Electrical Design Analysis

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.

This post is inspired by Problem 31, Chapter 14 (Synchronous Motor) of "Textbook-Reviewer in Electrical Engineering, 1st Edition" by Professional Electrical Engr. (PEE) Marcialito M. Valenzona. SITUATION A Y-configured, three-phase synchronous motor draws its rated current of 70 Amperes at a leading power factor of 0.8, with 6.6 kiloVolts at its terminals. If each phase has 2 ohms armature resistance and 20 ohms synchronous reactance, what is the induced electromotive force between the lines? ANALYSIS 1.) One-line diagram: o|---V,load,LN---R,ar,1ph---X,sy,1ph---EMF,load,LN---|> -V,load,LN-> + i,load-> (R,ar,1ph +j X,sy,1ph) + EMF,load,LN-> = 0 EMF,load,LN-> = V,load,LN-> - [ i,load-> (R,ar,1ph +j X,sy,1ph) ] EMF,load,LN-> = V,load,LN-> - [ i,load-> (Z,load,1ph) ] 2.) Armature resistance, 1-phase: R,ar,1ph = 2 ohms 3.) Synchronous reactance, 1-phase: X,sy,1ph = 20 ohms 4.) Load impedance, 1-phase: Z,load,1ph-> = R,ar,1p

The following items are examples listed in 2017 PEC Appendix D (Wiring Design Examples). Example D1: "Single Family Dwelling Unit, up to 50 square meters Floor Area with Load not exceeding 3,680 Volt-Amperes" Example D2: "Single Family Dwelling Unit, up to 150 square meters Floor Area (Not more than six circuits)" Example D3: "Single Family Dwelling Unit, up to 150 square meters Floor Area (More than six circuits)" Example D4: "Single Family Dwelling Unit, up to 150 square meters Floor Area, Optional Calculation" Example D5: "Single Family Dwelling Unit, more than 150 square meters Floor Area, 230 Volts, Single-Phase Service" Example D6: "Single Family Dwelling Unit, more than 150 square meters Floor Area, 400/230 Volts, Three-Phase, 4-Wire Service" Example D7: "Multi Family Dwelling Unit, 230 Volts, Single-Phase Service" Example D8: "Multi Family Dwelling Unit, 400/230 Volts, Three-Phase,

TREATING DESIGN ANALYSIS AS MANDATORY It is important to keep in mind that, before any actual electrical work is done, everything starts at the design stage. Considerations on SAFETY and QUALITY need to be included (and examined) on paper in order to avoid any mistakes that may result in costly back jobs, or worse, loss of life and property due to electrical fire. Therefore, electrical practitioners in the Philippines need to be aware of the following provisions in the Philippine Electrical Code (PEC) concerning design analysis. ===== 2017 PEC 1.0.1.5 (A) Mandatory Rules. [ 2017 NEC 90.5 (A) ] "Mandatory rules of this Code are those that identify actions that are specifically required or prohibited and are characterized by the use of the terms SHALL or SHALL NOT." ===== 2017 PEC 1.3.2.1 (F) Plan Requirements - Design Analysis . [ This PEC provision does not exist in NEC. ] "Design analysis SHALL be included on the drawings or SHALL be submitted on se

Consider this generator again from the previous scenario , but this time, with half of its rated capacity attached. This post examines what will happen to its internal EMF at half-load condition. SITUATION A compound DC generator rated 20 kiloWatts, with resistances in its armature winding at 0.05 ohm, series field winding at 0.025 ohm, and shunt field winding at 100 ohms, operates at HALF LOAD with 250 Volts at its terminals. With the machine configured as short shunt, what is the internal electromotive force produced by the generator? ANALYSIS 1.) Schematic diagram, short shunt compound: 2.) One-line diagram: o|---EMF---R,ar---|---R,se---V,L---|> | R,sh | v 3.) Power drawn by load: P,L = 10kW (HALF LOAD) 4.) Voltage at the generator terminals: V,t = 250V 5.) Resistances 5.1.) Armature winding resistance: R,ar = 0.05 ohm 5.2.) Series field winding resistance: R,se = 0.025 ohm 5.3.) S

Consider the generator from the previous scenario , but without any loads attached. This post examines what will happen to its internal EMF at no-load condition. SITUATION A compound DC generator rated 20 kiloWatts, with resistances in its armature winding at 0.05 ohm, series field winding at 0.025 ohm, and shunt field winding at 100 ohms, operates at NO LOAD with 250 Volts at its terminals. With the machine configured as short shunt, what is the internal electromotive force produced by the generator? ANALYSIS 1.) Schematic diagram, short shunt compound: 2.) One-line diagram: o|---EMF---R,ar---|---R,se---V,t---x | R,sh | v 3.) Power drawn by load: P,L = 0 (NO LOAD) 4.) Voltage at the generator terminals: V,t = 250V 5.) Resistances 5.1.) Armature winding resistance: R,ar = 0.05 ohm 5.2.) Series field winding resistance: R,se = 0.025 ohm 5.3.) Shunt field winding resistance: R,sh = 100 oh

SITUATION A compound DC generator rated 20 kiloWatts, with resistances in its armature winding at 0.05 ohm, series field winding at 0.025 ohm, and shunt field winding at 100 ohms, operates at full load with 250 Volts at its terminals. With the machine configured as short shunt, what is the internal electromotive force produced by the generator? ANALYSIS 1.) Schematic diagram, short shunt compound: 2.) One-line diagram: o|---EMF---R,ar---|---R,se---V,L---|> | R,sh | v 3.) Power drawn by load: P,L = 20kW (full load) 4.) Voltage at the generator terminals: V,t = 250V 5.) Resistances 5.1.) Armature winding resistance: R,ar = 0.05 ohm 5.2.) Series field winding resistance: R,se = 0.025 ohm 5.3.) Shunt field winding resistance: R,sh = 100 ohms 6.) Short shunt configuration 6.1.) Load current is same as series field current: i,L = i,se 6.2.) Load voltage is same as generator terminal vol

A SIMPLE COMPARISON OF SYNCHRONOUS AND INDUCTION MOTORS Synchronous Motor = stator windings are the same as those of induction motors = rotor windings are usually flexible copper coils connected to a separate DC exciter = the DC excitation supplied to the field coils produces the rotor magnetic flux = constant speed from no-load to full-load, but may stop when overloaded = works as power factor corrector while providing stable torque to drive loads Induction Motor = stator windings are the same as those of synchronous motors = rotor "windings" are usually rigid aluminum bars, skewed and short-circuited at each end = self-excited via induction like a transformer to produce rotor magnetic flux = variable speed based on frequency, but can be affected by harmonics = can provide high torque to accommodate temporary overloads Each has an advantage, depending on the desired application.