Technical Application Papers No.2: MV/LV transformer substations: theory and examples of short-circuit calculation

Overview

The paper first covers typical substation layouts, transformer selection and parallel operation, and how the utility's limits constrain MV protection. Typical transformer short-circuit voltages per IEC 60076-5 are 4% up to 630 kVA, 5% for 630-1250 kVA, 6% for 1250-2500 kVA, 7% for 2500-6300 kVA and 8% for 6300-25000 kVA. Paralleled transformers need matching vk%, the same vector group and compatible ratios, and paralleling raises the LV fault level and busbar and breaker duty.

An example shows the effect: with a network Sk of 750 MVA at 400 V, one 1600 kVA (vk 6%) transformer gives about 37 kA at the LV busbar, while two 800 kVA (vk 4%) units in parallel give about 55 kA. The calculation uses network, cable and transformer impedances, motor contribution and peak current. Motor contribution is commonly approximated as 4-6 times motor rated current, and IEC 60909 requires it to be considered when the sum of rated motor currents exceeds Ik/100.

Peak current is ip = k x sqrt(2) x Ik with k = 1.02 + 0.98 e^(-3R/X); for Ik = 33 kA and X/R = 6.6, k = 1.64 and ip = 76.6 kA. Breaker making capacity Icm must exceed the prospective peak current at the installation point, not only Icu above Ik. Breaker selection, switch-disconnector and RCD coordination and a complete MV/LV network study follow.

Annex A gives the transformer inrush current model as an exponentially decaying multiple of rated current, used to set MV protection above it. The annexes also include worked examples using symmetrical components and the power method.

Key points

1. IEC 60076-5 typical vk: <=630 kVA 4%; 630-1250 5%; 1250-2500 6%; 2500-6300 7%; 6300-25000 8%.
2. 750 MVA network at 400 V: one 1600 kVA (vk 6%) gives about 37 kA at the LV busbar; two 800 kVA (vk 4%) in parallel about 55 kA.
3. Paralleled transformers need matching vk%, same vector group and compatible ratios; paralleling raises LV fault level.
4. Motor contribution is about 4-6 times rated motor current; IEC 60909 requires it when the sum of rated motor currents exceeds Ik/100.
5. Peak current ip = k x sqrt(2) x Ik, k = 1.02 + 0.98 e^(-3R/X); Ik = 33 kA, X/R = 6.6 gives k = 1.64, ip = 76.6 kA.
6. Making capacity Icm must exceed the prospective peak current; Icu above Ik is not enough.
7. Annex A models transformer inrush as an exponentially decaying multiple of rated current; MV protection is set above it.

Using it in design

• Default transformer impedance from the IEC 60076-5 table when the user leaves vk blank.
• Compute busbar Ik for every operating mode (single transformer, transformers paralleled via bus-tie, generator only) and size breakers to the worst.
• Include motor (chiller/pump/fan) contribution when the sum of motor In exceeds Ik/100.
• Compute peak current from X/R and check Icm >= ip as a separate rule; warn when a bus-tie allows transformer paralleling that pushes Ik above downstream Icu.

Who should read it

Electrical designers doing MV/LV substation and short-circuit calculations.

Related standards

• IEC 60909-0
• IEC 60076-5
• IEC 60947-2

Publication details

Authors

ABB

Publisher

ABB

Year

2008

Document number

1SDC007101G0202 (QT2)

Document language

English

Source type

Manufacturer

Access

Free to read

Licence

Copyright retained; citation only — the publisher’s terms apply to the original document

Citation