Synchronous Machines in Power Systems and Drives 

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Most of the electrical power generators are threephase synchronous generators Synchronous motors are competitive in higher power ranges because of efficiency and lower costs Reluctance and permanent motors are popular at lower power ranges Synchronous generator in power systems 

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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

Stability Studies 

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Sub-transient time constant of machine is between 0.03 to 0.04 second, shorter than electromechanical oscillation Electromechanical oscillation frequency between synchronous 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 of electromechanical oscillation Slower acting component with longer time constants such as boilers and AGC response may need more time between 10sec to 2 min Different model shall fit into the different analysis

Basic Dynamics of Synchronous Generators 

Basic dynamic behavior of synchronous generator in transient situations: 

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voltage behind the transient reactance of a generator and network E’=Eth+ j Xt I, Xt = X’d + Xth jX’d

AC Network

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One-line diagram

jX’th

E’

E’th

Circuit Equivalent

take Thevenin’s voltage as reference phasor Eth= Eth ∠0, and E’=E ∠δ

Basic Dynamics of Synchronous Generators 

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electrical output power of the generator E ' Eth Pgen= R(E’I*) = pu sin δ Xt from the above equation, we can see that power transfer characteristic for the system is a sine wave with max value E’Eth/Xt the rotor motion without damping Pmech-Pgen= 2 H dω pu ωb dt replace the dω/dt with d2δ/dt2, we obtain swing equation

Pmech − Pgen 

2 H d 2δ = ωb dt 2

ω  dδ  pu or   = b 2H  dt  2

∫ (P

mech

− Pgen )dδ

If machine was to maintain synchronism, excursion of δ would be bounded and (dδ/dt) would have return to zero

Transient Power Angle Characteristics  δ max

∫δ

min

Pmech1

transient power angle characteristics ( Pmech − Pgen )dδ = ∫

δ ss

δ min

A2

( Pmech − Pgen )dδ + ∫

δ max

δ ss

( Pmech − Pgen )dδ

A2max

A1

Pmech1

δmax δSS

Pmech0

δ0 δ0 δSS δmax

π-δSS

Equal Criteria: A1 = A2 A1 < A2max A1 = A2max A1 > A2max

Stable Critically Stable Unstable

t0

t

Transient Power Curve And Dynamics 

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Without damping loss, rotor oscillates about δSS. Otherwise, δ eventually settles to δSS. The gain in rotor momentum could carry δ beyond critical angle π- δSS which Pmech1>Pgen and rotor accelerates to lose synchronism For purpose of determining the synchronism of the machine, area of A2max should be larger than area of A1 (area of A2max is from δSS to π- δSS) Transient power angle curve may be raised by increasing the excitation control of E’ 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 transient performance over the control of excitation system adverse impact of PSS: 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, to switch from motor notation to generator notation, it is required to invert the sign of all stator currents in the voltage equation, flux linkage equation, and torque equations.

Transient Model with d,q Field Windings 

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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 machine parameters directly obtained from standard tests, such as reactances and time constants For derivation of transient model equations, please see pp. 468-474 Transient model (without damper winding): state variable λd, λq 

stator winding equations

 dλq rs  − Ed' − λq  + vq = − '  + ωr λd Lq  ω  dt 

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rotor winding equations ' dE  Ld − L'd Ld ' q ' Tdo + ' Eq = E f +  ' dt Ld  Ld Torque

 ωr λd 

r vd = − s' Ld

 dλd  Eq'    ω − λd  + dt − ωr λq  

' '   − L L L dE q q q ' ' d  ωr λq + ' Ed = − E g − Tqo '   dt Lq  Lq 

'  3 P  λd Ed' λq Eq  1 1  + − ' − ' λd λq  Tem =  ' ' 2 2  ωr Lq ωr Ld  Ld Lq  

Transient Model with d,q Field Windings 

Simplification of transient model 

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in the transient analysis, damper windings are no longer active in transient stability prediction, rotor winding transient are dominant. first swing of the rotor would be the interval of interest, and the rotor transients vary at the rate of Tdo’ and Tqo’ the rotor transient would impact speed voltage term, ωrλd, ωrλq, and greater than that of dλq/dt, dλd/dt. Therefore, the effect of dλq/dt, dλd/dt could be neglected the transient model can be further simplified by neglecting dλq/dt, dλd/dt

Transient Model Equations 

Simplified transient model equations (ωr=ωe except for rotor mechanical dynamics) 

Stator winding equations (see pp.473 ) 

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inputs : vq , Eq' , vd , Ed'

vd = − rsid + xq' iq + Ed'

outputs : iq , id

Rotor winding equations 

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vq = − rsiq − xd' id + Eq'

Tdo'

dEq'

+ Eq' = E f − ( xd − xd' )id

dt ' ' dEd Tqo + Ed' = − E g + ( xq − xq' )iq dt Torque Equation:

Tem = −

inputs : E f , E g , id , iq outputs : Ed' , Eq'

3 P { Eq' iq + Ed' id + ( xq' − xd' )id iq } 2 2ωe

N.m

Transient Model Equations 

Rotor equations

dωrm J = Tem + Tmech − Tdamp (N.m) dt inputs : Tem + Tmech − Tdamp

d {(ωr − ωe ) / ωb } = Tem ( pu ) + Tmech ( pu ) − Tdamp ( pu ) (pu) 2H dt output : ωr dδ e = ωr − ωe , dt

P ωr = ωrm 2

Transient Model Block Diagram 

Stator Block

Transient Model Block Diagram 

Rotor Block

Transient Model Block Diagram 

Field voltage equation 

d axis

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q axis

Transient Model Block Diagram 

Overall block diagram

from excitation system

Synchronous machine model in Chap 7 

Overall block diagram

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 the transient synchronous machine model. The machine is connected to the following source: v1=1√2sin(120πt+0) pu v2=1√2sin(120πt-2π/3) pu v3=1√2sin(120πt+2π/3) pu

1.

With excitation reference voltage Ef= 1pu, Tmech = 1 pu (mechanical torque), apply three-phase bolted fault to ground at t=10 sec, fault clear at t=10.25 sec, observe and plot a. b. c. d. e. f.

vq, vd, iq, id, in one figure ia, ib, ic in one figure Pgen, Tem, δ, ω in one figure Qgen, If’, in one figure Show the critical fault clearing time and plot δ vs. time with stable and unstable conditions discuss what you see on the plots (ex. observe transient in field current and qd, abc current)

Project 10-1 Fault tests of Synchronous Machine 2.

With excitation reference voltage Ef = 1pu, Tmech = 1 pu (mechanical torque), apply single phase to ground fault on phase c at t=10 sec, fault clear at t=10.25 sec, observe and plot 1) 2) 3) 4) 5)

vq, vd, iq, id, in one figure ia, ib, ic in one figure Pgen, Tem, δ, ω in one figure Qgen, If’, in one figure 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 sec through the time when system becomes stable after the fault cleared

EXCITATION SYSTEMS 

Scheme of Excitation systems contains    

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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

rotating part

EXCITATION SYSTEMS 

Classification of excitation systems 

ac excitation (static) 

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field winding of alternator is on the same shaft as the rotor of the synchronous machine alternator’s stator and rectifier are stationary

EXCITATION SYSTEMS 

Classification of excitation systems 

ac excitation (rotary) 

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armature of alternator and rectifier are on the same shaft as the rotor of the synchronous machine alternator’s rotor field winding is stationary

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

EXCITATION SYSTEMS 

Overall scheme of excitation systems 

detector, regulator, exciter, stabilizer, diode bridge, power system stabilizer components

exciter

detector

diode bridge

regulator

COMPONENTS OF EXCITATION SYSTEMS 

voltage transducer and load compensation circuit 

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voltage transducer and rectifier are modeled by a single time constant with unity gain compensation of excitation voltage due to internal load is represented by RC+jXC compensator

transducer

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

stabilizer signal

stabilizer feedback signal

error amplifier with limiter transient gain reduction Tc (vref-vfb), (1/s) and feedback loop.

Case 2: Multi-machines System 

Inside generator model : 

exciter

Case 2: Multi-machines System 

Inside generator model : 

rotor block

ωr

Case 2: Multi-machines System 

Inside generator model : 

generator model Ef

1

Exc_sw(1) Exc_sw

in_Vref |Vt|

vqt

Sw

1 vdt 1 s

1/Tpdo(1)

sum

exciter

1/Tpdo xd(1)-xpd(1)

out_|Vt|

Eqp_ 3

2 out_|I|

Eqp out_Pgen id_

4 out_Qgen

VIPQ stator_wdg

Gain2

delta

2

5 out_delta

in_iqe iq

6

3 id

out_puslip

in_ide

7

qde2qdr

4

iq_

out_Tem in_Tmech

xq(1)-xpd(1)

Rotor

Gain 8 1/Tpqo(1) Sum

Gain1

1 s

out_Eqpe 9

Edp qdr2qde

out_Edpe

Case 2: Multi-machines System Overall model diagram

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ide1 U(E) Scope

U

y1

Selector1

Sbratio(1) Sys/Gen1VA_

T o Workspace Mux

Mux

vref(1)

Initialize and plot

Vref1

m2

Clock

T Eqpe1 Edpe1

T mech1

tmodel

Sys/Gen1VA iqe1 -K-

Sys/Gen2VA iqe2

-K-

U(E) Scope1

U

y2

Selector

Eqpe2

T o Workspace1 Mux

Mux_ 1.

vref(2)

T3

iqe4

Vref2

Edpe2

0

T1

T4

vde3

T mech(2) T mech2

T2

vqe3

Clock1

tmodel1

T5

ide4 ide2

network Sbratio(2) Sys/Gen2VA_

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 at t=7 sec, then a step change to 0.7pu at t=15 sec, finally a step change to 0.8 pu at 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.

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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 results similar to figure 10.24(a,b,c) and observe the interaction of generator 1 and 2 due to the change of electrical strength z14 and z24, report on the difference due to the change of electrical strength

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Tmech2 = 0.8pu and Tmech1 = 0pu, a fault current of iq4e-jid4e = -(2-j2) pu is to be introduced 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 figure 10.25(a,b,c) , plot all the bus voltages vs. time, and report the interaction of generator 1 and 2 due to the fault Observe how long the duration of the fault is so that the generator 2 will be out of synchronism?