Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12EE201 Lecture 38 P. 1In SSS, the complex power absorbed by a two-terminal device is a complex number defined by,S = VeffIeff*Ieff* is the complex conjugate of IeffExample: Calculate complex power if imtIti cos)( vmtVtv cos)(andComplex Power and its ComponentseffmeffmVVII 2;2 iveffeffaveIVP cosAverage powerEE201 Lecture 38 P. 2))((*ijeffvjeffeffeffeIeVIVS)(ivjeffeffeIVS aveiveffeffPIVS cosReComplex powerCharacteristics of complex power22QPIVSSQjPSeffeffComplex form“Apparent Power”Units: VAQ = reactive power = Im[S] Units: VAR(Volt-Amp-Reactive)EE201 Lecture 38 P. 3Calculation of power components for inductors+vL(t)-iL(t)Suppose iL(t) = tILsin2 90cos)( tItimL90LLLIjIIVL = ZLIL = jωLIL => ωLIL = VL 0oComplex power: SL = VLIL* = jVLIL = PL+jQLAverage power: PL = 0 {= VLILcos(0o- (-90o))}Reactive power: QL = VLILEE201 Lecture 38 P. 4Instantaneous power, pL(t)2 tLItLitWLLL222sin)(21)( LMAXLLLLLLLLLLQtWtQtWtIVtWtLItWtLItW)(22cos1)(22cos1)(22cos1)(2cos121)(22pL(t) = vL(t)iL(t)=ωL ILcos(ωt)x ILsin(ωt) pL(t)=VLIL sin(2ωt)Stored energy, WL(t)2EE201 Lecture 38 P. 5Calculation of power components for capacitors+vC(t)-iC(t)Suppose vC(t) = Vcsin(ωt)vC(t) = Vm cos(ωt-90o)VC = -jVC => VC -90IC = YCVC = jωCVC => ωCVC = IC 0oComplex power: SC = VCIC* = - jVCIC = PC + jQC Average power: PC = 0 {=VCICcos(-90o-0o)} Reactive power: QC = -VCIC2EE201 Lecture 38 P. 6Instantaneous power, pC(t)pC(t) = vC(t) iC(t)pC(t) = ωC Vccos(ωt) Vcsin(ωt)22pC(t) = VCICsin(2ωt)Stored energy, WC(t) CMAXcccCcCcCccCQtWtIVtWtCVtWtCVtWtCVtCvtW)(22cos1)(22cos1)(2cos121)(sin)(21)(22222EE201 Lecture 38 P. 7Example: Consider the circuit in the figure, where vin(t) = 100√2 cos(2000πt) V. The quantity 100√2 is a maximum value. Find the complex, average, reactive, and apparent power absorbed by the load. +-vin(t)iin(t)100Ω10kΩ16nFLoadZin(jω)SolutionStep 1. Compute Zin(jω).Zin(j2000π) = 100 +110-4+j2000π × 16 × 10-9= 5074 – j5000ΩEE201 Lecture 38 P. 8Step 2. Compute Iin. Iin =VinZin= 10 + j9.85 mAStep 3. Compute complex power absorbed by load. S = VeffI*eff = 100(10 – j9.85)10-3 = 1 – j0.985 VAStep 4. Compute average, reactive, and apparent power. Pav = Re[S] = 1 W(Reactive) Q = Im[S] = -0.985 var(Apparent) |S| = 1.404 VA(Average)EE201 Lecture 38 P. 9Conservation principles General Principle of Conservation of Power:In all circuits, instantaneous power is conserved. Sum of absorbed powers of all circuit elements is zeroPrinciple of Conservation of Complex Power in AC circuits:In AC circuits operating in steady state, complex power is conserved. Sum of all absorbed complex powers for all elements is zero.Note: conservation principle does not hold for apparent power.EE201 Lecture 38 P. 10Example: Calculate power delivered by the voltage source and the phasor current Iin if+-100VS1S2S3S4S5IinS1=360 + j160VA S2=360 - j120VAS3=420 + j540VA S4=130 + j80VAS5=40 – j100VASS = 1310 + j560 VA = 100 Iin*Pave = 1310 W ; Qs = 560 VAR|Ss| = 1425 VAIin= 13.1 – j5.6 AEE201 Lecture 38 P. 11Example: Consider a circuit which depicts a motor connected to a commercial power source. The motor absorbs 50 kW of average power and 37.5 kvar of reactive power, and has a terminal voltage Vm = 230 V. Find |Is|, the complex power delivered by the source Ss, and |Vs|.Rline = 0.5ΩIs+-Vs+Vm-Motor50kWSolutionStep 1. Find the apparent power |Sm| absorbed by the motor. SinceSm = Pm + jQm = 50 + j37.5 kVA|Sm| = 62.5 kVAEE201 Lecture 38 P. 12Step 2. Find |Is|. |Sm| = |VmI*s| = 230|Is|. |Is| = 271.7 A.Step 3. Compute the line loss.Pline = Rline|Is|2 = 36.92 kWStep 4. Compute the complex power delivered by the source. From the conservation of power, Ss = Sm + Sline = Sm + Pline = 50 + j37.5 + 36.92= 86.92 + j37.5 kVA Step 5. Compute |Vs|. |Vs| = |Ss| |Ss||I*s| |Is|= = 348.4
View Full Document