Electrical Design Analysis

SITUATION An impedance coil draws 10 Amps but dissipates only 250W when it is supplied with 220 Volts, 60 Hertz. How much power will it dissipate when it is supplied 110 Volts, 25 Hertz? ANALYSIS 1.) Existing supply at 220V 60Hz 1.1.) Apparent power: S<a = P + j Q S<a = S cos(a) + j S sin(a) S = V i = 220 * 10 = 2200VA 1.2.) Power factor angle, inductive: P = S cos(a) pf = cos(a) = P/S cos(a) = 250W / 2200VA = 0.1136 a = (+)arccos(0.1136) = 83.48deg -- Power positive angle means "lagging". 1.3.) Impedance, inductive: V = i Z Z = V/i = 220/10 = 22 ohms Z-> = R + j X Z-> = Z cos(a) + j Z sin(a) R = Z cos(a) R = 22 cos(83.48) = 2.5 ohms X = Z sin(a) X = 22 sin(83.48) = 21.86 ohms 1.4.) Inductance at 60Hz: X = 2 pi f L L = X / (2 pi f) = 21.86/(2 pi 60) L = 0.05799H 2.) New supply at 110V 25Hz 2.1.) Reactance at 25Hz: X = 2 pi f L X = (2 pi 25) * 0.05799H = 9.11 ohms 2.2.) Resistance remains the same: R = 2.5 ohm

"ELi the iCE man drinks ViR and eats PiE." 1.) ELi -- If inductive (L), current (i) lags voltage (EMF). 2.) iCE -- If capacitive (C), current (i) leads voltage (EMF). 3.) ViR Ohm's Law, V = i * R -- Voltage or EMF (V) is directly proportional to the current (i) through a constant resistance (R). 4.) PiE Electrical Power Equation, P = i * E -- Power (P) is the product of current (i) and voltage (EMF).

SITUATION A nominal 69-kiloVolt short transmission line has a length of 16 kilometers, with an impedance of (0.25 + j0.4375) ohm per kilometer. It supplies a 70MVA 3-phase load, with 80% lagging power factor, at 64 kiloVolts only. What is the line's transmission efficiency? ANALYSIS 1.) One line diagram: o|---V,src---Z,line---Z,load---|> V,src-> = i,line-> (Z,line-> + Z,load->) V,load-> = i,line-> Z,load-> V,src-> = i,line-> Z,line-> + V,load-> 2.) Line impedance: Z,line-> = (0.25 + j0.4375) * 16km Z,line-> = (4 + j7) ohms Z,line-> = 8.0623ohms < 60.2551deg 3.) Load voltage, line-line, as reference: V,load,LL-> = 64kV < 0deg 4.) Load voltage, line-neutral: V,load,LN-> = V,load,LL-> / sqrt(3) V,load,LN-> = (64kV < 0) / sqrt(3) V,load,LN-> = 36.95kV < 0 5.) Power factor angle: S<a = P + j Q S<a = S cos(a) + j S sin(a) P = S cos(a) cos(a) = P/S = pf = 0.8 a = (+)arcco

This post is inspired by Problem 21, Chapter 18 (Transmission Lines) of "Textbook-Reviewer in Electrical Engineering, 1st Edition" by Professional Electrical Engr. (PEE) Marcialito M. Valenzona. SITUATION A manufacturing plant is supplied by a 3-phase transmission line 10 km long. The plant receives 2.5MVA at 6kV with power factor 0.8 lagging. With a phase impedance of (0.30 + j0.40) ohm per kilometer, what is the sending voltage between the lines? ANALYSIS 1.) One line diagram: o|---V,src---Z,line---Z,load---|> V,src-> = i,line-> (Z,line-> + Z,load->) V,load-> = i,line-> Z,load-> V,src-> = i,line-> Z,line-> + V,load-> 2.) Line impedance: Z,line-> = (0.30 + j0.40) * 10km = (3 + j4) ohms |Z,line| = sqrt(3^2 + 4^2) = 5 ohms angle (b) = arctan(X/R) = arctan(4/2) = 53.13deg Z,line-> = |Z,line|<b = 5 ohms < 53.13deg 3.) Load voltage, line-line, as reference: V,load,LL-> = 6kV < 0deg 4.) Load volta

This post is inspired by Problem 36, Chapter 17 (Power Plant) of "Textbook-Reviewer in Electrical Engineering, 1st Edition" by Professional Electrical Engr. (PEE) Marcialito M. Valenzona. SITUATION An existing power plant has a maximum demand of 20MW. With a load factor of 55% and a capacity factor of 50%, what is its present reserve capacity? ANALYSIS 1.) Average demand: Load factor is the average demand over the maximum demand for a given operating period. LF = demand,avg / demand,max 0.55 = demand,avg / 20MW demand,avg = 20MW * 0.55 demand,avg = 11MW 2.) Maximum capacity: Capacity factor is the average demand felt by the plant over the plant's maximum installed capacity. CF = demand,avg / capacity,max 0.5 = 11MW / capacity,max capacity,max = 11MW / 0.5 capacity,max = 22MW 3.) Reserve capacity: Reserve capacity is the difference between maximum capacity and maximum demand. capacity,rsv = capacity,max - demand,max capacity,rsv = 22MW -

This post is inspired by Problem 27, Chapter 20 (Illumination) of "Textbook-Reviewer in Electrical Engineering, 1st Edition" by Professional Electrical Engr. (PEE) Marcialito M. Valenzona. SITUATION An industrial work area of 33 meters by 13 meters needs an illumination level of 72 lumens per square meter using 200W lamps with 2730 lumens of output per lamp. The area's coefficient of utilization is 40% and its maintenance factor is 71%. How many lamps does the area need? ANALYSIS 1.) Desired work area illumination: (72 lumens/m^2) * (33*13)m^2 = 30,888 lumens 2.) Lumens from sources: Coefficient of utilization is the lumens reaching the work area (diminished by absorption and reflection) over the lumens emitted by sources. 30,888 lumens / 0.4cu = 77,220 lumens 3.) Lumens when clean/new: Maintenance factor is the lumens emitted by source under normal working conditions (dust, dirt, smoke, etc) over the lumens emitted when everything is completely

This post is inspired by Example 10.2 of "Elements of Power System Analysis, 4th Edition" by William D. Stevenson, Jr. SITUATION A synchronous motor rated 30MVA, 13.2kV, with a SUBTRANSIENT reactance of 20%, draws 20MW with leading power factor of 80% from a feeder line with 10% reactance calculated from the machine base ratings. The line's terminal voltage is 12.8kV, and at the sending end is a synchronous generator also rated 30MVA, 13.2kV, with a SUBTRANSIENT reactance of 20%. When a 3-phase fault occurs at the motor terminals, what are the symmetrical fault currents at the generator, the motor and the fault point? ANALYSIS 1.) SOURCE SUB-CIRCUIT Base power, 3ph: S,base = 30MVA Base voltage, LL: V,base = 13.2kV 1.1.) Establishing mathematical relationships to avoid confusion: S,base = S,3ph V,base = V,LL V,LN = V,LL / sqrt(3) i,1ph = i,1L (assuming Y-config) i,base = i,1L S,3ph = 3 * V,LN * i,1ph S,3ph = 3 * ( V,LL/sqrt(3) ) * i,1L S,3ph

This post is inspired by the 2017 Philippine Electrical Code (PEC) Appendix D Example D13 (Available Short Circuit Current). The 2017 PEC approach is simple and straightforward, but it does not indicate the relationships of the parameters relative to each other. This calculation is geared towards providing clarity to the confused electrical practitioner. SITUATION A fault occurs down the line of a single-phase transformer rated 100kVA, 230 Volts, and 2.5% impedance. What is the maximum symmetrical fault current produced? ANALYSIS 1.) TRANSFORMER LOAD-SIDE SUB-CIRCUIT Base power, 1ph: S,base = 100kVA Base voltage, 1ph: V,base = 230V Impedance, given: Z = 2.5% = 0.025 pu 2.) METHOD USING ACTUAL VALUES 2.1.) Base impedance, 1ph: Z,base = (V,base)^2 / S,base Z,base = (230V)^2 / 100kVA Z,base = 0.529 ohm 2.2.) Actual impedance, 1ph: Z,1ph = Z,base * Z,pu Z,1ph = 0.529 ohm * 0.025 pu Z,1ph = 0.013225 ohm 2.3.) Use rated voltage as actual voltage.