Ac_machines_(alternators).pdf

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AC Machines Part 1 AC GE-NERATORS (AL; TEI~NATORS) '

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ESTABLISHED LEADER IN EE REVIEW,, ·,

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Rm. 867, Ground Floor, Isabel Bldg. F. Cayco corner Espail::1 Sts. Sampaloc, Manila Tel. No. 7J 17423

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AC MACliiNES Part I , AC GENERATORS (ALTERNATORS)

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CENERALTYPESO~ALTERNATOR

I.

Synchronous generator-- its speed is called synchronous speed and it is used in
2.

Induction generato1· (Asynchronous generator)- it is an induction motor which is run as a generator with a speed above the synchronous speed. Its power fac1or is leading and usually connected in parallel with a synchronous generator in order to supply I ighting and power loads.

3.

Inductor alternator-- it generates voltages. at higher frequencies (500 llZ to I 0,000 liZ). It is used to supply power to induction furnaces in orde;· to he,1t and melt alloys and steel.

Two Possible Constructions of an Alternator I. Stationary field and a revolving armature Stator_._ as field Rotor- as armature 1

Stationary armature and a revolving fkld Stator-- as ~mnature Rotor-·· as'field Note: The latter (2"d) arrangement Is mc~e preferable.

PRIME MOVERS FOR ALTERNATORS

A.

For large AC generators: I. Steam turbine . 2. Hydraulic tui·birye 3. Gas turbine "1. Internal combustion engine Frequency

ofGenera~ed

B. For small AC generators: I. Internal combustion engine

Voltage

f=PNs

120 where:

· r,

frequency, liZ or cps

P '-~no. of poles N5

'""

synchronous speed, rpm

Generated Voltage in an Alternator Em= N

f

x 10-R volts

where: Eave- average generated voltage, volts N -'no. of turns in coil ~- flux per pole, lines/pole 01 ma:-. wells/po!e t- time for flux to change by~, se;conds

• MUL TIVECTOR REVIEW AND TRAINING CENTER or

E.. ,c=

N~

volts (ifq, is in

weber~/pole)

~

e~

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Em~4fN<j>x 10- volts 8

For sinusoidal flux-density distribution, effective voltage is

E = 1.11 E.. ,c E = 4.44 f N·<jJ x 10·8 volts. ---·

2

10" 8 volts ____ .._

3

For a Jq, alternator, E, = 4.44 f N+ <jJ

X

where: E~

- generated voltage rer phase N$- no. of turns per phase

Note: Equations l, 2, and 3 are used for alternators having concentric armature wit1dings and full pitch ·' coils. '

ARMATURE WINDINGS FOR ALTERNATORS I

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Alternator windings are very similar to the de lap winding. Coils in ac 'armatures are joined together by merely connecting the proper coil ends in the correct sequence. .

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Two(2) (icncral Arrangements of Coils Employed:

I. 2.

Half-coiled (single layer)- one coil side/slot. Whole-coiled (double layer)- two ~bil sides/slot

Note: The whole-coiled winding is the more usual of the two arrangements. Coil Pitch- the distance between the two sides of a coil. It is usually express as a percent of full pitch. Coil pitch

Coil sides

A.

Full-pitch coil,-:- a coil having a distance between its two sides exactly equal to 180 electrical degrees.

Note: For a full-pitch coil, generated voltages in both coil sides are exactly in phase.

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MULTIVECTOR REVIEW AND TRAINING CENTER I

B.

Frnctionnl pitch coil · a coil having adistance between its two sides less than 180 electriCal degrees. Note : For a fractional pitch coil, ge·nerated voltages in the two(2) coil sides arc not in phase. , !

Pitch Factor·· the ratio of the voltage generated in ti1e fractional-pitch coil to the voltage generated in the full-pitch coil. Kr ==sink 2

where: · Kp =pitch factor in dcci111al p" =span of the coil in electrical degrees Por a full-pitch coil, Kr = I (unity) r:or a fractional-pitch coil, Kr < 1 Distribution Factor- the factor by which the generated voltage E must be multiplied becatlse the coils arc distributed in several slots under the poles instead of being concentrated in single slots under the poles. Kd "' sin n ( d~/2 ) nsin (d/2)

where: Kd = distribution factor, decimal n = no. of slots per pole per phase : d" = no. of electrical degrees per sl<)t

For concentric winding, Kd = I (unity) For distributed winding, Kd < 1 'corrected Voltage of an Alternator E+ = 4.44 K.r Kd f N+ ~ x 10" 8 volts

ALTERNATOit SCHEMATIC WIRINGbiAGRAM

a•

a

·I

+i DC source (Exciter)

DC field winding

c

'b

c

Armatme Windings I

3

To 3~ Load

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EQUIVALENT CIHCUIT ll,IAGRAMS

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For WYE connctted armature windings:

. IL -~~._·laT, .

.

VL V~;

~

btl

>

_IL~

VL

c

v~,·oo fJv~

1~,

=I+

E1

=

1J

E+

2. For DELTA connected armature windings:

tl

'

~)I

----.-----7 lr

Yr

SRr

~

j

V~,

~l

c

I

VL J,

~. c

vL = v~

IL

=

..f31,

E1.

= E~

Per Phase Equivalent Circuit Diagrai11: (Wye or Delta Connected) · E~ = V+ + Ia Z, Z,,;, R. -t j X,

v.

X,=

x.,

-->>

t XL

where : E+ =generated or open circuit or no-load voltage per phase Et =generated or open circuit or no-load line to line voltage V$ = terminal (output) voltage per phase VL = line to line terminal voltage 1$ = 1. = armature current per phase 1~, = line current Ra = Re = effective or ac resist'lnce df the armature per phase X,= synchronous reactance per phc:se XI. = armature leakage reactar!ce per phase

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added vectorially

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MUL TIVECTOR REVIEW AND TRAINING CENTER X., - n;actancc due to armature readion per phase Z,- synchronous impedance per phase Rr field resistance I r field ctlrrcnt . Vr~ vtiltagedrop across the field

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Alternator Voltage Regulation 0

o

V.R ..···

vN,.v.,"., --L.~ Vn.

X

100

or ~0 v.R. ~

E4 - V ~ v/

x 100

Factors affecting the drop or rise of the altemator terminal voltage: I. magnitude of the load . 2. actual over-all power factor of,the combined loads .Itffect of various types of load on the altcrn,ator terrninal voltage:

I.

Resistive loads (i.e. incandescent lamps, heating devices) or loads with unity powei factor. - S% to 20% drop in terminal voltage below its no-load value

2.

Inductive Loads (i:e. induction motors, electrical welders, fluorescent lighting) or lqads with lagging power factor. :l c 25% to 50% drop in terminal voltage below the no-load value.

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3. Capacitive loads (i.e. capacitor devices or special types of synchronous motor) or lo.dds with leading power

factor. · -tend to raise or increase the terminal voltage of the alternator above the no-load value. I {

Three (3) factors responsible for the change in voltage of an alternator: I. resistance drop .in the armature circuit (I.R.) 2. change in flux or armature-reaction effect (l.X,,) 3. armature reactance drop (!.XL) Alternator Phasor (Vector) Diagrams:

a. For lagging power factor

c. For leading power factor

b. For unity power factor

I.x, 0 0

1¢ = I. From the phasor diagrams: I. In complex form..

E~ =

V~ + I,Z,

2. In magnitude, E¢ = .Y(V¢ cos 8 + l,R.i +tV¢ sin 8 ± I.X,i where: · 8 ~power factor angle ohhe load EB - for lagging power fa;:tor 0 - for leading power factor

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MUL TIVECTOR REVIEW AND TRAINING CENTER Three (J) tests necessary to perform in order to obtain data for the calculation of the regulation of an alternator: I. J ~

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Mmatur\' de rcsJstnacc test Open-circuit test or no-load test Short-circuit test

The Armatun· DC Resistance Test

With the de field winding-open, measure the de resistance between each pair of terminals. The average of the three sets of resistance values is called R1•

r--------------. en On .

-

2

de f i ew.:cl).~3 ·. 1 d

DC Source

.

---

R1 u, R,2+R23+R,, 3

~J

. ;(i.e. Rp= Ydcrdg) - Adc rdg

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For Wye-eonnected armature windings, R'\Js = For Delta-connected armature windings.

• II igh current rheostat

~~

R~k = 3 ~~

To get the equivalent effective or ac resistance of the armature, usc a factor of 1.25 ito 1.75. Say,

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R. or Rc= 1.25

R"dc

The Open-Circuit Test or No-load Test

With the anm1ture-wlnding circuit open, the alternator is driven at synchronous speed. A de source is connected to the field.- making provision to adjust the field current so that, starting at zero. i,t may be raised until the ac voltage between any pair of terminals of the armature winding is somewhat above the rat'cd ' . voltage.

Open armature windings

--('MovO< p·

AC voltmeter

· . vo Itage per pI1ase, E-'-oc Open cJrcUJt

=

y 1f 1

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MULTIVECTOR REVIEW AND TRAINING CENTER The Short-Circuit Test

The alternator is driven at synchronous sped. Starting with a very low direct field current, progressively increase its value as the ac ammeters increase their deflections to rated current and above.

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r-

Highly Resistive F. R.

t

Shorted armature windings

DC Source

L--

~

_____./ ._

Average short circuit current per phase, I,c

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Synchronous impedance per phase, Z, a.

=

Eoc

('/\\

v.:__v

A,+ Az + A3

3

1sc

= -

For wye-connected armature windings Z

b.

::::;:

~

Eoc

'I sc

= Y/-{3 lsc

For delta-connected armature windings

,z =~=-_2j_ -~s

l,c

13

I sc

{j

Synchronous ~eactance per phase, X,=

f(Z}- (R.) 2

ALTERNATOR EFFICIENCY:

ll

p =

.:_Q_

Pin

P, x IO_Q___ x I 00 = p 0 + Total losses

Alternator losses inClude:

I.

Rotationallosses a. Friction and windage b. Brush friction at the field collector rings (often neglected; quite small) c. Ventilation to cool the machine (if r1ecessary) d. llysteresis and eddy currents in the stator

2.

Electrical losses a. Field winding · b.- Armature winding c. Brush contacts (often neglected; quite small)

3. 4.

Losses in the exciter used for field excitation Stray-load loss (negligible for small alte;nators)

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OPERATION OF ALTERNATORS IN PARALLEL

Rcq11isitcs for connecting alternators in parallel I. -, J.

Their voltages must be the same. Their wave forms (frequencies) must be the same. Their phase sequences must be the ~ame. If any ofthe above requisites is not met, there will be cross current between the alternator windings thru the common bus bars. With tile cross-current, the alternator cannot be synchronized.

Alternator Under Faults

I.

Three phase (3~) fault at the alternator terminals (either wye or delta connected alternator) Note: In a 3tp fault, the fault currents are symmetrical (balanced). I

sci. =

I 00 )

IL rntod (;;/ ;o

zl :...t

If alternator resistance is neglected. then I ' · "= I L rntcd ( 01I 00X ) 5CL to " 1

2.

Phase

to

phase fault at the alternator terminals (wye connected alternator)

In a phase to phase fault, the fault currents arc asymmetrical (unbalanced). One fault current is zero. The other two fault currcllts will have a magnitude of

...J3 I I+ rnted

(, ot 10

3.

x' 00+-YoXz o-

)

1

I

~ alternator resistance is neglected.

Line to ground or phase to ground fault (wye connected alternator) In a line to ground fault,. the fault currents arc asymmetrical (unbaia11ced). Two fault currents arc zero. The remaining one will have a magnitude of 100 3 [ l+rntcd( %XI+ o;.,X +%X" )] 2

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alternator resistance and neutral wire impedance arc neglected. '

4. Double line to ground fauit (wye connected alternator) In a double line to ground fault, the fault currents are asymmetrical (unbalanced). One fault current is zero. The other two fault'currcnts will have a magnitude of

J+ nllctf

[

0/ 0

/0

X

2

0/

where:

X

/0

Yo X 1 + %X 2 +

0 M

.,

I.e~.

=three phase fault line current rnted =rated line current of the alternator I~ rated= rated phase current of the alternator % z,~c percent positive phase sequence impedance % X 1= percent positive phasr sequence reactance % X2 ·= percent negative phase sequence reactance % Xo = percent zero phase sequence reactance I~.

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MUL TIVECTOR REVIEW AND TRAINING CENTER AC MACHINES PART 1- ALTERNATORS

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EXERCISES: 1. The no. of cycles generated in a 10 pole alternator in one revolution is _ _;__ A. 2 B. 5 C. 10 D. 20

2. What is the n1.,1mber of electrical degrees made per revolution for a special purpose 40 pole alternator? A. 28,800

B. 14,400

3. Alternators are usually designed to generate A. variable frequency B. definite currents

C. 7,200

C. definite frequencies

D. 1,600

· D. definite p.f.

4. A 4-pole, 3-phase, star-connected alternator armature has 12 slots with 24 conductors per slot and the flux per pole is 0.1 Wb sinusoidally distributed. Calculate the line emf generated at 50 Hz. A 1,066 v B. 3,198 v C. 1,846 v D. 5,538 v ,REE - May 2008 · 5. A three-phase wye-connected 50 HZ 2-pole synchronous machine has a st~tor with 2,000 turns of wire voltage of 5 KV? per phase. What rotor flux would. be required to produce a terminal (line to line) I A. 8.4 mWb B. 6.5 mWb C. 5.2 mWb D. 7.8 mWb REE - April 2006 i 6. ·A 3-phase, 8-pole generator is delta :connected. The terminal voltage is 2,400 v while the line current is 500 amperes. If the machine is converted tb wye, what will be the terminal voltage in volts? B. 4,157 v C. 4,600 v D. 3,800 v . 'A. 4,800 v ·REE - Sept. 2005

7. A 3-phase, 8-pole, 2,400-volt delta connected generator has a line current of 500 amperes. If converted in wye, what is the tolerable current? A. 371.5 a · B. 288.7 a C. 245.1 a

8. If the alternator winding has a fractional pitch of 4/5, the pitch factor kp is A. 0.833 B. 0.966 C. 0.972

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D.315.2A

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D. 0.951

9. A six pole, three phase alternator has 72 slots and a coil span of 1 to 10, w~at is the pitch factor? A. 0.924 B. 0.939 C. 0.966 i D. 0.985 10. Calculate the distribu.tion factor for a 36-slot. 4 pole three phase winding. A. 0.924 B. 0.951 C. 0.960

D. 0.975

11. A 144-slot stator has a whole-coiled 12 pole three phase winding. What' is the number of coils per phase and per group, respectively? ' A.12&4 B.24&4 C.36&4 D.48&4 12. The following information is given in connection wl.th an alternator: slots = 144; poles = 8; rpm = 900; turns per coil = 6; ~ =1.8 x 106 ; coil span = slots 1 to 16; winding is whole-coiled three-phase; winding connections = star. What is the voltage generated between terminals? A. 1,276 V B. 2,210 V C. 635 V D. 1,100V

AC Machines I P;Jn" 1 nl R

.. MUL TIVECTOR REVIEW AND TRAINING CENTER AC MACHINES PARTI-ALTERNATORS

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13.A part of an alternator winding consists of six coils in series, each coil having an e.m.f. of 10 V (r.m.s.) induced in it. The coils are placed in successive slots and between each slot and the next, there is an electrical phase displacement of 30°. Find the e.m.f. of the six coils in series. ' I A. 77.28 V B. 9.66 V . C. 38.64 V D. 19.32 V 1

14. A 4-pole alternator, on open circuit, generates 200 Vat 50 Hz when the field current is 4 A. Determine H1e generated e.m.f. at a speed of 1200 r.p.m. and a field current of 3 A, neglecting saturation in the iron parts A 40 V B. 240 V C. 60 V D. 120 V 15: A 3 ~, wye connected alternator having a sinu.soidal line potential with a 5% 5th harmonic content, 3% 7 1h harmonic content, 1.5% 11th harmonic content and 0.85% 13th harmonic content. What is the total harmonic distortion (THO) of the system line voltage? A. 11% B. 9% C. 6% '! D. 3%

~, Y.-connected alte,:nator will have a per phase no~inal impedance of _ __

16. A 100 MVA, 13.8 kV, 3 A. 3.5 0 .

B. 2.9 0

C. 1.9 0

D. 7.5 0

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REE -·March 1998 17. A generator is rated 100 MW, 13.8 kV and 90% power factor. The eff,ective resistance to ohmic resistance is 1.5. The ohmic resistance is obtained by connecting two terminals to a d.c. source. The current and voltage are 87.6 amperes and 6 volts, respectively. What is the:resistance per phase? A o.o617 n B. o.o513 o c. o.o6as n · o. o.342 o I

REE - April 2006 . · 18. A single phase alternator gives 250 amperes at 1,200 volts. What is the out~put of the machine in KVA? A. 300 B. 450 I C. 400 D. 350 I

REE - Sept. 2004 · 19. A 3,600v 500-KVA, 60 HZ, 3-phase Y-connected generator on test yields the following results: Mechanical and iron losses are 10 KW; field current at full-load 100% p.f. is 50 A; field current at fullload 80% p.f. is 70 A; resistance per phase of armature winding is 0.4 ohm. The exciter voltage is consta'nt at 120 volts and voltage control is done by means of rheostat. Determine the full-load armature current at 80% lagging p.f. A. 96.5 A B. 99.8 A C. 64.2 A D. 80.2 A REE -April 2006 :20. A single-phase generator delivers 80 A at 240 v and 75% lagging p.f. What kind of load is the generator supplying? B. resistive C. inductive D. reactive . · A capacitive REE- April 2005 ·21. A single-'-phase 2,400-volt synchronous· generator delivers 450 amperes at unity power factor. The synchronous impedance of the generator is 0.08 + j2.8 n. What is the regulation in percent? A 14.2 . B. 16.3 C. 20.1 D. 18.4 REE - May 2008 22. A three-phase wye connected wound rotor synchronous generator rated at 20 KVA, 240 V has a synchronous reactance cf 1.5 ohms/phase and an armature resistance of 0.6 ohm/phase. What is the percent voltage regulation at full-load with 80% lagging p. f.? A 38 B. 34 C. 47 D. 51 AC Machines I r1nr.,., ,,f n

MUL TIVECTOR REVIEW AND TRAINING CENTER AC MACHINES PART' I- ALTERNATORS

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REE- Sept. 2010 23. A 3 ~ wye connected, wound rotor synchronous generator rated at 10 kVA, 230 v has a synchronous reactance of 1.2 ohms/phase and armature resistance of 0.5 ohm/phase. What is the percent voltage regulation at full-load with 80% leading power factor? A -0.92% B. :-5.14% C. -3.08% D. -3.91% REE - May 2010 24. A three-phase wye~connected wound rotor synchronous generator rated at 10 kVA, 230 v has synchronous reactanc~ of 1.2 ohms per phase and armature resistance of 0.5 ohm per phase. What is the power factor such that the voltage regulation at full-load is zero? A 0.837 leading B. 0.894 leading C. 0.869 leading. D. 0.877 leading

25. A 1,000 kVA, 3,000 V, 50 Hz, 3-~ star-connected alternator has an armature effective resistance of 0.2 0 . .A field current of.40 A produces a short-circuit current of 200 A and an open-circuit e.m.f. of 1040 V (line value). Calculate the full load percentage regulation at a power factor of 0.8 lagging. A.-21.43% (B.21.43% C.·-24.31% D.24.31% 26. A 25 kVA alternator has a total loss of 2,000 watts when it delivers rated kVA to a l~ad at a power factor of 0. 76 .. Calculate its percent efficiency. A. 90.48% B. 92.32% C. 91.58% D. 89.12% 27. A 25 kVA, 220 Volt 3~ alternator delivers rated kVA at a power factor of 0.84. The effective ac resistance between armature winding terminals is 0.18 0. The field takes 9.3 amps at 115 volts. If friction and windage loss is 460 watts and the core loss is 610 watts. Calculate the percent efficiency. A. 87.82% B. 84.27% C. 86.41% D. 88.33% 28. Adjustment of field excitation on one of the two alternators operating in parallel will A. increase its load B.. change its frequency C. decrease its load D. change its power factor

REE - Sept. 2001 49. A generator is being synchronized to a large system. The actual system yoltage and frequency are · 13.7 kV and 60 t1z, respectively. The generator voltage and frequency are 13.6 kV and 60 hz, respectively. When the generator is switched to the system, choose which one happens. A. generator delivers MVAR B. generator takes MVAR C. generator delivers MW D. generator delivers both MW and MVAR REE - Sept. 2009 30. Two alternators operating in parallel supply 2,500 kW to a load at a power factor of 80% lagging. If one machine delivers 1,200 kW at a power factpr of 95% lagging, what is the ppwer supplied by the other machine? · A. 1,400 kW B. 1,600 kW C. 1,300 kW D. 1,500 kW

31. Two exactly similar turbo-alternators ·are rated 20 MW each. They are runn'.ing in parallel. The speedload characteristics of the driving turbines are such that the frequency of alternator one drops uniformly . from 50 Hz on no-load to 48 Hz on full-load, that of the alternator two from 50 Hz to 48.5 Hz. How will 'the two machines share a load of 30,000kW? · A. 14.56 MW, 15.44 MW B. 12.80 MW, 17.20 MW C. 17.47 MW, 12.53 MW D. 16.92 MW, 13.08 MW AC Machines I P;JQP :1nf fl

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MUL TIVECTOR REVIEW AND TRAINING CENTER AC MACHINES PART 1- ALTERNATORS

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,,-'____/ ........._

32. Each of two single-phase alternators has an armature winding whose\ resistance and synchronous reactance are respectively, 0.025 and 0.06 ohm. If the machines are operating without load, calculate the circulating current i.ri the windings if the alternators are paralleled, when the emfs are equal at 230 volts, but are dispfaced 30° from a position of phase opposition. A.915A B.1,830A C,.457.5A D.119A 33. In Prob. No. 32, when the voltages are in phase opposition, but one voltage is 230 and the other is 200 volts. A. 462 A B. 231 A C. 400 A D. 800 A I

34. A 600 kVA, 2,400 V, 3 ~ alternator has a Z!3ro phase sequence reactance of 12%, and a positive and negative phase sequence reactances of 8%. If the alternator resistance is negligible, determine the fault currents that the alternator can sustain if a 3 ~ occurs at its terminals. . A.1,042A 8.1,804A C.3,125A ·:, D.602A 35. In Prob. No. 34, determine the fault currents delivered if a phase to phase fault occurs at the alternator terminals. ' A.1,562A B.902A C.1,804A D.3,125A 36.A 3~ 11 kV, 10 MVA alternator has a sequence reactances of positive negative sequence, X2 j0.15 p.u. and zero sequence, Xo = j0.05 p.u: reactance. If the alternator is on no-load, calculate the line to ground and 3 A. 4499 A, 3499 A B. 2814 A, :2187 A C. 4499 A, 3030 A

=

~equence, X1 = j0.15 p.u., ~ith negligible neutral wire phase fault currents. : D. 1842 A, 854 A

SUPPLEMENTARY PROBLEMS 1. Calculate the average voltage generated in a six turn full-pitch coil of a 2t5 cycle alternator if the flux . . per pole is 7.2 X 105 m:axwells. A. 4 V B. 4.2 V C. 4.32 V '~ D. 4.8 V I

2. When the speed of an alternator is changed from 3,600 ·rpm to 1,800 rpm,· the generated emf I phase will become A. one half B. twice C. four times : D. one fourth {

3. An ac generator has wye connected armature windings with an effective resistance of 0.2 ohm/phase and synchronous reactance of 2.2 ohms/phase. If the terminal voltage is 500 volts when the load takes 50 amperes at unity p.f., what is the line to line open circuit emf? A. 866 volts B. 551.3 volts C. 288.7 volts ', D. NOT 4. The stator of a 3-phase, 20-pole alternator has 120 slots and· there are 4 conductors per slot accommodated in two layers. If the speed of the alternator is 300 r.p.m., calculate the e.m.f. induced per phase. Resultant flux in the air gap is 55 mWb per pole. Assume that the coil span is 150° electrical. A. 911 V B. 1822 V C. 455.5 V D. 1366.5 V 5. If in a 3-phase alternator, a field current of 50 A produces a full-load armature current of 200 A on short circuit and 1, 730 volts on open-circuit, then its synchronous impedance is A. 8.66 ohms B. 4 ohms C. 5 ohms D. 34.6 ohms AC Machines I PCl(lP <1 nf

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AC·MACHINES pARTI-AL TERNATORS 6. A 1500 kVA, 6,600 V, 3-phase, star connected alternator has an effecttve resistance of 0. 5 0 per phase and a synchronous reactance of 5 n per phase. Find the percentage change of voltage when the full-rated output at: power factor 0.8 lagging is switched off.. :1 A. 6.23% B. 12.47% C. 4.16% ~i D. 24.94% !· 7. Calculate the pitch factor for the armature windings of an alternator having 36 stator slots, 4 poles, coil span 1 to 8. A. 0.856 B. 0.940 C. 0.766 i D. 0.883 8. The winding of a 4-pole alternator having 36 slots and a coil span of 1 to ,8 is short-pitched by _ __ degrees. · A. 140 B. 80 C. 20 D. 40 I

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9. The harmonic which would be totally eliminated from the alternator emf usir,lg a fractional pitch of I

A.

B.

3rd

ih

1

c. s'h

D.

.~

5

is

gth

10. The disadvantage of a short-pitch coil is that---,-A. harmonics are introduced B. waveform becomes non-sinusoidal C. voltage around the coil is reduced D. both a and b 11. A 3 ~, 60 Hz, 6 pole alternator has an air yap flux which is sinusoidally distributed. The stator has 5 slots per pole per phase and each slot contains 4 conductors arranged in two layers. The coil pitch is 12 slots with normal excitation and rated speed, the voltage induced per conductor is 25 volts. The open-circuit emf of this Y-connected machine between lines when all conductors per phase are connected in series is nearest to ---A. 2730 volts · B. 2370 volts C. 4730 volts D. 4370 volts 12. The imaginary or fictitiqus part of synchronous reactance on alternator takes care of _ __ A. inductive reactance B. leakage reactance C. armature reaction D. copper losses I

13. Synchronous impedance test is taken on a 3-phase generator. Under short circuit condition, the currents in the three lines are 26.2 A, 23.7 A, and 27.4 A. What current should be used for the test? . A. 26.3 A B. 24.6 A C. 25.7 A D. 25.4 A 14. A 6.6 kV, 6 MVA, Y-con11ected, 3~ alternator has an effective ac resistance of 0.436 ohm between terminals. The field current of 103 Amp produces an open circuit emf of 6,600 Volt between terminals. The field current required to produce rated full-load current at short circuit condition is 82 Amp. Wha~ is the synchronous impedance per phase? ' A. 10 ohms B. 5.78 ohms C. 7.26 ohms D. 12.6 ohms 15. 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 B. operate the governor C. increase the exCitation D. decrease the excitation ·. 16. A 3 MVA, 50 Hz, 11 kV, 3 ~ wye-connected turbo-alternator when supplying 100 Amp at zero p. f. leading has a line to line voltage of 12,370 Vol~s. when the load is removed, the terminal voltage falls ' down to 11,000 Volts. The effective resistance of the stator is 0.4 ohm per phase. Find· the percent(%) regulation of this alternator when su'pplying full load at 80% p.f. lagging. A. 15.65% B. 16.5% C. 13.6% D. 14.5% AC Machines I P<JqP :0 nf R

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, MUL TIVECTOR REVIEW AND TRAINING CENTER AC MACHINES PARTI-AL TERNATORS

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17.A 3~, 1,200 kVA, 6.6 kV, wye-connectea turbo alternatorwith effective resistance of 0.4 ohm and reactance of 6 ohm per phase delivers full-load at 0.8 p.f. lagging and normal rated voltage and speed. The terminal voltage for the same excitation and load current of 0.80 p.f. leading is nearest to--,-A. 7,560 V . B. 7,900 V C. 4,560 V D. 7,360 V :18. Two 1200 kVA ac generators running in parallel supplies the following loads: 400 kW unity p.f., 500 kW at 0.90 p.f. lagging and 600 kW at 0.866 p.f. lagging. One machine is loaded to 1000 kW at 0.95 p.f. lagging. The p.f. of the other machine is nearest to _ __ A. 0.89 lagging B. 0.98 lagging C. 0.98 leading D. 0.89 leading 19. Two 3 synchronous generators connected in parallel are driven by waterwheel whose speed-load characteristics are as follows: The speed of the first falls uniformly from 624 rpm at no-load to 600 rpm at full-load at 1000 kW. The speed of the second waterwheel falls uniformly from 630 rpm at no-load to 600 rpm at 1000 kW loading. The output of each generator when the load is 1250 kVA, 0.80 p.f. are nearest to ___ . A. 500 kW each B. 445 kW and 555 kW C. 667 kW and 333 kW D. none of these 20. Two identical 2000 kVA alternators operate in paralleL The governor of the first machine is such U1at 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. What is the maximum unity power factor load that can be delivered Without overloading either machine? A. 3600 kW B. 6300 kW C. 4000 kW D. 1723 kW 21. A 3 star conne~ted 1000 kVA, 6.6 kV turbo synchronous alternator .with 20% reactance but of negligible resistance is supplying full-load at 80% p.f. lagging to a l.arge ·power system (infinite busbars). If steam supply is suddenly cuhoff, what is the new current? Neglecting losses. A. 77.58 A B. 57.58 A C. 47.58 A D. 67.58 A 22. A 3-phase, star-connected alternator has an armature resistance of 0.1 0 per phc.se. When excited to 86 line volts and short-circuited, the machine gave 200 A. What should be the e.m.f.(in line terms) the machine must be excited to, in order to maintain a terminal p.d. of 400 V with 100 A armature current at 0.8 p.f. lagging. A. 253 V B. 758 V C. 438 V D. 506 V 23. A 3~ Y-connected, 1000 kVA, 6.6 kV turbo-alternator has a synchronous reactance of 8.7 ohms per phase but negligibleresistance. It is supplying 87.5 Amp at 80% p.f. lagging to a large power system and the emf is increased by 25% (excitation was increased) the new value of armature current and p.f. are respectively. A. 182.48 A, 0.4761agging 8. 175.31 A and 0.567 lagging C. 190.82 A and 0.367 lagging D. 181.48 A and 0.467 lagging I

24. High speed alternators have a rotor construction A. similar to d. c. machines B. of non-salient type C. of salient type D. none of these

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i MUL TIVECTOR REVIEW AND TRAINING CENTER AC MACHINES PART 1- ALTERNATORS 25. A 1000 kVA, 3.3 kV, 60Hz, 4 pole, 3~ star connected, wound rotor synchronous alternator reactance of 1 ohm per phase and n€)gligible armatum resistance. The generator is operating at full-load, 0.80 p.f. lagging on an infinite .bus. If a disturbance causes the power angle to swing by 1° (mechanical). What is the synchronizing pbwer per phase and synchronizing torque respectively? · A. 133.6 kW and 708.9 N-m B. 148.5 kW and 508 N-m C. 190.8 kW and 367.6 N-m D. None of these 26 .. A 100 MVA, 2 pole, 60 Hz alternator has a moment of inertia of 50 x 103 kg-m 2 . What is the energy stored in the rotor at rated speed? A. 5533 MJ . , B. 3355 MJ C 3553 MJ D·. 5335 MJ 27. A 10 MVA, 10 kV, 3
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35. A three-phase Y -connected alternator delivers a unity power-factor load at 230 volts. If the synchronous-reactance voltage drop is 60 volts per phase, calculate the percent regulation, neglecting the resistance voltage drop. A. 9.74% B. 7.94% C. 12.74% D. 14.27% 36. In huge alternators, the moving part is A. brushes B. armature

c·. poles

D. none of these

37. In a 3-phase, 4-pole, star-connected alternator the e.m.f. per phase is 1,000 Vat a frequency of 50 Hz. Calculate the speed and line voltage.· A.750rpm,3000V B.1500rpm,1732V C.3000rpm,1414V D.1500rpm,2000V 38. The voltage generated across one inductor of a three phase alternator is 6 volts. There are 120 slots and 4 turns in every coil. The coil pitch is short by 2 slots. number of poles is 10 and frequency is 60 Hz. The armature winding correction factor of the alternator is nearest to _ _ _ _ . A. 0.957 B. 0.925 C. 0.987 D. 0.967 39. A 3-phase, star-connec.ted alternator is ra'ted at 2,000 kVA, 13.5 KV. The armature effective resistance and synchronous reactance are 1.3 0 c.nd 20 0 respectively per phase. Calculate the percentage regulation at full load for power factors of 0.8 lagging and 0.8 leading. A. 7.76%,-5.06% B. 15.52%,-10.12% C. 2.91%,-2.1% D. 6.56%,-4.73% 40. Two alternators running in parallel supply the following loads: (i) 1500 kW at 0.9 lagging p.f. (ii) 1000 kW at 0.8 lagging p.f. (iii) 800 kW at unity pl (iv) 600 kW at 0.8 leading p.f. The load on one machin& is adjusted to 2,000 kW at p.f. 0.95 lagging. Find the load and p.f. of the other machine. A. 1900 kW, 0.98 Jagging B. 950 kW, 0. 96' lagging D. 950 kW, 0.961eading C. 1900 kW, 0:98 leading REE - May 2009 41. A 30 kVA three-phase 230 v wye-connected synchronous generator has synchronous reactance of 0.8 ohm per phase. The armature resistance is negligible. What is the percent regulation on 25% full load at 80% power factor lagging? A. 6.6% B. 7.8% C. 6.2% D. 7.2%

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