ece 325 electric energy system components 6- three...
TRANSCRIPT
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Instructor:
Kai Sun
Fall 2015
ECE 325 – Electric Energy System Components6- Three-Phase Induction Motors
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Content
(Materials are from Chapters 13-15)
• Components and basic principles
• Selection and application
• Equivalent circuits
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• 3-phase, 60Hz, AC induction motor
• Rated power: 30.0 hp (=22.4kW)
• Service factor: 1.15 (i.e. 115% of
rated power for short-term use)
• Rated full-load current: 35.0A
• Rated voltage: 460V
• Rated speed: 1765rpm
• Continuous duty at ambient 40oC
• Full-load nominal efficiency: 93.0%
• Frame size: shaft height of 28/4=7
inches; body’s length of 6
• Insulation class: F (155oC for 20,000
hours lifetime)
• NEMA design: B (normal starting
torque combined with a low current)
• Start inrush kVA: G (5.60-6.29kVA)
http://www.pdhonline.org/courses/e156/e156content.pdf
Understanding the Nameplate
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Principle components
• Stator (stacked laminations)
• Rotor (stacked laminations)
− Wound rotor
− Squirrel-cage rotor
• Air gap (0.4mm-4mm)
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Stator
• Hollow, cylindrical core made up of stacked laminations.
• Stator winding is placed in the evenly spaced slots punched out of
the internal circumference of the laminations.
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Wound Rotor
• Has a 3-phase winding (usually in
Y-connection) uniformly distributed
in rotor slots, similar to the stator
winding
• Rotor winding terminals are
connected to three slip-rings which
revolve with the rotor.
• 3 stationary brushes connect the
rotor winding to external resistors
during the start-up period and are
short-circuited during normal
operations.
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Squirrel-Cage Rotor
• More adopted in induction motors
• Instead of a winding, copper bars are
pushed into the slots in the laminations
• Welded to two copper end-rings, all
bars are short-circuited at two ends, so
as to resemble a “squirrel cage”
• In small and medium motors, the
“cage” (bars and end-rings) are made of
aluminum.
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Principle of Operation: Faraday’s Law and Lorentz Force
• A magnet moving above a conducting ladder induces a voltage E=Blv and a current
I in the conductor underneath according to Faraday’s law (the right-hand rule).
• Cut by the moving flux, the current-carrying conductor experiences a Lorentz force
(the left-hand rule) always acting in a direction to drag it along the moving magnet
• When the system reaches a steady state, the ladder moves in the same direction as
the magnet but has a lower speed <v (why?)
• Roll up the ladder into a cylindrical squirrel-cage rotor and replace the magnet by a
stator an induction motor
Parallel conductors are short-circuited
by end-bars A and B
Conductors are hort-circuited
by end-rings A and B)
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Rotating magnetic field
• Consider a stator having 6 salient poles with Y-connected stator windings
carrying balanced 3-phase alternating currents.
Positive currents: always flowing
into terminals A, B and C( ) 10cos 10cos( )
( ) 10cos( 120 ) 10cos( 120 )
( ) 10cos( 120 ) 10cos( 120 )
a
b
c
i t t
i t t
i t t
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Rotating Magnetic Field: Instant 1
• =0: 6 poles together produce a magnetic field having
essentially one broad N pole and one broad S pole
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Rotating Magnetic Field: Instant 2
• 60o: the magnetic field moves CW by 60o; its angular speed equals
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Rotating Magnetic Field: Instant 3
• 120o: the magnetic field moves CW by 120o; its angular speed equals
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Rotating Magnetic Field
• For the stator producing a magnetic field having one pair of N-S poles
– The field rotates 360o during one cycle of the stator current.
– The speed of the rotating field is necessarily synchronized with the frequency f
of the source, so it is called synchronous speed =60f (r/min)
– With the phase sequence A-B-C, the magnetic field rotates CW.
– If we interchange any two of the three lines connected to the stator, the new
phase sequence will be A-C-B and the magnetic field will rotate CCW at the
synchronous speed
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Phase group• In practice, instead of using a single coil per pole, the
coil is sub-divided into 2 or more coils lodged and
staggered in adjacent slots connected in series, i.e. a
phase group
• Each phase group produces one N/S pole, so using
more phase groups allows us to increase the number
of poles (denoted by p)
• No. of groups = No. of phases No. of poles = 3 p
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Number of N-S poles and synchronous speed
• Synchronous speed ns=120f / p [r/min]
f : frequency of the source [Hz]
p: number of poles
If f=60Hz,
ns=120f / 4 =30f =1800r/min ns=120f / 8 =15f=900r/min
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Starting of an induction motor1. The magnetic field produced by the stator
current rotates CW at synchronous speed ns.
2. Relative to the magnetic field, the rotor rotates
CCW.
3. According to Faraday’s Law (right-hand rule),
the rotor windings (bars) have induced
voltages in the directions as indicated.
4. Large circulating currents are created in rotor
windings (bars) by the induced voltages
5. The currents have the maximum values at the
staring point because the rotor has the
maximum speed relative to the magnetic field.
6. The rotor windings rotates CW subjected to
Lorentz forces (left-hand rule) in the directions
as indicated
7. The rotor will accelerate until the total Lorentz
force equals the friction
8. Once the rotor starts rotating CW, its speed
relative to the filed will decrease, so its
winding currents and Lorentz forces will
decrease
Can the rotor reach ns?
• If it did, then its currents and Lorentz
forces would be zero and friction
would slow the rotor down, so the
answer is: No, the rotor’s speed n<ns at
a steady state.
• Since friction is very small, the rotor
speed n ns at no load conditions
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Slip
• Slip: is the difference between synchronous speed ns and rotor
speed n, expressed as a per unit or percent of synchronous speed.
f: frequency of the source connected to the stator [Hz]
f2: frequency of the voltage and current induced in the rotor [Hz]
ns=120f / p: synchronous speed [r/min]
n: rotor speed [r/min]
r=n/30, s=ns/30 : corresponding angular speeds of n and ns [rad/s]
2 (pu)s s r
s s
n n fs
n f
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Example 13-3
• A 0.5 hp, 6-pole induction motor is excited by a 3-phase, 60 Hz source. Calculate
the slip, and the frequency of the rotor current under the following conditions:
a. At standstill
b. Motor turning at 500r/min in the same direction as the revolving field
c. Motor turning at 500r/min in the opposite direction to the revolving field
d. Motor turning at 2000r/min in the same direction as the revolving field
Solution:
ns=120f/p=12060/6 = 1200 r/min
a. s= (ns-n)/ns=(1200-0)/1200 =1 pu
f2=sf=60Hz
b. s= (1200-500)/1200 = 0.583 pu
f2=sf=0.583 60Hz = 35 Hz
c. s= [1200- (-500)]/1200 =1.417 pu
f2=sf=1.417 60Hz = 85 Hz
d. s= [1200- 2000)]/1200 = -0.667 pu
f2=sf=-0.667 60Hz = -40 Hz
Slip Rotor speed
& freq
Operating mode
s=1n=0
f2=fTransformer (standstill)
0<s<10<n<ns
0<f2<fMotor
s>1n<0
f2>fBrake
s<0n>ns
f2<0
Generator (reversed
phase sequence A-C-B)
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Equivalent circuits: wound-rotor induction motor1. When the motor is at standstill (i.e. n=0, s=1), it acts exactly like a
conventional transformer. Assume a Y-connection for both the stator and rotor,
and a turns ratio of 1:1. Consider the per-phase equivalent circuit:
Eg: source voltage, line to neutral,
x1, r1: stator leakage reactance and winding resistance
x2, r2: rotor leakage reactance and winding resistance
Xm, Rm: magnetizing reactance and resistance modeling losses of iron,
windage and friction
Rx: external resistance, connecting one slip-ring to the neutral of the rotor
Note: Io may reach 40% of Ip due to the air gap between the stator and rotor (much bigger
than the air gap between the P and S windings of a transformer), so we cannot eliminate
the magnetizing branch.
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2. When the motor runs at a slip s, i.e. n=(1-s)ns
|E2|=|sE1|, |I2|=|I1|, jx2 jsx2 r2 and RX are the same
2 1 1 22
2 22 2 2 2 22 2
| | 0 | |= arctan
j | j | ( )
E sE sE sxI
R sx R sx RR sx
Note: the phasors on the primary side (E1 and I1) and the secondary side (E2 and I2)
cannot be drawn in one phasor diagram because they have different frequencies
RX
f sf
1 : 1
R2 =r2+RX
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• I1 and I2 have the same effective value even
though they have different frequencies
|I1|=|I2|
• sE1 and E2 have the same effective values
|sE1|=|E2|
• E1 leads I1 and E2 leads I2 both by
Phasor diagrams on the rotor and stator
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Simplified equivalent circuit: referred to the stator side
11 2
2 2j
sEI I
R sx
1 1
1
2 2 2/ j
E EI
R s x Z
def1
2 2 2
1
/ jE
Z R s xI
Note: The value of R2/s will vary from R2 to as the motor goes
from start-up (s=1) to ns (s=0)
jx =jx1+jx2
E1
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Eg
Ip
Io
r1jx1 jx2
R2
RmjXm
I1
2
1 sR
s
jx
2R
s
ImIf
• Active power supplied to the rotor:
• Total I2R losses in the rotor circuit
• Mechanical power developed by the motor
• Torque developed by the motor
• Total power absorbed by the motor:
• Stator iron losses
• Stator copper losses
2 2
2 21 1
| | | || | ( )
g g
m m
E E RS j I r jx
R X s
2
2 2 21 1 1
| || | | |
g
m
E RP I r I
R s
2
1 2| | /rP I R s
2
1 1| |jsP I r
2
1 2| |jr rP I R sP
2
1 2
1(1 ) | |m r jr r
sP P P P s I R
s
9.55 9.55 (1 ) 9.55
(1 )
m r r
s s
P P s PT
n n s n
2| | /f g mP E R
Note: The torque only depends on Pr
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Eg
Ip
Io
r1jx1 jx2
R2
RmjXm
I1
2
1 sR
s
ImIf
Active power flow
2
1 1| |jsP I r
2
1 2| |jrP I R
2| | /f g mP E R
2
1 2
1| |m
sP I R
s
/L eP P
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Eg
Ip
Io
r1jx1 jx2
R2
RmjXm
I1
2
1 sR
s
jx
2R
sIm
If
=accos(P/|S|)
=actan(x/r1)
Z1=r1+jx
Phasor diagram of the induction motor
S=P+jQ
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Breakdown (maximum) Torque
When |Z1|=|R2/s|, Pr and torque T both
reach their maximum values
I1
Z1I1
Eg
I1R2/s
1 2| | | |
cos2 2
gE I R
s
2
1| |b
Rs
Z
22
1 2
2
1
9.55 | |9.55 9.55 | | /
4 | | cos2
grb
s ss
EP I R sT
n nn Z
1
1
| || |
2 | | cos2
g
b
EI
Z
1 1
| || | cos
2 2
gEI Z
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Example 13-5• A 3-phase induction motor having a synchronous speed of 1200 r/min draws
80 kW from a 3-phase feeder. The copper losses and iron losses in the stator
amount to 5 kW. If the motor runs at 1152 r/min, calculate
a. The active power transmitted to the rotor
Pr=Pe – Pjs – Pf=80 – 5 =75 kW
b. The rotor I2R losses, i.e. Pjr
s=(ns-n)/ns=(1200-1152)/1200=0.04
Pjr=sPr=0.0475=3 kW
c. The mechanical power developed
Pm=Pr – Pjr =75 – 3 =72 kW
d. The mechanical power delivered to the load, knowing that the
windage and friction losses equal to 2 kW
PL=Pm - PV=72 – 2 =70 kW
e. The efficiency of the motor
=PL/Pe=70/80=87.5%
f. The torque developed by the motor
Tm=9.55 Pr/ns= 9.5575000/1200=597 Nm
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Two-wattmeter method to measure 3-phase active power
P1=|Vac||Ia|cos( - 30o)=ELILcos( - 30o)
P2=|Vbc||Ib|cos( + 30o)=ELILcos( + 30o)
3-phase load
in Y or
connection
3 13 1 23 2 3( )Q P PS jP P Pj
3 cos( 30 ) cos( 30 ) 3 cos( 30 ) cos( 30 )
2cos cos30 3 2sin sin 30
3 c s no si3 L L
L L L L
L L L L
L L
S E I j E I
E I j E I
EE IjI
Proof:
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a. Power supplied to the rotor
Pe=P1+P2=70 kW
b. Rotor I2R losses Pjr
r1=0.34/2=0.17
Pjs=3I12r1=37820.17=3.1 kW
Pr=Pe – Pjs – Pf=70 – 3.1 – 2 =64.9 kW
s=(ns-n)/ns=(1800-1763)/1800=0.0205
Pjr=sPr=0.020564.9=1.33 kW
1763 r/min
Example 13-7
c. Mechanical power supplied to the load
Pm=Pr – Pjr =64.9 – 1.33 =63.5 kW
PL=Pm - PV= 63.5 – 1.2 =62.3 kW
=62.3 1.34 = 83.5 hp
d. Efficiency
=PL/Pe=62.3/70=89%
e. Torque developed at 1763 r/m
Tm=9.55 Pr/ns= 9.55649000/1800
=344 Nm
A 3-phase induction motor having a nominal rating of 100 hp (~75 kW) and a
synchronous speed of 1800 r/min is connected to a 3-phase 600 V source.
Resistance between two stator terminals =0.34. Calculate
Two-wattmeter method
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no-load torque
Torque-Speed Curve 2
2
2
21
9.55 | |9.55
( )
gr
ss
s
s
E RPT
n R nZ n n
n n
Stable operating
Region (n, T)
2 2 21 1 1, | | , / ( )s
r g
s
n n R Rs P I I E Z
n s s
• The curve is nearly linear between no-load and full-load because s is small and
R2/s is big2 2
2
2 2
9.55 | | 9.55 | |( )
g g
s
s s
E ET n n s
R n R n 1 2 /gI sE R
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Eg=440/1.73 V
R2=1.2
Z1=1.5+6j
ns=1800 r/min2
2
221
9.55 | |
| | ( )
g
ss
s
E R
R n
n
T
nZ nn
Torque-Speed Curve
21
1
g
s
s
E
RI
n
nZ
n
Current-Speed Curve
2
2
2
1 2
9.55 | |
/
g
s
E R
Z R nT
s s
Torque-Slip Curve
The 5 hp induction motor
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Torque-Speed Curve: 5 hp motor
s I1 Pr T n
• When the motor is stalled, i.e. locked-rotor condition, the current is 5-6 times the
full-load current, making I2R losses 25-36 times higher than normal, so the rotor
must never remain locked for more than a few second
• Small motors (15 hp and less) develop their breakdown torque at about 80% of ns
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Torque-Speed Curve: 5000 hp motor
Big motors (1500+ hp) :
• Relatively low starting
(locked-rotor) torque
• Breakdown torque at
about 98% of ns
• Rated n is close to ns
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Effect of rotor resistance
R2
2.5R2
10R2
Rated Torque =15Nm
Rated torque =15Nm
Rated torque =15Nm
Current at the rated torque
Current at the rated torque
Current at the rated torque
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• When R2 increases, starting (locked-rotor) torque TLR increases, starting current
I1,LR decreases, and breakdown torque Tb remains the same
• Pros & Cons with a high rotor resistance R2
– It produces a high starting torque TLR and a relatively low starting current I1,LR
– However, at the rated torque, it produces a rapid fall-off in speed with
increasing load because the torque-speed curve becomes flat, and the motor
has high copper losses and low efficiency and tends to overheat
• Solution
– For a squirrel-cage induction motor, design the rotor bars in a special way so
that the rotor resistance R2 is high at starting and low under normal operations
– If the rotor resistance needs to be varied over a wide range, a wound-rotor
motor needs to be used.
Effects of rotor resistance
2 2
2
2
2
2
2
1 1
9.55 | | 9.55 | |
| | | |
g g
s s
LR
E E
Z n
RT R
R Z n
2
2
1
9.55 | |
4 | | cos2
g
b
sn Z
TE
2
2
1| |jr IP R1
1,
2
g
LR
E
ZI
R
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Effects of the terminal voltage: Problem 13-33• A 3-phase, 300kW, 2300V 60Hz, 1780 r/min induction motor is used to
drive a compressor. The motor has a full-load efficiency of 92% and power
factor of 86%. If the terminal voltage rises by 20% while the motor operates
at full load, how does each of the following change?
2
2
2
9.55 | |( )
g
FL s FL
s
ET n n
R n
2
22
1
9.55 | |
| |
g
LR
s
ET R
Z n
1,
1 2
g
LR
EI
Z R
2
2
1
9.55 | |
4 | | cos2
g
b
s
ET
n Z
Pm, T I1,LR
nFL Tb
I1,FL Average
motor
temperature
PF Flux per pole
Exciting
current
TLR Iron losses
Q=|I1|2x+|Eg|
2/Xm
Same
(Q)
About the
same
(Pf Pjs Pjr)
44%
20%
44%
About the
same (iron
windings )
20%
>20%
(considering
saturation)
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Asynchronous generatorConnect the 5 hp, 1800 r/min, 60Hz motor
to a 440 V, 3-phase line and drive it at a
speed of 1845 r/min
s=(ns-n)/ns=(1800-1845)/1800= -0.025
R2/s=1.2/(-0.025)=-48 <0
The negative resistance indicates that power
is flowing from the rotor to the stator
(asynchronous generator mode)
Power flow from the rotor to the stator:
|E|= 440/1.73=254 V
|I1|=|E|/ |-48+1.5+ j6|=254/46.88= 5.42A
Pr=|I1|2R2/s= -1410 W <0
1410W flows from the rotor to the stator
Mechanical power and torque input to its
shaft:
Pjr=|I1|2R2=35.2 W
Pm=Pr+Pjr=1410 + 35.2=1445W
T=9.55 Pm/n=22.3 Nm
Total active power delivered to the line:
Pjs=|I1|2r1=44.1 W
Pf =|E|2/Rm=71.1 W
Pe=Pr-Pjs-Pf= 1410-44.1-71.7=1294 W
P3=3Pe=3882W
Reactive power absorbed from the line:
Q3=(|I1|2x+|E|2/Xm) 3 =(176+586) 3
=2286 var
Complex power delivered to the line:
S3=P3 - Q3=3882 - j2286 VA
cos=86.2%
Efficiency of this asynchronous generator
=Pe/Pm=1294/1445=89.5%
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Tests to determine the equivalent circuit
• Estimate r1, r2, Xm, Rm and x (note: r2+RX=R2 where RX is the
external resistance)
1. No-load test
2. Locked-rotor test
Eg
Ip
Io
r1jx1 jx2
r2/s
RmjXm
I1
ImIf
RX/s
jx
R2/s
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No-load test
At no-load, slip s0
R2/s is high, I1<<Io
Steps:
1. Measure stator resistance RLL
between any two terminals
(assuming a Y connection)
r1=RLL/2
2. Run the motor at no-load using
rated line-to-line voltage ENL.
Measure no-load current INL
and 3-phase active power PNL
3NL NL NLS E I
2 2
NL NL NLQ S P
2 2
1 13 3NL f NL f NL NLP P I r P P I r
2 2
2
1
( / 3)
( ) / 3 3
NL NLm
f NL NL
E ER
P P P I r
(Ignoring PV)
2 2( / 3)
( ) / 3
NL NLm
NL NL
E EX
Q Q
44
Locked-rotor test
When the rotor is locked,
slip s=1, IP >>Io r2 =r2/sR2/s, neglect the magnetizing branch
Steps:
1. Apply reduced 3-phase voltage
ELR to the stator so that the stator
current Ip the rated value
2. Measure line-to-line voltage
ELR, current ILR and 3-phase
active power PLR
3LR LR LRS E I
2 2
LR LR LRQ S P
2 2 2
1 2 2 13 3 / (3 )LR LR LR LR LRP I r I r r P I r 23
LR
LR
Qx
I
46
Homework Assignment #5
• Read Chapters 13.0-13.16, 15.0-15.9
• Questions:
– 13-16, 13-23, 13-24, 13-28, 15-2, 15-3, 15-4, 15-5
• Due date:
– hand in your solution to Denis at MK 205 or by email
([email protected]) before the end of 11/4 (Wed)