# Induction Motors ```1/19/22, 11:29 AM
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AUDIENCE
Chapter 7: Induction Motors
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Chapter 7:
7-1.
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Induction Motors
A dc test is performed on a 460-V ∆-connected 100-hp induction motor. If VDC = 21 V and I DC = 72 A,
what is the stator resistance R1 ? Why is this so?
SOLUTION If this motor’s armature is connected in delta, then there will be two phases in parallel with one
phase between the lines tested.
VDC
R1
R1
R1
Therefore, the stator resistance R1 will be
VDC
I DC
R1
=
=
R1 (R1
R1
+ R1 ) 2
= R
+ (R1 + R1 ) 3 1
3 VDC
2 I DC
7-2.
3 21 V 
= 
 = 0.438 Ω
2  72 A 
A 220-V three-phase six-pole 50-Hz induction motor is running at a slip of 3.5 percent. Find:
(a) The speed of the magnetic fields in revolutions per minute
(b) The speed of the rotor in revolutions per minute
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1/19/22, 11:29 AM
(c) The slip speed of the rotor
The rotor frequency in hertz
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SOLUTION
(a)
The speed of the magnetic fields is
Hz ) = 1000 r/min
= 120P f e = 120(50
6
nsync
(b)
The speed of the rotor is
= (1 − s ) nsync = (1 − 0.035 )(1000 r/min ) = 965 r/min
nm
(c)
The slip speed of the rotor is
= snsync = (0.035)(1000 r/min) = 35 r/min
nslip
(d)
The rotor frequency is
fr
7-3.
nslip P
( 35 r/min )( 6 )
= 120 =
120
= 1.75 Hz
Answer the questions in Problem 7-2 for a 480-V three-phase four-pole 60-Hz induction motor running at
a slip of 0.025.
SOLUTION
(a)
The speed of the magnetic fields is
114
=
nsync
(b)
120 ( 60 Hz )
4
= 1800 r/min
The slip speed of the rotor is
= snsync = ( 0.025)(1800 r/min ) = 45 r/min
The rotor frequency is
fr
7-4.
=
= (1 − s ) nsync = (1 − 0.025)(1800 r/min ) = 1755 r/min
nslip
(d)
P
The speed of the rotor is
nm
(c)
120 f e
=
nslip P
120
=
( 45 r/min )( 4 )
= 1.5 Hz
120
A three-phase 60-Hz induction motor runs at 715 r/min at no load and at 670 r/min at full load.
(a) How many poles does this motor have?
(b) What is the slip at rated load?
(c) What is the speed at one-quarter of the rated load?
(d) What is the rotor’s electrical frequency at one-quarter of the rated load?
SOLUTION
(a)
This machine has 10 poles, which produces a synchronous speed of
nsync
(b)
120 f e
P
=
120 ( 60 Hz )
10
= 720 r/min
The slip at rated load is
s=
(c)
=
nsync
− nm
nsync
&times; 100% =
720 − 670
720
&times; 100% = 6.94%
The motor is operating in the linear region of its torque-speed curve, so the slip at &frac14; load will be
s
= 0 25(0 0694) = 0 0174
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s
0.25(0.0694)
0.0174
The resulting speed is
(d)
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The electrical frequency at &frac14; load is
fr
7-5.
Fullr/min
PDF) =Package
707 r/min
(1 − 0.0174 )(720
nm
= sf e = ( 0.0174 ) ( 60 Hz ) = 1.04 Hz
A 50-kW 440-V 50-Hz two-pole induction motor has a slip of 6 percent when operating at full-load
conditions. At full-load conditions, the friction and windage losses are 520 W, and the core losses are 500
W. Find the following values for full-load conditions:
(a) The shaft speed nm
(b) The output power in watts
(d) The induced torque τ ind in newton-meters
115
(e) The rotor frequency in hertz
SOLUTION
(a)
The synchronous speed of this machine is
=
nsync
120 f e
=
P
120 ( 50 Hz )
2
= 3000 r/min
Therefore, the shaft speed is
= (1 − s ) nsync = (1 − 0.06 )( 3000 r/min ) = 2820 r/min
nm
(b)
The output power in watts is 50 kW (stated in the problem).
(c)
=
POUT
50 kW
 2π rad  1 min 
=
ωm
(2820 r/min )
 1r
(d)
= POUT + PF&amp;W + Pcore + Pmisc = 50 kW + 520 W + 500 W = 51.2 kW
=
τ ind
Pconv
ωm
=
51.2 kW
2π rad  1 min 
(2820 r/min )


 1 r  60 s 
= 173.4 N ⋅ m
The rotor frequency is
fr
7-6.


 60 s 
The induced torque can be found as follows:
Pconv
(e)
= 169.3 N ⋅ m
= sf e = ( 0.06 )( 50 Hz ) = 3.00 Hz
A three-phase 60-Hz two-pole induction motor runs at a no-load speed of 3580 r/min and a full-load
speed of 3440 r/min. Calculate the slip and the electrical frequency of the rotor at no-load and full-load
conditions. What is the speed regulation of this motor [Equation (4-57)]?
SOLUTION The synchronous speed of this machine is 3600 r/min. The slip and electrical frequency at noload conditions is
nsync
s nl
=
f r ,nl
− n nl
nsync
3600 − 3580
&times; 100% =
3600
&times; 100% = 0.56%
= sf e = ( 0.0056 )( 60 Hz ) = 0.33 Hz
The slip and electrical frequency at full load conditions is
nsync
− n nl
100%
3600 − 3440
100%
4 44%
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1/19/22, 11:29 AM
sfl
=
f r ,fl
sync
nl
&times; 100% =
nsync
&times; 100% = 4.44%
3600
)( 60 Hz ) = 2.67
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The speed regulation is
SR
= n nl − nfl &times; 100% = 3580 − 3440 &times; 100% = 4.1%
nfl
3440
116
7-7.
A 208-V four-pole 60-Hz Y-connected wound-rotor induction motor is rated at 15 hp. Its equivalent
circuit components are
R1 = 0.220 Ω
R2 = 0.127 Ω
X 1 = 0.430 Ω
X 2 = 0.430 Ω
Pmech = 300 W
Pmisc ≈ 0
X M = 15.0 Ω
Pcore = 200 W
For a slip of 0.05, find
(a) The line current I L
(b) The stator copper losses PSCL
(c) The air-gap power PAG
(d) The power converted from electrical to mechanical form Pconv
τ ind
(e) The induced torque
(g) The overall machine efficiency
(h) The motor speed in revolutions per minute and radians per second
SOLUTION The equivalent circuit of this induction motor is shown below:
IA
+
R1
jX1
0.22 Ω
j0.43 Ω
j15
Vφ
jX2
j0.43
Ω
R2
0.127 Ω
1 − s 

 s 
R2 
jXM
Ω
2.413 Ω
-
(a)
The easiest way to find the line current (or armature current) is to get the equivalent impedance Z F
of the rotor circuit in parallel with jX M , and then calculate the current as the phase voltage divided by
the sum of the series impedances, as shown below.
IA
+
R1
jX1
jXF
0.22 Ω
j0.43 Ω
RF
Vφ
-
The equivalent impedance of the rotor circuit in parallel with jX M is:
ZF
=
1
1
1
jX
+Z
=
1
1
j15 Ω
1
= 2.337 + j0.803 = 2.47∠19&deg; Ω
+ 2 54 + j0 43
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1/19/22, 11:29 AM
jX M
j15 Ω
Z2
2.54 + j 0.43
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The phase voltage is 208/ 3 = 120 V, so line current I L is
(b)
= IA =
IL
R1 + jX 1 + RF + jX F
= I A = 42.3∠ − 25.7&deg; A
120∠0&deg; V
=
0.22 Ω + j 0.43 Ω + 2.337 Ω + j 0.803 Ω
The stator copper losses are
2
= 3I A2 R1 = 3 ( 42.3 A ) ( 0.22 Ω ) = 1180 W
PSCL
(c)
Vφ
IL
The air gap power is PAG
= 3I 2 2
2
(Note that 3I A RF is equal to 3I 2
2
R2
s
R2
s
= 3I A 2 R F
, since the only resistance in the original rotor circuit was R2 / s ,
and the resistance in the Thevenin equivalent circuit is RF . The power consumed by the Thevenin
equivalent circuit must be the same as the power consumed by the original circuit.)
= 3I 2 2
PAG
(d)
2
= 3I A2 RF = 3 ( 42.3 A ) ( 2.337 Ω ) = 12.54 kW
= (1 − s ) PAG = (1 − 0.05)(12.54 kW ) = 11.92 kW
The induced torque in the motor is
=
τ ind
PAG
ω
(f)
s
The power converted from electrical to mechanical form is
Pconv
(e)
R2
=
sync
12.54 kW



1 min
r  60 s 
= 66.5 N ⋅ m
The output power of this motor is
POUT
= Pconv − Pmech − Pcore − Pmisc = 11.92 kW − 300 W − 200 W − 0 W = 11.42 kW
The output speed is
nm
= (1 − s ) nsync = (1 − 0.05) (1800 r/min ) = 1710 r/min
=
τ
POUT
ωm
11.42 kW
=

(g)
1 r  60 s 
The overall efficiency is
η
=
η=
(h)
= 63.8 N ⋅ m
(1710 r/min ) 2π
POUT
PIN
&times; 100% =
POUT
3Vφ I A cos θ
11.42 kW
3 (120 V )( 42.3 A ) cos 25.7 &deg;
&times; 100%
&times; 100% = 83.2%
The motor speed in revolutions per minute is 1710 r/min. The motor speed in radians per second is
118
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