chapter11 ac power analysis.ppt
CHAPTER 11 AC POWER ANALYSIS,11.1 Instantaneous and Average Power11.2 Apparent Power and Power Factor11.3 Maximum Average Power Transfer11.4 Complex Power11.5 Conservation of AC Power 11.6 Power Factor Correction,1,Introduction,In this chapter,our goals and objectives include Determining the instantaneous power delivered to 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 it,2,Instantaneous and Average Power,1.Instantaneous power（瞬时功率）,3,The average power is the average of the instantaneous power over one period.,2.Average Power,Real POwer(平均功率、有功功率),4,Instantaneous power：瞬时功率,Average power：平均功率/有功功率,Power analysis to R:,5,Power analysis to L:,Instantaneous power：,Average power：,Stored magnetic energy：,Q is a measure of the energy exchange between the source and the reactive part of the load,6,Power analysis to C:,Instantaneous power：,Average power：,Stored electric energy：,7,inductor=90,capacitor=-90,P=0,8,Maximum Average Power Transfer,The current through the load is,9,Average power delivered to load is:,Our objective is to adjust the 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=0,10,For maximum average power transfer,the load impedance ZL must be equal to the complex conjugate of the Thevenin impedance Zth.,Setting RL=Rth and XL=-Xth in Eq.(1)gives us the maximum average power as,11,In 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 absorbedBy the passive linear network of Fig.11.1.,13,Solution:The instantaneous power is given by p=vi=1200cos(377t+450)cos(377t-100)Applying the trigonometric identity,Or p(t)=344.2+600cos(754t+350)W,The average power is,p(t)=600cos(754t+350)+cos550,14,Example：For the circuit shown in Fig.,find the average power supplied by the source and the average power absorbed by the resistor.,15,Solution:,The current through the resistor is,The average power supplied by the voltage source is,And the voltage across it is,16,Which is the same as the average power supplied.Zero average power is absorbed by the capacitor.,The average power absorbed by the resistor is,Example:Determine the power generated by each source and the average power absorbed by each passive element in the circuit Fig11.3.,17,Solution:We apply mesh analysis as shown in Fig 11.3.for mesh1,For mesh 2,18,For the voltage source,the current flowing from it isI2=10.5879.10A.And the voltage across it is 60300V,so that the average power is,V1=20I1+j10(I1-I2)=80+j10(4-4-j10.39)=183.9+j20=184.9846.210V,19,The average power supplied by the current source is,P1+P2+P3+P4+P5=-735.6+320+0+0+415.6=0,V3=-j5 I2=(5-900)(10.5879.10)=52.9(79.10-900),20,11.4 Effective or RMS Value,The effective value of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current.,The effective value of a periodic signal is its root mean square(rms)value.,21,The power factor is the ratio of the average power to the apparent power,which is dimensionless.,Apparent Power and Power Factor,1.Apparent power（视在功率）,The apparent power(in VA)is the product of the rms values of voltage and current.,1)pf is the cosine of the phase difference between voltage and current.2)pf is also the cosine of the angle of the load impedance.,22,Generally speaking,0cosj1,X 0,j 0,inductive,current lags voltage,X 0,j 0,capacitive,current leads voltage,Example：cosj=0.5(lagging)，j=60o cosj=0.5(leading)，j=-60o,Power factor,23,z=-36.9，capacitive pf=cos(36.9)=0.8（leading）,z=36.9，inductive pf=0.8（lagging）,pf is the cosine of the angle of the load impedance.,24,PD=1000W，U=220V，f=50Hz，C=30F，cosD=0.8(lags）,calculate the apparent power,the average power and power factor of the source.,solution：,pf is the cosine of the phase difference between voltage and current.,25,Complex power(in VA)is the product of the rms voltage phasor and the complex conjugate of the rms current phasor,its real part is real power P and its imaginary part is reactive power Q.,Complex Power,1.Complex Power（复数功率）,Reactive powerunit:乏(var),26,Relation among the average,reactive and apparent power:,Apparent power:S=UI,Average power:P=UIcosj,Reactive power:Q=UIsinj,27,Generally speaking：,28,Conservation of AC Power,Why?,29,Conservation of complex power：The complex,real,and reactive powers of the sources equal the respective sums of the complex,real and reactive powers of the individual loads.,Conservation of the AC power,30,Solution：,31,P and Q of R、L、C:,PR=UIcos=UIcos0=UI=I2R=U2/R,PL=UIcos=UIcos90=0,QR=UIsin=UIsin0=0,QL=UIsin=UIsin90=UI=U2/XL=I2XL0,PC=UIcos=UIcos(-90)=0,QC=UIsin=UIsin(-90)=-UI=U2/XC=I2XC0,32,Calculate the complex power of each branch:,Solution 1：,33,Solution 2：,34,35,36,37,38,39,40,41,42,43,Power Measurement,The wattmeter is the instrument for measuring the average power.,A current coil is connected in series with the load,the voltage coil is connected in parallel with the load.,positive：power absorb；negative：power supply,44,Solution：,P1=R1I12=100W，P2=R2I22=25W，P=P1+P2=125W,45,Check Your UnderstandingWhy power companies require that the customers power factor be high?,46,1.Improve pf:,(1)If Pf is higher,the load can get more real power to do work.,P=Scosj,cosj=1,P=S=75kW,cosj=0.7,P=0.7S=52.5kW,Motor:Empty load cosj=0.20.3 Full load cosj=0.70.85,Light:cosj=0.450.6,(2)If pf is high,the energy dissipated on the transmission line will be decreased.I=P/(Ucosj),47,2.How to correct pf:,48,3.How to determine C：,49,Another way:,50,Power-factor correction：current method,解：,51,pf=1，全补偿,pf=0.9（落后），欠补偿,pf=0.9（导前），过补偿,52,f=50Hz,U=380V,P=20kW,cosj1=0.6(lag)。Find the value of capacitance necessary to raise the pf to 0.9,E.,53,Pf should be corrected to be 0.85,calculate C.,54,AC CIRCUIT ANALYSIS,Knowledge points：1.Sinusoids and phasors.2.Phasor method to calculate ac circuits.3.Phasor diagram.4.Ac power calculation.,55,习题讨论,1、图示正弦稳态电路中，us1(t)=6sin2t V，us2(t)=4sin2t V，求 u(t)和各电源提供的功率。,56,u(t)=2.69sin(2t 21.8)V,57,6V电压源提供的功率,P1=0.563.76cos68.6=4.12W,4V电压源提供的功率,P2=0.542cos93.6=0.251W,58,=Z2I2=(R2+jX2)I2=4+j12,=71.6,z2=13.3,Z2=13.371.6=(4.20+j12.6),4.20I2=4,I=0.976A,59,P=UI1cos0,I1=10A,UAB=50V,UBF=100V,UBD=50V,XL=XC=8.66,60,S闭合时，P2=1000 600=400W,I3=I4,（由S闭合时的电路不能先求出I3或I4）,61,U3=80V,R1=6,R2=4,US=R1I+R2I2=10(6+4)=100V,S断开时，,10010cos()=600,=53.1,1000=6053.1+U336.9,L=1/C,62,G1C1+G1C2+G2C1=AG1C1,=0,G1G22C1C2=0,63,end,64,EXAMPLE11.7,11.7(a),300,1000var,1.For the power triangle in Fig.11.7(a),the apparent power is:,(a)2000 VA(b)1000 VAR(c)866 VAR(d)500 VAR,2.The average power absorbed by an inductor is zero.,(a)True(b)False,4.The instrument for measuring average power is the:(a)voltmeter(b)ammeter(c)wattmeter(d)varmeter(e)kilowatt-hour meter,3.A source is connected to three loads Z1,Z2,and Z3 in parallel.Which of these is not true?,(a)P=P1+P2+P3(b)Q=Q1+Q2+Q3(c)S=S1+S2+S3(d),11.7(b),500w,600,5.In the power triangle shown in Fig.11.7(b),the reactive power is:,1000 VAR leading 1000 VAR lagging 866 VAR leading 866 VAR lagging,6.Ihe Thevenin impedance of a network seen from the load terminals is 80+j55.For maximum power transfer.the load impedance must be:,(a)-80+j55(b)-80-j55 80+j55(d)80-j55,8.If the load impedance is 20-j20,the power factor is(a)/450(b)0(c)1(d)0.7071(e)none of these,7.The amplitude of the voltage available in the 60-Hz,120-V power outlet in your home is:,(a)110v(b)120V(c)170V(d)210V,9.A quantity that contains all the power information ina given load is the(a)power factor(b)apparent power(c)average power(d)reactive power(e)complex power,10.Reactive power is measured in:(a)watts(b)VA(c)VAR(d)none of these,Example：To a RC phase-shift circuit in Fig.(a),find the relationship between parameters R,C when lagging behind by/3.,We set R/XC=k,Solution：,Example：The load ZL absorbed 4W average power and 12 Var reactive power,and we know the phase difference of and is 300.Find ZL and I.,Solution：,+,+,R1=10,ZL,Because Q0,we know ZL is inductive load.,ZL=R+jXL,(1),We set,Because ZL is inductive load,(2),(1),(2),Analyze the circuit of Fig.11.46 to find the complex power absorbed by each of the five circuit elements.,EXAMPLE11.6,Solution：,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:,EXAMPLE:Consider the circuit of Fig.11.45.Specify the value of capacitance required to raise the PF of the total load connected to the source to 0.92 lagging if the capacitance is added(a)in series with the 100-mH inductor;(b)in parallel with the 100-mH inductor.,Zeff=j100+j300|200=237 54.25o.=PF=cos 54.25o=0.5843 lagging.,(a)Raise PF to 0.92 lagging with series capacitance,Znew=j100+jXC+j300|200=138.5+j(192.3+XC)W,Solution:,Solving,we find that XC=133.3=1/wC,so that C=7.501 mF,(b)Raise PF to 0.92 lagging with parallel capacitance,Znew=j100|jXC+j300|200=+138.5+j92.31 W=138.5+W,Solving,we find that XC=25 W=1/wC,so that C=40 mF,EXAMPLE11.10,1.find Vth,2.find Zth,3.find R and Pmax,Solution:,Answer:6.792,6.569W,
