I
ELECTRICAL MACHINES
REVIEWER
RICARDO V. CORREA
EE BSME
ASSOCIATE ELECTRICAl ENGINEER
-
MERRIAM & WEBSTER BOOKSTORE, INC.
Ma.nlla Philippines
COPYRIGHT 1985 BY
MERRIAM & WEBSTL•~I{, INC.
ISBN 971-30-0593-7
I
TABLE OF CONTENTS
PART-I
DIRECT CURRENT MACHINES
1. D. C GENERATOR & MOTOR PRINCIPLES
ALL RIGHTS RESERVED
J
i
2. DC DYNAMOS, CONSTRUCTION AND
ARMATURE WINDINGS
3. D. C GENERATOR CHARACTERISTICS
4. D. C MOTOR CQARACTERISTICS
5. EFFICIENCY, RATING, AN APPLICATION
OF DYNAMOS
3
10
22
34
50
PART- II
ALTERNATING CURRENT MACHINES
MERRIAM
WEBSTER
BOOKSTORE
METRO MANILA
PHILIPPINES
1.
2.
3.
4.
5.
6.
ALTERNATING CURRENT GENERATORS
TRANSFORMERS
POLYPHASE INDUCTION MOTORS
SINGLE- PHASE MOTORS
SYNCHRONOUS MOTORS
CONVERTERS AND RECTIFIERS
71
87
117
151
163
190
Dlf?ECT
CURRENT
2-1] THIRTY- SIX
MOTOI(
Df?Y CELLS Af?E CONNECTED
OF NINE CELLS
flATINt;
AND
GENEI(ATOfl
IS
IN FOUIC
14-5
AND
VOLTS
+
b) f?ECALCULATE
OF FWI?
CELLS
IN
SEI?IES
THE PROBLEM
I 45 VOLTS
-
I/CELL= 4- AMP
ILEQUIIZED:
0) Vi, Ir, 4 Pr Or THE COMBINATION
FIG. A.
IN
b)
Vr, Ir 'i- PT
IN FIG. B
=
(9)(1 45)
Vr
=
13 05' VOLTS
IT
=
(4) ( 4)
Pr
=
Vr rT
Pr
=
208 8 lv'ATTS
=
IG A~FE!?ES
=(13 05 )( 16)
b) Vr = (4-) (1 4'5)
[, ~ ~)(4-)
OF THE ENTITlE
PARALLEL GROUPS
-
Vr
A
FIGUfi'E
, L-Llllll-11
SOLUTION:
a) Vr
NINE
f Lt t ~ l~--B~-1
~ t ¥i I
3G DRY CELLS
=
fCATING
FOR
PEJL G'ICOUP
GIVEN
V/CELL
AND CUI?fZENT
AMP, llESPECT/VELY,
a) CALCULATE THE VOLTAGE 1 CL'/ZflENT, AND R')lvEJ(
COMBINATION·
PAICALLEL GIWUPS
PErc Gf?OUP IF THE VOLTAGE
IN ScfCIES
OF EACH CELL
PI?JNCIPLES
Vr
I
- - - - - -
-
-
-
-=..
-
-=..
-
IIIIIIII
FIGUilE
= 5·8 VOLTS
-
r
B
oG AMPERES
=
= (58)(36)
Pr== VrTr
Pr = 208· 8 \v'ATTS
2- 2] CALCULATE
THE AVEf?.AGE
25 X 10 6
IF IT CUTS
MAXWELLS
VOLTAGE
IN
Yto
GENE/lATED
GIVeN:
ip =
t
=
C'S' X 10
'/'10
6
AT
MAXIveUS
~ Yeo
SEC
f:AvE =
SEC
X IO_g
t
2·5
X
IO+G
X 10-B
9-w
Et~VE =
10 -S
EAve.= 2 VOLTS
t = V40 SEC
1._
X
SEC
Yeo
SOLUTION
EAYE =
_!
SEC
£-5 X lOG
AVEllAG'E VOLTAG'E GEi-JE/tATtD
AT
Yeo
=
+
J(t::Q UI flED.
l
t
IN A MOVING CONDUCTOr<
'/eo
SEC ; IN
I VOLT
_,
-'
X /0-8
DlrlECT
GENmATOR
CURRENT
f' -3] A SIX- POLE
D.C GENERATOR
504
CONNECTED
CONDUCTOilS
GENCfCATED
VOLTAGE
IN THIS
AND THE
MAX\v'ELLS
HAS AN
SPEED
Z= 504
CONDUCTOrtS
G PAJ?ALLL'L PATHS
==-
\vlNDJNG \viTH
6
llE<MilE/7:
xJ0- 8
6
8
)
ISOO
MODIFIED
SO HiAT
SPE:ED
PATHS
OF THE GENEf(ATOF? OF Pfl08LEM
\viNDJNG
IT HAS
SHOULD THE
Th/0
MACHINE
8E
DRIVEN
VOLTAGE
AS 8EFOI\E THE CHANGE, ASSUMING
fZE'MAIN
UN CH/It..J<;J:fJ ?
IS
INSTEAD OF SIX. AT lvHAT
IF IT IS TO DEVELOP THE SAME
ALL OTHEI?
CONDITIONS
TO
E'9=
VOLTS
ALL OTHEI? CONDITIONS TO
llPM= Eg X 0
4l
rlEMAIN UNCHANGED
== 250
~EQUIIZED•
SPEE17 IN I(PM.
X
60
X 2
10
X
504
GIVEN THE FOLLO\v'ING
PAfiT/CULAflS: NUMBEil O'F SLOTS IN AI?MATUI?E =5o i
NUMBER' OF CONDUCTOrz:S
GENEf?ATED
PET<: SLOT = 4 ;
.SPEED-= 1,200 .r.:PM; NUMBER' OF
PAI\ALLEL
A t=OU17 POle
U-C
MACHINE
PLUX PER FOLE = 2.62 X 106 MAX\v'EU.S
PATHS
IN ArtMATU17E
=Z
ZT = 4 CDHOUCTOllS
'P'"4 POLES
.SWf
S =55 SLOTS
ZT= Z20
ZJSLOT"' 4 CONDUCTORS/
7SLOT
E'9 ,~,
PX
X
Zr )( I\ PM
X
J0- 8
GO
(2l (eo)
2 P4fCALLEL FATtiS
:
P
l=g "' 1?30
VOLTS
!P
Z PAirALLEL PATtiS
VOLTS
X
CONDUCTO~
AllMATUilE
x
eo
x
10 8
P X I('PM
.e x Go x 10 8
6
X4
X
1500
Zr = Z70 CONOUCTOTCS
b) ZT/
OF
=
Z70 CONOUCTO/Z'i
SLOT
45
= G COf.JOUCTOo/SLOT
AND aJNDUCTOT'lS PEJr SLOT
lv'OULO BE
F/lOB. '2- 7 . JF Tlit:. WIND/IVG HAD
FOU~
GIVEN:
CJ, "' 4 PARALLEL PATHS
riWM
Uf../CHAN~Eil
IS rtEMAJNEV
f>IW5
#
b) L
=
SLOT
!SLOT
b) ZjSLDT
SOLUTION:
a) ZT = Eg
tt
X Q X GO
X
P
5"10 CDNDUCTOR:S
4S SLOT
E-7.
Z/
X
x 10 8
!(PM
;;: e50 X 4 X GO X 10 8
ZT = 5'40
X
1500
COt-1/JUCTOIC$
t-.----·----~--------------.a~~
·I
a
1·85 X 10
CONDUCTOI'(S
!·85 Xl0° x4
E:<j = ?
P X 'Zr >< RPM X IO -B
Zr = eso x
a) zT
CO/IJDVCTOI\S
~" (ZG£' X JOG)(4)(uo)(1ZOO)(I0- 8 )
SPEED; 1200 fCPIVI
l'lC.~UIREP
)( ~ .sLOTS
X
Zr = Eg x
MAX\VELLS
I?EQ'U.
Q X
MAX\v'ELLS
6
NECESSARY IN THE
I"ArtALLEL PATHS f'
OTHERS"
SOLUTION:
GIVEN:
a =
BY
1·85 X 10
f'-8] HO\v MANY TOTAL
VOLTAGE
PAilALl.C.L
0 X fOO
z/sLOT
X
f-5] CALCULATE THE
6
Q) f:q = if?
VOLTS
/SLOT
6
X
nvo
OF AIZMATUIZE CONUUCTmS :
8
llPM= GOO
X 10
Eg= ~50
OF AflMATl.JflE.
X 60
io 6
I 65 X
AT
SOLUTION:
LET Zr iS THE TOTAL NUMBER
x P xz xJ()B
OF'El7ATEV
IN EACH SLOT.
OF CONOLCTO!lS
4 POLES
P~
a=
a!( GO
Q "' Z PAR'ALLEL PATHS
p =· /!6£'
THe NUMBErt
Z50 VOLT-S lv'HEN
1·85 X /0 G MAX\v'ELLS , THE NL'M/31:7?
45, ANO THE AIIMATUilE \11/NDHJG HAS
SLOT= 15
<P X P X Z X I?PM X 10-s
JS
GIVEN:
tj =:
SOLUTION:
250
(b)
SLOTS IS
111:1 VOLTS
GENEIZATES
S = 1500 IZPM
GIVEM:
E9 ~
PATHS
PAJCALLEL
2-'5
MACHINE
FWX PE~ POLE
CALCULATE (a) TI-lE TOTAL NUM6fl(
J
PATHS,
4 X GO
((PM . IF THE
OF AJZMATUJ?E
GENEf?ATED VOLTAGE 1:'9
PARALLEL
E9 = 2.G2 X lOG X 4 X 220 X 1'200 X 10-B
C'J =
?-7] A fOUf?-POLE
(G) (60)
HAD FOU!r
2-5
THE GENERATEV VOLTAGE?
~=?
(1·6'5 X 10 )(6)(504)(1800)(10-
PRINCIPLES
SOWT/OIJ:
SOLUTION:
Eg = 250 VOLTS
AJ?MATUfi'E
15EEN
HAVE
GIVEN:
Q "'4 PA«ALLEL PATHS
6
e-t] THE
\vHAT \IIOULO
OF PllOB
\v'IND/NG
t800 ~PM.
IS
~ = 1·85 X 10 MAXh/E'LlS
1\EQUifZED:
CllRilENT GENEI?ATOil' AND MOTOTr
1!-6] IF THE Al'lMATUT?E
PATHS. CALCULATE THE
E 9 =~ X PXZ X ICPM
a x eo
P=G POLES
DlllECT
IF EACH POLE PJZODUCE"S J.B5 X 10
MACHINE
Gl VEN:
a~
ARMATURE
IN SIX PARALLEL
AI\'MATURE
PIZINCIPL~S
AND MOTOR
5
=
12 CONDUCTOI'(S /
/S'LOT
DIRECT- CURRENT
2-~ THE'
SPEED
GENEI?ATOfl
MOTOIC
AND
OF THE GENEIZATOI?
Of PirOB
1,550 RPM· (G)WHAT WiLL B"t: THE GENEIZATCD
POLE
IS MAINTAINED
AT THE SAME
;(b) 1V WHAT VALUE OF FLUX f'ET( POLE
GENEflATED VOLTAGE
IF THE
10
6
[.e.J
1
IN #
a)
f- 7,
QIVEN:
BE ADJUSTED
eso VOLTS(
tm~p
b)
7
i]j = Eij
p
=
GO
X
Z
f-10] WHAT 15 THE FflEQLIENCY
EIGHT- POLe
CONDUCT0f(5
750-flPM
ftPM
S=
X
10-B
b)
f=
F"'-8 POLES
s,_ 7so rcrM
c)
f=
b)
IZEQ
c) -F =
L,. LENbTtl (INCHES)
VOLTAGE GENEJlATED
5CO- fZPM
MACHINE
AMP.
B= 52/000
7
LINES/
/SQ.If.J.
MANY
FrrEQUENCY
POLES
IS "50 CPS
GIVEt,J:
CPS
f-=521 000 X 6 X90
EXCEflTED !JY Z:ACH
11;300,000
F
f = B X L xI
~
Z-1~WHAT
t;o CFS
=
LIES ON AN
x
WILL THE
T0/7GUE
CONDUCTOR'
c;NEN
== 4-1 7 DP.S
5"00
IN WHICH THE AflMATUflE
AT A SPUP OF 600 Rf>M
7
SOlUTfON;
+'=
DIA. = 9 IN(HfS
foo -f
flEQUiflW:
X
IZO
fZPM
NUtv113Ef? CF PGLE5
e=
30
L.
=
2'·48
X
9
IN'.
2
T= 0.932 L/LFT.
=
FORCE IN (LB)
RADI!JS
X 120
GOO
P= 6 POLES'
ll
IS g IN ?
T==FxJL
WHERE :
rl-
F ~ R"f'M
DEVELOP It= IT
T~fxR
SINCE
f
f-13
SOLUTION:
IZO
S= GOO RFM
PilOB.
!ZfQUWED.
ARE THEIIE IN A GfNEflATOI?
I
OF
AI7Jv1ATUTZE. THE DIAMETER OF Wrtl CH
TOJlGVJ:":?
SINCE
LBS'.
1/,300,000
1
WHEN OF'ERATING
= 2·48
120
g ~
10
= CUI(fZENT
SOLUTION:
CON PUC fOil
SINCE.
=
I
i(EQLIIil:ED
fO!lCE
4!? Cf'S
[7.
-f= 3D
WHE({E:
120
2-11] HO\v'
IS 52,000 LINES PER SQLI.4Rf
f3 = FLUX DENSITY
1
fREQlJENCY OF ALTEflNATING VOLT.&Gl:
D-C tvlOTOR: WHEN IT CA/lll1E5 A CU!lllENT OF 90 Atr1f'.
ANCl LIES IN A FIELD THE INTENSITY OF WHICH
120
flf'lfl
BY EACH CONDLICTOI?, 6 /N. LDNG, ON
I= 90
P= 10 POLES
s = 500
OF A
EXCEirrEO
GIVEN:
X 13£>0 X w-8
1'20
900 flFM
FWCE
4 X 270
gownoN:
a) -f= r x !{PM
P= G POLES
SCO Rf'lvJ
l= G INCHES
MACHIN!: 7 (c) A 10-POLE
GIVEN:
=
INCH.
A SIX-POLE 900-l(f'M MACHINE?
(D)
IZO
IZ.
t:50 X Z X GO
OF THE ALTEfZNATING
OF
x
=50
THE Ai'1'/IIATUilE
X 60
X
X 120
p
1!-13]C4LCULATE THE
'250 VOlTS/ fiND Ttl/:' VALVE. Of {P
X C1
-f
flPI.tl =
KPivl
~ = f.056 X 10 6 MAX\v'J::LLS
a)
F== 12 POLES
10-e
f''25 VOLTS
r9 =
1F
f'OLt 1
IN IHE AflMATURE
SOLUTJOIJ:
SPEED (1ZP11)=?
x
IS 150 Cf'S?
CONOUCTOf(S
+~ 5"0 CP.S'
8
X 13"50 x 4 X 270
f X
fg =
ROTATING IN A 12-PDLE MACHINE IF
AllMATURE
0 X 60
fWM .
b) FLux !'fr?
IN THE
lrEQ!lllrED
= IB5 X JOb
IS DECIZI:ASEP
GElJEfUTED VOL TAGt
(b) AN
FREQUENCY
MAX\v'ELLS i"
Eg = <P X P X Z X IZPM X 10-
ONLY THE Sft:ED OF THE-
1,3E>O
JS AN AIIMATU/lE
2-lr] AT WHAT SPEED
TiiE
SOLUTION:
GENERATOR'
a)
I·B5 X
IS TO llEMt\IN THE SAME
TO
If THE FLUX PEIZ
SHOULD THE EXCITATION
GIVEN:
THE SAME VALLI/"'
TO
DII7ECT- CUR17E.NT GENE/?ATOR AND MOTOR P/liNCIPLES
IS DECI{EASED
'2-7
VOLTAGE
i.e. J
VALUE 1
Pl71NCIPLES
7
X
FT
121~
DlflECT- CUfl17ENT
GENERATOR
MOT0/7
AND
DII?ECT-CUICI?ENT
Pf<INCIPLES
2-I~THE AI?IIJATUI?E OF A D-C MOTOR HAS 31 3LOTS hiiTH IG CONDUCTORS IN EACH SLOT- ONLY G8 PER' CENT OF TH/: CONDUCTOICS LIE
DIReCTLY UNDEf( THE POLE FACES, \VHEf<E THE FLUX DENSITY IS
LINES PErz SQUAR'c lf\JCH IF THE AR'MATUR'E-COIZE
AT 4-G,OOO
UNIFOfZM
IS 7!?5 JN ;
DIAMETEil
CONDUCTOI7
4-. 1?5 JN.
ITS LENGTH
AfCMATUI?E TENDING TO
!SLOT
TI-IEJ\.EFOJ(E:
L ""' 4-Z5 IN
F= lii;cat-Jp. x31
T
=
AirMATUfl'E I='ATHS
1?5 AMP
SLOT
AltMATUR'E
b) lVitQUE
'C-16] USING THE OATA
IF THE FLUX
IHE
IS INCICEASED
DENSITY
TO
1:5 =
I~
I= iO
r=
r= eeJ. 7-t
AMP.
0 95
LINf;f'/ :<
x3t
LB
T= G7 lB-fT.
/114
THBr
F= B><LxT
(5"8000)[7Z xG x0.705] (B)
ICONDUcrOilS
SINCE THmE AilE G PAITALLEL PATHS,
THE TOTAL CUI?flENT AIIUST THE
AllAIIATUilE
F
=
h= 57G AMF
x 16]
1050 L13 -f?f
7 e.5 IN X f f
~
TzTN.
95·96 X 6
_f1 Jl(. X 1?1
2
1~/K-
I?OO LB".
2-18] WHAT MUST 5E THE TOTAL AI?MATU({E- WINJ7JNG CUilRENT TN PI(OE'i.
THE
*
lV
REMAINING IS 9G /~
THE17.EFOtrE:
?-17 OtVLY WE TORQUE INCI?I:ASt:S
I =UJ71-1)(11,300,0oo)
1'200 LB-FT, WHILE:: THE fLUX
(5"8,000Xo.g,;;)(a)[7e XG X0-705}
Df:NSITY DfWF BY 1- 'f.
11'EQUII\EO:
I= 111 AMP.
TOTAL AI\MATIJTCE CU/Cf(J:NT (Ir)
IT= 111-
X
G fAf{ALLEL PATH
SOLUTION:
1\,300,000
F=T =
r
F = 1"371-4
X
e-17
IF THE TORQUE INCilEASES TO l,eoo LB-FT WHILE THE t=LUX DENSTTY DllOPS
!5Y "t rEI( CENT?
SINCE FWX DENSITY IS 17/fOP BY -"!-;I.
GIVEN:
ALL 1/ATA AI\E 11-IE SAMt IN 1"~015·
;<.
CAN CAilRY IS:
IT=
r""
= eel 71 LB
= (1-eoo) (11,"5oo ,ooo)
T= 1= x~r
BY 5 FEfl CENT WH1Lf:-
T=f'XYQ
SOLUTION:
X 0-70£>
I= 35 96 A1>1PI
5xLxr
"ur.
THEN:
TOI('QUE..:."'
7e. X 6
l/,300,000
11,300,000
t'5Y 5 /.
X
I
= G PARALLEL PATHS
SOLUTION'
LB-IW.
f= 46,000 X 095 X4-Z5 X40[068
/lEQUI17ED:
B"' tG,OOO
0
1D AMP.
FifOB. 1"-15
rrevuow
DIA."'" fl IN.
ln:QUIIlED:
't'T-1 L13" -FT
IS R'EVUCEP
GIVEN:
DAIA
ZT=
THUS:
OF PI?:QB. i?-15 , CALCULATE. THe TOI?QUE THAT WILL
l'E DEVELOPED
CU/?RENT
=
Al?l: EfffcriVE ONLY AI?E:
T == 1,050 l!> -FT.
If' l~h
T
l5UT TOTAL CONDUCTOR WHICH
B = 56,000 LIIIJES I "
IOTAL CLIJI.flENT (r)
=04-5·9)( ~)
IN (LB-f'f)
5xL
/IN
swTl(o68)(H.>5)(es)'
J
r= x 11,3oo ,ooo
=
POLE AIZC COVER'= 70-5 f.
rACES
F= Jt5""·9 LBS
(b) T::. r- x ra
0) FOflCE E.XCEI<TEV f5Y n-JE
IHUEFO/lf::
r
Zl
= G CONDI
/SLOT
/SLOT
11,;500,000
fll::QUif?E/7:
JN PAfl'ALLEL = 6 ; TOtrQUE = 1,050 l.i!- FT.
171VEN:
L= 811\1.
DILl= 7-c5 IN
OF CIKCUM-
DENSITY= 58,000 LINES PEf( SQVAI?E INCH j AllMATUrrE
5LOTS=7c
LIE ON THE Fl'JLE
L
nux
n:::rn:NCE ;
BUT ONLY GB /. OF THE CONDUCTOI<S
B = 4G,000 LINES I z
IJN
Afl'MATUIZE OF A D-C MOTOTr
CDirE. LENGTH= 8 iN. _j AftMATURE COIZ.E DJAMITEI? = f.'ltN. ; NUMB.EIT OF
a) F==BxLxi
11,300,000
= IG CONDUCTORS/
MUST THE
Pf(JNCIPLES
PEl? SLOT= 6 ; POLE AI?CS' COVCK 70-5 PETf CENT
CONVUCTOrlS
SOLUTION:
.SLOTS= 31
/SLOT
IN EACH
BY THE
C'-'17] WHAT TOTAL CUI?TrENT
AND MOTQI?
CAflflY 1 GIVEN THE FDLLOh!ING INFOIUIIATJON : AI?M4TUIZE SLOTS = 7e j
PRODUCE flOTATION ; (b) THE lVf(QUE IN POUND-FEET
GIVEN:
Z/
AND THE CUfC[(EiJT
J
IS 25 AMP, CALCULATE (o) THE RJTlCE EXCERTED
GENEJCATOTr
IT= 684 AMP-
1'200 L6-FT
10-5 IN X ·F"f/IZJN.
L6·
I.}
DII?ECT- CUfi'r?ENT
3-1] CALCULATE
DIRECT -CURRENT
DYNAMOS
THE COIL PITCHES
ANtl INDICATE THE" SLOTS
INTO \vHICH
THE
FlllST COILS SHOULD 6E PL.ACED FOil TtiE fOLLOWING ArtMA1UilE
WINDINGS i (a) 36 SLOTS, FOUit RJLES ~(b) 57 SLOTS, FOUR: PDLES
(C) 76 SLOTS , SIX FOLES j (d) 132 SLOT$, !:IGHT POLES ~ (e) 270 SLOTS )0 POLES
(f) e31" SLOTS
1
3-3] THE ARMATURE
WHAT CLIITI('ENT
SINCE THE WOI?KING EQUATION FOR
COIL PITCH OR COIL SPAN
::; 76 -
""6
±e
Ys=- 12 ~SLOT
d) Ys = 1~2 ~
Ys-=S -K
p
1 AND 13
WINOING
= e7 j SLOT
IZ
AN INTEQER.
~
=
1 AND ZB
Ys
=
ZO
"'1-
3-~
HOW
=
IX2
2 PARALLEL PATHS
11
= 350
lf'ATH
~
= 17£ AMP·
MANY
PAI<ALLEL
PATHS
( 0.) DUPLEX -LAP
AilE THERE IN THE WINDIN!iS OF ATrMATUICES
J=Ofl RJUI( POLES f' (b) TmPLE.X -LAP I"'OR' SIX
(c)DUPLEX-LAt' FO!r I':':IGHTPOLES? (d)Tr<IPLEX-LAP FOil IOPOI.ES?
FOT\ SIX A?LES?
SOLUTION:
a) PAITALLEL
3-~J HOW MANY PAI?ALLEL PATHS AflE THEIIE IN THE. ARMATUr?E
WINDINGS Of FROB. 3-1 IF (aUb) 'AND (c) A«E .SJt.i!PLEX-WAVE WOUNO AND
SIMPLEX- LAF' WOUND?
SOLUTION·.
p, = PLE.X )( 2
fl = PU:X X f"OLt':S
f't = 8 f'AR4LLEL
FATf'lS
FOR a), b) AND c) , WE PARALLEL
rz) f• = 1 x 10
f,
ARE THE SAMe
p, =I X e.
= 10 f'ARALU:L PATHS
f) P. =- 1 x re
p, = If PARALLEL
P, = ? PARALLEL PATHS
LAP- WOUND WI~DING THt=:
.e) fOI{
Fi
P•
X
POLJ:S
ru:.x x
=
2 X 8
IG
nex x rou::s
f't = <?4
POLt::.s
FArzALLEL F'ATH3
l"ARALLEL PATHS IS GIVfN AS
p, = FLEX x rou:.s
no
PATHS
= 4 XG
PAflALlfL PATHS
=
~
rn~LLEL
GUADllUFLEX-LAI" ,SIX I"OLE-S
fi =
C) FOR. DUPLEX ~LAP, 8 POLES
f"ATJ-IS
!0 FOLES
5 X 10
F't = 30
== 3 X 6
F• = 16
I
F"LEX X fOLES
f"AI{ALLEL PATHS
p, = FLt:X
THEREFORE :
=
=
= Z X 4
FOr< SIX f"OLES
"') x 8
TWPLEX -LAF
Pt
b) FATrALLEL PATHS FOre TRIF'l£X-lAF'
d) Fi = PLEX x fOLJ::S
WAVE WOUND WINDING IS GIVEN AS
d) FOI\
FATHS fVI{ DUPLEX-LAP
rOI( FOUR FOLES
p, = 8
nlfiZ!:FORf
SINCE THE PARAlleL PATHS FOR
FQR
= F'Li:X X 2
~ =
ASS'UME THAT THE
1
(e) QUADflUfl..EX- LAP
= It j SLOT I AND 15
PATHS
fAJ?ALLEL f'AT+IS
FOR WAVE;
f·
Af7E WOUND
POLL'S?
4
THE OTHE~ AJZ!O
NOT STATEC7
TH4T
57 - j__
+
b) YrATtt = .::.r_T- - -
12
I~~ SLOT 1 AND
Ys = 9 j SLOT 1 AND 10
b) Ys
IS LAP
TYPES OF W/ND!N(; IS Slt<lPLEX
a) Ys =~ -· K = J6
p
YrArH = SB 33 AMr
SINCE THE NUM13c(l OF FLEX IS
-f) Y.s= 23+ -§_
SUBHACTED TO MAK.E Ys
/PATH
WINDING IS WAVE.
IO
K= ANY PAliT OF 5/p THAT IS
= .:::3.:::_:50::..;A:.::Mc::.f'_.-=-:=::--:;:
1X 6 Ft\I?ALLEL PATHS
1/
SOLUTION:
e) Ys = KJQ..
Ys
a)
= 5SO AMF
b) CUflREI-.IT f'Eil PATH fl= TliE
Ys == 16 ~ SLOT I AND 17
F= NO. OF FOLES
FOre·.
a) CUIU?ENT PER FATH If IHE
WHERE:
5"" NO. OF $LOTS
PATH IF TliE WINDING IS (a) LAP? (b) WOUND?
lrfQUI RED:
t
IS
IN EACH
=6
f'OI.f.S
lr
c''J Y.s
OF A SIX-POLE GENEIZATOI? CATrmES A TOTAL OF 350 AMP
FLOWS
GIVEN:
IZ POLES .
SOLUTION:
DYNAMOS
;r'
PAR4LLEL
FAlliS
DIRECT- CUrrrrENT
DII?ECT- CUrrRENT
DYNAMOS
3-~ MAKt: 3/MPLE SKETCHES LIKE FIG· 39 SHOWING HOW LAP COILS ARE
CONNECTED TO THE COMMUTAT-OR
FOR SIMFLEX-, DUFLEX -~ TrnrLEX-:, AND
CONT. Or" FI?Of3.
DYNAMOS
3-5
c) TJ?IPLEX -LAP WINDING CONNECTED 10 THE
COMMUTATOR
QUADCWPLL'X- LAP WINDINGS.
COIL 2.
OIL3
CCMMUTATtm Sf:GME/JTS
COMMUTATOR'
U) SIMF'LEX -LAf' WIND)NGS
SEGMENTS
CONN!:CTED TO !HE COMMUTAT017
COILJ
COIL
d)QUADffUPLEX -LAF' WINDING
COMvECTtD TO THE COIVIMLJTATOR
o
I 111~~~~~1 Ill
SLOT 1
CQ'IAMUTATOil SEGMENTS
COMMVTATOrr
b) DUPLEX- LAF
WINDING
~ENTS
CONNt:CTED 10 THE COMMUTATOI? . '
12
I ·'
==.'
DIRECT- CUITI?ENT
Dll?ECT- CUI?l?ENT
DYNAMOS
3-~ MAl</: 3/MPLE SKETCHES LIKE FIG· 39 SHOWING HOW LAP COILS ARE
CONNECTED IV THE COMMUTAT-OR
QUADIWPL!:J<- LAP
FOR SIMFLEX-, DUFLEX -, TrCirLEX-:, AN17
DYNAMOS
CONT. Or' PI?OB. 3-5
C)TI?IPL.EX -LAP WINDING
CONNECTED TO THE COMMUTATOR
VVlNDINGS.
COIL 2
1-~,...-
COIL
SLOT C!
,
SLOT 3
CCI\1MUTATort SI:GMEIJTS
COMMUTATOR'
SEGMENT$
a) SIMF'LEX -LAf' WIND)NGS CONNt:CTI:D TO "THE Wv1MLITAT017
COIL 1
COIL
d)QUADfruF'LEX -LAP WINDING
COM.JECTI:D TO THE COMMtrrATOf\
o
! ~~~~~~~~~~~ ;
SLOT 1
CCMAIIUTATOIL SEGMENTS
COMMUTATOR' SEGMENTS
b) DUPLEX- LAF
WINDING
CONNECTED 1V THE COMMUTATOl? . '
12
l ·'
Dlr?ECT- CUI?f?ENT
UYNAMOS
Dll?'t=CT - CUTU7.ENT
THE FOLLOWING LAP
3-6] DETERMINE THE DCGIT'EES OF I?'EENTTZANCY FOR
(a) DUPLEX,
WINDINGS:
3G SEGMENTS :_ (b) DUPLEX , 35 SE'~MENTS ;(c) TR:If'LEX,
(d) TRIPLEX, liB SEGMENTS (e) GUADIWFt.EX, C?eG Sl::Ctv'\:ENTS.
117 SECMENTS
3~ DLTEfMNE
Trtlf'LEX, 117 SEGMENTS
SINCE THE !7EGI?EE OF IZEENTRANCY
I?"= Trt/PLEX =
3
OF LAP WINDINGS IS EQUAL TO THE
COMMON
FOUR Ft>LE..S ~
f(
FAC1VR' BETWEEN
= 3- TK!f'l..J:
J!2_
3
SINCf Yc::: C
Ri=ENTilANCE
WttER'E:
I?= 1 -SINGLE
AND THE PLEX Of THE WINDINGS
TltElJ,
lltl:RI:FORe j THE llf:fNTI?A NCY
lS
FATH
l:lRUSH ARMS 7
AND BY
EACH
IS WAVE 1 HOW
Bf?I)SH AR'M
(b)
ron:
DOUBLE.
h
o:
~
~
ore
£'2'J :!: l
Ciz
Yc.= 7G
G:J) FOrt
c''C7 SEGMENTS ,
%
Yc~
57
'+7
MUCfl CUR'TI'ENT IS HANDLeD
IF THERE AI\E (a.) FOUR'
3-9] IN tACH Of THE COMl51NATJONS
AfWUND
(b) TWO BRUSH AR:MS?
THE
COMMUTATOIZ
OF
F"n'OB". 5-8, TllACE TI-lE WINDING
ANIJ SHOW /HAT THt: F'fWPffl SEGMENT , AHEAD OF
SOWTION:
230 VOLT MDTOIT'
THEN:
a) rorz Yc = 37
!80 AMF
180
%R:USHARM.
f'OLES = 4
TRACING SECMENT
4je
1 -38 -75 ..... ONE 13Et!IND SeGMENT 1
TYF'E Or WIN!71NG =WAVE
FOI( Yc=38
90 Ait1_%1<USH All'M
tn:OQUIR"EI/ :
curm:ENT PElt FATrl ~ fEK BRUSH ARMS
"
(tV II
i'FATH
.:::: leo Alr1f'
z
SOLlJTION:
=
(a) f, .:: e FOil WAVf:
r;,:
=
100 A1...1r .
I/
, fATr:
=
90 AMFI
rATH
(l'I<USt! ARM
/F4Tt1
THEREFOR!: ; fOR
j
fOR Yc
90 AM%-rH
B"I~USH AR'M
4
=
/80
eh
/80 AMW
/BRV.SH ARM.
FOil. EVERY SET OF BRUSHES
lltERE IS ONLY
1- 39 -e .._ON!: AHEAD SI:GMENT 1
b) fOR Yc "' .46
1- 't7-9:5 -ONE l3EJ-IIND SEGMENT 1
= 47
1- tB -2 --+ONE AHEAD SEGMEUT 1
1!
e
c) FOR Yc= 70
1-77-153-<'.'?9 ..... ONE ISEHIND SEGMENT 1
d)
fOR Yc
=:
fi7
1-5B-IIS-17Z-2- ONEAHEAV 5t:GIIll::NT 1
I
BRUSHES ,
THEI\E Afl'E lWO SETS Of BIWSH
AIZM5'.
' ::'f-
EIGt-lr POLES
Yc == .f27! I
9!i .st:CMEJ.JTS JOUR FOLES
Yc= 1G
=-
Orr. f:!EHIND IHE FiftH ONE , IS I<EACI-+ED.
1
SINCE
Yc:::
)t
Yc = 93 ± l
GIVE:N:
50 tW
(C) FoR tc'3 St::GM!::NTS , SIX J=>OLES
'lc= 37 OR 58
o-~ THE TOTAL AI\MATUI\E CAflRIED l'JY A !50-HP FOUR POLE e30-VOLT MOTVIT'
If THE WINDING
FOil SIMI"'..t)( WAVE
~2
2 13 TtiE COMMOI--J FACTO/l
17 = 1 - SINGLE ffEENmANCE.
FOUR' f"'LES
OF POLES
-z-~
DUF'LEX LAF, 3S SECMENlS
1
±1
.5t:GMEN1S
IZffi-JTI?ANCE
IT'== t - eero
LAP , 36 SECMENTS
5Y EACH
SEGMENTS, E'IGHT POLES.
C =TOTAL NWI!'5EI? Of COMMUTATor?
e) QUAOI?UFLI:::X / 286 SEG'MENTS
THE llEENTII'ANCY IS 2 -IXlUBLE FOR
IS 180 AMP.
e·n
'lz
P- NO.
THEN:
b)
(c) ee9 SEt;ME.NlJ , SIX A:>LES ~(d)
(a) 7'5 SE"GMENT.S
Tl-IE NUMBEI( OF COMMUTATOR SEGMt:.NTS d) TIZ/f'LEX , /Ill SEb:IVlf:NTS
VUFLEX
FOR THE FVLLCW/NG WAVE-
Yc
SOLUTION:
c)
a) DUPLEX LAF , 3G SECMENTS
HIGHEST
PITCHES
ARIIIATU/lES : (o.) 75 SE(;MENTS, FOUR POLES ~ (b) 93 SEGMeNTS,
WOUND
SOLUTION:
THE COMMUTATOI?
DYNAMOS
IS
1
DYNAMOS
~~ECT-CU~RENT
CONT. OF PROB.
3-15
COIJT OF 3-17
SOLUTION:
T£7ACING
Cl) FOIT COIL PITCH
~orr
p
+
~ F017 COMMUTATOr?
=
I
P_lz
= o WINDING
Yc = IG Or?
THE 17ATIO
OF .SEGMtNTS 1V SLOTS
PITt~~!+
,
IS NOT
ELEMENT~
=
4fc
P, =- PLEX
f,
I!
=
B33- 33 AMF'
10 PATH
!PATH
·
/WINDING ELEMENT
= S3-3"3 AMP
e) cutmE.r-JT f'EK f5JWStl ARM
f p= 500, 000 WATTS
I!
= 8i5·33
!BflW'H ARM
1%
600 VOLT
COMBINATIONS OF SLOTS AND SEGMENTS: (a) 33 SLO'cy, 99 SI:GMEIJTS?
(b) 7G SLOJ:S' , 2e7 SEgMENTS? (C) .39 SLOTS, 77 SEGMENTS ? (d) 5"4 SLOTS,
I X 10 = 10 PARALLEL
PATH
/COIL
1 TlJflN/
5-1~ WILL THERE BE A ''DEAD" ELeMENT IN EACH OF TiiE FOLLOWING
X POLES
=
11JinJs;cmL
d) FULL-LOAD CUI"li7.EWT f"Eit CONDUCTOI?
17
I
Af?E;
PATH
/COIL
--. '3 WINDING fLEMENTJ'
A WHOLE NUMI3EI?-
33 !. 1
3
10 POLES
THE NUMBEJ"( OF PARALLEL
F'Eft~·WINVI/-.l~ ELEMENT
C)TUilNS
THERE A!CE
SINCE
108 SLOTS
r:LEMEIJTS I BECAUSE
1-S-9 ..:15--17
=C t
= 3"2t SEGMENTS
Yc =17
YES. THEITE WILL 131:. A DEAD
Ys = 4
Yc
b) WINDING CLEMENTS f"EI? COIL
1-18-2
=Jl..-j_
Tf(ACING;
Yc = IG
F017
1-17-:53
Ys= S -K
+
DYNAMOS
Dl/i'cCT-CUITITENT
It "" 833 ·3.3 AMP:
=l~·G7'
A~-1f
AMf.
216 SLGMENTS?
J-18] THE EMF GENER4TED IN EACH CONDUCTOI7 OF A SIX-POLe SIMPLEX-
SOLUTION:
Cl) 33 SL013', 99 SEGMENTS -
IJONE
TO
ij 76
SLOTS, 'lC7 SEGMt"NTS -
I
15fCAUSE THt: TlATIO OF SE"MEN"IS
SLOT IS A WHOLE NUMBEI?
YI:S, BECAUSE ..z.zJ_ l S NOT A WHOLE
76
NUMBel?
WINDING
IS 0 ·'tB VOLT. /F THE AICMATUR'f HAS 1-2 SLOTS
AND IZS COMMUTATVIT SEGMt:NTS . A~l'Y
S<ITttEm:ft'JRE THt:R!: IS NO Dr.AD EILMENT
GENERATOtr HAS AN AlrMATUfCE WITH
108 SLOTS 321- COMMUTATOIT SEGMENTS, AND A SIMPLI!X- LA f' WINDINB
WITH A TOTAL Of 618 CONDUCTORS. CALCULATE: (a) 71-lt: NVMISfll Of CONDUCTORS f'Eff. SLOT j (b) THE NUIIIBEIT OF WINDING eU:JviENT PER COIL ;
(c) THE NUMI3fiT Of TURNS PER WIND/I.JG t:LE:MEIJT j (d) THE FULLLOAD CUfCflENT FEll COI-41/UCTOf( i (e) THE. CURRENT P't:R BRUSH AI<'M.
ARE fOUir TURfJS
rm
10 -POLE
SOLUTION.·
F= G FVLr:J', SIMPLEX WAVE
ll!E 1DTAL NUMBER Of CONDUCTO!lS
;:,Fi~.
S =1-l SLOTS
C =ItS SECME!-JTS
Z1"" 't TURIJJ/
llJJUJS'/
=
/Wif.J/7/NG t:'.LI=MEIJT
GEf..!ErATt:l>
ISMr
>(
/WifJDING ELEMENT
+
125 SfG1r1tNT
ZT = 5VO CONDIJCIDIZ.f
IN eACH
VT = EMf
X TOT4L CONDUCTOiD'
COIJDUCTOI?
1
CONPUCTOrr = O·te VOLT
=
IZEQ'[7·
Tl:!ZMINAL VOLTAt9C Of THE
Vr
O·f8
= 21-0
X
500
VOLTS
G'fNE ItA TOR'
SOLUTION:
GIVEN :
F'=500 KW
Z= f}f8 CONDUCTOI?'S
a) CONDUCrortS
Z/
!::;= 6'00 VOLT
ISLOr
FOLcS = 10
=
f'ER SLOT
5-20]
GtB QJtJ!7UC[OIU
108 SLOTS
A 54-SLOT
ARMATUITE
216-SECMENT
A!rtv1ATUflE
HAS A FOUIC-POLE LAP
THAT IS EQUALIZED (00 PER CENT AT THE COMMUTATOfl .
CO~ECTIONS A[{E 11-lERt:. ? (b) TO WHAT TWO
o) flOW MANY EQUA LIZef(
SLOT = 108
==- G CONDUCTOIU';
/SLOT
C = 3t'f SEGMENTS
TY~ Of
Tl.lEI~E
ARMATURE -WINDING' t:"LEMENr, CALCULATE THE TEIZMINAL VOLTACE OF TH::';
GIVEN·.
d) 54 SLOTS, ~IG StGMt=NTS- NONE:. 6ECA~ 21G IS A WHOLE NOS.
600- VOLT
AIZMATVrlE
GENERATOll.
C) 39 SLOTS, 77 SEGME:NT..5- YES
3 -1~ A 500- KW
WAVE
SEGMENTS
iS THE Fll?'sT
EGURLIZEIZ
CONNECTED 7
WII-JDING= SIMf>L.f:X LAf'
,g
t9
r,~""'
DlflEIT- CUIU"CENT DYNAMOS
OF
CONT
01/?ECT- CUrmENT
;5-f~
F'IWB. 3-eo
GIVEN:
SOLlJTION:
S=54 SLOT.S
a) rOR: 1007- EGUALIZATION
= 2/C
<t F'OLEJ' , LAF' WINDING
p~
=L
Pf-z
C =t'IG SEGMEfJ'RS'
= 108
a) ftOW MAt-JY EGUAUZm CONNeCTIONS
ARE THEile ?
b)TO WH4T SE(;MEIJTS IS
f'AI?ALLEL
t:GUALJZEIZ'
MANY
EQlJALIZJ:ItS
IN
a) PAflALLEL PATHS FOR A
IN A SIX- POi.£ 50 Pflr CfNT
AlrE THEflE
A= 8 X 2
F()f'( fflOG-LtG
f'• - IG PAITALU::L
THE COIL ANO COMMUTATOR
SOLUTIOt-J:
p
6
FOrr FROG LEG WIIJDING
Y.r = IZ
Yc= C:t3
6 fVLE , 21" SEG'Mf:NTS
O·S (ZIG)
=
FOTC TR'IfLE.X WAVE ONLY
'lz
3G EQUALIZ!:RS
'/e = '288
G/2
'!:
3
=
'35' ore 97
~
FOTl PROB · 3-21 SIMILAR' TO THAT (;IVfN
FOR
EXAMFLE 9) p. 80·
GIVEN'
f'~G
c
='
fOLES
2JG' SEG'MENTS
SOLUTION:
TABLE OF WINDINGS
5G
FD~
EQUAL! ZERr:
ll-47-83-119 -155'-191
124-G0-96-132-168-204
12-48 -81-IZO -15G -192
t5-GI -97 -1~5 -IG9- 205
13-49-85-121-157-193126 -G2-9B-13"f--l70 -'206
1-57-73-109-145-181
It- so -BG-122-158 -19<1-
L'-38 -71'-110 -146-182
15'- 5'1-87 -123-159- 195 28-G4 -l00-136-17Z- 'Z08
3-39 -75 -II 1-147-183
liD- 5'2-86 -IZ"'-I~H9G ?9-GS'- lOl-!37-173-209
27- G3- 99-135 -171- 207
4-40 -7G-IIZ -148-1134 17- 53-69-/25-IGI-197 '30-GG-102-1~- 17+-210
S- tl- 77 -/13 -149-185' 18-.91- -.90-IZG-16'2-198 '31- b"T-/03-139 -17S- 211
6-42-78 -114-/50 -1~ 19 -55'-9l-IZ7-IG3-I9j 32-bB-104 -I'W-17Cl- 21Z
7-43-79-115-151-187 20-SG -92-128-164-200 33-G9-10.5 -Jil-177-'2/3.
8-44-80-116 -15? -lf.e 21- 51-93-IZ9-I65-20/ 34-70-IOG- lf2-17B-214
9- tS -81 -117 -153-18'3 22-58-94 -130-166-202
3~ -71-
PITCHES
Fl?OG-LEG AR'MATUflE
Ys- S = 72
!
FOlt 5'V /. EQUALIZATION
3- 22] MAKE A TAI5'LE
b) FVJ\ 8--FVLES, f/?:OG -LEG
P'ATHS
P• = ?..0 FAirALLI:l PATHS
:H~ DITEI?MINE
WHICH THET'{E ATl'E 2/{i SEbMEW
=
FlroG-
FIRSf
SOLUTION:
5Dj. E<WALIZATION
IN (O) A 10-FDLE
SOLUTION:
f88- SEGMENT SIX- R?LE
EQURLIZED ARMATUILE
AilE l'HEI?E
FrrOG-LEG WINDING?
= 10 X 2
EQUA Ll ZEfl CONNECTED 7
3-tO HOW
PATHS
(b) AN EIGHT- POLE
p, ~ POLrs X 2
b) 'f!WM st:G'MENT 1 TO :109
THE
'?
10- POLE FKOG- LEG WINDING
4fz
ILEGD·
HOW MANY
LE'G WINDING
DYNAMOS
107-·lt3 -179 -2/S
10- "''G'-·82 -118 -154-190 Z3- 5'3 -~~ -131-lfi7-2Q3 3c; -72-108 -I# -JB0-2/G
~~=~~~~~Da--------------------------------~------~~----------------------------.1
.?.C
.'I
WINDING.
FOK A 72-SLOT
DlllECT- CUili?ENT
4-1) EACH SHUNT- FIELD
GENEI'{ATOTC'
CHAIZACT!:I?ISTICS
COIL OF A D-C GENERATOR' PRODUCES
1UilNS. HOW 1/lANY 11JilNS
IS 15 AMF 7
AITE THErn:
= 2, 700
1·
2,700 AMP-
1111 THE COIL IF THE CUI?I?ENT
At.-"lP- TUrtlJS
1·5 AMr.
4-4] CALCULATE
NO LOAD VOLTAGE= 13!> VOLTS
FLU LOAD VOLTAGE=-120 VOLTS
Nl -
~700
NUMBEf{ OF TUR'NS 7
N =-
'2700
SERIES FIELD
WHICH
IN
EACH COIL IS WOUN/J
fo Vf( =-
HAS A
v..,~- VF-L
WITH 8 ~ TUTlNS . HOW MANY
COMFVUND GENER:ATOit
J.vrc= 12·"5" f.
REGULATION
X 100 j.
OF A 250-VOLT SHUNT GENEilATOI?
IS 6 f'EI? CENT·
CALCULATE. THE NO LOAD VOLTAGE.
GIVEN:
VOLTS
Y·rl.=
<;;I.
FULL LOAD VOLTAGI0=25o VOLTS
AMP- TURNS
l"Tl0£7UCEC1
BY
R.EG'D.
COIL?
CALCULATE. THE NO LOAD VOLTAGE..
SOLUTION:
SOLUTION"
LET: NI :IS THf: NUM.6EIZ OF
t V·R'. ~
1V 0-C LOAV
.AMF-TURNS
X 1001
VN·L -VFL·
VT-L
SINCE::
1]-IU.J:
f'"' e:I
VIJ.L."'
f. v.n:.
! - p
50,000
250
I=-
fiGUfZE
LONG SHUNT COvlPOUND GENETC'ATOTl'
w
v
""
-1-
Vn.
VI'{-L. =
200 Alf1P.
NI "'" B·:S TUITNS X ZOO AMP
4-G
1700 .AMP-TI.JR:NS
6
1.
J IF THE
2G5 VOLTS
NO-LOAD VOLTAGE OF A SfPAI?.ATELY eXCITED
IS 110 VOLTS
AT
1,350 171"~;1, WHAT WILL BE THE VOLTAGE
IS lNCI7l:ASl:D TO I,GOO RPM
(ASSUME CONsrAIJT
2.1
+ 250
X 2!>0
100
THEN:
=
X VF-L.
/00
E
=
lVI
120
l5Y EACH COIL 7
R't::Q'(l.
EtCH
"j.VR·= ~X 100
v,..L
'l~ TUilNS
8
Ns =
13'5 VOLTS AIJD
TitE EQUATION IS :
4--5] THE fi'EGULATION
VT=~50
VOLTAGES AilE
SOLUTION:
~IVt:N·.
P= 5"0 KW
%VOLTAG'E
CALCULATe THE
250 VOLT COMPOUND GEf.JERATOI1 (SHORT- SHUNT)
Afi't:: F'IIODUCEP
IZE.GULATION OF A SHUNT f:7EIVEI?ATOR:
rn:Q'o.
AMf'- TUI7:NS
1·5
AMP- TUI<NS
VOLTAGE
LET N = IS THE NUM5m OF TUIZNS
llEQ'D·
A SO- KW
CE/ff
SINCE:
N ~ I, 800 T1J RNS
t-{J
THE PER'
IN WHICH THE NO LOAD AND FULL LOAD
120 VOLTS , llESf"EC71VELY.
GIVEN:
SOLUTION:
GiVEN:
Nf
Dlf'?ECT-CUI'U?ENT GENE1?AT01? CHAflACTEmSTICS
?
SHUNT GE:t.JE17ATO(?
IF TtJE SFEe.P
IS OEC!lEASED TO 1,100 llPM?
FIELD EXCITATION)
.,--~'
DlflECf- C!HlflENT GENETlATOfl CHAI<ACTERISTJCS
DIRECT -CURRENT GENmATOO CHAI(ACTL"mSTIQ'
CONT. OF PR0f5.
CONT OF PIWB 4-8
13Y flATIO AND I?TWPOflTION:
+-"
GIVC:N:
Vu-~
SEPARATE"rl EXCITED SHUNT Gt:NtJlATOIJ
o)
R"FM
= !,GOO
v..o = -6oxryo)
riGLIIlE
SEPAR4TfLY EXCITED
rcf'IVl
~HUNT
trL-=
to8
OHMS
IF THE SHU.NT FIELD ftESISTANCc IS 20 OHMS, WHAT IS THE SE/liES-FlELD
CUILRENT AT FULL LOAD 7
AT S2
Vr = IS TI-lE VOLTAG!:
GIVEI-J:
FOR V•=!
V:5 = Vz 5s
V3"' IS THE VOLTAGE AT Sa
3Y 11ATIO AND rnorormo1-1:
VoJ-L = Vz
s;:
P~
s;
= 1"30 ·5" '/-.
Y!l
I~SV
= £39.7
FULL LOAD 7
SOLUTION:
LET:
Is= IS THE SEitiES flEW CURiltNT
13".50
LONG SHUNT
-
COMPOUND GENERATVR
IA =IS THE AIZMATU/Ct cuiTT?eNT
VOLTS'
f-8] A SHUNT GC:NER.ATOR HAS A NO-LOAD TEflWIINAL. VOLTAGE OF 270 AND
A VOLTAGE OF 24-0 WHEN IT
EXTERNAL CHARACTEiliSTICS
Yt= f50 v
s .
l.I.t.=I:
l7f
WHAT IS THE SE"ICIES -FIELD
VOLTS
IL·I50KW
'
wrmcNr AT
Va- 110 X 1"-00
= 150·5"
~ ]f
4
rrco'o.
1100
1~00
V-z = V!-1-L X S:o
-I,
/50 KW, -zso• VOLT
tlf= 20 OHMS
s2
= V3
1V vt::rf.!{MINE Vz
Yt
+ 270
"'t-10] A 150 -KW 250-VOLT COMPOUND GENErrATOR" IS CONNECTED LONG SHLWT.
GENEMTOIJ
SOLUTION:
1350
I 'lOA
y,.o = 2.'50 ~LTS
IS INCI?EASED 1D S, =!,GOO IZPI/I
AND IF THJ: SPEEV !.S fJECJ(EASED 1D
LET:
= tJ;O V
180.
WHAT WILL BE THE VOLTAGE IF 11-fE
= ,,,oo
IIZO
'270 -Voto
- - ==.
IOCJ
120
I?EQ'[7:
s~
120
.Z70-2.W
S-s =- I, 100 RPM
Sn:EIJ
b) ilL= y,.o
V.J-L- Y,.o
=
lBO
VH-L "'110 VOLTs AT 1"550 I"CF'M "'s,
S2.
-V•+o
DELIVERS 180 AMP. ASSUME A SmAI~HT -LINE
AND VITEfl'Mli-JE ntE VOLTAGE WtfEN THE CUI?IZENT
IS 120 AMP- WHAT IS Tilt EGUIVAU:NT LOAO RESISTANCE UNOC/"l 11-HS
SIIJCE THE CO/JNECTION IS LONG SHVNT
T..;=Is = IL +!f
BUT
IL =
...u,_.
\4
!L
=
GIVEN:
V120 ""
V= Z'l-0 VOLTS AT 100 Atllr
IS THE VOLTAGE WHEH lllc
CU~ItENT
Kt:QUIReD·.
Q) l7J:rEIZMINf WE VOLTAGt: WHEN
THE CURRENT IS 120 AMP:
lo) THt:. EQlJIVALI:IJT tOAD /Zt:SISTANCE
V~o ~ IS THt:
If= A_
= .fE2_
zo
IS 1'2.0 AMP·
VOLTAGE
t-11] If THE GENE!lATO!l OF F/W/5. 4-10 IS CONNECTED SHOflT-SHUNT, WHAT
IS THE FULl- LOAD Sf/liEf- FIELD CURRENT 7
fl':f.
WHEN THE
CUilRENT IS 180 AMP(,
IIri"
UNI/Eit TtiiS CONPITION·
24·
Is =f:lt-5" AMP.
ALSO THIS IS THE VALUE OF TA
If= Its· AMP
SOLUTION:
Le:T:
NO-LOAD VOLTAGE= ~70 VOLTS
1r = rooo ·Ht· 5
GOO Alv1f
nr
CONDITION 7
THEflt: FOflE ·.
w
250 v
= 150,000
I
0r/
~
-Is~IL
frA
TO fl-C
lOAD
25
FIGUflE
sHorn - SHUNT
COMPOUND GENE:I7ATOI?
DlflECT -CUI?I?ENT GENEI<ATOI? CHAT?ACTEniSTICS
COtvT OF f'IWB · 4-11
SOLUTION:
DlflECT -CU1?11ENT GENEI1ATOI? CHARACTE.R'ISTJCS
COhlT OF F'IW13 4-1+
THEN:
FIELD CUilllENT 1s eQUAL
FTWM THE FIGUT?E
Vr
GENEI?ATO!r , THE ''Is" OR S'Ef(l ES
= 150000
w
IL =IP +Ise
Ise =h -To
w
'l50V
THE LOAD CUTliCENT ''Ji.. ".
1s = GOO AMPERES
-t l'z
f'Eil POLE AN17 A SEIIIES- fieLD WITH
1$1!. =
IF Tti.E SHUNT- FIELD
AND JElliES-FIELD AI.1FJ:IZE- TUilNS .ARE , llESf'ECTIVELY,
I'ZOO ANI/ 196, CALCULATE THE F'OWfll
TEilMINAL VOLTAGe
DELIVERED TO A LOAD WHEN lHf
IN
flp =
i1s
~ ht
f(f
-TL
D
IrA
C
L
~
A
~
,..
SERIES -flEW AMP- TUR'NS= 19G" A/llP-TURNS
fVWEIZ DEliVERED TO THE LOA.II WHEN
LOtJG-SHUNT
THE TERMINAL VOLTAGE IS 230 VOLTS.
GENEI(ATOil
NTSHu~r
HOW MANY
AMPEilE-TUirNS Am: ffiODUCED BY EACH
4-15] DETEr?MINE
SE111ES-FIEL17 COIL OF A COMPOUND Gt:NERATOIT , GIVEN THE FOLLOWING
f'Af?TICLJLArtS:
!lATIN~= 100 Kw; FULL-LOAD VOLTS= 600; SEfi.IES-FIELD TUI?NS
f'f.l? COIL "" 8~-z
j
ll.I:SISTANCE
OF SEfW:S FIELD= O..OZS OHM j 171VE!ZTER
fCESISTANCE - O·OG8 OHM.
]?p
GIVEN:
fULL-LOAD VOLTAGE::. 600VOLTS
Tl+f.t-1:
IL
AMF-w-rffiS
1100 )1JR.1\1S
fL
I:JG AMP- WR+d.J-
Is= 43 .ss
=
y,., -It
Rs"'-= O·OZS ..Il.
4'5·!>!5 -1
!Zo = 0-0(88 ..fL
i2.
ID
!lf
OC LOAD
I1- = 12 5"5" AMP
AMf'.
1·5
SERIES -FlEW TUllNS f'ER COIL"'8
=
Ts:
=
~4
Q.()fi OHMS
F=-100 KW
t"I!Lb
Ns#I.INT FI£LP
Is
flo=
A
COMPOUND
SOLUTION:
FOR
:: (.11P)(O·Of)
0
D
~
ru:(io.
It "' 1
I.ns llJY
It>
SERIEf- FIELD TU/LNS= 4 Y2 TURNS
SHUNT-fiELD AI'IIP-TURNS~IZOO ,drvJP-TUITNS
= 1200
f'IGUfCE:
COMPOUND GENERATOR
f'AI?ALLEL
IP RP "" Ire Rre.
IS 230 ·
SHUNT- Flt:LD TUFlt.JS = IZOO TUR.NS
DC LOAD
3<0 AMP.
Vo ""Vse , 17EC.4Ust IT IS CO~I.JECTED
GIVEfJ:
If==
l<f
SINCE:
fER .POLE
TUilNS
TO
'0- 2t
e
4-12] A LONG- SHUNT COMPOUND GENEI!ATOI? HAS A SHUNT FIELD WITH
1,200 TUrlNS
flo
SOLUTION:
Is= Ji..
IN SHOrlT SHUNT COMPOUf.JD
nm:m
PL
ITEQUIRED:
IL Er
.,: (41 !;S") ( 230)
=
<:J7BG
=
WAITS _,..
rowm
DELIVE/Ll:D TO THE LOAt7
AM F.
f='IGUITE:
HOW MANY AMPERE- TUITNS AI7E ·
COMPOVND GENE!tATOR'
FTWDUCED BY EACH SEfl.IES FIELD
COIL OF A COMFVUND GENEI!ATOIZ.
SOLUTION:
4-14] 4 SHO!n- SHUNT COiv'IPOUND GENETlATOIZ HAS A FULL-LOAD CUTl/IENT
OF GO Alv'IF'. If THE SEfZ!ES- FIELD RESISTANCE
DIVERTER
CAR!ZIES
24 AMP, WHAT IS THE
GIVJ::N:
fULL- LOA o curm:ENT ,
TlESISTANCE 7
DIVcrc:TC.':fZ
THE SERIES- FIELD
lL=.fL =
V~;
RESIST;lNCE "i<o"
=
SII.JCE: JL., I~>+Isr
AMFEJ<E- TUI!NS.
iS 0 ·01 OHM AND A
DIVEfZTETl
fl~QUIILED:
r L =GO AMI'
U:T N-I s.r
JOO,OOOW
lt>= lL-Tse.
SUBSTJnJ11": : f t>
l?o=Ise.RsE
600 V
1P
flp = fS£ TU-e.
!L -J S:E.
h= IGG·G7 AMP.
= 0·0'1 OHMS
1D = 2.4 AMF
IZS·E.
2(
27
IIJ THE .EQ.
DIITECT- CUrmt:NT GENEITATOIT CHAfrACTEITISTICS
DlllECT-CUili7J:NT
CONT .OF PR05· f-J!l
4-17]
TO DE1EitMINE TsE :
WEN:
illSF
RDfL
~
A SHUNT- FIELD
+.ftp
BRUSHES, Or 0-02 OHM· WHEN THE AITMATUITE CUITIZENT IS 128 AM!"
INCLUDING
THE. tENERAIED
TO THE LOAD.
N.ts~; "" 103G ,4VIP-TURNS
ElvlF IS
2~'f.2
VOLTS. CALCULATE THE:. POWER DELIVEILED
Rs"
Tst: = (iG6.G1)(0 OC.B)
GIVEN:
(O.OGB+ 0·025)
= 77 .il..
l7sE= 0 DOB ..fl.
fl[
lse "" 1'21 ·87 AMf.
4-lG] A 5-KW 120-VOLT COMPOUND
TlESISTANCf OF 0-2"5 OHM, A SffZIES-fiELII
o o2 .n_
rzA =
GENtRATOI? HAS AN ARMATURE
ftc = 0 ·005 ..Q.
flESISTANCE Of 0-04 OHM, AND
128 AMP
A SHUI-JT- FIJ:'Lt:7 rn:SISTANCE OF 57· 5 OttMS. ASSUMING A LONG -SHLINT
IA
COf.JNECTION ANI/ A VOLTAGE
t'j= 23t Z VOLTS
THE GENERATE!/
DnOP AT THE BIZUSHES Of 2 VOLTS , CALCULATE
AT fULL -LOA!7.
EW!F
Vt~
--r..
~
5' KW
lf
120 VOLT
f'OWEfl. DEL IVEIZED TO THE LOAD
SOLUTION:
FfWM THE FIGUIZE .
!?2:. 0-04 ...0..
E~ == v~ +
ITt
ItA
Ve-e.= 2 VOLTS
J1EQUIRE'D:
LONG -SHUNT COMPOUIJD GENERATOrr
SOLUTION:
VOLTA{;E
r,.... lf+h
= 'IH7 + 'l·08G
Vt= TERMINAl VOLTAGE
Vsc = BlliJSH CONTACT VOLTAGE
Vt + IA (l?A
lL= FL
-Vt:
trrsE)
-+ Vac
TIL
=
5,000W
-)20V
E
F1-= I...Vt
= (lzs)(~3J)
If=
1000
77 ..{L
FL- = 28 -~ KW.
3 AMF
4-18] THE FOLLOWING INFOfl/IIIATION IS GIVEN
FOR A 300-KW 600-VOLT
t':-:t""
GENERATED
LONG-SHUNT
133.8 VOLTS
BY TltE .4flM4TUflE .
P; :'XJD KW , LONG SHUI.Jf FLAT COMf'OUND
V+.=_600 VOLTS
rlf = 7'5 OHMS
--
If
llf
,AMF.
GIVEN:
IL= 41·G7 AMP.
lf = y,.
R'f
Yt-= 231 VOLTS
= 115
E'~= 120 H·3·75G (0·23+0.04) + 2
THEN:
IA = lf
\4 ,:. '234 2- 1-ze (o. oz + o.oos)
.:::75 OHMS; Afllv1ATU17E
SINCE:
B't.ff
:=.128-3
Tt..
FLA1-ChllFOUN!7 ~ENEflATOT?. : SHUNT -FIELD RESISTANCE
f(ESISTANCE INCLUDING BflUSHES = 0-03 OHM ;
COivlMUT AT/NG -FIELD WINDING llESI STMJCf ~ 0·11 OHM WHJ:.N THE
MACH/1-JE I.S DELIVEfliNC RATED LOAD, CALCULATE THE VOLTAGE AND POWEll
!A = 4'3·75C ,4MF·
DfWf"'.
E'} ==
lL =lA-Ir
fA ( ((.. +l?c)
It=~= 231 V
Gef.JERATED VOLTA~E
E~- G~C:RATfiJ
t=lburlE:
SHOnT- SHUNT COMPOUND GENEJZATOfC
!?Ad O.f3 ..0..
flf= 575" ..fl...
LET:
=
flEQ/D:
BIVEN:
p~.
17ESISTANCE
POLE WINDING RESISTANCE OF 0·005 OHM, AND AN AIZMATU17E l?fSISTANCf,
Nse X Is£
= (8 '/z ) (121·87)
Ise = !Wk
R'se.
A SHORT- SHUNT COMPOUND GENEITATO 17 HAS
OF 77 OHMS, A SERIES- FIELD flESISTA 1-JCf OF O·OOB OHM, A COWIMUTATING -
IsE;R'se. =Ito (h-I sa)
l~E(Rs;e+RD)=
GENCI(ATOI? CHAilACTEiliSTICS
= IZOV
57· 5-0-.
2·066 AMP.
l?t:'Q'b:
l'lc>uc =-0.03 SL
CALCULATE THE VOLTAGE
l'lc"' 0 0/l.fL
GEN.EITATE!7 f'ft TilE ARMATURE
IZse"' 0-0IZ..O..
28
Gt:Nl=RAWR
fl""" 0·03G .fl...
--
•';)i)!
AND FOWErt
.
DIJ7ECT -CUJ7R'ENT GENEI7ATOJ7
DlllECT- CUI?I?ENT GENEJlATOI? PHAilACTEITISTICS
CHAITACTER/STICS
SOLUnON•
lL ~ PL
\h-
IL=
600
a)
w
v
f4-= IA z rz..
fc
Vr:
=
=
(0
=
0 ·00
A NO-LOAD
FIGUI'CE 18-2
4' fZ:::
AilE ALSO AllMATUfZE curz.rz.tNT
IT IS CONNECTED
IN SEJLIE'S
t-
f<J = C<JfA
fs
G:Z51- VOLTS
"1-1~ IN P/(05. 1-18, CALCULATE THE POWEll
(d) THE
SEIZIES
CHANGE
FIELD
!/:z
8
TU/lNS
FIELD HAS 800 TURNS PER
PEIC POLE. THE SHUtJT-
A NO- LOAD
VOLTAGE OF 230 AND A FULL- LOAD VOLTAGE
FIELD MUST
a) THE llESISTANCE
PflODUCE
22'5 AMf'Ef(E -TL'I?NS.
OF A DIVEIITEJ?. TO ACCOMPLISH THIS
b) THE TOTAL NUA<1C5Erz OF AMF'EitE --TUilNS
,AT NO-LOAD AND AT
PIWOUCEO 13Y EACH POLE
fULL- LOAO.
-rL
LOSSES 1111: (o.) AllMATUIZE ;
hf
W/ND/NG j (c) THE SEI'liES -fiELD WINDING j
l?o
1D LO.AD
!rf
DIVERTER ; (e) THE SHUNT- AELD Clll:CUIT-
SOWTION:
LET
fb= PI:JWE'IL LOSSES
fC,_ P'OWEJL
PsE:
PD
IN THE Af?tv1ATUfZE
LOSS'ES IN THE
C0Mtv1UTATING
POLE WINDING
.______________.___tj
F'OWE'R' LOSSES IN THE SERIES FIELD WINDING
LOSSES' AT TttE DIVi:RTER
fi{;UfZE
~ F'OWE!7.
3U
HAS
AilE 80 OHMS AND 0-07 OHM, ITESPEC-
R"ESISTANCES
Z30, THE SEI?IES
CALCULATE :
=(G25-4)(soa)/
/1000
= 31Y.t KW
COM FOUND (LONG SHUNT) GEI-.Jt:TrATOI?
IN Of'WE/l TO MAKE THE ~ENE!tATOJl FLAT- COMPOUND, SO THAT
IT WILL HAVE
508 ( 0.05 -t 0011 +0 00'1)
(b) THE COMMUTATINt; POLE
17-70.3 )<W
17-703 l<V= 17· 703 KW
VOLTAGE' OF 230. THE SHUNT
AND THE
AND SE171ES- FIELD
OF
Eg = Vt + IA (r&. +bC + f7.c -t-1le~)
E9=
POLE'
TIVCL Y-
BECAUSE
IHEN:
= ~00
F:l- r... ~ rT
~17-703 KV-~OOI<W=
-t-20] A 10- KW. 250 -VOLT
JL
IN ll"D
TO CHECK THIS·.
i«ltbc
03Ci )(0-012)
THE CUIT:fl.ENT FLOW
ll"'i
(508)(0-009)
IsE.- 381 AMF
llse
0-05G +0-0IZ
flp-tflsJO
rA
0-DI'l.fL
CALCULATE THE EQUIVALENT IZE'SISTANCE
.1-
p,... = 4800 WATT$
Vr= 4-572
Ise. = 4-512 V
!r"
Of Ttlf PAIZALLEL CONNECTION Of J7v
= (8)~ (75)
RsE
-1L
'508 AMP
llt'.<b-
PP ~ 581 WATTS
e) PSH =If • Rf
2
7S J'L
!Co r&e
=(1'27 )'' ( 0 -0"5&)
= 2859 IVATT.S
c) Fs~: = Ise Rse
BUT,· !so; = V,.
riGUI?E 18-1
+ 500
=
(581)"' (0-012)
f'sE -= 17"1-Z WATTS
d) Pp= Ip 2 (lp
FsE =
2
rA~Ir+h
rhj-
IP= -"().7]3r;
:£Q12J!.n. -= 127 AMf.
- (508) (0-011)
1J= 8 AMP-
I,..=
AT P.LIIZALLEL Cl[(CUIT OF l?o tj ILsE
b) Pc::: r.. 2 llA
lf= ~= 600V
= 8
POWEI'? LOSSES
FIELD
- (ooel(o.o3)
PA = 7742 WATT-S
500 AMP
llf
= SHUNT
Vr = VOLTAG"E
300,000
=
PsH
-I._
Eg = GENC/7ATED VOLTACE BY
THE AI<MATUf?E
P9= POWEIZ GENEflATED BY
THE AllMATUI?E
LET •
PI?OB. 4-19
CONT OF
CONT OF Ff<OB 4-18
31
DlflECT- CUIZrlENT GEN[IZATOrl CHAIT.ACT[f(ISTICS
Dlf?ECT- CUI"lJZENT
GENEflATOI?
CHAT7.ACTEI71STICS
,....~~-------------------------~~·cl:ll
4--21] A 3,000- KW
CONT Of
Prt05. 4-20
ARMATUfZE
lse.= Nis<
GIVEN:
f'L= 10 I<W
Vt ~ 250 VOL T5
Ise. =
V.,c= ??:f) VOLTS
NsE "' 6 '/• TU!li-/S
R:sE = 0·07 Jl..
EQUAL , TI-lE SEIZIES
-
b) TOTAL NUtvltmc OF AMP-llJfl.NS
LOAD ~
h=
If "'
LJ:T: ID= DIVEmEIL CUflfl.ENT
11>-GG
IL =LOAD
NI
NI
curm.ENT
If = 3HUIJT- fiELD CUflflENT
2?0 VOLTS TEflMINAL VOLTAGE
IL = PL = 10,000
Vt
2:'50
h= 40 AMP
w
v
IO,OOOW
= '1:3.f8 AMf.
~ = Z875 AtviP.
TliERE IN eACH
= Z300
Nir ==
IA '"" I-f + IL
r;7 X OF
h-=
UNDE/l EACH
=
CONDUCTO/lS PER. POLE f.ACE' ==
TOTAL ARMATUIZE CUflR.ENT
= AllMATUilE
CONDUCTORS PLR
=
r;ooo xc
EQUATING:
GOOOC"" 48000
c,
fBOOO
C=
a cowoucrorcs
GOOO
OF F'AIZALLEL PATHS
C = CONOUCfOJ"CS IN EACH POLE FACE
IT.:= 3 lXXJ, 000 W
500 VOLT
1#,000 AMP
FtELP
AT
rULL-LOAD
:: 3·125 +tO
IA = 43·125 AMP
3n
1-8,000
SJNC£ .
+ 2'2!ii
ZSZS AMP- TUTZNS
POLE={:'f'ZBH)(/1'2)
COIIIPENSATING WINDING AMPE/ZE-
f'i"" NO-
=
It
POLE fACE OF A
f'VLE
h
bOOOAMP = t28.57 AWl
11/PATH
= 111
II = AllMATUilE CUII.flENT fiR. PAW
lf'ATI-1
AT
=
Z;
= (Z3t0 CONO.) (O.G'f)
!I'OLE
50LUTfON:
IJsu x If
"'(800)( UJ75)
Nfr= IJIAT "'a.LD;I.P + Nf ~""'"'
rovr;:n:
COMF'ENSATJNG WINDING?
=
NI AT fULL-LOAD
3125 AMP
!PATH
AiZM.I\TUilE AMrE/lE CONDUCTOW'l
Zl
!/"OLE
NO-LOAD
--
I;.:
HOW MAIN CONOUCTOltS All.E
LET:
FlEX X POI.ES
t
"'(1 ) ( 11)
F,"' 14 FAIZALLEL F.LITHS
l7:EQUIIT.EI7:
AT NO LOAD
NI =- Z3Do AMP-TUI'lNS
If = Vt == :2 50 V
llf
80 _Q
f,
THf. ENTillE CITlCUMFEilr: NCE
80
IsE =SERIES -fiELD CU/lJlENT
lf=
1D
Z30V
SOlUTION:
AT
POt..E fACES
VOLTAGE AftE EGUAL TO ZW VOLTS
AT FULL -LOAD
KW
LAP-WOUND ARMATURE
b) If 1110-LOAD VOLTAGE ~ ruLL - LOAD
PIWDUCED 13Y EACH FDLE AT NO-
~,000
=Z310 CONDUCTO/CS
b
[lQ = 0·112 Otlt.1S
IZESISTANCE
Of A COMFTIJSATJNG WINDING?
P:. It POLE
llt> =Issfls£ = (2G.'f7)(0·07)
REQUIILEJ7:
a) ILo=DIVEilT.ETL
f'L =
lbf(Q =Is£ fl.sE
225 AMP- TURNS.
POLE FACE
Vt= 500 VOLT
a) SINCE YD =Yse
FIELD MUST
OF THE fNflllE Clff.CL'MFEflENCE . HOW MANY CONDIXTOilS Allf THEilE IN t-ACH
GIVEN:
Io = JG.GG AMP.
NO-LOAD ~ fUll-LOAD VOL TAt;c AilE
14-POLE GENETlATOrl HAS A LAP- WOUND
Z~.47 AMr.
=- 43.f2S -?IP-t7
so _n_
PfCODUCE
B·!:J~
IA = Io+!sE
IQ, lA -Is e.
NS~UIJT I'IEW ~ 800 TURNS
llf=
= 2?5 AMP_1JJR!IfS
N-se.
500-VOLT
2,3t0 CONDIXTOflS ··THE POLE FACeS COVEll G7 PElt CENT
WITH
33
.
5-1] A 250-VOLT SHU"IT MOTOIC
OHM. ASSUMING'
THE AflMATUflE
A 2- VOLT
CUflllniT IS 35-t AMP.
Lt:T Von.op= 15 THE VOLTAGE DllOP
IN THE ARMAllJflE ClflCUIT
T
0
VOLTS
..rt.
!(., 0-26
h.c "'· 2
YoRor = IA17A + V8 ...,..~ «>.rr.ocr
= (27.6:5)(0·~8) + 2
b
flA
VOLTS
~
IA"' 35-t Alv'IP
VoROr= 1?-5 VOLTS
17
flEQUIRED.
COUNTErC EMF?
fiG:
5-~ A 500-HP 600-VOLT COMPO!.}ND MOTOR: Of'EllATES AT A SPEED OF
495 RPM AT FULL-LOAD- If THE FLUX PEfC' POLE 15 91 X 10 6 MAXWELLS
AND THE AfC'MATU/lE T?E.S/STANCE 15 o.0/5 CALWLATE :(lfi) THE COUNTErl EMF;
(b) WE AllMATUilE CUftrlENT. (ASSUME A VALUE OF K~ 1·5 X 10- 7 AND A
5HUNT MOTO/l
SOLUTION:
LET Ec
=
13 THE COUNTE£'{
IMPRESSED
EMF
A/lMATU/ZE
OF A MOTOIT'
V~>c=
VA
AC!70SS
THE
TERMINALS
Ec., (~ -Vb.t.) --TArt,.
VOLTAGE ACflOSS
Afl/VIATU«E WINDING'
Ec= 218
AllMATUILE' CUI?J(ENT
8 VOLTS
Tf "'" SHUNT- FIELD CUI7.flENT
NOTE: THAT THIS COUNTE/l EMF CAN
NEVEI7 BE EQUAL TO NJD MUST
ALWAYS' BE
t:Mf IS
S=: 495 flPM AT fULI:-WAD
6u
~~
c
= :II X 106'
lZJ:
ll
MAXWELLS
/-3XI0-
E
7
fiG: COMFVUND MOTOll
Vs.c ~ 5 VOLTS
THEN:
Cl) Ec=l·3 XI0- 7 (9-IXI0{;)(4'35)
R!:'Q~D:
SHUNT
WHAT CUR/7.ENT
MOTOfl
HAS AN AJZMATUIZE RESISTANCE OF
WILL fLOW
227.5 VOLTS? (ASSUME
IN THE AllMATUIZJ: WHEN THE COUNTER
A 2-VOLT
BflUSH DflOP)
o) COUNTE/l t=MF Ec
1:>) AllMATUIZE CUrlflfNT IA
Ec~
b)
58!5.6 VOLTS
IA
= (v,.- Va-c )-l':c
RA
SiNCE Ec 15 A GENE/7.ATED VOLTA~,
"-L
=(G00-5)- 585 .c;
IT DePENDS UPUN TWO FACTORS
IZA=038.IL
t) FLUX PER POLE
Ec = 'C.?.7· 5 VOL To
Yb c = 2 VOLTS
Z) Sf'EED OF flOTATION (rz:PTVI)
lZEQ'D:
AfZMt\TUflC' CURRENT IA
E'c IS DIRfCTL Y
:JtiUNT MOTOIZ
7
I
B'Y OHMS LAW:
EC: K~S
WHER:E:
0·38
IA ~
IA rcA =(V.- \l?c) - Ec
K = JS THE fiWF'OilT/ONALJTY
?.7-1P3 AMP
J,. = (VA-Vb.c)-[c
CONSTANT
itA
·~~N-.'f
.<4
ii£W
-~
iii%1iU:t'S<
~
OUT IF Ec= 5'85 VOLTS 01-JL Y
FIWFD/ITIONAL TO
THE SPEED /"',r-FLUX f'EIZ POL[
IA = (240 -2) - 2Z7- 5
50WTION:
0·01'5
[A= GZ7 AMP·
IA = (GOO-I>) -585
THEflEFOIZE:
----<>-
OR !585 VOLTS
Ec= 5!75-G VOLTS
IF WE USE
SOLUTION:
V,. = 210 VOLT5
I
VA = 600 VOLT5
K-=
GIVEN:
l
D
c
fL= !'500 HP
Yft?LE:
IZA = 0·015 OHMS
LESS THAN , l}tc VOLTAGE
5'-2] A 2'1-0- VOLT
038 OriM
'----~+
GIVt:N:
E'c= ( 230-2 )- (55 ·t)(o. 2G)
i<A= AflMATUilE . flE515TANCE
DIWP OF 5 VOLTS.)
BflUSH
31NCE:
5RL.'5H CONTACT LOSSES
= IMF«ESSED
IA ""
ClflCUIT?
S'OLUTIOIIf:
·----ot-
~ 2~0
MOTor( CHArrACTE!liST/CS
5-3] IN PR:OB 2, WHAT IS THE TOTAL VOLTAGE DR:OP IN THE Af?MATUflE
HAS AN ARMATUilE ITESISTANCE OF 0.26
13/lUSH DfWP, CALCULATE THE COUNTEIT eMF WHEN
GIVEN.
VA
DlllECT- CUll RENT
MOTOR: CHARACTErliSTICS
DIRECT- CURRENT
0-015
IA= GG7 AMP
NOTf:
ANY
(JF
THE TWO ANSWE/7.5
IS CORILJ:Cf.
DII?ECT -CURI?ENT
MOTOR CHAI?ACTEltiSTICS
DlllECT -CUI?flENT
5-5] IF TiiE LOAD ON THE MOTOil OF Pfl.OB.t IS INCREASED SO 1HAT IHE
AI'U/lATUfZE Cl}flllE/1/T
!liSE'S TO
800 AMf. , AT
WHAT SPEED WILL THE MOTOR
CONT OF PfWB 5" -7
GIVEN:
INCI'lEASES ~ f"E«. CENT?
OPEI'l.ATE, ASSUMING THAT THE FLUX
PL= 10 HP
Vv·c + J,.llA ""Vr Ec ; 15UT E c ~ K~S
IA= 800 AMP·
KIP
SOLUTION:
(9.q113 XIO~) ( 1·3 X itf- )
7
Cl) CALC. THE COUNTER.. EMF
NOW
FDLE
b) THE F'OWER.. VEVElOfED f5Y
IS GIVEN .AS:
b) PM=Ec. x1;.
=-(52:?) (J<! -B)
THE MOTOFL IN WATTS~ Hf'.
!Jl-(9·1X 10~)(1·09)
p=
Ec = '522 VOLT.S
OEVEWF'EO BY THE MOTOR:
9%, THE TOTAL FLUX PER
Ptvt=
SOLUTION:
'31119 X IO~ MAXW,elLS.
fl,.,.= 772(; WAiTS X ~le7HP~=
THE MOTOfl
LOAD IN PIWB·4 IS flEMOVEO, THE SFEED 17.15E5 TO
560 flPtvl . WHAT IS THf ILEGULATION OF THE MOTOil?
7tG W.llTTS
f'M"" POWm DEVELOPED
THE
7726 WATTS
IN TffUvlS OF tiOICSE POWE/1.
LIT Ec = IS THE COUNTE/l EMF OF
5-~ WHEN THE
u
~
SHUNT MOTOIL
R:EQ'D:
S= 452 RPM
IS INCil-E-A SED
s
0
ri) (VA- Vru)- Ec~lAflA
Ec =-(VA- V8 .c) -lARA
t:c= (550-5) -(it.s)(J·55)
= 14 8 AMr.
Vee= 5 VOLTS
I.IJOTOIZ
51NCE TtfE FLUX
tc
+
p
E
IA
= (600-5)- (800)(0·015)
LEI S = IS THE SPEED OF THE
Ir.- 0I,-
S"' 1750 Jtf'M
R"' = 1·5'5 ..£L
5'= VA -IA!Uo -Vac
Sf'EED Or THE tviOTOfl ?
ITf
VA= 550 VOLTS
Tt-IEN SOLVEV FOIL Sf'EED :
rtEO'V:
10 .#f' X 0· 746 KW
.IW
P1-= 7·fC> KW
IAf'lA =\t.\-kiPS-Vsc.
rUJX INCIIf:AS'ES g J.
BY
P~,.
OR:
TI/1)5:
GIVEN:
MOTOR' CHAitACTEI71STICS
~y
f"1<1= IO·eb HP
MOTOfC
GIVt:N:
SI'I:ED AT FULL-lOAD "'4~5- trPM
S: AT NO LOAD -= 5'60 llPM
fCEQ'o:
'%
INCLUDING
Jl.t::(;ULATION Cff THE MOTOI7..
:tIt==
IS OPER:ATING UNDET? LOAD, THE AflMATUIZE TAKES
=
S'F·L
X
560-495
= 15 -13
IA
AM-P
Ec-= e.-so -510 (o.-t)
fc =
!"lEGb
THE
HAS AN AflMATUIZE
FULL LOAD : CALCULATE: (o) THE COUNTER
DEVELOPED
lf-8 AMF'· AT.
EMF PEVELOf'El7 I5Y THE
BY ltfE MOTOR'
t-IOR:SE FVWEJZ ·(ASSUME A S VOLT 1317..U5H DROP)
~
MOTOft?
F., "' fciA
fA "'V,., fA
VA=~
P.,= :o·t
=
8,280 W = Z'50 VOLTS
:56 AMP
~.·(::;.
x....,+J,_,_P_~
7-lG WATTS
74-G
IN WATTS,AND
VA
21S · G VOLTS , COUNTER EMf
== (ZI!HO) ( ~)
SOWTION:
StNC.E ·.
BllU5H
CONTACT VOLTAGE DfWP
0·"1' ..!l
HOIZSE FVWE!<. DEVELOPEP
.1.
fOWER
Ec- V.<~-IA (I?,., -t-a) INCLLIDING TTIE
)\ 100
OF 1-55 OHMS · IF THE AIUIIATUIZE TAI<ES
MOTOR i (b) THE
~36
IL.-~+a=
10-HP 1,750-flPM 550-VOLT SHUNT MO'lre
JtESISTANCE
IS DEVELOPJ:D BY
THEN:
PA· s2eo WATTS
100
t9!i
bl· fl..
f31WSHES, IS O·'T OHM, WHAT HOflSEPOWEK
GIVEN'·
J'H·L -
SF· I-
IN
A MOTOfC
THE MOTOrz?
50LUTION:
5-7] .A
5'-13] WHEN
8,280 WATTS AND ITS CLII{flENT IS J6 AMr. IF THE Afl'MATUILE -cm:CUIT flESISTAIJCf
--~·:-;..
f.W
~
DlflECT- CUfli?EfJT
5-~ A 5-HP
AT
?30-VOLT SHUNT MOTOIZ TAKES
FULL- LOAD. THE SHUNT- FIELD
AflMATUITE
RESISTANCE
5'-10] IN PI?OB 9
18 AMP WHEN OPEflATING
l?fSISTANCf
STAIZTE/t
IS liS OHMS AND THE
HP
1
-
SHUNT MOTOR'
VA= ?'50 VOLTS
IA = 18 AMP AT FULL LOAD
I'Cf= 115 OHMS
flA= 0-?5 OHMS
Ec~
A VALUE
71 VOLTS ,
ASSUMING THAT THIS IS
RATED VALUE.
1·5 TIMES ITS
l<a= IS THE' VALUE
LET
OF STAflTE{( 17ESISTOR
THE VALUE
OUT
OF CUT
IC.ESJSTOR
"r(' IS;
TO
I<: e. =
IT~
DC. SOURCe
Rf
1
IS 74 VOLTS AND THAT THE Afl/v1ATUIZE
EMF
JUMPS TO
IMMEDIATELY
WHEN
~--------.---------~+
IL
GIVEN·
fL.= 5
SECOND STUD
WHICH IS CUT OUT OF THE
SOLUriON:
THE llATED VALUE OF THE 3/A/ZTING INSTANT. (ASSUME A 5-VOLTS BRUSH
CONTACf).
TO THE
IN MINING
CUflflfNT
CUIZIZENT IS LIMITED TO 15 TIMES
AllMATUIZE
MOTDrr CHAllACTElliSTICS
CALCULATE THE ((ESISTANCE
WHEIJ THE COUt-JTER
DONE
CALCULATE THE VALUE OF THE
IS 0 25 OHMS
5TAIZTE!l llESISTOIZ IF THE
DlfZECT- CUI? RENT
MOTOI? CHAf?ACTITISTICS
_~
(v,- Ve-<-) - E'c
~
(?30-3) - 7t
-
flA + Ee-=7+ YoLO
fls"' 9 -08- 6'-04
1..
rc,
I( A +e,...o -
o ?5
fls= 3-04 otiMS
Z1-52
VB-C.= 3 VOLTS
rz~
fiGUI'(E
G. 04-
OHMS
llEtJ'D:
CI\LCULATE THE VALUE OF STARTING
, !.Hl] WHAT WOULD
IS LIMITED TO I 5 TIME5 THE flATED
VALU[ AT THE STAJ(TJNG INSTANT
"'(1·5) (IG-2'17)
SOLUTION:
LET IZ= IS THE VALUE OF STA{[TEfl
=
2'f. 32 AMP.
l4 ~ '1,;- Vac.
1HEN:
rlESISTOI? CONNECTED IN 5EfliES
ICA+ll= (VA-Vec)- y{
rr-=-(<'30-:5) -o
THtN
-
Tc,
rz..
Tl= 9-08
IA (IZA +ll) = (VA -Ve.c)- Ec
AT THE
~
9 I SL
5'-1~
I
THf VALUE OF STAilTEIZ
BUT COUNTEIT EMF" Ec" 15 ZrflO
ltEJISTANCE
151JT'
Ir-= 5 W X 7'1-G WADS
)W
f'- = 60
VA
AND AN 'AllMATURE
r<A= o o+
RsrA<~rEI'- -= . 0 ·GG _o _
V8 c
I
38
Jr THE
IT IS STAJ?:TED- (ASSVME A 3-VOLTS BIWSH
Tf-
~
Jc-17s
IC'f
h~
TO
:IOIJllCE
_n__
~
~-
f(E515TANCE OF 0-01 OHM
VOLTS
~
·r"
flE515TANCE OF
!N THE STATnEI? IS 0-GG OHM, CALCULATE THE CUilf(ENT JNPUT
llf= '5El 3 ...Q.
lb-ZI7 AMP.
HAS A SHUNT-FIELD
HP _: St!UNT fv10fOR
VA-~ 2'50
X 7'TG
Z50
Ir=
RESISTOR:
230-.3
TO Tf!E MOTOfl AT THE INSTANT
DROP)
GIVEN.
IA .=:- f.5 Ir-
lr"" !:J
~
A 60-HP 230-VOLT SHUNT MOTOR
OF ·:;e.:s OHMS
ILESISTOfl.
INSTANT OF STAIZTING
WITHOUT STAITIJNG
IA= 908 AMP
Jl"' 9."53- 0-25'
THUS:
STARTING; If
DlflECTL Y TO 1HE WJE
O·Z5'
Z'f33
CUflflENT.
CUfl((ENT UPUN
rw
0
r..
TO THe AflMATUI?E
f r- = IS TJ-IE rrATED AlrMATUflE
AllMATUR'E
MOTOI< l.'v f''FWf3. 9 WEfCE CONNECTED
i
!i WITHOUT 1-l STARTING fl.ESISTO{[?
SOLUTION-
1A=I5'Ir
IA
BE TtiE
I THE
ILESJSTOJ( IF THE A/lMATU!lE CU{[flJ:NT
3' 'v'Ol.T:'i'
--o
fiGURE
_ _ _ _ _ _ _ _ _g _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
39
__J
DIR'ECT- CUfi'fi'ENT
MOTOR' CHAilACTJ:IZISTIC.S
CONT OF PR'0/5. 5-12
l?EG'o:
O,STA{l.T
(V4 -Va.c)-y{ = IA (fZAtR.s)
CALCULATE THE CUIUZENT INPUT
IT IS
5-j4] WHAT H0/7SEPOWEfZ
AND:
TO THE tv'IOTOtt AT THE INSTANT
DIRECT -CU!rflENf MOTOfl CHAflACTErliSTICS
TA "' VA- Ya·c.
-rz..
Rs "'STAil:TE/Z flESISTANCE
IL = IS THE CUIZilENT INPUT
tfr : :Z11" >< fU'M
+ fts
IA==~
o.ot +O·GG
HP =
IA
HP = 19 HOrtSE POWER
zrr
~24.3
=
AMP.
THUS:
!L = IA +1-r-
A ?20-VOLT SHUNT IVIOTOI'( l-IAS AN AflMATUilE trESI5TANCE OF
AND A AElD liES/STANCE OF 110 OHMS. AT NO LOAD
THE AllMATUllE CURRENT IS G AMP AND THE SF'EED 1,800 IZI'M. ASSUME
5-15]
h-=- 321· 3 + G
lf= VA == Z30 v
JL.=: ~30·3
~
J4:
0·32 OHM
AMr:
THAT THE
It= G AMP.
FLUX
5-13] A FOUR- POLe
SHUNT MOTOR'
THE
FLUX
TORQUE
DENSITY
WITH ~ TOTAL OF
HAS AN ARMATIJ/ZE
POLE FACE HAS AN AIZEA OF Z5 Sa IN
AND
IS 50,000 LINES PJ::fl SGUAilE INCH, CALCULATE THE
IN POUND- FEET FOil AN AI!MATUITE CUlt/lENT OF 85 AMP. THE
AR-MATUR:E
GIVEN:
13UT
P=4 P'DLE
K -( ~~:3
Z ~ '282 CONDUCTOIZS
POLE fACf AREA= Z5 SQ. IN
nux DENSITY"
IA = 85 AMP.
4
5 = '50,000 UNESL
/SQ./N
)( ~
P Z) X {p X lA
WHffZE:
NO· Of f'OLES
l = NO· Of CONDUCTORS
Q=
Cl = 2
15 TtiE TO!UJUE
T= KiP IA -l15-FT
P'LEX X2 -
fOR lvAVE WIN CliNG
= 1 X 2.
SOLUTION'
F'Afl:ALLEL F'ATHS
1 [ x:
T =[o ~:3] 4
GIVEN:
VA= 220 VOLT, SHUNT IVIOTOT?
If= SHUNT- FIELD CU!tf?ENT
h IS G2 AMP.
SFL.-= SF'EED WtiEN
THEN.:
1~
IA=
=
I'Zf
'220 A
110 ..()_
lf = Z AMP.
G AMP
I~.-=
1L = IA +If
62 AMF'·
IA~G2-2
82
]
(so,oooxzs) (e5)
ltfQUIRED:
TA= GO AMP.
a) SPEED OF THE MOTOIZ WHEN
SINCE'·
THE IZATED LINe curm.ENT
S
IS G2 AMP.
sN.L·
MOTOIZ.
SOLUTION:
LEY:
=(VA -Vs.c) -(fA )(rt.Ay~ K
AT NO LOAD:
b) Sf'EED rlEGUtATION OF- ntE
=(220-2)-(G)(0·32x~ -- -.eqy(j)
AT J=ULL LOAP:
SF t. = (ZZO- 2)- (60)(0 32)/ -- -~.@
lo·I'·L.= IS THE ARMATURE
CU!r.riD.IT AT f=ULL LOAD
It.
= IS THE llATED LINE
CUR/lENJ.
,o1()
=- ':!!___
.LIT NO LOAD:
T= 70·3 LB-fT
--+V
flEGULATION Of=' THE
Sw.L."' 1800 [(PM
~ =·13 xA -fLUX/rou:
f'=
ttEG'D:
CALCULATE lVr<:GUE IN LI3-FT.
T=
Blr.USH DIWF') (b) THE SPEED
MOTOfl·
0 =NO OF FA!'rALLEL FATHS
ARMATUJZE WINDING-= WAVE
LET:
(ASSUME A 2-VOLT
LINE CUfUl.ENT IS G2 AMP
WHEIV THE IZATED
!!.A c 0· '52 ..n..
llf= 110 ..Q
Ve-e.-= 2 VOLTS
WiNDING IS WAVf.
WITH LOAD AND CALCULATE : (a) THE
DOES NOT VAftY
SPEED OF THE MOTOR'
?.BZ CONDUCTOrl5 · IF EACH
>< L"'"ZO X 70 3
'33,000
INSTANT IT IS STAIZTED
BUT:
X T
?f,,OOO
TO THE MOTOR , AT THE
SINCE:
DEVELOPJ:D 6Y THE MOTDI'C OF
1,420 IZP/'11 7
THE HOilSE POWEI? DEVELOPED 6Y WE MOTOR: OF
P/W13. # 13 fOR A 5f'EED OF 1420 ['{.P/'11 15 EQUAL TO;
STARTED.
SOLUTION:
OF
SOLUTION:
TH EIIEfOI?:E:
LET:
WILL BE
fOrt A ~FEED
F'IW/5. 13
41
/K~
DIRECT-CU1li7ENT MOTOR CHARACTETZISTICS
CONT OF PIZOB. 5-15
DIVIDING Sf.L. BY
a)
CONT OF PTZOB
51-l.L.
= 218-19-Zj~
Sf.L.
1800
b) Sf'fEV REGULATION
= S1-n -SF·'- x 1oo
Z+B- I 92/-i'W
::: IBOO- IG5G
216.08
=
Sn.
x 100 J(
550- VOLT
50-Hf'
INCLUDING
BIWStfES, OF
SHUNT MOTOR
IZESISTANCE StlOULD 13E
INSEI!TED
IN THf Af'lMATUilE Ct/1.CUIT TO OBTAIN A 20 f"EI'1. CENT SPEW
ITEDUCTIOI\1
WHEN THE
TORGUE.';; ASSUME THAT THETU:
IS OEVl:LOPJNG 70
A 240-VOLT
SE!l/ES MOTOR HAS AN AI7.1VJATUTtE RESISTANCE OF 0.-t?
OHM AND A SEflii:S -fiELD
R'ESJSTANCE Of 0·18 OHM JF THE SPEt:D IS 500 ICF'M
1HE LOAD flEDUCtS THE LINE CUIZ/Zl:NT TO 21 AMP? (ASSUME A :5-VOLT
VA
~
~
mn....._____...+
2'10 VOLTS
fZ.p 0·12 _a_
IL __.,_
T
0
I,=
t'4-
WHAT RESISTANCE ~,(~ :JHOULD
BE
Vs.c
It
c
E
Sf'EED /Z.EDUCTION
\"'\GUICE
AT IZATEO
IAR..I
=VA
LET S2
OF IZAT!:O TOilQUE
THUS
Ec == VA- TAI?A
TOflQUE
55"0- (75)(0.3<0)
51NCE, COUNTEil EMF IS D/ICfCTLY
THErz:E IS A TLEDUCT/ON OF 20
SPEfV
; THIS
WILL
THEJZEfOr?.E;
CUrm.ENr IN THE AIZMATUTi::E
J.
15
Tfif COUI-JTEfl_ fMf 15
42
REDUCED TO
THE
70 7•
"""3G
TO ?/ AMP'·
X 100;.
ONLY WHfN
lZ'f0-3) -3G (O·'h'+O·IB)--
K<l>
13'(
OF THE MOTO/l.
,•;
=
UI\II<NOWN Sf'EED
SINCE·
2'1
=
K4J
r<if!
S,
017.
!)E. 3 3 /•
IGATIO/ FROFOI?-TION
S2 _ 237-IH/0·4/Gl K<t"
500- 237 -'ZI·G
/K¥
S:z
6/
= soo[214·4
rLUX IS ALSO R:EDVCES BY 11 G7 i'o
THUS1
S, = (YA-VBc)- Iz (IZA+Rs)
Kcp
43
l
'ZIS" ·1j0·41!Pt j
::k:o 1194- llPivl
AND THE fWX IS Pf?OPORTIONAL
TO THE CUTtrl.ENT
T = 0·7 KP1A
P/[()f'Of{T/ONAL TO THE SPEED I Wltt:N
= IS
3'G AMP
= KiPIA
BECOMe
Ec-=- S 23 VOl T:3
0·<J-JG7
___ (!)
S,=(VA-Ysc)- I, (IZA+Rr)
THE curmENT IS Rt:DUCES Fl'lOM
SINCf..:
=
S2=<(Z10-3)-21(04Z+0·18)
I
THE TOit~UE OEVELOPE!7 15 ONLY 707.
-Ec
SE.'RIES MOTOR
SOLUTION.·
Sf"ECD IS flEDUCEO TO 80 7. FIWIIl 100 ~
LOAD j
'5 VOLTS
TO THE LINE CUTl/I.ENT TO 21 AMF" 7
IF THE TOilQUE DEVELOPED WHEN THE
WHEN THE MOTOII IS OPEI?ATED
~
5f'EE.D WHEN THE LOAD IZEDUCES
t:c ..,rgt.;)(o.,;)
Eo= 419-f VO!.TS
70/. Of RATED TOIWUE 7
SOLUTION:
~
WHAT WILL BE THE MOTOTr.
MOTO/t IS DEVELOF"ING
WHEN THE
3G AMP.
ITEdo:
INSEfiTED IN THE AITMATU/lE C//l.CUIT
TO OBTAIN 20 (.
s
u/'(
0
h= 2'1 AMP
u
RA
0
fl.~
:3, = 500 flPM
s
0
T
tL
f?s = 0 ·18 .J2..
-t
50 HP; SHUNT MOTOI?
550 VOLTS
rn:o'o
iN
i'5'
GIVEN:
PErt CEIJT OF rt4TED
IS NO FlUX CHANGE.
fu +e = 0·3G .fL
IA ~ 75 AMP
iNSEilTE D IN
THE AI'1.1Y\ATU/Z.E CITZCUIT
13/W.JH D/WP AND THAT THE FLUX IS PfWf'OIZTIONAL TO THE CUfUl..ENT)
GIVEN:
V,. =
TZESISTO~
(08).
WHEN THE CUfliZENT iS 30 AMP, WHAT WILL BE THE MOTO/l SPEED WHEN
0·3G OHM. Wl-ltN Of"E'fZATIN~ AT RATED LOAD AND
MOTOI?
THE VALUE Or
_Q ,
550-4J8.L(
o. "3G +fl.=
HAS AN Ail MATURE RESISTANCE,
75 AMP· WHAT
-0·3G
2-lt
=
(15)(0.7)
S·R·=B·71.
5-1~
SFEED, THE AftMATUfZE TAKES
PL~
rz.
f?At/l""
ltATED CUitiTE'NT
5-1~ A
/7.=. 2·S
TO THE ARMATUftE.
16!%
165G f'{f'M; SP!:ED AT
5-110
THE 17.fSISTO/t IS INSERTED
THUS:
IA (IZA+IZ) =VA- f.:
SF·L·
Sf.L. = (1800)(198·6)
lv10TOIZ CHAI?ACTEfl/STICS
DlllECT- CU/li?ENT
-@
MOTOR'
DIRECT- CL'flR'ENT
5-18] A ?0-HP 220-VOLT
SHUNT MOTOrt HAS
540-flPM
DIR'ECT -CUrrRENT
CHARACTEIZISTICS
AN AllMATlJilE
STAilTEI?
fCESISTANCE
IS
1·7 OHMS/ WHAT CUilflENT
MOTO{( TAKE AT THE 'INSTANT OF STAilTJNG
7
n:Ea'b:
TORGUE AT THE SECOND VALUE OF CUrt:/lENT "r2#
DOES THE
(ASSUME A 2-VOLT
SOLUTION.
BRUSH
T-z == ~ X <P"
T,
I, · ¢;-
OIWP)
-t
GIVEN:
F~=
eo
JL- T
HP, 5HUNT MOTOR
VA= 220 VOLT
llf
5 = 5'40 I"CPM
liA
0
Tz = 164
6v
Tz
ll
fl'sT=' 1·7 ..(l_
c
E
Yt>.c- 2 VOLT
I'Z4 = 0·24SL
FJGurzE:
TAKE AT THE INSTANT OF STAilT/NG
TUI'ZE CU/'Ul.ENT
IA = VA-Vac. -0
WHAT CU/ZilENT ODES THE MOTOI(
7
0·24
88
+ 1·7
IA
AT THE INSTANT OF STAIZnNG.
fc == COUNTER ctv'IF
=:
h= IA -rTf
TtJus·.
VA =
1f = i·i
IS
AT FULL L0,40
Al'-10
Of THE
flECULATION
2
tvf[TfOfZ.
1900= !i2b·1T __ -(})
AT
MOTO/l
~ !=LU)< IS D/lOP TO 88 ;1.
5N·L·=(VA -Va.c.,L.)- IA (flA+I<o)
- 5' VOLTS AT FULL LOAD
O·BB
rtEQb.
o-j WHAT
llEVELOPS
IGt LB-FT
INCI(EASES
OF TOfl.GUE WHEN THE CUI'tll.EIJT
THE CUIUl.ENT
50 fE/1.. CENT WITH A
Of ZO FEI'C CENT , CAlCULATE THE TORQUE AT THE SECOHO
Rs
CUtlRE{'JT
I
~
•+
INGrl.EASt=
VALUe OF
GIVEN:
T
0
SE:rCIES MOTOR
-r.· ~
/G"'- LI3-FT
I·='
9+ AMP.
r .. ~
iPz =
~
v
n.
t,
1·'2 P1
1·5'
c
E
FIG:
.l/·1
SEI'LIES MOTOR
NO LOAD
SINCE CUrc.rc.r:NT Dfl.OF TO 5 AMP
SJ'eeD WILL l1JE. MOTV/7..
0·88
s....... ,
'5 AMP- ,4ND DTWI' IN fLUX TO
EQUATf'
eg% OF THE AJLL LO,AD VALUE.
1.) CALCULATE THe
1-
s..,.L
S PEEP
1900
REGULA TIO'/J OF n-Jc MOTOR..
=
!=a>. (]) /t:a@
!;'fr;.t/o.ee tx.f
)"P
!;'23 . 7~
(o .ea)(~;23 71)
o.) SFL = (VA- Va< ) - I• (IV.+ n.r)
Kef
(!;'5'0-G) -
----@
s .. ,.= (Jsoo Y.>4!;.1)
SOLUTION:
=
54-§.1
Kt'l
0·88 K4>
ARMATURE CUR.IU:NT Dl'l"P TO
5t<L
pK
5>JL.=(S5'0-·Z) -5(0·9GIO)
OJ>EIZATE AT NO LOAD IF Tfti::
A SEITIE5
~
VJlc. =- 2 VOLTS .AT NO LOAD
AMI'
A BTWst-1 DTWP OF
VOLTS AT /'10 LOAD) (b) CALCULATE l"HE: FE/1..CEN
5F·L."' /,900 rtPtvf
Y~c
IF THE AfLMA --
OI'WP-5 TO 3 AMI" WITH A COfffiESF'ONDING Dfl.OP IN FLUX 10
R.s"' 0 -I;- .fl.
LAW.
NO LOAD
Of THE FuLL-LOAD VALUe 7 (ASSUME
IA= 22 AMr.
IL = l/3 -77 AMP.
1G7
94' AMP. IF THE LOAD
FL.UX
Ttlf FIGUI'l.t:
2'20V
IZf
-2D
fflOM
fU"SIS-
RPM WHEN THE AIZM4TUrtE CUTLfLE.NT IS 22 AMP
1,900
WILL THE MOTOrt OPERATE AT
11.4= O·BI!> .12..
~y CUIUZ.ENT k'IRCHOFfS
HAS AN AflMATUIZE
AND A St:RIES'- f=IELD 17.ESISTANCE /JF 0·15 OHM . THE
IS
VA= 550 VOLT ; LONG SHUNT MOTOR
h. = 11?.37 + I·+
If~
LB'-FT
GIVEN:
111· 37
TliEW
TA (fLo t- flsT) = (VA- Ve.c.) - fc
BUT Ec = 0 AT THE INSTANT OF STARTING
5
5f"EW
1·9t
h"' IS THE J-ilOWI!: CUfl.fl.ENT TAKE
FEI'l CENT
10 VOLTS
11.'"" (220- 2)-0
SOLUTION:
LIT
OF OB/r; OHM
TANCE
[.J·b"~9'fJ [ I·ZJh]
-;p;-
LONG- SHUIJT COMPOUNJJ MOTOR
(o.) AT WHAT Sf'EED
AND
fli::QUIIZW:
= 295
5'-Z2] A 550-VOLT
FULL LOAO SPEED
llf = 1!57 .fl..
CHAf'(ACTE!ZISTICS
CONT- OF Pll013. fi -21
AND A FIELD llE31STANCE OF 0·24 OHM AND 1'57 OHMS, IZESPECTIVELY.
IF THE
MOTOr?
S'•J.L
22 (OBIS +O·Ir;)
K~
45
= :22'17 J(f"M
AT NO LOAD
DlrlECT- CUrlilENT
CONT. OF f1W6.
DIRECT -CUIUlENT
5-22
1- 5 R-
b)
MOTOTl. CHAI'lACTI:Tl/STIC.S
CONT- OF Plt05·
:::: Vt-~.L - VF-~
CHARACTERISTICS
5-23
HENCE:
X 100
VF.L
=
MOTOfl
HP"' 2rt(rtPM)(T)
Z247- 1900 )( 100
:: 21t (1070)(9~·2)
33.000
1900
53000
+JP = 18-78 HP , OUTPUT
Of=
THE MOTOr{
1,5-fl·= IIPiJ.'
5-24] A 550-VOLT SE/ZIES MOTOfZ TAKES 111 AMP AND OPERATES AT
A IS -HP 440 VOLT COMPOUND
!)-23]
OF Z9;) OHMS
WHEN ITS SPEe:D
A SfiUNT -FIELD r!:ESISTANCf
HAS
'28·'3 Al.tlP AT rcATEV LOAD y.tHE'N 11 OPEil.ATES
AND TAKES
AT 1,150 RPM· WHAT WILL
MOTOrt
3E THE HOrcSEPOWET{ OUTPUT OF THE MOTOit
IS INCIZE.ASED
13Y
+
TO
f4
I L,~ '28 ·5 AMP.
s,9.r
= II !00 ltPM
L .• "'
31. !> AMf'·
4'
= 1,070
DC SOUR.CE
RPM
K<H.. = tl-r
HORSE POWER OUTPUT?
.Ai
{tf"IV1 = JIS'O
K~ =
SOLUTION:
==
JA
= 28·5 - H>
I!.-=
28 5 AMf'.
IL -If
AT h= :st·> AMP ~ 5·1070 ttPM
1 A.= IL -Tt
= I.t. =
5 =820
r.. =
'2.7 AMF
~-
s
liZ AMr.
0
v
RPA-1
R
c
O·ti;..Q.
E
1~..
DROrs TO 84 AMr.
~
IS I'ZEDUCEO BY IS 1.
SEitJt:S MDTOIL
:=
31-S -I·S
IS' GIVEN ,AS:
s
HF': 2Tr X ll:F't.-1 X T
THEf.J:
AT !A e
3~
SLJT:
5V6ST.
T==- Kqi IA
AMP-
K<P = 2·!>21-
~
RPM
I(~
5)C::.
T == K~ IA
~UT:
!=ROM U>-1
T=- (Z-':;4) (5.3)(1 1)
TH Efll: FO tu: :
ttP=
I= 92-20 t.B-FT
l
l<" (0·1!7)
h
=i?'t AMP.
!P
= O·lc; ~.
____ e~
·If!
sues T. E<? (!) _.. @)
zrr
[-'.17·4 l
[o-1s- ~J
[k..f r..J
33,000
tw= 2TT ( r~7·"1) (e-t)
KiJ.j =1·3 ----eq.(i)
S;<
f<.A
33000
11T (RPM)
AT
X'
(O·It;;)
.!."37·4
tff == Z1T S.x I
T = tJP (53000)
21f ( tJ20 ){11'2)
I-= Kef r .. (1·1)
3'3,000
ezo
=
[IA
SJC= sso-@t)(o-tc;)
SOLUTION:
l(~:: (7!> J-JP)('53000)
~ = FLUX WILL INCI!EASE I5Y 101.
~c.=- V,.. -
AND:
AT Io..= 112 AMP
IA = 33 LIMP.
IA = Z7 Al\1f'.
TORQUE' AT
rzs
= 5!)0 VOLT
t+P) ( 3.3 ,ooo)
K~ = ~·54 - - - - - er~~ (j)
I·S AMP.
IA
(n;
(ZIT") (11!:>0) (27)
== 'TfO v
-v..
-293Rf
ARMATURE CUI'Ut.ENT AT
IS :
(33,000)
2Tr (RPM)
If=
115 f'Ert CENT·
HP OUTPUT OF THE MOTOR?
THEir.EFO/lE :
= 10/. INCI'!:EASEV
=:
TlEOUCE.D BY
~EQ'o:
REG(D:
lf
VA
1L
llf = 293 SL
IS
GIVEN:
GIVEN:
VA= 440 VOLT
IS 0·1, OHM, CAlCULATE THE HORSEPOWER OUTPUT
WHEN THE CURRENT DROPS TO 84 AMr , ASSUMING
THAT THE FLUX
10 PErt CENT 1
PL= IS HP
Sj~ ...-
WHEN THE LOAD 15 75 HP· IF THE EFFECTIVE ARMATURE·-
IT.ESISTANCE
OF Tl1E MOTOft
IS 1,070 RPM, UNDER WHICH CONDITION IT T.AI<ES
3'1-·5 ANO AND THE A.UX
820 RPM
C.IrtCUIT
(o.t{;) (~3,coo)
t-IP.:: JiT. 3 tlf'
=7
SINCE·.
Ec== 1<!)15"
J
.If.'
't
-~
41
DII?ECT- CUfUlENT
Dii7ECT- CUfvt.ENT MOTOIZ CHARACTERISTICS
5"-25]
A
50-HP 230-VOLT SHUNT MOTOII HAS
CONT OF f'f7.06. 5-25
A FIELD IC:ESISTANCE
IN THE ClfZCUIT
AT 1/550 RPM· TO INCR.EASE THE SPEED Of THE MOTOit TO
If=~
I,GOO ICPM, A rtESISTANCE" OF 5."5 OHMS 15 ''CUT IN" THE FIELD
ltHEOSTAT i THE L.JNE CU/lJtEI>IT THEN
It-=
THE
FIELD
FOrt THE
AND THE FIELD fWE05fAf FOR THE
c) THE PElt CENT LOSSeS
AT 1,600- ltPM SPEaD.
I,GOO--ILPM SFEED;
FR
r~..~ ~o
~
llf .= 17.7 ..12.
IL,
REDUCE
Or
L:/50 ltPM
?..VfiEl.D IIHEOSTAT AT 1600 ll.f'M.
c)~ LOSSES IN THE. FIElD /<Y
Pt :::: "2989 WATTS
11 = ZO AMP.
rM
R.-=ll...JL
=-
vA rL
=
(:2~0) ( 181)
flEC.(V:
FM = 41,10.50 WATTS
POWER LOSS
j.
RHEOSTAT.
=
~ X 100
45,700
U.l
1.
,= 2,98']
PM= POWEll LOSSES IN THE MOTOR
POWEll LOSS AT FIELD
j{ LOSSeS =-
flJit:OSTAT
5
trA ~
lGt
c
E
IN THE
--o-
Fw= I,.,"" IT
= (20) 2 (1·2)
f'R = 480 WATTS
X 100
7· Z 1.
IS NOT DIVEIZTED
IN THE CIRCUIT
230
Rf
17-7
It= 11.99 4Mf.
I
A .,
:"6
IF THE
r--------r----o+
50LUTION:
41,C30
LOSS AT SHUNT FIELD
LOSS IN THIS fCESISTO!?
liES I STOlt. ''fl."
POWEFL LOSS IN TtiE FIELD IS
F"'~
LEr
OF PIWB. 22 TO 1,200 flPM,
TZ
j. LOSS=!:!_ x 100 j{
30LUTION:
If = :Y_t:._ ..,
j.
A R'ESISTANCE OF 1·2 OHMS IN THE AflMA-
GIVEN:
= (Z30)( 12 ·99)
b) f'OWCIC. LOSSES IN THE f!EL17
a) FIELD ILHEOSTAT
INSEf7.T
TUflE ClflCUIT. CALCULATE THE POWER
ITS PER:CENTAGE OF THE TOTAL POWER
PRH>!O==
100
CUflflENT IS 20 AMP.
Pt=- VA It
= I"OWfll
THE SPEED OF THE MOTOT?
IT IS NECESSAIIY TO
190 AIVIP.
INPUT FOIL THE 5f'EED
F't
""
f. LOSS
SHUNT MOTOR.
POWffl. LOSS IN THE FIELD AND
riELl7
= I,770 WATTS
5-2~ TO
!> ·3 .G..
rtEQ/D ·
a)
x
E
52 "'1600 frPM
.. ""
== f'~·u•o
f'M
~
5, == 1'350 11.PM
r~
1.
LOSS AT fiELD fCHE05TAT
j/, LOSS
c
=- 18! AMf'.
RrticOSTAT "'
= "f.OS
J.
2
F'f
X 100
4~700
F'f = If /If
= C10Y (17·7)
HP, sHuNT MOTOR.
r..,
::::. 1,770
p,...k = 530 WATIS
GIVEN:
LOSS AT FIELD = Pf x IOO
/0 AMP
2
IN THE FIELD AND IN THE FIElD RHEOSTAT
V,.= '230 VOLT
f
+ 5 ·~
ff.t~~~:e = 1f ITRt'I"O
= Qo)''(s-.3)
I, ;;·o ·- ll.PM SPEED ; (b) THE POWElL LOSSES 11-J
2"30 X 190
=:
17· 7
; AT h"' 190 AMF
f'M "" 13,700 WATTS
Z30
=
llf + fl.A
INCREASES TO 190 AMP. CALCULATE:
e<) THE POWEr( LOSS IN THE FlfLD AND ITS PERCENTAGE OF THE TOTAL
POWElL INFUT
c) PM =VAIL
b) FIElD IIHEOSTAT 15 CUT IN,·
OF 17.7 OHMS AND OrEr?.ATES AT FUI.L LOAD WHEN THE LINE CUflrz.ENT IS
1e1 AMP.
MOTOII CHAITACTWISTIC.S
w
-~~~~~~
'I·'}
EFFICIENCY, 11ATING
1
AND APPLICATIONS OF DYNAMOS
HAS AN EFFICIENCY OF 91 F"Ert CENT AT
6 -2] A 50 -I<W GENEl?ATOil
FULL LOAD CALCULATE THE INPUT IN I<ILOWAlTS AND THE POWEI?
LOSS.
GIVftJ:
EFFICIENCY flATING AND APPLICATICNS OF DYNAMOS
I
6-~ THE l?OTATIONAL LOSS IN A GENEflATOR WAS FOUND TO 13E 780 WATTS
WHEN
THE GENEflATED EMF WAS 1~2 VOLTS. DETeRMINE THE IWTATIONAL
lOSS FOR GENETtATEV VOLTAGES OF 138 ANO 12G VOLTS
GIVEN:
Pt..= ISO WATTS
Eff., 911. AT FULL LOAD
Sr.L. 011. flOTATIONAL LOSS::: 780 WATTS
~= 1~2
REQ'D:
CALCULATE THE IIJPUT IN KW Vvr POWEfZ LOSS
SOLUTION:
LOAD FVWEIZ
I
VOLTS
Q) AT Eg= 138 VOLTS
flEQ'p:
5·-P.l. =639)(t;.9)
OETeltr•1JNE THE nOTATIONAL LOSS
IS THE POWflt OUTPUT
= 81S·S
FOR GENErtATED VOLTAGES OF
SiNCE:
o) 138 VOLTS
l:ff =
f"oiiT
X 100 1.
b) 12,
f:N
)~"f-.835
... 7"f3 · "t WATTS
r"'
KW
1'52V
fVWfR LOSS'ES .. p,,.. - f"...,r
:=
VOLTS
s-:r.L = E9 x
IA = 780 WATTS
--;j'l.
=
= I'Z'
5 ·P.l· = (12& )(!; ·'3)
VOLTS
SOLUTION:
F, 01 = 150 KW )( 100 ;/
FiK
b) AT f9
WATTS
r~"'
(1~.835 -ISO) KW
5·9 AMP.
~-']THE OUTPUT TOflQUt: OF A MOTOR IS
.. lt.e,s; Kw.
G9.:Z l6-FT
WH~N
IT OPERATES
AT
950 I!PM. CALCULATE THE" LOSSES IN Tt1E MACHINE AND ITS EFFICIENCY IF,
G-~ A 15-HP MOTO/l OPmATES AT AN EFFICIENCY OF 87·'5 PEFZCENT AT
Ft/WEIZ LOSS IS APf"T'llXIMATELY ONE -FOURTH
FULL LOAD. IF THE ST/?AY
OF THE TOTAL LOSS
1
CALCULATe THf: COPPER LO.SS.
F~= IS HP
P'r =
MOTOTC
!11.
I
AT FULL LOAD
STRAY POWER. LOSS =
!4 TOTAL lOSS
P1 =
F,
P,.,.:: IS"
IS THE COPPCTt LOSS
HF X
746
=
POWER LOSS
JIF
11,1'30 WATTS
----
?
=
=~
T
-c;,:sz;
"') Err=?
£ff:::. p...,T
OUTf'UT
f'OWE/r
X IOO
J.
= '3,~ 2 S X IOOZ
10,'300
X 95"0 X 69·2
ffF.
3~.000
tiP"" 1'2·5" HP
P..ur
r....
I01 90o
P,NI'IIr
HP= 211' X 11PM X T
SPL:
P,N -
11 57ri WAITS
SOLUTION:
p.,UT =12·5)'11"' X
7-t' w
-w-
"' 9~2!0" WATTS
789 WATTS
"'-
50
~-
ftosses ~ ErPICIENCY
IS"'J<J- 39:1· 7G"
Pe<~"' J,I9'J.2t; ~ IZOO WATTS
w
=
=
Pu>$SU
ft.
11EG'D:
~,000
Pcu = Pr-
STIU!Y POWER lOSS
= TOrAL
o.)
69 · 2 L6 -FT
P,., = 10,900 WATTS
TWS:
0·97(;
:c
PT
S·f':L.=. S99·7SWATT5
0·875
p,N
~
= ~ (1£;99)
SOLUTION:
=
P...,r
/S"J9 WATTS
S·P.l· =-
CALCULATE" THE COPPER.. LOSS ?
f,"
PrN -
SIIJCE:
SpL =
=
SPEE/7 "' 9SO lr.PM
:: 1'2178'3 -111190
REQ'lJ:
u:.r: Pcu =
GIVEN:
iOrtQUE
GIVEN:
EFF= 87
UNDER. THIS CONCJITION , 1HE P(IWEfl. INPUT IS 10,900 WATTS.
51
=
er;.b
X
LffiCIENCY
r·~~-~l WHAT
1
APPLICATJONS
rtt.\TING, AND
OF DYNAMOS
SHOULD 13E THE rlJLL- LOAD HORSEPOWE!(
A
THAT DfVVeS
GIVEN:
r'L=
50 'r<W, GENERATOR
EFfc"' 89 9 /.
f"ovT
-4
~
= POWEll':
INPUT TO GENEflATOR:
f"1.1
=POWElL
OUTPUT 13Y MOTOR
5"0 KW
Pr =
Fr=
GENE11AT0fl:
FM
Pn-1- P""r
6+,10'1-.~
C) f'~.ess.,l'w. = .FiN~-
=
P... =
-50,000.
p.., =
p,.,M_ PouT.,.
6"1-,IO'f:G'- 5'"5,8/bG
€1Z5~ WATTS
f'ouTG
G.,-9] A 10-k'W 220-VOLT COMPOUND GENEIZATO!l IS OFERATEV AT NO LOAD AT
11-fc
PITOPEIZ. Afl.MATUR.c VOLTAGE
f'OWEIZ LOSS
THfnEFOR.E:
f"t.t= 55866
y/ X
SHUNT- fiELD
HP
= 7t. 89
AND SPEED. FfWM WHICH THE STRAY-
CALCULATIONS AIZE DE.TEfl.M/NED TO f5E 70/'i WATTS ·THE
liC:SISTANCE IS 110 OHMS, THE ARMATU/lE flESISTANCE IS 0-2GS
OHM , AND THE Sr:fliES- FIELD flESISTANCE
74G'P
f'M
= 5,8GG WATTS
= 55'",8fPIO- '5000
TltE MOTOJ"(
0·895
[ff9
"·
fr = It, 104. G WAITS
'50,000 'N
::::
f'Lo..,
GENEITATOIZ SET
P9 =PM
OUTFUT OF GENEflATOR. = OUTPUT OF
LET:
Pc.J=
b) TOTAL LOSSES IN MOTOR: -
Eff.,=8% /.
APPLICATIONS OF DYNAMOS
CONT OF PflOB. G-8
SINCE:
SOLUTioN:
PouT
IS 89.5
P..ur=
FIGURe
HP RATING OF A MOTOr?
J
= F,N
MOTOR
ILE'Q 'b:
Pc,::;;:;
RATING OF A lv10TOR
50-KW GENER.ATOR WHOSE EFfiCieNCY
fEfl CENT?
EFFICIENCY, ITATING, AND
BRUSH DROP
HP.
AND C4LCULATE
IS 0.035 OHM. ASSUME A 2- VOLT
THE FULL- LOAD EFFICIENCY.
GIVEN:
558GG WATTS
l<s
f'L: 10 k'W
6-8] If THE OVE!l-ALL EFFICIENCY
P/tOB. 7
OF THE
MOTOR-GENERATOI? St:T OF
EACH
ltf = 110 .lL
IN
~------~
OF MOTOR-
Pi~
~------~
Ve.c = 2. VOLTS
MOTOit
REQ D:
o) t::FF. OF THE MOTOR
TOTAL LOSSES
GENERATOR:
0)
Ef£..,
0 ,.,,._
Pu;..,
fiN -
CffM= 55,8GG
0i,IOt.G
IOOfo
Lff. ov ER.-il\A-
:Eff,. =
50,000 W X 100/.
FIWM
X 100/.
DETERMINE' flllST I. TO
CALCULATE TOTAL COff'Efl
LOSSES·
THUS
87- 15
7-
FA= AIZMATUflC: COPF'flt
P.. r + f'r L•<SE>
7
IL = PL
= 10,000 w
VA
'220 V
IL = 1'5· 45 AM F.
78;.
lf= VA = '220 V
rrf
?>4-,1 O<t-6
-110.n...
f's .•. ==
LOSS
SERieS riELD COF'PETl. LOSS
Ps.,...= SHUNT FIELD COPPETl LOSS
COF'PETL LOSSES AIZ'E:
FA= IA ~ 17..
FA
=.
(47·t5)~ (0·265)
f',. "' 59G. 7 \AI,IITTS
PsE"" IL ~
==
fls~
(t5 ·'fS) oz (0.03'S')
fSE= 72· 3 WATTS
If-= 2 AMP.
52
+ <f-5.'f-5
LET:
E.ff= - - f'Jl.OIS.~
== 2
I. = 47· 45 AMP.
SOLUTION:
P,NG!!NEil,AToR
= 55,SGG
SOLUTION:
7
Pour X 1001-
ISUT lt>ur""""'" ~
SET
IA= 1t .J-lL
SINCE':
fULL -LOAV EFFICIENCY
1.
FooT.,
I
= - - X 00 •
PtN M
/N THE MOTOR-
c) LOSSES 11\1 EACH MACHINE
X
FIG: COMPOUII!O GENERATOR
ttEG'o:
MOTOR - GI::IJcRATOrt SET
1
p,>IIT
DC LOACl
Rs = 0·035" ..lL
p.,M "' PiN,.
GENERATOR SET FTW/111 f"/W6 #- 7
Pm=
Rt
JlA = 0-265 JL
MACHINE.
GIVEN:
OVEI'1. ALL Eff = 787.
6)
TO
S·f.L.= 70'5 WATTS
IS 78 FER CENT, CALCULATE : (o) THE EFFICIENCY OF THE' MOTO/( ~
(b) THE TOTAL LOSSES IN THE' MOTOR- GENER:ATO/l SET (c) Tf'tl: LOSSES
"----<l+
VA =Z?.O VOLT
i3
EFFICff~CY;·I!tlTlliG.. ,_J.I.NV_
APPLICATIONS OF DYNAMOS
EFFICIENCY, flATING, AND APPLICATIONS OF DYNAMOS
CONT Or PROB . 6' -11
GIVEN:
CONT. OF PIWB. 6-9
If
2-co x e
f&• -= IA "
f'st+ == VA x
=
440 WATTS
P&H =
VA "' 2<;0 VOLT
= 94-9 WATTS
[A
t:Ff =
TOTAL LOSSJ:S::: S.P.L-+ FA +-!Set PsH
Yrs.c-
=- 47·l> X 2
10,000 W X 100%
10,000 -t- 181'1- + 9'1--9
=- 70'5+ 59b-7+72-3+4to
=1814 WATTS
Ef'F = 8:5· "37 /.
OCCURS WHEN THE SUM OF THE COPPEll LOSS IN THC. AllMATV/lE AND
IS EQUAL TO THE" SUM OF THE ST/?AY POWZ::Ir LOSS ANV
THE SHUNT- FIELD lOSS, CAlCULATE : (a) THE AIZMATU/lr: CUTUlENT; (b) llll:
LINE CUillrENT ; (c) THE KILOWATT OUTPUT ; (d) THE MAXIMUM EFFICIENCY.
= S-P·L + PsH
!f - 2 Atv·lf"' Flt.OM
PJtO~. ~ 9
;
= 701J +4f0
= llf!7
I'-""' '"S -t
WATTS
TI-IUS;
IA '2 f'lA
=
Pcu
lZ,o~ == "ffOW ~AT IA =1-0A
l"'fO)...
leu"' 'TOO WATTS,"-AflMATUitE COf"F'Efl.
IZA = 0-Zs ...J2...
LOSS
MAX· EFFICIENCY
AT
IZEQUII'{ED:
200-J-'ZG0-+100 =1A 2 flA
rOR THE
AflMATUitE CUTl.lrENT
IA :z.=
CONDITION OF MAXIMUM EJ=F.
!iGO
O.?r;
J A: ~
SOLUTION:
lo) II-= IA - [f
THAT VARY AS THE
SQUAfif OF THE LOAD.
f/'1.1CTION/'.,YWINOAGE LO~.S=- 200 WATTS
COflE LOSS = ZGO WATTS
SHUNT- FIELD LOSS:: 100 WATTS
SERIES FIELD
PciJ+~e
LOSSES
40 AMP, RATED CUR/lENT
61W5H CONTACT LOSSES:: 80 WATTS
6-1~ If THE MAXIMUM EFFICIENCY OF THE GENErl.ATOT{ OF PIW~· ~
SOLUTION:
=
StiUNT GENt:flATO/l
I
MAX-
EfFICIENCY OCCURS
2.2"!0
r A =17.33
WHEN
.4Mf'·
THE Af'flZOXIMATEV CONSTANT
lOSSES AI7.E fGUAL TO THOSE
h .. "' 59.8 AMP.
d.) TA • IS THE' AtzMAlVRE CIJI{IZENT )
c r.... r
lA (~+Jrse) = 114!7 w
IA 2 =
llt> w
lA =
J 3811D-G7
o-z~s
1A =
P1..osses =
;;:
=
=
6-12] A 25 -I<W
V.4l1..
=- (tt0)('$9-8)
WHEN
1000
+ o-o~;
p.,..r =
r.,uT. ')(
=
fFf
1!290 WATTS
-=
WHEN IT IS DE.LIVEfUNG
100:/.
f'5J% x
9!>.
HAS AN EFFICIENCY OF 85 f'EIUENT
ltATfD LOAD· IF THE SWAY- POWEll LOSS' IS ZO f'm. CENT
A LOAD
OF IS- KW, ASSUMING THAT THE STRAY -
CONSTANT AND THE OTHErl. LOS'SES VARY
l?s
GIVEN:
13,1?'
lltt; t lltS'
AT
POWER LOSS IS SUBSTAIJTIALL Y
AS THE SQUArz:E OF THE LOAD.
FIN
[rc:u + PseJ +[ FsH + s.n]
OF'Ef<.ATJNG
~ENEitATO,fl
OF THE fULL.-LOAD LOSS , CALCULATE THE EffiCieNCY OF THE GENERATO/l.
13.r;' Kw.
cl) Eff. =
61-8 AMP':
SE,fliES
roo
-
RATED lOAD" ZS KW
+~Z90
-I.-~=I._
fr. .
l:ff =85 'J. AT 11ATEO LOAD
z1.
S-f.L." 20'/. AT rULL-LOAD LOSS
0
L
flA
0
REG'o:
G-11] A 250- VOLT
SHUNT GENETlATOIZ
HAS A RATEP ARMATURE CURRENT
OF to AMP' AND THE fOLLOWilJG LOSSES AT FULL LOAD: FRICTION ANI7
WINDAGE= '200 WATTS ; CORE LOSS= 260 WATTS;
SHU~T-fiEL/l LOSS.:: 100 W.
ASSUMING THAT THE MAX/MUM EffiCIENCY OCCUllS
CONSTANT LOSSES
ARE EQUAL TO THOSE LOSSES ThAT VATl.Y
SG!UAflE Of THE LOAD, CALCULATI: TttE ATZMA TU/lf
CON17ITION OF MAXIMUM
I
i
I.
WHEN THE APPnOXIMATED
rFFICIENCY.
'54
CU/'lllE.NT
AS THE
FOil. THE
A
0
CALCULATE THE EffiCifNCY OF
THE GENEilATOT?. WHCN OELIVERJN~
fiG: SERIES GWcrtATO/'l
A LOAO OF 1; KW·
5rtUSH CONTACT LOSS "'80 WATTS ; AltMATVTZe COP'f>Eir LOSS = toO WATTS •
T
SOLUTION:
Lf"T:
F,= IS THE FULL LOAD f'OWER
F'e<~
LOSS
=
IS THE COPPER.
LOSS
ttf1!i"
IS THE EffiCIENCY WHEt-./
THE G£NfllATOI'{ IS PELIVflliNG
I~ i<\1\1
55
EFFICIENCY, llATING , AND APPLICATIONS OF DYNAMOS
EFFICIENCY, RATING, AND APPLICATIONS OF DYNAMOS
OF PR06. 6-12
CONT.
CONT. OF PfW8. 6 -13
SOLLJTION:
1!-IE LOAD IS DEC/tEASING FTWM
SOLUTION:
AT RATED LOAD:
2S KW UP TO It; t<W, THEilEfO!tE
DETEllMINE
PIN
= Pour
-X
fff
IA "~ ALSO DECREASED
THE
~
100/•
•
li;KW
1. LOAD
29."tl KW
Pr"' Pu·J- Four
= (z9.~1-2S)KW
f1=
f'cu
4·11 KW
= fT-
~o%
4"07. , THEN THE COI"F'Eit LOSS NOW
15 ONLY GO% AND IT WILL VAitY
r...
TO THE SQUAfZ-E. OF THE LOAD·
I,. =- z + <)0!3
THE COPPEll. LOSS NOW IS
S.P.L·
P~
= I·Z7
fffy
= ZD/. OF F'T
= (0.2)("1·41)
:=
==
0·8B'2. KW
f,~
+ f'c.4 + S ·P·L·
f.s 1-1
1'3 KvY
IS J(W + 1·27 KW +0..982 KW
LONG SHUNT COMF'OUN D GENERATOil : E = 220 VOLTS ; OUTPUT = ZO KW; STRAY-
!3RUSH DltOP== 3 VOLTS. CALCULATE
IZsE::
20 KW, OUTPUT POWEit
0
S.f.L::: 70S WATTS
= Q,'Z65
~s~:
L
RsH
RA
.n.
= o. o~s ..n._
0
A
0
V6 .c = 3 VOL"IS
EFFICIENCY
FltiCTION
+
.Arlc
X 100/.
Fe";
+ P,.
20,000
~osse.r
w
X
lOOt·
20,000 w + 4012-7
fff =
w
83-3%
r...
278·7 WATTS
WINDAGE == 200 WATTS ; SMUNT FIELD"' 100 WATTS ; COILE..:: Z~O
AT MAXIMUM
LOAD f"OWER.;
(c) THE
==
400 WAHS ·
EFFICIENCY, WHEN THE CONSTANT LOSSES
LOSSES THAT VAR.Y AS THE SQUArlf OF THE
LOAD, CALCULATE ~) THE , AllMATUR.E AND LINE
CUfmcNTS ; (_b) THE
MAXIMUM :EfFICIENCY.
GIVEN•
VT = ·2'='0 VOLT
IA = 10 .LI.Mr; FULL-LOAD
-IL
lsH
ARE
fRICTION -+ WINDAGE= 200 W
Fs~.:::
?
PauT
9Z·9
EQUAL TO THOSE
LOSSES'
FIG'· LON~ SHUNT COMpOUND GENEitATOR
fl.EG'o:
EFF. =
7 WATTS
6-1~ A 2SO- VOLT SHUNT GENERATOIZ HAS A FULL-LOAD ARMATU/lc
CVf?flENT OF 40 .AMF 1 UNDER WHICH CONDITION THE LOSSES AilE:
FOrt OPERATION
IZsH • 110 ...0...
X
X
"tOIZ.
WATTS ; BllUSH CONTACT =- 120 WATTS ; .AilMATU!tE COPPElL
r-----.,---------o +
-rL T
fT = 220 VOLT
RA
0·03!1 .1l ;
THE EfFICIENCY.
GIVEN:
rL. =
3
PDR.U.rH=
POWER LOSS= 705 WATTS ; llsw == 110 OHMS ; ll4 "'O·'ZG!; OHM ;
PT =
X CT
"'Va.c.
=
IS GIVEN IN CONNECTION WITH A
INFORMATION
""t"t0+278·7 +2<;;89 +70;;
"' 2 X 220
~ 'ttO WATTS
FBIWSH
+ F'sR.Usll + p.,.. + S ·f. L·
PT:: Psu
LOSSES AILE:
PsH ==IsH
X 100;1.
'fft 1!;'"' 87• "1!1 jl.
0-13] THE FOLLOWING
TOTAL LOSSES :
== IsH +lL
fOWEIZ.
KW
f"1'?
Z 589 WATTS
SPL == 705 WATTS
I A = 9 2 . 9 AMP.
f'eu =~·SZ8)( o.c)"
SINCE:
21?0 V
IsH= 2 AMf
=
I"e.~ -
90.9 AMP.
THE DEC/tEASED LOAD IS ABOUT
Pru = 3·528 KW
S.P.L :
lL.""
ET
220 v
IsH=--=-flst<
110 Jl.
"' "\'."11-0·882
S.f.l.
Er
25'1<W
85:f.-
r,w "'
FL
eo,ooow
lL=-=-
,
;1. LOAD:-- X lOO%
= ?5 KW X 1001.
IA ~ [ ·rtA + l!ss]
= (9:z.~)~ [o.zt~s +o.o3s]
Fcc..t =
FIRST lL;
res~
tIA
~
100 VVAITS
Fc"Re = 260 WATTS
SEE NEXT PAGI: I"OR: THE SOLUTION :
Pe~us+< c".rrAcT '='
Pc-u ""
'56,
FIG:
IZO WATTS
400 WAT15
"i7
Sl-tUI-JT GfNE"nATOft
+
EFFICIENCYI' llATING I AND
EFFICIENCY, RATING 1 AND APPLICATIONS OF DYNAMOS
CONT. OF Pl1.013
IT.EQUJ/lE[J:
CONf. OF Pl'l06. G-1S
6-14-
lst-~=- =
lsH
=
0-4 AMP·
TL := lA -
SOLUTION
lL
IA "' 4-0 AMP.
=400
lst<
MAJ<ING
TIA=- = IA z
("!!)z
1~.-=
HP
P~.
=(''1'1)(2!>0)
lvHEN THE CONSTANT LOSSES
F'L
= 11,750 \v'ATTS
IL.:::: """
EOUAL TO THOSE LOSSeS
ARE
TH,.AT VARY AS
THE
SQUARE
"'p,., = rA 2 RA
PccNSTANT
SINCE
CONSIDEITING THE CONSTANT
L0'5SES AfZ.E •
f'CDNS'l'A>Jf::
f'L
MAX· fff.
LOAD.
OF THf.
IsH=
c) fL "'Four
f"F+w ;
f'«>RE,
IA:=
100
CU/'W.ENT~
~
X
460
-
746
LONG SHUNT COMPOUI-JD MOTOR
0·88
X
AMP.
VT
fflOPOrlTJON
= <!-60
THE GENEIZATED VOLTAGE OF
z AMP:
THE MOTOR I 5 WHAT YOU CALL
COUNTEIZ EMr OR Ec
= II AMP.
lL= It.+ IsH
VOLTAGE AT A IZATEO
I...=
lr... = tG
II+ :Z
THE ONE
GIVEN
Ec = (Vr- Vs.c)- IA (l'lA +~)
CURRENTS
OF
AND CALCULATE THE EFFICIENCIES
CHANGC:S : LINE VOLTS= 4'60; SHUNT- FIELD RESISTANCE
=230
OHMS
ARMATURE ..P"c~-- INTEIZPOLE. RESISTANCE= 0·72 OHM ; SERIES- fiELD
flt:SISTANCE
~
D·lZ OHM j BRUSH DfWf'.:: 3 VOLTS.
GIVEN:
Bf.COME
STRAY
Ec= 4'f'7.7C VOLTS
THE ST/'lAY POll/ElL LOSS
ASSUMED
LINE VOLTAG'E = 4GO V
TO GENERATED VOLTAGE . SO IF
s.P.L
Ys.c
230 .fl.
-=
VOLTS
H·P. = 2~ HP
1088 WATTS
llA +lNTe.RPOf..E
=- 3
ASSUME:
~ Q.1'Z AMP.
Eff,BB(o AT FULL LOAO
Rse. -= 0 -IZ .fl.
58
THErte IS
IJECilEASED
THIS WILL
A REfEIZE.WCE GENETIATED
PO\v'W
A I 088 WATTS
LOSS.
AT IA"' II AMP. , THERE IS AN
POhlfR LOSSES :
TO BE
~
VOLTAGE WHICH HAS
Ec= (~-3)-11 (0·72-+0·12)
ARMATURE CUfUl.E:NTS= /1,22,3,, 1'1-, 5"S A/v'lF.
IZst~ =
I
l
i
41 AMr.
Ec.:= (4G0-3)- 1-+ (o.?z +1M)
OF THE MOTOR
II, 22, 53,44, AND SS AMP, ASSUMING THE FOLLO\VING
A:::
Ec = 4'20 VOLTS
Ec = IS THE COUNTErt EMF
fDTI AltMATURE
= "J-6; -2
THUS:
WHE~t::
TO
FROM EX. # 7
IA
I
LINE CUfl./lEIJT,
AMP, WHICH S·PL= 1088
WATTS
MOTDI'Z. IS GIVEN AS:
~o.Ts
MAJ<E A TABLE SIMILAR
TO THE GENERATED
VOLTAGE·
-230
THE GWEI'l.ATEV VOLTAGE AT THE
f%0
>
c
E
I~.-= 13 AMf'·
1.
+ 260
G-15] REfE'flfliNG TO EXAMPLE' 7, p. 192
u
rz
({,..
THEN DETEI'l.MINE WE GENErtATED
100 /.
11,7t;;O + 2(%o)+ 120
lllAX· Eff = 90·%
IA2.1ZA "'zoo+
X
AT : IA
= ----,--....,.--
f'sH
THEILEFOIZE :
IsH::
fco"STA'-'T
I 1,7!70
741;
I'Zsll
100/.
K +(2)(%0) + 120
feu =
MAX. Eff.
X
X
25
SL
= o.zs
5
0
'f60 X 0·88
h==~
1ZA
. Q511
A TABLE rOR DIFFEitEtJT
VALUES Or ARMATUftE
b) LOAD PO\v'ETI
FL == IL Vr
400
IL- T
0
D
I 11
c
COMPUTI: FIRST BEFORE
4'l·<f-0·f
=
\
SOLUTION:
h = 47 AMP : LOAD CURRENT
\v'AIT
Pc<.l
Z50V
11-ll.J s:
v) MAX. EFFICIENCY
fcu
--
VT
PO\v'ER.
MAKE: A TAf;LE SIMILAR TO
EXAMPLE # 7 , PAGE 192
!OOW
f'<>H
MAX. EFFICIENCY
AT
ltEQUIIlt:D:
a.) IA "'t7.'33 ~ 47·4 AI/IP.
a) AllMATUKE____.,.~J.- LINE CU/UlENT AT
b) LOAD
S OF DYNAMOS
INCJ{f.ASED
IS
f'rtOPOfl.TIONAL
AN INCREASeD
IN Ec
TtiE INCREASED Ec
= 4'+7·7~
4to
OR.
fc. = I· OGG VOL.T
IN THE GENEI'lAT'EV
VOLfAGE 1 STIZA'Y POIVER LOSS
ALSO JNCflEASEO OIL DECILEA:lED IN
59
eFFICIENCY, RATING, AND APPLICATIONS OF DYNAMOS
EFFICIENCY, RATING, AND APPLICATIONS OF DYNAMOS
TAI:5LE
CONT· OF PIWB- ;i 6-15'
nn:.rtE.FORE :
S.f'.L. =- (i.OGG )( 1088)
S-P-L-
LINE VOLTS
F;N :: VL >< I
= llbO WATTS
fu;=
PjiJ
AllMATUIZE LOSSES
PA"" TA z rz.A
f'A"' {Jl)"(0-7?)
P;~. =
LINE
4GO X 13
=
!;'380 \VATTS
F""'T:::
l"se = (ll):z(O·l?)
'57Gt;.r;;
BF:=
37G<;.r:;;)i<l X
StiUIJT- fiELD LOSSES
tll''= t;.0\0 1-tl'
= IS+l x YT
Ps+l = (2) (450)
PsH
CUIUr.et-JT Of
f's.c.=
IA
X
t:Ff
Yg.c
=
I
AttMATUIU'
+ 87
711: )/'("'
SHUNT riELD (Isu
TOTAL
INF"UT,\vATT5
1
~ x )oo/.
35
46
57
I
I
2
2
33
4+
2
55
"/-20
410-B
I 438·5 I 42'3-3
I
J4S"_
I
I
I
1112
17Bt
;;a
131
920
920
I
1088
I
IOG4-
1394
232
66
I
99
I
2!528
I
3.046
I
920
920
/'32
I C> B'
3.766
4,690
5',3~0
11,040
15,/80
21,160
26,220
~765· ;-
8$12
12)'31
17, 391-
21/5!'10
5r!P
114"
IG·3
23·3
28-BG
14214-5
P.N -fLosses)
I/'5G
__87__ ,3.8
0UTF'UTHF-(0uTPUT';ATT}
EFFICiENCY, f'EilCENT
G2·S7./. ln-1/.
X 100
179-':Jj.
I
8227.
I 5'2-11/.
~
G-lGJ A 20-HP MOTO[( HAS AN eFFICIENCY Or 885 PEflCENT AND OPEIZATES
+f4.c; +920 +33
NOTE:
PROCEDURES
DIFFEilENT VALUES OF THE flEMAINING
22,33,11,
'3E'E THE RESULTS' IN THE TABLE
PAGE. ...
SOLUTION:
PL = '20 HP , MOTOf(
/rtY ~5" AMr.
EFF :: 88· r;
t
"'
f' 0
/o
0
'200 HRS
0111/KIV-HR.
LET
PLo~>E.> =IS THE
a)
l
PC<JT
Eff.
20 ttP X 7%
f,,.,= . ''0·88!i
b) tNERGy LOSS f"E/l MONTtl
c.) COST Of ENERGY LOSS
P,N =
'/z 4/ jK\v-1-1)<.
61
(,()
p,., =
REQUII<r:D:
AT
f'Oh/ER LOSS
CLOss "" ENERGy LOSS FER- MONTH
Cl) POh/Eil LOSS
I
AT FULL LOAD . CALCULATE·
LOST AT 1 Yz CENTS FElL KILO\v'ATTHOUR..
GIVEN:
FOil CALCULATION
ARMATURE CUIW:E.NT . SUCH AS
ON THE NEXT
A VENTILATOfl
(a) THE POh/Efl LOSS j (b) THE ENEfl.GY LOST PEfl_ MOI-JTH 1 ASSUMING
OPERATION FOTr 200 HOURS DUfliNG THAT PERIOD ~ (c) THE COST OF THE
eNERGY
TO
l
//GO
(IL xvc)
CONTINUOUSLY TO Dl'l/Vf
SAME
I 24
I 2
I 22
x Vr)
920
61W5H CONTACT (vee xi") I 33
~-r;t+P
Pr == ?, 214 · 5' \v'ATTS
THE
I
(r,,' tt.)
SERIES
Eff =- GZ-97 /.
TOTAL LOS S'ES:
==- IIGO
13
"!47-7~
nr:w (rA.,rL.;E)
I HP
1/ AMF'S.
EFF = 37Gr;·t;
Pa.c. =-- 33 \vATIS
f'r
"tGO
2
I.A
STRAY FOIVE/t (5-P-L)
f'u-1
P,.,_:: (II) (3)
'tGO
I/
OUTPUT (
LOSS'ES
I ·4Go
ARMATURE
fFF/CiENCY FOIL ArtMATUrtE
f's~J = <:! '20 \vAnS ·
I 460
POlvCR LOSSES (\vATTS)
\vATT"S
OUTFLIT HOitSE f'O\VE.J<.:
f'sc. ::: 87 \v'ATT 5
460
SHUNT FIELD (IsH)
GENEf'lATEO VOLTS ( Ec
Pour = 'i>':380- :22•14·';;
5fl'liE5 FIELD LOSSES
l'i'
IJa.
curmENTS , AMPEilES:
L
f'our = f'nJ - F'; L..o>.>Er
87 \vATTS
~RUSH -CONTACT
OF PIWBLEWJ
-
IG, £?(;9 1\/ATTS
EFFICIENCY, fl.ATING , AND APPLICATIONS OF DYNAIIIOS
EFFICIENCY, ITATING, AND APPLICATIONS OF DYNAMOS
MOTOIZ
6-18] AN ENCLOSED
CONT OF FROB. 6-1G
c) COST=
$~
X 3!38 IW-Hft
= p,~ -
r~oss
PouT
I"LOSS
=.
COST==
t:JI X 7'f(;W
P1.-0s> = IG,859- 20
-v
.$
5.8 ; COST OF EIJER.G'l
L''SS
6-l~
LOAD \viTrl OUT EXCESSIVE HEATING , lvH.4T
MOllE
13E BIVEN Ttil': MOTOI<.?
SOLUTION:
50 HF'
:=
TLATING
6-1~ A
1
COST =
= G'T tW
LOADS : 105 Klv' AT 250 VOLT.S ; 35 Klv' AT liS VOLTS 13t::T\v'Et:N iHE !"051-
LOSS
$ G9·"
SOLUTION:
OE.LIVEllED 6Y THE GENEIZATOil; lb) THE CUI'l.fl.ENT IN THE POSITIVE
LINE~ (c) THE CUfl.ltENT IN THE. NEGATIVe LINE ; (d) THE CUrc.17.ENT AND
PER 'l~AR
ITS DIRECTION
IN THE NEUTJ(AL
Z50/115' VOLT , 3 -lv'IIZE GENE:f<ATOit
6S Klv AT Z30 VOLTS
(.,OST Or ENERGY ;
~S
~ X
AT liS VOLTS BETlV'EEN
THE F'OSITIV~_..V Nf.UT/lAL LINES
4 O.Oit;
Yfl
COST=-
7tG
_j
SAVING/
/YEAI1..
f' 1 ~ ==- I G ,30Gi \v'ATTS
t9·G'S
K\v'
A YE.Aft.
1-JE(;ATIVE
/rJ'
b) CVIlflENT
I· :58~ 1<1-/
~~;; t
3t;
12~
K\v
I
+ 'Z!>
IN THE F05)TIVE.
I+ 11 ~
=
Pa,.
=
:5 r;,ooo
II~Y
)'..
200 l-IIIS X 12 1\!JBNTflf
l+,•r; =
~
Mot-IT!+
f:~oss == l·'38G
tLoJJ
X '2.00
X
YEAn
-=- 3~20 k\.J-HR /.
I YEAll
Jl
I'
I no
=
\v
:=,
1t-
'283
+ 217
1~.~
If11r;
=
= r;;oo AMF
I+11r;
-I-11;
" 87- AMP1 TO\v'AIW THE
63
BECAUSE'
IS GrtEATE/t THAN 1-11>
• SEE THE Flt;UILE
(1
( ).'1
+ l-11,
l5ALANCE COIL
GS' ,000 \II
AMF'.
z ,.,
1tJ = 3'04- :217
30"1' AMP'·
2.93
IN NeGATIVE LINE'
I - "' I
ItJ =
lit; V
230V
I:t~D
v
a) CUI7.f1£NT) ITS DIRECTION 11-1
NEUTRAL WJE:
11
I
12
CURR:ENT IN NEGATIVE LINE.·
)It;
LINE:
EWEJZ,GY LOSS rt:rt YC.Af~:
c;a7 AMf.
I -11, ""' 2!i,000)v
f' LOff"' 1,:58'G VATTS
f'Lo.<>
'283 + 30"!-
I -II> "' 217 .AMf
!OT.AL K\v' =
P,N - PouT
::-
It- =
It=
NEV1/7AL LINES
a) TOTAL l(lv' LOAD DELIVE/lED
_j 19·71
FOSITIVE LINE
123<> +I+IIG'
SOWTIOIJ:
F._orJ = I<D,?OG- 14,920
El.Of[
IN
I+"'
c)
AT Jlc; VOl-TS , BEWEfN THE
25 K\v
= j ~9·G -~ 4.9·!9
:=:-
I
LOADS·.
AT EFf = 91· Sf.'
fiN=~
Eff
LINE.
GIVEN:
A YEAR.
COST=- 3326
0·91t;'
LINES. CALCULATE : (Cl)THE TOTAL KILOiv'ATT LOAD
N!:UTfZAL
FOLLO\v'ING
coST
f'L.Orf:::
X 1·28
2:50/HS -VOLT THflEE-1\/ffl.E GENERATOIL DeLIVERS THE FOLlOIIIING
TIVE AND
AT Efr..,. B8 s ;I.
COST-= -$ S .g x t-2
IZ.E0 V:
'
SAVINGS PER YEAtl.. IN
011-
=- 50
!UTIIJG' Of THE MOTVR.?
COST OF eNERGY
l:FF == 91 S ;I.
PLOss "
f'Efl. CENT
5l'8 Klv'-HIZS.
DATA IN F'fWI3 # IG
P,~=
SHOULD
1
I· '33~ X 200 H/U'.
GIVEN··
Y.
IS LOCATED \vHEilE IT IS
R.E0 D:
WIOTOIL Of Pl1013. IG HAD AN E-fF!CicNC'l' OF 91·<; F'ffi CENT?
20
28
rlATING
flATING
WHAT SAVING \vOULD BE MADE f'ER YEAil IN ENERGY COST IF lTIE
wt:::ru;y
CARRY
GIVEN:
b) ENERGY LOSS == K\v'~ 655 X -\;
ft-o<S"'
5"0 HP ./HE COVErt
CAPABLE t/F COOLING ITSELF EXTn:EMELY \VELL. IF ~STS SHO\V THAT IT CAN
I, 93g WATTS'= 1·939 Kh/
flOSS' =<
or
A IZATING
AI<.E flEMOVEO , AND THE MACHINE
PLATES
KV~HfC
HAS
,AT THf NEXT F'AGJ:.
•
EFFICIENCY, 17ATING, AND APPLICATIONS
EFFICIENCY, flATING', AND APPLICATIONS OF DYNAMOS
OF DYNAMOS
CONT OF P/ZOB#6-t?1
fiGUIZE
Of
FfWB. #"
G -19
If THE CUIZflENT IN POSITIVE
,. T
(+)
THUS,
\v'lflE IS EQUAL TO THE SLIM OF THE
K\v'+ 11 , = <tO Klv'
NEGATIVE ~
CUll/lENT IN THE
I+ ~ L
THeN,
.j.
IN
J<lv'230 =
1
H
SIMPLE
T\v'O-FtJU:
THILEE-\1/IfZ.E
6- 20] IF THE GENETZATGrl.
EXCEPT
THAT
THE T11/0
GENr:'itATOR:
FOR
UNDER \v'HICH
AflE BALANCED 1viTH RESPECT
SINCE THe T\\10 liS VOLTS
M "'
EQUAL J5l)T OPPOSITE
h "" p.. "" 125,000 \v'
=
p;..,I'VT
CUfllZENTS
IN THE
\v'lrtfS, CALCULATE THE
CALCULATf
LOADS
r. ..
X SOO
l:'FF
80 \v'ATTS
=
1< 230 = l<lv' WAD
G-23] A DYNAMOMETEr?
\v'INDING'S
HAS LO\v'- VOLTAGE
\\/HOSE
IN THE THTlEE \viJZES
DEL/VEfCS A WllflENT
POSITIVE .
CONDITION THE COreE
CALCULATE THE INPUT
!3ET\v'EEN
X
/007-
f.
RESISTANCES
INPUT/,;,...- HIGH- VOLTAGE
AilE ' rlES'fECT/V[L y
HIGH-VOLTAGE \v'INDING HAS
AS THE
LO\v'- VOL TAG'E
Of 0-'3 AMP
+
FI7./CTION
/o>AJ.--
I
0.1 OHM
TIMES
AT 180 VOLTS, UNDE/l \VHICH
+ \v'INDAGE
LOSS
CUlt RENT~ VOL TAGc.
SOLUTION
7'/z
AT THE NEXT PAGE
NEL'TIIAL Ll NE .
65
AS
\\/IN DING' . IF THE MACHINE
/'".v NEUTitAL LINt.
K\11_ 11 ,-:: i<\V tO,AD BETiv'EEN NEGATIVE
64
60-G
100 1.
II AMP X 12 VOLT
MANY CONOL!CT0/1S
LET:
THE THflEE
K
80\v'
[fF "" - l:i 2 lv'
AND 4 OHMS, AND THE
K\v'u~;"' K\v' LOAD
llEO'D:
=.:
Fo<tr
= (32 \\IATTS
OUTPUT
SOLUTION:
Ir+l =- It-> +I..,
Kill
A TOTAL LOAD OF 120 K\v'.
1\\V LOADS
TOTAL LOAD OF 120 K\v'
EFF =
EQUAL TO THE SUM OF THE
NEGATIVE AND NWTIZAL
G'IVEN.
I
DELIVERS
!VIllE IS
BATTERY
SOLUTION:
Fourr-ur == 0-IG
OF PflQB. 19
F!ZOIVI A 12 VOLTS
EFFICIENC'Y ?
5""1-"S-5" AMP
If THE CUflrlENT IN THE POSITIVE
I
IN
tJJflJ:CTION.
230. v
II AMP
ILEO'o:
THEY ARE
'500 VOLTS,
A 12- VOLT STO!ZAG:E
FIT.OM
GIVEN:
IT TAKE$
r.. =It us - L11s
r o , B'E:CAUSE
I 20 K\v'
eFFICIENCY.
THEfl.EFOR.J:::
THUS:
G-20 THE GENEIZATOIZ
THE.
II AMP
OUTPUT flAT/Nr7= 0-IG AMP: AT 'iOO VOLTS
PT"' TOTAL K\J LOAD
WILEE
f5ATTErtY. CALCULATE
=
HAS AN OUTPUT RATING OF 0.16 AMP AT
CONDITION. IT TAKES
Arl.E BAL,ANCED IN LOAD
NfUTfZ.AL CL'IlflENT
VL
LINES
"' 80 +40
Klv', 1 ,
G-ee] A DYNAMOMETER
LOA05
'10 K\v'
Klv'no = klv', 1,. + Klv_.,,
= 40 1-<hl
/IS"- VOLI
LET
I .. ~ IS THE LINE CUrtiZENT
L==
4 \(lv'_,.~ ~ -120-3Klv-
OF f'fl05. 19 DEliVEfl'S THE SAME TOTAL LOAD,
SOLUTJON:
I
Klv' 2 ; 0
Klv'
K\v'+II<> :. 80 K\v'
TO CHECK THIS :
'BUT=
2'30/115 ·VOLT Of'E!ZATION
TO EACH OTHETZ , CALCULATE THE LINE/ NEUTilAL CURilENTS.
IN =
K\v' liS= K\J"!>" + K\v'_u.,
to
~
K\V_ 11 ,
"Ol-T.S
f•eld
K\11
K\v+ 11 s- = 80 f<\v'.
}JEL'Tli.AL \v'lflL
THERE FOrtE:
Z30
+ 'TO
IS 35 \v'ATTS,
EFFICIENCY,
CONT. OF PfWB. #
RATING, AND APPLICATIO~S
l";NPVT
rz~ow == 0 .J. Sl.
AND THIS IS
HIGH VOLTAGE lv'I!JDING HAS 7 Y-z
v
35
7
FIGUflE:
2
4)
(
-
'/-z TIMES AS MANY
LOIV VOLTAGE
DELIVERS' A CU!Ul£NT
=
lls"
7'/~
AT 180 VOLTS.
+ Ffl/ICTION + \viNVAGE
TO SOURCE
'
HIGH VOLTAG!.'
180
LOSSES:
CUIZIZENT
7·S
"' 35 \v'ATTS
LET:
llp
LV "'iNPUT VOLTAGE
INPUT CUR.flENT
a) INFIJT
b) INF'lJT VOLTAGE'
CUfUI.ENT =
THE POiv'm
fZo+b
I'ZG \)IATTS
flp
= IS
IINI"UT
\JATTS
b) INPUT
f'<J "' f"O\vER OUTP'tJT
::::
\IOLTA~c=
VINPU;
==
5. 25 At>IIP.
H·Y
-71/-z.-
SHUNT ~OTOI(
G- zfj A 35- HP 250- VOLT
15/}USHES', OF 0-1'2!; OHM. THE
Ec
-7·f>
-
l:c
OF 0·73 OHM. CALCULATE
~=
.Z'f VOLTS
[ 230 + 18-T
THAT
TO
SHOULD
LIMIT
51:'
THE
PLACfD
IN flUSH
llESISTORS IN THE AUTOMATIC
OF A PLUGG!~JG
1~1
SERIES
6-
A
flESISTOil
AflMATU/t.E CURRENT TO
PLUGGED. ASSUME
Y1 -~ '250 VOLTS
R'.aH = 0-I'ZS .1L
Race
l-
2S
0·1'25 -0.7:5
J
..n...
FOfl THE MOTOIZ
OF PrWB. 24 , CALCULATE THE OH/'v11C VALUt: Or
ili:SISTOR THAT
TIMES
ITS
IVILL LIMIT THt INfWSH AilMATUFlE
IZLlTED VALUE.
SOLUTION.
LET
THAT Hie
fZp 13 = 15 THE VALUE OF DYNAMIC- BflAKING 17.ESSTOf(
l:'c
IZoa = - . - - - ilA+t.
Z 5TA
IS 80 PER. CENT OF ;HE IMP!lESSED VOLTAGE
35 HP
1·19
DYNAMIC- BllAKING
I'> TIMES ITS FULL-
GIVEN:
F'~.-~
~5]
CUfl/ZENT TO
\viTH THE ACCELERATION llESfST0R5
LOAD VALLIE AI THE INSTANT THE MOTOIL. IS
COUNTE/l EMF
rtr =
IS CONNECTED HAVE A TOTAL ILESISTANCE
THE OHMIC VALUE
-
= (o.a ) ( ?30)
= 184 VOlTS
HAS A FULL- LOAD AfllvlATUflc-
ACCELEflATING
STAftTE!l 10 \vHICH THE MACHINE
+ ;s
Cc = 80 /. OF VT
lBO
=
AND AN AIZMATVfl.e RESISTANCE , INCLUDING
OF 135 AMF
llESISTOfl
THE ARMATL'IZE ilfS/STOIL INCU'DING f5fWSHES
(\5)(135)
CUIVZI:t-JT llATING
PLUGGING
J _ ITA
VT + Cc
15 TA
~
Z4 VOLTS
PJ"' O·S AMF. X 180 V
= 90
IS THE
=
V INf'UT
OfLIYEflEIJ,
flESISTOfl
flacc = IS THE ACCElEflATIN(; IZE'>ISTOIZ
F'n<P<.!T
==
SOLUTION:
LlW?ITING
~OlUTION:
LV --= Z4 VOLTS
fZEQUI RED·
f'.,l
[{.,.
+ (os)
!HEN,
VOLTAGf 1\!INDING
TJ.lE MACHINE
a)
6-2'1-
2
CONDUCTORS AS THE LO\v VOLTA~e
AS MANY CONOUCTOilS
Of 0·5 AMI"
+ ~'"co><e.• + L ·<;
Sll'lCf HIGH VOLTA6't: "' leD VOLTS
K NIGH ~ 4 .i1.
COrtt
DYNA~OS
l'iMr'<JT = I'ZG \'IATTS
VOLTAGE OUTPUT
LOI~'
= f'OUT
f,Nr<Jt = '30\v +
LO\v VOLTA(;E IIJPUT
AS THE
APPLICATIONS OF
THUS:
GIVEN:
TIMES'
OF Pfl0f3. #
CONT.
G-25
DYNAMOTqll
HiGrl
EffiCIENCY, flATING, AND
DYNAMOS
OF
SEE FIGU/lf AI
IZ.:>cc
~
O-T5 l l
llo15=
ru~o'o·
b---------------~-
66
- 0 1'25
THJ:
NEXT F'A.;;E:
(2. <;;) (135)
CALCULATe TH!:
PLUGGING
(08)(230)
OHMIC VALUE
OF !HE
llo3
= 0-'t:i:'
OHMS
f[ESISTOfl..
,J
taz-
liiiilf:
!E ·11
--
~1
~
EFFICIENCY, 17.ATING:, AND APPLICATIONS
FIGURI:
OF Pll05.
*
OF DYNAMOS
G -25
!I
l
PART II
ALTERNATING CURRENT
MACHINES
I
.
l
,_w.~·-.--~~,,~.""""'"-_
II
!
_,.._
{,z.;
_,.,,,,._.,.,._..,__j
·--·--------------...J
69
-1
~
j
I
; !
ALTEflNAT/N/7- CVTlRENT
7-
~
A '5G- POLE AlTErZNATOI?
IS GENEIYATE0 7
GIVEN:
P~
3G POLE
S=
zoo
GENEfl.ATOilS
lS OPE11ATED AT 200 flPAIJ \vHAT FllEQUEI-JCY
SOWTJON:
f =
p )( fl.P'I\1
flf'M
\20
l'lw'o•
f=
310 X ZOO
FIZEQUI: NCY; f 7
I'ZO
f~ GO CYCLe
7-?J AT \v'HAT SPEW 5HO()LD A SIX- FVLE ALTEIZNATOIZ
DEVELOP "40 CYCLES 7
GIVEN
BE DRIVEN
TO
SOlUTION;
F~
G FOLE
f= p
f.= iO CYcLES
11-PM
X
1'20
lZEQ'[;.
= (1-0)(JW)
IZPM
5f'EED , ICPM 7
,,
G
ll.PM = 800 llPM
I
~-·~·
7-3]
A~J
FDLES
OOES IT
HAVE
It
SOLUTION:
p-= f
S"' 17(0' 5' IZP/>1
SO CYCLE'S
120
X
F'= SO x IZO
-
POLES, P=7
7-~
50 CYCLES 7
ll.f'M
fl.EQ'b.
IM
/ZPM - HOhl MANY
THE VOLTAGE IS GENE!l.ATED AT
GIVEN
f"'
m;. 5
AT A SF'EED OF
AL Tffl:NATO/l OPE17ATf5
F'=
--·-·=
CALCULATE
THE AVEIZAGE
PITCH COIL OF A 25 CYCLE
7-2 X 10 5" lv1AX\v'ELL $.
GIVEI-J.
34
171P ·G'
F'OLES
I
VOLTAGE GENERATED IN A SIX- TUIZN
ALTEI?.NATOTl IF Trlt= FLUX PET?. Pc7LE
FULLIS
SOLUJ/ON·
N= 0 TUICNS
f"' ;?<;
i!)l
'POLE
CYCLe
5
oo
7 Z X 10 MAX\v'ELLS
_t AYE
70
= "1--f Nip
.f-"V£
=("1-)(zs)(G)(nxJo") x 10-e
X
10 -lf
EAv~: = i·3Z VOLTS
lLfQ''u
AVEf(A;r
fAYE.
VOLMb'E GfNEfV.TE/7
"'" .'
71
i1
ALTERNATING- CU!lflENT
7-5] IN PJWB. 4, \v'HAT
FOtlM
IS
IS THE
SINUSOIDAL
ALTEflNATING- CUflflENT
GENErZATOR:S
EFFECTIVE
VOLTAGE
7-7] A 72-SLOT
If THE \v'AVE
STATOIC
HAS A HALF-COILED
\v'INDING . Ho\\1 MANY COILS Afl.E THt:fl.E
7
E"' IS THE
E= HI
!:-.= 1-11
effECTIVE
VOLTAGE
EavE.
4-~2
)f.
THE
IT iS A HALF COILED , TH!: COILS
FEJ< -FHASE
Cl)
HALf- COIL CD
OF PfWB. 4
ALTERNATOR
THE \\/IN DING
BET\v'EEN TERMINALS? (NEGLECT
COILS
CONNECTED
HAS
IS Y- CONNECIED, \vi-IAT
THE
SAMf
IN THE
A TOTAL OF 240 COILS
IS THE EFFECTIVE
PHASE
AND
%=
_72.:-.::.S.=.:LO.._T_ _ _ __
o/p =
1:2 COILS' I
/PHASE
b) COILS f1:IT. GflOUP
DISPLACEMENT OF SETl/ES-
b) NOTE
SOLUTION:
7p = COILS PElt.
%"'
B--..-,
Y- CONNECT!:D
THAT THEilE AIU: FOUR G/l:OUP.S
%=
LET·
A------------------1
N"' Z'W TUIINS
JS;
Z SLOT I
X 3 PHASE
I COIL
Cl) COIL-5 PE/l. FHASE'
VOLTAGE
POLE - PHASE GIWUP)
GIVEN:
AilE 7'1 SlOT A!>ID
J", 4 f'DLE
fLEl:¥'0·
7-~ IF
THR:EE -PHASE
S= 72 SLOTS
3 PHASE \v'INDING
E"'" <f.8 VOLTS
FOL!fl.-POLE
(a) PErl. PHASE: (b) PErl (;flOVP?
SIN~E , THEfl..E
GIVEN:
SOLUTION:
LET
GENEflATORS
11. COILS
4 GfWUPS
l'"t/ASC
C! = "3 COILS/
IG
IG!lOUP
COILS PER. GIWUP
REOL!IR.fD:
7-8] A
EfFECTIVE VOLTAGI: BET\vEfN
~
Tfrz:MINALS '' E"
7
1ft-SLOT STATOR
\VINI71NG. CALCULATE
lb)
SOLUTION:
Y-CONNECTION
\'IIINDING , SO
BET\vEEN
,J3
AilE
I
THE
EfFeCTIVE
rofl.
AND
evERY
f3Y
GrwuP llli:IZE
T\v'O COILS.
C
I'= 11. POLES
.:::.....----___.)
FIGUrlE
\VHOL!: COIL
1
THEN,
E.::: 4-'KfN <P X 10-
E
8
X
1:2-POLE
THilEE-l"HASE
PHASE
GfWUF'.
S= 144 SLOT
VOLTAGE
MULTIPLY
IS
COILED
NUME>ER OF COILS : (a) PEl?..
GIVE:W
IS A ~-PHASE
TERMINALS
PER'.
HAS A \\/HOLE
THE
0
X
Z
"'(4·4~) (zs)(24D) (~)( 7 2 X 10")
E.= (532) (z)
8
( I0- )
I
14'1 S'LOT
C; - - - - - - 1P
I SLOT I
X. 5 1"-HASE
/CoiL
3 I"HASE \v'INDING
lLEQ'D:
Cl "'"
IP
NUMBER. OF COILS
a) PElt l'"HASE
b)
!.) PC:R GrtOVP
THEn£
A
%=
7-9] A
1'> :1. SLOT I
/COIL
\vHOLE
%
=
G
FOIL
fER. GROUP
'1-B
1'2 f"'LES
"!- COIL~GflOUP
COIL \viNDING
~0- SLOT STATO/l HAS A THfl.EE- PHASE SIX- FDLE \vHOLE- COILED
\JINDING , EACH
CALCULATE :
COIL OF \v'HICH SPANS
(a) THe
PITCH
FltOI\1.. ,.S~QT, 1 TO SLOT 14 ·
rACT0/1: ; (bj ·~E
~
72
4f-8 COILS I
"~PHASE
COILS
SOLi.JrtON:
I
E= GG"t VOLTS
I
a) COl LS PEJL. PHASt:
73
••.11
D1Smtl3UTION rAC-TOR.
.:\..-::~,.. .~
'
Ili
I
1
AL1E'fZNATING- CI}IUlENT
ALTE/lNATING- CUflRENT GE'NER'ATOJZS
GENERATOIZS
CONT. Or PIZ0/3. # 7-10
CONT OF PI7.0B 'lF 7-9
GIVEN:
!i'LOT
S= 90 SLOT
P= G POLE:
=11-4
=8
P
I<.:J=
POLES
TUrtNs
SPAN FfWM SLOT 1 10 14
j COIL = e
Kd -=
~ = 1.8 X 10 6
rrrQ'o:
'KJ
COIL SPAN= SLOT 1 TO 1(0
(o) PITCH FACTO It "Kp"
\v'INDING = \v'HOLI: COILED,
(b) DISTR113l!TION FACTOrt ~Kclu
FIGUR.E.'
.,.
SOLUTION:
( lo)
==.
\v'HERE ·.
PHASE.
SPAN OF TltE COIL IN
11i::G!tEES
ELECTRICAL
= 1'3 , 13ECAUSt:
(Ct) THE COIL PITCH
THE SPAN
do=
IS FROM
SLOT 1 TO
P':: f!ATIO OF COIL PITCH
A\ 90 SLOT /
X 180"
Ys
=
do= - -
= SIN I:_
'2.
= SIN ~
'2
Kd'
7-lo] THE FOLLOiviNG
SLOTS = 1 <M ~
Al1ERNATOR: :
COIL= 10 ~
4>
INFOR:MATION
o::
\v'HOLE- COILED
IB x 10 6
:
SLOTh
POI..f-PHASE
(!> X I"Z'/t
;- ')( stn (it'/-z)
-
riG:
=
I'H- COILS
TUrtNS PER
= STAll •
c 6ET\vEEN
V OL.TAG
ii"HASE
AN
5LOT I
lf"OLE
r
•
IS"
=18
=
74
tA
48
l=jT
= G TUitNS I
!COIL
- 4-8 X (0 = 288
...!!!..___
8
x 18o
lc;o'
Kp = Stn - 2
n:rrMINALS
=
3 PHASC
n == -46
8-
= 900 i
f'Olt \v'YE. -CONNECTION
EjT = ~ ~
II"HASE
\viTH
orz:. \v'YE CONNECTI':'D
A lv'HOLE COIL \v'INCJING
COIL!
COIL SPAN = SLOTS 1 TO 160 ; \v'INDING =
THil.EE- PHASE' ~ \V!NDING CONNECTION
T
:5- f'HAS'E ALTEil.NATOil, STArt
SINCE 1HERE AltE 144 SLOTS , THEK[
THlJS·
'
0·957
IN CONNfCTION
CALCL1LATE •. (a) 11-ff VOLTAGE' PEfl. PHASE ; (_b) THE
8ET\v'EEN TE'Itlv11N4LS
IZO
rorz
Sin
POLeS -= 8 ~ llPM
VOLTAGE
110
TURN/
IS GIVEN
b)
{:= 8X900
o.r;n,;;
o-978
1275 VOLTS
ltPM
X
f= GO CYCLE
= o.t;
X 10-&
3tTIIIEEN Tt"IZMINALS
\IfiLL Oc A TOTAL OF lt't COILS
JG =
Kp: Sll\1 78.
Kr ==
-f= p
•
-s
n=- G. ')(3 -
THEN,
to'h
0.9G"<S
A
tfr == VOLTAGE
= 1'2 ISETIIIEEN SLOTS
90
1'3
•
P"==- X 180 ·= 1SG
15
l<p
13ET\IIEEN ADJACENT
IS"
cO
G Sin
=
EI<P = VOLTAGE PEI'l I"HASE
LET:
90 =- I!>" SLOTS I
~
/POLE
u~o·
go
~ET\v'EEN TEilMINALS
SOLUTION:
PER. POLE
6 POLE
Y s = - = 15'
------=--:---
r4 =-
b) VOLTAGe
SLOTS.
14
(nd'/'2.)
n x srnd'h
sin (6 x ro'/z)
CALCULAT!: THE VOLTAGE PBt PHASe
a)
WUM13ert OF ELI::CTRICAL
IJ!::Gtz.EI:S
•
e!)% = 'T·'T"l f Nt Kr K.l x 10· 8
Ef<r =- Ef.-t-'t[(;0)(1~8)(1.e XIO,;)
ltt'Glb:
\vHEQE:
n = NUM13ER OF SLOTS PErt
2
3 PHASE, STAR CONNJ:CTION
= 10
(o.%G)(O.%'G')
SIN (nd'/2)
n x s1n (d/z)
Kp =SIN-
-p'=.
KJ
90 SLOTS, G pOLES , SPAN 1 TO 1-t
18
SIN
SPEE'D = '300 RPM
\v'HOLE- COILED \viNDING
l
1130"
d"== -
GIVEN:
•
==
= 18
1so
•
=0· 9GG
75
= .)5 ( IZ7S')
=
21.08 VOLTS
II
I
\ ~
Al TE"f?NATING- CU/lRt:NT
ALTE:IZNATING- CUilllENT GENEilATOilS
7-1'2]VHAT
THE ALTEflNATOf(
SIX -
\v'OIJLO
VOLTAGE
Of
MW5.
10
If THE lviNDIN(; IS CONNECfEO
SLOT/
GIVEN:
;rou:
f
rtf'M=- 900
TUil.NS/cOIL
•
24-
= 24
T11"4
\JI-IOLE -COILED, 3- PHASE
Er =
180.
)<
= 11'2·5
\Uil.N/
:=
I FHAS!::
SIN
THEILE Al"l.E 144 COILS 15ECALISE
!He NUMBt::IZ Of SLOT IS 1"14
n
f
7-13] A
3
SIN
0
6
:1
¥
10-
1
8
'
]
X
ALTE11.NAIO/l HAS A IUITING OF 5,000 KVA AT !~,1.00
LINt" CUflflENT.
IL =~
5- rHASE ALTEflNAiOil
KY.A x.f3
I<VA = r;,ooo
==- 13,200 V=-
r;.z
'
!Lt::Gp:
FULL-LOAD CURil.EN\
l1.. =
KV
..
r;,ooo
lL ·
76
;:-oR. >-fHASr
~
DfWP
IS
i
I
j
wd"
itESISTANCE
DELIVERS A UNITY
SYNCHilONOUS- ll.EACTANCE
VOLfAGE
DTLOP.
SOLUTION:
Lf:T:
'% = TE/lM!N4L
FF =UNITY
Vr = 1!30 VOLTS
VOLTAGE
F"efl.
f'HAst:
REACTANCE VOLTAGE
OflOF'
.!:9 = G:ENEJ?ATI:.D VOLTAGE Of" AN
TXt..= t;,O VOLTS /Ff.JASE
VOLTAGE
X 100
PEl< f'"HASf / C4LCULATE THE FEll CENT
3- PHASE, Y-CONNECTED ALTEitNAT.OI?
CALCU LATf
<f-'"0
"TGO
= IG. 3 "j.
ALTE17.NATOI1.
AT Z30 VOLTS · IF THE
GO VOLTS
NeGLECTING' WE
At..n:g.NATorz
':/(j> =
fr.E~ULA TION
PE IZ.CENT
NEGLECTING' THE
JWA
~ X I3·Z
IL == ZJB -7
.. 7
POiv'Eil- FACTOIL LOAD
Y- CONNECTED
rlEQ~D:
SOLUTION:
&!VEN:
Y. V R.
GIVE I-I:
I
= 'T> CYCLE
IH17.EE- PHASE
rtEGULATION,
I 'tZC VOL.TS
1~0
JA
VOLTAGE
IO~e]
(. X 900
7- 1r;;
I;~~-
j. yR.
l
8 SIN 7·S/"2
•!
)( 100 i·
VF.L
PEfZ.CEt.JT VOLTAGE rtfGULATION
J<0·9S:~
TI·Hl.EE- F'HASE
AT
PE11.Cf.NT
THE
VF-L.
VN·L· -
'1o v. rz:. =
[l.EQ'D:
1:4 = ~['f.<f1"X'tS"X288XIBXIO'x0.831!;
=
460 VOLTS
SOLUTION:
VF.L-· =- 4GO VOLTS
/2
'JT= -f3["t·'MfN~ KpK.:I
o/,-
rfWM
AT NO LOAD· CALCULATE
GIVEN:
/ FtiASf
KJ. = 0·9S"G
VOLIS · CALCULATE THE f'ULL -LOAD
l:'T
>< 48-GefL/
SIN (8 X 7·S /t)
Kd=
\<M
= - = 48
110
535" VOLTS
RISES'
fl.EGULATION.
(nxd"jz)
P X rtPM
f=
TO
FULL-LOAD
.G&tt.
Kcl =-
X 13·2 KV
192 AMP
IL =
VOLTAGE OF AN ALTE/lNATOfZ
7-ls] THE
'288 TU~NS /PHASE
tJ
"r._''
K\v
f3
0-82 X
VN l. = 535 VOLTS
SOLUTION:
?JJ;ao
h=
5. 1 1"2
1-
="
G TURNS
n·
n "' --=---~,...---<$ fHASE)(CD ~OLE)
34'
ll'l· (;;
I:.z
E! ==?
14"~ SLOTS
13.2 VOLT ,
LINE CUICR.ENT
Z<j-
IZEQUIR.t:O:
K\v
TL = - - P'·F x Er >< .J3
1
d' ==~ =- ?·r;
VOLTAGE BET\v'E"EN TERMINALS
/FHASt:
SOLUTION:
rtEG D:
Kp:::- 0-8315
POLE
COIL!
1
0·8?.
p.f =
6
Kp = s1n
·\VIND!NG CONNECTION
f=
FOrt
DELIVERS A LOAD OF 3,GOO K\v
CALCULATE TI-lE LINt CUfZ17.ENT
LOAD:-:: 3,000 K\v
II-IUS;
COIL SFAN ~ S\.Oi 1 TO lG
=G
=
IS"
==
r'
=G
l·'t X lOG f\1\A)(.
=
n :o
AT A PO\vEIZ FACTOR OF 082
GIVEN:
fllOM I'TWB. 'It 10
S = 144- SLOT
f'
7-1'1] IF THE ALTEIZ.NATOIL OF PILOB. 13
IN
Of'EfV.TION ~
rou:
\P
131:T\vEEN TERMINALS
13E GENEIZATED
GE'NEflATOIZS
rz.
h ..
"iS
= 1~<'-79 VOL.T.S
~<>{'·"
fl.ESIST.4NCE
17TZ.OF . '' '/-
:2"50
I
~ IX"L
7
p<v
I
v/<j>
Zl9 AII\F.
77
=60 WLTS
J
= I.Ja.. 7'} VOLTS
~~
ALTEitNATING- CUfUlE"NT GENERATOIZS
CONT. OF F'Tl.0/5 #
ALTERNATING -CUTlllEIIIT
7-16
CONT OF F'fWI5. # 7- 18
r;-:----
,Jq_~·:n'3 +0) 2
E '3
1::.':1 =
Flfl.ST, DETERMINE
+(Go) ..
THE
OPEt-J-Ct!lCUIT VOLTAGE
t:'Ac/.
1"1"5"· 71.. VOLTS
1{5
."f. fC.EGULATlON -== E'<J- V<t>
FIG· 166. Of'EN-Cif1.CUIT__/ SHORT- Ct/UUlT
CVINeS' FOil AN ALTE!lNATOJZ
SOLUTION:
fq=~lY/cj> +IllY +ITXL?
">< 1001.
f,~~u
/{;
\14>
1-R"" 14-S"-72.' -I~Z.7'3
I?Z -7')
=
t>
--
1
-J5
~
!=i
tliT.OP
VOLTAGE
he=
PER. F'HASE IN FfW6. IG
IS
VOLTS"
!::!
t-!-
192' AMf (,if'FfWXI WI A TEL Y)
OF'EN CIRCUIT VOLTAGe
FACTOR
llf"WF 15
L
AT UNITy PO\vER
If THE
fZ.ESISTANCE
Zs =
VOLTAGE
r; VOLTS
I
TR
I
PftASOR.. DIAGRAM
+(!x'-Y
f<J = /Q3Z.79+ G") -z.
Zs
~(GO):
:~ R=- E9 - Yo$
ftA = 0
E"J = I 5"0 VOLTS
;t. f7.. ==
VOLT
Zs
TH/!EE- PI+ASE
AND
"TERMINALS
Xs. THE D-C 11ES'ISTANCE f5ETw'EeN
REQUirtED:
= 100
Ec.c =· '2;ZOO VOLT, Ti::lt.MINAL VOLTAGE
3 ~ f't+A S
c
A LTEIUMTOIZ
SYNC't"lf7.0NOU5 IMP""E17ANCE
_?""'«
!ZeACTANCE
Zs
D-C RESISTANCE , it-~.= 0.9 SL
Xs~
,.,
I
_,
/1,."<
700
<:::>
SOO
G
::>
fiJO !::
,;. IV
~
'100 ~
300 2
JU
200 lL
100
so 35 '10 ""'tS
2
+ x,"
z,"
-llAz
C>
SO SS 60 GS 70
J (G.G'I)"-(0.4s)"'"
Xs "' r;.c,
-+ Xs =?'
f'Ef(.CENT
"THE
OELIVEil.S
FACTOit OF
o.,;
liS
flEGULATION
fULL- LOAD
.fl..
GIVEN'
lv'HE"f./ THE
lLEOUlflEC7:
FltOM Ml015 · # 18
l<VA = 400
= 2200
VOLTS
Zs"' G.GI JL
CALCULATE
THE PERCENT REGULATION
SOLUTION:
l-J:T:
il.l"k:
IS f'ffl..CfNT lt.EGULATION
lt..= -FULL-LOAD LINe CVfW.E.I.JT
Xs=b"·b".ll..
h= VOLTAGE DJWP ACROSS
fZA=O·'I-CO.Il
EACj~ =
FOil PTW5 18
KILOVOLT- .4Mf'Erzc5 AT A FV\VEIZ
LAGGING.
THE
1'270 VOLTS
IZ-ESI ST 4NCE
Ex"' VOLTAGE. UfWf' ACROSS'
P·l" = o.r:; LAGGING
~~
900 1U
800 ~
~-·-
J ILA
Xs-= ~
~
2
ALTERNATOR
EAc
KVA
f
SYNCHil.ONOUS IMPEDANCE
IS 0-9 OHM.
GIVEN:
I
I
-4c; .1L
7-1~ CALCL'LATI:
OF FIG. 166 AilE fOI( A "100 -KVA , 2,2CXJ-
REACTANCE
~
X IOO
ALTEltNATOr<: , CALCULATE "THE.
THE SYNCHRONOUS
"'
~
131'·79
~--------------------------------------------------~
ASSUMING THAT TI-lE CURVES
1000
1!;0- 13?.79
;1.11.: 1£'.9/.
7-1~
1100
I
,...__.
v ~-
Zs:
IS EQUAL TO;
"2
12c»
J.
1 1/
S" 10 1$ 20 25
IIi!: RE"SISTA NCE rt:rt PHASe
x too/.
V<j>
1300
SINCE:
G.G'I OI-IMS"
=
-
,...,. ~
v
RA"' Rt
15"00
1100
flELV AMPERES' -If
192 A
=
I-'"
1...-c
.
9:J\J
v<P
'f5:r=fv4' +Err.)-...
TXL
jkl"'"
./
r- L~
'.f
J
_L
0
!'3c
1"270
SOLUTION:
~r/
0
v
l
200 - - - .•~.
!GO
_<n.r
n"
cl'""
:I:
80
V)
'10
~
LeY
1.::>~
::; 'f0
u 2ZO
Zs=---------~-----
VOLTS , CALCULATE THE F'Ctz.CENT flEGULATJON.
(:, I!EGULATION
1,'<.
-11:{
.;,
<:..">
1- 2
"THcR.EFOrtE:
7-17] IF THE. Rt:SISTAIJCE
§;
=:,
L.>
i{/(=9·74"/-
.(~tz
1-
= 1?70 VOLTS
= I:ZlO
tA</~
~~
.!:
Z'ZOO
THt:N r!WM riG. ""' AT
X /00
GENEflATOilS
"Tt!E.
1-9
rtEACTANCE
i
ALTfJ[NATING -CUrl/lENT GENEftATO!lS
ALTERNATING - CUfl.fl.ENT GENETlATO!ZS
CONT. OF PIWB #' 7-19
7-
SOLUTION:
f'HASOR
KVA
AS
h :=.-=--.[3 x EAc
400,000-VA
I l.
OETER.MINI: THE
•.
ANGLE
ER" lL R,. = lOS" X 0·4!1
C.:,=I0&Z+47)7.
LR = 47 VOLTS
1'270 X Q.(;
OF O·S
=7G2 VOLTS
'7- R
l':j-
il!llJUi
VOLTAGE BETivEfN STAWI? TEillv11NALS
lli~l!
== 1,300 VOLTS
~ fl.EGULAT ION
/OC(.
):<
CIRCUIT
=
FtlASOit DIAGRAM USING I AS
X
1270
tOO
ltEFErlENCE PHASOR
I 1.. = 1000 .ooo A1A
'1-
ri = "\-8.9(.
!1..=
,.,.
Of
fTWB. 19 FOil A
Clt.
2
2
2
Jl
I
Es = G93 VOLTS
= J z s flA"'
2
Xc;. :: J( 2.9'3)z_ I
f'·f =- O·B LEADING
Xs = 2.82
#" 19
voLrs
CR
FHASOit
{} = 3G.tJ1.
-v- x
"t R =- IG' ·It
lOGS" VOL lS
Cc:J == 1999 VOLTS
E'} -V·
o/. fl.=
1999 - 1"528
'/. rt =
Est>= r~ Xs
=(25 I) (z.8:2)
~ 710 VOLTS
X 100
"/.
FACTO/l
I
)<
100 /.
v~
11-J A LEADING
;i1·
+E,)"·
f~=1~328)(0·9)+'2SI]"l + [(l328)(0.G) 1- 710] "l
r.n.=
IL 17... t
Eo;._-= 708
/. R = i06'S- 1270
l'Z70
(7'"Z-G95) 2
E~ IS LESS TtiAN V
=
{3
E'3==kvcos~ + ER) 2 +(Vsm~
.fl..
ER = ~5"1 VOLTS
100
2500 = 1328 VOLTS
-
fill = ('2;1) (i)
7(;Z VOLTS
~ =JQOJG + 47)-.: +
FV\vEfl.
FOrt LEADING
FACTOR
%R= !:c:~-V
1"270 X 0·8 = IOIG' VOLTS
=
t71A~rlAM
P'O\vEfl
cos-• o .a
Vw 1170 X a.;;
2
Xe.
SOLUTION:
E:J"'
f3 X ZSI
VOLTS
EAC/f3 = t'Z70
Vx=
::l..n_
= '2 .99 .Q
Zsq = l:SOO
PO\v'ffl FACTOR
LcA17iNG'.
PTWB·
v, =
:2SI AMr. •·
"A• = : - : : : : 7-20] REPEAT THE: SOLUTION
=?'
SOlUTION:
Vs ~ IOIG" VOLTS
fT =
AT A
LAt;G/NG".
.[3 ( ?,?JJO V)
Ys= J6z10)~ -(7G?)?.
'+7
IS Z OHMS.
ALTEil.NATOfl
1000 KVA , 2300 VOLT, "3 -PHASE ALT.
+ (101{;+69!.)"
fof'£fJ
~~~~~~
GIVfr-J:
(Vs+Es)z
1590 ·8- I 'Z70
V5 "' { Y"- Yr< 2 -
l:.R=
STATOt< IE17.MINALS
fl.EGULATION OF TfiE
f~PE.N CI~C(.(IT
orr.
OF O·B
1,300. !HE EFFECTIVE
I!EQ'o·.
i- R =.
Vs = 1'270 x O·S = IOIG VOLTS
GIVf"N: fTWM
.8i:T\vEEN
IS
17.t "' 'Z Jl, ffft:CTIVE Jlt:SISTANCE
Es: G93 VOLTS
'0
SHOrcr:-
t:j"' 1890. ~ VOLTS
f.,,. l1..XL "'- 10!; ')( {;.(;
VR
--I
'- i
Cq= { (VR+EI< ) 7 +
STATOR TfltMtNALS
F'Eil CENT
1"0\VEfl. FACTO!!
cos-' o.G
53· rs. P·F.
=
6'ET\vEEN
(COrz.rtECTED) ltESISTANCE'
2?00.¥--
lOS" A~P.
:::
fJ. =
{t
X
2, 300- VOLT Al TEI7.NATOR IS
1.000- KVA
At-JD IS OPERATED AT
THE VOLTAGE
Es ~tx,
h=---,[3
A THflEE -PHASE'
llATED SffED 10 PTl.ODUCE 11.4TED
CVIZJZ.ENT. \IIITH TI1E SHOI7.T CIRCUIT /Z.Eiv10VEIJ ANO THE SAME EXCITATION,
PIAE>RAM USING I
IUofEI1ENCE fHASOrt.
TI-lE
2D
CI!(CtJITED
~1
1"529
;-o . s :/.
X 100
!ILTLI(NATING- CUI?flENT GENEilATDrZS
l ccj ft:tPEAT
ALTEflNAT/NG - CIHUZENT GENERATOilS
FOrL A FO\v'EfZ FACTO![ Or 0-8 LEADING.
PfWB. 21
CONT. OF Pfl03 #
SOLUTION:
fc;
C<:J=~+Ep.]"' +[VSih~-Es]-z
Ecj "' 1'311-
1. R =
A1 O·B LAGG!I.JG
2
f'OiliE'It t'ACTOI'l
3'1"£16 VOLTS
/'.fl.
I
DIAGRAM
!
Ec; =
VOLTS
Ec;
=
x 100 ;1.
E9 - V,p
f'HASO!t
+ E,;j 2 + [ VSINtl +f.: ] 2
rG =}kzGr;;r;)(o·a)+ islr+[(zG%)(o·G)i-loo?]
=)Urs2B)(o.s) + -z;1f + [Crszs)(o.G)-7s-ot
E':l
= j lV COS{).
7-23
l
I
I
-Yo~>
X 100
V.p
34bG -Z(i%
if.IL-=
Vt;
X
100
2G \>G
1314 -13ZB
;I.R=
X
1'528
100
~f7..=
. .v ,....
v
:1.1L=-11.
PHASOI'l
._./
DIAGRAM USING V
AS' THE
7-2-8 A 25- KVA
I'CEFEflENCE PHASOR.
11 OELIVErr.S
AND, OPEI[ATING
4,GOO-VOLT
GiVE
17.ATED
THflEE-PHASE ALTEflNATOfl
AT RATED Sf'EEP, "THE FffLD EXCITATION
A17.MATURE- \v'INDING CUIUlENT
IS SHOIZT-CI!ZCUITED,
IS ADJUSTED TO
THE SHO/ZT- Cti7.CUIT
IS THEN
tlEMOVED, AND, \\liTH THE SAME FIELD CUJZ/ZENT AND SPE'E.D , THE OPEN-CIIl.CUIT
VOLTAGE
5ET\v'EEN
EFFECTIVE · (A-C) llES'ISTANCE
1·2 OHMS.
CALCULATE
A PO\v'EIL FACTO![
IS FOUND TO /3E
STATOR: TEilMINALS
THE
PE/l F'liASE OF THE AIZMATUI'l.E
Pt:JLCENT
fl.EGOLATION
IL = --11;=3'...--,(4-'--,(;-00->f)
VOLTAGE 5ET\v'tEN STATOR.
fL
TERMINALS= 174-'t VOLT
o. 8
fElL CENT
,JS
Vq
flE(;U LATION ~
?
SOLUTION:
1744
~~*"'--···,_.._,..,
=
zr;<;G VOLTS
eR
=
f:~
= ItS.!; x 1·2.
.ER =
l:'s"'=-=:
PER. CENT
=
PoN
~ f',.oT
f"oN
=
2/)00 \v'ATTS
+
FLoo;SES
\'J,000 + 2,000
f' 1""" '21,000 \v'AHS
P0\1/ER rACTOil.= 0.7()
•I
IS
OF
EFFICIENCY.
Z 5 ~ KVA , ALTERNATDR
~oss
\vHEN
1•
llEGUifl.EV:
PouT
•
EFF."" - - X !00 1·1
r, ...
/'. EffiCIENCY=?
1'3,000
Po !AT
=
•/
~ Eff "' - - X 100 ;.
21,000
2 '5,000 X 0· 7G
% fFF =
f'our = 13000 \v'ATTS
90· 47;/.
= ' 1'2!:7 · c; AMP.
4600
V.p= - -
Lt\tGING
A TOTAL LOSS' OF Z,OOO lv'ATTS
GIVEN•
7-zs] A !>00 -KVA
I· '2 fl.
tU:QUifiED •
t: sf
\v'INOIN.G
1,000,0()0 -YA
P· F :
ITS
SOLUTION:
GIVEN:
=:
HAS
KILOVOLT- AMPERES TO A LOAD AT A i'VIv'Eit !".ACTOR.
THE
OF THE ALTEIZ.N.ATOR AT
OF 0·8 LAGGING.
1000-KVA, '10()()-VOLT,34> ALTEICNATO/l
ITA¢
1,74-t.
ALTI.:I?NATOR:
RATED
0· 7G. CALCULAIE
7-23] A 1,000 KVA
1
30/.
e"
I~R,.<t
IS'/ VOLTS
OUTPUT
ALTEilNATOit OPERATES AT I"ULL-LOAV
AT AN EFFICIENCY
OF 93.!;"
f'ER.CEN1.
KILOVOLT- AMF't=l'l.E.
IF THE LOAD PO\vEil. I"ACTOfl.
IS 0· 83, CALCULATE THE 1DTAL LOSS.
GIVEN'
i;OO KYA ALTER.NATOR
j'. EfF :-. 93. s •;.
rf-= 0·83
IZ-EQ'p .
-J3
TOTAL LOSSES= 1
1007 VOLTS
SOLUTION:
f<\v: 120
k\v; OUTf'UT
f,.., = F'ouT
T:ff
Po>~
=
Fl.bss~>s
=
FOivEIZ
<T20. :X 1000
\v
O·%S"
449,198 \v'ATTS
=- +T9 "'"- -tzo 1<.\v
f'Los>tes =
2c;J K\v ; IOTAL LOSSES
K\v' = 500 KVA X0·85
-
I
0,~
<>·~
i\3
J
ijl.'
=~
l:i··
:i
[
ALTEf(NATING- CUflfltNT GENEitATOn:S
7-26] THE.: D-C
A
750 !<VA
AIIMATL'f(f
\viNDING rtESISTANCE:
4:+00 VOLT THREE -PHASE
THE COPPErl LOSS
AL TEflNATJNG - CUrmENT GtNEflATO/lS
B'ET\v'EeN TEI1MINALS Of
ALTEilNATOfl
IS 0-9 OHM. CALCULATE
FULL LOAD.
IN THE lv/NDING AT
flA~
AIZE P'AJ7:ALLELED ~ (0) lv'HEN THE
NATORS
2
ALTEflNATOR
3'- FHASE
AT1.E: ,IIESf'ECTIVELY,
AIZE OPERATING
fl.A4> ~ 0-4~ _Q/PIIASE
13UT AilE
0-9 SL
2
COF"f'E!'L LOS'S = 1L ItA (3)
Cu.
COPFEIL LOSS IN THE \v'INDING
AT FULL-LOAD=?
LOSS'=
cu ·LOSS
(98·-tY'{o-<ts;)(3)
= 13,071 \-/ArTS
~
SOLUTION
13-1 K\v
FULL LOAD CUf7JlENT
DISPLACED 'SO" F!WM A f'OS'ITION OF FHAS'E Of'POSJTJON ; (b) lvHEN
OT\'lEfl. IS 200 VOLTS ;
(c) \vHEJ.J THE VOLTAGES AilE Z~O AND '200 , AND
DISPLACED 30" FJWM A POSITIOf-.1 OF f"HASE OPPOSITION'·
GIVEN:
T\110 SINGLE
Tt=o.oz;
(lo) V, = 230 /~ Voz= 200, BUT IN
I"'HASE AL TEIZNATOtzS
I"HASJ: OFF'OSITION
_o._
Xs, O·OG SL
IL~----.,f-3 X 4-, "100 V
'ZOOV
220- VOLT THREE- f"HME
I<ILOVOLT- AMPEilES
rLESIS'TANCE
FIELD
Zs
IS 0-18 OHM. THE
\IIIND/NG TERMINALS
9.3 AMI' AT Ill; VOLTS.
IF FRICTION
=
J lts
=
~ (0-025') 2 +(0·06) 'Z
AND \,III.JVAb'E LOSS lS
"'1'60 \v'ATTS AND THE CORE LOSS IS 610 \'IIATTS , CALCULATE Tf+E FEfl CENT
w. Loss "
25- KVA , 220-VOLT , :5-I'HASE ALT.
f'Oiv'Eft
( o) EMFS = 230,
FACTOIL; 0·84-
CIJ. LOSS
9·3 AMP
AT liS' VOLTS
=
G/0 \VATTS
=
X
Vf
'3·3 )( Ill>
FIELD LOSS' = 1070 \vATTS
TOTAL LOSSES
·;. EFFICIENCY= 1
•
1•
SOLUTION:
1
25,000 'riA
lv= - = - - -
.[3 )( 2.20 v
= 3302
\vATTS
~
::!__=SIN
2
;. nr = 86. t
;.
l~"(z;o)
v = z (z-so)(o.zs9)
V= 11'3 VOLTS
Ic
EFF = ' - - - - ' c - - - -
+ ?i30Z
'230V
TO DETERMINE V , USE COSINE LA\v'
lc
vz=
v
V
2
2
2
200 +230
= 40000
= 13,1.2tQ
2
'I,
-
2.[-zooxz;o] cosw·
+ 52'300
- 7967+
VOLTS
V=J 13;22t:;
V = lit; VOLTS
lc = .::!__
v
'22s
1. i!s
1)9 v
= -;---;----(0-065) Z ALTEII.NATORS
lc = 915'- 4
11:,1
I
Y,
V• = v2
(.zs.ooo)(o-a+) x IOO'l·
(1S,000)(0·8f)
PHASE Of"F'OSITION.
OF
BUT DISFUCE!7
-1
TOTAL LOSSeS= llG'2+1070+ "f6o +GIO
11EO/lJ:
(c) V,='Z30/V-z. ='2.00, BUT AILE
OIS'F"LACED 5o• rfWM A fOSITION
'. /
IIGZ \VATTS
X '251 AMP:
OPPOSITION
r L ...11A
FIELD LOSS = If
FJIICTIO~_...y\viNDAGE' LOSS= 1'60 \vATTS
COrlE LOSS
=
Tc = 230-6
30• fi10M A POSITION Or PHASE
""(3XGs-6'):z(o.o9)
fl.A = 0·18 .0..
If=-
3
(Z)(O·OGS)
Zs = O·OGr; SL
EFFICIENCY.
GIVEN:
+ Xs
Z3o-zoo
=
2
DELIVERS f7ATED
AT A PO\vEfl fACTOR Or 0·84. THE EFFECTIVE A-C
BCT\vEEN AftMATUil.E
TAKES
ALTEIZNATOfl
Ic.
2
30Vf
-z30V
~
SOLUTION:
95-4 AMP
Z~- KVA
A
IF THf: ALTER-
EMFS Al'lE EQUAL AT 230 VOLTS,
7!;0,000 VA
z~zfl
LOAD ,
THE VOLTAGES ArtE IN PHASE OFFOSITION, 3UT ONE VOLTAGE IS 230 AND THE ·
rz.e:Q/o:
lL"'
\viTHOUT
CALCULATE THI: CI11CULATING Ct)fU'LENT IN THE \VINDING'S
i?At = ~
,4100 VOLTS ,
HAS AN AJ'lMATUrzE \v'!NDING
AND SYNCHIWNOUS !IEACTANCE
0-0?S AND O·OG OHM . IF THE MACHINeS
GIVEI-J~
7~0-KVA
7-28] EACH OF T\vO SiNGLE-PHASE ALTE/lNATORS'
\¥HOSE flESISTANCE
AMP.
Ic=~
z Y.0-06!;'
Ic = 885' AMF'.
IL= G5·G AMP
flAI
/fHAS'E
..
~
.______________ _
='
zf7.o. -0-187 =
=0·09 .f.l
B4
85
:~·~' I
J
ul
ALTETZNATING- CUlli7ENT GENEllATOIZS
TI?AN5"FOITMms
7- '29] USING' THE GIVEN DATA OF PROI3- '2.8 , CALCULATE" 11-IE ANGLE
B'ET\vEEN T\v'O EQUAL
VOLTAGES
1.000 AMP \v'Ht:N THE ALTErzNATOrtS
CUiliZENT IS
8-1] A
OF 230 IF THe CIILCULATING
2;500/2"50- VOLT
1,100 TUflNS
AflE F'AIZALLI::Ll::D.
ON
THE HIGH SIDE.
OF THE lllON IS
GJVEI-I:
T\1/0 EQUAL VOLTAGES
V= 2 (t?>o) sm~
f7
6L'T:
v
IN THE COrtE
S"ECONC7AIZY
IN LINES
¢m ;
I
f'Ert SQUAIZ.f
TU/7.NS.
=-2 (Z30)
=
Sin-•
•
VOLT
T
1-Jp= I'ZOO TUIINS
"'" - '2. ('230)
fi "'
2 ( 1~-<t .)
,8 "'
5~-8· OUT OF fHASE·
l:s
Rf:QUIIlE/7:
o. 2B3
o) TOTAL fLUX (jl,
b) MAX
SECON!7AflY
FLUX DENSITY ii-J THE
J=IG[)ftf:."
b) t=LUX DENSITY ,8'"' fu
A
SOLUTION:
o.) Slt-.ICE THE MUTUAL FLUX
THE
PIIIMAIZ.Y
7l':l,470 MAX
IN
~ = __
9_l_N_•_
lv'INDINGS' AN[7
yB"" 79,94-IZ 80,000 MAX/JNz
IIIIND/NGS AflE THE
SECONOAR.Y
SAME,
c) SINCE·
Er= 4 ·'i'T
1-Jp .J."!Prn x /0-
8
!!L~..!:t
VOLTS'
Es
Ns
!Pm""
Ep
-
"'f·H IJpf. xJO
Ns =
8
NpEs
l':p
'2300
l·
~..,- (<~·+t)(i'ZW)(Go) no-•
}
dJ,. =
l
I
WINDING
COIZ..E IN LINES' PER SQlJAilf INCH
c) SE.CONI7AilY TlJIZNS Ns
fi-= '2. e
130 VOLTS
= 2.300 jt30
CI10S'S'-SECTIONAL Artf:A,.. 9 SQ. IN.
130
-e- = 1e.-r •
X 2:Zs
V = \000 >< 2 (O·OG!>).
V=
OF
FLUX
.f = GO CYCLE
FIGUJZ.E:
srn-e
FIWM THE f'lGUfl.E
Ic
E
Yo=Z50V
SOLUTIOtJ:
V=
CflOSS- SECTIONAL Ar~EA
IF THE NET
GIVEN·
l!EQ'CI:
ANGLE 15ET\IIEEN
DeNSITY
HAS
Tf?.ANSFORMEfl
SQ lN. , CALCULATE: (a) THe lDTAL
9
(b) THE MAXIMUM I=LUX
INCH; (c) THE f..IUM13Ert
Tc =1000 AMr.
Vo-+ V-z-= 2"ll0 VOLTS
OISTfZII3UTION
60 -CYCLE
N~ "'(1.200 )(z30J
'Z~OO
71">,470 Z 7-Z Xld' MAXIvi:LLS
Ns = 110 TUiliJS
f
·u
I
B-z] A corzE- TYPe: TRANsForzMm , FIG. rM _ 1s cONS"TIWCTED \v'ITH 0014IN.
THE
THICK
LAMINATIONS
MAXIMUM
FLUX
G· 2 X 10 s- MAX\v'ELLS
TttE
Alll Sf"ACES
HAVING A LINIFOflM
ANO
FLUX
DENSITY
\viDTH OF 2-?r; IN . IF
ArlE , IZESF'ECTIVELY,
AND 8.2 X 10 5 MAX\v'ELLS
f5ET\vEEN
LAMINATIONS
THE STACI<ED COflE , CALCULATE , THE NUMffEfZ.
THE
8()
f"Erl. SGUAilE INCH AND
OCWf"Y 8 PEIL CENT OF
OF
LAMINATIONS I!J
TIZ.ANSFOflMEil.
87
Jj
=I
TRANSFORMERS
TRANSFORMERS
CONT OF
6-t] CALCULATE
8-2
f'fW13' · #
A
OF
TeD
THE FULL LOAD PRIMARY
!;-KVA
2,400/120-
AND SECO),JDARY
CURREI-JTS
VOLT TllAI-JSFOilMEn.
STEEL COJtE
GIVEN'
f"'tiMARY COIL:
2/'fOO-VOLT
5-KVA
KVA
TrlANSFOJZt.1eR
Es
Is=
t:;OOOVA
=
110-'<1
rltQUIRED:
10 LOAD
Is= 41-67 AMP.
LOAD f"ll!MA/Z.Y /r/'v
FULL-
SECONDATZY
orr
CU/ZITENTS
Is
-=0..
Ip
SOLUTION:
!<VA
Ir = - t: r
COIL
FLUX
MUTUAL
Is= CLfr
S,OOO:VA
"' -'--2/too
Ir =
2-08 AMI':
s] THE
VOLTS
= 20 X 208
v-
Is= "ti-G AMF
FIG. IG9 : SIMf'LE TllANSfOI{fv!Eil !71AGILAM
GIVEN·
\-.!= 2-7'5 IN
rVm: G" 2 X 10 s- MAX.
f-lO· OF LAMINATIONS'= W
PER TUfZN
OF A 1>0- CYCLE
TRANSFORMER
IS
AND SECON17ARY
TUIT.NS ; (b) THt: FLUX
1-75"..l-f1.
VOLTS/TUJZN = G-68
0-014 )l<r.
~EQ'J7:
o.) NUMI5Eil OF LAMINATIONS IN
r;o- CYCLe
f =
2066" ?6 2100 TUJZNS
I
8-3] A 25"- CYCLE
TTt4NSf'OIZ.M!:"R
SECONDARY 1Ur?NS
IF THE MAXIMUM
HAS 2,250
PIVMAflY TURNS AND 11;0
VALUE OF THE
MUTUAL J=W)(
2-G4 x 10 6 ~AX\1/ELLS' , CALCULATE : (a) THE F''fZJMARY t4ND
VOLTAGES ;
(!o) THE RATIO
t::r ~
1£; CYCLE
Es
4f;O
=Gi;;8
&;.GB
X 10'" M4X.
rteQ'[7:
!:s
= "1-40
I.,)
ct
PRI·/'.,..Y SEC· INDUCED VOLTAGES
=
,
b) RATIO OF TRANSFOIZMATION
SOLUTION:
a) Er
8
='tt· 'l'f)(lc;o)(z5)(z.fi.ot x 10~) ( ,o-~)
Hs= IS"O TIJflNS
et)
X 10-
Q=
=
"t 'T'T Nr f
=
-'! <14 ( 1Zi;O)(z§ )( Z G"'t X 10') X 10-
q,.,., x
10 -a
88
8
~s=-
o) F'/riMAIZY /,.,.V SCCONOAflY TUTli.JS
Ns= ?:9 ;;::.,70 TUtU-IS
b) t=L UX IN THE COTZ.E
SECONDAilY
IP !;"94 X GfiOO VOLTS
r::~='f-4"4- IJsfcP.,
Nr= 2, zr;o TUflNS
IS
OF TnANSFOR.MATJON.
GIVEN:
¢!,.,= 2 ·G""t
v--
Np =
13,800/460 VOLT X 'FOrlMER
== 19G'-1 X 197
/"3,800
-(0-.G_8_-><r._j_TUIU-J
TllANSFOilMER
lt-.!DlJCfD
CO/IE:
Np =
fU'(/(7:
f~
IN THE
GIVEN:
2
liN.
TtiE
13,800 /t60- \/OL I
G-68 · CALCI...'LATE: (a) THE NUMBEJZ OF P/ZIMAIZY
t-
8-2 x \0~ MAXi
/J =-
B-
SOLUTION:
t= 0-011 IN.
b)
SOLUTION
Gl) Ep
-
IPm =
= c;.c;e,
RESPECT
TO PtliMAr'lY
13,800
Np
Np=
w;
El'
4-'Tt /Jpf x 10-e
- ("f. 4"4 )(2100)(~0 X 10- 8 )
~
tftm=
G.(;8
2-9G x lOb MAXWELLS
VOLTS
Er -
Es-
__
I II
!.
8-C] IF THE SECONVAflY
Nr
Ns
IS 2tO AMP , CALCULATE
GGOO
2'2£>0
'140
\.!;"0
J!;:l
LOAD CU/ZflENT OF THE TflANSFOilMEfl
THE
Ir=
8 AMF'
Z"fo[4GO_ l
I
f"fliMATZ.Y CU!l/7..ENT
SOLL1TJON:
Ir = Is [ kl=
Er J
IN P'TZ.0/3 !1
Ofl:
Ip: Is [
rr.aooJ
~;]
=
(_Z10)
lr -=
8 AMF
[ligo]
89
I,
I
II(ANSI OfZMERS
J11
1
11
l'r! lVI' II
h!AS TeSTED
11/J\IJo.l oltMII(
1;1 1\MI' AT
VOLTS
22(3
CALCULATE : (a) THE P/ZIMAf/:Y
3-2 AMI'
LINDElL LOAD AND \v'AS FOUt-JV To
h/HC.N THE Pfi.IMArtY CUI(JI.EIH MEASURED
lfJPUT VOLTAGE
;
(b) TH!:. RATIO
TllANSFOIZMATION.
OF
G'IVE.N:
Vp=
Vs Is
--
Ir
Is= G:Z. AMr:
Vs = 228 VOL1S
b)
fl.EGV:
Cl =
Vr
Is
=Vs
Ip
0
8-8] A 1/ZANSFDILMEfL
HAS
fiG. A) ENTIIZE PfliMARY
t;;:Z
'T\\/0
=
19-'t
= 19-1-
:
Q
= G900
Vs
230
=
0.
b)
Vp=
'900- (0-02-!i
vr
30:1
AND
VOLTA(;ES
OF 7rz.ANSF0fl./V1A TION
P!ZI.
SEC
l
BJ ·EJI~ r
EJQw EJff}v
0 V::=Ys
l
y.,
Yr=
10!15'5
= --
0'
7f.. /.
TAP
VOLTS
G38Z·\>
Vy
G38e-5
(l=-=---
2'50
41
AT 2 Y~ ;1. TAP
Yr = 6900- (0-07<;> X b900)
Vp = G555' VOLTS
Cl= Zlf-5:1
v,
S%7AP'
Cl
=
230
'Z7-7S: 1
AT 7'1. /. TAF
v
fiG. 1
01 =
Vs
a,~
riG-
= j:BOO
.±_
w: 1
210
-ff10
FIG 5
.,
I
l
03
=
z
o.,_ = 4-Boo
=
zo: 1
-zY, 'l·
r;; "f.
Vp= b'900- (0·10 xt;9oo)
t-r- - - " ' -
Yp~
G'ZlOV
Vs~'230V
L__ sr.
t::)
10;/, TAP
Q., 2.7: 1
FIG. 4
2400
0'\-= gjQQ
-240
a.,_ =
Oa = 10:1
90
Vp=~tiO
VOLTS
vr
10210
0=---=-Ys
230
Fl~-
5 VOLTS
230
1
AC· D)
Vpc. G900- (0-0t:i X6900)
0=-
X 690o) = rD7'27-
G727-s
Vs
rib'- C:) ~ /. TAF'
Vp
TAP
0= - = - - -
1
f"'ZIMArtY- TO- SECONl7AIZY
E.ACH CASE.
IN
zY-z. /.
Q = 29-25:
2,'1-00- VOLT PfZIMAfl.Y COILS
SEC.
=,:±
19·<)-
SECONDARY COILS'· INDICATE BY SKETCHES 'THE
/LAT/0
OF Trl.ANSFORMATION
~~;Owv
\\lAYS TO CONNECT THE TTLANS'FOIZME/L AND , FOIZ
EACH ONE, DETEft/VIINE THE
AND THE
FOil. A CONSTANT 2'~0 -VOLT
THAT MAY 61: IJSEV
AND THE I'ZATJO
SIDE.. D.ETErlMINE THe
SOLUTION:
Cl=--=2'28
32
SINCE:
VOLT
VOLTAGES
PfliMAI'lY
SEilVICE
ON Tl:l~ PTZIMAIZY
Is
Yr
SOLUTION·
2"!-0-
?Y2. AND JO PEIZCENT TAPS
CI=-=Ys
Ir
44-ZO
b) tlATIO OF IIZANSFO!ZMATION
FOUIL POSSIBLE
i
8-10] A 6,900/Z30-VOLT T/lANSFOI'lMEIZ IS PIZOVJDED h/ITH Z/'2 ,5,
lG1.l
-J
3.z
-= 228
G\) Pi'l1M41ZY INPUT VOLTAGE
i\v'O
I
'TflANS'FOIZMEJ?S
Vp= 1-TI7S ~ "1-'TZO VOLTS
lp= 3-2. AMP
a)
,i
480
c;: 1
91
AT 10/. TAP
I
I
".I
TI(ANSFORMEf?S
TrlANSFOfZMEflS
8 -11] THE NO- LOAD
TIZAI'-lSFORMEIZ
CALCULATE
AND FULL-LOAD
\VEllE FOUND
TO 5E
= 117
Vs~
CONT. OF PflOB. #
THE PEJZ. CENT flEGULATION.
VOLTS
SOWTIOt-J:
AT NO-LOAD
bj.
AT AJLL-LOAD
\13 VOLTS
17.e~ IS THI: eQUIVALENT
LET:
IN THE Pfl/MAI1.Y
YN·L- VFL
Vll =
X 100/.
117 - I 1'5
1. VR =
REGULATION = 7
l\3
/. YR = 3.5<j
RESISTANCE
Ze
·j.
Ze'= IS THE EQ. IMPEDANCE
IN PIZ.IMATZY SIDE
;I.
1
\1e'
flEGULATION
IS
2·G1
CALCULATE : (a) 'THE NO- LOAD
PErl. CENT·
b) THE
TRANSFOJZ.MEIT.
OF A 2,300/ '230- VOLT
l'
SOLUTION:
o.) ·;. V.ll::
111ANSFOllMEfL 2300/230 V
VN·l -YF·l
X 100 /.
~ .t;t "/.
7-VR.:~Gii(
VN·L·
1
!ZEQ D:
VN·L
a) NO- LOAD SEC. VOLTAGE
lo)
ItA TlO
=
X 2'50
100"/.
,.., "= 0·008 ..fL
,,e
=
FOLLO\VING
Oo == Yp
=
2500
23b'
fle'""
·~
I
a) fie }
llp=0·4Z
xr = o.
fts
AND SECO/JDA rzy
lo) Xe
c) Ze
12.
Ze~ =
Q
2
X.::+Xp
1·4' .sL
Xe'=
I.
8- 14] USING THE DATA
DrtOPS IN PfliMArtY
Of PfZOB. 13 , CALCULATE
AND SI:CONDAIZ'(
SOLUTION:
Re" .= 0
o.e IL
1000 =
Ts = 100,000
lf>O,OOO
Yp
==
41G·~ AMP:
240
Z400
IsRe"= (41.0.6'4!)(0·008)
Iy= 41·G7 AMI'
Isl1.e'= 3·5'3 VOLTS
!pRe'= (41·G7)(o.a)
o.ooes
IrRe'=
Is Y.e"= (4J6.10G)(0·014)
33.33 VOLTS
Is X~''=
s;-.93 VOLTS
lp Xo9' = f'ti·G7){1·4')
'
~'---~
008 Sl.
Xe"= 0·014Sl..
Xe'= 1·4-.CL
IN f1rlMA~Y ~
VOLTAGE
AT SECONDARY TEfl.M5 ·
VOLTAGE DIWP AT F'ILIMAILY SIDE:
X
THE FOLLD\v'ING
'TE.Jl.MS: (CI) Ine ~(b) TXe.
= D. 0038
Xs =
J(0·008)'+(0·011)-z.
1
Ip = KYA
SECONDARY TEIZMS
=J~ (Xe'') 2
0·8 ..Q
Sli<.:'E ·. rz.e'=
60CYCLE
c) Ze 1/
o.or-t.JL
Xe =- (10) 2 ( 0·0069) + o. 7'2.
llEQLI/T'l.f:V.'
2400(24<JY
I
IN PRIMAIT.Y
~ (C) Ze .
GIVEN:
100-KVA
2.40
=
Ze'' = O·OIG\ SL
b)Xe 1 =
1
Xp=O·?? j I'Zs= 0·0038; Xs=O·OOGB·
fZp=O·"t-2;
(b) Xe
~·75 :
GOCYCL[ lllANSFOflMEfl HASTHE
2,"1DOj240-YOLT
THE FOLLO\v'ING VALUES'
TE1'1.MS j (a) fl.e ;
100
xe~
Tle 1 = {Jo)z(o.oo:sa) +O·"i'Z
230
--
V,,,L
CONSTANTS:
CALCULATE
Xe"= O·OOG8i-0·7Z
0 = YF = 2.4'00
Ys
23G VOLTS
0.=
A 100-KVA
az
-n:rz.~s:
Pl'l.IMAftY
5lJT
If'
b) Xe''= 'Xs + ~
Ll=-10:1
+
""'
a} f'le'= a'2rzs + Rr
,,'ll:
9-13]
IN SECONDATZY
VF·L·
.f= GO CYCLE
b) IUrlNS
FOIL
.JL
a'
O·OO"o8 + O·+Z
SIDE.
SECONDARY VOLTAGE j
11JllNS flATIO.
GIVeN:
lfV\PEDACE
'-----.
I·GI
IDO
lle'~ Xe''/ze"= IS THE EO.
17.ESISTAIVCf. 1 flEACTAt-ICE , /
B-IZ] TME PERCENT
=
1
fOJZ. SECONDAitY 'Tf:l'lMS:
Cl) fle" = Rs + rt 1,
IN f"rziMAil Y SIDE
X 100 •
c) ze'= J(lle YtCxe') 2
""~ (0·8)'2+ (1 4)'2
SIDE
cG. ILEACTANCE
Xe'= IS THE
VF.L
tLcQUIRED:
PEtz.CENT
8- 13
SOLUTION:
GIVEN.'
Vs
OF A
SECONDARY VOLTAGES
117 AND 113 VOLTS / IZESPECTIVEL'(.
lp Xe'
92
=
£;8. 34 VOLTS
I
93
<i
<i
..
j
TIZAN3FOTZMEJZS
8-15] USING THE DATA OF PllOBS.
AT UNITY
ltEGULATIDIJ
1'3 AND 14, CALCULATE' THE PEI7. CEt-JT
230- VOLT TllANSFOfZMER
Psc=
51JT;
USING SECONDAflY SIDE i
YN.L.
= j"(v:-s-CO-S-fJ-+-1s_rl._e
VH.L
=J[240(1)+'3·'53Y+l24'0(0)+5·83Y
Vf'.L = 210 VOLTS
Is fie''= 3 ·'3'3 VOLT$'
Is Xe"= 5·85 VOLTS
vN·L· - v ... L
7
2
K )
(b) THE PER CENT
:-+:-(:-\ls-S-IN_fJ_+_I_sX:-e"-::-)z
VH.L. =
1. v.n
.,
X 100;.
=
2"t3·4- 210
240
I
100
Is.c.
8-IG] REPEAT
PrlOB. IS'
'•
VOLTAGE /Jfl.OF':
= 4.5t
AMr:
JXe'
= (4.54-)(I0./9)
lXe' = 129·6 VOLTS
a) lle, '/.e_,~Ze IN Fill TE!ZMS
'1-
l
Irle' = E'l·3't)(IO.J9)
IRe'= 4t. '22 VOLIS
m::o'o:
b)
.I
b) 'f. V:R AT f'·f'=0·107 LAGGING
2,300/230 -VOLT X-FOflMEIT
Ps.c. = 192 \v'ATTS
x
II
IZEGULATION Ai A LAGGING PO\vt::lt fACTOR OF 0·107 ·
Es.c."" 1:57 VOLTS
i- V·R· = 1·4 ;I.
Vr.L
CALCULATE IN FfliMARY TEIZMS: (a)fle,Ze~Xe;
GIVEN:
243 ·4 \JOLTS
THUS,
PEJZFOI'lME'D UF'UN A 10-KVA 2,'500/
\VITH THE FOLLO\v'ING ltESULTS : Es.c. = 1'57 \JOLTS;
1'32\vATTS;Isc= 'f.34 AMP.
10-KVA
f.r. =UNITY = 1
/. Vifl.. =
B-18] A SHORT- CIRCUIT -riST \VAS
PO\vEfl FACTOR.
SOLUTJON:
,I'
Tf(ANSFOIZMEITS
v... ~.. -
llEGULATION AT F.F-:0·701 LAGGING
•j.V·R=-
v~·L
•
XIOO/.
yi'·L·
FOil A PO\v't:l( FACTOR OF O.S LAGGINC.
SOU1TJON:
SOLUTION:
VN L.=
VNL.
FHASOfl DIAGflAM
O·B LAGGING FF.
AT
j te'tQ(0·6) +3·"53] +[ '2.10fO·G) + 5 ·8'5]
2
=
AT 0·9
LAGG/NS PO\vErz r.te'TOR
2
VOLTS
246·18
l
'
o)
Psc
'/. VR
246·16- 210
240
XIOO/.
Is
= 't·57 '/.
:!}
f
f
8-17] rtEPEAT Prl.Of5. I!; FOR A f"O\vE/l FACTOR OF 0· 8 LI:ADINS.
SOLUTION:
.I$
AT 0·8 LEA171NG' F-F
vN-L
=}[z4o(o.a) H33]Z +[:z4o(o.a)-5'·85]2
VN.L. = 2:59.,?5
o.f.
VI(.=
23~-
•
FHASO!l DIAGilAM AT A r.F.:0-8
)< 100 /.
~"10
LEADING USIIJG V$ AS fl.EFI:J'ZENCE
PHASOTZ
•(. V.IL: - 0·41 '/.
1
vfi.L. -::}r;;o(o.?o7)+ tt.n]' t['Z3oo(o-?o1)+1Z~·•J
Vtf.L. =
Tle'
Z-41.3 ·3
VOLTS
= 10·19 ll..
E<;;c
"Ze' "' - -
137
2423.3 - 2500
(: Vl1· =
=---
'2300
t.3t
Ze' = 31·5"&
;. y. fl. . .: 5·'36 /.
.n..
!7e" 1
Xe'= /(31.;6?-Qo.t9P
li,,
l1
Xt/:: t9.8' J1.
B-1~ THE
TEST
FOLLO\v1N6
PEitFOflMED
DAT4 \vErlE OBTAINED FIZOM A· SHOOT- CII?CVIT
UPUN A
5'0- KVA
2i500/ll!>- VOLT
Esc= 87 VOLTS ; Isc= 21.75' AMP: ;
IN PIZIMARY
TERMS ; (a) tZe, Ze, ANV
AT A PO\IIE'Il
FACTOT'{
Xe
~
60-CYCLE Trl.ANS -
Psc = 590 VATTS ·CALCULATE
(b) IHE f"Ert CE.N1 riEGULATION
OF o.S6G LAGGING.
GIVEN:
·I
II
50·1<V.4 , 2500/11!> VOLT 1 GO CYCLE X- FOrtMEfl.
Psc
:::
21.75' AMP.
t;'~O
\'IIATTS
4Xe
t~) rle, 'Ze,
IN Pl"llMAI'lY TEIZMS
b) 1- v.JZ. AT A f"0\1/Ert rACTOlZ
OF 0·86G
95
rl
I•
!te.c:{o:
Esc.=87 VOLTS
lse- =
94
X 100
Xe'= /ze' 2 -
rOI1MEil ;
VOLTS; USE 239 \)0L'T5
2:'40
l
== (<t.'S1Y
sc
lsc
o/. V.R·::
192
lle '= - T
2
LAGGING
TfCANSFORMEIC.S
CONT OF Prl.OB #
8-fl] FOQ THE TIZANSFOIZMEfl OF PflOB 19 , CALCULATE ; (a) THE KVA LOAD
b) /. V.R. AT A f.F:0-866 LAGGING
SOLUTION:
Q)
F5
I
(21-75}'
1-2\;
Xe =
= (21-7S)(l'Z.S)
= 27-'2.
\v'HEN THE COPPER
FV1v'E11. FACTOR
VOLTS
Cl) Wcu =
Yw'-.=
j[ 2YXJ(o-&rocs) + 21-2Y + [ 2300 (o.s)+B2·tY
b) Wcu= 714
4 OHMS
\VATTS AT P.F=0-8
F's.c
17.ATED 1-<VA = 50 KVA
V><-L- V'f"L
''/.
X 100.
;I. V.IT..:.
Z3G'5· 4-- 2300
CALC. THE KVA-LOAD OF o.) ~ lo)
.,
X I001·
'2300
o) Wcu = Ps-c
8-20] FOfZ THE TILANSFOflMETl OF PJWB _ 18, CALCULATE THE
COF'PEft LOSSES
IS : (o) 8 KVA ; (b) 10 1<\v' AT A PO\v'EIT. FACTOR Or 0·85
18
COf'PER LOSS(Wcu)
fp = -2'500 VOLTS
\vcu
230 VOLTS
Es =
1
fl.e = 10-'2 1L
=
lc 2 Re'
= (3 ·'1-B)-:z (10.2)
\lieu= 1'23'-£' \vATTS
fl.EQ'O:
b) AT
COPPElL LOSSES \v'HEJ.J
THe
LOAD
10 K\v' XIOOO
(0-85') ( 2500 V)
h.= S-11 AMr.
10 k\v' AT A f. F= 0-8\;
\'1/cu. = 1r.." lle
SOLUTION:
01) LOAD WIVZENT h
Ic
=
KVA x 100
Er
Vc4
2300
v·
-=
=(921) (50) 2
590
==) 30li;;-4
~:
~5""
r
K\v' = I-< VA X f'. r
= !;S X
K\v
o. f!
= "'l4-K\v
590
!
:1
i:
g- 22] TfiE EDDY- CUflflENT LOSS
I
IS 280 \vATTS · \v'H.<J \v'ILL B'E
IIJ A 21 300- VOLT 60 CYCLE TflAN SFOflM ER
THIS LOSS If= THE'. TRANSrOILMER
IS CONNECTED'
(Q) TO A 2,300-VOLT S'GI-GYCLE SOUILCE? (b) TO A ?,tOO-VOLT GO-CYCLE SOURCE
(c) TO A 2,200- VOLT
'
fi,
25-CYCLt: 50UflCE7
GIVEN:
EDDY
CL'rlfZEiJT LOSS=
,280 \v'ATTS
SOLt..'TION:
o) AT ?,'500 V, 50 CYCLE
llI
I
!I
I
!
Jr.. = 3-84- AMF'
KVA 2
\If
1
266 \v'ATTS
(£'0) 7 (714)
I~
Cs.tl) "( JO. z)
8 KVA x 100
=
1<\v'
lL=-~-~--
IS:
a) 8 KVA
io)
10
J
5'90[
v
LKVA n:t1ed J
THEN:
kVA l 2
[ rated KYA
=
0-8
= ----
KVA =
~22
~
i,
~'
GIVE}J:
-:z
KVA
SOLUTION:
;. vrz.= 2 81/.
OHMS
KVA
'r. F=
f__K~_:A_
Wcu. = F'sc X
S90 \v'ATTS
flEGUIRED:
VF-L
==f (<f)"- (1-zsrz'
FllOM F'TWB- #
b) AT 71't \v'ATTS / '
SO,
o;. V: f{-'<"
,.. ,;---;-:z
~z
-y Le - rce
\vliE~J THE LOAD
KVA = Gz.c; -f<VA
YN.L== 23GS-"l VOLTS
=
AILE 71'1- \VATTS.
KVA = ~ '3'306 ·8
\\IATTS
922
FROM fflOB . # 19
21-75"
X/= 3-7'3
LOSSES ARE 922 \v'ATTS ; (b) THE K\v' LOAD AT A
OF 0·8 \vHE:N TI-le COPPE11. LOSSES
GIVEN·
IXe' = (21- 75)( 'S-79) = 8'2-"'r VOLTS
87
Is.c
I
1
OHMS
Ze'= Es-c
Ze'=
I!1e
590
C
fie = 2
Is.c
fle' =
TfZANSFOfZMERS
8-19
Pe
= tBO [?300r
-,2500
f'e
=
ZBO \v'ATTS
c) AT 2,200- V, 21) -CICLE SOUflCE
Pe== 280 [
2200 l2
--J
2300
I
b) AT
2,400 V , GO CYCLe
Z400r
F'e= 280 [ -
2300
Pe = .3 O't. 8 'JATTS
9.£
Iiiii', I!
Pe== 25'6 \vAITS
--------------------9'7
Tf?ANSFOR'METlS
8-23] THE
CONT. OF Pfl.OB· #
LOSS IN A 6,600-VOLT 00-CYCLE: TfZANSFOR:MErl
liYSTETlESIS
<teD \v'J\TTS · \,!HAT
!S
T17ANSFOfl.MEflS
\v'ILL
W-CYCLE SOUI1CE ? (b) TO A b,GOO7
VOLT 'tO-CYCLE SOUflCE
(c) TO A 6,900-VOLT 40-CYCLE SDUIZCE:?
COI-11-JECTED:
IS
(a) TO A G)100- VOLT
~
FH ~
"tBO \VATTS , IN A 6,600- VOLT
[y.IX COStr- %TIZ. SIN-& ) 2
~-------­
'200
'J. flEG.= (1-18)(0.707) + 3-6(0-707) + ..!:.::.___:_:_ _,..:....__.:....__,_,__~
!~ r·G
1-80 [
<-00
•;. fl.EG
60 - CYCLE SOI.Hl.CE
c) AT ~900- VOLT "TO- CYCLE SOUilCE
0) AT G,']OO V, (;O CYCLE S'OLIIUE
P.,: FH, X
f_ E2
L"E:
]
I·G
Fo-~
6900]'~
r.. = 1ao [ (;Goo
51~
FH =
= f.
6600
)1·6
(~)O·GJ
111 •
!
F.. =-
.tI'
J
,'.~{
{;57 'w'ATTS
CALCULAIE THE PEfl. CENT
rMTEO KVA
eFFICIENCY
zs]
CALCULATE: (a) THE
ilEGULATION AT A f"O\V.EII
,
Poe
= '285
\v'ATTS
Esc = ISO VOLTS
OF Pfl05. 19,
Psc = 615 WATTS
'7. Til, MttJ THE "/.IX [}flOPS: (b) THE PEJZCENT
Es.c
'/'..fR= - -
"J. 1-t. = fouT
X
100/.
Fc11.1T
+
PLoSS'ES'
.
P..,.T,.
/007.
IS DELIVEitiNG
KVA
X
P·F
X
KVA X f'·F 1-
IOO"l·
PLoSS'ES
(50,000 VA)(0-8~)
X
/. 1-L= (50,000) ( O·B"'t}t-285
7. vt=
37.9
1007.
t2.1G
:z
SOLIJTION:
7.IIl=
h/HEI1E;
590W/IOOOW/KW
s;o
E's.c. = SHOllY ClfZCUIT VOLTAGE
= 87-VOLT
'1-h.::.:
AMP.
% EFFICIENCY =7
X 100'7.
ERATEII
ERA TED=
= 10-87
Isc
FACTO!( or 0.707 LAGGING.
rz.Ea'o:
-- X
101.5"\-vATTS; Isc~ IO-S7 A
AT A PO\vEit FACTO/I OF 0-84.
l:oc = 230 VOLTS
P's -c
=
Psc=
OPEN CII7.CUT TEST:
SOLUTION:
D) IUZ
:·I!
•,1'
50-KVA, 4,600/230 VOLT, 60 CYCLE TTl.ANSFOflMETl
FfWM THE SHOflT-Cifl.Cl!IT TeST
% rz
II' I
GIVEN·.
AT THE SAME
VOLTAGE
THE DATA
USING
)
lv'HEN THE Tfl.ANSF017.MEfl
SHOIZT CIRCUIT TEST:
8-
lt-.1 CONNECTION \viTH TESTS
50-KVA 4,G00/2"50-VOLT GO-CYCLE TflANSFOfl.MEri.:
St-!Orn CIRCUIT TfST ····· .... ·Esc= ISO VOLIS
b) AT 6,600 V, 4"0 CYCLE SOVItCE
f,]0·6
= r.. , [,,
INFOrlMATION IS GIVEN
OPEN Clll.CUIT TEST······· Eoc= '230 VOLTS ; Pee= 285 \VATTS
<fO
<JfJOl (1-0137) (l·Zlt;)
fH=
\IIATTS
[ ( 6900
H,
F'ErlFOflMED ON A
1
1
= 3· 'tO-t%
8-2G] THE FOLLO\viiJG
'I
II
1
11
FH"' 6!2 \vATTS
SOLUTION:
PH
+
7.f1Eb.: /.rrz COSfT +'7-IXSIN-6-
[(5 .r; )(o. 707)- (I·IB)(o.7D7)Y
GIVEN:
r,,,
8-2S
(b) '1. REGULATION AT A PO\IIE/l FACTOR OF 0-707 LAGGING
BE Tri!S LOSS \vHEN THE TI1ANSFO!rMEfl.
FfWM Pi7.06. 19
1- m =
)(100/.'
LET: '1_ = % EFFICIENCY
KVA
f'our
1-18 '7.
= 1"0\vER.
OUTPUT
1"1,. = FOil/ElL INPUT
[ATED VOLTAGE
1.
= 2500- VOLT
87
'7. J7= - -
2300
rx = ~ ;nz
;.rx
X 100:7.
=
2
-
;w:rz
8- Z7] FOil THe TRANSFORMER
J(.naf-0·18)'
EFFICIENCY \IIHE lr!E LOAD
100 KYA
fl'.EQL11/r.ED:
TI1A NSFOR Mt:n:
PERCEJ-IT EFFICieNCY=?
f"O\vtrl FACTOrt = D· 86
OTHEn: GIVEN 17ATA ,
I~~llf.J:'f."''~~-~
f=rrOM
.......
7
_..,..il
9'fl
XLW
AT A PO\v'ER FACTOR OF 0·86·
GIVEN··
-;z I X -= J. f 'l.
%flo=3-787.
OF Prl05 · t6, CALCUlATE THE P'CR CENT
IS GO I<VA
....,..,
2!
B
lII
1,1::11.·'
II!' I
Frr015. ZG
I
,m~ ~'·,
I
Cit/]
.I
1\1 1
1~1'1'
II!J!,'
Trl.4 NSFOfZMEJI:S
CONT OF PfZOB #
TflANSFOR:MEflS
B-27
8- 29] FOR
SOLUTION.·
=
TOTAL LOSSES'
""' ass. G + 28!;
LOSSES AT
= /170. G \v'ATTS
GO KVA
SJNCF AI
I
Fl101Vl
/.J,_ =
IS EGUAL TO THe
PouT
AT GO KVA IS tQUAL TO;
Wcu
Wcu. =Wcu
=:
so
Glt;
X
us.ses
Jt:
KYA Go
[KVAso
511000
]l
[W
/.1--t =
GO
97-78
\OOt
+- 1170-G"
~~
~
·~
Vi'
.....,
FTWf5. ZG
I
CALCULATE
=
LOAD I<YA
= 60
I
KVA
8-:soJ IHE
x1ooo x
IOOI'·
egs ·G)
\VATTS
( FllOtv) PIW/5. #
~
'27)
285 \v'ATTS (COrLE LOSS fTLOM l'IW/3 ~ ?(;)
fovT _ _
97.1;2
)<
!JO KVA
=
3i t<YA
+
X
t
I
100;1.
'·~
)
F,ossE.r
!
:)
X 100].
i
7.
ON
INFO/liV1ATION 15' GIVEN
A 25-KVA
IN CONNECTION \v'ITH TESTS
SHOrn- CIR.CUIT TES'T ·· ·· ·· · ·· · Esc=
(o)
0-86G
KVA
THE
PE/l CENT
AT A FVIvEfl fACTOIZ. OF 0-8
FVILMC.R.
EFFICIENCY
\1/HEN
THE
\v'HEN THE
EFFICIENCY
250 \VATTS
72 VOLTS; Fsc =380 \v'ATTS ~ I sc =10. t AM F.
REGULATION
LAGGING ; (b) THE f'E,z CENT
f'oc =
FOIZ. A PO\v'ER. FACTOil OF
EfFICIENCY
~ (c) THE
AT
KVA
PO\v'EIZ FACTO!l
1 '/-z TIMES
~(d) THE
IS 08.
GIVEN
ZS KVA , Z,400/Z·f0 VOLT , GO C"YCLE
OPEN Ci!LCUIT
TI<.ANSFOQ!IJE IL
TEST
SJ-IOOT Clfl.CUIT TEST
= 240 VOLTS'
Foe =
Esc= 72 VOLTS
Fsc
250 \v'ATTS
=
380 lv'ATTS
Isc = 10-t AMP.
~
H
ij
~--~-~~~._.~-----
lo
~
! ()
l
lZATEV
LOAD ON THE Tri.ANS-
IS A MAXIMUM
I
1;1'1;
2;W0/2'1-0-YOLT GO-CYCLE Tfl.ANSFOfl.MErl.·
OPEN -CirtCUIT TEST · · · · · Eoc = 2'tO VOLTS;
CALCULATE
.1,1
!II,
I
= c;,e. 04 ;;.
.J ~~:
=
I
1
(3'1-)(0-84)(1000) + 2 ( 285)
EFFICIENCY
FOLLO\v'ING
PE17.FOflMED
E0c
-·
IIi
ij.
.J
wooow
2~= ~000\V+MS~W+~SW
:ti~-lfilli<
I
PouT +2Wc
(34)(o.e4)
li
100/.
vVcv
··.~
It•'
f"ouT
\,
X
LOAD l<vA ~ flATW f<VA ~ We
ll1E FER cnJT
AT GO KVA
~it. =
1.q_"
PouT
SOLUTION
4-8 K\v AT ,4 fV\vEfl FACT0/1. OF 0 ·8.
THE LOAD IS
48 K\v
\<VA= - - - 0-8
f.h_=
iijl.'
\VATTS
a) AT MAXIMUM EFFICIENCY
w-1
50UJTfOfJ:
\lie=
KVA
'Z '1.1'>'1"
b) M,~XIMUIVI
!j
)\,
B- 28] FOfl. THE TrtANSFOrcMm OF
\\leu=
= 50
= 285
REGUI/lEO:
X D· 86
&
\1/HEN
1
MAXIMUM
-~-
EfFICIENCY
,I1
\v'HEN THE EfFICIENcy IS
= G0 /0ii{ X \000 VA
W!ll8l'ti?-'I!I'IUIIIIIIILL$U
1
o) LOAD KVA , AT A PF= 0 84
?.
I
IS MAXIMUM , AND
b) MAXIMUM Eff.
\1fcu = GI!; \IIATTS
f:>vr = 51 GOO \vATTS
~
\\/HEN THE EFF/Cif:NCY
II
\Vc
Wcu "" BBS-6' \v'ATTS
Pour!'IJT
1
PrlOB # ?G
511000 ?' 100
]~=
THE KVA LOAD
1
PJ1TED KVA
fouT + f,
COPn:rz. LOSS.
IIATED
26 , CALCULATE
GIVEN:
SHO/Z.T CIRCUIT lEST, THE SHOTl.T
ClllCUIT POh/Eil
TrzANSFOfZMER. OF P/ZOB
AT A PO\VEfl FACTOR. OF 0.84
THe MAXIMUM EFFICIENCY.
f'cu + f'c•RE
DElDlMINE FlfiST THE COPFER
THE
MAXIMUM
TI7ANSFOrlMEIZS
TR'ANSFORMEflS
CONT OF FTWB #' 8 -30
CONT. OF Prw3. #' 8-51
Pour =
SOLUTION:
a) (. n.EGULATION FOn A PO\v'E'fZ
FACTOn OF D·BGG
Esc
1. IZ= - Eraled
X
LAGGING
100 7.
'J., tfF
=
j
LOAD KVA = llATED KVA
=
b)rtATED KVA,
o) 3/1
d) 1/ 2
f2SO
i'S ~
3aQ
(J.5z)'
-
'bi-t.-~
+ .'J, I X ClJS i7
200
+ (2.S81D)(o.s)
z
+ (cz s&G)(o-BIDG) -u.szJ(os)]
200
.~z
(zoz:r)(o.e) x
+
·~
LtT:
=
IS TIMES flATED KVA
=
=
"580 \V (
2(0·2W)
97.Q1
'2
== (z) (a. zso)
=
l
OF 32
["11·475·+J.817
(. Yla = '?Jf:.o7
JK\,1-HQ" 100?.
'7.
O.!; !<W
GiVEN.
et) PruNSfOkMEo = r,~M [
1-
i--J
LOAD:: "52 K\v
LOSS OF 75
I
3/-t ITATED
r
'l
100 t.
44.'f75 K\v-H11.
=
K\v' AT A PO\vE/l F.ACTOfl OF 0·8. CALCtJL,ATE ~
AUTOTfl.A NSFDRME.fl.., 4000/2500-V
FOLLOiviNG LOADS :
')(.
(D) THE Tr<..ANSFOrl.MEO f'O\vER j (b) THo CONIJUCTEV 1"0\v'efl:.
/·,!;')Z
o.es
~ ENt:JlGY~LITf1JT +LOSSES']
8-32]AN AUTOTllANSFOtlMEfl, OESIGNED FOR 4.000-TO .?,300-VOLT OPE.Il.ATlON,
SuPPLIES A LOAD
lv'ATTS. CALCULATE
REG'o;
THE' ALL-DAY EFFICIENCY
1'/2 TIME'S THE fZATED
KILOVOLT -AMPERES
KILOVDLT-AMf'E'IlES' 1 F'OIVEIT. FACTOR= 10, 7 Hfl.; NO LOAD
102
1
I
b) CONOUCTEO f()\VErt
b)
Pco>'i>UCTEb
=
\1-l
1
1'3 .c; K\V'
= F,Nru;
-1'-rPAHSFo ..MED
32- i3.G
1
SOLUTION'
G Hll;
4 HR
FrAANS'F'oi!.MEP
32 K\v [1-
0) TflANSFO(lf.1ED POIVER.
HR j tlATED KILOVOLT- AMPERES , F:OivEIZ FACTorz=
KILOVOLT- AMF'EJIES I PO\v'I=IZ FACTOfl.= o.%'
=
1"-F ::0·8
8<51] A ~-KVA 'TI'lANSFOI?MER HAS A COrlE LOSS' OF 50 \v'ATTS AND A
FACTOR=
~ E:Hl:J7.GY OUTf'\!T
'TO'TAL = 'T4·47S 1<\'11-Hil
'i
s Hrz.;
b' K\v-Hfl
X JOO~
:; ENEJlGY IN PUT
(075)(3 KVA)(0·9S)(GHI'l.S)= 1'2 S25 K\v-tll'L
8!>S \v'ATIS
FULL-LOAD COPPE'fl
'Z VL.I
1<\v'.ti!Z.
~ fi-11:/ZG/ OLITP\JT
'Z '1.<1 =
/. I'J.d = JS THE ALL DAY Ef'f.
(1)(3 KVA)(0-9)(SHRS)= 13
100'2
KVA
!Jz. l'lATED
Z/'td"'
Q·5')(3 KvA)(O;Bs)(tfJit) = 7-6'13 K\v'-Hll
COFT'ER lOSS = '2 We
J
i
, -t HilS.
~ ENEl?.GY OUTPIJT:
?
KVA Z
Wcv =raTed Ww [ I · S -
I
= 1·817
ITATEV KVA / P.f= 1·0 , 7 Hf'lS.
(0·5)(.5 KVA)(1)(7 HILS)= IO·S K\v'-HR
"Z f],WI">(
AT A PO\vElt FACTOfl OF 0 8
PD\v'Efl
~ f.f.!ERGY LDSSCS' = 1·0~7+0.7<'
, c; HILS.
~~
~"
J
AT MAX t:FFICIENC'f' ~
2
b) 7. EFFICIENCY AT
(LOAD KVA,) (f. F)
-x 100
(LOAD KVAM)(PF)+ ZWc
(20 Z?)(OB)
VIZ= (i·S~)(0.8GG)
?. v.rz.. = t
O·'J
o. 95
SOLUTION:
,I
+('7.IX61NI7- ~ HlCOS-tJ-)'Z.
i
FF.=-0·9, 5HrzS
, f"·f=
e) I'O LOAD
d) MAXIMUM J::FFICIENCY,AT P:f= o.B
?·58G "2
= /. Ill COS-17
FOil THE
KVA
LOSSES = 30 \<1
[
-21141lS
1000\<1/
11<\-t
1
~ ~~u
1.0.97 K\v'- HR
~conE LOSSES-= 0·7Z 1<\v'- f.IR
a) 1·5 TIMES RATED KVA , P.F.= o.B5 2 HitS.
2
7.IZ -7.Trl."
~ (3) 2
i. IX=
\vcu
~COnE
LOAD KVA = 20· '27 KVA
;nx ~
t•
~C.. LOSSES =
ALL-DAY EFFICIJ:NcY FOfl !HE
I SZ /..
=
~w. LOS~ES= 0 O?r; K\-ll2(1·G)-z. + £(1 ) 2 +
+b(O·?Sf+ 7(o.r.)'"]
flEG'o·
9G. t£ /.
ZSKVA
'7.TX=
7. V. fl..
l"'l!LL LOAD COPPEJZ LOSS =75 \vATIS
100 /.
oox 1000\v +855 +250
380/IOOO K\v' X 1007.
7. Tf1.
== :SO \v'ATTS
FOLLOIVING LOADS :
Frtt1ed
'?.117.=
X
'Tf7ANSFO/l.ME11.
COf!J: LOSS
C) LOAD KVA AT MA)(. /':FFICIENCY
~XIOO/.
7. Ul.=
=
·
3 7.
30 X 1000 \y
7. Eff
x 100 'Z
PouT
PouT +We+ Woo
(. Eff, =
~ E.NEfl.{;)' LOSSES =~Cu. LOSSES+~ COR.E LOSS
GIVEN:
:S KVA
1TZ = - - X/00'/..
2400
=-
KVA )( o.a)
'50 Klv'
f'ouT =
n
'Z IZ
(zs
Cl
PeoNPucre»
4000
= - - = l;;r<j- : 1
~300
103
=
18 ."f i<\V
TRANSFOflMEITS
8 -33] FOR mOB. '3'2 , CALCULATE
Tr?ANSFORMCI(S
THE CUI?.I?.ENT IN : (a) THE Pill MArlY \v'INDING;
CONT OF PflOB.
(b) THE SECONDARY \v'INDINt;.
SOLUTION:
)
.!'.:.
b)
KVA
4000
v
Ip= /0·67
Is
d
-
Is = 17.4 AMF'
~Is=
=Q
Ip
A
2,400/2,200-VOLT
Vp= l5DOV
ILATING
AUTOTrZA~ISFOIZMEIZ
OF THE
DELIVERS
P.l=.,. 0·75
lr
Vp
"*""--~
LOAD
SOLUTION:
P·F. (Yp)
(0·75')( '2400)
I
=
'2
'}
'a=
END
Is
=
'f.58 AMP.
EITI~Efl
THIS'
SHOULD
CALCULATE
?300 /460
-VOLT
DISTRIBLITION TR.ANSFORMER
(a) MAKE A VI17.1NG
1Ht. KILOVOLT- AMf'E/lE
SHO\v'ING HO\v
LOAD OUTF'\JT.
\000
=
=
GS. 22 X 460
1000
30
!<VA
A 20
F'Ert.CENT
HAS A 5 AND A 10 PEfl CENT TAP FflOM ONE
TAP Ffl.OM THE DTHEIZ. IF 120 VOLTS
IS
ACI'ZOSS THE \viNDJN(; eNOS, CALCULATE THE VALUES OF ALL THE
THAT
END
MAY
13E OBTAINED
crET\vEEN
ANY T\v'O TAPS
AND FIWM
AND THE TAPS.
S KVA ,
iIt
·~
a)[4
'C_j
at
b)
1'20V
S ?. 'TAPS
V =(O·OS)(120)
V = G VOLTS
AT IS :Z TAP
V ~ (0.7 flzo)
V= 18 VOLTS
V"' 84 VOLTS
lo) KVA OUTPUT
Wo..\
AT 70 '7. TAP
V=(o.n;'XIto)
C!) \v'lfliiJG DIAGRAM
VOLT
1'1!
c)
RE!V'o:
<; 1-\VA Tfl.ANSFO/l MErt
I 4'60
DIAGRAM
13E DONE · (b) \/HEN USED 'TO TRANSFORM
GIVEN'
2300
IS TO BE
AUTOTR.ANSFORMER TO STEF' UP THE VOLTAtE rfl.OM
VOLTS.
x Vs raTed
SOLUTION:
I
2,7GO
KVA
s: 1
AlJTOTRANSFOIZMER
AND
VOLTAGES
o.7t;('2200)
TO
'%0 =
IMF'IlESSED
1'Z =- S<J. 85 AMP.
?,300
23DO
I<VA
= J>t.es- so.27
AS AN
h
KVA=-
Vs
90-S'
CONNECTED
b) I<VA OUTPUT
= IS THE LOAD CU/ZJlENT
8-3G] AN
f'·F ( Vs)
8-:55] A S-!<.VA,
h
Q = TflANSFOrlMATION flATIO
Is=Iz-lp
K\v'
=
Is= IS THE ClJ/ZiZEf.JT AT PoiNT D~C
'Ti-lviS;
50.27 AMf.
=
l
I
IL= G5·22 AMP.
fiGUflE
9o.s K\v'
p -
Ir
lc = 51-.35 + 10·97
Ip=IS THE CU/ZIZENT AT POINT A~B
I
K\v'
Ip= - - -
1 -
c
b
LOAD= 90.5' K\v
AMf.
h= Is+Ip
LET•
\
= 5't.3S"
Is= IL-Ip
-h
TflANSFOilMt:rt.
AUTOTILANSFOR.MER
S' (10·87)
BUT;
A LOAUOF 90·!;K\v'
IN t=ACH \VINDING SECTION
GIVEN:
2,400 j2, '200- VOLT
Is
Is= Ic-lp
AT A PO\vEfl. FACTO!'l. OF 0·7S. CALCULATE THE CUflfl.ENT
AND THE KVA
v
G\Ip
Is "'
J
8-34]
1-60
AMF.
SINCE
Is = Dip
10 AMf'.
5000 VA
27G0-'2300
a) \v'IRING DIAGflAM
= ()
=Q·7t)(ID)
lr "'
S'KVA
Ip-
Ir
32 KVA/0·8
Ip = ~ =
Gl
8-35'
=If:
SOLUTION:
105
TI?ANSFOIZMEI"lS
CONT OF ffWO #
d)
e) AT
AT 10"?. TAP
Rl·"·
'20 'Z TAP'
~
[j
24V
J_
y,.(o.zo)(tzo)
V=
V= 24 VOLTS
IZ VOLTS
h) AT 90
TAf
T
[Jr
~L
POTENTIAL TfLANSFOrtME'R
I
0
'20 )( [JI:fl-ECTION OF
iJ AT
I
ii
~I
!
v ... (p.9Q)(Izo)
V=(.o.g~;)(tzo)
V• 106 VOLTS
V= lit VOLTS
IS CONNECTED 'TO A
S -AMPErtE
OF 4.102 AMP. IVHAT IS THE
UNE CUI'l.ll.ENT.
VOLTS
2360
=
VliNE
9S ;4 TAf'
V=(q·e)(lzo)
15'0-YOLT
:I
:LO\v'- SEC VOLTAGE
FIG:
B-10] A POLARITY TEST
FOllMER. · IF THE INFUT
fl.EADI~G
IF :
SUBTRACTIVE.
(a) THE
POTENTIAL Tll.ANSFORMEIZ.
IS FErlFOflMED
VOLTAGE
FOLAIZITY
UFUN A
LISO /liS' -VOLT TR.ANS-
IS 116 , CALCULATE' 'THe
IS ADDITIVE .)
VOLTIVIETEI1
(b) /tiE PDLAIZITY IS
'
·~
''~
[,'
r,_ 1r--- ----~1"171MAIZY
TURNS
1
LINE__./ 1t
CUI7JlENTS
I
I
~SECONDARY
- j_
- -_________
- - --- - -- J - JI
I
.t
l
·~
COIL
(MANY TLlRNS)
i
IHAN \11
v,
.~
a) ADDITIVE POLARITY
b) SL'BTfZACTIYE FOLARITY
SOLUTION:
LET: V-= IS THE VOLTMETEIT /lEADING (DEFLECTION)
Yt= IS' THE LINE' VOLTAGE
Q = TILAf.JSfORtv'\ATION RATIO
FIGURE:
CUI'l.l'l.ENT TRANSfORMf.R
SOLUTION:
BO
h
=
=
ADDITIVE POLAfliTY
1\(;'0
lL = -
h
Q)
X OEfLECT10N OF AMMI!TER
5'
IG
C\= - - = 10:1
II!>
V =
X
4·~:?
7'5.9t AMP.
y,
Ia) SlJ 6TftACT lYE F'OLA RIT'I
Y= y,_y~ ; Ys = V'/a
V= 111>-J..I.£
+ Ys · Vs =v,
/j"'l
Y =
It(;+~
V =
tiG t- II·G
10
Q
V = IIG -'11·10
V = 104. 'f VOLTS
10
V = 1'27- G VOLTS
106
107
'
I.
IIi
VOLI METER.
I
Y=(p.7S)(tzo)
V= 90 V
108V
A DEFLECTION
IS 118 VOLTS , CALCULATE
~-------,
VOL'TMETER
1
'---------' 1
INDICATeS
IS USED \IIITH A 1!>0-\/0LT
DEFLECTION
I
fEIUENT TA I'
GURI'/.f:NT 'TrtANSFORMEfl.
,,II
,,
SOLUTION:
VLINE =.
V= 9G VOLTS
AMMETfTt Ttf~T
20:1
VOLTMETE/7. . IF THE. lNS'TIWMENT
THE LINE VOL TAI7f.
20 X 118
V=(o.f)(tzo)
8-37] AN 80: 5'
8-39] A
I~
[_j
9) AT BO',Z
I
S- 36
l.,,,l
TRANSFOI?MER.S
f
,,,!I I·
TfZANSFOr<MERS
Tf?ANSFV!lMERS
CONT OF Pfl015
8-'1-1] THE FOLLO\viNG INFORMATION IS GIVEN FOrt T\110 TrlANSFOil.MERS THAT
AflE CONNeCTED
IN
PARALLEL:
TfZANSFOfZMErl 1
e
TRANSFOflMEfl
!ZATIN G = 10 KVA
RATING
4-,GOO /250 VOLTS
4 ,'9-8 !>I ~30 VDLlS
Ze = O.IG OHM, IN SECONDARY 'TfllMS
Ze"' 0.22 OHM ,IN St:C TERMS
THE SECOND,AITY
CALCULATE
=
Ir
(20-19.\; )( 230)
Ic = -:---:-::--::---;---::-::---::
+ Cl 2 Ze,
(o·IG)( 20)+~·Z2)(J9.~)
Ic =
Clz ==
liS
Ic =
4GOO = 20:1
Q, == ~
\
"
I
.1':
·.II
7·"9-9
KVA 2 == KVz (h)
KVA 7 = B· 4"2 KVA
= (4.G)(2·51G)
KVA 1 = 11·57 K\/.4
KVA 1
DATA OF F'rl.OB. -4-1
8-43]i.JSING !He
THAT \vlll
1!:> .35 AMP.
"~II
l
'•·
8-"H'] IF THE TAP OF TflANSFOtl.MER 2 IS SHIFTED SO THAT ITS VOLTAGE
IS !HE SAME. AS THAT OF TflANSFORMEf"l L i.e., 4,G00/230'VOLTS,
THE LOAD ASSUMED
rOR. /t TDTA L
LOA tJ OF
1?.5 KVA ·
SOLUTION:
(CI,-Clt.)Vs + OrZe.Tr
Iz == - ' - - - - - - - -
Q?.r;)(IOOO)
Ir= - - - 230
C\,Ze,
h = 7(; AMP.
I,==
6Y EACH OF THE T\v'O TRANSFO/lMEI'ZS
'20 KVA.
-------·1
r--~r.
I
GIVEN:
',
II=
v%
SOLi.JTION:
KVA LOA£7
=
I<VA,
= (; .<;
TRANSFORMEIZS IN f'ArtALLEL
lr=fr+Iz
r,::
_ Ze 2
kVAz
= (2;o)(47·B2)
IOOO
i<VAz = II i<VA LOAD
1000
KVA
LOAD
FOLLO\v'ING INFORMATION
o.22 J
r'Z rlli'iG
12------- e-~.(1)
108
Jz
= 1'.31-- 1·37!; Tz
h
=c·-4·54
2•'51!;
I2
= I·!B AMP.
IS GIVEN FOR 1\v'O TRANSFORMERS
IN PARALlel:
lTtANSFOfZMCR 2
RATING= 75 KVA
rtATING = SO KVA
2,400/240 VOLTS
2;'t00 240 VOLTS
I
le= 2.22
Ir-h
I. = 4.31'-h
Ze,
1·375
47.82 AMF.
(19 S)(o.z2)
TllANSt=ORMER 1
13LIT:
----
I,=
(?O)(O·lG) +Q9.~;)(o.zz)
AMP·
CONNECTED
SINCE.·
=
KVA,
8-tt] THE
20,000 VA
4-GOO V
IT= 4-.34- AMf'.
T"'
r,
+
~--:--:--:----;c----:-;:--~
TOTAL
YL
lz
+[czo)(O·IG)(7b"))
Ze,
f.czo-1<:~ s)(-z;o) + (-zo)(O.IG)(?co)]
-
------"----'--~'---.::____
+~z
(230) ( 28.176)
kVA,/ KVA-.. LOAD
J,
(zo- !%)('Z3o)
h -
1'2.::
G\ 1Ze, +Clz Ze,
(ZO)(O·IG)
llcQ/o:
l
(O,-Cl 2 )Vs +(Cl,Z2Ir)
lr == 28. 17G
"TOTAL LOAD= ?0 K\v
Ir=
FOil A TOTAL LOAD OF
i'
4',4-8!:> = 1'3-5; 1
CALCULATE
THE I<ILOVOLT- AMPET(E LOAD
CALCULATE
BY eACH TflANSFOflMI:!ZS
13E CARRIED
F0/'1. UNEGUAL flATIOS OF TICANSFDR.MAT!ON
RATING
1\
= (t.G)( 183)
=(1.85)(1·375)
l, = 2.516 AMP.
CIRCULATING ClJI1flENT AT NO LOAD.
Vs
= .(o,
;:,. .__-Qz)
__:.__
Q, Ze,
8-'1-2
INA, = KV1 (J,)
7- S KVA
SOLUTION:
Ic
#
1t1ETlEFORE :
CALCULATE
FORMER
THE
Ze == 4·1S
KILOVOLT- AMFEI7.E
FOR. A IOTAL
LOAD OF
LOAD
1-?S
109
CARRIED
KVA
13Y
eACH Tfl.A"-15-
Iii
TflANSFOfZMER.S
TRANSFOI1MEIZS
•
CONT. OF FfWI3 . .#
CONT.
GIVEN:
I TO'TA~ = I
T,
+ I z
I
lET'
l':cp =
5' 2 ·08
f.ff7
Ze-=-f.22Sl.
1 -z.
I
113- 15 AMP.
=.
'T'Z.
I?.AT/1-.JG -= t;O KVA
SfNCI:::
Ze = 4-15 .sL
I
rn:o/o:
LOAD CAf11ZJED 13Y eACH
I
= 3'3. 9"1- A MP
l
KVA, =(t-4 KV)(33-0t)
KVA, = 131-"tG'
.SOLUTION'
'2"\'00
·.~
.l
tl
THEitE.FOR.E;
TrLANSFOIZMEJLS
TO'!A l ::::.
l
I. : 1·97 I..:
I,= 0-a?)(ta-ls)
2'400/240 VOL.TS
I
~
sz.o8 AMP
KVA-.
=
43-SG
E1... = 10
Es
F'ER. Pt+,A.st BASIS
TOTAL \<;VA = KVA,
I:z
Ze,
81-!;
EcP
1:L
8-40
Irs
10
IHn.EE 50:1
STEP-DO\VN
THE
ti--'c'
D== '50:1
THE SECDNOAflY
LINE' VOlTAGE
Sl'EF'-DO\v'N X-rORMER
=
12£? KVA
1-1371-z -------- e'b(l)
LINE VOLTAGE
SOLUTION:
.
.
r--~-----------'··- ~I
ARE CONNECTED
A- y
FOR STEPPING
?,500- VOLT THREE PHASE SOUI'lCE · CAlCULATE THE SECONDARY
LINE VOLT AGE.
~---------
'Cei<O<JV
I
-;:-r
-[5
Tftl:'11.EFOILE:
E'L AT f'lliMAILY
LINE VOLTAGI::.
TIMES
E.ps
= J3
XC\
IS EQUAL TO
1'52,000 VOLTS
PHASE VOLTAGE, OR
{3 E4> -
E<ts =
J3
FOil. \v'Y E
J:4,6
EL =- E¢
=
:r
(3 o)
2S'4Sl VOLTS
SINCE E~ =Eq,
rOIL
DELTA
',!,
ELs
a= EL
fs
IN TERMS
\vYE -- CONNECTION
() "'10: 1
VOLTAGE
SiNCE;
CONNECTION
5
FIGUIT.E:
AT. DELTA CONNECTED TRANSFOILMER.
"'t.t_.t-.
Et~_E<j>
TRANSFORMATION RATIO
[\ ==
Ej
'2
DELTA
VOLTAGE
rMASE
' '';;;>.,
LU
AT \vYE CONNECTED TRANSFORMER
THE
U~l
DElTA
I
ll"'
FOIL
·-
\v'YE.
'EL"' LINE
E4=
···,:.· '· ~
E¢;:
~,!
r
LEr
TRANSFORMERS
§
[:!!.
f~
ILEQ/o·.
SECON.;oAR.Y
Y~Ll.
.i\___1
'Ec
· ••
t KVA,
2-~2
8- '\-!;] 'THfl.EE 10:1
AilE CONNECIED
TflANSFOfZ.MEflS
132,000 -VOLT THILEE-F'HASE' TilANSMISSJOIJ• VOLTAGE.
~IVEN:
+ 43·S
AT
VOL..TS'
Ec= 1"52,000 VOLT, 5-F"i+ASE
Ze,
TOTAL KVA
39,7~0
Ek.f
THl'S
CALCULATE
81-S KVA
f3 (?300)
SECON[/ARY SIDE
STEPPING' 00\v'N
% "\'3-1> KVA
I•
<f-.1!;
THE
P fiAS'E VOLTAGE
lD CHetK:
I , = - - x Iz
UP
=a
_:_t_
KVA-.:: (2.4)(\fHT;)
TI-IIJS;
J, =
tL = 10[3 E<l>
f.:L = LIN£. VOLTAGE
}.., = - -
2100/210 VOLTS
1'2SOCO ,
'0 -45
OF Pll013.
SOLLITION:
5?-0B = 1·97Iz +Iz
RATING== 7S KVA
kVA
II,
B-44
0=
=
2~4-3
VOLTS; LINE VOLTAGE
,!'
AT SECONDAR.Y TER.IVIINALS
OF I"HASE VOL1AG'E
OF
DELTA TIZANSFOilMER.
Et/£
~s
FOIL
; lO
111
:IIl11 ·t.
/"
,,
TflANSFOflMEflS
8-'t~ THREE TftANSFOrlMEfZS, CONNECTED A ON !HE PfZIMARY SIDE AND
Y ON THE SECONDM1Y SIDE, STEP THE VOLTAGE IJO\VN FfZOM
f60 VOLTS
AI\ID
OF TfZANSFOfitvlATIOJ\1 OF tACH TITANS'FORMER;
CALCULATE : (o.) THE IWITIO
(b) THE KJLOVO!X- AMF'ETZE
13,200 TO
A 7SO -l<YA O·B fD\v'ER fACTOfZ.. LOAD·
DELIVEfL
AND
I<ILO\vATI LOAD ON
EACH lfiANSfOilMER;
I
PRIMARY
\v'IRES ;
LiNE
(C:) THE
EL"'I3.21<V
d) Wfl.f'lENT
f
1
TRANSFORMEI'Z \-/IND!NG
__K_VA...f./_><__FO_It_M_e_n._
P4>-
13UT
Eq,
Ccf>=EI.
13 · 2
!I
-f) CVfl.RENT
·~·
LOAD
~~
I f.FooOB
fl
=
IN EACH
PIZIMAf(.Y
lTIAN~fOfl.MEIZ
= 18· 9'1- AMP.
-!(;V
.J3 lp4>
f5 (18.94)
\viNDINt;;
/<VA /x'FORt.II.ER.
Ir<~> =
lpL"'
750 KVA
E~
'2!00 I<.VA
Ip.p =
Ir~.- ==- "5Z·!3 AMP.
lpq,
=:.
1'3. 2-J.N
18 ·'34 AMp
£,_
I
v
8-i9] AN OPEN- Ll
---Ir
--
t>--~
DELIA
GIVEN
b)
Siff'-DO\VIJ mANSFOrtMEil
13,200/460 VOLTS
t=OR.MER?
/ TILANSf'Ofl.I\1E:ft
LOAD KVA : 7t;O
fQ\vfll. FACTOil =0.8
=
c=:
illlANSl'ORMl:R
KWL;
lL. =
Is =
1p~
Ipq,
/TV,AflSFOI1MEit
-
X'FORME.Il
ll
PIIIMAJtY
II
~ f'rtASE CVRilE.NT
IN FIWN.lll.Y
/
KVA"jX'Ft>ltMER.
SQU)T/ON'
E<i>s
0) f'U\110 Of T/LANSi=DIIIVlATION
Er
()=- =
Es
lL
E<!>
--
(2,;0 KVA )(IOOO)
4-GO
Ets/(5
/-13
E
L
b)
KV,4 '-/TftANSf=OfU·1ERJ'
)'
:I
FIG: OPEN DELTA BANK
AT Of'ei.J DELTA
·b) \<VAL
IsL -= -[3
suT
Is~=
lsL =
KVAL •
-cL
-~
r;o WA
=
---
O.'fGO.)i;V'
1'30·45 AMP.
=
Is 4 =
{3
130. "1-i>
"'"
.J3
7S. 3 Alv1f'
I
!
I,
= lsq
X
l:q>
= (]!;;;)(O."fGO K.V)
ls.p
I Sc
Is40
49·7 :1
I,
/")( filfl.MEil.
\I-IUS;
a'"'{3l4wJ
-Is
SOLUTION·
a)
'I'
l
h"' 941· 33 AMF.
\' 13,?00l
u=
=
A
D
o) Is41"'?
.
G) CUrtfl-Et./T DELIVERED TO THE LOAD
IL = - - ' = - - - -
0
)
'200 K\v'
II
L
f(:fQb:
x n=
/X FoRMER,
= tf50)L0 .e
::::
II
OOES EACH TRANSFOn.Mt=ll CAn.rty 7
lOAD== fPO KVA AT 4GO V
CtJIZIIENT IN THE SECONJ/ArtY \VINDING
II
IN THE SECONDARY OF EACH TfZANS-
(b) \v'IIAT KILOVOLT- AMPEftE LOAD
!<VA
= i<VA;,
k>
Dl::LIVERED TO THE LOAD
CURI'LEI\IT
FL0\1/S
LOAD OF 0 KVA
OPEN OELTA X'FORMt:llS
3
=: ~r;o
KVA ON I':ACtl
A BALANCED 1H17.EE- PrtASE
GIVEN:
7SO t<VA
LET:
KVA/
DELIVERS
__,Jo;L
klv' LOAD IN EACH X'FORMEit
KVA/
\<VALL
Q= TRAf.JSFOfl..MA1tON llATIO
!3AfJK
AT "'GO VOLTS. (tl) \v'HAT CUf'W.ENT
\vYE
FIGURE
!I'" Ill
'1'1:
lf\US;
.#! -
'1,1
:r
I~ =h =94'1.5'5 ~942 AMP
IN DELTA Ofl rrtl1v1ARY SIDE
-zr;;o~
1r4>=
TL
IN EACH SECO~IDATtY
i
i/
f'lliMArtY TRANSf'OrtMI:.ll. \"INDING.
~
e) cumm.JT
IN THE PTZIMAR.Y l:I).JE lvlflES
1
IN cACM SECOIJDAR'/ Tl"lAtJS-
CURRENT
r--::--r,-
CONT· OF PROB # 6 -47
I
(c) Trtc CURRENT DfLIVEI?.ED TO THE LOAD; (d) THE. CUrtRENT IN THE
POI'lMEIL \v'JNI7/f.iG$ ; (f') THE CUilflfNT IN C'ACtf
TfZANSfOfiMERS
i:;
113
34.7 kVA
il : ~I
lTlANSFOTZMEfl.S'
8-S~
1\VO TltANSf=OilMEI1S
ARE CONNECTED OPEN A AND DELIVER A
SOLUTION:
ON EACH TflANSFO/lME/l ;
(b) THE PO\v'Eit DELIVERW
a) CU11.nENT IN EACH X'FOrtMEIL
13Y EACH
\v'INDING'
OF THE T\vO TrzANS'fOilMEflS j (c)Tt+E AVAILABLe T/lANSfOILMI:Il KILOVOLTUNIT OF THE SAME
AMF'EIZ.ES
Jr A THifl.O
IS ADDeD
TO FOR/vi A 4
LOA 17 KVA
;b) .FJ?\.v'ER JZ~LI.'!l;TU;D. ~y .J;;.Aetl OF
= BG ·7
I)VA
kVAfT =- KVA PE:R. TRANSFO/lMEfl..
fT==
D.ELIVCR.ED TO T+IE LOAD
1=orz. A TH1f1.D
x'FOJIME!l
Q)
KVA LOAD ON EACH Trl.4NSFOIIMEJZ
kVA
KVAjT
8G·7 KVA
= ff
KVA/ == KVATOT4L.
/T
'l.
l~
{~
tf.J EACH )( 'rortMEJ1.
ON THE SEC· SIDE
(Is)
\\>0 K.VA
\(\v' X 1000
--,--:---:=
Is -
'2
P·F.(Ys)-[3
\.::V"i~
:·,;·
Is=
1!;0 l'VA
=
7!;;: INA !lATIN&
130,000
Co .a~D<O)(<rGo5C.Js)
Is=
0
\BB·'f AMP.
p, = so O!S (3ri.B7 -go)
8-5~]
Pz
=
t;;O COS (7!.rNJ7+30)
F~
=
19·6't K\v'
C) I<VA
-[3
==
=3
SCOTT CONNECTION IS USED 10 TRANSfORM FROM 460 VOLTS
THE
THitEE-PHASE
TO II!; VOLTS
T\v'O-FHASf.Jr THE LOAD IS
4-0
li.VA,CALCULATC.:
(_o.) TttE CUI"U7..ENT IN THE MAIN AND TEASeR.. f"RIMAR.IcS j (6) THE
f'OR. OEL-TA ( t;,) BA N K
kVAA
SO I<VA
KVAjT
\Vlt-IDING
f't = 49 ..:;'<f- 1<:\v'
TO Be AI7DCD.
130 KW
D.BGG
=
&s.~ AMf.
b) CUrl.ll.E.NT
cos-' o.g
fr= :;G.e?
f'z = POiv'EIL AT TILANSFOILMErz. '2
J<YA 3 "' KVA
Ip=
gur:
POiv'ER. Of Tf1ANSFORMJ:Jl. 1
TOTAL KVA =
Tt+E
P, = KVAfr COS ( -().-30)
AND
P2. = KVAfr COS (-e- t30)
LET:
ON Tl+E. Prtl· SIDE (Ip)
130 1<\v' X 1000
FOIL OPEN DELTA
SOLUTION:
C) KV.A RATING tiF EACH >(RJ/l.MEIL
Ir = -0-·-BG_r;_x_~_s_oo_
Tr<,ANSFOR:ME~
· T\v'O
PO\vER F'ACTOIL =0·8
p, =
lUTING AS ONE OJ= THE OTHEfl. T\v'O
BANK
GIVeN
~
~ 8-51)
(COI-JT. OF FJWI3.
"'
LOAD Or 86 · 7 KVA AT A POVEIL FACTO!t OF 0.8 · CALCULATE : (a) THE ,,KILOVOLTAMPeRE LOAD
r
TrZANSFOflMER:S
AND TEASEIL SECONDARIES; (c) TttE I<ILOVOLT-
CUrt!l'EIJT
IIJ iHE MAIN
AMPE!t.E
I'LATING OF t:ACH
OF THF- Trz.ANSJ=DRMERS'.
I<.VAjy
:I
!.::VAL> =(3)(£;'D)
KVAA = l!;O I<.VA
\It; VOLTS
a-s1] THE P.iZIMAIZrEs
o'F
Tl~o 1-~ANsro~MErz.s:-~l'?~.'~NN:E~J?,:TO.
A 2,3oo-
VOLT T\v'O F'H.ZI'SE ''SOCJm'E · i"~-S'fCDNDAfzlfs. AR.E cONNECTED SCOTT,
T , AND
DELIVER
FACTOCL. LOAD.
\1/INDING
\v'INDI/JG
4-6'0- VOLT
130 K\v' THReE PHASE O·BGG LAGGING
PO\v'E[(
CALCULATE: (0) Tl-tE CUfl.IGENT IN EACH TII.ANSFOIZMJ:Tl.
ON THE
0/J
A
L·C'·, IN
Pfl.IMAILY SIVE ~ ( 6) THe CUrz.iLENT IIJ EACH T/lANS'FO/ZMER
THe SECOI-JDAfl.Y SIDE ; (c) THE I<ILOVDL-T- AMfER.E
OF EACH OF ntE T\v'O TflANSFOJLMEflS'.
llATJNG
~
2 FttASE
:z
1EilMINALS
'<1:
~
1:::!
[___jV~T
~
~
<t:
f
4GOV
~T!:ASER.
t<1
GIVEN.
SCOTT CONNECTIOIJ TllAf.JSI''ORMETl. (SECONDARY)
Pc=
t;o
K\v' ,
"160
vo~ T
, 3
q,
1"0\v'EIL t=ACTOI'l = O·BGG LAGGING
TITANSFOJIMJ::IL
FIGUTZ.c:
GIVEN:
40 !'VA , 460 jnr;
scorr , r, TltAf.lSFOIZMEfl
II
114
l\
115
li.
.•• 11'
1flANSFORMEI7..S
SOLUTION:
SINCE THE T\vO SECOfJ.DArz.IES
Cl) Tt!E CU!Wei\JT IN THE M.41N
i<.VA
I"''iT =
Df':LIVEilS'
ffl.l. \v'INDINGS
fOfL A "6- PHASE
>t
INPlir
'lOOOO
1... -~,l(==
_!_]
398j
MOTORS
AT WHAT SPEED WILl A 14- POLE
Of'f:RATE IF WE SUP IS 0·09?
=
II!;
17t AMr.
f'OLE = 14
INDUCTION M01Vrt
LET:
-f
=
s
= 0-09
I
~0 CYCLE
Ns "' SYNCH!lONOUS SPEED
Nrt =
~
c) KVA 1?-ATING OF MAIN Tn.ANSf'O~Mt:.rt.r.·
GO- CYCLE
GIVEN:
?.0 KVA eACH,
IDO!J
J3 E<t>
INDUCTION
,~,
CONT OF PfWB · 4F 8 -!;2
Af.JD TEASER
'
f'OLYPHASE
.,
¥,f, I.
S =
ILEQ'rJ.
rtoTOIZ. SPEED
SLIP
SPEED (rlYM)
4-60 ,(r;;o.z
4-0000
J5 ( 4100)
1M~T
=
!;0. 'Z AMP
f:(VAM""
KVAM =
2b KVA
(12o)(Go)
l20t
AT MAif-J
Ns=-p
KVA OF TEASER. X'FOILMER.
b) Clifl/ZE'fJT IN M A I N /
14-
Ns == 511-. Z9 ILPM
4GO X \>0-2
--rt::hSEfL. S'E.CON!7AILY
T
-M1~S£c) =
SOU!TION:
1000
\(VAT""
?.0 )(VA X 1000
,\'.VAT=
NIL= Ns ( 1-S)
1000
Z~
= (Sf~. '29) ( J- 0·09)
I' VA
J-...
N11- = 1-GB R.PM
lis v
~J
THE
39Bj
IS
NAME- PLATE
SPEeD
o.) lliE.
b)
es-CYCLE
THE SPEED AT
7ZO RPM· IF
CALCULATE :
OF A
NO-LOAD
INIJUCTlON
I\IJOTO!t
IS 7-tS rz.F'M,
SLIP;
THE PE!l. CENT ltESULATION
GIVEN:
.f=
II
e5 CYCLE
N.L =
I'
1
7ZO !rPM
~·I
NO-l.OAO SPEt:V
= 7"1'!>
ItPM
SOWTJON:
P
~ 4 POL!=$
= (JZO) (t'£;)
7ZO
THUS;
N.s
= 1~0 f'
= (lt:O )(t !;)
F
Ns""'
o.)
NJ\..
s
1'
7SO
= 11.0
Tt..PM;
- Nl'= N.r Ns
s ...
rt.PM
7~0-
1e0
=: - - - -
750
CJ. 04 J--...
II
ttG
117
i
.~ fi
POLYPHASE
CONT. Of PIW6.
N~.
POLYPHASE INDI)CTlON
,'1
MOTORS
I'~
CONT· Of PfW!3 # 4
f
IZEQ~p:
N... L.
1- R = lv'H·L -
b)
I
INDUCTION MOTORS
Np.L. = ~JJL
.J
NJ'.L.
a) Roron. SPET::V
WHERE:
b)
:,.'/
Nlf.l·
=
NO-l-OAD SPEf.l7
Np.L.
=
j=ULL- LOAD SPEED
7 'j-s; - 720
'/. lL =
iz.EvOLVING FleLO SPEED
SOLUTION:
a) NR
Ns
=
(I-s)
(!'20) (60)
X 100 /.
Ns =
7'20
;. fl.= .5-47/. /<>w~..
N~t-
= 7?0 !Z.Pl\11
10
=(no) ( 1- o.075)
N R = GGG llP/11 ./~.
_E._]
.39!Jj
FOR. HOW MANY
IF THE: NAME
IS A SO- CYCLE
POLES
SPEED
PLATE
fl.EVOLVIN~ FIELD= l'ls -NR
b)
MOTOR WOIJNIJ
INDUCTION
IS 460 RPM?
"= 720- GG!d
=
GIVt:N:
f=
!70 CYCLe
Ns= +GO rt.PM
5 :
rz.eo/v:
A
TtiiZ.EE- PHASE
3EJ9j MOTOR
NUM3Ert
OF pOLES (P)
IZO~
1\ls=--
f'
r == (lto)(so)
tGD
MACHINE~ THE
POLES
MtJST BE
IF NOTUv1AL
(a.) TftE fWTOIL IS AT J1.EST ;
ilOTOIL
IS 17llNEN
on.
11- POLI::S
~0- CYCLE
A THrr.EE -PHASI::
FULL-LOAD
THE:
THE
AND ffiEGUt:NCY
IS Af'PLIED TO THE STATon.
(b) THE IWTOfl.-. SLIP
BY ANOTffEil.. MACHINE
f
./(kola .
SLIP
!0-POLE
tt.E.VO\...VING
INDUCTION MOTOR
OF 0-07!;. WI+AT
rc.oTOit. ltELATIVE TO THE
R..EVOLV/N@
INDUCTION MOTOTl.
=GO CYCLE
FIELD
IS lllE SPEED
!l.EVOLVING FIELD
!Z.ELATIVE
?
To TttE STATOR ?
OF
J-IAS
Z 30 VOLTS
SOLLITION;
Q.l
E.rt_
=
S
XEs,_
Ea 1L/ = 230 X
,<j>
_, S =I
0·7ri
=
, SINCE MOTO!L IS AT flEST
llt·!,; VOLTS
Cf!. =(f) (171?-!;;)= 17?.!; VOLTS
S=
f=
IN A
FIELD.
CIVE.N:
f::.
IS 0-04 _;
AT BOO !l.PM
"G/VEI-J:
t: =
a)
b)
IN
AS
P = G POI,ES
P = l'l
399j A
WOUIJD_ !Z.OTOrt •
AND ITS IlOTOR.
fl.DTOR CONDUCTOM
VOLTAGE
FJWM THAT OF THE
OPPOSITE
THLIS~
_!]
MAIJY
AS
VOLTAGE
SLIP RINGS
3-cP
AN EVEN NUMBeR
PER-CENT
THE
/?l!l.ECTION
USEP IN A
7S
AtLE
230-VOLT
JN DELTA
STATon. CONDUCTOns · CALCULATE
W!+I:N
IS NO AVO NUM6Efl. OF POI..E.S
TI!ffl.E
GO- CYCLE
SETWCEN
(c) THE
=13
Sf X- FOLE
HAS ITS STATOR CONNeCTED
STArt· n/EILE
SOLUTION:
SINCE
S4- IU' M)'aov.z. .
SO CYCLE , 3 PHASE.
VOLTAGE BE.TWEE"N STATOIL TETUI11Nl\L.J'==
0·07S
V=
10 POL~-$
V=
118
f3
(17Z·S)
299 VOLTSJ'a.u.
119
,fj
ER
POLYPHASE INDUCTlON
I
MOTOITS
-~
N2. 5
CONT. OF PIWB
::,}
=S
a) .J:r
~ ROTOR "fllfQUE NC"i
X .f
SPEED
t;;O CYCLES/-..
tr= f.'.<J-
Pfl.OB. #" G
RPM ;
IS 895
-t =
f3
X
P=
S=
s
f730
POLES
8
!'20
f
p
(1-zo)("o)
Ns =
= l-2'00 RI"M
G
1-Jil
a)
I "ZOO- (-BOO)
Ns
(1'20) (GO)
NK
€·9
Ns =
Ns -
=
p ·""' !"Zo-f
BOO ft.PWJ
l'ls =
WATT.S
SOLUTION:
Ell-j~
i:IL = 11·9!; VOLTS /lfMf4..
c) Nrt =
AMp.
= eo, Boo
P,~pur
"THE ROTOIL VOLTA&E $!:TWEEN 1E/Z.MIN41.S
SQUHUU:L-CAG:E J./VI.
bO CYCLE
lt.. = G'-t
~-9 VOLTS
fJ'. =
(d) Ton.Que:(e) EFFICIENCY:
NR= 850 RPM
' ·,~
..[5 x
IF THE NO-LOAD
flEGULATION
FACTOI7.;
Z20 VOLTS; 3-PHASE
=
f
= (D.O"f- )(ln?.t;)
EJl. =
(C) POWEIZ
FOWEIL = i?f> HP
CPS.,/GM<£.
f:p../<j! ~ S X .f:BILA,
Ep..jq. =
MOTORS
GIVEN:
(0·01-)(~o)
1.) .fr =
CONT. OF
INDUCTION
CALCUL.ATE: (a) SLIP~ (b) PER CENT
= I X 60
fr=
A?LYPHASE
s
I"ZOO
8
900 RPM
Ns - NR.
= ----Ns
s =
= l·b'7
(l'ZO)(Go)
"" ·----
900-8.50
=
900
0-078 f().t<d.
TI-IEl-1 _;
.fr = Q·tn)(t;;o)
.f.r=
Nu., - NF.t
b) 7. n. =
NF·L = NR
JOO CI'J'pf..__
NF·L
89!; - 8.30
X 100
-~-·--
fJt./4> = J'
=
X
£.sn...jq
830
J
(Hn )(11t-6)
i::IL/ =
'ZS?· J;"
4
c,)
YOLTJ'
:EJt. BETWEEN rto:otL TE1U¥11NALJ'
.EJL
= [3
=[:3
IZ =
7·83 /.
f
7
f'~
f
PF= - FA
E~'-f¢
i,:
WI-!EC7,E:
PT : TrwE powen: (WATTS)
PA = AFPARENT r~awErL (VA)
X t87·J;
PA;[-3 EI
~
f:..rt. = "ftJg ')(. !>00 VOL. T.J'
J3
(220)(1?1)
.., ~f'
fAc• 21,.38/VA
G
J
39!7
rz,'<-<y'b
~,:y
rHEri.EfOJC.E
THE NAME
THE
PLATE
OF A SGUII1.rtEL- CAGE
FOLLOW/1-JG tNFO!LMATION :
GO CYCLE-S , 630 RPM , b4
20,900 WATTS
WHEN
25 Hf' , 220
INDUCTION MOTOR
HAS
VOLTS, THtl.E.E PHASE ,
AMP PER. LINE • IF THE MOTOIL TAl<ES
~'·r· =·
2D,eoo
24,387
p, ~ '20,800 w
PF = 0·55"3
orEMTING AT' -FULl-lOAD, CALCULATE.:
120
PR
F·f
121
POLYPHASE
INDUCTION
I
MOTOIZS
IIRII!I
POLYPHASE
l
CONT OF f"IWB.
d)
I;
#-,_~
~;,
CONT-
fZPD = ROTOR
J
FVYVER
b)
tJEVELOFt::D
NR = 1720 ILPM
NR.
7·04 X (2>;)(74-G)
830
1 =
(1'20)(1!>0)
1'-ls =
p
4
!800 llPM
IS8 FT- L5-
e\
f
f:Ff' ("?_)
'
11111111!
l'ZO-f'
= --- = ----
Ns
=: ------~
T-
OVTPtJ T
X
SLIP(S)=
Ns-NR
=
0-0"r4-
100 j.
S
fiNt'tlT
(25) ( 7 4-G)
MOTOflS
OF PfWB # 7]
7-0'1- X fl.PD
T=
INDUCTION
X 100%
lJNDE(l
1800-1720
N_,,
1800
:
CONDITION IHE fLOTOfl_ CUfW.ENT IS~
TI-lle>
'20,600
'1 =
A
7 l
STATOR AND A
NUM6ER Or CONDUCTORS
AND fl.EACfANCE
AT STANDSTILL
RESPECTIVELY.
IF THE VOLTAGE
AND
(o.)
TtlE.
FREQUENCY
AT STARTING
IS
j(p~,._
MOTOR. .HAS .4 Y-
A- CONNECTED
ON
_; (b) WHt:N TtiE SPC!=D
n-tc
lS
GIVEN·-
7
r0-18 ]2
-i
,1""'-'J..
I.S -3 AMP.
TR"'
110
.-
fWTOJL CU1ll1ENT:
IS L720 RPM.
+(XBP-)
[o.oH + (o 7!7f
PE/1. PHASE,
IMPRESSED ON THE STATDfZ
r
110/[3
="'
rLESIS'TANCE
AND 0-7'Si OHM
fJO CYCLES, CALCULATE
r~
fWTOR, WITH TtfE
EACH· 11-IE fWTO!l
ARE 0-18
EBIC/ [3
IR :::
Ja#;L.
INDUCTION
WOUND-IWTOIT
fOUR-POLE
399j CONNECTED
SAME
•;
99-G'G' /•
3~
PR06. 7 , CALCULATE
IN
PErt . PHASE
EQUIVALENT
RESiSTANCE
LOAD.
THE
Or
THe
,,,
P= <t,- POLE , WOUND- Tl.OTOrL
Y- CONNECTED STATOR ,
SOLUTION:
INDUCTION MOTOR
Ll- CONN!':CTfD IWTOCL
THt= EQUIVALENT
fl"-/q,
=
0-IB ..JL
Xeryq,
=
a. 79 JL
=
f- =
cBR
n"~/q,
110 VOLTS
o)
IR
=
:'
PHASE OF THE LOAD
I
=
4
0·18 [ I- 0-040-04-<t
3-9 OHMS
1
J
I
,[
JaJJJ'.
,I,
1,,,,,I
'Ill' I'
Ee"-/.f3
1111'11'1'
j(RR)7. +(:>(BRr
110
!,1'
1'!!'
,,l.~j,
9l
v
{3 x Jco-ta) +(0-7<;)t
1
II(
= 82-34 AMP. ,f,._..,__
1
A (00- CYCLE
3 ~j STATOR
LAYEIL
a)
'230- VOLT fNDUCTJON
THAT
CONTAif\/S
WINOJNO.
THE:.
CALCULATE
VOLTAGE
MOTOn.
A SIX'- POLE
HAS A
!.J
THE
twTOil..
90-SLOT
Y- CONNECTeD
DOUBLE-
ACM!:!S ONE COIL
GROUP OF THE
SPEED ,4ND t'"f!.EQ IJENCY.
123
I''
l :i
,I
J
WINDING.
122
:~
IS;
-s
[ ~]
RK
=
Re.ycf>
I
=
G"O CYCLES
SOLUTION:
R.ESISTANCE
INDUCTION
POLYPHASE
CONT. OF
MOTORS
I"Tl.DI3. # 9]
I
i
,:;:
10
l
A S-HP f2fJWEltE
-f= roo CYCLE
CONNECTED , DOU5LE LAYER. WINDING'
SOLUTION:
o.)
VOLTAG'c ACfl.OSS' ONE COIL GR.OUP OF THE WINDING
i=sR
:230 v'
E
/4>
SINCE
VOLTAG'E ACROSS ONE. COIL GROUP
V;
1:52·8'
EsP..jcj,
-
/COlt. QRCVP-
=G
c;;
zz.
!o VOLTS _ft:l.+14.
!S
o.) Ffl./CTJON ~
HOTOR
COPI"EJ:l.
S =
Ns
N R. =
=
OUTPVT
!j1
TORQUE.
k)
PEK CENT
~) WAD
WATTS
E:FF!Cit:.NCy
fOWe.tz FACTOR
GIVEN:
!; HP
P= 4
_:,
GO
J (I-
=:
P-t- 1~0-LOAD
NO- LOAO COPPE~ LOSS = -:.5 X Ir~.L r
= (.3)
P.F- W• ~
rn>tl LOS-I'
=
X
COPPER LOSS
RAe:-
(b'·Z) 2 (o ..:s)
= 31-·G'
1-J.r ( I-s)
X" 3.£" WATTS
¢ ] 0 - 3,!;"
,I
£.75 WATTS.!~.
0-0'28)
I
lo)
STATOR
COPPER LOSS
=3
-=( 3
xf
=
=(0·028) ( (;o)
-h-= I· 7
11:-IDUCTJDN MOTOR
CYCLE
0.) fRJCTJON~WINDABE~ IHON LOSS'EJ'
1200 RPM
= S'
3 -f'HAS"f!
POL.ES
== (:....rz_o_)..:...(G_o_)
N R =- ll GG' RPM f C~->W. .
-fr
UNDER LOAD;
(0
(!tOO
11
UNVER lOAP;
fN
0·028 SLJP jcoJL GMI)P
p
Ns =
3,C:SO ~I :=II<'::; !1.PM=1,7JD
SOL.LITION:
(131.!3 )~ COlL GROUP)
= _1-z_o_+_
Pt=
ANP I!WN LOSSES';
LOSS
f.IORSEPOWER
.f.=
22.1:3
WINDAGE
f)
ER = .S X &~t/q,
S=
220;
e) ROTOR OLITPUT
THE KOTOI<. SPEED ,J.fJD FREGUENCY
SINCE:
Vt..=
r·;
A -C !l.ESISTANCE OF .STAWfl.. PDR PHASE = 0·5 ..!l...
b) STATOR. COPPER LOSS'
c) !1.0TOfl. POWER INPVT ~
HP =
/o)
= 220 _; P-~;.= 310..: I= "·2
VNL
CALCIJLATE:
a)
132-8 VOLTS
G A?l..ES, TIIERE ,<1RE ALSO r; COIL .GROUP
AILE
TtfEfl.E
Tl-fi.JS j
=
= --
I
TEST:
EFFECflVE
P= 6' POLES
Y-
AND THE FOLI.OW!NC DATA
WAS TESTED
LOAD TEST:
SL.OT= 90
Ttll'lEE -PMASE
OBTAINED:
NO- LO.AD
= Z 30 VOLT-S
:E.BR
MOTOrtS
FOUR- POLf 60- CYCLE
VOLT
399j INDUCTION MOTOR.
GIVEN:
INDL'CTION
POLYPHASE
X
rL :z
l(
lJNDER LOAD
RAe
!IIUiil
I!
) ( 11·3)::1. X D. 3
l/!; WATTS,ft><u.
~ !· 8 CFU' p;
:/a.t...
11:'11.
.,1,·'
1,'1
124
1
125
11'
1
1
1'
l ,,, .
11111':.1'1.''
II:.
'
'
pOLYPHASE
CONT-
INDUCTION
MOTORS
OF Pfl.05. #- 10]
c) fl.PI
=
Pt -
POLYPHASE
i>
h)
WHt=:RE:
P.- =
=:590 WATTS
RPl = 32rJO WATfS )IQM4..
ROTOR COPPER LOSS l)NVER
RCL = Ik
2
S=
S=
mus:
e)
+
= .3,097
LOAD
'j.
-= (3260) ( 0 -0!;)
RCL
=
=
3~097 WATTS)'~.
p.f.
11
J
·:;a
A
dF
OUTPUT
ROTOR 0UTPtiT
7<fG
.3~097
w
74G-W-ji-H-'
ttP o\JTPVT
<j-.1!;
!-1-P. /a-.
TOR&>UE
I=
!'<PI
7·0+ :><
T
7·0'1- :><
T
I.Z· 7t;
Ns
L
(vA)
w
Pr=
.J3 (uo)(ll·!>)
= O·B•t-7
=GO'IO
THE MOTOR
OF TI+E MTDR
ROTOR
.
r~~.
.
- -:---:-:
BLOCKED- ROTOR
VII
X O-e5"
JEST WAS"
Pt.:RFORMED UPUN THE MOTOR OF.
CALCULATE :
WERE OBTAIN e);> :
(~) TI-lE EQUIVALENT
Vs~t =
48;
RESISTANCE
PEK PtfASE
R., ; (1J nte E=QUJVAC..ei-IT RESlSfANCE
PER PHASE..
R~ ; (c) THE
REAcrANCE
PER
PHASE
X sF..
EQUIVALENT
BLOCJ<ElJ_
•
GIVEN:
Vsp.. = 4-9 VOl-TS
I-= IS AMP
Pt."'
'})
rA
3<;;s;o
PttoB· 10/ AN17 THE FOLLOWING" D./'TA
I = 19" , f'-t:
Hf'OUTPtiT =
/cuw..
S-t .gs; /.
P-.r (w)
p. F- =
RP! - RCL
3280- !lb5
lfORSE fOWER
X JOG;:
3/;; :;o
l'6WEI< FACTOR
P·f' =
!G.5 WATTS/~.
ROTOR OVTPUT
.3, 097
""'l.. =
l:-f'F
...t_) LOA!J
!fWO
RCL
i:
II
WATTS
RPM
O-DS
ltOTOH OUTPUT
f)
= 3Ci!;O -11!,; -!G3- 27!;
fou-r
=
Ns
FRICTION, WINPAGE~ !RON LOSSES
ft'Wfll. OUTPUT IS Al.SO EQUAL TO ROTOR OUTP~T :
!BOO -1710
Ns- NR
il
100 /•
P<>o"T
'{_ =
= IBOO
o;
X
p,.,
= RPI xs
RR
(IW)(Go)
N s ""
p...,()T
= --
Pm - STAT;7R. COPPER LOSS -
!1.PI = 30_!;0 - .390
d)
I(_
EFFICIENCY UNDER LOAD
I
+STATOR COPPER LOSS
'27!; +II!;
PER CENT
'(. EfT.
f?L = TOTAL FOWER UNDER LOAD
PUlls$=
MOTORS
CONT· OF PIWB- # 10]
Pt.oGsEs
Pl-OSS= P~.\0/+IRO... LOSS'
liJDUCTIOI-J
F'ROM
GIO
PROB -#- 16 :
t=:f'f'EcTIVE
a-C ftESIJ'TANCE::: 0·5
32'6'0
1900
llll'r'lll,
La-)=T./a...... .
II
'·
126
127
I
INDUCTION
R:JLYPHASE
-
CDNT-
Of PFWB. #
MOTOR.S
I
,,
!fi.
.\'
11]
I'
.SOLUTION:
[
a)
GlOW
f't
Re
=
J:<e
~J
CALCULATE
399
wHEN
THE STA>ZTII-JG
lT 1s- srAtnED
GIVEN:
Vs~<
3 (18)Z
=. o.G'f!7
OHtv'S
:,,_i·ll
;
(PEK PHASE)
b)
RR
=
Re -Rae
=
o.~e7-
=
RIZ
tI,
Xe =
48 VOLTS
6'10
Rae=
0-3 .D..
STATOR CURRENT WITH
VI'.ATED
J3
IllR
POWER
(18)
INrtlT
VP.ATeo
l
1·4
XsR= - - = - Z
2
.Jl.
lill
STATOR
#a.m. .
'lll1i
[re]
Yar.
l
'2'20
~
COPPER
II
"' f12.f> A
ROTOR BLOCKE.iJ
WITH HIE
f
I·+JL
o. 7
l
J..Q_RJ
i-8
J-!;;"4_Q
Xe
]'2 c610)\ =
LOSS
~
3
x
J
l'Z
39.9j WHICH
THE RESULTS
11,1314 WATTS
TBK :1. X Rae
WILL BE
I, I'•
li'If!.lt
,,1,1
==
OF F'JL08 · II , CALCULATE THE SPEED AT
TI+E TORQUE
I
Ill
2 ( pBRJ
'.'IIIII.'
II'
= (3) (8?-5)'2 (o.3)
USING
I
ROTOR. J3LOCKEQ
[•
220
= ~(H>4)"-(0-G~7)?
=
.
Ns = 1800 RPM
Vsp_
413
J3
Xs~t
I
SOLUTION:
Vs~~,
Ze
Xe=
-
w
PeR=
Ze t- Re""
ife
Xe
PRO~
l8 AMP.
---
BUI:
OF THE MOTOR OF
voLTs _; (L>J AT 11o voLTs
ri<ICTiON ,WINDAQE~ IIZON LOSSES= 27!:7 J<W
0-327 OHMS
j
zzo
o.3
0-)
c)
=
lB11..=
I,
TORQUE
(Q) AT
MOT017.S
#- 10
rRoM
=--
3XT.BI\z.
INDUCTION
POLYPHASE
A MAXIMUJVI.
RP1
=
GI'2Y -G
~~~~~~.·
111.1
WATTS
P,NPur - STATOR COPPER. LOSS'- F._, W +!RoN LOSS
11·
,1'1,)1
J!r:
1
-SOLUTION:
f?.JI..
RPJ
SLIP AT /VIAXJMUIVl TORGUE = - -
==-
11;814 -
GIZ£" · 6 -
G413. 1-
WATTS
'27~
'1'111
XgR
{}. ..521
s,..,_T
= --o.7
Sm.r
AT
'220
=
1800
lrll,,l/11
VOLTS
= o. <1-6'8
/,,,
7-0t X RPI
lSTARTI"-'"
Ns
=
Ns
[<.PM (FROM # 10)
7-04 X
RPMMr =
ij{'".'i\'1;r
N.r ( J- S...,t)
G;413 ·4
1800
l,fl,.·'.
ljl
,111
'I
l
Tsr. =
= (1800)( 1- 0·4-68)
RPM l'lt = gs;y.r;
~
?S.I
FT-LB/ruw..
9,!;9 RPM.#;
i'l' '
li'!ill'il;.
0-tW...
:I'll
~ ~ :f
129
128
lil1
1;!
'1,.
•111'II'
II
1
'
ft)l.YPHASE
f'OL.YPHASE
INDUCTJOH MOTOflS
CONT. OF PM6. # 1"1-]
CONT. OF PMB. #: 15)
SOLUTioN:
b)
TORt.WF. AT
STARTING'
STATOR CIJ!.Ulf.NT
WITH
VAATEP
--
(
VBR
1!0 VOL lS'
I~-.)
: : :- - - -
(
48
I~>. R!<. [~]
2
HPOUTPUT=
ROTOR BLOCKeD
liO
rOR tiP OUTPUT
.
18
)
t3U1~
CJ!IR
IR=
WrW UOTOR 13l.OCKeO
f'OWEIZ INPUT
{(?Y+ Xs~
STATOR
,.
vAAlE.P lj -z( PsllJ1 = [''o
Jt ((;101,1
Ys11.
1'$
.3203. (;
"'(3) ( 11· z~)
=
l~3J·
f!J·lr; )~ + (O·!:if
lo.o~
WA TT.J'
==
COPPER LOSJ'
2
(
3 X lsR,<t X
1ll == 7!:i · b'Z AMP.
Rc.c
11J-US'
o.~;)
'
!fP
</- WATTS
OUTPIJT =
/397-2
=:
(?t'·(;~l(o.n;)
[/- 0
I+P
= 21·8~ tff'/~
wAITS'
fOR TORQUE:
,AT
IJO VOLTS:
T
= 7·0'/- ( ftP
7·04 :>< RPJ
=
TsTA~<n"'G
1397-
TO</- :X
=-
Tsr.
~:4G)
Ns
TST. =
I'ZOf
Ns= - -
p
1800
n-L6.
t;·5"
z
(l'ZD) (GO)
= - -t - -
N.s == 1800 RPIVl
.f"a..w-
NJt..= Ns {1-.J')
1800(1-0·0r;)
j_!_J
A
399j A
FOUR- FOLE
fWTOfL
REACTANCE
250-VOLT
RESISTAHCE
60-CYCLE
OF O·lb OHM
OF OS OtiM Pt:R PHASE.
AND THE TOIZQUE
POWEI< OUTPUT
INOUCT!ON
MOTOR HAS
P=
4
EBR =
f'O~ES
=
.RPM
CALCUL.ATC
FOR A
SUP OF
THE
S
tfORSE-
PER CENT.
T =
7·04
[
21·85 .X 74G
1710
T = C1·2 1=-T-LB fa.w....
t+P OUTPUT AND
230 VOLT
f =- G'O CYCLE
Rryif o-1~ .JL
X!yq, == o .g ..n._
s
1710
PEP. PHASE AND A ROTOR
P-!O'cv'o:
GIVEN:
tJR =
TORQUE
~~-
130
131
·OJ;
o.or;
71-ID
!<PI = 3?.03.(;' -· 1£31·1 - 27,;
KPl
jS== o.or;
2
230
.!R=
l
:::
S
7'TG
"l-1-Z!; AMI'·
=
INDUCTION MOTOfZ..S
J
J
tr
POLYPHASE
IS"
l
399j
THE FOLLOWING VATA
UPUN
LOAD TEST:
=
CONT- OF PROB. t1: 1;]
Z 30 _.: P-t:""' 8,35"0 ; I
= 23 _;
+ WINDAGE
(c) P.OWR
POWt=l<
IW.PUT
UNDER L040
(d)
d)
ROTOR COPPER
ROTOR COPPE-R LOSS
UNDER LOAD
120f
(i'lO)(c<;O)
Ns=-- = - - - p
G
(e) HORSEPOWER. O.IJTPLIT _;(-f.) LOAD TORGUE _,
LOSS UNDE!< LOAD
EFFICIENcy __; (h) FVWER FACTOR.
('}) LOAD
LOS~-STATOR IP..Or-1
#awJ-.
RP1 = 80!;1- WATTS
~).STATOR. IRON LOSS ; (b) ROTOR CDPPER. LOSS JJNI7ER
CALCULATE:
WAU ~
STATOR
P..Pf = 6'3!;0- Ill- 18!;
f7.PM = I, 1"10
210 WATTS
LOSS' :
ft-
C) ROTOR POII)ER INPUI =
-i-1 0 ; I= 8 .r;
fOff'"J:CTtVe RESISTANCE OF STATOR PER PHASE;::: 0·14- OHM
FR"TION
INDUCTION. MOTORS
INDUCTION MOTOR:
fi. :
V»t. = 2.30 ~
VL
FROM TESTS PERR'.?RMf:D
WERE OBTAINED
230-VOLT Sl>C-POJ.E
A fO-IfP
NO- LOAD TEST :
FVLYPHASE
lNVIJCTION MOTOR
Ns=
1'200 RPM
GivEN:
F-Ff'ECTIVE.. RESI.ITA-t.JCE Or .sTA-TOR
HP""- 10
P=
f'E~ PHASt:: :::: 0
r;; POLI5S
E= 2.30 VOI.-"):5'
J=RICTION
5=
·I:\" ..!L
+ W!.t.JDABJ;
Ns -NR
1'200-1140
Ns
/ZOO
S = 0-0f;
SLIP
LOSS= '210 W
rlCL "' 1-<.PJ x t:;
NO- LOAD TEST:
L.OAO TEST:
Vtil- "" '2~0 VOJ...TS'
Pt =
I =
"='
V1.. = '230 VOLTS
(80!;"4-)(o.ot;)
=
40~
/am_
WATTS
w
"'flO WAITS
Pt ==
e3s;o
g.~;;
I=
:23 4MP-
MilP.
J-lCL
:
fZPD
=-
/~1, HORSC'.PD!IJER OUTPUT
~-,
746
RI>M== 1,110 RPM
f\f'D :::. RPI- RCL -(F-+VII.LOSJ'J
= 8054-
SOLUTION:
STATOR COPPER LOS'S AT NO-LOA!?
0-1</-
Rst"f'E'criii!O A-c-
=
2
llJiJS, STATOR COPPER
=
= 3 (I NY"2. R~;pF(A-c)
0.) STATOR
-403- '2/0
= 7441 WATTS
7441 w
H P OUTPUT=
0-07 ...t'LJfHASE
f)
WATTS
LOAD TO!'<<¥ tJE :
RPlJ
T= 7·Dt x
IRON LOSS= P;,N-l- -F.-+
w.
LOSS- STATOR
c.,. LOSS
T
~
N~<.
7441
7·0'7- X
1140
1-10 _ 210 - IS
STATO!Z
LOAi:l J::f'F!CtE.NCY:
STATOR. LOSS UNDI:::R LOAD
l=.fF =
=
3
(tt.}~l:;Ff'. Ae)
(z:;)z(0-07)
_
3
=
ill WATTJ' /~.
132
fa-.
T= 4!0 f'T-l.B
IRON LOSS'= ff1!;; WATTS.f~.
'}J
b)
= !3-97 J-fp /cw....
w-/I+ p
74-(;
LOSS AT NO-LOAD IS
::: .0 (g.s;)"L(0-07)
lf;
RPD
RPD
~ .. T
7
X
IOO
LMP
't't I
8.3S"O
1.
•
x
Job
cf=r= e9.1 /.' f~
133
J.
LOSS
pOLYPHASE
CONY.
INVUCTION
MOTOflS
POlYPHASE
OF pnoa. :fF 1~)
h)
17
RPI
=
+
IS
TOTAL STATOR LOSS
P
STARTING CURRENT OF
+
IMPflESSCD · WtMT VOLTAGE. StiOULV Bt! .APPLIED AT THE
tP
It; H-P .J 3
,f::
0·91
IF ~ CURRENT
IS Nai TO EXCEED GO AMP?
tr;IVEN:
IB.G +Ill
..)3 (:u;o) (23)
P·F. =
/
A I{; -HP 440-VOLT THRE.E -
JS J3'Z AMP WHEN IZATED ~LTAGC!
INDUCTION MOTOR
STARTINt:; INSTANT
Et.IH
80~4
f;F. =
THE
1-00] PHASE
f'OWER. -FACTOR
P.F.
J
INDLICTION MOTOn.S
,fa..w...
= '1-'TO
INVlJCTION MOTOR
VOL-T.S
IsT == 13£ AMI>·
SOLUTION:
l
~
JF 11-IE STARTIN~ TORU\UE OF
.,.OOJ
MOTOR
RATED
IS
2
Y-z
VOlTAGE
TORQUE
11\/HEI-I
TIIVIES
A 7·5-HP
4~-VOLT 1_730-RPM
FULl- LOAD TORGUE
ITS RATEV
IS IMPRESSED, CALCUlATE
230 VOLTS
IS APPLIED
THE STARTING'
132
AT THE INSTANT QF I~E
=
V
p.~:.a'v=
cwe:IJ:
CA!..CVLATE TI-J.E STARTING
Ht>= 7-t;
TO~QtJE
vot:~s
18
J .A
itX>
3 -HP
THE
Y-
WHt.N
RATEV
WHEN
THE Y-~
Tf'·l·
.SOLUTION:
THfl.EE- PHASE
METHOO · IF Tf+E
N,."' l73o RPM
Y-z
ZOO VOLTS ~~-
<?08- VOLT
.1
Tsr =
2
~'HCJ
(
(b'O) ( "l-40)
v
MOTOP.. IS STAKTED.
E = 410
= 1r~r
Iso
WHSN
VOLTAGE
SWITCH
Zi~ (TN)l ~~r
IS USED?
3 tW , 34' MOTOR
~08
f...:=
VOLT
1sT'=' 5"1- AMP.
BiJT:
T F.L.= 7· o+
x
Tfl. = 7·04 X [
Tf'L
=
I<PD
SOLi..JTION:
NR
(7·5)(74-fD)]
T.4T 2.sov
-?'l-77 f"T-LS
V.p
r y
= (-zhJ(tt.n)lz;;J
-=
.[3
AT WYE
1730
-zgo
TAus"v
i=:L.
Vt; =
TtfU~
.,.._.
IS· g; FT-1.8 /,
LY-.d
134
:ZOB
f-3
j
]y_~
=1ST (
~~)
= ~4 [
208
=
IS STAl'UEIJ
/-nJ
..zoa
31-2 AMf
/<.WYL.
135
BY
Ci,IRRENT IS J;"f' AMf
IS APPLIED/ WHAT CURRENT WILL
GIVEN:
T t>t (no)
MOTOR
STARTING
FLoW
rot YPHASE
191 WHEN
"'ffOj
AN INOVCTION
JNDUCTJON MOTOitS
THE LINE ... Tt+E STARTJN(; TOJ2.QUE
TORQVE ANP Tf-IE
IS STARTeD
FUlL- LOAD
CONDITION.
'
.
CONT. OF PIWB. #
F/R.ST FOR !0 POLES
AND
1-zo f
Ns = - -
Y-4 !VIeT-HOD? (b) Wi-IAT WILL
BY TI-lE:
Ns =
UNDER T#IS
TY-1.1
-
1· 7!1
(J3f
I
-
Y-A
-
8 Ir=.L.
[
lY-6 ::: +-b'~
STARTINt;
I
TF.t
I
J
v;,n
-v--
II=L
"f.b("
Iv-c
TWO 60- CYCLE
CONNECTED
A~E
FOL!.-OW!NG
DOES
EACtl
a.)
11- PoL-ES
-:500 RPM/,
V'OM.<..
METHoD IS USED To OBTAIN
100- CYCLE
POLES
IN
.SYNCHRONOIJ.I' SPEEDS :
?
EACI-1 WIND/N{;
CONSE6lUEJ..JT
/'20
P=
POLE CONNEOTI15N
P=
-t-
G
POLES
l-200
/an.J'
= '2·{;7 TF.L • .;-~.
J\llOTORS
1-fAV!NG
10 ,4ND 14
Pot.ES ARE
P=
goo RPM
(J-zo)(Go)
:300
g
POLE.!' fcu.w..
SYNCHR.Ot-IOU.J' ..SPISEDS
FOR GOO RPM
P=
flC.G'P:
GIVEN:
fOJ'.J'/8 l E
60 CYCL-E
!O
Tt+E.
PC>l.-E.S
p.,_- !4 POLES
IJ..r FO /':.
GOO
J-2
POLE..!'/cuv~.
MOTOR.
fOR
P=
136
02o)(,;;o)
<K;"O RP/11
(lZo)(Go)
=/(;
137
IS TJ-IE.
!+AVE?
(izo)(Go)
RPIVJ
Ti-lE
/ZOO RPIVI,
STANOAJ<C
(Ia) HOW MANY f'OLE.S
!"OR /'200 FZPM
lFL
FI?SSIBI.E 7
p, =
104
SOLUTION:
Vj{?,]
[-v
IN CONCATENATION. WHAT
f=
=
10+14
23] THE CONSEQVEI-JT- POL-E
fCVj
FOR
22. ]
100 j
~·
900 RPM/ 600 P..PM / AND 4£;0 RPM. (0..) WHAT
{3-
==:.
Ns =
I'JUMJ3E.R OF
In P'a......
= 'T-IO'l
S"l4- RPM~.-~
fQ.ft.z..
CURRENT ON THE liNe SIDE
'(-6
=
}4
(rzo)(6o}
,I
= o.gg3
TF-L.
STARTING CVRR.ENT
(120)(60)
rOR IHE CoJV1BINATION OF
== l·7!> TF.L.. [
=-
/a.m..
'Z
2
Ty_A
720 RPM
=
Ns
t vq,v. . ]
v;_; J
I· 75 T F·L
a.) Ty_e.
IO
11- ft)LES
FOR
SOLUT!OIJ:
O-zo)(6D)
=----
p
?
it>)
?2]
SOLUTION:
VALVES _. WI+EN
STARTING" CURR~NT ON TtfE. LIN!: SIDE
i3E THE
nAtED
MOTOR. ST,t!RTI~G TORGUE
CURRENT _. IN TE:RMS' OF
1ltE MACtiiNE
1·7S TIMES f:=ULL-LOAI7
CURRE~T JS 8 TIMES
STARTING
WILL BE THE
VALUE • (o..) WHAT
STA~TJNG
IS
MOTORS
{\
DIReCTL'Y FF.OIIil
IS STARTEV
INDUCTION
POLYPHASE
,'!~
MOTO/t
POLES'Ja..tU.
.,
,i#
,f.
POLYPHASE
CONT. OF PROS. 1F
INDLICTION
POlYPHASE
MOTOTl.S
INDUCTION
MOTOT'Z.S
J
CONT. OF PR06 . ll' 2"
23]
SOWTION:
h)'
I· S'IX_/ EIB*'IT POI..E.S
~.
l'Z ~ IQ pol-ES
=~
SLIP AT MAXIMUM TORG>UE
j
X sit= "1-RR
XsR
_foR
SMT
~ IT IS DESIRED TO OPERATE A THREE -PHASE
tOo]
CYCLE
MOTOR FROM
~; -CYCI..E
A
4i0-VOI..T
bO-
i~
1
s..,1 == o. -zG"
SOURCE. WliAT VOL.TA(;E
'I
Slt01JLD
BE. APPLIED
OeNSITY
TO TI+E M.ACI-IINE
IS TO f5E MAINTAINED
IF THE AIR-G'AP FL.UX
Ns == (1'20) (so)
.f =
N11. =: ll'Zt;
flO CYCLE
fSDUP.ce
=
'21#
27 ]
~~IVEN y~TED
=
[
"'tOO
1
IS
I
'I'I1,
1
I
RPM ;fa#~-.
A SQUIRREL- CAGE MOTOIL
IOP.QIJE
WHEN TtiE
(a) THE
PER CE/-JT TAP
L13-FT OF ST47<TING
By THE
WJLL BE DEVELOPED
A
(25) ( <H-O)
= 18.3· 33,
FOUR-POLE
1lt
~
~IVEN:
1i
DEVE:LOP ToRGUE
VOLTS ,.f~.
;-a-CYCLE
=
90 LI3-FT
REACTANCE
G /VE.N:
P== 4 .POLES
f=
tl,ii'
INDUCTION MOTOR
PER PHASE
THAT /.S
a) STARTING TORQUE
T "sJ., =
I
11o
XsR =
+ RR
[_€£] z
.c;n
[:r
Tt;sj
=
80
T,.q
=
135·2 Ll3-FT /aw.r.
80 FEI<_ CENT TAP
TB<>l.
Bo [ ~ ]z
T 11o1. =
zo-t. 8
~'1·1.1
I'•l[·l
I'
I''
·I;~
II''III
fl~ ~r·
I"
111
1
1
'!:1~II·
~1 ~1.
,In,
,li!!li
·~
RE.QUtt<ED:
A1
111AX!MUM TOI:laUE?
AT tO>!; PERCENT TAP
b) STARTING TOR(ftJE 4T
5"0 CYCLE
l 1j
Iii'
111'1
THFtEE-PHASE
WILL !He MOTOR DEVELOP
1
,
1
11
FOUR TIMES THE ROIOR I<ESISTMJCE. l"l:.P. PHASE . AT WHAT
Nil.
1
'111
liiiW
fDO
400] HAS A 61..0CJ<ED- ROTOR
SPEED
~
W:
JS USeD ?(b) THE eo PER CE.f.-IT TAP
SOLUTioN:
l
~
REDUCED -VOL.TAGE
IS USED?
v =
2G
1
PER CENT TAP ON A COIVIPENS,ATOIZ.
f;O
WHAT STARTINt;' TOR~VE
6'J;
80
DEVELOPS
USED TO SrART THE MAO/H-IE
METHOD·
v =
y
~
CYCLE
SOl.UTIOIJ:
,j
I
'
\
N11. = Ns Cr-s) = Js;oo(I-O-'Z,!;)
E'::440V~5£P
-fzs
I
1500 RPM
+
.AT ITS NORMAt VALUE.?
GIVEN:
l
I
=-
= --
L5 -FT.fa.u.
MAXIMUM TORQUE
,,1111
',I,
~I
1.11"[1.·
I.'
'II'·
~~
II
~~
138
139
ill
11{11·1
L
i.
. I.
POLYPHASE
EXAMPLE 1] CALCULATE TilE
FDLE
(o..)
INDUCT/ON
A 60-CYCLE
MOTOfl.S : /LWSTrzATIVE
INDUCTION
MOTOR
SYI-JC~fWNOUS
WHEN SUPPLIED
-.3]
EXAMPLE
SPEEDS OF AN c/GiiT-
MOTOR.
WJTH POWER. FTZ.OM :
=
(o..) t=OfZ. 100 CYCLES :
(~ "
Of{
=
es
(c) FOR
PER CENT SLIP::=
,f"'-·
~
(r.<o)(so)
8
7
so
"""'.R...
O-zo) (..ZG") = 37S
t~oov.z
9
-t]
EX4MPLE
RPM_;.
MOTOR
SLIP
7-
SLIP =
CALCULATE
IF THE
=
D!.
,•
CYCLES:
fl.PIYlsyw =
RPIVJ
SYN
JS
X 100
lt"TDR
SYN
t000- 91DO
X
1000
100
4 /. /"""'"'".
THI:: SPEED
S
SUP
1:11
-RPM
tz.PM
CyCLES' :
5"0
llf'MsYN
= 900 RVM
SO-CYCLE INDUCTION
CENT SLif'·
~ = (12o)(.so)
p
c;
tooo R.PM
SOLUTION:
8
F'E:R
OF A
60- CYCLE
1"1--f'OLE
O·OS.
SOLUTION:
I:':XAMPLE
INDUCTION
z']
CALCULATE
Tl-lf:
SYNCHitONOUS
HAVII-JG : (a) FOUR
tv10TOrt.
11111,1'
1"20~
SPEEDS OF GO-CYCLE.
POLES
l!l
SOUJTION:
R.PMsyN =
(1-zo) ( 6o)
ILLUSTrl.ATIVE EXAMPLES
11-IE ROTOR SPEED OF A SIX -f'VLE
SO\JitCe.
ILPMsYIJ =
/lllOTOR.S:
IS 3Go RPM. CALCULATE THE
W-CYCl..E SOUTZ.CE ~(c) A 2!;-CYCL-E
SOUR.CE':; (h) A
INDUCTION
POLYPHASE
EXAMPLES
Ii
Ill!
RPMgcTOft = - - ( 1- S)
p
~ (b) SIX ft)LE.S;
(c) 10 FDLES' .
(!'ZO)(Go) (
=
SOLUTION:
14"
=
RPM ~DTO)(
1
1- 0-0EJ
i!JJI,.I.·
11;11!
4"68 Rf'M ;/~.
I1·111II'
(o.) t=OR
fOUR fbLES :
120 f
fl.PMSYN == - -
p
RPMsyN=
=
~_:_:'--
f
RPM j'....,._
I, BOO
(r-zo)(100)
G
10
..6 ,4/JLJ A
!N
HAS
lfALF AS
~PM //~.
ROTOR
VO...TAGE
(c.) !='OR
INDVCTI ON
TED
SPEeP OF
/,:zoo
A THREE-PHASE
WOUND- ROTOR
(b) fOR SIX POLES:
RPMsy>l=
POLES:
ROTOR
P.PM,.
VOLTAGE PER
PER
1l=RMINALS;
MOTVR HAS
Tt+AT
MANY TURNS
1_,110
PHASE
60-CYCLE
A STATOR
IS CONNECTED
AS THE
CALCULATE :
PHASE
SIX-f'OLE
!':BR ;
STA10~.
2'20-VOLT
THAT
IS CONNE.C-
STAR. IHE
FOR
ROTOJ:Z
A ROTOR
(a) THE SLIP _; (b) THE dLDCJ<EV-
(c) THE ROTOR GeNERATED
c~; (d) THE J:ZGTOR
VOI..IAGE
!3ETWEEN
(e) T+-IE. J:.!.OTOIZ FREQUENCY.
(1-zo)(~o)
ILPM SyN
1
:,111r
1:·~,~:1
Iii
1'1'11
~
i;
11•'1
TZ..P~fa..w.
(o-)
I
11,
!
Ns -Nn.
S =
.I I'
Ns
I
1'1
Ns= (1-zo)c;; (60) =
\ZOO RPM
==. 0-071; ~
S=
I
140
II
11111:1:
I
SOLLJT 101-J :
IO
= no
r1,r~·
II
EXAMPLE.~]
(1-zo) (lDo)
. __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ___J
141
lilj;.l
,,
.
CONT- OF C)(.
h)
=
f.l!R
'Z
FOWJ:R
= llO VOLTS
I:::)'.. :
£R
=
DEVELOPED
!:~;
13Y WE
a.)
l b)
THE P-O TOR COPPf:R
P-OTOR.
IWTOfZ
ttoRSE.f'oWER (1 ffp = 746' WATIS) .
IN
ls-JJ
"- r RR
XTK
8-'Z'!; VOLTS
RPI == 3 ( 77·5 )
l3E1WE.EN TERMINALS:
.[3- )(
0.1
t
X --
0.07!;
RPl -= Z4,000 WATT-S
8·'ZS"
l4-~~
LOSS _: (c) TI-lE P.OTOR
\VATIS .-· (9) Tf+E. ROTOR PQWE.R
):t)WER INPUT:
fl.PI=~
=fiXER
=
EXAMPLES
SOLI.JTIOI--J:
0 -0"1!i X 110
ROTOR. YOLTAGE
ILLUSTRATIVE
BY Tl-1£: ROTOR/ IN
DEVELOPED
X i:.gl:!.
S
.£11. ==
d)
ROTOR POWER INPUT ;
'2'ZO X_!__
TI-IJ:: ROTOR GE~JERATEP VOLTAGE I"J::.R l"tiASE
MOTORS:
EXAMPLE 7] USJN' THE DATA OF EXAMPLES !i AND 6 , CALCVLATE : (C\) THE
VOt..TAGE PER f>HASE:
Bl..OO<EP-P.OTOR
INDUCTION
FOLYPHASE
1;]
t=:!?R.
c)
MOTOfl.S ~ ILLUSTRATIVE EXAMPLES
INDUCTION
POLYPHASE
b)
P,OTOR
COPPER LOSS:
VOt..TS
RCL= P.PT X S
e)
ROTOR
f'f\.E.QUENCY:
-fr=
RCL
.f
S X
.ft.= 0-07r;;
tr =
= .Zi,OOO X 0-075
I<CL
c)
X GO
1--(0TOK
~OTOR
CVP.P.EIJT
I, 800 WATT.S
lOwER
4-5" CYCLE$'
RPD
CXAMf'LE G] USING Tt+E
=
DATA OF EXAMPLE
IF RR= 0·1 Otii'IJ
ANP
!;'",
==
P-PP=
CALCUL4TE JllE
3
KPD ~ (u-.1 f-JORSE.PoWERJ
'22,200
---)(8R
74G W/HP.
z.
Ri"D
- · ·~--:j(o!/o.07!i)'Z +co-~)"
EXAMPLE
IR
Jv_ =
!10
OF
----------------------~33
r
-1-
(o~-;;y­
_l_!£
I
,...
l
1R = 77- 5" AWIP.
8]
CALCULATE
EXAMPLE
29- B H)>.
THE TORQUE
/"-<u.
=
RPI
7·04-
T = 7·04 x
T=
140-9
x--
RPMsyN
24,000
\,ZOO
LB-FT ./~.
_j
142
DEVELOPED
7-
T
--·-
~
SOLUTJOIJ:
1.4-z
I
.w..
RYIJ""
JIO
IR =
1- o-o7~
P.PP= 22,200 WATTS
d)
/e'}Y +
I~IZ ( J~S)
3 x (n-s)-zx 0-1 x
.el!k.
R -
2
X IR
0·07!':
XaR= O·!i OHM.
SOLUTION;
T-
DEVE.l.Or'E-!J: (IN WATTS)
143
BY Tt+E. MOTOR
I
A?LYPHASE
UAMPLE
INDUCTION
.9] CALCULATC:
WIOTOftS ·.
ILLUSTRATIVE
POLYPHASE
(a) THE MA)(IMUM TORQUE WAT CAN 6E. Df:VELOP.et>
i'Y lHE MOTOR OF I:XAMPLt: 6 ; (!.) THE SP!:EO
WILL
EXAMPLES
AT WHICI4 THIS TORQUE
OCCUR.
INDUCTION
MOTORS : IlLUSTRATIVE
EXAMPLE lil A S-l-IP GO-CYCLE
INDUCTION
EIGHT -POLE
liS -VOLT
~
EX- 6
-= 110
EaR
LOAO TEST:
Rp. = 0-1 OHM
N.s
=
V-C STATOR • RESISTANCE
2
E 811.
Z Xs11
L
.
EFriCIEr-JCY ; (d) TI-lE
PEl{ CENT
THE
SOLUTION:
o) RPI.,t = 3
J [
-(uo)"
-2 XO·!i
"" 5
GWEN
~r;.wo
TMAK ""
7-0+ X
J
=
A-C RESISTANCE: OF STATOR
WATrS
0-118
R,._c-= - -
=
T..,..,JC
N~
(SINCE
THE
!,too
OF
ABOUT
AC RESISTANCE
TtlE EFFE:CTIVE
IS USUAl-LY
RPM~o~"" Ns ( 1- Slht)
Nrt.
10] CALCULATE
I
'
=
0-1)
zoo (,_0-S
~GO ~Pt.ll
S..,t =
;
COPPEg LOSS = 3 )(
= I.ZOO
IHL z
= 3 X (10) X 0-08
Rro.
= 24 WATTS
X0-8
FRICTIOIJ
+
WII.JDAGE
+
IRON LOSSES
= (R + P.)- NO-LO.AO COPPJ::RLOSS
.fa.....
==
THE STARTINS TORGUE
R~:t
EeR."
...
z
_..
Ts-t
=
t.
X
1
)~
+~._0-!>
14,000 W4TTS
(ns- 4ts-) - 24
WATTS
(IT
IS
PERMISSIBLE TO CHA~GE THIS LOSS !::NTIRELY TO Tt-IE
WITH t-JO
RR
APPRECIABLE
PORTION
t:J-?ROR , ALTHOUGJ-1
IS ACTUALLY
A
LOAD COPPErt LOSS IN STATOR =
T!-IE. FRICTION
AND
PART OF TlfE ROTOR Loss.)
;5
X
ll
7
X RA-e
= .3 X (27-5)Zx 0-08
179 WATTS
X 0-l
lVTAL STATOIZ LOSS UNDER LOAD=
r- +W+J· LOSSES
+ LOAD COPPERLOSS
7-04 X
1st =
]
+ XsR
(110)~
Tst:
BY
STATOR
RPlst :::: -3 ( 1 )Z
,p.J
=
DEVELOPED
WINDAf;E
SOLUTION:
RPI st = 3 [
X RA-e.
2
)(I!R
EXAMPLE G.
RPfst
HIGI-IER 'ltlAN
I· 2!; .)
'276.
Of
JS
MULTIPLIED BY A rACTOR
t;>l3 L/3 -FT./'-...
NR
MOTOR
IS
= 0-08 ..fL
X 1·25"
D-C RESISTANCE. , IT
NO- LOAD
b)
PER t'HASE
'2
oG,3oo
7-0'1- X
(lo) TtiE TORGUE ; (c) THE
FOWER fACTOR OF Tt-+E MOTOR FoR
SOLUTION:
RPI .,.,.
T.,...,.
BETWEeN TERMINALS :::0·1'28 OHM
VALUES.
LOAD
t::FFECTIYE.
RPT »11 =
Vl=ll&"; F\=3,140:P-z=I.S70;1L=:27-3~RPM 11 or011 =810
CALCULATE: (o.) TI+E fiORS'EA:>WER OIJTPUT ;
~PM
1.200
Vt-~1..= 1115~ P,:: 72S ~ Pz=- -425' ~ J.,L:: 10
tJO-LOAD TEST:
VOLTS
XaR-== O·S Ol-lt-1
THE
DATA WE.J<E
OBTAINED:
GIVEN~
exAMPLE
Tt-IREE PHASE
f'VI..LOWIN~
WAS TESTED , AND THE
MOTOR
PROBLEMS
27(;
~Plst
N,
7-0+ X _ 11-.000
I. ZOO
179
:::: 455" WATTS
!Z-PI
UIJDER LOAD
= (3,1"10+ I, ~70)- 4!;;!;
=
4,2:!05' WATTS
82 LB- FT ,f',.__.
144
+
145
~
I:
POLYPHASE'
INDUCTION
MOTOI1.S ·.
ILWST/lATIVE
Pfl.06LEMS
CONT OF EXAMPuJ 4t II
5=
Ns -1-.JR
.
=300 ~PM
Q.JO
.900
RCL
RCL
= tl_pf xs
= 4, 25!:> X 0·10
RCL =
ROTOR
4-ZG WATTS
= 3:2;
= 4,2!>5- 42G
=
AND 11-!E
REACTANCE
OF TI-lE
REACTANCE
D,R ANV Xs£?
OF nlE
~
b)
THE
b)
c)
RPI
Ns
4,:2!;5
900
c)
T= 3~~2 L.ll-FT
WATTS LOSSES'
/.. cFF
d)
X
=
+ 42G
Xe =
0·4"5:2 01-lM/a.u...
R"'
= Re -
=
x JOo
FO~
X'g~<_""'
0. :21~ OHM /a.w.
(1- s..,tJ
== ~s (1- ~ J
( 1- -O·IOG)
0-ZI"
RPM= Ns
T-HE biVEN LOAD:
)(1111.
900
-1,710
D-86{;
IISX'27·3
146
0·43::2
-= ----z
e
P== [3 ET cos-G-o ; COS 8-=P·F.
P·F· =
)(.,.
=
t=ROM:
.J3 )(
RsrAT<>R
0· 1010 OHWI/,._
d)
P.F.=
X 32
Re ·z
J~·47) 2 -tP-IfljQ) 2
)(~ll_
J
81·3 /..
f'OWEK FACTOK OF THE M01VR
.j3
x.. =
100
WATTS OUTPUT+WATTS LOSSES
4,710
ZG
=
0-47 OHM
Xe :::: ~ :Ce z. -
1{,.._
-----------4!;5"
OHM~.
D·I6G- 0-08
PER CENT erFICIEf-.ICY:
( 1-
~(32)2
=
VsR./J3
Ze=
Ze=
1;.13
I -
z.
Ii!.R
TOS<!QVE :
f. a=F=
570
X1BR
Re = 0-JB<;
T= 7-04 X - - -
c)
R"' · (b) THE. !=Ql.IIVALEIJT
RPM
RESISTANCE
,4ND
(d) lllE SPEED AT 'NHICj.j· MAXIMUM
PeR
Re =
3
:5,829
T= 7-0t X
MOTOR.
Xe ; (c) THE MfOR
MOTOR
+W= - - = - - 74-G
7+G
-H-P=
WAS PERFORMED ON THE
!=OLLOWING DATA WERE OBTAINED
WILL OCCUR.
o)
PQ06LEMS
P, = 1,+30 ~ Pz""' - BGO · CALCULATE : (o.) THE
RESISTANCE
3, n'Z9 WATTS
OUTP.VT:
RPP
ll,
SOl.J)TION:
POWER OUTPUT= RPI -RCL
J.IOASEPOWER
EXAMPLE
t=GU/VAI.ENT
1VRGVE
RPD
a.)
MOfOR OF
V8R = fG ~ "IsR
900- 6/0
S==
INDIJCTION MOTOIZS: ILLUSTR.ATIVE
EXAMPLE 12) A BLOC I< ED- Il..OTOR Tt:ST
120 X <00
Ns=--8--
Ns
FI?LYPHASE
=
4£;B /..._
147
POLYPHASE
INDUCTION
MOTOR.$:
PflOBLE.MS
ILLUSTI1ATIVE
EXAMPLE 13] CALCULATE THE STARTING TORQUE OF THE MOTOR
-EXAiv?PLES
II
II>JDUCTION
POLYPHASE
11>.1
CONT Or
MOTOrlS ·. /LLUSTflATJVE PROBLEMS
EX. 14-)
AND lZ.
b) PE~
209
== - -
CENT I
~3"
SOLUTION:
EX· II /
rROM
IRON LOSS'= '2.7t;; WATTS
o/
532 1•
X 100 :::
101
PfR CENT T:
2'27
X
lt.JO == 45
!
X
STATOR CURRENT AT II!; VOLTS Wj ROTOR 13LOC:<ED
II!>
=
X 32
2G
I~
AMP.
It-/
POWER INPUT AT II~ VOLTS Wj ROTOR BLOCI<ED
= (
I U ~0 WATTS
X ';;10 =
AT
MOTVR
'220
2
== 3 X (14Z } X
STATOR. COPPER LOSS'
=
RPI=
0·08
OF
4,830 WATTS
-27~-
Tl-1-E.
ISO
IS
THE.
1V
TERMS
7·04- X
EXAMPLE 1-B Tf!E J=OLLOWINb'
SO-HP
o.)
900
RESf'ECTIVE.LY:
ARE
3DG LB-FT
STARTING
AI-JD
CVFtRE.NT
I, I <DO-P-PM
+t0
AT
STARTING'
Pt=RCENTAGDS OF Tt+E
IN
ii-IFORMATION
4'TO- VOLT
AND CURRENT
AMP, ANO BLOCkED- POT"OR CURREIJT
(o-) THE
(c) THE.
STARTING CURRENT
CURRENT Ot-.1 T-HE
MA~NETIZING
f'ER CENT
lv)OTOR.
LINE SIPE
CUU!<!!:t-JT..S
IN
IMoroiZ ""
VOLTS
TORQUE
3G2
IS QIVI=.N
IN
CONNECTION
c)
INDUCTION MOTOR: -FULL-LOAD
ARE
'227
AND CURRENT
RATED
AT
AT
IIIIi
TI-lE
1!;"0= 97-!> AMP.
0- ~r; )( '37.Ji =
=
VALUES
~:5·4- AMP-
t17-J:
I'ER CENT MOTOR CVII.REt-JT = - - X 100
Zli-4
=
</-'TO VOLTS
2!;;4- VOLTS. (.J,)
VALVES ARE
x
LB- FT liND G~ AMP;
3G9/.
t;_;.4
LINE CURRE-NT = - - X IOO
PER caJT
AMP, RESPECTIVELY. (o.) CALCULATE THE
AIJD TORQVE
0-ti!;
zt;-'T
Wl+AT
=
2-+0
;1.
CALCULATED
(o.)?
eXAMPlE It;] TWO GO- CYCLE.S'
SOWTiOJ--l:
<A)
NATION
.3G2
12!;4
X
=
MACHINE
440
COMBINATION
2
tlAS
LOAD·
t::IGf.fT
IVIOTOP.S
ARE
IF MACl-liNC:
COI-JNECTED
I
POLE.S, CAL.CIJLATE
IF THE SLIP
HAS
4-40J
IS 0·075-
102 LB-FT
I '20
f.
RPM=--
p, + P-z
RPM =
<:;+8
"RPM
148
=
X ( 1-s,)
1'20 xr;;o
x (1-o.o7s)
"l-7~
149
IN CONCATE-
SIJ(- POLES
11-IE SPEED
SOWTION:
2!;41 z
3010 X (-
DRIVE A
2£;4
== 209 AMP
T:zs-1
TO
i[
AND LINE
OF ItAlED AWIPE.I<E S.
SOWTION"·
b) 1l.INJ:
A
THREE -PHASE MOTOR.
G,OSA
Tst=- 47·.5 LB-Fif,.__.
WITH
A COMPENSATOR AKE USED
CALC.ULATE:
IVIOTOR ; (b) Tl-IE
AJ.TT"OT.RAI-lSFClRMERS;
G;,os;;q. WATTS
T.st =
TORQL.!f:
'2(;.4
AMP.
ON
220- VOLT
10-HP
COMPI:.NSATOR., IJE~LECTtNG THE
THE
CUAAENTS , IN
4,flo0
A
CURRENT IS
VOLTS
DELIVERED
PER CENT TAP.S
WITH
CONNECTION
RATED
II!;" )-:z
---:zG
RF!= I!_IGO
fi!i
EXAMPLE 15] Tl-IE
AIJV
Of= 1-I+E
POLYPHASE
J IN
EXAMPLE 17
BE Ofl.IVEN
SAME
INDUCTION
EXAMPLE
IF eACI-I
VAWE
MOTOitS
IUl'STnATIVE
16 , AT WHAT
MOTOR
OTHER SPEEDS
IS OPERATED
SINGLE- PHASE MOTORS
Pfl06LEMS
CAN THE LOAD
S'f:PARATELY, ASSUMINt; T»E
~_l
CALCULATE
"1-tlj
PHASE
THE
'THAT HAS THE
DATA: E = liS; I=-+·" ;
OF SLIP?
Ef=FICIENCY OF 60
POWER -FACTOR OF A SINGLE-
FULL-LOAD
MOTOR
l-IP=
Y-r ;
FOLLOWINI9
NAME~F'LATE
llll!i
,II
,]I
~PM= /,7-10 . .ASSUME AN
1:11,
PER CeNT·
1.'
SOLUTION:
I
GIVE)'/:
ll?O.X GO
fl..PMc;"'
G
X
(1- 0·07!;)
P-EQ'V:
,.I
·:,
'II
l:.=/IS VOLTS
I==- ""f-.6 AMF:
ll.PMc; = I, 110
,·
FULL- LOAD POWI::R FACTOR
il',
Y
1-lP=: 4 =0-2& HP
llPMg ==
RPMe -=
l'lO X ~0
8
832
111
11,1,
RPM= 17"')-0
. )( (1- 0·07f;)
EFF.
=
lllil
11
60j.
11
I,
SOLUTION:
Ef'F =:.
il,,
f<>UT
,·;
- - )( 100(.
P,tJ
~liT= -
F.wpor
#f'"
X T"f&;
II'
PouT
E.FF
I
4
Pou-r=
P,N =
:il
I
IIi,!
WI
/JW
18~-1;; WATTS
1!'1
1
'II'
11'11
IIIII
=
)9(0._!;
:::: 310. &3 WATTS
O.t;;
:11'1'
~
'
.~~~
~
fiN= 310·63 W
II"'\
il
Ill
·'11111,1
'l'li,.J,.
,,1,'111
,I
iJjl'
11'
FIN
f'·F = . VA
.310-8.3
== (irs) (4.t;)
P.F-== O-!i87
..sl
1-t~
#.:;o..w.
1
'II II
I
!1,1
I~
!!,,
r
:J'II
WHAT WILL BE A WATTMETE:rt fl.EGISTEfl IN THE MOTOn. OF
PJW6. 'Z
?
:'·1·ll
' II'
:1(',.
SOLUTION:
I,,
··1·'
WATTMETER KEADINC
=::
.310·83 ~ .311 WATTS
I'
ill
.11'
1 .1
II
:'[1:1:,
I'll
1
'li.
I·
l~I.:J
150
151 :
l:il
IIIIi'
11,1•1
'1': ,11
1
dl\:
I
~-
SINGLE - PHASE.
SINGLE -PHASE
+ ]
CALCULATE
1-1-7
THE
MOTORS
PHASE MOTOTZ l"HAT HAS
AND A FULL- LO,AJ
RPM
FrtoM
;:\ G'O-CYCi.E
ANO SLIP OF A SINCLE -
[lE(;"lJJ.ATION
A IJO- LOAD
OF
'5PEED
~J
A
Y:z0 -HP
+t7j WHEN
HP=
REG'tll.ATIOH
SPEED F-t.'·
l,l'lO HPtvl
Yeo
F.FFICIENCY
?
N"" tSS"O rtPM
RE<l'"O:
1)9.5" RPM
TZ.EG '():
GIVEN;
SOLIR.Ct:.
SPEEOti.L.:
SHADeD-PoLE MOTOn. TAKt:S 120 WATIS
Of"E.rt..ATING AT FULL-LOAV. WHAT IS ITS EFF!Cif'..NcY?
SPEED OF U9Si
UZCI ftPM WHEN OPERATED
GIVEN:
tSSO-fl.PM
MOTORS
~
P :::- I"ZO WATTS
SLIP
III,
.SOt.UTION:
F"ou-r =
SCLUfiON:
• IJP:zo
I
"Y.HP
X. 71-ID'
= 37.3 WATTS
'I
1
111 I1:1
7./l.
/. rt.
N N.~ -
= -
"'F-l
wzo
Jl9!;"-
=
~ iOO/.
NF.L
EFF =
)( 100/.
IIZO
1. rt::: fi-7/. )t-.
~~
2 :
EFF
iHE FVLl- LOAD
CALCULATE
Pot!T
=p,.,-
j':f'l=
(o.) IN POIJN 0- FEET ;
TOTZ.QUE
OF THE MOTOit IN Pll06·
(lo) IN OIJNCt= - ftJCtl£.5.
~J
WHAT
.37·3
w
I:ZO
w
I
1:,1
·il
x
~I
I
11!
II,
lOOt
I•
I,:
= 51·08 /.
PE~ CENT
OF THE
RATED TO/lQUE
DOES A
STAflTING
TOttQUE
IS
r;.o
1
GIVEN:
/+
~PM=I740
Y-z'"
Hp =
''II
t+P
,;l!li:l
N = 3,'200 RPM
SOLUTION:
T ==
E:ZW x HP
L8-FT
RPM
!>'250 X
j
4
T==
li
:,
1
Hf
f1JLL-L0.417 TORQUE=-- X JO' OZ-IN-
ltPM
lf".l
X 10
fi
= 0-04
!0~
---
XJO
T:::: 14'3·b'8 0~-Jt-l /,_
a.ow..
' l,il''l'
'i
1'2·5 OZ-IN·
5"
=
OZ-IN
1-2-t;
40 ~~
:II:,,
1
'!!'
~ OF fLATED TOfLQ iJf::
l7"t0
1
11
G
3200
OZ-JN
T.F·k =
X
1
hi\111!
SOLUTION:
rtPM
Y+ wr
II
'1'1'
oF n-IE. IZATEV TOTZ.al..lE
17"1-0
T= - -
1'1
1
""-----
l-IP
,il,!l·
AVE. STAil.TINC TOKQLIE.:: 5·0 OZ-JN.
Ct.EQ..-0:
;~
T =0·7!:;-t LB-FT ~•'LWI-·
b)
OZ-IN?
Fll.OM PnDS · 2
HP=
o.)
~.!> HP
SHAVEV- POLE MOT0/2. DEVELOP AT STAI'lTlNC,
3,200 R-PM
IF THE AVERAGE
GIVEN:
•
X JOO /.
OZ-J-N
/a.wz.
1il'
J
TG.!__ X 100
TFL
X lOoj
.
' II1,!. ·
'I
11!
1
·~1·
l'
I
',Jill
152
153
!jlr,
I,Ji,
,,,,
;.l·li\.1
~-
SINGLE- PHASE
11
J
A
Y2 -HP
1-17j CENT
SINGLE- PHASE
MOTOfl.S
/,7SO-fl.PM C4FAC!TOR
MOTOR
DEVELOF'S 5SO peR
CONT· OF PrtOB. #' l'Z]
STARTING TORQUE· WHAT IS THE START!N& TORGUE IN
POUND -FEET
SOLUTION:
ANJJ OUNCE -INCHeS?
a)
!'ZOf
= -----
SPEED
F
G/VE.N:
HP=
N
=
/z
1
WHEn...e;::
HP
f'=
1,750 tl.PM
X =OEVC.LOPS
FOP. STARliNG TORC:VUE
FULL-LOAD
i+P
TO~GVE =
--
}( 10
G
;; ZGO X
TFL =
/·!;
1)
LH- FT
7.
AT:
f=
f' ==
s;. -zs t.s -FT /<JtM..
STARTING TORGUE
X
10
f =
b
0?-fN.
417]
fOR. lfOW MANY
AT:
(o.)
FltOM A
X TFL
=
~.9:l/.J('28S·71"1-)
too(.
__ 1~] CALCULATE
IDt.> /.
ffij
Oi!-IN· /~MU-.
/000
POLES
8~0 ltPM FROM
1,!;-CYCLE
GO
,;ru,..._,
RPM
CYCLE
p=
(1'20){,0)
11.30
P=
G·37 ~ ~ Pol.ES/a.tw...
zs~·?l"!Di!-IN.
IS A
A
MOTOR
l'HE PER CENT SLIP !"on.. EACH OF THE MOTORS
OF Pfi..OB. l'Z.
SOI..UTION:
aj
l
= 1_!30
s
1750
Tsr = f.
Ts-r=
c) AT
IN OZ-IN.
jl., 1-1-r
?.!;)
l;4b'D
P = 2 POLES
jOOj.
=
TF-L=
I'Z
2J;; CYCL.f:
(1~0)(
_(!:;s-o 'f. ) ( 1-s)
-
TH =
S= I ;'I-50 RPM
X Tt=-L
JOOJ.
FOJ'<.
<D·2S ~ G POl.ES,f'a.u.
p"'
LB-FT
{T.cT) = - - - - , -
Tsr
CYCLE
!;;O
/7!;0
Tf'-L "'
STA~TING: TOR41VE
3b'O fU>M
(1-zo)(so)
F=----9GO
OZ-JN.
Yz Iff'
IS THt: FRE.GIJENCY
s =
f=
IN LB-fT.
RPM
oR
NO. OF POLES
f-=
3J;;O (. STAI'tTING TO~IJE
AT:
SOLUTION:
MOTOfl.S
AT:
WOUND T»AT OPERATES
!;;O-CYClE .SOURCE.
SOURCE?' (c) Ll.30 Il.PM
~
t;;O-CYCLE
=
=
t:: POLES
SO CYCLE
N~t=
(J..) 1,4!;0 tl.PM
FMM A
p
-f:
S=
~%0
Ns-N~t.
IJ.s
SoUn.cE?
RPM
_, am
Jzo
Ns=
1000- '!JGO
5=
1000
s =
154
0·04
= 4/. .l'a.t4..
155
f
(I"Xo)(rt>)
f..ls=--= - - - p
c;;
IOOO /l..PM
"'"Iii
·'Ill
I
I,
MOTO fLS
SINGLE - PHASE
ill,.
SINGLE- PHASe /VlOTO!Z.S
:!1.1
111!:'1·
CONT· OF PIW6 #
1)
13]
AT:
CONT. OF PfWB · -# 1-1-]
P=
2 POLES'
f-=
'2J; CYCLE
SOLI.JTJON:
SINcE:
NR = 14>"0 ILPM
=
I 'I
a ~C
C =
Wt+E)l.E:
rzo -f
(l-zo) ('u;)
Ns = - - = ,_______:._ _
p
:2
Ns
'I'
;111,1,
Q
= llATIO OF Ttz.ANSFOfLMATION
c
=
CAPAC!TOfL VALUE
USED IN
I
,I
AUTOTRAN.SFOn.t-1ER
1>00 /LPM
.I
~
~I
TI-IETLEFORE:
1!1
T!-HJS:
Gt== G-8
S=
,,,I
Jr;;00-14!;0
C= f1?1-9co
11
~ tB!;:,-J--1[ /fow,w_.
I
~~-
~~
GO CYCLES
NR = 1130 I"ZPM
A
TWO- POLE
HAS
A
<dO-CYCLE
SUBSYNCHROIIIOIJS
IU7r017..
WITH
HY.STE[!ES IS
MOTOn
3'2 TEETH (POLES). AT WHAT SPEED
(120)(60)
!"ZOO IZ-PM
~
.£
II
~~
I'
'ij
f'OLE TEETH= 52 f'OL~S
·;
0·058.3% _f;-[1.5 , • .?~.
s
,,h,,
i'l 1
f=GO CYCLE
I"ZOO
I
+t7j
IN
A lV\10-- VAI,.UE
JLATIOS
CAPACITO/l.
ON, Ti-lE STARTING
6"·8 r,\-ND l-!:7
VALUES
A-ND A 4 -.A-If 0\PACITOR
~·~
MOTOR.-
AND !WNNJIJG
.32
Ii ~ . I'
SOLUTION:
A f?_E USED
J'20f
IF THE STEP-UP
N = ---
p
POSITIONS AILE
RESPECTIVELY/ CALCULATE THE EFFECTIVE
OF THE C4P,ACITANQ
VlEWl=.D
FftOM
N
=
II:
(1-t.O)( GO)
fZ.E.G"D:
SPE£0
11-] AN AUTOTTIAJ-.ISF01l.MfR
~
:Ill
GlVEN"-
!"200- 1130
,II
11
WILL IT OPEMTE·
Ns=
1:
11
·11.11
P=t::PoLEs
+=
'I'
1
C= (1D·fl)2. (4 ..uf]
J!;OO
S= 0-0333 ~ 6-35/.
c)AT:
AT PR.IMA/LY SlOE
= -----
~~
~I~!:.~
'~. I'
:22!; RPM.
WE l"fUI>MILY
~
II. .
SlOE.
J
ff?.J
GIVEN:
1"1
A UTOT!tANSFOf!..ME/7:.
~-
G·B
l~) THE
poSITION= I·];
MOTOIL Y!EI...DE:O
.?,500-rz.PM fl.Ef'IJLSlON
W"HEIJ
STARTED :
WATTS.
E"= liS VO[..T.S.i
CALCULATE:
oF THE STAI<..TING TORQVE TO THE f=Ult-LOAD
(Ia) THE STAlL T1 NG
P~IMAR.;>' SIDE
OTL AT
STARTING POINT.
•
ti-P::::. !13 Hf' ,
E-=
]. =
156
t<-EPULSJON
11S VOLT
ll-7 AMP.
Y!CifOR
T= 4.'2 l.-13 -F
P"' 'J2Cl WATrJ'
N-=
?{;"CJD RPM
' i[l'·,
li~I'
I•:!
I1'1
1·1:,
jl~
~0
PoWElL FACTOR.
~
b"JVE.N:
)::Fff.CT!VE VALVES OF G4PACITANCE
A<OM
ll&" VOLT
ll.ATIO
T0T1..QI..JE: ;
-lZE~/D:
VIEWED
-HP
1= 11-7 AMP; T= 4-2 LB-F-f; P= el2o
STAILTING fOS!TfCN=
R~NN!NG
1/
3
11-iE FOLLOWING TE.ST DATA
C= 4AF
"
A
'li
(TsrJ
'
~~
11'I,,' 11'I'1
l'ilt
,111:1:
157
'1,~
' I
\.'!ii'l_
1'·1
SINGLE- PHASE
MOTOTZS
SIN~LE
CONY. Or PROS . iF 17]
CONT· OF PfL08. #
50LVTION:
a)
CALCULATE FIRST
~
~"ZSO X HP
RPM
=
F-1.-
PotrrptJT
18G-5 WATTS
ilATIO= (!;
STAR..TING
~ IN
Pf.lLFOILMING
STArtT
THE
FOLLOWING
p =- .3!0
(1.) POWEll.
~
gzow
,_. ;.
I
~-
-J.O"
I
p.f'.
TOJl.QUE
T=
VOL.T AMp.
T=
T=
OUTPUT
UPUN A
A!l.JL'STEO
E=
fLPM= IJ25 · CALCULATE ·.
!'-r- HP
CAPACITOR.-
TO fl.ATED
VALUE
1
IJ£; VOLTS ~I: 3 ·8 AMP;
(o) ErFJCIENCY;
(c) TOfLQ.Uc IN POUND -INCtlES.
GIVEN:
3JOW
(11!;)(.3·8) VA
J
CAPACITOR STAll.T /V10T0f7...
E = 115 VOLTS
=
/1 ctMA..
JN ftJlJNV-INCHES:
!;2.1;7) X HP
L6 -P=F X 12 INj~
RPM
5150 X
;4
X
1'2
l7L"~
.9-13
WAS
/"tWw..
UNDER. LOAD FIZOM
20
A UNIVERSAL
JIG- VOLT
TOOl<
28'1' WATTS_, DEVELOf'EV 3.2 OZ-FT OF TOfl.QVc ,AND RAI-l
AT fD,OOO
MOTOn.
LB-JN.
11-Bj
GO-CYCLE
fl.PM.
l=t:IWErt.
EACH
IJ.JPUT
OPEll.ATEU
SOLJn.CE, UNOEn. WHICH
WHEN fC.ESISTANCE
THE SPEEU lU
4,000
5AME TORGUE, THE MO-p:;>R.
V<t- l+P
/OJ«..
VA
REACTIVE
llEOIJCE
l+P =
1.
•
W
VYATT
BR..AKE -LOAD TE'ST
;I
6"0·ICO
100
FVWEIL FACTOfl.:
u)
FACTOR
(J1,!;)(11·7) VA
0. G'84" ,{1a....:.
:=::.
J
X
310
p. t= = 0. 70'3
VOLT AMP.
FACTOfL _;
J
=
t4v...
DATA WEllE OBTAIN EO :
WATTS ;
EFF-
ll%-1,;
F-F-=- = - -
-·--
WIT!~ ITS
MOTOTZ.
=
WATTS
1'-f.=
-11-j
I=:FF·
0-7 LB-FT
POWJ':R.
P-r=
ff.
/
b)
-f. 2 Lf3-FT
Tl"l
/o)
746' W
tiP X - - Hp
L6-FT
TG-T
flAT/0 =
= -I
LB-I=T
2WO
TF-L.= 0-7
310 WATTS
"t
1
/ 3)
Tn = ( :;zso) (
=
To FULL-LOAD TORQUE:
THE FULL -LOAD TOfl.QUf:
MOTOITS
19]
a) Pu·"ruT
flATIO OF STAIZTING TORQUE
- PHASE
~~0
OPefl.ATING
WATTS.
tl.PM
VOLTAG'E
W,4S ADDEU IN SETZIES
WHILE
WAS
cALCULATE
90 VOLTS ANO THE
THE .E:FF/CIENC:Y LINDER.
CONDITION.
GIVEN:
REa>'o:
lJNIVE!l.SAl lVlOTOlt
!LPM = 172!;
EFFICIENcy Uf.IDEn.
E= 11!7 VOLT
f=
SOLUTION:
(a.) EFFICif:.NO':
t:.ff
=
PouT ruT
x too%
FIN=
T =
J=,4CH OPeJ?.ATING
fDO CYCL-E
ze+
CONDITION
wArrs
3·'2 OZ-FT
N= GOOD CtPM
FINI"UT
158
159
TO
OEVEI.OPING THE
3.9 AMP·
P = .310 WATTS
A
CONIJIT/ON IT
SINGLE- PHAS!=
SINGLE - PHASE MOTOrtS
CONT-
EXAMPLE 1] CALCULATE
zo]
OF PfWI3. #
MOTORS : ILLUSTRATIVE
AND OUNCE
SOLUTION:
THE
FULL-LOAD TORGUE, IN
Or A
INCHES
PROBLEMS
f's-o
HP
l,t;OO- RP/Vl
POUNIJ -rEf:T
SHADED -POLE
MOTOR
IJNDEIL- LOAD CONDITION
SOLUTION:
SINCE:
x;.rx Iii>
g290 X HP
T=
OZ-J#
f'T
X-
0~-FT
rr.PM
T
S 1 '2b0
T=
fZPM
X
IG
X
T""-
3-2 x r;;ooo
HP=
Tf+IJS;
p.,~-.
s-zs-o
= t+P
X
IG
HP
10:
T== - - X 10
--m
Pcu-r = o. 22857 fl-Y
0-07 li:S-FT ,{/a-..
t?· 2'2B57 tiP
=
7 ~4- w
X
0-0'l
X
1,<;00
HP~
:72'30
LB-F-T
RPM
SZ<;O X I~ X fW
T=
S,2.SO X HP
I=-----
fi)U
tlPM
OZ-IN
RPM
;x 74G
0-0'2
T- - - X JOe:;== 13-4-
-
~P
v:;oo
OZ-11-Jji~.
f'.,.,r = -170._!; WATTS
Tl-lER.Ef'OfLe:
a;
EFF
PouT
= -P,H
lTt>-J;
Err=
!.>)
fi:>R
S:==
GO
MAXIMUM
w
I'Z
XIOo%
Z94 W
EFF =
EXAMPLE 2]
•
X IODj.
.
-2!;;0
+-IP=
0?-FT
SOL!JTIOIJ:
(3·2) (<fOOD)
!+r(.!;:Z!;D) (lG)
O·l!:>t:-<t
HI x
=
HI>
0- l_\;-2-} ffp.
T= - -
RPM
7<!-6"
""/w
113·7 w.4rrs
2t;;O WATT-I'
~FF
=
G
X 10
0·05
=1,550
T= 52· 3 OZ-IN
c;.
X 10
(FULL-LOAD TORQUEJ
/13·7 WATIS
Tl}-Ej/.:6 t"'' IZE:
Wr·=
HP
MAX· TORQUE== 39 oz-IN-
tl-PM
=
'/zo
STARTING ToRQUE= 12. OZ-f}-1.
!Ii-zSO X li<i x HP
p,u-r=
WITH RESPECT
RPM= J_t;;t;O
WAiTS
"T = TI-ft: SAME JN (0.)
T ==
ARE THESE TOP.GUES
t;IVEJ..i:
-tooo Jl-PM
l't~-t=
PERCEI-JTASES
WHAT
TO THE PULL- LOAD TOP.GIJE?
.,
J'~
j.
OZ-!1~-
A '/20-HP
I,SSO -RPM SHADED-POLe MOIOK HAS A
TORQUe OF 38 OZ-IN- ANO A STARTING TORQUE OF
,..,...,.._
<f-5-!;/. ~
160
X 100"1
I•
I
L
38
PERCE~T
MAX. TORQUE
PEI(C~NS
STAP:riNt; TORGIIJE=-- X 100= 57"/
= -
32.;
X 100 =JIB •1
I·
"
32-3
/.
161
~>.i
J'c
SINGLE -PHASE MOTORS:
EXAMFLE 3] FOR HOW JVlANY
IF IT OPEfZATES
WOUND
100
CYCL.E
/S A SPUT-PtiASE
POLES
t7~0
AT
SYNCHnDNOUS
ILLUSTRATIVE" PR061J:MS
RPM AT FULL LOAD
l
___!_ AT WHAT SPEED
MOTOR
MOTORS
WILL A SYIVCHfWNOVS MOTOR. OPeRATE
SOUR_Ct=:?
GIVEN:
SOLUTiON:
REO'!?:
P=- 10 POt.E.S
THIS
OF
MACHINE OBVIOUSLY
1,800 RP/Vl.
HAS
G>'/'ICH~ONOUS
A
SPEED
-Fa=
TfiEJZE}"O.RE
P=
P==
1'20 X
-f
+
I,BOO
POLt=:S
TRANSrORMER, AS SHOWN
VOl.TAGE
!$
ON STAI-\TING
/JO
AIZE THE E'f=FECT/VE
Sr::CONDAAY
IS CONNECTED
IN THE
VOLTAGE
FIG-
IS 700
TO AN AUTO-
IF THE
ANO 400 VOLTS
CAPAC!fANCES
b)
PR1fv1ARY
ON RJJNNIN(;'/ WHAT
12o.f,
Or THE CAPACITOR
J.zo-t,.
S:z
4'~.. _
(l'<?o)(<Do)
= '-----'-/0
= --
p
7~0 R.FM
/a<-44.
IF TI+E
PoSIT/oN.?
IN EITHEF
(r-:zo) (.!:;o)
6'00 fLPM
5,._.::
c)
SOLUTION:
ON
S3
t'<!of.,.
=p
s3 =
(lw)(zs)
== ;;.___ __
10
300 11.-PM /,._
THE STARTING POSITION,
1700)
C= G X ( - .
1
lfQ
_gj
= 24-3,u.C
HOW
MANY POLES VOES
A SYNCHJZONOUS lv10TOrL !-lAVE
THE
RUNNING:
'f7Bj OPErr.ATES
AT 'ZOO
fZ.PM
wttEN CONNECTED
POSITION/
700 \<~
C== 6 X (-
GIVEN:
<tOo) =
5= too·rtPM
18-1- .,t~f
-
Ollr TYPE CAPACITOR
f,;
GO CYCI..E
ll-EQ'a:
A LifO-TMN s fO!<MER
NQ _ OF
rou:.r "'?
50L.UT!ON:
·--··
ttof
CEI-JTJZ.IfllGAL SWITCH
(i-zo)(<Do)
P.::: - - "'" - - - S
'2.00
--AUXILIARY WINDING
P=
p=
R~:
7'l.OO
zoo
-31;; PO t.t::.J' ;f_
404W...
VI/IRING DIAGMI\1 OF 2-VALUE CAPAC/fOR SPLIT-PHA<;E MOTOR.,
iJS!Nt; AN AUTOTMNSFogfV)Ef.<
162
IF IT
I
SOURCE?
ON
:-?
= -P- = -'---ro
a) s,
CAPACITOR
/ S2 / S_3
SOWTIOtJ:
s, =
~AMPLE 4] A G-Af
S,
!:>O CYCLE
fz:: 60 CYCI..E
h"'" i'5 CYCLE
i20X GO
tBOO
tF
ANV A
CAI~CJT.OR.
IT
1-78j HAS 10 ft:ILES AND IS CONNECTED TO A EO-CYCLE $0Ufl.CE ~
A 6'0- C'YCLE SOUR.CE? A '25 -CYCLE SOURCE!
FR.OM A
1
163
TO A
60- CYCLE
sYNCHRONOUS
~~
MOTOR$
SYNCHI!ONOVS
MOTOrtS
I
I
A 10-FOI..E
17~ AND
MANY
25-CYCLE.
IS VIUVEN
POI..I!S
IS
ALTERN,<\TOIZ.
i3Y , A
60- CYCLE
ARE THE/1.E
IN THE
Ollti=CTLY COUPLED
TO,
I
CONT· OF PIWB # 13]
SYNCHiWNOUS MOTOR. . HOW
SOLUTION:
LATTE[l.?
~ POWErt.
INPUT TO THE MOTOit
(Pm.):
i'l
GIVEN:
PouT
=o
PA
'
10 f'Dl.E$'
fA~~
~.t;
f..,"'
GO CYCLE
Of::'LE
p,"'
•
Povr = 50JW X 7<t~ 111/HP. :: 37,'300 WATIS
57300
F,» ==
IZ.E.Gi/:
NR.. OF FOLE.S OF S>'Nct+I'!Ot-JDU$ MoTOR
(P..,)
b)
SINCE THE. ALTErtNATO/l.
= 4-0,989 WATTS
0·91
f'tw =
SOLi.JTIO;..i:
40. '389 I<I'J l'am.. .
~KVAR
EAC-H OT-HEIL.
TI+Ert£f'Oit(? ;
l<W
SA=
s,.,
K'N
J:>.f. =
··-
5,..== SPeED Or AI.-iE;JtNAT<?/l.
I.ZO .fA
SA~:
G] THE
10
:500 ILf'M
=SM
\IVHAT
s"'"" 1'2.o+..,
/20
PM -
PM -== Q.zo)(Ga)
=
;300
21' PO!.E.S
POWER. INPUT
'T7<Jj SYI.ICHRONOUS
TtiUSj
PM ==
kW
=
PF.
= !;'f.N Kvv.;_.
Q1!o)(?~J
=---·
~;,.
S.., =
I<VA
J
~
o.e
KVA=
S101: SPEED OF WE MOTOft-
Ttl\::IJ:
.
KVA
WHI::'RS:
1~; THE
TO THE
CU/7.JU:.NT
0/I?..E.CT
FIELD
0-C
VOLTS?
= IS' KVA
PINPLIT
300
Ptt·mtT(v.c)=
.f'.....w..
(A-C INpUT)
It; kVA
G /.. C>F
'"' 0.9 kfA"' 900 YA
1'7~
O·fJ
I
I
l
(b)
S'O-ttP
A
OF
91
SYNCtlfWNOUS
PCIL CENT
CALCIJL<\TE: (o-j TH£:
I..EAOIN@.
THE
MOTOfL HAS A FIJll-LOAO
AIJD Of'E_RA'TE£
kiLOVOLT- AMPEi<E
11-lPIJT
,&T A
JbL\7!::1<
To n+E
E.FFIC/ENC'Y
Fl?WEtL rACTon.. OF
[LEQ'D ·
0-C
CURRENT
lNPIJT Of' THE MOTDR:
IIIOIOP..
SOLUTION:
'7
lv.c
GIVE.f.J:
P=
!?O ttr ;
FUl-L- L.OAD
P. F
!=fTlCIENCY
=
91 ~/.
t V<:..
Vvc
SY~ICHIWNOUS MOTO/i!.
1 De =-
900VA
/20
7· c; A !VJ V. /a..w .
=_:~~=-,t-r.~_r;;_____
164
/,l;-kVA (INPUT]
A -C
THAT E-XCITES THE F/Et..D
GIVEN.
7'ZOO
OF A
IS 6 PE.!l CENT OF THE f1.4TEO
/VIOTO!l
0-C VDLTtiGE S = 12.0 VOLTS
__£]
OfL
INPUT TO THe MOTOf'L
KVA
IS CoUPLED To THE.
S>'NCHil-01-KJIJ..<; MO!Ofl. , TI-lE SAME SPF-ED WILL
OCCVit TO
•
x 100/.
E'""F- = - -
165
v
INPUT·
AT
$YNCHfWNOUS
7]
f79j
!t-IE
STAfl.TING
MOTOfl.
!S
MOTOfl.S
OF A
TORGUE
25 -HP
1~0 PErt CENT OF ITS
THE STARTING
IN
TOR.QUE
SYNCHR.ONOUS
720-fl./'M SYNCHI7..0NDlJ.51
rtATED TOJLQUE. CALCULATE
POUND- FEET.
II] IT IS OESIIi.ED TO COIUl.ECT. THE 2,400 -KVA
fflJ
POWER -FACTOit LOAD 1"-1 A
ltATING OF
s.;.. no
!;'Z!;'O X ltP
LA(;.
18~-3
TsT.""
T~>T
-t7 ""' 49.
LB-FT.
=
= 273. 4S LB-FT }f,.,..._.
AT
IN
A
!;000-KVA
---5Y~CH~~US
CUrut.E."'T
AILE
I 824
~
kVA.fa..w.
479j
KVA
IN
!?E THE RATING
r;"ROI3. II , !F
OF THE SYNCHRONOUS
THE OVEf'l-At.L
POWER t=ACTO!t.
GIVeJ..l:
160 J<:.W
.P.I=.
<0,600 VOLTS
!LEQ'o:
LOAD Ctlrt.rz.ENT
POWf.fl FACTO/L
(IL)
KW~
,8::::.
cos·'o.9z
~=
23-07°
I+K~A~Ii
I
WAR=Nf'.W
I
J?EQ"p:
-f?: ( 10 .G
SOLUTION:
K.V)
"\-37· 4 AMP· ,fa<M..
f(W
f'·f· = - - ~VA
f'.F.
P=1.400 !CVA
MOTOF{
SDOO KVA
=
)WAs.,
l l
( P·F)
SOLUTiO/-.~:
MOTO!t.
IS TO BE 0·92?
15<b'O
=0·9Z
I<VA ll.ATING OF S)'NCI-IRONOVJ"
1L=
"''4
16'0 K.W. CALCULATE
1-z] WHAT SHOUI.[J
It ==
1<-w
f1-=o"f9.ib'~
ANIJ pOWElL FACTOfl·
GIVEN:
PA = s,ooo
Kill:= 1!><00
"'::'-,-----.-v
=(2"1-00)(Sli.J 1-9·"1-G)
=
CONUENSER
1\J PllT I<.VA
UNITY 1-'0WI;ft. t=ACTOR:
THILEE-PHASE
160 KW
= -5'000
- -KVA
kVAJ<..:::- ;<.W X TAN '23-07"
=
KY.A""':= NE.WkVA- kV4~ AT P·F=0·9Z
=::
KVAsM
166
(i_>;Go)(TAN z;.tJ?')
KVAJZ == ~64 ·"t-' Z t;;C?f /(VAk
0·05Z /a.w.
=
i6'Z.</ 11(;;'0
'
1:;
!
F!GUJLE:
tVA X P-F.
= 1!160 KW
f.!? (IS'Z-3)
b,6'00-VOLT SY~CHROt-IOl.IS CONVETt:TE.R.
-t=tiLL
"'*"" •
= (2400 )(0-G!#)
!JEW KVA
1/ =
DC SOURCE
cos-• o.as-
-&=
TOTAL IC.W LOAD
FU~L-LOAU LOSSES
':=
I,'
I'
1',
SOLUTION:
TlfER.E f=OILE..:
f'L
P.F-=0-Ili!;
PF=I
0-Gf? 1.4GCII.JG
NEW VALUE Of' WA
LB-FT
x es
TZO
TFI-=
FtJLL-LOAD
2,400 KYA
11.PM
.!7'2SO
TFL =
Tl+l=
2400 KVA
.,...__
IZ.E.Q'o:
fUILL·LOAD TOil.QUE =-
-t79'J
LATTER.
P.f'=
SOLUTION:
TtiE
0.~£- LAG(; lNG-
UNITY BY THE INSTALLATION
3-<f> SOU/l.CE
fA =
R-PM
TsT= l!:>Oj. Tr·L·
10l
n;E
(;IVEIJ:
SYt-!CtlMNOIJS MOTOfl
}
PLANT TO
OF A SYNCHMNOUS CONVERTERS· CALCUlATE THE KILOVOLT-AMPERE
GIVEI-J:
P= 2G HF
lv10TOfl.S
GG"'-
I<.VA_,/'Ci-oi4
(KvA
167
OF
~NCH/?..ONOtJS MoroR)
t
lllbiAf
f"F&i
MOTORS
SYNCHfWNOUS
1'31
ELECTRICAL SYSTEM
AN
'f9lj
POWETl. FACTOR.
SER
IS INSTALLED
HAS A
OF O·G7.
FOlt
SYNCHTWNOVS
LOAD OF
5,000 KW AT A LAGGING
IF A 3,000 KVA
SYNCHRONOUS CONPEN-
pOWER FACTOR.. C01l.RECTJON
CONY.
OF PfWJ3.
MOTOR.S
# 15)
O
PUR POSES ,
I<.W= 'fiOOD
M
V
CALCULATE : (a) TttE OVErt-ALL POWER fACTOR. ; (b) THE TOTAL kiLOVOLT-
AMPERES
N~
ON THE SYSTeM.
171VEN:
LOAO
= 5000
KyAR
=!i5"f{>
I<VAs.c.= 30001<VA
kW
Pf = 0·67 LAGGING
p
KVA OF SYNCtiR..ONOIJS CONDEnSER= 3000 !<VA
KVA
~
~
3. PHASE
a) OVErt- ALL FUWEfl FACTOT?.
1
LOAD
5000
KW
1
SOURCE
P.J== o.G7
LAGGING
ij
D-C SOUflCE
5YNCHIWN0Ll
WNDENSER
~
1-79
FIGURE:
= cos-'
NEW P.F.::
0-~91
NDIJS
o-G7
ITS
r;ooo
I<VA"' KW
p.f
TOTAL I<VA OF THE SYSTEM= Sb"OB KVA.J'a.....
PLANT
HAS A lOAD OF
800 /<W AT A
OF '2.00 k'W
W
AND ALSO SERVE
MOTOR HAS
TO, 0-92.
(sooo)
CDfl.fl.ECT THE OVER-ALL
ASSL'MIM~
THAT TJ+t: SYCHfW.
tz.AT!NC
AND THE PowEfl.. FACTOft.
IT WILL OPEfV\TE.
= 5540
MN = MP -NP
P·F=0·9f J.AC.
o
TAt--1 47.93"
,
:;,
>
f"·F=0-8
I
LAGGIN~
KVArl.
(Fi?.OM PHASOrl. DIAGn.AM)
= 55'10 -3000
1\1\N =
zgqo
kVAK
TOTAL KVA= [KW z
=
TOTAL
I<. VA=
DC SOURCE
t
I( VAll.
z
J (s-roo)' + (z910)"
Wll?.tt-J(; VJAGn:AM:
r;;G08 I<VA
168
A
AN EFFICIENCY OF 9! FEfl.. CENT ., Or=Tefl..MINI':
I<ILOVOLT- AMPERE INPUT
WHICH
POWEfl.
MOTOR. OF SUFFICIENT CAPACITY TO l7El..IVER A
~-rHASE~--~--------~~ BOOKW
LOAD
SOURCE
74-G3 KVA
J<W TAN -6-
=
/a.m.
0·(07
KVA (OP) =
KVArL =
AT
5G09 I<:VA
KYAT
FACTOR OF 0-8 LAGGJN(; · IT IS DESil!EIJ TO PURCHASE
LOAD
47·93.
f)-=
5000 KW
- - = ----
PL..AI-lT POWER-FACTOR
SOLUTION:
KWT
NEW P·F- =
AN iNDUSTRIAL
j
SYNCHMNOUS
-tJ
OIAG'Il.At.ll
PHASOI?.
169
II
SYNCHRoNOUS /VIOTOfl...S
SYNCHfl.ONOUS MOT0fl.S
tij
CONT OF PfWI3.
CONT. OF PIWB :IF 14]
ON = 800 + 2~0
ltll.IS;
GWf!H:
f<W LOAD= 800 KW
NR
_I<W LOAD OF SYNCtHWNOUS MOTOrt"' 'tOO l<W { OIJTP.UT)
T4N
ON
r-Fz = 0-92
=
NR
EffM= 91/.
iT~
ON X TAN ~...
=(1020 )(T4N '2.3-07J
tu=:a ...u:
Nil.=
434 kVA/l.
iZS =
NS- NR
leVA INPUT fl.A71NG
b) POWER. FACTOrt AT WHICH IT Will Ol'ert.ATE
GOO- 434
f7.S =
SYNCI'IIlONOUS MOTOIL
a)
KW LOAD= 800
~..1 '-...,~.,
M
oQ,._
INPUT~
KVA
WH£flE
1'1
};
·~
T
j(f>sf' +(ns)Z
PS=Mf.J=
KVA INPUT=
.... v
,
/cPG /CVAR
Ia)
"VA JNPIJT
.z-zo
i'-w
J (:?:zoJ'+(IGb)"'
=
SYNCHRONOUS IW'IOll:>R:
II
27S-fD ~ -27C: I<.VA-fruu._
,,
POWER. -FA&TOil .
F·t=~ =
IC.VA
KVA
KW
MP-= NS= GOO KVAR
f. F. 1 = 0-8 LA"It.JG
a)
= 1020
(N THf FIGU11.£,
SI/'ICE
PHASOrt
PF =
2ZO
'Z'f-(;
D-797
LE"ADINb" TO
Tt-l£ V0t..TA8E,I',..,...
SOLUTION:
= COS-
LET ~.
-6-z =
~w
1
co§'
0-8
= .:5(; -87 •
~
A
0· 9t
=
::.,.19)
!"-ACTOrL.
23 ·07"
eoo
OP..:--= - - =IOOOWA
P-F.
o.e
FACTORY TAkES
~Set/
I::VA
KW (OUTPUT)
9t
(a)
= (BOO) (TAN o5t.0-97")
Jlllf'=-
IS TO
A
BE
MOTOn.
IIAV/N(; AN INPUT Tt4TINI?,
INSTALLED TO CA/lR.Y
OF
AN AODITIONAL LOAD OF
ANV ALSO TO
IMPMVE
ntE POWErt FACTOfl..
MOTOfl. eFFICIENCY
OF 92-4
PErt CENT/ CALCULATE':
T+IE OVER-ALL
F.AClVR ;
INA AT 0-G LAG(;INb POWER
A LOAD OF 2,400
SYNCHI7.0NOU$
/_ZOO
ASSUMINI!j'
MP= OM )( TAN
A
J<lLOVO(.T- AMPERE WAD; (io) Tl+E OVER-ALL POWER
(C) THE SYNCHRONOUS
/VJOTOfl-
POWER. -rACIDR..
GOO I<.VA!L
t;IVEN:
ON=' OM+ MN
.&.17:
MN::
LOAD= 2/400 kVA
J(W LOAD OI.JTPUT OF SrN· M070ft
f;Ff'l.
MN=
MN=
'2.00J<.W
p F1
=
0 · G "LAGGING
S)'NUI/7.0!-JOIJS' M07DTL
Ef=l=. tvroro~ = 92 . "t (.
0-91
'2.'2.0 K.W
170
INPUT
nATINf;
= l,b-00 JCVA
ADDIDOfJAL La40 "' !ZOO J<W ( 0UTP1JT)
171
SYNCHRONOUS
CONT. Or"" f'l1.0l5.
SYNCHfWNOUS
MOTORS
CONT.
15'
I'
[lJ
0
1
3- PHASE
~·ic~
OF
MOTORS
Pfl.OB .
2,100 !'VA
t
LOAD
P.f..o<
LAC.
I
0 k' ':' \
1
.-. .....
~'""
~:.·
---- . •·,-
I
•
V
'\
IZOOKW
OUTPUT
•
-t.c
1500 I<VA SYNC· MOTOR
FIGURE:
""~
P-C SOURCE
Wlfl.ING
'V
KYA
~J./ASOR.
V!A{;n.AM
DIAGRAM
SOLl.I'TION:
LET:
s
p
KVAr = OVE'/l ALL KVA LOAD
P.F.r =
OVER ALL POWEll. FACTOR
t'-f. M =
S>'NCHRONOU$ MOTOR PO!.'Je/'l rACTOfl..
-
f1.S
PS
= TAN'ff3
I<W
=
OUTf'IJT
(l.S
kW
=-0·924
EFF.
k'WAUO. :::::
IWO
RS=JPR."-PS"'
!"ls=[(rs;oo)''-- (1239) ..
KW INPUT:
.ADDITIONAL
OR
TO
12.'99 f(W.
= ?SO tVA It
DETERMINI': THE VALUE OF
NR== MP-RS
NR
IN k.VAR
_j SINCE Mf': NS"" 1920 )<.VAR
NR= 19'20- 7Si0
TOTAL hW LOAD=
KWLoAo
+
NR= 1170 kVAR
KWADP.
13tJT:
KW LOAD= INA
>(
a}
P. F
KW LOAD: l"t"''-0 KW
-e;
fJ-2
=
cos-'o.c; =
+ l'l99 = 2739
S'.3.J3.
KVA
LOAD:
rROM TIUANGLE ONR
= (-z4oo) (o.c;)
TOTAL KW LOAD= 1"1'40
OVER -ALL
KVAT =OR=
I<VII =ON
JOR"'-+ NR ..
KVAT =/('2739)'Z +(1170}-z
kVAT =
2978·<+ X
2980 KYA
= p. F. T
k>) OVER- ALL f>OWER
FACTO/l
f'rl. =::1500 J<VA (GIVE;!-J)
f'5 = MN = 1,~99 KW ( COMPL'TEO)
f'-F.T =
i<WT LOA/7
ON
=
KVAT LOAD
MF
--
OM
=TAN
2739
iTJ
2378·1-
MP == 1440 (TAN J;3.J3•)
Mf':. 19ZO I<VAR
172
P.Fr
==
0·919 /~.
173
-Ofl
,/'t:>AM-.
SYNCHRONOUS
CONT-
SYNCHQONOUS
MOIDRS
CONT. OF Prl.OB. IG'j
OF Pn.D6- 15]
tr. == cos·'o-b = 53.13 •
-tr..,= cos· 0·9= ?G-84
C) POWEQ FACTOR OF SYtJCHR..ONOUS MOTOR
PS
p. /': SM
::
LJ
1233 k.W
R
h
u;oo KVA
PR
F-f-SM:: 0-86b LEADING
KVA LOAV =
B'E
r lt9~"~~~ s
INCQEASEIJ 8Y THE ADDITION Of= A
MUST ..BE 11:S
OPEfl.ATE
AND
FVWER FACTOR..
WHAT
ISOO /<VA
'
''
''
''
4-!:Z,,_,goopJ ' ', .... ~
N'
•V
SYNCHrtOt-lOIJS MOWit
THAT TAKES 450 KW. AT WHAT POWEtL rACTOJL MUST TI-llS
OVER-ALL
900 JC.W
0-G
~ A FACTOil.Y LOAD OF 900 KW AT 0.6 POWER. FACTOR LAGGING IS 10
f!Sj
MOTORS
KILOVOLT-AMPEil.E
MOWn.
1/-IPtiT IF Tfll:.
IS TO BE 0-9 LAG'GJNGr
1<-VAR
~IVEN:
LOAD=900 KW
I'
P.f'1= Q· G LAGGING
FIGUil,E:
KVA PHASOR OIAG'KAM
KiN OF SYNCHIWNOLJS tv!OTO/l= <fSO K.W
FROM TRIAN.9lE OMP:
P.l'_,=:: 0-9 LAGGING
3-PHASE----·-a
-
----
SOtJI1.CE
MP
-- =
OM
!VIP
900 KW
LOAD
I"'P
Pf.-:0-G
=
=
TA-N -B-1
(900 KW)(TA.; .,r.H3')
1'200 KYAR
LAt;GINt
FflOM TRIANG't.E. ONR:
NR.= {900+4-SO) I'W
X TAN-&-.._
~R: (r3S"O ~W){TAIJ ~S-el'!-")
D·C SOURCE
NR.;=
G'_!;"f- 1<-VAK
RS: MP- IJR
rJr;"URE:
RS-=> 1'200-.,;-s4
WllWJG DIAG'RAM
RS =
~G
ICVAR.
R~an:
OJ POWER
h)
t'V~
p;,n,,J.R
nh•pu···
MLJS'T TillS' MOTOR OPEIZATE (P.F3 )
~-!OTOR._
,cF
-FROM Tll.JAN$l~
ps
I' c3 = Po-e,:[:;, FACfQR OF Tt+E WIOTOit
-fT 3 =
f',;ii\1ER
f'- 2
OVER ALL
l___________~
OF T+fE
-6-o = TAt-!-' ~
SOU1") 101'·'
LET
a.) POWE)l. FACTOR
"'
FACTO!'- 4t-.16U:: Or
i73""
WE MOTOR
nkJS_;
f'I'WER. FAC.TOI\: ANG).E,
PRS;
MOTOfl..
PS-tvH.Jc:4,!;'0 KW
TAt-J-1 S'f-G
==
"T!;O
.5'0-s-1'
f'.F = COS s-t>.j;'f3
P-F~ = 0-fir." Lf:A.Pti-JG..f"ow.,_.
_"::.'~:--.
---------f
7/
---1.
l7S
;:1'11',:
,',ll·i
1
SYNOifWNOL'S
MOTOR$
COt~T OF PILOB. libj
b)
CO NT-
KVA
INf'LIT =
[~w 2 ~~
=
V (%-o)"+(~...;;-f
,.----··
!Ni'IJT =·
t:VA
707-
5<f-
1
I''1
OF PU.OS · 17]
Ill:i1'. I'•. i
SOL.UTJON:
1/'IPUI OF MOlVR
kVA
'l!:lj:\i
11 1':1
SYNCHRONOUS MOTORS
----
1
t1i =
cos-• o. 707 =
-e-2 =
COS
~
4·s-.,
0-937==- 20-<f-£;
II111_
~''''!,'
.
lil'l',ll:'ll:l
I(WLO.tlp= KVAibAO X
COS-e.;
::01!1
= (3r;;oo)(0.707)
jc•.VA
'II'
,/. r:.·l''l.··l·r
OR
OM
f!;O
!(\AI
KVA =
Pre:,
707· \;"·f I<Vo/.-._.
o.a.;co
= '254!;; KW
rl,'
t'"li
1;)1
t<VAR= Mt'= KVA SIN 8-1
1,!111
1'··'''1
... ('u;oo)(o-7o7)
17l A MAf.JVFACTOfl.ING
!-lAS A L..OAV OF .3,600 K.VA AT A
PLANT
. 17.-:Jj .POWER FACTOR OF 0· 707 ·
.I
.~
!;;CO- HP SYNOIII.ONOL!S lvlOTOR
EFFICIENCY oF go PER cEIJT
IIVIPMVE
T!+E.
KILOVOLT- AtvlPERE
THI:. /"OWER
IS 1/>JS'TAL.U::v
!"OWER FAC1GJR
Tf+E OYER-ALL
AND
,II ...
1ll'i'
!NF'iJT
TO
AN.O
0.937
1s
LAG'GI~
uSED Aws-o
1 11
'11'
Z.5-""f-&"" j<:.VAP.
'1
''
1
'1'1 '
fifL,;ff
'11'1111''
I<W fNPJJT OF sYNCI-IROIJOUS MOTOR.: Mf.J = PS
HAVING AIJ
w
I(W INPUT=
MOTOR
1- I 4 . 4-<t
1
·~
!1l1
1·'1"1'
0.9
=
'hi·
111.,)1
ll
(r;;oo wr-) ( o.74G l'vJjtlr)
l-AGGING. CALCULATE
SY~lCHROf-.J.O!J.S
1-':ATir-.IG' OF TI+E
j.1f'::.
'll:!li.!'i
1
11 llj
!<~'
'11,,1
1'11!1.''.1
!l11i,:
FACTOR
AT WHICH
IT IVJUST OPERATE.
TOTAL ICW LOAD= 2S4s.:z +414-44
=
GIVEN:
;Z9S9. G4 J£W
t.A~G/1-JG
:'t.ill''i
:= :1;;00
-~..,I
~:=J
-£J-C SOUitCE
I
V
NR.
T
~
s
KVA
PHASOR DIACR.AW'I
=:
TOTAL ll'N X IAN -9-_,
NR
!(EGO:
KATING' OF -S:>'NCHRONOUS ~OTOR
f'AcroR AT
wjc
IT oi11JS'T
r,= ""'"".
'I~ :il I
I I
llli·j!
l .li.:l.i':.:'l l
l
I 'Ill
111'1!,1;
' ·' '1!1
i':l'lll!!l.l
!!!li'
"II"·111
'
1
1
'!'l·j·
li!'l!
i:lrll
-
WIRING DIAGRAM
Of'ER4Tf!
II
I'
_________,.,,I
176
I
'I
p
~/
POWER
N
iIi
L..4°~GING
MOTOR
J
I
~1Rt
p. F-0-707
~i'YNct-tRO~
.b)
M
t' I
!IOO:::p
!NI'UT
,
',
~~~oi<VA
I I
l<VA
·~·-··~ ..
\~
_,flr-.
i
----r---~
o)
rl'i ll~'1l.1 ,1
HP
414•'1-'tf.<W
----··-·--·t__
fiG'UI<E :
I
1
P.F z::: (). 937" LAGG INQ
SOVRGE
'jl<l:
1
!'1
"I'll;'
1,1'11"''
1'1 1
·'l·.u.'',
OUTPUT OF ~)INCH RONC!IJS /YlOTOK
.3 -F'I-IASE:o---
j
1
'11':!
,:1.!.',11
r! ,.;1
LOAD= 5,Coo k:VA
P-F. 1 = D-707
l
il~'l.1
l
l'!llli
Ntl. = (295"'3-£"1-)(TAI-1 20·4S•)
Nil. :
1103.4
KVAR
l.l11,
!!ii,!!
rl1
RS .= NS- NR; Oft
1
!1
··111'·
1
il!l:
lr.
RS: MP- NR
RS:: '.?545 ·'Z -110.5 ·4
RS= l"t'tl·fl I<VAR
1
1
~l~l:
1
177
:
1111\
rll1
:I ill'
'lli\1 '-
..,,
SYNCHfLONOUS
SYNCHRONOUS
MOTOfLS
CONT. OF Pfl.06. 18]
SOLl'TlOIJ:
CONT OF P/WB . 17]
a)
MOTORS
_,
KVA
INF't.JT OF SYNCHI'W NOUS MOTOR. :
KVA
INPUT=
j
RS ,Z
INPUT=
KW LOAD
PCN>U.
ISOO )(.VA
0
1'3·9!>
KW LOAD "" KVA COS-(f,
= (.:3000) (o. n)
j(IH1·8) 2 +@-11·41}'
=:
KVA
+
RS -z.
o. 7'2 -::
ff, = CDS
~I GO
==
I<W ""OM
€00 #!"' X 0·71-G f',W~
KW OF MOTOR (JWPIJT)=
b)
o.e9s
FVWE'IL FACTOR OF WE MOTOR.
orz
-8-.3 "" TAN_, RS
cos-'
f'IZ
~
=
_, 1411·8
-1
TAN - - -
414.44
OJ: COS
~) OVE~
414·1'4
1s;-0o
-ALL KW
=
IU-Ifft>TAL
= 2f0l>O .KW
= CDJr3·!Jb'•
o:: COS-& 3
p.f", == 0.27G LEAIJJN<9
0
18l THE AVERAGE INPUT TO A IVIANViACTORING PLANT IS 3_DOO
HAVING
Al'J
i=ACTOR
OJ'= 0·7?
11\JPL'T OF
1,300
LAGGING.
kVA
A
SyNCHl<Ol'JOUS
IS THEN
INSJALlED
KVA
+!700
=ON =OM +M~
M
N
--•--~·-.-!-1
I<W:ZJ6'0
't<=::::\
~~.
--i
POWeR
KWLoAo -t KWMOrOK
'21~0
TI+US ~
4-7~ A
INPJJT:
KWTOT4L.
-e.~= 73 ·':I G ~
P.F3
&00 l<.W :: f'S = MH
KWMOTDII=
ps
PS
,_
~R
AT
I
'-I
MOTOR
AND
OVEK-ALL
INPUT; (c)
-n+t::
OVEI'Z- ALL
DELIVERS
I
I
OVER-ALL kiLOVOLT-AMPERE /NPJ.Jf.'
f'MM
iRIAN<;LE OMP j
LOAD= 3{)00 KVA
MP = OM TAN-&,
PF, = 0 7?
MP= (21GO) (TAN 43 .95°)
LAGGING
J<VA INPUT OF S)'NCHILDNOUS MOTOR. :1,.300 !<.VA
POWEfL OUTPIJT OF SYN
!::ft'ICIENCY =
SOUCUE
s
b)
PLAI-JT POWEk FACTOR.
GIVEN:
~
I<: VA INPUT
INPUT; (b) Tt-!E OVJ::R-At..L kiL0\/0LT-AII1f'cRE
kiLOWATT
"-"'""" .
~00
I
I-RS= 11~ J<.VAii.'
A WAD OF GOO t+p. AT AIJ ISFFICIENCY OF f'19.h- PE)Z CENT CALCIJLATE:
Tt+E::
•V
69.[;
l
MOTOR
-
l'i!P= '2082. ZS' KVAR =HS
FROM
! •'''"""'
"~~~12
~
LAGGING
TJI.I ANGLE
Pll.S ;
I:::VAR INPUT OF 5Yi'JCHRONOUS MOTOR = P.S
RS =
fl.S=
J P~
2
-
f"S .,
ro; ----
I'Z.S-=:
~
,.z
(,
ooJ - \soo)
1,'200 I:.VAR.
NK =
Z082. '2 !>- 1ZOO
NR~=8BZ.'25
KvAg
FIGURE:
178
2
N~= MP-RS
D-C SOL'rtCE
KVA I N P U T + - - - -
-= 1,300
Jc:w l-Ll"o
1000 HP.
1-
SYNCU·~~
:, WI-I ERE. OM I&" THE
179
SYNCt-ll'lONOUS
CONT.
fo!JOTOR$
SYNCHfi.ONOUS
Oi= Pfl.OB- 18]
CONi- OF PrtOB. 1ij SOLUTION:
!<Wtr =
KVA,.-
=
j iZWT z + N!Z""'
J(Z'6fl0)-z+(862-Z!>)
KVAr =
c)
280'2 ·!; KVA
cos~=
COS{J-..,_ =
I<W LOAD
KVA LOAD= DP= - - P-F1
7
I,'H-0 KW
0-G
J'aNW...
KWT
2€>60
I::VAT
2802-£
KVA LOA!J = '2,400 I<:VA
)8, = {}-, - f}-7 = !;3.)3·- 25 -8-t~~=
27-3"
f'. F, = 0-94j X 0-9S_,fa...t.
KYAT =OR= ~VA LOAD COS'{S 1
(PLAIJT OVER- ALL POWER f'ACTOR)
=
(z-roo) (cos
KVAT -== 2133
~J
11-iE
180
FACTOrt OF 0-G
tzONOUS
0-8
PLANT IS
INDUSTRIAL
TO AN
LAQGING'.
FACTOR.
MAINS
LEADING
POWER
Ai-JO HAVE IT COfW.ECT
0-9 ·. WHAT
TO
SYNCH[1.0NOUS
AT A
SHOULD
27·"3·)
k VA
1_440 k'W AT A POWER
DESinEo 10 CONNECT
IT IS
THAT DPE!ZATES
MOTOTl
TO TI-lE POWER.
POWE.R.
TI+E
INPVT
MOTOI<.S
FP.OM
A SYNCi-lFACTOR. OF
PR =
ff< =
THE OVEn-ALL
BE THE POWE"R 11-JPVT
TKIANc;LE
OPP.:
PR = KVA INPUT Of= MOTOR
TO
f'R:
lv1DTDIZ_7
KVA
LOAD )(
S'"';1'
(~400)( Sit-.} '27·3 ")
OR I<VAR OF TRIANGLE OPR
IIOO KVA
BIJT
K VA
OF
Ll
r'KS.
'IVfN:
LOAD
=
PS
1440 I<W
P F-.1 =
0-G LAGGING
P-F-,
0.9
P·F3
=
j:S =
O,
'20]
!NPUT TO THE MOTOR.
1-80 j
FOR WE
OUTPUT
INDU$TIUAL
OF A
= £;3.13
cos-• 0-9
&3=
cos-'o.s =36·£?7·
ANV AN
OPERATIIJG AT
EFFICIENCY
OF
SIIJf;f;:
v
EFf· =
PooT
p,...j
Pour =(880){0·£774)
four = 7f;9 .1::-W
INf'tiT (MOTOR)
I
{).3
p
.....j--
(l+W + 709) kw
!+P oLJTPJJrc,...,roR> =
KW INPUT {MOTOR.)
s
~P OJJTPVT
J<VA PHASOR. DIAGRAM
(NaT TO SCALE)
180
. G l<w/Hr
0 74
= 29GI
181
A LEADING'
97·"1- PEt{ C'EIJT
A:JWEP. rACTOTt TO UNITY.
SOLUTION:
N
M
OF O,S
MOTD{L
Tf-4AT WILL COflRECT ntE OVER-ALL
= 2b -81-
f71 =
0.
OJS -(73
PIZOB- 19, CALCULATE THE HP
PLANT OF
SYNCHRONOUS
POWElL FACTOR.
_,
= COS 0-G
X
PS = ~go I<W ~C!M/z..
tz.EG>/D:
SOLUTION.
kVA COS 93 = )?f{
Rr = 6Joo)(0·8)
0-8 LEAOIIJG
fbWER
= k:W INPUT OF SYNCHROt-JOV-S MOTOR
H-p f~-
SYNCHRONOUS
MOTOftS ·.
EXAMPLE 1] AN INDUSTRIAL
AVERAGE
ILLUSTRATIVE PIWBLEMS
PLANT
HAS A
CAlCULATE : (o.) THE I<ILOVOLT-AMPERJ:
fOR· AN OVeR- ALL
CONDJ::NS'ER
TOTAL
LOAD OF
1_500 KVA AT AN
OF O·G LAGGING. i>JEGL.ECTING ALL LOSSES,
POWER FACTOR
INPUT TO A
SYNCHRONOUS
POWER FACTO!< OF UNITY; (b) THE
SYNCHRONOUS
EMMPLE 2]
~
TO
FVWEI~
1. NEGLECTING
KILOVOLT- AMPERES
11
~ ~
~
r
0
§~ :
~6
O·G IF
LAGGING
:I:
LOAD
I')
ALL LOSSES , CALCULATE:
Or THE
r
i
11
I: i,~o:;;VA
T-
I
LAGGING
D-C SOUR.CE
DC SOLIP.CE
SYNCHIWI>KJUS CONDENSER
'
~INPUT=
'
'
~
$3°
1,200 KVA
lv1
I
TOTAL KVA
I
I
/
I<VA LOAD= I, 500
KVAc 1,!>00
AND WIRING DIAGRAMS FOR SOLUTION OF E.lc'-1
KVA / / WIRING
FIG.
I
I
I
OM= 900 KW
AN OVER-ALL POWE~ FACTOR OF UNITY , THE
/VIP= 1,200 J<VA
WILL ltAVE' 70 COVNTt:!.RACT
THE VERTICAL CO~PONE~T OF 1,500 I<.V.A I Wl-IICH IS
ll-IE
=
SYNCHRONOUS
I, SOO X 0·6
~VA INPUT=
1.200 KVA
o)
1-1) TOTAL. kiLOWATT LOAD
TOTAL LOAD= I(YAt.Mp
CONDeNSER CORRECTION
X
COS-9-
= /500 .X O•G
TOTAL
MN
= MP -Nr ::-
MN
=
I.Z00-750
4-!:iO KVA
kVA OF THE
TOTAL kVA=
==
TOTAL LOAD .:. <300 kW
182
IS Nr~ 750 I<VA
SOLUTION:
INA= 1,500 X SIN !;;3•
!<.VA
I<VA Pl-lASOfl DIAGrzA/vl
OP ::: 1,500 KVA
0.) KVA lt.JPIJT TO A SYNCHRONOUS CONDENSER:
CONDeNSE~
p
DIAt;R.AMS FOIL SOLUTION OF EX· 2
R.EFFERING 10 THE
SYIJCffROIJOUS
KVA
PF ANGLE
SOLUTIOf-1:
FO~
CONDENSEfl
.-v
I
I
I
J<.VA
= 750
INPUT
0
I
riG:
SYNCHRONOUS
t~\,
' 'yTOTAL KW= 900
~
PLANT; lb) THE. OVEI<-
t..IL <>-------------;>---7----t- PLANT
LOAD
~li.
PLANT
I
AVAILABLE
&<b
I:' <OO ~A
"- !;\..
jS
POWEP. FACTOR OF THE
FACTOR.
a:2
~!\! 1 ~
CONDENSER.
COIZIZECT THE LAGGING
EXAMPLE
(_o) TilE TOTAL
ALL
KILOWATT LOAD.
ILLUS'TflATIVE f7WI3LEM.S
A 7'50 -l<VA SYNCHR.Ot-.IOUS
.ANO IS USED
PLANT IN
MOTOfl.S' :
TOTAL KVA
f'LANT:
J (0111) +(NM)
J(900)'Z+(4b"O)-z
2
2
= 4b"O KYA/a.o.c..
183
I
1111111
!
~
MOTORS : ILWSTRATIVE
SYNC!-IRONOI)S
b)
Tt-IE
CONT. OF EX.#
z]
CONT- OF PR06. #
S'YNCHRONOVS
PROBLEMS
ANGLE
OVER-ALL A?WER FACfOR
=-
OVER-ALL P·F· = ON
SOWTION:
NOM=' 18" (SINCE COS 18"= 0·9Si)
MP:: OP X SIN 48 •
900
OM
$]
MOTORS: ILlUSTRATIVE PrtOBLEMS
l,OOG'
MP= 2AOO X 0· 74~
= 0·89<; LAGCJN(;
~VA
MP= 1,7fJ3
EX4!Y1PL!:. 3] IT iS DESIRED
[7ENSER TO
FACTOR
CORRECT TilE
IN AN
LOAD
2,100- kVA
INDUSTRIAL
0-6"7- LAGGING
PLANT TO
OM= 2,400 X 0·67
POW.ER-
OM= fi>De INA
0·95 LAGGJNt;.
LOSSES , CALCULAT~ : (o.) TI-lE KILOVOLT -AMPe[<E
NEGLECTING
KJLOVOL'T -AMf'ERES
(b) 11-f£ TOTAL
~
:<!
"'-
~~:
~~
MN
.,
- - = TAN 18
SYNCHRONOUS CONDENSER ~
INPUT RAT/Nt; OF 11ft: REGLJJREV
~
OM= OP X COS48'"
TO PURCHASE A SYNCHRONOUS CON-
OM
~ ~
1
!Q<>-
0·2)2!;'
Tltcp.Ef:ORE
OF THE PLANT.
MN-== 0/Vl X0·52fi
~ I~ I:~KVA
-~~~c
MN"" I,G08 X0·32G
MN= 572'2. KVA
0~~
w
=
0.) J<VA
~OAD
SYI-JCHRONOUS
I NPIJT RATING OF
SYNCtlROI\IOr.lS CONDENSeR RATINb':::.
~
D-C SOIJRCE'
b)
TOfAL
=
I, 783 - !;22
""
tZIQ/ KVA (USE A
KVA OF Tf.}E
CONDENSER;
MP-MN
1,2SO KVA /VIACH/NE)
PLANT:
TOTAL kVA == [(OM)"2+(_MN)Z
SYNCHRONOUS CONDENSER
',
_,;><
',,
I
o~
= ~ (t<DDB)-z+ (szz)?.
INPtJ'T-::: !_1(;/
TOTAL KVA ==
OVt::R-ALL I'OWCR FACTOR(=0-95)
',
• , 19 ~s:;;::=
.ANbLE
<~
.
1,10.90 I<'VA fa.......
v
t'XAMPLE
-i}
.AN INfJUSTRIAL PLANT
HAS AN AVERAGE LOAD OF 900-
KVV AT A f~OWER FACTOR OF D·G LAGGING. A sYNCI-IMNOUS /YIOTOR
IS TO
BE !1-JSVIt.LEO
THE OVER-ALL
MATES
FIG:
KVA ANV WIP..!NC
DIAGRAMS FOP. SOLUTION OF EX· -5
WILL BE
TO DRIVE A
POWER FACTOR
INLJJCATE THAT
ABOUT
THE'. INPUT TO TtlE
(SINCE
184
AND TO RAISE
PRELI/VJJNARY ESTI-
SYNCHRONOUS MOTOR
1</LOVOL.T-AMPEI~E
FACTOR AT WHJCJ-1 IT WILl Of'ERA.TE.
Of':::: 'l ,'tOO kVA
ANGLE ft'M= 48"
0·9e · IF
2!;"0 I<W, CALCULATE : (o) lTG
INPUT RATING .: (b) TI-lE FOWER
rtEFERR/NC TO THE f'IG·
D-C ~NERATOR
TO
COS -t9"= 0·10'7)
185
SYNCHIWNOUS
CONT. OF
EX-# Al
11'"1]
MOTORS: ILLUSTRATIVE
PROBLEMS
SYNCHI1.0NOUS lllOTOftS : IUUSTilATIVE P11.03LEMS
SOLUTION:
COI•JT. OF EX.
+]
c.:
900 KW
PS =MN = ZSO KW
y:
0 ·tO _I"· I=· L.4GGINI':
ON= OM +MN
6~
(J)s~---
~ ~. .- _ _ J' -
T
-PLANT LOAD
ON=- 300
=1.150 J<W
+ 2!10
NR:; ON X TAN 23°
----~
"" 1,/J;"O X 0·+2-T
488 KVA
t.JR=
D-C SplJfl.CE
RS= NS -NP.
= 1,200-4-89
KVA SYNCHrz.ONOUS MOTOJ<
RS= 712 KVA
~'
{;;-___-SYNCHRONOUS MOTO
a)
R
INPUT
' ' ';<ORIGINAL
'·
r.'l
kW LOAD
PR
N
ANHE
~Y
SYIIJCffRDNOUS
tJ.
KVA LOAD
FIG:
PR=
~TOR:
K.VA IN PUT
~
MOTOR INPUT
7
(Z'W)-z+(7fZ)-z
PR== 754 INA fi/t~--..
b)
KW INPUT
KVA Af.JP WIRlN~
=SYNCHRONOUS
PR = ~(PS) 2:t(l-i:S)
P·F. ANGLE
p
KVA JNP!JT RATING :
POWER FACTOR OF SYNCI-IRONOUS MOTOR.·
s
P.F-
= COS
ANt::LE SPR
DJAGRAM.S FOR SOLUTION OF EX·i2li>O
P·r.= -752}-
REFEP,RING TO THE FIG:
PS
= --
PR
= 0·332 LeADIN~
/a-.
OM= 900 !CW
ANGL-E POIVl= .>0"
ANGLE
RON
=
(SINGe COS &3·==
z.;• (SINCE COS ~~·= 0.9z)
OM
- - =COS 5"3
OP
THERERJRE;
OP ::::
MP
o ..:=;)
•
= O·G
t:XAMPLI:.
3,000
l<VA
MOTOF-Z
1£
LINE
900
-- =
O·G
LSOO kVA
.r;J
AT A
TO
SHAFT
RATING OF
LOAD
THI: AVERAGE
INPUT
POWER rACTOR OJ= 0-72 LAt;GJNt;. A SYNCHROt-011$
BE 11-iSTALLEO FOR THE PIJRF'OSE
OP
1,300 ~VA.
IS ABOUT GOO HF'
MP == NS =
I, !;OO X 0-8
=
I,ZOO KVA
186
A NEW
ASSUMING
THAT THE SYNCHRONOIJS-MOWR
,AT AN EFFICIENCY OF 89·5 PER. CBNL ANO
l<llDVOL T -AMPERE INPUT , cALCULATE
(o.} Tf-IE OVER-ALL J(ILOVOL.T-AMPERE LOAD; (b) THE OVE.R-ALL POWI::R
FACTOR.
T-I+ER!=.FORE;
Of=" OPf':.Jl.ATJNG
AND IMPROVING THE PLANT POWER FACTOR, Wm+ A
THAT IT IS OPERATED AT RATt=D
SIN 53°:::: 0·8
TO A MANUFACTURING PLANT IS
187
SYNCHRONOUS MOTORS : IUUSTRATJVE. PROBLEMS
CONT·
OF
3-PHASE
P·F=0·9S
ex ..# ~J
CONT· OF EX #
SOLUTION=
l_
soo:C_e
______
o
l
LAGGJN~
SYNCHRONOUS IVIOTORS : ILLUSTRATIVE
!i]
PM ~ SN
__..~l-:3-ooo lNA
0·17. PF
Pl'W6LEMS
=
Of' X SIN
=
3000 X O·G9~
"t+"
PM= 2,0f3E KVA
LA~~ING
PLANT
NR= SN -RS
LOAD
=
"2,08!;"- !,200
NR= BB5" KVA
0-C SOIJRCf
OM= 3,000 >< COS'i-"1- •
OM= 2,160 I<W
ON= OM +-MN =<- Z,IIDO+ SOO
SYNCHR~NOlJS MOTOR
/
e,.x,
''
Ok
ON= 2,6GO KW
1Nf'UT
•
''
0.) OVER-ALL KVA LOAD:
I'"
''
KVA LOAD
..... . . . . . - ",,..,
'-·
= J(ON)"'- + (NRY
= {Cz,ta6'0)
=::
SYNCHRONOUS MOTOR:
I • I
p
fit;:
KVA INPlJT
kW INPUT
s
ICVA ANJ) WIRING DIAG-RAMS fOR SOLUTION
OF EX.!;
1:1)
OV.E'R-ALL
POWER
SINCE;
44o
KW INPUTS:::
AND
(SINCE
ON
?,6GO
OR
2.eoo
p.j=.==---=--
P·f·
=
0·9S
COS '14• =O·TZ)
HOIU'EPOWER
X 0·74-G
fff'JC/ENCY
ps
=
GOO X 0·7-+G
rs
=
!;oo Kw
0·89!;'
SYNCHRONOUS MOTOR INPUT= PK:: /,300 i'VA
f{S
=~ (PR}'"-
(PS).,.
=l(l,300)?.~00)
2
RS= IJZOO I<VA
188
/a..u..
FACTOP-:
OP= Z,OOO J<VA
POM=
+(BBSY
2,600 KVA
REFERRING' TO T!tE J=lG:
,4NBLE
2
189
Jtl-1'1d·
CONVERTERS
1l
IT IS
DESJflED
St!j r=r"'oM
A
250
VOL.TG
DiflECT CURRENT
CONVERTER . WHAT
A -C
VOLTA~!::.
J IF
3
.!>41
j
THE
B4
USED 7
!3E
MUST
TO OBTAIN
Sli-J9LE- PHASE
CONVERTERS
RECTIFIERS
AND
CALCULATE
SOLUTION?
= ,r;
=
EAc
PROB. f:
OF
AND OPERATES
THE GLIP RING
HAS AN EFFICIENCY
OF
AT A poWER FACTOR OF 0-92,
CURRENT.
GIVEN:
zr;o
Eve
. EA-c
CONVERTEIZ
Pl=R CEI-JT
RECTIFIERS
AND
DATA
PQOB # 2
FROM
/o
.fi
EFf'-= 84P.f= 0.92
1710-78 ~ 177 VOLTS
REG>'o:
2
l·
A
StiJ IS
OPERATED
LOSSES
(o.) THE
AND
FROM
UNITY POWER
SLIP R.JNG CUTW.EIJT (1
CALCULATE
OF ? J<W ANO
A -C SOURCE . NEGLECTING
220- VOL-T
A
ASSUMING
D-C OUTPUT
HAS A RATING
CONVERIER
S/Nf:LE- PHASE
Ac)
SOLUTION:
FACTOR, CALCULATE :
PotiTPUT
E.ff.
VOLTAGE AND CURRENT; (h) Tl-IE SLIP- RING
ftt-~f'\JT
CURRENT
PINMlT =
-
-2000 W
0·84
= f?:38J WATTS
GIVEN:
1<-ATING
E.Ac =
'2 KW
f'Ac =
220 VOLT
!=."" =
P-F= V"IITY
EFf.
=
100
J.
f:,.~c
LOSSES
,BECAUSE
ARE /JE~LECTED
SINCE
=
1A-c =
EAc ==
E DC-
fvc =
Epc
v
"LI-e=
J2
E.Ac
r=z (ezo)
311 VOLTS
#"a.w... .
PRATING
' ~rc. --
AMP
/'W'Z.
Lcc=
Powe.- Ra+o'"S
E';~c
2000
PtNPVI
EAc AT P.F-==0·92
'238 I W
202.4 v
F'rtOM
SIN~LE -PHASE
240 -VOLT
A
D-C
INVERTED CONVERTER.
SOURCE.
OF 91 PER. CENT
CALCULATE:
ALTERNATING
ft.,...__.
(o.) THE
AND
ASS'UIV!ING
A LOAD
OPERATES
A FULL-LOAD
A::>WER.. FACTOR
OF
SLIP-RING VOt...TAGE; (b) THE
CURRENT OUTPUT; (c) TI-lE
DIRECT- CURRENT INPUT
GIVEN:
b) SLif- RING CURRENT
Lc=
7-5- KW
0·85,
31/
G-43
A
EFFICIENCY
2000W
=
J
5-tlj
El>c.
Ivc
'20:?- 4' VOLTS
1Ac= II-7G4- AMP-
41pc. =
'2'20 x o .gz
THUS;
SOLUTION:
o.)
X I'·F
fAc
w
220
7·!; KW
SIN"'-.E-PHASE lf..JVERTE.D CONVER.lER
l:.pe =
240 VOLT
RILL-LOAD EFFICIENcY
LOAD
POWE.IZ FACiOR
=:
=
0.91 OR 91 j.
0· 9.3
lAc.= 9-1 AMP.
. 190
191
fJ
~
CONVETUEfLS
CONT
OF Pit0l3-
flECTIFIERS
AND
CONVERTERS
1-]
AND
fl.ECTIFIERS
CONT. OF PROS. # 5]
SOWT!ON:
b)
o..) SLIP-RING VOlTAGE:
E~c
.SINCE
'240
E AC-ff- =.jZ
EAc
N2. OF TAPS
N.f!.
= IG9. 7 VOLTS /a......
G
FOR
cVE~Y
FOR AN £1Gi-IT- PoLE SIX -PHASE CONVERTER
PHAst:: , THE.Rt=.
G TAPSjp
X
e
:A/}Z Of' PDLE..S
24 TARS'.f'c.t-wz. .
pRATING>
=
c
P-F
X
.l-Ac
1Ac.
10 ]
w
7,r;;oo
lAc=
c) THE
8
OF TAPS= -
b) ll-IE AL.TI:/2..NATING CURRENT OUTPUr:
r.. c
ARE. G TARS' FOR
OF POLES, THERER?RE
PAIR
5"41] AND
l69-7XO-f?,3
AND
VI RECT- CURReNT INPUT:
Toe
GOO VOLTS.
THE
HAS
n+E
A D-C RATING OF 120 kW
AN r='FFIC/ENCY OF 91- PER CENT
ASSUMING
0- 9S,
rACTOR IS
PDWER
CUKRENT OUTPUT~ (_lo)
;f'a.m..
S3. ~E AMP.
=
A THREE- PHASE CONVERTER
CAI,.CULATE: (o) THE 0/R.ECT-
SL!P-R.INS VOLTAGE; (c) THE SLIP-
Tl-IE. kiLOVOL.T- AMPERE
fliNG
CURRENT. (d) ALSO
DE.TER/Y11NE
LOAD
ON
TRAN6FOR/VIERS'.
EACH
OF THE
FINPLIT
=
GIVEN:
E.f.>C
PourPtJT
P,~I'UT =
P1N. =
~vc""
~241
240
3- Pi-lASE
w
·8
CONVERTER
D-C I<ATIN(;
0.91
.EFF.
8.<?41 8
lpc =
7,.!:'00
= IZO KW
E vc -=:.- GOO VOLTS
WAIT$
EfF = 94- /o
f.F. = 0·9!;
w
v
SOLUTION:
34.34- ~ 3"1--4 AMP /cw.....
Cl) DIRECf OJRI..(E/-.JT OUTPUT
f;
]
I+OW
MANY
S41j WINDIN(;
PHASE
AilE
TAPS
TO THE
CONVERTER
SLIP
7
BROUGHT
fl..INGS
OUT
OF:
FROM
(o.) A
(h) AN EiGHT- POLE
THE ARMATURE
SIX- POLE
Ivc=
p /'-A TINt;
2pc.
THREE
Sill- PHASE CONVER-
}pc =
(Ivc)
1'20,000
TER.?
w
000V
'L>c
SOLUTION:
o) SINCE TI-IEP.E. ARE
OF
POLES
IN
3
TAPS
3
PHASE
G
POLE.S
rDR
EVERY
PAll<
=
N.il. Of
TAPS
=
9
TAPS //
Epc
~
X 0-B~G
'J't?
X
2
VOLTAGE (E:.AC)
6) SLIP- RING
SYSTEM,
f'Ac =
I'J<I.. OF TAPS
ZOO AMP. ,/Mw..
3 TAPS/pAIR OF POLES
~Ac. =
c;oo
Vl
X O·SGG
.l"a..u.
E,...c_ =
192
3 f>7 VOL.TS /~.
193
CONVEnTER.S
CONT
CONT. OF PrtOB. #
SUP-RING
T~c
I~tc =
KVA
200
IAc = - - - - - - {3 EAc
0 9'1-
ON
IA c
KW
3 X P-F
I.ZO
3
);VA I
-,
I TIZ4t'ISFORMEJ( --
7
"j
54-1
A
Of
1,000- kW
~00
THREE- PHASE
44- .g
POWER FACTOR
IS
0.96 /
OUTPUT;
(Jo) THE
A-C
AIVIPERE
LOAD DEUVEIZ.ED
DELIVETLS
IS
FULL LOAD
(C) THE
X
EFF
.J3 X
kVAfT =
0-9G XO
'J+
3G9·+ ;:;} 369 f<:VA
SYNCHRONOUS
541] CUR.RENT·
CONVE.RTER
POWE/1.
foe = 600 VOLTS
IF Tt!E
94-/.,
(C)
Pf: = 0·9G
CONVEil.TErt
ASSUMING
FACTO((. , Df".TERMI"IE
TtiE A-C
SLIP-fCING
12 17..1NGS.
SY~CH~NOUS
( T oc)
Dlru':CT
rowt.R
=----I X lOG W
Ipc=
GOO
FROM TABLE-#- 10 / PAGE 49/
V
o)
1vc. ~~ I· GG7
CONVE~TER
CUrtf'lENT = '000 ,LIMP
SOLUTION:
toe.
A !VIP 11
//(!...,...
-FOR
THJLEE
RIN9S
lAc
='
0 · 943 T
1Ac
=
o -945 x
1Ac = 17/. S'
194
DC
s;oo
AMP.
I
__j
L __,
500- AMP
DIRECT-
EFFICIENCY AND UNITY
CUfUUO.Ni
(o) THIZ.EE 17..1N9S ; (b) SIX ll./1'-IGS ;
J.iAS
CONVE/lTEfC
DEUVER.S
100 PERCENT
G/VEf.J:
SOLUTION:
CUR/7...t:NT OUTPUT
,fa,....,.
i<ILOVOLT-
Tr<ANSFOR.ME/L.
RATING=: 1000 -kW ; 3 -PHASE
lpc
X P.F
1000 ,ketAl
=
91- PER CENT AIJD THE
CURReNT;
BY t::ACH
J3
kV,LijT
DETERMINE' (~) THE Otrl..ECT CUR.R.E~H
SUP-fLING
BY EACH TRAf-iS'POrUviETl..S
DELIVERED
KVAh
9l A
D.) DIRECT
LOAD
jOOO /M
GI\/EI-J:
Ef=F==
KVA
kYA ,1
-:7a.,..._.
CONVeRTER
VOLTS. IF THE EFFICIENCY
c)
0·94-
AMP J~
1,71?·+
1Ac=
f(W
X 0·9); X
w
= -J3;:::3:--X-O,:..G-=-I-='2-x_6_oo
__
x~0--9-+_x_o=-.-96
t.f'F.
X
Ef=F. X PF.
1000,000
E.Aet-1 OF THE TKANSFORMER.S
KV,A;
==
/TRANSFORMER
X
EAc = 0.012 Epc
.:BUT
212.7 AMP.J"""'"•.
LOAD
(lAc)
CURRENT
p
== - - -
ErT
rtECTIPIEfZ.S
AND
7]
b) SLIP- 11.11-JG
CUIZfU:NT
IAc = - -
d)
CONVERTERS
f?.ECTlFlEn.S
e]
OF PROS. -#
C) THE
AND
195
j',.,__
CONVt:tzTErlS
CON f. OF Pf7.013 · *-
AND
IZC:CTIFIEIZ$
CONVEfl.TElZS
flECTIFIEfl.S
AND
CONl~ OF PROB · #' lQ]
9]
IAc == 0-472 It:>c
IAc
=:.
~~
h) r~ULL- LOAD DlilECT- CUn.n.ENT OUTPUT (I De)
SIX IZING"S
b) FOR
Toe = J:_RATINC:_
0-"'1-72 X "!;00
Epc
IA<. = 23G A~1F,f;;......-....
I
_
I)C-
---
/2QV
C) t=On. 12 fUNGS
lvc. =
I,4c = 0 25G Ivc
11'1,
JO,OOOW
83-33 AMP
,fCIM4...
]Ac= 0-236 X 500
C) A -C VOLTS
1Ac = 118 AMP. /-....
10
JA
LABORATORY
CONVEI1TER
HAS THE
S"41j 10-I<W SIX- PHASE , 120 VOLTS
DETEf7MINE
TH~
A-C
FOLLOWING:
A -C VOLTS
0/RECT-Clll"lflENT
Al7JACEJ\IT
SLIP-RINC CURRENT, A9SUMJNC..:
AND UWTY
o. 35t
f;.c
O. 3!?4 X /20
f?.ATING:
(d)
d)
A-C
I
TAc =
AN EFFICIENCY OF 92 PCR.
IAc
~
GIVEN·
=
/0 KW ,
SIX~ PHASE.
IAc =
PRATING
----------G X EFf X EAc:X P.F
10,000
w
0 X0-92
X
/200
r:rf ==
o-sz
P-F.""
X
I
OR)
= GO CYCLeS
RPM=
4?·4
i
!=OR Six FtlASE
4?· 72 X 42· 8 AMP
.Ere ""' 120 VOLTS
f
~·
42.48 VOLTS /a.-w..
SLIP fl.ING CUR.fl.ENT
THE
POWEll.. fACTOR. .
17.ATJI-J~
I
Eve
i'
i!!
OUTPUT;
SUP fl.INCS _:
=
fAc ~
(o) THE NUMBER. OF POLES IN
BEHVEEN
EAc'=
0-C . 60 CYCLES / 11300 RPM.
MACHINE; (b) THE: Fl!Ll-L0-10
(c) THE
CENT
THE:.
FOLLOWING
BETWEeN ADJACENT SUP RINGS
FfWM
UNITY
TABLE
Tf\c.
10, PAGE
49)
0· 472. 1 pc
------
:EFF
SOLUTION:
0· '1-72 )( 83-53
1Ac
a) NUt118Efl. or r"'OLES :
LET
P IS THE NUM8Ef~ OF POLE.-5'
SINCE:
rtPM
!2-0f
= ---·····F
P~
p""
J'20F
i? i
lri
1Ac
--~~-JI
WHAT
$4-2
OF THE
SHOULD
THlt.EE
BE. THE
0-92
4-2 78 Z 4-'.Z-8 A~P/a..u...
RATING Of= eACH
KILOVOLT-AMPEil.E
mA/'..1..5F017..ME(LS
USED
IN
PIWS.
10?
120 X G·~­
--,e(x)-
p ~ --:f
i 't}J- r· t.._,~, ·;~.-~
196
SEE
SOLUTION AT THE NEXT P.A49E
___j
197
~1 1 1
1111
CONVEIZTETZ-S
CONVERTERS
AND IZECTIFlEilS
D
CONT OF Prt013 ~ I
Ill
AND 11ECTIFJER.S
CONT OF PfWS. #< 12] SOLUTION:
b) DIIU:CT
SOLUTION:
KW
=
KVA/
ITl<ANSf=Orl.Mt:R
P.F. X EFF X 5
=
Epc.
3,000~000
.
] De=
I X0-92 X.3
KVAfT -==
pi'U>Til'IG
Ivc==
10 KW
KVA/r
CURRENT OUTPUT (I 0 c)
w
750
Ioc "" 1;000 AMP j!~.
3.fi2 KVA .fa..a.
C) A-C VOLTAGE BETWEEN Ati..JACENT SUP-ft/NGS
12.
l
HIGH -VOLTAGE
Al7.E
LINE
ON TH~ LOW SIDE
IN
A
4
(o.) MAkE
BANK
ON THE
A
(b) Tl-IE Dln..ECT
9G
TH(tEE -PHASE
CUMENT
SUP RIN&;$' ~
AND
OUTP!JT;
0.9~
(c) Tf-JE
DIAG/t.Aivl
AN
R.ATIO Of TflANSFOtz.MATION
Or
EACH
FOIL SIX Pt1ASt:
j
E.Ac = 0·3G4-X 750
SHOWING ALL
.f-.Ac = 2(;b VOL-T$ Q\-C)Jaw....
!.,
ErFICIENCY AND
J)
17..t=SPECTIVELY 1 CALCULATE:
A-C VOLTAGE
J:'/ICH TRAN.SrOIZMER. SECONDARY COlL
VOLTAGE OF
I
lffi
BETWEEN
W
-- '2
X EA
IT s<oc. coiL-
(cl) THE VOLTAGE OF EACH TfLANSFOIT./VIER
(e) HIE A-C SUP fUNG
COIL ;
f.Ac = o.;s::;4 EPc
23_DOO-VOLT
TILANSFORME:R.S
HIGH SIDE AND DIAMETTUCAL
WINDING
PE/7.. CEI-.JT
SIX- PHASE S!:JNCH/7..0-
OF TH/7..EE
FOfL THE PROBLEM- ,ASSUMING
POWE/7.. FACTOR OF
SECONDARY
IS FED FJWM A
THrLOUGH
CONNECTED
CONNECTIONS
AD.JACEt-lf
0-C GO CYCL.E.
750 VOLT
CONVEIZ.TE/l
5"42J NOUS
TJ-IAT
TABLE 10, PAGE 491
FI'WM
A 5,000 I<W
CUfULENT ; (-f) THE
(Tl-flS
FIWM T-HE G'EOMETRY Of=
FOLLOWS
THE SIX PHA£E
OF THE THREE TRANSFOfLMERS.,
Tf-tE
c
HEXAGON _., IN
ACROSS TttE DIAGONAL IS
VOLTAGE
GIVEN:
flATING
=.3,000 kW
:=:
=
7£"0 VOLTS
V,!
IT se:c.
Eve
-f
=
GO CYCLI:_.S
5 PHASE SOURCE
I:.Ff= 9G
P·F.
:= 2~,000
VOLTS
= 0-95
e)
3-PHASI:
LOAD
SOURCE
SIDE
4- CON NECTElJ
J
=
2
:::;
~60 VOLTS/A<J.f'.
l(
2G'S
co1L.
THE A-C SLIP
Inc=
CONVER-TER-
7sov. D-C-
"---DIA~RAM
198
rUNG
0·4-'72 1vc._
EFF x P·F-
9
CUTUU::NT
.
100
X c.PF.
OR,
I
II
I
I,.c=
POWE.IZ/rJ<A~fORMI::R
VOl-TS/rRAN.\"FORl'<1ER X P·f. X EFF.
lf\-c
=
lAc"'
3,000,000/3 X'FORME.PS
.\?30
X 0-9<C
X
0·.95
2PG'S ~ 2,070 Alv'\P/"""'",_
L
199
I
(IAc)
rR.OM TABLE 10 AT
ANV
DC SOURCE
WINDING
VOLTS/SIDE
X
/p
SOLUTION:
a)
2
WHICH
I ..
P.F= I
CONVERTERS
OF Pfl.OB.
CONT.
CONVEflTEJlS AND flECTIFIETZS
flECT/F/Efl.S
AND
E_] THE THREE
. 5'42 j CONVETZTE.Jl
::jj:: 1e]
f)
TfZANSFOfltv'lATION
(a)
fl.ATIO
Ep
?3.000
Es
~0
Y
lS
0.= 43·4-: 1
OF
Pfl.l
.A
I
THfl.EE -PHASE
ON THE
PIUMARY SIDE AND
SIDE· /F THE SLIP-tl.ING CURRENT
PlllMAR.Y COIL
2/.7: I ~ (b) THE
THE
IN .1
CONNECTED
CALCULATE : (a) THE CUfl.TtENT
AMP,
FOfl.MER
THAT FEED
Tfl.ANSFORMEQS
AflE
ON THE SECONDARY
"1'7J.~
a=-=--
SEC:
IIIII
I
IN EACH
JS
TRANS-
IF l11E /7.ATIO OF TIV>NSF011..MATION
CURn.ENT ON TfiE
LINE
PltiMAil.Y SIVE
TR.,4NSF011.MEfl.S.
GIVEN:
13l
A
512] AT
<f-000- KW
UNITY
POWErl FACTOrl
12-F'HASE
AND
Dlrz.ECT
CONVERTER
AN EFFICIENCY
(a) THE A-C VOLTAGe
CALCULATE:
(b) 1HE
1,5"00 -VOLT
CVrmENT OUTPUT;.
6E1WEEN
(c) Tl+E
OF
3 -PHASE
QP'EilATES
9G.S PETtCE:NT.
4DJACENT
CONVEI"lTEf7..
ltlrt.EE Trl,ANSFORMER
.6 ON THE Pfl.IMArtY
SUP RINGS;
Y ON THE
a.= 27: I
A-C SLIP !ZING CUIII<ENT.
G/Vr::W
lAc== 47f·f'
rt,ATING'
= 4;000 KW
E.oc =
1,500 VOLTS
SECONDARY
AMP·
IIIII I
!
EFF. == 96.~ /.
IU
~
12- PHASE CONVERTER
SOLUTION:
a.)
TABLE 10, PAGE 491
A-C VOLTAGE
BETWEEN AD..JACENT SLIP RINGS
E.Ac
= o. 182
EAc ==
i'73
~ ~
~
!:!
0..
II..
,:j
~
I
It)
EL
fAc=3~7 V
lL
EDC
t1)
FR./MAllY (
SECONVAn..y
X ~S"OO
VOLTS
r:::!
-TAc,.4SJ<i'A~
-TAc:
E..4c == 0.182
~
~71SA
~
f'.F. =UNITY
FROM
-'IAc::
EL
!/)
~
>
z.
8
II
. Ill
I
I'I
~
I
(WYE)
FIGURE:
./a.,.,_.
I
I'·
I
lo)
DII7.ECT-CUfULENT
Ioc=
SOLUTION:
OUTPUT
4:. 000,000
vv
o.)
I se-c.
IAL
1/'P.t.
Ir
~=-
SLIP IIING CUfl.RENT
IAc=
JN EACH Tfl.ANSFORMEit (Jp)
PRIMARY CUrU7.ENT
I.SOO V
lvc = c,GGG ·G7 AMP.
C) A-C
1
1"1
1:111
Ir=
0· 23G loc
EFF X p.p
I,o~c.
a.
47/·!; A
=
21·7
Ip =- ZJ. 73 Afv'lf'.
(ALSO THESE VAWE
0·23G X I?GGG. fi7
IAc
= --o-.-9-G_S'_X_/_ _
lAC == G 92 AMP J'CMM-.
la)
SfNCf:
IL =
{3 IrHASl!
1~.- == .[.3 X 2l· 7.3
I L == 67·10~ A~P.j'a..,_ .
200
201
IS A PHASE
GUMENT IN DELTA, PJ?.IMATl.Y)
~ .D. S>'STJ:!M
CONVERTERS
18] IF THE SLIP-RING
stZ}
LATE :
CONVETlTEf?_S
AND fl.ECTIFJEIZS
IN Pf?_OB. 17
VOL-TAGE
IS 3G7 VOLTS
1
CALCU-
(a) THE D-C VOLTAGE ; (b) THE VOLTAGE pEn. Tf/..ANSFOQMER.
SECONDARY
~
COIL
if')
CONVErl.T
Ffl..IMAI<.Y LII-JE VOLTAGE;.
WE
21] (a) WHAT IS
512j \)SE-D
OPEI1ATE
THE
TWO
IN
FIWM
AND f?_ECTIFIERS
SMALLEST
NUMBErL
OF POLES
/V!ACHINE-S
SYNCHRONOUS
GO TO SO CYCLES
IF IT
THAT CAN 13E
IS DESIR-ED
(b) AT WHAT SPEED
?
TO
WILL THE SET
7
GIVI'::I-J:
e._,.c
=
SOLUTION:
5G7 VOLTS
,,)
a-) ~
UGE DATA FfWM PfWI3. 17
SOLUTION:
a-)
=
-f.,
THE !7-C VOLTAGE
~
I
f=OR gyNCt-lfWNOL'S FnEQUENCY
CONVfllTEil
p,
WHErLf:
{Eve):
-f, =
FOil. 3- PHASE CONYErLTE.R,
GO CYCLE
-\\ = SO CYCLE
E,4c
= O·<iii'Z
Eoc
--r-14US ;
TJ-IEnEFOil.E:
1=-Dc= ~
0-<012
~99-G7 ~ 600 VOLTS
Epc =
1?)
GO
12
A
-=-=10
Pz.
50
3G7V
EAc
..fa..w..
-FTWM
FIG·
f'OIL
IN
PfWB,
#- 1 HAS 12 POLES, Wi-JILE
MACI-JINI:
#
6)
VOLTAGE PE11. T/l.ANSl"OIUVIfOR. SE.CoNDArty COIL
'I~.-
.f3
e
HAS
1'20 fl
flPM = - - -
# 17
RPM=
Eq>
RPM=
120 X f.i>O
-
10 POLES
120
=
p,
W)IE. 9J'STEMS
l=t. =
iii
MACI-IINc
+.,_
---
p.,_
120X !?0
---10
12.
GOO RPM.fa.u
,,, '"
II
"" T 4>
SINCE:
22
E.._=E.Ac
r:.<t =
E... c.
fAc
'1rsec.
A SIX- POLE THilEf -PHASE WOUND- fWTOiL MOTOR IS TO BE USED
AS AN INDUCTION FREQUENCY CONVERTER . THE STATOR IS
CONNeCTED
f-3
VOLTAGE/TitANS· SEC. COIL
J
S-tiJ
..J3 -
21-2 VOL.TSJaw.,_
.3C.7
-S-3
TO A GO-CYCLE
DJI1ECTION, MUST
(a) 120 CYCLES
1
(b) 100
ATWHAT
SOU/lCE.
THE ROTOR
BE DRIVEN
CYCLES
7
I
I-
SPEED, AND IN WHAT
IF IT IS TO DEVELOP.
(C) 180 CYCLES
(d) W
CYCLES
7
GIVEN:
SIX-POLl: , Tt-I~FHASc, WOUND-rWTOn. MOTOR
f= GO CYCLES
c)
przrMAfl..Y LINE VOLTAGJ'::
$1-
=
SOLUTION:
0-
X 212
El :::: '21·7 X 21'2
el-
SYNCHgoNOUS' SPEE.O = Ns =
ltof
-p1'20 X60
= f,G'OO VOLTS#'~-
G
Ns = 1;200 RPM
202
203
!Ill~
I i',
t''"
!,IIIli
CONVEflTEfZS
CONVERTE/7.S
CONT
OF PIWB #
17.ECTlFIE:R.S
AND
~ IF TH£ VOLTAGE
<?2]
fcoNv. -
f
5+2]
BETWEEN
SUP
fliNGS
IJ..J PIZ0/3· t?2
STANDSTILL , CALCULATE THE SLIP-fl.ING VOLTAGES
OIFFEILENT
I± RPMcmJVEI<TEI'- )
RPM s-yt.JcH
(
AND TlECTIFIER0
IS
120 AT
Ill
illlllf
1
FOIL THE
FflEQUENC/ES.
11111
IIIII I
·SOLUTION :
NOTE:
USED PLUS SIGN
.1\GAINST
WHEN
flEVOLV/1\JG
THE
Tr+E ROTOR- TUTLNS
ALSO
IS GREATEt<.
FfZ.EQUENcY
120 CYCLES
II'' '
MINUS SIGN
E= 120X (
IN THE SAME 0/fl.ECT/ON.
IF THe f=-TLEQUENCY
THE SYNCHTLONOUS
Af!JD
FIELD
I
~) AT
WHEN THE rWTOfZ IS ORJVEN
THAN
b)
AT
fZPM
r
l
-fcoNv.
100)
J
-1
X
c)
RPMsvr-Jc.
AT
CONV
Jf~O)
l
j
- I
d)
1,200
AT 100 CYCLE
RPMcotJv =
AT
l4J
X 1200
800 fLPM
I
I
(THE SAlliE AS IN
P.J
II
A TUNGATl flECT!FIETl CHARGES
512
AND G;·<J- VOLTS. ASSUMING
THE
HALF-WAVE
THE
CURRENT
RECTIFIER
A STORAGE
Bt\TTcRY
SINOSOIDAL OUTPUT
ANO A TUBE
AND VOLTAGE RATINGS
AT 5 AMP
VARIATiONS
f
FOfL
Dfl.OP c>F 10 VOLTS, CALCULATf
OF THE TRANSFORMER SECON-
IIIII
~
DARY.
180 CYCLE
RPMco~o~v
~I
E = 120 X ( : ) = 80 VOLTS
'II
J
100
RPMcoNv = [ CO -I
_,60 VOLTS
AT 1D CYCLES
1,'200 fZPM (Off'OSITE TO THE REVOLVING
coNV =
j =
E= 12'0 X( GO
rl 120
r;o
=
200 vDLT.S
lBO CYCLES
fsYNC·
FiELD)
J
if'
- <00
AT IW CYCLE
riPM
C
=2'TO VOLTS
E=- 120 X ( - - }
RPJYl CONV. =
b)
c;o"J
100 CYCLE$
USED P'WS SIGN.
II+US;
o.)
120\
GIVEN-'
l
180
j
= [ GO -I
Ioc =
X /,200
!;
Eve ==
I'''
Alv'lf'·
G'· <J- VOLTS
TUBE VOLTAGE 0/lOP = 10 VOLTS
RPM wtJv = 2400 RPM
p.e.e>..-o:
d)
AT
10 CYCLE
r
IWM
CONY
=
Il
I -
40 l
GO
JX
1;2 00
.;oLUTJON:
TKE
RPM c.Nv =
llil
L A c / V.qc
400 RPM
( IT SHOULD
BE THE SAME
DIILECTION TO THE
ILEVDLVING FIELD)
RECTIF-IER
VALUE
il l
UNID/l<CCTIONAL LOAD CUIZJl.ENT /f..l /> 1-fALF WAVE
FLOWS
DURING
OF
EACH CYCLE / ITS AVERAGE
IS
l
loc = 2
It>c ::::
204
Yz
2
I,
1T'
X-
0· ~IB
Im
~
]..,., =
MAX. VALUE OF TI+E S/NI::.
WAVE.
205
''I
CONT.
CONVERTErlS AND 17ECTIFIERS
AND RECTIFIEflS
CONVEfZTEJ"lS
z"'j
OF Pfl.OB. #
'28] A HALF-WAVE
5"43] 10 AMP
ALSO TttE
eFFECTIVE VALUE OF SUCH CU!<flE.NT IN
A
VOLTAGES
RECTIFIEr?. IS
HALF WAVE
Irn
I
I.o~c =
J2
=
OF THE
VOLTS. IF THE
LECTING
2
D-C LOAD OF
NEG-
CUPll.E/'IT.
GIVEN:
It>c
Epe
Ipc
2X0·3~
A
AND GECONDAfl.Y
ARE , fl.ESFECT/VELY, 11'0 AND G?,
TflANSFORMEn..
THE EXCITING
THEREFDRE j
IAc=
DELIVERS
PfliiVlARY
CALCULAIE: (o.) THE TUBE DfWP; (!o) THE P/liMAflY CUT7.RE.NT,
Im
[2
X
RECTIGON flECTIFIER
AT 20
5
~ __2__x_o-.-3-18
= 10 AMP
= 20 VOLTS
f:p =
liE VOLT-S
EAC= G2 VOLTS
TAc= Y.8G AMP. ,.:/'"'''··
TOTAL
p.(. OUTPUT VOLTAGE =
SOLUTION:
o.)
Eoc + TUBE DROP;
JT IS ALSO EQUAL TO
Epc
== 0·318 Em
~
TUBE
E;~c
l=m= MAX. VALUE OF TtlE SINE
WAVE
E..,
tpc +TO
[2
X 0·318
E.Ac X
.f2
X 0·318
TD.= - - - - - -
0·318 X
Jz
TD=
!..)
G·4 +IO
-o. 316
xrz-·
7.9 VOLTS
PRIMARY
(TUBE !JROPJ
CURRENT
EAc
IAcPRI.=
:E"Ac =' 3~ · "1-7 VOLTS ;(C~M>.
X IAcsEc·
t:..p
BUT:
.Ipc
~ JF THE PRIMARY OF THE T/lANSFOrtMER
s1-3j
CONI-JECTED
AMPERE
TO
IN PfWB. 24
IS
A !18-VOLT soUn.CE ,COMPUTE THE VOLT-
INPUT, NI::G;LECTJN(; THE
EXCITING CUri.I?.ENT.
VA
Z X0·318
lAc sec=
,-s: 7?,
IAcPRI.
EAc X TAc
= 36"·47
VA:::
IAcs,.c =
=
10
_Z_X_0-·5_1_8
AII'1V
THEREfVRE:
SOLUTION:
VA ==
- 20
Epc
EAc == ~
EAc ==-
=
(T.IJ.) VOLT-$
D11..0P
IACpl(j.=
X 7·8Ci
28ft;·7 VA/~.
=
2'3
J ASSUMING
543j TIVE
115" y
X 1!0·73 AMP.
8·48 ~ 8.t; AMP Ja-.
A n..ECTANGULAf( WAVE FORM , CALCULATE TttE EFFEC-
ANODE
JtECTIFIEit
G2V
CUfCRENT
FOn... A CJ-C
IN A TWO
ANODe
GIVEN:
Ioc =
~--· ,_--:--j
206
IVJE/lCURY- ArLC
LOAD Of !10 AMf'.
SOLUTION:
IA·~F= 0·707 Ipc
!=iO AMP.
= 0·707 X
ReQo:
EFFECTIVE ANODE CURRENT
207
I;O
== 35.35 AMPf'a.u.
.,n':111
1
CONVERTERS
AND flECTIFIEnS
CONVEflTElZS
AND
,;1
RECTIFIER-S
ill,, II
30] A TWO- ANODE
~43j 15 VOLTS
MEilCUIIY- AftC
A D-C LOAD
DELIVER$
ANO
THE MAXIMtJM
POTENTIAL
RECTIFIER
DlFFE.RENCE
HAS AN ARC DROP OF
AT
CONT. OF PfWB
120 VOLTS. CALCULATE!
BETWEEN
c)
CATHODE AND
#
31]
=12
/'l ANODES ~ ~
AT
r;,OOO,OOD
ANODE.
IAE.ff-
GIVEN:
A~C OIZOP
32] A T\NO-ANODE
Em = .)2 X !'20
= 169
IS
s::t3) S'-913
MAX· E
KW
CONNECTED
VOLTS
AND A
AMP
1
;fa.w..
MERCUI1Y-AnC rz.ECTIFlt:JZ
AT 115' VOLTS· IF THE
1--,~00
TO A
RECTANGULAR
= Eve +
£,..,
CUR/tENT, CALCULATE:
+
1109
fLATING OF EACH
120
= 001 · 4
1'20 VOLTS
MAXIMUM VALUE. OF SINUSOIVAL VOLTAGE
=
1,111i
,.
IAcFF
SOLUTION:
E
m
x
II ••
== IS VOLTS
Epc "'
MA)(·
2;100
1
(C) THE
= c89 VOl:TS_fa.u.
DELIVERS A LOAD
PJZIMAflY
OF TIZANSF0/7.MC:R.
OF
1
11
1111:
IS
-VOLT .SOUfl.CE, THE TUBE DMP IS IS VOLlS,
IS ASSUMED FOil. THE AI-JODE
WAVESHAPE
(o) THE ANOOE CtJR11ENT; (h) TI-lE VOLTAGE
OF THE lflANSFORMER. SECONDARY WINDINGS~
PR.IMA11.Y CUfiiZENT, NEGLECTING THE EXCITING CUflnt:.NT;
(d) THE I<ILOVOLT- AMPERE
IZATING
(e)
RATINC OF THE TRANSFOfl.tv?Et:\ SECONDARY.
i:l
!'I
OF THE TRANSFORMER Pl'liMARY _;
:1
~
S'!-3
j
ASSUMING
TIVE
KILOVOLT- AMPERE
WAVE FOfl.MS. COMPUTE TI-lE EFFEC-
fl.ECT.ANGULAfl
ANODE
THE
FOR THE FOLLOWING
CURRENTS
D-C LOADS
IIIIi
GIVEN.'
lWO AI-JOVE
/VlEilCVIlY -Ail.C ftECTIFIERS ; (o.) 75 .kW'
230 VOLTS , THREE ANOJJES ~ ( 6) t 500 k.W 1 1,200 VOLTS , SIX
IN THE GIVEN
MULTIANODE
KW LOAD= E -98 KW
Eoc
ANOD!::S _; (c) 5>000 KW • 2~ 400 VOLTS , 12 ANODES.
=
E A-cr =
sownoN:
~
0-J IA-~
~rc
liS VOLTS
"LGOO VOLTS
TUBE DROP (TD):.: l'i VOLTS
=-ff- _ WHERE a/)
II)C
'Iuc =
7'S,OOO W
.::o IS THE IJUMBER Of A/-JODE.S
.
"'-' 3210-09 At<W
SOWTION:
0-)
ANODE CUfl.flENT (I,; "'-FF)
230V
1.vc
IAtoFF "" ~
fOil.
3
A-NODES ;
J:pc
J5
IA-EFF=
b) AT
1
'f>
~ ~ =2
FO/l. TWO /NODES
=-~
5210-0'3
=---]3
SIX AJ..JODE.S' ;
'1J =G
G-9B X 1000
186 5 AMPfa.w..
-
w
r
IH> X .fZ
IE
IA
I:FF
= 56-8 AMP
l'a..u.
b) VOLTAGE RATING OF EACH OF THE TnANSFOil.MEIT
IA.Eff'
1A~ff'
=
=
1,5'00,000
w
S.fCONDAflY WINDINGS
xfG
J::pc
= VL +
510·0 .AMP~'
.va.....
C.pc
-=
1,200
203
ARC DTWP
II~+ I~
=:
209
(50 VOLT'-S'
II
WNVEflTETlS
CONT
AND
Of Pfl.OB · # 32]
ALSO;
Eoc
130
v
9J .
=if X.f2 ><
2
= -11'
=
VSin
1/
(130)
2
J2
THE
rlcCTJFIER.S
A THREE-ANODE MER.CUfO'-Afl.C llECTIF!ER. SUPPLIES A 2.;0-K1
.!>1'3j
250-VOLT LOAD AND
WHOSE
1T
eo
2
AND
AflE IN DELTA.
PRIMA fliES
VOLTS,
JS ENE!lGJZ.ED
FrZOM A 4,GOO -VOLT
SOURCE TtHI.OUGH A THREE- PHASE TflANSFOR/111':R
PHASE
AI1C OflOP OF
LOAD CURRENT ~(b) THE VOLTAGE
CALCULATE: (a) THE P-C
KILOVOLT- .AMF'ERE flAYING
ASSUMING AN
Or !';L\Cf.l Tfl.ANSFOf7.MER SECONC7ARY;
(C) THE CUT'lfl.ENT AND KILOVOLT -AMPERE RATING
go•
SIN
V = 144. 4- VOLT-Sfa.w..
C)
AND
33l
11-IIU:E-PHAS!::
t
X~ X V SIN -
n
CONVEIZTERS
RECT!FfEfl.S
OF TI-lE THREE -
TRANSFORMER.
GIVE'N:
25;0 KW
PfZIMATLY CURRENT:
II
E:: '230 VOLTS
Tp =
I
).)1-
X
AEFF
..5G·8 X 14+·4
Tp =
d)
KVA
f.l!; AMP /a.w.
OF THE TrzANSFORMER PfUMAfl.Y:
P.ATIN(;
J'li
KVA P =
EAc = 4,G'OO VOLT 3-PtlASE .SOURCE
4;GOO
E,..cp
(Afl.C Dfl.OP) = 20 VOLTS
T.O.
il
SOLUTION:
a)
D-C
LOAD CURRENT
250,000 w
Ioc = ----'----
x lAc XV
230
v
/000
.[2
KVA p =
1000
KVA p ==
e)
KVA
RATING
It>c=
X 3G·8 )( 144 ·4
7.f>Jt;
KVA/aw2..
OF THE TRANSFOfZMER SECONDARY:
b)
1087 AMP.
VOLTA{;E..ft"'J- kVA RATING OP EACH TfZ.ANSFOfZMER
SECONDARY
ii!i'i'
Epc
KVA Sf'c
=
,n
~
TOTAL p..c VOLTAG;E CEpe)-= 230 + 20
Il>c V
1000
E17c ==
~
Tr
[Z V
2~0
=:
VOLTS
iT
SIN ~
!;.98 X 1000
Ioc ==
}pc:::
II!;
!;2
V
= Eoc
x 1f
fZ X OSIW'l/;
rOR 3 ANOOE.$'
AMP.
y ::::
"WAsEc =
-' ~ = 0
-J2
X ;z X li<t·4
1000
250 IT
.j2
X
3 SIN _!80: x-fi.3
_;:rt
V = 213 .7§ VOLTS /tM>t..
KVA St:c =
IO·G'2 kVA
kVAsec
l(VAsec
J3
}pc V
; IL>c = 1087 ;AMP
1000
~ X !087 X 213·75'
= -'--------
\000
.- 40'Z ·44
210
211
kVA
,fla..w_.
I
I
CONVERTERS
PROB lF- .3:5]
CONT. OF
!
CONVE17TEFS
AND RECTIFIERS
CONT OF PR0/3 · -Jlo 31-]
C) CUTULENT AND I<VA RATING OF T+-IE THREE-PHASE
D-C CURRENT:
TrtANSf=OJLMEfl
r,.~.ef=F =
Ioc.
IilePF -
-1087
-
ff
..J5 = fD27. G ,AMP
<f,fDOO
29 .1c; .AMP. f4tt4-.
Ip:::.
KVA
=
J.oc
kYAT
Ivc,
=
Ioc
=
!:1000 l<YA X JG
!]30 V
X G
Z/9t;; AMP .f,_..
FOWER:
29·1G X 1-.600
Tp EAcp
1000
.JCf
= ~ = {<if
G27·f> X 213·75
XV
E11c
Ioc
BtJr:
I;~.EFF
TAeFF
Tp =-
AND RECTI F/ERS
1000
POWER
(t<\\1):::
EJ>c X Ioc
1000
134 !<VA ..f'a.-w..
k VA.::
1'23/ X ZJg!;
11.:1
IOOO
~ A SIX
MERCUllY-.Arte RECTIFIER IS SERVED BY A
ANODE
S43j
THILEE- PHASE
AND
13,800/930- l)i3GO VOLT$. ASSUMING
ANODE
HAVING A RATING OF S,OOOKVA
Tfl_.tlNSFOJLMER
CALCULATE : (£)) THE D-C LOAD
(b) THE
pOWER
b) TI+E
= 2,700 J<:W
A 25-VOLT An..c OfWP,
VOLTAGE , CUflRENT AND
POWER ;
GIVEN:
2,Jg~;;
Tpc
IA EI'F
IAEFF
;<t
f39G
AMP/""'-,..
Y= 930 VOLTS
i 3, 900 VOLTS
TD. (AI<C DROP)= 25 VOLTS
kVA =
!>000
SOLUTION:
0)
TOTAL DC
VOLTAG'c
= -} fZ
G
r::;-
V SJN
Ever=- X ~2 X.930
.
1t
=l=vcrarAL =
Epc ==
fpc
=
Eoc=
8>cToTAL--
I'Zs-tt: -
~
1T'
lBO
X SIN- X G
rr
1'25"5·94 ~ I'Z!;G VOLTS
TD.
25
lt31 VOLTS
212
A
.[6
SIX ANODE ~ ~ =G
==
tl:
ANOJJE CURRENT:
CUP:RENT.
EAc
/a.w..
213
1111
:I
CONVERTERS .ANV RECTIFIERS: IllUSTRATIVE PROBLEMS
,l:XAMPLE 1] A
SIN~LE- PHASE
1·5'-I<W
/<T FULL LOAD FROM
CONVERTEP.
A 2.30-VOLT .A-C SOURCE.
LOSSES_. ANO ASSUMING
OPERATES
NEGLECTIN~
rACTOR Or UNITY, CALCULATE:
A POWER
(o.) THE D-C VOLTAGE ANO CURRENT ; (b) THE A-C IN-PUT
CURRENT-
EXAMPLE
.5]
A 2!7
GIVEN:
A?WER RATING= Z!1 -KW
E:pc
VOLTAGe ANO CUP-RENT
= .[.2 l=Ac
=
J;:
Gl) A-C /NPtJT VOLTAGES
1_5"00
EPC
b) rULl
0 ·GI2 Euc
FOR THReE PHASe
-
141 VOLTS/a-.
LOAD
CURRENT
D-C OUTPUT (l"")
F
lpc = - E
_ PAc
(EAc)
O·<llt X t'.30
EAc=
Tpc = · 4-<CZ .AMP /a.m.
A-C INPUT
=
EAc =
= --32g
lPc= - -
= 230 VOLTS
501-lJTION:
'X ~30
EAc
Pee
b)
IIi
.3- P~ASI': CONVERTER
Epe = .3ZG VOLT$ /£~.1M..
~
A-C
PER UNE.
SOLUTION:
I
111
RINGS; (b) THE rULL-L~,dO 1:1-C OUTPUT; (c) THE FLILL-L..OAO
EAc = 230 VOLT
EDC
CONVERTER HAs A D-i:: OUTPUT
2.30 VOLTS. ~SUM lNG: Nt~·blt;JB-t:J: ,· tOSSES AND UNITY
!"'WER FACTO~, CALCyLATE: (o.) Tl-IE A-C INFr.JT VOLTAGES BETWEEN
Pp..ATING =I·!> KW
a) 0-C
-J<w,' lliREE -PHAS.!::
PROBLEMS
VOLTASE OF
INPUT
GIVt::N:
~ECTIF!ERS : ILLUSTRATIVE
CONVEIUERS AND
!
l>c
1,500
=
2£i,OOCI
230
= tOB-7 AMP,I...._
I Ac---==?.30
EAc
c) FUU-LOAV A-c 1'-IP'UT PER. LINE (IAc)
I.Ac -= ~ · S' z A lv1 P j'a.«.
0·~-t-5 I!>c
1Ae =
:s
OPERATED
INVI=!ZfEV, L·e, IT COt-!VERTS
DIRECT TO ALTERNAT!Nt:'
CIJRRENT·
IF THE V-C INPUT VOLTAGe
IS
EXAMPLE
TAGE ;
z]
A
Z·\;-KW SING:LE -PHASE
CONVeRTER
2.30, CALCULATE: (o.) THE .&t-C VOL-
(b) Tl-IE ALTERNATlN~ AND lJI~ECT CURRENTS AT FUU.. Lad!l
EXAMPLE 4] A, LS'OO- KW SIX -PHASE
t-fAS
SOLUTIOI'J:
o.)
A FULL-lOAD
VOLTAGE
A.-C VOLTAGE (EAc) :
SYNcHRONOUS CONVERTER.
D-C VOLTAGE OF C>OQ. DETERMINE :(a) THI= A-C
SETWEEN
RINt;S ; (lo) THE D-C OUTPUT~ (C) THE A-C
INPUT PER LINE.
cpc
'230
[2
J2
EAc = - - = - t::.A-c ::
(FoR THREe PHASE)
lAc= 0·~+3 X 108·7
TAc = IOZ-3 AMP- ,!'........_.
pOWER. AATIN(;
IG2·!< VOLTS ,/..-...
G- PHASE
.b) ALTERNATING/ D!~ECT CUH!<ENTS AT FIJLL LOAD:
PAc
IAc
= -t=Ac
- ="
Ire
=-
2,500
--
)t;'Z ·fi
= Is; ·4 AMP·
E~c
_
fpc
-
2, S:.OO
230 ::: I0-9AMP.
''214
=
= l-£>00
I
kW'
I
I
SYNCH~NoUS CONVE.RTEP.
rDOO VOLTS
SOWTION:
a) A-C
Ppc
!
GIVEf.J:
VOLTAGE. BETWEeN
P.INGG:
EAc
=
0-3S"f-
EA<:
=
21Z·4. VOLTS/~.
xr;;oo
t2J5
;;~.
(FeR S PHASE)
II
COIJVERTERS AND P.l:CT!FIERS : ILLUSTRATIVE
CONT
CONVERTEP.S
PROBLEMS
!:.)(AMPLE
OF EX· 1t 4]
b) THE
D-C OUTPUT:
=-
Ioc
p
=
Iuc
CONNECTE.D
1,!700,000
c) A-C INPUT
FACTOR OF
(FOR SIX-PHASE)
VERTER
OPERATES
i,200-VOLT
AT FULL LOAD
ON THE
f'tJLL WAD
St:CONOARY
fiT A POWER
liltf
o.gz AND AN cf=FJCIENCY OF- 93 PERCENT ,AND 0\I..CIJLATE:
Iii I[
13Y WE TRANSFOR-
CURRENT DELIVERED
SfO_CONDARIE.S TO TflE SLIP RINbS;
TIDN
OF EACH
TRANSFORMER
12-PHASE SYNCHRONOUS
AT AN
OPERATE.S fiT
(d) 11-+E. JI.AT!O OF IF.AI.JSFOR/11141111
== 1)80 AMP. /am..
A 5,000-KW
ON n+E PRJIVIAR.Y SIDE ,AND Y
MER
WIRES
s]
£,.
RINGS ; (c) THE. ALTERNATING
IAc.= 0·47:::' X 2,5DO
IAc
IN
(D) Ttfl':. 0-C OUTPUT ; (b) 11-lE. THREE- PHASE A-C VO(.TAGE .AT THE. SLIP
PER LJNE :
lAc= 0-472 X Ior
A 2!>0-ICv\l 1HREC -PHASE SYNCHRONOUS CONVE.JUE.R HAS A
VOL-TAGE Of ~40 VOLTS· IT RE.CE.IVE.S ITS A-C POWER
ASSUME Tt+AT TI+E MACHINE
SIPE.
'2SOOAMP /~.
ILLUSTRATIVE PROBLEMS
A ?.300- VOLT SOUil.J;E TJ+ROGtl ,A SA~I( OF 'WREC.. TRANSf=ORME.RS
FROM
= ---'-bOO
Epc
t=.J(AMPLE
~J
0-C OUTPUT
RE.CTIFJE~S:
AND
CON-
EFFICIENCY OF 9G PER CfNT
ON mE
TRANSFORMER. ; (e) THE CURRENT
PRIMARY COILS; (.f) TflE a.JRIZENT
IN EACH OF THE
IN EACH OF THE LINE
PRIMARY SIPE.
SOLUTION:
a) IHE
(I De)
0-C OUTPUT
~
AND A POWER FACTOR OF 0·95. CALCULATE : (o.) THE A-C VOl.TA~E
15ETWEEN
IN
EACH
RING:S ; (b) THE D-C OUTPUT; (C) THI: ALTERNATING CURRENT
Of THE
1vc =
2SO ,00~
:: I Ot'Z AMP.
2"1-0
II
'
!2 WIRES.
b) THE. THREE ·-PHASE
GIVEN:
= 5',000 J<.W
fAATJNj;
c I>C =
P.F
.EAc =
SYNC. CONVE.RTER
eFF::: 9G
A-C VOLTAGE
E.Ac = O·GI'l. E.pc
1, 200 VOLTS
12 PHASE
t:Ac =
1-
c)
= o.gr,
AT THE £LIP RINGS:
(FOR .3 -PHASE)
O·IDI2 X 2.40
147 vOL-TS
ALTEIZNATINC- CURReNT DEL.IVEJZElJ BY TJ..IE TRANSFORMER
SECONDARY
SOLUTION:
o-)
A -C
VOLTAGt=:
BETWEEN
RINGS:
Et~e
= 0· 192
~c
= 0·162 X 1;100
lAc =
(FOR JZ -PHASE)
Eoc
1/01-2
X 0·94.b
0-92
THE T.t'.ANSFORMER
x o.gz;
= tiS'O
AMP
SE.CONOAI<Y CURRENT CAN
A~O
BE
CALCUL-ATED AS FOL..LDWS
CAe::: 216-"T VOLTS,f',.....__
b)
D-C
J AC
OUTPUT:
lvc
=
'0;,000,000
1,'200
= 4,1G7 AMr /~
C) ALTE..RNATIN~ CURRENT IN
(rof.l
::. 0·:25G X"!', IG7 =
IAc(EFF= 9G7. /
lAc
=1,080
P.f.=O·!J!;)
=:
.J3 X
p
EAc X
PF
lAc ==
lAc
1-t- PHASE)
984 AMP
d)
ilAT/0
=
fi-
EFF
X 147 X: 0·92 X 0.96
I, 150 AtviP.
OF TRAtiSFOfZ/VlAT/ON
OF EACH TMNSFORME.Jt:
98
= o.9G +xo.gs
VOlTS
PER. TRANSFORMER
CDJL SECONDAR.Y =
AMf _/M/4-.
216
X
'2!70,000
.
EACH OF THE 11 WIRES:
IAc(EfF= IOOj. _ / ' P·F.=l):: O·Z3fd lpe.
I,.
217
147
-,f3"
== 85
YOLTS
CONVEkTEP-..S
CONVe.RTEgs AND RECTIFIERS; lLWSTRATIVE PROBLEMS
'Z,:500
a.=8S
e)
b)
=27~:t
IN t:ACI-I OF THE 1.../NE WIRES
CIJJI.R.E.HT
ON
c)
IN
X
o.;,4 X 7,;0
E,ac
=
2G~ VOL.TS
G =
73·~ AMP·
(POl< SIX-PHASE)
Et>c
SECONDA~Y:
VOLTAGE OF EACH TRANSFoRMER
T+lE
(THIS
It..= 4t·G"
= 0-3s-4
=
VOLTS
PRIMARY SlJJE:
PR/MAJ.{Y
t,Ac
CAc
tJ~;o
= - - = 1-Z·G" .AMP.
rr.l·a:lll
"2.7·1
/X-FOf<lVl!SR..,
SOLUTION
THE SIX-PHASE. VOLTAGE SETWE:E.N ,4UJACENT SLIP R-INGS:
THt= CURRENT IN EACH OF TtfE TRANSt=ORIVIER PRIM/IRY:
CURRE:JiT/
.f)
7]
CONT· OF EX· #
COHf. OF EXAMPI.JE #- G]
i ILLUSTRATIVE PROBLEMS
ANO Rt::CTIF!ER.S
PER TRANSFORMER SI::COH.DAIZY COIL= :2 X 2t;r; =~VOLTS
FROM TI-lE GEOMJ:OTRY
FOLLOWS
WHICH
d)
THE VOLTAGE
Tf-fE SIX'-PHASE HE.XAGON/
OF
ACROSS A OIAGONAL
= 'l X
VOLTS PER SIVE)
THE RATIO OF TRANSFOR!ViATION OF EACH TRA"ISFORMER:
THE PI<IMAl<.Y LINE CURRENT CAN ALSO l:lE
il
~
AS FOLLOW$:
CALCULAT!::.D
:@
13,200
0..=
550
~
ji
ill[
= '!.4 .g: 1
I
p
~~
!
PRIMARY
I
h.
.ja
=::
e)
ELX Pf X f':.fF
X
THE ALTERNATING
2[;0,000
- J3 X '2,300 X 0·9'l.X0·93
~
2,GG7 X O. 472
IAc ""
i
73 ·S'
CW<.RC.NT VI::UVf:llt=P
/>MP. ./'t:tw~.
==
0·9GX0-9J;;
TI-lE TRANSFoRMERS
CONVE.I<TEI<
IN VEl...TA
ON
'2,000- K.W
IS FE.D BY
LOAD
OF
VOLTS.
AT
AN
THE THREE -PHASE
EFFICIENCY
OF 9G
PER CEI-JT
VOLTAGE.
DIRI:..CT-
BE. TWEEN
OF EACH TRANSFORMER
VOLTAGE
DEUVERl=P
(·f) TlfE CURRENT IN
('}) TltE CUJ<RENT
lAc
AND A
=
BY f:ACH
S30 X 0·9G
0-9s-
X
POWER. !=ACTOR
lAc =
ADJACENT SLIP RINGS~
-f.)
1,380 AMP.
CURREf.Jf IN EACH TRANSFORME)Z COIL PJZIMARY
SE.CON!JA~>'; (d) f#E. IVITICl
TIJ.ANSt=OR.ME.R
POWER. PER TRANSf=ORMER
2,000,000 /3
CURRENT OUTPUT ;
OF THE. TRANSFORMER
EACH. 01=' THE
/11-JD E'.ff.'=: IOOJ.)
CAN ALSO B£
VOLTS PER TIZANSfi?RIVlER X I'F X r=FF
IS
AT FULL -
1,390
=- - -
OF EACH TRAI-JSFOJ<MEJZ ; (e) THE AL TER.NATING
OF TRANSFORMATION
CURI<.E:.NT
PRIMARY LINE VOLTAGE
POWER fACTOR =I
SECONDARY CURJ?.E.NT
CALCULATED AS FO/..LOWS:
SYNCHMNOtlS
AND DIAMETRICAL ON THESE-
o.9S, AND CALCULATE : (o.) THE
'WE
SIX'- PHASE
ASSUME THAT THE. MACHINE OPJ::KATES
(b) TffE SIX-PHASE
(c)
D-C
A BANK OF TRANSrORMJ::}:(S CONNECTED
TtfE PRIMARY SIDE
CONDARY SIDE.
13,200
750 -VOLT
:!Ill
1,380 AMP·
(Tt-1-C EQUATiON IN TABLe 10 IS FOR
EXAMPLE 7] A
BY E.AC!-1 OF T+fE
llZANS'FOR.IVIER SE.COND.Il/ZIE.S:
2.4-9
SECONPAP.IES j
!J)
PRIMARY COILS;
PRIM,ARY
= r;;s. Ei
AMP.
L.liJE= CUI{RENT = .!;,!;-~
THE PRIMARY LINE CURRI::.NT
CA-N A/..SO
xS3
==
9G AMP
8E CALCULATE)) ;IS ~OLlOIVS:
HJ EACI-l OF THE Lli-JE WIR.ES ON THE PRIMARY SIVE.
p
I
SOL..UTION:
PRIMARY
h
=
[3 x
EL
x
PF
x EFF
I
()) l>IRE.CT -Cli!<P.ENT OUTPUT
IDe
=
2.,000,000
7130
=
'2,<;,~7 AIVJP.
'I
.
=
IL--·------------------------~--------------~
'2,000,000
.{3 X 13;200 X0-910 X0-9{;
9~
218
AMP.
219
CONVERTEI7S
e]
l':XAMPLE
WHAT
13E USED
IN T\VO
CONVERT
f="~M
THE
AND
SPEED
ARE
THE FEWEST
MACHINES
SYNCHRONOUS
OF- POLES THAT CAN
NUMBI::J"S
IF IT IS
DESIRED
GO TO 2S CYCLES? (b) 4-0 TO GO CYCleS
(0.)
O'F THE
eACH
SET IN.
':3]
CONT OF EX.
120 )(GO
HAS
1'2
4
CASE .
1-2.
Pz
f,
25
S
MACHINE 2
MACHINE
CAN OBVIOUSLY
(i..e. ,!?) ,
IT
WHILE
FOLLOW-s
MACHINE
THE SET
PPMcoNv)
I ± ---·--
\
GO
1
f (
o
+coNv ~
+,
POLES
= U300 RPM
RPMSYN =
a)
p,
THIS MEANS THAT IF
.
SHOULD HAVE
MACHINe
21 POLES.
J MUST HAVE
WILL THEREFORE
AT f
1
10
+ _R_P_M_cc_,.,_v)
I, BOO
THE SPEED
POLES/
90
RPMcONv= (
01=
\
GO -I ) 1,800
== 900 RPM (OPPOSITE 01 RECTION)
II
b) AT
1
= lt;"O CYCL.ES
-fcoNv
1'2.0X2!;
l'l.OXGO
---·-
24
F'
30 CYCLES
CDNV =:
90 = 60(1
6E
1'2.0 -f
RPM= - -
RFfVlsyN
NUMBER OF POLES
MUST f-!.AVE
2
MACI-H~E
5" POt.ES . SINCE. A
AN ODD
NOT HAVE
THAT
PROBLEMS
SOLUTION :
TO
? DETERMINE
SOLUTION'
a.)
AND RCCTIFJER.S: ILLUSTRATIVE
CONVERTERS"
rlECTIFIEIZS ; lLWSTn.ATIVE PROBLEM$
10
Rl)f•l]
COIN=
(
150
)
1:0
-I
1~;:1
It!'
I
1,800= ?,700 RPM(OPPOSITE DIRECT/ON)
I:ZPM = 300
c)
b)
t-IAVE
4p,
-t.
40
- = - =-=
Pz
GO
G
·fz
"'!- POlES
WHILE
AT fu:n<V :::
THEREFORE , MACHINE 1 MUST
MACHINE 2
50 CYCLf.S
£;0"' GO
TffE SPEt'D WILL BE
40
RPWl= - - - -
"!-
s:;o) 1,800 =300
11.0
'
I<P1·1wNv= ( 1 - \
GO
x r;;o
d)
G
fiT RPMco,.,v
~
~XAMPLE
9]
.A FOUR- POLE
USED .AS
THE. ROTOI~
IF THE
l3E
OUTPUT
GO- CYCLE
WOUND-ROTOR
DRIVEN, AND IN
FI~EQUENCY
WHAT DIRECTION
MOTOR
IS TO
f
MUST IT ROTATE,
IS TO 13E : (a.) 90 CYCLES
P.OWR
FREGIJENC)"
(o) 2,200 RPM AGAINST THE FIELD'S DIP.ECTION
THE SAME
( AGAINSf THE FIE.L[) DIRECTION)
?,"200 RWl
-
c-aHV -
1
2,'200)
+ -l,fiQQ
13.3 !_~
'17
CPS rt;~
Lf'
A FREQUENCY CONVERTER. AT WHAT SPEED MUST
(C) SO CYCLES 7 WHAT WILL BE Tt-IE
IS·.
]<.Pf\11 (SAME DIRECTION)
.
I, '2.00
·f CONV -·_eo
o_ ( ,
BE
~A>Nv)
RPM
1,1300
I
J'ZO X
RPM =
(I--
MUST !-lAVE C POLES·
?
?(b)
AT RPIIlcot.n =
IF THE SPEED
fcoNv"'
\.
f (flNV =
GIVeN:
EXAMPLE
OF
10
DV..OP
10l
~
AMP
CU!Z!Zf'ot>!T
CW
A RECTIGON
AT
(SAME. DIRECTION)
400 \
1,800)
4G%
CPS//
-~
// ~.
8AHER.Y
'
CHARGER
12 VOLTS. ASSUMING
VA!l!ATIDHS
DELIVERS
SINOSOIDAL
fOR Ti-lE H/ILr- WAVE
10 VOLTS, CALCULATE
L.::.::_:THE" TJl-:;~~RME~
220
&o( 1_
(e) 1-QO RPM IN
DIRECTION AS Tf-tE FIELD?
P::: 4 POLES
f= (;0 CYCLES
"'tOO RPM
ISO CYClES
A D-C LOAD
VOL.TAG'E
RECTIFieR
ANP
AND A TUBE.
THE CURRENT ·AND VOLTAGE RAT!Nb"S
----------------J
SECONDAR,_Y_.
221
I"
I
CONV!=.RTERS AND RECTIFIERS: ILWSTRATIVE PROBLEMS
CONT. OF EX. 10]
EXAMPLe ll] THE TJZANSPORMER
SOLUTION
SINCE THE UNJDI/.ZECTIONAL LOAD CLJRRI=NT TN A HALr-WAYE
!<ECT!FIE)Z FLOWS
DllRING:
AYEAAGE'
IS
VAWE
1oc .:: -
l
WHERE
lm
ONE -HALF
'2
X- ]
11'
'2
IS THE MAXIMUM
I,o~c ==-
111
OF eACH cYCLE , ITS
IF TtlE
D-C
LOAD
THe
VALUE
VALUE
[2
X
OF TI-lE SINE WAVE
0) TIJSE. P~OP (TD):
Vpc
VAc. =
II'!\
{2
~.
--~-
=
..[2
1.m
2
:ZX0-318
lAc=
TtlE TOTAL
TI-lE
[2
b) t:f-FE.cTlVE
D-C OUTPUT VOLTAGE , WHICH IS n-IE SUM OF
ANV THE. WBE pgop, IS ECVUAL TO
Vm
0.318 Vm
I~ THE MA)(lMUiv'l
BliT THE
t::rf'ECTJVE
ll+E. TRAN.SFORME.R
OF T»E. CYCLE , J-5
VALUE
VALUE
X
ff
X 0·318) - 1!:
TRAtJS:FO~MEg
2 X0•3'/t1
2 X 0-3'18
lAc.s= 7·8!; AMPf-...
c) 11tE TFlAm'FORME.R J>R!M.AF~Y
OF TI-ll: S!NE WAVE·
OF THE. YOLTAb=E ACROSS
SECON/JARY, APPEARING
SOT+f -HAL YE..S
T;~cr
Vs:
=-
T.4cp
=
1Acp
=- 3
Yp
= -lf
CURRENT:
>< 1Ac_..
S'Z
Vm
VAc.
Sf:CDNPARY CURRE.NT:
=---s_____
}pc
T;.ci"
!J-C LOAD VOLrs
WRERE
X 0·318
ID::: 8·4 VOLTS _faw7.
'2 X 0·"518
15·7 AMP
Vpc =
;i¥i
0·318
=:. - - - - - -
TO ::: (,!;2
!0
>(
IS+ T·V.
!:i'2
Ioc
=
CU~RENT-
EXC!TIN(;;
IT FOL-LOWS, THt..RErORE, THAT
IAc
AN}) A .!;'2-VOLT SECOI-JOARY.
IS VOLTS , CALCULAIE : (a) THE
IS
VOLTAGE
OF SUCH A CURRENT IN A
(S
J
OF A .t; -AMP TLJNGA'R BATIEIZY
PRIMARY
TIJSE DROP (TV) ,; (b) Tf+E EFFECT/'/£ TRANf::'FORMER SECONDARY
CURJU::NT ; (c) THE. TRANSFORMER PR!MA~Y CURRENT, NE.GL.~CT/NG
= 0.318lm
A II~ -VOLT
CHARGER HAS
50WTIOI-J:
AI.So , TilE eFFECTIVE
1-lAl...F-WAYE. RECTIFIER.
CONVERTERS AND RECTIFIERS; ILW9Tl<ATIVE PROBlEMS
l'i'S x
7-e&
·&!;' AM~.._.
IT FOLLOWS, THEREFORE, THAT
VDc
YAc::::
J2
X 0-318
EXAMPl-E
rz]
VAc
=
rzJ.Z x+
10
0·318
ASSUJ\11/N~
48·'2l VOLTS_..?',._.
222
A
THE
2,300/230-li!;-VOI.-T TIMNSFOR./VJER
'TWO-ANODE
lb'-VOLT
L..OAD EFFECTIVE
IN
VAc =
A )0-kVA
TO A
CONNECTED
I.S
MERCURY-ARC RECTIFIER·
ARC O!.WP, CALCUL..ATE :(a) THE.. FULL-
..1NODE CURRENT; (b) THE JJIRECT CURRENT
LOAD ; (c) THE
D-C LOAD VOLTAGE ; (cl) n+E J<W LO;Ab
OUTPUT ; (e) 11-IE FULL- LOAD
PRIMAI~Y CUI<RENT.
223
~I
CONVERTERS ANIJ RECTIFIERS: JLWSTJ~ATIVE
CONT· OF EX. l.Z]
CONVERTERS AND RECTIFIERS; ILLUSTI~TIYE PROBLEMS
PR0t3LEMS'
eX'AMPLE 13] Tl-fE ll- COIJNECTED
SOLUTION
FORME!<
CL) FULL- LOAD
EFFECfiVE .ANODE CUAAENT:
VERS
II&""
0·707 loc
43·!;'
lpc:::. - <;;I ·S' AMP
o.707
=>
=
11-JERE/'VR£
(Eve)
TI-t!=. D-C LOAV VOLTA.9E.:
SINCE H-IE.. VOL.TA~E
.J2
En1 ::
Em =
OF
pgiMARY; (_e) TtiE kYA
a)
X III;
foe
VOLTS
+ I!>
DELI-
AN
Vvc
=-=rr
3
11-
b)
= 311'{ZX t3§
::
XO·SGG
DIRECT- CURRENT
REQUIRED FD!<:
~
IN EACH TAAI-!Sf=ORMER
OF T-HE TRANSt:Oiz..MER.
-= (2:20 +I,!;)""- J2
Ji,c :::
'2
ANODE (TI<AN Sf=ORIVJER SECON/:1ARY)
CURP.ENT
RATIN<;;
1T'
V SIN-3
Zoo VOLTS
11-J T+-IE. LOAD; {1uc)
100,000
Z'20
=
I
2.00 AMP.
·
C) EFFECTIVE ANOPE CURRENT (
Tf+EREFVRE
SECOI~D,ARy)
4-SS
Epc = (
.Epc =
~ ~
~~
.Eve =
1A Efr -= -{5
- - ==
) - 1!;
d)· f.!TE.CTIVE
JG.''2.G;) -I!;
X
J<w
PRIMARY
=S·"'!-g
e)
IN E.KH TRANSf=ORI'v115R PRIMARY:
iOO
2,300
llf!=.. KVA RAT/Nt;; OF TH!:: TRANSFDRMER:
I
CURREN'r:
S!l-.ICE n+E. RATIO OF -rnANSFORMATJON SETWEEN
AI-JIJ EACH
2.(():3. X
2103 AMP
lp -= 1'2. '.) AMP.
ourrvr = BB.s;1,000
)( Gt·~
e) THE FULL-LOAD
CURRENT
lp ""
f3fi.J;; VOLTS
d) THE. l<.W LOAD OUTPUT:
J'RIMAJ<Y
AT '2.'20 VOLTS. ASSUMING
SOLUTION:
11-1-E LOAD AND ARC DROP l.S
:=.
2,300- VOLT THREE -
.SECONDARY IS CONNEC-
(a.) THE VOLTAGE ACROSS
.SECONDARY; (b) THE DIRECT CURt<.ENT IN
CUJ<RE:NT; (d) "THE EFFECTIVE
y
BUT Tl-tE AVE.RABE.. D-C VOLTAGe
f=Av!O
100 KW
WAVE IS SINOSOIDAL /ITS
J G '2 · t;;
CONNECTED
T#E LOAD ; (c) THE EFt=ECTIVE
VAWI::: IS
MAXIMUM
Y-
A THREE -PHASE TRANS-
pgoM A
A THIZEE -ANODE MERCURY-AIZC RECTifiER, WHICH
A V-C LOAD
t::ACH TRANS't'ORMER
THE DIRECT- CUXRENT 11-J WE LOAb:
1A£fl'
c)
TO
PRIMARY OF
POWER
ARC !JROP OF I!; VO!..TS , CALCUL.ATE :
::: 1-3·5 AMP
b)
PHASE SOURCE , AND THE
TED
!0,000/'2
IA'"FF """
WITH
IS SUPPLIED
J3_X_200
X_
4!;!;
__
_
!<VA-~;.=__:...
1,000
HALt= Or THE SECONDARY IS
KYAt =
.1,300
ls:t' .:s; KVA
C\..= - - - "" 'ZO: 1
!IS
THEREFORE
Gl·~
Jp
"'-;zo
=
224
3 ·07,t; AMP
225
1
CONVEI<TERS AND RECTIFIEfl.S;
~XAM;LE
I! L~AD AT
t<fj
A
PRIMARIES
A "'1;000-KW
OF WE THREE -PHASE
.
THAT SERVES THE RECTIFI El< A IJO LOAD ARE CONNECTED
TRANSFO!<IYIER
IN ;::";
D-C. IF THE
VOLTS
Pf20BLEMS
tvlfi?CURY -ARC RECTIFIER Df'LIVERS
SIX-'-AIJODE
:2/400
tLWSTPATIVJ::
.AI·I}) TO A
GG',OOO -VOLT T-HI<EE- PHA£'E SOURCE·, CALCULATE THI::
FOLLOW/N~ , ASSUMJN(;
ARC Ot<OP : (o..) Tt+E DIRECT- CUR-
A '2.0- VOLT
RE:NT IN THE LOAD -~ (b) THE EffECTIVE
CUP-RENT; (c) TH-E. VOLTAGE
SECONDAKY COILS /
(cl)
ANODE (TRANSroP-MER)
.AC~SS EACH
Tf-IE
01= THE TI'/ANSFOKMER
I<VA P.ATING OF HIE TAANSFOR!VIER.
SOL.UTIOI--1:
o)
THE
Dll<cCT- CURRENT IN 1/-IE LOAD:
4_DOO
Jvc=
= J,GrD7 AMP
2-4
b) THE EfFECTIVe ANODE CUI<RHIT:
JA=
c)
1,6G7
VOLTAGE ACROSS EACH OF T!+E TRANSFORJVIEP.. SE:.C. COILS
Vvc = "2,400
y =
6)
=-.:;go AMP
JG
~.VA I<ATIN(;i
+ 20
G
=--;:;:;;:
'l.;'l"ZO 11
GJZ X o.;;
II
JZ
OF T+-'E TI<ANSRJRMER:
7,?:.00 kVA
2:2G
G
= 1,790 VOLTS
fG X 1,790 )( I,GID'7
KVAt == - - - - - - 1,000
k'VA-l :=
rr
V SIN-