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

2. Four configurations
We will study only this one, the DFIG. 2

3. Basic concepts Rotor is wound: it has 3 windings.AC DC
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. 3

4. Basic Concepts
rotor
(fs: 60 Hz,
p: # of pole pairs)
Balanced voltages applied to stator windings provides a rotating magnetic field of speed
which 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 4

5. Basic concepts Mechanical rad/sec We can manipulate to get:
The induced rotor voltages have frequency of :
Substitution into slip expression above yields:
Observe three modes of operation:
ωm< ωs ωr>0s>0Subsynchronous operation
ωm= ωs ωr=0s=0Synchronous operation
ωm>ωs ωr<0s<0Supersynchronous operation 5

6. Per-phase steady-state model
STATOR VOLTAGE EQUATION:
at fs
=stator voltage with frequency fs
These quantities are referred to stator side.
= emf in the stator windings with frequency fs
= stator current with frequency fs
=stator resistance
=stator leakage reactance
ROTOR VOLTAGE EQUATION:
at fr
These quantities are referred to rotor side, indicated by prime notation.
=rotor voltage with frequency fr
=induced emf in the rotor windings with frequency fr
=induced rotor current with frequency fs
=rotor resistance
=rotor leakage reactance= 6

7. Referring quantities
Application of Faraday’s Law allows the stator back emf and the induced rotor voltage to be expressed as:
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, respectively
φm : magnetizing flux
Solve both relations for φm and equate:
But recall:
The ratio Ks/Kr is normally very close to 1, therefore
Define the effective turns ratio:
Define the induced rotor voltage referred to the stator side: 7

8. Referring quantities We just derived that: (*)
At a locked rotor condition (s=1), the device is simply a static transformer, and we have:
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:
Rotor quantities are referred to the stator-side, indicated by unprimed quantities. Is Rs
jωsLσs
jωrLσr Rr Ir 3
This is locked rotor condition (s=1), therefore ωr=ωs and Ers=Es Es Vs
Ers Vr
We can account for other slip conditions using ωr=sωs and from (*), aE’rs=sEs. 8

9. Referring quantities Is Rs jωsLσs jsωsLσr Rr Ir 3 Es Vs Ers=sEs Vr
Now write the rotor-side voltage equation (referred to stator):
Divide by s
and we get the following circuit: Is Rs
jωsLσs
jωsLσr
Rr/s Ir
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. 3 Es Vs Es
Vr/s 9

10. Referring quantities Is Rs jωsLσs jωsLσr Rr/s Ir 3 3 3 Es Vs jωsLm
Vr/s 10

11. Power relations Is Rs jωsLσs jωsLσr Rr/s Ir 3 3 3 Es Vs jωsLm Vr/s
We modify the above circuit slightly in order to clearly separate slip-dependent terms from loss terms:
Change the circuit accordingly…. 11

12. Power relations Vr(1-s)/s Is Rs jωsLσs jωsLσr Rr Ir 3 3 + - Rr(1-s)/sEs Vs
jωsLm Vr
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:
Power balance relation:
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.
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. 12

13. Power relations Pmech Veq= Vr(1-s)/s Is Rs jωsLσs jωsLσr Rr Ir 3 3 + -
Req= Rr(1-s)/s 3 Es Vs
jωsLm Vr
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.
If Pmech>0the machine is delivering power through the shaft: MOTOR!
If Pmech<0the machine is receiving power through the shaft: GEN!
If 0<s<1Req term is positive Veq term is positiveSupplying P to cct
If 0>s>-1Req term is negative Veq term is negativeConsuming P from cct. 13

14. A first torque expression
Veq=
Vr(1-s)/s Is Rs
jωsLσs
jωsLσr Rr Ir 3 3
+ -
Req= Rr(1-s)/s 3 Es Vs
jωsLm Vr
(p: # of pole pairs)
Recall from slide 5:
and
Therefore: 14

15. A second (equivalent) torque expression
Veq=
Vr(1-s)/s Is Rs
jωsLσs
jωsLσr Rr Ir 3 3
+ -
Req= Rr(1-s)/s 3 Es Vs
jωsLm Vr
Stator power:
Stator voltage:
Substitute Vs into Ps:
The middle two terms are purely imaginary, therefore:
First term is purely real, only the second term contains real and imaginary, therefore: 15

16. A second (equivalent) torque expression
Veq=
Vr(1-s)/s Is Rs
jωsLσs
jωsLσr Rr Ir 3 3
+ -
Req= Rr(1-s)/s 3 Es Vs
jωsLm Vr
Rotor power:
Rotor voltage:
Substitute Vr into Pr:
The middle two terms are purely imaginary, therefore:
First term is purely real, only the second term contains real and imaginary, therefore: 16

17. A second (equivalent) torque expression
Now substitute Ps and Pr into the power balance equation:
Observe we have loss terms added and subtracted in the above, so they go away.
Now consider what happens when you take the real part of a vector multiplied by j (or rotated by 90 degrees): ja a
Observe that
Re(ja) = - Im(a)
Im(a)
Re(ja)
Therefore: 17

18. A second (equivalent) torque expression
Let’s consider another vector identity: taking imaginary part of a conjugated vector:
Observe that
Im(a*) = - Im(a) a
Im(a)
Im(a*) a*
Therefore:
Recall:
Recall:
Therefore: 18

19. Two equivalent torque expressions
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…. 19

20. Additional equivalent torque expressions
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):
Mutual
inductances
Stator winding
Rotor winding
Self inductances
ASIDE: Each self inductance is comprised of mutual and leakage according to:
Therefore:
From stator winding equation:
From rotor winding equation:
Choose one of these equations and substitute into torque expression #2…. 20

21. Additional equivalent torque expressions
From stator winding equation:
From rotor winding equation:
Substitute into torque expression #2….
Using stator winding equation:
Using rotor winding equation:
Purely real
Purely real 21

22. Airgap and slip power
On slides 15 and 16, we derived the following relations for the power into the stator and rotor respectively:
Subtracting losses from both sides, we obtain:
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:
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:
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:
The term 3Im{} in the slip power expression is Pairgap. Therefore: 22

23. Airgap and slip power So we just proved that: where
Our power balance relation states:
Therefore:
Substituting
we obtain
Recall: 
Substituting:   23

24. Approximate relations between active powers
On slides 15 and 16, we derived the following relations for the power into the stator and rotor respectively:
If we neglect the stator losses (3RSIs2) and rotor losses (3RrIr2):
Bring out front the “s” in the rotor power expression and use Re{ja}=-Im(a) (both):
Use Im(a*) = - Im(a) on the rotor power expression
The term 3Im{} in the rotor power expression is PS. Therefore:
Recall the power balance relation:
Neglecting losses:
Substituting Pr expression:
Recall:     24

25. Active power relations - summary
Exact
Both
Approximate 25

26. Power balance With losses Without losses Pairgap Pslip Pairgap Pslip
Pgrid Ps Pr
Pgrid Ps Pr
Ploss,r
Ploss,s
Pmech
Pmech
These figures assume proper sign convention (power flowing to the rotor is positive). 26

27. Generator modes 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)
<0 (mch receives mech pwr)
<0 (mch delivers power via stator)
<0 (mch delivers power via rotor)
3. Generator
s>0, ωm<ωs
(subsynchrnsm)
4. Motor
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<0Ps<0 and Tem<0, s<0 ωr<0Pr<0
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. 27

28. Generator modes Recall the approximate relation
Operation must have |s|<1, so rotor power is always smaller than stator power.
Mode 2
Pm= Pmech
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.
Mode 3
These figures show actual flow direction for generator operation.
They also neglect losses. 28

29. A question on rating Without losses Pairgap Pslip Pgrid Ps Pr
This figure assumes proper sign convention (power flowing to the rotor or into the stator is positive).
A question on rating
Without losses
Pairgap
Pslip
Pgrid Ps Pr
Assume an operating condition such that Pmech=PWTrating. Then
Pmech
For example, consider Pmech=PWTrating=-2 MW.
In supersynchronous mode, with s=-0.3,
Therefore stator winding must be rated for MW.
But in the subsynchronous mode, s=+0.3, then
Question: Does this mean that the stator of a 2 MW turbine must be rated for ?
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: 29

30. Question on sign of losses
Question: Since stator losses (3RSIs2) and rotor losses (3RrIr2) 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?
Motor operation:
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.
Motor operation:
Generator operation:
50 = = 30

31. Per-unitization
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 impedance:
Base inductance:
Base flux:
Base speed:
Three-phase power base:
Base torque: 31

32. Per-unitization – stator side
Once all base quantities are obtained, then per-unitization is easy:
Stator voltage in pu:
Stator current in pu:
Stator flux in pu:
Stator active power in pu:
Stator reactive power in pu:
As usual, only the magnitude is transformed (angle remains unchanged). 32

33. Per-unitization – rotor side
Rotor voltage 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).
Rotor current in pu:
Rotor flux in pu:
Rotor active power in pu:
Rotor reactive power in pu:
As usual, only the magnitude is transformed (angle remains unchanged). 33

34. Per-unitization – torque, speed, R, L
Torque in pu:
On the rotor side, we use the same base quantities as on the stator side (with actual quantities referred to the stator side).
Speed in pu:
Resistances in pu:
Inductance in pu:
Note that the resistances and inductances when expressed in pu are lower case.
As usual, only the magnitude is transformed (angle remains unchanged). 34

35. Voltage equations expressed in per unit
From slides 15, 16, we obtain voltage equations for stator and rotor circuits:
which we rearrange by collecting terms in jωs:
From slide 20, we obtain the equations for stator and rotor flux linkages:
(*)
We recognize the flux linkage expressions in the voltage equations. Therefore:
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: 35

36. Voltage equations expressed in per unit
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:
Now consider the flux linkage equations:
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:
Per-unitize one of the torque equations (#2)
as follows: 
Per-unitize the power expressions to obtain: 36

37. Homework #3 Homework #3: This homework is due Monday, March 26.
A. Using previous relations provided in these slides, derive the following torque expressions.
(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
The stator, Qs
The rotor, Qr
C. For each DFIG condition below, compute Pairgap and Pslip and draw the power flows similar to slide 28.
Pmech=-1 MW with s=+0.30 (subsynchronous operation).
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. 37

38. Homework u (or a) Rs Lσs Lm R’r Lσr Rr Ls Lr Vbase Ibase lσs lm rr lσr38

39. Phasor diagrams for generator operation
We have developed the following relations:
(1) Stator voltage equation
(2) Rotor voltage equation
(3) Stator winding flux equation
(4) Rotor winding flux 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: ….
Vs - Isrs
Isrs Vs Is
Ir=λs/lm – ls Is/lm
lm Is
λs/lm
– ls Is/lm
λr=lm Is+lr Ir
λs= -j(Vs – Isrs)
lr Ir 39

40. Phasor diagrams for generator operation
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:
Vr=Irrr+jsλr, s<0
super-syn
jsλr, s>0, sub-sync
Vr=Irrr+jsλr, s>0
Vs - Isrs
Isrs Vs
Irrr Is
Ir=λs/lm – ls Is/lm
lm Is
jsλr, s<0
λs/lm
– ls Is/Lm
Observe that the angle of Vr is heavily influenced by the sign of s.
λr=lm Is+Lr Ir
λs= -j(Vs – Isrs)
lr Ir 40

41. Question: How to know quadrant of Is?
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
Machine
Observe that current angle is always negative of impedance angle, θi=-θ
Real pwr
Reactive pwr
P>0 motor
R>0
Q>0 absorbing
X>0 Z
Real pwr
Reactive pwr
P<0
gen
R<0
Q>0 absorbing
X>0 Z V V I I
Lag
Lag
Real pwr
Reactive pwr
P>0 motor
R>0
Q<0 supplying
X<0 I
Real pwr
Reactive pwr
P<0
gen
R<0
Q<0 supplying
X<0 I V V Z Z
Lead
Lead 41

42. Example Problem
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:
(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
(a) Synchronous speed
(b) Line-to-neutral voltage
(c) Line current
(a) Synchronous speed:
Alternatively, the synchronous speed was given as 1500 rpm, therefore: 
(b) Line-to-neutral voltage:
(c) Line current:
(d) Stator flux 42

43. Example Problem
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:
(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
(a) Synchronous speed
(b) Line-to-neutral voltage
(c) Line current
(e) Rotor current
This is the referred rotor current!
We can obtain the actual rotor current from a (or u) =0.34:
This phasor is at the rotor frequency, of
fr=sfs=-0.25(50)=-12.5 Hz
(f) Rotor flux
(g) Rotor voltage
Actual rotor voltage: 43

44. Example Problem
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:
(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
(a) Synchronous speed
(b) Line-to-neutral voltage
(c) Line current
(h) Rotor real power
(i) Rotor reactive power
(j) Total real power generated
Comments:
Pm must be larger in magnitude to supply losses
This wind turbine’s rating should be 2.55 MW.
The DFIG stator winding is rated for 2MW. 44

45. Example Problem
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:
(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
(a) Synchronous speed
(b) Line-to-neutral voltage
(c) Line current
(k) Tem 45

46. Wind turbine control levels
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 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. 46

47. Level 1 control
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.
This (open-loop) control not heavily used for DFIGs 47

48. Level 1 control
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. 48

49. Level 1 control
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. 49

50. Level 1 control: Qs equation
From voltage equation (slide 35):
If we neglect drop across the stator resistance (it is typically very small), then:
Substitute into the stator reactive power equation:
Use Im(ja)=Re(a):
From previous slide, note that ɣi is the angle by which Is leads λs , i.e.,
Substituting:
Final equation for Qs: 50

51. Level 1 control: Tem equation
From HW3 (see slide 37):
Again (from phasor diagram), note that ɣi is the angle by which Is leads λs , i.e.,
Substituting:
Final torque equation: 51

52. Level 1 control: Is equation
From phasor diagram:
But recall our Qs and Tem equations:
Substituting into Is equation:
Recall from slide 50: 
Substituting into Is equation: 52

53. Level 1 control: λr equation
From slide 20:
Using these relations, together with:
we may derive: 53

54. Level 1 control: λr equation
Now use the rotor flux equation derived on the previous slide
together with the rotor voltage equation (slide 35):
Neglecting the voltage drop in the rotor resistance, we may derive: 54

55. Level 1 control: summary
Also, we have stator and rotor powers as a function of Tem:  55

56. Level 1 control: magnitudes
Magnitudes are attractive because then we can plot them.
And this shows that these terms are functions of our desired reference quantities.
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. 56

57. Level 1 control Fixed Qs=0 Fixed Tem=-1
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. 57

58. Level 1 control
Fixed torque implies fixed rotor current if stator flux is fixed.
Because Tem=Pmechp/ωm, Pmech must decrease as ωm increases.
Fixed Tem=-1
Ir is independent of ωm for fixed torque but increases as Qs moves from + (absorbing) to – (supplying). 58

59. Level 1 control
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)
Add them to obtain magnetizing current
Very close to zero since Ls~Lm.
Magnetizing component.
Qs=0 (no stator reactive power):
Magnetized from rotor current
0<Qs<3Vs2/Lsωs (reactive power into stator, abs)
Magnetized from both currents.
Qs=3Vs2/Lsωs (reactive power into stator, abs)
Magnetized from stator current.
Qs<0 (reactive power from stator, sup):
Magnetized from both currents. 59

60. Level 1 control Fixed Qs=0 Fixed Tem=-1
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.
Remember: ωm=(1-s)ωs,
ωr=sωs. 60

61. Level 1 control Fixed Qs=0 Fixed Tem=-1
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.
Remember: ωm=(1-s)ωs,
ωr=sωs. 61

62. Level 1 control Fixed Qs=0 Fixed Tem=-1
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) 62

63. Generator modes
Mode 2
Pm= Pmech
Mode 3 63

64. Representing RSC with impedance
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: Is Rs
jωsLσs
jsωsLσr Rr Ir 3 3
Req 3 3 Es Vs
Ers=sEs Vr 3
jωrLeq =jsωs Leq
In slide 9:
Now:
Divide by s
Divide by s 64

65. Representing RSC with impedanceIs Rs
jωsLσs
jωsLσr
Rr/s Ir 3 3
Req/s 3 Vm Vs
jωsLm
Vr/s 3
jωs Leq
Equivalent RSC impedance is:
Represent it in the circuit with:
Let’s assume the DFIG operates at unity power factor. Then Qs=0, and for Vs=Vs/_0°,
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-RsIs2= (Vs-RsIs)Is, i.e. use phasor notation for the current instead of just magnitude. 65

66. Representing RSC with impedanceIs Rs
jωsLσs
jωsLσr
Rr/s Ir 3 3
Req/s 3 Vm Vs
jωsLm
Vr/s 3
jωs Leq
From slide 25, we know for the model (with losses) that
Equating the two airgap expressions:
Rewriting, we find a quadratic in Is:
Obtain roots:
Could be positive (motor) or negative (generator)
With stator current calculated, we can use the circuit to find Vr and Ir…. 66

67. Representing RSC with impedanceIs Rs
jωsLσs
jωsLσr
Rr/s Ir 3 3 Im
Req/s 3 Vm Vs
jωsLm
Vr/s 3
jωs Leq=jXeq/s
(Xeq= ωrLeq)
From KVL we can compute Vm:
Then compute the magnetizing current Im:
Then compute the rotor current Ir:
Then compute the rotor voltage Vr:
We can now obtain Zeq/s or Zeq:
where Ir is computed from above relations. 67

68. Representing RSC with impedance
Tem is increasing here.
Req>0rotor delivers active power to the converter.
Req<0converter transfers active power to rotor. 68

69. Homework #4
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? 69

70. Homework #4 70

71. Homework #4 Converter equivalent impedance at 1500 rpm:
So 1500 rpm is synchronous speed! 71

72. Homework #4
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. 72

73. SCIG Torque-slip characteristic
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
Subsynchronous
Supersynchronous
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. 73

74. Comparison of equivalent circuits: SCIG vs DFIGRs
jωsLσs
jωsLσr
Rr/s Ir 3 3 Im 3 Vm Vs
jωsLm
SCIG
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. Is Rs
jωsLσs
jωsLσr
Rr/s Ir 3 3 Im
Req/s 3 Vm Vs
jωsLm
Vr/s 3
jωs Leq=jXeq/s
DFIG
(Xeq= ωrLeq)
Where do we see rotor losses in these circuits? … (next slide) 74

75. Comparison of equivalent circuits: SCIG vs DFIGRs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3 Im 3 Vm Vs
jωsLm
SCIG
Split up the R/s terms in each circuit as R+R(1-s)/s and the rotor losses become immediately apparent. Is Rs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
DFIG
(Xeq= ωrLeq)
Where do we see mechanical power in these circuits? … (next slide) 75

76. Comparison of equivalent circuits: SCIG vs DFIGRs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3 Im 3 Vm Vs
jωsLm
SCIG
The mechanical power is represented by the slip-dependent resistances. Is Rs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
DFIG
(Xeq= ωrLeq)
But what do the other two terms in the DFIG circuit represent? … (next slide) 76

77. Comparison of equivalent circuits: SCIG vs DFIGRs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3 Im 3 Vm Vs
jωsLm
SCIG
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). Is Rs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
DFIG
(Xeq= ωrLeq)
How to compute torque in for these machines? … (next two slides) 77

78. Comparison of equivalent circuits: SCIG vs DFIG78

79. Torque equation for SCIGIs Rs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3 Im 3 Vm Vs
jωsLm
SCIG
Note that the “s” on the denominator provides that Pmech is positive for s>0, motor action, and negative for s<0, generator action.
How to obtain Ir? …. (next slide) 79

80. Torque equation for SCIGIs
Zs=Rs+jXσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3 Im 3 Vm Vs
Zm=jωsLm
Find Thevenin looking in here.
Comment: Zm>>ZS, so Vth≈Vs, Zth=Zs is not a bad approximation. Is
Zth
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Vth 80

81. SCIG Torque-slip characteristic
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
Subsynchronous
Supersynchronous
Now let’s take a look at the torque-speed curves for the DFIG…. (next slide) 81

82. Torque equation for DFIGIs Rs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
How to obtain Ir? …. (next slide) 82

83. Comparison of equivalent circuits: SCIG vs DFIGRs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
Zm=jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
Find Thevenin looking in here.
Comment: Zm>>ZS, so Vth≈Vs, Zth=Zs is not a bad approximation. Is
Zth
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s
Vth
Req
Vr/s 3
jωs Leq=jXeq/s 83

84. Torque-slip characteristic for DFIG
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):
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~ωm3. But Pmech=Temωm therefore Tem~ ωm2.
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): 84

85. Torque-slip characteristic for DFIG
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. 85

86. Efficiency
Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG. Is Rs
jωsLσs
jωsLσr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
Zm=jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
At 1750, the slip is s=( )/1500=
From your homework, you should compute that Is= amperes
Ir= amperes, Req= ohms, Xeq= ohms.
The mechanical power supplied to the generator 86

87. Efficiency
Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG. Is Rs
jωsLσs
jωsLσr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
Zm=jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
From your homework, you should compute that Is= amperes
Ir= amperes, Req= ohms, Xeq= ohms.
The rotor power is
This power is negative (because Req is negative); it is supersynchronous, therefore it is flowing out of the rotor to the RSC. 87

88. Efficiency
Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG. Is Rs
jωsLσs
jωsLσr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
Zm=jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
From your homework, you should compute that Is= amperes
Ir= amperes, Req= ohms, Xeq= ohms.
The rotor and stator winding losses are 88

89. Efficiency
Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG. Is Rs
jωsLσs
jωsLσr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
Zm=jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
From your homework, you should compute that Is= amperes
Ir= amperes, Req= ohms, Xeq= ohms.
The stator active power is 89

90. Efficiency
Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG. Is Rs
jωsLσs
jωsLσr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
Zm=jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
From your homework, you should compute that Is= amperes
Ir= amperes, Req= ohms, Xeq= ohms.
The total power delivered to the grid is
The difference between Pm and Pg is the losses on the stator and rotor windings:
Efficiency is: 90

91. DFIG for non-unity power factor
“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 Large
Generator Interconnect Agreement (LGIA) if the Transmission Provider shows, in the system impact study that they are needed to ensure the safety or reliability of the
transmission 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 (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 (Aug. 12, 2005), 112 FERC ¶ 61,173 (2005).
“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.”
E. Camm and C. Edwards, “Reactive Compensation Systems for Large Wind Farms,” IEEE Transmission and Distribution Conference and Exposition, 2008. 91

92. DFIG for non-unity power factor
E. Camm and C. Edwards, “Reactive Compensation Systems for Large Wind Farms,” IEEE Transmission and Distribution Conference and Exposition, 2008. 92

93. DFIG for non-unity power factor
“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. 93

94. DFIG for non-unity power factorIs Rs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
Define: φ as power factor angle: -180<φ<180
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.
Identify the current phasor as
Therefore:
Recalling
, we may write
We have just made the numerator positive for all values of φ. 94

95. DFIG for non-unity power factorIs Rs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
jωsLm
Req
Vr/s 3
jωs Leq=jXeq/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:
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: 95

96. DFIG for non-unity power factorIs Rs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
Define the magnetizing current factor:
From the circuit, KCL requires:
But the magnetizing current is entirely imaginary: or
Substitution of the Im expression into the rotor current expression yields:
If the machine is absorbing Q, then
. Substituting into Ir: 96

97. DFIG for non-unity power factorIs Rs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
Assume the machine is operated at rated power, Ps,rated, and recall
Recall from slide 94:
and the substitute into previous expression :
Factor out the Ps,rated/3Vs….(next slide): 97

98. DFIG for non-unity power factorIs Rs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
Factor out the -Ps,rated/3Vs 
Combine terms with “j” 
Simplify  98

99. DFIG for non-unity power factorIs Rs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
So this is for absorbing (underexcited operation)
If we repeat the exercise for supplying (overexcited operation), we will obtain this:
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) 99

100. DFIG for non-unity power factorIs Rs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
Rotor current for rated stator active power and reactive power generation
It is interesting to see the relative magnitude between Ir and Is. Again, from slide 94:
100

101. DFIG for non-unity power factorIs Rs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
Rotor current for rated stator active power and reactive power generation
Rotor current for rated stator active power and stator unity power factor
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
101

102. DFIG for non-unity power factorIs Rs
jωsLσs
jωsLσr Rr Ir
Rr(1-s)/s 3 3
Req(1-s)/s Im 3 Vm Vs
jωsLm
Req
Vr/s 3
jωs Leq=jXeq/s
Krs factor for a given stator power factor at rated stator active power.
Krs factor for unity 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.
102

103. DFIG for non-unity power factor
Krs factor for a given stator power factor at rated stator active power.
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.
103

104. DFIG for non-unity power factor
Ratio of Krs factor for a given power factor to Krs factor for unity power factor, for rated stator active power.
104