(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 2 ppt
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 2 ppt
...
()()
EQ
11 5 0 V 0 .14 0 0.5 32 8.7 36 87 A
PS S
Zj.
′
=+ =∠°+ + Ω ∠− °VV I
11 8.8 1. 4 V
P
′
=∠°
V
11 8.8 -11 5
VR 10 0% 3.3%
11 5
=×=
(2) 1. 0 PF:
()()
EQ
11 5 0 V 0 .14 0 0.5 32 8.7 0 A
PS ... be
(1. 2 T)(0.04 m)(0.04 m) 0.0 019 2 WbBA
φ
== =
(c) The total reluctance of the core is:
28
11 6.3 -11 5
VR 10 0% 1. 1%
11 5
=×=
(3) 0.8 PF Leadin...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 8 ppt
... generator is
15 1
()
11
1
21
2
384 V
sin sin sin sin 36.4 31. 1
4 41. 6 V
A
A
E
E
δδ
−−
== −°=−°
The new armature current is
2
2
480 0 V 4 41. 6 31. 1 V
22 7 24 .1 A
1. 1
A
A
S
jX ... a
10 % decrease,
2
11 , 907 V
A
E = , and
11
1
22
2
13 ,23 0 V
sin sin sin sin 27 .9 31. 3
11 ,907 V
A
A
E
E
δδ
−−
== °=°
Therefore, the new arm...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 1 pot
... 0.7 24 9.74
0.733 25 5.08
0.767 25 9 .2
0.8 26 3.74
0.833 26 7.6
0.867 27 0.8
0.9 27 3.6
0.933 27 6 .14
0.966 27 8
1 27 9.74
1. 033 2 81. 48
1. 067 28 2.94
1. 1 28 4 .28
1. 133 28 5.48
1. 167 ...
1. 133 28 5.48
1. 167 28 6.54
1. 2 287.3
1. 23 3 28 7.86
1. 26 7 28 8.36
1. 3 28 8. 82
1. 333 28 9 .2
1. 367 28...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 3 ppsx
...
()
2
2
base
base
base
15 kV
1. 125
20 0 MVA
V
Z
S
== =Ω
so
(
)
(
)
EQ
0. 0 12 1. 125 0. 013 5 R =Ω=Ω
()( )
EQ
0.05 1. 125 0.0563 X =Ω=Ω
49
625 0 VA
52 .1 A
12 0 V
S
S
S
I
V
== =
2 -16 . A ...
(
)
(
)
-1
11
tan 12 0 kW tan cos 0.8 90 kvarQP
θ
== =
2
50 kWP =
(
)
(
)
-1
22
tan 50 kW tan cos 0.9 24 .2 kvarQP
θ
== =
TOT
17 0 kW P =...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 4 pot
...
2
rms
315 15
sin sin
22 664 3 3
M
V
V
ππ π π
π
=−−−
22
rms
315 313 3
sin sin
23 4 3 3 23 422
MM
VV
V
πππ π
ππ
=−−=−−−
22
rms
313 33 3
0.8407
23 422 23 4
MM
M
VV
VV
ππ
ππ
...
()()
()()
EQ
1. 513 1. 134 3.36
0. 010 0.040 0.00 723 0.04 82 0.040 0 .17 0
1. 513 1. 134 3.36
jj
Zj j j
jj
+−
=+ + + ++ +
++−
EQ
0. 010 0.040 0.0078...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 5 ppsx
...
12 /T−
12 /T
c b SCR
3
SCR
5
SCR
5
-30
°
12 /T 12 /3T
a b SCR
1
SCR
5
SCR
1
30°
12 /3T
12 /5T
a c SCR
1
SCR
6
SCR
6
90
°
12 /5T
12 /7T
b c SCR
2
SCR
6
SCR
2
15 0°
12 /7T
... voltage will be:
86
1
22 2
1
M
V
B
RC
ω
=
+
and
2
222
1
M
RC V
B
RC
ω
ω
−
=
+
Therefore, the forced solution to the equation is
()
,
22 2 22 2
s...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 6 pps
...
()()()()
13 28 0 0 .15 2 51 36.87 A 1. 1 2 51 36.87 A
A
j=∠°+ Ω∠− °+ Ω ∠− °E
15 37 7.4 V
A
=∠°E
(c) The equivalent open-circuit terminal voltage corresponding to an
A
E of 15 37 volts is
()
,oc
3 15 27 ...
()()
OUT
10 00 kVA 0.8 800 kWP ==
(
)
(
)
2
2
CU
3 3 2 51 A 0 .15 28 .4 kW
AA
PIR== Ω=
F&W
24 kWP =
11 4
OUT
IN
554 kW
10 0% 10 0% 90.5%...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 7 doc
...
AAASA
RjX
φ
=+ +EV I I
()( )()( )
13 28 0 0 .15 2 51 36.87 A 1. 1 2 51 36.87 A
A
jE =∠°+ Ω∠°+ Ω∠ °
12 17 11 .5 V
A
E =∠°
The voltage regulation would be:
12 17 13 28
VR 10 0% 8.4%
13 28
−
=×=−
(d) If the ... MW/Hz
2 Hz
PA
S ==
3 MW
1. 676 MW/Hz
1. 79 Hz
PB
S ==
3 MW
1. 9 61 MW/Hz
1. 53 Hz
PC
S ==
11 6
()()()()
13 28 0 0 .15 2 51 0 A...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 9 pot
...
(
)
(
)
,nl
13 28 0 V 2. 8 10 .5 0 A 13 28 .3 1. 27 V
A
j=∠°− Ω ∠°= ∠−°E
()( )
,half
13 28 0 V 1. 5 93.6 0 A 13 54 11 .2 V
A
j=∠°− Ω ∠°=∠−°E
(
)
(
)
,full
13 28 0 V 2. 8 19 8 0 A 14 39 22 .7 V
A
j=∠°− ... )
()
OUT
load
21 0 00 hp 746 W/hp
12 4,700 N m
1 min 2 rad
12 00 r/min
60 s 1 r
m
P
τ
π
ω
== = ⋅
()
()
load
525 2 21 0 00 hp
525 2
91, 91...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 10 pps
... RjXX
+−+
=
++ −+
()
22 2 2
11 11 1 1 1 1
TH
2
2
11
MMM MMM
M
RX X RX X RX j R X X X X X
Z
RXX
−+++ ++
=
++
() ()
22 22
1 111
TH TH TH
22
22
11 11
MMMM
MM
RX R X X X X X
ZRjX ... is
()
2
1
TH
2
2
11
M
M
RX
R
RXX
=
++
. If
1M
XR>> , then
()()
22
2
11 1
MM
RXX XX++ ≈+
,
so
2
TH 1
1
M
M
X
RR
XX
≈
+
The Thevenin re...