EENG350 Lecture Notes Ch5

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EENG350 Lecture Notes Ch5 Powered By Docstoc
					                                          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
                              22
                          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 je 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   cos1 (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 cosa  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  1140 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  103

                                    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 jZ     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  jZ
                                            I 1
                                                 Z        Z e jZ  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  sin1 (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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