03 frequency

Frequency Regulation of Synchronous Machines

POWER SYSTEM DYNAMICS AND CONTROL (EEEN40550)

Prof. Federico Milano

Email: [email protected]

Tel.: 01 716 1844

Room 157a - Electricity Research Centre

School of Electrical, Electronic and Communications Engineering

University College Dublin

Dublin, Ireland

Dublin, 2015 Frequency Regulation of Synchronous Machines - 1

Time Frames of Synchronous Machine Transients

Electrical transients short circuits 0.001 s÷ 0.1 s

Torsional oscillations sub-synchronous resonance 0.005 s÷ 0.5 s

Rotor fluxes subtransient (damper winding) 0.005 s÷ 0.3 s

transient (field) 0.2 s÷ 20 s

Electromechanical oscillations OMIB 0.1 s÷ 5 s

Two machines 2÷ 30 s

V/Q control primary 0.2÷ 5 s

secondary 0.2÷ 100 s

f/P control primary 1÷ 20 s

secondary 10÷ 500 s


Interaction between f/P and V/Q controls

f/P control

valves

TurbineΩ, δ

Pe

PeGrid

V/Q controlfield voltage

MachineV,Q

• The V/Q control affects the f/P control much more than what the f/P control

affects the V/Q one.

f/P V/Q


Interaction between f/P and V/Q controls

• In fact, we have:

Pe =EV

Xsin δ

E ∝ ΩΦmax ≈ constant (if Ω ↑, Φ ↓)

• Hence the f/P control does not affect too much the V/Q control.

• The f/P control is also much slower than the V/Q one.


Non-interacting f/P and V/Q control

• The system is nonlinear, hence a fully non-interacting control cannot be obtained.

• However, let’s consider the linearisation at an operating point.

• We need the model of the system (all quantities are assumed in pu):

T

τm

τe

vf+−

ω

Gv

τm − τe = M dωdt

τe = τe(vf , ω)

v = v(vf , ω)


Non-interacting f/P and V/Q control

• General Scheme:

ωref

vref

εω

εv

Reg. System

τm

vf

ω

v+

+

−

−

→ If we linearise the model of the system, we can obtain an independent control.


Principle of Non-Interoperant Control

∆ωref

∆vref

∆ω

∆v+

+

−

−

R P

w

∆ωref

∆vref

∆ω

∆v

RI

RII

PI

PII

+

+

−

−


Linearization of the system at the O.P.

⇒ ∆τm −∆τe = Md∆ω

dt(∗)

⇒∆τe =∂τe∂ω

∣∣∣∣0

∆ω +∂τe∂vf

∣∣∣∣0

∆vf (∗∗)

⇒ ∆v =∂v

∂ω

∣∣∣∣0

∆ω +∂v

∂vf

∣∣∣∣0

∆vf (∗ ∗ ∗)

Substituting (∗∗) in (∗):

∆τm =∂τe∂ω

∣∣∣∣0

∆ω +∂τe∂vf

∣∣∣∣0

∆vf + Ms∆ω︸︷︷︸

s stands for Laplace operator

(∗ ∗ ∗ ∗)


Linearization of the system at the O.P.

• The outputs are ∆ω and ∆v, while the inputs are ∆τm and ∆vf

• Hence we rewrite (∗ ∗ ∗) and (∗ ∗ ∗ ∗):

∆ω =∆τm

Ms+ ∂τe∂ω

∣∣0

−

∂τe∂vf

∣∣0

Ms+ ∂τe∂ω

∣∣0

∆vf

⇒ ∆ω = p11∆τm + p12∆vf

∆v = ∂v

∂ω

∣∣0

Ms+ ∂τe∂ω

∣∣0

∆τm+ ∂v

∂vf

∣∣0−

∂v

∂ω

∣∣0

∂τe∂vf

∣∣0

Ms+ ∂τe∂ω

∣∣0

∆vf

⇒ ∆v = p21∆τm + p22∆vf


Synthesis of the Regulator

u ε y

R P

+

−

• We know P .

• Moreover: y = Dε, where D = PR

we impose that:

p11 p12

p21 p22

r11 r12

r21 r22

=

d11 0

0 d22

Hence, D must be diagonal.


Synthesis of the Regulator:

• We obtain:

p11r11 + p12r21 = d1

p11r12 + p12r22 = 0

p21r11 + p22r21 = 0

p21r21 + p22r22 = d2

• Unknown quantities are r11, r12, r21, r22.

• d1 and d2 can be imposed by assuming the desired behaviour of the system.


Some simplifications:

• The elements of P can be simplified:

∂v

∂ω≈ 0 ⇒ p21 ≈ 0

p22 ≈∂v

∂vf

∣∣∣∣0

• Moreover:

∂τe∂ω≈ 0 ⇒ p11 ∼=

1

Ms

p12 ∼=∂τe∂vf

∣∣∣∣0

1

Ms


Synthesis of the Regulator:

• Applying the simplifications above, we obtain:

R⇒

r21 = 0

r22 = d2

p22

r11 = d1

p11

r12 = −p12

p11

d2

p22= −p12

p11r22


Resulting Control Scheme

∆εω

∆εv

d1p11

r22

−p12p11

∆τm

∂τe∂vf

∣∣∣∣0

∆vf

1

Ms

∂v

∂vf

∣∣∣∣0

∆ω

∆v

R P

+

++

−

Observe that: −p12p11

=

−(−∂τe∂vf

∣∣∣∣0

·1

Ms)

1Ms

=∂τe∂vf

∣∣∣∣0


Non-interacting f/P and V/Q controls

• Previous scheme is valid only for a particular O.P.

• To work for a general O.P., the control scheme must be adaptive

• We can obtain some results using a state-space approach:

x = Ax+Bu

y = Cx+Du

where u = [τm, vf ]T , x = [δ, ω]T , y = [ω, v]T


f/P Regulation of Synchronous Machines

• General scheme of the turbine governor and of the turbine/machine:

Ωref

Wa

Pm

Pe

We

ΩSpeed

Control

Valve

PositionTurbine

Network

Inertia

Damping

+

+

−

−−

We = perturbation within the network (e.g., load variation)

Wa = perturbation within the mechanical system


f/P Regulation of Synchronous Machines

• Let’s consider a linearisation around an operating point.

∆Ωref

∆Ω

∆Ω∆εΩ ∆β ∆A ∆Pm

∆Pg

∆PL ∆Pc

Gr(s) Gv(s) Ga(s)

Gg(s)

Gc(s)

1

sM

+ +

−−

−

−

Pm − (Pe + Pg) = M dΩdt

→ ∆Pm − (∆Pe +∆Pg) = M d∆Ω

dt

where Pe = Pc + PL ∆Ω(s) = 1sM [∆Pm − (∆Pe +∆Pp)]


f/P regulation of Synchronous Machines:

• Where:

∆β = angular position for the valve regulation

∆A = valve position

Gc(s) = transfer function that takes into account the dependence of the load on frequency

Gg(s) = transfer function that takes into account mechanical losses, damping etc.

Gr(s) = ∆β/∆εΩ

Gv(s) = ∆A/∆β

Ga(s) = ∆Pm/∆A


f/P regulation of Synchronous Machines:

• We define Pe = Pc + PL ⇒ ∆Pe = ∆Pc +∆PL

PL = load variation that does not depend on Ω

Pc = load variation that depends on Ω

• For the sake of simplicity, let assume:

Gg(s) = 0

Gc(s) = 0


Simplified f/P Control Scheme

∆Ωref ∆Ω∆εΩ ∆Pm

∆PL

Gf (s)1

sM

++

−−

where:

Gf (s) , Gr(s)Gv(s)Ga(s)

• We want a control scheme with the following features:

- fast

- stable

- zero static error


Synthesis of the f/P Control:

• There are various possibilities for Gf (s):

1.KI

s, (integrator)

2. KI1 + sT2

s, (PI)

3. KIT11 + sT2

1 + sT1, (lead-lag) T1 > T2 or T1 < T2

• In the following slides, we will consider each case.


Synthesis of the f/P Control Scheme

1) Integrator KI/s then: G(s) = Gf (s)1

sM = KI

s2M

KI < 0

ℑ

ℜ

KI > 0

ℑ

ℜ

|G| unstable!

−2 (= −40dB/decade)

ωs

⇒

Ω

t



2) PI regulator: G(s) = KI1+sT2

s2M

|G|−2 (= −40dB/decade)

−1 (= −20dB/decade)

1T2

νTωs

stable if νT ≫1T2

hence:

KI

s2M sT2

∣∣s=νT

= 1

ℑ

ℜ



3) lead-lag controller. G(s) = KI1+sT2

sM(1+sT1)

T1 > T2

− 1T1

− 1T2

ℑ

ℜ

T2 > T1

− 1T1

− 1T2

ℑ

ℜ

T1 ≃ 10÷ 15 s

T2 ≃ 3÷ 5 s

T2 ≃ 10÷ 15 s

T1 ≃ 3÷ 5 s



• Bode’s diagram:

|Gf (jωs)|

−1

−1

−2

∠Gf (jωs)

1T1

1T2

ωs

−90

−180

bpbt

=T2

T1

Gf (s) = KIT11 + sT2

1 + sT1

−

(∆Ω

∆Pm

)

∆Ωref=0

(0)

=1

KIT1

−

(∆Ω

∆Pm

)

∆Ωref=0

(∞)

=1

KIT2


Static and Dynamic Droops

• Let define the static droop bp on machine bases:

bp =Pn

Ωn·

1

KIT1⇒ −

(∆Ω/Ωn

∆Pm/Pn

)

∆Ωref=0

(0) = bp

• Let define the dynamic droop bt on machine bases:

bt =Pn

Ωn·

1

KIT2⇒ −

(∆Ω/Ωn

∆Pm/Pn

)

∆Ωref=0

(∞) = b2


Droop Change of Bases

• The value of the droop depends on the power and frequency bases.

• If we use different bases, say, Pn and Ωn, the droop bp can be obtained as:

bp =Pn

Ωn

·Ωn

Pn· bp

• Then, assuming that the frequency is unique for all the AC network:

bp =Pn

Pn· bp

• Note that, machines tend to have similar bp (on machine bases).


Static Error of the Lead-Lag control (I)

•

∆Ω

∆PL=

− 1sM

1 + 1sMKI

1+sT2

1+sT1

Note that: limt→∞ ∆Ω(t) = lims→0 s∆Ω(s)

• If ∆PL is a step of magnitude A, e.g.,

A

⇒ ∆PL(s) =A

s

then: limt→∞ ∆Ω(t) = lims→0

− 1sM

1 + 1sMKI

1+sT2

1+sT1

sA

s=−A

KIT16= 0

⇒ The static error is not zero!


Static Error of the Lead-Lag Control (II)

∆Ω, p

∆Ω(t)

PL(t)

t

A

−A

KIT1

• Observe that in this simulation ∆Ωref = 0, as it is in practise for the f/P primary

control.

• If the system is islanded we need to use scheme(2).

• For interconnected networks, we use (3) for all machines and a secondary control to

obtain ∆Ω = 0 in steady state.


Static Error of the Lead-Lag Control (III)

• An interesting consequence of the non-zero static error of the primary frequency

regulator is the bimodal PDF of the frequency.

49.80 49.85 49.90 49.95 50.00 50.05 50.10 50.15 50.20

Frequency [Hz]

0

2

4

6

8

10

12

14

PD

F

PDF distribution of the freqeuncy of the Irish system (∼ 6M samples, one week).


Steady-State Characteristic of the f/P Control

• We have, in general:

∆Ωref ∆Ω∆ε ∆β ∆A ∆Pm

∆Pe

Gr(s) Gv(s) Ga(s)1

sM

+

+

−

−

where:

Gf (s) = Gr(s)Gv(s)Ga(s) = KIT11 + sT2

1 + sT1

• In steady-state:

∆Pm = Gf (0)(∆Ωref −∆Ω)

∆Pm =Pn

Ωn

1

bp(∆Ωref −∆Ω)


Steady-State Characteristic of f/P Control

Ω

Ω∗

Ωo

PmP ∗ Po

bp 6= 0

−Pn

Ωnbp

special cases:

Ω

Ωo

Pm

bp → 0

Ω

Po Pm

bp →∞


Change of the Operating Point

• Let’s assume we set a different set point for the mechanical power:

∆Pm =Pn

Ωn

1

bp(∆Ωref −∆Ω) +

[∆P ref

m

]← change of O.P.

Ω

PmPm,o Pm,o +∆Pm

∆P refm


Effect of Gc(s) and Gg(s)

∆Ωref∆Ω

Gf (s)

Gc(s) +Gg(s)

∆Pm

∆Pe

1

sM

+ +

−

−

−

• Let’s check if the effect of Gc(s) and Gg(s) can be neglected.

• Without Gc(s) and Gg(s), we have:

∆Ω

∆Pe=

− 1sM

1 + 1sMGf (s)

=−1

Gf (s) + sM



• With Gc(s) and Gg(s), we have:

∆Ω

∆Pe=

−1

Gf (s) +Gc(s) +Gg(s) + sM

• In steady-state (s→ 0), we have:

−1

Gf (0) +Gc(0) +Gg(0)

• Sometimes, Gf (0) is defined as regulating energy, then Gf (0) +Gc(0) +Gg(0) is

the total regulating energy.

• In general Gc(0) and Gg(0)≪ Gf (0)



• Moreover, observe that:

1

Gf (0)>

1

Gf (0) +Gc(0) +Gg(0)

so, the static error is a bit smaller if we consider Gc and Gg .

• From the dynamic point of view, Gc(s) is actually very negligible.

• Less intuitive is to neglect Gg(0)→ it takes into account load inertia.

• We can take it into account by defining:

M∗ = (1.15÷ 1.20) M


Parallel of Synchronous Machines

• Let’s assume that the system frequency is unique (no electromechanical transients).

∆Ωref

∆Ωref

∆Ω

Gf,1(s)

Gf,2(s)

∆Pm,1

∆Pm,2

∆Pm

∆Pe

1

sM

M = M1 +M2

+

+

++

+ −

−

−

→ Let’s define a unique Gf (s)



∆Ωref∆Ω

Gf (s)∆Pm

∆Pe

1

sM

+ +

−

• Where we define:

Gf (s) = Gf,1(s) +Gf,2(s)

• If we have n machines:

M =n∑

i=1

Mi

∆Pm =n∑

i=1

∆Pm,i

Gf (s) =n∑

i=1

Gf,i(s)



⇒ Let’s compute bp, T1, T2 for the resulting Gf (s)

• We know bp,i, T1,i, T2,i of the n machines:

Gf (s) =

n∑

i=1

Pn,i

Ωn,i·

1

bp,i·1 + sT2,i

1 + sT1,i

• Since T2 = 3÷ 5 s and T1 = 10÷ 15 s, let assume that T2,i and T1,i are equal to

mean values, say T2m and T1m.

• Then Ωn is the same for all machines.

• Hence:

Pn

bp=

n∑

i=1

Pn,i

bp,i⇒ bp =

Pnn∑

i=1

Pn,i

bp,i

where Pn =

n∑

i=1

Pn,i



• So the resulting total transfer function Gf (s) is:

Gf (s) =Pn

Ωn·1

bp·1 + sT2m

1 + sT1m

• In the same way, we can define bt for the interconnected system.

• Observe that connecting several machines in parallel is “good” for the system as:

M increases → ∆Ω decreases



• Let’s see the effect of the bp,i on the overall system.

• bp,i can be zero or not zero:

Gf,i(s)

KI

1 + sT2m

s⇒ bp,i = 0

Pn,i

Ωn

·1

bp,i·1 + sT2m

1 + sT1m

⇒ bp,i 6= 0


Example of the Parallel of 3 Machines

• Let’s consider the following example:

bp,1 = 0

bp,2 6= 0bp,3 6= 0

Grid

ΩΩΩ

50 Hz∆Pe

Pm,1 Pm,2 Pm,3

P om,1 P o

m,2P om,3P o

m,1 +∆Pe



• To balance the power, we have

P oe = P o

m,1 + P om2 + P o

m,3

• Since bp,1 = 0, any variation ∆Pe is provided by machine 1.

• This explains why no machine in an interconnected system can have bp = 0.

• Actually, in interconnected systems, the bp,i are more or less all the same so that all

machines contribute to ∆Pe proportionally to their power rate.

• ... but if all bp,i 6= 0, then we have that ∆Ω(t)∣∣t→∞

6= 0.



• Let’s now assume bp,1 6= 0, bp,2 6= 0, bp,3 6= 0.

ΩΩΩ

50 Hz

∆Ω

Pm1P om,1 P o

m,1 +∆Pm,1 Pm,2P om,2 P o

m,2 +∆Pm,2 Pm,3P om,3 P o

m,3 +∆Pm,3

∆Pe = ∆Pm,1 +∆Pm,2 +∆Pm,3

⇓

Pe = Pm,1 + Pm,2 + Pm,3

but: ∆Ω 6= 0



• If ∆Ωref = 0, then: ∆Pm,i = −Gf,i(s)∆Ω

• In steady-state: ∆Pm,i = −Pn,i

Ωn

1

bp,i∆Ω

• Moreover, since1

sMis an integrator:

n∑

i=1

∆Pm,i = ∆Pe = −Pn

Ωn

1

bp∆Ω

→ ∆Ω = −bpPn

Ωn

∆Pe = static frequency error

Moreover: ∆Pm,i =Pn,i

Pn

·bp

bp,i·∆Pe ⇒

∆Pm,i ∝ Pn,i

∆Pm,i ∝1

bp,i


Steady-State Error of the f/P Control

• Assume:

Gf (s) =Pn

Ωn

1

bp

1 + sT2,m

1 + sT1,m=

Pn

Ωn

1 + sT2,m

bp + sT2,mbt

where we used the identity:bpbt

=T2,m

T1,m

• Then, in steady-state and assuming ∆Ωref = 0:

⇒∆Ω/Ωn

∆PL/Pn(0) = bp

• Note that the total droop bp is a measure of the steady-state frequency error.

• bp ∈ [0, 5]%

• bt ∈ [25, 40]%


Synthesis of the f/P control

• We have Gf = GrGvGa ⇒ Gr =Gf

GvGa

• Gv can be assumed to be a constant (we neglect delays)

• Ga depends on the type of turbine:

1. thermoelectric

2. hydroelectric

• Once we know Gv and Ga, we impose the resulting Gf (s) and determine the

transfer function of Gr(s)


Model of Gv(s)

• The valve transfer function can be assumed as follows:

Gv(s) =∆A

∆β≈

Kv

1 + sTv≈ Kv

where Tv ∈ [0.3, 0.5] s.

• The time constant Tv is “small” with respect to the timer frame of the f/P regulator.

• Hence let assume that

Gv(s) ≈ Kv = constant


Model of Ga(s) for Hydro Plants

reservoir

wall

tunnelpipe

Peltonturbine

400÷ 500 m 700 m

Pm = ηρHQ

η = efficiency

ρ = specific weight (of the water)

H = height (m)

Q = m3/s volumetric flow



• If ∆η = ∆ρ = 0 ⇒∆Pm

P om

=∆H

Ho+

∆Q

Qo

→ Injector model:

Ac

A

12

pipe

H1 = 0 , H2 = H

v =Q

A

[m

s

]

mgH1 +1

2mv21 =mgH2 +

1

2mv22

⇒Q2

1

A2=

Q22

A2c

+ 2gH

since Ac ≫ A ⇒ H =

(QA

)2

2g⇒ Q = A

√

2gH



• Hence we have:∆Q

Qo=

∆A

Ao+

1

2

∆H

Ho

• We still need the relation between ∆Q and ∆H . It depends on the dynamics of the

water in the tunnel, wall and pipe.

• Analogy with electrical systems:

H

ztunnel zpipe

Cwell

→ ∆Q

∆H

+

−

Q ⇒ I (current)

H ⇒ V (voltage)



• We neglect ztunnel and Cwell. Let’s define zpipe(s).

H

LP

α Hv −H(∗) Hv −H = LP sinα+

Pv − P

ρ

water weight: ρAPLP sinα

⇒ force balance: ρAPLP sinα+ (Pv − P )Ac =ρAPLP

g

d(

QAc

)

dt(∗∗)

⇒ Let’s divide by ρAc and use (∗), hence:

LP

g

1

Ac

dQ

dt= Hv −H


Model of Ga(s) for Hydro-Plants

⇒We define JP ,LP

SAP= pipe inertance

⇒ So: −∆H = JPd∆Q

dt⇒ zpipe(s) = sJP

The pipe behaves similarly to a coil (inertance⇒inductance)

Inertance: measure of the pressure gradient in a fluid to cause a change in flow rate

with time [Pam−3s2]



• In summary we have:

∆Pm

P om

=∆A

Ao+

∆H

Ho

∆H = −sJP∆Q

∆Q

Qo=

∆A

Ao+

1

2

∆H

Ho

⇒ Let’s define Tw =JPQ

o

Ho[s]⇒

∆H

Ho=−sTw

1 + sTw2

∆A

Ao

∆Q

Qo=

1

1 + sTw2

∆A

Ao

⇒ Ga(s) =∆Pm

∆A



⇒∆Pm

P om

=

(1

1 + sTw/2−

sTw

1 + sTw/2

)∆A

Ao

⇒∆Pm

∆A=

P om

Ao

1 + sTw

1 + sTw/2≡ Ga(s)

⇒ Bode’s Diagram:

1Tw

2Tw

|Ga(jωs)|

∠Ga(jωs)0

−90

−180



• Observe that, actually, the model of Ga(s) is much more complex:

Ga(s) =P om

Ao

1−aPgAP

Qo

Hotanh

sLP

aP

1 +aPgAP

Q0

H0

1

2tanh

sLP

aP

where aP = wave proportion speed within the pipe≈ 1000 m/s

⇒ the resonance frequency is: νr =π

2LP

aP

≈ 3 Hz



• Bode diagram of the precise Ga(s)

νR 2νR 3νR

2P

m

A= 2Ka

P

m

A= Ka

0

−90

−180


Synthesis of Gr(s) for Hydro Plants

• Despite the complexity of the model of Ga, for the synthesis of Gr(s) we assume:

Ga(s) ≈ Ka (and Gv(s) ≈ Kv)

• Hence we have:

Gr(s) =Pn

Ωn·1

bp

1 + sT2

1 + sT1︸︷︷︸

Gf (s)

·1

Gv(0)Ga(0)

• There are two kinds of regulators:

1. Accelerometer

2. Transient Feedback


Accelerometer-based Regulator

∆Ωref

∆Ω

KT (1 + sTn)

∆εΩ

KA

Kp

KSM

s

∆β++

−−

Amplifier

Sensor

Feedback

oleo pneumaticdevice

⇒ servomotor

⇒∆β

∆εΩ= KT (1+sTn)

KAKSM

s

1 +KAKSM

s Kp

=KT

Kp

1 + sTns

KAKpKSM+ 1

=KT

Kp

1 + sTn

1 + sTReg

where

TReg =1

KAKpKSMǫ [10÷ 15] s ⇒ T1

Tn ∈ [3, 5] s ⇒ T2,KT

Kp=

Pn

Ωn

1

bp

1

Gv(s)Ga(s)


Transient Feedback Regulator

∆Ωref

∆Ω

KT KA

Kp

KpsTd

1 + sTd

KSM

s

∆β++

−−

we have:

Gr(s) =KT

Kp

1 + sTd

(1 + sT ′

Reg)(1 + sTs)

where

TReg =Td(Kp +KT )

Kp

Ts is very small∼ 0.1 seconds.

T2 , Td

T1 , T ′

Reg

KT

Kp, Pn

Ωn

1bp

1Gr(0)Ga(0)


Behavior of the regulators of a Hydro Plant

• In steady state, t→∞, s→ 0, both regulators behave in the same way.

To obtainbp = 0 ⇒ KP = 0

bp →∞ ⇒ KT = 0

• During a transient, since Ts is “small”, the behaviour of the two regulators is very

similar.

• In practise, the two regulators are similar and both are rather slow.


Effect of Varying the Reference of β

• Let’s assume we add an additional signal to vary the reference angle β.

∆Ωref∆Ω∆ε

KT (1 + sTn) KA

Kp

KSM

s

∆β

∆βref

Gv Ga1

sM

∆A ∆Pm

∆Pe

+ +

+− −

−

−

• In steady state, we have:

∆β = ∆βref +KT

Kp∆ε = ∆βref +GR(0)∆ε

• Moreover:

∆β

∆βref=

KAKsM

s Kp

1 +KAKSMKp

s

=KAKSMKp

s+KAKSMKp⇒

∆β

∆βref

∣∣∣∣s=0

= 1


Effect of varying the reference of β

⇒ ∆β = ∆βref +KT

KP(∆Ωref −∆Ω)

∆Pm = Ga(0)Gv(0)∆β⇒ in steady state:

∆Pm = Ga(0)Gv(0)∆βref +Ga(0)Gv(0)KT

KP(∆Ωref −∆Ω)

⇒ so we can vary the reference of the mechanical power by imposing:

∆P refm = Ga(0)Gv(0)∆βref



• During a transient, we have:

∆β = ∆βref +GR(0)∆εΩ

accelerometer:→∆β

∆βref≈

1

1 + sT1where T1 ∈ [10, 15]s

transient feedback:→∆β

∆βref=

KP

KTGR(s) ≈

1

(1 + sT1)(1 + sTs), T1 > Ts

→ During the transient, the transient feedback regulation is faster, as its response

depends on Ts.



• If in the acceleromter-based regulator we have the signal ∆βref :

∆Ωref

∆Ω

KT (1 + sTn) KA

Kp

KSM

s

∆βref

∆βref∗

+

+ +

−− −

−

∆βref∗

= ∆βref Kp

KT (1 + sTn)

hence:∆β

∆βref∗

= GR(s)Kp

KT (1 + sTn)=

KT

Kp

1 + sT2

1 + sT1

Kp

KT

1

1 + sTn=

1

1 + sT1

for the transient feedback regulator we have:∆β

∆βref∗

∼=1

1 + sTs


Model of Ga(s) for a Thermo Plant

AUX ECO EVA RH

AP MP BPBP G

PG

Ps

PA

PB PB

PM

QS

QA

Q′

A

QSP

P : Pressure Q: Volumetric flow


Model of Ga(s) for a Thermo Plants

• If there was only the turbine:

Ω∆PA =

(PA

Qs

)

nom

∆Qs

and

∆Qs = h′

A∆A

• The re-heater (RH) introduces a delay in the fluid passing from AP and MP.


Model of Ga(s) for thermo plants

• For the complete model, we have:

Ga(s) =∆Pm

∆A=

∆Pm

∆Qs

∆Qs

∆A

with:∆Pm

∆Qs=

∆PA +∆PM +∆PB

∆QS

assume: α =PA

Pm

∼= 0.3

and assume:

∆PM +∆PB =

(PM + PB

QS

)

nom

1

1 + sTR∆QS

with: TR ∈ [10, 15] s.


Model of Ga(s) for Thermo plants

• So, we obtain:

∆Pm

∆Qs=(Pm

Qs

)

nom

(

α+1− α

1 + sTR

)

• and, finally:

∆Pm

∆Qs=

Pn

Qsn

1 + αsTR

1 + sTR

• So:

Ga(s) =∆Pm

∆Qs·∆Qs

∆A=

Pn

Qsn

1 + αsTR

1 + sTRh′

A = K ′

A

1 + sαTR

1 + sTR

observe that αTR < TR, so the situation is better than what we had for the hydro

plant.


Model of Ga(s) for a Thermo Plant

• If TR = 15 sec, αTR∼= 5 sec, so we do not face too many control issues.

K′

A

αK′

A1

TR

1

αTR0

−90

−1



• Actually assuming ∆PA =(PA

QS

)

nom∆Qs is too simplistic.

• There are delays in the turbine:

∆PA =(PA

QS

)

nom

1

1 + sTA∆QS TA ∈ [0.1, 0.5] s

• Hence:

∆PM +∆PB =[(

Pm

QS

)

nom

+(

PB

QS

)

n

1

1 + sTB

] 1

(1 + sTR)(1 + sTA)∆Qs

(We neglect the delay in MP because it is really small).

⇒ GA(s) = k′A1 + α′TRs

(1 + sTR)(1 + sTA)where α′ ∼= 0.8α



• For slow transient, we have to take into account the boiler dynamic.

• We have:

∆QS =RvhA

Rv +Rs∆A+∆PG

∆Qi −∆Qs = CGd

dt∆PG

(assume ∆Qi = 0 and obtain ∆PG)

→We obtain:

GA(s) =K ′

AsTG(1 + αTRS)

(1 + sTG)(1 + sTR)

where: TG = CG(Rv +Rs) ∈ [50, 500] s

50 s: once through boiler

500 s: drum boiler


Synthesis of the f/P Regulator

• Again, we want: Gf (s) =Pn

Ωn

1

bp

1 + sT2

1 + sT1= Gr(s)Gv(s)Ga(s)

• Since we know Gv(s) and Ga(s), we have:

Gr(s) =Pn

Ωn

1

bp

1 + sT2

1 + sT1

1 + sTR

1 + sαTR

1

Gv(0)Ga(0)

where: T1 = 15 s, T2 = 5 s, αTR ≈ T2 and TR ≈ T1

hence: Gr(s) =Pn

Ωn

1

bp

1

Gv(0)Ga(0)≈ constant

• Observe that Gr(s) depends on the operating point!


Effect of the Reference Signal ∆βref

∆Ωref ∆ΩGr(s)

∆Pm

∆Pe

∆β

∆βref

(∂Pm

∂β

)

0

1 + sαTR

1 + sTR

1

sM+

+

+

+ −

−

→ ∆βref comes from the II f/P regulation.

→ In steady state, we have:

∆Pm =

(∂Pm

∂β

)

0

∆βref +

(∂Pm

∂β

)

0

Pn

Ωn

1

bp

1

(∂Pm

∂β

)

0

(∆Ωref −∆Ω)


Auxiliary Power Regulator

• With the current regulator scheme, we canot obtain a constant frequency regulator.

• Let’s add a PI controller before ∆βref :

∆P ref

∆Pe

∆εΩ

Pn

Ωn

1

bp

Kp1 + sTP

sTP

∆βref

frequency bias

++

−

where∆P ref comes from the secondary f/P regulator

∆Pe is the power produced by the plant ( = ∆PL )


Auxiliary Power Regulator

⇒ hence we have ∆Pe = ∆Pm − sM∆Ω

⇒ In steady state:

∆Pm =(

∂Pm

∂β

)

0∆βref + Pn

Ωn

1bp(∆Ωref −∆Ω)

⇒ −∆Pe +∆P ref + Pn

Ωn

1bp(∆Ωref −∆Ω) = 0

⇒ ∆Pe = ∆P ref + Pn

Ωn

1bp(∆Ωref −∆Ω) = ∆Pm

⇒ In such a way, ∆P ref forces to produce the required power.

∆P ref

· to obtain ⇒GR(0) = 0

frequency bias = 0· to obtain ⇒ disconnect

∆P ref

∆Pe


Regulation: Boiler Follows Turbine

Boiler

Turbine

Ω

PS

Reg I f/P

Ω

⇒ In this way the boiler follows the control on the turbine. It is fast but I have to open the

valve several times.

⇒ There is also a control of the type: turbine-follows-boiler. It is slow but does not

stimulate too much the valve.


Co-ordinated Control

• This control has all the positives of the “boiler-follows-turbine” and the

“turbine-follows-boiler”.

∆P ref

Pn

Ωn

1

bp

Kp1 + sTP

sTP

Gr(s)

(∂Pm

∂β

)

0

1 + sαTR

1 + sTR

∆Pm

∆Pe

∆Pe

1

sM

∆Ωref

∆Ω

+

++

+++

−

−

−

⇒∆P ref

GP (s)Aux steam

boiler

∆Ps+

−

This is the boiler-follows-turbine. To obtain the turbine-follows-boiler→ disconnect the

feedback on ∆Ω


Secondary f/P Control

• We have seen that using the primary f/P regulation, the frequency static error is not

zero.

• We used an external signal ( ∆βref or ∆P ref ) to obtian ∆Ω = 0 in steady state.

• For example, if the load increases:

∆Ω

t

∆εΩ 6= 0


Secondary f/P Control

• The secondary f/P control is ”unique” for the system (or area).

→ The National Control Center co-ordinates the Regional Control Centers.

+

−

∆Ωref

∆Ω

∆εΩReg II

yR

∆Pb,i

Real− timeDispatch

∆P refi

• ∆P refi are the signals that go to each power plant (center into the primary f/P

control of each generator).

• ∆Pb,i are the real-time power dispatches→ comes from short-term market (or from

III f/P control).


Effect of Secondary f/P Control

⇒ The static error of the secondary f/P control must be zero:

t

∆Ω 0.2÷ 5s 2÷ 50s

∆εΩ 6= 0

I f/P control

II f/P control∆εΩ = 0

→ In steady-state, we have: ∆εΩ = 0, ∆Ωref = 0, ∆Ω = 0


Scheme of the Seconday f/P Control

∆Ωref

∆Ωref

∆Ω

∆Ω

G0(s)1

R

∆yR

R1

Ri

Rn

∆Pb,i

∆Pd,i

Fpi(s)∆P

′′

m,i

∆P′′

m,i

Gf,i(s)∆P

′

m,i

∆Pm,i

∆Pe

1

sM

+

+

+ +

+

+

+

−

−

−

∆Pm,i = ∆P′

m,i + ∆P′′

m,i

I f/P reg. II f/P reg.


Scheme of the secondary f/P Control

• We have:

∆P refi = Ri∆yR +∆Pb,i (∗)

∆Pd,i = ∆Pb,i + Ri(∆Pd −∆Pb)

• Where Ri = Ri/R, R =n∑

i=1

Ri

• ∆Pd −∆Pb is the power imbalance due to a wrong load forecast.

• Ri can depend also on economical factors (market dispatch).

• Then we have: ∆P refi = ∆Pb,i + Ri(∆Pd −∆Pb) (∗∗)



• Imposing that (∗) and (∗∗) are equal, we obtain:

∆Pb,i + Ri(∆Pd −∆Pb) = Ri∆yR +∆Pb,i

⇒R1(∆Pd −∆Pb) = R1∆yR

R2(∆Pd −∆Pb) = R2∆yR

....

⇒(∑

Ri)(∆Pd −∆Pb) = (∑

Ri)∆yR

but, by definition,∑

Ri = 1 and∑

Ri = R ⇒ ∆yR =∆Pd −∆Pb

R



• From (∗) we obtain:

∆yR =∆Pd,i −∆Pb,i

Ri

• hence:∆Pd −∆Pb

R=

∆Pd,i −∆Pb,i

Ri

• and finally:

Ri

R= Ri =

∆Pd,i −∆Pb,i

∆Pd −∆Pb

• Conclusion: The frequency error is shared proportionally by all power plants involved in

the secondary f/P control.


Simplified Secondary f/P Scheme

∆Ωref ∆ΩGM (s)

K0

s

∆Pm

∆Pe

+

+

−

−

where

GM (s) =∆Ω

∆Pm −∆Pe=

1

E

1 + sT1

s2

ν2

0

+ 2ξ sν0

+ 1⇒ synthesis of the I f/P control

and where we assumed RiFp,i(s) = 1


Synthesis of GM(s)

→ Observe that:

∆Pm −∆PL∆Ω

1

sM

Gf (s)

+−

where Gf (s) =∑

Gfi(s) , M =∑

Mi

Gf (s) = E1 + sT2

1 + sT1

⇒ GM (s) =1

sM

1 + 1sMGf (s)

=1 + sT1

sM + s2MT1 + E(1 + sT2)


Synthesis of GM(s)

• Hence, the closed-loop transfer function of the machine with its primary frequency

regulation becomes:

GM (s) =1 + sT1

sM + s2MT1 + E(1 + sT2)

=1

E

1 + sT1

s(ME + T2) + s2MT1

E + 1

=1

E

1 + sT1

s2

ν2

0

+ 2ζ sν0

+ 1

⇒ hence:

ν20 = EMT1

2ζν0

= ME + T2


Stability of the Secondary f/P Scheme

• Let’s consider:

∆Ωref ∆ΩGM (s)

K0

s

∆Pm

∆Pe

+

+

−

−

• We can expect that1

T1< ν0, hence:

small K0

medium K0

big K0

−1

0

−2

ν0

1T1


Stability of the Secondary f/P Control Scheme

• So we cannot use “big” K0

• The Bode cut-frequency must be ν′′c < ν0

• Observe that the cut-frequency of the primary f/P control is ν′c ≈ ν0

so we have:

• ν′c ≈ 0.3 Hz

• ν′′c ≈ 0.03 Hz

• The secondary f/P control must be slower than the primary f/P control.


Regulation Band

• Even though the secondary f/P control is relatively slow, some kinds of plants can be

even slower and cannot follow the ∆P ref .

• Thus we define a “regulation band”:

0÷ 100%→ hydro

10%→ thermo

0%→ nuclear

(Pelton→ 100%

Francis→ 50%)

• So far most renewable energy sources do not participate to the f/P control→ This

must change in the near future.


Alternative Secondary f/P Control Scheme: AGC

• The AGC (Automatic Generation Control) is used in USA

• It is heavily based on telecommunication systems.

• It co-ordinates power exchanges between different system operators.

∆Ωref

∆Ω

∆Ω

∆εΩB0

Bi

∆Pe1

∆Pei

∆Pei

∆Pei

∆Pen

∆Pe

∆Pd

∆Pb1

R

∆yR

R1

Ri

Rn

∆Pbi

∆Pdi∆P ref

i

Ki

s

∆Ωrefi Gfi(s)

1

sMi Reg f/P I

H

K

+

+

+

+

+

+

+

+ +

−

−

−

−

−

−


AGC Scheme

• We have:

H© ∆Pd = ∆Pe +B0∆εΩ

K© ∆Pdi −∆Pei +Bi∆ǫΩ = 0 (⇒ in steady state)

⇒∑

∆Pdi = ∆Pd

⇒∑

∆Pei = ∆Pe

⇒ ∆Pd −∆Pe +n∑

i

Bi∆εΩ = 0

⇒ 0 = (B0 +n∑

i

Bi)∆εΩ

Since B0 +∑

Bi 6= 0 ⇒ ∆εΩ = 0 in steady state

Moreover if ∆εΩ = 0 ⇒ ∆Pdi−∆Pei = 0⇒ Each generator produces

exactly its reference power


Effect of the Transient Response of Synchronous Machines

• Even though the system frequency in steady-state is unique, during a transient, each

machine has its own oscillations.

∆Ω

t

∆PL

• Hence, we need to measure several frequencies in the system and compute a mean

value (⇒ COI ), otherwise the frequency control could fail.


Effect of Local Frequency Measurements

ΩrefΩref

ΩA

PAB

Grid A Grid B

R R

• In this case, if we measure only ΩA and the two systems seperate (PAB = 0)

→ the regulator on Grid B does not operate properly.


Frequency Regulation of Interconnected Systems

• Let’s consider the two systems above and their primary and secondary f/P control.

B1

∆ΩA ∆ΩA

∆ΩA

−K

s

−K

sGfA(s)

∆PmA

∆PeA

∆εA 1

sMA

K

s

∆PAB∆PAB

B2

∆ΩB

∆ΩB ∆ΩB

GfB(s)∆PmB

∆PeB

∆εB

1

sMB

Secondary f/P Control Primary f/P Control Interconnected System

++

+ +

+

++

+

− −

−

− −

−

−


Comparison of f/P control and system modes

⇒ To define the oscillation mode of the system, let’s put ∆PmA = ∆PeA = ∆PmB = ∆PeB = 0

∆εA = −K

s(∆ΩA −∆ΩB) (∗)

∆εB = +K

s(∆ΩA −∆ΩB) (∗∗)

s2MAMB x© ∆ΩA −∆ΩB =

1

sMA

∆εA −1

sMB

∆εB

⇒ (∆ΩA −∆ΩB)s2MAMB = sMB (−K

S(∆ΩA −∆ΩB))

︸︷︷︸

∗

−sMA (+K

S(∆ΩA −∆ΩB))

︸︷︷︸

∗∗

⇒ λ1,2 = ±j

√

K

MAB

, with MAB =MAMB

MA +MB

λ1,2 ≈ 1 Hz ⇒ ν′c ∼ 0.3 Hz ⇒ ν′′c ∼ 0.03 Hz ⇒ fully decoupled!


03 frequency

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