A.C. Generators or Alternators Most of the electrical power used aboard
Navy ships and aircraft as well as in civilian applications is ac.
As a result, the ac the most important means of generator producing is electrical power. Ac generators, generally called alternators, vary greatly in size depending upon the load to which they supply power.
Regardless
of size, all electrical generators, whether dc or ac, depend upon the principle of magnetic induction. An emf is induced in a coil as a result of (1) a coil cutting through a magnetic field, or (2) a magnetic field cutting through a coil. ROTATING-ARMATURE ALTERNATORS The rotating-armature alternator is similar in construction to the dc generator in that the armature rotates in a stationary magnetic field the generated ac is brought to the load unchanged by means of slip rings. The rotating armature is found only in alternators of low power rating and generally is not used to supply electric power in large quantities.
ROTATING-FIELD ALTERNATORS The rotating-field alternator has a stationary armature winding and a rotating-field winding. The advantage of having a stationary armature winding is that the generated voltage can be connected directly to the load A rotating armature requires slip rings and brushes to
conduct the current from the armature to the load. The armature, brushes, and slip rings are difficult to insulate, and arc-overs and short circuits can result at high voltages. For this reason, high-voltage alternators are usually of the rotating-field type.
Since the voltage applied to the rotating field is low
voltage dc, the problem of high voltage arc-over at the slip rings does not exist. The stationary armature, or stator, of this type of alternator holds the windings that are cut by the rotating magnetic field. The
stator consists of a laminated iron core with the armature windings embedded in this core as shown in figure. The core is secured to the stator frame.
FUNCTIONS OF ALTERNATOR COMPONENTS A typical rotating-field ac generator consists of an alternator and a smaller dc generator built into a single unit. The output of the alternator section supplies alternating voltage to the load.
The only purpose for the dc generator is to the supply the direct current required to maintain alternator field. This dc generator is referred to as the exciter.
PRIME MOVERS source of mechanical power to turn their rotors. Two classes: 1. high-speed --- Steam and gas turbines 2. low-speed ---- while internal-combustion engines, water, and electric motors ALTERNATOR ROTORS Types of rotors used in rotating-field alternators. 1. Smooth cylindrical (or turbine-driven) - for high speed - 2 or 4 poles 2. Salient (or projecting) - for low to medium speed - With 6 or more poles
Smooth-cylindrical
Salient - pole
36-pole rotor
4-pole rotor
ALTERNATOR CHARACTERISTICS AND LIMITATIONS Alternators
are rated according to the voltage they are designed to produce and the maximum current they are capable of providing. The maximum current that can be supplied by an alternator depends upon the maximum heating loss that can be sustained in the armature. This heating loss (which is an I2R power loss) acts to heat the conductors, and if excessive, destroys the insulation. Thus, alternators are rated in terms of this current and in terms of the voltage output — the alternator rating in small units is in volt- amperes; in large units it is kilovolt-amperes
SINGLE-PHASE ALTERNATORS A generator that produces a single, continuously alternating voltage. The stator (armature) windings are connected in series. They are most often used when the loads being driven are relatively light.
TWO-PHASE ALTERNATORS A two-phase
alternator is designed to produce two completely separate voltages. Each voltage, by itself, may be considered as a single-phase voltage. Note that the windings of the two phases are physically at right angles (90º ) to each other. The graph shows the two phases to be 90º apart, with A leading B.
THREE-PHASE ALTERNATOR Has
three single-phase windings spaced such that the voltage induced in any one phase is displaced by 120º from the other two.
Three-Phase Connections The stator coils of three-phase alternators may be joined together in either wye or delta connections, as shown in the figure. With these connections only three wires come out of the alternator
FREQUENCY The output frequency of alternator voltage depends upon the speed of rotation of the rotor and the number of poles.
f
NP 120
where, N - speed in rpm, P - number of poles
Note: A machine that runs at a fixed or constant speed is called a SYNCHRONOUS MACHINE Synchronous speed, NS, is computed as 120 f
N S
P
Synchronous machine can be operated as a synchronous generator (alternator) or as a synchronous motor.
INDUCED or GENERATED EMF Let Z - no. of conductors per phase T – no. of turns per phase f – frequency, in Hz - flux/pole in Weber kd – distribution factor kp – pitch or coil span factor E - rms value of generated emf per phase
E 4.44k p kd fT 2.22k p kd fZ
ARMATURE WINDINGS
•The single layer winding will have one coil side per slot, while double layer winding will have two coil sides/slot. •If one slot per pole or slots equal to number of
poles are employed, then concentrated winding is obtained. Such windings give maximum induced emfs. •If the conductors are placed in several slots under one pole, the winding is known as distributed winding.
•When the two coil sides forming a complete coil of a winding are 180 electrical space degrees apart, the winding is known as the full pitch winding. •When the coil span of the winding is less than 180 electrical space degrees i.e. the two coil sides forming a complete coil are less than 180 electrical space degrees apart, the winding is known as fractional pitch winding. • In this type of winding, the induced emfs in the two coil sides is not in phase, so the resultant emf, which is equal to the vector sum of induced emfs in the coil sides, is less than their arithmetic sum and so the emf induced in short pitch coil is less than that in full pitch coil under the same conditions.
Pitch or chording Factor, kp
vector sum of induced emf's per coil cos 2 arithmeticsum of induced emf's per coil where, -- angle (in electrical degrees)by which the coil span kp
falls short of full pitch
Es
Es
Algebraic sum 2Es
E Es
Es
Vector sum
Distribution or Breadth Factor, Kd
kd
vector sum of coil emf ' s with distributed winding arithmetic sum of coil emf ' s with concentrated winding
Where
n – no. of slots per pole per phase 180 no. of slots per pole
sin n
n sin
2
2
Algebraic sum
Vector sum
1. Calculate the pitch factor for the under-given windings: a) 36 stator slots, 4 poles, coil span 1-8; b) 72 stator slots, 6 poles, coil span 1-10; c)96 stator slots, 6 poles, coil span 1-12. 2. Calculate the distribution factor for a 36 -slot, 4- pole, single layer three-phase winding. 3. Find the value of k for an alternator with 9 slots per pole d for the following cases: a)one winding in all the slots, b)one winding using only the first 2/3 of the slots per pole, c)three equal windings placed sequentially in 60° group. 4. A three phase, 16-pole alternator has a star-connected winding with 144 slots, 10 conductors per slot. The sinusoidally distributed flux per pole is 0.03Wb and the speed is 375rpm. Find the frequency; and the phase and line emf. Assume full-pitched coil.
5. Find the no-load phase and line voltage of a starconnected 3Ф, 6-pole alternator which runs at 1200rpm, having flux per pole of 0.1Wb. Its stator has 54 slots having double layer winding. Each coil has 8 turns and the coil is chorded by 1 slot. 6. The stator of a 3 phase, 16-pole alternator has 144 slots there are 4 conductors percalculate slot. If the speedand of the alternator is 375 rpm, the induced emf per phase. Resultant flux in the air gap is 0.05Wb/pole. Assume the coil span is 150°.
Alternator on Load •Fig. (1) shows Y-connected alternator supplying inductive load (laggingp.f.). •When the load on the alternator is increased (i.e., armature current Ia is increased), the field excitation and speed being kept constant, the terminal voltage V (phase value) of the alternator decreases.
•This is due to (1) Voltage drop IaRa where Ra is the armature resistance per phase. Fig. (1) (2) Voltage drop IaXL where XL is the armature leakage reactance per phase. (3) Voltage drop because of armature reaction.
(1) Armature Resistance (Ra) Since the armature or stator winding has some resistance, there will be an IaRa drop when current (Ia) flows through it. The armature resistance per phase is generally small so that IaRa drop is negligible for all practical purposes. (2) Armature Leakage Reactance (XL) When current flows through the armature
winding, flux is set up and a part of it does not cross the air-gap and links the coil sides as shown in Fig. (2). This leakage flux alternates with current and gives the winding self-inductance. This is called armature leakage reactance. Therefore, there will be IaXL drop which is also effective in reducing the terminal voltage. Fig.(2)
(3) Armature reaction As in dc generators, armature reaction is the effect of armature flux on the main field flux. Its effect is of the nature of an inductive reactance. Therefore, armature reaction effect is accounted for by assuming the presence of a fictitious reactance XAR in the armature winding. The quantity X is called reactance of armature reaction. The value of XAR AR is such that Ia XAR represents the voltage drop due to armature reaction.
For the same field excitation, terminal voltage decreases from its no-load value EΦ to V (for lagging pf) due to: 1. Drop due to armature resistance, IR a 2. Drop due to leakage reactance, IX L 3. Drop due to armature reaction, IX ar
EΦ E
θ I
V
NOTATIONS Ra → effective value of the armature resistance per phase XL → leakage reactance per phase XAR → reactance due to armature reaction per phase Xs → synchronous reactance per phase Xs=XL+XAR Zs → synchronous impedance per phase I=Ia →RMS armature current per phase V → terminal voltage per phase EΦ →RMS induced or no load emf per phase E → RMS Load induced e.m.f. It is the induced e.m.f. after allowingfor armature reaction. It isequal to phasor difference of E Φ and IaXAR
EQUIVALENT CIRCUIT
EΦ
E V I a ( Ra jX L ) E E I a ( jX AR )
EΦ
E V I a Z S V I a ( Ra jX S ) X S X L X AR Z=R +j X s
a
s
Phasor Diagram of a Loaded Alternator Fig. (4) shows the phasor diagram of an alternator for the usual case of inductive load. Fig.(4)
The armature current Ia lags the terminal voltage V by p.f. angle φ The phasor sum of V and drops IaRa and IaXL gives the load induced voltage E. It is the induced e.m.f. after allowing for armature reaction.
The phasor sum of E and IaX AR gives the no-load e.m.f. E 0 Note that in drawing the phasor diagram either the terminal voltage (V) or armature current (Ia) may be taken as the reference phasor.
A phasor diagram of a synchronous generator with a unity power factor (resistive load)
Leading power factor (capacitive load).
Voltage Regulation The voltage regulation of an alternator is defined as the change in terminal voltage from no-load to full-load (the speed and field excitation being constant)divided by full-load voltage.
%VR No load voltage Full load voltage x100 EO V x100 Full load voltage V
Note that (E0−V) is the arithmetic difference and not the phasor difference.
For leading load p.f., the no-load voltage is less than the full-load voltage. Hence voltage regulation is negative in this case. The effects of different load power factors on the change in the terminal voltage with changes of load on the alternator are shown in Fig. (5). Since the regulation of an alternator depends on the load and the load power factor, it is necessary to mention power factor while expressing regulation.
Fig. (5)
Determination of the parameters of the equivalent circuit from test data • The equivalent circuit of a synchronous generator that has been derived contains three quantities that must be determined in order to completely describe the behaviour of a synchronous generator: – The saturation characteristic: relationship between If and (and therefore between If and Eo) – The synchronous reactance, Xs – The armature resistance, Ra •
• The above three quantities could be determined by performing the following three tests: – Open-circuit test – Short-circuit test – DC test
Open-circuit test • The generator is run at the rated speed • The terminals are disconnected from all loads, and the field current is set to zero. • Then the field current is gradually increased in steps, and the terminal voltage is measured at each step along the way. • It is thus possible to obtain an open-circuit characteristic of a generator (EOC or Vt versus If) from this information If
+ Vdc
Vt
Short-circuit test •
Adjust the field current to zero and short-circuit the terminals of the generator through a set of ammeters.
•
Record the armature current Isc as the field current is increased.
•
Such a plot is called short-circuit characteristic. If +
VDC
A
A
Isc
DC Test – This test is conducted to measure-winding resistance of a synchronous generator when it is at rest and the field winding is open. The resistance is measured between two lines at a time and the average of the three resistance readings is taken to be the measured value of the resistance, RL, from line to line. – If the stator is Y-connected, the per phase stator DC resistance is
Rdc
RL 2
– If the stator is delta-connected, the per phase stator DC resistance is
Rdc
3 2
RL
Due to skin effect, use: Ra = 1.5Rdc
Determination of X s
The internal machine impedance is EOC ( ) ZS I SC ( ) EOC ( L )
EOC ( L )
ZS I
if ;
3
ZS I
SC ( L )
3
if
SC ( L )
Then the synchronous reactance Xs could be obtained using
X s Z s2 Ra2 Where,
Ra
is known from the DC test.
Example : A 200 kVA, 480 V, 50 Hz, Y-connected synchronous generator
with a rated field current of 5 A was tested and the following data were obtained: 1. VT,OC = 540 V at the rated IF. 2. IL,SC = 300 A at the rated IF. 3. When a DC voltage of 10 V was applied to two of the terminals, a current of 25 A was measured. Find the generator’s model at the rated conditions (i.e., the armature resistance and the synchronous reactance). Since the generator is Y-connected, a DC voltage was applied between its t w o phases. Therefore:
RL
VDC 10 0.4 I DC 25
1 Ra 1.5 Rdc 1.5 0.4 0.3 2
The internal generated voltage at the rated field current is
V 540 E A V ,OC T 3 3
311.8V
The synchronous reactance at the rated field current is …
Xs Z R 2 s
2 a
311.82 300
2
0.32 0.995
The equivalent circuit
Example :: A 480 V, 60 Hz, Y-connected six-pole synchronous generator has a per-phase synchronous reactance of 1.0 . Its full-load armature current is 60 A at 0.8 PF lagging. Its friction and windage losses are 1.5 kW and core losses are 1.0 kW at 60 Hz at full load. Assume that the armature resistance (and, therefore, the I2R losses) can be ignored. The field current has been adjusted such that the no-load terminal voltage is 480 V.
a. What is the speed of rotation of this generator? b. What terminal voltage the generator if lagging; 1. It is is the loaded with the ratedofcurrent at 0.8 PF 2. It is loaded with the rated current at 1.0 PF; 3. It is loaded with the rated current at 0.8 PF leading. c. What is the efficiency of this generator (ignoring the unknown electrical losses) when it is operating at the rated current and 0.8 PF lagging? d. How much shaft torque must be applied by the prime mover at the full load? e. What is the voltage regulation of this generator at 0.8 PF lagging? at 1.0 PF? at 0.8 PF leading?
Since the generator is Y-connected, its phase voltage is
V VT
3 277 V
At no load, the armature current IA = 0 and the internal generated voltage is EA = 277 V and it is constant since the field current was initially adjusted that way.
a. The speed of rotation of a synchronous generator is
120 nm f e P which is
m
1200 60
120
60 rpm 1200 6
2 125.7 rad s
b.1. For the generator at the rated current and the 0.8 pf lagging, the phasor diagram is shown. The phase voltage is at 00, the magnitude of Eɸ is 277 V,
jX S I A j 1 60
36.87 60 53.13
Two unknown quantities are the V
magnitude of and the angle of Eɸ. From the phasor diagram:
E2 V X S I A sin X S I A cos 2
2
2
V E2 X S I A cos X S I A sin 236.8V Since the generator is Y-connected,
VT 3V 410 V
b.2. For the generator at the rated current and the 1.0 PF, the phasor diagram is shown. Then:
V E2 X S I A cos X S I A sin 270.4V 2
VT 3V 468.4 V
b.3. For the generator at the rated current and the 0.8 PF leading, the phasor diagram is shown. Then:
V E2 X S I A cos X S I A sin 308.8V 2
VT 3V 535 V
c. The output power of the generator at 60 A and 0.8 PF lagging is
Pout 3V AI cos
3 236.8 60 0.8
34.1 kW
The mechanical input power is given by elecl oss Pin Pout P The efficiency is
corel oss mechl oss P 0 1.0 1.5 36.6 kW P 34.1
Pout 100 % Pin
34.1
36.6
100%
93.2%
d. The input torque of the generator is P
in app m
36.6
125.7
291.2 N - m
e. The voltage regulation of the generator is
Lagging PF:
VR
Unity PF:
VR
Lagging PF:
VR
480 410 410
100% 17.1%
480 468 468 480 535
535
100% 2.6%
100%
10.3%
Practice Problem A 381-V, 60-Hz, Y-Connected synchronous generator, having the synchronous reactance of 0.8 ohm and negligible armature resistance, is operating alone. 1. Determine the induced emf
a. If load current is 100A at 0.8 PF lagging b. If load current is 100A at 0.8 PF leading c. If load current is 100A at unity PF 2. Calculate the real and reactive power delivered in each case. 3. Calculate the %VR at each case.
PARALLEL OPERATION OF ALTERNATORS Alternators are connected in parallel to (1) increase the output capacity of a system beyond that of a single unit, (2) serve as additional reserve power for expected demands, or (3) permit shutting down one machine and cutting in a standby machine without interrupting power distribution.
The
machines must be synchronized as closely as possible before connecting them together. The generators are synchronized when the following conditions are set: 1. Equal terminal voltages. This is obtained by adjustment of the incoming generator’s field strength. 2. Equal frequency. This is obtained by adjustment of the incoming generator’s prime-mover speed. 3. Phase voltages in proper phase relation.
