BASICS OF INDUCTION

MACHINES

Dr. Monalisa Pattnaik

Department of Electrical Engineering,

National Institute of Technology, Rourkela,
Odisha, India, 769008.
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TYPES OF WIND TURBINE GENERATORS (WTG)

Electric Generators
used in WT

Asynchronous Synchronous
Generators Generators

Doubly-fed Squirrel Cage Permanent

Induction Induction Wound Rotor
Magnet Rotor
Generator Generator

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Principle of operation:

 When the balanced stator windings, wound for p pole pairs and displaced
in space by 120 electrical degrees
 excited by a balanced three-phase sinusoidal supply, the current in the three
phases of the stator winding (also displaced in time-phase by 120 electrical
degrees)
 creates a rotating magnetic field of constant strength. The sinusoidally
distributed magnetic field moves around the air gap at a speed given by:
120𝑓𝑓
𝑁𝑁𝑠𝑠 =
𝑝𝑝
 which is known as the synchronous speed of the induction machine.
 The rotating air-gap field induces emf in the rotor windings, and these
emfs in turn produce currents in the short-circuited rotor bars or windings.
 The interaction between these rotor currents and the air-gap flux creates a
torque which, by Lenz's law, acts in the direction of the rotating field.
 For the production of torque, a speed difference between the rotor and the
air-gap field is necessary, so that the rotor flux linkage can change.
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 A very important variable in the performance analysis of an induction
machine is its per-unit slip s, which is expressed :
𝑁𝑁𝑠𝑠 − 𝑁𝑁𝑟𝑟
𝑠𝑠 =
𝑁𝑁𝑠𝑠
 where 𝑁𝑁𝑟𝑟 is the rotor speed in rps. For motoring operation, 𝑁𝑁𝑟𝑟 < 𝑁𝑁𝑠𝑠 ,
i.e., 0 < s < l.
 As the relative speed between the rotor and the air-gap field is (𝑁𝑁𝑠𝑠 − 𝑁𝑁𝑟𝑟 ),
current will be induced in the rotor at the slip frequency sf
 The slip frequency rotor currents create a rotor mmf, the space fundamental
component of which moves past the rotor at the slip speed 𝑁𝑁𝑠𝑠 − 𝑁𝑁𝑟𝑟 .
 The rotor-speed being 𝑁𝑁𝑟𝑟 , the rotor mmf and the air-gap field are stationary
with respect to each other.
 The torque magnitude depends on the space angle δ, also called the torque
angle, between these two waves in addition to their magnitudes.
 The rotor mmf lags behind the air-gap field. It can be shown that this torque
angle δ equals 90°+𝜙𝜙2 , 𝜙𝜙2 being the rotor power-factor angle at the slip
frequency sf.

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Steady-state Equivalent Circuit Model:
 The resultant air-gap flux ∅𝑚𝑚 is set up by the combined action of the stator
and the rotor mmfs.
 This synchronously rotating flux induces a counter emf 𝐸𝐸1 in the stator phase
winding.
 The stator terminal voltage 𝑉𝑉1 differs from this counter emf 𝐸𝐸1 by the stator
leakage impedance drop.
 The stator emf phasor equation is then:𝑉𝑉1 = 𝐸𝐸1 + 𝐼𝐼1 𝑅𝑅𝑠𝑠 + 𝑗𝑗𝑋𝑋𝑙𝑙𝑙𝑙
 where, 𝐸𝐸1 = 2π𝑓𝑓𝑓𝑓𝑤𝑤1 𝑇𝑇1 ∅𝑚𝑚

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The stator circuit model:
 Just as in the case of a transformer primary winding, the stator winding
current 𝐼𝐼1 can be resolved into an exciting component 𝐼𝐼𝑂𝑂 and compensating
load component 𝐼𝐼2′ .
 The load component current 𝐼𝐼2′ counteracts the rotor mmf, thereby
demanding power from the source.
 The exciting component 𝐼𝐼𝑂𝑂 can be resolved into a core-loss component 𝐼𝐼𝑐𝑐 , in
phase with the stator induced emf 𝐸𝐸1 and a magnetizing component 𝐼𝐼𝑚𝑚 ,
lagging behind the induced emf 𝐸𝐸1 by 90°.
 It is this magnetizing component which sets up the air-gap flux ∅𝑚𝑚 .
 The stator emf phasor equation is then:𝑉𝑉1 = 𝐸𝐸1 + 𝐼𝐼1 𝑅𝑅𝑠𝑠 + 𝑗𝑗𝑋𝑋𝑙𝑙𝑙𝑙
 where, 𝐸𝐸1 = 2π𝑓𝑓𝑓𝑓𝑤𝑤1 𝑇𝑇1 ∅𝑚𝑚
The rotor circuit model:
 The air-gap flux wave induces a slip frequency rotor voltage 𝐸𝐸2𝑠𝑠 given
by: 𝐸𝐸2𝑠𝑠 = 2π𝑠𝑠𝑓𝑓𝑓𝑓𝑤𝑤2 𝑇𝑇2 ∅𝑚𝑚

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 With respect to Fig., the rotor current per phase may be expressed as:
𝐸𝐸2𝑠𝑠
𝐼𝐼2𝑠𝑠 =
𝑅𝑅𝑟𝑟 + 𝑗𝑗𝑋𝑋𝑙𝑙𝑟𝑟
 It may be noted here that the current 𝐼𝐼2𝑠𝑠 flows at slip frequency. Combining
𝑠𝑠𝐸𝐸1
above Eqns: 𝐼𝐼2𝑠𝑠 =
𝑚𝑚 (𝑅𝑅 +𝑗𝑗𝑋𝑋 )
1 𝑟𝑟 𝑙𝑙𝑙𝑙
 where 𝑚𝑚1 is the ratio of the effective stator turns (𝑘𝑘𝑤𝑤1 𝑇𝑇1 ) to the effective rotor
turns (𝑘𝑘𝑤𝑤2 𝑇𝑇2 ). 𝐼𝐼2𝑠𝑠 can also be rewritten in the form:
𝐼𝐼2𝑠𝑠 𝐸𝐸1
= 𝑅𝑅′
𝑚𝑚1 𝑟𝑟 +𝑗𝑗𝑋𝑋 ′
𝑠𝑠 𝑙𝑙𝑙𝑙

Steady-state per-phase circuit models of a three-phase induction motor: (a)

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rotor circuit model at slip frequency, (b) rotor circuit model at stator frequency
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 where the stator-referred rotor resistance 𝑅𝑅𝑟𝑟′ = 𝑚𝑚2 𝑅𝑅𝑟𝑟 and the stator-referred
′
rotor leakage reactance 𝑋𝑋𝑙𝑙𝑟𝑟 = 𝑚𝑚2 𝑋𝑋𝑙𝑙𝑟𝑟 .
 From the mmf balance between the ideal stator circuit and the rotor, the
𝐼𝐼
stator-referred rotor current 𝐼𝐼2′ = 2𝑠𝑠 .
𝑚𝑚1
 𝐼𝐼2′ arises out of the line-frequency voltage 𝐸𝐸1 in a circuit with impedance
𝑅𝑅𝑟𝑟′ ′
( + 𝑗𝑗𝑋𝑋𝑙𝑙𝑙𝑙
𝑠𝑠
)
 The actual rotor circuit in Fig. (a) may be replaced by the stator-referred
equivalent circuit shown in Fig. (b) at the line frequency.

(c) complete stator-referred T-form equivalent circuit 8

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 In both the circuits, the current maintains the same phase relation with
respect to the corresponding input voltage. The power-factor angle of the
rotor circuit is given by:
′
𝑠𝑠𝑠𝑠𝑙𝑙𝑙𝑙 𝑋𝑋𝑙𝑙𝑙𝑙
∅2 = tan−1 = tan−1 𝑅𝑅′𝑟𝑟�
𝑅𝑅𝑟𝑟
𝑠𝑠
 Since the input terminal voltage in Fig. (b) is the same as the output terminal
voltage, the two circuits may be connected in tandem to yield a complete
equivalent circuit for the induction motor, as shown in Fig. (c).
 This circuit is generally known as the T-form steady-state equivalent circuit.

(c) complete stator-referred T-form equivalent circuit 9

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Power and Torque relations:
 The equivalent circuit shows that the total power transferred across the air
gap, i.e. the power input to the rotor, is:
′ ′2 2 ′
′
𝐼𝐼2 𝑅𝑅𝑟𝑟 𝑠𝑠𝐸𝐸1 𝑅𝑅𝑟𝑟
𝑃𝑃𝑎𝑎𝑎𝑎 = 3𝐸𝐸1 𝐼𝐼2 cos ∅2 = 3 = 3 ′2 ′ 2
𝑠𝑠 𝑅𝑅𝑟𝑟 + 𝑠𝑠 2 𝑋𝑋𝑙𝑙𝑟𝑟
 Under actual operating conditions, the electrical power consumed in the rotor
circuit is the rotor copper loss, given by:
2 ′2 ′
𝑃𝑃𝑐𝑐𝑐𝑐𝑐 = 3𝐼𝐼2 𝑅𝑅𝑟𝑟 = 3𝐼𝐼2 𝑅𝑅𝑟𝑟 = 𝑠𝑠𝑃𝑃𝑎𝑎𝑎𝑎
 The effective mechanical power is then given as:
2 1 − 𝑠𝑠
𝑃𝑃𝑚𝑚 = 𝑃𝑃𝑎𝑎𝑎𝑎 − 𝑃𝑃𝑐𝑐𝑐𝑐2 = 3𝐼𝐼2′ 𝑅𝑅𝑟𝑟′ = (1 − 𝑠𝑠)𝑃𝑃𝑎𝑎𝑎𝑎
𝑠𝑠
 The power expressions show that:𝑃𝑃𝑎𝑎𝑎𝑎 : 𝑃𝑃𝑐𝑐𝑐𝑐2 : 𝑃𝑃𝑚𝑚 = 1: 𝑠𝑠: 1 − 𝑠𝑠
 The electromagnetic torque developed by an induction motor is:
′ 2 ′
𝑃𝑃𝑚𝑚 𝐼𝐼2 𝑅𝑅𝑟𝑟
1 𝑃𝑃𝑎𝑎𝑎𝑎
𝑇𝑇𝑒𝑒 = =3 =
2𝜋𝜋𝑁𝑁𝑟𝑟 𝑠𝑠 2𝜋𝜋𝑁𝑁𝑠𝑠 2𝜋𝜋𝑁𝑁𝑠𝑠
 The power flow across the air gap is a measure of the electromagnetic torque
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generated.
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(a) stator-referred T-form equivalent circuit (b) Phasor diagram

 If the power components in Eqns. and are to be emphasized in the equivalent

circuit of Fig. (c).
 The power dissipation in the slip-dependent resistance is the developed
mechanical power per phase.
 Fig. (b) presents the corresponding phasor diagram.
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Torque-speed characteristics:
 Previous torque equation is not convenient for examining the torque-speed
characteristics of an induction machine, as 𝐼𝐼2′ itself depends on s.
 Thevenin's theorem permits the replacement of the circuit in Fig. (c) by that in
Fig. 3.4 with an equivalent voltage source 𝑉𝑉𝑇𝑇𝑇 .
 With no significant loss in accuracy, 𝑅𝑅𝑐𝑐 of Fig. (c) is ignored for the sake of
simplicity.

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 The 𝑉𝑉𝑇𝑇𝑇 and new impedance elements in Thevenin's equivalent circuit are:
𝐸𝐸1
𝐼𝐼2′ =
𝑅𝑅𝑟𝑟′ ′
+ 𝑗𝑗𝑋𝑋
𝑠𝑠 𝑙𝑙𝑙𝑙
𝑘𝑘𝑘𝑘1
= ′
𝑅𝑅𝑟𝑟 ′
+ 𝑗𝑗𝑋𝑋
𝑃𝑃𝑚𝑚
2
𝐼𝐼2′ 𝑅𝑅𝑟𝑟′ 1 𝑠𝑠 𝑙𝑙𝑙𝑙
𝑇𝑇𝑒𝑒 = =3
2𝜋𝜋𝑁𝑁𝑟𝑟 𝑠𝑠 2𝜋𝜋𝑁𝑁𝑠𝑠

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Torque-speed characteristics:
 A negative value of the slip implies running the machine above
synchronous speed in the direction of the rotating field.
 As the torque direction is simultaneously reversed (opposite to the
direction of the rotating field), the machine has to be driven by a source
of mechanical power to counteract the opposing torque.
 In the process the machine acts as a generator feeding power to the
source.
 For s > 1, the machine runs in a direction opposite to that of the rotating
field and the internal torque.
 In order to sustain this condition, the machine should also be driven by
a mechanical power source.
 This mode of operating the induction machine is known as plugging,
and is equivalent to an electrical braking method.

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Torque and power relations:
 For a constant input voltage, if the ratio of the total rotor circuit
resistance to the slip, i.e., ( 𝑅𝑅𝑟𝑟 + 𝑅𝑅𝑥𝑥 ) /s, remains constant, the
developed torque 𝑇𝑇𝑒𝑒 would also remain constant.
 As this makes the effective circuit impedance constant, the stator
input current also remains constant.

where 𝑅𝑅𝑥𝑥′ is the stator-referred value of the rotor external resistance 𝑅𝑅𝑥𝑥

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Induction Machine Power-speed Characteristics

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Modified Equivalent Circuit of IM (Inverse-г model) :
 No significant error is introduced in the prediction of the performance of
the machine if the shunt resistance, representing the iron loss in the
stator, is ignored in favour of simplicity of manipulating the equations.
 Applying Kirchhoff's circuit laws to the equivalent circuit, the following
equations can be written:
 𝑉𝑉1 = 𝐸𝐸1 + 𝐼𝐼1 𝑅𝑅𝑠𝑠 + 𝑗𝑗𝑋𝑋𝑙𝑙𝑙𝑙
= 𝐼𝐼1 𝑅𝑅𝑠𝑠 + 𝑗𝑗𝑋𝑋𝑙𝑙𝑙𝑙 +𝑗𝑗(𝐼𝐼1 − 𝐼𝐼2′ )𝑋𝑋𝑚𝑚
′
′ 𝑅𝑅𝑟𝑟 ′
0=𝐼𝐼2 + 𝑗𝑗𝑋𝑋𝑙𝑙𝑙𝑙 -𝑗𝑗(𝐼𝐼1 − 𝐼𝐼2′ )𝑋𝑋𝑚𝑚
𝑠𝑠
 Upon rearranging the terms and putting
′
 𝑋𝑋𝑠𝑠 = 𝑋𝑋𝑙𝑙𝑠𝑠 + 𝑋𝑋𝑚𝑚 ,𝑋𝑋𝑟𝑟′ = 𝑋𝑋𝑙𝑙𝑙𝑙 + 𝑋𝑋𝑚𝑚
 𝑉𝑉1 = 𝐼𝐼1 𝑅𝑅𝑠𝑠 + 𝑗𝑗𝑋𝑋𝑠𝑠 − 𝑗𝑗𝐼𝐼2′ 𝑋𝑋𝑚𝑚
= 𝐼𝐼1 𝑅𝑅𝑠𝑠 + 𝑗𝑗𝑋𝑋𝑙𝑙𝑙𝑙 +𝑗𝑗(𝐼𝐼1 − 𝐼𝐼2′ )𝑋𝑋𝑚𝑚
′
′ 𝑅𝑅𝑟𝑟
0=𝐼𝐼2 + 𝑗𝑗𝑋𝑋𝑟𝑟′ − 𝑗𝑗𝑗𝑗1 𝑋𝑋𝑚𝑚
𝑠𝑠
 Here 𝑋𝑋𝑠𝑠 and 𝑋𝑋𝑟𝑟′ may be recognized as the self-inductance of the stator
winding and the stator-referred self-inductance of the rotor winding
respectively.
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Modified Equivalent Circuit of IM (Inverse-г model) :
 Let the stator-referred rotor current 𝐼𝐼2′ be replaced by a new variable 𝐼𝐼2𝑒𝑒 related
to 𝐼𝐼2′ by the following relation:
𝑋𝑋𝑚𝑚
 𝐼𝐼2′ = 𝐼𝐼2𝑒𝑒 → 𝑉𝑉1 = 𝐼𝐼1 𝑅𝑅𝑠𝑠 + 𝑗𝑗𝑋𝑋𝑠𝑠 − 𝑗𝑗𝑗𝑗2′ 𝑋𝑋𝑚𝑚
𝑋𝑋𝑟𝑟′
𝑋𝑋 2
 𝑉𝑉1 = 𝐼𝐼1 𝑅𝑅𝑠𝑠 + 𝑗𝑗𝑋𝑋𝜎𝜎 + 𝑗𝑗(𝐼𝐼1 − 𝐼𝐼2𝑒𝑒 ) ′𝑚𝑚
𝑋𝑋𝑟𝑟

Stator-referred equivalent circuit: Inverse-г model

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𝑋𝑋𝜎𝜎 is the familiar expression for the transient reactance of the induction machine.
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Modified Equivalent Circuit of IM (Modified-T form) :
 The circuit models of synchronous machines and dc machines contain a
rotational emf term.
 Even the generalized theory of electrical machines indicates the
presence of a rotational emf in the rotor circuit of an induction machine.
 The conventional circuit model of an induction motor shows the
presence of a negative resistance in the rotor circuit, implying generating
action at a speed above the synchronous speed.

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Modified Equivalent Circuit of IM (Modified-T form) :
 However, the absence of any rotational emf in the rotor branch complicates
the explanation of the behavior of the rotor of an induction machine that
serves as a source of power during its generating action, particularly, in the
self-excited mode.
 The concept of a negative resistance seems to offer a computational
advantage rather than a convincing explanation.

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Modified Equivalent Circuit of IM (Modified-T form) :
 However, the absence of any rotational emf in the rotor branch
complicates the explanation of the behavior of the rotor of an induction
machine that serves as a source of power during its generating action,
particularly, in the self-excited mode.
 The concept of a negative resistance seems to offer a computational
advantage rather than a convincing explanation.

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Effect of Rotor-injected Emf-Slip Power Recovery Scheme:

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Effect of Rotor-injected Emf-Slip Power Recovery Scheme:

 Fig. shown below as the modified version of the per-phase stator-referred

conventional equivalent circuit of the induction motor with injected emf in the
rotor.
 Such an induction machine is also known as a doubly fed induction machine
(DFIM) because of the two power sources employed.
 Since motoring convention has been followed, 𝑃𝑃𝑎𝑎𝑎𝑎 and 𝑃𝑃𝑚𝑚 will be negative in the
generating mode.
 𝑃𝑃2 has been considered positive for the power absorbed by the auxiliary
source, i.e., for the power flowing out of the slip-ring terminals.

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Mode I-Sub-synchronous motoring operation:
 In this mode, 0 < s < 1 and 𝑃𝑃𝑚𝑚 > 0. Consequently, as 𝑃𝑃𝑚𝑚 = (1 − 𝑠𝑠)𝑃𝑃𝑎𝑎𝑎𝑎 ,
𝑃𝑃𝑎𝑎𝑎𝑎 > 𝑃𝑃𝑚𝑚 , 𝑃𝑃𝑐𝑐𝑐𝑐2 + 𝑃𝑃2 = 𝑠𝑠𝑃𝑃𝑎𝑎𝑎𝑎 , is positive.
 For constant-torque operation, 𝑃𝑃𝑎𝑎𝑎𝑎 is constant.
 An increase in 𝑃𝑃𝑐𝑐𝑐𝑐2 +𝑃𝑃2 , 𝑃𝑃2 raises the value of s, which implies a drop
in the speed.

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Mode II-Super-synchronous motoring operation:
 In the super-synchronous region, the rotor speed is greater than the
synchronous speed, i.e., s < 0 and 𝑃𝑃𝑚𝑚 > 0.
 As 𝑃𝑃𝑚𝑚 = (1 − 𝑠𝑠)𝑃𝑃𝑎𝑎𝑎𝑎 , 𝑃𝑃𝑎𝑎𝑎𝑎 < 𝑃𝑃𝑚𝑚 and positive, 𝑃𝑃𝑐𝑐𝑐𝑐2 + 𝑃𝑃2 = 𝑠𝑠𝑃𝑃𝑎𝑎𝑎𝑎 , must be
negative.
 As 𝑃𝑃𝑐𝑐𝑐𝑐2 is always +ve, 𝑃𝑃2 must be negative. Therefore, power must be
fed into the slip-ring terminals from the auxiliary source.
 For constant-torque operation, 𝑃𝑃𝑎𝑎𝑎𝑎 is constant.
 An increase in the input to the rotor, 𝑃𝑃𝑚𝑚 can be increased, with increase
in machine speed.

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Mode III-Sub-synchronous generating operation:
 For generating action 𝑃𝑃𝑚𝑚 is negative, and as the rotor speed is less than
the synchronous speed ( 0 < s < 1), 𝑃𝑃𝑎𝑎𝑎𝑎 is negative.
 As 𝑃𝑃𝑚𝑚 = (1 − 𝑠𝑠)𝑃𝑃𝑎𝑎𝑎𝑎 , 𝑃𝑃𝑎𝑎𝑎𝑎 > 𝑃𝑃𝑚𝑚 , 𝑃𝑃𝑐𝑐𝑐𝑐2 + 𝑃𝑃2 = 𝑠𝑠𝑃𝑃𝑎𝑎𝑎𝑎 , is negative.
 As 𝑃𝑃𝑐𝑐𝑐𝑐2 is always +ve, 𝑃𝑃2 should be made sufficiently negative by
injecting power into the rotor circuit in order to make the rotor electrical
power 𝑠𝑠𝑃𝑃𝑎𝑎𝑎𝑎 negative.
 The net electrical power flowing into the grid is, therefore, 𝑃𝑃1 − 𝑃𝑃2 .

27
 INDUCTION MACHINES
Mode IV-Super-synchronous generating operation:
 For generating action 𝑃𝑃𝑚𝑚 is negative, and as the rotor speed is more than
the synchronous speed s < 0, 𝑃𝑃𝑎𝑎𝑎𝑎 is negative.
 As 𝑃𝑃𝑚𝑚 = (1 − 𝑠𝑠)𝑃𝑃𝑎𝑎𝑎𝑎 , 𝑃𝑃𝑎𝑎𝑎𝑎 < 𝑃𝑃𝑚𝑚 , 𝑃𝑃𝑐𝑐𝑐𝑐2 + 𝑃𝑃2 = 𝑠𝑠𝑃𝑃𝑎𝑎𝑎𝑎 , is positive.
 The remaining surplus energy 𝑠𝑠𝑃𝑃𝑎𝑎𝑎𝑎 is returned via the rotor circuit to the
grid after providing for the secondary losses 𝑃𝑃𝑐𝑐𝑐𝑐2 ·
 Mathematically, 𝑠𝑠𝑃𝑃𝑎𝑎𝑎𝑎 > 0, both s and 𝑃𝑃𝑎𝑎𝑎𝑎 being negative.

28
 INDUCTION MACHINES
Double-output System with a Current Converter:
 The provision for bidirectional flow of power through the rotor circuit can
be achieved by the use of a slip-ring induction motor with an ac-dc-ac
converter connected between the slip-ring terminals and the utility grid.
 Figure shown below presents the main components of the solid-state
system for the controlled flow of slip power at variable speed through
current converters.

29
 INDUCTION MACHINES
Double-output System with a Current Converter:
 The intermediate smoothing reactor is needed to maintain current continuity and
reduce ripples in the link circuit.
 For the transfer of electrical power from the rotor circuit to the supply, converters
I and II are operated, respectively, in the rectification and inversion modes.
 On the other hand, for power flow in the reverse direction, converter-II acts as a
rectifier and converter-I as an inverter.
 The step-down transformer between converter-II and the supply extends the
control range of the firing delay angle of α2 converter-II.
 The firing delay angle α1 of converter-I on the rotor side controls the phase
difference between the injected rotor phase voltage and the rotor current, while
the delay angle α2 of converter-II on the line side dictates the injected voltage
into the rotor circuit.
 Line-commutated converters cannot generate leading VAR, and so, for
maximization of the power output, α1 should be set at 00 (rectification mode) in
the super-synchronous region to draw power out of the rotor.
 At 1800 (inversion mode) in the sub-synchronous region to inject power at the
slip frequency into the rotor circuit.
 Power flow characteristics can be studied using ac as well as dc equivalent
circuits.
30
 INDUCTION MACHINES
Double-output System with a Current Converter:
 A reasonable estimate for the required variation in α2 as a function of slip can be
obtained by considering the dc voltage balance between the two sides of the
smoothing reactor in the antiparallel bridge network.
 To derive the equivalent circuits and analyze this system, we make the following
assumptions:
(a)The magnetizing current and iron loss are neglected.
(b)The inverter-side transformer is assumed to be ideal.
(c)Commutation of the switching devices is assumed to be instantaneous and
the device losses are neglected.
(d) The harmonic effects are ignored.

(i) AC equivalent circuit: Neglecting stator and rotor leakage impedance drops,
the average voltage output of converter-I at a slip s is given by

where 𝑉𝑉2 is the slip-ring voltage (per phase) at standstill.

For converter-II, the average voltage output is given by

𝑚𝑚2 being the turn ratio of the step-down transformer between converter-II and the
supply.
31
 INDUCTION MACHINES
Double-output System with a Current Converter:
 For dc voltage balance, neglecting the resistance drop in the dc-link smoothing
inductor
 This yields

 Where 𝑉𝑉2′ is the stator-referred, slip-ring open-circuit voltage.

 Above equation can be used for the evaluation of 𝛼𝛼2 .
 Under the assumption of negligible stator impedance drops, 𝑉𝑉2′ equals 𝑉𝑉1 and Eqn
becomes

 In fact, the presence of the dc-link circuit resistance demands

 Neglecting the losses in the semiconductor switches, the slip power, i e., the rotor
electrical power, is partly dissipated in the dc-link and rotor resistances, and the
rest is fed back to the supply system through converter-II.
 The power input to converter-II from dc-link side (inverter operation) is:

32
DOUBLE-OUTPUT SYSTEM WITH A CURRENT CONVERTER:

33
 INDUCTION MACHINES
Double-output System with a Current Converter:
 For 1200 conduction of each device in the converters, the fundamental rms
component 𝐼𝐼2 of the ac-side current of the converters rotor or transformer
secondary) and the dc-link current are related by 𝐼𝐼2 = 𝐼𝐼𝑑𝑑 6 /𝜋𝜋
3𝑉𝑉
 Power fed back to the supply by the rotor is obtained as:𝑃𝑃2 = − 𝑚𝑚 1 𝐼𝐼2 cos 𝛼𝛼2
2
3𝑚𝑚1
 Referring the current 𝐼𝐼2′ to the stator side, we get:𝑃𝑃2 = 𝑉𝑉1 𝐼𝐼2′ cos ∅2
𝑚𝑚2
 where ∅2 is the supplementary of the delay angle 𝛼𝛼2 of converter, ∅2 = 𝜋𝜋 − 𝛼𝛼2
 The rms value of the quasi-square wave rotor current is: 𝐼𝐼𝑟𝑟 = 𝐼𝐼𝑑𝑑 2/3
 The total secondary circuit copper loss:
 𝑃𝑃𝑐𝑐𝑐𝑐2 = 3𝐼𝐼2 2 𝑅𝑅𝑟𝑟 + 𝐼𝐼𝑑𝑑 2 𝑅𝑅𝑑𝑑 = 3𝐼𝐼2 2 (𝑅𝑅𝑟𝑟 + 0.5 𝑅𝑅𝑑𝑑 )
 The rotor rrns current consists of the fundamental rms component and the higher
harmonic rms components.
 Assuming that the torque is produced by the fundamental component of the rotor
current, mechanical power can be expressed as:
1−𝑠𝑠 2 3𝑉𝑉1
 𝑃𝑃𝑚𝑚 = (𝑃𝑃𝑐𝑐𝑐𝑐2 𝑑𝑑𝑑𝑑𝑑𝑑 𝑡𝑡𝑡𝑡 𝐼𝐼2 + 𝑃𝑃2 ) = 3 𝐼𝐼2 (𝑅𝑅𝑟𝑟 + 0.5 𝑅𝑅𝑑𝑑 ) + 𝐼𝐼 cos ∅2
𝑠𝑠 𝑚𝑚2 2
 The air-gap power is:𝑃𝑃𝑎𝑎𝑎𝑎 = 𝑃𝑃𝑐𝑐𝑐𝑐2 + 𝑃𝑃2 + 𝑃𝑃𝑚𝑚

34
 INDUCTION MACHINES
Double-output System with a Current Converter:
 Substituting the expressions for 𝑃𝑃𝑐𝑐𝑐𝑐2 , 𝑃𝑃2 , 𝑃𝑃𝑚𝑚 and referring all the quantities to the
stator side, we get :

35
 INDUCTION MACHINES
Double-output System with a Current Converter:
 Substituting the expressions for 𝑃𝑃𝑐𝑐𝑐𝑐2 , 𝑃𝑃2 , 𝑃𝑃𝑚𝑚 and referring all the quantities to the
stator side, we get :

36
DOUBLE-OUTPUT SYSTEM WITH A CURRENT CONVERTER:
The dc equivalent circuit:

 The ac side (up to the rotor-side converter) represents the per-phase equivalent
circuit of the induction machine referred to the rotor.
 The dc side of the equivalent circuit (to the right of the rotor-side converter)
consists of the series resistance of the smoothing reactor and a voltage source
representing the line-side converter.
 It is convenient to refer the complete equivalent circuit to the dc side.
 First the ac-side resistances 𝑠𝑠𝑠𝑠𝑠𝑠′ and 𝑅𝑅𝑟𝑟 are converted to their dc equivalents.

37
DOUBLE-OUTPUT SYSTEM WITH A CURRENT CONVERTER:
The dc equivalent circuit:
 To find out the dc equivalent resistance, recall that the input current of a phase-
controlled converter has a quasi-square waveform as shown in Fig.
 In this figure, the dc-side current is assumed to be ripple-free. The commutation
overlap effect is also neglected.
 The rms value of the ac line current (in terms of the de current) is given by
𝐼𝐼𝑟𝑟 = 𝐼𝐼𝑑𝑑 2/3
 Now a resistance 𝑅𝑅𝑑𝑑𝑐𝑐 connected to the dc side of the phase-controlled converter-I
will be called the dc equivalent of the ac-side resistance.
 if the power dissipated in 𝑅𝑅𝑑𝑑𝑐𝑐 equals the total power dissipated in all three
phases.
 The ohmic power dissipated in all the three phases of the induction motor is:
 𝑃𝑃𝑐𝑐𝑐𝑐2 = 3𝐼𝐼𝑟𝑟 2 ( 𝑠𝑠 𝑅𝑅𝑠𝑠′2 + 𝑅𝑅𝑟𝑟 )=2𝐼𝐼𝑑𝑑 2 ( 𝑠𝑠 𝑅𝑅𝑠𝑠′2 + 𝑅𝑅𝑟𝑟 )
 Therefore, the equivalent dc resistance of the induction motor when viewed from
the dc-link side of converter-I is
 𝑅𝑅𝑑𝑑𝑑𝑑 = 2( 𝑠𝑠 𝑅𝑅𝑠𝑠′2 + 𝑅𝑅𝑟𝑟 )
 The resistance portion of the ac side
equivalent circuit is transferred to the dc side

38
DOUBLE-OUTPUT SYSTEM WITH A CURRENT CONVERTER:
The dc equivalent circuit:
 The rest of the ac circuit along with the rotor-side converter-I can be represented
by a dc voltage source 𝑉𝑉𝑑𝑑1 in series with an equivalent internal resistance 𝑅𝑅𝑑𝑑1 .
 This takes into account the reduction in the mean output voltage of converter-I
caused by the induction motor reactance.
 The values of 𝑉𝑉𝑑𝑑1 and 𝑅𝑅𝑑𝑑1 in terms of the ac-side circuit parameters are given by:

39
DOUBLE-OUTPUT SYSTEM WITH A CURRENT CONVERTER:
The dc equivalent circuit:
 Being an inductive phenomenon, 𝑅𝑅𝑑𝑑1 does not represent a power loss
component.
 Figure below shows the complete equivalent circuit referred to the dc side.
 Note that the entire slip power, i.e., the rotor-side electrical power, is supplied to
the right of the dotted line. From the figure,
 The condition 𝐼𝐼𝑑𝑑 > 0 is required because the rotor-side and the line-side
converters allow only unidirectional current.

40
DOUBLE-OUTPUT SYSTEM WITH A CURRENT CONVERTER:
The dc equivalent circuit:
 Substituting the expressions for 𝑉𝑉𝑑𝑑1 > 𝑉𝑉𝑑𝑑2 , and 𝑅𝑅𝑑𝑑1 in Eqn and arranging the
terms, we get

41
DOUBLE-OUTPUT SYSTEM WITH A CURRENT CONVERTER:
The dc equivalent circuit:
 The approximate rms value of the fundamental component of the stator current
for level dc-link current 𝐼𝐼𝑑𝑑 is:

 The power input to the utility grid through the supply-side converter-II of the de
link is

 As the motoring convention has been followed, the net electrical power output for
the generating operation,

42
DOUBLE-OUTPUT SYSTEM WITH A CURRENT CONVERTER:
Reactive Power and Harmonics:
 The grid-connected induction generator draws its excitation from the power line
to set up its rotating magnetic field for regeneration and thus always demands
lagging reactive power.
 Such reactive power demand may adversely affect the network voltage level
particularly in weak public utility networks-and increase system losses.
 For large wind turbines driving induction generators, the voltage fluctuation and
the flickering arising from power output variation may exceed the statutory limits
of the utility system.
 By applying the definition of reactive power 𝑄𝑄 = 𝑉𝑉1 𝐼𝐼1 sin 𝜑𝜑1 , to the T-circuit model
of an induction motor the reactive power in one phase of the motor under the
approximation becomes 𝐼𝐼1 ≈ 𝐼𝐼2′
 The reactive power consumed by the electrical machine thus comprises a
constant value, which is the magnetizing volt-ampere, and a parabolic value
dependent on the dc-link current.

43
DOUBLE-OUTPUT SYSTEM WITH A CURRENT CONVERTER:
Reactive Power and Harmonics:
 A three-phase naturally commutated bridge converter on the rotor side is
not capable of generating leading VAR.
 The lagging reactive power requirement is then transferred from the
supply through the stator side of the machine, thus reducing the stator
active power output for the same current loading.
 On the other hand, if the rotor-side converter is made a forced-
commutated converter and its firing angle is made greater than 180° for
operation below the rated speed and less than zero for operation above
the rated speed, the reactive power demand of the machine can be met
by the rotor-side converter.
 For controlled converter II, the phase angle between the fundamental
alternating current and the ac sinusoidal current is equal to the firing
delay angle 𝛼𝛼2 .
 As 𝛼𝛼2 is always less than 180°, the fundamental lagging reactive power
requirement of converter-II comes from the electrical source through the
step-down transformer, and varies with the operating point.

44
DOUBLE-OUTPUT SYSTEM WITH A CURRENT CONVERTER:
Reactive Power and Harmonics:
 Here too, if forced commutation is employed, unity or leading power-
factor operation in order to improve the overall power factor of the
system is possible.
 Whatever may be the firing strategy, the current-fed dc-link converter
system requires an expensive choke and an extra commutation circuit for
operation at synchronous speed (if it lies within the operating speed
range).
 It also results in poor power factor at low-slip speeds. Besides, a current-
fed de link generates rectangular current waves, which inject low-order
harmonics into the supply side of converter-II, which are difficult to
eliminate.
 The low-order harmonics from converter-I injected into the rotor mmf
produce variable-frequency stator current harmonics and torque
harmonics.

45
DOUBLE-OUTPUT SYSTEM WITH A VOLTAGE CONVERTER:
 The drawbacks of naturally commutated or line-commutated converters and
low-frequency forced-commutated converters can be overcome by the use of
dual PWM voltage-fed, current-regulated converters, connected back to back,
in the rotor circuit, as shown in Fig.

46
DOUBLE-OUTPUT SYSTEM WITH A VOLTAGE CONVERTER:
 PWM converters with dc voltage link circuits offer the
following characteristics:

 (a) Realization of the field-oriented control principle for decoupled control of

the generator's active and reactive power.
 (b) Low distortion in stator, rotor, and supply currents, owing to the shift of the
harmonic spectra from lower to higher order, requiring a small-sized filter for
attenuation of higher harmonics.
 (c) Improvement in the overall system power factor through the control of the
displacement factor between the voltage and current of the supply-side
converter II.
 (d) Operation at synchronous speed with direct current injected into the rotor
from the dc voltage link circuit.

47
PROBLEM

48
PROBLEM

49
PROBLEM

50
PROBLEM

51
PROBLEM

52
PROBLEM

53
 The adoption of a suitable control strategy in the operation of bidirectional
converters to control the flow of slip power in the double-output system can
help to reduce the reactive power demand of the generator from the utility
system.
 However, squirrel cage type generators require a different approach. Direct
control of reactive power demand can be achieved by using a bank of
capacitors, or other VAR compensators, located either centrally or near the
wind farms.
 The network structure and the location of wind farms dictate the choice of the
system.
 VAR compensators improve voltage stability, increase network
capability, and reduce losses.
 Various types of VAR compensators, with various features suitable for different
applications, are in use:

 Switched capacitor scheme (TSC)

 Thyristor controlled reactor (TCR)

 Static VAR compensator (SVC)

54
 Switched capacitor scheme (TSC)
 The switched capacitor scheme comprises a bank of parallel capacitors which
are switched on and off by contactors in response to preset voltage levels.
 The speed of response is limited by the contactor closing time and is suitable
for the discreet and slow control of the system voltage.
 For faster control, thyristor pairs are used to switch the capacitors (TSC) as
shown in Fig.
 Continuous control is not possible with the TSC scheme, as the capacitor
would remain in the circuit for a full cycle before the thyristor switches off when
the current reaches zero.
 In practice, current-limiting reactors are used
in series with the capacitor banks to limit the current that
may arise owing to the difference between the supply
and capacitor voltages at the switching-on instant.
 In a three-phase system, capacitor banks are usually
delta-connected.

55
 Thyristor controlled reactor (TCR)
 Continuous control of effective reactive power is possible if thyristor-phase-
controlled reactors (TCRs), shown in Fig., are used in parallel with fixed
capacitor banks rated at the full-load reactive power demand of the induction
generator.
 Variable VAR is realized by varying the firing angle between 900 and 180°.
 The excess reactive power from the capacitor bank at reduced load is
absorbed by the reactor when the delay angle approaches 900.
 In a three-phase system, the thyristor-controlled
reactors are normally delta-connected to avoid triple
harmonic components in the compensating line currents.
 Owing to the high cost of the system, it can be cost
effectively employed at large wind farms rather than
at each wind turbine site.

56
 Static VAR compensator (SVC)
 The recent trend in reactive power control is based on the forced-commutated
voltage source PWM converter. The basic circuit configuration is shown in Fig.
 It is the static realization of the synchronous condenser. Inductors are included
in series with the ac supply and a capacitor on the dc side.
 The dc capacitor is of considerably lower rating (typically 20%) and of small
physical dimension compared to the ac capacitors that are used with
conventional reactive power controllers.
 The main feature of this VAR compensator is that the converter can generate
or absorb reactive power by controlling the switching pattern of the devices
with gate turn-off capability, such as the GTO thyristor and IGBT.

57
 Static VAR compensator (SVC)
 With a charged capacitor, the inverter produces a set of balanced voltages at
the output terminal, which are controlled to be in phase with the corresponding
ac system voltage, so that only reactive current can flow between the
converter and the system.
 The basic principles of the control of reactive power flow are similar to those
of the rotating synchronous condenser, and are shown in the phasor diagram
of Fig.(b), for the current direction assumed in Fig. (a), for the fundamental
component.
 When the inverter output voltage 𝑉𝑉𝑟𝑟 is above the system voltage 𝑉𝑉𝑠𝑠 , 90°
lagging reactive current flows to the ac system and the converter therefore
acts as a reactive power generator for the system.
 In other words, the converter appears as a capacitive load drawing leading
current from the system.
 If the inverter output voltage is decreased below that of the ac system, the
reactive current will flow from the system to the converter, and thus the
converter will appear as an inductive load on the system absorbing reactive
power.

58
 Static VAR compensator (SVC)
 Since only the reactive power is involved, ideally the de capacitor should
retain its charge.
 The inverter simply helps connect different phases for the exchange of
reactive currents between them.
 In a practical inverter, the devices are not lossless, and the inverter output
voltage is made to lag behind the ac system voltage in case of leading
VAR, and lead slightly for lagging VAR, so that the system can supply the
small losses and the capacitor voltage can be maintained at the desired
level.
 This principle can be used to increase or decrease the capacitor voltage,
thereby controlling the inverter output voltage, for controlling the VAR
generation or absorption.
 This static VAR generator can provide fast and continuous control of
reactive power.

59
PROBLEM

60
PROBLEM

61