188
te
le
lu io
RL H.
(c (d
If
Solution:
sin (wt 50
For
15 At
60
v2
~d
or
j27r
2.305
47.66 100
id
v2
-2.965
47.66
(max.)
(0
Maximum de 161. 161.66 66<' <' or -18. -18.34 34°. °. 100Y 100Y2s 2sin in
61.6 61.66° 6°
100Y 100Y2s 2sin in -18. -18.34 34
±44.5 .6 °.
100V2sin71.66° (d
For
wt 7r
-27r60
100\1"2
(1
E-l.0461
7r
rad.
18 Similarly, 1.
ycle
311"
ater
2.940 and 5.
ycle later:
1111"
wt
2.817
(5 time constant
10.2
ne at
on ct
ou
cl X;
ke 19%, X~ 26 (a
th
er an
Xd
us aine
ated 0%
short-circui
and (c breaker. Solution:
100.000
Base curren
-:--3 (b
c)
3207.5
16,882
er co
10.3
3207.5
2,467
3207.5
jO.19
J2
18
16,882 23,874
ec ed
he
ne at
find (a high voltag id an low-voltage side. Solution: (0.19
0.10) 240
Base li»
Ii>
18
-j3.448
240. 3207."
es
be
ob
0.
ii
.448
\1
3.44
(b
40.6
.5
3207.5
11,060
with
60-Hz
0.20
400
dire tl across th terminal
th ge erat r.
Solution: head
500
1.0 I"
1.0 jO.16
jO.20
jO.16
per unit or 5.
it
0.15 and 0.35, transforme is 0.10 pe unit 0.9 1.0
the solution
0.8
mp in
lt
in
su ra ie th
Solution: jO.l0
MOTOR
P: faul poin
0.9 I"
(0.8
0.81
jO.12
jO.25
jO.6)(j0.15)
0.81
jO.12
jO.6)(j0.45)
1.17
jO.36
.4
j3.24
pe unit
in
19 1.1i-jO.36
I"
j3.34
jO.35
I"
I"
I"
pe
j6.58 pe
it
it
(b
jO.6)(jO.1) jO.25
th
I; I::'
0.
j6.58 pe
jO.146 curr
so rc an
jO.6 jO.6
;~::
it
jO.146per unit
jO.60 0.96 jO.08
I" By replacing I' by
jO.35
jO.08 pe
0.96
then applyi
th
j6.58)
(-
;~:!~
principl of superpositio 0.48- j3.24
j6.58)
(-
it
pe unit
j3.34 pe
10.6
tam
Solution:
we
P: fault point
BaseZ XL Xsc th
0.~82 0.023 0.1152 96 ...L
0.1152 20 pe
it
0.20 pe unit 0. 30 pe unit
it
19 j1.30
or
2000
7.05
7,00
0.48
10.7
bus
included in
Zbus.
ef ul
and (2 ve en gl ct d,
Solution: @,
1"
-j8.33 pe unit
jO.12 0.12
2,
0.25
jO.2
0.25
pe unit
-3.75 pe unit
ns
nc an
nt
19 bt
10.17,
fro
generato
CD-~
im edance matrix in th calculatio s. -,....---------r--Q)
Solution:
-j2.5
-j5
Thevenin Network (Admittances marked in pe
bus
unit)
[-j~; -1;5 j2
j397.5
th im edance
ne de
re j5
D.21
-102
D.21
j397.5
j397.5
D.23
j397.5 02 -J
-)4.
jO.2465 jO.2465
I"12
0.515
jO.2
j397.5
-j2.58
0.515 pe 0.592
ni
er ni
pe unit
jO.2
-j2.43
per unit
jO.1195 jO.2465 jO.1006
19 10.17
ct
bus Zbus
Prob. 10.8. Solution: j5 bus
j2.5
-j8.5 -0.4167
-j5.4167 j3.3333
U-1L
Ybus
-j6.1154]
where
0.4167
-I
-1).1667]
O.0833 jO.0769 jO.0692
0.6154
jO.1846 jO.l006
jO.I63S]
Hence, Zbus
Ybus
-1
U- L-
it
=jO.1l95
10.10 If
10.5
ag
(2), Zbus
10.7. to llustrat
ou calculat ons.
Solution: jO.2436
-j4.105
195
[ ~
;~:i~!!] jO.1456
0.2444LQ:.] 0.3662LQ:. 0.4023LQ:.
-j1.992
lOA.
10.11
se equivalent circuits base on Zbus to illustrate your calculations Solution: 1.0~
jO.1338
-j7.474
19
1.
Vd
(1.0 L .
10.12
j3.753
jO.ll11
jO.1111
0.5830LJ!::_ 0.5037 LJ!::_
jO.0664
. (
lOA.
Zbus
@@, 10.11.
Solution:
I" Zkk, new Zth,53
1.0~
where
Zkk,ne Z55
Zb-
Zb
Z55
Hence, Zth,53
10.1301
0.0603]
jO.2970
kk, new
I"
jO.097
1.
jO.2970
-)3.367
per unit
19
(a
bu
Zbus,
(D-@
bus and line @-@. Solution: Reactanc
diagram:
-j
Ybus jO
j26.88
-j42.60 j15.72
j26.88 j15.72 -j46.60
(b
Zbus
(2
bus
rOO1357
jO.1234
jO.1278 jO.1246 jO.1492
jO.1143] jO.1266 jO.1222
@,
(c
1.0LJE
I"
jO.1492
(d
i]
: ]
;~:i;~~
LK Q)-@ jO.0372
-j3.857
.!
it
19 @-@
jX43
j2.846
jO.0638
10.14
actor, 90 betwee ;'
ef icienc
with
he mach ne
(a
er
alcu at
and ng
pr
P. Simplify
en
Q. (d
ea
ar (b
Solution:
MOTORS
0.9 0.2 o r n t r ru p n g
c ur re n
207.2 kVA
625 207.2
se
1.5X
625
r;
2.4
jO.905/3 jO.1203
jO.2
-j8.315
JO.1203
19 rom th generator: -)8.315
jO.905/3
-j5.0oo
jO.2
rom each motor -j1.105 ault at
(b
Thru
75
(gen. only)
Thru B:
-j5.0+2(-j1.105)
-j7.210
1084
ault at
(c
75 ru
(gen. only)
166.2
(on motor)
(d
Thru
3(166.2)
ru Ma imum currents to
493.
166.2
interrupte
an
10
A, respectively.
10.15
(a
ated to Instru tor:
tt
ti
th
tu
ld
re te
th
hort
ra
reaker applications should re ie Applicatio Guid fo AC High-Voltage Circuit Breaker Rate on Symmetrical Current Basis AN C3 .010-1979,Amer ic atio ta ar Instit te York This licati is ls IE td 320-1979.
Solution: (a
olta
ta lo ch rate rt circ it curr ecre se an th al of th curr ar 38 1.65 1.65
he rate
short-circ it curren at 34
in re se
23.0kV 22,000 kV is
22,000 38 34.5
36,300
ra in
