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1、CHAPTER 11 AC POWER ANALYSIS11.1InstantaneousandAveragePower11.2ApparentPowerandPowerFactor11.3MaximumAveragePowerTransfer11.4ComplexPower11.5ConservationofACPower11.6PowerFactorCorrection1IntroductionIn this chapter, our goals and objectives include Determining the instantaneous power delivered to
2、an element Defining the average power supplied by a sinusoidal source Using complex power to identify average and reactive power Identifying the power factor of a given load, and learning means of improving it2 Instantaneous and Average Power 1. Instantaneous power （瞬时功率 ）3i tOupUIcos - - UIcos(2 t
3、1 )Theaveragepoweristheaverageoftheinstantaneouspoweroveroneperiod.2. Average Power , Real POwer(平均功率、有功功率平均功率、有功功率) 4inductor =90capacitor =- -90P=0RX|Z| P+- -RjX+- -8 Maximum Average Power TransferThe current through the load is ZLZth9Average power delivered to load is:Our objective is to adjust t
4、he load parameters RL and XL so that P is maximum. To do this we set P/ RL and P/ XL equal to zero. We obtain P/ RL=0 P/ XL=010For maximum average power transfer, the load impedance ZLmust be equal to the complex conjugate of the Thevenin impedance Zth.Setting RL = Rth and XL = -Xth in Eq. (1) gives
5、 us the maximum average power as 11In a situation in which the load is purely real, the condition for maximum power transfer is obtained from Eq. (2) by setting XL = 0; that is,12 Example： Given that v(t)=120cos(377t+450）V and i(t)=10cos(377t-100)AFind the instantaneous power and the average power a
6、bsorbedBy the passive linear network of Fig.11.1.13Solution:The instantaneous power is given by p=vi=1200cos(377t+450)cos(377t-100)Applying the trigonometric identityOr p(t)=344.2+600cos(754t+350)WThe average power isp(t)=600cos(754t+350)+cos55014Example： For the circuit shown in Fig.,find the avera
7、ge power supplied by the source and the average power absorbed by the resistor.15Solution:The current through the resistor isThe average power supplied by the voltage source isAnd the voltage across it is16Which is the same as the average power supplied. Zero average power is absorbed by the capacit
8、or.The average power absorbed by the resistor isExample:Determine the power generated by each source and the average power absorbed by each passive element in the circuit Fig11.3.17Solution:We apply mesh analysis as shown in Fig 11.3.for mesh1, For mesh 2,18For the voltage source ,the current flowin
9、g from it isI2=10.5879.10A. And the voltage across it is 60300V, so that the average power isV1=20I1+j10(I1-I2)=80+j10(4-4-j10.39) =183.9+j20=184.9846.210V19The average power supplied by the current source isP1+P2+P3+P4+P5=-735.6+320+0+0+415.6=0V3=-j5 I2=(5-900)(10.5879.10 )=52.9(79.10-900)20Thepowe
10、rfactoristheratiooftheaveragepowertotheapparentpower,whichisdimensionless.Apparent Power and Power Factor1.Apparentpower（视在功率（视在功率）The apparent power (in VA) is the product of the rms values of voltage and current.1)pfisthecosineofthephasedifferencebetweenvoltageandcurrent.2)pfisalsothecosineofthean
11、gleoftheloadimpedance.22Generallyspeaking,0cos 1X 0, 0,inductive,currentlagsvoltageX 0, 0,capacitive,currentleadsvoltageExample：cos =0.5(lagging)， =60ocos =0.5(leading)， =-60ocos 1,resistive0，reactivePowerfactor23z= -36.9， capacitive pf=cos(36.9)=0.8（ leading ）z=36.9， inductive pf=0.8（ lagging ）pfis
12、thecosineoftheangleoftheloadimpedance.24PD=1000W，U=220V，f =50Hz，C =30 F，cos D=0.8(lags）,calculatetheapparentpower,theaveragepowerandpowerfactorofthesource.+_DCsolution：pfisthecosineofthephasedifferencebetweenvoltageandcurrent.25Complex power (in VA) is the product of the rms voltage phasor and the c
13、omplex conjugate of the rms current phasor, its real part is real power P and its imaginary part is reactive power Q.ComplexPower1.ComplexPower（复数功率 ）load+_Reactivepowerunit:乏乏(var)26Relationamongtheaverage,reactiveandapparentpower:Apparentpower:S=UI RX+_+_+_ SPQPowertriangle |Z|RXImpedancetriangle
14、UURUXVoltagetriangleAveragepower:P=UIcos Reactivepower:Q=UIsin 27Generallyspeaking：+_+_+_28 Conservation of AC PowerWhy?S0iuC+- -PC=UIcos =UIcos(- -90 )=0QC =UIsin =UIsin(- -90 )=- -UI=U2/XC=I2XC0, we know ZL is inductive load.ZL=R+jXL(1)We setBecause ZL is inductive load,(2)(1)(2)Analyze the circui
15、t of Fig. 11.46 to find the complex power absorbed by each of the five circuit elements.EXAMPLE11.6Solution：EXAMPLE11.8: The load in Fig. 11.44 draws 10 kVA at PF =0.8 lagging. If |IL |=40 A rms, what must be the value of C to cause the source to operate at PF =0.9 lagging?Solution:RLXL LGB EXAMPLE:
16、ConsiderthecircuitofFig.11.45.SpecifythevalueofcapacitancerequiredtoraisethePFofthetotalloadconnectedtothesourceto0.92laggingifthecapacitanceisadded(a)in series withthe100-mHinductor;(b)in parallel withthe100-mHinductor.Zeff=j100+j300|200=237 54.25o. =PF=cos54.25o=0.5843lagging.(a)RaisePFto0.92laggi
17、ngwithseriescapacitanceZnew=j100+jXC+j300|200=138.5+j(192.3+XC) Solution:Solving,wefindthatXC=133.3 =1/wC,sothatC=7.501mF(b)RaisePFto0.92laggingwithparallelcapacitanceZnew = j100 | jXC + j300 | 200 = +138.5 + j92.31 W= 138.5 + W Solving,wefindthatXC=25 =1/ C,sothatC=40 FRX EXAMPLE11.101.findVth2.findZth3.findRandPmaxSolution:Answer:6.792 ,6.569W
