Electrical Design Analysis

Get it on  Huawei AppGallery now! (Keyword: "ElectricalDEAN") ========== Electrical DEAN for Mobile Devices is an electrical design analysis app based on the Philippine Electrical Code, the SI Modernized Metric System, and equivalent provisions from the National Electrical Code (NFPA 70). Electrical design analysis is primarily about detailed calculations of wire gauges, conduit sizes, protective device ratings, fault currents, voltage drops and other technical matters necessary for the safe and proper operation of electrical systems. The Electrical DEAN mobile app is intended as an educational tool for those who are new in the electrical trade, and as a research tool for veterans who need quick calculations to compare with their own electrical designs. Currently, it covers the fundamentals of electrical design analysis: conductor and conduit data, fault currents and voltage drops (1-phase and 3-phase), and general-purpose circuit sizing (circui...

SITUATION ​​ A conductor of a certain length and cross-sectional area has a resistance of 10 ohms. If the conductor's length is tripled and ​its ​cross-sectional area​ is doubled, what is its new resistance? ANALYSIS 1.) Resistivity equation: Resistance = resistivity constant * (Length / cross-sectional Area) R = r * (L / A) ​2.) Resistance, old: R,1 = r * (L,1 / A,1) R,1 = 10 ohms 10 = r * (L,1 / A,1) ​3.) Parameters changed:​ ​L,2 = 3 L,1 A,2 = 2 A,1 ​ ​4.) Resistance, new: R,2 = r * (L,2 / A,2) R,2 = r * ​[ ​(3​ ​L,1​)​ / ​(​2​ ​A,1) R,2 = [ r * (L,1 / A,1) ] * 1.5 R,2 = 10 * 1.5 R,2 = 15​ ohms​ CONCLUSION ​With the conductor's length tripled and cross-sectional area doubled​, the conductor's resistance becomes 15 ohms.​ Resistance is directly proportional to length and inversely proportional to​ cross-sectional area, according to the resistivity equation.

SITUATION A defective battery is found to have excessive heat losses during quality control tests. How its internal resistance determined to verify if it matches with its original design specification? ANALYSIS 1.) One-line diagram: o|---EMF,batt---R,batt---R,load---|> EMF,batt = battery internal electromotive force R,batt = battery internal resistance R,load = dummy load resistance 2.) Ohm's Law: V,load = i * R,load i = V,load / R,load 3.) KVL: (-EMF,batt) + i * (R,batt + R,load) = 0 (-EMF,batt) + (V,load / R,load) * (R,batt + R,load) = 0 (-EMF,batt) + (V,load / R,load) * R,batt + V,load = 0 V,load (R,batt / R,load) = EMF,batt - V,load R,batt / R,load = (EMF,batt / V,load) - 1 R,batt = [ (EMF,batt / V,load) - 1 ] * R,load At no load: V,NL = EMF,batt At full load: V,FL = V,load R,batt = [ (V,NL / V,FL) - 1 ] * R,load CONCLUSION The internal resistance of a battery can be determined by measuring its no-load voltage, then placing a known resi...

This post presents a simple circuit that can be thought of as a basic model for wire insulation. When insulators weaken, high voltages can push stray currents out, hence the need for insulation resistance tests. SITUATION A 40-picoFarad capacitor is connected in series with 40-ohm resistor. If the supply is 220 Volts at 60 Hertz, what is the current flowing through the circuit? ANALYSIS 1.) Capacitive reactance, vector: X-> = 1 / (jwC) = -j / (2 pi f C) 2.) Impedance, vector: Z-> = R-> + X-> Z-> = 40 + [ -j / (2 pi 60 * 40p) ] Z-> = (40 - j 66.315M) ohms Z-> = 66.315M ohms < (-89.999965) 3.) Current, vector: i-> = V-> / Z-> i-> = 220V<0 / [ 66.315M ohms < (-89.999965) ] i-> = 3.3175uA < 89.999965 CONCLUSION The current is 3.3176 microAmperes, leading.

This post presents a simple circuit that can be thought of as a basic model for conductor cables in large wiring installations, such as in big residential apartments, tall commercial buildings, or wide industrial complexes, where short-circuit or ground fault currents may not be enough to kick a circuit breaker open. In such cases, the fault condition will simply become what is known as a "phantom load". It unnecessarily heats up the cable and contributes to the energy consumption even when electrical appliances are unplugged. As cable length increases, conductor resistance and reactance become more evident, hence the need to consider them carefully in wiring design calculations. SITUATION A 0.0001-henry inductor is connected in series with 40-ohm resistor. If the supply is 220 Volts at 60 Hertz, what is the current flowing through the circuit? ANALYSIS 1.) Inductive reactance, vector: X-> = jwL = j 2 pi f L 2.) Impedance, vector: Z-> = R-> + X-...