Lecture Electric circuit theory: Three-phase circuits include all of the following content: Introduction, three-phase source, three-phase load, three-phase circuit analysis, power in three-phase circuits.
Nguyễn Công Phương Electric Circuit Theory Three-phase Circuits 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 Three-phase Circuits - sites.google.com/site/ncpdhbkhn Three-phase Circuits Introduction Three-phase Source Three-phase Load Three-phase Circuit Analysis Power in Three-phase Circuits Three-phase Circuits - sites.google.com/site/ncpdhbkhn Introduction (1) • Polyphase: – Phase: branch, circuit or winding – Poly: many • Three-phase: three phases • Advantages: – – – – Machine: less space & less cost Transmission & distribution: less conducting material Power delivered to a three-phase load is always constant Single phase source from three-phase source Three-phase Circuits - sites.google.com/site/ncpdhbkhn Three-phase Source (1) A vAA’ vBB’ vCC’ C’ B’ t B S C Stator A’ v AA ' = Vm sin ωt vBB ' = Vm sin(ωt − 120o ) vCC ' = Vm sin(ωt + 120o ) Three-phase Circuits - sites.google.com/site/ncpdhbkhn Three-phase Source (2) v AA ' = Vm sin ωt vAA’ vBB’ vCC’ vBB ' = Vm sin(ωt − 120o ) t vCC ' = Vm sin(ωt + 120 ) o v AA ' + vBB ' + vCC ' = = Vm (sin ωt + sin ωt cos120o − cos ωt sin120o + sin ωt cos120o + cos ωt sin120o ) = Vm (sin ωt + sin ωt cos120o ) −1 = Vm sin ωt + 2sin ωt = v AA ' + vBB ' + vCC ' = VCC ' ω 120o 120o VBB ' Three-phase Circuits - sites.google.com/site/ncpdhbkhn VAA' 120o Three-phase Source (3) v AA ' = Vm sin ωt vAA’ vBB’ vCC’ vBB ' = Vm sin(ωt − 120o ) t vCC ' = Vm sin(ωt + 120 ) o v AA ' + vBB ' + vCC ' = VCC ' Symmetrical (balanced) Three-phase Source: ω - Same magnitude 120o 120o VBB ' VAA ' 120o - Same frequency - Displaced from each other by 120o Three-phase Circuits - sites.google.com/site/ncpdhbkhn Three-phase Source (4) A C C’ N Load C B’ A’ Star connection C A’ A Load A B C’ B’ B’ Load B C B Delta connection B C’ A’ A Three-phase Circuits - sites.google.com/site/ncpdhbkhn Three-phase Source (5) B’ A C C’ N C B’ A’ B C’ A’ A B A A B B C C N Three-phase Circuits - sites.google.com/site/ncpdhbkhn Three-phase Source (6) iA vAB vCA vBC A B’ C vA iB iBC B vB iC vC iC vC B iB C N A’ vB C’ iCA iA A vA iAB vAB, vBC, vCA: line voltages vA, vB, vC: line voltages/phase voltages vA, vB, vC: phase voltages iAB, iBC, iCA: line currents iA, iB, iC: line currents/phase currents iA, iB, iC: phase currents Three-phase Circuits - sites.google.com/site/ncpdhbkhn 10 Three-phase Circuit Analysis (9) ICA VCA = = I AB × 120o Z∆ I A = I AB − ICA VCA A C VBC ICA IA VAB +– VAB I AB = Z∆ V I BC = BC = I AB × − 120o Z∆ Z∆ Z∆ B IC c IBC Z∆ IAB b IB → I A = I AB (1 0o − 120o ) = I ab (1 + 0.50 − j0.87) = I ab I B = I AB a − 30o − 150 o IC = I AB 90o Three-phase Circuits - sites.google.com/site/ncpdhbkhn 26 Three-phase Circuit Analysis (10) A VCA Method VAB VBC Z∆ c IBC IC IAB Z∆ b IB IC VC a Z∆ B +– C ICA IA IA VA c ZY A C +– N a ZN VB B Three-phase Circuits - sites.google.com/site/ncpdhbkhn ZY IB n IN ZY b 27 Three-phase Circuit Analysis • • • Y–Y, Y–∆, ∆–∆, ∆–Y kinds of three-phase circuit: balanced & unbalanced Balanced three-phase circuit: – Balanced three-phase source and balanced three-phase load – Balanced three-phase source: same magnitude, same frequency, displaced from each other by 120o – Balanced three-phase load: three identical loads • Unbalanced three-phase circuit: – Unbalanced three-phase source and/or unbalanced three-phase load • To solve a balanced one: – Exploit the symmetry of a balanced three-phase circuit, or – Treat it like a normal three-source circuit • To solve an unbalanced one: – Treat it like a normal three-source circuit Three-phase Circuits - sites.google.com/site/ncpdhbkhn 28 Three-phase Circuit Analysis (11) VAB = V 0o VCA − 120 VCA = V 120 o a IA VAB o VBC C KVL for AaNbBA: +– VBC = V A ZY ZY c N ZY b IB B IC o VAB V = ZY I A − ZY I B = VAB → I A − I B = ZY ZY I B = I A × − 120o → I A − I B = I A (1 − − 120o ) = I A 30o → IA = V 3ZY − 30 o V IB = 3Z Y − 150 o IC = = I A × − 120o Three-phase Circuits - sites.google.com/site/ncpdhbkhn V 3ZY 90o = I A × 120o 29 Three-phase Circuit Analysis (12) A VCA a IA VAB Method VBC ZY c b IA VA c C N a ZN VB B Three-phase Circuits - sites.google.com/site/ncpdhbkhn ZY ZY A +– ZY IC IC VC N IB B +– C ZY IB n IN ZY b 30 Three-phase Circuit Analysis • • • Y–Y, Y–∆, ∆–∆, ∆–Y kinds of three-phase circuit: balanced & unbalanced Balanced three-phase circuit: – Balanced three-phase source and balanced three-phase load – Balanced three-phase source: same magnitude, same frequency, displaced from each other by 120o – Balanced three-phase load: three identical loads • Unbalanced three-phase circuit: – Unbalanced three-phase source and/or unbalanced three-phase load • To solve a balanced one: – Exploit the symmetry of a balanced three-phase circuit, or – Treat it like a normal three-source circuit • To solve an unbalanced one: – Treat it like a normal three-source circuit Three-phase Circuits - sites.google.com/site/ncpdhbkhn 31 Three-phase Circuit Analysis • • • Y–Y, Y–∆, ∆–∆, ∆–Y kinds of three-phase circuit: balanced & unbalanced Balanced three-phase circuit: – Balanced three-phase source and balanced three-phase load – Balanced three-phase source: same magnitude, same frequency, displaced from each other by 120o – Balanced three-phase load: three identical loads • Unbalanced three-phase circuit: – Unbalanced three-phase source and/or unbalanced three-phase load • To solve a balanced one: – Exploit the symmetry of a balanced three-phase circuit, or – Treat it like a normal three-source circuit • To solve an unbalanced one: – Treat it like a normal three-source circuit Three-phase Circuits - sites.google.com/site/ncpdhbkhn 32 Ex Three-phase Circuit Analysis (13) ICc Solve for currents 220 0o 220 − 120o 220 120o + + 20 j10 − j10 Vn = 1 1 + + + 20 j10 − j10 + j = 57.46 − 122o V C 220 o V N 220 120o V +– Suppose VN = IAa B c A 1+ j2 Ω InN 220 − 120o V IBb a n j10 Ω 220 0o − Vn 220 o − 57.46 − 122o → I Aa = = 12.76 11o A = 20 20 o o 220 − 120 − 57.46 − 122o 220 − 120 − Vn = 16.26 150.7 o A = I Bb = j10 j10 220 120 o − Vn 220 120o − 57.46 − 122 o ICc = = = 25.21 − 161.6o A − j10 − j10 o 57.46 − 122 Vn = 25,70 174, 6o A I nN = = + j2 + j2 Three-phase Circuits - sites.google.com/site/ncpdhbkhn 33 Ex Three-phase Circuit Analysis (14) ICc Solve for currents − j10I purple + j10( I purple + Iaqua ) = 20Iaqua + j10( Iaqua + I purple ) = = −220 − 120o + 220 0o I purple = −19.05 + j 43.21A I aqua = 38.11A 220 o V 220 120o V B c A N o a n +– = −220 − 120 + 220 120 o C IAa 220 − 120o V IBb j10 Ω I Aa = I aqua = 38.11A ICc = I purple = −19.05 + j 43.21A I Bb = −I purple − I aqua = −57.16 + j 43.21A Three-phase Circuits - sites.google.com/site/ncpdhbkhn 34 Three-phase Circuit Analysis (15) Ex VA VA = 220 V; VB = 220 − 120 V; VC = 220 120 V o o Z L = 5Ω; Z1 = 10Ω; Z = j 20Ω; Z3 = − j 30Ω; o N VB VC find currents ? ZL – + – + – + ZL ZL A Z1 Z2 B Z3 C (2 Z L + Z1 )I red − Z1I green − Z L Iblue = VA − VB − Z1I red + ( Z1 + Z + Z3 )I green − Z 3I blue = − Z L I red − Z3I green + (2 Z L + Z ) Iblue = VB − VC I red = 13.57 + j 0.48 A → I green = −6.89 − j12.59 A I = 2.06 − j11.02 A blue → currents Three-phase Circuits - sites.google.com/site/ncpdhbkhn 35 Three-phase Circuit Analysis (16) Ex VB VA = 220 0o V; VB = 220 − 120o V; VC = 220 120o V VA Z2 –+ Z1 = 10Ω; Z = j 20Ω; Z = − j 30Ω; find currents? Z1 VC Z3 − Z 2I green = − VB ( Z1 + Z )Iblue − Z2 Iblue + ( Z + Z )I green = − VC Three-phase Circuits - sites.google.com/site/ncpdhbkhn 36 Three-phase Circuits Introduction Three-phase Source Three-phase Load Three-phase Circuit Analysis Power in Three-phase Circuits Three-phase Circuits - sites.google.com/site/ncpdhbkhn 37 Power in Three-phase Circuits (1) ZY = Z φ A v AN = V sin ωt iA = I sin(ωt − φ ) vBN = V sin(ωt − 120o ) iB = I sin(ωt − φ − 120o ) vCN = V sin(ωt + 120 ) iC = I sin(ωt − φ + 120 ) o B N ZY o ptotal = pa + pb + pc = v AN iA + v BN iB + vCN iC ZY ZY C = 2VI [sin ωt sin(ωt − φ ) + sin(ωt − 120o ) sin(ωt − φ − 120o ) + + sin(ωt + 120 o )sin(ωt − φ + 120o )] sin A sin B = [cos( A − B) − cos( A + B)] → ptotal = 3VI cos φ Three-phase Circuits - sites.google.com/site/ncpdhbkhn 38 Power in Three-phase Circuits (2) ptotal = 3VI cos φ Pp = VI cos φ S p = VI Q p = VI sin φ S p = Pp + jQ p = VIˆ Three-phase Circuits - sites.google.com/site/ncpdhbkhn 39 Ex Power in Three-phase Circuits (3) A three-phase balanced Y–Y system has a phase voltage of 220 V The total real power absorbed by the load is 2400 W, the power factor angle of the load is 20o Find the line current? ptotal 2400 Pp = = = 800 W = V p I p cos 20o = 220 I p × 0.94 3 800 → Ip = = 3.87 A 0.94 × 220 → I L = I p = 3.87 A Three-phase Circuits - sites.google.com/site/ncpdhbkhn 40 ... 210o Three-phase Circuits - sites.google.com/site/ncpdhbkhn 12 Three-phase Circuits Introduction Three-phase Source Three-phase Load Three-phase Circuit Analysis Power in Three-phase Circuits Three-phase. .. ZB Three-phase Circuits - sites.google.com/site/ncpdhbkhn 14 Three-phase Circuits Introduction Three-phase Source Three-phase Load Three-phase Circuit Analysis Power in Three-phase Circuits Three-phase. .. sites.google.com/site/ncpdhbkhn Three-phase Circuits Introduction Three-phase Source Three-phase Load Three-phase Circuit Analysis Power in Three-phase Circuits Three-phase Circuits - sites.google.com/site/ncpdhbkhn