. Load Frequency Control for Interconnected Systems Using Optimal Controller

ABSTRACT

The need for satisfactory operation of power stations running in parallel and

the relation between system frequency and the speed of the motors has led to the

requirement of close regulation of power system frequency. Power systems are

frequently subjected to varying load demands. The perturbation in generated power

must match the load perturbations if exact nominal state is to be maintained. A

mismatch in the real power affects primarily the system frequency. For an efficient

and successful power system operation in the wake of area load changes and

abnormal conditions, such mismatches have to be corrected via supplementary

control.

In this project work, a detailed investigation on load frequency control problem for

both the isolated power system and interconnected power system has been carried

out. In case of interconnected power system, a two area system model is taken into

consideration for simplicity. Conventional Transfer function approach and State

Space approach are adopted to analyze the dynamic performance of the system.

The response obtained by the two approaches are verified by using MATLAB.

Firstly the system studies have been carried without proportional feed

back controllers, later the proportional plus integral strategy is implemented to

obtain an improved response for the system. Also the effect of + 50% variation in

system parameters from their nominal values on the dynamic performance of the

system has been studied by obtaining the response plots of frequency deviation of

disturbed area.

Finally, the techniques of Optimal control theory are applied to develop an optimal

feed back controller for enhancing the system dynamic performance of both

Isolated and Interconnected power systems. Numerical examples have been

considered to demonstrate the effectiveness of optimal controller over the PI

controller and the results are presented and duly discussed. ka

1

CONTENTSAbstract

1. Introduction1.1 Introduction ……11.2 Load Frequency Problem ……21.3 Literary review ……3

2. Load Frequency Control of Isolated Power Systems2.1 Introduction …….42.2 Modeling of Power system components ..…..4 2.2.1 Modeling of speed governing system …….4

2.2.2 Modeling of turbine ..…..82.2.3 Modeling of Generator-Load ..…..9 2.2.4 Block Diagram of an Isolated Power system ……11

2.3 Dynamic Response without feedback PI Control ……12 2.3.1 Transfer Function Approach ……13

2.3.2 State Space Approach ……16 2.4 Dynamic Response with PI Control ……25 2.4.1 Control strategy ……25

2.4.2 Transfer Function Approach ……27 2.4.3 State Space Approach ……30

2.5 Case Studies ……36 2.6 Discussions ……46 3. Load Frequency Control of Interconnected Power Systems3.1 Introduction …….473.2 Modeling of Multi Area Power Systems ……48

2

3.3 Modeling of Two Area Systems ……503.4 Dynamic Response of Two Area Systems

with PI control …….52 3.4.1 Area Control Error …….52 3.4.2 State Space Approach …….553.5 Case Studies …….643.6 Discussions ……..734. Load Frequency Control for Interconnected Systems using Optimal Controller4.1 Introduction …….74 4.2 Optimal Control Theory …….74 4.2.1 System State x …….75 4.2.2 System Cost C …….75

4.2.3 Optimal Controller .……76 4.2.4 Calculation of the Optimal controller K .……76 4.2.5 Snag of Optimal Control

…….78 4.3 Application of Optimal Control to an

Isolated Power System …….79 4.3.1 Isolated Power System with

Reheater constraint …….83 4.3.2 Isolated Power System with

Reheater constraint using Optimal controller …….85

4.4 Application of Optimal Control toInterconnected Systems ……89

4.5 Discussions …….95 5. Conclusions

…….96

3

References …….99

4

INTRODUCTION The continuous growth in size and complexity of electric power

systems along with increase in power demands has motivated the power

control engineers to put their best efforts in the area of Power System

Control. The operation of an interconnected power system usually leads to

improved system security and economy of operation. In addition, the

interconnection permits the utilities to take the advantage of the most

economical transfer of power. The benefits have been recognized from

beginning and interconnections continue to grow. The various areas or

power pools are interconnected through tie-lines. These tie-lines are utilized

5

for contractual energy exchange between areas and provide inter-area

support in case of abnormal conditions.

1.1 Introduction

Normally, the power systems operate in nominal system state which is

characterized by constant system frequency and voltage profile with certain

specified system reliability. The change in frequency and voltage from their

nominal values change when there is any mismatch in real and reactive

power generations and demands. It can be proved by sensitivity analysis that

a mismatch in the real power balance affects primarily the system frequency,

but leaves the bus voltage essentially unaffected. Also a mismatch in the

reactive power balance affects only the bus voltage magnitudes, but leaves

the system frequency essentially unaffected.

Automatic generation control (AGC) of interconnected power systems is

defined as the regulation of power output of generators within a prescribed

area, in response to change in system frequency, tie line loading, or the

relation of these to each other, so as to maintain scheduled system frequency

and/or established interchange with other areas within predetermined limits.

Over the years, many automatic generation control (AGC) schemes have

been suggested to deal this problem efficiently and effectively. The main

requirement of AGC is to ensure that:

1) Frequency of various bus voltages and currents are maintained at near

specified nominal values.

2) Tie line power flows among the interconnected areas are maintained at

specified levels

3) Total power requirement on the system as a whole is shared by

individual generators economically in optimum fashion.

1.2 Load Frequency Problem

6

~ ~S1 S2

Load Load

Consider two machines S1 and S2 running in parallel .

Fig 1 Two Plants connected through a tie-lineThe possibility of sharing the load by the two machines is as follows: Say,

there are two stations S1 and S2 interconnected through a tie-line. If the

change in load is either at S1 or S2 and if the generation of S1 alone is

regulated to adjust this change so as to have constant frequency, the method

of regulation is called Flat Frequency Regulation. Under such situations

station S2 is said to be operating on base load. the major draw back of flat

frequency regulation is that S1 must absorb all load changes for entire system

thereby the tie line between the two stations would have to absorb all load

changes at station S2 since the generator at S2 would maintain its output

constant.

The other possibility of sharing the change in load is that both S1 and S2

would regulate their generations to maintain the frequency constant. This is

known as Parallel frequency Regulation.

The third possibility is that the change in a particular area is taken care of by

the generator in that area thereby the tie-line loading remains constant.

This method of regulating the generation for keeping constant frequency is

known as Flat-Tie line loading Control. This arrangement has the advantage

that load swings on station S1 and the tie line would be reduced as compared

with the flat frequency regulation.

The application of modern control theory to AGC problem of

interconnected power system has been the subject wide range of applications

over the past three and half decades. Among the various types of automatic

7

generation controllers, the most widely employed are the conventional

proportional integral (PI) controller and the state feedback controllers based

on optimal control theory to achieve better system dynamic performance.

8

LOAD FREQUENCY CONTROL OF AN ISOLATED POWER SYSTEM

2.1 Introduction

The main objective for the load frequency control is to exert

control of frequency and at the same time exchange of real power via the

Tie-lines. The change in frequency and tie-line real power are sensed which

is a measure of the change in rotor angle δ, i.e. the error ∆δ to be corrected.

The error signals i.e. ∆f and ∆Ptie are amplified mixed and transformed into a

9

real power command signal ∆PC which is sent to the prime mover to call for

an increment in the torque. The prime mover therefore brings in the

generator output by an amount ∆PG which will change the values of ∆f and

∆Ptie. The process continues till deviation ∆f and ∆Ptie are well below the

specified tolerances.

2.2 Modeling of Power System Components

Modeling of different power system components i.e. Speed

governing system, Turbine, Generator-load are described and the various

block diagrams representing the components are presented in this section.

2.2.1 Modeling of Speed Governing System

The schematic diagram of speed governing system which controls the real

power flow in the power system is shown in fig 2.1.

A

BC D

El1 l3l2 l4

Speed changer

Lower

Raise

Speed governor

High pressure oil

Pilot valve

XDirection of positive movement

steam

To turbine

10

Fig:2.1 Speed Governing System

The Speed Governing System consists of the following parts:-

1. Speed Governor: This is a fly-ball type of speed governor and

constitutes the heart of the system as it senses the change in speed or

frequency. With the increase in speed the fly ball move outwards and the

point B on linkage mechanism moves downwards and vice versa.

2. Linkage Mechanism: ABC and CDE are the rigid links pivoted at B

and D respectively. The mechanism provides a movement to the control

valve in the proportion to change in speed. Link4 (l4) provides a feed back

from the steam valve movement.

3. Hydraulic Amplifier: This consists of the main piston and pilot valve.

Low power level pilot valve movement is converted into high power level

piston valve movement which is necessary to open or close the steam valve

against high pressure steam.

4. Speed Changer: The speed changer provides a steady state power

output setting for the turbine. The downward movement of the speed

changer opens the upper pilot valve so that more steam is admitted to the

turbine under steady condition. The reverse happens when the speed changer

moves upward.

Consider the steady state condition by assuming that linkage

mechanism is stationary, pilot valve closed, steam valve opened by definite

magnitude, the turbine output balances the generator output and the

turbine/generator is running at a normal speed or at a normal frequency

f° ,the generator output PGO and let the steam valve setting corresponding to

these conditions be XE.

11

Let the point A of the speed changer lower down by an amount

∆XA as a result the commanded increase in power ∆PC then ∆XA = K1∆PC.

The movement of linkage point A causes small position changes ∆XC and

∆X D of the linkage points C and D. With the movement of D upwards by

∆X D high pressure oil flows into the hydraulic amplifier from the top of the

main piston thereby the steam valve will move downwards a small distance

∆XE which results in increased turbine torque and hence power increase,

∆PG. This further results in increase in speed and hence the frequency of

generation. This increase in frequency ∆f causes the link point B to move

downward a small distance ∆XB proportional to ∆f. Assume the

movements are positive if the points move downwards.

Two factors contribute to the movement of C:

i)Increase in frequency causes B to move by ∆XB when the frequency

changes by ∆f as then the fly-ball moves outward and B is lowered by ∆XB .

Therefore, this contribution is positive and is given by K1∆f.

ii) The lowering of the speed changer by an amount ∆XA lifts the

point C upwards by an amount proportional to ∆XA, i.e. K2∆PC.

∆XC = K1∆f - K2∆PC …………………..2.1

Where K1 and K2 are the positive constants depends upon the

length of the linkage arms AB and BC and upon the proportional constants

of the speed changer and the speed governor.

The movement of D is contributed by the movement of C and

E. Since C and E move downwards when D moves upwards,

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therefore, ∆X D= K3∆XC + K4∆XE ……………………2.2

Where K3 and K4 are positive constants depend upon the length of the

linkage CD and DE

Let the oil flow into the hydraulic cylinder is proportional to position

∆X D

of the pilot valve, the value of ∆XE is given by

∆XE=K5 -(∆X D )dt ……………………2.3

Where the constant K5 depends upon the fluid pressure and the

geometries of the orifice and the cylinder.

Taking Laplace transforms to equations 2.1, 2.2 & 2.3

∆XC (s)= K1∆F(s) - K2∆PC(s)

∆X D (s)= K3∆XC (s)+ K4∆XE (s)

∆XE (s)= -K5∆X D (s) /s

= -K5 (K3.( K1∆F(s) - K2∆PC (s))+ K4∆XE (s)) s

Eliminating the variables ∆XC and ∆X D ,

∆XE (s) = K2K3 ∆PC(s) - - K3 K1∆F(s) K4+s/K5

∆XE(s)= K G [∆PC (s) -∆F(s)/R] ……………….2.4 1+sTG

Where R = K2/K1 → speed regulation of governor.

K G = K2K3/K4 → gain of speed governor.

TG = 1/K4K5 → time constant of speed governor.

The above equation 2.4 can be represented as a block diagram shown in

fig2.2

13

Fig. 2.2 Block diagram of speed governing system for steam turbine

2.2.2 Modeling of Turbine

The turbine power increment ∆P T depends entirely upon the

valve power increment ∆Pv and the response characteristics of the turbine. A

non-reheat turbine with a single gain factor K T and a single time constant T T

is considered and in the crudest model representation of the turbine the

transfer function is given as

G T (s)= ∆P T (s) = K T ….……………..2.5

∆XE (s) 1+sT T

The above transfer function is represented in the form of Block diagram

along with the governor as shown in fig 2.3

Fig.2.3 Block diagram of power control mechanism of turbine

KG

1+sTG

+-

1/R

∆PC ∆XE(s)

∆F(s)

KG 1+sTG

+-

1/R

∆PC ∆XE(s)

∆F(s)

KT 1+sTT

14

∆PT(s)

2

2.2.3 Modeling of Generator-Load

The model gives relation between the change in

frequency as a result of change in generation when the load changes by a

small amount.

Let ∆PD be the change in load demand, as a result the generation also swings

by an amount ∆PG. The net power surplus at the busbar is ∆PG-∆PD and this

power will be absorbed by the system in two ways

a) Rate of increase of stored kinetic energy in generator rotor

Let Wo be the Kinetic Energy before change in load occurs when

the frequency is f o.

Let ∆f be the change in frequency.

Let W be the Kinetic Energy when the frequency is ∆f+ f o.

As K.E. is proportional to square of the speed of the generator

W= W° f° +∆f f° → W=W° (1+2∆f/f°) (neglecting the higher terms)…………2.6

By Differentiating equation 2.6,

dW = 2W° . d (∆f) ………………2.7 dt f° dt

b) The load on the motors increases with increase in speed. The load

on the system being mostly motor load the rate of change of load w.r.t.

frequency can be regarded as nearly constant for small changes in

frequency,

i.e. D=∂PD where D, Load Frequency Constant, can be obtained empirically ∂f

Therefore, the net power surplus at the busbar is given by

∆PG-∆PD = 2W° . d (∆f) +D ∆f ………………2.8 f° dt

15

Let H be the inertia constant of the generator in MW-sec/MVA

Let P be the rating in MVA, then W° =H*P.

Substituting W° in equation 2.8,

∆PG-∆PD =2HP. d (∆f) +D ∆f ……………….2.9 f° dt

Dividing equation 2.9 throughout by P,

∆PG (p.u.)-∆PD (p.u.) = 2H . s∆F(s) +D(s)∆F(s) f°

= ∆F(s){D(s)+2Hs/f°}

i.e. ∆F(s) = ∆PG (s)-∆PD (s) D(s)+2Hs/f°

Or ∆F(s) = [∆PG (s)-∆PD (s) ] KP ……………..2.10 1+sTP

Where Tp=2H/Df° →power system time constant

Kp=1/D →power system gain

The transfer function in equation 2.10 is represented in the form of a

block diagram as shown in fig 2.4

+-

∆PG

∆F(s)

∆PD

KP 1+sTP

16

Fig2.4 Block diagram of Generator- Load model

2.2.4 Block diagram of an Isolated Power System

The models of speed governor, turbine, generator-load

are combined to represent complete block diagram of an isolated power

system for Load Frequency Control and is represented in fig 2.5

Fig 2.5 Block Diagram of an Isolated Power System for load frequency control

From the block diagram in fig 2.5, the change in ∆F is due to

1) either change in speed changer setting(∆Pc)

2) or Load Demand(∆PD)

+- +-

1/R

∆PC(S)

∆PD(S)

∆F(S) KP 1+sTP

KT 1+sTT

KG 1+sTG

17

Consider a fixed setting in speed changer i.e. ∆Pc =0, which is known

as free governor operation, and a sudden change in load (i.e. step disturbance

in load) and so ∆PD(s)= ∆PD/s

By representing the block diagram in fig 2.5 in the form of a

transfer function ∆F(s) ,

∆F(s) = KP * (∆PD /s) ………2.11 (1+sTp)+ K G KT KP/R

(1+sTG)(1+sTT)

2.3 Dynamic Response without Feedback PI Control

To obtain the dynamic response giving change in frequency as a

function of the time for a step change in load, there are two different

approaches

1) Transfer Function approach

2) State Space approach

2.3.1 Transfer Function Approach

To obtain transient response or frequency response analysis of

single input and single output linear systems, the conventional transfer

function representation forms a useful model. The transfer function of a

linear time-invariant system is defined as the ratio of the laplace transform

of the output variable to the laplace transform of the input variable under the

18

assumption that all initial conditions are zero. The highest power of the

complex variable s in the denominator of the transfer function determines

the order of the system.

When the transfer function of a physical system is determined,

the system can be represented by a block. It describes the input and output

behaviour of the system and does not give any information concerning the

internal structure of the system.

Recalling the equation 2.11,

∆F(s) = KP * ( ∆PD/s) ……………2.12 (1+sTp)+ K G KT KP/R

(1+sTG)(1+sTT)

By simplifying the equation 2.12 and assuming K G KT ~ 1

∆F(s) = R (1+sTG) (1+sTT) 1+{(1+sTG) (1+sTT) (2Hs+D) R}

= R(1+s2TT TG+s(TT+ TG)

1+[{1+ s2TT TG+s(TT+ TG)}{2Hs+D}R]

∆F(s) = s2 TTTG+s(TT+TG)+1 s3TTTG2H+s2{TTTGD+2HTT+2HTG}+s{DTT+2H+DTG}+

{D+1/R}

To obtain the response of isolated power system without feedback by the

transfer function approach using Matlab programming. The following is the

program to use in MATLAB

Program For Transfer Function Model(Protfsa.m):

19

T=input('TOTAL RATED CAPACITY in MW :');

R=input('ENTER THE SPEED REGULATION GOVERNOR(R)in

Hz/puMw:’);

F=input('ENTER SYSTEM FREQUENCY(F) in Hz :');

H=input('ENTER INERTIA CONSTANT(H) in sec :');

D=input('ENTER THE LOAD FREQUENCY CONSTANT(D)in puMW/Hz:');

TG=input('ENTER THE TIME CONSTANT OF GOVERNOR in sec :');

TT=input('ENTER THE TIME CONSTANT OF TURBINE in sec :'); Pp=input('ENTER THE CHANGE IN LOAD DEMAND IN TERMS OF %

INCREASE:');

dP=Pp*0.01;

R=R/F;

Num=[TG*TT TT+TG 1]*-dP;

Den=[TT*TG*H (TT*TG*D+2*H*TT+2*H*TG) (D*TT+2*H+D*TG)

(D+1/R)];

Tf=tf(Num,Den);

step(Num,Den);

Tf

>> Protfsa

TOTAL RATED CAPACITY in MW :250

ENTER THE SPEED REGULATION GOVERNOR(R) in Hz/puMw :3

ENTER SYSTEM FREQUENCY(F) in Hz :60

20

Fig 2.6 Response plot for an isolated power system without PI feedback control by Transfer Function approach

ENTER INERTIA CONSTANT(H) in sec :5

ENTER THE LOAD FREQUENCY CONSTANT(D) in puMw/Hz :0.8

ENTER THE TIME CONSTANT OF GOVERNOR in sec :0.2

ENTER THE TIME CONSTANT OF TURBINE in sec :0.5

ENTER THE CHANGE IN LOAD DEMAND IN TERMS OF % INCREASE :20

Transfer function:

-0.02 s^2 - 0.14 s - 0.2

-----------------------------------

0.5 s^3 + 7.08 s^2 + 10.56 s + 20.8

2.3.2 State Space Approach

The state variable approach is a powerful technique for the analysis and

design of control systems and which has a lot of advantages over transfer

21

functional approach. The state variable analysis can be applied to multi input

and multi output systems.

The transient analysis can be carried with initial conditions and can be

carried on multiple input and multiple output systems. In this method of

analysis, it is not necessary that the state variables represent physical

quantities of the system. The variables that do not represent physical

quantities and those that are neither measurable nor observable may be

chosen as state variables.

The state model of a system consist of state equation output equation.

(t)= Ax(t)+Bu(t) ………..state equation

y(t) = Cx(t)+Du(t) ……….output equation

where

A→ n x n Sate Distribution Matrix

B→ n x m Control Distribution Matrix

C→ Output matrix

D→ Transition Matrix

Analysis of State Space Equation:

22

Fig 2.7 Block diagram of an isolated power system for analyzing state space equations

The state space equations can be derived by considering

the intermediate states and from the block diagram

x1=∆f, change in frequency

x2=∆PG, change in power generation

x3=∆XG, change in governor position

x1=( x2-u) 1 …………….2.13 2Hs+D

x2= x3/(1+sTT) …………….2.14

x3=- x1 1 ……………2.15R (1+sTG )

from equation 2.14 x2+TTs x2= x3

→ x2+ 2 TT = x3 ……………2.16

KG

1+sTG

KT

1+sTT

1 2Hs+D

+- +-

1/R

u

∆PD(S)

∆F(S)∆XG

x2x3

x1

23

-D/2H 1/2H 0 0 -1/TT 1/TT

-1/RTG 0 -1/TG

-1/2H 0 0

B=

from equation 2.15, - x3R- x3R TGs= x1

→ - x3R- 3RTG = x1

……………..2.17

from equation 2.13, x1*2Hs+ x1D= x2-u

→ 1 2H+x1D= x2-u …………….2.18

By solving the equations 2.16, 2.17 and 2.18,

1 =(- x1D- x2-u)/2H ……………….2.19 2 =(- x2+ x3)/ TT

………………2.20

3 = (-x1- x3R)/(R* TG) ………………2.21

Writing the equations 2.19, 2.20 and 2.21 in Matrix form ,

where A=

C=(1 0 0) and D= ( 0 )

1

2

3

= -D/2H 1/2H 0

0 -1/TT 1/TT

-1/RTG 0 -1/TG

x1

x2

x3

+ u

-1/2H

0

0

24

To obtain the response of isolated power system without feedback by the

state space approach using Matlab programming. The following is the

program to use in MATLAB.

Program for State Space Model(Prosssa.m)


R=input('ENTER THE SPEED REGULATION GOVERNOR(R) in Hz/puMw:');



D=input('ENTER THE LOAD FREQUENCY CONSTANT(D) in puMW/Hz: ');

TG=input('ENTER THE TIME CONSTANT OF GOVERNOR in sec : ');

TT=input('ENTER THE TIME CONSTANT OF TURBINE in sec :');

Pp=input('ENTER THE CHANGE IN LOAD DEMAND IN TERMS OF %

INCREASE:');

dP=Pp*0.01;

R=R/F;

A=[-D/(2*H) 1/(2*H) 0 ;0 -1/TT 1/TT ;-1/(R*TG) 0 -1/TG];

B=[-dP/(2*H) ; 0 ; 0 ];

C=[1 0 0 ]; D=[0 ];

fprintf('THE MATRICES A,B,C,D ARE:- ');

A

B

C

D

step(A,B,C,D);

25

>> Prosssa


ENTER THE SPEED REGULATION GOVERNOR(R) in Hz/puMw :3



ENTER THE LOAD FREQUENCY CONSTANT(D) in puMw/Hz :0.8



ENTER THE CHANGE IN LOAD DEMAND IN TERMS OF % INCREASE :20

THE MATRICES A,B,C,D ARE:-

A =

-0.0800 0.1000 0

0 -2.0000 2.0000

-100.0000 0 -5.0000

B =

-0.0200

26

0

0

C =

1 0 0

D =

0

Fig 2.8 Response plot of Isolated Power System without PI feedback control by State Space Approach

27

2.4 Dynamic Response with PI Control

With the primary LFC loop, a change in the system load will

result in a steady state frequency deviation depending on the governor

speed regulation. In order to reduce the frequency deviation to zero, a reset

action is to be provided. The reset action can be achieved by introducing an

integral controller to act on the reference setting to change the speed set

point. The integral controller increases the order of the system by one.

2.4.1 Control Strategy

Uncontrolled system is subject to steady state errors and so

control strategy is required. The control specifications are

1) Control loop must be characterized by a sufficient degree of stability

2) Following a step load change, the frequency error should return to

zero. This is referred to as isosynchronous control. Magnitude of transient

frequency must be minimized.

3) Integral of the frequency error should be minimized.

4) The individual generators of the control area should divide the total

load for optimum frequency.

28

Let ∆Pc be the negative feed-back signal drawn from frequency

deviation. Suppose, if it was not an integral feedback, i.e. if ∆Pc=-K1∆F(s)

where k1→gain for proportion control

Recollecting the system response from equation 2.11

∆F(s) = KP * ( ∆PD /s) (1+sTp)+ K G KT KP/R

(1+sTG)(1+sTT)

→ ∆F(s)= - KP(1+sTG)(1+sTT) ∆PD(s) {assuming KG KT~ 1} ….2.22

(1+sTG)(1+sTT) (1+sTP)+(K1 +1/R)KP

i.e. Steady state frequency error,

∆FSS = Lt ∆F(s)= -KP ∆PD S→0 1+(K1+1/R)KP

i.e. As ∆Fss→0, K1→∞, which is not suitable for proportional control and so

cannot be recommended for control

So, if ∆Pc(t)=-KI ∫ ∆f(t)dt

∆Pc(s)=-KI∆F(s)/s

So , ∆F(s)= -KP∆PD(s) s(1+sTP)+(K2+s/R)

∆FSS= Lt s. ∆F(s) =0s→0

29

Thus by using Integral control strategy, steady state error can be

eliminated(i.e.∆Fss=0). Thus PI controller when introduced improves the

transient performance and ensures better stability.

The complete block diagram with integral control for an isolated power

system is shown in figure 2.9

Fig 2.9 Block Diagram of an Isolated Power System with PI controller

The dynamic response for a step change in load, with integral

control can be obtained by two different approaches

2.4.2 Transfer Function Approach

From the above block diagram shown in fig 2.9 the system equation can be

written as

∆F(s)= - KP * ∆PD ………….2.23 (1+sTP)+( 1 + KI ) * KP o

R s ( 1+sTG)(1+sTT)

But D=1/Kp and Tp=2H/D

Hence on simplifying, equation 2.23 becomes

KG

1+sTG

KT

1+sTT

KP

1+sTP+- +-

1/R

∆PC(S)

∆PD(S)

∆F(S)

KI/s

30

∆F(s)= Rs( 1+sTG) (1+sT T) s[(1+sTG)(1+sTT) (1+sTp)+s/R+KI

→∆F(s)= s3 TTTG+s2 (TT+TG)+s s4TTTG2H+s3{TTTGD+2HTT+2HTG}+s2{DTT+2H+DTG}+s{D+1/R}

+KI

From the above equation, the response in the transfer function

approach can be obtained by a program using MATLAB.

Program for Transfer Function Model(Protfsawi.m)

T=input('TOTAL RATED CAPACITY :');

R=input('ENTER THE SPEED REGULATION GOVERNOR(R) in HZ/puMW:');



D=input('ENTER THE LOAD FREQUENCY CONSTANT in puMw/Hz :');

Tg=input('ENTER THE TIME CONSTANT OF GOVERNOR in sec :');

Tt=input('ENTER THE TIME CONSTANT OF TURBINE in sec :');

Ki=input('ENTER THE INTEGRAL GAIN :');

Pp=input('ENTER THE CHANGE IN LOAD DEMAND IN TERMS OF

%INCREASE:');

dP=Pp*0.01;

R=R/F;

Num=[0 Tg*Tt (Tg+Tt) 1 0]*(-dP);

Den=[Tg*Tt*2*H Tg*Tt*D+Tg*2*H+Tt*2*H)(2*H+Tg*D+Tt*D)(D+1/R)

Ki];

31

Tf=tf(Num,Den);

Tf

step(Num,Den);

>> Protfsawi

TOTAL RATED CAPACITY :250

ENTER THE SPEED REGULATION GOVERNOR(R) in puMW/Hz:3



ENTER THE LOAD FREQUENCY CONSTANT in puMw/Hz :0.8



ENTER THE INTEGRAL GAIN :7

ENTER THE CHANGE IN LOAD DEMAND IN TERMS OF % INCREASE:20

Transfer function:

-0.02 s^3 - 0.14 s^2 - 0.2 s

---------------------------------------

s^4 + 7.08 s^3 + 10.56 s^2 + 20.8 s + 7

32


The block diagram with different states required for analyzing

the state equations is shown in fig 2.11

KG

1+sTG

KT

1+sTT

KP

1+sTP+- +-

1/R

∆PC(S)

∆PD(S)

∆F(S)

KI/s

x4 x

3

x2

x1

33

Fig 2.10 Response Plot of an Isolated Power System with PI control using Transfer function Approach

Fig 2.11Block Diagram with PI Control in State Space Representation

1=( x2/2H)-( x1D/2H)-(u/2H) …………………..2.24

2=-( x2/TT)+(x3/ TT) …………………..2.25

x3=( x4-1/R) 1/(1+sTG) → x3 (1+s TG)= - x4 – x1/R

i.e. 3= -( x4/ TG) –( x1/R TG)-( x3/ TG) ……………………2.26

x4= x1Ki/s

i.e. 4= x1Ki ……………………2.27

So the standard form of equations are

To obtain the response of isolated power system with PI control by the

state space approach using Matlab programming, the following numerical

example is illustrated

1

2

3

4

=

-D/2H 1/2H 0 0

0 -1/TT 1/TT 0

-1/RTG 0 -1/TG 1/TG

Ki 0 0 0

+

-1/2H 0

0

0

u x1

x2

x3

x4

34

Program for State Space Model(Prosssawi.m):-

T=input('TOTAL RATED CAPACITY in MW :’);

R=input('ENTER THE SPEED REGULATION GOVERNOR in Hz/puMw :');


H=input('ENTER INERTIA CONSTANT(H)in sec :');






%INCREASE:');

dP=Pp*0.01;

R=R/F;

A=[-D/(2*H) 1/(2*H) 0 0;0 -1/Tt 1/Tt 0;-1/(R*Tg) 0 -1/Tg -1/Tg;Ki 0 0 0];

B=[-dP/(2*H) ; 0 ; 0 ; 0 ];

C=[1 0 0 0];

D=[0 ];

disp('THE MATRICES A,B,C,D ARE AS FOLLOWS:-');

A

B

C

D

step(A,B,C,D);

>> prosssawi

35


ENTER THE SPEED REGULATION GOVERNOR(R) in Hz/puMW :3


ENTER INERTIA CONSTANT(H)in sec :5

ENTER THE LOAD FREQUENCY CONSTANT in puMw/Hz :0.8





THE MATRICES A,B,C,D ARE AS FOLLOWS:-

A =

-0.0800 0.1000 0 0

0 -2.0000 2.0000 0

-100.0000 0 -5.0000 -5.0000

7.0000 0 0 0

B =

-0.0200

36

0

0

0

C =

1 0 0 0

D =

0

Fig 2.12 Response plot for an isolated Power System with PI control using State Space Approach

37

.

2.5 Case Studies

There are several parameters which effect the dynamic response of a

system and they are

Case 1 : Governor Speed Regulation

Case 2: Time Constant of the Govenor

Case 3: Time Constant of the Turbine

Case 4: Integral Gain

Case 5: Load Disturbance

Different Case Studies have been carried by varying the above parameters.

The variations are depicted by using MATLAB programming for an Isolated

Power System with PI controller. The dynamic response for different cases

is obtained by considering the example with ± 50% changes in the parameter

values.

38

Program for Case studies in Singe area (prosacs.m):-


F=input('ENTER SYSTEM FREQUENCY(F)in Hz :');


D=input('ENTER THE LOAD FREQUENCY CONSTANT in puMw/Hz:');

disp('ENTER THE PARAMETER IN WHCICH THE VARIATION IS TO BE SOUGHT:')

disp('1.CHANGE IN SPEED REGULATION(R)\n');

disp('2.CHANGE IN GOVERNOR TIME CONSTANT(Tg)\n');

disp('3.CHANGE IN TURBINE TIME CONSTANT(Tt)\n');

disp('4.CHANGE IN INTEGRAL GAIN(Ki)\n');

disp('5.CHANGE IN LOAD DEMAND(dP)\n\n');

value=input('ENTER A VALUE :');

switch(value)

case {1}

Tg=input('ENTER THE TIME CONSTANT OF GOVERNOR IN SEC :');



Pp=input('ENTER THE CHANGE IN LOAD DEMAND IN TERMS OF % INCREASE:');

R=input('ENTER THE SPEED REGULATION GOVERNOR(R) in Hz/puMW');

dP=Pp*0.01;

r=R/F;

A1=[-D/(2*H) 1/(2*H) 0 0;0 -1/Tt 1/Tt 0;-1/(r*Tg) 0 -1/Tg -1/Tg; Ki 0 0 0];

A2=[-D/(2*H) 1/(2*H) 0 0;0 -1/Tt 1/Tt 0;-1/(r*1.5*Tg) 0 -1/Tg -1/Tg;Ki 0 0 0];

39

A3=[-D/(2*H) 1/(2*H) 0 0;0 -1/Tt 1/Tt 0;-1/(r*0.5*Tg) 0 -1/Tg -1/Tg;Ki 0 0 0];

case {2}



Pp1=input('ENTER THE CHANGE IN LOAD DEMAND IN TERMS OF %NCREASE:');

R=input('ENTER THE SPEED REGULATION GOVERNOR(R) in Hz/puMW:');


dP1=Pp1*0.01;r=R/F;

A1=[-D/(2*H) 1/(2*H) 0 0;0 -1/Tt 1/Tt 0;-1/(r*Tg) 0 -1/Tg -1/Tg;Ki 0 0 0];

A2=[-D/(2*H) 1/(2*H) 0 0;0 -1/Tt 1/Tt 0;-1/(r*Tg*1.5) 0 -1/(Tg*1.5) -1/(Tg*1.5);Ki 0 0 0];

A3=[-D/(2*H) 1/(2*H) 0 0;0 -1/Tt 1/Tt 0;-1/(r*Tg*0.5) 0 -1/(Tg*0.5) -1/(Tg*0.5);Ki 0 0 0];

case {3}


R=input('ENTER THE SPEED REGULATION GOVERNOR(R) in HZ/puMW :');


Pp1=input('ENTER THE CHANGE IN LOAD DEMAND IN TERMS OF % INCREASE:');


dP1=Pp1*0.01; r=R/F;


A2=[-D/(2*H) 1/(2*H) 0 0;0 -1/(Tt*1.5) 1/(Tt*1.5) 0;-1/(r*Tg) 0 -1/(Tg) -1/(Tg);Ki 0 0 0];

A3=[-D/(2*H) 1/(2*H) 0 0;0 -1/(Tt*0.5) 1/(Tt*0.5) 0;-1/(r*Tg) 0 -1/(Tg) -1/(Tg);Ki 0 0 0];

40

case {4}

R=input('ENTER THE SPEED REGULATION GOVERNOR(R) in puMW/Hz :');




%INCREASE:');


dP=Pp*0.01; r=R/F;


A2=[-D/(2*H) 1/(2*H) 0 0;0 -1/Tt 1/Tt 0;-1/(r*Tg) 0 -1/Tg -1/Tg;Ki*1.5 0 0 0];

A3=[-D/(2*H) 1/(2*H) 0 0;0 -1/Tt 1/Tt 0;-1/(r*Tg) 0 -1/Tg -1/Tg;Ki*0.5 0 0 0];

case {5}

R=input('ENTER THE SPEED REGULATION GOVERNOR(R) in puMW/Hz :');





dP1=Pp1*0.01;

dP2=Pp1*0.5*0.01;

dP3=Pp1*1.5*0.01;

r=R/F;

41

A=[-D/(2*H) 1/(2*H) 0 0;0 -1/Tt 1/Tt 0;-1/(r*Tg) 0 -1/Tg -1/Tg;Ki 0 0 0];

B=[-dP1/(2*H) -dP2/(2*H) -dP3/(2*H); 0 0 0 ; 0 0 0; 0 0 0];

C=[1 0 0 0];

D=[0 0 0];

step(A,B,C,D);

break;

otherwise ,

end

B=[-dP/(2*H); 0; 0; 0];

C=[1 0 0 0]; D=[0];

sys1=ss(A1,B,C,D);

sys2=ss(A2,B,C,D);

sys3=ss(A3,B,C,D);

step(sys1,sys2,sys3);

Dynamic Response after execution is as follows:

>> prosacs


ENTER SYSTEM FREQUENCY(F)in Hz :60


ENTER THE LOAD FREQUENCY CONSTANT in Hz/puMw :0.8

ENTER THE PARAMETER IN WHCICH THE VARIATION IS TO BE SOUGHT:

1.CHANGE IN SPEED REGULATION(R)\n

42

2.CHANGE IN GOVERNOR TIME CONSTANT(Tg)\n

3.CHANGE IN TURBINE TIME CONSTANT(Tt)\n

4.CHANGE IN INTEGRAL GAIN(Ki)\n

5.CHANGE IN LOAD DEMAND(dP)\n\n

ENTER A VALUE :1

ENTER THE TIME CONSTANT OF GOVERNOR IN SEC :0.2





43

Fig2.13 Response plot of an Isolated Power System with PI controller with variation in Governor Speed Regulation

ENTER A VALUE :2






44

Fig2.14 Response plot of an Isolated Power System with PI controller with variation in Time Constant of Governor

ENTER A VALUE :3






45

Fig2.15 Response plot of an Isolated Power System with PI controller with variation in Time Constant of Turbine

ENTER A VALUE :4

ENTER THE SPEED REGULATION GOVERNOR(R) in Hz/puMW:3





46

Fig2.16 Response plot of an Isolated Power System with PI controller with variation in Integral Gain

ENTER A VALUE :5






47

Fig2.17 Response plot of an Isolated Power System with PI controller with variation in Load Disturbance

2.4 Discussions

48

The response of an isolated power system without integral feedback control when subjected to load change are shown in figs 2.6&2.8. From these figs. it is clear that response for the isolated power system obtained by the transfer function and state space approaches. Also the settling time is 8 secs and the steady state frequency deviation is .0095Hz/pu.

The response of an Isolated Power System with PI controller when subjected to a load change are shown in figs 2.10&,2.12. From these figures, it is clear that the settling time is 15 seconds and the steady state frequency deviation is zero. Hence it can concluded that the feed back control reduces the steady state deviation to zero.

The response for the variation in different parameters are obtained in section 2.5. The response for + 50% variation in speed regulation of governor is shown in figure 2.13. From this figure, it can be observed that when R is increased the settling time decreases but frequency deviations increases and vice versa.

The response for + 50% variation in Time constant of governor is shown in figure 2.14. From this figure, it is clear that when TG is decreased the settling time decreases and frequency deviations decreases and vice versa.

The response for + 50% variation in Time Constant of Turbine is shown in figure 2.15. From this figure, it is evident that when TT is increased the settling time decreases and frequency deviations decreases and vice versa.

The response for + 50% variation in Integral Gain is shown in figure 2.16. From this figure, it is evident that when Ki is increased the settling time increases and frequency deviations increases.

The response for + 50% variation in Load disturbance is shown in figure 2.17. From this figure, it can be observed that when dP is increased the frequency deviation increases with no considerable change in settling time.

49

CHAPTER-3

50

LOAD FREQUENCY CONTROL OF INTERCONNECTED SYSTEMS

3.1 Introduction

All power systems today are tied together with neighboring

areas and the problem of load-frequency control becomes a joint

undertaking. By considering a practical system with a number of generator

stations and loads, it is possible to divide an extended power system into sub

areas in which the generators are tightly coupled together so as to form a

coherent group. Such a coherent area is called a control area in which the

frequency is assumed to be the same throughout in static as well as dynamic

conditions. The important advantages can be derived by pool operation are

a) to improve system security and economy of operation

b) the interconnection permits the utilities to make economy transfers

The basic operating principles for interconnecting systems are

a) under normal operating conditions each pool member or control area

should strive to carry its own load, except such scheduled portions of the

other members’ loads as have been mutually agreed upon.

b) Each control area must agree upon adopting regulating and control

strategies and equipment that are mutually beneficial under both normal and

abnormal situations. The advantages belonging to a pool are particularly

evident under emergency conditions.

51

The Problems of frequency control of interconnected areas are more

important than those of isolated areas. The objective of load frequency

control of interconnected power systems is two fold: minimizing the

transient error deviations in both frequency and tie line power and ensuring

zero steady state errors of these two quantities. By a simple proportional

integral control law the above mentioned objective is achieved. This chapter

presents proportional control for minimizing the transient error and the

integral control for zero steady state error.

3.2 Modeling of Multi area Power Systems

In an isolated control area case, the incremental power (∆PG-∆PD) was

accounted by the rate of increase of stored kinetic energy in or out of an

area. Changes in the Tie-line power flows also affect the power balance in

corresponding areas.

Consider any area i in an n area power system corresponding to

change in load demand ∆PD.

Let ∆ Ptie, i be the tie line schedule deviation

∆ Ptie, , i = ∆ Ptie, , ij ………………..3.1

where ∆ Ptie,, ij is the change in tie line power flow over the line connecting

areas i and j

Ptie, , ij= Vi Vj sin[(δi + ∆ δi )- (δj + ∆ δj ) ] ……………….3.2 Xij

where Vi,Vj → voltage magnitudes at the tie-line ends in areas i and j

respectively

Xij → Reactance of the same tie-line

δi , δj → Nominal bus voltage phase angles

52

∆ δi , ∆ δj → Changes in the phase angles

Expanding equation 3.2,

∆ Ptie, , ij = Vi Vj sin[(δi - δj )+ Vi Vj cos (δi ° - δj ° )(∆ δi -∆ δj ) …….3.3 Xij Xij

but we have Pmax = Vi Vj / Xij

So from the Equation 3.3

∆ Ptie, , ij = Pmax sin(δi - δj )+ Pmax cos (δi ° - δj ° )(∆ δi -∆ δj )

= Ptie,°, ij + Pmax cos (δi ° - δj ° )(∆ δi -∆ δj )

Synchronizing Coefficient Tij is given as

Tij = Ptie, , ij - Ptie,° , ij = Pmax cos (δi - δj )

(∆δi -∆ δj)

So, ∆ Ptie, ij = Tij (∆δi - ∆ δj ) ….……….3.4

where ∆ Ptie, , ij = Ptie,, ij- Ptie °, ij

For Frequency Deviation (∆f i) in the ith area ,

Ptie, ij = 2Π Tij ( ∆ f i dt - ∆ f j dt ) …………..3.5

By applying d/dt to the equation 3.5

d (∆Ptie,, ij ) = 2Π Tij (∆f i -∆ f j) …………….3.6 dt

Applying Laplace Transform to the Equation 3.6

53

∆ Ptie,, i(s) = (2Π/s) Tij {∆ Fi (s)-∆ F j (s)} ……………..3.7

Thus for the case of a multi-area interconnected systems,

∆PGi -∆PDi = Di + 2Hi - d (∆fi ) + ∆Ptie, i fs dt

Taking Laplace transform on both side of equation and rearranging terms,

∆ Fi (s)= KPi [ ∆PGi(s)-∆PDi(s)- Ptie,, i(s)] ……………..3.8 (1+sTPi)

3.3 Modeling of Two-Area Systems

In fact, a Two Area system is a case of multi area system,

where the areas are connected through the tie line power exchange. The

reasons for choosing the two-area system as the object of our study are

1. It is the simplest form of multi area systems.

2. Because the few papers published on multiple-area control have

limited their analysis to two area-systems and the results can be compared.

3. before understanding the larger systems, it is of paramount

importance that the two area case must be thorough commanded.

From Equation 3.8 we have

∆F1(s)= KP1 [ ∆PG1(s)-∆PD1(s)- Ptie, , 1(s)] (for area1 i.e. i=1 )………..3.9 (1+sTP1)

∆F2(s)= KP2 [ ∆PG2(s)-∆PD2(s)- Ptie,, 2(s)] (for area2 i.e. i=2 )………..3.10 (1+sTP2)

from Equation 3.7 assigning i=1 and j=2

54

KG

1+sTG1

KT

1+sTT1

1 2H1s+D1

-+ -

1/R1

∆PL1(S)

KG

1+sTG

2

KT

1+sTT2

- ++-

1/R2

1 2H2s+D2

∆PL2(S)

T12/S+-

-

-

∆ Ptie, 1(s) = 2 Π T 12 [∆F1 (s)- ∆F2(s)] ……………..3.11s

∆ Ptie,, 2(s) = -2 Π T 12 a 12 [∆F1 (s)- ∆F2(s)] …………….3.12 s

where a12 is the area-size ratio coefficient and it is the negative

ratio between the rated Megawatts of areas 1 and 2, respectively. .ie.a12 ~ -

Pr1/ Pr2 . It must be include in the block diagram because the term ∆ P tie,, 1

represents the tie-line power exchange out from area i expressed in per unit

of area rating Pri . We have the following relation for the two tie line powers

∆ Ptie, 2 ~ a12 ∆ Ptie, 1

Combining the block diagrams of two single areas (fig

2.4) with synchronizing coefficient and area-size ratio coefficient, the block

diagram for uncontrolled two area system is shown in fig 3.1

Fig 3.1 Block Diagram of two-area system without PI feedback control

Dynamic Response of Two-Area System with PI Control

55

The feedback control without PI controller is not used, as it is

not satisfying the minimum specifications which are

1) the static frequency error following a step load change must be zero

2) the transient frequency swings should be normal.

Hence the control strategy is used to meet the above

requirements and is of linear and integral form i.e. the Proportional Integral

(PI) control is used to study the dynamic response of the 2 area system.

3.4.1 Area Control Error(ACE)

By using the integral control, let the speed changer be

commanded by a signal obtained by first amplifying and then integrating the

frequency error, i.e. ∆Pref ∆ -Ki ∫ ∆f dt.

Where the unit for Ki is puMw/Hz/sec. The signal fed to the integrator is

referred to as area control error (ACE), i.e. ACE ∆ ∆f

Integral control will give rise to zero static frequency error

following a step load change because as long as an error remains, the

integrator output will increase, causing the speed changer to move. The

integrator output, and thus the speed changer position, attains a constant

value only when the frequency error has been reduced to zero.

In case of two area systems, the tie-line bias control is based on

the principle that all operating members must contribute their share to

frequency control in addition to taking care of their own net exchange

56

KG

1+sTG1

KT

1+sTT1

1 2H1s+D1

-+-

1/R1

∆PL1(S)

KG

1+sTG2

KT

1+sTT2- +-

1/R2

1 2H2s+D2

∆PL2(S)

T12+-

+-

b2

b1

--

K2/s

K1/s

+

+-

+

1/s

1/s

In applying the reset control method to the two area system, the control error

for each area consists of a linear combination of frequency and tie-line error

i.e. ACE1= ∆ Ptie , 1 +b1f1 …….. …..………..3.13

Where the constant b1 is called area frequency bias constant

Applying Laplace transform to equation 3.19

ACE1(s)= ∆ Ptie,, 1 (s)+b1F1(s) …………………..3.14

Similarly for control area-2

ACE2(s)= ∆ Ptie , 2 (s)+b2F2(s) ..………………...3.15

Combining the basic blocks of two diagrams with ∆PC1(s) and ∆PC2(s)

generated by integrals of respective ACEs and employing the block diagram

from the fig 3.1 a new block diagram is obtained which is shown in fig 3.2

57

fig 3.2 Block diagram of Two-area system with PI controller

Let the step changes in loads ∆PD1 and ∆PD2 be simultaneously

applied in areas 1 and areas2, respectively. When Steady state conditions are

reached, the output signals of all integrating blocks will become constant

and in order for this to be so, their input signals must be zero.

So from fig 3.6,

∆ Ptie , 1 +b1f1 =0 ……… …………3.16

and ∆ Ptie , 2 +b2f2 =0 ……..……..……3.17

∆f1-∆f2=0(input of integrating block-(2ΠT12/s)…………3.18

From equations 3.10 and 3.11 we have

∆ Ptie, 1 = - T12 = - 1 = constant …………3.19

∆ Ptie, 2 -T21 a12

Hence the Equations 3.16, 3.17, 3.18 and 3.19 are simultaneously satisfied for

∆ Ptie, 1 = ∆ Ptie, 2 =0

∆f1 =∆f2 =0

Thus under steady state condition change in tie line power and frequency of

each area is zero. This has been achieved by integration of ACEs in the feed

back loops of each area.

58

The dynamic performance of two area system with proportional

Integral Controller is obtained first by using the state space approach.


The steady state space analysis is carried out by using the block

diagram(fig 3.3) and developing the state space equations.

KG

1+sTG1

KT

1+sTT1

1 2H1s+D1

-+-

1/R1

∆PL1(S)

KG

1+sTG2

KT

1+sTT2- +-

1/R2

1 2H2s+D2

∆PL2(S)

T12 +-

+-

b2

b1

--

K2/s

K1/s

+

+-

+

1/s

1/s

x1x2x3x4

x5

x6x7x8

x9

59

Fig 3.3 Block diagram of Two-Area system with PI Controller to represent the states

So from the block diagram 3.7, we have the equations

x1=( x2-d1- x9)(1/2 H1s+ D1) → 12 H1+ x1 D1= x2- d1- x9

→ 1= -( x1 D1/2 H1)+( x2/2 H1) -( x9/2 H1) -( d1/2 H1) ……………….3.20

x2= x3/(1+s TT1)

→ 2=-( x2/ TT1)+( x3/ TT1) …………………..3.21

x3=( x4- x1/ R1).1/(1+s TG1)

→ 3=-( x1/ R1 TG1)-( x3/ TG1)+( x4/ TG1) ….………………3.22

x4=-( b1 x1+ x9).Ki1/s

→ 4=- Ki1 b1 x1- Ki1 x9 ………………….3.23

x5=( x5+ x7- d2)/(2 H2s+ D2)

→ 5=-( x5 D2/2 H2)+( x6/2 H2)+( x9/2 H2)-( d2/2 H2) …………………3.24

x6= x7/(1+s TT2)

→ 6=-( x6/ TT2)+( x7/ TT2) ………………….3.25

x7=( x8- x4/ R2).1/(1+s TG2)

→ 7=-( x5/ R2 TG2)-( x7/ TG2)+( x8/ TG2) …………………..3.26

8=- Ki2 b2 x5+ Ki2 x9 …………………..3.27

60

x9=( x1/s) T12-( x5/s) T12

→ 9= x1 T12- x5 T12 ..…………………3.28

So combining equations from 3.26-3.34 and writing in standard matrix form,

we have

1 -D1/2 H1 1/2 H1 0 0 0 0 0 0 -1/2 H1 x1

2 0 -1/ TT1 1/ TT1 0 0 0 0 0 0 x2

3 -1/ R1 TG1 0 -1/ TG1 1/ TG1 0 0 0 0 0 x3

4 -K i1 b1 0 0 0 0 0 0 0 0 x4

5 0 0 0 0 - D2/2H2 1/2H2 0 0 1/2H2 x5

6 0 0 0 0 0 -1/ TT2 1/ TT2 0 0 x6

7 0 0 0 0 -1/ R2TG2 0 -1/ TG2 1/ TG2 0 x7

8 0 0 0 0 -K 2 b2 0 0 0 K2 x8

9 T12 0 0 0 - T12 0 0 0 0 x9

- d1/2H1 0 1 1 0 0 0 0 0 0 0 0 x1

0 0 2 0 0 0 0 1 0 0 0 0 x7

0 0 3 0 0 0 0 0 0 0 0 0 x9

0 0

and B= 0 - d2/2H2 which is of the standard form

0 0 =Ax+Bu

0 0 =Cx+Du

0 0

61

=+

=

0 0

A Solution to the above system of equations represented is obtained

by writing a MATLAB program and applied to numerical example, which is

illustrated as

Program for Two-Area with PI Controller

(prossmapi.m):-


F=input('ENTER THE SYSTEM FREQUENCY in Hz :');

R1=input('ENTER THE GOVERNOR SPEED REGULATION OF AREA1 in

Hz/puMw :');


Hz/puMw :');

H1=input('ENTER THE INERTIA CONSTANT OF AREA 1 in secs :');


Tg1=input('ENTER THE TIME CONSTANT OF GOVERNOR OF AREA1 in

sec:');

Tg2=input('ENTER THE TIME CONSTANT OF GOVERNOR OF AREA 2 in

sec :');

Tt1=input('ENTER THE TIME CONSTANT OF TURBINE OF AREA 1 in

sec :');

Tt2=input('ENTER THE TIME CONSTANT OF TURBINE OF AREA 2 in

sec :');

D1=input('ENTER THE LOAD FREQUENCY CONSTANT OF AREA 1 in

puMw/Hz :');

D2=input('ENTER THE LOAD FREQUENCY CONSTANT OF AREA 2 in

puMw/Hz :');

62

b1=input('ENTER THE FREQUENCY BIAS CONSTANT OF AREA 1 in

puMw/Hz :');

b2=input('ENTER THE FREQUENCY BIAS CONSTANT OF AREA 2 in

puMw/Hz :');

Ki1=input('ENTER THE INTEGRAL GAIN OF AREA 1 :');


T12=input('ENTER THE SYNCHRONISING POWER COEFFICIENT in

puMw: ');

Pp=input('ENTER THE CHANGE IN LOAD IN TERMS OF PERCENTAGE

INCREASE:');

dP=Pp*0.01;

r1=R1/F;

r2=R2/F;

A=[-D1/(2*H1) 1/(2*H1) 0 0 0 0 0 0 -1/(2*H1);0 -1/Tt1 1/Tt1

0 0 0 0 0 0;-1/(r1*Tg1) 0 -1/Tg1 1/Tg1 0 0 0 0 0;-Ki1*b1 0 0

0 0 0 0 0 -Ki1;0 0 0 0 -D2/(2*H2) 1/(2*H2) 0 0 1/(2*H2);0 0

0 0 0 -1/Tt2 1/Tt2 0 0; 0 0 0 0 -1/(r2*Tg2) 0 -1/Tg2 1/Tg2

0;0 0 0 0 -Ki2*b2 0 0 0 Ki2;T12 0 0 0 -T12 0 0 0 0];

B=[-dP/(2*H1) ;0 ;0 ;0 ;0 ;0 ;0; 0;0 ];

C=[1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0];

D=[0 ;0 ];

step(A,B,C,D);

63

The execution is as follows:-

>>prossmapi


ENTER THE SYSTEM FREQUENCY in Hz :60

ENTER THE GOVERNOR SPEED REGULATION OF AREA 1 in Hz/puM:2.4

ENTER THE GOVERNOR SPEED REGULATION OF AREA2 in Hz/puMw:2.4

ENTER THE INERTIA CONSTANT OF AREA 1 in secs :5


ENTER THE TIME CONSTANT OF GOVERNOR OF AREA1 in sec :0.08

ENTER THE TIME CONSTANT OF GOVERNOR OF AREA 2 in sec :0.08

ENTER THE TIME CONSTANT OF TURBINE OF AREA 1 in sec :0.3


ENTER THE LOAD FREQUENCY CONSTANT OF AREA1 in puMw/Hz :0.00833

ENTER THE LOAD FREQUENCY CONSTANT OF AREA 2 in puMw/Hz :0.00833

ENTER THE FREQUENCY BIAS CONSTANT OF AREA 1 in puMw/Hz :0.425


ENTER THE INTEGRAL GAIN OF AREA 1 :1


ENTER THE SYNCHRONISING POWER COEFFICIENT in puMw :0.545

ENTER THE CHANGE IN LOAD IN TERMS OF PERCENTAGE INCREASE:20

64

Fig 3.4 Response plot of Two-Area system with PI controller with a load disturbance in area-1

65

Case Studies

There are several parameters which effect the dynamic response

of a system. They are Governor Speed Regulation, Time Constant of the

Governor, Time Constant of the Turbine, Integral Gain, Load Disturbance,

Frequency Bias Constant and Synchronizing Coefficient. The first four case

studies have been discussed in section 2.5. These case studies can be applied

in two-area systems also, but the case studies that have been made by the

two area system are

case 1: Variation in Frequency Bias Constant

case 2: Variation in Synchronizing Coefficient

case 3: Variation in Load disturbance

Different Case Studies are depicted by using MATLAB programming

for an Two area Power System with PI controller. The dynamic response for

different cases with ± 50% changes in the parameter values are obtained by

considering the numerical example. The MATLAB programming for this

case is as follows.

Program for Case Studies in Multi Area

System(prossmacs.m)


F=input('ENTER THE SYSTEM FREQUENCY in Hz :’);


Hz/puMw:');

66


Hz/pumw:');



Tg1=input('ENTER THE TIME CONSTANT OF GOVERNOR OF AREA1 in

sec:');

Tg2=input('ENTER THE TIME CONSTANT OF GOVERNOR OF AREA2 in sec :’);

Tt1=input('ENTER THE TIME CONSTANT OF TURBINE OF AREA1 in

sec :');

Tt2=input('ENTER THE TIME CONSTANT OF TURBINE OF AREA2 in

sec :');

D1=input('ENTER THE LOAD FREQUENCY CONSTANT OF AREA1 in

puMw/Hz');

D2=input('ENTER THE LOAD FREQUENCY CONSTANT OF AREA2 in

puMw/Hz');

b1=input('ENTER THE FREQUENCY BIAS CONSTANT OF AREA1 in

puMw/Hz');

b2=input('ENTER THE FREQUENCY BIAS CONSTANT OF AREA2 in

puMw/Hz');



T12=input('ENTER THE SYNCHRONISING POWER COEFFICIENT in puMw

:');

Pp=input('ENTER THE CHANGE IN LOAD IN TERMS OF PERCENTAGE

INCREASE:');

67

disp('ENTER THE PARAMETER IN WHICH THE VARIATION IS TO BE

sOUGHT')

disp('1.CHANGE IN FREQUENCY BIAS CONSTANT(b)');

disp('2.CHANGE IN SYNCHRONISING COEFFICIENT (T12)');

disp('3.CHANGE IN LOAD DEMAND(dP)');

dP=Pp*0.01;

r1=R1/F;

r2=R2/F;

value=input('ENTER A VALUE :');

switch(value)

case {1}

A=[-D1/(2*H1) 1/(2*H1) 0 0 0 0 0 0 -1/(2*H1);0

-1/Tt1 1/Tt1 0 0 0 0 0 0;-1/(r1*Tg1) 0 -1/Tg1 1/Tg1 0 0 0 0

0;-Ki1*b1 0 0 0 0 0 0 0 -Ki1;0 0 0 0 -D2/(2*H2) 1/(2*H2) 0 0

1/(2*H2);0 0 0 0 0 -1/Tt2 1/Tt2 0 0; 0 0 0 0 -1/(r2*Tg2) 0 -

1/Tg2 1/Tg2 0;0 0 0 0 -Ki2*b2 0 0 0 Ki2;T12 0 0 0 -T12 0 0

0 0];

A1=[-D1/(2*H1) 1/(2*H1) 0 0 0 0 0 0 -1/(2*H1);0

-1/Tt1 1/Tt1 0 0 0 0 0 0;-1/(r1*Tg1) 0 -1/Tg1 1/Tg1 0 0 0 0

0;-Ki1*0.5*b1 0 0 0 0 0 0 0 -Ki1;0 0 0 0 -D2/(2*H2) 1/(2*H2)

0 0 1/(2*H2);0 0 0 0 0 -1/Tt2 1/Tt2 0 0; 0 0 0 0 -1/(r2*Tg2)

0 -1/Tg2 1/Tg2 0;0 0 0 0 -Ki2*0.5*b2 0 0 0 Ki2;T12 0 0 0 -

T12 0 0 0 0];

A2=[-D1/(2*H1) 1/(2*H1) 0 0 0 0 0 0 -1/(2*H1);0

-1/Tt1 1/Tt1 0 0 0 0 0 0;-1/(r1*Tg1) 0 -1/Tg1 1/Tg1 0 0 0 0

0;-Ki1*1.5*b1 0 0 0 0 0 0 0 -Ki1;0 0 0 0 -D2/(2*H2) 1/(2*H2)

68

0 0 1/(2*H2);0 0 0 0 0 -1/Tt2 1/Tt2 0 0; 0 0 0 0 -1/(r2*Tg2)

0 -1/Tg2 1/Tg2 0;0 0 0 0 -Ki2*1.5*b2 0 0 0 Ki2;T12 0 0 0 -

T12 0 0 0 0];

case {2}

A=[-D1/(2*H1) 1/(2*H1) 0 0 0 0 0 0 -1/(2*H1);0

-1/Tt1 1/Tt1 0 0 0 0 0 0;-1/(r1*Tg1) 0 -1/Tg1 1/Tg1 0 0 0 0

0;-Ki1*b1 0 0 0 0 0 0 0 -Ki1;0 0 0 0 -D2/(2*H2) 1/(2*H2) 0 0

1/(2*H2);0 0 0 0 0 -1/Tt2 1/Tt2 0 0; 0 0 0 0 -1/(r2*Tg2) 0 -

1/Tg2 1/Tg2 0;0 0 0 0 -Ki2*b2 0 0 0 Ki2;T12 0 0 0 -T12 0 0

0 0];

A1=[-D1/(2*H1) 1/(2*H1) 0 0 0 0 0 0 -1/(2*H1);0

-1/Tt1 1/Tt1 0 0 0 0 0 0;-1/(r1*Tg1) 0 -1/Tg1 1/Tg1 0 0 0 0

0;-Ki1*b1 0 0 0 0 0 0 0 -Ki1;0 0 0 0 -D2/(2*H2) 1/(2*H2) 0 0

1/(2*H2);0 0 0 0 0 -1/Tt2 1/Tt2 0 0; 0 0 0 0 -1/(r2*Tg2) 0 -

1/Tg2 1/Tg2 0;0 0 0 0 -Ki2*b2 0 0 0 Ki2;T12*0.5 0 0 0 -

T12*0.5 0 0 0 0];

A2=[-D1/(2*H1) 1/(2*H1) 0 0 0 0 0 0 -1/(2*H1);0

-1/Tt1 1/Tt1 0 0 0 0 0 0;-1/(r1*Tg1) 0 -1/Tg1 1/Tg1 0 0 0 0

0;-Ki1*b1 0 0 0 0 0 0 0 -Ki1;0 0 0 0 -D2/(2*H2) 1/(2*H2) 0 0

1/(2*H2);0 0 0 0 0 -1/Tt2 1/Tt2 0 0; 0 0 0 0 -1/(r2*Tg2) 0 -

1/Tg2 1/Tg2 0;0 0 0 0 -Ki2*b2 0 0 0 Ki2;T12*1.5 0 0 0 -

T12*1.5 0 0 0 0];

case{3}

dP2=Pp*0.01*0.5;

dP3=Pp*0.01*1.5;

A=[-D1/(2*H1) 1/(2*H1) 0 0 0 0 0 0 -1/(2*H1);0

-1/Tt1 1/Tt1 0 0 0 0 0 0;-1/(r1*Tg1) 0 -1/Tg1 1/Tg1 0 0 0 0

0;-Ki1*b1 0 0 0 0 0 0 0 -Ki1;0 0 0 0 -D2/(2*H2) 1/(2*H2) 0 0

1/(2*H2);0 0 0 0 0 -1/Tt2 1/Tt2 0 0; 0 0 0 0 -1/(r2*Tg2) 0 -

69

1/Tg2 1/Tg2 0;0 0 0 0 -Ki2*b2 0 0 0 Ki2;T12 0 0 0 -T12 0 0

0 0];

B=[-dP/(2*H1) ;0 ;0 ;0 ;0 ;0 ;0; 0;0 ];

C=[1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0];

D=[0 ;0 ];

B1=[-dP2/(2*H1) ;0 ;0 ;0 ;0 ;0 ;0; 0;0 ];

B2=[-dP3/(2*H1) ;0 ;0 ;0 ;0 ;0 ;0; 0;0 ];

sys1=ss(A,B,C,D);

sys2=ss(A,B1,C,D);

sys3=ss(A,B2,C,D);


otherwise,

end;

B=[-dP/(2*H1) ;0 ;0 ;0 ;0 ;0 ;0; 0;0 ];

C=[1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0];

D=[0 ;0 ];

sys1=ss(A,B,C,D);

sys2=ss(A1,B,C,D);

sys3=ss(A2,B,C,D);


70

The execution of the program is as follows

>>promacs



ENTER THE GOVERNOR SPEED REGULATION OF AREA 1 in Hz/puMw :2.4

ENTER THE GOVERNOR SPEED REGULATION OF AREA 2 in hz/puMw :2.4


ENTER THE INERTIA CONSTANT OF AREA 2:in secs :5












ENTER THE CHANGE IN LOAD IN TERMS OF % INCREASE:20

ENTER THE PARAMETER IN WHCICH THE VARIATION IS TO BE SOUGHT:

1.CHANGE IN FREQUENCY BIAS CONSTANT(b)

71

2.CHANGE IN SYNCHRONISING COEFFICIENT (T12)

3.CHANGE IN LOAD DEMAND(dP)

ENTER A VALUE :1

Fig 3.5 Response plot in Area-1 with variation in Frequency Bias Constant

72

fig 3.12 response plot of in area-2 with variation in frequency bias constant


>> ENTER A VALUE :2;

73

Fig 3.7 Response plot in Area-1 with variation in Synchronizing Coefficient


ENTER A VALUE :3

74

Fig 3.9 Response plot in Area-2 with variation in Load Disturbance in Area-1

Fig 3.10 Response plot in Area-2 with variation in Load Disturbance in Area-1

75

Discussions

The response of a Two Area Power System with PI controller when subjected to a load change in area-1 is shown in figs 3.4. From these figures, it is clear that the settling time for both the areas is 250 seconds and the steady state frequency deviation is zero. But the peak value of the transient response in area-1 is0.0095 Hz/pu and in area-2 is 0.0062 Hz/pu. Hence it can be concluded that the feed back control reduces the steady state deviation to zero.

The response for the variation in different parameters is obtained in section 3.5. The response for + 50% variation in Frequency Bias Constant is shown in figure 3.5 for area 1 and in fig 3.6 for area-2 with a load disturbance in area-1. From these figures, it can be observed that when b is increased the settling time decreases and frequency deviations increases for both the areas and vice versa.

The response for + 50% variation in Synchronizing Coefficient is shown in figure 3.7 for area-1 and in fig 3.8 for area-2 with a load disturbance area-1. From these figures, it is evident that when T12 is increased the settling time decreases where as the frequency deviations increases and vice versa.

The response for + 50% variation in Load disturbance in area-1is shown in figure 3.9 for area-1 and in fig 3.10 for area-2. From these figures, it is evident that when dP is increased the frequency deviation increases with no considerable change in settling time for both the areas.

76

LOAD FREQUENCY CONTROL OF INTERCONNECTED SYSTEMS USING OPTIMAL CONTROLLER

4.1 Introduction

The classical design techniques used so far utilize the plant

output for feed back with a dynamic controller. In this chapter, Modern

control designs that require the use of all state variables to form a linear

static controller are employed. Optimal Control is a branch of modern

control theory that deals with designing controls for dynamic systems by

minimizing a performance index. This is also called as linear quadratic

regulator (LQR) problem. The object of the optimal controller design is to

determine the optimal control vector uopt(x,t) which can transfer the system

from its initial state to the final state such that a given performance index is

minimized. The performance index used in optimal control design is known

as the Quadratic Performance Index and is based on minimum-error and

minimum-energy criteria.

4.2 Optimal Control Theory

The problem in optimal control theory can be explained as follows:

Given the linear time invariant system represented by the state

variable differential equation

(t)=Ax(t)+Bu(t) …………………4.1

where x→ n X 1 state vector

u→ m X 1 control vector

78

A→ n X n State Distribution matrix

B→ n X m Control Distribution Matrix

The objective is to find the control vector uopt which minimizes the

cost function

C= ½ ∫ (x´Qx+u´Ru) dt …………………4.2

where Q → n X n positive semi-definite symmetric state cost weighting

matrix

R → m X m positive definite symmetric control cost weighting matrix

x´ and u´ → transpose of x and u respectively.

4.2.1 System State x:

If the system is linear and time invariant , the state x can be

represented in the form of equation 4.1 with the state variables x1,x2,…..,xn

are the components of the state vector x. these state variables are the

minimum number of variables containing sufficient information about the

previous state of the system, assuming the control inputs are known. The

state variables are not purely mathematical but have true physical meaning.

4.2.2 System Cost C:

The performance of the system is specified in terms of a cost that is to

be minimized by the optimal controller. The components Q and R can be

chosen mathematically by the way the system is to be performed. There are

two cases in choosing Q and R which are

1) If R=0 but require Q≠0 then it means there is no charge for the control

effort used but the state for being nonzero is penalized. Here the best control

79

strategy would be in the form of infinite impulses. This control would drive

the state to zero in the shortest possible time with the greatest effort.

2) If Q=0 and R≠0 then the control effort is penalized but do not charge for

the trajectory the state x follows. In this case , the best control is to use u=0

i.e. not to provide any control effort at all.

These two cases are the extreme cases, but they emphasize the

importance in choosing the components of Q and R.

4.2.3 Optimal Controller:

The optimal controller that minimizes the cost C of the system

in state variable form is a function of the present states of the system

weighted by the components of a constant gain matrix K of dimension m

X n:

u(t)= -Kx(t) …………………4.3

This optimal gain matrix is determined by solving the

differential equation, the matrix Riccati equation. For the infinite time

problem, the Riccati equation has a steady solution. Since the gain matrix is

a constant, the optimally controlled system can be expressed in the closed

form. x=Acx

Ac ~ A-BK

4.2.4 Calculating the value of optimal controller K:

Consider the plant described from the equations 4.1 and 4.2

The objective function is to: minimize C= ½ ∫ (x´Qx+u´Ru) dt

with the constraint: Ax(t)+Bu(t)=

To obtain the formal solution, Lagrange

80

multipliers method is applied. The constraint problem is solved by

augmenting the equation 4.1 into equation 4.2 using an n-vector of Lagrange

multipliers, λ . The problem reduces to the minimization of the following

unconstrained function. L (x,λ,u,t) = [x´Qx+u´Ru]+λ[Ax+Bu-

] …………………4.4

The optimal values(denoted by opt) are found by partial differentiating

the Lagrangian function w.r.t. λ, u, x and equating them to zero

i.e. ∂L /∂λ = Axopt+Buopt- opt=0

→ opt = Axopt+Buopt ………………………..4.5

∂L/∂u = 2Ruopt+λ´ B =0

→ uopt= ½R-1λ B ………………………..4.6

∂L/∂x = 2xoptQ+λ’´ +λ´ A=0

→ λ’ = 2Qxopt – A´λ ………………………..4.7

Assuming there exists a symmetric, time-varying positive

definite matrix P(t) satisfying

λ = 2P(t)xopt ……………………….4.8

substituting equation 4.3 into equation 4.6 gives the optimal closed-loop control law

uopt (t)= -R-1B´ P(t)xopt …………………. ……4.9

by derivating equation 4.8,

λ’ = 2(P’xopt+px’opt) ………………………..4.10

Finally equating 4.7 with 4.10,

P’(t)= -P(t)A-A´P(t)-Q+P(t)BR-1B´P(t) ……………………….4.11

The above equation is referred to as the matrix Riccati

Equation. The boundary condition for equation 4.11 is P(tf) =0. Therefore,

81

Equation 4.11 must be integrated backward in time. Since a numerical

solution is performed forward in time, a dummy variable τ = tf-t is replaced

for time t. Once the solution to equation 4.11 is obtained the solution to the

state equation 4.5 in conjunction with the optimal control equation 4.9 is

obtained.

The optimal controller gain is a time varying state-variable feed

back. the time –varying optimal gain K(t) is replaced by its seady value, as

its inconvenient to implement time varying feedback. For linear time-

invariant systems, since p’=0, when the process is of infinite duration, that is

tf→∞ ,Equation 4.11 reduces to the algebraic Ricatti equation

PA+A´P+Q-PBR-1B´P=0……………………….4.12

4.2.5 Snag of optimal Control

The Optimal control determined by the above equations is quite

often impractical due to the following reasons:-

1) The optimal control is a function of all the states of the system. In

practice, all the states may not be available. The inaccessible states or

missing states are required to be estimated

2) It may not be economical to transfer all the information over long

distances.

3) The control which is a function of the states in turn is dependent on

the load demand. Accurate prediction of the load demand may be essential

for realizing the optimal controller.

4) The optimal control is also dependent on the weighting matrices

which are not unique.

82

4.3 Application of Optimal Control to an Isolated Power System

The optimal controller is applied to isolated power

system so as to enhance the dynamic performance of the system. In case of

an Isolated Power System the order of matrices Q and R is 4 x 4 and 1 x 1

respectively.

The dynamic response to an Isolated Power System with PI controller

and with Optimal Controller is illustrated by the following MATLAB

program, considering the same example in section 2.4

Program with Optimal Controller to Isolated power system

(Prosawioc.m)

T=input('TOTAL RATED CAPACITY :');

R=input('ENTER THE SPEED REGULATION GOVERNOR(R) in

puMW/Hz:');



D=input('ENTER THE LOAD FREQUENCY CONSTANT in puMW/HZ :');



83


Pp=input('ENTER THE CHANGE IN LOAD DEMAND IN TERMS OF %

INCREASE:');

dP=Pp*0.01;

R=R/F;

A=[-D/(2*H) 1/(2*H) 0 0;0 -1/Tt 1/Tt 0;-1/(R*Tg) 0 -1/Tg -

1/Tg;Ki 0 0 0];

B=[-dP/(2*H) ; 0 ; 0 ; 0 ];

C=[1 0 0 0];

D=[0 ];

sys1=ss(A,B,C,D);

disp('ENTER THE STATE COST WEIGHTING MATRIX(Q)IN n X n

FORM:');

for i=1:4

for j=1:4

temp=input('');

Q(i,j)=temp(1);

end

end

Q

R=input('ENTER THE CONTROL COST WEIGHTING MATRIX IN m X m

FORM:');

Bc=B/dP;

84

[K,P]=lqr2(A,Bc,Q,R);

disp('THE RICCATI EQUATION SOLUTION (P) IS :'); P

disp('THE OPTIMAL GAIN MATRIX (K) IS :'); K

Ac=A-Bc*K;

sys2=ss(Ac,B,C,D);

step(sys1,sys2);

The Execution is as follows

>> prosaoc





ENTER THE LOAD FREQUENCY CONSTANT in puMW/Hz :0.8





ENTER THE STATE COST WEIGHTING MATRIX(Q)IN n X n FORM:

20 0 0 0

0 15 0 0

0 0 10 0

0 0 0 5

Q =

20 0 0 0

85

0 15 0 0

0 0 10 0

0 0 0 5

ENTER THE CONTROL COST WEIGHTING MATRIX IN m X m FORM:0.15

THE RICCATI EQUATION SOLUTION (P) IS :

P =

149.0273 -3.7741 -6.7772 9.2148

-3.7741 3.3239 0.6093 0.0968

-6.7772 0.6093 0.9375 -0.0661

9.2148 0.0968 -0.0661 7.3990

THE OPTIMAL GAIN MATRIX (K) IS :

K =

-99.3515 2.5161 4.5181 -6.1432

86

Fig 4.1 Response Plot of an Isolated power system with PI and Optimal Controller

4.3.1 Isolated Power System with Reheater Constraint:

A case study has been carried out using the optimal

controller i.e. to a thermal system in real terms. A simplified model of

turbine is shown in fig2.3 i.e. a non-reheat turbine. Practically, to increase

the overall efficiency the turbine is often divided into two or several stages.

Between the stages the steam is reheated in reheaters. Considering the two

stage turbine, The reheater represents a delay Tr , and the Total Turbine

model is represented by the transfer function as

87

GT(s) = (1-s Kr ) KT (1+s Tr) (1+s TT )

where Tr → reheating Time Constant

Kr → Gain of reheats in Thermal unit

The transfer function corresponding to the Reheater constraint

is added to the Block diagram in fig2.5 and the new Block diagram is

represented below

Fig 4.2 Block diagram of an Isolated power system with Reheater constraint

Analysis of state space Equations:

The state space analysis is carried to the model shown in fig5.2 and the state equations are

x1=( x2- d)(1/2Hs+ D) → 12 H+ x1 D= x2- d

→ 1= -( x1 D/2H)+( x2/2H) -( d/2H) …………………..4.13

x2= (1-s Kr)x3 → x2+ 2 Tr = x3- 3Kr

(1+s Tr)

→ 2= -( x2/ Tr)+ (x3/ Tr)-( x3’ Kr/ Tr) ..………………..4.14

KG

1+sTG

KT

1+sTT

1 2Hs+D

+- +-

1/R

∆PC(S)

∆PD(S)

∆F(S)

KI/s

1-s Kr

1+sTr

x1

x2

x3

x4

x5

88

x3= x4/(1+sTT )

→ 3=-( x3/ TT)+( x4/ TT) ……….…………4.15

x4=( x5- x1/R) 1/(1+sTG) → x3 (1+s TG)= - x4 – x1/R

→ 4= -( x4/ TG) –( x1/R TG)-( x5/ TG) …………………4.16

x5= x1KI/s

i.e. 5= x1KI …………………4.17

The above equations 4.13 to 4.17 can be written in standard form as

1 -D/2H 1/2H 0 0 0 x1 -dP/2H

2 0 -1/ Tr (1+ Kr)/( TrTT) -(Kr/ TrTT ) 0 x2 0

3 = 0 0 -1/TT 1/TT 0 x3 + 0

4 -1/RTG 0 0 -1/ TG 1/TG x4 0

5 KI 0 0 0 0 x5 0

and the output equation is

C=[ 1 0 0 0 0] and D=[0]

4.3.2 Isolated Power System with Reheater Constraint using Optimal

Controller :

The application of optimal control is done to attain the

frequency a steady state value after a disturbance.

The application of optimal controller is illustrated through MATLAB

programming and the dynamic response is obtained for the example and is

as follows

89

u

Program for Isolated Power System with Reheater Constraint (prosahc.m)


R=input('ENTER THE SPEED REGULATION GOVERNOR IN Hz/puMW(R) :');

F=input('ENTER SYSTEM FREQUENCY(F)in Hz :');





Kr=input('ENTER THE GAIN OF REHEATS IN THERMAL UNIT :');

Tr=input('ENTER THE REHEAT TIME CONSTANT in sec :');

Ki=input('ENTER THE PROPORTIONAL INTEGRAL GAIN :');


dP1=Pp1*0.01;

r=R/F;

A=[-D/(2*H) 1/(2*H) 0 0 0;0 -1/Tr 1/Tr+(Kr/(Tr*Tt)) -Kr/(Tr*Tt) 0;0 0 -1/Tt 1/Tt 0;-1/(r*Tg) 0 0 -1/Tg -1/Tg;Ki 0 0 0 0];

B=[-dP1/(2*H) ; 0 ; 0 ; 0;0 ];

C=[1 0 0 0 0];

D=[0 ];

sys1=ss(A,B,C,D);

disp('ENTER THE STATE COST WEIGHTING MATRIX(Q)IN n X n FORM:');

for i=1:5

for j=1:5

temp=input('');

Q(i,j)=temp(1);

end

90

end

Q

R=input('ENTER THE CONTROL COST WEIGHTING MATRIX IN m X m FORM:');

Bc=B/dP1;




Ac=A-Bc*K;

sys2=ss(Ac,B,C,D);

step(sys1,sys2);

The execution is as follows

>> prosahcoc


ENTER THE SPEED REGULATION GOVERNOR IN puMW/Hz(R) :3

ENTER SYSTEM FREQUENCY(F)in Hz :60


ENTER THE LOAD FREQUENCY CONSTANT in Hz/puMw :0.8



ENTER THE GAIN OF REHEATS IN THERMAL UNIT :0.5

ENTER THE REHEAT TIME CONSTANT in sec :10

ENTER THE PROPORTIONAL INTEGRAL GAIN :7



20 0 0 0 0

0 15 0 0 0

91

0 0 10 0 0

0 0 0 5 0

0 0 0 0 0

Q =

20 0 0 0 0

0 15 0 0 0

0 0 10 0 0

0 0 0 5 0

0 0 0 0 0

ENTER THE CONTROL COST WEIGHTING MATRIX IN m X m FORM:0.15


P =

112.1477 -4.1513 -4.5221 -5.7077 5.3851

-4.1513 65.1044 5.3891 0.6167 -0.2692

-4.5221 5.3891 2.6981 0.6112 -0.2339

-5.7077 0.6167 0.6112 0.7725 -0.2900

5.3851 -0.2692 -0.2339 -0.2900 0.3334


K =

-74.7652 2.7675 3.0148 3.8051 -3.5900

92

Fig 4.3 response plot of an isolated power system with heat constraint and with PI and with optimal controller

From fig 4.3, it is observed that without optimal control the steady

state value of frequency deviation is increasing and with the optimal control,

the steady state value of frequency deviation is zero with minimum peak

value of transient frequency.

4.5 Application of Optimal Control to Interconnected Systems

93

KG

1+sTG1

KT

1+sTT1

1 2H1s+D1

-+-

1/R1

∆PL1(S)

KG

1+sTG2

KT

1+sTT2- +-

1/R2

1 2H2s+D2

∆PL2(S)

T12+-

+-

b2

b1

--

K2/s

K1/s

+

+-

+

1/s

1/s

The Theory of Optimal control can be applied to two area

systems also to enhance the dynamic performance of the system. The block

diagram of two area system is shown in figure 4.4

fig 4.4 Block diagram of Two-area system with PI controller

The block diagram and the state space Equations with PI controller

are same and also can be applied with optimal controller. In this chapter only

the State Cost Weighting Matrix(Q), of order n X n , and Control Cost

Weighting Matrix(R), of order m X m are introduced to enhance the

response of the system. In a Two Area Power System the values of n and m

are 9 and 1 respectively.

The optimal control theory is applied to a example(which was

discussed in section 3.4) and is illustrated in the form of MATLAB

programming

94

Program for Two-Area System with optimal control (Promaoc.m)


F=input('ENTER THE SYSTEM FREQUENCY in Hz :');

R1=input('ENTER GOVERNOR SPEED REGULATION OF AREA1 in Hz/puMw :');

R2=input('ENTER GOVERNOR SPEED REGULATION OF AREA2 in Hz/puMw :');


H2=input('ENTER THE INERTIA CONSTANT OF AREA 2:in secs :');

Tg1=input('ENTER THE TIME CONSTANT OF GOVERNOR OF AREA 1 in secs :');

Tg2=input('ENTER THE TIME CONSTANT OF GOVERNOR OF AREA2 in secs :');

Tt1=input('ENTER THE TIME CONSTANT OF TURBINE OF AREA 1 in secs :');

Tt2=input('ENTER THE TIME CONSTANT OF TURBINE OF AREA 2 in secs :');

D1=input('ENTER LOAD FREQUENCY CONSTANT OF AREA1 in puMw/Hz :');

D2=input('ENTER LOAD FREQUENCY CONSTANT OF AREA 2 in puMw/Hz :');

b1=input('ENTER THE FREQUENCY BIAS CONSTANT OF AREA1 in puMw/Hz:');

b2=input('ENTER THE FREQUENCY BIAS CONSTANT OF AREA2 in puMw/Hz:');



T12=input('ENTER THE SYNCHRONISING POWER COEFFICIENT in puMw :');

Pp=input('ENTER THE CHANGE IN LOAD IN TERMS OF PERCENTAGE INCREASE:');

dP=Pp*0.01;

95

r1=R1/F;

r2=R2/F;

A=[-D1/(2*H1) 1/(2*H1) 0 0 0 0 0 0 -1/(2*H1);0 -1/Tt1 1/Tt1 0 0 0 0 0 0;-1/(r1*Tg1) 0 -1/Tg1 1/Tg1 0 0 0 0 0;-Ki1*b1 0 0 0 0 0 0 0 -Ki1;0 0 0 0 -D2/(2*H2) 1/(2*H2) 0 0 1/(2*H2);0 0 0 0 0 -1/Tt2 1/Tt2 0 0; 0 0 0 0 -1/(r2*Tg2) 0 -1/Tg2 1/Tg2 0;0 0 0 0 -Ki2*b2 0 0 0 Ki2;T12 0 0 0 -T12 0 0 0 0];

B=[-dP/(2*H1) ;0 ;0 ;0 ;0 ;0 ;0; 0;0 ];

C=[1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0];

D=[0 ;0 ];

sys1=ss(A,B,C,D);

disp('ENTER THE STATE COST WEIGHTING MATRIX(Q)IN n X n FORM:');

for i=1:9

fprintf('row %d :',i);

for j=1:9

temp=input('');

Q(i,j)=temp(1);

end

end

Q

R=input('ENTER THE CONTROL COST WEIGHTING MATRIX IN m X m FORM:');

Bc=B/dP;




Ac=A-Bc*K;

sys2=ss(Ac,B,C,D);

step(sys1,sys2);

>> prossmaoc

96






ENTER THE INERTIA CONSTANT OF AREA 2:in secs :5


ENTER THE TIME CONSTANT OF GOVERNOR OF AREA2 in sec :0.08

ENTER THE TIME CONSTANT OF TURBINE OF AREA1 in sec :0.3

ENTER THE TIME CONSTANT OF TURBINE OF AREA2 in sec :0.3

ENTER THE LOAD FREQUENCY CONSTANT OF AREA1 in puMw/Hz :0.00833







ENTER THE CHANGE IN LOAD IN TERMS OF PERCENTAGE INCREASE:20


row1 : 1 0 0 0 0 0 0 0 0

row 2 :0 1.545 0 0 0 -0.545 0 0 0

row 3 :0 0 1 0 0 0 0 0 0

row 4 :0 0 0 0 0 0 0 0 0

row 5 :0 0 0 0 0 0 0 0 0

row 6 :0 -0.545 0 0 0 1.545 0 0 0

row 7 :0 0 0 0 0 0 1 0 0

row 8 :0 0 0 0 0 0 0 0 0

row 9 :0 0 0 0 0 0 0 0 0

Q =

97

1 0 0 0 0 0 0 0 0

0 1.5450 0 0 0 -0.5450 0 0 0

0 0 1.0000 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 -0.5450 0 0 0 1.5450 0 0 0

0 0 0 0 0 0 1.0000 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

ENTER THE CONTROL COST WEIGHTING MATRIX IN m X m FORM:1


P =

231.66 1.25 -0.84 -9.231 -38.615 0.134 0.114 1.534 -1.188

1.250 0.26 0.051 -0.0491 0.3975 -0.074 -0.015 -0.0158 -0.046

-0.84 0.051 0.053 0.0341 0.1773 -0.0144 -0.003 -0.0070 -0.008

-9.23 -0.049 0.0341 0.5981 1.5160 -0.0047 -0.004 -0.2622 0.189

-38.6 0.397 0.1773 1.5160 463.207 5.8628 -0.007 -18.421 0.162

0.13 -0.074 -0.0144 -0.004 5.8628 0.4076 0.085 -0.2317 0.229

0.114 -0.015 -0.003 -0.0043 -0.0073 0.0858 0.0629 0.0009 0.060

1.534 -0.015 -0.007 -0.2622 -18.421 -0.2317 0.0009 0.9570 -0.338

-1.18 -0.046 -0.008 0.1897 4.162 0.2291 0.0607 -0.338 20.13


K =

-23.16 -0.125 0.0847 0.923 3.8615 -0.0135 -0.0114 -0.153 0.118

98

Fig 4.4 Response plot of area-1 with a step load disturbance in area-1

with PI and Optimal controller

99

Fig 4.5 Response plot of area-2 with a step load disturbance in area-1

with PI and Optimal controller

o Discussions

The response of an Isolated Power System when subjected to load change with PI and optimal controller is shown in fig 4.1 From this fig it is clear that the settling time without optimal controller is 14 secs and with optimal controller is just 6 secs. It can also be observed that the peak value of transient frequency with and without optimal controller is 0.0145Hz/pu and 0.0016 Hz/pu.

The response of an Isolated Power System with Reheater Constraint when subjected to load change with PI and optimal controller is shown in fig 4.2 From this figure, it is evident that a steady state value of frequency deviation without optimal control is not reached where as when optimal control is applied, a steady state value is reached within 2.2 seconds.

The response of an Two Area Power System when subjected to a load change in area-1 is shown in figs 4.4 and 4.5. From these figures, it is clear that there is no considerable change the settling time for both the areas for with PI and optimal controller (i.e. 250 secs) but a vast difference in the peak value of transient frequency deviation is observed. i.e. the value for without optimal controller is 0.009 Hz/pu and for with optimal control is 0.045Hz/pu. for area-1.

100

CONCLUSIONS

An exhaustive study on load frequency control problem of both

isolated and interconnected power systems has been carried out. Necessary

computer programs have been developed to carry out these studies by

Transfer function approach, State space approach and finally verified by

using Matlab.

The techniques of PI control and optimal control have been

employed to enhance the dynamic performance of both the isolated and

interconnected power systems. The dynamic response for a step load change

and response plots for variation in the system parameters from their nominal

values have been presented.

The results of the response plots obtained for the isolated power

system without PI controller, with PI and optimal controller are presented in

a tabular form below.

System Maximum Frequency Deviation in Hz/pu

Settling Time in seconds

Steady State error In Hz/pu

Isolated Power System without PI feedback Control 0.0138 7.5 0.01

Isolated Power System with PI controller 0.0142 14 0

Isolated Power System with Optimal controller 0.0012 6.5 0

Isolated Power System with Reheater using PI control ∞ ∞ ∞

Isolated Power System with Reheater using Optimal control 0.0012 3.7 0

Table 5.1 Results of response plots for step load change in isolated power system

102

From the table 5.1, It is observed that the dynamic response with PI

controller decreases the steady state frequency deviation to zero compared to

the system without PI controller whereas the settling time increases. With

the optimal control the maximum value of frequency deviation is decreased

for the system in addition to decrease in settling time. In case of Isolated

Power System with Reheater, the frequency deviation is drastically

increasing even with PI controller without any settling time. By adopting

Optimal Controller, the steady state frequency error is zero with a decrease

in settling time.

The results of the response plots for various case studies (i.e.,50%

increase in parameter values) carried out for the isolated power systems are

presented in a tabular form below.

Table 5.2 Results of response plots of Different Case Studies for isolated power system

Case Max Frequency Deviation Settling TimeIncrease in Governor Speed Regulation(R) Increases Decreases

Increase in Time Constant of Governor(TG) Increases Increases

Increase in Time Constant of Turbine (TT) Decreases Decreases

Increase in Proportional Integral Gain (KI)

Increases Increases

From the table 5.2, It is clear that for isolated power systems for

normal operation settling time should be low for which the Governor speed

regulation should be high, the Governor time Constant should be small, the

Turbine time constant should be high and the Proportional Integral gain

should be low.

103

The results of the response plots, with PI and optimal controller obtained, for

the two area power system are presented in a tabular form below

System Maximum Frequency Deviation in Hz/pu

Settling Timein seconds

Steady State error in Hz/pu

Two Area Power System with PI controller (Area-1) 0.0095 260 0

Two Area Power System with PI controller (Area-2) 0.0064 260 0

Two Area Power System with Optimal Controller(Area-1) 0.0045 260 0

Two Area Power System with Optimal Controller(Area-2) 0.0035 260 0

Table 5.3 Results of the response plots of the two area system From table 5.3, it is clear that there is no change in settling time in

case of two area-systems with PI controller and Optimal controller, but the

maximum frequency deviation decreases incase of optimal controller.

The results of response plots for various case studies (i.e.,50%

increase in parameter values) carried out for two area systems are presented

in a tabular form below

Table5.4 Results of response plots for Different Case Studies for Two Area system Case Max Frequency Deviation Settling TimeIncrease in Frequency Bias Constant(b) Decreases Decreases

Increase in Synchronizing Torque Coefficient(T12)

Increases No considerable change

From table 5.4, it can be inferred that for normal operation of

two area power systems, the frequency bias constant should be high and the

synchronizing torque coefficient should be low.

104

In totality, it can be concluded that optimal controller gives a

better performance in terms of settling time as well as frequency deviation

compared to PI controller not only in the case of isolated system but also for

interconnected system.

REFERENCES

1) O.I.Elgerd and C.E.Fosha, “Optimum Megawatt-Frequency

Control of Multiarea Electric Energy Systems”, and “The Megawatt-

Frequency Problem: A New Approach Via Optimal Control Theory,” IEEE

Trans. Power System Apparatus and Systems, vol. PAS-89, No.4, April

1970 pp. 556-576.

2) N.N.Benjamin and W.C.Chan , “Multilevel Load Frequency Control

of Interconnected Power Systems”, IEE vol. 125, No. 6 , June -1978.

3) M.M.Adibi, J.N.Borkoski, R.J.Kkafka, and T.L.Volkman, “Frequency

Response of Prime Movers during Restoration”, IEEE Trans. Power System

Apparatus and Systems, vol. PAS-14, No.2, May 1999 pp. 751-756.

4) Ibraheem and P Kumar, “Study of Dynamic Performance of Power

Systems with Asynchronous Tie-Lines Considering Parameter

Uncertainties”, IE(I) Journal –EL , vol. 85, June-2004.

5) Hadi Saadat , “ Power system analysis ”, first edition, Tata Mc Graw

Hill publications.

6) Olle l. Elgerd , “ Electric energy systems theory”, second

Edition(1983), TMH publications.

7) C.L.Wadhwa, “Electric power systems” third edition., Tata Mc Graw-

Hill Publications

105

8) P.S.R. Murthy, “Power system operation and control”, First Edition,

TMH publications

9) I.J.Nagarath and J.P. Kothari, “Modern power system analysis”. third

edition, Tata Mc Graw Hill Publications.

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