Electrical Design Analysis

SITUATION In 2017 PEC Appendix D Example D14 Steps 10.2.d and 10.3.d, computing the fault currents yields almost 19,000 Amperes for point "b" and 11,000 Amperes for point "c". However, this same example as presented in the previous scenario results in fault currents of only around 17,000 Amperes for point "b" and 10,000 Amperes for point "c". Why are they different? ANALYSIS 1.) PROCEDURE IN 2017 PEC EXAMPLE D14 By inspection, it can be verified that all values in 2017 PEC D14 and in the previous scenario are almost identical, except for the new motor per-unit values used in the impedance diagrams of 2017 PEC D14 Steps 10.2.b and 10.3.b. 1.1.) 2017 PEC D14 Step 8 Motor Contribution In this step, the motor per-unit impedance is initially estimated at 0.25 pu (equivalent to a 25% subtransient reactance neglecting resistance), after considering the recommendation in IEEE Std 141-1993 Section 4.5.4.1. Conversion to the common

Below is a modified version of Example D14 (Simplified Fault Current Calculation) in Appendix D of the 2017 Philippine Electrical Code (PEC). This post is almost identical to the previous scenario , except that this time, an alternative common base value is used. The goal here is to show that regardless of the common base values utilized and the new per-unit values computed, the actual values remain practically the same. SITUATION An industrial complex receives 230 V, 60 Hz from a bank of three single-phase distribution transformers interconnected into a three-phase configuration. Each distribution transformer is rated 100 kVA, and the entire bank has an impedance of 5%. The transformer bank taps into a 34.5 kV supply with a 1,000 MVA short-circuit capacity. Using the per-unit method, what maximum symmetrical fault currents may occur in each of the fault points "a", "b", and "c"? From these fault currents, what are the minimum symmetrical kiloAmper

This post is inspired by the 2017 Philippine Electrical Code (PEC) Appendix D Example D14 (Simplified Fault Current Calculation). While the 2017 PEC solution is straightforward, it does not explain the essence of base values and their role in per-unit conversions. This approach aims to provide clarity and hopefully enlighten confused electrical practitioners that the Per-Unit Method is a handy tool in analyzing electrical circuits of any size, from the smallest residential loops to the largest transmission networks. SITUATION An industrial complex receives 230 V, 60 Hz from a bank of three single-phase distribution transformers interconnected into a three-phase configuration. Each distribution transformer is rated 100 kVA, and the entire bank has an impedance of 5%. The transformer bank taps into a 34.5 kV supply with a 1,000 MVA short-circuit capacity. Using the per-unit method, what maximum symmetrical fault currents may occur in each of the fault points "a", "