Aps Chap10(SynPwrSys)
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Transcript of Aps Chap10(SynPwrSys)
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Synchronous Machines in Power Systemsand Drives
Most of the electrical power generators are three-phase synchronous generators
Synchronous motors are competitive in higherpower ranges because of efficiency and lower costs
Reluctance and permanent motors are popular atlower power ranges
Synchronous generator in power systems transient stability study: maintain synchronism from
large oscillations caused by a transient disturbance dynamic stability study: small signal behavior and
stability about some operating point
long-term dynamic energy balance study: dynamics of
slower acting components
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Stability Studies
Sub-transient time constant of machine is between0.03 to 0.04 second, shorter thanelectromechanical oscillation
Electromechanical oscillation frequency betweensynchronous generators in a power system lies
between 0.5 to 3 Hz (0.33 to 2 second) Transient time constant of machine is between 0.5
to 10 second which is longer than the period ofelectromechanical oscillation
Slower acting component with longer timeconstants such as boilers and AGC response mayneed more time between 10sec to 2 min
Different model shall fit into the different analysis
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Basic Dynamics of Synchronous Generators
Basic dynamic behavior of synchronous generatorin transient situations:
voltage behind the transient reactance of a generatorand network
E=Eth+ j Xt I, Xt = Xd + Xth
One-line diagram Circuit Equivalent
take Thevenins voltage as reference phasor
Eth= Eth0, and E=E
ACNetwork
E
jXd jXth
Eth
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Basic Dynamics of Synchronous Generators
electrical output power of the generator
Pgen= R(EI*) = pu
from the above equation, we can see that power transfercharacteristic for the system is a sine wave with max value EEth/Xt
the rotor motion without damping
Pmech-Pgen= pu
replace the d/dt with d2/dt2, we obtain swing equation
If machine was to maintain synchronism, excursion of would bebounded and (d/dt) would have return to zero
sin'
t
th
X
EE
dt
dH
b
2
( ) =
=
dPPHpudt
dHPP genmech
b
b
genmech2dt
dor
22
2
2
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Transient Power Angle Characteristics
Pmech1
Equal Criteria: A1 = A2
A1 < A2max Stable
A1 = A2max Critically StableA1 > A2max Unstable
A1
A2
0 SS max
Pmech0
-SS
A2max
t0
Pmech1
0
t
max
SS
transient power angle characteristics
+=max
min
max
min
)()()(
ss
ss
dPPdPPdPP genmechgenmechgenmech
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Transient Power Curve And Dynamics
Without damping loss, rotor oscillates about SS. Otherwise, eventually settlesto SS.
The gain in rotor momentum could carry beyond critical angle - SS whichPmech1>Pgen and rotor accelerates to lose synchronism
For purpose of determining the synchronism of the machine, area of A2maxshould be larger than area of A1 (area of A2max is from SS to - SS)
Transient power angle curve may be raised by increasing the excitation controlofE
As a need to give a high speed control of E would introduce a negativedamping and adversely affect the dynamic stability, see [103]
The power system stabilizer (PSS) is introduced to obtain a better transient
performance over the control of excitation system adverse impact ofPSS: interaction of PSS and torsional mode of turbine shaft
gives rise to sub-synchronous oscillations
Transient stability study is mainly concerned in the synchronous generator, toswitch from motor notation to generator notation, it is required to invert the
sign of all stator currents in the voltage equation, flux linkage equation, andtorque equations.
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Transient Model with d,q Field Windings
In this chapter, there are two more models to express the synchronous
machine dynamics: transient model and sub-transient model The model difference between transient model in this chapter and that
in chapter 7 is that transient model in this chapter uses more machineparameters directly obtained from standard tests, such as reactancesand time constants
For derivation of transient model equations, please see pp. 468-474
Transient model (without damper winding): state variable d, q stator winding equations
rotor winding equations
Torque
qrd
d
q
d
sddr
q
qd
q
sq
dt
dE
L
rv
dt
dE
L
rv
+
=++
=
'
'
'
'
qr
q
qq
gd
q
qdqodr
d
ddfq
d
dq
doL
LLEE
L
L
dt
dET
L
LLEE
L
L
dt
dET
=+
+=+
'
'
'
'
''
'
''
'
'
'
+= qdqddr
qq
qr
ddem LLL
E
L
EP
T
'''
'
'
' 11
22
3
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Transient Model with d,q Field Windings
Simplification of transient model
in the transient analysis, damper windings are no longeractive
in transient stability prediction, rotor winding transientare dominant.
first swing of the rotor would be the interval of interest,and the rotor transients vary at the rate ofTdo and Tqo
the rotor transient would impact speed voltage term,
r
d,
r
q, and greater than that of d
q/dt, d
d/dt.
Therefore, the effect of dq/dt, dd/dt could beneglected
the transient model can be further simplified byneglecting dq/dt, dd/dt
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Transient Model Equations
Simplified transient model equations (r=e except forrotor mechanical dynamics)
Stator winding equations (see pp.473)
Rotor winding equations
Torque Equation:
dqdqqdsd
ddqqqddqsq
iiEixirv
EvEvEixirv
,:outputs
,,,:inputs
''
''''
++=
+=
''''''
''
'
'
,:outputs)(
,,,:inputs)(
qdqqqgdd
qo
qdgfdddfq
q
do
EEixxEEdtdET
iiEEixxEEdt
dET
+=+
=+
{ } N.m)(22
3 ''''qddqddqq
e
em iixxiEiEP
T ++=
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Transient Model Equations
Rotor equations
( ){ }
rmer
pudamppumechpuember
dampmechem
dampmechemrm
P
TTTdt
dH
TTT
TTTdt
dJ
2,
dt
d:output
(pu)/
2
:inputs
(N.m)
re
r
)()()(
==
+=
+
+=
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Transient Model Block Diagram
Stator Block
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Transient Model Block Diagram
Rotor Block
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Transient Model Block Diagram
Field voltageequation
d axis
q axis
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Transient Model Block Diagram
Overall block diagramfrom excitationsystem
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Synchronous machine model in Chap 7
Overall block diagram
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Project 10-1 Fault tests of Synchronous Machine
(Homework): You are given a synchronous machine model with the
machine parameters given in Table 10.7 Set 1 to construct thetransient synchronous machine model. The machine is connected tothe following source:v1=12sin(120t+0) puv2=12sin(120t-2/3) puv3=12sin(120t+2/3) pu
1. With excitation reference voltage Ef= 1pu, Tmech = 1 pu(mechanical torque), apply three-phase bolted fault toground at t=10 sec, fault clear at t=10.25 sec, observeand plot
a. vq, vd, iq, id, in one figureb.
ia, ib, ic in one figurec. Pgen, Tem, , in one figured. Qgen, If, in one figuree. Show the critical fault clearing time and plot vs. time with
stable and unstable conditionsf. discuss what you see on the plots (ex. observe transient in field
current and qd, abc current)
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2. With excitation reference voltage Ef= 1pu, Tmech = 1 pu(mechanical torque), apply single phase to ground fault onphase c at t=10 sec, fault clear at t=10.25 sec, observeand plot1) vq, vd, iq, id, in one figure2) ia, ib, ic in one figure3) Pgen, Tem, , in one figure4) Qgen, If, in one figure5) discuss what you see on the plots (ex. observe transient in field
current and qd, abc current)
Suggestion:the figure time scale can be shown starting from t=9 secthrough the time when system becomes stable after thefault cleared
Project 10-1 Fault tests of Synchronous Machine
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EXCITATION SYSTEMS
Scheme of Excitation systems contains
pilot exciter main exciter to provide field winding voltage/current of synchronous machine slip rings (optional) automatic voltage regulator
Classification of excitation systems dc excitation
primary excitation power is from dc generator whose field winding is on the same shaft
as rotor of synchronous generator
rotatingpart
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EXCITATION SYSTEMS
Classification of excitation systems ac excitation (static)
field winding of alternator is on the same shaft as the rotor of thesynchronous machine
alternators stator and rectifier are stationary
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EXCITATION SYSTEMS
Classification of excitation systems ac excitation (rotary)
armature of alternator and rectifier are on the same shaft as therotor of the synchronous machine
alternators rotor field winding is stationary
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EXCITATION SYSTEMS
Classification of excitation systems ac excitation (from ac bus)
pilot exciter function is replaced by ac bus voltage
use controllable rectifier to adjust dc excitation
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EXCITATION SYSTEMS
Overall scheme of excitation systems detector, regulator, exciter, stabilizer, diode bridge, power
system stabilizer components
detector
regulator
exciter
diode bridge
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COMPONENTS OF EXCITATION SYSTEMS
voltage transducer and load compensation circuit voltage transducer and rectifier are modeled by a
single time constant with unity gain
compensation of excitation voltage due to internalload is represented by RC+jXC
compensatortransducer
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COMPONENTS OF EXCITATION SYSTEMS
voltage regulator consists of an error amplifier with limiter transient gain reduction can be achieved by adding a zero-pole
compensator
zero-pole (lead-lag) compensator
error amplifier with limiter
transient gainreduction Tc
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COMPONENTS OF EXCITATION SYSTEMS
Exciter output signal from regulator must be
amplified by the exciter before it isused to excite the field winding ofthe synchronous machine
the resistance and inductance of thearmature winding of exciter is
neglected due to the small numberof turns voltage of the field winding and
armature winding in exciter are:
field current can be expressed interms of saturation function Se andarmature voltage vx
( )
exciterofvoltagearmatureis
,,
x
xfx
ff
fff
v
iifv
dt
idriv =+=
xevB
exexe
ag
xf ASvSR
vi expwhere, =+=
saturation part
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COMPONENTS OF EXCITATION SYSTEMS
Stabilizer provide more phase margin in the open-loop frequency response ofregulator/exciter loop (add zero to increase stability) transient gain reduction (to counter negative damping) can be achieved by
adding a zero-pole compensator with a proper value of TF or in TC and TB
regulator
exciter
stabilizer
PSS
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COMPONENTS OF EXCITATION SYSTEMS
Exciter
substitute if in vfequation and we get vfpu equation
(1)
transfer function of the exciter, integrate (1)
( )
( )agbase
base
f
E
xpu
xpuf
E
xpu
Expuxpuepu
base
f
Efpu
RRR
rK
dv
vd
where
dt
dvvvS
R
rKv
===
+
+=
,,
( )
( )
+=
+
+=
dtvvSR
rKdtvv
dtdt
dvdtvvSRrKdtv
xpuxpuepu
base
f
Efpu
E
xpu
xpu
Expuxpuepu
base
f
Efpu
1
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COMPONENTS OF EXCITATION SYSTEMS
Exciter
transfer function of the exciter
block diagram of the exciter
KE
( )
( )
+=
+=
dtvvSRrKvv
dtvvSR
rKdtvv
xpuxpuepu
base
f
Efpu
E
xpu
xpuxpuepu
base
f
Efpu
E
xpu
1
1
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COMPONENTS OF EXCITATION SYSTEMS
Diode bridge (optional)
mode 1: dc voltage output: Vd=Vdo-RCId mode 2: dc voltage output:
end of mode 3:
range of three modes of a diode bridge rectifier
Sdo VVwhere
33 =
2
3
21
2
3
=
S
dCdod
V
IXVV
Cessd LVII /3/2 2 ==
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COMPENSATION OF EXCITATION SYSTEMS
Why to we need the Power System Stabilizer (PSS) As a need to give a high speed control of E would introduce a negative damping and
adversely affect the dynamic stability, see [103]
The power system stabilizer (PSS) is introduced to obtain a better transientperformance over the control of excitation system
adverse impact ofPSS: interaction of PSS and torsional mode of turbine shaft givesrise to sub-synchronous oscillations
Instability problem of exciter even the amplifier gain KA is small, AVR step response would be likely cause system
unstable
Solution to the instability of exciter introduce a controller which add a zero to AVR open loop transfer function
How to add a zero to AVR open loop transfer function? add a rate feedback to the control system by properly adjust KF and F model of rate feedback regulator exciter
stabilizer
PSS
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COMPENSATION OF EXCITATION SYSTEMS
Power system stabilizer (PSS)
filter to suppress the frequency component in the input signal that couldexcite undesirable interactions wash-out circuit for reset action to eliminate steady offset two phase (lead-lag) compensator to make phase compensation (phase
margin), compensation center frequency at , limiter to prevent output of PSS from driving exciter into heavy saturation
stabilizer
PSS
212/1 TT 432/1 TT
preventsaturation compensate
frequency bandwidtheliminatedc offset
suppressundesired frequency
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SIMULATION OF EXCITATION SYSTEMS
Overall scheme ofsimplified excitation systems regulator, exciter, stabilizer components
regulator
exciter
stabilizer
VF
PSS
Project 10 2 Excitation tests of Synchronous
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Project 10-2 Excitation tests of SynchronousMachine
(Homework):You are given a synchronous machine model
with the machine parameters given in Table 10.7 Set 1 toconstruct the transient synchronous machine model withthe excitation system. The machine is connected to thefollowing source:v1=12sin(120t+0) puv2=12sin(120t-2/3) puv3=12sin(120t+2/3) pu
1. With excitation reference voltage Vref= 1pu, Tmech = 1 pu(mechanical torque), change Vref= 0.5pu at t=10 sec,observe and plot
a. vq, vd, iq, id, in one figureb. ia, ib, ic in one figurec. Pgen, Tem, , in one figured. Qgen, If, in one figure
e. discuss what you see on the plots (ex. observe transient in fieldcurrent and qd, abc current)
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2. With excitation reference voltage Vref= 1pu, Tmech = 1 pu(mechanical torque), change Vref= 1.5pu at t=10 sec,observe and plot
a. vq, vd, iq, id, in one figureb. ia, ib, ic in one figure
c. Pgen, Tem,, in one figured. Qgen, If, in one figure
e. discuss what you see on the plots (ex. observe transient infield current and qd, abc current)
Suggestion:
the figure time scale can be shown starting from t=9 secthrough the time when system becomes stable after thefault cleared
Project 10-2 Fault tests of Synchronous Machine
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Case 1: Transient Models (single machine)
Case: one machine is connected to a simple externalnetwork:Vz = (re+ j xe) IZ
Such a phasor quantity could be expressed in qdcomponents ofsynchronous reference frame: Vz = vqz
e jvdz
e and Iz = iqe j id
e
To incorporate with generator side parameter in rotor frame,bus voltage of synchronous reference frame should be
transformed from synchronous frame into rotor frame bymultiplying e-j
The rotor frame voltage can be expressed as:vq
r jvdr = e-j (vqz
e j vdze)= (re+ j xe) e
-j (iqe j id
e )= (re+ j xe) (iq
r j idr)
re jxe
+- VZ
I
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Case 1: Transient Models (single machine)
The external line drop ofre+ j xe can be directlyadded to the stator winding voltage equations.
The infinite bus voltages in phasor quantity shouldbe expressed in qd quantities and the synchronousframe needs to be transformed into rotor frame
vqr jvd
r = e-j (vqe j vd
e)
''
''
)()(
)()(
dqeqdesd
qdedqesq
Eixxirrv
Eixxirrv
++++=
+++=
0,2,0~
~2,
~2
=====
eda
eq
oaa
a
e
d
e
qa
e
d
e
q
vVvVVfor
IjiiVjvv steady state to qd
from synchronousframe to rotor frame
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Case 1: Transient Models (single machine)
Stator module with external network stator resistor: rs+re stator reactance: xd+xe, xq+xe
( ) ( )( )eqedesZ xxxxrrD ++++=
''2
11
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Case 1: Transient Models (single machine)
qd synchronous to rotor frame module transform bus voltage of synchronous reference to rotor
reference value
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Case 1: Transient Models (single machine)
Overall synchronous generator transient model
rotor winding
rotor winding
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Case 2: Multi-machines System
Main interests in the study of multi-machine
examine the interactions between generators
transients of the electro-mechanical oscillations
check whether the generators will maintain in synchronism
Case study, two machines interconnected with externalbuses
Four bus test systemr14
jx14
I4
r24jx24
I2
gen1
gen2
r34jx34
load4
I1
3
41
2 bus
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Case 2: Multi-machines System
Model setup: network is expressed by [ ie ]=[ Y ][ Ve ] in synchronous reference frame
machine model is in rotor reference frame
needs a module between stator and network to convert quantities (v, i) ofsynchronous frame to rotor frame
network matrix should have voltages (Eqd, Vqd) or injected currents (iqd) as
inputs and associated currents iqd as outputs to feed the inputs ofgenerator model or voltages (vqd) as output of injected bus (load bus)
Eqpe1
Eqpe2
Edpe2
Edpe1
iqe1
ide1
iqe2
ide2
1.
vqe3
0
vde3
network
iqe4
ide4T5
T4
T3
T2
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Case 2: Multi-machines System
Incorporate stator voltage equation into network equation stator voltage equation is in rotor frame
network equation is in synchronous frame and expressed in phasor form
to incorporate these two sets of equations together, transform statorequation in synchronous frame whose q-axis is aligned with referencephasor
stator voltage in synchronous frame and phasor form:
the fixed stator impedance of (rs+jxd) can now easily added into Zbus or
Ybus of the network matrix can be obtained from simulation of rotor field winding equation if
Thevenin equivalent circuit is used
'E~
I~
)(V~
)())(()(
'
'''
++=
++=
ds
dqje
deqds
ed
eq
jxr
jEEejiijxrjvv
'~E
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Case 2: Multi-machines System
Incorporate stator voltage equation into networkequation combine stator admittance with network admittance
Y11=(g14+ggen1)+j(b14+bgen1) Y14=Y41=-(g14+jb14)
Y22
=(g24
+ggen2
)+j(b24
+bgen2
) Y24
=Y42
=-(g24
+jb24
)
g14jb14
I4
g24jb24
I2
gen1
gen2
g34jb34
load4
I1
3
41
2
ggen1jb
gen1
ggen2jbgen2
Eq1-jEd1
Eq2-jEd2
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Case 2: Multi-machines System
Bus admittance matrix of the network including transient admittance oftwo generators
=
ed
eq
e
d
e
q
ed
eq
ed
eq
ed
eq
e
d
e
q
ed
eq
ed
eq
jvvjvv
jEE
jEE
YYYY
YYYY
YYYY
YYYY
jiijii
jii
jii
44
33
'2
'2
'1
'1
44434241
34333231
24232221
14131211
44
33
22
11
choose bus 4 voltage as output and bus 4 injecting current as input forload or fault current
=
ed
eq
ed
eq
ed
eq
ed
eq
ed
eq
ed
eq
ed
eq
ed
eq
jii
jvv
jEEjEE
jvv
jii
jiijii
44
33
'2
'2
'1
'1
44
33
22
11
gyratedY
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Case 2: Multi-machines System
Model setup: network is expressed in [ ie ]=[ Y ][ Ve ], Y is complex
matrix with Gij+jBij (conductance and susceptance), ie is
matrix with iqe+jid
e, Ve is matrix with Vqe+jVd
e
need to separate xq+jyd components into q, d
components method to separate complex quantities
(iq - jid)=(G + jB)(vq - jvd) into q, d quantities:
matrix gyration to reform input and output components
base VA ratio of network and generator
=
d
q
d
q
v
v
GB
BG
i
i
gen
sys
gen
net
S
S
i
i=
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Case 2: Multi-machines System
Overall model diagram
Edpe1
Eqpe1
Eqpe2
Edpe2
ide1
iqe1
iqe2
ide2
1.
vqe3
0
vde3
tmodel1
tmodel
network
Initialize
and plot
m2
iqe4
ide4
vref(2)
Vref2
vref(1)
Vref1
y2
To Workspace1
y1
To Workspace
Tmech(2)
Tmech2
Tmech1
T5
T4
T3
T2
T1
T
Sbratio(2)
Sys/Gen2VA_
-K-
Sys/Gen2VA
Sbratio(1)
Sys/Gen1VA_
-K-
Sys/Gen1VA
UU(E)
Selector1
UU(E)
SelectorScope1
Scope
Mux Mux_
Mux Mux
Clock1
Clock
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Case 2: Multi-machines System
Model setup: Network module
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Case 2: Multi-machines System
Inside generator model : generator model
Eqp_
|Vt|
iq_
id_
iq
id
delta
Ef
vqt
vdt
9
out_Edpe
8
out_Eqpe
7
out_Tem
6
out_puslip
5
out_delta
4
out_Qgen
3
out_Pgen
2
out_|I|
1
out_|Vt|
sum
stator_wdg
qdr2qde
qde2qdr
exciter
VIPQ
Sw
Sum
Rotor
xd(1)-xpd(1)
Gain2
1/Tpqo(1)
Gain1
xq(1)-xpd(1)
Gain
Exc_sw(1)
Exc_sw
1
s
Eqp
1
s
Edp
1/Tpdo(1)
1/Tpdo
4
in_Tmech
3
in_ide
2
in_iqe
1
in_Vref
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Case 2: Multi-machines System
Inside generator model:
stator module inputs: Eq, Ed
, iq, id not Eq, Ed
, vq, vd stator module outputs: vq, vd
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Case 2: Multi-machines System
Inside generator model :
excitation system: Ef=> (vref-vfb), (1/s) andfeedback loop.
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Case 2: Multi-machines System
Inside generator model : exciter
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Case 2: Multi-machines System
Inside generator model :
rotor blockr
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Case 2: Multi-machines System
Inside generator model : generator model
Eqp_
|Vt|
iq_
id_
iq
id
delta
Ef
vqt
vdt
9
out_Edpe
8
out_Eqpe
7
out_Tem
6
out_puslip
5
out_delta
4
out_Qgen
3
out_Pgen
2
out_|I|
1
out_|Vt|
sum
stator_wdg
qdr2qde
qde2qdr
exciter
VIPQ
Sw
Sum
Rotor
xd(1)-xpd(1)
Gain2
1/Tpqo(1)
Gain1
xq(1)-xpd(1)
Gain
Exc_sw(1)
Exc_sw
1
s
Eqp
1
s
Edp
1/Tpdo(1)
1/Tpdo
4
in_Tmech
3
in_ide
2
in_iqe
1
in_Vref
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Case 2: Multi-machines System
Overall model diagram
Edpe1
Eqpe1
Eqpe2
Edpe2
ide1
iqe1
iqe2
ide2
1.
vqe3
0
vde3
tmodel1
tmodel
network
Initialize
and plot
m2
iqe4
ide4
vref(2)
Vref2
vref(1)
Vref1
y2
To Workspace1
y1
To Workspace
Tmech(2)
Tmech2
Tmech1
T5
T4
T3
T2
T1
T
Sbratio(2)
Sys/Gen2VA_
-K-
Sys/Gen2VA
Sbratio(1)
Sys/Gen1VA_
-K-
Sys/Gen1VA
UU(E)
Selector1
UU(E)
SelectorScope1
Scope
Mux Mux_
Mux Mux
Clock1
Clock
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Project 10-3 Multi-synchronous machines Project
Read carefully on project 2 in 10.9.2: multi-machines system Use the simulation model (machine parameters are in Set 1 of TABLE 10.7) to
run the simulation as follow: run the simulation to create plots as figure 10.24 (a), (b), and (c). In this case,
step changes in torque is applied at generator 2. As you can see in the figure,machine originally operate in Tmech = 0.8pu, a step change in torque to 0.9pu att=7 sec, then a step change to 0.7pu at t=15 sec, finally a step change to 0.8 puat t=22 sec. Use the line impedances (in pu) as follow: z14 = 0.004+j0.1, z24 =0.004+j0.1, z34 = 0.008+j0.3, y40=1.2-j0.6, report and comment on the figures.
run the similar simulation as above but increase the line impedance of z14 =0.016+j0.4, z24 = 0.016+j0.4 (decrease the electrical strength), plot resultssimilar to figure 10.24(a,b,c) and observe the interaction of generator 1 and 2due to the change of electrical strength z14 and z24, report on the difference dueto the change of electrical strength
Tmech2 = 0.8pu and Tmech1 = 0pu, a fault current of iq4e-jid4
e = -(2-j2) pu is to beintroduced at t=5 sec.
the fault duration is 0.15 seconds. Use the line impedances (in pu) as follow: z14= 0.004+j0.1, z24 = 0.004+j0.1, z34 = 0.008+j0.3, plot results similar to figure10.25(a,b,c) , plot all the bus voltages vs. time, and report the interaction ofgenerator 1 and 2 due to the faultObserve how long the duration of the fault is so that the generator 2 will be outof synchronism?