Sizing calculations for 20/3.3 kV, 12.5 MVA transformer feeder cable

20kV transformer feeder cable

Typical calculations for a 20kV transformer feeder cable are presented below. After correct cable voltage classification the following considerations apply:

1. Current carrying capacity
2. Short circuit rating
3. Voltage drop
4. Earth loop impedance

Consider a 20/3.3 kV, 12.5 MVA transformer to be fed by direct buried, 3 core XLPE, SWA, PVC, copper conductor cable.

1. Cable current carrying capacity

Transformer full load current is calculated by:

I_full-load = 12.5 × 10⁶ / 1.73 × 20 × 10³ = 361 A

Don’t forget derating factors… Manufacturers provide data sheets for cables including appropriate derating factors based upon IEC 60287 (Table 1). For a ground temperature at depth of laying of 20°C, the derating factor is 0.97.

Therefore subsequent current rating of cable to be 361 / 0.97 × 0.84 = 443 A

From manufacturers tables selected cable size is 240 mm².

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2. Short circuit rating

The maximum system fault level in this application is 8.41 kA. From standard IEC 60364-5-54 (Electrical Installations In Buildings – Earthing arrangements, protective conductors and protective bonding conductors):

I_sc = K × A / √t

Where:

• K – constant, 143 for XLPE cable
• A – cable cross-section, 240 mm² based on current carrying capacity
• t – short circuit duration, for MV cables use 1 second

I_sc = 240 × 143 / √1 = 34.4 kA

Note tables are conservative and assume a fully loaded cable. At the initiation of the fault conductor temperature is 90°C and at the end of the fault conductor temperature is 250°C.

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3. Voltage drop (Vd)

Consider a 100 m route length of cable with resistance, R = 0.0982 Ω/km and inductive reactance, X_L = 0.097 Ω/km. At full load current, I_fl = 361 A at 0.85 power factor the cable voltage drop over a 100 m cable length,

V_d = I_fl × X_L × sinφ + I_fl × R × cosφ volts

V_d = (361 × 0.097 × 0.53 + 361 × 0.0982 × 0.85) × 100 / 1000
V_d = (18.56 + 30.13) ×100 / 1000
V_d = 4.87 V
V_d = 0.042%

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4. Earth loop impedance

For building services work it is important with small cross-section wiring and low fault levels to ensure that sufficient earth fault current flows to trip the MCB or fuse protection. For distribution power networks with more sophisticated protection the check is still necessary and allows the calculation of the likely touch voltages arising from the earth fault.

This in turn can then be checked against the allowable fault duration to avoid danger. See this article for a consideration of the design criteria associated with touch and step potentials.

Criteria #1 – Consider the earthing resistance at the source substation 0.5 .

Criteria #2 – The source substation 20 kV neutral is approximately 10 km from the 100 m cable under consideration. In addition parallel copper conductor earth cable is run to supplement and improve power cable armour resistance values from equipment back to the primary substation infeed neutral.

Criteria #3 – The combined resistance of the 100 m, 240 mm², cable armour (0.028 Ω / 100 m) and in parallel 2 × 95 mm² copper supplementary earth cables (0.00965 Ω /100 m) = 7.18 × 10^-3 Ω.

Criteria #4 – Consider the earthing resistance at the cable fault to be 0.5.

Criteria #5 – The effective earth circuit is shown in Figure 2. The effective primary substation neutral-to-fault cable resistance 0.15.

Criteria #6 – The maximum earth fault current at 20 kV has to be determined. Sometimes this is limited by a neutral earthing resistor and the maximum limited current may be taken for calculation. Maximum earth fault current for this calculation is 1000 A.

For a fault to earth at the end of the 100 m cable, 10 km from the primary power infeed the fault current,

I_f = (1000 × 0.15) / (1 + 0.15) = 131 A

Therefore touch voltage to earth at the cable fault 131 × 0.5 = 65.3 V

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APPENDIX – Derating factors based on IEC 60287

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Reference // Transmission and Distribution Electrical Engineering by Dr C. R. Bayliss CEng FIET and B. J. Hardy ACGI CEng FIET