8
Load Frequency ControlII
OBJECTIVES
After reading this chapter, you should be able to:
 develop the block diagram models for a twoarea power system
 observe the steady state and dynamic analysis of a twoarea power system with and without integral control
 develop the dynamicstate variable model for singlearea, twoarea, and threearea power system networks
8.1 INTRODUCTION
An extended power system can be divided into a number of load frequency control (LFC) areas, which are interconnected by tie lines. Such an operation is called a pool operation. A power pool is an interconnection of the power systems of individual utilities. Each power system operates independently within its own jurisdiction, but there are contractual agreements regarding internal system exchanges of power through the tie lines and other agreements dealing with operating procedures to maintain system frequency. There are also agreements relating to operational procedures to be followed in the event of major faults or emergencies. The basic principle of a pool operation in the normal steady state provides:
 Maintaining of scheduled interchanges of tieline power: The interconnected areas share their reserve power to handle anticipated load peaks and unanticipated generator outages.
 Absorption of own load change by each area: The interconnected areas can tolerate larger load changes with smaller frequency deviations than the isolated power system areas.
For analyzing the dynamics of the LFC of an narea power system, primarily consider twoarea systems.
Two control areas 1 and 2 are connected by a single tie line as shown in Fig. 8.1.
FIG. 8.1 Two control areas interconnected through a single tie line
Here, the control objective is to regulate the frequency of each area and to simultaneously regulate the power flow through the tie line according to an interarea power agreement.
In the case of an isolated control area, the zero steadystate error in frequency (i.e., Δf_{steady state} = 0) can be obtained by using a proportional plus integral controller, whereas in twocontrol area case, proportional plus integral controller will be installed to give zero steadystate error in a tieline power flow (i.e., ΔP_{TL} = 0) in addition to zero steadystate error in frequency.
For the sake of convenience, each control area can be represented by an equivalent turbine, generator, and governor system.
In the case of a single control area, the incremental power (ΔP_{G}−ΔP_{D}) was considered by the rate of increase of stored KE and increase in area load caused by the increase in frequency.
In a twoarea case, the tieline power must be accounted for the incremental power balance equation of each area, since there is power f low in or out of the area through the tie line.
Power flow out of Control area1 can be expressed as
where ∣E_{1}∣ and ∣E_{2}∣ are voltage magnitudes of Area1 and Area2, respectively, δ_{1} and δ_{2} are the power angles of equivalent machines of their respective areas, and X_{TL} is the tieline reactance.
If there is change in load demands of two areas, there will be incremental changes in power angles (Δδ_{1} and Δδ_{2}). Then, the change in the tieline power is
Therefore, change in incremental tieline power can be expressed as
where
T_{12} is known as the synchronizing coefficient or the stiffness coefficient of the tieline.
Equation (8.3) can be written as
where Static transmission capacity of the tie line.
Consider the change in frequency as
In other words,
Hence, the changes in power angles for Areas1 and 2 are
and
Since the incremental power angles are related in terms of integrals of incremental frequencies, Equation (8.2) can be modified as
Δf_{1} and Δf_{2} are the incremental frequency changes of Areas1 and 2, respectively. Similarly, the incremental tieline power out of Area2 is
where
Dividing Equation (8.6) by Equation (8.3), we get
Therefore, T_{21} = a_{12}T_{12}
and hence ∆P_{TL2} = a_{12}∆P_{TL1} (8.7)
From Equation (7.25) (LFC1), surplus power in p.u. is
For a twoarea case, the surplus power can be expressed in p.u. as
Taking Laplace transform on both sides of Equation (8.8), we get
Rearranging the above equation as follows, we get
where
By comparing Equation (8.9) with singlearea Equation (7.26), the only additional term is the appearance of signal ∆P_{TL1} (S)
Equation (8.9), can be represented in a block diagram model as shown in Fig. 8.2. Taking Laplace transformation on both sides of Equation (8.4), we get
FIG. 8.2 Block diagram representation of Equation (8.9) (for Control area1)
FIG. 8.3 Block diagram representation of Equations (8.10) and (8.11)
For Control area2, we have
The block diagram representation of Equations (8.10) and (8.11) is shown in Fig. 8.3.
8.2 COMPOSITE BLOCK DIAGRAM OF A TWOAREA CASE
By the combination of basic block diagrams of Control area1 and Control area2 and with the use of Figs. 8.2 and 8.3, the composite block diagram of a twoarea system can be modeled as shown in Fig. 8.4.
8.3 RESPONSE OF A TWOAREA SYSTEM—UNCONTROLLED CASE
For an uncontrolled case, ∆P_{c1} = ∆P_{c2} = 0, i.e., the speedchanger positions are fixed.
8.3.1 Static response
In this section, the changes or deviations, which result in the frequency and tieline power under steadystate conditions following sudden step changes in the loads in the two areas, are determined.
FIG. 8.4 Block diagram representation of a twoarea system with an LFC
Let ∆P_{D1}, ∆P_{D2} be sudden (incremental) step changes in the loads of Control area1 and Control area2, simultaneously.
∆P_{G1}, ∆P_{G2} are the incremental changes in the generation in Area1 and Area2 as a result of the load changes.
Δf is the static change in frequency. This will be the same for both the areas and ∆P_{TL1} is the static change in the tieline power transmitted from Area1 to Area2. Since only the static changes are being determined, the incremental changes in generation can be determined by the static loop gains. So, we have
and for static changes (8.13)
For the two areas, the dynamics are described by:
and
Under steadystate conditions, we have
After substituting Equations (8.12), (8.13), and (8.16) in Equations (8.14) and (8.15), we get
and
Since ∆P_{TL2} = −a_{12}∆P_{TL1} and ∆f_{1} = ∆f_{2} = ∆f, from Equation (8.17), we have
Substituting ∆P_{TL1} from Equation (8.18(a)) in Equation (8.18), we get
Substituting Δf from Equation (8.18(b)) in Equation (8.18 (a)), we get
Equations (8.18(b)) and (8.18(c)) are modified as
Tieline frequency,
Tieline power,
where
Equations (8.19) and (8.20) give the values of the static changes in frequency and tieline power, respectively, as a result of sudden stepload changes in the two areas. It can be observed that the frequency and tieline power deviations do not reduce to zero in an uncontrolled case.
Consider two identical areas,
B_{1} = B_{2} = B, β_{1} = β_{2} = β, R_{1} = R_{2} = R and a_{12} = +1
Hence, from Equations (8.19) and (8.20), we have
and
If a sudden load change occurs only in Area2 (i.e.,∆P_{D1} = 0), then we have
and
Equations (8.23) and (8.24) illustrate the advantages of pool operation (i.e., grid operation) as follows:
 Equations (8.19) represents the change in frequency according to the change in load in either of a twoarea system interconnected by a tie line. When considering that those two areas are identical, Equation (8.19) becomes Equation (8.21). Hence, it is concluded that if a load disturbance occurs in only one of the areas (i.e., ∆P_{D1} = 0 or ∆P_{D2} = 0), the change in frequency (Δ f) is only half of the steadystate error, which would have occurred with no interconnection (i.e., an isolated case). Thus, with several systems interconnected, the steadystate frequency error would be reduced.
 Half of the added load (in Area2) is supplied by Area1 through the tie line.
The above two advantages represent the necessity of interconnecting the systems.
8.3.2 Dynamic response
To describe the dynamic response of the twoarea system as shown in Fig. 8.4, a system of seventhorder differential equations is required. The solution of these equations would be tedious. However, some important characteristics can be brought out by an analysis rendered simple by the following assumptions. A power system of two identical control areas is considered for the analysis:
 τ_{gt} = τ_{t} = 0 for both the areas.
 The damping constants of two areas are neglected,
i.e., B_{1} = B_{2} = 0
By virtue of the second assumption, Equations (8.14) and (8.15) become
Taking Laplace transformation on both sides of Equations (8.25) and (8.26) and by rearrangement, we get
From the block diagram of Fig. 8.4, the following equations can be obtained:
(, since two control areas are identical)
By solving Equations (8.27)–(8.30), we get
From the above equation, the following observations can be made:
(i) The denominator is of the form:
where
and α and ω^{2} are both real and positive. Hence, it can be concluded from the roots of characteristic equation that the time response is stable and damped.
The three conditions are:
If α = ω_{n}, system is critically damped
α > ω_{n}, system becomes overdamped
where α = damping factor or decrement of attenuation
ω_{0} = damped angular frequency
Since parameter α also depends on B, but in practice, therefore, the effect of coefficient B is neglected on damping.
(ii) After a disturbance, the change in tieline power oscillates at the damped angular frequency.
(iii) The damping of the tieline power variation is strongly dependent upon the parameter α, which is equal to . Since f ^{0} and H are essentially constant, the damping is a function of the R parameters. If the R value is low, damping becomes strong and vice versa.
The transient change in the tieline power will be of undamped oscillations of frequency, ω_{o} = ω.
If R = ∞, i.e., if the speed governor is not present (α = 0), the variation in frequency deviation and the tieline power would be as shown in Fig. 8.5.
It can be seen that the steadystate frequency deviation is the same for both the areas and does not vanish. The tieline power deviation also does not become zero.
Although the above approximate analysis has confirmed stability, it has been found through more accurate analyses that with certain parameter combinations, the system becomes unstable.
FIG. 8.5 Frequency deviation and tieline power change following a stepload change in Area2 (two areas are identical)
Example 8.1: A twoidentical area power system has the following parameters (Fig. 8.6(a)):
Power system gain constant, K_{ps} = 105
Power system time constant, τ_{ps} = 22 s
Speed regulation, R = 2.5
Normal frequency, f ^{0} = 50 Hz
Governor time constant, τ_{sg} = 0.3 s
Turbine time constant, τ_{t} = 0.5 s
Integration time constant, k_{i} = 0.15
Bias parameter, b = 0.326
2πT_{12} = 0.08
Plot the change in the tieline power and change in frequency of controlarea 1 if there exists a stepload change of 2% in Area1 (Fig. 8.6(b)).
FIG. 8.6 (a) Simulation block diagram for a twoidentical area system of Example 8.1; (b) frequency and tieline response for Example 8.1
Example 8.2: A twoarea power system has the following parameters (Fig. 8.7(a)):
For Area1:
Power system gain constant, K_{ps} = 120
Power system time constant, Ƭ_{ps} = 20 s
Speed regulation, R = 2.5
Normal frequency, f ^{0} = 50 Hz
Governor time constant, Ƭ_{sg} = 0.2 s
Turbine time constant, Ƭ_{t} = 0.4 s
Integration time constant, k_{i} = 0.1
Bias parameter, b = 0.425
For Area2:
Power system gain constant, K_{ps} = 100
Power system time constant, Ƭ_{ps} = 22 s
Speed regulation, R = 3
Normal frequency, f ^{0}= 50 Hz
Governor time constant, Ƭ_{sg} = 0.3 s
Turbine time constant, Ƭ_{t} = 0.5 s
Integration time constant, k_{i} = 0.15
Bias parameter, b = 0.326
2πT_{12} = 0.08
Plot the change in the tieline power and change in frequency of Controlarea 1 if there exists a stepload change of 2% in Area1 (Fig. 8.7(b)).
FIG. 8.7 (a) Simulation block diagram of Example 8.2; (b) Frequency and tieline power response of Example 8.2
Example 8.3: Determine the frequency of oscillations of the tieline power deviation for a twoidenticalarea system given the following data:
R = 3.0 Hz / p.u.; H = 5 s; f ^{0} = 60 Hz
The tieline has a capacity of 0.1 p.u. and is operating at a power angle of 45°.
Solution:
The synchronizingpower coefficient of the line is given by
T^{0}^{12} = P_{m} cos δ_{12} = 0.1 × cos 45° = 0.0707 p.u.
Hence, the frequency of oscillations is given by
8.4 AREA CONTROL ERROR —TWOAREA CASE
In a singlearea case, ACE is the change in frequency. The steadystate error in frequency will become zero (i.e., Δf_{ss} = 0) when ACE is used in the integralcontrol loop.
In a twoarea case, ACE is the linear combination of the change in frequency and change in tieline power. In this case to make the steadystate tieline power zero (i.e., ΔP_{TL} = 0), another integralcontrol loop for each area must be introduced in addition to the integral frequency loop to integrate the incremental tieline power signal and feed it back to the speedchanger.
Thus, for Control area1, we have
where b_{1} = constant = area frequency bias. Taking Laplace transform on both sides of Equation (8.33), we get
Similarly, for Control area2, we have
8.5 COMPOSITE BLOCK DIAGRAM OF A TWOAREA SYSTEM (CONTROLLED CASE)
By the combination of basic block diagrams of Control area1 and Control area2 and with the use of Figs. 8.2 and 8.3, the composite block diagram of a twoarea system can be modeled as shown in Fig. 8.4. Figure 8.8 can be obtained by the addition of integrals of ACE_{1} and ACE_{2} to the block diagram shown in Fig. 8.4. It represents the composite block diagram of a twoarea system with integralcontrol loops. Here, the control signals ∆P_{c1}(s) and ∆P_{c2}(s) are generated by the integrals of ACE_{1} and ACE_{2}. These control errors are obtained through the signals representing the changes in the tieline power and local frequency bias.
8.5.1 Tieline bias control
The speedchanger command signals will be obtained from the block diagram shown in Fig. 8.6 as
and
The constants K_{I1} and K_{I2} are the gains of the integrators. The first terms on the righthand side of Equations (8.36) and (8.37) constitute and are known as tieline bias controls. It is observed that for decreases in both frequency and tieline power, the speedchanger position decreases and hence the power generation should decrease, i.e., if the ACE is negative, then the area should increase its generation.
So, the righthand side terms of Equations (8.36) and (8.37) are assigned a negative sign.
8.5.2 Steadystate response
That the control strategy, described in the previous section, eliminates the steadystate frequency and tieline power deviations that follow a stepload change, can be proved as follows:
FIG. 8.8 Twoarea system with integral control
Let the step changes in loads ∆P_{D1} and ∆P_{D2} simultaneously occur in Control area1 and Control area2, respectively, or in either area. A new static equilibrium state, i.e., steadystate condition is reached such that the output signal of all integrating blocks will become constant. In this case, the speedchanger command signals ∆P_{c1} and ∆P_{c2} have reached constant values. This obviously requires that both the integrands (input signals) in Equations (8.36) and (8.37) be zero.
Input of integrating block is
Input of integrating block is
and input of integrating block is
Equations (8.38) and (8.39) are simultaneously satisfied only for ∆P_{TL1(ss)} = ∆P_{TL2(ss)} = 0 and ∆f_{1(ss)} = ∆f_{2(ss)} = 0.
Thus, under a steadystate condition, change in tieline power and change in frequency of each area will become zero. To achieve this, ACEs in the feedback loops of each area are integrated.
The requirements for integral control action are:
 ACE must be equal to zero at least one time in all 10minute periods.
 Average deviation of ACE from zero must be within specified limits based on a percentage of system generation for all 10minute periods.
The performance criteria also apply to disturbance conditions, and it is required that:
 ACE must return to zero within 10minute periods.
 Corrective control action must be forthcoming within 1 minute of a disturbance.
8.5.3 Dynamic response
The determination of the dynamic response of the twoarea model shown in Fig. 8.6 is more difficult. This is due to the fact that the system of equations to be solved is of the order of nine. Therefore, actual solution is not attempted. But the results obtained from an approximate analysis of a twoidenticalarea power system for three different values of the parameter ‘b’, are presented in Figs. 8.9(a), (b), and (c).
The graphs of Fig. 8.9(a) correspond to the case of b = 0. It can be seen that the tieline power deviation reduces to zero while the frequency does not.
The graphs of Fig. 8.9(b) correspond to the other extreme case of b = ∞. Now, the frequency error vanishes. But, the tieline power does not vanish.
The graphs of Fig. 8.9(c) show an intermediate case wherein both the frequency and the tieline power errors decrease to zero. This is the desired case.
Therefore, it can be concluded that the stability is not always guaranteed. Hence, there is a need for proper parameter selection and adjustment of their values.
8.6 OPTIMUM PARAMETER ADJUSTMENT
The graphs given in Fig. 8.9(c) stress the need for proper parameter settings. The choice of b and K_{I} constants affects the transient response to load changes. The frequency bias b should be high enough such that each area adequately contributes to frequency control. It is proved that choosing b = β gives satisfactory performance of the interconnected system.
The integrator gain K_{I} should not be too high, otherwise, instability may result. Also the time interval at which LFC signals are dispatched, two or more seconds, should be low enough so that LFC does not attempt to follow random or spurious load changes.
FIG. 8.9 Approximate dynamic response of twoidenticalarea power systems with three different values of b parameters
First, a set of parameters, which ensure stability of the control, is selected. For example, b_{1} and b_{2} cannot both be zero, i.e., one of them should be chosen for the control strategy. Later, the values of these parameters are adjusted so that a best or an optimum response is obtained. In other words, the values of parameters, which give rise to an optimum response, are to be determined.
The procedure is as follows:
The popular error criterion, known as the integral of the squared errors (ISE), is chosen for the control parameters ∆f_{1}, ∆f_{2}, and ∆P_{TL1}. For a twoarea system, the ISE criterion function C would be
where α_{1}, α_{2}, and α_{3} are the weight factors, which provide appropriate importance, i.e., weightage to the errors ∆P_{TL1}, Δf_{1}, and Δf_{2}. There is no need to choose ∆P_{TL2}, since ∆P_{TL2} = a_{12}∆P_{TL1}.
Since Δf_{1} and Δf_{2} behave in a similar manner, we need to consider only one of them. So, let us consider Δf_{1} only. This is a parameter selection. Then, α_{3} = 0. Also, let α_{2} = α. Since, we are interested only in the relative magnitudes of C for parameter setting, we can set α_{1} = 1.
With these, Equation (7.41) reduces to
For a twoarea system, ∆P_{TL2} and Δf_{1} would be functions of the integrator gain constants K_{I1} and K_{I2} as well as the frequency bias parameters b_{1} and b_{2}.
The procedure for obtaining the optimum parameter values would be as follows:
First, a convenient and suitable value is chosen for the weight factor ‘α’. Then, for different assumed values of K_{I1}, K_{I2}, b_{1}, and b_{2}, the values of ∆P_{TL1} and Δf_{1} are determined at different instants of time. With these values and α, the value of C is computed using Equation (8.42). The set of values of K_{I1}, K_{I2}, b_{1}, and b_{2} for which C is a minimum is the optimum one.
If we consider the two identical areas, then the number of parameters reduces to two, viz., K_{I1} = K_{I2} = K_{1} and b_{1} = b_{2} = b. In this case, values of C for different values of K_{I} and b can be plotted as shown in Fig. 8.10. As can be seen, the variation of C with K_{I} for different fixed values of b is plotted to get a family of curves called constantb contours. For illustration, only three curves are shown in Fig. 8.10. In practice, a number of curves have to be determined and drawn. It can be seen that C is minimum for b = 0.2 and K_{I} = 1.0.
In this case, the optimum control strategy would, therefore, be
and
In practice, the frequency and tieline power deviations are measured at fixed intervals of time in a sampledata fashion. The sampling rate (the rate at which the frequency deviation and tieline power deviation samples are measured) should be sufficiently high to avoid errors due to sampling.
Note: LFC provides enough control during normal changes in load and frequency, i.e., changes that are not too large. During emergencies, when large imbalances between generation and load occur, LFC is bypassed and other emergency controls are applied, which is beyond the scope of this book.
FIG. 8.10 Constant bcontours of the ISE criterion function C
8.7 LOAD FREQUENCY AND ECONOMIC DISPATCH CONTROLS
Economic load dispatch and LFC play a vital role in modern power system. In LFC, zero steadystate frequency error and a fast, dynamic response were achieved by integral controller action. But this control is independent of economic dispatch, i.e., there is no control over the economic loadings of various generating units of the control area.
Some control over loading of individual units can be exercised by adjusting the gain factors (K_{I}) of the integral signal of the ACE as fed to the individual units. But this is not a satisfactory solution.
A suitable and satisfactory solution is obtained by using independent controls of load frequency and economic dispatch. The load frequency controller provides a fastacting control and regulates the system around an operating point, whereas the economic dispatch controller provides a slowacting control, which adjusts the speedchanger settings every minute in accordance with a command signal generated by the central economic dispatch computer.
EDC—economic dispatch controller
CEDC—central economic dispatch computer
The speedchanger setting is changed in accordance with the economic dispatch error signal, (i.e., P_{G (desired)} − P_{G (actual)}) conveniently modified by the signal ∫ ACE dt at that instant of time. The central economic dispatch computer (CEDC) provides the signal P_{G (desired)}, and this signal is transmitted to the local economic dispatch controller (EDC). The system they operate with economic dispatch error is only for very short periods of time before it is readily used (Fig. 8.11).
This tertiary control can be implemented by using EDC and EDC works on the cost characteristics of various generating units in the area. The speedchanger settings are once again operated in accordance with an economic dispatch computer program.
FIG. 8.11 Load frequency and economic dispatch control of the control area of a power system
The CEDCs are provided at a central control center. The variable part of the load is carried by units that are controlled from the central control center. Mediumsized fossil fuel units and hydrounits are used for control. During peak load hours, lesser efficient units, such as gasturbine units or diesel units, are employed in addition; generators operating at partial output (with spinning reserve) and standby generators provide a reserve margin.
The central control center monitors information including area frequency, outputs of generating units, and tieline power flows to interconnected areas. This information is used by ALFC in order to maintain area frequency at its scheduled value and net tieline power flow out of the area at its shedding value. Raise and lower reference power signals are dispatched to the turbine governors of controlled units.
Economic dispatch is coordinated with LFC such that the reference power signals dispatched to controlled units move the units toward their economic loading and satisfy LFC objectives.
8.8 DESIGN OF AUTOMATIC GENERATION CONTROL USING THE KALMAN METHOD
A modern gigawatt generator with its multistage reheat turbine, including its automatic load frequency control (ALFC) and automatic voltage regulator (AVR) controllers, is characterized by an impressive complexity. When all its nonnegligibility dynamics are taken into account, including crosscoupling between control channels, the overall dynamic model may be of the twentieth order.
The dimensionality barrier can be overcome by means of computeraided optimal control design methods originated by Kalman. A computeroriented technique called optimum linear regulator (OLR) design has proven to be particularly useful in this regard.
The OLR design results in a controller that minimizes both transient variable excursions and control efforts. In terms of power system, this means optimally damped oscillation with minimum wear and tear of control valves.
OLR can be designed using the following steps:
 Casting the system dynamic model in statevariable form and introducing appropriate control forces.
 Choosing an integralsquarederror control index, the minimization of which is the control goal.
 Finding the structure of the optimal controller that will minimize the chosen control index.
8.9 DYNAMICSTATEVARIABLE MODEL
The LFC methods discussed so far are not entirely satisfactory. In order to have more satisfactory control methods, optimal control theory has to be used. For this purpose, the power system model must be in a statevariable model.
8.9.1 Model of singlearea dynamic system in a statevariable form
From the block diagram of an uncontrolled singlearea system shown in Fig. 8.12, we get the following ‘sdomain’ equations:
In time domain, the above equations can be expressed as
FIG. 8.12 Statespace model of a singlearea system
Let us choose the state variables , input, u = ΔP_{C}, and disturbance, d = ΔP_{D}
The above equations are written in a statevariable form:
The state equation would then be written in a more general form as
where x is the ndimensional state vector, u the mdimensional controlforce vector = [u] = [ΔP_{c}], and p the disturbance force vector = [p] = [ΔP_{D}].
8.9.2 Optimum control index (I)
The optimum linear regulator design is based on the integralsquarederror index of the form.
where q’s and r’s are positive penalty factors. Let us consider index I for a singlearea system as
Here, the idea of optimum control is to minimize the index (I). Consider the term q_{3} (Δ f)^{2}, squaring of frequency error will contribute to ‘I’ independent of its sign. If Δf is doubled, its contribution to ‘I’ will quadruple. The integral causes Δf to add to ‘I’ during its entire duration. The penalty factors q_{i} distribute the penalty weight among the statevariable errors. If the error in a particular variable is of little significance, we simply set its penalty factor to zero.
Similarly, in the case of controlforce increments, the penalty factors r_{i} distribute the penalties among the m control force. (Here, none of r’s are set to zero.) If they are set to zero, then the control force will assume an infinite magnitude without affecting ‘I’. An infinite control force could do its correcting job in zero time. This would obviously be a very unrealistic regulator.
If all q’s and r’s constitute the diagonal elements of the two penalty matrices, then we have
IndexI in a compact form is
8.9.3 Optimum control problem and strategy
The task that an OLR or an optimum controller must perform is the fulfillment of the optimum control problem, which can be stated as follows: consider a system that is initially under steadystate condition. If it is disturbed by a set of step type of disturbances, it goes through a transient state first. It is required that, after the expiry of the transient period, it should return to the original or new prescribed steadystate condition. The problem is to determine the set of control forces which will not only take the system to the original or new prescribed steady state, but will also do so by simultaneously minimizing the chosen control criterion function or optimum control index (I).
The value of which fulfills the above optimum control requirement, is the desired optimum control strategy. An optimum controller or OLR is the one that carries out the above strategy.
8.9.4 Dynamic equations of a twoarea system
From the block diagram of uncontrolled twoarea systems shown in Fig. 8.4, get the following ‘sdomain’ equations:
where X_{E1}(s) and X_{E2}(s) are the Laplace transforms of the movements of the main positions in the speedgoverning mechanisms of the two areas.
By taking inverse Laplace transform for the above equations, we get a set of seven differential equations. These are the timedomain equations, which describe the smalldisturbance dynamic behavior of the power system.
Consider the first equation,
Taking the inverse Laplace transform of the above equation, we get
In a similar way, the remaining equations can be rearranged and an inverse Laplace transform is found. Then, the entire set of differential equations is
8.9.4.1 State variables and statevariable model
The state variables are a minimum number of those variables, which contain sufficient information about the past history with which all future states of the system can be determined for known control inputs. For the twoarea system under consideration, the state variables would be Δf_{1}, Δf_{2}, ∆X_{E1}, ∆X_{E2}, ∆P_{sg1}, ∆P_{sg2} and ∆P_{TL1}; seven in number. Denoting the above variables by x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, and x_{7} and arranging them in a column vector as
where is called a state vector.
The control variables ∆P_{c1} and ∆P_{c2} are denoted by the symbols u_{1} and u_{2}, respectively, as
where ū is called the control vector or the controlforce vector.
The disturbance variables ∆P_{D1} and ∆P_{D2}, since they create perturbations in the system, are denoted by p_{1} and p_{2}, respectively, as
where is called the disturbance vector.
The above state equations can be written in a matrix form as
where ; i = 1, 2, 3, … 7.
The above matrix equation can be written in the vector form as
where [A] is called the system matrix, [B] the input distribution matrix, and [J] the disturbance distribution matrix.
In the present case, their dimensions are (7 × 7), (7 × 2), and (7 × 2), respectively. Equation (8.45) is a shorthand form of Equation (8.44), and Equation (8.44) constitutes the dynamic ‘statevariable model’ of the considered twoarea system.
The differential equations can be put in the above form only if they are linear. If the differential equations are nonlinear, then they can be expressed in the more general form as
8.9.5 Statevariable model for a threearea power system
The block diagram representation of this model is shown in Fig. 8.13.
FIG. 8.13 Threearea model
From the block diagram, the following equations are written as
Taking the inverse Laplace transform for the above equations, which we get in a similar way, the remaining equations can be rearranged and the inverse Laplace transform can be found. Then, the entire set of differential equations is
and
and
and
The above equations are written in a vector form as shown below:
where
where X is called a state vector.
The control variables ∆P_{c1}, ∆P_{c2}, and ∆P_{c1} are denoted by the symbols u_{1}, u_{2}, and u_{3}, respectively, as
where u is called the control vector or the controlforce vector.
The disturbance variables ∆P_{D1}, ∆P_{D2}, and ∆P_{D1}, since they create perturbations in the system, are denoted by p_{1}, p_{2}, and p_{3}, respectively, as
where is called the disturbance vector.
where 
m 
= 
2π(T^{0}_{12} + T^{0}_{13}) 

n 
= 
2π(T^{0}_{21} + T^{0}_{23}) 

o 
= 
2π(T^{0}_{31} + T^{0}_{32}) 

p 
= 
−2πT^{0}_{12} 

q 
= 
−2πT^{0}_{13} 

r 
= 
−2πT^{0}_{21} 

s 
= 
−2πT^{0}_{23} 

t 
= 
−2πT^{0}_{31} 

ū 
= 
−2πT^{0}_{32} 
8.9.6 Advantages of statevariable model
The statevariable modeling of a power system offers the following advantages:
 Modern control theory is based upon this standard form.
 By arranging system parameters into matrices [A], [B], and [J], a very organized methodology of solving system equations, either analytically or by computer, is developed. This is important for large systems where a lack of organization easily results in errors.
Example 8.4: Two interconnected Area1 and Area2 have the capacity of 2,000 and 500 MW, respectively. The incremental regulation and damping torque coefficient for each area on its own base are 0.2 p.u. and 0.8 p.u., respectively. Find the steadystate change in system frequency from a nominal frequency of 50 Hz and the change in steadystate tieline power following a 750 MW change in the load of Area1.
Solution:
Rated capacity of Area1 = P_{1(rated)} = 2,000 MW
Rated capacity of Area2 = P_{2(rated)} = 500 MW
Speed regulation, R = 0.2 p.u.
Nominal frequency, f = 50 Hz
Change in load power of Area1, ΔP_{1} = 75 MW
Speed regulation, R = 0.2 = 0.2 p.u. × 50 = 10 Hz/p.u. MW
Damping torque coefficient, B = 0.8 p.u. MW/p.u. Hz
Change in load of Area1, ∆P_{D1} = 75 MW
p.u. change in load of Area1
p.u. change in load of Area2
Steadystate change in system frequency,
where
Steadystate change in tieline power following load change in Area1:
Example 8.5: Solve Example 8.4, without governor control action.
Solution:
Without the governor control action, R = 0
Steadystate change in tieline power following load change in Area1:
It is observed from the result that the power flow through the tie line is the same in both the cases of with governor action and without governor action, since it does not depend on speed regulation R.
Example 8.6: Find the nature of dynamic response if the two areas of the above problem are of uncontrolled type, following a disturbance in either area in the form of a step change in electric load. The inertia constant of the system is given as H = 3 s and assume that the tie line has a capacity of 0.09 p.u. and is operating at a power angle of 30^{o} before the step change in load.
Solution:
Given:
Speed regulation, R = 0.2 p.u. = 0.2 × 50 = 10 Hz/p.u. MW
Damping coefficient, B = 0.8 p.u. MW/p.u. Hz
Inertia constant, H = 3 s
Nominal frequency, f ^{0} = 50 Hz
Tieline capacity,
From the theory of dynamic response, we know that
It is observed that the damped oscillation type of dynamic response has resulted since α < ω_{n}:
∴ Damped angular frequency
∴ Damped frequency = f_{d}
Example 8.7: Two control areas have the following characteristics:
Area1: 
Speed regulation = 0.02 p.u. 

Damping coefficient = 0.8 p.u. 

Rated MVA = 1,500 
Area2: 
Speed regulation = 0.025 p.u. 

Damping coefficient = 0.9 p.u. 

Rated MVA = 500 
Determine the steadystate frequency change and the changed frequency following a load change of 120 MW, which occurs in Area1. Also find the tieline power flow change.
Solution:
Given R_{1} = 0.1 p.u.; R_{2} = 0.098 p.u.
B_{1} = 0.8 p.u.; B_{2} = 0.9 p.u.
P_{1 rated} = 1,500 MVA; P_{2 rated} = 1,500 MVA
Change in load of Area1,
∆P_{D1} = 120 MW, ∆P_{D2} = 0
p.u. change in load of Area1
∴ Steadystate frequency change,
i.e., Steadystate change in frequency, ∆f_{ss} 
= 
0.0012415 × 50 

= 
0.062 Hz 
∴ New value of frequency, f = f^{0} − ∆f_{ss} 
= 
50 − 0.062 

= 
49.937 Hz 
Steadystate change in tieline power
Example 8.8: In Example 8.6, if the disturbance also occurs in Area2, which results in a change in load by 75 MW, determine the frequency and tieline power changes.
Solution:
Change in load of Area1, ∆P_{D1} = 120 MW
p.u. change in load of Area1
Change in load of Area2, ∆P_{D2} = 75 MW
p.u. change in load of Area2
Steadystate frequency change,
∴ Steadystate frequency change = 0.002 × 50 = 0.1 Hz
∴ New value of frequency = f ^{0} − Δf_{ss} = 50 − 0.1 = 49.899 Hz
Steadystate change in tieline power,
Example 8.9: Two areas of a power system network are interconnected by a tie line, whose capacity is 250 MW, operating at a power angle of 45^{o}. If each area has a capacity of 2,000 MW and the equal speedregulation coefficiency of 3 Hz/p.u. MW, determine the frequency of oscillation of the power for a step change in load. Assume that both areas have the same inertia constants of H = 4 s. If a stepload change of 100 MW occurs in one of the areas, determine the change in tieline power.
Solution:
Given:
Tieline capacity, P_{tie(max)} = 250 MW
Power angle of two areas, (δ^{0}_{1} − δ^{0}_{2}) = 457°
Capacity of each area, P_{rated} = 2,000 MW
Speedregulation coefficient = R_{1} = R_{2} = R = 3Hz/p.u. MW
Inertia constant, H = 4 s
Since, α < ω_{n}, the dynamic response will be of a damped oscillation type.
Damped angular frequency,
∴ Frequency of oscillation,
If a stepload change of 100 MW occurs in any one of the areas, the total load change will be shared equally by both areas since the two areas are equal, i.e., a power of will flow from the other area into the area where a load change occurs.
Example 8.10: Two power stations A and B of capacities 75 and 200 MW, respectively, are operating in parallel and are interconnected by a short transmission line. The generators of stations A and B have speed regulations of 4% and 2%, respectively. Calculate the output of each station and the load on the interconnection if
 the load on each station is 100 MW,
 the loads on respective bus bars are 50 and 150 MW, and
 the load is 130 MW at Station A bus bar only.
Solution:
Given:
Capacity of StationA = 75 MW
Capacity of StationB = 200 MW
Speed regulation of StationA generator, R_{A} = 4%
Speed regulation of StationB generator, R_{B} = 2%
(a) If the load on each station = 100 MW
Speed regulation
From Equations (8.48) and (8.49), we have
0.000533P_{1} = 0.0001P_{2}
5.33P_{1} = P_{2} (8.50)
P_{1} + P_{2} = 200
Substituting Equation (8.50) in Equation (8.47), we get
The power generations and tieline power are indicated in Fig. 8.14(a).
(b) If the load on respective bus bars are 50 and 150 MW, then we have
i.e., P_{1} + P_{2} = 50 + 150 = 200 MW
5.33 P_{1} = P_{2}
P_{1} + 5.33 P_{1} = 200
⇒ 6.33P_{1} = 200
P_{1} = 31.6 MW
∴ P_{2} = 200 − 31.60 = 168.4 MW
The power generations and tieline power are indicated in Fig. 8.14(b).
FIG. 8.14 (a) Illustration for Example 8.10; (b) illustration for Example 8.10; (c) illustration for Example 8.10
(c) If the load is 130 MW at A only, then we have
P_{1} + P_{2} = 130
5.33P_{1} = P_{2}
∴ P_{1} = 5.33 P_{1} = 130
6.33P_{1} = 130
⇒ P_{1} = 20.537 MW
∴ P_{2} = 130 − P_{1} = 109.462 MW
The power generations and tieline power are indicated in Fig. 8.14(c).
Example 8.11: The two control areas of capacity 2,000 and 8,000 MW are interconnected through a tie line. The parameters of each area based on its own capacity base are R = 1 Hz/p.u. MW and B = 0.02 p.u. MW/Hz. If the control area2 experiences an increment in load of 180 MW, determine the static frequency drop and the tieline power.
Solution:
Capacity of Area1 = 2,000 MW
Capacity of Area2 = 8,000 MW
Taking 8,000 MW as base,
∴ Speed regulation of Area1,
Damping coeffi cient of Area1,
Speed regulation of Area2, R_{2} = 1 Hz / p.u. MW
Damping coefficient of Area2, B_{2} = 0.02 p.u. MW/ Hz
Given an increment of Area2 in load,
∴ Static change in frequency,
Static change in tieline power,
Note: Here, a_{12} value determination is not required since values of R_{1}, B_{1}, and β_{1} are obtained according to the base values.
Alternate method:
Find Then, obtain the ∆f_{(ss)} and ∆P_{tie(ss)} values.
Here, there is no need to obtain, R_{1}, B_{1}, R_{2}, and B_{2} separately.
Example 8.12: Two generating stations A and B having capacities 500 and 800 MW, respectively, are interconnected by a short line. The percentage speed regulations from noload to full load of the two stations are 2 and 3, respectively. Find the power generation at each station and power transfer through the line if the load on the bus of each station is 200 MW.
Solution:
Given data:
Capacity of StationA = 500 MW
Capacity of StationB = 800 MW
Percentage speed regulation of StationA = 2% = 0.02
Percentage speed regulation of StationB = 3% = 0.03
Load on bus of each station = P_{DA} = P_{DB} = 200 MW
Total load, P_{D} = 400 MW
Speed regulation of StationA:
Speed regulation of StationB:
Let P_{GA} be the power generation of StationA and P_{GB} the power generation of StationB:
P_{GB} = Total load − P_{GA} = (400 − P_{GA})
⇒ 0.002P_{GA} = 0.001875 (400 − P_{GA})
= 0.75 = 193.55 MW
(0.002 + 0.001875) P_{GA} = 0.75
⇒ 
P_{GA} 
= 
193.55 MW 

P_{GB} 
= 
206.45 MW 

P_{GA} 
= 
193.55 MW 
∴ 
P_{GB} 
= 
206.45 MW 
The power transfer through the line from StationB to stationA
= P_{GB} − (load at bus bar of B)
= 206.45 − 200
= 6.45 MW
Example 8.13: Two control areas of 1,000 and 2,000 MW capacities are interconnected by a tie line. The speed regulations of the two areas, respectively, are 4 Hz/p.u. MW and 2.5 Hz/p.u. MW. Consider a 2% change in load occurs for 2% change in frequency in each area. Find steadystate change in frequency and tieline power of 10 MW change in load occurs in both areas.
Solution:
Capacity of Area1 = 1,000 MW
Capacity of Area2 = 2,000 MW
Speed regulation of Area1, R_{1} = 4 Hz/p.u. MW (on 1,000MW base)
Speed regulation of Area2, R_{2} = 2 Hz/p.u. MW
Let us choose 2,000 MW as base, 2% change in load for 2% change in frequency
Damping coeffi cient of Area1,
Similarly, damping coefficient of Area2 on 2,000MW base
Speed regulation of Area  1on 2,000MW base = R_{1}
Speed regulation of Area2, R_{2} = 2 Hz/p.u. MW
If a 10MW change in load occurs in Area1, then we have
Steadystate change in frequency,
Steadystate change in frequency,
or Δf_{(ss)} = −0.007633 × 50 = 0.38 Hz
Steadystate change in tieline power:
i.e., the power transfer of 7.938 MW is from Area2 to Area1.
If a 10MW change in load occurs in Area2, then we have
∴ Steadystate change in frequency,
Steadystate change in tieline power:
i.e., A power of 2.061 MW is transferred from Area1 to Area2.
Example 8.14: Two similar areas of equal capacity of 5,000 MW, speed regulation R = 3 Hz/p.u. MW, and H = 5 s are connected by a tie line with a capacity of 500 MW, and are operating at a power angle of 45^{o}. For the above system, the frequency is 50 Hz; find:
 The frequency of oscillation of the system.
 The steadystate change in the tieline power if a step change of 100 MW load occurs in Area2.
 The frequency of oscillation of the system in the speedgovernor loop is open.
Solution:
Given:
Capacity of each control area = P_{1(rated)} P_{2(rated)} = 500 MW
Speed regulation, R = 2 Hz/p.u. MW
Inertia constant, H = 5 s
Power angle = 45^{o}
Supply frequency, f ^{0} = 50 Hz
(a) Stiffness coefficient,
Since α < ω_{n}, damped oscillations will be present.
∴ Damped angular frequency,
(b) Since the two areas are similar, each area will supply half of the increased load:
ΔP_{tie} = 50 MW from Area1 to Area2.
If the speedgovernor loop is open, then
Damped angular frequency,
KEY NOTES
 An extended power system can be divided into a number of LFC areas, which are interconnected by tie lines. Such an operation is called a pool operation.
The basic principle of a pool operation in the normal steady state provides:
 Maintaining of scheduled interchanges of tieline power.
 Absorption of own load change by each area.
 The advantages of a pool operation are as follows:
 Half of the added load (in Area2) is supplied by Area1 through the tie line.
 The frequency drop would be only half of that which would occur if the areas were operating without interconnection.
 The speedchanger command signals will be:
and
The constants K_{12} and K_{12} are the gains of the integrators. The first terms on the righthand side of the above equations constitute what is known as a tieline bias control.
 The load frequency controller provides a fastacting control and regulates the system around an operating point, whereas the EDC provides a slowacting control, which adjusts the speedchanger settings every minute in accordance with a command signal generated by the CEDC.
SHORT QUESTIONS AND ANSWERS
 What are the advantages of a pool operation?
The advantages of a pool operation (i.e., grid operation) are:
 Half of the added load (in Area2) is supplied by Area1 through the tie line.
 The frequency drop would be only half of that which would occur if the areas were operating without interconnection.
 Without speedchanger position control, can the static frequency deviation be zero?
No, the static frequency deviation cannot be zero.
 State the additional requirement of the control strategy as compared to the singlearea control.
The tieline power deviation due to a stepload change should decrease to zero.
 Write down the expressions for the ACEs.
The ACE of Areas1 and 2 are:
ACE_{1} (S) = ∆P_{TL1} (S) + b_{1}∆F_{1}(S).
ACE_{2} (S) = ∆P_{TL2} (S) + b_{2}∆F_{2}(S).
 What is the criterion used for obtaining optimum values for the control parameters?
Integral of the sum of the squared error criterion is the required criterion.
 Give the error criterion function for the twoarea system.
 What is the order of differential equation to describe the dynamic response of a twoarea system in an uncontrolled case?
It is required for a system of seventhorder differential equations to describe the dynamic response of a twoarea system. The solution of these equations would be tedious.
 What is the difference of ACE in singlearea and twoarea power systems?
In a singlearea case, ACE is the change in frequency. The steadystate error in frequency will become zero (i.e., Δf_{ss} = 0) when ACE is used in an integralcontrol loop.
In a twoarea case, ACE is the linear combination of the change in frequency and change in tieline power. In this case to make the steadystate tieline power zero (i.e., ΔP_{TL} = 0), another integralcontrol loop for each area must be introduced in addition to the integral frequency loop to integrate the incremental tieline power signal and feed it back to the speedchanger.
 What is the main difference of load frequency and economic dispatch controls?
The load frequency controller provides a fastacting control and regulates the system around an operating point, whereas the EDC provides a slowacting control, which adjusts the speedchanger settings every minute in accordance with a command signal generated by the CEDC.
 What are the steps required for designing an optimum linear regulator?
An optimum linear regulator can be designed using the following steps:
 Casting the system dynamic model in a statevariable form and introducing appropriate control forces.
 Choosing an integralsquarederror control index, the minimization of which is the control goal.
 Finding the structure of the optimal controller that will minimize the chosen control index.
MULTIPLECHOICE QUESTIONS
 Changes in load division between AC generators operation in parallel are accomplished by:
 Adjusting the generator voltage regulators.
 Changing energy input to the prime movers of the generators.
 Lowering the system frequency.
 Increasing the system frequency.
 When the energy input to the prime mover of a synchronous AC generator operating in parallel with other AC generators is increased, the rotor of the generator will:
 Increase in average speed.
 Retard with respect to the statorrevolving field.
 Advance with respect to the statorrevolving field.
 None of these.
 When two or more systems operate on an interconnected basis, each system:
 Can depend on the other system for its reserve requirements.
 Should provide for its own reserve capacity requirements.
 Should operate in a ‘flat frequency’ mode.
 When an interconnected power system operates with a tieline bias, they will respond to:
 Frequency changes only.
 Both frequency and tieline load changes.
 Tieline load changes only.
 In a twoarea case, ACE is:
 Change in frequency.
 Change in tieline power.
 Linear combination of both (a) and (b).
 None of the above.
 An extended power system can be divided into a number of LFC areas, which are interconnected by tie lines. Such an operator is called
 Pool operation.
 Bank operation.
 (a) and (b).
 None.
 For the static response of a twoarea system,
 ∆P_{ref1} = ∆_{ref2}.
 ∆P_{ref1} = 0.
 ∆P_{ref2} = 0.
 Both (b) and (c).
 Area of frequency response characteristic ‘β’ is:
 1/R.
 B.
 B + 1/R.
 B  1/R.
 The tieline power equation is ΔP_{12} = _____
 T (Δδ_{1} + Δδ_{2}).
 T/(Δδ_{1} + Δδ_{2}).
 T/(Δδ_{1}  Δδ_{2}).
 T(Δδ1  Δδ2).
 The unit of synchronizing coefficients ‘T ’ is:
 MWs.
 MW/s.
 MWrad.
 MW/rad.
 For a twoarea system, Δf is related to increased step load M_{1} and M_{2} with area frequency response characteristics β_{1} and β_{2} is:
 M_{1} + M_{2}/β_{1} + β_{2}.
 (M_{1} + M_{2}) (β_{1} + β_{2}).
 (M_{1} + M_{2})/(β_{1} + β_{2}).
 None of these.
 Tieline power flow for the above question (11) is ΔP_{12} = _____
 (β_{1} M_{2} + β_{2} M_{1})/β_{1} + β_{2}.
 (β_{1} M_{2}  β_{2} M_{1})/β_{1} + β_{2}.
 (β_{1} M_{1}  β_{2} M_{2})/β_{1} + β_{2}.
 None of these.
 Advantage of a pool operation is:
 Added load can be shared by two areas.
 Frequency drop reduces.
 Both (A) and (B).
 None of these.
 Damping of frequency oscillations for a twoarea system is more with:
 LowR.
 HighR.
 R = α.
 None of these.
 ACE equation for a general power system with tieline bias control is:
 ΔP_{ij} + B_{i} Δf_{i}.
 ΔP_{ij}  B_{i} Δf_{i}.
 ΔP_{ij} /B_{i} Δf_{i}.
 None of these.
 For a twoarea system Δf, ΔP_{L}, R_{1}, R_{2}, and D are related as Δf = _____
 ΔP_{L} / R_{1} + R_{2}.
 ΔP_{L} / (1/R_{1} + R_{2} + B).
 ΔP_{L} / (B + R_{1} + 1/R_{2}).
 None of these.
 If the two areas are identical, then we have:
 Δf_{1} = 1/Δf_{2}.
 Δf_{1} Δf_{2} = 2.
 Δf_{1} = Δf_{2}.
 None of these.
 Tieline between two areas usually will be a _____ line.
 HVDC.
 HVAC.
 Normal AC.
 None of these.
 Dynamic response of a twoarea system can be represented by a _____ order transfer function.
 Third.
 Second.
 First.
 Zero.
 Control of ALFC loop of a multiarea system is achieved by using _____ mathematical technique.
 Root locus.
 Bode plots.
 State variable.
 Nyquist plots.
REVIEW QUESTIONS
 Obtain the mathematical modeling of the line power in an interconnected system and its block diagram.
 Obtain the block diagram of a twoarea system.
 Explain how the control scheme results in zero tieline power deviations and zerofrequency deviations under steadystate conditions, following a stepload change in one of the areas of a twoarea system.
 Deduce the expression for staticerror frequency and tieline power in an identical twoarea system.
 Explain about the optimal twoarea LFC.
 What is meant by tieline bias control?
 Derive the expression for incremental tieline power of an area in an uncontrolled twoarea system under dynamic state for a stepload change in either area.
 Draw the block diagram for a twoarea LFC with integral controller blocks and explain each block.
 What are the differences between uncontrolled, controlled, and tieline bias LFC of a twoarea system.
 Explain the method involved in optimum parameter adjustment for a twoarea system.
 Explain the combined operation of an LFC and an ELDC system.
PROBLEMS
 Two interconnected areas 1 and 2 have the capacity of 250 and 600 MW, respectively. The incremental regulation and damping torque coefficient for each area on its own base are 0.3 and 0.07 p.u. respectively. Find the steadystate change in system frequency from a nominal frequency of 50 Hz and the change in steadystate tieline power following a 850 MW change in the load of Area1.
 Two control areas of 1,500 and 2,500 MW capacities are interconnected by a tie line. The speed regulations of the two areas, respectively, are 3 and 1.5 Hz/p.u. MW. Consider that a 2% change in load occurs for a 2% change in frequency in each area. Find the steadystate change in the frequency and the tieline power of 20 MW change in load occurring in both areas.
 Find the nature of dynamic response if the two areas of the above problem are of uncontrolled type, following a disturbance in either area in the form of a step change in an electric load. The inertia constant of the system is given as H = 2 s and assume that the tie line has a capacity of 0.08 p.u. and is operating at a power angle of 35^{o} before the step change in load.