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CHAPTER 4
                              SYNCHRONOUS MACHINE

4.1 INTRODUCTION

•   Synchronous machine is designed to be operating at synchronous speed, nsync.

•   The rate rotation of the magnetic fields in the synchronous machine is given as:
                                 120 f e
                            nm =
                                    P
    Where,     nm      =       rate rotation of synchronous machine’s magnetic field, rpm
               fe      =       electrical frequency/frequency supply,Hz
               P       =       number of poles in the machine.

•   The synchronous machine can be used to operate as:
       [a] Synchronous generator              [b] Synchronous motor

•   Used principally in large power applications because of their
       - high operating efficiency,
       - reliability and
       - controllable power factor .

•   Rotates at constant speed in the steady state.

•   The rotating air gap field and the rotor rotate at the same speed.

•   Applications:
       - Used for pumps in generating stations
       - Electric clock
       - Timers
       - Mills
       - Refineries
       - Assist in power factor correction and etc

•   It can draw either leading or lagging reactive current from the ac supply system.

•   It is a doubly excited machine:

       - Rotor poles are excited by a DC current
       - Stator are connected to the ac supply

•   The air gap flux is the resultant of the fluxes due to both rotor and stator.




                                                                                        88
Figure 4.1

4.2 SYNCHRONOUS GENERATOR

Construction

•    Synchronous generator is also known as alternator. This machine consists of two
     main parts:
     i. Field winding (rotor winding)
        - winding that produce the main magnetic field in the machine.
    ii Armature winding (stator winding)
        - winding where the main voltage is induced.

•   Main construction:
    - Rotor
    - Stator

•   Stator
    - has a 3 phase distributed windings(AC supply)– similar to induction machine.
    - Stator winding is sometimes called the armature winding.

•   Rotor
    - has a winding(DC winding) called the field winding.
    - Field winding is normally fed from an external dc source through slip rings and
brushes.
    - Rotor can be divided into two groups:


                                                                                        89
•   High speed machines with cylindrical (or non salient pole) rotors
              •   Low speed machines with salient pole rotors




                                       Figure 4.2

•   Two types of rotor:

       [a] Salient pole rotor – magnetic pole stick out from the surface of the rotor
           • Its rotor poles projecting out from the rotor core.
           • Is use for low-speed hydroelectric generator.
           • Need large number of poles to accumulate in projecting on a rotor (large
               diameter but small length).
           • Almost universal adapt.
           • Has non uniform air-gap.




                      Figure 5.3: Two poles salient pole rotor

       [b] Non salient or cylindrical pole rotor – magnetic pole constructed flush with
          the surface of the rotor.


                                                                                          90
•   Has its rotor in cylindrical form with dc field winding embedded in the
               rotor slots.
           •   Has uniform air-gap.
           •   Provide greater mechanical strength.
           •   Per-unit more accurate dynamic balancing.
           •   For use in high speed turbo generator.
           •   2 / most 4 poles machine use.
           •   Simple to model & analyze.




                      Figure 4.4: Two poles round (cylindrical) rotor

•   Two type of armature winding:
      [a] Single layer winding.
      [b] Double layer winding.

Generated voltage

•   The magnitude of the voltage induced in the given stator phase is given as:

               E A = 2πN C φf       or      E A = Kφω
       where φ = flux in the machine
       f = frequency of the machine
       ω = speed rotation of the machine
       K = Constant representing the construction of the machine

•   From the equation, it can concluded that
       (i) EA proportional to flux and speed
       (ii) Flux proportional to field current, If
       (iii) Thus EA also proportional to If


                                                                                         91
The internal generated voltage EA Vs field current If plot is shown below:

                     EA




                                                                        I
                                      Figure 4.5: Magnetization F
                                                                curve

Equivalent Circuit of a Synchronous Generator

                                 IF    RF                          IA
                                                        jXS   RA
                             +                                              +

                                              +
                      VF                LF         EA                       VΦ
                                                                            V
                      (dc)
                                               -

                             -                                              -
                  Per-phase equivalent circuit of synchronous generator


            E A∠δ = Vφ ∠00 + I A∠ ± θ 0 ( RA + jX S )

EA = internal generated or generated emf per phase
IA = armature current
Vθ = per phase terminal voltage
Xs = synchronous reactance




                                                                                 92
Zs

                                     Ra             Xs
                                                                           Direction Ia out from
                                         Ia                                the generator
                                                                           because generator




                                -
                                    EG           1.0k Vt
                                                 1.0m            load      supply power to the
                                                                           load

                                +




Phasor Diagram of The Synchronous Generator

The load for synchronous machine may be of three types:
           (i)       Pure resistive load (unity power factor)
           (ii)      Inductive load (lagging power factor)
           (iii)     Capacitive load (leading power factor)

(i) Synchronous generator with pure resistive load (unity power factor) :

            Ra             Xs

        I
   -




                    1.0k




        EG         1.0k
                   1.0m                  load
   +




                                         EA
                                                         jXSIA          Unity pf

 IA                                  Vφ          IARA
The equation:

       E A = Vφ ∠0 0 + I A ∠0 0 ( R A + jX S )




                                                                                           93
(ii) Synchronous generator with inductive load (lagging power factor):

            Ra                  Xs




                         1.0k
                        1.0m
                   I
    -




         EG            1.0k
                       1.0m
                                             load
    +




                                                           EA
                                                                         Lagging pf
                  θ                                             jXSIA
                                                   Vφ
              IA                                        IARA
The equation:

        E A = Vφ ∠0 0 + I A ∠ − θ ( R A + jX S )

(iii) Synchronous generator with capacitive load (leading power factor):

             Ra                 Xs
                          1.0k
                         1.0u




              I
-




         EG                                  load
                       1.0k
                       1.0m
    +




                                     EA
                                                        jXSIA
    IA                                                            Leading pf
                                                        IARA
                                           Vφ
The equation:

        E A = Vφ ∠0 0 + I A ∠ + θ ( R A + jX S )

NOTE:
AT NO LOAD CONDITION, IA = 0 A, THUS EA = Vφ∟00 VOLT.
Example 1


                                                                                      94
A 3-phase Y-connected synchronous generator supplies a load of 10MW at power factor
0.85 lagging and the terminal voltage is 11kV. The armature resistance is 0.1 ohm/phase
and synchronous reactance of 0.66 ohm/phase. Calculate the armature current, the
internal generated voltage and voltage regulation. Draw the phasor diagram.

Solution


                    Ia



   0.1Ω
                         1, 0k
                            0




                                 Vt
 j0.66Ω                                    load
             -




                    Eg
             +




        Power factor = 0.85 (lagging)

        θ = cos-1 PF
          = cos-1 0.85
          = 31.790

        Pout = 10MW
              = √3 VLIL cos θ
        IL = IA = Pout / (√3 VL cos θ)
           = 10M / (√3 x 11k x 0.85)
           = 617.5 A

        EA = Vφ∟00 + (IA∟-θ)(RA + jXS)
           = (11k/√3 )∟00 + (IA∟-31.790) (0.1 + j0.66)
          = 6624.07∟2.720 V



                                                  EA    =6624.07V

                                                                Lagging pf
     θ
                         2.72º                         jXSIA
           31.79º                     Vφ
   IA                                        IR
                                     A A
Power flow diagram and Torque in Synchronous Generator



                                                                                     95
Pin=Input torque x gen. speed

                                                        in r/sec

                                                    Pin=Pout + Total Losses



                                                                                     Pout=√ 3VTILcosθ
              Pin=τ appω m

                                                                   2                    Cop.Loss=3(IA)2RA
                                                                   I R losses
                                                       Core        (copper losses)
                                      Friction
                          Stray                        losses
                                      and
                          losses      windage
                                      losses


                                                                                           Can be calculated
                                                      Usually given in                     when RA is given
             If not given in the                      the question
             question, Pstray = 0                                                          and IA is known
                                                                                           If the question don’t
                                                                                           mentioned about RA,
                                                                                           the copper losses = 0



       Synch Generator
       Pin   = Pout + P iron loss + P copper loss
             = Tmwr


                                                                         P copper loss(3Ia2Ra)

P in                          Pm                                                 Pout


                          P iron loss, Pμ (stray+friction+windage+core+etc)




For delta connection,




                                                                                                 96
Vφ = VL
IA = IL / √3
= Pout / (3 Vφ cosθ)
= S / 3 Vφ

For Y connection,

Vφ = VL / √3
IA = IL
= Pout / (√3 VL cosθ) or          Pout / (3 Vφ cosθ)
= S / √3 VL           or          S / 3 Vφ


● Real output power can be determined using the following equation:

           Pout = 3VT I L cos θ             OR         P =3Vφ I A cos θ
                                                        out




● Since XS >> RA, then RA can be ignored. The new phasor diagram will be:

                                                                EA       c

                                                                     θ       EAsinδ =
                                                        jXSIA
                    δ                                                        XSIAcosθ
                                                       Vφ
                O            γ
                        θ                                   a            b
                        IA




● From the pahsor diagram, it can be seen that IA cosθ can be represented as:

                                           E A sin δ
                             I A cos θ =
                                              XS



● Insert IA cosθ in the real output power equation will give:
                                      3Vφ E A sin δ
                                 P=
                                           XS




                                                                                        97
● At maximum condition, δ = 900, thus the equation will be:

                                         3Vφ E A
                              Pmax =
                                           XS

● Other equation for induced torque for synchronous generator:

                                        3Vφ E A sin δ
                              τ ind =
                                          ωm X S




4.2.6   Efficiency and Voltage Regulation

● The voltage regulation of synchronous generator is given as:

                      E A − VPHASE
               VR =                × 100%
                         VPHASE


● The efficiency of the synchronous generator is given as:

                   Pout
              η=        × 100%
                   Pin


                        Pout
              η=                  × 100%
                   Pout + Plosses


Example 2

A 480V, 60Hz, delta-connected, 4 pole synchronous generator has the OCC shown in
figure 1. This generator has a synchronous reactance of 0.1 ohm per phase, and armature
resistance of 0.015 ohm per phase. At full-load, the machine supplies 1200 A at 0.8 PF
lagging. Under full load conditions, the friction and windage losses are 40kW and the
core losses are 30 kW
i) What is the speed rotation of the magnetic field in rpm?
ii) How much is the field current must be supplied to the generator to make the terminal
voltage 480V at no load?
iii) If the generator is now connected to the load and the load draws 1200 A at 0.8 PF
lagging, how much field current will be required to keep the terminal voltage to 480V?
What is the voltage regulation of this generator?
iv) How much power is now generator is supplying? How much power is supplied to the
generator by the prime mover? What is the generator’s efficiency?


                                                                                      98
v) If the generator is now connected to a load drawing 1200 A at 0.8 PF leading, how
much field current will be required to keep VT = 480 V and what is generator’s voltage
regulation?




                                           Figure 1


Solution

For delta connection:

Vφ = VL
IA = IL / √3

i) The speed rotation of magnetic field:
        nm = (120fe )/ P
           = (120 x 60) / 4
           = 1800 rpm

ii) If at no load condition.

At no load, IA = 0 A. Thus,


                                                                                         99
EA = Vφ∟00 Volt.
          = 480 ∟00 V.

Refer to OCC, when EA = 480 V, If = 4.5 A.

iii) If at load (0.8 power factor lagging). IL = 1200 A.

       PF= cos θ = 0.8 (lagging)

       θ = cos-1 0.8
         = 36.870

       IA = IL / √3
          = 1200 / √3
          = 692.82 A.

       Vφ = VL = 480 V

For lagging load, the internal generated voltage is given as:

       EA = Vφ∟00 + (IA∟-θ) (RA + jXS)
          = 480∟00 + (692.82∟-36.870) (0.015 + j0.1)
          = 529.88 + j49.19
          = 532.16 ∟5.300 V

From OCC, when EA = 532.16 V, If = 5.7 A.

The voltage regulation, VR:

       VR = [ (EA - Vφ) / Vφ ]x 100%
         = [(532.16 – 480) / 480] x 100%
         = 10.83%

iv) The output power:

       PF= cos θ = 0.8 (lagging)

       Pout = √3 VL IL cosθ
            = √3 (480) (1200) (0.8)
            = 798.129kW

The input power:

       PIN = Pout + Pstray + Pf&w + Pcore + Pelect




                                                                100
Pelect = 3 IA2RA
              = 3 (692.82)2(0.015)
              = 21.6kW

        PIN = 798.129k + 0 + 40k + 30k + 21.6k
           = 889.729kW

The generator’s efficiency:

        η = (Pout / PIN ) x 100%
          = (798.129k / 889.729k) x 100%
          = 89.7%

v) If at load 0.8 leading.

        IL = 1200 A

        IA = IL/√3
           = 1200 / √3
           = 692.82 A.

        PF = cos θ = 0.8 (leading)

        θ = cos-1 0.8
          = 36.870

        EA = Vφ∟00 + (IA ∟+θ)(RA + jXS)
           = 480∟00 + (692.82∟+36.870)(0.015 + j0.1)
           = 446.74 + j61.66
           = 450.98∟7.860

Refer to OCC, when EA = 450.98 V, If = 4A

The generator’s voltage regulation:

        VR = [(EA - Vφ) / Vφ] x 100%
          = [(450.98 – 480) / 480] x 100%
          = 6.06 %




Measuring Synchronous Generator Model Parameters




                                                       101
● 3 quantities that to be determined,
            Relationship between field current, If and EA
            Synchronous reactance, XS
            Armature resistance, RA
● 2 tests to be conducted,
            open circuit test – terminal of generator is open-circuited, generator run
               at rated speed, If is gradually increased in steps, and terminal voltage is
               measured. Produced open-circuit characteristic (OCC) graph.

                                               Air-gap line
                                    VT (V)


                                                     Open-circuit
                                                  characteristic (OCC)




                                                               If (A)

            short circuit test – terminal of generator is short-circuited through an
             ammeter, IA or IL is measured as If is increased. Produced short-circuit
             characteristic (SCC) graph.

                              IA (A)            Short-circuit
                                             characteristic (SCC)




                                                                        If (A)
● From the short circuit test, the armature current, IA is given as:
             IA = (EA/ (RA + jXS)

● The magnitude of IA is given by:

               |IA| = |EA| / √(RA2 + jXs2)

● The internal machine impedance is given as:

               ZS = √(RA2 + jXS 2 ) = EA / IA




                                                                                        102
● Since Xs >>RA, the equation reduce to:

                                   Xs = Vφoc / IA
                                   = EA/IA

5.2.8 The Synchronous Generator Operating Alone


                                                        E’A
                                                                                                         E’A
                                                        EA
                                     jXSI’A              jXSIA                         δ δ’               EA

           θ         δ δ’            V’φ           Vφ
      IA       I’A                                                         IA    I’A              V’φ    Vφ
                                                                        Unity PF
                                                                        +P or resistive load added, Vφ
 Lagging PF                                                             and VT ↓
                                                                        VR = small +ve
+Q or inductive load added,
Vφ and VT ↓↓
VR = large +ve




                             E’A           jXSIA         Leading PF
     I’A                                                 -Q or capacitive load
IA                             EA           jXSI’A
                                                         added, Vφ and VT ↑↑
                δ       δ’            Vφ       V’φ       VR = -ve
General conclusions from synchronous generator behavior are
          – If lagging loads (+Q or inductive load) are added, Vφ and VT decrease
              significantly but voltage regulation VR is positive large.
          – If unity power factor loads (no reactive load) are added to a generator,
              there is a slight decrease in Vφ and VT and VR is positive small
          – If leading loads (-Q or capacitive power loads) are added, Vφ and VT will
              rise and VR is negative.




5.3 SYNCHRONOUS MOTOR



                                                                                                         103
● Synchronous motor converts electrical power to mechanical power.

● The equivalent circuit of synchronous motor:
                              IF   RF                            IA
                                                      jXS   RA
                          +                                              +

                                             +
                   VF               LF           EA                       V
                                             -                           VΦ
                   (dc)


                          -                                              -


● The difference between synchronous generator equivalent circuits and the
  synchronous generator equivalent circuit is the direction of IA.


            Vφ ∠0 0 = E A ∠ + I A ∠±θ ( R A + jX S )
                           δ


The phasor diagram of synchronous motor with various power factor are shown below:

i) Unity Power Factor




                               E A = Vφ ∠0 0 − I A ∠0 0 ( R A + jX S )




ii) Lagging Power Factor


                                                                                 104
E A = Vφ ∠0 0 − I A ∠ − θ 0 ( R A + jX S )

iii) Leading Power Factor




                                 E A = Vφ ∠0 0 − I A ∠ + θ 0 ( R A + jX S )


Power flow for synch motor is reverse from synch motor

       Synch Motor
       Pin   = Pout + P iron loss + P copper loss
             = Tmwr

                                           P iron loss, Pμ (stray+friction+windage+core+etc)


P in                          Pm                                              Pout


                          P copper loss(3Ia2Ra)

Efficiency and Voltage Regulation

The voltage regulation of synchronous motor is given as:

                          VPHASE − E A
                  VR =                 ×100%
                              EA
                                                                                           105
The efficiency of the synchronous motor is given as:

                    Pout
               η=        × 100%
                    Pin


                         Pout
               η=                  × 100%
                    Pout + Plosses



Example 3

A three phase, Y connected synchronous generator is rated 120 kVA, 1.5 kV, 50 Hz, 0.75
pf lagging. Its synchronous inductance is 2.0mH and effective resistance is 2.5 Ω.
(i)     Determine the voltage regulation at this frequency.
(ii)    Determine the rated voltage and apparent power if the supply frequency is going
        to be twice.
(iii)   Determine the voltage regulation if the frequency is increased to 120% of the
        original frequency.

Solution

(i)

                  2.0 mH           2.5 ohm
           +




                           DC M1




      Ea                   1,0m
                           1,0k              120kVA
                                             1.5kV
           -




                                             50HZ




Z S = RS + jX S                               X S = 2πfLS
                                             = 2π (50)(2m)
                                             = 0.628Ω
Therefore




                                                                                   106
S = 3V L I L
                                                            S
Z S = 2.5 + j 0.628                                  IL =
                                                            3VL
= 2.578∠14.1°
                                                         120k
                                                     =
                                                         3 x1.5k
                                                                     cos φ = 0.75
                                                     = 46.188∠ − φ
                                                                     φ = 41.41°

I L = 46.188∠ − 41.41°
E g = V∠0° + I a Z S
  1.5k
=      ∠0° + 46.188∠ − 41.41°(2.578∠14.1°)
    3
= 971.825 − j 54.63
= 973.36∠ − 3.22 0 A

       E g − VT          973.36 − 866.025
VR =                 =                    = 12.39%
          VT                 866.025

(ii) f = 100 H z

Rated Voltage:
50 H Z → 1.5k
                          100
100 H Z → 1.5k ⇒              x1.5k = 3kV
                           50
Apparent Power
50 H Z → 120kVA
100 H Z → 240kVA

(iii) f = 120%of 50 H Z

f = 60 H Z


    New X S = 2πfLS = 2π (60)(2m) = 0.754

Z S = 2.5 + j 0.754 = 2.61∠16.78°

E g = VT = I a Z s
VT = ? at 60Hz



                                                                               107
Rated Voltage:

50 H Z → 1.5k
                         60
60 H Z → 1.8k ⇒             x1.5k = 1.8kV
                         50

Apparent Power

50 H Z → 120kVA                                                     144k
                                                             IL =
                60                                                 3 x1.8k
60 H Z →           x120 = 144kVA
                50                                           = 46.188 A

E g = VT + I a Z s
        1.8k
  =      + ( 46.188∠ − 41.41°)(2.61∠16.78°)
      3
  = 1148.81 − j 50.24V
  = 1149.91∠ − 2.5 0 V

                                    1.8k
                        1149.91 −
         E g − VT                     3
VR =                =                      = 10.65%
               VT           1.8k
                               3

Example 4

A 2300V, 120hp, 50Hz, eight poles, Y-connected synchronous motor has a synchronous
inductance of 6.63mH/phase and armature resistance of 1Ω/phase at rated power factor of
0.85 leading. At full load, the efficiency is 90 percent. Find the following quantities for
this machine when it is operating at full load.

(i)        Draw a phasor diagram to represent back emf, supply voltage and armature
           current.

(ii)       What is its voltage regulation?

(iii)      Output power.

(iv)       Input power.

(v)        Developed mechanical power.

(vi).      Draw the power flow diagram.


                                                                                       108
Solution

1hp = 746W
2300V, 120hp=89.52kW, 50Hz, 8 poles, efficiency = 90%
Y connected(IΦ=IL, VΦ=VL/√3)

        Xs       = 2πfL=2π(50)(6.63m)=2.08Ω
        Z        = Ra+jXs = 1 + j2.08 Ω
(i)
               Pout         Pout 89.52kW
        η=          , Pin =     =        = 99.47kW
               Pin           η      0.9
        Pin = 3VφIφ cos θ
                Pin           99.47 k
        Iφ =           =                  = 29.38∠31.79° A
             3Vφ cos θ 3(2300 / 3 )(0.85)
        Iφ = 24.97 + j15.48 A


                       2300
        E = V − IZ =             − (24.97 + j15.48)(1 + j 2.08) = 1335.13 − j 67.42V = 1336.83∠ − 2.89°V
                         3


                                                              θ = 31.79°
                                                              δ = −2.89°




                Vt − E       2300 / 3 − 1336.83
(ii)    VR =             =                      = −0.0067 x100% = −0.67%
                  E               1336.83
(iii)   Pout = 120hp = 89.52kW

                Pout 89.52kW
(iv)    Pin =       =        = 99.47kW
                 η      0 .9
(v)     Pmech                =          Pin-Pcl
                             =          99.47kW - 3Ia2R
                             =          99.47kW - 3(29.38)2(1)
                             =          96.88kW

(vi)                                       P iron loss, Pμ (stray+friction+windage+core+etc)

P in                     Pm                                                                           Pout

                             P copper loss




                                                                                                           109
Steady-state Synchronous Motor Operation

Torque Speed Characteristic Curve

 Maximum torque            τind
 when δ=90°
                    τpullout


                                                                  nnl − n fl
                                                           SR =                X 100% = 0%
                                                                     n fl
                     τrated


                                                  nsync            nm

1)    Speed of the motor will be constant (because it’s locked to the electrical
      frequency). The result torque speed characteristic is shown in the figure above.

2)    The speed is constant from no load torque until max torque, thus SR=0%.


Effect Load Changes on Synchronous Motor
                                                                                     3Vφ E A sin δ
                                                               P = 3Vφ I A cos θ =
                                                                                          XS




                                                                                               110
IA1 I
       A2
            IA3                    Vφ
                    IA4
                                                     ∝P1
                                                            ∝P2
                                          EA1                     ∝P3
                                                                        ∝P4
                                         EA2
                                        EA3
                                   EA4



When the load is increased:

        1) θ is change from leading to lagging.
        2) jXSIA is increased, thus IA is increased too.
        3) Torque angle, δ, is increased
        4) |EA| is constant

Effect Field Current Changes on Synchronous Generator

 ∝P (=constant)

                  IA4
                  IA3
                  IA2                          Vφ

                  IA1                                         ∝P (=constant)

                                 EA1          EA2 EA3 EA4
When the field current is increased:




                                                                               111
I     V
      V
  δ
I δ   E
           E
                  1) |EA| increased
                  2) Torque angle, δ, is decreased
                  3) IA first is decreased and then increased.
                  4) θ changed from lagging to leading.
                  5) Real power supply is constant
                  6) VL is constant.


          Synchronous V Curve


           IA

                                                     P = P2
                                                     P = P1
                  Lagging
                  power                             Leading
                  factor                            power
                                                    factor
                               PF = 1.0

                                                           IF
          Starting Synchronous Motor

          3 methods to start a synchronous motor:
                 i) Reduce electrical frequency, fe
                 ii) Use external prime mover
                 iii) Use damper windings or amortisseur windings.


          4.4 RELATIONSHIP BETWEEN SYNCHRONOUS MOTOR AND
          SYNCHRONOUS GENERATOR

                 P  Q         Supply Q             EAcosδ > Vφ
                                                                     Consume Q           EAcosδ < Vφ

           Supply P
                                                              E                          E
                Generator                                                        I
                                                    δ V                              δ           V
           EA leads Vφ
                                                I

           Consume P
                                           I          V                                      V
                 Motor
                                                δ                                    δ    112
           EA lags Vφ                                            E               I           E
4.5 INDUSTRIAL APPLICATION

The three phase synchronous motor is used when a prime mover having a constant speed
from a no-load condition to full load is required. Such as fans, air compressors and
pumps.

Also used to drive mechanical load and also to correct the power factor.

Only used as a correct power factor of an industrial power system, as a bank capacitor
used for power factor correction, also called a synchronous capacitor.

Rating up to 10hp are usually started directly across the rated three-phase voltage.
Synchronous motor of larger sizes are started through a starting compensator or an
automatic starter.




Tutorial 4

1. A 90kVar, 445 V, 60 Hz 3-phase synchronous motor is delta connected. The motor
   has a synchronous inductance 1.3 mH/phase and armature resistance of 4 ohm/phase.
   Calculate the back emf and power angle if the motor is operating at 0.85 lagging
   power factor.

2. A 2400 V, 60 kW, 50 Hz, 6 poles, delta-connected synchronous motor has a
   synchronous reactance of 4 Ω/phase and armature resistance of 2 Ω/phase. At full
   load, the efficiency is 92 %. Find the followings for this machine when it is operating
   at full load at rated power factor 0.85 lagging.

   (i) Phasor diagram to represent back emf, supply voltage and armature current.

   (ii) Voltage regulation.


                                                                                      113
(iii) Input power

   (iv)Developed mechanical power

   (v) Power flow diagram

3. A 3-phase, 50 Hz, Y-connected synchronous generator supplies a load of 10MW. The
   terminal voltage is 22 kV at power factor 0.85 lagging . The armature resistance is 0.2
   ohm/phase and synchronous inductance is 0.3 mH/phase. Calculate the line value of
   emf generated.

4. A hydraulic turbine turning at 200 rpm is connected to a synchronous generator. If the
   induced voltage has a frequency of 60 Hz, how many poles does the rotor have.

5. A 2300 V 3-phase, star connected synchronous motor has an armature resistance of
   0.2 ohm/phase and a synchronous reactance of 2.2 ohm/phase. The motor is operating
   on 0.5 power factor leading with a line current of 200A. Determine the value of
   generated or counter emf per phase. Also draw the phasor diagram. From the phasor
   diagram, discuss what happen to the counter emf if the power factor is increased to
   0.8 leading. No need to recalculate the new emf being generated.

6. A 200kVA, 600 V, 50 Hz three-phase synchronous generator is Y- connected. The
   generator has a synchronous reactance 0.1 ohm/phase and armature resistance of 2
   ohm/phase. Calculate the voltage regulation if the generator is operating at 0.75
   leading power factor. Draw the phasor diagram.

7. A 2000V, 500hp, 3-phase Y-connected synchronous motor has a resistance of 0.3
   ohm/phase and synchronous reactance of 3 ohm/phase respectively. Determine the
   induced emf per phase if the motor works on full-load with an efficiency of 92% and
   p.f = 0.8 leading.




                                                                                      114

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Chapter 4 synchronous machine

  • 1. CHAPTER 4 SYNCHRONOUS MACHINE 4.1 INTRODUCTION • Synchronous machine is designed to be operating at synchronous speed, nsync. • The rate rotation of the magnetic fields in the synchronous machine is given as: 120 f e nm = P Where, nm = rate rotation of synchronous machine’s magnetic field, rpm fe = electrical frequency/frequency supply,Hz P = number of poles in the machine. • The synchronous machine can be used to operate as: [a] Synchronous generator [b] Synchronous motor • Used principally in large power applications because of their - high operating efficiency, - reliability and - controllable power factor . • Rotates at constant speed in the steady state. • The rotating air gap field and the rotor rotate at the same speed. • Applications: - Used for pumps in generating stations - Electric clock - Timers - Mills - Refineries - Assist in power factor correction and etc • It can draw either leading or lagging reactive current from the ac supply system. • It is a doubly excited machine: - Rotor poles are excited by a DC current - Stator are connected to the ac supply • The air gap flux is the resultant of the fluxes due to both rotor and stator. 88
  • 2. Figure 4.1 4.2 SYNCHRONOUS GENERATOR Construction • Synchronous generator is also known as alternator. This machine consists of two main parts: i. Field winding (rotor winding) - winding that produce the main magnetic field in the machine. ii Armature winding (stator winding) - winding where the main voltage is induced. • Main construction: - Rotor - Stator • Stator - has a 3 phase distributed windings(AC supply)– similar to induction machine. - Stator winding is sometimes called the armature winding. • Rotor - has a winding(DC winding) called the field winding. - Field winding is normally fed from an external dc source through slip rings and brushes. - Rotor can be divided into two groups: 89
  • 3. High speed machines with cylindrical (or non salient pole) rotors • Low speed machines with salient pole rotors Figure 4.2 • Two types of rotor: [a] Salient pole rotor – magnetic pole stick out from the surface of the rotor • Its rotor poles projecting out from the rotor core. • Is use for low-speed hydroelectric generator. • Need large number of poles to accumulate in projecting on a rotor (large diameter but small length). • Almost universal adapt. • Has non uniform air-gap. Figure 5.3: Two poles salient pole rotor [b] Non salient or cylindrical pole rotor – magnetic pole constructed flush with the surface of the rotor. 90
  • 4. Has its rotor in cylindrical form with dc field winding embedded in the rotor slots. • Has uniform air-gap. • Provide greater mechanical strength. • Per-unit more accurate dynamic balancing. • For use in high speed turbo generator. • 2 / most 4 poles machine use. • Simple to model & analyze. Figure 4.4: Two poles round (cylindrical) rotor • Two type of armature winding: [a] Single layer winding. [b] Double layer winding. Generated voltage • The magnitude of the voltage induced in the given stator phase is given as: E A = 2πN C φf or E A = Kφω where φ = flux in the machine f = frequency of the machine ω = speed rotation of the machine K = Constant representing the construction of the machine • From the equation, it can concluded that (i) EA proportional to flux and speed (ii) Flux proportional to field current, If (iii) Thus EA also proportional to If 91
  • 5. The internal generated voltage EA Vs field current If plot is shown below: EA I Figure 4.5: Magnetization F curve Equivalent Circuit of a Synchronous Generator IF RF IA jXS RA + + + VF LF EA VΦ V (dc) - - - Per-phase equivalent circuit of synchronous generator E A∠δ = Vφ ∠00 + I A∠ ± θ 0 ( RA + jX S ) EA = internal generated or generated emf per phase IA = armature current Vθ = per phase terminal voltage Xs = synchronous reactance 92
  • 6. Zs Ra Xs Direction Ia out from Ia the generator because generator - EG 1.0k Vt 1.0m load supply power to the load + Phasor Diagram of The Synchronous Generator The load for synchronous machine may be of three types: (i) Pure resistive load (unity power factor) (ii) Inductive load (lagging power factor) (iii) Capacitive load (leading power factor) (i) Synchronous generator with pure resistive load (unity power factor) : Ra Xs I - 1.0k EG 1.0k 1.0m load + EA jXSIA Unity pf IA Vφ IARA The equation: E A = Vφ ∠0 0 + I A ∠0 0 ( R A + jX S ) 93
  • 7. (ii) Synchronous generator with inductive load (lagging power factor): Ra Xs 1.0k 1.0m I - EG 1.0k 1.0m load + EA Lagging pf θ jXSIA Vφ IA IARA The equation: E A = Vφ ∠0 0 + I A ∠ − θ ( R A + jX S ) (iii) Synchronous generator with capacitive load (leading power factor): Ra Xs 1.0k 1.0u I - EG load 1.0k 1.0m + EA jXSIA IA Leading pf IARA Vφ The equation: E A = Vφ ∠0 0 + I A ∠ + θ ( R A + jX S ) NOTE: AT NO LOAD CONDITION, IA = 0 A, THUS EA = Vφ∟00 VOLT. Example 1 94
  • 8. A 3-phase Y-connected synchronous generator supplies a load of 10MW at power factor 0.85 lagging and the terminal voltage is 11kV. The armature resistance is 0.1 ohm/phase and synchronous reactance of 0.66 ohm/phase. Calculate the armature current, the internal generated voltage and voltage regulation. Draw the phasor diagram. Solution Ia 0.1Ω 1, 0k 0 Vt j0.66Ω load - Eg + Power factor = 0.85 (lagging) θ = cos-1 PF = cos-1 0.85 = 31.790 Pout = 10MW = √3 VLIL cos θ IL = IA = Pout / (√3 VL cos θ) = 10M / (√3 x 11k x 0.85) = 617.5 A EA = Vφ∟00 + (IA∟-θ)(RA + jXS) = (11k/√3 )∟00 + (IA∟-31.790) (0.1 + j0.66) = 6624.07∟2.720 V EA =6624.07V Lagging pf θ 2.72º jXSIA 31.79º Vφ IA IR A A Power flow diagram and Torque in Synchronous Generator 95
  • 9. Pin=Input torque x gen. speed in r/sec Pin=Pout + Total Losses Pout=√ 3VTILcosθ Pin=τ appω m 2 Cop.Loss=3(IA)2RA I R losses Core (copper losses) Friction Stray losses and losses windage losses Can be calculated Usually given in when RA is given If not given in the the question question, Pstray = 0 and IA is known If the question don’t mentioned about RA, the copper losses = 0 Synch Generator Pin = Pout + P iron loss + P copper loss = Tmwr P copper loss(3Ia2Ra) P in Pm Pout P iron loss, Pμ (stray+friction+windage+core+etc) For delta connection, 96
  • 10. Vφ = VL IA = IL / √3 = Pout / (3 Vφ cosθ) = S / 3 Vφ For Y connection, Vφ = VL / √3 IA = IL = Pout / (√3 VL cosθ) or Pout / (3 Vφ cosθ) = S / √3 VL or S / 3 Vφ ● Real output power can be determined using the following equation: Pout = 3VT I L cos θ OR P =3Vφ I A cos θ out ● Since XS >> RA, then RA can be ignored. The new phasor diagram will be: EA c θ EAsinδ = jXSIA δ XSIAcosθ Vφ O γ θ a b IA ● From the pahsor diagram, it can be seen that IA cosθ can be represented as: E A sin δ I A cos θ = XS ● Insert IA cosθ in the real output power equation will give: 3Vφ E A sin δ P= XS 97
  • 11. ● At maximum condition, δ = 900, thus the equation will be: 3Vφ E A Pmax = XS ● Other equation for induced torque for synchronous generator: 3Vφ E A sin δ τ ind = ωm X S 4.2.6 Efficiency and Voltage Regulation ● The voltage regulation of synchronous generator is given as: E A − VPHASE VR = × 100% VPHASE ● The efficiency of the synchronous generator is given as: Pout η= × 100% Pin Pout η= × 100% Pout + Plosses Example 2 A 480V, 60Hz, delta-connected, 4 pole synchronous generator has the OCC shown in figure 1. This generator has a synchronous reactance of 0.1 ohm per phase, and armature resistance of 0.015 ohm per phase. At full-load, the machine supplies 1200 A at 0.8 PF lagging. Under full load conditions, the friction and windage losses are 40kW and the core losses are 30 kW i) What is the speed rotation of the magnetic field in rpm? ii) How much is the field current must be supplied to the generator to make the terminal voltage 480V at no load? iii) If the generator is now connected to the load and the load draws 1200 A at 0.8 PF lagging, how much field current will be required to keep the terminal voltage to 480V? What is the voltage regulation of this generator? iv) How much power is now generator is supplying? How much power is supplied to the generator by the prime mover? What is the generator’s efficiency? 98
  • 12. v) If the generator is now connected to a load drawing 1200 A at 0.8 PF leading, how much field current will be required to keep VT = 480 V and what is generator’s voltage regulation? Figure 1 Solution For delta connection: Vφ = VL IA = IL / √3 i) The speed rotation of magnetic field: nm = (120fe )/ P = (120 x 60) / 4 = 1800 rpm ii) If at no load condition. At no load, IA = 0 A. Thus, 99
  • 13. EA = Vφ∟00 Volt. = 480 ∟00 V. Refer to OCC, when EA = 480 V, If = 4.5 A. iii) If at load (0.8 power factor lagging). IL = 1200 A. PF= cos θ = 0.8 (lagging) θ = cos-1 0.8 = 36.870 IA = IL / √3 = 1200 / √3 = 692.82 A. Vφ = VL = 480 V For lagging load, the internal generated voltage is given as: EA = Vφ∟00 + (IA∟-θ) (RA + jXS) = 480∟00 + (692.82∟-36.870) (0.015 + j0.1) = 529.88 + j49.19 = 532.16 ∟5.300 V From OCC, when EA = 532.16 V, If = 5.7 A. The voltage regulation, VR: VR = [ (EA - Vφ) / Vφ ]x 100% = [(532.16 – 480) / 480] x 100% = 10.83% iv) The output power: PF= cos θ = 0.8 (lagging) Pout = √3 VL IL cosθ = √3 (480) (1200) (0.8) = 798.129kW The input power: PIN = Pout + Pstray + Pf&w + Pcore + Pelect 100
  • 14. Pelect = 3 IA2RA = 3 (692.82)2(0.015) = 21.6kW PIN = 798.129k + 0 + 40k + 30k + 21.6k = 889.729kW The generator’s efficiency: η = (Pout / PIN ) x 100% = (798.129k / 889.729k) x 100% = 89.7% v) If at load 0.8 leading. IL = 1200 A IA = IL/√3 = 1200 / √3 = 692.82 A. PF = cos θ = 0.8 (leading) θ = cos-1 0.8 = 36.870 EA = Vφ∟00 + (IA ∟+θ)(RA + jXS) = 480∟00 + (692.82∟+36.870)(0.015 + j0.1) = 446.74 + j61.66 = 450.98∟7.860 Refer to OCC, when EA = 450.98 V, If = 4A The generator’s voltage regulation: VR = [(EA - Vφ) / Vφ] x 100% = [(450.98 – 480) / 480] x 100% = 6.06 % Measuring Synchronous Generator Model Parameters 101
  • 15. ● 3 quantities that to be determined,  Relationship between field current, If and EA  Synchronous reactance, XS  Armature resistance, RA ● 2 tests to be conducted,  open circuit test – terminal of generator is open-circuited, generator run at rated speed, If is gradually increased in steps, and terminal voltage is measured. Produced open-circuit characteristic (OCC) graph. Air-gap line VT (V) Open-circuit characteristic (OCC) If (A)  short circuit test – terminal of generator is short-circuited through an ammeter, IA or IL is measured as If is increased. Produced short-circuit characteristic (SCC) graph. IA (A) Short-circuit characteristic (SCC) If (A) ● From the short circuit test, the armature current, IA is given as: IA = (EA/ (RA + jXS) ● The magnitude of IA is given by: |IA| = |EA| / √(RA2 + jXs2) ● The internal machine impedance is given as: ZS = √(RA2 + jXS 2 ) = EA / IA 102
  • 16. ● Since Xs >>RA, the equation reduce to: Xs = Vφoc / IA = EA/IA 5.2.8 The Synchronous Generator Operating Alone E’A E’A EA jXSI’A jXSIA δ δ’ EA θ δ δ’ V’φ Vφ IA I’A IA I’A V’φ Vφ Unity PF +P or resistive load added, Vφ Lagging PF and VT ↓ VR = small +ve +Q or inductive load added, Vφ and VT ↓↓ VR = large +ve E’A jXSIA Leading PF I’A -Q or capacitive load IA EA jXSI’A added, Vφ and VT ↑↑ δ δ’ Vφ V’φ VR = -ve General conclusions from synchronous generator behavior are – If lagging loads (+Q or inductive load) are added, Vφ and VT decrease significantly but voltage regulation VR is positive large. – If unity power factor loads (no reactive load) are added to a generator, there is a slight decrease in Vφ and VT and VR is positive small – If leading loads (-Q or capacitive power loads) are added, Vφ and VT will rise and VR is negative. 5.3 SYNCHRONOUS MOTOR 103
  • 17. ● Synchronous motor converts electrical power to mechanical power. ● The equivalent circuit of synchronous motor: IF RF IA jXS RA + + + VF LF EA V - VΦ (dc) - - ● The difference between synchronous generator equivalent circuits and the synchronous generator equivalent circuit is the direction of IA. Vφ ∠0 0 = E A ∠ + I A ∠±θ ( R A + jX S ) δ The phasor diagram of synchronous motor with various power factor are shown below: i) Unity Power Factor E A = Vφ ∠0 0 − I A ∠0 0 ( R A + jX S ) ii) Lagging Power Factor 104
  • 18. E A = Vφ ∠0 0 − I A ∠ − θ 0 ( R A + jX S ) iii) Leading Power Factor E A = Vφ ∠0 0 − I A ∠ + θ 0 ( R A + jX S ) Power flow for synch motor is reverse from synch motor Synch Motor Pin = Pout + P iron loss + P copper loss = Tmwr P iron loss, Pμ (stray+friction+windage+core+etc) P in Pm Pout P copper loss(3Ia2Ra) Efficiency and Voltage Regulation The voltage regulation of synchronous motor is given as: VPHASE − E A VR = ×100% EA 105
  • 19. The efficiency of the synchronous motor is given as: Pout η= × 100% Pin Pout η= × 100% Pout + Plosses Example 3 A three phase, Y connected synchronous generator is rated 120 kVA, 1.5 kV, 50 Hz, 0.75 pf lagging. Its synchronous inductance is 2.0mH and effective resistance is 2.5 Ω. (i) Determine the voltage regulation at this frequency. (ii) Determine the rated voltage and apparent power if the supply frequency is going to be twice. (iii) Determine the voltage regulation if the frequency is increased to 120% of the original frequency. Solution (i) 2.0 mH 2.5 ohm + DC M1 Ea 1,0m 1,0k 120kVA 1.5kV - 50HZ Z S = RS + jX S X S = 2πfLS = 2π (50)(2m) = 0.628Ω Therefore 106
  • 20. S = 3V L I L S Z S = 2.5 + j 0.628 IL = 3VL = 2.578∠14.1° 120k = 3 x1.5k cos φ = 0.75 = 46.188∠ − φ φ = 41.41° I L = 46.188∠ − 41.41° E g = V∠0° + I a Z S 1.5k = ∠0° + 46.188∠ − 41.41°(2.578∠14.1°) 3 = 971.825 − j 54.63 = 973.36∠ − 3.22 0 A E g − VT 973.36 − 866.025 VR = = = 12.39% VT 866.025 (ii) f = 100 H z Rated Voltage: 50 H Z → 1.5k 100 100 H Z → 1.5k ⇒ x1.5k = 3kV 50 Apparent Power 50 H Z → 120kVA 100 H Z → 240kVA (iii) f = 120%of 50 H Z f = 60 H Z New X S = 2πfLS = 2π (60)(2m) = 0.754 Z S = 2.5 + j 0.754 = 2.61∠16.78° E g = VT = I a Z s VT = ? at 60Hz 107
  • 21. Rated Voltage: 50 H Z → 1.5k 60 60 H Z → 1.8k ⇒ x1.5k = 1.8kV 50 Apparent Power 50 H Z → 120kVA 144k IL = 60 3 x1.8k 60 H Z → x120 = 144kVA 50 = 46.188 A E g = VT + I a Z s 1.8k = + ( 46.188∠ − 41.41°)(2.61∠16.78°) 3 = 1148.81 − j 50.24V = 1149.91∠ − 2.5 0 V 1.8k 1149.91 − E g − VT 3 VR = = = 10.65% VT 1.8k 3 Example 4 A 2300V, 120hp, 50Hz, eight poles, Y-connected synchronous motor has a synchronous inductance of 6.63mH/phase and armature resistance of 1Ω/phase at rated power factor of 0.85 leading. At full load, the efficiency is 90 percent. Find the following quantities for this machine when it is operating at full load. (i) Draw a phasor diagram to represent back emf, supply voltage and armature current. (ii) What is its voltage regulation? (iii) Output power. (iv) Input power. (v) Developed mechanical power. (vi). Draw the power flow diagram. 108
  • 22. Solution 1hp = 746W 2300V, 120hp=89.52kW, 50Hz, 8 poles, efficiency = 90% Y connected(IΦ=IL, VΦ=VL/√3) Xs = 2πfL=2π(50)(6.63m)=2.08Ω Z = Ra+jXs = 1 + j2.08 Ω (i) Pout Pout 89.52kW η= , Pin = = = 99.47kW Pin η 0.9 Pin = 3VφIφ cos θ Pin 99.47 k Iφ = = = 29.38∠31.79° A 3Vφ cos θ 3(2300 / 3 )(0.85) Iφ = 24.97 + j15.48 A 2300 E = V − IZ = − (24.97 + j15.48)(1 + j 2.08) = 1335.13 − j 67.42V = 1336.83∠ − 2.89°V 3 θ = 31.79° δ = −2.89° Vt − E 2300 / 3 − 1336.83 (ii) VR = = = −0.0067 x100% = −0.67% E 1336.83 (iii) Pout = 120hp = 89.52kW Pout 89.52kW (iv) Pin = = = 99.47kW η 0 .9 (v) Pmech = Pin-Pcl = 99.47kW - 3Ia2R = 99.47kW - 3(29.38)2(1) = 96.88kW (vi) P iron loss, Pμ (stray+friction+windage+core+etc) P in Pm Pout P copper loss 109
  • 23. Steady-state Synchronous Motor Operation Torque Speed Characteristic Curve Maximum torque τind when δ=90° τpullout nnl − n fl SR = X 100% = 0% n fl τrated nsync nm 1) Speed of the motor will be constant (because it’s locked to the electrical frequency). The result torque speed characteristic is shown in the figure above. 2) The speed is constant from no load torque until max torque, thus SR=0%. Effect Load Changes on Synchronous Motor 3Vφ E A sin δ P = 3Vφ I A cos θ = XS 110
  • 24. IA1 I A2 IA3 Vφ IA4 ∝P1 ∝P2 EA1 ∝P3 ∝P4 EA2 EA3 EA4 When the load is increased: 1) θ is change from leading to lagging. 2) jXSIA is increased, thus IA is increased too. 3) Torque angle, δ, is increased 4) |EA| is constant Effect Field Current Changes on Synchronous Generator ∝P (=constant) IA4 IA3 IA2 Vφ IA1 ∝P (=constant) EA1 EA2 EA3 EA4 When the field current is increased: 111
  • 25. I V V δ I δ E E 1) |EA| increased 2) Torque angle, δ, is decreased 3) IA first is decreased and then increased. 4) θ changed from lagging to leading. 5) Real power supply is constant 6) VL is constant. Synchronous V Curve IA P = P2 P = P1 Lagging power Leading factor power factor PF = 1.0 IF Starting Synchronous Motor 3 methods to start a synchronous motor: i) Reduce electrical frequency, fe ii) Use external prime mover iii) Use damper windings or amortisseur windings. 4.4 RELATIONSHIP BETWEEN SYNCHRONOUS MOTOR AND SYNCHRONOUS GENERATOR P Q Supply Q EAcosδ > Vφ Consume Q EAcosδ < Vφ Supply P E E Generator I δ V δ V EA leads Vφ I Consume P I V V Motor δ δ 112 EA lags Vφ E I E
  • 26. 4.5 INDUSTRIAL APPLICATION The three phase synchronous motor is used when a prime mover having a constant speed from a no-load condition to full load is required. Such as fans, air compressors and pumps. Also used to drive mechanical load and also to correct the power factor. Only used as a correct power factor of an industrial power system, as a bank capacitor used for power factor correction, also called a synchronous capacitor. Rating up to 10hp are usually started directly across the rated three-phase voltage. Synchronous motor of larger sizes are started through a starting compensator or an automatic starter. Tutorial 4 1. A 90kVar, 445 V, 60 Hz 3-phase synchronous motor is delta connected. The motor has a synchronous inductance 1.3 mH/phase and armature resistance of 4 ohm/phase. Calculate the back emf and power angle if the motor is operating at 0.85 lagging power factor. 2. A 2400 V, 60 kW, 50 Hz, 6 poles, delta-connected synchronous motor has a synchronous reactance of 4 Ω/phase and armature resistance of 2 Ω/phase. At full load, the efficiency is 92 %. Find the followings for this machine when it is operating at full load at rated power factor 0.85 lagging. (i) Phasor diagram to represent back emf, supply voltage and armature current. (ii) Voltage regulation. 113
  • 27. (iii) Input power (iv)Developed mechanical power (v) Power flow diagram 3. A 3-phase, 50 Hz, Y-connected synchronous generator supplies a load of 10MW. The terminal voltage is 22 kV at power factor 0.85 lagging . The armature resistance is 0.2 ohm/phase and synchronous inductance is 0.3 mH/phase. Calculate the line value of emf generated. 4. A hydraulic turbine turning at 200 rpm is connected to a synchronous generator. If the induced voltage has a frequency of 60 Hz, how many poles does the rotor have. 5. A 2300 V 3-phase, star connected synchronous motor has an armature resistance of 0.2 ohm/phase and a synchronous reactance of 2.2 ohm/phase. The motor is operating on 0.5 power factor leading with a line current of 200A. Determine the value of generated or counter emf per phase. Also draw the phasor diagram. From the phasor diagram, discuss what happen to the counter emf if the power factor is increased to 0.8 leading. No need to recalculate the new emf being generated. 6. A 200kVA, 600 V, 50 Hz three-phase synchronous generator is Y- connected. The generator has a synchronous reactance 0.1 ohm/phase and armature resistance of 2 ohm/phase. Calculate the voltage regulation if the generator is operating at 0.75 leading power factor. Draw the phasor diagram. 7. A 2000V, 500hp, 3-phase Y-connected synchronous motor has a resistance of 0.3 ohm/phase and synchronous reactance of 3 ohm/phase respectively. Determine the induced emf per phase if the motor works on full-load with an efficiency of 92% and p.f = 0.8 leading. 114