Machine Addiii
Machine Addiii
Machine Addiii
MACHINE
ECE 3303
Chapter One
1. Magnet and Electromagnetic Principles
Magnet is a material which produces a magnetic field.
Magnetic field is invisible & responsible for the most notable
property of a magnet.
Notable Properties of a magnet are:
It has a force that pulls ferromagnetic materials such as iron
It has a force that attracts or repels other magnets.
Magnets are categorized in to two main groups. They are:
1) Permanent magnet
2) Electromagnet(Man-made)
1.1) Permanent magnet
Permanent magnet is an object made from a material that
is magnetized and creates its own persistent magnetic field.
They are found in a natural state in two main types:
Magnetite also called "iron oxide", (Fe3O4) &
Lodestone also called "leading stone“.
Field or flux
line
Current-carrying
conductor
Figure 1.2 Direction of magnetic field Figure 1.3 Ampere’s right hand rule
around a current-carrying conductor. showing the direction of field
Ampere's right-hand rule : If we grasp the conductor with our
right hand, the thumb pointing in the direction of the current, & our
fingers will point in the direction of magnetic field.
To determine the direction of the magnetic field in a cylindrical
coil of many turns with right hand is by grasp the coil with our right
hand as follow;
with the fingers pointing in the direction of current &
the thumb will point in the direction flux & north pole.
1) Magnetic Flux
Overall strength of a magnet measured by total magnetic
flux.
Magnetic flux- number of magnetic field lines passing
through an area.
symbol - 𝛷(phi)
unit - Weber (Wb).
2. Magnetic Lines of Force- is the "quantity of magnetism"
which exists in a magnetic field.
Solution
𝛷= 0.032 x 10-3 Wb
πD 2 π(0.01) 2
A= = = 78.53 × 10 − 6 m 2
4 4
Φ 0.032 × 10 −3
B= = − 6
= 0.407 T
A 78.53 × 10
This magnetic flux density exists only at the immediate end of the
magnet. As we move away from the end of the magnet, the
magnetic flux density decreases.
4. Magnetomotive Force(MMF):
Fm = NI 1.2
Fm = NI = 1000 × 20 × 10 −3 = 20At
Φ 6 × 10 −6
B= = = 76 mT
A ( π / 4)(1 × 10 − 2 )2
5. Magnetic Reluctance
Reluctance- ratio of MMF to magnetic flux.
Fm
= ℜm 1.3
Φ
Φ = magnetic flux, Wb
ℜm= reluctance of the magnetic circuit. At/Wb
Fm = ℜmΦ
𝛷 is directly proportional to the MMF.
Equation 1.3 represents Ohm's law of magnetic circuits.
Proportionality factor (ℜm) is called reluctance of magnetic
circuit.
Reluctance of a homogeneous magnetic element(circuit)
may be expressed in terms of its physical dimensions and
magnetic property as follows:
l
ℜm = 1.4
μA
−7
µ 0 = 4π × 10 H / m
µ 1.25 × 10−3
For 1.0 T: µ r = = −
= 994.72
µ0 4 π × 10 7
Cast steel has at least 1000 times more ability to set up magnetic flux lines
than nonmagnetic materials.
2. MAGNETIC CIRCUITS
2.1. Electric Circuit Analogs
In our discussion so far, we note the following analogous
relationships between magnetic quantities and electric
quantities:
Electric circuit Magnetic circuit
+ Rag
+
E NI
-
-
N = 550 turns
l1 = 20 cm = 20 ×10-2 m
l2 = 12 cm = 12 ×10-2 m
A = 4 cm2 = 4 ×10-4 m2
φ =560 ×10-6 Wb
560 × 10−6
B= = 140 × 10− 2 = 1.4T
4 × 10− 4
2200 × 0.64
I = = 2.56 A
550
EXAMPLE6: Figure 1.10 is built up of iron of square cross-section, 3 cm
side. Each air gap is 2 mm wide. Each of the coils is wound with 1000
turns and the exciting current is 1A. The relative permeability of part A
and part B may be taken as 1000 and 1200 respectively. Calculate:
(i) reluctance of part A, (ii) reluctance of part B, (iii) reluctance of two
air gaps,
(iv) total reluctance of the complete magnetic circuit,
(v) the mmf, (vi) total flux, and (vii) flux density. Leakage and fringing
may be neglected.
17 cm PART A
I 2 mm
N N
10 cm
PART B
20 cm
lA
Reluctance for part A, ℜA =
µ0 µ r A
lA = 20 − (1.5+1.5) = 17 cm = 0.17m
0.17
ℜA =
4π × 10 − 7 × 1000 × 9 × 10 − 4
= 15.03 × 10 4 At / Wb
lB
ii. Reluctance of part B, ℜ B =
µ0 µ r A B
Iafe Ibcd
Rg
Rf Rc
+ +
- -
Re Ig Rd
a) Magnetic circuit (b) equivalent magnetic circuit (c) analogous electric circuit
Figure 1-11 Magnetic circuit with center leg
The flux that is produced by the MMF in the center leg divides
into two parts, one going in the path afe and the other in the
path bcd.
The MMF drop around afe must be equal to the MMF drop
around bcd. This can be stated more precisely as:
= Iafe (Ra + Rf + Re )
EXAMPLE 7: In Figure 1.12, the following dimensions apply:
lg = lf = lc = 12 cm
la = lb = le = ld = 14 cm
Aa = Ab = Ac = Ad = Ae = Af = 1 cm2
Ag = 3 cm2
The material is sheet steel. The flux density in the center leg is 0.9 T.
Calculate the MMF required to produce this flux density.
Figure 1.12: Magnetic circuit with center leg & equivalent magnetic circuit
Solution
The total flux in the center leg is 0.9 × 3 × 10-4 = 2.7 × 10-4 Wb. The flux
divides into two parts, the left-hand path through afe and the right-hand
path through bcd. The flux density in path g is Bg = 0.9 T and therefore
Hg = 320 At/m. The flux density in section a is
−4
2.7 × 10
Ba = −4
= 1. 35 T
2 × 1 × 10
and therefore
Ha = 950 At/rn
Ha = Hb = Hc =Hd =He= Hf
Therefore,
NI = Hglg + Ha (la + lf + le )
= 320 × 12 × 10-2 + 950 ( 14 + 12 + 14 ) × 10-2
= 38.4 + 380 = 418.4 At
EXAMPLE 8: In Figure 1.12 we cut an air gap in the center leg, and
the air gap is 1.5 mm wide. All other dimensions remain
unchanged and the flux density in the center leg is still 0.9 T.
Find the number of ampere-turns on the center leg required to
produce this flux density.
Solution
NI is now in series with two reluctances in the center path, the air
gap and the steel in leg g.
(NI) - (MMF drop in air gap) - (MMF drop in section g )
= MMF drop in section b + c + d
= MMF drop in section a + f + e
In the center leg, the flux density is still fixed at 0.9 T. Therefore.
Bg = 0.9 T
H in the center steel section is still, Hg = 320 At/m, as before.
Therefore, MMF drop in leg g = 320 (12 - 0.15) × 10 -2 = 37.92 At
The MMF drop across the air gap is found from
Fmgap = Hgaplgap
For air: µ = µ0 = 4π×10-7 Wb/(At/m) or H/m
Therefore, 0.9 5
H gap = −7
= 7. 16 × 10 At / m
4π × 10
MMF drop across the path afe is still 380 At, as before,
NI − (37.92 + 1074) = 380 At
NI = 1491.91 At
Inductance:
From Faraday’s law , changes in magnetic flux create a voltage e, called
the electromotive force (emf), across the coil equal to
dφ
e = N 1.16
dt
d Ni N 2 di di
e=N = =L
dt ℜ ℜ dt dt
N2
L = heneries
ℜ
CHAPTER TWO
2. TRANSFORMERS
2.1. INTRODUCTION
Transformer is a static device that transfers electrical energy
from one electrical circuit to another electrical circuit through the
medium of magnetic field without a change in the frequency.
Primary winding - electric circuit which receives energy from the supply mains &
Secondary winding - circuit which delivers electrical energy to the load.
If the secondary winding has more turns than the primary winding, then the
secondary voltage is higher than the primary voltage and the transformer is called a
step-up transformer.
When the secondary winding has less turns than the primary windings then the
secondary voltage is lower than the primary voltage and the transformer is called
step down transformer.
Shell-type Transformer.
In core form transformer - primary and secondary windings are wound
In shell type transformer - the primary and secondary windings pass
inside the steel magnetic circuit (core) which forms a shell around the
windings.
Figure 2.1: Core-type & shell-type transformer construction
Figure 2.2: Transformer Core Lamination Types
Core-type transformer core is forming from
two “L” stampings or
“U" stampings with "I" end closing stampings.
V1 P N1 N2 S
Example 2.1: A single phase transformer has 350 primary and 1050
secondary turns. The net cross-sectional area of the core is 55 cm2. If
the primary winding be connected to a 400 V, 50 Hz single phase
supply. Calculate
(i) the maximum value of flux density in the core and
(ii) the voltage induced in the secondary winding.
Solution:
i) E1 = 4.44f φm N1 = 4.44f Bm A i N1
Maximum value of flux density in the core,
400
Bm =
4.44 × 50 × 55 × 10 − 4 × 350
= 0.93 T ( Wb / m 2 )
ii) For an ideal transformer,
E1 N1
=
E2 N2
Vs , nl − Vs , fl V p a − Vs , fl
=
VR ⋅100%
= ⋅100% (2.7)
Vs , fl Vs , fl
Pout Pout
η = ⋅100% = ⋅100% (2.8)
Pin Pout + Ploss
efficiency is η
Vs I s cos θ
=
The transformer ⋅100% (2.10)
PCu + Pcore + Vs I s cos θ
Example 2.3: A single phase transformer working at unity power factor
has an efficiency of 90% at both one half load & at the full load of
500W. Determine the efficiency at 75 % of full load.
Solution: Efficiency of the transformer at full-load = 0.9
Output at full load = 500 W
Let the iron losses of the transformer be = x watts
and the total copper losses at full load be = y watts
Then, the total losses at full load = x + y
Hence,
500
0.9 =
500 + x + y
(2.12)
(2.13)
Cont…d
The Y-Y connection has two very serious problems:
1. If loads on one of the transformer circuits are unbalanced, the
voltages on the phases of the transformer can become
severely unbalanced.
2. The third harmonic issue.
(2.14)
One problem associated with this connection is that the secondary voltage is
shifted by 300 with respect to the primary voltage. This can cause problems
when paralleling 3-phase transformers since transformers secondary voltages
must be in-phase to be paralleled. Therefore, we must pay attention to these
shifts.
Cont…d
3. ∆ -Y connection:
The primary voltage on each
phase of the transformer is
(2.17)
(2.18)
(2.19)
(2.20)
(2.22)
Phase Phase
Connection Line Voltage Line Current
Voltage Current
Star VP = VL ÷ √3 VL = √3 × VP IP = IL IL = IP
Delta VP = VL VL = VP IP = IL ÷ √3 IL = √3 × IP
Example 2.4: An 11000/415V, delta-star transformer feeds power to a
30 kW, 415V, 3-phase induction motor having an efficiency of 90% and
full-load pf 0.833. Calculate the transformer rating and phase and line
currents on both high and low voltage sides.
Solution 30
Transformer kVA rating = = 40KVA
0.9 x 0.833
Total load in VA 40,000
Line current on l.v. side of transformer = = = 55.65A
3 x line voltage 3 x 415
.
For star connected l.v. winding, phase current in l.v. winding = line
current on l.v side = 55.65A.
Line current on HV, side of transformer = 40,000 = 2.1A
3 x 11000
For delta connected HV winding, phase current in HV winding:
phase current in = 1 (line current on h.v. side) = 1 x 2.1 = 1.212A
HV winding 3 3
Chapter Three
3. Three-Phase Induction Machines
3.1. Electromechanical Conversion
Types of electrical Machines
DC Machines AC Machines
Asynchronous (Induction)
DC Gen DC Mot Synchronous Machines
Machines
Michael Faraday
Passing a current through a conductor in a fixed magnetic field creates
a force which causes the conductor to move through the field.
The force created by the current known as the Lorentz force acts
between the current carrying conductor and the magnetic field.
The magnitude of the force acting on the conductor is given by:
F = BLI
Where; F - force on the conductor,
L - length of the conductor &
I - current flowing through conductor.
Generator Action
Moving a conductor through a magnetic field, or moving the magnetic field
relative to the conductor, causes a current to flow in the conductor.
The magnitude of the EMF generated in this way is given by:
E = BLv
Where; E - generator EMF &
v is the velocity of the conductor through the field.
Electrical Machines
Majority of electrical machines sold today are still based on the Lorentz force.
Single turn coil carrying electrical current rotates in a magnetic field between
the two poles of a magnet.
Cont…d
Operational principle,
of induction machines
construction &
included in this chapter.
characteristic features
Induction (Asynchronous) Machines
Introduction
§The induction machine is the most rugged and the most widely used machine in
industry.
§It has a stator and a rotor separated each other by an air gap.
..\..\..\Animates & Figs\Figure 3.1x.pptx
generator are not satisfactory for most applications. Thus induction machine is
extensively used as a motor in many applications.
§Of all the ac motors the poly-phase induction motor is the one which is
extensively used for various kinds of industrial drives.
IM has the following main advantages & some disadvantages.
Advantages
It has very simple and extremely rugged, almost unbreakable construction
(especially squirrel cage type).
Its cost is low and it is very reliable.
It has sufficiently high efficiency. In normal running condition, no brushes
are needed, hence frictional losses are reduced.
It has a reasonably good power factor.
It requires minimum maintenance.
It starts up from rest and needs no extra starting motor and has not to be
synchronized.
Its starting arrangement is simple especially –for squirrel-cage type motor.
Disadvantage
Its speed cannot be varied without sacrificing some of its efficiency.
Its speed decreases with increase in load.
Its starting torque is somewhat lesser to that of a dc shunt motor.
The induction motor is used in various sizes:
Large three-phase induction motors are used in pumps, fans,
compressors, paper mills, textile mills and so forth.
Small single-phase induction motors are used in many household
appliances such as juice mixers, washing machines & refrigerators.
Two-phase induction motors are used primarily as servomotors in
a control system.
The linear version of the induction machine has been developed
primarily for use in transportation systems.
Rotor cage
Schematic Diagram
No slip rings, brush gear, short circuiting devices, rotor terminals for starting rheostats are required. The
It has better space factor for rotor slots, a shorter overhang and consequently a smaller copper loss.
It has bare end rings, a larger space for fans and thus the cooling conditions are better.
It has smaller rotor overhang leakage which gives a better power factor and greater pull out torque and
overload capacity.
The greatest disadvantage of squirrel cage rotor is that it is not possible to insert resistance in the rotor
The cage rotor motor has a smaller starting torque and larger starting current as compared with wound
rotor motor.
Principle Operation of IM:
a) 3-phase supply is feeding to stator windings then a magnetic flux
generated in the stator winding.
b) The flux rotate through the air gap cuts the rotor conductors. (Fig. a)
c) Due to the relative speed between the rotating flux and the stationary
conductors in rotor slot, an emf is induced in the rotor. (Fig. b)
Cont…d
I1 Io
' 2
Ic Im
V1 E1 Rc Xm
Pag
The mechanical speed ωmech is related to the synchronous speed by
Where;
3
Thevenin Equivalent circuit of IM:
R1 X1
I1 Io
'
V1 Xm
R''
Pag
Vth
Pag
120 × 50
N = N S (1 − S) = (1 − 0.03) = 970 r.p.m.
6
Example 3.3: A 4-pole, 50 Hz, 3-phase induction motor has a rotor
resistance of 0.024Ω per phase and standstill reactance of 0.6 Ω per
phase. Determine the speed at which the maximum torque is developed.
Solution
Rotor resistance per phase, R2=0.024Ω
Rotor standstill reactance per phase, X2=0.6Ω
N = N S (1 − ST max )
120 × 50
= (1 − 0.04 )
4
= 1440 rpm
Example 3.4: A 3φ, 460V, 1740rpm, 60Hz, 4-pole wound-rotor induction
motor has the following parameters per phase:
R1=0.25Ω R2’=0.2Ω
X1=X2’=0.5Ω Xm=30Ω
The rotation losses are 1700 W. with the rotor terminals short-circuited, find
a) i) Starting current when started direct on full-voltage
ii) Starting torque
b) i) Full-load slip
ii) Full-load current
iii) Ratio of starting current to full-load current
iv) Full-load power factor
v) Full-load torque
vi) Internal efficiency and the motor efficiency at full-load
Solution
460
a) V1 = = 265.6 V / phase
3
At start S=1. The input impedance is
r2' '
jX m + jx 2
s
j30(0.2 + j0.5)
Z1 = r1 + x1 + ' == 0 . 25 + j0 . 5 +
0.2 + j(30.5)
r2
s
(
+ j xm + x 2 '
) = 1.08∠66 Ω
265.6
I st = °
= 245.9 ∠ − 66°A
1.08∠66
2πN s 2π × 1800
ωsyn = = = 188.5 rad / s
60 60
V1 jX m 266.5 × j 30.0
Vth = = = 261.3V
r1 + j ( xm + x1 ) 0.25 + j 30.5
2
3 Vth R2'
Tst = ⋅ .
ω syn ( Rth + R2 ) + ( X th + X 2 ) S
' 2 2
3 261.3 2 0.2
= × × =
188.5 ( 0.24 + 0.2 ) + ( 0.49 + 0.5 )
2 2
1
= 185.2 N .m
b) At full-load Air gap power:
N s − N 1800 − 1740
i) S = = = 0.0333 vi) Pag = Tωsyn = 163.11 × 188.5 = 30,746.2 W
Ns 1800
ii) R2' 0.2 Rotor copper loss:
= = 6.01 Ω P2 = sPag = 0.0333 × 30,746.2 = 1023.9 W
s 0.0333
j30(6.01 + j0.5) Pmech = (1 − s )Pag = (1 − 0.0333)× 30,746.2 = 29,722.3W
Z1 = 0.25 + j0.5 +
6.01 + j(30.5)
= 0.25 + j0.5 + 5.598 + j1.596 Pout = Pmech − Prot = 29 ,722.3 − 1700 = 28 ,022.3 W
= 6.2123∠19.7 Ω Pin = 3V1 I 1 cos θ 1 = 3 × 265.6 × 42.754 × 0.94 = 32 ,022.4 W
265.6
I FL = °
= 42.754 ∠ − 19.7°A
6.2123∠19.7
Pout 28,022.3
I st 245.9 ηmotor = = × 100 = 87.5%
iii) = = 5.75 Pin 32,022.4
I FL 42.754
= 163.11 N.m
Example 3.5: A 3φ, 15 hp, 460V, 4-pole, 60Hz, 1728 rpm
induction motor delivers full output power to a load connected
to its shaft. The windage and friction loss of the motor is 750 W.
Determine the
a) Mechanical power developed.
b) Air gap power
c) Rotor copper loss.
Solution
a) Full-load shaft power, Pshaft = 15 × 746 = 11,190 W
Mechanical power developed, Pmech = Pshaft + Pf&w
= 11,190 + 750 = 11,940 W
120 × f 120 × 60
b) Synchronous speed, Ns = = = 1800 rpm
P 4
Ns − N 1800 − 1728
Slip, S = = = 0.04
Ns 1800
Pmech 11,940
Air gap power, Pag = = = 12 ,437.5 W
1 − S 1 − 0.04
3.Inter-poles
Modern direct current machines are provided with interpoles with
windings on them in order to improve commutation under loaded
conditions to insure sparkles operation of machine.
They are arranged midway between the mains poles and are bolted
to the yoke.
B) Armature
It is the rotating part (rotor) of the dc machine
It is a cylindrical body, which rotates between the magnetic poles.
It is separated from the field system by an air gap.
The armature consists of:
Armature core with slots and
Armature winding accommodated in slots
C) Commutator
The commutator is mounted on the rotor of a dc
machine and it performs with help of brushes a
mechanical rectification of power from
ac to dc in case of generators and
dc to ac in case of motors.
BRUSHES AND BRUSH HOLDER
Brushes collect the current from the rotating commutator or
to lead the current to it.
Brushes are made up of carbon & graphite, so that while in
contact with the commutator it is not rigid...\..\..\Animates &
Figs\Brush.pptx
Brushes are accommodated in the brush holder.
4.2) PRINCIPLE OPERATION OF DC GENERATOR
The basic essential parts of an
electrical generator are:
Magnetic Field &
Conductors
Whenever a conductor cuts magnetic
flux, dynamically induced emf is produced
in the conductor. (Faraday’s law).
This emf cause a current to flow if the
conductor is closed.
Fig.4.1: Schematic diagram
of a simple machine
Right-hand Rule
According to Faraday’s laws of
electromagnetic induction, an emf will
be induced in the rotating coil and is
given by
Generating of ac voltage
The 0° position of the coil is defined as in (a) where the
coil sides move parallel to the flux lines.
Basic Principles:
Conductor move parallelly to lines of force 0 electrom. induction.
Conductor move perpendicularly to lines of force max. ele.mag.indu
4.3) TYPES OF DC Generators
Field winding & Interconnected in various ways
to provide a wide variety of
Armature winding performance characteristics
Ea3
Ea2
Ea1
Ear
If1 If2 If3 If4 If5 If Ish
Some residual magnetism must exist in the magnetic circuit of the generator.
Assume that the field circuit is initially disconnected from the armature circuit and
the armature is driven at a certain speed, Ear will appear across the armature
terminals because of the residual magnetism in the machine. If the field circuit is
connected to the armature circuit, a current will flow in the field winding. If the mmf
of this field current aids the residual magnetism, a current If1 will flow in the field
circuit. With If1 following in the field circuit, the generated voltage is but the terminal
voltage is Vt = Ifl RF < Eal.
4.4) Emf Equation of DC Generator
Eg = emf generated in one of the parallel path
dφ
Average emf generated / conductor = , volt
dt
Flux cut / conductor in one revolution, dφ = φP ,Wb
1. Armature Reaction
2. Commutation
IL / Ia
Figure : External characteristics of separately excited generator
External characteristics clearly indicate that the terminal voltage falls
as load on the generator increase.
VL II Over-compounded
VL
I
s
I Level (flat) compounded
tic
VNL
ris
II
te
s
ac
tic
ar
tic is r
Ch
III
ris te
te ac
s
n
io
ac ar
III Under-compounded
at
C h
iz
et
Ch al
n
rn
ag
ar
te
M
In
al
IV
rn
te
Ex
Differential compounded
IFL IL
_
Figure : connection diagram of
series-wound motor
2. Shunt wound motor
Important relationships
Ish IL VL
+ I sh =
Field R sh
Rheostat Ia
I L = I sh + I a
Eb VL
Eb = VL − I a Ra
Shunt
Winding
Pdrawn = VL ⋅ I L
Pdev = Eb ⋅ I a
_
Figure : connection diagram of shunt-wound
motor
3. Compound wound motor Important relationships
I se = I L
Ish IL
+
Ia Eb + I a Ra V − I se Rse
Ise I sh = = L
Rsh Rsh
Eb VL
Shunt Series I L = I sh + I a
Winding Winding
Eb = V L − I a Ra − I L Rse
_ Pdrawn = VL ⋅ I L
(a) short-shunt compound motor
Ish
Pdev = Eb ⋅ I a
IL Important relationships
+
Ia Ise I a = I se
E + I (R + R se ) V L
Shunt Eb Series VL I sh = b a a =
Winding Winding
R sh R sh
I L = I sh + I L
_
E b = V L − I a (Ra + R se )
(b) long -shunt compound motor
Figure : connection diagram of compound- Pdrel = VL ⋅ I L
wound motor Pdev = Eb ⋅ I a
4.6.2. DIRECTION OF ROTATION
The direction of rotation of a motor can be reversed by:
Reversing current through either armature winding or field coils.
If current through both is reversed, the motor will continue to rotate in the
same direction as before.
2.6.3. BACK EMF
The direction of back emf is opposes the applied voltage.
Since the back emf is induced due to the generator action , the
magnitude of it is given by the generated emf in a generator:
+
φZN P
Ra Eb Eb = × volts,
60 a
V +
Eb
V Ia V = Eb+Ia Ra
-
60 PΦNZI a PΦ I a Z
Torque, Ta = . = 0.159. [ N .m]
2π 60 a N a
For a particular dc motor; P, Z & a are fixed. Hence,
Ta ∝ φ I a
For dc shunt motor, the flux per pole is practically constant, hence
the torque developed is directly proportion to the armature current,
i.e.
Ta ∝ I a (for dc shunt motor)
E b = V − Ia R a
T, N.m
N
I, A
N
N, rpm
T, N.m
N, rpm
T
Ia, A
V − Ia R a a V − Ia R a
N= ⋅ = K r.p.s.
Zφ P φ
Machine :
Stator of the three-phase synchronous machine has a three-
induction machine.
ac supply system.
The rotor has a field winding, which carries direct current
np
f = Where; n is the rotor speed in rpm
120 P is the number of poles
E f = 4.44 fφ f NK w Where
φf is the flux per pole due to the excitation
Current If
E f ∝ nφ f N is the number of turns in each phase
Kw is the winding factor
5.3.1. Equivalent Circuit of 3-phase synchronous generator