Solved Questions for
Assignment - Electric
Machinery | EEL 4205
Electrical and Electronics Engineering
University of Central Florida (UCF)
14 pag.
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)
5-1.
At a location in Europe. it is necessary to supply 300 liW of 60LH7. power. The only power sources
available opcrate at 50 W L . It is decided to gcncralc the powcr by meinns of a noto or-generator bet
consisting oP a aynclironoi~smotor driving a hynchronous gc~ierator. T-Tow many poles should each of the
t~\!o~nachineshave in order to convert SO-I-Tz powcr to 00-I-lz powcr?
501 L I IOA Thc spccd of a synchronou\ maitunc I \ rclatcd to it\ f~~ccluc~icy
b) thc ccluat~on
T o makc a SO H/ '~nd a 60 HL michine have the aarnc rnwhsnical \pwd
together. we see that
$0
that the) can he coupled
Therefore, a 10-pole s):nchronous motor Inust be coupleci lo a 12-pole synchronous yenerator tu accomplish
this frecluency con\;ersio~~.
5-2.
A 1300-V 1000-kVA 0.8-PF-lagping 60-Hz two-polc Y-connectetl synchronous generator has a
synchronous reactance of 1. I 12 arld an arlnatiire resistance oi' 0.15 (2. At 60 HL. its friction and windage
losses are 2 3 kQ', and its core losses are 18 kW. The iielt! circuit has 21 dc \:oltage 01' 200 V, and the
maximum I, is 1 0 A. T h e resistance of the field circuit is adjustable over the range from 20 to 200 I1.
The OCC of this generator is shown in Figure P5- I .
((I!
How much field current is required to ~rlakc\/,
load'?
equal to 2300 V when the gerreralor is runni~lgat no
Ih) What i \ the intc>rnalgenerated ~ o l t a g coP thi\ m a c h ~ n cat ratcd conditio~ls?
( c ) How 111uch field current is rrquirecl to make V , equal to 23UO V when tlie generator is sunning at rated
condit~ons'?
(dl How niuch power and torque must the generator's priirle rnover be capable of' supplying'.'
( c ) C t ) l ~ \ t r ~a~capability
it
curbe for th15 generator.
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)
0.0
1.0
2.0
3.0
3.0
5.0
60
Field current, A
8.0
7.0
9.0
10.0
1C the no-load [erminal voltage is 2300 V. Lhc required field currcni can bc read directly Irom the
open-circuit characteristic. It i h d.15 A.
in)
This generator is Y-connected, so I ! = I , > . :It rated conditions, [lie line anil phase current
generator IS
(12)
P
1 , = I I =----T
3
- 1000 k\JA
V
-
v'?(2.100 V )
= 25 1 ;\ at
2111
ill
this
angle of -.;6.87^
The phase voltage of this niachine is V , = 1/, /J?= 1328 V . The il~tel.nalgeneraretl voltage of the machine
is
E , = kTqD
+ RH,I,,+ j X , 1 ,,
E, := 1328~0"-t-(0.15~ ) ( 2 5 1 L-36.87" A ) t ,!(I. l R)(23 lL-36.87^ A )
E , = 1537L7.4" V
(c)
Thc equivalcnl opcn-circuit [crininal voltage coi-rcsl.roiidinsio an
L;', ol'
1537 volt,<is
From the OCC, the required field current is 5.0 A
(LZI The ~nputpowa- to [hi\ gencr~itoi-I \ ccli~alto thc oi~tputpouel plu\ lo,\c\ Thc I ,i~edoiltpul po\cer I \
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)
Therdhre the prime mover ruust be capable olsupplying 175 kW. Since the generator is a two-pole 60 H L
machine. to must be turning at 3000 rlmin. 'I'lre required tolque is
( e ) The rotor cul.rcnl liinit of llle capability curve would hc drawn Irorn an orig111of
The radius of the rotor currenl limit is
Thc stator ciirrcnt l~mitir a circle at [lie origiii 01'radiu\
A MATLAB program that plots this capability diagram is shown below:
% M - f i l e : prob5-2.m
% M-file t o d i s p l a y a capabil.i.ty curve f o r a
% synchronous g e n e r a t o r .
% C a l c u l a t e t h e waveforms f o r t i m e s £:on:
Q = -4810;
DE = 5567;
S = 1000:
0 t o 1/30 s
% Get p o i n t s f o r s r a t o r c u r r e n t l i m i t
theta = -95:1:95;
% A1:gle i n d e g r e e s
rad = t h e t a * p i / 180;
% Angle i r : r a d i a n s
s -c u r v e = S . * ( cos ( r a d j -i. j * s i n ( r a d ) ) ;
% Get p o i n t s f o r r o t o r c u r r e n t l i m i t
o r i g = j*Q;
t h e t a = 75:1:105;
% Angle i n d e g r e e s
s a d = t h e t a * p i / 18O;
% Angle i n r a d i a n s
r-curve = o r i g + DE . * i cos ( r a d ) + j * s i n ( r a d ) ) ;
% P l o t t h e c a p a b i l i t y diagram
f i g u r e ( 1) ;
p l o t ( r e a l i s -c u r v e ) , imay ( s - r u r v e ) , b l , ' L i n e w i d t h ' , 2 . 0 ) ;
hold o n ;
p l o t ( r e a l ( r-c u r v e ) , imag ir--:urve i , ' X - - ' , ~ L i n z w i d t h,' 2 . 0 );
% ~ d x
d and
y axes
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)
% Set titles and axes
title ('\bfSynchronousGenerator Capability Diagram'!;
xlabel('\bfPower (kW)' ) ;
ylabel ( ' \~,bf~eactive
Power i k V P K ) ' ;
axis ( [ -1500 1500 -1500 150Cl j ;
axis square;
hold off;
The rc\ultlng capal3illty d~agramis sho\+n below.
Synchtonous Generator Capab~lttyDtayram
1500
5-3.
Assu~nethat the field current uf the generator i n Problem 5-2 ha\ heen adjusted to a value of 4.5 A.
ctr)
R h a t will the terminal \oltagc or t h ~ $generator bc
~~nped'incc
of 2 0 L 3 0 ' L2 ?
I [ 11 is
connected to a 3-connected load w ~ t ha n
( 0 ) Sketch the phaso~d~agraniof t h ~ sgener'ltor
(c)
What is the cfliciency o r tht: generator a[ thcsc conditions'?
(rl) Now assurnc that another identical A-conncctecl load is lo bz paralleled with thc fir\[ one. W'hvr
happens to 111cphasor diagram h)l- the gcnerator?
i c ) What
IS
the new terminal boltage after the load has been added?
(f) What must be clone to restore the terlninal voltage to its original value?
If the field current i\ 4.5 A, the opm-circuit terrn111:il koltage
voltage in the gerierator will be 3385 /\&::1 1377 V
ju)
~ l 1 1 1be
about 2585 V. and the phase
The load is A-connected with threz impct1;inces of 10L30" L2. From the Y-A transform, t h i b load is
ecluivalerit to a Y-connected load with three irilpedances of 6.667L30L R . The resulting per-phase
equivalent circuit is shown below:
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)
Thc magnitude or the phase cut rent f l o ~ ~ In
n gthi4 gc11c1a t o ~14
Therefore, the tnagnitude of the phaw voltage is
arid the terminal voltage is
( b ) Armature curt-cnt is I , = 1 SbL - 30' A . ancl the pliilzt. ~oltage1s V ,= 114010- V . T h e r e h e. the
intcnwl generated voltage
i4
The rebulting phaaor- cliagr~im14 \howri heIo\v (riot to scale).
= 1777ih.8' V
E
4
((.)
'T'he efficiency of the ~eiieratorunder these conditions ?can he found as I'ollows:
I:
,,..,= 3 Lt0 I,
cos O =
3(1210 V ) ( I 86 ~ ) ( 0 . 8=)554 kW
PbTr~,,
= ( ass umcd 0 )
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)
id)
When the new load is kidded. the total current Tlo\l: increases at the sarnc phase angle. Tlicrelbre,
j X , I , increases in lerigth at the same angle. while the tnagnirude of E , [nust remain constant. Therefore,
E, "swings" out along tlle arc of constant m'igrutucle L I ~ L Itlie
I neb\ , j X , I , fit, exactly betureri V, ancl
E4
E',
( e ) The ne\v impedance per phase will be IialC of [he old value. so Z = 3 . 3 3 3 ~ 3 0 "R . 'I'he magnitude of
the phase current flowing in this generator is
Tlirrefore, the ~nagnitudeoS the phase ~ o l t a g r1s
V , = 1,Z = ( 3 3 5 A)(3.333 R ) = l l 17 V
and the trrirlinal vol~agris
If)
5-4.
To restore the ter.mlnrtl voltage to 1t\ 01~glnallaIuc. Incrca\e the llelcl c~irrmtI,
Assume that the field current of tlie gelieruLor ill Problelri 5-2 is adjusted to achieve rated voltage (7300 V)
at 1 1 1 1 load coridilioris in each of the cluestions belo\v.
juj What is the efficiency of the generator at rated load'?
(hl What ia thc \oilage rcgulatlon of thc gencralor if it i\ lo,tcled to ra~edkilo\oltaniperc w~lliO 8-PFlagging loada?
(c) What is the voltage regulation of the gellaator if it is loaded to rated kilovoltalnperes with 0.8-PFleading loads4?
!dl What is the voltage regulation of thc gmerntor ir i l is loacled LO rated kilovol~ampcrzswith unity-powerfactor loads'!
(e) U$e MATLAB to plot the terminal loltagc of the gcncrato~'15 a f ~ ~ n ~ ol
t ~lo'id
o n Tor
faclora.
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)
,ill
thrce po\\u
This generator is Y-connected. so I ,
generator IS
(0)
1n. ,
1L
I-'
-
I , . ,At rated conditions. the line and pliase curreht in this
1000 k V A
= 25 1 4 at an angle of -36 $7'
~;(230OV)
-
---- ;
2,
1
,
-
,..
Thc phasevoltage o f tliis ~nacl~ine
is I ? , 1; / J3
1328 V . The internal generated voltagt. or thc ~rr~chine
is
.E,=\'~+R,,I,.tjX,T,
E, = 1 3 2 8 ~ 0+" (0.15 S Z. ) ( ~ ~- I36.87"
L
,:I)+ , j ( ~I .S Z ) ( XI L- 36.87" A )
E, = 1537L7.4' V
,
The input power to this generator is eql~alto the output power plus losses. The rated output power is
P,,,.
= ( I 000 k V ~ j ( 0 . 6:=) 800 kW
(O) If the generator is loaded to rated kV.4 with lagging loads. the phase voliage is F', = 132X~.!03V and
E,
the internal generated voltage is
= 1537/7.4" V . Tllerefore. the phase voltage at no-load would be
V, = 1537L0" V . The voltage reguln~ionwoulcl be:
(
If he generator is loaded
L O raiecl
k\'A with leadins loads, thc phaac voltagc is
y , 132XLO" V and
::::
the internal generated voltage is
E,
= V,
+ /<,I,, + j X , 1 ,
E,, = 1328LOG+(0.15 IZ)(251~36.87"A ) + j ( l . I i2)(251L3(5.87" A)
E, = 1217L1 1.5'' V
The voltage regulation would be:
VR =
((1)
V
1217-1328
x 1004.i = -8.4%
1-328
II' the generator 1s loadd to r'ited kVA '11
1318L0° V and the internal generated \oltage I <
UIIIIY
power factor. the phase voltage 1s
-
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)
Thc ~olragercgulation nould be:
(e) For this problcni. we will assume that the tcrrninal volrage is >~cljustedto 1300 V at no loact
conditions, arid see \+#hathappelis to the voltage as load ilicrcases at 0.8 lagging, i~nity.ancl 0.3 leading
power fcictors. Note that the rnaxiliurri cur~.entuill be 25 I ;\ in any case. 11phasor diagrarri representing
the situation at lagging po\ver factor is shown below:
E ?t
A pllasor tliagr:ralrl represeti~ingthe siti~atiollat leacling powel- PLictoris sho\i..ll below:
By thc Pytliagorcan Theorcm,
A ph;~sordiagram representing the situation at unity power factor is shown below:
EA
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)
B y the Pythagorean Theorem.
The M4'rl.A13 program i< shown helow rakes adkantage o f rhic f x t .
%
%
%
%
M-file: prob5-4e.m
M-file to calculate and plot the terminal voltage
of a synchronous generator as a function of load
for power factors of 0.9 lagging, 1.0, acd 0.6 leading.
% Oefine values for this generator
EA = 1329;
% Internal g e n voltage
I = 0:2.51:251;
% Current values (A)
K = 0.15;
% R iohmsi
X = 1.10;
-% XS ! ~ h m s i
% Calculate the voltage for the l a g y l ~ yPF case
VP-lag = sqrt( E A ^ ~- (X.*I.*0.6 - E.*I.*C.G'.A2)
- R.*I.*C.8 - X.*I.*0.6;
VT-lag
= VP-lag . * sqrt ( 3 ) ;
% Calculate the voltage for the leading PF case
VP-lead = sqrc( E A ^ ~- (X.*I.*0.8+ R.*I.*C.6;.^2 )
- R.*I.*0.8 + X.*I.*0.6;
VT-lead = VP-lead . * sqrt (3);
% Calculate the voltage for the unity PF ease
VP-unity = sqrt( E A * ~- (X.*I).*2 j ;
VT-unity = VP-unity . * sqrt ( 3 ) ;
% Plot ~ h e
terminal voltaye versus load
plot(I,abs(VT-lag),'b-','LireWidth1,2.Cj;
hold on;
ploti1,abs iVT-unity) , 'k-'LiceWidth',2.0)
;
plot(I,abs(VT-lead) , I s - . ' , ' L i n e W i d t h 1 , 2;. 0 )
title (l\bfTerminal Voltage Versus Load');
xlabel ( \bfLoad (A! ' ; ;
ylabel ( '~,bfTerminal Volzage (V! ' ) ;
legendi'0.8 PF l a g g i n g l , ' l . O
P F 1 , ' Q . PY
8 leading'! ;
axis ( [0 250 1500 2500: ) ;
grid on;
hold off;
I ,
The resulting plot is shown below:
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)
..
...
Terrn~nal'/olta@? Versus Load
?SOL1
ZCrJD
iaoo
lid0
-
' 50?
I
J
EU
10'
150
200
-.
‘1s~
1 oad (A)
5-5.
Assume that the field currmt of the genel.ator in Problern 5-7 has been adjusted so that it supplies rated
voltage wlle11 loaded with rated current at unity power factor. (You [nay ignore the ef:fects of R , when
answering these cluestions.)
((1)
What. is the tol.c/ue angle 60f tllr generator when supplying rated current at unit): power kictor?
(b) When (his gener:r[ol- is running a[ full load wilh ~lni[ypo\vel. I'ac[oi-. how close is it t c ~[he static stability
limit of [he machine?
S 0 1 , ~1 !O\
((I)
The torque aangle can be Sour~db l calculatiilg F , , .
E,
-
V,+ R,41,
+ jX,I,4
E , = 1 3 2 8 ~ 0 ' + ( 0 . 1 5hlj(25 ILO" A ) + j ( 1 . I ~t ) ( 2 i 1 ~ 0 A" )
E, = 13C)3L11.4" V
Thus the torque angle d'= I 1.4".
jh) The static s~abilitylin111wcurs at (5 = 9OC. This generator is a very lory \?a): firon1 thal linlil. If' we
ig11o1.e the internal resistance of tlle grneraioi.. the outp~itpower will be given by
p=
3v, F:,
sin 6
X.5
ant1 the output power is proportional to sin 6. Since sin I I .4" = 0.198 . ant1
stabilily lirrlit is about 5 tirnes the current ouiput power of [lie generator.
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)
bin
90" =: I .(K) , the staiic
5-23.
Two identical 600-kVA -180-V s y ~ ~ c h r o ~ ~generators
ous
are connectetl in parallel to supply a load. The
prime movers of the two generators happen to ha\:e different speed droop characteristics. When the field
currents of the two generators are eqr.~al.one delivers 300 A at (1.9 PF lagging, while the other ddi\lers 300
A at 0.72 PF lagging.
((I) What are the real power arid the reactive power supplicvl by each gencratc~sto the loacl'!
(11) What is the overall power factor ol' the loacl"
1
( c ) In what direction must the field current on each 2encr:ltor he ~~djusted
in order for them to operctte
same power factor'!
(0)
the
The real and rcacti\:e po\vcrs :ire
Q? =
v/? V, I , sin 0 = A ( 4 8 0 Vj(200 Ajsin cos-' (0.72) = 115 kVAR
f h ) The overall power racto~.can be lound from thc LO(SI
P,,,, = 4 + P 2 =299 kW
+
l?Ok\?'
=
1 . ~ 1 1 and
rcact~vcPO\\CI \ ~ ~ p plol ~~hc:
~ load.
d
319L\V
The otel.all power factor i \
PF = cos t a n
,
!
Q,,,,.
---
= 0.850 lagging
PT,
( c ) The iield cul-ren~of gcncratol I should be i~lcreasetl.a ~ i dthe lield current of genel-atol-2 should be
si~iiultaneously decreased.
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)
5-25.
A generating station for a power system consists of four 120-MVA 15-hV (1.85-I'F--laggingsynchronous
generators with identical speed drc~op characteristics operating ill parallel. The governors on the
gcncrators' prime rnova-s are adji~stcclto prcducc a 2 - H L cirop FI-omno load to full load. Three of thcsc
generalors arc each supplying a stcacly 7 5 h,lW 211 a Srccluency of 60 Hz, while the lijurlh generator (cillled
the swing generutor) Ilandlcs 2111 incremental load changes on the systcrn while mainlaining the system's
frequency at 60 Hz.
(a) At B given instant, the total system loads are 200 hlW at a frequency of 60 tlx. What are thJ no-load
frequencies of each of the system's generators'!
(19)
If (he system load rises to 790 MW ancl the gencri~tor':,governor set points tlo not chul~ge.\vha~will the
new syslern frcquency be:'
(c) To what frequency niu\t the no-lo'td l'reqi~enc)of (he \wing gcncriltor he adjustcd
systern frecl~lencyto 61) H/.?
In
order to restore the
(dl II' the system is operating a1 the co~~ditions
desc~.ibedin part ( c j ; what would happen if the swing
generator were tripped off'tl'le line (disc'onnwted 1.1-o~n
the power line)?
The tull-load poMel of there generator\ I \ (1 30 M VA ) (0.85) = 102 M W and thr droop t~ noload to full-load is 3 HL Thcrclorc. rhc ,lope. of the po\~c~-l~eclucncq
cLirLe rol thew ro111~encl-'ilor\I \
(u)
If generators 1, 7 , ancl 3 are supplying 7 5 MW each. then generator 4 must he supplying 75 MW. The IIOload Srecluency of the l'irsl lliree ge~lcrutorsis
T11c no-load frcquency or thc li)urth generator is
Pi = .',~r (hii- f;!.,)
Ihl
The setpoinls of generators I . 7, 3: ant1 4 d c ~not change, so the llew xystern irequenc) will he
The go\lernor setpoinls of the swing generator nus st be increased until the system liequenc) rises back
)
to 60 Hz. /It 60 H L . the 0tht.r three generators will be suppiling 7 5 MW e~tcli.s o tllc swing generalor must
supply 290 h l w - 3 ( 7 5 hlWi = 65 MW ill 60 l i ~ .Tl~erel'ort.,the su.ing generator's setpoints lnusl be set
to
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)
--
(d) If [he swing generator [rips off the line. [he other three generators woulti have to upply all 390 M W
of [he load. Therefore, the system frecluency will become
p,,,,,
=.~,.;(.~.,!-.f~ys~+1.2(fnl?-f~.)+~~i~.'j,fE,~-.t;,)
290 MW = (34)(61.21 - ji;.;,)
. , +- (34)(6?.21 -
fY5)
+ (34)(61.11- j ),
>)5
8.529~
186.63-31',,,,,
,f;), = 59.37 IIr
Each generator will S L I ~ P I96.7
~ M W to
5-27.
L ~ loads.
C
A 25-bIVA [hree-phase 13.8-kV two-pole 60-Hz synchronous generator was tested by Ihe open-circui~[es~.
and its air-gap voltage was extrapolated with the following rzsults:
Open-circuit test
The short-circuit test was then p e r l o r ~ n ~wd~ t hthe lollou~llgrcs~llt\
Sllort-circuit test
Field cul.rellt. 12
Armature current, h
1
1
310
1040
1
1
365
1 190
1
1
360
1240
1
1
475
1550
1
570
(
1 SX5
The armature resistance is 0.24 il per phase
(01 Find the unsatul.ated synchronous reactance of this generalor in ohms per phase and in per-unit.
( b ) Find the approxitnate satul-ated synchronuus reactance
answer both in ohms per phase and in per-unir.
.Y,ctt
a field currelit of 380 A. Express the
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)
( c ) Find tlie approximate saturated synclirono~~s
reactance at a fielcl current of 375 A. Express the answel
both in ohms per phase and in per-unit.
(d) Find the short-circuit ratio For this generator.
'The unsaturated synchronous reactance of this gener;itor is the same at any field current, sq we will
look at it at a field c~lrrentof 380 A. Tlie extrapolated air-gap vollagc at this point i h 38.3 kV, and the
Since this gcna-ator is Y - c o n n a t d . the phase voltagc is
sIio~.t-ci~-cuitcurrent is 1230 A.
r
ITc, = 18.3 kVlti3 = 10,566V and the armature current is I, = t 240 A . 'lherefore, the mnsclruroted
(a)
synchronous reactance is
The base impedance of t h ~ \generator is
Therefore. the per-unit unsaturated qnchronous reactance is
(6) Tlie saturated synchronous reacrance at. a field current of 380 A can be found from the OCC and the
SCC. The OCC voltage at I,,. = 380 A is 14. I kV, and tlie short-circuit. current is 1240 A. Since rhis
generator is Y-connected, tlie corresponding phase voltage is ITc2
= 14. I k ~ / &= 8 141 V and rhe artnature
current is 1 , = 1 230 A
. 'Therefore. the .sc~turtlrc.dsynchronous raictancc is
and the per-unit i~nsaturatedsy~~chronous
reactance i s
The saturated syncl~ronousrcuctancc at a field current of 375 ;2 can hc found Irorn ttlc OCC anti [he
SCC. Thc: OC:C voltage :it I, = 475 ;2 is 15.2 LV. iuid the short-circuit curl-ent is 1550 A. Since this
(r)
7
generator is Y-connected, the correspc)nding pliase voltage is V, = 15.2 k ~ / \ , ' 3= 8776 V ancl the armature
current is I , = 1550 A . Therefore. the .strtriraterl syncllronous reactance is
and the per-iinit unsaturatetl synchronous reactance is
id) The rated voltage of this generator is 13.8 kV. which req~iiresa field current of -365 A . The rated line
and armature current of this generator is
The field current required to produce a short-circuit current of 10465 A is about 330 A. Therefore. the
short-circuit ratio or this generator is
SCK =
365 A
= 1.14
320 A
-
Document shared on www.docsity.com
Downloaded by: anasmaraaba (als.maraaba@gmail.com)