Calculating And Measuring Power In Three Phase Circuits

Transcription

PDHonline Course E344 (6 PDH)Calculating and Measuring Powerin Three Phase CircuitsJoseph E. Fleckenstein, P.E.2013PDH Online PDH Center5272 Meadow Estates DriveFairfax, VA 22030-6658Phone & Fax: 703-988-0088www.PDHonline.orgwww.PDHcenter.comAn Approved Continuing Education Provider

www.PDHcenter.comPDH Course E344www.PDHonline.orgCalculating and Measuring Power in Three PhaseCircuitsJoseph E. FleckensteinTABLE OF CONTENTSSection – DescriptionPage1.Introduction 52.Calculating Power in Single Phase Circuits 63.2A.Single Phase Voltage and Single Phase Current 62B.Single Phase Power 82C.Kilowatts, Lumens, BTU’s and Horsepower 13Calculating Power in Balanced Three Phase Circuits 143A.General 143B.Calculating Power in Balanced Three Phase Wye Circuits 183B1.General 193B2. Calculating Power in Balanced Three Phase Wye Circuits - Resistive Loads 203B2a.Using Phase Parameters 203B2b.Using Line Parameters 203B3. Calculating Power in Balanced Three Phase Wye Circuits – Inductive or Capacitive Loads 213B3a.Using Phase Parameters 213B3b.Using Line Parameters 213C.Calculating Power in Balanced Delta Three Phase Circuits 213C1. General 213C2. Calculating Power in Balanced Three Phase Delta Circuits - Resistive Loads 223C2a.Using Phase Parameters 223C2b.Using Line Parameters 223C3. Calculating Power in Balanced Three Phase Delta Circuits – Inductive or Capacitive 233C3a.Using Phase Parameters 233C3b.Using Line Parameters 234.Calculating Power in Unbalanced Three Phase Circuits 234A.General 234B.Calculating Power in Unbalanced Three Phase Wye Circuits 23 Joseph E. FleckensteinPage 2 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.org4B1. General 234B2. Calculating Power in Unbalanced Three Phase Wye Circuits - Resistive Loads 244B2a.Using Phase Parameters 244B2b.Using Line Parameters 254B3. Calculating Power in Unbalanced Three Phase Wye Circuits – Inductive or Capacitive Loads 264B3a.Using Phase Parameters 264B3b.Using Line Parameters 274C.Calculating Power in Unbalanced Three Phase Delta Circuits 294C1. General 294C2. Calculating Power in Unbalanced Three Phase Delta Circuits - Resistive Loads 294C2a.Using Phase Parameters 294C2b.Using Line Parameters 304C3. Calculating Power in Unbalanced Three Phase Delta Circuits – Inductive or Capacitive 304C3a.Using Phase Parameters 304C3b.Using Line Parameters 314D.4D1.4D2.4D3.5.General 35Using Phase Data 36Using Line Data 36Measuring Power in a Single Phase Circuit 365A.6.Calculating Power in a Three Phase Circuit with Mixed Wye and Delta Loads 35General 36Measuring Power in Three Phase Three Wire Circuits 376A.General 386B.Using One Single Phase Wattmeter to Measure Power in a Three Phase Three WireCircuit 406E.Using a Three Phase Wattmeter to Measure Power in a Three Wire Three PhaseCircuit 467.Measuring Power in a Three Phase Four Wire Circuit 487A.General 487B.Using One Single Phase Wattmeter to Measure Power in a Three Phase Four WireCircuit 487C.Using Three Single Phase Wattmeters to Measure Power in a Three Phase FourWire Circuit 497D.Using a Three Phase Wattmeter to Measure Power in a Three Phase Four WireCircuit 508.Power Analyzers and Power Quality Meters 51 Joseph E. FleckensteinPage 3 of 88

www.PDHcenter.com9.PDH Course E344www.PDHonline.orgSummary of Symbols and Equations 539A.Symbols 539B.Equations 5410. References 58Appendix A - Demonstration that Equations 105 is the equivalent of Equation 106 60Appendix B- Table of Computed Values of Instantaneous Power for Example 1 62Appendix C - Demonstration that Total Power of an Unbalanced Delta Circuit isEquivalent to that of an Assumed Wye Circuit 64Appendix D – Proof of Two Wattmeter and Three Wattmeter Methods. 70Appendix E - Table of Typical Three Phase Power Values vs. ωt for One Cycle 79Appendix F – Equations for Calculating Currents 82 Joseph E. FleckensteinPage 4 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.orgCalculating and Measuring Power in Three PhaseCircuits1.IntroductionThe generation and transmission of electricity is commonly accomplished bymeans of three phase circuits. Although electrical service to residentialbuildings in the USA are exclusively by single phase circuits, electricalservice to commercial and industrial users are usually by means of three phasecircuits. The distribution of electrical power within these buildings is mostlyaccomplished by means of three phase circuits. And, all larger motors arealmost exclusively of the three phase type. In short, three phase circuits arethe most practical means of dealing with larger amounts of electrical power.When planning a new three phase circuit, it very often becomes necessary foran engineer to determine the power those circuits will require. Likewise, theneed may arise to determine the power consumption of an existinginstallation. Sometimes, these values may be established by calculation. Onthe other hand, it sometimes becomes necessary to take measurements inorder to determine the power that is being used by an existing installation.In the case of an existing installation, calculations and measurements go handin hand. If measurements are to be made, a person must first know whatmeasurements are to be made and, secondly, how are field measurementsconverted to actual power values. A number of conditions have a bearing onthe subject. There are delta connected loads and wye connected loads; thereare three phase three wire circuits and three phase four wire circuits. And, ofcourse, there are balanced loads and unbalanced loads. This spectrum ofpossible scenarios can be confusing to a person who may not regularly dealwith three phase power.This course treats the subject of three phase power in detail and in a mannerthat a reader, well experienced in three phase circuits or otherwise, will find Joseph E. FleckensteinPage 5 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.orgeasy to follow. The course considers all of the possible types of three phasecircuits - balanced, unbalanced, three wire, four wire, wye and delta circuits.The course is arranged so that power values may be determined fromcalculations or measurements and from either phase parameters or lineparameters.Numerous diagrams and examples are used in the course to explain theconcepts of the course in simple and unambiguously terms. In the spirit ofpresenting subject matter in easily understood concepts the course avoids theuse of complex variables and polar notation. Rather the more simple methodof vector algebra is used.2.Calculating Power in Single Phase CircuitsTo better understand three phase power, a person would be well advised tofirst review and understand the principles applicable to single phase power.After all, a three phase circuit is essentially a combination of three separatesingle phase circuits which happen to have peaks and valleys separated by aperiod of time. Following is a brief review of the principles involved in singlephase power.2A.Single Phase Voltage and Single Phase CurrentIn general, the instantaneous voltage in a single phase AC electrical circuitwith respect to time can be expressed by the relationship,vi (vPK) sin ωt Equation 101where,vi instantaneous value of voltage (volts) ivPK peak value of instantaneous voltage v (volts)ω 2πf (radians)f frequency (hz)t time (sec)Similarly, instantaneous current can be expressed as,ii iPK sin (ωt θSP) Equation 102 Joseph E. FleckensteinPage 6 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.orgwhere,ii instantaneous value of current (amps)iPK peak value of instantaneous current ii (amps)θSP angle of lead or angle of lag (radians) of current with respect to voltagein a single phase circuit. (The subscript “SP” designates “single phase.”)for a lagging power factor, θSP 0for a leading power factor, θSP 0The trace of thevoltage and current ina typical electricalcircuit with a laggingpower factor is shownin Fig. 1. In a circuitthat is said to have alagging current it isthe current that lagsvoltage.The trace of thevoltage and current ina typical electricalcircuit with a leadingpower factor is shownin Fig. 2.Commonly, voltageand current areexpressed as afunction of time inrms values, Joseph E. FleckensteinPage 7 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.orgV(t) V sin (ωt) Equation 103I(t) I sin (ωt θSP) Equation 104where,V rms voltageI rms current(Note: Equations 101 and 102 are accurate expressions for instantaneousvoltage and instantaneous current with respect to time. Equation 101 can beused to compute the precise value of voltage at a selected value of time (t).Likewise, Equation 102 will compute the precise value of current at a selectedvalue of time (t). On the other hand Equations 103 and 104, while convenientfor calculating power and other electrical properties of a circuit, do notprovide precise values of voltage and current at selected values of time.Equations 103 and 104 are not used in this course for calculatinginstantaneous electrical properties.)The commonly recognized relationships between instantaneous values of rmsvoltage and rms current are,V (1/vi iPK, andI (1/i PK2B.Single Phase PowerPower is defined as the rate of flow of energy with time. In the MKS (MeterKilogram-Second) system of units, which today is more commonly called theSI (Système International) system, the unit of the flow of electrical energy iscalled the watt. One unit of energy in the SI system is the joule. One jouleflowing for one second is one watt. In other words, 1 watt-second 1 joule.Of course, the more commonly used measure of electrical energy is thekilowatt-hour which would be the energy equivalent of one kilowatt flowingfor one hour. The watt has been the standard unit for the measure of electricalpower and it appears the use will continue well into the foreseeable future.However, some of the other units of measure associated with the use ofelectrical power are changing. Common uses for electrical power have been Joseph E. FleckensteinPage 8 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.orgfor lighting, to drive electrical motors and for heating. All of these usages arebeing subject to changes in the units used to describe the particular end use.Instantaneous power in a single phase circuit is given by the expression,iiiP v i (watts)From Equation 101,vi (vPK) sin ωtand from Equation 102,ii iPK sin (ωt θSP)Therefore,iP [(vPK) sin ωt] [iPK sin (ωt θSP)], oriP (vPK)(iPK) (sin ωt) [sin (ωt θSP)]Since, sin (ωt θSP) sin ωt cos θSP cos ωt sin θSPiP (vPK)(iPK) (sin ωt) [sin ωt cos θSP cos ωt sin θSP]iP (vPK)(iPK) (cos θSP) (sin2 ωt) (vPK)(iPK) (sin θSP) (sin ωt) (cos ωt) Equation 105iThe equation for instantaneous power (P ) is often presented in a differentform. One common version is:iP [(vPK)(iPK)/2] cos θSP – [(vPK)(iPK)/2] cos (2ωt θSP) Equation106(Reference #1) As demonstrated in Appendix A, Equations 105 and 106 areequivalent equations although each is stated in a different form.Example 1It will be informative to plot the computed values of Equation 105 for onecomplete cycle (ωt 0º to ωt 360º) of the variable ωt for a typicalsingle phase application. The plot provides a pictorial view of theinstantaneous value of power as a function of time. In the way ofillustration, consider a case having the following parameters:Vrms 480 VAC, single phasevPK 480 678.82 VACIrms 10 ampsiPK 10 14.14 amps Joseph E. FleckensteinPage 9 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.orgPower Factor 0.70, lagging & cos θ 0.70, and θ –45.57ºEquation 105 states,iP (vPK)(iPK)(cos θSP)(sin2 ωt) (vPK)(iPK)(sin θSP)(sin ωt)(cos ωt)Let,2A (vPK)(iPK) cos θ sin ωt, andB (vPK)(iPK) sin θ sin ωt cos ωtiP A B2A (480) (10) (0.7) sin ωtA (6720.00)2sin ωt, andB (480) (10) (sin –45.57º)sin ωt cos ωtB (9600) (–.7141) sin ωt cosωtThe computed valuesof Function A andFunction B in thisexample are containedin Appendix B. Ascaled plot of thevalues of Function Ais shown in Fig. 3. Ascaled plot of thevalues of Function Bis shown in Fig. 4 anda plot of Function Aplus Function B isshown in Fig. 5. Joseph E. FleckensteinPage 10 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.orgIt is interesting tonote that in Fig. 5,which describespower flow in atypical single phasecircuit with a powerfactor of 0.70, powerflow has two largepeaks in the positivedirection and twosmaller valleys in thenegative direction. This power flow will be compared below to power flow ina three phase circuit which, as demonstrated below, is drastically different.According to Equation 105, the instantaneous power in a single phase circuiti2is: P (vPK)(iPK) (cos θSP) (sin ωt) (vPK)(iPK) (sin θSP) (sin ωt) (cos ωt)Generally speaking the average power is of the most interest. To determinethe average power a single cycle of ωt is considered, i.e. from the value ofωt 0 to ωt 360º. To determine average power, the net power over a singlecycle is determined and then divided by the time of one period.Let T the time for a single cycle (from ωt 0º to ωt 360º). The net powerin the period is determined from integration of the equation for instantaneouspower (Eq. 101) with respect to t. Let,iP A(t) B(t), wherebyA(t) (vPK)(iPK) cos θ sin2ωt, andB(t) (vPK)(iPK)sin θ sin ωt cos ωt A(t) (vPK)(iPK) cos θ sin2ωt dt (vPK)(iPK) cos θ sin2ωt dt A(t) (vPK)(iPK) cos θ [t/2 – (sin 2 ωt)/4ω], evaluated from t 0 to t T. Joseph E. FleckensteinPage 11 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.orgThe integral is evaluated from t 0 to t T, (with value “T” occurring atωt 360º) A(t) [(vPK)(iPK) cos θ]{[T/2 – (sin 2 ωT)/4ω] – [0/2 – (sin 0)/4ω]} A(t) [(vPK)(iPK) cos θ] (T/2)The average value of A(t) throughout the period is:P [(vPK)(iPK) cos θ] [T/2] / T [(vPK)(iPK) cos θ] / 2It will be apparent that the contribution of B(t) to the value of the integral ofiP throughout the period t 0 to t T is B(t) 0. This is the case since halfof the function is positive during the period under consideration and half isnegative, the net result being that the total contribution throughout the periodis zero.Voltages and currents are commonly used in rms (root mean squared) values.By definition,V vPK /, andI iPK /And,P [(vPK)(iPK) cos θ] /2Substituting rms values for peak values of voltage and current gives:P VI cos θ Equation 107Where,P power (watts)V potential (rms voltage)I current (rms amperage)cos θ power factor(The valid range of θ for a single phase circuit is from –90º 90º, and thepower factor will always be between 0 and 1.0.)Equation 107 is the commonly recognized equation for calculating power in asingle phase circuit. Joseph E. FleckensteinPage 12 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.org2C. Kilowatts, Lumens, BTU’s and HorsepowerIt is fair to say that a lage usage of three phase electrical power can beattributed to lighting, electric motors and heating.Electrical lighting, and especially incandescent lighting, has long been definedby the amount of power to activate the light, i.e. by the wattage. This practiceis changing as more efficient light sources have become available. More andmore the various types of lights are being defined by the amount of lightproduced rather by the amount of electrical power the light requires. One unitthat seems to be gaining in acceptance is the ‘lumens’, which is a metric unit.The incandescent light which remains the most common means of providingillumination, at around 17 lumens per watt, is one of the least efficient meansof converting electrical energy into light. There are of course other popularunits of measure for defining the amount of light a device produces. In anyevent the unit of the electrical power to activate a light will most certainlyremain the “watt.”When describing heat, the BTU (British Thermal Unit) has long been used inthe United States and in Commonwealth Countries. In the United States, theBTU will probably continue is use for the foreseeable future althoughoverseas the BTU is being replaced by the “calorie” as well as other units.Nevertheless, electric resistance heating is mostly classified by its kilowattrating.Another big change occurring in the United States is in the way electricalmotors are defined. Traditionally the size of a motor has been defined by“horsepower.” Because of globalization and increased international trade, theterm “horsepower” is being replaced by the “watt.” In electrical units, 1horsepower 746 watts. For example, a 1 kilowatt motor with an efficiencyof 74.6% would require 1 kilowatt (1,000 watts) of electrical energy tooperate and would produce 1 kilowatt X 0.746 kilowatts of output power(brake power), or 746 watts of output which would be the equivalent to 1 Joseph E. FleckensteinPage 13 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.orghorsepower of output. A 1 kilowatt motor with an efficiency of 80% wouldproduce 0.80 X 1 kilowatt of output power or (800/746) 1.07 HP.3.CalculatingPower in BalancedThree Phase Circuits3A.GeneralBy definition, a threephase circuit consists ofthree separate circuits.A typical trace of thethree voltages withrespect to time wouldbe similar to thatrepresented in Fig. 6. The phases are commonly identified as Phase A, PhaseB and Phase C and the common sequence is A-B-C. Each of the phases is120 from one another in a time plot.There are two basic types of three phase circuits: delta and wye. A typicaldelta circuit is shown in Fig. 7 and a typical wye circuit is shown in Fig. 8. Ina delta circuit, phasevoltage equals linevoltage but phasecurrent is not equal toline current. In a wyecircuit, line currentequals phase currentbut phase voltage isnot equal to linevoltage. So, in orderto perform correctpower calculations,one must make the Joseph E. FleckensteinPage 14 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.orgdistinction betweenline parameters andphase parameters andthen apply theappropriatecalculation.A wye circuit couldbe the three-wire orfour-wire type. If awye circuit is thefour wire type, afourth conductor isextended from thewye neutral point to ground. A three wire wye circuit is absent a conductorextending from the neutral point to ground.To calculate the power of a three phase circuit, specific parameters must beknown. Voltage must be known. Also, currents and the leads or lags ofcurrents with respect to voltages must be known. Unless care is taken to usethe appropriate equation, a correct answer cannot be expected.When computing three phase power, whether for a wye circuit or a deltacircuit, it is particularly important that a person clearly understands thesignificance of current lead and current lag as well as the bearing that lead/laghas on the determination of power. Power factor in a single phase circuit is aclearly understood condition. More specifically, power factor in a single phasecircuit is the cosine of the angle between current and voltage. A leadingcurrent with a lead angle of, say, 30º has a power factor equal to the cosine of30º or 0.866. A lagging current of 30º would likewise have a power factor of0.866. In three phase circuits, the matter of power factor is somewhat morecomplicated.A motor with a nameplate that states the motor power factor is giving thepower factor of the phases. In other words, a motor power factor is the cosine Joseph E. FleckensteinPage 15 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.orgof the angle between the phase current and the phase voltage whether themotor is a wye wound motor or a delta wound motor. The most common typeof three phase motor is the induction motor which always has a lagging powerfactor. Equation 12-1 in Appendix 12 gives the relationship between phaselead/lag and line lead/lag for balanced wye circuits and balanced deltacircuits. Equation 12-1 can be helpful when a person needs to determine linecurrents and the associated lead/lag of the line currents in order to performpower calculations.With a wye circuit, line currents and line voltages of an existing installationcan generally be read with ease, but phase voltages for a three wire circuitmay be difficult to read. In a delta circuit line voltages, which are the same asphase voltages, can be conveniently read but phase currents may be difficultto read. For example, to measure the phase currents of a delta wound motorwould require removing the terminal box cover and moving existingconnections. Particularly with high voltage motors, an effort of this type maybe especially hazardous and undesirable.By definition a balanced circuit has line voltages, line currents, phasevoltages, phase currents and the lead/lag of the currents that are all identical.According to Equation 107, the power of a single phase circuit is described bythe relationship,P VI cos θ, where the parameters are the single phase parameters.It follows then that the power for a balanced three phase circuit is,P 3VPIP cos θP (Reference 1)where,P power (watts)VP phase voltage (rms)IP phase current (rms)cos θP power factor of phaseθP angle of lead or angle of lag (radians or degrees) (phase current withrespect to phase voltage) (The subscript ‘P’ designates ‘phase’.)for a lagging power factor θP 0for a leading power factor, θP 0 Joseph E. FleckensteinPage 16 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.orgAs in a single phase circuit, instantaneous power of each of the phases of athree phase is given by Equation 105.Example 2In above Example 1, the values of instantaneous power are plotted for onecomplete cycle in a typical single phase application. As will bedemonstrated, a plot of three phase power throughout the same period hasa very different appearance. Consider a case having the followingparameters which are typically characteristic of a three phase motor:Vrms 480 VAC, three phaseVPK 480VACIrms 10 amps (phase current)IPK 10ampsPower Factor 0.7, laggingTherefore,cos θ 0.70, andθ –45.57ºEquation 105 statesthat for each of thethree phases,iP VPKIPK cos θ2sin ωt VPKIPK sinθ sin ωt cos ωt .Equation 105It may be seen thatthe power factor isthe same as thatassumed in Example1. In this laterexample the user is athree phase motor. Asummary of the power computations of instantaneous power vs. ωt forExample 2 are tabulated in Appendix E. A plot of the values of power forthe assumed parameters for one complete cycle is shown in Fig. 9. It issignificant to note that the value of total instantaneous power throughout Joseph E. FleckensteinPage 17 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.orgthe cycle is constant. In other words, the sum of the powers of the threephases is calculated to be exactly a constant value, in this instance10,080.00 watts. The computations of Example 1 and Example 2demonstrate one of the significant differences between a single phasemotor and a three phase motor. The power of the single phase motor in asingle cycle has two sharp peaks and two sharp valleys per cycle whereasthe power of the three phase motor is exactly constant for all values of ωt.As mentioned, the nameplate on a three phase motor will bear the powerfactor which is applicable to the phase currents and phase voltages. Thenameplate will also state the full load amperage, as ‘FLA’, which is themagnitude of the line current. This practice is seemingly inconsistent, butnonetheless it has become standard practice.It is usually a good idea to establish the sequence of a circuit underconsideration. Knowing the phase sequence is especially important whendealing with three phase motors. For this purpose a phase sequence metercan come in handy.The Model PRT200 by Extech Instrumentsis a non-contact phase sequence meter Joseph E. FleckensteinPage 18 of 88

www.PDHcenter.com3B.PDH Course E344www.PDHonline.orgCalculating Power in Balanced Three Phase Wye Circuits3B1. GeneralIn a balanced three phase wye circuit the phase currents of all three phases areof the same magnitude as the line currents, and all three currents are at thesame lead/lag angle. The phase voltages are at a magnitude that is a fixed ratioof the line voltages.In the typical three phase wye circuit as represented in Fig. 8 the line currentin Conductor A is the current of Phase A-D. The line current in Conductor Bis the current of Phase B-D, and the line current in Conductor C is the currentof Phase B-D. For a wye circuit the magnitude of the phase voltages are afixed ratio of the line voltages and are given by the expression,VL ) VP Equation 109or,VP [1/)] VLwhere,VL line voltage (rms)VP phase voltage (rms)The lead/lag of the line currents in a wye circuit are related to the lead/lag ofthe phase currents by the general expression,θP θL – 30º (Reference: Equation 12-1 in Appendix F.)Where,θP lead/lag of phase current with respect to phase voltageθL lead/lag of line current with respect to line voltageMore specifically,θP-A/AD θL-A/CA – 30ºθP-B/BD θL-B/AB – 30ºθP-C/CD θL-C/BC – 30º Joseph E. FleckensteinPage 19 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.org3B2. Calculating Power in Balanced Three Phase Wye Circuits Resistive Loads3B2a. Using Phase ParametersAccording to above Equation 108,P 3VPIP cos θPIn a resistive circuit the power factor is unity, or cos θP 1.0.Thus,P 3VPIP Equation 3B2a3B2b. Using Line ParametersFrom Equation 109,P 3VPIP cos θPFrom Equation 108,VL ) VP, orVP [1/)] VLIn a balanced wye circuit the magnitude of the line current is equal to themagnitude of the phase current (although the lead/lag of the current withrespect to phase voltage is different from the lead/lag of the current withrespect to line voltage).IP ILFor a balanced or unbalanced wye circuit the current lead/lag are related bythe expression:θP θL – 30º Equation 12Since cos θP 1.0 for resistive loads, θP 0ºθP 0º θL – 30ºθL 30ºThus,The net power for all three phases becomes,P 3VPIP cos θP 3{[1/)] VL} IL cos (θL – 30º)P VL IL cos (θL – 30º) VL IL cos (30º – 30º) VL IL cos (0º) VL IL (1)P VL IL Equation 3B2b Joseph E. FleckensteinPage 20 of 88

www.PDHcenter.comPDH Course E344www.PDHonline.org3B3. Calculating Power in Balanced Three Phase Wye Circuits –Inductive or Capacitive Loads3B3a. Using Phase ParametersFrom Equation 109,P 3VPIP cos θP Equation 3B3a3B3b. Using Line ParametersFrom Equation 109,P 3VPIP cos θPFrom Equation 12, θP θL – 30ºVP [1/)] VLIn a balanced wye circuit the magnitude of the line current is equal to themagnitude of the phase current.IP ILP 3VPIP cos θPFrom Equation 12θP θL 30ºP 3 [1/)] VL IL cos (θL – 30º)P ) VL IL cos (θL – 30º) Equation 3B3bNote: Equation 3B3b can be used for any balanced three phase load whetherthe user is wye, delta or a mix. The reasons for this condition are explainedbelow.3C.Calculating Power in Balanced Delta Three Phase Circuits3C1. GeneralIn a balanced three phase delta circuit the phase voltages are the same as theline voltages and the line currents are a fixed ratio of the phase currents.In the representation of a typical delta circuit in Fig. 7 li

Calculating Power in Unbalanced Three Phase Wye Circuits - Resistive Loads_24 4B2a. Using Phase Parameters _24 4B2b. Using Line Parameters _25 4B3. Calculating Power in Unbalanced Three Phase