Induction Motor Drive

Induction Motor Drive • Why induction motor (IM)? – – – – Robust; No brushes. No contacts on rotor shaft High Power/Weight, Lower Cost/Power ratios E...
Author: Derick Rogers
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Induction Motor Drive • Why induction motor (IM)? – – – –

Robust; No brushes. No contacts on rotor shaft High Power/Weight, Lower Cost/Power ratios Easy to manufacture Almost maintenance-free, except for bearing and other “external” mechanical parts

• Disadvantages

– Essentially a “fixed-speed” machine – Speed is determined by the supply frequency – To vary its speed need a variable frequency supply

• Motivation for variable-speed AC drives – – – –

Inverter configuration improved Fast switching, high power switches Sophisticated control strategy Microprocessor/DSP implementation

• Applications

– Conveyer line (belt) drives, Roller table, Paper mills, Traction, Electric vehicles, Elevators, pulleys, Airconditioning and any industrial process that requires variable-speed operation.

• The state-of-the-art in IM drives is such that most of the DC drives will be replaced with IM in very near future. Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Torque production (1) • Only “squirrel-cage” IM (SCIM) is considered in this module • Neglecting all harmonics, the stator establishes a spatially distributed magnetic flux density in the air-gap that rotate at a synchronous speed, ω1 :

ω1 =

ωe p

where ωe : supply frequency (in Hz) p: pole pairs (p=1for 2 pole motor, p=2 for 4 pole motor etc) • If the rotor is initially stationary, its conductor is subjected to a sweeping magnetic field, inducing rotor current at synchronous speed. • If the rotor is rotating at synchronous speed (i.e. equals to f1), then the rotor experience no induction. No current is induced in the rotor. Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Torque production (2) • At any other rotor speed, say wm, the speed differential ωi-ω2 creates slip. Per-unit slip is defined as:

ω − ωm s= 1 ; ω1

ω1 =

ωe p

where : ωe : supply frequency

ω m : rotor frequency p : pole pair

• Slip frequency is defined as: ω2=ω1-ωm. • When rotor is rotating at ωm., rotor current at slip frequency will be induced. • The interaction between rotor current and air-gap flux produces torque.

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Single-phase Equivalent Circuit (SPEC) 1:nS R1

V1

L1

L2

Rm

Lm

Vm

STATOR SIDE

V2 = nSVm

R2

ROTOR SIDE

R1 : Stator resistance L1 : Stator leakage inductance R2 : Rotor resistance L2 : Rotor leakage inductance Lm : Magnetising inductance v1 : Supply voltage (phase voltage)

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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SPEC, referred to stator I1

V1

R1

L1

I2

Rm

L2

Lm

R2 S

• From previous diagram, SPEC is a dual frequency circuit. On the stator is ω1 and on the rotor ωm • Difficult to do calculations. • We can make the circuit a single frequency type, by referring the quantities to the stator

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Rotor current If E1 is the back EMF in the stator phase, then the back EMF in an equivalent rotor phase with the same effective turns ratio will be E 2 where : E 2 = sE1 At standstill, i.e when ω m = 0, E 2 = sE1 = 1E1 = E1 At synchronous speed, i.e whenω m = 1, E 2 = (0) E1 = 0 Hence the current in the rotor phase, E2 sE1 = R2 + jsX 2 R2 + jsX 2 E1 = R2 + jX 2 s Note that the quantities are now referred to the stator, but with the rotor resistance alteration. I2 =

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Performance calculation using SPEC I1

R1

V1

L1

Rm

I2

L2

Lm

R2 S

Pin = 3V1 I1 cos φ

Input Power :

Note : V1 and I1 must be phase voltage and current Stator copper loss : Core loss : Power across the air - gap : Rotor copper loss :

Pls = 3I12 R1 3V12 Plc = Rm 3 I 2 2 R2 Pg = s = Pin − Pls − Plc Plr = 3I 2 2 R2

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Performance calculation (2) Gross output power : Po = Pg − Plr

2 I R (1 − s ) 3I 2 2 R2 3 = Pg (1 − s ) = − 3I 2 2 = 2 2 s s

Power at the shaft : Psh = Po − PFW ; PFW : friction and windage loss. Developed (electromagnetic) torque : Po 3I 2 2 R2 (1 − s ) Te = = sω m ωm Since

ω1 − ω m s= ω1

ω m = (1 − s )ω1 ,

3 I 2 2 R2 ∴ Te = sω1 But ω1 = Then,

ωe p

; ω e is the supply frequency.

3 pI 2 2 R2 Te = sω e Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Example calculation •

A single phase equivalent circuit of a 6-pole SCIM that operates from a 220V line voltage at 60Hz is given below. Calculate the stator current, output power, torque and efficiency at a slip of 2.5%. The fixed winding and friction losses is 350W. Neglect the core loss.

I1

R1

X1

0.2Ω

0.5Ω

V1

I2

X2 0.2Ω

Xm

R2

20Ω

0.1Ω

V1 = 220V line-to line ÷ 3 220V = 127V 3 = 2.5% = 0.025

=

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Calculation (solution) X 1 = 0.5Ω, X 2 = 0.2Ω, X m = 20Ω R2 + jX 2 s 0 .1 + j 0 .2 0 . 025 = 0.2 + j 0.5 + j 20 = 4.2∠20o Ω 0 .1 + j 0.2 + j 20 0.025 V 220 3 o I1 = 1 = = ∠ − A 30 . 0 20 Z in 4.2∠20o

Z in = ( R1 + jX 1 ) + jX m //

Pin = 3V1I1 cos φ = 3(220

3)(30)(cos 20o )

= 10,758W Pls = 3I12 R1 = 3(30 2 )(0.2) = 540W Power transferred to rotor (neglecting core loss) Pg = Pin − Pls = 10,758 − 540 = 10,216W Gross power Po = Pg (1 − s ) = 10,216(1 − 0.025) = 9,961W Power at the shaft Psh = Po − PFW = 9,961 − 350 = 9,611W Output power 9611 = = 89.3% Input power 10758 Electromagnetic Torque

Efficiency =

P Te = o =

ωm

Po 9611 = (ωe1 p) (1 − s ) 2π (60 / 3)(1 − 0.025)

= 78.4 N .m Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Starting current •

For the previous example, Calculate the stating current when motor is first switched on to rated applied voltage.

Solution : At standstill, s = 1 Z in = = 0.2 + j 0.5 + j 20

0.1 + j 0.2 0.1 + j 0.2 + j 20

= 0.76Ω V 220 3 I1 = 1 = = 167 A Z in 0.76

Note that the starting current is about 5 times than full load current. This is common for induction motors.Care should be taken when starting induction motors.

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Approximate SPEC +

R1

L1

L2 R2 s

LM

V1 -

Since L m is large, the circuit above can be drawn I2 =

R R1 + 2 s

2

V1 + ω12 ( L1 + L2 )2

Power at the rotor (per phase), Po = I 2 2

R2 s

Electromagnetic (developed) torque, Te =

3Po

ω1

=

3R2V12 R sω1 R1 + 2 s

2

+ ω12 ( L1 + L2 )2

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Single (fixed supply) frequency characteristics For a give frequency ω1 , the torque (versus slip) characteristics can be shown as below. Note that : ω1 − ωm s= ; at standsill s = 1, at sync speed, s = 0.

ω1

TORQUE(+)

ωm

ωe

PLUGGING

MOTORING

Tem (max torque or pull-out torque)

ωe

ωe

ωm

ωm GENERATING

Tes (starting torque) 0 unity slip (standstill)

rated slip

zero slip ω e (sync.speed) SPEED

SLIP,s TORQUE(-)

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Single frequency characteristics CURRENT

TORQUE

operating point (rated torque)

EFFICIENCY POWER FACTOR

rated current rated slip

1.0

Standstill

SLIP 0

synchronous speed

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Single frequency characteristic •

As slip is increased from zero (synchronous), the torque rapidly reaches the maximum. Then it decreases to standstill when the slip is unity.



At synchronous speed, torque is almost zero.



At standstill, torque is not too high, but the current is very high. Thus the VA requirement of the IM is several times than the full load. Not economic to operate at this condition.



Only at “low slip”, the motor current is low and efficiency and power factor are high.

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Typical IM Drive System − IDC VDC



Modulation Index,

BLOCK DIAGRAM Supply

Rectifier and Filter 3-phase Inverter

n

CIRCUIT Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

IM 16

Variable speed characteristics •

For variable speed operation, the supply is an inverter.



The frequency of the fundamental AC voltage will determine the speed of IM. To vary the speed of IM, the inverter fundamental frequency need to be changed.



The inverter output frequency must be kept close to the required motor speed. This is necessary as the IM operates under low slip conditions.



To maintain constant torque, the slip frequency has to be maintained over the range of supply frequencies.

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Variable voltage, variable frequency (VVVF) operation • In order for maximum torque production, motor flux should be maintained at its rated value. Φ = Φ m sin ω1t But the back emf is : dΦ e1 = N = Nω1Φ m cos ω1t dt In RMS, 1 E1 = Nω11Φ m = 4.44 f1 N1Φ m 2 E1 or = 4.44 N1Φ m f1 Therefore, in order to maintain the motor flux, the ( E1 f1 ) ratio has to be kept constant.

This is popularly known as the constant Volt/Hertz operation Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Constant Torque region •

Hence for VVVF operation, there is a need to control the fundamental voltage of the inverter if its frequency (and therefore the frequency of the IM) need to be varied.



To vary the fundamental component of the inverter, the MODULATION INDEX can be changed.



The rated supply frequency is normally used as the base speed



At frequencies below the base speed, the supply magnitude need to be reduced so as to maintain a constant Volt/Hertz.



The motor is operated at rated slip at all supply frequencies. Hence a “constant torque” region is obtained.

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Constant Torque Region f1 f 2

TORQUE(+)

f1 > f 2 > f 3 > f 4 > f 5 f3

f4

f5

rated torque

0 rated slip

SLIP,s

TORQUE(+)

SPEED

rated torque

0

SLIP,s

SPEED

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Constant Power region • Above base speed, the stator voltage reaches the rated value and the motor enters a constant power region. • In this region, the air-gap flux decreases. This is due to increase in frequency frequency while maintaining fixed voltage. • However, the stator current is maintained constant by increasing the slip. This is equivalent to field weakening mode of a separately excited DC motor.

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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Constant Power region

TORQUE(+)

rated torque

0

SLIP,s

Base speed

TORQUE(+) CONSTANT TORQUE REGION

0

SLIP,s

SPEED

"FIELD WEAKENING"

Base speed

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

SPEED

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VVVF Summary

CONSTANT TORQUE

CONSTANT POWER

Electromagnetic torque,eT Terminal (supply) voltage, V1 slip frequency,fs slip,s 0

Base speed

Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

SPEED

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Examples •

A three-phase 4-pole, 10 horsepower, 460V rms/60Hz (line-to line) runs at full-load speed of 1746 rpm. The motor is fed from an inverter. The flux is made to be constsnt. Plot the torque-speed graphs for the following frequency: 60Hz, 45 Hz, 30Hz, 15Hz.

• A three-phase induction motor is using a three-phase VSI for VVVF operation. The IM has the following rated parameters: • • • •

voltage: frequency: slip (p.u) pole pair

415V (RMS) 50Hz 5% 2

– If the inverter gives 415V (RMS) with modulation index of 0.8, calculate the required modulation index if the motor need to be operated at rotor mechanical speed of 10Hz. Dr. Zainal salam; Power Electronics and Drives (Version 2),2002, UTMJB

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