LibraryReferenceShort-circuit currents

Peak current and thermal equivalent current

A short-circuit current starts with a direct-current (DC) component that decays with the network's time constant. The first peak of the current, $i_{p}$ , sets the current a circuit breaker must make. The heat of the current over the fault's duration, expressed as the thermal equivalent current $I_{th}$ , sets the short-time rating. IEC 60909-0:2016 calculates both from the initial symmetrical current $I_{k}^{''}$ (Short-circuit currents by IEC 60909-0) and the R/X ratio at the fault location.

Read first: Short-circuit currents by IEC 60909-0

1The peak short-circuit current

The peak factor κ expresses the first peak as a multiple of the symmetrical peak $2 I_{k}^{''}$ :

i_{p} = κ 2 I_{k}^{''}

(1)

IEC 60909-0:2016 §8.1.1, Eq. 54

κ = 1.02 + 0.98 e^{- 3 R / X}

(2)

IEC 60909-0:2016 §8.1.1, Eq. 55

κ approaches 2 for a network of reactance alone ( $R / X \to 0$ ) and falls as the resistance rises (Figure 1).

Method B in a meshed network

The design treats the network at a substation bus as meshed and takes κ by method B. Method B takes the R/X ratio of the impedance at the fault location, calculates $κ_{b}$ by Eq. (2), and multiplies it by 1.15:

κ = 1.15 κ_{b} \leq 2.0

(3)

IEC 60909-0:2016 §8.1.2, Eq. 59

The limit is 2.0 in high-voltage networks and 1.8 in low-voltage networks. The design applies the high-voltage limit at the high-voltage (HV) and medium-voltage (MV) buses. It takes $I_{k}^{''}$ as the largest of the fault types' currents at the bus. At the default project's buses, the R/X ratios are 0.0778 at the HV bus and 0.0330 on a MV section.

Ratio R/X0.250 · κ by method B, Eq. (3): 1.705 · κb, Eq. (2): 1.483

Figure 1.The peak factor κ falls as the ratio R/X rises. Method B multiplies it by 1.15 and holds it at its limit of 2.0 for low ratios. The design service calculated the values by Eq. (2) and Eq. (3).

where
$i_{p}$	peak short-circuit current	kA
$κ$	peak factor
$κ_{b}$	peak factor by Eq. (2) from the R/X ratio at the fault location
$R / X$	ratio of the resistance to the reactance of the positive-sequence impedance at the fault location
$I_{k}^{''}$	initial symmetrical short-circuit current	kA

2The thermal equivalent current

The thermal equivalent current is the current of constant value that heats a conductor as much as the short-circuit current does over the duration $T_{k}$ . The factor $m$ adds the heat of the DC component to that of the alternating-current (AC) component, expressed by $n$ :

I_{th} = I_{k}^{''} m + n

(4)

IEC 60909-0:2016 §10, Eq. 100

The design takes $n$ = 1, for a fault far from a generator. $m$ follows from κ by Eq. (2), the frequency and the duration. $m$ is largest for a short fault and falls as the fault lasts longer (Figure 2). The design takes the duration as 1 s at the HV bus (3 s at an HV level of 33 kV) and 3 s at the MV bus.

Duration1.60 s · κ 1.908, MV bus: 1.032 · κ 1.796, HV bus: 1.014

Figure 2.The ratio

I_{th} / I_{k}^{''}

falls towards 1 as the fault lasts longer, and is higher for a higher κ, at 50 Hz. The curves take the κ of the default project's buses by Eq. (2). The design service calculated the values by Eq. (4).

where
$I_{th}$	thermal equivalent short-circuit current	kA
$m$	factor for the heat effect of the DC component
$n$	factor for the heat effect of the AC component, 1 far from a generator
$T_{k}$	duration of the short-circuit current	s

3The switchgear rating checks

The design checks the switchgear at the HV bus and at the MV bus against the maximum case. For a fault far from a generator, the symmetrical breaking current equals the initial symmetrical current:

I_{b} = I_{k}^{''}

(5)

IEC 60909-0:2016 §9.1

The breaking current is checked against the rated short-circuit breaking current of the breaker (IEC 62271-100:2008 §4.101). The peak current is checked against the rated short-circuit making current, 2.5 times the breaking current at 50 Hz and 2.6 times at 60 Hz (IEC 62271-100:2008 §4.103). The thermal equivalent current is checked against the rated short-time current for its duration. The rated current at the HV bus is the level's design fault level, and at the MV bus the rating of the MV switchgear. The design also checks the earth mat's design current against the HV earth-fault current.

Example 1Peak and thermal currents of the default 220 kV project

Given the maximum case of the default 220 kV / 33 kV project at 50 Hz (Short-circuit currents by IEC 60909-0, Example 1).

Quantity	HV bus	MV section
Ratio R/X at the bus	0.0778	0.0330
Largest current $I_{k}^{''}$ (three-phase at HV, three-phase at MV)	21.98 kA	24.25 kA
Duration $T_{k}$	1 s	3 s
Rated short-circuit current	50 kA	31.5 kA

Peak factor by Eq. (2) and Eq. (3). At the HV bus, $κ_{b}$ = 1.796, so the factor of method B is 2.065, limited to 2.000. On a MV section, $κ_{b}$ = 1.908 and the factor is 2.194, limited to 2.000:

κ_{b} = 1.02 + 0.98 e^{- 3 \times 0.0778} = 1.796

1.15 \times 1.796 = 2.065

Peak current by Eq. (1):

i_{p, HV} = 2.000 \times 2 \times 21.98 = 62.17 kA

i_{p, MV} = 2.000 \times 2 \times 24.25 = 68.58 kA

Thermal equivalent current by Eq. (4), with $m$ = 0.044 at the HV bus and 0.034 on a MV section:

I_{th, HV} = 21.98 \times 1.044 = 22.46 kA

I_{th, MV} = 24.25 \times 1.034 = 24.66 kA

Rating checks by Eq. (5), with the making factor 2.5 at 50 Hz:

Check	HV calculated	HV rating	HV result	MV calculated	MV rating	MV result
Breaking current $I_{b}$	21.98 kA	50 kA	Pass	24.25 kA	31.5 kA	Pass
Making current $i_{p}$	62.17 kA	125 kA	Pass	68.58 kA	78.75 kA	Pass
Short-time current $I_{th}$	22.46 kA	50 kA	Pass	24.66 kA	31.5 kA	Pass

Result The ratings of the default project cover the calculated currents. The making current is the check nearest its rating at both buses.

Sources

[1]IEC 60909-0:2016, Short-circuit currents in three-phase AC systems, Part 0: Calculation of currents. §8.1.1, Eq. 54; §8.1.1, Eq. 55; §8.1.2, Eq. 59; §10, Eq. 100; §9.1.

[2]IEC 62271-100:2008, High-voltage switchgear and controlgear.

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