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11/30/2012 Overview • • • • • • • 32-Synchronous Generators Part 1 text: 7.7 to 7.9 ECEGR 450 Electromechanical Energy Conversion Excitation Induced frequency Induced EMF Equivalent Circuit Armature Reaction Power Relationship Approximate Power Relationship 2 Dr. Louie Excitation Excitation • Synchronous generators (motors) require revolving magnetic field a b c Permanent magnet Field winding (dc) stator i • Exciter: supplies current to field winding (if ) commutator N DC generator Brushless generator Power rating: <3% of generator rating N if shaft Pilot exciter field coil S slip rings dc generator • Field current is related to fp by kf Voltage: 125 to 600VDC Automatically controlled (terminal voltage magnitude, reactive power) 3 Dr. Louie field coil S stator 4 Dr. Louie Brushless Excitation Induced Frequency • 2-pole synchronous generator • Balanced three-phase voltage induced a b c stator i 3-phase rotor N N if x shaft Pilot exciter rectifier x xx S x S stator x 5 x x Two-pole machine: electrical frequency = mechanical frequency x N x x Dr. Louie x x x field coil S Stationary field AC generator x x field coil x Tm, wm x x Dr. Louie 6 1 11/30/2012 Induced Frequency Induced EMF • 4-pole synchronous generator • Each coil “sees” two Norths and two Souths per rotation • We next examine the induced emf in a synchronous generator • Flux linking a single stator coil (fc): Two electrical sinewaves for each mechanical rotation • In general: fc fpkp cos(wt) • Induced voltage in an Nc turn coil is: f P N P f m m 2 120 ec Ncfpwkp sin(wt) S N N S • Maximum induced voltage: Em Nckpfpw Tm, wm 7 Dr. Louie Induced EMF Induced EMF • RMS value of the induced emf: Em Nckpfpw | Ec | • If a generator has “a” parallel paths and P poles, then the emf per phase is: N P f m 120 1 Em 4.44fNckpfp 2 | Ea | • Induced voltage in a phase group, accounting for the number of coils in series, pitch factor and the distribution factor, is: | Epg | nk dEc 4.44nNckpk dffp Ne P 4.44nNck wffp a PnNck w a Ne: effective turns per phase • We can then write: | Epg | 4.44nNck wffp kw 8 Dr. Louie | Ea | 4.44Neffp kpk d (winding factor) 9 Dr. Louie Dr. Louie Example Example Consider a 16-pole, 144-slot, three phase Yconnected synchronous generator. Each coil has 10 turns, and the flux per pole is 0.025Wb. The rotor rotates at 375 rpm. The pitch factor is 0.94, and the distribution factor is 0.96. There are two parallel paths. Compute: frequency of the induced voltage RMS value of the per-phase voltage RMS value of the line-line voltage Dr. Louie 10 11 Consider a 16-pole, 144-slot, three phase Yconnected synchronous generator. Each coil has 10 turns, and the flux per pole is 0.025Wb. The rotor rotates at 375 rpm. The pitch factor is 0.94, and the distribution factor is 0.96. There are two parallel paths. Induced frequency: f 375 16 50 Hz 120 Dr. Louie 12 2 11/30/2012 Example Example Consider a 16-pole, 144-slot, three phase Yconnected synchronous generator. Each coil has 10 turns, and the flux per pole is 0.025Wb. The rotor rotates at 375 rpm. The pitch factor is 0.94, and the distribution factor is 0.96. There are two parallel paths. Induced RMS phase voltage: Computing generator parameters: PnNck w 16 3 10 0.902 216.48 a 2 | Ea | 4.44fNefp 1201.5V Ne k w kpk d 0.902 n Consider a 16-pole, 144-slot, three phase Yconnected synchronous generator. Each coil has 10 turns, and the flux per pole is 0.025Wb. The rotor rotates at 375 rpm. The pitch factor is 0.94, and the distribution factor is 0.96. 144 3 coils/pole/phase 16 3 | EL | 1201.5 3 2081V 13 Dr. Louie Equivalent Circuit Equivalent Circuit • Generator terminal voltage (Va) of a synchronous generator depends upon the load Terminal voltage may be greater or smaller than induced emf Will be higher when the power factor is leading Assumes generator is not grid-connected Ra: per-phase armature resistance (Ohm) Xa: armature leakage reactance (Ohm) Rf Armature resistance voltage drop Armature leakage reactance voltage drop Armature reaction + 15 Dr. Louie Equivalent Circuit jXf armature circuit vf Ra fp if - Ea jXa Ia Va Ea Ia(R a jXa ) Va Dr. Louie 16 Equivalent Circuit Phasor diagrams (compare magnitude of Ea, Va) Draw the phasor diagram for a synchronous generator with a leading PF Ea Ia(R a jXa ) Va Ea Ea IaRa • Equivalent circuit ignoring armature reaction field circuit • Terminal voltage is affected by: Ia 14 Dr. Louie jIaXa Ia fPF V a IaRa jIaXa Va Unity power factor Lagging power factor Va: reference Ia: in phase Va (unity PF) IaRa: in phase with Ia jIaXa: 90o out of phase from Ia E a > Va Va: reference Ia: lags Va (by fPF) IaRa: in phase with Ia jIaXa: 90o out of phase from Ia Ea > Va Dr. Louie Va Leading power factor 17 Dr. Louie 18 3 11/30/2012 Equivalent Circuit Armature Reaction Draw the equivalent circuit for a leading PF Consider a load with unity power factor Ea lags fp by 90 degrees Ia is in phase with Va fe = fp + far Ea Ia • fe: effective flux per pole • far: armature reaction flux per pole jIaXa IaRa fPF fe induces an emf the armature winding Va • Ear armature reaction emf • Ear lags far by 90 degrees Leading power factor Va: reference Ia: leads Va (by fPF) IaRa: in phase with Ia jIaXa: 90o out of phase from Ia Ea < Va Possible for induced voltage to be larger than terminal voltage Ee = Ea + Ear • Effective per-phase emf Ee = Va + Ia(Ra + jXa) 19 Dr. Louie 20 Dr. Louie Armature Reaction Armature Reaction • Armature reaction can be modeled by placing a voltage source in series with the induced emf • Equivalent circuit • Phasor Diagram (unity PF load) Armature reaction reduces the effective flux per pole Terminal voltage is smaller than the induced voltage Ea Rf + - Nf vf Ear + Ea - if - + Ra + Ee - Ia fe = fp + far + Va - 21 Ea fe = fp + far Ia IaRa 22 Dr. Louie Armature Reaction • Phasor Diagram (leading PF load) Armature reaction increases the effective flux per pole Terminal voltage is greater than the induced voltage Ear Ee jIaXa Ee Ea Ear • Phasor Diagram (lagging PF load) Armature reaction reduces the effective flux per pole Terminal voltage is smaller than the induced voltage far fPF Va IaRa Va Unity power factor Armature Reaction fe fp Ea Ear + Ee = Ear Ia(R a jXa ) Va Ee Ea + Ear Dr. Louie Ee fe Ia Ea Ear + Ia(R a jXa ) Va fp Ear far jXa far jIaXa fe = fp + far fp fe Ia Ea fPF Ear Ee jIaXa IaRa Va Leading power factor Lagging power factor Ea Ear + Ee = Ear Ia(R a jXa ) Va Ea Ear + Ee = Ear Ia(R a jXa ) Va Ee Ea Ear Ee Ea Ear Dr. Louie 23 Dr. Louie 24 4 11/30/2012 Armature Reaction Equivalent Circuit • Per phase equivalent circuit: • Note: if must be adjusted with changing load to keep terminal voltage constant • Armature reaction can be expressed as a magnetizing reactance Xs is used instead of Ear, Xa Ear = -jIaXm (emf uses active sign convetion) • Xm and Xa can be combined together as the synchronous reactance, Xs Rf Xs = Xm + Xa + • Synchronous impedance of the generator is Zs = Ra + jXs vf 4 1.06 d Ia Ia fPF Ea IaRa jIaXs Lagging power factor Unity power factor Ea fPF Va Ea as a function of power factor. Terminal voltage held constant at 10kV. Induced Voltage (V) Ea Va x 10 lagging 1.04 1.02 1 0.98 leading 0.96 0 jIaXs IaRa d 0.2 d is measured from Va to Ea d is positive for generators Va Leading power factor 26 Equivalent Circuit d: angle between Ea and Va (induced voltage and terminal voltage), known as the power angle or torque angle Ia Va Ia Dr. Louie • Phasor diagrams of new per-phase circuit jIaXs jXs Ea Ia(R a jXs ) Va Equivalent Circuit IaRa Ea if - 25 Dr. Louie d Ra jXf Dr. Louie 27 0.4 0.6 Power Factor 0.8 28 Dr. Louie Example 1 Example A synchronous generator has a per-phase synchronous impedance of 0.2 + j4. The generator supplies a per-phase load current of 100A at a lagging power factor of 0.866 lagging. The perphase terminal voltage is 10kV. • Per phase armature current: Ia 100 30 A • Solving the circuit: Ea 10.21.8 kV • Power angle: 1.8 degrees Compute the per-phase induced voltage. Compute the power angle. 0.2 Ea j4 100 kV Ea Ia(R a jXs ) Va Dr. Louie 29 Dr. Louie 30 5 11/30/2012 Voltage Regulation Power Relationships • Similar to dc generators, the voltage regulation of a synchronous generator is: VR • Mechanical power supplied to the shaft of a synchronous generator by the prime mover | Ea | | Va | 100 | Va | Ea: induced emf, also the no-load terminal voltage Va: terminal voltage at full load (V) steam turbine combustion turbine dc motor others 31 Dr. Louie 32 Dr. Louie Power Relationships Power Relationships • Mechanical power in: • Power output: Po 3 | Va || Ia | cos fPF 3 Re{VaIa} Pinm Tsws Requires knowledge (usually computation) of armature current Ts: shaft torque (Nm) w s: shaft speed (rad/s) • Desired to have an equivalent expression of generator power output without having to compute armature current • Total power in: Pin Tsws v fif • Electrical power out: Po 3 | Va || Ia | cos fPF 3 Re{VaIa} • Copper2 losses: Pcu 3 | Ia | R a 33 Dr. Louie Power Expressions Power Expressions • From the equivalent circuit: Ia Continuing: Ea Va Ea Va R a jXs Zs Pd 3 Re{ • Power developed: * Po 3 Re{VaIa} 3 Re{ 3 Re{ VaEa | Va |2 } Z*s VaE*aZs | Va |2 Zs V E* Z | V |2 R | V |2 X } 3 Re{ a a 2 s a 2 a j a 2 s } | Zs |2 | Zs |2 | Zs | | Zs | | Zs | Zs | Zs | z | Zs | z | Z | Zs 1 1 Z*s s 2 z Z*s Z*s Z*s | Zs |2 2z | Zs | | Zs |2 Dr. Louie VaE*aZs | Va |2 R a | V |2 X j a 2 s} | Zs |2 | Zs |2 | Zs | 3 Re{ VaE*aZs | Va |2 R a } | Zs |2 | Zs |2 3 Re{ VaE*a (R a jXs ) | Va |2 R a } | Zs |2 | Zs |2 Note: Ea | Ea | d Va | Va | 0 | Va | |V| |E |(cos d jsin d )(R a jXs ) | Va |2 R a 3 Re{ a a } | Zs |2 | Zs |2 Above expansion uses: Z*s | Zs | z 34 Dr. Louie Recall that dividing by a phasor means dividing by the magnitude and subtracting the angle 35 3 | V|a|Ea| | V |2 R (R a cos d Xs sin d ) a 2 a | Zs |2 | Zs | Dr. Louie Important result 36 6 11/30/2012 Power Relationships Power Relationship • Power balance equation: • Generator efficiency: Pin Tsws if v f 3 | Va || Ia | cos fPF 3 | Ia |2 R a if v f Pr Psl Pr: rotational losses (W) Psl: stray load losses (W) 3 | Va || Ia | cos fPF 3 | Va || Ia | cos fPF 3 | Ia |2 R a Pc • For maximum efficiency: • Constant losses grouped as: 3 | Ia |2 R a Pc Pc if vf Pr Psl 37 Dr. Louie 38 Dr. Louie Power Relationship Approximate Power Relationship • Armature resistance is small • Common to ignore it Pd = Re{EaIa*} power (p.u) Po jXs Ea Ia d Va Ia 0 50 100 d (deg.) 150 fPF Ea jIaXs Va Example lagging PF load 200 Armature copper losses negligible 39 Dr. Louie Dr. Louie Approximate Power Relationship Approximate Power Relationship Computing the real power output: • Synchronous generator power output (approximate) Ea | Ea | d | Ea | cos d j | Ea | sin d Ia | Ia | fPF | Ia | cos fPF j | Ia | sin fPF Euler’s Identity Va | Va | 0 | Va | j0 Ia d Ea Va | Ea | cos d | Va | j | Ea | sin d 0 jXs jXs jXs | Ea | sin d | E | cos d | Va | j a Xs Xs | Ia | cos fPF Ia fPF Po 3 | Va || Ia | cos fPF Ea jIaXs 3 | Va || Ea | sin d Xs 3 | Va || Ea | sin d Xs • Assumes: Va Example lagging PF load Va Ea jIaXs | Ea | sin d (equating real parts) Xs Po 3 | Va || Ia | cos fPF 40 Armature resistance is zero Constant speed Constant field current Cylindrical rotor Important result Dr. Louie 41 Dr. Louie 42 7 11/30/2012 Approximate Power Relationship Power Relationship • Power-angle relationship: Po • Torque developed (approximate): 3 | Va || Ea | sin d Xs Td • Maximum power: 3 | Va || Ea | Pdm Xs Po (approximate) Pd ws 3 | Va || Ea | sin d Xs • Maximum torque (approximate): Tdm power (p.u) 3 | Va || Ea | Xsws • Maximum power and torque occur at d = 90o 0 50 100 d (deg.) 150 200 Dr. Louie 43 Dr. Louie Example 44 Example A 2-pole synchronous generator has a per-phase terminal voltage of 7.5 kV, a per-phase induced voltage of 7.9 kV and a synchronous reactance of 1W. If the power angle is 15 degrees, compute the total real power delivered to the load. Assume the rotational losses are 1MW. A 2-pole synchronous generator has a per-phase terminal voltage of 7.5 kV, a per-phase induced voltage of 7.9 kV and a synchronous reactance of 1W. If the power angle is 15 degrees, compute the total real power delivered to the load. Assume the rotational losses are 1MW. Po 3 | Va || Ea | sin d 46MW Xs Rotational losses are not electric, so we do not need to subtract them. Dr. Louie 45 Dr. Louie Power Expressions Power Relationship Summary Pin = Tsws +vfif (total input power) Several different forms of round-rotor power output: vfif (field winding loss) Po 3 | Va || Ia | cos fPF =3Re{VaIa} 46 Pinm=Tsws (input mechanical power) 3 | Ea || Va | 3 | Va |2 R a (R a cos d X s sin d) | Zs |2 | Zs |2 Pr + Psl (rotational and stray load loss) Pd 3Re{EaI*a} developed electrical power 3 | Va || Ea | sin d Po (valid only if R a can be ignored) Xs Pcu 3 | I2a | R a (copper loss in armature) Po 3 | Va || Ia | cos fPF =3Re{VaIa} (output electrical power) Dr. Louie 47 3 | Ea || Va | 3 | Va |2 R a (R a cos d X s sin d) | Zs |2 | Zs |2 Dr. Louie 48 8 11/30/2012 Power Relationship Example Summary Pin = Tsws +vfif = 44.21MW Let: vf=400V if = 250A Pr + Psl= 2MW Zs = 0.2 + j4W d = 30o Va = 10kV |Ea| = 11kV Po • Exciters are used to supply DC current to the rotor of synchronous generators • Frequency of induced voltage increases with the number of poles for a fixed mechanical speed • Leakage reactance and armature reaction can be combined into Xs, the synchronous reactance • Approximate power delivered by a synchronous generator is: Vfif = 0.1MW Pinm=Tsws = 44.11MW Pr + Psl = 2MW Pd 3Re{EaI*a} 42.11MW Pcu 3 | I2a | R a 1.14MW Po 3 | Va || Ia | cos fPF 3 | Ea || Va | 3 | Va |2 R a (R a cos d Xs sin d) 40.97MW | Zs |2 | Zs |2 Dr. Louie 49 3 | Va || Ea | sin d Xs Dr. Louie 50 9

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