Electrical-Engineering-portal.com-Dry Transformer Percent Impedance Definition

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  • 7/28/2019 Electrical-Engineering-portal.com-Dry Transformer Percent Impedance Definition

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    Dry Transf ormer Percent Impedance Def inition (on photo dry type transf ormer by Engineeringcompany B&S, Ukraine)

    http://electrical- engineering- portal.com/dry- transformer- percent- impedance- definition April 12, 2013

    Dry Transformer Percent Impedance Definition

    Edvard

    Introduction

    The percent impedance is the percent voltage required to circulate rated current f low through

    http://electrical-engineering-portal.com/dry-transformer-percent-impedance-definition
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    one transformer winding when another winding is short-circuited at the rated voltage tap atrated f requency.

    %Z is related to the short circuit capacity of the transformerduring short circuit conditions.

    For a two winding transformer with a 5% impedance, it would require 5% input voltage appliedon t he high voltage winding to draw 100% rated current on t he secondary winding when thesecondary winding is short-circuited.

    If 100% rated voltage is applied to the high voltage winding, approximately 20Xratedcurrent would flow in the secondary winding when the secondary winding is short -circuited.

    Impedance Levels

    Based kVA Minimum Impedance, %

    0 150 Manufacturers standard

    151 300 4

    301 600 5

    601 2,500 6

    2,501 5,000 6.5

    5,001 7,500 7.5

    7,501 10,000 8.5

    Above 10,000 9.5

    Important Notes

    1. The impedance of a two-winding transformer shall not vary from the guaranteed value bymore that 7.5%

    2. The impedance of a transformer having three or more windings or having zig-zag windings shall not vary from the guaranteed value by more than 10%

    3. The impedance of an auto-transformer shall not vary from the guaranteed value bymore than 10%

    4. The difference of impedances between transformers of the same design shall notexceed 10% of the guaranteed values

    5. Dif ferences of impedance between auto-transformers of t he same design shall notexceed 10% of the guaranteed values

    Impedance vs. Percent Impedance

    Impedance is def ined, in the Standard Handbook for Electrical Engineers, as the apparentresistance of an alternating current circuit or path the vector sum of the resistance andreactance of the path. Impedance may be comprised of resistance, capacitive reactance and

    http://electrical-engineering-portal.com/transformer-heat-copper-and-iron-losseshttp://electrical-engineering-portal.com/power-transformer-construction-windings
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    inductive reactance, and is expressed in ohms.

    From the perspective of a load, the tot al input impedance may include the impedance of theupstream generator, transformer, line reactor and conductors.

    The power system impedance is useful fo r estimating the available short circuit current.

    Sample calculations for a three phase transformer rated 500kVA, 4160:480, 60Hz, 6%impedance:

    Transformer reactance Xt = (kV2/MVA) x %Z/100 = (0.482 / 0.5) x 0.06 = 0.027648 ohms

    Approximate available short circuit current= 480/(1.732 x 0.027648) = 10,023.7 amps

    Effective Percent Impedance

    Effective impedance is the relative impedance of a reactor or transformer under actualoperating condit ions. Since smaller (kVA) loads have higher impedance and thus draw lowercurrent than larger (kVA) loads, the internal ohms of a reactor or transformer represent asmaller percentage o f the load impedance for a small (kVA) load t han for a large load.

    The value in ohms will cause a lower voltage drop when less than rated reactor or transformercurrent is f lowing. If the load is only one half the rated current, then the voltage drop across theimpedance will be onehalf of the rated voltage drop.

    Sample calculations for a three phase transformer rated 500kVA, 4160:480, 60Hz, 6%impedance:

    Transformer reactance Xt = (kV2/MVA) x %Z/100 = (0.482 / 0.5) x 0.06 = 0.027648 ohms

    Rated secondary current= 500,000 / (480 x 1.732) = 601.4 ampsActual Load current= 300 ampsVoltage drop at actual load= 300 x 1.732 x 0.027648 = 14.36 volts (14.36 / 480 = 0.0299, or3%of 480 volts)Effective percent impedance = 6% x (300 / 601.4) = 2.99%

    Transformer Percentage Impedance (VIDEO)

    Cant see this video? Clickhere to watch it on Youtube.

    Resource: Substation Comissioning Course Dry Type Transformer

    http://www.youtube.com/watch?v=iU0JmFub7xAhttp://electrical-engineering-portal.com/short-circuit-currents