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Synchronous Machines The cylindrical-rotor construction is used for two- and four-pole turbine generators. The salient-pole construction is best adapted to multipolar, slow-speed, hydroelectric generators and to most synchronous motors. The dc power required for excitation is supplied by the excitation system. In older machines, the excitation current was typically supplied through slip rings from a dc machine, the exciter, which was often mounted on the same shaft as the synchronous machine. In modern systems, the excitation is supplied from ac exciters and solid-state rectifiers (either simple diode bridges or phase-controlled rectifiers). When a synchronous generator (G) is connected to a large interconnected system containing many other synchronous generators, the voltage and frequency at its armature terminals are fixed by the system. G cannot significantly affect the system voltage or frequency. It is usual to represent the remainder of the system as a constant-frequency, constant-voltage source, referred to as an infinite bus. Torque in a synchronous machine: P 2 T R F f sin RF 22 R : resultant air-gap flux per pole. F f : mmf of dc field winding. RF : torque angle, angle between magnetic axes of R and F f . Torque-angle characteristic; rotor mmf and resultant airgap flux are assumed constant. Positive values of torque represent generator action, corresponding to positive values of RF for which the rotor mmf wave leads the resultant air-gap flux. loss of synchronism (or pulling out of step): An increase in prime-mover torque will result in a corresponding increase in the torque angle. When RF becomes 90°, the electromechanical torque reaches its maximum value, known as the pull-out torque. Any further increase in prime-mover torque cannot be balanced by a corresponding increase in synchronous electromechanical torque. The result is that synchronism will no longer be maintained and the rotor will speed up. Under these conditions, the generator is disconnected from the external electrical system by the automatic operation of circuit breakers, and the prime mover is quickly shut down to prevent dangerous overspeed. Synchronous Machine Inductances Flux linkage of phase-a: a L aaia L abib L acic L af i f Stator-to rotor mutual inductances: L af L fa Laf cos me Rotor rotating at synchronous speed s , m s t 0 0 : rotor position at t 0. me ( P / 2)m et e0 Flux linkage of phase-a due to field: L af L fa Laf cos(e t e 0 ) af Laf I f cos(et e0 ) Stator self-inductances: L aa L bb L cc Laa Laa 0 Lal Laa 0 : self-inductance due to fundamental air-gap flux, Lal : armature leakage flux Mutual inductances between armature phases: 1 L ab L ba L ac L ca L bc L cb Laa cos(120 ) Laa 0 2 Resultant phase-a flux linkage: 1 a ( Laa 0 Lal )ia Laa 0 (ib ic ) af 2 Balanced currents ia ib ic 0 ib ic ia 3 a ( Laa 0 Lal )ia af Therefore, 2 Lsia af Ls : synchronous inductance Equivalent Circuits Terminal voltage (phase-a). d a di va Ra ia Ra ia Ls a eaf dt dt eaf : voltage generated by field winding flux (excitation voltage) d af eaf e Laf I f sin(et e 0 ) dt e Laf I f rms value Eaf 2 ˆ e Laf I f As an rms phasor Eaf j e je 0 2 Phase-a terminal voltage eqn. in phasor form: ˆ ˆ ˆ ˆ Va Ra I a jX s I a Eaf (motor reference direction for current) X s Ls : synchronous reactance. ˆ ˆ ˆ ˆ Eaf Ra I a jX s I a Va (generator reference direction for current) (a) motor reference direction (b) generator reference direction. 3 X s e Ls e Lal e Laa 0 X al X A 2 X al : armature leakage reactance X A : armature reaction reactance ˆ ER : air-gap voltage Example 5.1 A 60-Hz, three-phase synchronous motor is observed to have a terminal voltage of 460 V (line-line) and a terminal current of 120 A at a power factor of 0.95 lagging. The field-current under this operating condition is 47 A. The machine synchronous reactance is equal to 1.68 Ω (0.794 per unit on a 460-V, 100-kVA, 3-phase base). Assume the armature resistance to be negligible. Calculate (a) the generated voltage Eaf in volts, (b) the magnitude of the field-to-armature mutual inductance Laf, and (c) the electrical power input to the motor in kW and in horsepower. For motor reference direction for the current ˆ ˆ ˆ Va Eaf jX s I a ˆ ˆ ˆ Eaf Va jX s I a ˆ 460 265.6 V line-to-neutral I cos1 (0.95) 18.2 ( sign is for lagging p.f.) Va ˆ a 3 ˆ I a 120 e j18.2 A ˆ Eaf 265.6 j1.68(120 e j18.2 ) 278.8e j 43.4 V line-to-neutral The field-to-armature mutual inductance 2 Eaf 2 278.8 Laf 22.3 mH e I f 120 47 The three-phase power input to the motor, P 3Va I a cosa 3 265.6 120 0.95 90.83 kW 122 hp in Example 5.2 Assuming the input power and terminal voltage for the motor of Example 5.1 remain constant, calculate (a) the phase angle of the generated voltage and (b) the field current required to achieve unity power factor at the motor terminals. a. Unity power factor current in phase with voltage Pin 90.83 ˆ Ia 114 A I a 1140 A 3Va 3 265.6 ˆ ˆ ˆ Eaf Va jX s I a 265.6 j1.68 114 327.45e j 35.8 V line-to-neutral 35.8 2 Eaf 2 327.45 b. If 55.2 A e Laf 120 22.3 103 OPEN- AND SHORT-CIRCUIT CHARACTERISTICS Open-circuit Saturation Characteristic open-circuit characteristic: The air-gap line represents the machine open-circuit voltage characteristic corresponding to unsaturated operation. Deviation from this curve is a measure of the degree of saturation in the machine. The open-circuit characteristic can be used to measure the field-to-armature mutual inductance Laf. Example 5.3 An open-circuit test performed on a three-phase, 60-Hz synchronous generator shows that the rated open-circuit voltage of 13.8 kV is produced by a field current of 318 A. Extrapolation of the air-gap line from a complete set of measurements on the machine shows that the field current corresponding to 13.8 kV on the air-gap line is 263 A. Calculate the saturated and unsaturated values of Laf. 13.8 2 Eaf 2 7.97 103 Eaf 7.97 kV Saturated value of Laf 94 mH 3 e I f 120 318 2 7.97 103 Unsaturated value of Laf 114 mH 120 263 The mechanical power required to drive the synchronous machine during the open-circuit test is used to measure the no-load rotational losses. no-load rotational losses friction and windage losses + core loss (due to flux at no load) The friction and windage losses at synchronous speed are constant. The open-circuit core loss is a function of the flux (which is proportional to the open-circuit voltage). open-circuit core-loss curve Short-circuit Characteristic Open- and short-circuit characteristics SCC: SCC is obtained with the machine driven at synchronous speed. The field current is increased and the corresponding armature current values are measured. The machine operates in unsaturated condition. The short- circuit armature current is directly proportional to the field current over the range from zero to above rated armature current. At If = Of , Ia,sc = O b line-to-neutral voltage corresponding to Oa = Va,ag V unsaturated synchronous reactance X s ,u a ,ag I a ,sc ER Eaf RaIa jXalIa jXsIa Ia Phasor diagram for short-circuit conditions. Armature short-circuited ˆ ˆ ˆ Va 0 Eaf I a ( Ra jX s ) Air-gap voltage: ˆ ˆ ER I a ( Ra jX al ) The saturated value of the synchronous reactance at rated voltage Va,rated is Va ,rated Xs Ia The short-circuit ratio (SCR): Of SCR Of Of : field current required for rated voltage on open circuit. Of : field current required for rated armature current on short circuit. 1 SCR ( X s ) pu Va ,rated Of Ia I Xs . From the short-circuit characteristic: a ,rated slope of SCC Of I a Of Of Va ,rated Of Va ,rated Of Va ,rated Xs Of 1 Xs . . Z base ( X s ) pu Of I a ,rated I a ,rated Of I a ,rated Z base Of SCR Example 5.4 The following data are taken from the open- and short-circuit characteristics of a 45-kVA, three-phase, Y- connected, 220-V (line-to-line), six-pole, 60-Hz synchronous machine. From the open-circuit characteristic: Line-to-line voltage = 220 V Field current = 2.84 A From the short-circuit characteristic: Armature current, A 118 152 Field current, A 2.20 2.84 From the air-gap line: Field current = 2.20 A Line-to-line voltage = 202 V Compute the unsaturated value of the synchronous reactance, its saturated value at rated voltage in accordance and the short-circuit ratio. Express the synchronous reactance in ohms per phase and in per unit on the machine rating as a base. Solution: air-gap line: If = 2.20 A Va,ag = 202 / 3 = 116.7 V for the same field current the armature current on short circuit is Ia,sc = 118 A Xs,u = 116.7 / 118 = 0.987 Ω / phase. rated armature current 45000 I a ,rated 118 A 3 220 Ia,sc = 1.0 per unit Va,ag = 202 / 220 = 0.92 per unit Xs,u = 0.92 / 1.00 = 0.92 per unit saturated synchronous reactance: Va ,rated 220 / 3 Xs 0.836 /phase Ia 152 In per unit I a 152 /118 1.29 Xs = 1.00 / 1.29 = 0.775 per unit the short-circuit ratio SCR = 2.84 / 2.20 = 1.29 Xs = 1 / SCR = 1 / 1.29 = 0.775 per unit STEADY-STATE POWER-ANGLE CHARACTERISTICS Synchronous machine is connected to an external system represented as an impedance in series with a voltage source ˆ E1 E1 e j ˆ E2 E2 X Z R jX Z e jZ Z tan 1 R ˆ The power P2 delivered through the impedance to the load-end voltage source E 2 : ˆ ˆ j P2 E2 I cos ˆ E E2 E1 e E2 E1 e j ( Z ) E2 e jZ I 1 Z Z e jZ Z Z ˆ ˆ E1 E I I e j I cos jI sin Re( I ) I cos cos( Z ) 2 cos Z Z Z E1 E2 E2 P2 cos( Z ) 2 cos Z Z Z R E1 E2 E2R E E E 2R cos Z P2 cos( Z ) 2 2 1 2 sin( Z ) 2 2 Z Z Z Z Z where Z 90 Z ˆ Power P1 at source end E1 of the impedance: E1 E2 E2R P 1 sin( Z ) 1 2 Z Z If the resistance R is negligible, Z 90 Z 0 E1 E2 P P2 1 sin the power-angle characteristic for a synchronous machine, : power angle Z Maximum power transfer occurs when 90 E1E2 P1,max P2,max Z ˆ ˆ ˆ ˆ ˆ ˆ If > 0 E1 leads E 2 and the power flows from source E1 to source E 2 . If < 0 E 2 leads E1 and the ˆ ˆ power flows from source E 2 to source E1 . ˆ For a synchronous machine with generated voltage Eaf and synchronous reactance Xs connected to a ˆ system whose Thevenin equivalent is a voltage source VEQ in series with a reactive impedance j XEQ, Eaf VEQ P sin X s X EQ power transferred from the synchronous machine to the system. is the phase ˆ ˆ angle of Eaf with respect to VEQ . Example 5.6 A three-phase, 75-MVA, 13.8-kV synchronous generator with saturated synchronous reactance Xs = 1.35 per unit and unsaturated synchronous reactance Xs,u = 1.56 per unit is connected to an external system with equivalent reactance XEQ = 0.23 per unit and voltage VEQ 1.0 per unit, both on the generator base. It achieves rated open-circuit voltage at a field current of 297 amperes. (a) Find the maximum power Pmax (in MW and per unit) that can be supplied to the external system if the internal voltage of the generator is held equal to 1.0 per unit. (b) Using MATLAB, plot the terminal voltage of the generator as the generator output is varied from zero to Pmax under the conditions of part (a). (c) Now assume that the generator is equipped with an automatic voltage regulator which controls the field current to maintain constant terminal voltage. If the generator is loaded to its rated value, calculate the corresponding power angle, per-unit internal voltage, and field current. Using MATLAB, plot per- unit Ear as a function of per-unit power. Solution: Eaf VEQ 1 Pmax Eaf 1.0 pu, VEQ 1.0 pu Pmax 0.633 pu 47.5 MW X s X EQ 1.35 0.23 the generator terminal current, ˆ ˆ Eaf VEQ Eaf e j VEQ e j 1 ˆ Ia j ( X s X EQ ) j ( X s X EQ ) j1.58 The generator terminal voltage ˆ ˆ ˆ e j 1 j Va VEQ jX EQ I a 1.0 j 0.23 0.8544 0.1456 e j1.58 generator loaded to its rated value P = 1.0 per unit VaVEQ P sin t 4.35sin t ˆ t : angle of the terminal voltage with respect to VEQ X EQ 1.0 t sin1 (1 / 4.35) 13.3 Generator current, ˆ Va e j VEQ e j13.3 1 ˆ Ia 1.007e j 6.65 jX EQ j 0.23 ˆ ˆ ˆ Eaf VEQ j ( X s X EQ ) I a 1.78e j 62.7 Eaf 1.78 I f 1.78 297 529 A Example 5.7 A 2000-hp, 2300-V, unity-power-factor, three-phase, Y-connected, 30-pole, 60-Hz synchronous motor has a synchronous reactance of 1.95 Ω/phase. For this problem all losses may be neglected. (a) Compute the maximum power and torque which this motor can deliver if it is supplied with power directly from a 60-Hz, 2300-V infinite bus. Assume its field excitation is maintained constant at the value which would result in unity power factor at rated load. (b) Instead of the infinite bus of part (a), suppose that the motor is supplied with power from a three-phase, Y- connected, 2300-V, 1500-kVA, two-pole, 3600 r/min turbine generator whose synchronous reactance is 2.65 Ω/phase. The generator is driven at rated speed, and the field excitations of generator and motor are adjusted so that the motor runs at unity power factor and rated terminal voltage at full load. Calculate the maximum power and torque which could be supplied corresponding to these values of field excitation. Solution This machine is of the salient-pole type. We will solve the problem by simple cylindrical-rotor theory. Rated kVA = 2000 × 0.746 = 1492 kVA, three-phase = 497 kVA/phase 2300 Rated voltage = 1328 V line-to-neutral 3 497000 Rated current = 374 A 1328 From the phasor diagram, Eafm Va2 ( I a X sm )2 1515 V 1328 1515 Va Eafm Pmax 1032 kW/phase 3096 kW 3-phase X sm 1.95 In per unit, Pmax = 3096/1492 = 2.07 per unit 2 2 P 3096 103 Synchronous speed, s e 2 60 8 rad/sec Tmax max 123.2 kN-m P 30 s 8 Va = 1328 V, Eafm =1515 V Eafg Va2 ( I a X sg )2 1657 V Eafg Eafm 1657 1515 1638 Pmax 546 kW/phase 1638 kW 3-phase. In per unit, Pmax 1.10 X sg X sm 4.60 1492 Pmax 1638 103 Tmax 65.2 kN-m s 8 STEADY-STATE OPERATING CHARACTERISTICS Consider a synchronous generator delivering power at constant frequency and rated terminal voltage to a load whose power factor is constant. The curve showing the field current required to maintain rated terminal voltage as the constant-power-factor load is varied is known as a compounding curve. Synchronous generators are usually rated in terms of the maximum apparent power (kVA or MVA) load at a specific voltage and power factor (often 80, 85, or 90 percent lagging) which they can carry continuously without overheating. The real power output of the generator is usually limited to a value within the apparent power rating by the capability of its prime mover. When the real-power loading and voltage are fixed, the allowable reactive-power loading is limited by either armature- or field-winding heating. capability curves for a large, hydrogen-cooled turbine generator: Capability curves of an 0.85 power factor, 0.80 short-circuit ratio, hydrogen-cooled turbine generator. Base MVA is rated MVA at 0.5 psig hydrogen. For a given real-power loading, the power factor at which a synchronous machine operates, and hence its armature current, can be controlled by adjusting its field excitation. The curve showing the relation between armature current and field current at a constant terminal voltage and with a constant real power is known as a V curve.

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posted: | 6/26/2012 |

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