The flux produced by field winding Flux and MMF Phasors The flux produced by field winding MMF is sinusoidal MMF Flux is sinusoidal 1 3 3 1 No of conductors 2 4 4 2 Therefore, Induced armature voltage is sinusoidal.

Consider cylindrical rotor alternator operation. Flux and MMF Phasors Consider cylindrical rotor alternator operation. Case 1: No Load Operation Alternator is rotated at syn speed by prime mover Field wdg is excited Armature voltage is induced which is given by Va0=Vt0=Ef=4.44 f Tph Φf Kw Ef= No load voltage, Excitation voltage or Excitation emf. Tph= Turns per phase of arm wdg. Φf = Flux per pole produced by field winding Kw= Winding Factor

Generated emf lags the flux by 900. No Load Voltage, Excitation Voltage Ef Ef Φf Axis of Field 900 Φf Ff Alternator phasor diagram at no load

Case 2: Unity Power Factor Load B1 Y2 Axis of Field R1 R2 N S Ff ω Φf B2 Y1

Case 2: Unity Power Factor Load N pole rotate anticlockwise Conductor R1 moves clockwise. Axis of armature winding R1R2 B1 Y2 S Fa Axis of Field Φa R1 R2 N S Ff ω Φf B2 N Y1 Emf is induced with dot polarity in R1 (RHR) In R2 cross polarity

Case 2: Unity Power Factor Load N pole rotate anticlockwise Conductor R1 moves clockwise. Axis of armature winding R1R2 B1 Y2 S Resultant air gap flux Φr Fa Axis of Field Axis of Field Φa R1 R2 N S Ff ω Φf B2 N Y1 Emf is induced with dot polarity in R1 (RHR) In R2 cross polarity

Case 2: Unity Power Factor Load Ef ω Axis of armature winding R1R2 Ia ω Resultant air gap flux Φr Fa Te Axis of Field Axis of Field Fr Φa Time-phasor diagram of Ef and Ia at upf Ff ω Φf Electromagnetic torque Te is towards resultant mmf or flux. Space-phasor diagram of mmf and flux

Case 2: Unity Power Factor Load It is a Rotor which has to move clockwise due to Te Ef ω S Ia Attraction Φr Fa Te Fr Φa N S Ff Φf Repultion N Combined Space and Time phasor diagram at upf

Case 2: Unity Power Factor Load It is a Rotor which has to move clockwise due to Te Ef Te and ω are in opposite direction Generator Operation ω S Ia Field poles are leading to arm poles w r t Te. Attraction Φr Fa Te Fr Φa S N Ff Arm reaction mmf is cross-magnetizing at upf. Φf Repultion N Combined Space and Time phasor diagram at upf

Combined Space and Time phasor diagram at zero pf lagging Case 3: Zero Power Factor Lagging Load Ef Arm reaction mmf is de-magnetizing at zero pf lagging. Ff -Fa=Fr ω Φf - Φa= Φr Te Φr Ff Φf Fr Φa Fa Ia Combined Space and Time phasor diagram at zero pf lagging

Combined Space and Time phasor diagram at zero pf leading Case 4: Zero Power Factor Leading Load Ef Arm reaction mmf is magnetizing at zero pf leading. Ff +Fa=Fr ω Φf + Φa= Φr Ia Fa Φa Te Φr Ff Fr Φf Combined Space and Time phasor diagram at zero pf leading

Combined Space and Time Phasor Diagram at Lagging Power Factor load Case 5: Lagging Power Factor Load Load with lagging pf is common load Ef Ff +Fa=Fr Φf + Φa= Φr ω Φ Ia Fr Φr Te Fa ψ Φa Ff Φf Combined Space and Time Phasor Diagram at Lagging Power Factor load

Case 5: Lagging Power Factor Load Two mmfs are sinusoidaly distributed along the air gap periphery. The relative velocity between the two mmfs is zero. Ψ=90+Φ Due to uniform air gap Φ Ia Φr Φa Φf Fa ψ Φa Φr Ff Φf ψ

Case 5: Lagging Power Factor Load Er= Air gap voltage Ef Er Ψ=90+Φ Φ 900 Ia Φr Er Fa ψ Φa Φr Ff Φf

Case 5: Lagging Power Factor Load Er= Air gap voltage jIaXm jIaxl From air gap voltage, subtract leakage impedance drop Ef Er Ψ=90+Φ jIaxl and Iara Φ Iara Ia Φr Terminal voltage or bus voltage is obtained Fa ψ Φa Vt Ff Φf In air gap voltage, add Mutual Inductance drop, jIaXm No load voltage Ef is obtained

Case 5: Lagging Power Factor Load jIaXm jIaxl Ef Er Iara Ef Vt Φr Φf Φa Vt Φf Φa Φr Er

Case 5: Lagging Power Factor Load jIaXs Xm+xl =Xs jIaXm jIaxl Ef Er Ψ=90+Φ Φ Iara Ia Vt

Case 5: Lagging Power Factor Load jIaXs Xm+xl =Xs Thus voltage equation of cyl. rotor generator is Ef Ef=Vt+Iara+jIaXs Ψ=90+Φ δ For generator Ef leads Vt Φ Iara θ Ψ=90+ δ+θ for generator Ia The angle between Vt and Ia is θ, power factor angle Vt The angle between Ef and Ia is Φ=δ+θ, internal displacement angle or internal power factor angle The angle between Vt and Ef is δ, power angle, torque angle or load angle

Case 5: Lagging Power Factor Load Ef ¯ ¯ jIaXs Iara Ef=Vt+Iara+jIaXs ¯ Ψ=90+Φ Actualy variables are phasors Vt ¯ Ψ=90+ δ+θ for generator δ θ Ef leads to Vt for Generator Ia

Case 5: Lagging Power Factor Load jIaXs Ef Ef=Vt+Iara+jIaXs Ψ=90+Φ Vt If resistance is neglected Ψ=(90+ δ+θ) <180 for generator δ Ef=Vt+jIaXs θ Ia Ef cosδ > Vt Over-excited P=+ ve Supplying Q=+ ve Supplying

If resistance is neglected Ψ=90+ δ+θ for generator δ Unity Power Factor jIaXs Ef=Vt+Iara+jIaXs Ψ=90+Φ Ef Vt If resistance is neglected Ψ=90+ δ+θ for generator δ Ef=Vt+jIaXs Ia Ef cosδ = Vt Nominal-excited P=+ ve Supplying Q=0

If resistance is neglected Ef δ Leading Power Factor q-axis Ef=Vt+Iara+jIaXs Ψ=90+Φ jIaXs Vt If resistance is neglected Ef δ Ef=Vt+jIaXs Ia θ Ef cosδ < Vt Under-excited P=+ ve Supplying d-axis Q= - ve Absorbing

Cylindrical Rotor Synchronous Motor For motor arm current is opposite wrt generator Ia= -Ia Axis of armature winding R1R2 Consider lagging current for Gen B1 Y2 N Axis of Field Ia R1 Φf R2 N S -Ia Φr B2 S Y1 Voltage applied with cross polarity to R1 (RHR) and to R2 with dot polarity

Cylindrical Rotor Synchronous Motor It is a Rotor which has to move anticlockwise Axis of armature winding R1R2 N Repultion Axis of Field Ia Φf ω N S -Ia Φr Attraction S

Cylindrical Rotor Synchronous Motor It is a Rotor which has to move anticlockwise Te and ω are in same direction Electromagnetic torque Te is towards resultant mmf or flux. Axis of armature winding R1R2 N Repultion Te is from N of field pole to S of arm pole Axis of Field Ia Φf ω S N Te -Ia Φr Field poles are DRAGGED behind the resultant air gap flux or by arm poles. Attraction S Field poles are lagging to arm poles. This is a motor operation

Cylindrical Rotor Synchronous Motor Vt Thus voltage equation of cyl. rotor motor is jIaXs Iara Vt=Ef+Iara+jIaXs Ψ=90+Φ For motor Vt leads to Ef Ef Ψ=90+ θ-δ for motor δ The angle between Vt and Ia is θ, power factor angle θ Ia The angle between Ef and Ia is Φ=θ-δ, internal displacement angle or internal power factor angle The angle between Vt and Ef is δ, power angle, torque angle or load angle

Cylindrical Rotor Synchronous Motor P=+ ve Absorbing Ef cosδ < Vt Q=+ ve Absorbing Under-excited jIaXs Vt Vt=Ef+Iara+jIaXs Vt Ef jIaXs If resistance is neglected δ Ef Vt=Ef+jIaXs δ θ Ia Ψ=(90+ θ – δ) <180 for motor θ Ia Phasor Diagram at Lagging Power Factor

Cylindrical Rotor Synchronous Motor Unity Power Factor Vt=Ef+Iara+jIaXs jIaXs Vt If resistance is neglected Ef δ Vt=Ef+jIaXs Ia Ψ=(90+ θ – δ) <180 for motor Ef cosδ = Vt Nominal-excited P=+ ve Absorbing Q=0

Cylindrical Rotor Synchronous Motor Leading Power Factor Ef jIaXs Vt=Ef+Iara+jIaXs Vt If resistance is neglected δ Vt=Ef+jIaXs Ia θ Ef cosδ > Vt Over-excited P=+ ve Absorbing Q= - ve Supplying

Salient Pole Synchronous Generator In cyl rotor, air gap is uniform. The arm flux is independent of spatial orientation wrt field poles. In salient pole, air gap is not uniform. The reluctance along d axis is much smaller than q axis. The arm flux is greater along d axis than along q axis. Resolve arm mmf along d axis and along q axis.

Salient Pole Synchronous Generator ARMATURE MMF D - Axis ARMATURE FIELD Q - Axis Arm wdg along Q-axis Arm wdg along D-axis

Salient Pole Synchronous Generator In cyl rotor, air gap is uniform. The arm flux is independent of spatial orientation wrt field poles. In salient pole, air gap is not uniform. The reluctance along d axis is much smaller than q axis. The arm flux is greater along d axis than along q axis. Resolve arm mmf along d axis and along q axis. So two mmf along d axis and one mmf along q axis

Salient Pole Synchronous Generator Consider arm current Ia lagging to Ef by 900. Q D D Ef Ff -Fa=Fr Φf - Φa= Φr ω Φf Φa Φr Te Φr Ff Φf Fr Φa Fa Ia N S Combined Space and Time phasor diagram Demagnetizing action Resultant flux decreases.

Salient Pole Synchronous Generator Consider arm current Ia leading to Ef by 900. Q D D Ef Ff +Fa=Fr Φf + Φa= Φr ω Φr Φf Te Φa Fa Ia Φa Φr Ff Φf Fr N S Combined Space and Time phasor diagram Magnetizing action Resultant flux increases.

Salient Pole Synchronous Generator Consider arm current Ia is in phase with Ef. Q D D Ef Ff +Fa=Fr ω Φf +Φa= Φr Φf Ia Φa Fa Saddle shape Te Φa Ff Φf Fa N S Combined Space and Time phasor diagram Along q axis, air gap is max. More reluctance, less flux

Salient Pole Synchronous Generator Consider arm current Ia is in phase with Ef. Q D D Ef ω Φf +Φa= Φr Φf Ia Φa Fa Saddle shape Te Φa Ff Φf Fa N S Combined Space and Time phasor diagram Saddle flux consists of Fundamental & 3rd harmonic component

Salient Pole Synchronous Generator Consider arm current Ia is in phase with Ef. Thus distorted resultant flux is obtained Q D D Ef ω Φf +Φa= Φr Φf Ia Φa Fa Saddle shape Te Φa Ff Φf Fa N S Combined Space and Time phasor diagram So emf consists of fundamental & 3rd harmonic component

Salient Pole Synchronous Generator Thus it can be concluded that If Ia lags Ef by 900, then there is demagnetizing action If Ia leads Ef by 900, then there is magnetizing action If Ia is in phase with Ef, then resultant flux is distorted containing 3rd harmonic flux. In actual practice, Ia lags Ef due to industrial and domestic load. Arm mmf Fa produces effect on both axes d and q.

Salient Pole Synchronous Generator Id=IasinΦ Iq=IacosΦ Fd=FasinΦ Fq=FacosΦ Two MMFs along d axis Ef One MMFs along q axis The voltage drop due to Id =jIdXd Iq Ia Fq Φ Fa jIdXd= jIdXmd+ jIdxdl Φa ψ The voltage drop due to Iq =jIqXq Ff Φf Fd Id jIqXq= jIqXmq+ jIqxql

Salient Pole Synchronous Generator Id=IasinΦ Iq=IacosΦ Fd=FasinΦ Fq=FacosΦ Two MMFs along d axis Ef Er jIaxl One MMFs along q axis Iara The voltage drop due to Id =jIdXd Iq Vt Ia Fr Fq Φ Fa jIdXd= jIdXmd+ jIdxdl Φa ψ The voltage drop due to Iq =jIqXq Ff Φf Fd Id jIqXq= jIqXmq+ jIqxql

Salient Pole Synchronous Generator jIqXmq Ef jIdXmd jIaXm jIdXd jIqxql jIdxdl q Er jIaxl Iara Id IdXd Vt Iq Ia jIqXq ψ Id IqXq d Iq

Salient Pole Synchronous Generator jIqXmq jIqxql Ef jIdXmd jIdxdl q Iara Vt Iq Ia ψ Id d jIdXd= jIdXmd+ jIdxdl jIqXq= jIqXmq+ jIqxql

Salient Pole Synchronous Generator jIqXmq jIqxql jIqXq Ef jIdXd Voltage equation of salient pole Syn Gen. jIdXmd Ef=Vt +Iara +jIdXd jIdxdl +jIqXq q Iara δ Vt Iq θ Ψ=(90+δ+θ) <180 Ia ψ Id d jIdXd= jIdXmd+ jIdxdl Simplified Phasor Diagram for Lagging Power Factor jIqXq= jIqXmq+ jIqxql

Salient Pole Synchronous Generator Phasor Diagram for Unity Power Factor Voltage equation of salient pole Syn Gen. jIqXq Ef jIdXd Ef=Vt +Iara +jIdXd +jIqXq q Ef leads to Vt for generator Iara δ Vt Ia Iq θ=0 Ψ=(90+δ+θ) <180 ψ Id d

Salient Pole Synchronous Generator Phasor Diagram for Leading Power Factor Voltage equation of salient pole Syn Gen. jIqXq Ef jIdXd Ef=Vt +Iara +jIdXd +jIqXq Iara Vt q δ Ia Iq θ Ψ=(90+[δ-θ]) <180 ψ d Id

Salient Pole Synchronous Generator Phasor Diagram for Leading Power Factor Voltage equation of salient pole Syn Gen. Iara Ef=Vt +Iara +jIdXd +jIqXq jIdXd Ef jIqXq Vt q δ Ia Iq θ Ψ=(90-[θ- δ]) <180 ψ d Id

Salient Pole Synchronous Motor Iara jIqXq Voltage equation of salient pole Syn Motor Vt jIdXd Ia=-Ia, wrt Gen Ef Vt=Ef +Iara +jIdXd +jIqXq q δ Iq Vt leads to Ef by δ,for motor θ Ia ψ Consider Lagging Power Factor operation Id d Phasor Diagram for Lagging Power Factor Ψ=90+(θ-δ) <180

Salient Pole Synchronous Motor jIqXq Voltage equation of salient pole Syn Motor Iara Ef jIdXd Vt Ia=-Ia, wrt Gen δ Vt=Ef +Iara +jIdXd +jIqXq q Vt leads to Ef by δ,for motor Ia Iq θ=0 ψ Consider Unity Power Factor operation d Id Phasor Diagram for Unity Power Factor Ψ=(90-δ) <180

Salient Pole Synchronous Motor jIqXq Ef Voltage equation of salient pole Syn Motor Iara jIdXd Vt Ia=-Ia, wrt Gen δ q Vt=Ef +Iara +jIdXd +jIqXq Iq Ia Vt leads to Ef by δ,for motor θ ψ Consider Leading Power Factor operation d Id Phasor Diagram for Leading Power Factor Ψ=(90-θ-δ) <180

Salient Pole Synchronous Machines Phasor Diagram of Saturated Machines Ef jIaXq Saturation means more flux. More flux is along d axis. Iara Under saturation reactance is negligible. q Vt δ So under saturation Xd is negligible. Iq θ Ia Value of Xq is fixed ψ Id d Voltage eqn for Generator Ef=Vt +Iara +jIaXq Phasor Diagram is Similar to Cyl. Rotor Generator

Salient Pole Synchronous Machines Phasor Diagram of Saturated Machines Vt jIaXq Saturation means more flux. More flux is along d axis. Iara Under saturation reactance is negligible. Ef q δ So under saturation Xd is negligible. θ Iq Ia Value of Xq is fixed ψ Id d Voltage eqn for Motor Vt=Ef +Iara +jIaXq Phasor Diagram is Similar to Cyl. Rotor Motor

Analysis of Phasor Diagram In the problem, generally Vt or Vb or Va, Ia, power factor angle θ, Xd, Xq are given. But for calculating Id and Iq, the power angle δ must be known. For this purpose, analysis of phasor diagram is required. First consider the phasor diagram of salient pole syn generator at lagging power factor.

Analysis of Phasor Diagram First draw a perpendicular line from tip of Iara for GEN. Since it is perpendicular to Iara, it is a reactance drop jIaX jIqXq From tip of Iara,draw perpendicular line on ob Ef jIdXd b Ef’ Extend Ia line jIaX jIaXq Consider triangle oa’b and acb δ+θ a c Vt Therefore bac= V V δ+θ Iara a’ δ x ab=IaX ac=ab cos(δ+θ) Iq θ Ia IqXq=IaX cos(δ+θ) ab=IaXq o IqXq=IqX Id Xq=X (1)

Ef’=ob=Vt +Iara +jIaXq =[Mag] δ Thus δ is calculated jIqXq The δ can also be calculated from Δoa’b Ef jIdXd b Ef’ jIaXq δ+θ a c Vt Iara a’ δ x Iq θ Ia o Id

Now obtain Ef - Ef’ jIqXq d Ef jIdXd b Ef’ jIaXq δ+θ a c =+ve So Ef >Ef’ Vt Iara a’ δ Phasor diagram is correct. x Iq θ Ia Ef =Ef ’+bd o Id

Consider Syn Gen with Leading Power Factor First draw a perpendicular line from tip of Iara for (GEN) Since it is perpendicular to Iara, it is a reactance drop jIaX a Iara d jIaX a’ jIdXd Vt ab=jIaX c Ef b jIqXq From tip of Iara,draw perpendicular line on ob Ef’ e δ Iq Extend Ia line e’ Ia Draw line parallel to ae θ Id o

Consider Syn Gen with Leading Power Factor Consider triangles abd and obe Angle dab=θ- δ θ- δ a Iara d ad=ab cos(θ-δ) jIaX a’ jIdXd IqXq=IaX cos(θ-δ) Vt c IqXq=IqX Ef b jIqXq Xq=X ……(1) Ef’ e δ Iq Ef’=Vt+Iara+jIaXq =[Mag] δ e’ Ia θ Thus δ is calculated Id o

The δ can also be calculated from Δoeb θ- δ a Iara d jIaX a’ jIdXd Vt c Now obtain Ef - Ef’ Ef b jIqXq Ef’ e δ Iq e’ Ia θ =Negative Id o So Ef’ >Ef Phasor diagram is not correct. Ef =Ef ’- bc

Now consider the phasor diagram of salient pole syn MOTOR at lagging power factor. θ-δ Iara jIqXq d First draw a perpendicular line from base of Iara for MOTOR. c jIdXd Vt jIaX Extend Ia line b Ef’ bc is perpendicular to Ia, so must be reactive drop IaX Ef e δ a Iq Consider Δ oab and bcd θ Ia bcd =θ-δ Id o bc=IaX cd=bc cos(θ-δ) =IaX cos(θ-δ) IqXq =IqX Xq =X…………(1)

Now consider the phasor diagram of salient pole syn MOTOR at lagging power factor. θ-δ Iara jIqXq d Ef’= Vt- Iara- jIaXq c jIdXd Vt =[Mag] δ jIaX Thus δ is calculated b Ef’ Ef The δ can also be calculated from Δoab e δ a’ a Iq θ Ia Id o

Now consider the phasor diagram of salient pole syn MOTOR at lagging power factor. θ-δ Iara Now obtain Ef’- Ef jIqXq d c jIdXd Vt jIaX b Ef’ Ef =Positive e δ a’ a Ef’ >Ef Phasor diagram is correct. Iq θ Ia Ef =Ef ’- be Id o

Now consider the phasor diagram of salient pole syn MOTOR at leading power factor. First draw a perpendicular line from base of Iara for MOTOR. jIqXq d Ef jIdXd c and extend Ia line jIaX Iara Ef’ bc is perpendicular to Ia, so must be reactive drop IaX θ+δ a’ Vt e b From base of Iara,draw perpendicular line on od a δ Ia Iq θ Consider Δ bce and oac cbe =θ+δ o Id bc=IaX, be=bc cos(θ+δ) =IaX cos(θ+δ) IqXq =IqX Xq =X…………(1)

Now consider the phasor diagram of salient pole syn MOTOR at leading power factor. jIqXq d Ef’= Vt- Iara- jIaXq Ef jIdXd c δ jIaX =[Mag] Iara Ef’ Thus δ is calculated θ+δ a’ Vt e b The δ can also be calculated from Δoac a δ Ia Iq θ o Id

Now consider the phasor diagram of salient pole syn MOTOR at leading power factor. Now obtain Ef - Ef’ jIqXq d Ef jIdXd c jIaX Iara Ef’ θ+δ a’ Vt e b =Positive a δ Ef >Ef ’ Phasor diagram is correct. Ia Iq θ Ef =Ef ’+cd o Id

Example A salient pole synchronous generator has the following per unit parameters: Xd=1.2, Xq=0.8, ra=0.025 Compute the excitation voltage Ef on a per unit basis, when the generator is delivering rated kVA at rated voltage and at power factor of (a) 0.8 lagging and (b) 0.8 leading

Solution The voltage equation for salient pole syn generator is: Ef=Vt +Iara Ef is always along q axis +jIdXd +jIqXq jIqXq First calculate δ Ef jIdXd Ef’=Vt +Iara +jIaXq (a) With Vt as a reference phasor, Vt =Rated voltage Vt =1.00+j0.00 Vt Iara For rated kVA, Ia= Rated value δ Iq θ Ia Ia =1.00 -36.9 For 0.8 pf lagging =0.8 - j0.6 Iara =(0.8 - j0.6)(0.25) Id =(0.02 - j0.015)

Solution jIaXq = j(0.8 - j0.6)(0.8) = (0.48 +j0.64) Ef’=Vt +Iara +jIaXq jIqXq =(1.00+j0.00)+ (0.02 - j0.015)+ Ef jIdXd (0.48 +j0.64) = 1. 5 +j0.625 = 1. 625 22.62 Vt Iara δ=22.620 and δ+θ=22.62+36.9=59.52 δ Iq θ Ia Id = Iasin(δ+θ) = 1.00 sin(59.52) =0.862 Id Iq = Iacos(δ+θ) =0.507

Ef= Ef’ +Id (Xd-Xq) = 1.9698 22.62 (b) Phasor diagram for leading pf θ=+36.9 Ia=0.8 + j0.6 Iara=0.02 + j0.015 jIqXq jIaXq=-0.48+ j0.64 Ef jIdXd Ef’=Vt +Iara +jIaXq jIaXq 50.500 =0.549+j0.655 =0.849 Ef’ δ= 50.500 δ Vt Iara Ia δ-θ=13.6 Iq θ Id=0.235 Ef= Ef’ +Id (Xd-Xq) Id 50.500 = 0.943

W I S H Y O U A L L T H E B E S T