CHAPTER III
ALTERNATORS
3.1. Fundamentals of AC Generator (Alternator)
Alternating current generators or alternators (as they are usually called) operate on the same
fundamental principles of electromagnetic induction as DC generators. They also consist of an armature
winding and a magnetic field. But there is one important difference between the two. Whereas in DC
generators, the armature rotates and the field system is stationary, the arrangement in alternators is just the
reverse of it. In their case, standard construction consists of armature winding mounted on a stationary element
called stator and field windings on a rotating element called rotor. The details of construction are shown in
figure 1.
Figure 1 (p.1402, Theraja, B.L. and Theraja, A.K. A Textbook of Electrical Technology)
The stator consists of a cast iron frame, which supports the armature core, having slots on its
inner periphery for housing the armature conductors. The rotor is like a flywheel having alternate N and S poles
fixed to its outer rim. The magnetic poles are excited (or magnetized) from direct current supplied by a dc
source at 125 to 600 volts. In most cases, necessary exciting (or magnetizing) current is obtained from a small
dc shunt generator which is belted or mounted on the shaft of the alternator itself. Because the field magnets are
rotating, this current is supplied through two slip rings. As the exciting voltage is relatively small, the slip rings
and brush gear are of light construction. Recently, brushless excitation systems have been developed in which a
3-phase ac exciter and a group of rectifiers supply dc to the alternator. Hence, brushes, slip rings and
commutator are eliminated.
When the rotor rotates, the stator conductors (being stationary) are cut by the magnetic flux,
hence they have induced emf produced in them. Because the magnetic poles are alternately N and S, they
induced an emf and hence current in the armature conductors, which first flows in one direction and then in the
other. Hence, an alternating emf is produced in the stator conductors (i) whose frequency depends on the
Number of N and S poles moving past a conductor in one second and (ii) whose direction is given by Fleming’s
Right-hand rule.
36
3.2. Generated EMF Per Phase (Design Equation)
Ep
2.22k p kb f Z
n
2
sin
kb
f
kp
1800
slots per pole
PN
120
sin 900
pitch
n sin
n
2
number of slots per pole per phase
pitch
coil span in number of slots
slots per pole
Where: Ep = generated emf per phase (volt)
kp = pitch factor or coil span factor
f = frequency (Hertz)
Kb = belt factor or distribution factor
= flux per pole (weber)
Z = number of conductor per phase
P = number of poles
N = speed of rotor’s rotation (rpm)
If flux is given in unit maxwells or lines, the above formula reduces to:
EP
Sample Problem 1:
What is the pitch factor?
2.22k p kb f Z x 10
8
A six-pole, 3-phase AC generator has 72 slots, the coil span is 12.
Answer :
coil span
slots per pole
pitch
12
1
72 / 6
pitch factor
sin 900 x pitch
sin 900 x 1
1
Sample Problem 2:
A 400kVA, 6.9kV, 60 cycle Y-connected alternator has the following
data: poles = 6; conductors per slot = 12; total slots = 54; winding = 2 layer lap; pitch = 8/9;
flux per pole = 4 x 107 maxwells. Determine the (a) belt factor (b) pitch factor (c) conductors in series per phase
(d) terminal voltage.
Answer :
1800
slots per pole
A.
kb
1800
54 / 6
3(20 0 )
2
n
sin
2
sin
n sin
200
3sin
2
B. k p
2
200
n
slots / pole / phase
54
6(3)
3
0.96
sin[(900 )( pitch) sin[(900 )(8 / 9)] 0.985
12 conductors
648
x 54 slots 648 conductors
Z
216 conductors / phase
slot
3
2.22k p kb f Z x 10 8 2.22(0.985)(0.96)(60)(4 x107 )(216) x10 8 10,882.42 volts / phase
C. ZTOTAL
D. EP
EL
3EP
3(10,882.42) 18,849 volts
37
3.3. Frequency of the Generated EMF
PN
120
f
where: f = frequency (Hertz)
P = number of poles
N = speed of rotor’s rotation (rpm)
3.4. Generated EMF Per Phase (Electric Circuit Relationship)
Case 1: If load has Unity Power Factor
EP
VP
I P Ra
2
IP X a
2
tan
1
IP X a
VP I P Ra
VP = load voltage per phase (volt)
Where: EP = generated emf per phase (volt)
IP = armature current per phase (ampere)
Ra = armature winding resistance per phase
Xa = inductive reactance of armature winding per phase
between EP and IP
Case 2: If load has a Lagging Power Factor
38
EP
VP cos
I P Ra
2
VP sin
IP X a
2
tan
1
IP X a
2
tan
1
VP sin
VP cos
IP X a
I P Ra
VP sin
IP X a
VP cos
I P Ra
Case 3: If load has a Leading Power Factor
EP
VP cos
I P Ra
Note:
VP sin
EP
Using Complex Numbers:
Where: +
2
VP 0
f power factor is leading
P and VP
0
IP
Ra
jX a
-
Sample Problem:
A single phase AC generator has an armature winding resistance and
reactance of 0.2 ohm and 2.0 ohms respectively. At rated load, the current delivered is 50A at 500V. Determine
the generated emf when the load power factor is (a) Unity (b) 0.866 lagging (c) 0.707 leading.
Answer :
cos 1 1.0 00
A.
I P Ra
2
EP
VP cos
VP sin
EP
500(cos 00 ) (50)(0.2)
2
IP X a
2
500(sin 00 ) (50)(2.0)
2
519.71 volts
cos 1 0.866 300
B.
VP cos
EP
500(cos 300 ) (50)(0.2)
cos 1 0.707
C.
I P Ra
2
EP
VP sin
2
IP X a
2
500(sin 300 ) (50)(2.0)
2
564.59 volts
450
I P Ra
2
EP
VP cos
VP sin
EP
500(cos 450 ) (50)(0.2)
2
IP X a
2
500(sin 450 ) (50)(2.0)
2
443.24 volts
3.5. Test on Alternators
3.5.1. Short-Circuit Test – The armature terminals are short circuited while a small field current
is applied. Adjustments maybe made such that IP is approximately equal to the rated load armature current per
phase.
39
I SC ( wye )
A
I SC ( delta )
A
3
where: A = ammeter reading during the test
3.5.2. Open-Circuit Test – The armature terminals are open circuited and the field current is
made equal to the field current during the short circuit test.
EOC ( wye )
V
3
EOC ( delta )
V
where: V = voltmeter reading during the test
3.6. DC Resistance Test
Using an ohmmeter, the DC resistance per phase is to be determined.
Ra ( wye )
1
Rt
2
Ra ( delta )
3
Rt
2
Where: Ra = armature DC resistance per phase
Rt = DC resistance of the alternator between any two terminals (ohmmeter reading during the test)
40
Remember: In practice, it is customary to multiply the DC resistance Ra by a factor of 1.25 to
1.75 to obtain a value more nearly equal to the armature resistance when it carries AC current.
Alternator parameters derived from test
EOC
I SC
Zs
( Z s ) 2 ( Ra ) 2
Xs
Where: Xs = synchronous reactance per phase
Ra = armature resistance per phase
ISC = short circuit armature current per phase
Zs = synchronous impedance per phase
EOC = open circuit voltage per phase
Sample Problem:
A generator is rated 100MW, 13.8kV and 90% power factor. A wyeconnected alternator was tested for its effective resistance. The ratio of the effective resistance to ohmic
resistance was previously determined to be 1.35. A 12-V battery was connected across two terminals and the
ammeter reads 120A. What is the per phase effective resistance of the alternator?
Answer :
Rt
EDC
I DC
Reffective
12
120
0.1
1.35Ra
Rt
2
Ra
0.1
0.05
2
1.35(0.05) 0.0675
3.7. Per Unit Values of the Alternator Parameters
The Per Unit value of any quantity is defined as the ratio of the quantity to its base value
expressed as decimal.
Single-Phase System
I base
Sbase
Ebase
Z base
Ebase
I base
Ebase 2
Sbase
Three-Phase System
I base
Sbase
3Ebase (line )
R pu
% IR
X pu
% IX
Ra
Z base
Xa
Z base
Z base
Sbase Ra
Ebase 2
Sbase X a
Ebase 2
Ebase (line ) 2
Sbase
Pcopper ( rated )
Sbase
I base 2 X a
Sbase
Where: Sbase = rated apparent power of the alternator (volt-ampere)
Ebase = rated line to line voltage of the alternator (volt)
Ibase = rated line current of the alternator (ampere)
Rpu = per unit resistance or percentage resistance
Xpu = per unit reactance or percentage reactance
Sample Problem 1:
phase alternator.
Determine the per phase nominal impedance of a 30MVA, 15kV three-
Answer :
Z phase
Ebase ( line ) 2
Sbase
(15, 000)2
30 x106
7.5
41
Sample Problem 2:
A generator is rated 100MW, 13.8kV and 90% power factor. The
effective resistance to ohmic resistance ratio is 1.5. The ohmic resistance is obtained by connecting two
terminals of the wye-connected windings to a dc source of 6V. The current drawn is 87.6A. What is the percent
effective resistance?
Answer :
EDC
I DC
6
87.6
Ra ( effective )
1.5Ra
Rt
Sbase Ra
Ebase 2
% IR
0.0685 ohm
Ra
Rt
2
0.0685
2
0.03425 ohm
1.5(0.03425) 0.051375 ohm
Pbase
pf
Ra
Ebase 2
100 x106
0.9
0.051375
x100%
(13.8 x103 ) 2
% IR 3%
3.8. Voltage Regulation
Percent voltage regulation is the percentage rise in the terminal voltage of a generator when its
load is removed.
EP VP
x100%
VP
%VR
Where:
%VR
(cos
+%IX = if pf is lagging
% IR) 2 (sin
% IX ) 2 1
-%IX = if pf is leading
Sample Problem:
A 2.5MVA, 6.6kV alternator is operating at full load and 0.8 pf
lagging. This machine has a synchronous reactance per phase of 10.4 ohms and a negligible resistance.
Calculate the percentage voltage regulation.
Answer :
I
EP
S
3E
2.5 x106
3(6, 600)
I P Ra ) 2 (VP sin
(VP cos
cos 1 0.8 36.8690 ; VP
218.69 A;
6, 600
3
3810.5 V
I P X a )2
[3810.5(cos 36.8690 ) 218.69(0)]2 [3810.5(sin 36.8690 ) 218.69(10.4)]2
EP
5485.66 V
%VR
5485.66 3810.5
x100%
3810.5
43.96%
3.9. Power Developed in the Armature Per Phase
On a large synchronous generator, except in very small ones, Ra << Xs making
Pdelivered = Pdeveloped.
Pd
EPVP
sin
XS
Pd (max)
EPVP
XS
Where: Pd = power developed per phase (watt)
Pd(max)
EP = generated emf per phase (volt)
VP = terminal voltage per phase (volt)
XS = synchronous reactance per phase (ohm)
P and VP
900
0
)
42
Sample Problem:
An 11,000V three-phase wye-connected turbo alternator has a
synchronous reactance of 6 ohms and a negligible resistance per phase. When the field current is 8A, the open
circuit voltage is 12,000V. Determine the (a) maximum power developed (b) armature current when it
developed maximum power.
Answer :
A. Pd (max)
ELVL
XS
12, 000(11, 000)
6
900
B. At max . power developed ,
VP 00
EP
12, 000
900
3
IZ a
j 6928.203 6350.85 I ( j 6)
I
22 MW
I
11, 000 0
0 I ( j 6)
3
6350.85 j 6928.203
j6
1566 A
3.10. Power Losses and Efficiency
Pout
Pout Plosses
Pout (1
)
Pout (3
)
Pf
Plosses
VL I L cos
Pcu (1
3VL I L cos
2
I f Rf
I fVf
Pcu
Pfw
)
I p 2 Ra
Pcu (3
)
Pcore
Pf
Pv
Pstray load
3I p 2 Ra
Vf 2
Rf
Where: VL = line to line voltage
IP = phase current
IL = line current
cos
Pcu = armature winding copper losses
Pfw = friction and windage losses
Pcore = core losses or iron losses = eddy currents and hysteresis losses
Pstray load = stray load losses
Pv = ventilation losses
Pf = field winding losses
If = field current or exciter current
Rf = field resistance
Ra = armature resistance per phase
Vf = voltage across the field winding
Remember:
Friction and windage loss, field loss, core loss, stray load and ventilation losses
are constant losses, unless otherwise specified.
Armature copper losses at any size of load
Pcux
( x) 2 PcuFL
Where: x = percentage loading on the alternator
Pcux = copper losses at x% load
PcuFL = full load copper losses
Sample Problem:
Determine the efficiency of a 1500kVA, 2300V, 3-phase, Y-connected
alternator, which operates at rated output with a power factor of 80%. The dc armature resistance between
terminals is 0.08 ohm. The field takes 70A at 120V from the exciter equipment. Friction and windage losses
are 15kW, iron loss is 35kW and stray load losses are 1.5kW. Assume the effective armature winding resistance
is 1.5 times the DC value.
43
Answer :
1500 x103
3(2300)
I
S
3VL
Ra
1.5Rdc
1.5(0.04) 0.06
Pa
3I P 2 Ra
3(376.53) 2 (0.06)
Plosses
Pa
Pf
Pout
Pout Plosses
Pcore
376.53 A
Pfw
Rt
2
Rdc
25.52 kW
Pstray load
0.08
2
Pf
0.04
Vf I f
120(70) 8.4 kW
25.52 8.4 35 15 1.5 85.42 kW
1500(0.8)
x100%
1500(0.8) 85.42
93.35%
3.11. Governor’s Speed Regulation (GSR)
%GSR
N NL N FL
x100%
N FL
%GSR
f NL
f FL
f FL
x100%
NFL = full load speed
Where: NNL = no load speed
FNL = no load frequency
fFL = full load frequency
3.12. Characteristic-Triangle and Parallel Operation of Alternators
Note:
generator decreases.
Normally, as the ac generator load increases, the operating frequency of the said
f
P
f NL f FL
PFL
f
P
%GSR( f FL )(100)
PFL
GD
Where:
PFL = rating of the alternator
GD = governor’s droop
Parallel Operation of Alternators:
In order to increase the capacity of a system serving
different loads, a number of generators are connected in parallel. Here are the basic requirements:
1. equal magnitude of terminal voltage
2. the same operating frequencies
3. the same phase sequence
44
CASE 1:
load, etc.)
With external characteristics given (kW rating, voltage rating, speed regulation, bus
f
Pa
(GSR ) a xf FLa
PFLa
(GSR )b xf FLb
PFLb
f
Pb
Before bus load changes:
Pt
Pa
Pb
Pb
Pb
After bus load changes:
PaNEW
Pa
f NEW
f
Pa
PbNEW
f
PtNEW
Pt
PtNEW
P
P
PaNEW
Pa
PbNEW
Pb
Where: +
f = frequency prior to the change in bus load
fNEW = frequency after the change in bus load
-
s an increase in bus load
Sample Problem:
Two alternators are driven by shunt motors. The shunt motors have
speed-load droop characteristics of 3% and 4% respectively. The alternators are in parallel and each carrying
50kW. There is no automatic speed-load control. An additional 50kW is switched on. What are the resulting
loads of the alternators assuming that the speed-load control of each is not adjusted?
Answer :
N N NL N FL
P
PFL
P1
P
N (50)
0.03 N FL
P1
P2
GSRxN NL
PFL
1666.67
N
N FL
P
( N )( PFL )
GSRxN FL
Eq.1
P2
N (50)
0.04 N FL
1250
N
N FL
Eq.2
Eq.3
substitute Eq.1 and Eq.2 in Eq.3 : 50 1667.67
N
N FL
1250
N
N FL
45
N
N FL
0.0171428
P1
50
P1
50 1666.67(0.0171428) 78.571 kW
P2
50
P2
50 1250(0.0171428) 71.428 kW
CASE 2:
With load sharing conditions given (load carried by each generator, operating power
factor of each alternator, current delivered, etc.)
Pt
Pa
Pb
It
Ia
t
a
Ib
Where: P = real power (watt)
Note:
St
b
t
Sa
a
Sb
b
S = apparent power (volt-ampere)
- = if operating at a lagging pf
The (+) sign or (-
t
the equation.
is dependent after the result of simplifying the right side of
Effect of Change in Excitation:
For two alternators (A and B) connected in parallel, if the
excitation of alternator A is increased, its operating power factor will decrease thus making its kVAR share to
increase while maintaining its kW share. Similarly, the operating power factor of alternator B will increase, thus
making its kVAR share to decrease while still carrying the same kW share.
Sample Problem:
Two alternators A and B are operating in parallel supplying a load
drawing 1000kVA at 0.80 power factor lagging. Alternator A contributes 500kVA at 0.60 power factor lagging.
For alternator B, determine the (a) kW, kVAR and kVA share (b) operating power factor.
t
cos 1 0.8 36.8690
ST
SA
SB
500
SB
SB
cos
ST
j 200 538.516
By inspection :
PB 500kW ; QB
pf B
A
cos 1 0.6 53.130
S A 1000
36.8690 500
53.130
21.80
200 kVAR; S B
538.516kVA
0
B
cos(21.8 ) 0.928 lagging
46
Sample Question:
The power factor of a generator is 75%. The operator is ordered to
increase the power factor to 80%. What will he do?
a. increase the voltage
b. operate the governor
c. increase the excitation
d. decrease the excitation
Problem Exercises:
1. What is the speed of a 100kW, 230V, three-phase, four-pole, 60Hz alternator?
a. 450 rpm
b. 900 rpm
c. 1200 rpm
d. 1800 rpm
2. A 6-pole, 3-phase, 60-cycle alternator has 12 slots per pole and four conductors per slot. The winding
is 5/6 pitch. There are 2,500,000 maxwells (0.025 weber) entering the armature from each north pole, and this
flux is sinusoidally distributed along the air gap. The armature coils are all connected in series. The winding is
wye-connected. Determine the open circuit emf of the alternator.
a. 532.1V
b. 504.2V
c. 512.4V
d. 572.4V
3. A 1200kVA, 6600V, 3-phase Y-connected alternator has an effective resistance of 0.40 ohm and a
reactance of 6 ohms per phase. It delivers full load current at 0.80 lagging power factor at rated voltage. What
will be the terminal voltage for the same excitation and load current, if the power factor is 0.80 leading?
a. 4560V
b. 9878V
c. 7898V
d. 4250V
4. To get the armature resistance of a 100MW, 13.8kV and 90% power factor generator, two terminals
are connected to a DC source. The measured current and voltage are 87.6A and 6V respectively. What is the
DC resistance per phase?
a. 0.0223 ohm
b. 0.0342 ohm
c. 0.0685 ohm
d. 0.0617 ohm
5. A 3-phase, 11kV wye-connected synchronous alternator has a synchronous reactance of 8 ohms per
phase but negligible resistance. If the excitation is such that the open circuit voltage is 14kV, determine the
power factor at the maximum output.
a. 0.786
b. 0.772
c. 0.793
d. 0.708
6. A 3-phase AC generator is supplying power to a load of 3200kW at 2300V and a power factor of
60%. Assume that the loss of the line, the generator armature and the load is equal to 10% of the load, what
would be the savings in watts if the power factor were raised to 80%.
a. 100,000
b. 140,000
c. 80,000
d. 230,000
7. A 3-phase turbo alternator has a reactance of 15 ohms and a negligible resistance. The machine
draws an armature current of 250A at 0.8 pf lagging when running on 12,000V infinite bus bars. If the steam
admission is constant but the emf is raised by 20%, calculate the new operating power factor.
a. 0.547
b. 0.586
c. 0.509
d. 0.575
8. A standby Diesel generator set will have the following loads: Inductive load drawing 50kVA at 0.8
power factor. Lighting load drawing 20kVA at 1.0 power factor. At what power factor will the generator
operate.
a. 0.855 lagging
b. 0.872 lagging
c. 0.821 lagging
d. 0.894 lagging
9. A 250V, 30Hz generator supplies power to a parallel circuit consisting of a 20Hp motor whose
efficiency is 90% at 0.80 pf lagging and a second load that draws an apparent power of 7kVA at unity pf.
Determine the system reactive power.
a. 23.582 kVAR
b. 12.435 kVAR
c. 10.341 kVAR
d. 20.384 kVAR
10. There are two alternators, 100kW, 3-phase in parallel are driven by shunt motors whose speed-load
characteristics are as follows: Alternator 1, no-load speed 600 rpm and the full load speed 530 rpm, while
alternator 2, no-load speed 590 rpm and full-load speed 550 rpm. For what load will the alternator divide the
load equally.
a. 62.45 kW
b. 68.78 kW
c. 67.54 kW
d. 64.67 kW
References:
Theraja, B.L. and Theraja, A.K. A Textbook of Electrical Technology in S.I. Units Volume II (AC and
DC Machines). MultiColour Illustrative edition. 2007.
Siskind, Charles S. Electrical Machines – Direct and Alternating Current. 2nd Edition. McGraw-Hill
Book Company, New York. 1979.
Rojas, Romeo Jr. A. Complete Electrical Engineering Formulas and Principles. Benchmark Publishing,
Republic of the Philippines. 2004.
47