-voltage engineers. Results are calculations from the published methods shown and depend entirely on the data you enter. They are not a substitute for a system study, manufacturer data, the applicable standards and Engineering Recommendations, or your organisation’s safety rules. Never use a result to justify working on or near live HV equipment.

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