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AC MACHINES
(ALTERNATORS)
M E Z Z A N IN E F L O O R , D O N A A M P A R O B U IL D IN G
C O R N E R E S P A N A & G. M. T O L E N T IN O S T R E E T S
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AC MACHINES ( ALTERNATOR) REVIEW LECTURE
ALTERNATORS An electrical machine converting mechanical energy to AC electrical energy.
GENERAL TYPES OF ALTERNATOR:
A) Synchronous generator - a generator called synchronous because it is driven at constant speed (synchronous speed ) and it is used in almost all type o f applications.
B) Induction generator - it is an induction motor which run as a generator with a speed above synchronous speed. Its p.f. is usually leading and connected in parallel with a synchronous generator in order to supply power for lighting loads.
C) Induction alternator - it generates voltage at higher frequency ( 500 Hz to 10,000 Hz ) it is used to supply power to induction furnace in order to heat and melt the metal.
TWO TYPES OF ALTERNATOR :
- Revolving Armature Type
- Revolving Field Type a. Stator - Stationary armature b. Rotor - Rotating field poles
A d v a n ta g e s o f S ta tio n a r y ,A r m a tu r e
- Output current is delivered to the load without passing it through brush contacts
- Easier to insulate stationary armature winding for high ac voltage
- Armature winding can be easily braced to prevent deformation which could be produced by mechanical stresses. A. Easily insulated slip rings which transfers low voltage, low power dc field circuit.
Two types of rotor used :
a. Salient pole construction - used for low speed alternator, large diameter than axial length , having more poles ( from 6 to 40 poles)
b. Smooth cylindrical construction - used for high speed alternator, diameter is less than axial length, having less number of poles (2 or 4 pole)
PRIMEMOVERS FOR ALTERNATOR
A) F or large AC g en erato r B) For Sm all generator
- Steam turbine ' 1. Use internal combustion engines 2. Gas turbine
- Hydraulic turbine
- Internal Combustion Engine
F re q ue ncy o lG e n e ra te d E m f
Where : P = number o f field poles Ns - synchronous speed, rpm
f -
PN 120
— Hz
Alternator Equivalent Circuit:
From Exciter Circuit
For star connected armature windings: Ej, — J?> E* I* = I,. = I
For Delta connected armature windings:
El = E* (^) la “ I* S
v + = v L
E= K+ 7 „ (Zs)
Equivalent Circuit Per Phase
t_JL where : Zs = Rc + j
Vector Diagram
/
la^1 t 1 _va _
4 Notations: R* = effective armature (stator) resistance per phase X s = synchronous reactance per phase = Xar + XL Zs = synchronous impedance per phase X L - leakage reactance per phase Xar = reactance due to armature reaction EcJ) = generated or no-load voltage per phase E l - generated or no-load line-to-line voltage V L —Line-to-line terminal voltage IL = Line current lo ~ Ia ~ armature current or phase current V f - exciter voltage R f = field resistance If = field current
Alternator on load, the variation of terminal voltage is due to the following reasons:
- Voltage drop due to armature resistance
- Voltage drop due to armature leakage reactance
- Voltage drop due to armature reaction ( major reason)
Vector diagram:
( a ) if lagging (b )i Heading (c ) if unity
ALTERNATOR VOLTAGE REGULATION : E - V X 100% = - t ---- t-x 100% Ka
in complex fo rm E = Vj>+ Ia ( Z s ) in magnitude
Where: © is lag’g. p.f. 0 f o r leading p.f. and 0 = 0 for unity
4 Notations: R* = effective armature (stator) resistance per phase X s = synchronous reactance per phase = Xar 4- XL Z s ~ synchronous impedance per phase X L = leakage reactance per phase Xar = reactance due to armature reaction Ecj) = generated or no-load voltage per phase E l “ generated or no-load line-to-line voltage V L = Line-to-line terminal voltage IL = Line current la? ~ h = armature current or phase current Vf = exciter voltage R f = field resistance If ~ field current
Alternator on load, the variation of terminal voltage is due to the following reasons:
- Voltage drop due to armature resistance
- Voltage drop due to armature leakage reactance
- Voltage drop due to armature reaction ( major reason )
Vector diagram:
@ i f lagging (b ) if leading ( c ) if unity
ALTERNATOR VOLTAGE REGULATION :
% VR= .... VrL x 1 0 0 % = 100%
in complex fo rm E = V 4 + I a ( Z , ) in m agnitude
Where: © is lag’g. p.f. 0 f o r leading p.f. and 0 = 0 for unity
LABORATORY TEST FOR ALTERNATORS :
5
a. DC Resistance Test While Alternator is at rest, and dc field is open, measure the dc resistance between each terminals. The average o f three sets of resistance is called R(. p , " I ----------------------------- f x
Field is / Open / Rf
_ /?, H- R2 + R
b. Open Circuit Test or No load Test The alternator is driven at synchronous speed, the field current (excitation) is adjusted from a low value up to that sufficient for voltage reasonably beyond the rated voltage.
E o c - open ckt. e m f per phase c __ V.
c. Short Circuit Test The alternator is driven at synchronous speed (rated frequency), the field current (excitation) is adjusted to give 150 % to 200 % of rated current to How.
I sc ~ short ckt. current per phase is c --! i + l2 ± h -
z s =
Open circuit voltage per phase _ ^ o c Short circuit current per phase I sc
MAXIMUM POWER OUTPUT E£R ALTERNATOR
n EV watts. ,. P„,ax = —-------— If Rc IS neglected A , Where :
E = em f generated per phase-
V = terminal voltage per phase
- Xs = synchronous reactance per phase
Zs = synchronous impedance per phase
_SYNCHRONIZING POWER (P™) of ALTERNATOR :_*
Because o f some disturbance on the system, the power angle changes by ij/ (as shown Va = V 0 ), w a
the machine to developed an additional power, thereby keeping in synchronism. This additional power is knowa ‘ \svnch\ onizingpow er *.
Consider the vector diagram with lagging p.f. load
From the vector diagram E / a - V J 0° i 0 , ‘■ w #, ' ----- —-----Ec / a - (j) - V, / - (]) J z sr t
I a - ~ ~^1 [ Eo cos (a - (|)) - cos i|) ] + — j — [ E0 sin ( a - <|>) + V* sin (|> ] Z s
Pcj = ( Rectangular ) o f V0 Ia * after simplifying :
p d = _ [ £ o cos ^ _ V^cos ( <|) ~ a ) ] Z s When a becomes a + S :
Pd = Ejl [ Ec cos — cos ( (|) + a + 8 ) ] Z s A n d since PSYn = ?d “ Pd n P (^) s y n ~------ Eoy< --— f> r I (^) sin co (^) sin (cj)• /j. (^) +. (^) a )\ , o + 2 cos ( (j) + / j (^) a ) sinN. v ‘ —] — S , watts -----
E V ,
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ALTERNATOR REVIEW PROBLEMS
1. A 60 cycle alternator has 2 poles. What is the speed of alternator?
A. 3600 rpm C. 1500 rpm
B. 1800 rpm D. 1200 rpm
2. A 60 cycle alternator has a speed of 120 rpm. How many poles has it?
A. 24 poles C. 36 pole
B. 20 pole D. 60 pole
3. The effective voltage for a 5-turn coil on a 6 pole generator is 10.6 volts. The flux
per pole is 0.00795 weber. Calculate the frequency of the machine.
A. 30 Hz C. 60 Hz
B. 50 Hz D 40 Hz
4. A 3-phase, 4 pole machine has 36 slots. The winding is double layered and short
pitched being made up to 6 turns per coil whose pitch is 140° electrical.
Determine the winding factor.
0.902 C. 0.
B. 0.96 D. 0.
5. A 72 slots three phase stator armature is wound for 6 poles, using double layer
lap coils having 20 turns per coil with a 5/6 pitch. The flux per pole is 4.8x
lines, and the rotor speed is 1200 rpm. Calculate the generated EMF per phase.
A. 6000 volts C. 6480 volts
B. 5680 volts D. 6000 volts
6. A generator is rated 100 MW, 13.8 kV and 90 % power factor. The effective
. resistance to ohmic resistance is 1.5. The ohmic resistance is obtained by
connecting two terminals to a 6 volts d.c. source. If the current 43.8 Amperes,
what is the percent resistance per phase?
A. 1.5% C. 4.5%
B. 6% D. 6.9%
7. A 150 kVA, 1000 volts three phase wye-connected alternator has an open circuit
' emf of 1000 volts. When an alternator is short circuited at the same excitation, the
armature current is 460 amperes. What is the synchronous impedance?
A. 1.26 ohms C. 2.17 ohms
B. 2.23 ohm£ D. 3.23 ohms
8. A 3-phase, 60 Hz wye connected round rotor synchronous generator rated 10
kVA, 230 volts has a synchronous reactance of 1.2 ohm per phase and an
armature resistance of 0.5 ohm per phase. Calculate the percent regulation at full
load with 0.8 lagging power factor.
A. 12.8% C. 21.8%
B. 28.1% D. 18.2%
AC Machines (Alternator) Supplementary Problems
- In large capacity ac generator, the moving part ja A. Brushes { C) Poles B. Armature D. Frame 1
- Calculate the electrical displacement between adjacent slots of a 16 pole alternator with 144 slots. A. 30° C. 60° B. 20° D. 180°
- The induced em f in each coil side of a certain alternator winding is calculated as 10 volts, if the measured voltage across the coil is 19.32 volts, the winding is short by how many electrical degrees? A. 30° C. 60° B. 50° D. 20°
- For a 3-phase winding with 6 slots per pole per phase and with coil span of 15 s lo t, the value of pitch factor is A. 0.851 C. 0. B. 0.966 D. 1.
- Damper or “Amortisseur” windings are used in alternators t o ...... A. achieve synchronism C. reduced windage losses B.prevent hunting D start up
- What is the frequency of the generated emf of a 10 pole alternator driven at 720 rpm? A. 50 hz C. 60 hz B. 55 hz D. 25 hz
- A 6 pole, three phase AC generator has 72 slots, the coil span is 12 slots. What is the pitch factor? (jy - 100 C. 0. B' 0.939 D. 0.
- A 3-<t», 60 hz , 10-pole alternator has 120 slots and 4 turns in every coil. The coil pitch is short by 2 slots. The winding distribution factor of the alternator is ...... A. 0.925 C. 0. B. 0.967 ; 0.
- ■A three phase alternator has three armature coils each rated 1200 volts and 120 amperes. W hat is the kVA rating of this unit? A. 288. - C. 432 B. 249.4 /f*h 144
- An alternator oft open circuit generates 360 Volts at 60 Hz when the field current is 3. Amp: Neglecting saturation, determine the generated emf at a frequency of 30 Hz and ‘ a field current of 3.0 Amp. A. 240 Volts C. 160 Volts B. 300 Volts D. 150 Volts.
11 Determine the distribution factor for a 3- phase winding in which there are 18 slots per pole. A. 0.966 -CL 0. B. 0.957 D. 0.
- Calculate generated emf per phase of a 10 pole, 3-phase, 60 Hz alternator with 5 slot per pole per phase and 4 conductors per slots in two layers. The winding is short by 3 slots & the flux per pole is 0.12 W eber A. 2908 Volts C. 720 Volts B. 1000 Volts D. 3200 volts
- A 100 MVA, 13.8 kV, 3, Y-connected alternator will have a per phase nominal impedance o f ...... A. 7.5 ohms d ) 1.9 ohms B. 2.9 ohms u. 3.5 ohms
- A generator is rated 100 MW, 13.8 kV and 90 % power factor. The effective resistance to ohmic resistance is 1.5. The ohmic resistance is obtained by connecting two terminals to a d.c.source. The current & voltage are 87.6 Amperes and 6 Volts respectively. W hat is the percent resistance? Ai 3 C. 4. B. 2 D. 1
- A wye-connected alternator was tested for its effective resistance. The ratio of the effective resistance to ohmic resistance was previously determine to be 1.35. A 12-V battery was connected across two terminals and the ammeter read 120 amperes. What is the per phase effective resistance of thf«.alternator? A. 0.135 ohms 0.0675 ohms B. 0.117 ohms D. 0.0234 ohms
- Dirt accumulation in generator can cause all of the following except A. low power factor C. poor voltage regulation B. flashover D. overheating
.17. A generator is rated 100 MW, 13.8 kV and 90 % power factor. The effective resistance to ohmic resistance is 1.5. The ohmic resistance is obtained by connecting two terminals to a d.c. source. The current & voltage are 87.6 Amperes and 6 Volts respectively. What is the resistance per phase? A. 0.0513 ohm C. 0.0685 ohm B. 0.342 ohm D. 0.0617 ohm
- A generator is rated 100 MW at 22 kV, and 90% power factor. The effective ac resistance to ohmic resistance is 1.35. The resistance is obtain by connecting two terminals to a 12 volts dc source. If the current is 120 Amperes, What is the percent effective resistance per phase? A. 1.5% C. 4.5% B. 6% D. 6.9%
- A dc resistance test is conducted on a 3-phase 200 kva, 240 volts, delta connected generator. When the two terminals are connected from 6 volts dc source, the ammeter inserted in the line reads 150 amperes. The effective resistance to ohmic resistance is 1.25. What is the percent resistance? A. 7.5% C. 26%
29 A 25 kVA 220 volt three phase alternator delivers rated kva at 0.84 power factor. The effective ac resistance between the armature winding terminals is 0.18. The field takes 9.3 amperes at 115 volts. If the friction and windage loss is 460 watts and the core loss is 610 watts. What is the Hp delivered by the prime mover?
- A generator is being synchronized to a large system. The actual system voltage and frequency are 13.7 kV and 60 Hz respectively. The generator voltage and frequency are 13.9 kV and 60 Hz respectively when the generator is switched to the system. Choose which one happens. A. Generator delivers both MW and MVAR (C. .Generator delivers MVAR B. Generator delivers MW u. Generator takes in MW
- It is never advisable to connect a stationary alternator to live bus-bars because A. It is likely to run as synchronous motor It will get short circuited C. It will decrease bus bar voltage though momentarilly D. It will disturb generated emf of other alternator connected in parallel
- An alternator is operating with 100 Amp field current. If the field current is increased to 125 Amp with the same electrical load on the machines, it will A. be less likely to go out of synchronism C.j operates at new power factor angle B. be more likely to go out of synchronism ''”D. overheat
- A generator is being synchronized to a large system. The actual system voltage and frequency are 13.8 kV and 60 Hz respectively. The generator voltage and frequency are 13.8 kV and 60.1 Hz respectively when the generator is switched to the system. Choose which one happens. ■A. Generator delivers both MW and MVAR C. Generator delivers MVAR B. Generator delivers MW D. Generator takes in MW
- The driving power from the prime mover driving an alternator is lost but the alternator remain connected to the supply (i.e. its power circuit breaker failed to trip) and its excitation is on. The alternator will .... A. behaves as an induction motor B. stop & get burnt C. behaves as synchronous motor but will rotate in reverse direction. D. behaves as synchronous motor and rotate at same direction.
- Increasing the speed of alternator, what will happen? A Decrease MW C. Increase MVAR B. Increase frequency D. Decrease MVAR
- The power factor of an alternator is 75 %. The operator is ordered to increase the power factor to 80 %. What shall he do? A. increase the voltage C. operate the governor B. increase the excitation D. decrease the excitation
A. 32.
B. 31.
C. 28.
D. 33.
- In a large alternator, dampers A. reduce frequency fluctuations B. reduce voltage fluctuation
C. increase stability D. none of the above
- Two 1500 kVA ac generators running in parallel supplies the total load of 2,250 kW at 0.90 p.f. lagging. One machine is loaded to 1000 kW at 0.85 p.f. lagging. What is the kVA and p.f. of the other machine?
B. 470 & 0.96 leading >• _ D. 1250 & 0.78 leading
- Two 60Hz 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 50 kW. There is no automatic speed-load control. An additional 50 kW load is switched-on. What are the resulting loads of the alternators assuming that the speed-load control of eachis not adjusted? A. 78.57 kW & 71.43 kW C. 78 kW & 75kW
B. 82.51 k W & 67.49 kW D. 80.31 kW & 69.69 kW
- Two identical 2000 kW alternator operates in parallel. The governor of the first machine is such that the frequency drops uniformly from 50 Hz at no-load to 48 Hz at full-load. The corresponding uniform speed drop of the second machine is 50 Hz to 47.5 Hz, W hat is the maximum load that can be delivered without overloading either machine? A. 3600 kW C. 6300 kW B. 4000 kW D. 3800 kW
- Two-3cj) alternators operating in parallel delivers power at a line potential of 2200 Volts to an inductive load of 150 kW at 80% p.f. If the armature currents of the two alternators be equal one be operating at unity p.f. How much power does each alternator supply to the load? A. 117.2 kW for unity p.f.& 32.8 kW C. 112.7 kW for unity p.f. & 38.2 kW B both 75 kW one at unity p.f. D. 100 kW for unity &50 kW
- A wye-connected turbo alternator having s synchronous reactance of 0.5 and negligible resistance is connected to a large power system having a busbar voltage of 13.8 kV supplying a load of 15000 kVA at 0 8 lagging power factor. If the steam supply is cut-off, what current will the machine carry assuming negligible losses? A. 256.55 A C. 525.62 A •B. 384.26 A D. 627.55 A
- An ac generator 1500 kVA capacity has a full load efficiency of 91% at 100% power factor and 86% efficiency at 80% power factor lagging. What engine drive in Hp is required if the genset is to supply entirely incandescent lamp? A. 3125 C. 1980 B. 2525. D. 2210
- A 50 MVA, 13.8 kV, 3<j), Y-connected alternator will have a per phase nominal impedance o f ...... A. 7.5 ohms C. 3.8 ohms B. 2.9 ohm 9 - D. 15 ohms
- A three phase salient pole, 440 volts, wye connected synchronous generator operates at a power angle of 20° while developing power of 36 kW. The machine constant are Xd=2Xq=5 ohms and Ra is negligible. Calculate the percent voltagfe regulation
A. 1750 & 0.8 lagging (^) 1335 & 0.94 lagging
A. 70%
B. 144%
C. 78%
D. 54%
