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