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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
Referring quantities
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.
(*)
Referring quantities
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.
Referring quantities
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
A second (equivalent) torque expression
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
A second (equivalent) torque expression
17
*2 Re33 rsmsrrr IILjsIRP
Now substitute Ps and Pr into the power balance equation:
rlossslossrsmech PPPPP ,,
*2 Re33 srmssss IILjIRP
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
A second (equivalent) torque expression
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 srmssss IILjIRP
*2 Re33 rsmsrrr IILjsIRP
*2 Re33 srmssssairgap IILjIRPP
*2 Re33 rsmsrrrslip IILjsIRPP
airgapslip sPP
rlossslossrsmech PPPPP ,,
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 ,,
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).
As usual, only the magnitude is transformed (angle remains unchanged).
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:
As usual, only the magnitude is transformed (angle remains unchanged).
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
Real pwr Reactive pwr
P>0 motor
R>0
Q<0 supplying
X<0
Real pwr Reactive pwr
P<0 genR<0
Q>0 absorbing
X>0
Real pwr Reactive pwr
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
(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
(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
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
44
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:
(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
(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
rlossslossrsmech PPPPP ,,
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
(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
(k) Tem *Im3 rs
s
mem I
L
LpT
kNmTem 9.125.164.18079028.1Im10587.2
105.223 *
3
3
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:
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
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.
emr
r Tp
P
Remember: ωm=(1-s)ωs, ωr=sωs.
Level 1 control
61
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.
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
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)
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 )()(
Representing RSC with impedance
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
Representing RSC with impedance
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….
Representing RSC with impedance
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)
Representing RSC with impedance
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)
Comparison of equivalent circuits: SCIG vs DFIG
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
Comparison of equivalent circuits: SCIG vs DFIG
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
Comparison of equivalent circuits: SCIG vs DFIG
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
Comparison of equivalent circuits: SCIG vs DFIG
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)
Comparison of equivalent circuits: SCIG vs DFIG
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
Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG.
kW
RIP eqrr
29.204
)05375.0()6.1125(3
32
2
The rotor power is
33
Req
jωs Leq=jXeq/s
33
3333
jωsLσrRs jωsLσsIs Ir
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
Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG.
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
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.
kW
RIP ssslosses
07.9
)00265.0()2.1068(3
32
2,
Efficiency
89
Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG.
The stator active power is
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.
kWIVP ssss 64.1276)180cos(2.10683
690cos3
Efficiency
90
Consider our HW assignment, at a speed of 1750 rpm and unity power factor. Compute the efficiency of the DFIG.
The total power delivered to the grid is
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.
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.”
DFIG for non-unity power factor
92
E. Camm and C. Edwards, “Reactive Compensation Systems for Large Wind Farms,” IEEE Transmission and Distribution Conference and Exposition, 2008.
DFIG for non-unity power factor
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.
DFIG for non-unity power factor
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.
DFIG for non-unity power factor
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
DFIG for non-unity power factor
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
DFIG for non-unity power factor
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):
DFIG for non-unity power factor
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
DFIG for non-unity power factor
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)
DFIG for non-unity power factor
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
DFIG for non-unity power factor
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
DFIG for non-unity power factor
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
DFIG for non-unity power factor
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.
DFIG for non-unity power factor
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
,