Electrical Design Analysis

This post is inspired by the 2017 Philippine Electrical Code (PEC) Appendix D Example D11 (Voltage Regulators, Single-Phase). SITUATION A single-phase automatic voltage regulator (AVR) rated 5 kVA, 60 Hertz, 165-280 Volts input, 230 Volts output needs a feeder circuit from the main service panel. The AVR is installed in an air-conditioned data center with raised floors, and the conduit for electric conductors shall be run underneath. The wires to be used are made of copper conductors, with insulation rated for 75 degC operating temperature and of type THW (Thermoplastic, Heat-resistant, for Wet location). The raceway for the circuit is a rigid PVC Schedule 80 conduit. What size of 1ph circuit breaker, wires, and raceway are needed? ANALYSIS 1.) ONE-LINE DIAGRAM i,avr,1ph ---> o|---V,msp,1ph---V,avr,1ph---|> 2.) CIRCUIT CALCULATIONS 2.1.) Kirchhoff's Voltage Law, 1ph: -V,msp,1ph + V,avr,1ph = 0 V,avr,1ph = V,msp,1ph 2.2.) Power equation,

RACEWAY SIZES (CONDUIT AND TUBING) 1.) Raceway Sizing There are three options for sizing raceways according to 2017 PEC Table 10.1.1.1 [ or 2017 NEC Chapter 9 Table 1 ], based on cross-sectional areas: A.) 53% raceway area for 1 conductor (a multiconductor cable is treated as a single conductor [Note 9]) B.) 31% raceway area for 2 conductors C.) 40% raceway area for over 2 conductors 1.1.) Choosing Between Alternatives For future expansion purposes, option (A) is far from practical. This may have use in very specific installations that are assumed to be not needing any expansions, but upgrades will definitely happen and option (A) will eventually become a bottleneck. Option (B) for 2 conductors having a smaller area than either option (A) for 1 conductor or option (C) for over 2 conductors may seem nonsensical, but it actually does make sense. Having 2 conductors usually means 2 live wires supplying loads that may require an equipment grounding conductor (EGC) later on

WIRE AMPACITIES, INSULATORS, AND IMPEDANCES Below are some useful tables pertaining to ampacities, insulators, and impedances of various wires. These are derived from the 2017 Philippine Electrical Code [ or the 2017 National Electrical Code ], but are rearranged in a manner that is easier to interpret and understand. 1.) Allowable Ampacities of Insulated Conductors in Raceway. The areas, ampacities, and insulation types of the following tables are based on 2017 PEC Table 3.10.2.6(B)(16) [ or 2017 NEC Table 310.15(B)(16) ]. 1.1.) Copper Wire Ampacities --------------------------------------- |  Copper  |  Insulation Max Op. Temp | --------------------------------------- |   Area   | 60degC | 75degC | 90degC | |   mm^2   |  Amps  |  Amps  |  Amps  | --------------------------------------- |    2.0   |    15  |    20  |    25  | |    3.5   |    20  |    25  |    30  | |    5.5   |    30  |    35  |    40  | |    8.0   |    40  |    50  |    55  | |   14.0   |    55  |    65

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