ENGR_400_Module_7_Assignment
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Mechanical Engineering
Date
Jan 9, 2024
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docx
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Jason L. Wilson
12/1/2023
7.3 - Chapter Assignment
ENGR 400
Problem
12.10
The speed of a 12-pole, 480 V, 60 Hz three-phase induction motor at full load is 560 rpm. The motor has the following parameters:
R₁ = 0.1Ω; R
′ = 0.5 Ω; X + X ′ = 5 Ω
a.
Compute the developed torque.
η
s
=
120
(
60
)
12
η
s
=
600
rpm
S
=
600
−
560
600
S
=
0.067
T
=
3
(
60
)
2
π
(
600
)
(
(
480
√
3
)
2
(
0.1
+
0.5
0.067
)
2
+
5
2
)
(
0.5
0.067
)
T
=
0.048
(
934.42
)(
7.46
)
T
=
334.60
N
−
M
b.
While the torque is unchanged, the voltage is changed to reduce the speed of the motor to 520 rpm. Compute the new voltage.
S
=
600
−
520
600
S
=
0.133
334.60
=
3
(
60
)
2
π
(
600
)
(
(
V
L
√
3
)
2
(
0.1
+
0.5
0.133
)
2
+
5
2
)
(
0.5
0.133
)
334.60
=
(
0.179
)
(
(
V
L
√
3
)
2
39.89
)
Jason L. Wilson
12/1/2023
(
V
L
√
3
)
2
=
334.60
(
39.89
)
0.179
(
V
L
√
3
)
2
=
74565.32
V
L
=
√
74565.32
V
L
=
273.07
∗
√
3
V
L
=
472.97
V
Problem
12.16
A wind turbine has 50 m long blades. The far stream wind speed is 15 m/s and the coefficient of performance Cp is 30% at the operational pitch angle. The induction generator is six-pole, 60 Hz machine and is rotating at 1260 rpm. The mechanical efficiency of the turbine is 85% and the electrical efficiency of the generator is 90%. Compute the following:
a.
Power captured by the blade
P
=
1
2
(
1.23
)(
π
∗
50
2
)(
15
3
)(
0.3
)
P
=
4890576.189
W
P
=
4.89
MW
b.
Slip of the generator
S
=
(
120
(
60
)
6
)
−
1260
120
(
60
)
6
S
=−
0.05
c.
Power delivered to the stator
P
s
=
0.85
(
4.89
)
P
s
=
4.15
MW
d.
Output electric power of the wind turbine
P
out
=
4.15
(
0.9
)
P
out
=
3.74
MW
e.
Rotor Copper Loss
P
g
=
3.74
1
−
0.05
Jason L. Wilson
12/1/2023
P
g
=
3.94
P
cu
(
R
)
=
0.05
(
3.94
)
P
cu
(
R
)
=
0.197
MW
f.
Stator Losses
P
cu
(
S
)
=
4.15
−
3.94
P
cu
(
S
)
=
0.21
MW
g.
Total efficiency of the system
η
=
3.74
4.89
∗
100
η
=
76.5%
Problem
12.19
A synchronous generator is connected directly to an infinite bus. The voltage of the infinite bus
is 15 kV. The excitation of the generator is adjusted until the equivalent field voltage Ef is 14 kV.
The synchronous reactance of the machine is 5 Ω. Compute the following:
a.
Pullover power (maximum power)
P
max
=
(
14
∗
15
)
sin
(
90
)
5
P
max
=
42
MW
b.
Equivalent excitation voltage that increases the pullover power by 20%
P
=
1.2
(
42
)
P
=
50.4
MW
E
f
=
50.4
(
5
)
15sin
(
90
)
E
f
=
16.8
kV
Problem
12.25
An industrial load is connected across a three-phase, Y-connected source of 4.5 KV (line-to-line). The real power of the load is 160 kW, and its power factor is 0.8 lagging:
a.
Compute the reactive power of the load. S
=
160
(
0.8
)
S
=
200
k
θ
=
cos
−
1
(
0.8
)
θ
=
36.86
°
sin
(
θ
)
=
0.60
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Related Questions
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Consider the equations of motion of an armature controlled DC motor given by
Jw(t)+bw (t) Ki (t)
=
di (l)
L + Ri (t) + Kw (t)
=
-Tfriction - Tload (t)
Vin (t),
dt
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When the input voltage and torque are constant, the transient response of this system has the form of the homogeneous solution.
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notation.
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Friction torque, Tfric = 0.003 N-m
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Armature resistance, R = 0.35
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QUESTION 8
Consider the equations of motion of an armature controlled DC motor given by
Jw(t)+bw (t) Ki(t) = -Tfriction - Tload (t)
di (t)
+ Ri (t) + Kw (t) =
Vin (t),
dt
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At steady-state operation, the motor equations satisfy
bo) -Ki =-T -T.
fric
load
Ri +Kov_-
55
55
The stall torque is the load torque required stead-state speed of the motor is zero.
The stall current is the armature current when the motor speed is zero.
The no load speed is the speed of motor when the load torque is zero.
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What is stall torque for this motor in N-m? Keep 3 significant figures, and do not include units. Use Scientific notation.
The constants are given below.
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с
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Unit
Remarks
Inertia
36
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Damping
20
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Train mechanisms only
Load Torque
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kN-m
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I
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I
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Saved
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T, =5.37s
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T2 =5.64s
3
T3 =5.31s
4
T4
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5
T5 5.73s
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