Bài giảng Electric circuit theor - Chapter X: AC Power Analysis - Nguyễn Công Phương

Nguy ễn Công Ph ươ ng Electric Circuit Theory AC Power Analysis Contents I. Basic Elements Of Electrical Circuits II. Basic Laws III. Electrical Circuit Analysis IV. Circuit Theorems V. Active Circuits VI. Capacitor And Inductor VII. First Order Circuits VIII.Second Order Circuits IX. Sinusoidal Steady State Analysis X. AC Power Analysis XI. Three-phase Circuits XII. Magnetically Coupled Circuits XIII.Frequency Response XIV.The Laplace Transform XV. Two-port Networks AC Power Analysis - sites.google.com/site/ncpdhbkhn 2 AC Power Analysis 1. Instantaneous and Average Power 2. Maximum Average Power Transfer 3. RMS Value 4. Apparent Power and Power Factor 5. Complex Power 6. Power Factor Improvement AC Power Analysis - sites.google.com/site/ncpdhbkhn 3 Instantaneous Power (1) pt()= vtit ()() =ω + φ vtV()m sin( t v ) =ω + φ it() Im sin( t i ) →=ωφ + ωφ + ptVI( )mm sin( t v )sin( t i ) V I =m m [cos(φφ −− ) cos(2 ωφφt ++ )] 2 vi vi VI VI =mmcos(φφ −− ) mm cos(2 ωφφt ++ ) 2vi 2 vi AC Power Analysis - sites.google.com/site/ncpdhbkhn 4 Instantaneous Power (2) VI VI p() t=mm cos(φφ −− ) mm cos(2 ωφφ t ++ ) 2vi 2 vi p(t) Vm I m 2 V I m m cos(φ− φ ) 2 v i 0 t AC Power Analysis - sites.google.com/site/ncpdhbkhn 5 Average Power (1) 1 T P= ∫ p( t ) dt T 0 VI VI p() t=mm cos(φφ −− ) mm cos(2 ωφφ t ++ ) 2vi 2 vi →=1φφ − 111T − T ωφφ ++ PVImmvicos( )∫ dtVI mm ∫ cos(2 t vi ) dt 2T0 2 T 0 The average of a sinusoid over its period is zero 1 →P = V I cos(φ − φ ) 2 mm v i AC Power Analysis - sites.google.com/site/ncpdhbkhn 6 Average Power (2) = φ V Vm v →* = φ− φ VI Vm I m v i = φ →* = −φ I Im i I Im i * = φφ−= φφ −+ φφ − VI Vm I m vimmVIcos( vi ) jVI mm sin( vi ) 1 P= V I cos(φ − φ ) 2 mm v i 1 →P = Re(VI * ) 2 AC Power Analysis - sites.google.com/site/ncpdhbkhn 7 Average Power (3) 1 1 P=Re(VI * ) = V I cos(φ − φ ) 2 2 mm v i φ= φ 1 1 1 2 v i : PVIVIIR=cos(0) = = 2mm 2 mm 2 m o 1 φ− φ = ± 90 : P= V I cos(90o ) = 0 v i 2 m m AC Power Analysis - sites.google.com/site/ncpdhbkhn 8 Average Power (4) • Ex.: v(t) = 150sin(314 t – 30 o) V i(t) = 10sin(314 t + 45 o) A Find P? AC Power Analysis - sites.google.com/site/ncpdhbkhn 9 AC Power Analysis 1. Instantaneous and Average Power 2. Maximum Average Power Transfer 3. RMS Value 4. Apparent Power and Power Factor 5. Complex Power 6. Power Factor Improvement AC Power Analysis - sites.google.com/site/ncpdhbkhn 10 Maximum Average Power Transfer (1) = 1 2 IL PIRL Lm L 2 + E E V Z I = eq →I = eq L L + Lm + – ZZeq L ZZeq L = + ZeqR eq jX eq = + ZLR L jX L Z I → += + ++ eq L ZZeq LR eq jX eq R L jX L + + = ++ + – V Z (Req R L )( j X eq X L ) Eeq L – →+= +2 + + 2 ZZeq L(RRXX eq L )( eq L ) AC Power Analysis - sites.google.com/site/ncpdhbkhn 11 Maximum Average Power Transfer (2) 1 I P= I2 R L L2 Lm L + Eeq V Z I = L Lm + – Zeq Z L += +2 + + 2 Zeq Z L(RR eq L )( X eq X L ) 2 1 EeqR L →P = × L +2 + + 2 2(RReq L )( X eq X L ) Z IL  ∂ eq PL =  0 ∂ + +  RL PL is maximum if:  E – V Z ∂P eq L  L = 0 – ∂  X L AC Power Analysis - sites.google.com/site/ncpdhbkhn 12 Maximum Average Power Transfer (3) 2 1 EeqR L P = × L +2 + + 2 2(RReq L )( X eq X L ) +  ∂P ∂P 2 RL( X eq X L ) L = 0 →L =E = 0 ∂ ∂Xeq [( RR +++ )(2 XX )] 2 2  X L L eqL eq L  2 2 ∂P ∂P 2 (RR+++− )( XX )2( RRR + )  L = 0 →L = E eq L eq L Leq L = 0 ∂ ∂Req 2[( RRXX +++ )(2 )] 2 2  RL L eqL eqL X= − X = −  L eq XL X eq →  →  =2 + + 2 R= R RL R eq( X eq X L )  L eq = * For maximum average power transfer, the load impedance must ZZL eq be equal to the complex conjugate of the equivalent impedance AC Power Analysis - sites.google.com/site/ncpdhbkhn 13 Maximum Average Power Transfer (4) 2 1 EeqR L P = × 2 L +2 + + 2 2(RRXXeq L )( eq L ) E →P = eq XX= − Lmax  L eq 8Req  = RRL eq AC Power Analysis - sites.google.com/site/ncpdhbkhn 14 Maximum Average Power Transfer (5) For maximum average power transfer, the load impedance must = * be equal to the complex conjugate of the equivalent impedance ZZL eq If ZL = RL ? → XL = 0 ∂P L =→=0RRXX2 + ( + ) 2 ∂ L eq eq L RL →=2 + 2 = RRXL eq eqZ eq AC Power Analysis - sites.google.com/site/ncpdhbkhn 15 Ex. 1 Maximum Average Power Transfer (6) E = 20 − 45o V; J = 5 60o A + Z Z =12 Ω ; Z = −j16 Ω 1 1 3 – Z2 Z3 Determine the load impedance Z2 that maximize the average power. What is E J the maximum average power? Z 1 Zeq Zeq I2 Z3 + – Eeq Z2 ZZ 12(− j 16) Z =1 3 = =7.68 −j 5.76 Ω eq + − ZZ1 3 12j 16 * → = + Ω ZZ= Z2 7.68j 5.76 2 eq AC Power Analysis - sites.google.com/site/ncpdhbkhn 16 Ex. 1 Maximum Average Power Transfer (7) E = 20 − 45o V; J = 5 60o A + Z Z =12 Ω ; Z = −j16 Ω 1 1 3 – Z2 Z3 Determine the load impedance Z2 that maximize the average power. What is E J the maximum average power? a E − J + + Z1 Z1 = = – I Eeq V a Eeq Z3 Zeq 2 1+ 1 Z Z E – J 1 3 + o – 20 − 45 Eeq Z2 − 5 60 o = 12 = 54.38 −140.4o V 1+ 1 2 − 2 E 12j 16 2 eq Eeq 54.38 P = →E = 54.38V →=P = = 48.13W 2 max eq 2max × 8R 8Req 8 7.68 eq AC Power Analysis - sites.google.com/site/ncpdhbkhn 17 Maximum Average Power Transfer (8) 1. Find the Thevenin equivalent + Z1 a. Zeq – Z b. E Z2 3 eq E J = * 2. ZL Z eq 2 Eeq 3. P = Z1 Z max 8R eq eq Z3 = − Ω 1a.Zeq 7.68j 5.76 a 1b.E = 54.38 −140.4o V eq + + Z1 = + Ω – 2.Z2 7.68j 5.76 Eeq Z3 E J = – 3.P2 max 48.13W AC Power Analysis - sites.google.com/site/ncpdhbkhn 18 Ex. 2 Maximum Average Power Transfer (9) 16 0o V + – Determine the load impedance Z that maximize the j6Ω average power. What is the maximum average power? −j8 Ω V 2I1 = open-circuit a Zeq b I short-circuit Ic 4Ω I1 Z Ω o o j6 16 0 V 2 30 A c + – −j8 Ω 1. Find the Thevenin equivalent 2I1 a b a. Zeq + b. Eeq Ic 4Ω I * 1 Voc 2. Z= Z 2 30o A L eq – 2 c E = eq 3. Pmax 8Req AC Power Analysis - sites.google.com/site/ncpdhbkhn 19 Ex. 2 Maximum Average Power Transfer (10) 16 0o V + – Determine the load impedance Z that maximize the j6Ω average power. What is the maximum average power? −j8 Ω V E 2I1 Z =open-circuit = eq a eq + b Ishort-circuit J eq I 4Ω I c (VV− )166 −+j II + 4 = 0 1 Voc c b 2 1 2 30o A = − – c Voc V b V c → =−+ + 1. Find the Thevenin equivalent Voc 16j 6 I2 4 I 1 a. Z = o eq I1 2 30 b. PEeq = = =× o = * I2 Ic 2 I 1 2 2 30 2. ZL Z eq → =− + ×× o + × o 2 Voc 16j 6 2 2 30 4 2 30 E 3. P = eq o max = 32.53 130.4 V 8Req AC Power Analysis - sites.google.com/site/ncpdhbkhn 20 Ex. 2 Maximum Average Power Transfer (11) 16 0o V + – Determine the load impedance Z that maximize the j6Ω average power. What is the maximum average power? −j8 Ω V E 2I1 =open-circuit = eq a Zeq b Ishort-circuit J eq Ic 4Ω I1 Z Ω o o j6 16 0 V 2 30 A c + – −j8 Ω 1. Find the Thevenin equivalent 2I1 a b a. Zeq b. E I P eq Ω c 4 I1 Isc 2. Z= Z * 2 30o A L eq 2 c E = eq 3. Pmax 8Req AC Power Analysis - sites.google.com/site/ncpdhbkhn 21 Ex. 2 Maximum Average Power Transfer (12) 16 0o V + – Determine the load impedance Z that maximize the j6Ω average power. What is the maximum average power? −j8 Ω V E 2I1 =open-circuit = eq a Zeq b Ishort-circuit J eq I 4Ω I c − o + =→ = o − 1 Isc I1 2 30Isc 0 I sc 2 30 I1 o 2 30 A c + = o j6I2 4 I 1 16 0 1. Find the Thevenin equivalent − + o − = I2 I 1 2 30Ic 0 a. Zeq →I − I + 2 30o − 2I = 0 b. 2 1 1 PEeq → − = o = * 3I1 I 2 2 30 2. ZL Z eq → = − 2 I1 0.67j 0.41 A E 3. P = eq → = o − − = o max Isc 2 30 (0.67j 0.41) 1.76 52.9 A 8Req AC Power Analysis - sites.google.com/site/ncpdhbkhn 22 Ex. 2 Maximum Average Power Transfer (13) 16 0o V + – Determine the load impedance Z that maximize the j6Ω average power. What is the maximum average power? −j8 Ω V E 2I1 =open-circuit = eq a Zeq b Ishort-circuit J eq Ic 4Ω I1 Z V = 32.53 130.4o V o oc 2 30 A c = o Isc 1.76 52.9 A 1. Find the Thevenin equivalent o 32.53 130.4 a.PZeq →Z = =1.67 +j 19.0 Ω eq o b. 1.76 52.9 PEeq 2. = * →=Z 1.67 −j 19.0 Ω PZL Z eq 2 2 32.53 Eeq P = = 79.2 W 3. P = max × P max 8 1.67 8Req AC Power Analysis - sites.google.com/site/ncpdhbkhn 23 AC Power Analysis 1. Instantaneous and Average Power 2. Maximum Average Power Transfer 3. RMS Value 4. Apparent Power and Power Factor 5. Complex Power 6. Power Factor Improvement AC Power Analysis - sites.google.com/site/ncpdhbkhn 24 RMS Value (1) i(t) T T + 1 R →P = i2 Rdt = i 2 dt – ∫ ∫ e(t) R T0 T 0 (AC) T → = 1 2 Ieff ∫ idt I T 0 + → = 2 E – R P I R (DC) T = 1 2 Veff ∫ v dt T 0 I is the effective/RMS value of i(t) T = = 1 2 Xeff X rms ∫ x dt T 0 AC Power Analysis - sites.google.com/site/ncpdhbkhn 25 RMS Value (2) T = 1 2 Irms idt 1T 1 T ∫0 → =2 = ω 2 T Irms ∫ idt ∫ [ Im sin tdt ] T0 T 0 = ω it( ) Im sin t T − ω = 12 1 cos 2 t ∫ Im dt T 0 2 2 IT I =m∫ dt = m 2T 0 2 I V I = m V = m rms 2 rms 2 AC Power Analysis - sites.google.com/site/ncpdhbkhn 26 AC Power Analysis 1. Instantaneous and Average Power 2. Maximum Average Power Transfer 3. RMS Value 4. Apparent Power and Power Factor 5. Complex Power 6. Power Factor Improvement AC Power Analysis - sites.google.com/site/ncpdhbkhn 27 Apparent Power 1 P= V I cos(φ − φ ) 2 mm v i V V = m P= V I cos(φ − φ ) rms 2 rms rms v i I I = m rms 2 S = Vrms Irms AC Power Analysis - sites.google.com/site/ncpdhbkhn 28 Power Factor =φ − φ P Vrms I rmscos( v i ) = S Vrms I rms P pf = =cos(φ − φ ) S v i •φ −=→ φ =→== v i 0pf 1 PSVIrms rms •φ −=± φ o → =→= v i 90pf 0 P 0 AC Power Analysis - sites.google.com/site/ncpdhbkhn 29 AC Power Analysis 1. Instantaneous and Average Power 2. Maximum Average Power Transfer 3. RMS Value 4. Apparent Power and Power Factor 5. Complex Power 6. Power Factor Improvement AC Power Analysis - sites.google.com/site/ncpdhbkhn 30 Complex Power (1) I Z +V – 1 S= VI * 2 = φ V Vm v = φ →* = −φ I Im iI I m i 1 1 1 →S = V I φφ−=VIcos( φφ −+ ) jVI sin( φφ − ) 2 m m vi2 mm vi 2 mm vi AC Power Analysis - sites.google.com/site/ncpdhbkhn 31 Complex Power (2) 1 S= VI * 2 V= ZI  * * φ −φ 2 1V  1 VV 1 (Vm v)(V m v ) V SV= = = = rms    * * * →  2Z  2 Z 2 Z Z  =1* = 1 φ −φ =1 2 = 2 S ZII Z ()Im i()I m iZI m Z I rms  2 2 2 Z =R + jX  P=Re(S ) = RI 2 →=+S (R jX ) I2 = RI 2 + jXI 2 →  rms rms rms rms = = 2  QIm(S ) XI rms AC Power Analysis - sites.google.com/site/ncpdhbkhn 32 Complex Power (3) 1 1 V 2 S= VI * =+P jQ = VI φ− φ = V I φ− φ =ZI 2 = rms 2 2 m m v i rms rms v i rms Z* == =2 + 2 SVIPQS rms rms = =φ −= φ 2 PRe(S ) S cos(v i ) RI rms = =φ −= φ 2 QIm(S ) S sin(v i ) XI rms P pf = =cos(φ − φ ) S v i AC Power Analysis - sites.google.com/site/ncpdhbkhn 33 Complex Power (4) • Ex.: v(t) = 150sin(314 t – 30 o) V i(t) = 10sin(314 t + 45 o) A Find Vrms , Irms , S, S, P, Q, pf ? AC Power Analysis - sites.google.com/site/ncpdhbkhn 34 AC Power Analysis 1. Instantaneous and Average Power 2. Maximum Average Power Transfer 3. RMS Value 4. Apparent Power and Power Factor 5. Complex Power 6. Power Factor Improvement AC Power Analysis - sites.google.com/site/ncpdhbkhn 35 Power Factor Improvement (1) pf = cos φ > →φ < φ R pf2 pf 1 21 + E – L IC IL E φ φ 2 1 I I R + C E – IL I L I L C φ< φ → = = 2 1 CP? ( const) AC Power Analysis - sites.google.com/site/ncpdhbkhn 36 Power Factor Improvement (2) Q =φ = φ 1 ∆Q QP1tan 12 , QP tan 2 S1 ∆ = − QQQ1 2 φ Q2 φ 2 1 E 2 ∆Q ∆Q =rms = ω CE 2 →C = rms ωE 2 P X C rms QQP−tanφ − P tan φ I C =12 = 1 2 R ω2 ω 2 + C E E – rms rms E φ− φ L tan1 tan 2 I I = P L C ω 2 Erms AC Power Analysis - sites.google.com/site/ncpdhbkhn 37 Ex Power Factor Improvement (3) A load is connected to a 220 V (rms), 50 Hz power line. This load absorbs a power of 1000kW. Its power factor is 0.8. Find the capacitor necessary to raise the pf to 0.9? tanφ− tan φ C= P 1 2 ω 2 Erms =→φ =→= φo → φ = pf 10.8 cos 11 0.8 36.9 tan 1 0.75 =→φ =→= φo → φ = pf 20.9 cos 22 0.9 25.8 tan 2 0.48 0.75− 0.48 →=×C 1000 103 = 0.0178F 314× (220) 2 AC Power Analysis - sites.google.com/site/ncpdhbkhn 38