Frequency Control


Power Balance



Primary Control

- cinetics of the rotating masses

dv    
linear motion:   F  =  m · a  =   m · v   ;       v   =  
 dt
 
rotation: Tq  =  J · ω   ;       ω   =   2 π f


Tq : mechanical shaft torque   [ Nm = Ws ]

J : total moment of inertia of the rotating masses   [ kg m2 ]

ω  : angular acceleration   [ 1 / s2 ] ;     ω : angular velocity   [ 1 / s ]


-- torque balance of rotating turbine/generator shaft


1
ω   =     =     · ( TqT − TqG )
dtJ

TqT : turbine torque (accelerates)

TqG : generator torque (retards)


-- mechanical power balance


with   P  =  ω · Tq

ω   ω   =   J -1 · ( ω TqT − ω TqG )   =   J -1 · ( PT − PG )   =   J -1 ·  Pacc


 PT(t)  −  PG(t)  =  Pacc(t)   :   accelerating power ( =  dEkin / dt   with  Ekin  =  J · ω2 / 2 )


ω ω P0Pacc(t) 1Pacc(t)
·   =    ·    =    · 
ω0     ω0 ω02 JP0 TAP0

 TA  =  ω02 J / P0  ≈  8 ... 20 s  :     accelerating time constant



- electrical power balance


 PG(t)  =  PL(t)  +  ΔP

 ΔP :   disturbance of power balance (load increase, generator outage or tie line outage) at t = t0

 ΔPG(t)  =  PG(t)  −  P0  =  PL(t)  −  P0  +  ΔP

 ΔPG(t)  =  ΔPL(t)  +  ΔP



- turbine controller (frequency or speed control)


dΔPT1   · ( −ΔPT(t)  −  KTC · ( f (t)  −  f0 )  )        simplified turbine model
(first order delay)
  =  
dt     TT

 ΔPT(t)  =  PT(t)  −  P0   ;     Δf (t)  =  f (t)  −  f0


final state


dΔPT / dt  =  0   =>   ΔPT  =  − KTC · Δf


ΔPTKTC · f0ΔfΔf
  =   −    ·    =   −   kTC  · 
P0P0f0f0


- frequency dependence of load


PL(t)  =  P0  +  KL · Δf (t)


ΔPL(t)  =  PL(t) − P0  =  KL · Δf (t)


ΔPLKL · f0 ΔfΔf
  =    ·    =   kL  · 
P0P0 f0f0



- states of primary control


-- initial state (before disturbance)

 ΔP  =  0  for  t  <  t0

 =>   PT(t)  =  PG(t)  =  PL(t)  =  P0   ;     ΔPT(t)  =  ΔPG(t)  =  ΔPL(t)  =  0


-- disturbance at t = t0

 ΔP  ≠  0  for  t  ≥  t0


-- immedeately after disturbance at t  =  t0

 Δf  =  0  ;    ΔPT  =  0  ;    ΔPL  =  0   but   ΔPG  =  ΔP  =  − Pacc

ω ω 1Pacc(t0) 1ΔP
·   =    ·    =   −    ·          initial slope only depends on TA and ΔP
ω0     ω0 TAP0 TAP0


-- final state (after primary control)

ω   =  0   =>   Pacc  =  ΔPT  −  ΔPG  =  0

 − KTC · Δf  −  KL · Δf  −  ΔP  =  0   =>   Δf  =  −  ΔP / ( KTC  +  KL )  =  −  ΔP / KS

 KS  =  KTC  +  KL

  Δf ΔPP0 ΔP1  
    =   −    ·    =   −    · 
  f0 P0f0  ·  KS P0kS


Secondary Control


 ΔPSC(t)  =  − KSC · Δf (t) dt


dΔPT1   · ( −ΔPT(t)  −  KTC · Δf (t)  −  KSC · Δf (t) dt  )       turbine with
secondary control
  =  
dt     TT

- final state

 Δf  =  0   =>   frequency recovered to normal ;   primary control relieved


Demonstration


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Interchange Control

- Generator Outage in System 2

  ΔP2   =>   Δf


- Primary Control

  ΔPPC1   =   − KS1 · Δf  ;       ΔPPC2   =   − KS2 · Δf  ;       KSi  =   KLi  +  KTCi

  ΔP2   =   ΔPPC1  +  ΔPPC2       outage compensated by primary control in both systems

  ΔP12   =   ΔPPC1   =   ΔP2  −  ΔPPC2   =   − ΔP21       power exchange


- Secondary Control   ( here: interchange control )

  ace1( t )   =   β1 · Δf  +  ΔP12   :     area control error

  ΔPSC1   =   − KSC1 · ( β1 Δf + ΔP12 ) dt

  ΔPSC2   =   − KSC2 · ( β2 Δf + ΔP21 ) dt


  adjust     β1  ≈   KS1  ;       β2  ≈   KS2


  ace1( t )   =   β1 · Δf  +  ΔP12   =   β1 · Δf  −  KS1 · Δf   ≈   0  
                  =>   secondary control inactive in system 1


  ace2( t )   =   β2 · Δf  +  ΔP21   =   β2 · Δf  −  KS2 · Δf  −  ΔP2   ≈   − ΔP2  
                  =>   secondary control active in system 2


- Final State

 Δf   =   0     frequency recovered to normal

 ΔPPC1   =   0  ;     ΔPPC2   =   0     primary control relieved in both systems

 ΔPSC1   =   0  ;     ΔPSC2   =   ΔP2     deficit completely covered by secondary control of system 2

 ΔP12   =   0     power exchange returned to schedule


Demonstration


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