Premium Membership

Get access to premium HV/MV/LV technical articles, electrical guides, studies and much more! Apply 20% OFF Code: EEP09PE
Home / Technical Articles / 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:

Cable sizing for 20/3.3 kV, 12.5 MVA transformer feeder
Cable sizing for 20/3.3 kV, 12.5 MVA transformer feeder (photo credit:
  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:

Ifull-load = 12.5 × 106 / 1.73 × 20 × 10= 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.

The group derating factor based upon 3 cables laid in trench at 0.45 m centres is 0.84. Ground thermal resistivity taken as the normal of 1.2°Cm/W for a UK installation and 1.00 rating factor. Cable installation depth to be 0.8 m and 1.00 rating factor.

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

From manufacturers tables selected cable size is 240 mm2.

Go back to contents ↑

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):

Isc = K × A / √t


  • 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

Isc = 240 × 143 / √1 = 34.4 kA

From manufacturers tables and/or Figures 1 (a–c) for working voltages up to and including 19000 / 33000 XLPE based insulated cable the selected 240 mm2 cable is just capable of this 1 second short circuit rating.

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.

(a) Paper-, (b) PVC- and (c) XLPE-insulated copper conductor cable short circuit ratings
Figure 1 – (a) Paper-, (b) PVC- and (c) XLPE-insulated copper conductor cable short circuit ratings

Go back to contents ↑

3. Voltage drop (Vd)

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

Vd = Ifl × XL × sinφ + Ifl × R × cosφ volts

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

Important Notes //

  1. At 20 kV the voltage drop is negligible over such a short length of cable.
  2. IEE Wiring Regulations require a voltage drop for any particular cable run to be such that the total voltage drop in the circuit of which the cable forms part does not exceed 2½% of the nominal supply voltage, i.e. 10.4 volts for a three phase 415 V supply and 6 volts for a single phase 240 V supply.
  3. Industrial plant users may use different specifications and apply 5% (or even 10%) under no load to full load conditions and perhaps 20% at motor terminals under motor starting conditions.
  4. Manufacturers data for building services installations is often expressed in terms of voltage drop (volts) for a current of 1 ampere for a 1 meter run of a particular cable size.

Go back to contents ↑

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.

For this example assume power and supplementary earth copper cables and armour over the 10 km distance have a combined effective resistance of 0.143.

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.

Calculation example – earth loop impedance
Figure 2 – Calculation example – earth loop impedance

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,

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

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

Go back to contents ↑

APPENDIX – Derating factors based on IEC 60287

Derating factors based on IEC 60287
Derating factors based on IEC 60287

Go back to contents ↑

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

Premium Membership

Premium membership gives you an access to specialized technical articles and extra premium content (electrical guides and software).
Get Premium Now ⚡

Edvard Csanyi

Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry facilities. Professional in AutoCAD programming.


    Aug 15, 2018

    I have 2MVA 11/0.433KV GMT.I want to connect XLPE/SWA/PVC copper cable. How many cables can i use per phase. What calculation is used

  2. Shoaib
    Jul 27, 2018

    Dear sir,

    Please tell me the between KW and KVA for example 1 machine has 30kva load capacity how we have to select the cable.kindly explain

    Thanks & regards,
    Shoaib Noor.

  3. javed
    Jun 24, 2018

    but how to calculate cable size for low voltage side

  4. vijay manikanta
    Dec 15, 2017

    Can we take parallel runs into account for calculation in the calculation of short circuit current capacity?

    In my case a feeder has to be selected for feeding 33KV, 31.5 ka, 1 sec rated panel, for a load of 20MW.

    I have selected 2R, 33KV, 300SQMM, AL Conductor, XLPE Insulated cable to be installed in RCC trenches, over 2000m length.
    As per manufacturer catalog 300sqmm Al, XLPE cable can cater 28 ka, as two runs are considered, I expect it can take care 31.5ka for 1sec.

    It is requested to advice on the above.

  5. Aye Kyaw Kyaw
    May 08, 2017

    For current 443 A, why you selected 240 sqmm? Can I see the manufacturer’s table that you referred? According to my knowledge, 150 sqmm of MV Cu/XLPE cable can carry about 500 A at free air from many cable makers.

Leave a Comment

Tell us what you're thinking... we care about your opinion!

Subscribe to Weekly Digest

Get email alert whenever we publish new electrical guides and articles.