Transcript PPT

Double-fed electric machines –
steady state analysis
Set 2
J. McCalley
Example Problem
The 2 MW DFIG (rated stator power) given by the data for HW Prob 4 (and on slide 43 of DFIG set
1) is delivering, from the stator, a load of 2 MW at rated voltage with zero stator reactive power in a
50Hz grid. The slip is s=-0.25 (super-synchronous). Compute: (i) Rotor reactive power
(a) Synchronous speed
(b) Line-to-neutral voltage
(c) Line current
(d) Stator flux
(e) Rotor current
(j) Total real power generated
L
(f) Rotor flux
(k) Tem using T  3 p L Im I 
(g) Rotor voltage
(l) Pmech using Ps+Pr-Pr,loss-Ps,loss
R (1  s )
1 s
(h) Rotor real power & using P  3 I  s   3 s  ReV I 
m
em
s
*
r
s
mech
2
r
r
r
*
r
(a) Synchronous speed: s  2f s  2 (50)  314.16 rad/sec
Alternatively, the synchronous speed was given (table) as 1500 rpm, therefore:
1500rev 2rad min
s 
 157.08rad / sec  s  p s  2(157.08)  314.16rad / sec
min
rev 60 sec
690
0  398.40 volts
*
*
3
6
 Ps    2 10 
*

  
  1673.4180 amps
P

j
0

3
V
I

I

(c) Line current: s
s
s s
 3V s   3  398.40 
(b) Line-to-neutral voltage:
(d) Stator flux
Vs 
V s  I s Rs  js  s
(V s  I s Rs ) 398.40  (1673.4180)2.6 10 3 
 s 

 1.28  90webers
js
j314.16
2
Example Problem
The 2 MW DFIG (rated stator power) given by the data for HW Prob 4 (and on slide 43 of DFIG set
1) is delivering, from the stator, a load of 2 MW at rated voltage with zero stator reactive power in a
50Hz grid. The slip is s=-0.25 (super-synchronous). Compute: (i) Rotor reactive power
(e) Rotor current
(j) Total real power generated
L
(f) Rotor flux
(k) Tem using T  3 p L Im I 
(g) Rotor voltage
(l) Pmech using Ps+Pr-Pr,loss-Ps,loss
R (1  s )
1 s
(h) Rotor real power & using P  3 I  s   3 s  ReV I 
(a) Synchronous speed
(b) Line-to-neutral voltage
(c) Line current
(d) Stator flux
(e) Rotor current
Ir 
 s  Ls I s
Lm
m
em
s
*
r
s
2
r
mech
r
r
*
r
 s  Ls I s  Lm I r
1.28  90  2.587 103 (1673.8180)

 1807.4  16.5amps
3
2.5 10
This is the referred rotor current!
phasor is at rotor frqncy, of f =sf =-0.25(50)=
We can obtain actual rotor current from a (or u) =0.34: This
-12.5 Hz. Our development (slide 8 of DFIG data
r
s
accounts for the rotor circuit being at
I r  a I r  (0.34)1807.4 16.5  614.5 16.5amps set#1)
ω =sω by dividing out the “s”, resulting in an
r
s
effective rotor frequency (for purposes of circuit
analysis) of ωs. The negative frequency indicates
rotating magnetic field from rotor is in opposite
direction to rotational direction of rotor.
(f) Rotor flux  r  Lm I s  Lr I r
 r  2.5 103 1673.8180  2.587 103 1807.4  16.5  1.358  77.4 weber
This term, with λ represents the drop across L and L as
(g) Rotor voltage V r  I r Rr  jss  r
observed in   L ( I  I )  L I
r
3
r
m
m
s
r
r
σr
r
V r  (1807.4  16.5)2.9 10  j(0.25)(314.16)(1.358  77.4)  102.2  165.9volts
V r 102.2  165.9

Vr 
 300.6  165.9
Actual rotor voltage:
a
0.34
3
Example Problem
The 2 MW DFIG (rated stator power) given by the data for HW Prob 4 (and on slide 43 of DFIG set
1) is delivering, from the stator, a load of 2 MW at rated voltage with zero stator reactive power in a
50Hz grid. The slip is s=-0.25 (super-synchronous). Compute: (i) Rotor reactive power
(a) Synchronous speed
(b) Line-to-neutral voltage
(c) Line current
(d) Stator flux
(e) Rotor current
(j) Total real power generated
L
(f) Rotor flux
(k) Tem using T  3 p L Im I 
(g) Rotor voltage
(l) Pmech using Ps+Pr-Pr,loss-Ps,loss
R (1  s )
1 s
(h) Rotor real power & using P  3 I  s   3 s  ReV I 
m
em
s
*
r
s
mech

(h) Rotor real power Pr  3 Re V r I r
*
2
r
r
r
*
r

Pr  3 Re102.2  165.9  (1807.4  16.5)*  0.477 MW

(i) Rotor reactive power Qr  3 Im V r I r
*



Qr  3 Im 102.2  165.9  (1807.4  16.5)*  23.4kVAR
(j) Total real power generated
Comments:
Ps  Pr  2  0.477  2.477MW 1. P must be larger in magnitude to supply losses
m
Pmech  Ps  Pr  Ploss, s  Ploss,r
2. The slip, at -0.25, is large; this is consistent with
the stator winding being at its 2MW rating.
3. Qs=0, Qr=23.4kVARMagnetization is via rotor 4
Example Problem
The 2 MW DFIG (rated stator power) given by the data for HW Prob 4 (and on slide 43 of DFIG set
1) is delivering, from the stator, a load of 2 MW at rated voltage with zero stator reactive power in a
50Hz grid. The slip is s=-0.25 (super-synchronous). Compute: (i) Rotor reactive power
(e) Rotor current
(j) Total real power generated
L
(f) Rotor flux
(k) Tem using T  3 p L Im I 
(g) Rotor voltage
(l) Pmech using Ps+Pr-Pr,loss-Ps,loss
R (1  s )
1 s
(h) Rotor real power & using P  3 I  s   3 s  ReV I 
(a) Synchronous speed
(b) Line-to-neutral voltage
(c) Line current
(d) Stator flux
Tem  3 p
(k) Tem
m
em
s
*
r
s
mech
2
r
r
r
*
r
 
Lm
*
Im  s I r
Ls


2.5 103
*


Tem  3  2
Im
1
.
28

90

1807
.
4


16
.
5

 12.9kNm
3
2.587 10
(l) Pmech
Pmech  Ps  Pr  3 I s Rs  3 I r Rr
2
2
 2 E 6  0.477E 6  3(1673.4)2 (0.0026)  3(1807.4)2 (0.0029)
 2 E 6  0.477E 6  0.021842E 6  0.02842E 6  2.527 MW
2
2
Alternatively, under a slight
Pmech  Ps  3 I s R  Pr  3 I r Rr
rearrangement (for reasons
that will be clear on next slide)

E6
0
.021842
.477
E
6
 0
.02842
E6
2

E6
0


Pairg ap
Pslip
 2.0218E 6  0.50542  2.527 MW
5
Example Problem
The 2 MW DFIG (rated stator power) given by the data for HW Prob 4 (and on slide 43 of DFIG set
1) is delivering, from the stator, a load of 2 MW at rated voltage with zero stator reactive power in a
50Hz grid. The slip is s=-0.25 (super-synchronous). Compute: (i) Rotor reactive power
(a) Synchronous speed
(b) Line-to-neutral voltage
(c) Line current
(d) Stator flux
(e) Rotor current
(j) Total real power generated
L
(f) Rotor flux
(k) Tem using T  3 p L Im I 
(g) Rotor voltage
(l) Pmech using Ps+Pr-Pr,loss-Ps,loss
R (1  s )
1 s
(h) Rotor real power & using P  3 I  s   3 s  ReV I 
m
em
s
*
r
s
mech
2
r
r
r
*
r
(l) Pmech
Now let’s compute it using
 
 R (1  s )   1  s 
*
Pmech  3 I r2  r
  3
 Re V r I r 
s

  s 
 0.0029(1.25)   1.25 
 3(1807.4)2 (0.0029)
  3
 Re(102.2  165.9)(1807.4  16.5)
  0.25    0.25 
 0.1421E 6  2.385E 6  2.527E 6  2.527 MW
Relate this calculation to the one on the previous slide:
 
 R (1  s )   1  s 
*
Pmech  3 I r2  r
  3
 Re V r I r
From previous slide, Pslip=-0.50542 MW. So:
s

  s 
1 s 
Pmech  
Pslip 
1 s  2
1 s 
1 s 
*

 3 I r Rr  3 Re V r I r  
Ploss,r  Pr   
Pslip 
 s 
 s 
 s 
 s 
 1  .25 
 
( 0.50542)  2.527 MW
  .25 
6

 
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
I
characterize the circuit as an impedance
Machine
Z=R+jX=Z/_θ, as indicated. Then we may
express the current according to
Z
V
R  jX 
Z 
I  I i 
V
V
V

  
Z Z Z
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
Real pwr Reactive pwr
P>0
motor
R>0
Q<0
supplying
X<0
Z
Real pwr Reactive pwr
V
I
Lag
I
P<0
gen
R<0
Q>0
absorbing
X>0
Z
Lead
V
I
Lag
Real pwr Reactive pwr
V
Z
P<0
gen
R<0
Q<0
supplying
X<0
I
V
Z
Lead
7
Phasor diagrams for generator operation
We have developed the following (per-unit) relations:
V s , pu  I s , pu rs  j  s , pu
(1) Stator voltage equation
V r , pu  I r , pu rr  js r , pu
 s , pu  ls I s , pu  lm I r , pu
 r , pu  lm I s , pu  lr I r , pu
(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 (in this case, we assume lag).
Step 3: Use (1) to draw the stator flux phasor λs:  s , pu   j (V s , pu  I s , pu rs )
Step 4: Use (3) to draw the rotor current phasor Ir: I r , pu   s , pu / lm  ls I s , pu / lm
Step 5: Use (4) to draw the rotor flux phasor λr:  r , pu  lm I s , pu  lr I r , pu
Step 6: ….
Vs - Isrs
-Isrs
Vs
lm Is
Ir=λs/lm – ls Is/lm
Is
λs/lm
– ls Is/lm
λr=lm Is+lr Ir
λs= -j(Vs – Isrs)
lr Ir
8
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:  s , pu   j (V s , pu  I s , pu rs )
Step 4: Use (3) to draw the rotor current phasor Ir: I r , pu   s , pu / lm  ls I s , pu / lm
Step 5: Use (4) to draw the rotor flux phasor λr:  r , pu  lm I s , pu  lr I r , pu
Step 6: Use (2) to draw the rotor voltage phasor Vr: V r , pu  I r , pu rr  js r , pu
Vr=Irrr+jsλr, s<0
super-syn
lm Is
Vr=Irrr+jsλr, s>0
Irrr
Is
Ir=λs/lm – ls Is/lm
jsλr, s<0
λs/lm
Observe that the angle of Vr is
heavily influenced by the sign of s.
jsλr, s>0, sub-sync
Vs - Isrs
Isrs
Vs
– ls Is/Lm
λr=lm Is+Lr Ir
λs= -j(Vs – Isrs)
lr Ir
9
Wind turbine control levels
Rotor-side converter (RSC) is
controlled so that it provides
independent control of Tem
and Qs. We will study the
steady-state actions of this
particular control function.
Level I: Regulates power flow
between grid and generator.
Level 2 computes reference
quantities for Level 1. Level
2 performs MPPT for
conventional designs, in
which case Level 3 is
absent. Level 3 provides
the ability of the WT to
provide ancillary services.
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.
Source: G. Abad, J. Lopez, M. Rodriguez, L. Marroyo, & G Iwanski, “Doubly fed induction machine: modeling and control for wind energy generation,” Wiley 2011.
10
Level II: MPPT Approach #2 Concept
2. MPPT with optimal tip speed ratio:
“In this method, the maximum power operation of the wind
turbine is achieved by keeping the tip speed ratio to its optimal
value λT,opt. The generator speed ωm is controlled by the power
converters and will be equal to its reference in steady state, at
which the MPPT is achieved.”
The block diagram above shows that if we
have a measured vw, then we can compute
the optimal speed ωm* from knowledge of
the optimal tip speed ratio λT,opt. The optimal
tip speed ratio λT,opt is obtained for a given
pitch from the CP plot (performance
coefficient) to the left, as a function of pitch.
The particular value of λT,opt shown is for a
pitch of θ=3 degrees.
λT,opt
11
Level II: MPPT #3 Concept
Popt=Koptωm3
since Popt=Toptωm
Topt=Koptωm2
Source: O. Anaya-Lara, N.
Jenkins, J. Ekanayake, P.
Cartwright, and M. Hughes, “Wind
energy generation: modeling and
control,” Wiley, 2009.
12
Level II: MPPT #3 Concept
Source: O. Anaya-Lara, N.
Jenkins, J. Ekanayake, P.
Cartwright, and M. Hughes, “Wind
energy generation: modeling and
control,” Wiley, 2009.c
13
Level II: Summary of MPPT #1, 2, 3 Concepts
This leads to three possible maximum power point tracking (MPPT) schemes.
1. MPPT with turbine power profile:
“The wind speed is measured in real-time by a wind speed sensor.
According to the MPPT profile provided by the manufacturer, the power
reference Pm* is generated and sent to the generator control system,
which compares the power reference with the measured power Pm from
the generator to produce the control signals for the power converters.
Through the control of power converters and generator, the mechanical
power Pm of the generator will be equal to its reference in steady state,
at which the maximum power operation is achieved. It is noted that the
power losses of the gearbox and drive train in the above analysis are
neglected and, therefore, the mechanical power of the generator Pm is
equal to the mechanical power PM produced by the turbine.”
2. MPPT with optimal tip speed ratio:
“In this method, the maximum power operation of the wind turbine is
achieved by keeping the tip speed ratio to its optimal value λT,opt. The
generator speed ωm is controlled by the power converters and will be
equal to its reference in steady state, at which the MPPT is achieved.”
3. MPPT with optimal torque control:
“The maximum power operation can also be achieved with optimal
torque control according to equation (2.8), where the turbine
mechanical torque TM is a quadratic function of the turbine speed ωM.
For a given gear ratio and with the mechanical power losses of the
gearbox and drive train neglected, the turbine mechanical torque TM
and speed ωM can be easily converted to the generator mechanical
torque Tm and speed ωm respectively. Figure 2-21 shows the principle
of the MPPT scheme with optimal torque control, where the generator
speed ωm is measured and used to compute the desired torque
reference Tm*. The coefficient for the optimal torque Kopt can be
calculated according to the rated parameters of the generator.
Through the feedback control, the generator toque Tm will be equal to
its reference Tm* in steady state, and the MPPT is realized. It is noted
that there is no need to use the wind speed sensors in this scheme. ”
rT: rotor radius (blade length)
Source:B. Wu, Y. Lang, N. Zargari, and S. Kouro, “Power conversion
and control of wind energy systems,” Wiley, 2011.
14
2012 Tutorial paper on WT Control for APC
ACTIVE POWER
CONTROL
(APC)
A focus on level 3
control (but it also
does good job
covering basics)
http://www.nrel.gov/docs/fy12osti/54605.pdf
Review section II for exam 1.
15
Level 1 control
We achieve Tem
and Qs control
objectives by
controlling rotorside voltage.
DC bus voltage is
controlled by grid-side
converter (GSC) to a predetermined value for
proper operation of both
GSC and RSC.
We control rotor
voltage to achieve a
specified torque and
stator reactive power.
Source: G. Abad, J. Lopez, M. Rodriguez, L. Marroyo, & G Iwanski, “Doubly fed induction machine: modeling and control for wind energy generation,” Wiley 2011.
16
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 will also examine the stator flux, stator current,
rotor current, rotor flux, and stator real power, as shown in the diagram below.
A given.
What results
from our
control.
So our problem is this:
Given desired targets T*em
and Q*s, compute the
required Vr.
The control
variable to
achieve our
targets.
Targets that we
want to achieve.
Source: G. Abad, J. Lopez, M. Rodriguez, L. Marroyo, & G Iwanski, “Doubly fed induction machine: modeling and control for wind energy generation,” Wiley 2011.
17
Level 1 control
We draw the phasor diagram with stator flux as 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 diagram.
It is operating as a motor (stator current is
almost in phase with stator voltage), and
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).
Note that ɣi is the angle
by which Is leads λs.
Source: G. Abad, J. Lopez, M. Rodriguez, L. Marroyo, & G Iwanski, “Doubly fed induction machine: modeling and control for wind energy generation,” Wiley 2011.
18
Level 1 control: Qs equation
From voltage equation (slide 40 of DFIG Set #1): V s  I s Rs  js  s
If we neglect drop across the stator resistance (it is typically very small), then:
V s  js  s
 

Substitute into the stator reactive power equation: Qs  3 Im V s I s  3 Im js  s I s

Use Im(ja)=Re(a): Qs  3 Re s  s I s
*

*
*

From previous slide, note that ɣi is the angle by which Is leads λs , i.e.,
 s  s 0;
Substituting:
I s  I s  i
Qs  3 Res s 0I s    i   3s s I s Re   i 
 3s s I s Recos  i  j sin  i   3s s I s cos  i
Final equation for Qs: Qs  3s s I s cos  i
Source: G. Abad, J. Lopez, M. Rodriguez, L. Marroyo, & G Iwanski, “Doubly fed induction machine: modeling and control for wind energy generation,” Wiley 2011.
19
Level 1 control: Tem equation
From HW3 (see slide 42 of DFIG Set #1): T  3 p Im I 
em
*
s
s
Again (from phasor diagram), note that ɣi is the angle by which Is leads λs , i.e.,
 s  s 0;
Substituting:
I s  I s  i
Tem  3 p Ims 0 I s  i   3 ps I s Im i 
 3 ps I s Imcos  i  j sin  i   3 ps I s sin  i
Final torque equation:
Tem  3 ps I s sin  i
20
Level 1 control: Is equation
From phasor diagram:
I s  I s cos  i  jI s sin  i
But recall our Qs and Tem equations:
Qs  3s s I s cos  i
I s cos  i 
I s sin  i 
Tem  3 ps I s sin  i
Qs
3 s s
Tem
3 p s
Substituting into Is equation:
Is 
Qs
3s s
j
Tem
3 ps
V s  js  s  Vs  s s
Q
T
Substituting into Is equation: I s  s  j s em
3Vs
3 pVs
Recall from slide 19:
So… given the targeted torque and stator reactive power, we
know the necessary stator current. But what we need
(because we can control it) is the rotor voltage…
21
Level 1 control: λr equation
From slide 22 of DFIG Set #1:
Solve for Is, Ir in
Ls I s  Lm I r terms of λs, λr.
s 
 r  Lm I s  Lr I r
L
1
s  m r
Ls
Ls Lr
L
1
Ir  m s 
r
Ls Lr
Lr
Is 
L2m
  1
Ls Lr
Using these relations, together with:
V s  js  s
Qs
sTem
Is 
j
3Vs
3 pVs
we may derive:
 Vs Lr Qs Ls Lr 
r  


s Lm 3Vs Lm 
 V 1 Qs Ls 
Ir   s


s Lm 3Vs Lm 
 sTem Ls Lr 
j

3
pV
L
s
m 

Suggest to do
this on your own
before Exam1.
 T L 
j  s em s 
 3 pVs Lm 
22
Level 1 control: λr equation
Recall the rotor flux equation derived on the previous slide
together with the rotor voltage equation (slide 40 of DFIG Set #1):
V L
Q Ls Lr 
r   s r  s

s Lm 3Vs Lm 
  T L L 
j  s em s r 
 3 pVs Lm 
Neglecting the voltage drop in the rotor resistance,
and equating:
r 
V L
Vr
Q Ls Lr 
 s r  s


jss s Lm 3Vs Lm 
V r  I r Rr  jss  r
V r  js s  r   r 
Vr
js s
 T L L 
j  s em s r 
 3 pVs Lm 
Solving for Vr , switching order of terms, and simplifying:
 Vs Lr Qs Ls Lr 
  sTem Ls Lr 
V r  js s 

 ( js s ) j 



L
3
V
L
3
pV
L
s
m 
s
m 
 s m

  sTem Ls Lr 
 Vs Lr Qs Ls Lr 
 s s 
 js s 



3
pV
L

L
3
V
L
s
m 
s
m 

 s m
Then, using r
 ss
sTem Ls Lr 
V r  r 
 3 pVs
Lm
 Vs Lr Qs Ls Lr 

  jr 


L
3
V
L
s
m 

 s m
23
Level 1 control: summary
Qs
T
 j s em
3Vs
3 pVs
 Vs Lr Qs Ls Lr   sTem Ls Lr 
r  

  j


L
3
V
L
3
pV
L
s
m 
s
m 
 s m

 Vs 1 Qs Ls   sTem Ls 
Ir  

  j


L
3
V
L
3
pV
L
s
m
s
m
 s m

 T L L 
V L
Q Ls Lr 
V r  r  s em s r   jr  s r  s

3
pV
L

L
3
V
L
s
m 
s
m 

 s m
Is 
 s  V s / js
Also, we have stator and rotor powers as a function of Tem (see slide 28 of
DFIG Set #1):
Ps 
s
p
Tem  Pr 
 r
Tem
p
24
Level 1 control: magnetization
 Vs 1 Qs Ls   sTem Ls 
Ir  

  j


L
3
V
L
3
pV
L
s
m
s
m
 s m

Q
T
I s  s  j s em
3Vs
3 pVs


 Vs 1 Qs Ls Qs    sTem Ls  sTem 
Im  



  j


L
3
V
L
3
V
3
pV
L
3
pV
m
m
s
m
s 
 s
s  
s
stator
rotor

+
Recalling that the stator voltage is at 90 degrees, we know that the inductive part
of Im is the real part. This is consistent with the fact that the imaginary part of the
above equation is almost zero, since LS=Lm+Lσs , and Lσs is small, so that Ls≈Lm.
Thus, the magnetizing current is provided by the real part of the above. There are
three “extreme” situations of interest:
1. Magnetized by rotor: When Qs=0, Im is provided by Vs/(ωsLm).
2. Magnetized by stator: When Qs>0 and Vs/(ωsLm)=QsLs/(3VsLm), Im is provided by
Qs/3Vs.
3. Capacitive operation: Here the real term in the above Im expression must be
Vs 1 Qs Ls Qs
V 1 Qs Ls  Qs
negative. This occurs when Qs<0 and:


0 s



L
3
V
L
3
V

L
3
V
L
3Vs
m
m
s
m
s
m
In this case, rotor current is
s
s
s
stator
rotor
increased. BUT….it is possible to do
25
it with DFIG!!!! (SCIG cannot).
Level 1 control: comments
Source: G. Abad, J. Lopez, M. Rodriguez, L. Marroyo, & G Iwanski,
“Doubly fed induction machine: modeling and control for wind
energy generation,” Wiley 2011.
26
Level 1 control: magnitudes
2
2
Q 
 T 
I s2   s    s em 
 3Vs 
 3 pVs 
2
2




V L
Q Ls Lr
sTem Ls Lr
2r   s r  s

 

  s Lm 3Vs Lm   3 pVs Lm 
V
s  s
s
2


 T
V
Q
L
1
I r2   s
 s s    s em
  s Lm 3Vs Lm 
 3 pVs
2



T

L
L
2  Vs
s r
Vr2  r2  s em


r 

3
pV
L
s
m


 s
Pr 
r
p
Tem
Ls 

Lm 
 f Is (Vs , Qs , Tem )
 f r (Vs , Qs , Tem )
 f s (Vs )
2
Q Ls Lr 
Lr
 s

Lm 3Vs Lm 
 f Ir (Vs , Qs , Tem )
2
 fVr (Vs , Qs , Tem , r )
 f Pr (Tem , r )
And this shows that these terms are functions of our desired reference quantities.
The last two relations are given as a function of ωr, but it may be more
intuitive to express them as a function of slip, where we can use ωr =sωs
27
Level 1 control: magnitudes
2
2
Q 
 T 
I s2   s    s em 
 3Vs 
 3 pVs 
2
2




V L
Q Ls Lr
sTem Ls Lr
2r   s r  s

 

  s Lm 3Vs Lm   3 pVs Lm 
V
s  s
s
2
2




V
Q
L

T
L
1
I r2   s
 s s    s em s 
  s Lm 3Vs Lm 
 3 pVs Lm 
2
2





T

L
L
V
L
Q

L
L
2
2
s r
s r
Vr2  s s   s em
 s s   s r  s


3
pV
L

L
3
V
L
s
m 
s
m 

 s m
s
Pr  s Tem
p
 f Is (Vs , Qs , Tem )
 f r (Vs , Qs , Tem )
 f s (Vs )
 f Ir (Vs , Qs , Tem )
 fVr (Vs , Qs , Tem ,m )
 f Pr (Tem ,m )
You can think of the rotor speed using 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.
28
Level 1 control
2
Q 
 T 
I s2   s    s em 
 3Vs 
 3 pVs 
2
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 (and thus Is)
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.
29
Level 1 control
2
V 1
 T L 
Q L 
I r2   s
 s s    s em s 
  s Lm 3Vs Lm 
 3 pVs Lm 
2
Fixed torque implies fixed rotor
current if stator flux is fixed.
Tem
 
Lm
*
Im  s I r
Ls
L
 3 p m s I r sin s   I r 
Ls
Tem  3 p
Therefore fixed torque implies fixed
rotor current if λssin(θλs – θIr) is fixed.
Fixed Tem=-1
If IS (last slide) and Ir are both constant, how can Pmech change?
(Tem=Pmechp/ωm, Pmech must decrease as ωm increases).
Answer: Vr is changingSee next slide.
Last comment here: Ir is independent of ωm for fixed torque but increases as Qs moves
from + (absorbing) to – (supplying). This is consistent with our observation on slide 59,
30
which indicated rotor current would increase for capacitive (supplying) operation.
Level 1 control
2
Ls Lr 
Qs Ls Lr 
2  T
2  Vs Lr


Vr2  s s   s em

s


s

 L

3
pV
L
3
V
L
s
m 
s
m 

 s m
0.5
0
Fixed Qs=0
-0.5
2
Fixed Tem=-1
• Vr is decreasing with ωm to ωm=ωs and then increases 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.
31