(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 1 pot
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 1 pptx
... will be
19 6
16 00 16 20 16 40 16 60 16 80 17 00 17 20 17 40 17 60 17 80 18 00
0
50
10 0
15 0
200
250
300
350
400
450
n
m
τ
ind
Induction Motor Torque-Speed Characteristic
R2 = 0.0059 ohms
R2 = 0 .17 2 ohms
...
Ω
+
-
V
φ
I
A,
start
R
1
jX
1
R
F
jX
F
The equivalent impedance of the rotor circuit in parallel with
M
jX at starting conditions (s = 1. 0) is:
,start
2...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 2 ppt
... = 11 5 V, and the internal generated voltage
,1Ao
E for a current of 58 A and a speed of
o
n =
12 00 r/min is
,1Ao
E
= 13 4 V.
()
,1
2
21
1,2
10 7.4 V 13 4 V
10 50 r/min 13 26 r/min
99 .1 V 11 5 ...
()
(
)
(
)
240 V 54.3 A 0.44 216 .1 V
ATAA S
EVIRR=− + = − Ω=
219
The resulting plot is shown below:
0 20 40 60 80 10 0 12 0
800
850
900
950
10 00
10 50
11 00
1...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 3 docx
...
current');
axis ([0 5 0 15 0]);
set(gca,'YTick',[0 10 20 30 40 50 60 70 80 90 10 0 11 0 12 0 13 0 14 0
15 0]')
set(gca,'XTick',[0 0.5 1. 0 1. 5 2.0 2.5 3.0 3.5 4.0 4.5 ... efficiency is
OUT
IN
11 ,19 0 W
10 0% 10 0% 79.4%
14 ,10 0 W
P
P
η
=× = × =
(e) The no-load
A
E will be 230 V, so the no-load speed will be
()
230 V
18 00 r/min 18...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 4 pptx
... 60
1. 170 2.3 31
1. 282 2.40 60
B
jj
Zj
jj
+
==+Ω
++
(a) The input current is
1
11
I
0.5 0.5
FB
RjX Z Z
=
++ +
V
()( )( )
1
120 0 V
I4.0359.0 A
1. 80 2.40 0.5 25. 91 43.83 0.5 1. 170 2.331jj ...
voltage of 11 6 V, as shown below.
2 71
()()
1. 282 2.40 60
1. 185 2.332
1. 282 2.40 60
B
jj
Zj
jj
+
==+Ω
++
(a) The input current is
1
11
I
0.5 0.5
FB
RjX Z Z...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 5 ppsx
... 19 0∠0° V,
1
11
I
0.5 0.5
FB
RjX Z Z
=
++ +
V
()( )( )
1
190 0 V
I 11 .2 27.0 A
1. 40 1. 90 0.5 26.59 9.69 0.5 0.7 41 1.870jjj
∠°
==∠−°
++ ++ +
()
()( )
2
2
AG, 1
0.5 11 .2 A 13 .29 16 67 W
FF
PIR== ... factor is
()()
915 .00.962 0.9 51 ===
dpw
kkk
276
1. 4 Ω j1.9 Ω
+
-
V = 220∠0° V
I
1
R
1
jX
1
s
R
2
5.0
j0.5X
2
j0.5X
M
j1.90 Ω
j30 Ω
s
R
−2
5.0
2
jX...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 6 ppt
... ?
FIGURE P1 13
The core of Problem 1 17 .
t (ms)
12 345678
0
φ
(Wb)
0. 010
0.005
–0.005
– 0. 010
FIGURE P1 12
Plot of flux
as a function of time for Problem 1 16 .
cha65239_ch 01. qxd 10 /23/2003 ... 3 10 .
cha65239_ch03.qxd 10 /30/2003 1: 15 PM Page 226
342
0 0 .1
Open-circuit voltage, V
Field current, A
12 00
11 00
10 00
900
800
700
600
500
400
300
200
10 0
0
0.2 0.3 0.4...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 7 ppt
... Subfractional Horsepower Electric Motors. New York: McGraw-
Hill, 19 70.
4. Werninck, E. H. (ed.). Electric Motor Handbook. London: McGraw-Hill, 19 78.
cha65239_ch10.qxd 11 /14 /03 12 : 41 PM Page 680
DC MOTORS ... Kingsley, Jr. Electric Machinery. New York: McGraw-Hill, 19 52.
2. National Electrical Manufacturers Association. Motors and Generators, Publication No. MG1-
19 93. W...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 1 pot
... 276 .14
0.966 278
1 279.74
1. 033 2 81. 48
1. 067 282.94
1. 1 284.28
1. 133 285.48
1. 167 286.54
1. 2 287.3
1. 233 287.86
1. 267 288.36
1. 3 288.82
1. 333 289.2
1. 367 289.375
1. 4 ... 0.233 10 4.4
0.267 11 8.86
0.3 13 2.86
0.333 14 6.46
0.367 15 9.78
0.4 17 2 .18
0.433 18 3.98
0.467 19 5.04
0.5 205 .18
0.533 214 .52
0.567...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 2 ppt
... Lagging:
()()
EQ
11 5 0 V 0 .14 0 0.532 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 0.532 8.7 0 ... is
10 6.2 10 0
VR 10 0% 10 0% 6.2%
10 0
PS
S
VaV
aV
−−
=×= ×=
The input power to this transformer is
()() ()
IN
cos 10 6.2 V 11 .1 A cos 2.6 41. 0
PP
PVI
θ
=...
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 3 ppsx
... its ratings be?
54
11 30 V
11 30
1. 00 A
SC
EQ
SC
V
Z
I
== = Ω
()()
11
260 W
cos cos 76.7
11 30 V 1. 00 A
SC
SC SC
P
VI
θ
−−
== =°
11 30 76.7 260 11 00
EQ EQ EQ
ZRjX j=+ = ...
()
2
2
base
base
base
15 kV
1. 125
200 MVA
V
Z
S
== =Ω
so
(
)
(
)
EQ
0. 012 1. 125 0. 013 5 R =Ω=Ω
()( )
EQ
0.05 1. 125 0.0563 X =Ω=Ω
49
6250 VA
52 .1 A
12 0 V
S
S...
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