9.unit3new

R Ramanujam

UNIT III – MODELLING OF POLY PHASE INDUCTION

MACHINES (9 hours)

Induction machines – Equivalent circuit – Complete speed-torque

characteristics - Voltage and torque equations in static and rotating

reference frames – Analysis of steady state and dynamic operations –

Induction machine dynamics during starting and braking, accelerating

time, under normal conditions - Computer simulation

REFERENCE:

1. Paul C.Krause, OlegWasyzczuk, Scott D.Sudhoff “Analysis of

Electric Machinery and Drive Systems” IEEE Press, Second Edition,

2002.

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UNIT 3 except VRM,

UNIT 5

UNIT 5

UNIT 4

UNIT 2

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IM-1

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For proof see the document

titled “Proof of the

inductance for single-phase”

dated 16/11/2013 which

follows

Winding

symmetrical

to the left of

the line

θ

Fig.1

Sinusoidally

distributed

winding

n

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1. Proof of the inductance for single-phase magnetizing inductance

Added 15/11/2013 SIM-1.1

Ref: Ned Mohan, Advanced Course in Electric Drives, 2001

The field intensity, flux density and MMF at any angle θ from a chosen

reference is given by (See Fig.1)

( )

(1)

( )

(2)

( ) ( )

(3)

1

where, H, B and F are magnetic field intensity,

flux density and magnetomotive force (MMF).

Definition of single-phase stator magnetizing inductance

(4)

(5)

Note: = Ls,self = Lasasetc. Lm1-ph = Lms

1 The proofs of expreesions (1), (2) and (3) are given later.

c

b

a

g

θ

g

dθ

r dθ

Fig.2

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SIM 1.2

Expression Lm-1ph

It is obtained by equating energy stored

in the volume shown to

See Fig.2. In this figure “r” is the mean radius.

Energy density at an angle θ from

from a-axis

( ) ( )

( ) (6)

Energy stored in the elemental

volume at θ is

( ) (7)

Using the expression (2) for flux density

energy stored in the elemental

volume

(

)

( ) .

/ (

)

(

)

( ) (8)

Total energy stored in the volume is obtained by integrating Eq.(8)

around the stator periphery:

∫ .

/ (

)

(

)

( )

r

SIM 1.3

(

)

(

)

(9)

This should be equal to

.

Therefore,

(

)

(

) (10)

2. Stator-to-stator mutual inductance

This is obtained by exciting phase a by ia and measuring λb. Then

(11)

The only flux linking the phase b winding is the flux produced by ia. But

the phase b winding is displaced from phase a winding by120º.

Therefore

λb due to ia = λa-magnetizing (cos 120º)

(16)

But from Eqs.(4) and (5)

(17)

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SIM 1.4

Therefore, from Eqs.(110, (16) and (17) we see that

(18)

Identical expressions hold good for mutuals between phases b and c and

phases c and a.

3. Per-phase magnetizing inductance Lm

In this case all the three phases carry currents (compare with single-

phase case where only one phase is excited). Therefore,

(19)

If the currents are balanced

ia + ib + ic = 0 (20)

Using Eq.(20) in Eq.(19) we get

( )

(21)

Therefore,

(

)

(

) (22)

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SIM 1.5

Note: The single-phase magnetizing inductance does not include the

effect of mutual coupling from other two phases, whereas per-phase

magnetizing inductance does. This explains the factor 3/2.

4. Stator self inductance when all three phases are excited

Rotor is still open.

(23)

Or

(24)

The proofs for expressions (1), (2) and (3) follow after the introduction

of sinusoidally distributed winding and expression for conductor density.

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SIM 1.6

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SIM 1.7

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SIM 1.8

SIM 1.8

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SIM1.9

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SIM - 2

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SIM-3

Lms = (Ns/Nr)2 Lmr

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SIM - 4

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Page nos. SIM-7 and SIM-8 not required

SIM - 9

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TL is positive for load on

shaft

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FOR PROOF OF THESE EXPRESSIONS SEE

DOCUMENT TITLED “EXPRESSIONS FOR TORQUE”

dated 13/112013 WHICH FOLLOWS

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EXPRESSIONS FOR TORQUE

13/112013 T1 of 2

Recap: Power flow in a induction motor p.P-1of 1

We will use a simple and intuitive approach given by C.

Concordia1 to derive the expression for torque developed by the

motor.

Power input to the motor is given by

Pin = vasias+ vbsibs + vcsics (1)

In terms of q, d, 0 variables the power input is

Pin =(3/2)( vdsids+ vqsiqs + v0si0s) (2)

The relevant voltage equations for the stator quantities in terms

of the transformed variables (similar to balanced transmission

line see p.RTF-7) is given by the phasor diagram (given in same

page):

vqs = rs iqs + p λqs + ω λds

vds = rs ids + p λds - ω λqs

v0s = rs i0s + p λ0s (3)

Using Eq.(3) in Eq.(2) and simplifying we get

Pin =(3/2)[(rsids+pλds -λqs )ids+(rsiqs +pλqs+λds )iqs+(r0si0s+pλ0s)i0s]

1 C. Concordia, Synchronous Machines, Wiley, New York, 1951

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T2 of 2

Pin = (3/2)[(rsids2+rsiqs

2+rosios

2)+(pλdsids+pλqsiqs+pλ0si0s)+(λdsiqs-λqsids)]

(4)

Here comes the intuition of Concordia: The interpretation of

above equation is as follows.

The first term on the right hand side represents the stator ohmic

loss.

The second term on the right hand side represents rate of

change of magnetic energy.

The third term should therefore, represent the power converted

from electrical to mechanical, i.e, power transferred across the

air-gap to the rotor. Hence, interpretation of the equation itself

is

Power input = Stator ohmic losses

+ rate of change of stator magnetic energy

+ Power transferred across the air-gap.

Therefore,

Torque developed = Power transferred across air-gap

Mechanical speed

i.e, Te = (3/2)(λds iqs - λqs ids )/mech

= (3/2)(P /2) (λds iqs - λqs ids) (5)

where mech is the speed of the ARF on the mechanical side and

P is the number of poles.

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Q.E.D

OTHER EXPRESSIONS FOR TORQUE

14/11/2013

OT1 of 4

Other expressions for torque as given by P.C. Krause are:

.

/ .

/ (

) (1)

.

/ .

/ (

) (2)

The primes in the above expressions denote quantities referred

to the stator.

We will derive these expressions from Eq.(5) of the section

titled EXPRESSION FOR TORQUE dated 13/11/2013.

We will ignore the zero sequence flux linkages and currents.

The flux linkage equations are (change of notation M→LM)

( ) (3)

( ) (4)

( ) (5)

( ) (6)

Inserting the RHS Eqs.(3) and (4) in the factor (λds iqs - λqs ids)

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OT2 of 4

we see that

λds iqs - λqs ids

= ,( ) - [( ) ]

Expanding the square brackets and simplifying we get

λds iqs - λqs ids = (

) (7)

Inserting Eq.(7) in Eq.(5) of the section titled EXPRESSION

FOR TORQUE dated 13/11/2013, we get Eq.(1).

Eq.(2) is slightly more complicated. The essence of the

derivation is to replace the stator flux linkages and currents in

the factor (λds iqs - λqs ids) by corresponding primed rotor

quantities.

Defining

( ) (8)

we see from Eqs.(3) and (5) that

(9)

(10)

From Eqs.(9) and (10) we get

(11)

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OT3 of 4

Similarly,

(12)

Hence,

λds iqs - λqs ids

= ( ) (

)

= λdrʹ iqs - λqrʹ ids - (

) (13)

This expression is the result of replacing stator flux linkages by

the corresponding primed rotor flux linkages.

Now let us do the same thing for stator currents.

From Eqs.(5) and (6)

,

( ) - (14)

[

( ) ] (15)

Consider the factor (λdrʹ iqs - λqrʹ ids) in Eq.(13). Using the

expressions (14) and (15) for the currents, this factor becomes

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OT4 of 4

λdrʹ iqs - λqrʹ ids

=

[

( ) ]

,

( ) -

(16)

The above expression can be simplified as

λdrʹ iqs - λqrʹ ids =( )

(

)

which can be further simplified as

λdrʹ iqs - λqrʹ ids (

)

(

)

(17)

Consider the second term of Eq.(17). Using Eqs.(5) and (6) to

replace the primed rotor flux linkages we get

[ ( )] , ( )-

= ( ) (18)

Inserting Eq.(18) in Eq.(17) and the result in Eq.(13) we get

λds iqs - λqs ids =

(19)

Inserting Eq.(19) in Eq.(5) of of the section titled EXPRESSION

FOR TORQUE dated 13/11/2013, we get Eq.(2).

R Ramanujam

SIM – 12

COMPUTER SIMULATION OF INDUCTION MOTOR DYNAMICS

IN ARBITRARY REFERENCE FRAME

Introduction

For computer simulation, the terminal voltages of the stator

and stator-referred rotor voltages are taken as the input.

The computed quantities are currents, flux linkages,

torque developed by the motor and speed of the motor.

Required equations

1. Flux linkage equations:

( ) (CS-1)

( ) (CS-2)

(CS-3)

( ) (CS-4)

( ) (CS-5)

(CS-6)

Eqns.(CS-2) and (CS-5) can be put in the matrix form as

λqs (Lls + LM) LM iqs

= (CS-2A)

λqrʹ LM (Llrʹ +LM) iqrʹ

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SIM-12.1

Or in a compact form

[λqsrʹ] = [Lqsrʹ] [ iqsrʹ] (CS-2A)

Similarly, Eqns.(CS-1) and (CS-4) can be put in the matrix

form as

λds (Lls + LM) LM ids

= (CS-1A)

λdrʹ LM (Llrʹ +LM) iqrʹ

Or in a compact form

[λdsrʹ] = [Ldsrʹ] [ iqsrʹ] (CS-1A)

Note that [Lqsrʹ] = [Ldsrʹ]

2. Voltage equations:

For the purpose of computer simulation the voltage equations

given in p.SIM-5 are re-written as

pλqs = vqs - rs iqs - ω λds (CS-7)

pλds = vds - rs ids + ω λqs (CS-8)

pλ0s = v0s - rs i0s (CS-9)

pλqrʹ = vqrʹ - rrʹiqrʹ - (ω-ωr)λdrʹ (CS-10)

pλqrʹ = vdrʹ - rrʹidrʹ + (ω- ωr)λqs (CS-11)

pλ0rʹ = v0rʹ - rrʹi0rʹ (CS-12)

3. Torque equation (see Eq.(5)of p.T2-2): Te = (3/2)(P /2) (λds iqs - λqs ids) (CS-13)

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SIM-12.2

4. Mechanical (acceleration) equation:

(CS-14)

which can be re-written as

( ) (CS-14A)

State variables

We will choose the flux linkages λds, λqs etc. and ωmr as state

variables.

Knowns at t = 0

Apart from motor electrical and mechanical data of the motor

The state variables [λqd0s(0)], [λqd0rʹ(0)]

The currents [iqd0s(0)], [iqd0rʹ(0)]

Motor speed ωr(0) (or equivalently slip(0) and position

θr(0)

Speed of the arbitrary reference frame, ω and position θ(0)

We can conveniently choose θr(0) = θ(0) = 0 which means that

they are aligned along the stationary reference frame initially.

Inputs

Stator and rotor voltages in abc coordinates

Load torque versus speed curve

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SIM-12.3

Stepwise computations

1. Set t = 0

2. Evaluate the elements of transformation matrices [Ks] and

[Kr] which are functions of θ and β = θ- θr evaluated at t

respectively.

3. Transform the stator and referred voltages, currents and

flux linkages to arbitrary reference frame.

4. Compute the right-hand side of Eqs.(CS-7) to (CS-12) at

time t. For the first time step the initial values must be

taken for computation.

5. Integrate the first-order differential equations (CS-7) to

(CS-12) using a suitable numerical integration method

(fourth order Runge-Kutta method or modified Euler

method) to obtain the flux linkages [λqd0s], [λqd0rʹ]

at time t+Δt. For the first time step this will give flux

linkages at Δt. Note that the time is incremented at this step

and also, that the electrical system equations are solved

(this step) before mechanical system equation (Step 9).

6. Compute currents at t+Δt using Eqns.(CS-1A), (CS-2A),

(CS-3) and (CS-6):

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SIM-12.4

ids (Lls + LM) LM -1 λds

= (CS-1B)

idrʹ LM (Llrʹ +LM) λdrʹ

iqs (Lls + LM) LM -1 λqs

= (CS-2B)

iqrʹ LM (Llrʹ +LM) λqrʹ

i0s = λ0s/Lls (CS-3A)

i0rʹ = λ0rʹ/Llrʹ (CS-6A)

If necessary for output/plot purposes compute the abcs and

abcr’ currents using the respective inverse transformation

matrices.

7. Compute the electrical torque developed by the motor Te

using Eq.(CS-13) with flux linkages from Step 5 and

currents from Step 6.

8. Compute load torque knowing the motor speed ωr at time t .

9. Using the value of the electrical torque Te computed in

Step 7 and load torque computed in Step 8 compute the

right-hand side of Eq.(CS-14A) and integrate the same

using a suitable numerical integration method (fourth order

Runge-Kutta method or modified Euler method) to obtain

the motor speed ωr.

[vabcs]

[iabcs]

[vqd0s]

[vqd0s]

[vqd0rʹ] [vabcrʹ]

[iqd0rʹ] [iabcrʹ]

ω

ωr

Te

TL

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SIM-12.5 10. Increment time: t = t+Δt

11. Check whether maximum time is reached. If yes stop. Else go to

next step.

12. Compute the latest values of θ and β:

θ(t+Δt) = θ(t) + ωΔt

β(t+Δt) = β(t) + (ω – ωr)Δt

13. Go to Step 2.

Simulation Diagram

[ks]

[ks]-1

Computation of

Flux linkages

(Step 5)

Currents (Step 6)

Torque(Step 7)

[kr]

[kr]-1

Solution of

mechanical

system equation

for speed (Step 9)

PAGES SIM-13 AND SIM-14 ARE NOT REQUIRED

AND HENCE, ABSENT

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SIM-15

R Ramanujam

SIM-15.1

Typical simulation results and discussions

Torque-speed

characteristics during

free-acceleration (motor

starting)- 3 h.p motor

Torque-speed

characteristics during

free-acceleration (motor

starting)- 2250 h.p motor

These variations are at stator frequency;

why?

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SIM-15.2

Observations:

1. 3 hp motor is a high-slip machine rated torque

developed at a speed considerably less than synchronous

speed. 2250 hp motor is low-slip machine rated torque

developed closer to synchronous speed.

2. For the large-hp (2250 hp) machine the rotor speed

overshoots synchronous speed. Speed response is

characterized by decaying oscillations about the final

operating point. In the case low-hp (3 hp) machine speed is

highly damped and final operating point reached is attained

without any oscillations.

Variables observed from different reference frames

Results for free-acceleration characteristics

1. Stationary reference

q axis stator voltage (top) and current; d

axis quantities similar with phase

changed. All in p.u

Torque (top) and speed in p.u.

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SIM-15.3

2. Reference frame fixed in the rotor 3. Synchronously rotating

ref.frame

The torque and speed curves will be identical to those obtained for

stationary reference frame. Why?



changed. All in p.u. The variations are at

slip frequency which varies from a high

value initially and becomes small as the

rotor speed approaches steady state.



changed. All in p.u. The quantities are

unidirectional.

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SIM-16 Added 2/2/2014

BALANCED STEADY STATE OPERATION

1. Voltage Equation

Recall stator voltage eqns in ARF:

;

For balanced steady-state operation in synchronously rotating

reference frame (SRF) zero-sequence quantities are zero and pλ

terms are also zero. SRF is also sometimes known as Kronʹs

reference frame.

Using upper case for steady-state the voltage eqns become

* (

)+

(20) i.e

(

)

Similarly,

(

)

Forming the phasor

we get

(

) (

)

[(

) (

)]

From eqns.(6) and (7) p.T-4 of 13, Unit 1- Part 2 notes for

steady state we see that the above eqn. becomes (after cancelling

√2 factor throughout, see Eq.(15) of p.T-8 of 13))

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SIM-17 Added 2/2/2014

( ) (21)

The rotor voltage eqns in ARF:

( )

(22)

where s = slip in p.u of ω

( )

(23)

For balanced steady-state operation in the SRF above eqns.

become

(

) (24)

(

) (25)

Forming the phasor and relating it to phasor in SRF we get

( )

Or

( ) (26)

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SIM-17.1

Note:

To an observer on the stator (stationary reference frame)

the currents, voltages and flux linkages (which are at slip

frequency) in the rotor appear to have a frequency equal to

synchronous frequency.

For example, the rotating mmf produced by slip frequency

rotor currents rotate at slip speed ωe – ωr w.r.t to rotor. But

rotor carries the rotating mmf at the speed ωr w.r.t

stationary reference frame. Hence, w.r.t to stationary

reference frame the rotating mmf of the rotor rotates at

ωe-ωr+ωr = ωe

As a corollary, for an observer on the synchronously

rotating reference frame, the rotor quantities appear to be

constants (since w.r.t stationary reference frame they vary

at frequency ωe).

⁄

= 0 for balanced

conditions

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SIM-18

2. Equivalent Circuit

The equivalent circuit for steady-state operation from Eq.(26) is

shown below:

3. Expression for torque in terms of current phasors

In terms of qd0 quantities in synchronously rotating reference

frame (SRF), the instantaneous power is given by

(

)

For steady-state balanced conditions

(

) (27)

where * denotes conjugation and

etc.

In SRF we have the following voltage equations (from Unit1-

Part2)

vqse = rs iqs

e + pλqse + ωe λds

e

vdse = rs ids

e + pλdse - ωe λqs

e

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SIM-19

For steady-state balanced conditions (setting transformer emf

terms to zero) the voltage equations are

Vqse = rs Iqs

e + ωe Λdse (28)

Vdse = rs Ids

e – ωe Λqse (29)

The flux linkage equations are:

Λqse = Lls Iqs

e + M(Iqse+Iqrʹ

e) (30)

Λdse = Lls Iqs

e + M(Idse+Idrʹ

e) (31)

Forming the phasor

we get

(

)

Substituting for the flux linkages from eqns.(30) and (31) we get

(

) (32)

where XM= jωeM.

Therefore,

(

)

| |

|

|

| |

( )

(33)

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SIM-20

In Eq.(33), the first term is proportional to stator copper loss, the

second term is proportional to reactive power “lost” in the

magnetic field and the last term is proportional to power

transferred across the air-gap.

Actual power transferred across the air-gap is given by

{

( )

} (34)

Hence, torque is given by

{

( )

} (35)

But

√ and

√ (36)

(see p.T-8 of 13, Eq.(15))

Using Eq.(36) in Eq.(35) we get

{

( )

} (37)

Pin,stator

Pcu losses,stator

Pair-gap

Pcu losses,rotor

Pe,rotor

Plosses,friction and windage

Pshaft

a

b

Pair-gap Is

Fig.1

Fig. 2 Note: Xs and

Xr are leakage

reactances.

Ir Pin, stator

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PF-1 of 1

POWER FLOW IN AN INDUCTION MOTOR

19/11/2013

See Fig.1(no-load core losses not included).

The symbols are self-explanatory.

Note: Ignoring friction and windage losses, Pe,rotor is the power

available at the shaft. Hence,

The per-phase equivalent circuit of a squirrel-cage induction

motor is given in Fig. 2

Note: The rotor electrical parameters and current are referred to stator (prime is absent-sorry!)

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OR-1 of 2

OPERATING REGIONS OF AN INDUCTION MACHINE Ref: 1. R Krishnan, Electric Motor Drives, Modelling, Analysis and Control,

Prentice-Hall, 2001

(2) V.Venikov, Transient processes in Electrical Power Systems, MIR Publishers,

Moscow, 1977

Normal operation: motor, positive slip

Braking: rotor rotates in a direction opposite to that of the stator

(rotating) magnetic field; slip >1. The braking action brings the

motor to standstill.

ωe

ωr

ωe

ωr

ωr ωe

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OR-2 of 2

Generating: the rotor is driven ahead of stator magnetic field;

slip is <0

direction of induced emf phase reversed

torque – slip reversed in sign

breakdown torque much higher (see dashed lines in the

figure) because mutual flux linkages are strengthened due

to reduction in motor impedance (boost in magnetizing

current and hence, an increase in mutual flux linkages and

torque)

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ST-1 of 2

EXPRESSION DESCRIBING TORQUE-SLIP

CHARACTERISTIC DURING STEADY-STATE

From Fig.2 (see p.PF-1 of 1)

(per phase) (1)

(per phase) (2)

( )

(3)

Hence, the torque developed is

(4)

where is the angular speed of the rotor on the mechanical

side. The angular speed on the electrical and mechanical sides

are related as

(5)

But

( ) (6)

Using Eq.(6) in Eq.(5) and the resulting expression in Eq.(4) we

get

( ) (7)

From Eqs.(1), (3), (4) and (7) we get

.

/

(8)

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ST-2 of 2

We will use Eq.(1). To calculate current Ir we can simplify

matters by applying Thevenin’s theorem to the to find the

equivalent to the left of the line ab in Fig.2 (p. PF-1 of 1).

Thevenin impedance:

( ) (9a)

(say) (9b)

Thevenin voltage (voltage divider)

,( ) ( ) -

(10)

With the circuit to the left of the line ab in Fig.2 replaced by

Thevenin voltage behind Thevenin impedance, we have for the

current

[.

/

( ) ]

(11)

From Eqs.(1), (8), (10) and (11) we get, considering all the three

phases

( ) .

/ .

ω /

[.

/

( ) ],( ) ( ) - (12)

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