induction mach slides

8/18/2019 Induction Mach Slides

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Department of Electrical andDepartment of Electrical and

Computer EngineeringComputer Engineering

EE20AEE20A -- Electromechanical EnergyElectromechanical Energy

ConversionConversion

Induction MachineInduction Machine


2/35

Principle of Operation

• The stator coils, when energised, create arotating magnetic field.

• Rotating magnetic field cuts through therotor inducing a voltage in the rotor bars.

• This voltage creates its own magnetic fieldin the rotor.

• The rotor magnetic field will attempt to lineup with the stator magnetic field.

• The stator magnetic field is rotating, therotor magnetic field trying to line up withthe stator magnetic field causes the rotor torotate.

• The rotor magnetic field, never catches up,

but follows slightly behind.

Principle of OperationThe stator coils, when energised, create a rotating magnetic field.

Rotating magnetic field cuts through the rotor inducing a voltage in the rotor bars.

This voltage creates its own magnetic field in the rotor.

The rotor magnetic field will attempt to line up with the stator magnetic field.

The stator magnetic field is rotating, the rotor magnetic field trying to line up with the stator magnetic field causes the rotor to rotate.

The rotor magnetic field, never catches up, but follows slightly behind.


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Motor AnalysisMotor Analysis

• Slip is the difference betweenthe speed of the stator magneticfield and the speed of the rotor

• SLIP,S, = (NS - N) / NS

• When motor is stationary, it

behaves like a transformer

• At a given Speed, flux cutting

rate is reduced => thereby

reducing output voltage by a

factor of the slip.

Motor AnalysisMotor Analysis

Slip is the difference between the speed of the stator magnetic field and the speed of the rotor

SLIP,S, = (NS - N) / NS


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AnalysisAnalysis

Xm

I NL

Rs

jXs

Vph

IIN

I1

I2

RO

a : 1IO I

m

V1 V2

Rr

jXr

Per Phase Equivalent Circuit


5/35

AnalysisAnalysis

Xm

I NL

Rs

jXs

Vph

IIN

I1

I2

RO

a : 1IO I

m

V1 V2

Rr

s jX

r



6/35

AnalysisAnalysis

Xm

I NL

Rs

jXs

Vph

IIN

I1

RO

IO Im R

r

s

jXr

Pair gap



7/35

Power per PhasePower per Phase

• Total Torque =

(3Pmech_gross- PF&W)/ωm

• Pag = I12R r ̀ /s

• Pcu = sPag

• Pmech_gross = (1-s)Pag

Xm

I NL

Rs

jXs

Vph

IIN

I1

RO

IO I

m Rr

s

jXr

Pair gap



Total Torque =

(3Pmech_gross- PF&W)/ωm


8/35


mo NL III +=

Pag = Power across the air gap

′′ +++

=

s

R xIP r

2

1ag

′=


9/35


P mech_gross = (1-s) Pag per phase

r 2

1cu R IProtor,inlossesCu ′=

m

W&Fmech_gross

ω

P-Px3 TorqueTotal =

[ ] ⎥⎦⎤⎢⎣

⎡ += ′s

s)-(1 R R IP r r 2

1ag


10/35


[ ] [ ]s

s)-(1R IR IP r

2

1r

2

1ag += ′

Pcu_losses_in_rotor Pmech_gross

PPagag :: PPcucu :: PPmechmech = 1:s:(1= 1:s:(1--s)s)


11/35


[ ]

[ ] r 2

1

r 2

1 _per_phasemech_gross

R Is

s)-(1

s

R Is)-(1P

′

′

⎥⎦

⎤⎢⎣

⎡=

⎥⎦

⎤⎢⎣

⎡=

Slip is variable and affects only rotor circuit

Ignoring Stator values

r r

ph1

X j

s

R V I

′′ +

=


12/35


[ ]

[ ][ ]

[ ]

2

r

2

r

r

2

ph

m

m

2

r

2

r

r

2

ph

r

2

1mech_gross

X

s

R

R .V x

ns2

s)-(1

2 ω

ω

phase perPowerTorque

Xs

R

R .V x

ss)-(1

s

s)-(1 R IP

′′

′

′′

′

′

+⎥⎦

⎤⎢⎣

⎡Π⇒

Π=

=

+⎥⎦

⎤⎢⎣

⎡=

=

n


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TorqueTorque

Simple Algebraic manipulations yield

( )

⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡⇒

⎪

⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪

⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ⎥⎦⎤⎢⎣

⎡=

′

′

′

′

′

′

′

′

′

22

r

r

2

r

r 2

ph

r

2

r

2

2

r

r

2

r

2

ph

mech_gross

sX

R

s.

X

R .V

xnsX2

s)-(1

XsX

R

s.R .V xns2s)-(1 T

π

π


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TorqueTorque

⎥⎦

⎤

⎢⎣

⎡

+⎥⎥⎦

⎤

⎢⎢⎣

⎡⇒

⎥

⎥

⎦

⎤

⎢

⎢

⎣

⎡

+⎥⎦

⎤⎢⎣

⎡=

=

′

′

′

′

22r

2

ph

2

22

ph

r

mech_gross

r

r

s

.s x

nX2

V.s)-(1

s

.s.V x

nsX2

s)-(1 TgetThen we

X

R let Now

2

α

α

π

α

λ

π

α


15/35

TorqueTorque

22

r s

2

phmech_gross

s

s

s

ss x

Xn2V T

:getn weforngsubstitutiBy

s)-(1nn

n

n-n sslip,But

+=

=⇒

=

′ α π

Since the above calculations was derives as power per phase,then the total torque for all three phases would be three times

the gross mechanical torque for each phase calculated above.


16/35

22

r s

2 ph

22

r s

2

ph

s

s. .kTorqueTotalThen

Xn2

3V kLet

s

s x

Xn2

V x3TorqueTotal

+=

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+=

′

′

α

α

π

α π

TorqueTorque

The maximum torque is obtained when:

α sorX

R sslip,

r

r ==′

′


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Torque CharacteristicsTorque Characteristics


18/35

SpeedSpeed--Torque characteristicsTorque characteristicsModifications in the design of the squirrel-cage motors

permit a certain amount of control of the starting current and

torque characteristics.

These designs have been categorised by NEMA Standards

(MG1-1.16) into four main classifications:

1. Normal-torque, normal-starting current motors (Design A)

2. Normal-torque, low-starting current motors (Design B)

3. High-torque, low-starting-current, double-wound-rotor

motors (Design C)

4. High-slip motors (Design D)


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Design A MotorDesign A Motor

• Hp range 0.5 – 500 hp.

• Starting current 6 to 10 times full-load current.• Good running efficiency (87% - 89%).

• Good power factor (87% - 89%).

• Low rated slip (3 –5 %).

• Starting torque is about 150% of full load torque.• Maximum torque is over 200% but less than 225% of full-

load torque.

• Typical applications – constant speed applications where high

starting torque is not needed and high starting torque is tolerated.


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Design B MotorDesign B Motor•Hp range – 0.5 to 500 hp

•Higher reactance than the Design A motor, obtained by means ofdeep, narrow rotor bars.

•The starting current is held to about 5 times the full-load current.

•This motor allows full-voltage starting.

•The starting torque, slip and efficiency are nearly the same as forthe Design A motor.

•Power factor and maximum torque are little lower than class A,

•Design B is standard in 1 to 250 hp drip-proof motors and in

totally enclosed, fan-cooled motors, up to approximately 100 hp.•Typical applications – constant speed applications where high

starting torque is not needed and high starting torque is tolerated.

•Unsuitable for applications where there is a high load peak


21/35

Design C MotorDesign C Motor

•Hp range – 3 to 200 hp

•This type of motor has a "double-layer" or double squirrel-cagewinding.

•It combines high starting torque with low starting current.

•Two windings are applied to the rotor, an outer winding having

high resistance and low reactance and an inner winding havinglow resistance and high reactance.

•Operation is such that the reactance of both windings decrease

as rotor frequency decreases and speed increases.

•On starting a much larger induced currents flow in the outer

winding than in the inner winding, because at low rotor speeds

the inner-winding reactance is quite high.


22/35

Design C MotorDesign C Motor

•As the rotor speed increases, the reactance of the inner winding

drops and combined with the low inner-winding resistance, permits the major portion of the rotor current to appear in the

inner winding.

•Starting current about 5 times full load current.

•The starting torque is rather high (200% - 250%).•Full-load torque is the same as that for both A and B designs.

•The maximum torque is lower than the starting torque,

maximum torque (180-225%).

•Typical applications – constant speed loads requiring fairly

high starting torque and lower starting currents.


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Design D MotorDesign D Motor

•Produces a very high starting torque-approximately 275% of

full-load torque.

•It has low starting current,

•High slip (7-16%),

•Low efficiency.

•Torque changes with load

•Typical applications- used for high inertia loads

The above classification is for squirrel cage induction motor


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Wound RotorWound Rotor

•Hp 0.5 to 5000hp

•Starting torque up to 300%•Maximum torque 225 to 275% of full load torque

•Starting current may be as low as 1.5 times starting current

•Slip (3 - 50%)

•Power factor high•Typical applications – for high starting torque loads where very

low starting current is required or where torque must be applied

very gradually and where speed control is needed.


25/35

Current Effects on the MotorCurrent Effects on the Motor

•Induction motor current consists of reactive (magnetizing) and

real (torque) components.

•The current component that produces torque (does useful work)is almost in phase with voltage, and has a high power factor close

to 100%

•The magnetizing current would be purely inductive, except that

the winding has some small resistance, and it lags the voltage by

nearly 90°.

•The magnetizing current has a very low power factor, close to

zero.•The magnetic field is nearly constant from no load to full load

and beyond, so the magnetizing portion of the total current is

approximately the same for all loads.

•The torque current increases as the load increases


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•At full load, the torque current is higher than the magnetizing

current.

•For a typical motor, the power factor of the resulting current is between 85% and 90%.

•As the load is reduced, the torque current decreases, but the

magnetizing current remains about the same so the resulting

current has a lower power factor.

•The smaller the load, the lower the load current and the lower

the power factor. Low power factor at low loading occurs

because the magnetizing remains approximately the same at noload as at full load

Current Effects on the MotorCurrent Effects on the Motor


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Methods to vary speed of theMethods to vary speed of the

Induction MotorInduction MotorAn induction motor is a constant-speed device. Its speed depends on

the number of poles in the stator, assuming that the voltage and

frequency of the supply to the motor remain constant.

•One method is to change the number of poles in the stator,

for example, reconnecting a 4-pole winding so that it

becomes a 2-pole winding will double the speed. This methodcan give specific alternate speeds but not gradual speed

changes.

•Another method is to vary the line voltage this method is notthe best since torque is proportional to the square of the

voltage, so reducing the line voltage rapidly reduces the

available torque causing the motor to stall


28/35

Methods to vary speed of theMethods to vary speed of the

Induction MotorInduction Motor•Sometimes it is desirable to have a high starting torque or to have

a constant horsepower output over a given speed range. These and

other modifications can be obtained by varying the ratio of voltageto frequency as required. Some controllers are designed to provide

constant torque up to 60 Hz and constant hp above 60 Hz to

provide higher speeds without overloading the motor.

•An excellent way to vary the speed of a squirrel-cage induction

motor is to vary the frequency of the applied voltage. To maintain

a constant torque, the ratio of voltage to frequency must be kept

constant, so the voltage must be varied simultaneously with thefrequency. Modern adjustable frequency controls perform this

function. At constant torque, the horsepower output increases

directly as the speed increases.


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NO LOAD TESTNO LOAD TEST

W a t t m

e t e r

AC

I

V

Current

Coil

V o l t a g e

C o i l

Xm

I NL

Rs

jXs

Vph

IIN

I1

RO

IO Im Rr

s

jXr

Pair gap



30/35


n - ns = 0 ‘No load Speed ≈ Synchronous Speed’

i.e. no power transfer which implies that Torque = 0

⇒I1 = 0 & T = 0

Power Consumed = Core Losses + Friction &

Windage

Measure Vph , IIN and Wph

⇒ ( Infinite Impedance ) since I1 = 0

E

sr

R ′


31/35

• INL = I0 – jIm

=⏐

INL ⏐ ( cos ∅NL - jsin ∅NL )

• cos∅NL = Wph

Vph ⏐ INL ⏐

• Ro = Vph Xm = Vph

I0 Im



32/35

Lock Rotor TestLock Rotor Test

W a t t m

e t e r

AC

I

V

Current

Coil

V o l t a g e

C o i l

Xm

I NL

Rs jXs

Vph

IIN

I1

RO

IO Im R

r

jXr

Pair gap

s


33/35

Lock Rotor TestLock Rotor Test• In the Lock Rotor test, No Load Speed, n = 0

⇒ Slip, s = ns – 0 = 1, s = 1

ns

• Then Rr⇒⇒

Rr

s

•Apply Voltage to Variac, VLR = (10% - 25% ) Vph

• Since INL


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•Zeq = VLR / ILR

•cos∅LR

= WLR

VLR ⏐ ILR ⏐

• Zeq

=⏐

Zeq

⏐

{cos∅

LR

- jsin∅

LR

}

=⏐

Zeq⏐ cos∅LR - ⏐Zeq⏐ jsin ∅LR

Rs+ Rr Xs + Xr


35/35

•At Standstill Under d.c. conditions = 0

⇒ X= L

⇒ X = 0

•R1 & R2 can be measured using an ohmmeter over two stator

windings, which gives a value of Rs

• Rr = Zeq cos∅LR - Rs


Rs jXs

Ohmmeter

R1

R2

STATOR ROTOR

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