aps chap10(synpwrsys)

7/28/2019 Aps Chap10(SynPwrSys)

1/55

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


2/55

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


3/55

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


4/55

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


5/55

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


6/55

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.


7/55

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


8/55

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


9/55

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 ++=


10/55

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

)()()(

==

+=

+

+=


11/55

Transient Model Block Diagram

Stator Block


12/55


Rotor Block


13/55


Field voltageequation

d axis

q axis


14/55


Overall block diagramfrom excitationsystem


15/55

Synchronous machine model in Chap 7

Overall block diagram


16/55

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)


17/55

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



18/55

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


19/55

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


20/55

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


21/55

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


22/55

EXCITATION SYSTEMS

Overall scheme of excitation systems detector, regulator, exciter, stabilizer, diode bridge, power

system stabilizer components

detector

regulator

exciter

diode bridge


23/55

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


24/55


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


25/55


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


26/55


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


27/55


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


28/55


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


29/55


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 ==


30/55

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


31/55

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


32/55

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


33/55

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)


34/55

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



35/55

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


36/55


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


37/55


Stator module with external network stator resistor: rs+re stator reactance: xd+xe, xq+xe

( ) ( )( )eqedesZ xxxxrrD ++++=

''2

11


38/55


qd synchronous to rotor frame module transform bus voltage of synchronous reference to rotor

reference value


39/55


Overall synchronous generator transient model

rotor winding

rotor winding


40/55

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


41/55


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


42/55


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


43/55


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


44/55


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


45/55


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=


46/55


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


47/55


Model setup: Network module


48/55


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


49/55


Inside generator model:

stator module inputs: Eq, Ed

, iq, id not Eq, Ed

, vq, vd stator module outputs: vq, vd


50/55


Inside generator model :

excitation system: Ef=> (vref-vfb), (1/s) andfeedback loop.


51/55


Inside generator model : exciter


52/55


Inside generator model :

rotor blockr


53/55


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


54/55


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


55/55

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?

aps chap10(synpwrsys)

Documents