Protection of capacitor banks by fuses during energization and operation

Stress during bank energization & operation

Stress specific to the protection of capacitor banks by fuses, which is addressed in IEC 60549, can be divided into two types: Stress during bank energization (the inrush current, which is very high, can cause the fuses to age or blow) and Stress during operation (the presence of harmonics may lead to excessive temperature rises).

The types of stress also vary depending on the type of configuration: single bank or multiple step bank.

Contents:

1. Temperature rise
2. Inrush current peak
   1. Single capacitor bank
      • Practical example for a single capacitor bank
   2. Multiple step capacitor bank
      • Practical example for a multiple step capacitor bank
3. Appendix 1: Cold resistance of Fusarc CF fuses
4. Appendix 2: Time/current characteristics in the Fusarc CF range of fuses

1. Temperature rise

If capacitors are used, because of the harmonics, which cause additional temperature rise, a common rule for all equipment is to derate the rated current by a factor of 30 to 40 %.

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2. Inrush current peak

2.1 Single capacitor bank

This type of circuit can be illustrated as shown in the diagram in Figure 1, in which:

• L = generator inductance
• R1 = fuse resistance (see appendix 1)
• R2 = resistance of the upstream circuit, calculated based on UN, Isc and cos φ.

When starter D closes, the transient current I_T of the load C is applied, where R = R1 + R2, according to the equation:

V is the voltage in A on closing. The orders of magnitude of L, R and C mean that the following terms can be ignored: R² / 4L² on the basis of which:

This transient current is superimposed onto the 50 Hz wave, which results in the current wave form illustrated in figure 2.

The current is at maximum value when V reaches the voltage peak, i.e.:

As with transformers, the fuse selection can be validated by checking the relationships between the peak current and the rated current IT / IN as well as the rise time constant. Therefore, if:

the I_T / I_N ratio can be expressed as:

and

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Practical example for a single capacitor bank

• U_N = 10 kV
• I_N bank = 35 A
• Isc = 40 kA (cos φ = 0.1)
• C = 19.3 × 10^-6 F

By means of the calculation:

• L = 0.46 × 10^-3 H
• R₂ = 14.5 × 10^-3 Ω.

In appendix 1:

• R1 = 13 × 10^-3 Ω for a fuse rated at 63A / 12kV
• or R = 27.5 × 10^-3 Ω

on the basis of which I_T = 1670 A and τ = 33.5 × 10^-3 s.

When recorded on the curves for point (B), this point confirms that a rating of 125 A is indeed correct.

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2.2 Multiple step capacitor bank

When the bank in position n is switched on, supposing that the (n-1) other banks have already been switched on, the oscillatory load will be identical. However, in this case, the other banks connected in parallel will act as additional sources of very low internal impedance.

This internal impedance (inductance Li in figure 3) comprises stray inductances from busbars and cables (order of magnitude 0.4 μH/meter) and may also comprise an interference suppressor choke designed to limit the inrush current.

In effect, this current not only poses a danger to the fuse elements, it also compromises the electrical endurance of the switchgear and may even compromise the service life of the capacitors themselves.

As the inrush current flowing via L may be negligible, the inrush current is calculated by applying the following formula:

Where

Although appearing to be similar to the calculation for the single bank, this calculation is in fact slightly more complicated. You must:

1. Select a fuse link according to the thermal criteria:
2. Calculate:
   
3. Calculate:
   
4. Record the point (I_T, τ_T ) on the time/current characteristics,
5. If necessary, select a different rating, calculate a new R_E value and start the verification process again.

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Practical example for a multiple step capacitor bank

• 3 banks:
• U_N = 10 kV
• I_N bank = 35 A
• Isc = 40 kA (cos φ = 0.1)
• Cables = 5 rn or Li = 2 μH

When the third step closes:

• Ci = 19.3 × 10^-6 F and L_s = 1 μH
• Ri = 5 × 10^-3 Ω and R_s = 2.5 x 10^-3 Ω

On the basis of which:

• C_E = 12.9 μF
• L_E = 3 × 10^-6 H
• R_E = 7.5 10^-3 Ω
• I_T = 16900 A
• τ = 0.8 × 10^-3 s

The time/current characteristics do not provide specific data for pre-arc durations of less than a millisecond. A value of I²t constant can be estimated in this zone. If the minimum value of the fuse assembly is assumed to be 125 A (defined in the previous example), this results in the following values:

The stress applied to the fuse assembly rated at 125 A is:

0.8 × 10^-3 × (16900)² = 228 × 103 A² s

and even 200 A is unsuitable!

This type of bank cannot be protected in this way using for example Fusarc-CF fuses. In some impossible cases, there is a solution that consists of protecting all three banks with a single common fuse (see figure 4).

With this type of diagram, two cases should be considered:

Case #1

The three banks cannot be powered up simultaneously. In this case, each time a bank is powered up, the fuse interprets this as the activation of a single bank. A fuse with a rating of 125 A can withstand the inrush current (see the example above).

Case #2

All three banks can be powered up simultaneously. The system becomes equivalent to a single bank with triple power, i.e.:

• L = 0.46 × 10^-3 H
• C = 57.9 × 10^-6 F
• I_T = 2900A

Point C, appendix 2, shows that 200 A is suitable in the majority of cases. Paradoxically, in this case, it is possible to protect all three banks globally using one fuse despite the fact that protecting each individually is impossible.

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Appendix 1: Cold resistance of Fusarc CF fuses

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Appendix 2: Time/current characteristics in the Fusarc CF range of fuses (Schneider Electric)

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Reference // Design and use of MV current-limiting fuses by O. Bouilliez and J.C. Perez Quesada