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...

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...

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.

MOTOR EFFICIENCY DISSECTED ANALYSIS 1.) Law of Conservation of Energy (ideal): W,in = W,out 2.) First Law of Thermodynamics (practical): Q = W + dU Total energy = Work done + change in Internal energy W,in = W,out + W,losses W,in > W,out Energy input is always greater than energy output because of energy losses. 3.) Utilizing energy through time: (W,in / t) = (W,out / t) + (W,losses / t) 4.) Power and energy relationship: P = W / t P,in = W,in / t P,out = W,out / t P,losses = W,losses / t 5.) Power flow equation: P,in = P,out + P,losses P,in > P,out Power input is always greater than power output because of power losses. 6.) Usable power and energy: efficiency = (P,out / P,in) * 100 efficiency = (W,out / W,in) * 100 SUMMARY Motor efficiency is all about how well a motor converts energy input into energy output, given imperfections like friction and waste heat.

SITUATION Lamp 1 and Lamp 2, operating in parallel across 230 Volts, dissipate rated power at 100 Watts and 60 Watts respectively. How much power will each lamp dissipate if they are connected in series? ANALYSIS 1.) Parallel operation 1.1.) Power: P,1 = 100W P,2 = 60W 1.2.) Resistance: R,1 = (V)^2 / P,1 R,1 = (230V)^2 / 100W = 529 ohms R,2 = (V)^2 / P,2 R,2 = (230V)^2 / 60W = 881.67 ohms 2.) Series operation 2.1.) Current: V = i * (R,1 + R,2) i = V / (R,1 + R,2) i = 230V / [(529 + 881.67) ohms] i = 0.16A 2.2.) Power: P,1 = i^2 * R,1 P,1 = (0.16A)^2 * 529 ohms = 13.54W P,2 = i^2 * R,2 P,2 = (0.16A)^2 * 881.67 ohms = 22.57W CONCLUSION When connected in series, Lamp 1 dissipates around 13.5 Watts while Lamp 2 dissipates around 22.6 Watts. Both outputs decrease significantly, but Lamp 2 now has a higher power output than Lamp 1.

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...

SITUATION An induction motor, rated 240 Volts, draws a current of 150 Amps and brings the system power factor down to 70.7%. What capacitor size is needed to raise system power factor to 95%? ANALYSIS 1.) Complex power equations, vector: S-> = P-> + Q-> S<a = P + jQ S<a = S cos(a) + j S sin(a) 2.) System power factor angle: P = S cos(a) pf = P/S = cos(a) = 0.707 a = arccos(0.707) = 45deg 3.) Motor apparent power, scalar: S = Vi = 240V * 150A = 36kVA 4.) Motor real and reactive power: P,load = S cos(a) P,load = 36k cos(45) = 25,455.84W Q,load = S sin(a) Q,load = 36k sin(45) = 25,455.84VAr 5.) System power factor angle after capacitor correction: pf = cos(a) = 0.95 a = arccos(0.95) = 18.19deg 6.) Real power, unaffected: P,load = 25,455.84W 7.) Desired system apparent power: P = S cos(a) 25,455.84 = S cos(18.19) S = 25,455.84 / cos(18.19) S = 26,794.87VA 8.) Desired system reactive power: Q,sys = S sin(a) Q,sys = 26,794....