J. McCalley Double-fed electric machines – steady state analysis.

J. McCalley

Double-fed electric machines – steady state analysis

Four configurations

2

We will study only this one, the DFIG.

Basic concepts

3

ACDC

DCAC

DFIG

Rotor

Power Grid

DC Link

Rotor is wound: it has 3 windings.Stator has three windings.Induction machine looks like a transformer with a rotating secondary (rotor).In DFIG, we will inject a voltage control signal via that converter.

Basic Concepts

4

p

fn ss

60Balanced voltages applied to stator windings provides a

rotating magnetic field of speedwhich induces an emf in the rotor windings according to•eind=induced emf in one conductor of rotor•v=velocity of conductor relative to stator flux rotation•B=stator magnetic flux density vector•L=length of conductor in direction of wire

LBveind )(

rotor

(fs: 60 Hz, p: # of pole pairs)

Basic concepts

5

rpm 60

;

;rad/sec 377 ;

p

fn

n

nn

psslip

ss

s

ms

mmss

ms

We can manipulate to get:

)1(

)1(

s

snn

sm

sm

The induced rotor voltages have frequency of : msr

Substitution into slip expression above yields: srsrs

r sffss

Observe three modes of operation:

ωm< ωs ωr>0s>0Subsynchronous operation

ωm= ωs ωr=0s=0Synchronous operation

ωm>ωs ωr<0s<0Supersynchronous operation

Mechanical rad/sec

Per-phase steady-state model

6

sssss IjXREV )( STATOR VOLTAGE EQUATION:

at fs

rrrrsr IXjREV )( ROTOR VOLTAGE EQUATION: at fr

=stator voltage with frequency fssV

sE

sI

sR

sX

= emf in the stator windings with frequency fs

= stator current with frequency fs

=stator resistance

=stator leakage reactance

=rotor voltage with frequency frrV

rsE

rI

rR

rX

=induced emf in the rotor windings with frequency fr

=induced rotor current with frequency fs

=rotor resistance

=rotor leakage reactance=

These quantities are referred to rotor side, indicated by prime notation.

These quantities are referred to stator side.

rrL

Referring quantities

7

Solve both relations for φm and equate:

But recall:

Application of Faraday’s Law allows the stator back emf and the induced rotor voltage to be expressed as:

mrrrrs

mssss

fNKE

fNKE

2

2

Ks, Kr: stator and rotor winding factors, respectively, which combine the pitch and distribution factors. Ns, Nr: number of turns of stator & rotor, respectively.fs, fr, frequency of stator & rotor quantities, respectivelyφm : magnetizing flux

rrr

sss

rs

s

rrr

rs

sss

s

m fNK

fNK

E

E

fNK

E

fNK

E

22

sr sff sNK

NK

sfNK

fNK

E

E

rr

ss

srr

sss

rs

s

The ratio Ks/Kr is normally very close to 1, therefore sN

N

E

E

r

s

rs

s

r

s

N

Na Define the effective turns ratio:

a

EsE

s

a

E

E srs

rs

s

Define the induced rotor voltage referred to the stator side:a

EE s

rs


8

We just derived that:a

EsE s

rs

At a locked rotor condition (s=1), the device is simply a static transformer, and we have:

aII

aEE

rr

rsrs

/

rsss

rs EaEa

EE

This tells us it we want to move a voltage from rotor side to stator side, we multiply it by a=Ns/Nr. We can obtain similar relationships for currents and impedances, and so we define the rotor quantities referred to the stator according to:

2

2

aLL

aRR

rr

rr

33

33

3333

RrjωrLσr

ErsEs

Rs jωsLσsIs Ir

Vs Vr

Rotor quantities are referred to the stator-side, indicated by unprimed quantities.

This is locked rotor condition (s=1), therefore ωr=ωs and Ers=Es

We can account for other slip conditions using ωr=sωs and from (*), aE’rs=sEs.

(*)


9

33

33

3333

RrjsωsLσr

Ers=sEsEs

Rs jωsLσsIs Ir

Vs Vr

Now write the rotor-side voltage equation (referred to stator):

rrsrsr ILjsREsV )(

Divide by srrs

rs

r ILjs

RE

s

V)( and we get the following circuit:

33

33

3333

Rr/sjωsLσr

EsEs

Rs jωsLσsIs Ir

Vs Vr/s

The voltage on both sides of the xfmr is the same, therefore, we may eliminate the xfmr. .We represent a magnetizing inductance jωsLm in its place.


10

33

3333

Rr/sjωsLσr

jωsLmEs

Rs jωsLσsIs Ir

Vs Vr/s

Power relations

11

33

3333

Rr/sjωsLσr

jωsLmEs

Rs jωsLσsIs Ir

Vs Vr/s

s

sRR

s

sRR

s

sR

s

sRsRR

s

R rr

rrrrrrr )1(

s

sVV

s

sVV

s

sV

s

sVsVV

s

V rr

rrrrrrr )1(

Change the circuit accordingly….

We modify the above circuit slightly in order to clearly separate slip-dependent terms from loss terms:

Power relations

12

33

3333

RrjωsLσr

jωsLmEs

Rs jωsLσsIs Ir

VsVr

It is possible to prove that the mechanical power out of the machine is the power associated with the slip-dependent terms R2(1-s)/s and Vr(1-s)/s. To do so, use:

rlossslossrsmech PPPPP ,,

where Ps and Pr are powers entering the machine through the stator & rotor windings, respectively, and Ploss,s and Ploss,r are the stator and rotor winding losses, respectively.Expressing the right-hand-terms of the power balance relation in terms of the above circuit parameters leads one to identify the slip-dependent terms as Pmech.

+ -Rr(1-s)/s

Vr(1-s)/s

Knowing that the slip-dependent terms are those responsible for mechanical power, we may obtain the power expressions from the circuit, as on the next slide.

Power balance relation:

Power relations

13

33

3333

RrjωsLσr

jωsLmEs

Rs jωsLσsIs Ir

VsVr

*2

*2

*2

Re1

3)1(

3

1Re3

)1(3

Re33

rrr

r

rrr

r

reqeqrmech

IVs

s

s

sRI

Is

sV

s

sRI

IVRIP

+ -Req=

Rr(1-s)/s

Veq=Vr(1-s)/s

If Pmech>0the machine is delivering power through the shaft: MOTOR!If Pmech<0the machine is receiving power through the shaft: GEN!

Rotor current (Ir) direction is out of positive side of voltage source; therefore it supplies power to circuit. But a normal (positive) resistance Req

always consumes power. So these two terms should be opposite sign. Defining Pmech>0 (see below) as motor mode implies Req term should be added and Veq term should be subtracted.

Pmech

If 0<s<1Req term is positive Veq term is positiveSupplying P to cctIf 0>s>-1Req term is negative Veq term is negativeConsuming P from cct.

A first torque expression

14

*2 Re1

3)1(

3 rrr

rmech IVs

s

s

sRIP

*2 Re1

3)1(

3 rrm

rr

mmmechem

memmemmech IV

s

sp

s

sRI

ppPT

pTTP

33

3333

RrjωsLσr

jωsLmEs

Rs jωsLσsIs Ir

VsVr

+ -Req=

Rr(1-s)/s

Veq=Vr(1-s)/s

(p: # of pole pairs)

Recall from slide 5: ;s

rs

s

msm ss

1)1(

Therefore:

*2

*2

Re33

Re33

rrrr

rr

rrr

m

mr

mrr

mem

IVpRIp

IVpR

Ip

T

r

m

r

s

s

m

s

s

1

and

ivrrrr

rr

em IVpRIp

T

cos33 2

A second (equivalent) torque expression

15

*Re3 sss IVP

33

3333

RrjωsLσr

jωsLmEs

Rs jωsLσsIs Ir

VsVr

+ -Req=

Rr(1-s)/s

Veq=Vr(1-s)/s

Stator power:

msrssssss LjIILjRIV Stator voltage:

Substitute Vs into Ps: *222

****

**

Re3

Re3

Re3Re3

srmssmssssss

srmsssmssssssss

smsrssssssss

IILjILjILjIR

IILjIILjIILjIIR

ILjIILjRIIVP

The middle two terms are purely imaginary, therefore:

*2Re3 srmssss IILjIRP First term is purely real, only the second term contains real and imaginary, therefore:

*2 Re33 srmssss IILjIRP


16

*Re3 rrr IVP

33

3333

RrjωsLσr

jωsLmEs

Rs jωsLσsIs Ir

VsVr

+ -Req=

Rr(1-s)/s

Veq=Vr(1-s)/s

Rotor power:

msrsrsrrr

msrsrsr

rr

msrsrsrrrrr

LjsIILjsRIV

LjIILjs

RI

s

V

LjIILjs

sRRI

s

sVV

11

Rotor voltage:

Substitute Vr into Pr: *222

****

**

Re3

Re3

Re3Re3

rsmsrmsrrsrr

rsmsrrmsrrrsrrr

rmsrsrsrrrrr

IILjsILjsILjsIR

IILjsIILjsIILjsIIR

ILjsIILjsRIIVP

The middle two terms are purely imaginary, therefore:

*2Re3 rsmsrrr IILjsIRP First term is purely real, only the second term contains real and imaginary, therefore:

*2 Re33 rsmsrrr IILjsIRP


17


Now substitute Ps and Pr into the power balance equation:



rlossslossrsmsrrsrmsssmech PPIILjsIRIILjIRP ,,*2*2 Re33Re33

Observe we have loss terms added and subtracted in the above, so they go away.

** Re3Re3 rsmssrmsmech IILjsIILjP Now consider what happens when you take the real part of a vector multiplied by j (or rotated by 90 degrees):

Re(ja)

a

ja

Im(a)Observe that Re(ja) = - Im(a)

Therefore: ** Im3Im3 rsmssrmsmech IILsIILP


18

Let’s consider another vector identity: taking imaginary part of a conjugated vector:

** Im3Im3 rsmssrmsmech IILsIILP

Im(a*)

a

a*

Im(a)

Observe that Im(a*) = - Im(a)

Therefore:

sIIL

IIsIIL

IILsIIL

IILsIILP

rsms

rsrsms

rsmsrsms

rsmssrmsmech

1Im3

ImIm3

Im3Im3

Im3)(Im3

*

**

**

***

Recall: )1( ssm

*Im3 rsmmmech IILP Therefore:

mmechem

pPT

Recall:

*Im3 rsmem IIpLT

Two equivalent torque expressions

19

*Im3 rsmem IIpLT ivrrrr

rr

em IVpRIp

T

cos33 2

Torque expression #1: Need rotor speed, rotor voltage and rotor current

Torque expression #2: Need stator current and rotor current

A third set of equivalent torque expressions follow….

Additional equivalent torque expressions

20

If we assume the magnetic core of the stator and rotor is linear, then we may express flux linkage phasors of each winding (stator winding and rotor winding, respectively):

rmsss ILIL rrsmr ILIL

Self inductances

Mutual inductances

Stator winding Rotor winding

ASIDE: Each self inductance is comprised of mutual and leakage according to:

rmrsms LLLLLL ;

Therefore:

ssrsm

rmsssms

ILIIL

ILILIL

)(

rrrsm

rrrmsmr

ILIIL

ILILIL

)(

From stator winding equation:

;;m

sssr

s

rmss L

ILI

L

ILI

From rotor winding equation:

m

rrrs

r

smrr L

ILI

L

ILI

;

Choose one of these equations and substitute into torque expression #2….

*Im3 rsmem IIpLT

Additional equivalent torque expressions

21

From stator winding equation:

s

rmss L

ILI

From rotor winding equation:

r

smrr L

ILI

Substitute into torque expression #2….

*Im3 rsmem IIpLT

*

2*

**

*

Im3

Im3

Im3

Im3

rss

m

rmrss

m

rrmrss

m

rs

rmsmem

IL

Lp

ILIL

Lp

IILIL

Lp

IL

ILpLT

Using stator winding equation:

Purely real

*

2*

**

*

Im3

Im3

Im3

Im3

rsr

m

smrsr

m

ssmrsr

m

r

smrsmem

IL

Lp

ILIL

Lp

IILIL

Lp

L

ILIpLT

Using rotor winding equation:

Purely real

Airgap and slip powerOn slides 15 and 16, we derived the following relations for the power into the stator and rotor respectively:

*2 Re33 srmssss IILjIRP *2 Re33 rsmsrrr IILjsIRP

Subtracting losses from both sides, we obtain:

*2 Re33 srmssss IILjIRP *2 Re33 rsmsrrr IILjsIRP This quantity is the power that flows from the stator terminals to the rotor (negative for generator operation). In other words, it is the power across the airgap. Therefore:

*2 Re33 srmssssairgap IILjIRPP

This quantity is the power that is transferred from the grid to the rotor through the converter (negative when it is into the grid). It is called the slip power. Therefore:

*2 Re33 rsmsrrrslip IILjsIRPP

Bring out front the “s” in the slip power expression and use Re{ja}=-Im(a) (both):

Use Im(a*) = -Im(a) on slip expression:

22

*2 Im33 srmssssairgap IILIRPP *2 Im33 rsmsrrrslip IILsIRPP

rsmsrrrslip IILsIRPP *2 Im33

The term 3Im{} in the slip power expression is Pairgap. Therefore:

*2 Im33 srmssssairgap IILIRPP

airgapslip sPP

Airgap and slip power

23

So we just proved that:

Our power balance relation states:

Recall: s

ms

1 airgaps

mmech PP

airgapsm

airgaps

m

mmechem P

ppP

pPT

slips

em Ps

pT

s

rs

slipsr

sem P

pT

slipr

em Pp

T

airgapslip sPP where

*2 Re33 srmssssairgap IILjIRPP *2 Re33 rsmsrrrslip IILjsIRPP

slipairgap P

rlossr

P

slosssrlossslossrsmech PPPPPPPPP ,,,,

Therefore: slipairgapmech PPP

Substituting airgapslip sPP we obtain airgapairgapairgapmech PssPPP 1

Substituting: slipairgapairgapslip Ps

PsPP1

Approximate relations between active powers

24

On slides 15 and 16, we derived the following relations for the power into the stator and rotor respectively:

*2 Re33 srmssss IILjIRP *2 Re33 rsmsrrr IILjsIRP

If we neglect the stator losses (3RSIs2) and rotor losses (3RrIr

2):

*Re3 srmss IILjP *Re3 rsmsr IILjsP Bring out front the “s” in the rotor power expression and use Re{ja}=-Im(a) (both): *Im3 srmss IILP *Im3 rsmsr IILsP

Use Im(a*) = - Im(a) on the rotor power expression

*Im3 srmss IILP rsmsr IILsP *Im3

The term 3Im{} in the rotor power expression is PS. Therefore: sr sPP Recall the power balance relation: rlossslossrsmech PPPPP ,,

Neglecting losses:rsmech PPP

Substituting Pr expression: sssmech PssPPP )1(

Recall: s

ms

1 ss

mmech PP

ssm

ss

m

mmechem P

ppP

pPT

rs

em Ps

pT

s

rs

rsr

sem P

pT

rr

em Pp

T

Active power relations - summary

25

sr s

Exact ApproximateBoth



*2 Re33 srmssssairgap IILjIRPP

*2 Re33 rsmsrrrslip IILjsIRPP

airgapslip sPP


slipairgapmech PPP

airgapmech PsP 1

s

ms

1

airgaps

mmech PP

airgaps

em Pp

T

mmechem

pPT

slipr

em Pp

T

*Re3 srmss IILjP

*Re3 rsmsr IILjsP

*Re3 srmssairgap IILjPP

*Re3 rsmsrslip IILjsPP

sr sPP

rsmech PPP

smech PsP )1(

ss

mmech PP

ss

em Pp

T

rr

em Pp

T

Power balance

26

Ps

Ploss,s

PairgapPslip Pr

Ploss,r

slipairgap P

rlossr

P

slosssrlossslossrsmech PPPPPPPPP ,,,,

These figures assume proper sign convention (power flowing to the rotor is positive).

PgridPs

PairgapPslip PrPgrid

With losses Without losses

Pmech Pmech

Generator modes

27

ems

s Tp

P

emr

r Tp

P

Mode Slip and speed Pmech Ps Pr

1. Motor(Tem>0)

s<0, ωm>ωs

(suprsynchrnsm)>0 (mch delivers mech pwr)

>0 (mch receives power via stator)

>0 (mch receives power via rotor)

2. Generator(Tem<0)

s<0, ωm>ωs

(suprsynchrnsm)<0 (mch receives mech pwr)

<0 (mch delivers power via stator)

<0 (mch delivers power via rotor)

3. Generator(Tem<0)

s>0, ωm<ωs

(subsynchrnsm)<0 (mch receives mech pwr)

<0 (mch delivers power via stator)

>0 (mch receives power via rotor)

4. Motor(Tem>0)

s>0, ωm<ωs

(subsynchrnsm)>0 (mch delivers mech pwr)

>0 (mch receives power via stator)

<0 (mch delivers power via rotor)

sr s

Focusing on the generator modes, we observe the standard induction machine generating mode, supersynchronism, where ωm>ωs (mode 2). We also observe a subsynchronous mode (mode 3), where ωm<ωs, which is available to the DGIG as a result of the machine receiving power from the grid via the rotor circuit.

For each mode, we may use the three relations to track the sign Ps, ωr, and Pr from the signs of Tem and s. For example, for mode 2, Tem<0Ps<0 and Tem<0, s<0 ωr<0Pr<0

Generator modes

28

sm

sm

These figures show actual flow direction for generator operation.They also neglect losses.

Pm= Pmech

Mode 2

Mode 3

Recall the approximate relation

sr sPP Operation must have |s|<1, so rotor power is always smaller than stator power.

In fact, DFIGS always run within about-0.3<s<0.3.

Therefore, the rating of the PE converter circuit need be only about 30% of the stator winding rating.

A question on rating

29

This figure assumes proper sign convention (power flowing to the rotor or into the stator is positive).

PsPairgap

Pslip PrPgrid

Without losses

Pmech

rsmech PPP

s

PP mechs

1

rsg PPP

s

sPP mechr

1Assume an operating condition such that Pmech=PWTrating. Then

WTratinggmech PPP

s

PP WTratings

1 s

sPP WTratingr

1For example, consider Pmech=PWTrating=-2 MW. In supersynchronous mode, with s=-0.3,

MW.5385.13.01

2

sP Therefore stator winding must be rated for 1.5385 MW.

But in the subsynchronous mode, s=+0.3, then MW8571.23.01

2

sP

Question: Does this mean that the stator of a 2 MW turbine must be rated for 2.8571?

Answer: No. In subsynchronous mode, the mechanical power from the generator shaft is lower that that in the supersynchronous mode. If Pmech increases beyond a certain level, then machine speed increases into the supersynchronous mode. So above situation never occurs. We can obtain the maximum power in subsynchronous mode as:

MW0769.1)3.01(5385.1)1( sPP smech

Question on sign of losses

30

Question: Since stator losses (3RSIs2) and rotor losses (3RrIr

2) are always positive, and since we get sign changes with the numerical values of Pmech, Ps, and (sometimes) Pr, do the loss terms in the above equation need to have different signs for motor operation than for generator operation? That is, do we need to do the following?


rlossslossrsmech PPPPP ,, Motor operation:

rlossslossrsmech PPPPP ,, Generator operation:

Answer: No. Our original equation applies for both motor & generator operation.

Remember: Pmech is positive for motor operation; Ps, and Pr are positive when flowing into the device from the grid. It may help to think about the equation in two different, but equivalent forms.

rlosssloss

Input

rs

Output

mech PPPPP ,,

Motor operation:

rlosssloss

Input

mech

Output

rs PPPPP ,,

Generator operation:

50 = 45 +10 - 3 - 2 - 50 = - 55 + 3 + 2

Per-unitization

31

In general, per-unitization enables inclusion of DFIGs within a system model.It also facilitates identification of inappropriate data. Finally, a per-unitized voltage provides the ability to know how far it is from its nominal value (usually also the “normal” value) without knowing that nominal value.

The procedure is to choose three base quantities and compute other necessary base quantities. We will choose our base quantities as •rated rms line-to-neutral stator voltage, Vbase=|Vs|rated (rms volts); •rated rms stator line current, Ibase=|Is|rated (rms amperes)•rated stator synchronous frequency, ωbase= ωs,rated (rad/sec))Then we compute:

base

basebase I

VZ • Base impedance:

base

basebase I

L

• Base inductance:

base

basebase

V

• Base flux:

tVtdt

dv :ionJustificat

basebasebase IVS 3• Three-phase power base: basem

basebase

ST

,• Base torque:

pbase

basem

,• Base speed:

Per-unitization – stator side

32

Once all base quantities are obtained, then per-unitization is easy:

base

spus V

VV ,

• Stator voltage in pu:

base

spus I

II ,

• Stator current in pu:

base

spus

,• Stator flux in pu:

base

sspus S

IVP

*

,

Re3

• Stator active power in pu:

base

ss

pus S

IVQ

*

,

Im3

• Stator reactive power in pu:

As usual, only the magnitude is transformed (angle remains unchanged).

Per-unitization – rotor side

33

base

rpur V

VV ,

• Rotor voltage in pu:

base

rpur I

II ,

• Rotor current in pu:

base

rpur

,• Rotor flux in pu:

base

rrpur S

IVP

*

,

Re3

• Rotor active power in pu:

base

rr

pur S

IVQ

*

,

Im3

• Rotor reactive power in pu:

For the rotor side, we use the same base quantities as on the stator side (with actual quantities referred to the stator side).


Per-unitization – torque, speed, R, L

34

base

empuem T

TT ,

• Torque in pu:

basem

mpum

,,

• Speed in pu:

base

rpu Z

Rr • Resistances in pu:

basepu L

Ll • Inductance in pu:


Note that the resistances and inductances when expressed in pu are lower case.

On the rotor side, we use the same base quantities as on the stator side (with actual quantities referred to the stator side).

Voltage equations expressed in per unit

35

From slides 15, 16, we obtain voltage equations for stator and rotor circuits:

msrssssss LjIILjRIV msrsrsrrr LjsIILjsRIV

ssrsms ILIIL )(rrrsmr ILIIL )(

From slide 20, we obtain the equations for stator and rotor flux linkages:

which we rearrange by collecting terms in jωs: mrsssssss LIIILjRIV mrsrrsrrr LIIILjsRIV

We recognize the flux linkage expressions in the voltage equations. Therefore:

sssss jRIV rsrrr jsRIV

Now we can replace voltages, currents, and flux linkages with the product of their per-unit value and their base quantity, then the base quantities can be used to per-unitize the resistances and frequency to obtain:

pusspuspus jrIV ,,, purrpurpur jsrIV ,,,

(*)

Voltage equations expressed in per unit

36

sssss jRIV rsrrr jsRIV

Replace voltages, currents, flux linkages with the product of their pu value and their base quantity, then base quantities are used to per-unitize resistances and frequency to obtain:

pusspuspus jrIV ,,, purrpurpur jsrIV ,,,

Now consider the flux linkage equations:

rrsmr ILIL rmsss ILIL Replace currents and flux linkages with the product of their pu value and their base quantity, then base quantities are used to per-unitize inductances to obtain:

purrpusmpur IlIl ,,, purmpusspus IlIl ,,, Per-unitize one of the torque equations (#2) *Im3 rrem IpT as follows:

p

IVST

basem

basebase

basem

basebase /

3

,,

*

,,

*

,

,

*

, Im3/

Im3

/ 3

Im3purpur

base

r

basembase

r

basem

basebase

rrpuem I

I

I

Vp

IVIp

T

Per-unitize the power expressions to obtain:

)sin();cos(

)sin();cos(

,,,,,,

,,,,,,

ivpurpurpurivpurpurpur

ivpuspuspusivpuspuspus

IVQIVP

IVQIVP

Homework #3

37

Homework #3: This homework is due Monday, March 26.

A. Using previous relations provided in these slides, derive the following torque expressions. ssem IpT ,Im3.1 *

*,Im3.2 rrem IpT

srsr

mem LL

LpT

,Im3.3 * (and identify σ)

B. Use Q = 3Im{V I*} and the equivalent circuit to derive reactive power expressions, in terms of Is and Ir for

1. The stator, Qs

2. The rotor, Qr

C. For each DFIG condition below, compute Pairgap and Pslip and draw the power flows similar to slide 28.1.Pmech=-1 MW with s=+0.30 (subsynchronous operation). 2.Pmech=-1MW with s=-0.30 (supersynchronous operation).D. Complete the table on the next slide (the boxed section) by computing the per-unit values of the indicated five resistances/inductances for the 2 MW machine.

Homework

38

u (or a)Rs

Lσs

Lm

R’r

Lσr

Rr

Lσr

Ls

Lr

Vbase

Ibase

Rs

lσs

lm

rr

lσr

Phasor diagrams for generator operation

39

We have developed the following relations:

pusspuspus jrIV ,,,

purrpurpur jsrIV ,,,

purmpusspus ILIL ,,, (3) Stator winding flux equation

(4) Rotor winding flux equation

(2) Rotor voltage equation

(1) Stator voltage equation

Draw phasor diagram per below (CCW rotation is pos angle):Step 1: Draw Vs as reference (0°). Step 2: For gen, Qs>0, lag; for gen Qs<0, lead. Draw Is phasor.Step 3: Use (1) to draw the stator flux phasor λs: Step 4: Use (3) to draw the rotor current phasor Ir:Step 5: Use (4) to draw the rotor flux phasor λr:Step 6: ….

)( ,,, spuspuspus rIVj mpussmpuspur lIllI // ,,,

purrpusmpur IlIl ,,,

Vs

Is

Isrs

Vs - Isrs

λs= -j(Vs – Isrs)

– ls Is/lmλs/lm

Ir=λs/lm – ls Is/lmlm Is

lr Ir

λr=lm Is+lr Ir

purrpusmpur IlIl ,,,

Phasor diagrams for generator operation

40

Draw phasor diagram per below (CCW rotation is pos angle):Step 1: Draw Vs as reference (0°). Step 2: For gen, Qs>0, lag; for gen, Qs<0, lead. Draw Is phasor.Step 3: Use (1) to draw the stator flux phasor λs: Step 4: Use (3) to draw the rotor current phasor Ir:Step 5: Use (4) to draw the rotor flux phasor λr:Step 6: Use (2) to draw the rotor voltage phasor Vr:

)( ,,, spuspuspus rIVj

purrpurpur jsrIV ,,,

Vs

Is

Isrs

Vs - Isrs

λs= -j(Vs – Isrs)

– ls Is/Lmλs/lm

Ir=λs/lm – ls Is/lmlm Is

lr Ir

λr=lm Is+Lr Ir

jsλr, s>0, sub-sync

Irrr

Vr=Irrr+jsλr, s>0

jsλr, s<0

Vr=Irrr+jsλr, s<0super-syn

Observe that the angle of Vr is heavily influenced by the sign of s.

mpussmpuspur lIllI // ,,, purrpusmpur IlIl ,,,

Question: How to know quadrant of Is?

41

Consider the circuit below, which is analogous to our stator winding circuit.At any operating condition, we may characterize the circuit as an impedance Z=R+jX=Z/_θ, as indicated. Then we may express the current according to

Z

jXR

ZV

I

Z

V

Z

V

Z

VII i

Real pwr Reactive pwr

P>0 motor

R>0

Q>0 absorbing

X>0

Observe that current angle is always negative of impedance angle, θi=-θ


P>0 motor

R>0

Q<0 supplying

X<0


P<0 genR<0

Q>0 absorbing

X>0


P<0 genR<0

Q<0 supplying

X<0

Z

I

I

Z

Z

I

Z

I

V

V

V

V

Lag

Lead

Machine

Lag

Lead

Example Problem

42

(a) Synchronous speed(b) Line-to-neutral voltage(c) Line current

(d) Stator flux(e) Rotor current(f) Rotor flux(g) Rotor voltage

(h) Rotor real power(i) Rotor reactive power(j) Total real power generated(k) Tem

(b) Line-to-neutral voltage: volts04.39803

690sV

(c) Line current: amps 1804.167304.3983

102

330

*6*

*

s

sssss V

PIIVjP

(d) Stator flux

webers9028.1

16.314

106.2)1804.1673(04.398)( 3

jj

RIV

s

ssss

sssss jRIV

(a) Synchronous speed: rad/sec 16.314)50(22 ss f

Alternatively, the synchronous speed was given as 1500 rpm, therefore:

sec/08.157sec60

min2

min

1500rad

rev

radrevs

sec/16.314)08.157(2 radp ss

The 2 MW DFIG given by the data on slide 38 is delivering, from the stator, rated load (2 MW) at rated voltage with zero stator reactive power in a 50 Hz grid. The slip is s=-0.25 (super-synchronous). Compute:

Example Problem

43




(e) Rotor current

(f) Rotor flux

(g) Rotor voltage

rmsss ILIL

amps5.164.1807105.2

)1808.1673(10587.29028.13

3

m

sssr L

ILI

This is the referred rotor current! We can obtain the actual rotor current from a (or u) =0.34:

amps5.165.6145.164.1807)34.0( rr IaI This phasor is at the rotor frequency, of fr=sfs=-0.25(50)=-12.5 Hz

rrsmr ILIL weber4.77358.15.164.180710587.21808.1673105.2 33

r

volts9.1652.102)4.77358.1)(16.314)(25.0(109.2)5.164.1807( 3 jV r

rsrrr jsRIV

Actual rotor voltage:

9.1656.30034.0

9.1652.102

a

VV r

r


Example Problem

44





(h) Rotor real power

(i) Rotor reactive power

(j) Total real power generated

*Re3 rrr IVP

MW 55.0)5.164.1807(9.1652.102Re3 * rP

*Im3 rrr IVQ

kVAR4.23)5.164.1807(9.1652.102Im3 * rQ

MW55.255.02 rs PP


Comments:1.Pm must be larger in magnitude to supply losses

2. This wind turbine’s rating should be 2.55 MW.3. The DFIG stator winding is rated for 2MW.

Example Problem

45




(k) Tem *Im3 rs

s

mem I

L

LpT

kNmTem 9.125.164.18079028.1Im10587.2

105.223 *

3

3


Wind turbine control levels

46

Level I: Regulates power flow between grid and generator.

Level II: Controls the amount of energy extracted from the wind by wind turbine rotor.

Level III: Responds to wind-farm or grid-central control commands for MW dispatch, voltage, frequency, or inertial control.

Rotor-side converter (RSC) is controlled so that it provides independent control of Tem and Qs. Let’s study the steady-state actions of this particular control function.

Level 1 control

47

This (open-loop) control not heavily used for DFIGs

Assume DC bus voltage is controlled by grid-side converter (GSC) to a pre-determined value for proper operation of both GSC and RSC.

We achieve control objectives by controlling rotor-side voltage.

We control rotor voltage to achieve a specified torque and stator reactive power.

Level 1 control

48

Our objective here is, for a fixed stator voltage (fixed by the grid), and a desired torque Tem,ref and a desired stator reactive power Qs,ref, we want to determine the rotor voltage to make it so. We are also interested in the stator flux, stator current, rotor current, and rotor flux, and stator real power, as shown in the diagram below.

Level 1 control

49

We draw the phasor diagram with stator flux as the reference (0 degrees). Here, the stator flux, denoted by ψs (instead of λs), is specified as the reference. We have identified particular angles in this phasor diagram. It is operating as a motor (current is almost in phase with voltage), and the stator is absorbing reactive power (Is has a negative angle relative to Vs, so Zmotor=Vs/Is has a positive angle, indicating it is inductive and therefore absorbing.

Level 1 control: Qs equation

50

From voltage equation (slide 35): sssss jRIV

If we neglect drop across the stator resistance (it is typically very small), then:

sss jV Substitute into the stator reactive power equation: ** Im3Im3 ssssss IjIVQ Use Im(ja)=Re(a): *Re3 ssss IQ

From previous slide, note that ɣi is the angle by which Is leads λs , i.e.,

issss II ;0

Substituting:

isssiisss

isssissss

IjI

IIQ

cos3sincosRe3

Re30Re3

Final equation for Qs: issss IQ cos3

Level 1 control: Tem equation

51

From HW3 (see slide 37):

Again (from phasor diagram), note that ɣi is the angle by which Is leads λs , i.e.,

issss II ;0

Substituting:

Final torque equation:

ssem IpT ,Im3 *

issiiss

ississem

IpjIp

IpIpT

sin3sincosIm3

Im30Im3

issem IpT sin3

Level 1 control: Is equation

52

From phasor diagram:

But recall our Qs and Tem equations:

Substituting into Is equation:

isiss jIII sincos

issem IpT sin3

issss IQ cos3ss

sis

QI

3cos

s

emis p

TI

3sin

s

em

ss

ss p

Tj

QI

33

Recall from slide 50: sss jV sssV

Substituting into Is equation: s

ems

s

ss pV

Tj

V

QI

33

Level 1 control: λr equation

53

Using these relations, together with:

From slide 20:

rmsss ILIL

rrsmr ILIL

rrs

ms

ss LL

L

LI

1

rr

srs

mr LLL

LI

1

rs

m

LL

L2

1

sss jV s

ems

s

ss pV

Tj

V

QI

33

we may derive:

m

rs

s

ems

m

rs

s

s

m

r

s

sr L

LL

pV

Tj

L

LL

V

Q

L

LV

33

m

s

s

ems

m

s

s

s

ms

sr L

L

pV

Tj

L

L

V

Q

L

VI

33

1

Level 1 control: λr equation

54

m

rs

s

s

m

r

s

sr

m

rs

s

emsrr L

LL

V

Q

L

LVj

L

LL

pV

TV

33

Neglecting the voltage drop in the rotor resistance, we may derive:

Now use the rotor flux equation derived on the previous slidetogether with the rotor voltage equation (slide 35):

m

rs

s

ems

m

rs

s

s

m

r

s

sr L

LL

pV

Tj

L

LL

V

Q

L

LV

33 rsrrr jsRIV

Level 1 control: summary

55

m

rs

s

s

m

r

s

sr

m

rs

s

emsrr L

LL

V

Q

L

LVj

L

LL

pV

TV

33

m

rs

s

ems

m

rs

s

s

m

r

s

sr L

LL

pV

Tj

L

LL

V

Q

L

LV

33

m

s

s

ems

m

s

s

s

ms

sr L

L

pV

Tj

L

L

V

Q

L

VI

33

1

s

ems

s

ss pV

Tj

V

QI

33

Also, we have stator and rotor powers as a function of Tem:

s

ss

V

ems

s Tp

P

emr

r Tp

P

Level 1 control: magnitudes

56

2

2

2

22

33

m

rs

s

s

m

r

s

sr

m

rs

s

emsrr L

LL

V

Q

L

LV

L

LL

pV

TV

22

2

33

m

rs

s

ems

m

rs

s

s

m

r

s

sr L

LL

pV

T

L

LL

V

Q

L

LV

22

2

33

1

m

s

s

ems

m

s

s

s

ms

sr L

L

pV

T

L

L

V

Q

L

VI

22

2

33

s

ems

s

ss pV

T

V

QI

And this shows that these terms are functions of our desired reference quantities.

),,( emssIs TQVf

),,( emssr TQVf

),,( emssIr TQVf

),,,( remssVr TQVf

Magnitudes are attractive because then we can plot them.

The above relations are given as a function of ωr, but it may be more intuitive to plot them as a function of rotor speed, ωm, where we can compute ωr =sωm/(1-s). You can think of the rotor speed as ωm=(1-s) ωs which shows that for low positive slips, rotor speed is just below synchronous speed, and for low negative slips, rotor speed is just above synchronous speed.

s

ss

V

)( ss Vf

emr

r Tp

P

),(Pr remTf

Level 1 control

57

Fixed Qs=0 Fixed Tem=-1

22

2

33

s

ems

s

ss pV

T

V

QI

• Is is independent of ωm but increases with |Tem| and with |Qs|• Is is the same independent of whether machine is absorbing or supplying vars.• Above equation indicates Is should be the same for Tem=1, Tem=-1. However,

above equation neglected stator resistance Rs. Assuming fixed Vs, in motor mode (Tem=1), Rs causes voltage across rotor circuit to be less, and so Ir must be greater to deliver same torque. In gen mode, Rs causes voltage across rotor circuit to be more, and so Ir must be less to deliver same torque.

Level 1 control

58

22

2

33

1

m

s

s

ems

m

s

s

s

ms

sr L

L

pV

T

L

L

V

Q

L

VI

Fixed Tem=-1

• Ir is independent of ωm for fixed torque but increases as Qs moves from + (absorbing) to – (supplying).

*Im3 rss

mem I

L

LpT

Fixed torque implies fixed rotor current if stator flux is fixed.

Because Tem=Pmechp/ωm, Pmech must decrease as ωm increases.

Level 1 control

59

Both rotor current and stator current equations have real part determined by Qs and imaginary part determined by Tem (Vs is at 90° so real part of currents is in quadrature with Vs)

0<Qs<3Vs2/Lsωs (reactive power into stator, abs)

Magnetized from rotor currentQs=0 (no stator reactive power):

s

s

V

Q

3 Magnetized from stator current.

Magnetized from both currents.

s

ems

s

ss pV

Tj

V

QI

33

m

s

s

ems

m

s

s

s

ms

sr L

L

pV

Tj

L

L

V

Q

L

VI

33

1

m

s

s

ems

s

ems

m

s

s

s

ms

s

s

s

m

s

s

ems

m

s

s

s

ms

s

s

ems

s

srsm

L

L

pV

T

pV

Tj

L

L

V

Q

L

V

V

Q

L

L

pV

Tj

L

L

V

Q

L

V

pV

Tj

V

QIII

333

1

3

33

1

33

Very close to zero since Ls~Lm.Magnetizing component.

ms

s

L

V 1

Qs<0 (reactive power from stator, sup):

Qs=3Vs2/Lsωs (reactive power into stator, abs)

Magnetized from both currents.

Add them to obtain magnetizing current

Level 1 control

60


• Pr linearly decreases w/ ωm for –Tem (gen) and linearly increases w/ ωm for +Tem(mot). • Pr is independent of whether machine is absorbing or supplying vars.

emr

r Tp

P

Remember: ωm=(1-s)ωs, ωr=sωs.

Level 1 control

61


• Vr is linearly decreasing with ωm to ωm=ωs and then linearly increasing with ωm.

• Vr depends mainly on speed of machine.• Vr does not change much with Tem or with Qs because VsLr/ωsLm tends to dominate.

2

2

2

22

33

m

rs

s

s

m

r

s

sr

m

rs

s

emsrr L

LL

V

Q

L

LV

L

LL

pV

TV

Remember: ωm=(1-s)ωs, ωr=sωs.

Level 1 control

62


• Efficiency increases with ωm under all conditions (see next slide):• In the subsynchronous mode, stator windings carry |Pmech|+|Pr|.• In the supersynchronous mode, stator windings carry |Pmech|-|Pr|.

• Efficiency decreases as |Qs| increases (most efficient for unity power factor).• More efficient when absorbing (magnetized from stator) than supplying

(magnetized from rotor)

Generator modes

63

sm

sm

Pm= Pmech

Mode 2

Mode 3

Representing RSC with impedance

64

It can be convenient in analyzing the steady-state performance of the DFIG to represent the RSC as an equivalent impedance, as indicated in the below figure.We can follow our earlier development (see slide 9), but with our RSC equivalent impedance represented:

33

33

3333

RrjsωsLσr

Ers=sEsEs

Rs jωsLσsIs Ir

Vs Vr

rrsrsr ILjsREsV )(

Divide by s

rrsr

sr ILj

s

RE

s

V)(

33

Req

jωrLeq

=jsωs Leq

In slide 9:

rrsrseqseqr ILjsREsLjsRI )()(

Divide by s

rrsr

seqseqr ILj

s

RE

s

LjsRI)(

)(

Now:

rrsr

seqseq

r ILjs

RELj

s

RI )()(


65

33

Req/s

jωs Leq

eqreqeq LjRZ

33

3333

Rr/sjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/s

Equivalent RSC impedance is:

eqseqeqreqeq Ljs

R

s

Lj

s

R

s

Z

Represent it in the circuit with:

Let’s assume the DFIG operates at unity power factor. Then Qs=0, and for Vs=Vs/_0°,

ssssssssslosssairgap IIRVIRIVPPP 33 2,

Question: Do we need to specify motor or generator operation in the above equation?Answer: Not for the relation Pairgap= Ps-Ploss,s (see slide 30). For motor op, Ps>0 and losses subtract so that Pairgap is smaller than Ps, consistent with the fact that power flows from stator to rotor. For gen op, Ps<0 and losses add so that Pairgap is larger than Ps, consistent with the fact that power flows from rotor to stator.However, the relation on the right assumes that Is is a magnitude (positive), and so it is correct for motor op. For gen op, we must use a negative magnitude to get the sign of VsIs correct. We could correct this by writing the RHS as VsIs-RsIs

2= (Vs-RsIs)Is, i.e. use phasor notation for the current instead of just magnitude.

Vm


66

33

Req/s

jωs Leq

33

3333

Rr/sjωsLσr

jωsLmVm

Rs jωsLσsIs Ir

Vs Vr/s

ssssslosssairgap IIRVPPP 3,

From slide 25, we know for the model (with losses) that ems

airgapairgaps

em Tp

PPp

T

Equating the two airgap expressions: ssssems IIRVTp

3

Rewriting, we find a quadratic in Is: 03

2 ems

ssss Tp

IVIR

Obtain roots:

s

emss

ss

s R

Tp

RVV

I2

342

Could be positive (motor) or negative (generator)

With stator current calculated, we can use the circuit to find Vr and Ir….


67

33

Req/s

jωs Leq=jXeq/s

33

3333

Rr/sjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

sssssm LjRIVV From KVL we can compute Vm:

Then compute the magnetizing current Im:

Im

ms

sssss

ms

mm Lj

LjRIV

Lj

VI

Then compute the rotor current Ir:

sms

ssssssmr I

Lj

LjRIVIII

Then compute the rotor voltage Vr:

rsr

rsssss

rsr

rmr

Ljs

RILjRIV

Ljs

RIVsV

/

We can now obtain Zeq/s or Zeq:

r

rsr

rm

r

reqs

eqeq I

LjsR

IV

I

sVLj

s

RsZ

/

/

r

rsrrm

r

reqseqeq I

LjsRIVs

I

VLjsRZ

where Ir is computed from above relations.

(Xeq= ωrLeq)


68

*Im3 rss

mem I

L

LpT

Tem is increasing here.

Req<0converter transfers active power to rotor.

Req>0rotor delivers active power to the converter.

Homework #4

69

Consider a 1.5 MW, 690 v, 50 Hz 1750 rpm DFIG wind energy system. The parameters of the generator are given on the next slide. The generator operates with a maximum power point tracking (MPPT) system so that its mechanical torque Tem is proportional to the square of the rotor speed. The stator power factor is unity. For each of the following speeds: 1750, 1650, 1500, 1350, and 1200 rpm, compute:•Slip•Tem (kN-m)•Vr (volts)•Ir (amps)•Req (ohms)•Xeq (ohms)

What kind of machine is this at 1500 rpm?

Homework #4

70

Homework #4

71

Converter equivalent impedance at 1500 rpm:

sec/0796.157sec60

min2

min

1500rad

rev

radrevm

sec/1592.3140796.157*2 radp mm

Hzf mm 502/1592.3142/ So 1500 rpm is synchronous speed!

Homework #4

72

There is another solution which has very large current and is clearly not realistic.

Be careful here because this solution assumed the direction of current Ir opposite to what we have assumed.

Observe that slip=0. This implies that a DC current flows through the rotor circuit from the converter and the rotor leakage reactance and equivalent reactance are zero. The DFIG is operating like a synchronous machine where the rotor flux is produced by a DC current through a DC exciter.

SCIG Torque-slip characteristic

73

You may recall, from EE 303 or your undergraduate course on electric machines that the torque-slip characteristic of the squirrel-cage induction generator (SCIG) appears as below. One observes that the SCIG operates as a generator only when it is in supersynchronous mode and a motor only when it is in subsynchronous mode.

Motoring

Generating

Let’s see how we obtain this curve for SCIG, and let’s also compare what we do to what we need to do to obtain the analogous curves for the DFIG.

Subsynchronous Supersynchronous

Comparison of equivalent circuits: SCIG vs DFIG

74

33

Req/s

jωs Leq=jXeq/s

33

3333

Rr/sjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

Im

(Xeq= ωrLeq)

33

3333

Rr/sjωsLσr

jωsLm

Rs jωsLσsIs Ir

VsVm

Im

SCIG

DFIG

The difference between the machines in terms of steady-state models is the ability to electrically absorb or supply complex power S via the rotor.

Where do we see rotor losses in these circuits? … (next slide)


75

33

Req

jωs Leq=jXeq/s

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

Im

(Xeq= ωrLeq)

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

VsVm

Im

SCIG

DFIG

Split up the R/s terms in each circuit as R+R(1-s)/s and the rotor losses become immediately apparent.

Where do we see mechanical power in these circuits? … (next slide)

Rr(1-s)/s

Req(1-s)/s

Rr(1-s)/s


76

33

Req

jωs Leq=jXeq/s

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

Im

(Xeq= ωrLeq)

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

VsVm

Im

SCIG

DFIG

The mechanical power is represented by the slip-dependent resistances.

But what do the other two terms in the DFIG circuit represent? … (next slide)

Rr(1-s)/s

Req(1-s)/s

Rr(1-s)/s


77

33

Req

jωs Leq=jXeq/s

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

Im

(Xeq= ωrLeq)

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

VsVm

Im

SCIG

DFIG

These terms represent the real and reactive power exchange between the rotor and the RSC. As we saw on slide 68, these terms, Req and Xeq can be pos (rotor transfers power to RSC) or neg (RSC transfers power to rotor).

How to compute torque in for these machines? … (next two slides)

Rr(1-s)/s

Req(1-s)/s

Rr(1-s)/s


78

Torque equation for SCIG

79

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

VsVm

Im

SCIG

Rr(1-s)/s

s

sRIP rrmech

)1(3 2

Note that the “s” on the denominator provides that Pmech is positive for s>0, motor action, and negative for s<0, generator action.

s

RI

p

s

sRI

s

p

s

sRI

pP

pPT

rr

s

rr

s

rr

mmech

mmech

mem

22

2

3)1(

)1(3

)1(3

1

How to obtain Ir? …. (next slide)

Torque equation for SCIG

80

33

3333

RrjωsLσr

Zm=jωsLm

Zs=Rs+jXσsIs Ir

VsVm

Im

Rr(1-s)/s

Find Thevenin looking in here.ms

msth ZZ

ZVV

ms

msth ZZ

ZZZ

rrth

thr jXsRZ

VI

)/(

3333

RrjωsLσrZthIs Ir

Vth

Rr(1-s)/s

22

2 /33

rthr

th

srthrr

sem

XXsR

R

sRpV

s

RI

pT

Comment: Zm>>ZS, so Vth≈Vs, Zth=Zs is not a bad approximation.

SCIG Torque-slip characteristic

81

You may recall, from EE 303 or your undergraduate course on electric machines that the torque-slip characteristic of the squirrel-cage induction generator (SCIG) appears as below. One observes that the SCIG operates as a generator only when it is in supersynchronous mode and a motor only when it is in subsynchronous mode.

Motoring

Generating

Now let’s take a look at the torque-speed curves for the DFIG…. (next slide)

Subsynchronous Supersynchronous

Torque equation for DFIG

82

33

Req

jωs Leq=jXeq/s

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

s

sRRIP eqrrmech

)1)((3 2

s

RRI

p

s

sRRI

s

p

s

sRRI

pP

pPT

eqrr

s

eqrr

s

eqrr

mmech

mmech

mem

22

2

3)1)((

)1(3

)1)((3

1

How to obtain Ir? …. (next slide)


83

33

Req

jωs Leq=jXeq/s

33

3333

RrjωsLσrRs jωsLσsIs Ir

Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

Find Thevenin looking in here.

33

Req

jωs Leq=jXeq/s3333

RrjωsLσrIs Ir

Vth Vr/s

Req(1-s)/s

Rr(1-s)/sZth

Zm=jωsLm

ms

msth ZZ

ZZZ

ms

msth ZZ

ZVV

)/(/)( sXXjsRRZ

VI

eqreqrth

thr

22

22 /)(3

3

s

XXX

s

RRR

sRRpV

s

RRI

pT

eqrth

eqrth

seqrtheqrr

sem

Comment: Zm>>ZS, so Vth≈Vs, Zth=Zs is not a bad approximation.

Torque-slip characteristic for DFIG

84

So how do we obtain the torque-slip characteristic for the DFIG?

1. Develop values of Zeq for various values of torque-speed control point (slides 66-67):

s

emss

ss

s R

Tp

RVV

I2

342

sssssm LjRIVV

s

ms

ssssssmr I

Lj

LjRIVIII

r

rsrrm

r

reqseqeq I

LjsRIVs

I

VLjsRZ

Aside: The above points result from the turbine control characteristic. This characteristic originates from the maximum power extracted from the wind, which is given by the power curve, described by Pmech~ωm

3. But Pmech=Temωm therefore Tem~ ωm

2.

2. For each value of Zeq, express Tem as a function of s (or ωm= ωs(1-s)) for various values of s. torque-speed control point (slides 66-67):

22

22 /)(3

3

s

XXX

s

RRR

sRRpV

s

RRI

pT

eqrth

eqrth

seqrtheqrr

sem

Torque-slip characteristic for DFIG

85

The sign of Req and Xeq are for rotor current direction defined out of the rotor. These signs reverse for rotor current direction into the rotor as we have done.

Efficiency

86

Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG.

kW

ssRRIP reqrmech

1500

)1667.0/()01667.01)(00263.005375.0()6.1125(3

/)1)((32

2

At 1750, the slip is s=(1500-1750)/1500=-0.1667

The mechanical power supplied to the generator

33

Req

jωs Leq=jXeq/s

33

3333

jωsLσrRs jωsLσsIs Ir

Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

Zm=jωsLm

From your homework, you should compute that Is=1068.2 amperesIr=1125.6 amperes, Req=0.05375 ohms, Xeq=0.02751 ohms.

Efficiency

87


kW

RIP eqrr

29.204

)05375.0()6.1125(3

32

2

The rotor power is

33

Req

jωs Leq=jXeq/s

33

3333


Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

Zm=jωsLm

This power is negative (because Req is negative); it is supersynchronous, therefore it is flowing out of the rotor to the RSC.

From your homework, you should compute that Is=1068.2 amperesIr=1125.6 amperes, Req=-0.05375 ohms, Xeq=-0.02751 ohms.

Efficiency

88


kW

RIP rrrlosses

0.10

)00263.0()6.1125(3

32

2,

The rotor and stator winding losses are

33

Req

jωs Leq=jXeq/s

33

3333


Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

Zm=jωsLm


kW

RIP ssslosses

07.9

)00265.0()2.1068(3

32

2,

Efficiency

89


The stator active power is

33

Req

jωs Leq=jXeq/s

33

3333


Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

Zm=jωsLm


kWIVP ssss 64.1276)180cos(2.10683

690cos3

Efficiency

90


The total power delivered to the grid is

33

Req

jωs Leq=jXeq/s

33

3333


Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

Zm=jωsLm


93.148029.20464.1276|||| rsg PPP

The difference between Pm and Pg is the losses on the stator and rotor windings:

07.1993.14801500|||| ,, slossesrlossesgm PPPP

Efficiency is:%7.98

1500

93.1480

m

g

P

P

DFIG for non-unity power factor

91

“FERC 661-A [1] specifies that large wind farms must maintain a power factor within the range of 0.95 leading to 0.95 lagging, measured at the POI as defined in the LargeGenerator Interconnect Agreement (LGIA) if the Transmission Provider shows, in the system impact study that they are needed to ensure the safety or reliability of thetransmission system..”[1] Order for Wind Energy, Order No. 661-A, 18 CFR Part 35 (December 12, 2005). See also Interconnection for Wind Energy, Order No. 661, 70 FR 34993 (June 16, 2005), FERC Stats. & Regs. ¶ 31,186 (2005) (Final Rule); see also Order Granting Extension of Effective Date and Extending Compliance Date, 70 FR 47093 (Aug. 12, 2005), 112 FERC ¶ 61,173 (2005).

E. Camm and C. Edwards, “Reactive Compensation Systems for Large Wind Farms,” IEEE Transmission and Distribution Conference and Exposition, 2008.

“The Electrical System Operator (IESO) of Ontario essentially requires reactive power capabilities for large wind farms that are equivalent to that for synchronous generators,taking into consideration an equivalent impedance between the generator terminals and the POI [2]. The requirements include:… Supplying full active power continuously while operating at a generator terminal voltage ranging from 0.95 pu to 1.05 pu of the generator’s rated terminal voltage.”

“The Alberta Electric System Operator’s requirements [4] include: The wind farm’s continuous reactive capability shall meet or exceed 0.9 power factor (pf) lagging to 0.95 pf leading at the collector bus based on the wind farm aggregated MW output.”


92

E. Camm and C. Edwards, “Reactive Compensation Systems for Large Wind Farms,” IEEE Transmission and Distribution Conference and Exposition, 2008.


93

“Along with the evolution of wind turbine technology, technical standards of wind generation interconnections become more restrictive. For example, unity power factor has been required for wind generation interconnections in many utilities or control areas in earlier years. Recently, the more restrict requirement with 0.95 lead and lag power factor has been under discussion since the DFIG and full converter wind turbine technology has become mainstream of wind generation interconnection requests.”

I. Green and Y. Zhang, “California ISO experience with wind farm modeling,” IEEE Power and Energy Society General Meeting, 2011.


94

33

Req

jωs Leq=jXeq/s

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

cos3cos3

s

sssss V

PIIVP

Define: φ as power factor angle: -180<φ<180

Identify the current phasor as)sin(cos jII ss

Therefore: )cos

sin1(

3)sin(cos

cos3

j

V

Pj

V

PI

s

s

s

ss

Recalling 222 cos1sin1cossin , we may write

)cos

cos11(

3

2

jV

PI

s

ss

Ps is negative for gen; then cosφ is also negative; Ps is positive for motor; then cosφ is also positive; so Is is always positive.

We have just made the numerator positive for all values of φ.


95

33

Req

jωs Leq=jXeq/s

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

The sign of Ps determines the sign of the real part of the current. Ps is negative if machine is in generating mode (supplying real power).In this case, cosφ is negative because φ is in quadrant 2 or 3.If machine is supplying Q, then sign of Qs should be negative, sign of Im{Is

*} should be negative, and therefore sign of Im{Is} should be positive. Given cos φ is negative:

)cos

cos11(

3

2

jV

PI

s

ss

If machine is absorbing Q, then sign of Qs should be positive, sign of Im{Is*} should

be positive, and therefore sign of Im{Is} should be negative. Given cos φ is negative:

)cos

cos11(

3

2

jV

PI

s

ss

)cos

cos11(

3

2

jV

PI

s

ss


96

33

Req

jωs Leq=jXeq/s

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

Define the magnetizing current factor: ratedsmmrateds

mm IKI

I

IK ,

,

From the circuit, KCL requires: smr III

But the magnetizing current is entirely imaginary: mm jII or ratedsmm IjKI ,

Substitution of the Im expression into the rotor current expression yields:

sratedsmr IIjKI ,

If the machine is absorbing Q, then . Substituting into Ir:

)cos

cos11(

3

2

,

jV

PIjKI

s

sratedsmr

)cos

cos11(

3

2

jV

PI

s

ss


97

33

Req

jωs Leq=jXeq/s

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

)cos

cos11(

3

2

,

jV

PIjKI

s

sratedsmr

Assume the machine is operated at rated power, Ps,rated, and recall

cos3cos3,

,s

ratedsrateds

s

ss V

PI

V

PI Recall from slide 94:

)cos

cos11(

3

2,

,

jV

PIjKI

s

ratedsratedsmr

and the substitute into previous expression :

)cos

cos11(

3cos3

2,,

jV

P

V

PjKI

s

rateds

s

ratedsmr

Factor out the Ps,rated/3Vs….(next slide):


98

33

Req

jωs Leq=jXeq/s

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

)cos

cos11(

3cos3

2,,

jV

P

V

PjKI

s

rateds

s

ratedsmr

Factor out the -Ps,rated/3Vs

)

cos

cos11(

cos3

2,

j

jK

V

PI m

s

ratedsr

Combine terms with “j”

cos

cos1

cos1

3

2, m

s

ratedsr

Kj

V

PI

Simplify

cos

cos11

3

2, m

s

ratedsr

Kj

V

PI


99

33

Req

jωs Leq=jXeq/s

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

So this is for absorbing (underexcited operation)

cos

cos11

3

2, m

s

ratedsr

Kj

V

PI

If we repeat the exercise for supplying (overexcited operation), we will obtain this:

cos

cos11

3

2, m

s

ratedsr

Kj

V

PI

The difference in sign on the square root term indicates higher rotor current is required for overexcited operation than for underexcited operation. No big surprise there!

And so the rotor winding should be rated for the overexcited operation, at rated stator active power output. This would be…. (next slide)


100

33

Req

jωs Leq=jXeq/s

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

Rotor current for rated stator active power and reactive power generation

22

,

cos

cos11

3

m

s

ratedsr

K

V

PI

It is interesting to see the relative magnitude between Ir and Is. Again, from slide 94: cos3cos3

,,

s

ratedsrateds

s

ss V

PI

V

PI

22

,

,, cos

cos11

3

cos3

m

s

rateds

rateds

s

rateds

rrs

K

V

P

P

V

I

IK

22

cos

cos11cos

mrs

KK


101

33

Req

jωs Leq=jXeq/s

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

Rotor current for rated stator active power and reactive power generation

22

,

cos

cos11

3

m

s

ratedsr

K

V

PI

Rotor current for rated stator active power and stator unity power factor

2,0.1 13 m

s

ratedspfr K

V

PI

Ratio of rotor current required for a given stator power factor when supplying Q and that required for unity stator power factor, all at rated stator active power

2

22

0.1 1

cos

cos11

m

m

pfr

r

K

K

I

I


102

33

Req

jωs Leq=jXeq/s

33

3333

RrjωsLσr

jωsLm

Rs jωsLσsIs Ir

Vs Vr/sVm

ImReq(1-s)/s

Rr(1-s)/s

Krs factor for a given stator power factor at rated stator active power.

Ratio of Krs factor for a given power factor to Krs factor for unity power factor, for rated stator active power.

2

22

0.1 1

cos

cos11

cosm

m

pfrs

rs

K

K

K

K

22

cos

cos11cos

mrs

KK

Krs factor for unity stator power factor at rated stator active power.

20.1 1 mpfrs KK


103

Krs factor for a given stator power factor at rated stator active power.

22

, cos

cos11cos

m

rateds

rrs

K

I

IK

ratedsmmrateds

mm IKI

I

IK ,

,

The prime notation on the Krs

at the top of the graph indicates the values have been referred to the rotor for an “a” of about 0.3.


104

Ratio of Krs factor for a given power factor to Krs factor for unity power factor, for rated stator active power.

2

22

0.1 1

cos

cos11

cosm

m

pfrs

rs

K

K

K

K

rateds

rrs I

IK

,

J. McCalley Double-fed electric machines – steady state analysis.

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Transcript of J. McCalley Double-fed electric machines – steady state analysis.