Course: EE3003D ELECTRICAL MACHINES II
Semester: Monsoon/ 2022-’23
Topic 2
Alternator
Alternating current generator
Constructional Requirements
Field system
System of conductors or coils
Relative motion between the two
Stationary and rotating field systems
Advantages of rotating field system
Lesser amount of power transfer to the rotating system
Easy to provide insulation to the system of coils
Stationary system insulation is free from mechanical stresses
Easy to provide cooling to the system of coils
Frequency of induced EMF: f = PN/120 Hz
Need of standard frequency
Speeds of operation for 50Hz
P
2
4
6
N rpm 3000 1500 1000
8
10
12
16
20
24
30
40
48
50
60
80
750
600
500
375
300
250
200
150
125
120
100
75
Synchronous machine
Synchronous generator
Synchronous motor
Synchronous speed
Stator construction
Stator core
3 phase stator winding like 3 phase induction machine
Rotor construction
Smooth cylindrical type (non-salient pole) for high speed
Salient pole type for medium and low speed
DC field system
Non-salient pole synchronous machine
Salient pole synchronous machine
P Ananthakrishnan, EED, NIT Calicut
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Course: EE3003D ELECTRICAL MACHINES II
Salient pole synchronous machine
Semester: Monsoon/ 2022-’23
Non-salient pole synchronous machine
Prime movers for synchronous generators
Steam turbines
Good efficiency at high speed, 3000rpm
Horizontal mounting
Cylindrical rotor of small diameter and large length
Hydraulic turbines:
Salient pole synchronous machines
Pelton wheel: head ≥ 400m, high speed, 1500rpm, horizontal mounting
Francis turbine: head 400-100m, 500-50rpm, vertical mounting
Kaplan turbine: head ≤ 80m, 500-50rpm, vertical mounting
Internal combustion engines
Diesel or gas fuelled captive power plants
Horizontal mounting
Salient pole synchronous machines
Three phase winding
Winding details
Number of poles = P
Number of slots = S
Number of slots per pole per phase, m = S/ (3P)
P Ananthakrishnan, EED, NIT Calicut
2
Course: EE3003D ELECTRICAL MACHINES II
Semester: Monsoon/ 2022-’23
Slot angle, β = 180º/ (S/P)
Short chording angle = α
Coil span = 180º - α
Pitch factor/ chording factor/ coil span factor = kp
Distribution factor = kd
Winding factor = kw = kp kd
kp = phasor sum of emfs induced in 2 sides of a coil/ arithmetic sum of the same
= 2E Cos (α/2)/ 2E = Cos (α/2),
where E is the emf induced in a coil side and ER is the phasor sum of emfs in 2 coil sides.
kd = phasor sum of emfs induced in coils distributed in m slots/ arithmetic sum of the same
= 2 r Sin (m β/2)/ [m 2 r Sin (β/2)] = Sin (m β/2)/ [m Sin (β/2)], where
AC, CD, DE, etc. are the emf phasors of m different coils under one pole pitch,
AB is the phasor sum of emfs induced in coils distributed in m slots and
r is the radius of a circle to which AC, CD, DE, etc. and AB will fall as chords.
P Ananthakrishnan, EED, NIT Calicut
3
Course: EE3003D ELECTRICAL MACHINES II
Semester: Monsoon/ 2022-’23
Presence of harmonics
Presence of harmonics in the magnetic flux waveform gets truly reflected in the emf waveform.
Then the slot angle for the nth harmonic frequency waveform becomes nβ and
The short chording angle for the nth harmonic frequency waveform becomes nα.
As a consequence, kpn = Cos (nα/2), kdn = Sin (m nβ/2)/ [m Sin (nβ/2)] and kwn = kpn kdn.
EMF Equation
Let, P is the number of poles, Z is the number of conductors per phase, T is the number of turns per
phase, ∅ is the flux per pole (Wb), N is the speed of rotation (rpm), f is the frequency (Hz) and kf is the
form factor (= 1.11 for sinusoidal).
P Ananthakrishnan, EED, NIT Calicut
4
Course: EE3003D ELECTRICAL MACHINES II
Semester: Monsoon/ 2022-’23
Average value of emf induced per conductor = d∅/ dt = P∅/ (60/N) = PN∅/ 60 = (2 ∅ f) V
Average value of emf induced per phase = (2 kp kd Z ∅ f) V = (4 kp kd T ∅ f) V
RMS value of emf induced per phase, E = (4.44 kp kd ∅ f T) V. Hence,
RMS value of fundamental emf per phase, E1 = (4.44 kp1 kd1 ∅1 f1 T) V,
RMS value of nth harmonic emf per phase, En = (4.44 kpn kdn ∅n fn T) V and
RMS value of the resultant emf per phase, E = [E12 + E22 + E32 + ….]1/2 V.
Open Circuit Characteristic
Open circuit characteristic (OCC) or no load characteristic of an alternator is the graph between the
induced emf, E and the field (excitation) current, If at a constant speed. Circuit diagram for conducting
an open circuit (OC) test on the machine and a sample OCC are shown below.
P Ananthakrishnan, EED, NIT Calicut
5
Course: EE3003D ELECTRICAL MACHINES II
Semester: Monsoon/ 2022-’23
Rating of Alternators/ Synchronous machines
Alternators/ Synchronous machines have their apparent power (kVA/ MVA) as their power rating like
other ac machines.
Examples
1. A 3-phase, 10-pole, 50Hz, star connected alternator with 3 slots per pole per phase and 6
conductors per slot has a double layer winding with a coil span of 140º electrical. If its flux per pole
has a fundamental component of 0.06 Wb and a 20% 3rd harmonic component, calculate the phase
and line values of its no-load voltage.
P = 10, f = 50 Hz, m = 3, conductors/ slot = 3, coil span = 140º, ∅1 = 0.06 Wb, ∅3 = 20% of ∅1.
β = 180º/ (S/P) = 180/ (3 x 3) = 20º; α = 180º-140º = 40º; T = 6 x 3 x 3 x 10/ (2 x 3) = 90.
kp1 = Cos (α/2) = 0.9397; kp3 = Cos (3α/2) = 0.5;
kd1 = Sin (m β/2)/ [m Sin (β/2)] = 0.9598; kd3 = Sin (m 3β/2)/ [m Sin (3β/2)] = 0.6667.
E1 = (4.44 kp1 kd1 ∅1 f1 T) = 4.44 x 0.9397 x 0.9598 x 0.06 x 50 x 90 = 1081.23 V
E3 = (4.44 kp3 kd3 ∅3 f3 T) = 4.44 x 0.5 x 0.6667 x (0.06 x 0.2/ 3) x 3 x 50 x 90 = 79.92V
Eph = [E12 + E32]1/2 = 1084.17 V; El = √3 x 1084.17 = 1877.84 V.
Alternator on load
The terminal voltage, V will be same as the generated emf, E when the alternator is under no-load
condition. The generated emf, E will depend upon the speed of the prime mover and the excitation
current. When the alternator is loaded by way of connecting a 3-phase load across its terminals, it
starts delivering current and power to the load. As a consequence, the terminal voltage shows a
variation from the generated emf, E. Factors causing this voltage variation are (i) winding resistance,
R, (ii) leakage reactance, Xl and (iii) armature reaction, its quantification being different for non-salient
pole and salient pole machines.
Winding resistance, R
Leakage reactance, Xl
Armature reaction
Cross-magnetising
Demagnetising
Magnetising
Dependence of armature reaction on amount and nature of current
P Ananthakrishnan, EED, NIT Calicut
6
Course: EE3003D ELECTRICAL MACHINES II
Semester: Monsoon/ 2022-’23
Unity power factor: purely cross-magnetising if saturation is negligible.
Zero power factor lagging: purely demagnetising
Zero power factor leading: purely magnetising
P Ananthakrishnan, EED, NIT Calicut
7
Course: EE3003D ELECTRICAL MACHINES II
Semester: Monsoon/ 2022-’23
Lagging power factor other than ZPF: cross-magnetising and demagnetising
Leading power factor other than ZPF: cross-magnetising and magnetising
Representation of Cylindrical type Alternator/ Synchronous machine
Armature reaction reactance, Xa.
Per phase representation of the machine.
A per phase representation of a non-salient pole alternator or synchronous machine is possible as
shown below, under balanced operating conditions.
Where, XS which is the sum of leakage reactance, Xl and armature reaction reactance, Xa is known as
synchronous reactance.
The impedance comprising of R and XS is known as synchronous impedance, ZS. Then, the per phase
synchronous impedance, ZS = R + j XS = IZSI∠ , where IZSI = (R2 + XS2)1/2 and
angle,
= tan-1 (XS/ R). This
is known as the impedance angle of the machine which has a greater role in its performance
as a synchronous machine. Accordingly, the network equation of the machine operating as an
alternator or synchronous generator will be E = V + IZS and that of the machine operating as a
synchronous motor will be E = V - IZS.
Phasor diagrams
Phasor diagram is the pictorial representation of different phasors as they appear in the network
equation. The network equation of an alternator or synchronous generator given as E = V + IZS can be
written as
E = V + IZS = V + I (R + j XS) = V + IR + j IXS = V + IR + j IXl + j IXa.
P Ananthakrishnan, EED, NIT Calicut
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Course: EE3003D ELECTRICAL MACHINES II
Semester: Monsoon/ 2022-’23
Then, the phasor diagram of an alternator or synchronous generator for lagging power factor load will
be as shown below. As the terminal voltage, V is not at all alterable in the case of a synchronous
machine operating on infinite bus, it is taken as the reference for the phasor diagrams of alternators
and synchronous machines every time. Phasor diagrams for upf and leading power factor loads are to
be drawn in a similar fashion.
Note: E0 and E in the diagram
are to be taken as E and Ei
respectively.
.
Examples
2. A 3-phase, star connected alternator, the effective resistance and synchronous reactance of which
are 0.1
per phase and 0.7
per phase respectively supplies a load of 10MW at a power factor of
0.8 lagging and a terminal voltage of 11kV. Calculate the line value of the emf generated.
Po = 10MW, Cos ∅ = 0.8 (lagging), VL = 11000V, ZS = (0.1+j0.7) / phase.
Vph = VL/ √3 = 11000/ √3 = 6350.85∠0º V = (6350.85+j0)V
IIphI = Po/ (3 x Vph x Cos ∅) = 10 x 106/ (3 x 6350.85 x 0.8) = 656.08A
Iph = 656.08 (0.8-j0.6)A
Eph = Vph + Iph R + j Iph XS = 6686.94∠2.81ºV
IELI = √3 IEphI = 11.582kV.
3. A 3-phase, 1000kVA, 6.6kV, star connected alternator, the effective resistance and synchronous
reactance of which are 0.4 per phase and 6 per phase respectively delivers full load current at a
power factor of 0.9 leading and rated voltage. Estimate the terminal voltage for the same excitation
and load current at a power factor of 0.9 lagging.
P Ananthakrishnan, EED, NIT Calicut
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Course: EE3003D ELECTRICAL MACHINES II
Semester: Monsoon/ 2022-’23
S = 1000kVA, VLrated = 6600V, ZS = (0.4+j6) / phase.
V1 = Vrated = 6600/ √3 = 3810.51∠0º V.
II1I = IIratedI = 1000 x 103/ (3 x 3810.51) = 87.48A; Cos ∅1 = 0.9 (leading).
I1 = 87.48 (0.9+j0.44)A.
E1 = V1 + I1 ZS = 3643.86∠7.69ºV.
For a power factor of 0.9 lagging,
IE2I = IE1I = 3643.86V, II2I = II1I = 87.48A and I2 = 87.48 (0.9-j0.44)A.
If V2 = is the new reference phasor, V2 = IV2I∠0º V,
Equating the magnitudes of RHS and LHS of E2 = (IV2I+j0) + I2 ZS and solving for IV2I,
IV2I = 3352.66V and IV2LI = √3 x 3352.66 = 5806.98V.
Load Angle
With reference to the phasor diagram shown above,
E = Ea + j Er = IEI∠ , where Ea = E Cos , Er = E Sin , IEI = (Ea2 + Er2)0.5 and
= tan-1 (Er/ Ea).
IEI depends on the excitation condition at constant speed or frequency as per the EMF equation and
will be seen later to depend on the real power handled by the generator. As seen from the examples
and the phasor diagrams, generated emf (E) always leads the terminal voltage (V) by an angle
in the
case of an alternator or synchronous generator. Or phasor E leads phasor V for generator operation in
line with the fact that generated emf (E) is the initial cause to make the voltage (V) available across the
terminals here.
The direction of generated emf (E) which causes the terminal voltage (V) to be established is along the
inter polar axis (quadrature axis or q-axis), the pole axis being the direct axis or d-axis. On no load, in
the absence of any voltage drop, phasor E coincides with phasor V or the angle
is zero. On load
condition, a rotating magnetic field is set up by the armature current, the relative speed between this
field axis and the main field axis being zero on all steady state operating conditions. Then the phasor
V, the direction of which represents the rotating magnetic field axis always lags the phasor E, the
direction of which represents the main field axis (quadrature axis or q-axis) by the angle , which is
non zero. Hence
is known as the load angle, the value of which increases with real power and it is
positive for generator operation.
P Ananthakrishnan, EED, NIT Calicut
10
Course: EE3003D ELECTRICAL MACHINES II
Semester: Monsoon/ 2022-’23
Voltage Regulation
As seen above, the terminal voltage of alternators changes with load condition which includes the
magnitude of the current and its power factor. On loading, the magnitude of the terminal voltage
decreases from its no load value for upf and lagging power factors while it increases for leading power
factors. The voltage regulation which is a performance index of the alternator indicating the level of
voltage drop occurring on different loading conditions is defined as the change in terminal voltage from
no load to load condition expressed as a fraction or percentage of terminal voltage, the speed and
excitation remaining the same.
% voltage regulation = (IEI – IVI) x 100/ IVI, at constant speed and excitation.
Regulation characteristic, which is a plot of voltage regulation vs power factor at constant current,
takes the shape shown below for an alternator.
As mentioned above, voltage regulation is seen to be +ve for upf and lagging power factors while it is
more or less -ve for leading power factors.
P Ananthakrishnan, EED, NIT Calicut
11
Course: EE3003D ELECTRICAL MACHINES II
Semester: Monsoon/ 2022-’23
Exercise
1. A 3-phase, 10-pole, 50Hz, star connected alternator with 2 slots per pole per phase and 4
conductors per slot has a double layer winding with a coil span of 150º. If its flux per pole has a
fundamental component of 0.12 Wb and a 20% 3rd harmonic component, calculate the phase and
line values of its no-load voltage.
2. A 3-phase, 1500kVA, 6.6kV, star connected alternator, the effective resistance and synchronous
reactance of which are 0.4 per phase and 6 per phase respectively delivers full load current at a
power factor of 0.8 lagging and rated voltage. Estimate the terminal voltage for the same excitation
and load current at a power factor of 0.8 leading.
3. A 3-phase, 22kVA, 415V, star connected alternator, the effective resistance and synchronous
reactance of which are 0.5
per phase and 4.2
per phase respectively, delivers rated kVA at a
power factor of 0.8 lagging and rated voltage. Its field winding is supplied at 3.2A and 230V. If its
mechanical and iron losses are 310W and 480W respectively, calculate (i) the full load efficiency
and (ii) the terminal voltage when the load is thrown off.
P Ananthakrishnan, EED, NIT Calicut
12