Electrical Design Analysis

SITUATION It is desired to improve the power factor of a 230V, 60 Hz, 3-phase system from 0.7 to 0.92 lagging. A wye-connected capacitor bank with capacitance of 10 microfarad per phase is to be used. What is the kW of the load? ANALYSIS 1.) Capacitor reactance, 1ph: 1.1.) Scalar: X,c,1ph = 1 / (2 pi f C) X,c,1ph = 1 / (2 pi 60 * 10uF) X,c,1ph = 265.2582 ohms 1.2.) Vector: X,c,1ph-> = 265.2582 ohms / j X,c,1ph-> = -j 265.2582 ohms X,c,1ph-> = 265.2582 ohms < (-90 deg) 2.) Capacitor power, vector: 2.1) Real power The capacitor is purely a reactive device. Since there is no resistance, no power dissipation happens. Therefore, capacitor real power does not exist. P,c-> = 0 Watts 2.2.) Reactive power, 3ph: Q,c,1ph-> = V,c,LN-> (i,c,1ph->)* Q,c,3ph-> = 3 Q,c,1ph-> Q,c,3ph-> = 3 V,c,LN-> (i,c,1ph->)* -- Conjugate (i->)* means "to reverse angle sign". Q,c,3ph-> = 3 V,c,LN-> (V,c,LN-> / X,c,1ph

Below is a modified version of Example D15 (Voltage Drop Calculation) in Appendix D of the 2017 Philippine Electrical Code (PEC). This post is a single-phase version of the three-phase example in the previous scenario . The same methods in understanding schematics and applying Kirchhoff's Laws are used, with minor adjustments in the loop diagrams and voltage calculations. SITUATION A single-phase transformer supplies 230 Volts to a structure via two 250 mm^2 THWN copper conductors in steel conduit 15.2 meters long, where the main service panel draws a total of 295 Amperes. One of the circuits in the main service panel feeds a single-phase motor via two 5.5 mm^2 THWN copper conductors in steel conduit 30.5 meters long. This electric motor is rated 60 Hertz, 230 Volts, and draws a full load current of 20 Amperes at 80% power factor. How much voltage drops are felt at the main service panel and at the motor terminals, coming from the transformer? ANALYSIS 1.) ONE-LI

This post is inspired by the 2017 Philippine Electrical Code (PEC) Appendix D Example D15 (Voltage Drop Calculation). The 2017 PEC solution is straightforward, but it forces the use of daunting formulas that are easy to forget, or worse, mix up values in the process. This take on the problem removes the necessity of memorizing formulas, by guiding the confused electrical practitioner to understand the schematics, then apply the principles of KVL and KCL instead. SITUATION A three-phase transformer bank supplies 230 Volts line-line to a structure via three 250 mm^2 THWN copper conductors in steel conduit 15.2 meters long, where the main service panel draws a total of 295 Amperes. One of the circuits in the main service panel feeds a three-phase motor via three 5.5 mm^2 THWN copper conductors in steel conduit 30.5 meters long. This electric motor is rated 60 Hertz, 230 Volts line-line, and draws a full load current of 20 Amperes at 80% power factor. How much voltage drops are fel