Through

  • May 2020
  • PDF

This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA


Overview

Download & View Through as PDF for free.

More details

  • Words: 366
  • Pages: 2
Through-Fault Current Calculator: A Resource for Selecting a Substation Transformer Protective Device One criterion in selecting a distribution substation transformer protective device is its ability to protect the transformer from secondary-side limited faults . . . or “through-faults.” These faults are difficult to detect by the overcurrent relay of the line-terminal circuit breaker, because the magnitude of the fault current is relatively low—being limited by the impedance of the transformer. These faults are a challenge to clear as well, because of their high transient recovery voltage. That’s why a stand-alone protective device such as an S&C Circuit-Switcher is typically applied at each transformer. S&C has tested each of its distribution transformer protective devices—Series 2000, Mark V, and Mark VI CircuitSwitchers, and Trans-Rupter II® Transformer Protector—specifically for this duty, and assigned a secondary-fault interrupting rating to each device. This formula can be used to determine the secondary-fault interrupting rating required for the substation transformer protective device, to properly protect a particular size and impedance of transformer: I = (57.8P) / [(%Z)E] The formula assumes an infinite (zero-impedance) source.

Through-Fault Current Calculator

Adobe Flash® Player Plug-in Required. Where: I = Inherent secondary-fault current, amperes P = Transformer self-cooled three-phase base rating, kVA E = System phase-to-phase voltage, kV %Z = Percent transformer primary-to-secondary impedance, referred A protective device is appropriate for the application if its secondary-fault interrupting rating is equal to or greater than the value for “I” calculated above. EXAMPLE: The inherent secondary-fault current for a 20/40/50-MVA, 115-kV, 8%-impedance transformer would be: I = [(57.8) (20 000 kVA)] / [(8) (115 kV)] = 1257 A

Check out these other resources for selecting a substation transformer protective device: “Selection Guide for S&C Substation Transformer Protective Devices” The selector table in this publication lets you compare the features and capabilities of S&C’s substation transformer protective devices, so you can select the appropriate device for your application. “Primary-Side Transformer Protection”. This paper offers a complete overview of transformer protection, including typical protection schemes, a discussion of transient recovery voltage and why secondary-side faults are difficult to interrupt, and criteria for selecting a device appropriate to a particular application.

©2009 S&C Electric Company All rights reserved. Site Map | Privacy Statement

Related Documents

Through
June 2020 28
Through
May 2020 11
Walk Through
June 2020 16
Walk Through
May 2020 25
Walk Through
August 2019 30
Breaking Through
December 2019 37