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# ControlSystems ME478

Lafayette College

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This 58 page Class Notes was uploaded by Izabella Jerde on Tuesday October 13, 2015. The Class Notes belongs to ME478 at Lafayette College taught by KarlSeeler in Fall. Since its upload, it has received 45 views. For similar materials see /class/222311/me478-lafayette-college in Mechanical Engineering at Lafayette College.

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5 Dr Seeler Department of Mechanical Engineering Fall 2004 Lafayette College ME 478 Automatic Controls Systems and Mechatronics Alternating Current Phasors Complex Impedance and Three Phase Current INTRODUCTION Alternating current AC is driven by sinusoidal voltages Analysis of AC circuits considers only the steadystate response not the transient response Since AC input voltage is sinusoidal and we are interested in the circuit s steadystate response we can use the tools we developed in ME 352 for frequency response Recall from ME 352 that the steadystate response of a linear system to a sinusoidal input will be sinusoidal at the same frequency as the input but possibly shi ed in time The amplitude of the output sinusoid is the product of the amplitude of the input sinusoid and the magnitude of the transfer function evaluated by substituting for s the imaginary number j times the input frequency s jm Asin03t AIGGwIsincot gt Block Diagram of the SteadyState Response of a Linear System to a Sinusoidal Input In AC circuits the frequency is generally xed at 60 Hz or 377 E We will not be sec concerned with the response of a system over a range of frequencies just at one frequency Further the input sinusoid is always voltage and always assumed to have zero phase shift Consequently the frequency response relationship can be fully described by the amplitude of the input A and the amplitude and phase shi of the output AIG and d A quantity with a magnitude and direction is a vector We will review the mathematics needed to work with vectors in a complex plane We will then introduce a vector representation tailored to AC circuit analysis Phasors are similar to the vectors we worked with in ME 352 but rather than using the amplitude of the sinusoid as the magnitude of the vector the magnitude of a phasor is the root mean square RM S or effective amplitude of the sinusoid Three phase current widely used in industry will be presented in the time domain and then represented using phasors Analysis of AC circuits is performed using complex impedances based on the frequency response equation The com lex impedance method lends itself to graphical vector construction Finally we will investigate power AC systems Review of Complex Exponentials Euler s Sine and Cosine Formulae Before we begin working with sinusoidally varying voltages and currents we will review prerequisite material from ME 352 Recall that Euler discovered that a complex exponential e 9 is a unit vector in a complex plane e e cos jsin Expressed graphically eJ6 cos jsin Re Complex Exponential Unit Vector The projection of a complex exponential unit vector 9399 onto the real axis yields the cosine of the angle 9 and the projection onto the imaginary axis yields the imaginary unit vector j times the sine of the angle 9 The projections of the complex exponential unit vector onto the real and imaginary axes are expressed in Mathcad and other engineering software as scalar functions Ree cos and Ime sine Note that the function Imel yields the magnitude of the projection onto the imaginary axis as scalar not an imaginary num er Euler s sine and cosine formulae also yield sin and cos 18 16 18 7 le cos and sin Euler s cosine and sine formulae expressed graphically as vector constructions Euler s Cosine Formula Euler s Sine Formula Finally recall that the angle 9 can be time varying With the form 9 mt d Where m is the circular frequency in E and d is a constant angle called the phase angle sec MPORTAN T Everything multiplied by the imaginary number j in the exponent of a complex exponential is part of the expression for the angle of the complex unit vector 18 mum ee FREQUENCY RESPONSE Derivation of the Frequency Response Equation The derivation of the frequency response equation requires the use of Laplace transforms partial fraction expansion and a fact from complex analysis For our derivation we Will use a rst order system With the transfer function GS 7 Outputs 7 L 7 Inputs 7 s 2 We will apply the inputt A sinnt to the transfer function The Laplace transform of the input is A03 s203 g A sin03t 2 Multiplying the transfer function Gs by the Laplace transform of the input to obtain the Laplace transform of the output yields InputsGs A0 Outputs s2032 s2 It is essential to remember that frequency response is the steadystate response of the system to the sinusoidal input The frequency 03 is a constant We need to perform the inverse Laplace transform on Outputs to return to the time domain and then take the limit as t gt 00 Outputs is not in the Laplace transform table so we will need to perform partial fraction expansion to produce terms which are in the table Normally we would keep the denominator factor 52 02 because sine and cosine are in Laplace transform tables However in this case we will factor the denominator into the product of first order factors for a reason which will be apparent later in the derivation The partial fraction expansion is O quot ls2llsizlzlmll l famllfiwllsle Determine the numerator constants B1 B2 and C by multiplying both sides by the denominator of the corresponding term canceling identical factors and evaluating using the pole of that term Constant B1 lw ll lmwlsfgmjltsjmgt gmjltsjwgt ltsjm lltsi 3 mgtll lB1 33Jltsjwgt ltswgt B1 B sjn S JOJ Sfijwmsizljw jb 39o l il lzg l jm mbl l jmm 1 Hence the constant B1 is B1 ll Q2l joi2izll j Evaluating the transfer function 5711 5711 l G s s2 fors jn yields 1 G Jw D2 The constant B1 can be written as A 1 A Bl lz jllml l2 JG W Using the same procedure B2 and C are found to be Bz rstatrium 4122 where C is a real number The partial function expansion is Outputs B 1 B C 1 1 2 SD S JOJ s2 Which becomes more intimidating when the complex constants B1 and B2 are substituted in A 1 A 1 A 1 Outputs 2 jGJmmEGJmsjm4mz Is2j The fact that B1 and B2 are complex constants does not affect the inverse Laplace transformation Constants either real or complex can be factored out of the inverse transform because it is a linear operator Outputt g 391Outputs 9 391B1 1 J B2 s 103 s Jon s 2 l l l Outputtg 1Bl 9 1B2 Z 1C S 103 8 103 s 2 OutputtB1391 1 B2391 1 C39l 1 SD S JOJ s2 All of the inverse transforms are of the form Therefore Outputt Ble j Bzejw39 Ce Before dealing with the constants B1 B2 and C let us take the limits of the exponentials as t gt 00 We will start with the limit of en l 1 lim 6 216 zoo 0 taco e 00 What this is telling us is that the homogeneous response of the system decays to zero in steadystate Regardless of the order of the system we will always have the same result Only the particular or forced response remains in steadystate Now consider the limit 739 t 739 11m e w 9 W tgtoo Does e jwo decay to zero No Recall that ejl is the complex exponential unit vector where the angle of the unit vector is the term in the exponent multiplied by j The complex exponential unit vector 61quot t has a time varying angle oat This unit vector rotates about the 2n origin of its complex plane complet1ng a rotation of 211 rad1ans as t increases by At The 0 negative sign is does not indicate decay it inverts the sign of the angle of the unit vector Consequently the limit lim e39j m does not yield a nal value The limit lim e39j m represents a taco tgtoo unit vector which are accumulated an in nite angle rotating about the origin of a complex plane However we cannot distinguish between 00 and 00 271 Therefore there is no nal value for either ej t or e jl Hence as t gt 00 the steadystate output is Outputtmadym Ble jw39 Bzejw39 NO Substituting in the complex constants B1 and B2 OUtplntsteadyrstate G JYDWj w Gj 063 t J J Because we know that the steadystate response of the system the particular solution of the system equation for the input will be a sinusoid since we are forcing the system with a sinusoid our objective is to eliminate the complex terms and yield a real trigonometric function We will need to use either Euler s sine or cosine function 61 61 cos 2 61 64 s1n 4 2 G003 is the system s transfer function Gs evaluated for s joa where n is the frequency of the forcing function The relationship between G003 and Gjoa can be seen when both are expressed vectors in the form of their magnitude times a complex exponential unit vector 1 I1Iej 1 5w Jw2mgl quot Vw24 G joa 1 New w 21 eijltlez w 1 jm2 Jmejmnil77 03 4 0 4 G003 and Gjoa have the same magnitude and differ only in the sign of their angle Let us generalize our notation by de ning their magnitude 1 Im24 IGj03I E and angle 4 mm zNjoa mm 04m1 tan 1 Using this notation with complex exponential unit vectors Goes Gltjoae and G joa Gltjoae We can now rearrange the steadystate output expression into a form which can be simpli ed using one of Euler s equations OutputGLteadymm 2 C Ijwe39j t 2 Gjwej t Outputtedyse ileomle jte W ileomlejtejw39 2 2 ontplntsteadyrstate JGjmlej ejm Neilwt I ejwt ejwt OUtplntsteadyrstate AGJml Outputtsteadyisme AGjoasinoat 4 This is the Frequency Response Equation we worked with in ME 352 As you recall from ME 352 I is the phase shift Ifcl is negative then the response follows or lags behind the input A system is classi ed as a lag if its phase angle is negative Another term used is phase lag to describe a negative phase angle eg The system has a phase lag of 35 means I 35 Note the use of degrees to report the phase angle This is convention because degrees are easier to visualize than radians If I is positive the response precedes or leads the input The system is classi ed as a lead due to its phase lead Phase leads may seem to violate the principle of cause and effect since the effect output appears to occur before the cause input This is not the case because the frequency response equation only applies in steadystate The input has been applied long enough for the system to get into a rhythm or the swing of things if you prefer The output is not being caused by the cycle that is immediately following it but by the cumulative effect of all the cycles that preceded it during the transient portion of the response Although the derivation of the frequency response equation is involved its application is straight forward Example Given the system equation 2 2 6v8d 21620i dt dt dt determine the steadystate response of the output yt when the input Xt 4sin3t 1 Create the transfer function Gs VS 2 Z 2d V 6v 9 8d 216 20i dt dt dt 2sIs 6Is 8s2Vs 6sVs 20Vs Is 2s 6 Gs Vs 8s26s20 2 Evaluate G003 where n is the frequency of the forcing function in radians per second In this example 03 3 2j36 6j6 6j6 G J 8j326j320 7220j18 52j18 10 3 Determine the magnitude and phase angle of G03 Gj3 l6j6l 6262 015 52j18 2 2 39 52 18 2Gj3 26j6 2 52j18 tan lgj tan l 079 281 202 rad 4 Evaluate the frequency response equation The magnitude of the input sine is A 4 Outputtsteadyrstate AGjm5inmt Outputtsleadystate E itSS 4015 sin3t 202 Outputtsteadystate E itSS 060 sin3t 202 Alternating current circuit analysis is a frequency response analysis The input is always a sinusoidally varying voltage vt V0 sinnt Note that the input voltage is assumed to have zero phase shift Often the output variable of interest is the current drawn by the system The phase angle 4 of the current drawn by a system it I0 sinnt is measured relative to the input voltage Current phase leads or lags are created by capacitive or inductive energy storage within a circuit For many of our analyses we will find it convenient to represent the circuit being driven by the source as a single complex impedance load Complex impedances are developed below They will allow us to work with phasors vectors in a complex plane as a convenient graphical representation of sinusoidally varying voltages and currents ALT ERNATING CURRENT Alternating current AC is the worldwide standard for the distribution of electric power The United States and Canada use a frequency of 60 hertz cyclessec Elsewhere AC frequency is 50 Hz When AC current was developed in the United States in the 1890 s the frequency was 25 Hz Why we now use 60 Hz and the rest of the world uses 50 Hz is a good question We will have to use a mix of units for frequency We will be able to perform some calculations in Hz but others w111 requ1re c1rcular frequency w1th units of i We also m1x sec units using both radians and degrees One number that is useful to keep in mind is 60 Hz 377 rad sec I The residential service voltage in North America is 110 to 120 VAC It is a range and does vary with the load on a system and the distance from the electrical distribution line Typically the voltage of 115 VAC is assumed in an analysis The residential voltage elsewhere 11 in the world ranges from 220 to 240 VAC Since lower voltage is safer why is North American voltage lower than elsewhere Increased current means larger diameter wire and more copper Copper was cheap in North America when standards were set in the late 1800 s European standards re ected the higher cost of copper there Resistive energy loss during transmission of electric power is a function of the current ow E Ri2 Lost Since electrical power P vi if you increase voltage you may decrease current and deliver the same power Consequently the voltage of AC power is stepped up for transmission and stepped down for use decreasing voltage and increasing current Although it is possible to step up and step down DC voltages of large power ows it is more expensive DC electrical energy must be converted to mechanical energy by a DC motor which then drives a DC generator to produce DC electrical power at a different voltage Paired DC motors and DC generators are expensive to purchase and to operate since they need maintenance AC voltages can be stepped up or down using transformers which are far less expensive to build and require little maintenance We use a various voltages above for industrial motors Large motors use voltages greater than 110 120 VAC for a number of reasons including economy The hazard posed by higher voltages is not an issue because large industrial motors are hardwired by electricians The following are the typical voltages 208VAC 230 VAC 460 VAC 575 VAC 2300 VAC In case you are wondering an induction motor using 2300 VAC pulls only 49 Amps to produce 200 hp The same type of motor designed to run at 230 VAC would need 490 Amps and a lot more copper to produce 200 hp Industrial motors also use variable frequency AC current for speed control AC is supplied to the speed control circuitry at 60 Hz The AC is converted to a direct current DC and then to the needed frequency using circuitry known as an inverter ROOT MEAN SQUARE RMS 0r EFFECTIVE VALUES In some AC analyses a single value is used to represent an average or effective voltage or current The effective value corresponds to the constant or DC voltage and current which would deliver the same power The average value of a sinusoid with no vertical offset such as vt V0 sinnt shown below is zero because of the symmetry of the sine wave VO nT n1T n2T Two Cycles of a Sinusoidal Voltage with Amplitude V0 Consequently a straight average of vtVUsinmt over a period is not a useful calculation Squaring vt before calculating the average eliminates the problem of negative values over half the period but also inordinately increases the contribution of large numbers to the calculation A way of compensating is to take the square root of the average of squared function This yields the rootmean square or RMS value Expressed mathematically the rootmean square RMS value of the vt is l T 2 VRMS dt vRMS Him 511mmZ dt We will not evaluate this expression because there it is far more convenient to integrate complex exponentials Recall Euler s expressions for sin9 and cos9 e 7e lie 16 le sin9 8T and cos9 i 2 If we integrate over a full cycle it makes no difference if we use cosine or sine Using a cosine function vt VD cosmt o l3 Expressing it using complex exponentials mm 4mm Vt V0 cos03t 4 V0 Those suffering from postcalculus stress syndrome feel free to skip from here to the results below You are not responsible for the following integration T T jwt 711mm 2 VRMS JV0sinmt zdt VO dt 0 0 V V5Tejltwt eijltwtmgt2dt RMS 4T 0 Expanding the integral helm MilWWdtzhejzwm 26jwt eejvx erj2ox dt 0 0 T T J39ejwt 64mm 2 dt J39ej2wt 26jwt erjwt 642mm dt 0 0 j ej2wt 26jwt eijwt 642mm dt ej2wt 260 642mm dt 0 0 T T T T RelMW 2e0 e jzlwl dt Iepwmdt J lmmdt J 2e dt 0 0 0 0 d d quot Evaluating the first integral term us1ng e 1 11 e and recognlzlng that the elrcular frequency x x 21 T T 0 E 411 612an engomjj 1ej4n2 em 14 Complex eXponentials are vectors The complex exponentials ejllmzl and em are the same vector since 0 211 411 and any other integer multiple of 211 are the same angle Therefore jej22quott dt 4lej4n2 em 4161392 em 0 0 TC TC and Ieejzquott dt 4 ej4n2 642 4 671392 1 642 0 leaving the only term in the integral to be T I I T j e WW 2e0 e 12quot dt j 2e dt 2T 0 2T 0 0 Hence Here is the result The effective or rootmeansquare RMS value of vt V0 sinnt 0707 v0 RMS Ilt nT n1T n2T Sinusoidal and RMS Voltages Note that the amplitude or peak value V0 of the sine can be calculated from its rootmean square value vn JEVRMS 141 vRMS IMPORTANT Voltages reported as VAC are rootmean square RM S values not the amplitude of the sinusoidal voltage THREE PHASE ALTERNATl39NG CURRENT Electric power is delivered to industrial users as three phase alternating current which consists of three sinusoidally varying voltages spaced in time by phase angles that differ by 120 degrees radians Each phase is transmitted on a separate line The 1nd1v1dual lines can be used as single phase power or two or three of the lines can be used together as polyphase power Two phases are delivered to residential electrical services Most small motors in home appliances or hand tools run on single phase 120 VAC which is the voltage drop between one and neutral with is at ground potential voltage Large motors such central air conditioning compressors and force air system fan and large resistive loads such as electric stoves and electric dryers use a voltage drop between the two phases Similarly industrial e uipment can be either one two or three phase depending on the power demand and the available supply 16 The gure below shows three phase 120 VAC sine waves where VAC is the rootmean squared RMS voltage The peak voltage is 120 VAC 170 volts Unfortunately it is o en dif th to visualize the relationship between trigonometric functions For example consider the signals at bt and ct below Assume the phase angle 1 of at is zero at VD sinmt Do signals bt and ct have positive or negative phase angles oh and on 200 I I 39 am I00 135 v 00 7100 l 39I I quot N r 200 I I I I I I o 0002 0004 0000 0003 001 00m 00 0010 I It is impossible to tell the sign of a phase angle a steadystate time domain plot of two or more sinusoids Signal bt could lag at by 120 or lead at 240 Likewise ct could lag at by 240 or lead at 120 When we work with three phase alternating current we will assume that voltage at has a phase angle of zero bt lags at 120 by and AC voltage ct lag at by 240 bt VD sinmt7120 ct VD sinmt 7 240 IMPORTANT We conventionally mix angular units when working with AC We MUST use 03 EE and E rad when we evaluate these expressions However the phase angle 1 is sec usually expressed in degrees when we report our results since degrees are easier to visualize than are radians Al ebraic expressions of these trigonometric functions can also be dif cult to visualize For example consider the voltages created by the differences atbt btct and ctat at 7 bt vu meaty sinmt7120 bt7 ct vu 51mm 120 7 mom 7 240 at7 at v sinmt7 sinmt 7 240 The algebraic sum of two sinusoids with the same frequency will be a sinusoid at that frequency What are the magnitudes and phase shi s relative to at bt and ct w atbl quot br7cr 0 clzrl uu 39 39 l l r r l l 0 0002 0004 0006 0008 00l 0012 0014 001 1 The magnitude of atbt btct and ctat may be a surprise It is a common error to presume that the amplitude of is V0 but comparing the plots of at and atbt we see that the amplitude of atbt is greater Do you remember the trigonometric formulae you learned in high school but never used No one does but now is when you could nally use them However with apologies to your high school math teacher it is easier to nd the magnitude sinmt7sinmt7120 as a vector construction using complex exponential unit vectors and Euler s sine formula than it would be to use trigonometric formulae We wish to determine the magnitude of at 7 bt The rst step is to create complex exponentials which represent at and bt The vectors must ve the correct amplitude and phase angle 1 Since we are evaluating steadystate relationships that must hold for any arbitrary time t we are free to choose the time t at which we will draw the vector diagrams The most convenient time is t 0 Phase at has a phase angle of zero so it plots on the positive real axis Positive angles are counterclockwise Phase bt has a negative phase angle since it peaks later in time than does at 00 M alt0gt IVoleJ Re bo Ivole39j120 Complex Exponential Vectors which represent at and bt at t 0 We now subtract the vectors to determine the magnitude of their difference First we invert vector bO j120 j60 m bogt voe Voe 00 aOVoeJ IVol Re 420quot bOV0e j120 Then we add vectors a0 and b0 N o g 160 bO Voe qul 30 Voej0 Re Now the hard part We need to nd the magnitude and phase angle of a0 7 b0 We know the lengths of two of the sides of the triangle formed by a0 bO and a0 7 bO and we know the angle between a0 and b0 There is surely a sophisticated formula which we could hunt for in a math handbook that would yield magnitude and phase angle of a0 7 bO but it is usually faster to use what we know then search for what we may not be able to nd We can create a right triangle with a0 7 bO as its hypotenuse using simple trig V0sin60 The Pythagorean theorem is used to calculated the magnitude of the vector a0 7b0 a07b0 lVUlcos60 zlv lsin60 z Factoring IVU out of the radical W lVUWlcos60 zsin60 z 20 a0 b0 v0 1 cos 60 2 sin60 2 Expanding the first term in the radical 1 cos 60 2 1 2 cos 60 cos2 60 Re arranging a0 b0 v01 2cos60 cos2 60 sin2 60 Using the trigonometric relationship from a unit circle cos2 0c sin2 or 1 a0 b0 v0 12cos60 1 Yields the result a0 b0V0 Line to Line Voltage This is a useful relationship to remember The line to line voltage in this case at bt in a three phase machine is 6 times the voltage of a single phase relative to the reference voltage neutral with ground potential Returning to the time domain plots note that it is difficult to visualize the phase angle of atbt relative to at from these plots Also note that the phase angles of the signals atbt btct and ctat differ from each other and from at bt and ct The phase angle of a0 b0 relative to a0 is easily derived from the vector construction from trigonometry lVolsin60 Re Phase Angle 421 q tanil lv lsin60 tanil sin60 tanilpg j ou ab lVUlcos60 lcos60 15 PHASORS The complex plane vector construction we created to determine the magnitude and phase angle of the voltage atbt is the basis of a phasor diagram The only difference is that the VECTOR MAGNITUDES USED IN PHASOR DIAGRAMS ARE THE ROOTMEAN SQUARE RMS MAGNITUDES strictly for convenience in subsequent calculations The advantage of using the RMS value is that calculation using phasors yield the effective values of voltage or current which is most o en what one needs The term phasor connotes that the angles of the vector construction represent the phase angles of the vector quantities relative to the input voltage THE INPUT VOLTAGE IS ALWAYS ASSUMED TO HAVE A PHASE ANGLE OF ZERO and lies on the positive real axis of phasor diagram Three phase AC is shown in the diagram below as both phasors and as sinusoids The three phasors rotate counterclockwise as time progresses There is one odd aspect of phasor diagrams The phasors represent RMS values of sines but the construction is used to calculate the contributions of the three phases at an instant arbitrarily assigned t 0 But RMS values are effective or average values over an entire cycle This makes no difference since RMS values and peak values of sines are proportional as we derived Phase a increasing time gt Three Phase AC Current shown as Phasors and Sinusoids 23 Wye Y and Delta A Three Phase Connections We will now more formally consider three phase electrical current using phasors Three phase lines can be connected in a Y also spelled as Wye connection or a delta connection The names refer to the con guration of the connections shown below where the lines are phasors which represent the voltage of each of the three phases The voltage drop in the connection is between one phase and neutral which is a common return line used to complete each of the three circuits Consequently a wye connection requires four lines The voltage drop in the delta connection is between two of the phases The vector calculation we performed above found the linetoline voltage between phases at and bt This is a delta connection We determined that the voltage drop between the two phases was larger than the voltage of either phase to neutral by a factor of J3 The delta connection needs only three lines and has a larger voltage drop across a load than does the wye connection for the same phase voltages If the delta connection needs fewer wires and provides a larger voltage drop why use a wye connection to drive a three phase motor Many three phase motors in fact use both types of connections to maximize performance by varying torque or power or to provide two speeds where a wye connection is used for low speed and a delta connection is used for high speeds Motors operated this way are called single winding 7 two speed motors b neutral Y or Wye Connection Delta Connection The Two Types of Three Phase Connections A wye connection is easier to understand and handle analytically A schematic of a three phase wye connected generator is shown below The three AC sources a b and c shown in the schematic are in fact three sets of armature windings within one generator The a b and c windings are spaced around the stationary frame of the generator called the stator such that when the generator s rotor carrying the field winding a DC electromagnet passes over the armature windings in succession the voltages induced in the three windings peak at successively later time This yields three sinusoid which have phase shi s of 120 AC Source C A Three Phase Wye Connection Generator re is an odd aspect of the Wye connection generator schematic Notice that the AC sources are marked with a positive and negative polarity What do polarity markings mean if AC voltage is a sinusoid symmetric about zero The polarity of an AC source refers to the phase shift Reversing the source polarity it is equivalent to inverting the sine wave which is equivalent to a phase shi of180 The neutral line is needed to complete each of the three circuits If you have done any home wiring you know that neutral and ground are the same terminals in a residential electric service Grounding the neutral line assigns it a voltage of zero If we consider just one phase we can visualize the positive terminal of the phase swingng through the sinusoid voltage alternately pushing current into the neutral line and pulling out of the neutral line As it happens when the three phases are connected to the same neutral line if they have a balanced load such as a three phase motor which pulls the same amount of power from each line the neutral line will carry no current At any instant the net current being pushed into the neutral line equals the net current being pulled out of it You will see schematics in which the neutral line is omitted However it must be present for safety It will carry current in the event the load becomes unbalanced It is conventional to analyze three phase circuits by considering just one of the phases and then tripling the resulting current and power calculated For example a three phase generator driving a three motor is shown in the schematic below neutral Three Phase Generator and Motor in Wye Connections Any of the three phases can be considered in isolation but it is convenient to work With phase a since the voltage supplied by phase a has a phase shi of zero neutral Phase 21 windings of motor Phase 3 windings of generator Phase a Windings of a Generator and a Motor in Wye Connections The schematic becomes a familiar ME 352 rst order system When it is redrawn R L D neutral Phase a Windings of a Wye Connection 26 An alternative way to draW the Wye generator 7 Wye motor schematic is to rearrange as Phase a Windings of a Generator and a Motor in Wye Connections Delta Connected Machines Delta connected machines do not have a neutral line 3 phase delta connected generator 3 phase delta connected motor Three Phase Generator and Motor in Delta Connections It is dif cult to visualize the equivalent circuit of a single phase in delta connections since the three phases are referenced to each other rather than to neutral as they are in the Wye 27 connection Fortunately it is easy to convert a schematic from a delta connection to a wye connection for analysis In fact we developed the method when we calculated the line to line voltage between at and bt Our result rewritten with subscripts ab for line a to line b and an for line a to neutral Vab Van van IMPORTANT We must know if a reported terminal voltage is linetoline vab or lineto neutral van in order to create the equivalent single phase circuit In industry the term line voltage means linetoline The term phase voltage means linetoneutral Vline Vphasng Example 1 The phase voltage of a delta connected motor is 460 VAC Determine the equivalent wye connected motor for analysis 460 VAC 460 VAC 460 VAC C We are given the phase voltage or linetoneutral is 460 VAC The voltage dropping across the windings of the motor is the linetoline voltage Consequently we must increase the voltage applied to the equivalent wye connected motor Vline VphaseN3 460 VAC 798 VAC The equivalent wye connected motor is 798 VAC 798 VAC 798 VAC b a neu1ra Equivalent Wye Connected Motor with LinetoNeutral Voltages Example 2 The line voltage of a delta connected motor is 460 VAC Determine the equivalent Wye connected motor for analys1s 460 VAC 0 I b a 460 VAC 460 VAC In this case the applied voltage of 460 VAC is reported as linetoline The voltage Which drops across the windings of a motor in a delta connection is the linetoline voltage We do not need to adjust the magnitude of the applied voltage if we show the voltage applied to equivalent Wye connected motor as linetoneutral 29 450 VAC 460 VAC 460 VAC a neutral Equivalent Wye Connected Motor with LinetoNeutral Voltages We do need to adjust the voltage applied to the equivalent Wye connected motor if we show the voltage as linetoline A linetoline voltage drop is greater than a linetoneutral voltage drop so we must increase the applied voltage linetoline voltage to have the correct linetoneutral voltage vim vphase J5 460 VAC 798 VAC 798 VAC C I b a 798 VAC 798 VAC neutral Equivalent Wye Connected Motor with LinetoLine Voltages COMPLEX IMPEDANCE and COMPLEX ADMITTANCE 30 The complex impedance method can reduce the effort required to derive the transfer function of a dynamic system The essence of the complex impedance method is to l transform the elemental equation using the Laplace transform and then 2 rearrange the equation into the ratio of the across variable over the through variable The example to remember is the complex impedance of a resistor because it is the source of the nomenclature 9vt itR Vs IsR VS Is R Capital Z is used as the symbol for complex impedance The term complex impedance refers to the transformation to the complex domain and to the generalized resistance or impedance to power ow in a system since resistance impedes the ow of current in a circuit In the case of the resistor the resistance in the time domain equation is the same parameter as the complex impedance in the complex domain However the elemental equations of energy storage elements are transformed from differential equations to algebraic equations The transformation of differentiation with respect to time leads to the Laplace variable s being included in the complex impedance The complex impedance ofa capacitor is d t it 9 0 KS CsVs Vs l Is Cs C The complex impedance of an inductor is d 39 t vt 53 L Vs LsIs VS Is LSEZ L 31 We can apply the reduction techniques for linear graphs or circuits which use time domain elemental equations and yield differential system equations directly to reductions which use complex impedances and yield transfer functions Reductions using complex impedances can be more algorithmic because Laplace transforms of differential elemental equations are algebraic equations allowing separation and collection of the input and output variables The reduction linear graph or a circuit in which the elemental equations are expressed as complex impedances can be performed as a graphical technique The method is to successively combine elements in either parallel and or in series and replace them with single equivalent elements The process is repeated until the network is reduced to the source element and a single complex impedance When a system is reduced to a source and a single equivalent impedance the equivalent impedance is called the driving point impedance since it is the impedance the source driving the system feels A driving point impedance is used to determine the power the source must provide to drive the system We derived the relationships to replace resistors in parallel or in series with single equivalent resistances in ME 352 The derivation is repeated below using the complex impedances of the resistors as a review Recall that the equivalent element carries the same through variable ow and has the same across variable drop across it as the portion of the network it replaces Reduction of Impedances in Series to a Single Equivalent Impedance Impedances in Series Single Equivalent Impedance KEY R Z Eqmv 1 2 3 1 3 Continuity 115 12 S E 15 15 Compatibility V s Viz s V23 5 Vus Viz 5 V23 5 V S R E z R E z 13 R E z Elemental 11 S 1 1 12 S 2 2 15 5qu 5qu Reduction Vus ms ms 2 15 was zR212lts ZEWVKS lesZZls ZEqvaS Zl Z2 15 32 AL Equiv enes This reduction is applicable to any complex impedance not just the impedance of resistors It can also be generalized to any number of compleX impedances in series Complex impedances in series Z Z1 Z2 Zn The reduction of compleX impedances in parallel to a single equivalent impedance is as straight forward It is generally more time consuming though because the result must be placed over a common denominator and inverted to be useful Reduction of Impedances in Parallel to a Single Equivalent Impedance Impedances in Parallel Single Equivalent Impedance Z 1 Z Equiv Z2 Continuity 11S 12 S E 15 15 Compatibility V125 Vlz S V12 S V12 S VuS V S V S R Ez LR EZ LR EZ Elemental 118 1 1 I S 2 2 15 Equiv 1aqu Reduction 11s12s Is Vlzs V12S VIZS V12 siij V12 8 Z1 Z2 Equiv Z1 Z2 ZEquiv 33 Equiv 1 2 parallel This reduction can also be generalized to any type and number of complex impedances in parallel l l 1 Complex lmpedances In series Zparallel Z1 Z2 Zn The dif culty comes in inverting the expression to nd Zparallel rather than The result parallel for two elements in parallel is easy enough to invert l ii l Z2 Z1 Z1Z2 Zmlll ZIZ2 Zparallel Z1 Z2 Zparallel ZIZZ Z1Z2 Z1Z2 p Z1 Z2 The result for three elements in parallel is more dif cult to invert 1 i L i 1 i i i Zparallel Z1 Z2 Z3 Zparallel Z1Z2Z3 Z1Z2Z3 Z1Z2 Z3 1 Z2Z3Z1Z3Z1Z2 Z 2 Zparallel ZIZZZS pamllel Z2Z3 Z1Z3 ZIZZ There is a hidden cost to using the complex impedance technique Our objective is to create a transfer function We need a proper fraction to create a transfer function If Z1 Z2 and Z3 are ratios then menel is an improper fraction which must be cleaned up The algebra can be very time consuming and error prone It can get so involved that if there are four or more elements in parallel then it is common to use complex admittances rather than complex impedances to simplify the reduction Complex Admittance Complex Admittance is the inverse of complex impedance The term admittance applies to the ow of the through variable Higher admittance for a given across variable drop results in higher ow through the element Admittance is the inverse of impedance If VS Z E Impedance then its inverse is 18 i E Y E Admittance Is Vs Z 34 Capital Y is the symbol for complex admittance Again recalling that impedance has the functional form of resistance VtitR R ZR E 1t 15 Where voltage is an across variable and current is a through variable 1 7 ls Y 7 R zR Vs Physically the complex admittance of a resistor is its conductance i R he units of conductance are Qquot Which are called siemens in SI and mhos by most engineers no kidding Admittance can be very convenient since we o en Wish to calculate the through variable current ow from an across variable voltage source to size system components Because admittance is the inverse of impedance the relationships for combining parallel and series admittance elements are the inverse of impedance Speci cally it is easy to combine adrnittances in parallel Reduction of Admittances in Parallel to a Single Equivalent Admittance Single Equivalent Admittance v1 Y Admittances in Parallel Equiv Con nuity Illts12lts21s 15 Compatibility Vlzs 25 Vl25 VHS Elemental 115 1 1 Y 125 7 1 7 1 Y 15 7 1 7 l 7 i Y ulV ms R Z Vzs R Z Z Vus Rm ZEW 1 35 Reduction 11s12s Is Y1V12s Y2V12s YEquiVVlzs Y1 Y2 V12S YEquivV12S Y Y1Y2 Equiv arallel The drawback using admittance is working with elements in series since admittance is the inverse of impedance There is one important restriction on the use of complex admittance Both the complex r and J quot of 39 J39 39J 39 elements are physically realizable They can be expressed using either differential or integral causality However there are system transfer functions for which the admittance is not physically realizable 1 Example A springmassdamper system driven by a force source is shown as a schematic and a linear graph below Derive the transfer function for the velocity of the mass using complex impedances or complex admittances Spring Mass Damper System Linear Graph Because this system has three elements in parallel it will be easier to work with complex admittances than with complex impedances Equation List Continuity Ft FM t FK t Fb t Ft FMt FKt Fbt Fs FM 5 FR 5 E S Compatibility v1gt v1gt E Vt trivial Vt Vs Elemental equations expressed as complex admittances m 9FMtJLM dt FM s sMVs YM 5M Lug 7 FKs 7 K 9i dt if 9mm SFK S KWS VS 7 YR 7 Fbs 9m 0 7 9bvt F45 7 bws Vs 7 Yb 7 Reduction The three parallel elements expressed as complex admittances can be combined into a single admittance With a surnma 39 9 YMYKYb Ydn E Y Vs mg pmm vmg pu mt Transfer functions are Gum um Gs Fs is the input and Vs is the output so Inputs we Will need to invert this relationship GS FS deiving oint l l l Gs Yg yigs YM YK Yb sM b om S Unfortunately as noted above for the transfer function to be useful it must be a proper fraction and often most of the work in a complex impedance or admittance problem is in clearing improper fractions S S Gs s2MKsb Ms2bsK sM b S COMPLEX IMPEDANCES AS PHASORS AND AC CIRCUIT CALCULATIONS We will use phasors for AC circuit calculations We will represent complex impedances and complex admittances as vectors so that we can perform graphical computations To eliminate possible confusion with admittance jY is not used to identify the imaginary vertical axis in a complex plane Unfortunately the veltical imaginary axis is identi ed as jX which takes some getting use to or alternatively by its units which also take some getting use to The units imaginary axis are jQ and are called reactance to sound similar but distinctive from the units of the real which are resistance Although reactance appears as an imaginary resistance physically reactance represents energy stored in inductance or capacitance Complex impedances and admittances can be used to derive transfer functions for transient analyses eg a step response or frequency response analyses Phasors are only used with the frequency response analyses and specifically AC circuit calculations A complex 39 J or J 39 is J as a phasor by evaluating the impedance or r 1 rad adm1ttance us1ng s 103 where 03 1s the frequency of the AC c1rcu1t 377 and then sec plotting the result as a vector In order to make the result easy to plot the imaginary number j is cleared from the denominator if necessary by multiplying the phasor by 4 yielding the J product l in the denominator and j in the numerator Impedances and Admittances of Electrical Elements Resistance Capacitance Inductance V5 Vs 1 W REZ E LsEZ Impedance 15 C 15 C5 C 15 L 15 1 1 15 1 15 1 1 GEY csEY EY Admittance S R Z R R VS Z0 0 VS LS Z L 1 Phasors 39 tne 39 39 39 quot Electrical Elements Resistance Capacitance Inductance VJ 112zc VJ L JEZc MZJLmEZL 1Jm 1Jm JCm Cm 1Jm 19m iGLEYR 1Jf jcmigyc 1J9JLEYL Vm R ZR Vm ZC Vm JLm Lm ZL jQ reactance Q resistance 1 L C ij 03C Impedances of Electrical Elements Plotted as Phasors Before using phasors we Will review vector arithmetic in a complex plane Complex numbers are represented as complex exponential unit vectors scaled by the magnitude of the number Vector addition and subtraction are performed as they are in a real plane Multiplication and division are performed using operations derived from the laws of exponentials Given complex numbers KAe 1 and Ble1B Their product equals the product of their magnitudes at an angle equal to the sum of their angles KEIAe BIe1B gt K lAllBle e1B gt K IA BIeJM XE ABza3 Their ratio equals the ratio of their magnitudes at an angle equal to the difference of their angles EJAISW g m i eiw IAIZQ E lBle1B B lBl e1B A Vector Arithmetic in a Complex Plane Vector Addition Vector Subtraction Vector Multiplication XE ABza3 Example Phasor calculation of current in an RC circuit Example A 120 VAC circuit is shown below Determine the voltage across the capacitor and express the input and output voltages as phasors Where R 10K Q and C 1 HF 41 We Will use phasors to calculate the current ow in an RC circuit driving by an 120 VAC source The input frequency w is 60 Hz or 377 sec Asinnt AGjwsinwt 1 Frequency Response Block Diagram We are told that the input voltage is 120 V AC Is that the peak voltage or the root meansquare RMS effective voltage Unless otherwise indicated AC voltages are RMS values AC circuit calculations are frequency response calculations Where the amplitude of the variables are expressed as RMS values rather than peak values The rst step is to derive the transfer function of the circuit The transfer function Gs evaluated at the excitation frequency by substituting for s jm is an operator since it operates on the input to yield the output We Will call the transfer function evaluated at the excitation frequency the frequency response operator We can use the frequency response operator either analytically or graphically in a phasor diagram to calculate the frequency response of the circuit Vinpul RMS vulgagllcomlm AC Frequency Response Operator The rst step is to derive a transfer function to use in the frequency response equation Equation List Loops vlg EvvlZ ng Nodes dvZg Elements vlz R1R 1C C 42 Energy E System 1 Ec ECEV2 Reduction Input Vlg Output Vzg VV12V2g VRiRV2g VRiRV2g VRicv2g dV2g VRC dt V2g System Equatlon UnitCheck V it R C i 1 H i 1 M M Rod v vXi tiv MMM dt 25 i V t Form the Transfer Function dV2 9v RCg 9523 VsRCsVzgsVzgs Outputs Vz g S VS RC5 1V2g S Inputs Vs RCs1 GS Calculate the frequency response operator the transfer function evaluated at the excitation frequency and represent it as a phasor The complex impedance is the transfer function evaluated at AC frequency 377 rad The magnitude of the complex impedance is sec 1 1 L 1 lGltJmgtl ch1 10103Q110 6Fj3771 377j1039256 The angle 0fthe phasor is 4Gjoa 4 41 z377j1 0 tan391 131rad 75 RCJOJ1 1 30377 1 1j377 AC Frequency Response Operator of the RC Circuit Expressed as a Phasor Calculate the Output Voltage Vzg We now use the AC frequency response operator to calculate the output variable The analytical technique uses the frequency response equation th lvllGjmlsinwt The steadystate response to sinusoidal input is vZg tss 120 VAC0256sin377t775 vzgtss 307 VACsin377t775 IMPORTANT Note the convention to use mixed units The phase shi is in degrees and ra the frequency IS 1n This expressron cannot be evaluated unless the units match sec The graphic method uses a phasor diagram of the voltage input and the frequency response opera or 120VRMS V input 750 Re 997 0 1 60377 1 377 Frequency Response Operator Phasor Diagram Showing AC Input Voltage and the Frequency Response Operator Multiplying the vectors vinpm and G0 377 using the relationship KEA BZXB yields cmw l120025640775 30747 75 Q 120 Vinpm 75 Re 29 output Phasor Diagram Showing Input and Output Voltages AC Calculations using Complex Impedance and Phasors When we connect a machine to a power line we know the line voltage We do not know how much current the machine will draw from the line To size the conductors and to calculate the power drawn by the machine we need to calculate the current 45 Example The schematic of a motor driven by 120 VAC is shown below Determine the current drawn bythe motor L 0001 H and R 01 Q R 2 L n neutral he approach we developed in ME 352 of 1 deriving a system equation 2 taking its Laplace transform 3 creating the transfer function and then 4 calculating the frequency response always works However if we need a transfer function we save effort by expressing the elemental equations as Laplace transforms rather than as differential equations AC circuit calculations are o en easier to work by taking the elemental equations one step further and expressing them as complex impedances Impedance E Z 2 15 If the schematic is simple as this one is then it is possible to determine the driving point impedance used to determine the current provided by a voltage source either analytically or graphica y We will derive the transfer function two difference ways First we will reduce the circuit to a driving point impedance analytically The second derivation will use phasors to sum the complex impedances of the resistor and capacitor of the circuit The two derivations will yield the same result Equation List Continuity it iR t 2 it 9 iR t 1s IR s 1R t 1L t 3 0 9amp1 0 1RS1LS Compatibility vt v12 t v2n t 9 vt 9 V12 t V2n t VS 112 S V2quot 5 46 Elemental Inductor V2n t L V2n t V2n s leL s V 2quot S Ls E ZL complex impedance of inductor IL 5 Resistor V12 t RiR 9Vlz t 9R iR V12 s RIR s V R E ZR complex impedance ofresistor IR 5 1 2 Energy I IL EL1L Reduction 1 Direct reduction without simpli cation of the schematic by the use of equivalent impedances Input Vs Output Is Vs Vus v2quot s Vs RIRs LsILs Vs RIs les W3 2 Zd RLs Is P Evaluate the driving point impedance for the values R 01 Q and L 0001 H VQWEZWRMm APQMKQWHDGWEQAMQpSWQ lD sec Reduction 2 Simpli cation of the schematic by the use of equivalent impedances 391 neutral The complex impedances of the inductor and resistor are in series and can be added to form an equivalent single impedance ZLZRsLREZdp The subscript dp on the equivalent impedance stands for driving point and is the impedance of the system as seen by the source driving the system Z dp neutral Having reduced the system to the voltage source and a single impedance E4 sLR 15 p This yields the same driving point impedance When evaluated for R 01 Q and L 0001 H W Ezdp 11ij zap 0lQj0377Q 10m Reduction 3 Vector summation of the series impedances We perform the summation ZLZR sLREde graphically The resultant vector is the driving point impedance Recall that resistance inductance and capacitance are represented by phasors on the positive real axis the positive imaginary axis and the negative imaginary axis respectively jQ reactance The drian point impedance is a vector inclined 75 from the positive real axis With a magnitude of039 j reactance ZR R 2 M 2 resistance 49 Current Calculation We have the driving point impedance VS Is P where Vs is the source voltage and Is is the current drawn from the source by the system We know the source voltage is 120 VAC and we wish to calculate the current E 15 Z dp Using the results of Reductions 1 and 2 Reductions l and 2 yielded the driving point impedance in Cartesian coordinates de 0lQj0377Q Dividing the input voltage by the driving point impedance yields 120 VAC 1 39m J 01oj0377o We need the current in the form of a single complex number not the ratio of a real and a complex number We have two options for clearing fractions The rst technique is to multiply the numerator and denominator by the complex conjugate of the denominator 120 VAC 019 03779 IJw 0 lQj0377Q 0 lQ j0377Q Expanding and collecting like terms 120VQ j452VQ K1092 0019 0377 Q 120VQ 39452VQ 1Jm J2 0152 Q 1j03 789 j297789A j297A 50 The second technique of clearing fractions is to express the numerator and denominator as the products of a magnitude and an exponential unit vector and then divide them 120 VAC INIejth lNl WW 1003 0IQj0377Q Dejl D M The input voltage is ALWAYS assumed to have zero phase shift It lies on the positive real axis so no calculations are needed to express it as a magnitude time a complex unit vector 120 VACEIZO vRMS Nej N 120 VRMSejO The denominator is a complex number We need to determine its magnitude and phase angle 0IQj0377Q Dej rgt D 019 0377 22 039Q 0377 tan 1 75 D 01 J 0 IQj0377 2 039 gem We can now calculate the current drawn by the system 120 VAC 120 VRMSejO 120 vRMS Kw 0IQj0377Q 039 gem 039 Q 475 1j03 308ARMS e Using the results of Reduction 3 We have the driving point impedance expressed as a phasor which allows us to perform the calculation of the current 1003 by vector division in a complex plane Vjoa I D J Z dp I120VRM5 Vsource Re We see that the current lsource follows or lags the input voltage Vsome This results from the inductance of the circuit A magnetic eld exists around a moving charge or current There is magnetic energy stored in the eld The eld cannot reach its nal state instantaneously without an in nite power ow Consequently there is a time dependent growth of the eld and of the current The voltage applied to the circuit rises before the current ow through the circuit AC Power Calculation The apparen power also called total power is the product of current and voltage phasors Apparent Power V IVe v1e lVllllem39ew VIe quotvquot Since the AC source voltage has no phase shi V 0 Apparent Power V I VIe quot Apparent Power is expressed in units of volts amperes 2 VA IMPORTANT Apparent Power 2 Volts amps 2 VA I watts Huh Isn t a wattiv Well no not in AC circuits Total PowerEi v where total power includes power that is either dissipated as heat or can be transformed or transduced This power is expressed in watts Total power also includes power which ows 52 into inductive and capacitive energy storage Because the AC source cycles continuously Im the inductive and capacitive energy ows in and then back out each cycle and then back out the other part of the cycle This power is not watts Phasors to make this easier to visualize Vsouree 1 71 0 Re HANG Isource Vsource Power tvv components of Apparent Power vector projected on the real and imaginary axis are power or active power and reactive powe jo Reactive Power var p Active Power watts Power VA Vector Decomposition of Apparent Power into Active Power and Reactive Power e real component P is active power Active power is power that is either dissipated as heat in the system as heat transduced from electrical to mechanical power in a 53 motor or transformed from one Ac eurrent and voltage to a another m an eleetneal transformer Excludlng the amount drssrpated as heat aetwe power ls what we ean use as meehameal power Aetwe powerhas umts ofwatts The lmaglnary eomponent of the apparent power veetor jQ ls ealled reactlve power Reaetwe power represents the flow of energy rnto erther the magneue eld of an rnduetanee or the eleetne eld of a eapaertanee Reaeuve power ls real power but rt ls not eleetrre elds of an Ac erreurt or maehrne flows m and then back out eaeh eyele of the usoldal eurrent Reaetwe power ls genulne power Reaetwe power ls neeessary for a t rt run r le t e motor39s torque ls the result of the magnetre elds of the stauonary androtatrng eomponents attempung to allgn Reaetwe powerls lmaglna quot m the sense that rt eannot be taken out of the motor as meehameal power on the output shaft Reaeuve power has umts ofvar lower ease whreh stands for uoltsampsreaedve Eleetne motors speerfreatrons lrst a power faetorquot The power faetor ls eose the t n r 100cos6 to elrmrnate the deermal fraeuon The produet of the power faetor and the apparent ower veetor ls the projection of the apparent power veetor onto the real axls The real eomponent of apparent power ls the aeuve power Example AC Ppwer Calculatinns Example 1 A general purposequot slngle phase Ac 10 hp motor and rts speerfreatrons are shown below caleulate the motors power Baldnr single Phase AC 10 Hp General Purpnse Mntnr CL3712T Baldor Single Phase AC Motor CL3712T Power Factor We are given the voltage the Full Load Amps current and the motor s power factor Calculate the apparent power Apparent Power 230 VAC41A 9 430 VA Calculate the power Multiply the apparent power by the power factor yields the projection of the apparent power vector on to the real axis Apparent PowerPower Factor 9 430 VA093 8 770 W Power Reduce by the full load efficiency Note that the Full Load Efficiency is calculated after projecting the apparent power vector onto the real axis Calculated PowerEf ciency 8 770 W084 7370 W Available Power The motor is rated as 746 KW so our calculations are accurate with the precision of the data To represent these results as phasors we need the angle between the apparent power vector and the real axis 55 9 005 1 power meter cos 1 2 215 100 100 jg Reactive Power var r P 87 kW p Active Power watts Apparent Power VA Example 2 A general purpose three phase delta connected 100 hp motor and its speci cations are shown below Baldor Single Phase AC Delta Connected 100 hp General Purpose Motor M2583T Baldor Three Phase Delta Connected AC Motor M2583T 1 40C AMBCONT IMPORTANT NOTES 1 The Full Load Amps current given in a specification of a three phase motor is for one leg or electrical supply line of the three phases 2 The rated voltage given in the specification or on the nameplate of a three phase motor is the linetoline voltage Linetoline voltage is measured between two of the three phases ie vab vbc vca The voltage needed for the power calculation is the linetoneutral voltage The linetoline voltage is a factor 5 greater than the linetoneutral voltage L J3 Repeating the above calculations using the data for the three phase motor V rated Vlinertorneutral Calculate the apparent power for line a J3 Apparent Power 230 VAC238 A 316 kVA Calculate the power Multiply the apparent power by the power factor to yield the projection of the active power vector on to the real aXis Apparent PowerPower Factor 316 kVA083 262 kW Active Power jQ Reactive Power var I 262 kW P 34quot Power watts 0quot I Q I 9 76D4 E jQ v Apparent Power VA Reduce the active power by the full load efficiency to the sha power Active PowerEf ciency 262 kW094l 246 kW Sha power from one phase The average sha power from the other two phases is the same The average sha powers over a cycle add as scalars so the sha power is three times the average sha power from one phase 3246 kW 738 kW The rated power is 746 kW Our calculation within approximately one percent You will see the power calculations for three phase motors performed using a formula which simpli es the calculation but removes some of the meaning Rated Power kW Rated VAC Full LoadAmp Power Factor Ef ciency To limit errors it is important to see that the factor J5 results from correcting from lineto line voltage to linetoneutral voltage and from multiplying by three to sum the average sha power from all three of the phases 58 Figures drawn during preparation of powerpoint lecture to be added to text follow

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