!"!#$%! '(%$")'(%$" #$%*+$, $$ -'(%$" !./ 012314 $$5 '6*6+ $"78,%$ 9 :!%!
!"# %&'&" !&#(
'6*6+ $"78,%$ 9 :!%!
';<14 =>?1@ %)*+,-./0 1 2,*+/3/- +/4/ 56/673 ! *)-89/50:2/- *)4/- +,6/3; <0:/ 2,*+/3/- 56/673 409,<,-82/4)-8/- 5,*<)3 6)8/-8/- 1 =/5/
•
>,*+/3/- 56/673 6)34030 4/30 1 ./06, 2,*+/3/- / ?/; < ?< 4/- @ ?@
•
A0:/ 409,<,-82/- 4)-8/- 5,*<)3 6)8/-8/- 1 =/5/; */2/ 405630<,50 /3,5 +/4/ */50-8?*/50-8 =/5/ <)3<)4/ BCDDE
•
,>AB1@14 %?1?C@ <14 73D?@3E>D3 !@>D :1D1 •
•
•
•
F/4/ 6B ! G,25 3)5,:6/-5 *)*+,-./0 /3/9 ./-8 5/*/ 4)-8/- /3/9 G,25 ./-8 409/50:2/- 7:)9 2,*+/3/- /?/ F/4/ 6C ! G,25 3)5,:6/-5 *)*+,-./0 /3/9 ./-8 5/*/ 4)-8/- /3/9 G,25 ./-8 409/50:2/- 7:)9 2,*+/3/- @?@ F/4/ 61 ! G,25 3)5,:6/-5 *)*+,-./0 /3/9 ./-8 5/*/ 4)-8/- /359 G,25 ./-8 409/50:2/- 7:)9 2,*+/3/-
:F>GD @;D>F?14 H142 <3IAE>FG14 CF;J G>AB1@14 1)1K G>AB1@14 G>AB1@14 E)E <14 G>AB1@14 G>AB1@14 L)L
nS
120 f =
p
!
f = frekuensi; p = jumlah jumlah kutup
'6*6+ $"78,%$ 9 :!%!
';<14 =>?1@ %)*+,-./0 1 2,*+/3/- +/4/ 56/673 ! *)-89/50:2/- *)4/- +,6/3; <0:/ 2,*+/3/- 56/673 409,<,-82/4)-8/- 5,*<)3 6)8/-8/- 1 =/5/
•
>,*+/3/- 56/673 6)34030 4/30 1 ./06, 2,*+/3/- / ?/; < ?< 4/- @ ?@
•
A0:/ 409,<,-82/- 4)-8/- 5,*<)3 6)8/-8/- 1 =/5/; */2/ 405630<,50 /3,5 +/4/ */50-8?*/50-8 =/5/ <)3<)4/ BCDDE
•
,>AB1@14 %?1?C@ <14 73D?@3E>D3 !@>D :1D1 •
•
•
•
F/4/ 6B ! G,25 3)5,:6/-5 *)*+,-./0 /3/9 ./-8 5/*/ 4)-8/- /3/9 G,25 ./-8 409/50:2/- 7:)9 2,*+/3/- /?/ F/4/ 6C ! G,25 3)5,:6/-5 *)*+,-./0 /3/9 ./-8 5/*/ 4)-8/- /3/9 G,25 ./-8 409/50:2/- 7:)9 2,*+/3/- @?@ F/4/ 61 ! G,25 3)5,:6/-5 *)*+,-./0 /3/9 ./-8 5/*/ 4)-8/- /359 G,25 ./-8 409/50:2/- 7:)9 2,*+/3/-
:F>GD @;D>F?14 H142 <3IAE>FG14 CF;J G>AB1@14 1)1K G>AB1@14 G>AB1@14 E)E <14 G>AB1@14 G>AB1@14 L)L
nS
120 f =
p
!
f = frekuensi; p = jumlah jumlah kutup
'6*6+ $"78,%$ 9 :!%!
!41F3D1 %;L1@1 M;G?C@ ) N
: O P>GD •
S76/50 ./-8 408,-/2/- ,-6,2 *)-./6/2/- +750R= /6/, -)8/R=-./ /3,5 ./-8 *)-8/:03 +/4/ 2,*+/3/ / ?/; < ? < 4/- @ T@ ./06, K
? 9/38/ +750R= ! <0:/ 6/-4/ 50:/-8 IUJ 6)3:)6/2 +/4/ +/-82/: 27-4,2673 I/; <; @J ? 9/38/ -)8/R= ! <0:/ 6/-4/ RR2 IEJ 6)3:)6/2 +/4/ +/-82/: 27-4,2673 •
•
V/*
BE
&3/9 G,25 ./-8 40R*<,:2/- 7:)9 /3,5 ./-8 *)-8/:03 4/:/* 5,/6, :0-82/3/- ! 5)5,/0 4)-8/- +)3+,6/5)23,+ I 4/+/6 40:09/6 +/4/ 8/*
CE
>)<)5/3/- G,25 ./-8 40R*<,:2/--./ ! 5)
'6*6+ $"78,%$ 9 :!%!
R1G?> ? O ?N
R1G?> ? O ?9
!41F3D1 %;L1@1 M;G?C@ ) Q
R1G?> ? O ?Q
R1G?> ? O ?S
'6*6+ $"78,%$ 9 :!%! •
•
!41F3D1 %;L1@1 '1?;A1IG1 ) N
%05/:2/- G,25 ./-8 409/50:2/- 7:)9 2,*+/3/- / T/; < T<; 4/- @ T@ */50-8? */50-8 /4/:/9K F a
=
F a
F b
F b cos(! " 120
0
=
)
F c
=
F c cos(! " 240
0
)
cos !
>/3)-/ /*+:06,4/ G,25 #H/#; #H<# 4/- #H@#; <)3,
=
F max
cos! t
F b
=
F max cos(! t ! 120
F c
=
F max cos(! t ! 240
0
)
0
)
'6*6+ $"78,%$ 9 :!%! •
H:,25 3)5,:6/-6 IHBJ /4/:/9 Y,*:/9 2)R8/ G,25 6)35)<,6; 4/- *)3,+/2/=,-850 6)*+/6 IZJ 4/- =,-850 L/26, I6J K
F 1 (! , t ) = F m cos !.cos" t + F m cos(! " 120
Transformasi trigometri F 1 (! , t ) = 1 2
!41F3D1 %;L1@1 '1?;A1IG1 ) Q
1 2
F m cos(! + " t " 480
karena : F 1 (! , t ) =
1 2 3 2
0
1 2
).cos(" t " 120 0 ) + F m cos(! " 240 0 ).cos(" t " 240 0 )
cos# . cos$ = 12 cos(# + $ ) + 21 cos(# " $ )
#
F m cos(! + " t ) +
0
F m cos(! " " t ) + 12 F m cos(! + " t " 240
0
) + 12 F m cos(! " " t ) +
) + 21 F m cos(! " " t )
F m cos(! + " t ) +
F m cos(! " " t ) $
1 2
F m cos(! + " t " 240
0
) + 21 F m cos(! + " t " 480 0 ) = 0, maka
merupakan gelombang berjalan
'6*6+ $"78,%$ 9 :!%!
=@34D3B ,;@T1 ) N
=@3D3B G;@T1 AC?C@ 34<>GD3 K
BE
!,*<)3 6)8/-8/- 1 =/5/ 409,<,-82/+/4/ 2,*+/3/- 56/673 ! *)-0*<,:2/*)4/- +,6/3 4)-8/- 2)@)+/6/ns
Q5 1E
120 f =
p
';<14 B>?1@ D?1?C@ /2/- A;AC?C42 E1?142 GC4<>G?C@ B1<1 @C?C@ &20AB1@14 @C?C@ 1G14 IAE>F 21H1 2;@1G F3D?@3G E 2 s
=
4.44 f 2 N 2! m
(untuk satu fasa)
[C5 K 6)8/-8/- 0-4,250 +/4/ 5//6 37673 <)3+,6/3 XE \E ]E
>/3)-/ G>AB1@14 @C?C@ A;@>B1G14 @142G1314 ?;@?>?>B ; 88: I[J /2/- A;42J1D3FG14 1@>D -$/ &4/-./ /3,5 I#J 40 4/:/* *)4/- */8-)6 ! A;43AE>FG14 21H1 -:/ B1<1 @C?C@ A0:/ GCB;F A>F1 ./-8 409/50:2/- CF;J 21H1 -:/ B1<1 @C?C@ L>G>B E;D1@ ,-6,2 A;A3G>F GCB;F E;E14; @C?C@ 1G14 E;@B>?1@ D;1@1J <;4214 A;<14 B>?1@ D?1?C@
'6*6+ $"78,%$ 9 :!%!
=@34D3B ,;@T1 ) Q
^E
>/3)-/ /4/-./ 6)8/-8/- 0-4,250 ./-8 R*<,: +/4/ 2,*+/3/- 37673 /20?1@ D?1?C@ -4D/ <;4214 G;L;B1?14 E;@B>?1@ @C?C@ -4 @/
QE
F)3<)4//- 2)@)+/6/- /-6/3/ -3 4/- -5 405)<,6 DF3B -%/ 40-./6/2/- 4)-8/- K S
=
ns ! nr
x 100%
ns
_E
A0:/ -3 ` -5; 6)8/-8/- R4/2 /2/- 6)30-4,250 4/- /3,5 R4/2 *)-8/:03 +/4/ 2,*+/3/- Y/-82/3 37673; 4)-8/- 4)*020/- R4/2 *)-89/50:2/- 27+):E ,CB;F AC?C@ 1G14 <3IAE>FG14 E3F1 4 @ F;E3J G;L3F <1@3 4D
BDE (0:09/6 4/30 @/3/ 2)3Y/-./; *7673 0-4,250 405)<,6 Y,8/ 5)
'6*6+ $"78,%$ 9 :!%! •
•
%F3B ) N
A)3,E>4214 U@;G>;4D3 <;4214 DF3B 4/+/6 40:09/6 5)
A0:/ U N O U@;G>;4D3 T1F1 Y T1F1
•
•
•
•
ns
=
f 2
=
120 f 1
p
F/4/ @C?C@ E;@F1G> J>E>4214 Z
!
p(ns ! nr ) 120
f 1
=
=
p x ns 120
p x ns 120
x
(ns ! nr )
ns
=
f 1 x S
F/4/ 5//6 56/36 ! ` BDD a b = C ` = B ()-8/- 4)*020/- 6)3:09/6 A E;@B>?1@K U@;G>;4D3 B1<1 D?1?C@ D1A1 <;4214 U@;G>;4D3 B1<1 @C?C@ (/:/* G;1<114 @C?C@ E;@B>?1@ ; U@;G>;4D3 1@>D @C?C@ <3B;421@>J3 CF;J DF3B I= C ` 5= BJ
'6*6+ $"78,%$ 9 :!%!
%F3B ) Q
>/3)-/ ?;214214 34<>GD3 4/- @;1G?14D3 G>AB1@14 @C?C@ *)3,+/2/- U>42D3 U@;G>;4D3; */2/ 9/38/-./ ?>@>? B>F1 <3B;421@>J3 CF;J DF3B E E 2 s
=
=
=
!
4.44 f 2 N 2! m 4.44
sf 1 N 2! m
!
E 2
=
4.44 f 1 N 2! m
sE 2
X 2 s
=
" s L2 s
=
2# f 2 L2 s
(Q ` ?;214214 34<>GD3 +/4/ D11? D?1@? I40/*J (QD ` ?;214214 34<>GD3 +/4/ D11? AC?C@ E;@B>?1@ X 2 s
=
2 ! f 2 L2 s
X 2 s
=
sX 2
=
2 ! sf 1 L2 s
[QD ` @;1G?14D3 +/4/ D11? @C?C@ E;@B>?1@ [D ` @;1G?14D3 +/4/ D11? D?1@? -<31A/
'6*6+ $"78,%$ 9 :!%!
+142G1314 +C?C@
!)6):/9 40?1@ ?;214214 34<>GD3 @C?C@ -( QD/ <14 @;1G?14D3 @C?C@ -[QD/ ?>@>? <3B;421@>J3 CF;J DF3BK 1@>D @C?C@ *)-Y/40 K I 2
=
E 2 s Z 2 s
E 2 s
=
( R ) 2
2
+
( X ) 2s
2
SE 2
=
( R ) 2
2
+
( SX )
2
2
atau I 2
E 2
=
( ) R2
2
S
+
R2
karena :
S
( X )
2
2
= R
2
+ R
2
" 1 ! S % $ ' # S &
*/2/ @142G1314 @C?C@ 4/+/6 Y,8/ 408/*
x R2
2
x R2 $
I 2 I 2
=
daya yang hilang berupa panas
" 1 ! S % ' daya keluar rotor yang diubah menjadi daya mekanik # S & =
'6*6+ $"78,%$ 9 :!%!
+142G1314 (G3\1F;4 ) N
,;@T1 AC?C@ 34<>GD3 D;B;@I T>21 G;@T1 ?@14DUC@A1?C@ 1<1F1J E;@<1D1@G14 B@34D3B 34<>GD3 ;F;G?@CA124;? E c:)9 2/3)-/ 06, *7673 0-4,250 4/+/6 40/-88/+ 5)4<;@ H142 E;@B>?1@
d)2673 40/83/* 5)
'6*6+ $"78,%$ 9 :!%!
+142G1314 (G3\1F;4 ) Q
+142G1314 ;G3\1F;4 *7673 0-4,250 40:,2052/- 5)
M;G?C@ <312@1A ,-6,2 3/-82/0/- )20W/:)- 5)
'6*6+ $"78,%$ 9 :!%!
,CB;F - * / ) N
(/30 3/-82/0/- )20W/:)-; 1@>D $Q] 1<1F1J K E 1
'
I 2 =
(a
cos ! =
2
R2
2
S
) ( a X ) +
tahanan impedansi
2
P
dan
2
T
=
(a
2
R2
2
R2
P =
) ( a X ) +
A0:/ eB ` "B f Y gB 40/-88/+ 2)@0:; [B 9/*+03 5/*/ 4)-8/- dB
!
P3 fasa
=
3 E 1 I '2 cos "
3
=
E 1 I 2 cos " !
T : kopel
S
2
S
T x !
! P : daya
2
a
=
2
2
! : kecepatan sudut
2
3
T =
!
2
2 V 1
Sa R2
(a R ) 2
2
2 +
2
S
(a X ) 2
2
....... (1)
2
h)-6,2/- 9/38/ ! /8/3 9/38/ h */250*,* i ! j/38/ ! ,-6,2 *)-4/+/62/- h */250*,* /4/:/9; <0:/ 4hk4! ` D (/30 40l)3)-50/50 4hk4! ` D ; 40+)37:)9 9/38/ h */250*,* +/4/ 5//6 ! ` m "CkgC NN ICJ 2
T maks
=±
3V 1 2
2! a X 2
........ (3)
'6*6+ $"78,%$ 9 :!%!
,CB;F - * / ) Q
(/30 2)R8/ +)35/*//- 6)35)<,6 40 /6/5 4/+/6 406/302 <)<)3/+/ 2)50*+,:/- K •
•
•
•
•
F)35E IBJ ! <0:/ 402)6/9,0 A B1<1 D11? D?1@? - % O N / 31F1J <;4214 A;AE>1? + Q O [Q j/38/ GCB;F A1GD3A>A <1B1? <3>E1J 4)-8/- A;421?>@ J1@21 [ Q /6/, D>AE;@ ?;214214 M NK <1B1? <3F3J1? B1<1 B;@D5 -9/ F)35/*//- IBJ 4/- ICJ *)-,-Y,22/- +Q I<1G A;42>E1J J1@21 GCB;F A1GD3A>A; A;F134G14 J14H1 A;42>E1J J1@21 % B1<1 D11? GCB;F A1GD3A>A ?;@T1<35 =;@>E1J14 +Q <1F1A J>E>42144H1 <;4214 GCB;F -*/ <14 DF3B -%/ 4/+/6 40:09/6 4/30 2,3W/ <)302,6 0-0
'6*6+ $"78,%$ 9 :!%!
71H1 - = /
()-8/- *)*+)39/R2/- AC<;F @142G1314 402)6/9,0 G @C?C@ -?;@<1B1? B1<1 L;F1J ><1@1/ K P2
'
=
3 E 1 I 2 cos !
( ) '
=
3 I 2
2
a
2
(
R2
+ R
2
( )) 1!S S
( ) '
=
2
3 I 2
a
71H1 G;F>1@ @C?C@ -<1H1 A;G143G B1<1 @C?C@ P2
?;@A1D>G @>23 2;D;@ <14 14234 K
+>23 ?;AE121 @C?C@ Z = L> O 9 -$Q]/Q 1Q+Q
( ) '
=
3 I 2
2
2
( ) R2 S
2
a R2
(! ) 1 S S
T1<3 =Q Z =A Z =L> O N Z -N)%/ Z %
()-8/- 4)*020/- 40+)37:)9 @/3/ *)-8906,-8 ./-8 :)<09 @)+/h ./06, 4/./ 2):,/3 37673 4/+/6 Y,8/ 40+)37:)9 4/30 4/./ */5,2 37673 402,3/-80 3,80 6)*
=
P2 ! Pcu
'6*6+ $"78,%$ 9 :!%!
.C4?CJ ) N1
!,/6, *7673 0-4,250; BDDD j+; CCDD W7:6; C\ @+5; BC 2,6,+; 1 =/5/ 9,<,-8/- <0-6/-8 ; *)*+,-./0 4/6/ 5)
"/-82/0/- )20W/:)- I5/6, =/5/J +/4/ 8/*
V 1 = V t ! I 0 ( R1 + jX 1 ) (/- 5)@/3/ +)-4)2/6/-; 2)<)5/3/MN <1B1? <3?>F3DG14 K V 1
= V
! I 0
t
(R
diketahui V t 530*/L/35/04M./977E@7*NEE %)50-?*)50- O056302 ?
2
1
2
+ X
1
2200 =
3
) dan = 1270
volt
'6*6+ $"78,%$ 9 :!%!
.1?1?14
Maka pada beban nol diketahui , maka pada rangkaian ekivalen V 1 = V t ! ( R1 + jX 1 ) I 1 " V t volt , diketahui dan
pada kondisi berbeban :
(
)
'
V 1 = R1 + R2' + jX 1 + jX 2' I 2
V 1
'
# I 2 =
daya keluar (P2 ) " Pmekanik (Pm )
( R
1 +
R2' + jX 1 + jX 2'
)
faktor kerja (pf) dalam kondisi berbeban :
2
" Paktif = arus tahanan
2 $1 ! s ' 1 # P2(3! ) = 3 I 2' x R2' & hp ) watt " 1 watt = % s ( 745.7
P3! = 3P1!
kecepatan sinkron n S : ns
=
120 f p
rpm (rotation per minute)
ns ! n r
# n r = n s (1 ! s) rpm
ns
Efisiensi (" ) =
Kopel (Tr ) : Tr
=
Pm !
=
Pm
2" f r
=
Pm
2" ( f r (rpm) / 6 0)
newton-meter " (
1 0.046
lb - 3.2808399 feet)
daya masuk (P1 ) = Pm + P3! !rugi!rugi + P0 = P2 + P3! !rugi!rugi 2
" P3! !rugi!rugi = 3 I 2' ( R1 + R2' ) dan P0 daya pada inti (rugi-rugi dari Rm )
Pin"daya aktif masuk Qin"daya reaktif "masuk
Q1 = Qinti + Q reaktansi = P0 tan! 0 + 3 I 2'
! ! 0 = cos"1 pf 0 # "1 Q1 & pf = cos % tan ( P1 ' $
kecepatan sinkron n r : s=
dari segi tiga daya ! tan ! =
P2 P1
2
( X
1+
X 2'
)
'6*6+ $"78,%$ 9 :!%! Maka pada beban nol diketahui I 0 V 1 = 1270 ! 75.1
(0.102)
2
+
=
.C4?CJ)N1
75.1 dan cos ! 0
(0.313)
2
=
=
0.053 lag
1245 volt
pada kondisi berbeban : I '2
V 1
=
( R
1
+
R2'
2
) ( X +
1
+
X 2'
)
2
V 1
=
" a 2 R2 % R + $ 1 S '& #
2 +
1245
=
( 0.102
+
0.0992 0.018)
2
+
( 0.313
+
0.313)
2
( X
1+
=
a 2 X 2
)
2
220 amper
sehingga daya keluar (P2 ), merupakan daya mekanik (Pm ) : P2
=
( )
2
2
0.018 3 I 2' R'2 ( 1!S S ) = 3 ( 220) x 0.0992 ( 1!0.018 ) = 789 kW = 1060 hp
Kecepatan sinkron (n s ) : ns
=
120 f
=
p
120 x 25 12
=
250 rpm
Kecepatan rotor (n r ) : nr
=
ns (1 ! S ) = 250(1 ! 0.018) = 242 rpm
Kopel (Tr ) : T r
=
P ! r
=
789000 2" f r
=
78900 = 30.600 newton-meter = 22.600 lb-ft 2" (242 ) 60
'6*6+ $"78,%$ 9 :!%!
.C4?CJ)N1
Daya masuk stator (P1 ) : P1 = P2 + Pcu
=
( )(
)
78900 + 3 I 2' R1 + R2'
+
P0
2
P1 = 789000 + 3(220) (0.102 + 0.992) + 15200 = 833 kW
Daya reaktif pada beban nol (Q 0 ) : Q0
=
P0 tan ! 0
=
(
)
15200 tan cos!1 0.053
=
248 kVar
maka daya reaktif pada keadaan berbeban (Q) : Q = Q0 + Qcu
=
(
28400 + 3( I '2 )2 X 1 + X '2
)
2
Q = 248000 + 3(220) (0.313 + 0.313) = 375 kVar
Efisiensi (! ) : ! =
P2 P1
=
789 833
=
0.945
Faktor kerja (pf) :
(
pf = cos tan !1 QP
)
=
(
cos tan !1 375 833
)
=
0.912
'6*6+ $"78,%$ 9 :!%!
.C4?CJ ) NE
o02/ +/4/ @7-679?B/; 402)9)-4/20 *7673 *)*+,-./0 27+): */250*,* +/4/ 5//6 56/36; <)3/+/2/9 6/9/-/- :,/3 ./-8 40+)3:,2/- ,-6,2 */25,4 6)35)<,6 4/- <)3/+/ <)5/3 27+): */250*,* 6)35)<,6E (/- <0:/ 5)6):/9 56/36 6/9/-/- :,/3 0-0 4090:/-82/-; 6)-6,2/- 5:0+ 2)R2/ 27+): */250*,* 4/<)3/+/ 2)@)+/6/--./E =;A;L1J14 K
F/4/ D11? GCB;F A1GD3A>AK DF3B *)*+,-./0 9,<,-8/- 5)
Bila R1 dan X1 diabaikan , maka S m
=±
R2 X 2
Bila R1 dan X1 tidak diabaikan , maka '
S m
R2
=
2 R1 +
(
' X 1 + X 2
)
2
bila hendak dipasangkan tahanan luar pada saat start, yaitu S m
=±
1.0 , maka: '
1.0
R2
=
( 0.102)
2
+
( 0.313
+
0.313 X )
2
!
'
R2
=
0.632 ohm - perfasa
'6*6+ $"78,%$ 9 :!%! Jadi tahanan luar R luar
=
.C4?CJ ) NE
0.632 ! 0.0992 = 0.533 ohm - perfasa
(harga ini adalah harga tahanan luar rangkaian rotor yang dipindahkan ke rangkaian stator) Arus rotor pada keadaan start : '
I 2
V 1
=
( R
' 1 + R2
2
) ( X +
' 1 + X 2
)
2
1245
=
( 0.102
+
0.632)
2
+
( 0.313
+
0.313)
2
=
1290 amper
Jadi Torsi (T) : T =
12 4! (25)
(3)(1290)2 (0.632) = 121.000 N-m = 89.00 lb-ft
Bila kemudian tahanan luar dihilangkan, kopel maksimum terjadi pada saat slip : S m
=
0.0992 2
(0.102)
+
(0.313 + 0.313)
Kecepatan rotor (n r ) : nr
=
250(1 ! 0.157) = 211 rpm
2
=
0.157
'6*6+ $"78,%$ 9 :!%! T m =
±
P=
2
3 V 1 2! a
T m
3 =
2
X 2 3
=
2a
!
2! a
2
2
X 2
2
X 2
2
2
I 2' Z 2'
2
I 2' Z 2'
2 '
! Z 2 2
Z 2'
2
2
(a R / s) (a X ) 2
=
+
2
2
(a R ) (a X ) 2
=
2
+
2
3
2
(
(
2
'
=
3 I 2
2
( a R ) 2
T =
=
'
3 I 2
2
(
(
a 2 R2
2
R2
+
'
3 I 2
2
2
2
=
2
( a R ) 2
2
'
3 I 2
2
2a
2
X 2
+
2
( a X )
2a
2
2
2
2
X 2
)
Q
2
2
! Sm = 1 = '
3 I 2 =
2
( a R ) 2
2 x 2! f r
R2 X 2
(
3 1290
2
=
2
) ( 0.632) 4! ( 25)
3 1290
) (a X ) +-, 2
X 2
2
2
2
" a 2 R2 %2 2 2 =$ ' + ( a X 2 ) ! s = 1 # s &
2
2
( a R )
2! a =
2
2! a
!
=
X 2
'
P
+
2
3 I 2
2
2
2
2a
2
2
I 2' * a 2 R2 ) 2 a X 2
P=
.C4?CJ ) NE
= 10048.1962
2
) (0.632) ! f r
4! f s
=
sf s
'6*6+ $"78,%$ 9 :!%!
.C4?CJ ) Q
!,/6, *7673 0-4,250; 1 =/5/ 9,<,-8/- <0-6/-8 ; CCD W7:6 I6)8/-8/- Y/:/?Y/:/J; BD 9+; ]D @+5; ] 2,6,+; *)*+,-./0 27-56/-6/ 5)
"/-82/0/- )20W/:)- *7673 4/+/6 408/*
=
Z '2 * jX m Z '2 + jX m
=
( R
2 +
)
jX '2 jX m
R2 + jX '2 + jX m
! R2 =
R2' s
" 0.144 % + j 0.209 ' ( j 13.25) $ # 0.02 & Z f = = 5.41 + j 3.61 = 6.75(32.40 " 0.144 % + j 0.209 ' + j 13.25 $ # 0.02 & V 1 =
220
=
127 volt (tegangan jala-jala)
'6*6+ $"78,%$ 9 :!%!
.C4?CJ ) Q Arus stator (I1 ) : I 1
127!0 =
=
6.75!32.4 Faktor daya cos -32.4
=
0.884
Kecepatan sinkron (n s ) : 120 f
ns
=
nr
=
120 x 60 =
=
6 P Kecepatan rotor (n r ) :
1200 rpm
1200 rpm =
=
60
20 rps
1200(1 ! 0.02) 1176 rpm =
Daya yang dipindahkan pada "air gap" = daya masuk rotor (P2 ) : P2
=
( )
3 I '2
2
R2
2
=
3 ( I 1 ) R f
2
=
3 (18.8) ( 5.41)
=
5740 wat t
S Daya mekanik pada rotor (termasuk juga rugi gesr + angin + besi) ; (Pin-m ) : Pin!m
=
(1 ! S ) P2 (1 ! 0.02) 5740 =
=
6530 watt
18.8! " 32.4
'6*6+ $"78,%$ 9 :!%!
.C4?CJ ) Q
Daya mekanik (keluar), Pout !m Pout !m Pin!m ! Prugi!rugi =
=
5630 ! 403 5230 watt =
Kopel (T) : T
P =
P =
2" f R
!
R
5230
P
=
=
2" f s (1 ! S )
=
2" *20(1 ! 0.02)
42.5 N-m
=
7.0 hp
[p50)-50 40906,-8 5)
Efisiensi (! ) : !
5230
Pout =
=
Pin
=
6060
0.863
",80 B; 6)*
`I1JIBQEQJCIDEC_XJ
` 1BC
L/q
",80 C; 6)*
`I1JIDEDCJI\^XDJ
` BB\
L/q
",80 /-80- f 8)5)3 f <)50 EE 402)6;E
` XD1
L/q
",80 676/:
` Q1D
L/q
(/./ 2):,/3 NEF7,6?*
` \C1D
L/q
71H1 A1D>G -=34/ O <1H1 G;F>1@ _ @>23)@>23
O `V`V
a1b
'6*6+ $"78,%$ 9 :!%!
'
I 2
V 1
=
! a R2 $ # R1 + & S % " 2
7312@1A #342G1@14 ) N (/30 *74): 3/-82/0/- *7673 0-4,250; *)-,-Y,22/-
.... (1)
2 +
( X
1
+
a
a X 2
)
2
(/30 +)35/*//- IBJ 4/- ICJ; 4/+/6 406,:05 K
a
X 1 + a X 2
sin! =
! a R2 $ # R1 + & S " % 2
.... (2)
2 +
(
a
X 1 + a X 2
)
2
'
I 2
=
V 1 2
X 1 + a X 2
sin !
'6*6+ $"78,%$ 9 :!%!
7312@1A #342G1@14 ) Q
o02/ 3)/26/-50 40/-88/+ 27-56/-; 4)*020/- +,:/ +,6/3/- 4/- 6)8/-8/- Id BJ; +)35/*//- 40 /6/5 *)3,+/2/- +)35/*//- +7:/3 4)-8/- 8/305 6)-8/9 5)<)5/3K V 1
()-8/- <)3,
2
( a R ( )) 2
2
X 1 + a X 2
1!S S
j/38/ 50- Z /2/- <)3,
r-6,2 *)-88/*
'6*6+ $"78,%$ 9 :!%!
7312@1A #342G1@14 ) 9
N5 =;42>G>@14 E;E14 4CF
(/30 +)-8,2,3/- <)G>@14 <;4214 @C?C@ <3?1J14 (/30 +)-8,2,3/- 37673 406/9/- /2/- 40+)37:)9 9/38/ /3,5 # 05 4/- =/2673 2)3Y/ Z95 4/4)-8/- 4)*020/- 4/+/6 406)-6,2/- 6)*+/6 2)4,4,2/- RR2 ! ` B o02/ 2)4,/ RR2 6):/9 402)6/9,0 2)4,4,2/--./; 4/+/6:/9 40<,/6 40/83/* :0-82/3/- 4)-8/@/3/ 5)
h0R2 +767-8 /-6/3/ 8/305 5,*<, 4)-8/- 8/305 6)35)<,6 +/4/ 4 *)3,+/2/- +,5/6 :0-82/3/-
8E ()-8/- 4)*020/- 40/83/* :0-82/3/- 4/+/6 40:,2052/-
'6*6+ $"78,%$ 9 :!%!
7312@1A #342G1@14 ) S ?
4/30 40/83/* :0-82/6/- 4/+/6 +,:/ 40906,-8 K
?
(/./ */5,2
?
(/./ 2):,/3 37673 ` 4/./ *)2/-02
?
o,*:/9 3,80
?
27+):
?
%05/:2/- 206/ 9)-4/2 *)-8/-/:05/ 5,/6, *7673 4)-8/- 27-4050 <)
?
%/2/ 8/305 P( 4/+/6 40-./6/2/- 5)
?
!)4/-82/- 4/./ */5,2 /4/:/9 K d Bs#B @75 ZB IdB 27-56/-J
?
o/40 4)-8/- *)*
?
F0- ` P(s 52/:/ /3,5sdB
?
V/305 P( 2)*,40/-; 4/+/6 40-./6/2/- 5)
'6*6+ $"78,%$ 9 :!%! ? ? ? ? ?
?
?
7312@1A #342G1@14 ) c
V/305 [( 2)*,40/-; 4/+/6 40-./6/2/- 5); 5)4/-82/- Y,*:/9 3,80 6)*/3)-/ 06, 5)R/+ RR2 +/4/ 8/305 &A 2) 8/305 *)-4/6/3 I8/305 ./-8 40<,/6 5)+)3R 6)35)<,6 +/4/ 4J +/4/ 40/83/* :0-82/3/- +37.)250 4/+/6 40-./6/2/- 5)
'6*6+ $"78,%$ 9 :!%! ! rugi
tembaga stator =
! rugi
tembaga rotor =
( I )
2
hs
( ) I hs
2
7312@1A #342G1@14 ) `
x R1 2
x a x R 2
Maka akan didapat perbandingan : BG
! rugi
tembaga rotor
! rugi
tembaga stator
=
GK
Akhirnya daya keluar rotor = daya mekanik dapat ditentukan dari :
(CD - ED - HE - HI ) x skala arus x V1
Pm
=
Pin " ! rugi
Pm
=
CI skala arus x V1
=
'6*6+ $"78,%$ 9 :!%!
+C?C@ 0;F3?14
?
?
%7673 0-4,250 Y)-05 0-0 A;AB>4H13 @C?C@ <;4214 E;F3?14 G>AB1@14 I21 U1D1 D1A1 D;B;@I G>AB1@14 D?1?C@ >,*+/3/- 56/673 4/- 37673 Y,8/ *)*+,-./0 Y,*:/9 2,6,+ ./-8 5/*/
?
!)+)3R 6)3:09/6 +/4/ 8/*1@ D1AB13 J1@21 ?;@?;4?>K <1B1? A;AE>1? GCB;F A>F1 A;4L1B13 J1@21 GCB;F A1GD3A>A4H1
?
>7+): *,:/ ./-8 <)5/3 *)*/-8 <3B;@F>G14 B1<1 a1G?> D?1@?
?
%7673 0-4,250 4)-8/- 37673 <):06/- A;A>42G34G14 B;41AE1J14 -B;421?>@/ ?1J1414 F>1@
?
h/9/-/- :,/3 ./-8 4/+/6 40/6,3 0-0; <3J>E>42G14 G; @C?C@ A;F1F>3 L34L34 ; 5):/0- ,-6,2 A;42J1D3FG14 GCB;F A>F1 H142 E;D1@ ; 6/9/-/- :,/3 6/40 <3B;@F>G14 >4?>G A;AE1?1D3 1@>D A>F1 H142 E;D1@ B1<1 D11? D?1@?
?
(05/*+0-8 06, 4)-8/- A;42>E1J ?1J1414 F>1@K G;L;B1?14 AC?C@ <1B1? <31?>@
