Part 7 · Chapter 33

Load Frequency Control and Automatic Generation Control

Frequency is the one number every machine on the grid agrees about, and it moves the instant generation and load stop matching — so the whole apparatus of governors, droop, tie-line bias and area control error exists to turn that single measurement into the right instruction for the right generator.

Electric Power Systems Prof. Mithun Mondal Reading time ≈ 48 min
i What you'll learn
  • Why an imbalance between generation and load appears immediately as a rate of change of frequency, and how the swing equation of Chapter 27 linearises into the generator–load block.
  • How the speed governor's droop \(R\) makes stable load sharing possible, and why it guarantees a residual frequency error.
  • The area frequency response characteristic \(\beta = D + 1/R\) and the one-line result \(\Delta f = -\Delta P_L/\beta\).
  • Why only integral action in the supplementary loop can drive \(\Delta f\) to zero, and how its gain is chosen.
  • How a tie line couples two areas through the synchronizing coefficient \(T_{12}\), and what the linearised tie-line power looks like.
  • The area control error \(\mathrm{ACE}_i = \Delta P_{\text{tie},i} + B_i\,\Delta f\), and why the bias setting \(B_i = \beta_i\) makes control non-interactive.
  • How AGC distributes its correction using the participation factors that Chapter 31's economic dispatch supplies.
Section 33-1

Frequency as the Balance Signal

Chapters 31 and 32 decided which units should run and at what output, minimising fuel cost over a day. Both calculations assumed the demand was known. It never is. Load changes second by second — a rolling mill starts, a city's air conditioners cycle, a cloud crosses a solar farm — and no schedule computed an hour ahead can track it. Something must close the gap continuously, and it must do so without any central computer being in the loop for the first few seconds.

The mechanism is already present in the physics. Electrical energy cannot be stored in the network, so at every instant the mechanical power delivered by all the turbines must equal the electrical power consumed by all the loads plus the losses. If it does not, the difference has only one place to go: the kinetic energy of the rotating masses. An excess of load draws energy out of the rotors, they slow down, and the frequency falls. An excess of generation pushes energy in, and the frequency rises.

This makes frequency an exceptionally convenient measurement. Because all synchronous machines in an interconnected system are locked to a common electrical speed (Chapter 26), frequency is a single global number. Measure it anywhere — at a generator terminal, at a distribution substation, at a domestic socket — and you learn the same thing about the balance of the entire interconnection. No communication network is needed to distribute it; the grid distributes it itself, at the speed of the synchronizing torques.

Reactive power behaves in exactly the opposite way, and it is worth fixing the contrast now because Chapter 34 is built on it. Reactive power does not travel far: the voltage at a bus is set mostly by the reactive balance in its own neighbourhood, so voltage is a local quantity and there are as many voltage problems as there are buses. Frequency is one problem for the whole system. The two control tasks are therefore organised completely differently, and they are almost perfectly decoupled — a fact that lets us treat active power and frequency in this chapter and reactive power and voltage in the next.

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Why frequency control exists
Active-power imbalance changes the stored kinetic energy of the system's rotors, and therefore the frequency. Frequency is the direct, instantaneous, system-wide indicator of that imbalance.

Holding frequency close to nominal is not cosmetic. Turbine blades have resonances that forbid prolonged operation more than about \(2\%\) off nominal; induction motors change speed and slip; synchronous clocks and any process timed from the mains drift; and a deep enough excursion trips under-frequency load shedding and then the generators themselves.

The control problem then decomposes naturally by time scale. In the first ten seconds the primary control — the speed governor on each turbine — acts alone, arrests the frequency excursion and settles the system at a new frequency slightly away from nominal. Over the next few minutes the secondary control, which is what "automatic generation control" (AGC) means in practice, adjusts the governor reference settings to return frequency to nominal and to restore the scheduled interchange between neighbouring systems. Over tens of minutes the tertiary layer re-dispatches units economically, which is where the machinery of Chapter 31 re-enters. This chapter builds the first two layers from their component models.

Section 33-2

The Generator–Load Model

Start from the swing equation of Chapter 27, written for a single machine on its own rating as a base:

Swing equation, Chapter 27
\[ \frac{2H}{\omega_s}\,\frac{d^2\delta}{dt^2} = P_m - P_e \qquad\text{(per unit)} \]

For stability studies we followed the rotor angle. For frequency control we want the speed itself. Since \(d\delta/dt\) is the deviation of rotor electrical speed from synchronous, and speed in electrical radians is \(2\pi f\), differentiating once fewer times gives an equation directly in the frequency deviation \(\Delta f = f - f_0\):

Swing equation in terms of frequency deviation
\[ \frac{2H}{f_0}\,\frac{d(\Delta f)}{dt} = \Delta P_m - \Delta P_e \]

Here \(\Delta P_m\) is the change in mechanical power input and \(\Delta P_e\) the change in electrical power output, both in per unit on the chosen base; \(H\) is the inertia constant in seconds on the same base. For a whole control area, \(H\) is the MVA-weighted sum of the individual machine inertia constants referred to a common base, exactly as in Chapter 4 — the area behaves like one large machine as far as the common frequency is concerned.

The electrical load must now be modelled with a little more care than "a fixed number of megawatts". Part of any composite load is frequency-sensitive: induction motors driving pumps and fans slow down when the frequency falls, and the mechanical power they demand falls with them, roughly as the cube of speed for a fan. Resistive load — heating, incandescent lighting — is indifferent to frequency. Linearising the composite characteristic about the operating point,

Frequency-dependent load
\[ \Delta P_e = \underbrace{\Delta P_L}_{\text{genuine load change}} + \underbrace{D\,\Delta f}_{\text{frequency sensitivity}}, \qquad D = \frac{\partial P_L}{\partial f}\bigg|_{f_0} \]

The coefficient \(D\) is the load damping constant. It is positive: a fall in frequency reduces the load, which helps. Its usual magnitude is such that a \(1\%\) change in frequency produces a \(1\%\) to \(2\%\) change in the motor part of the load. Throughout this chapter \(D\) is quoted in MW/Hz, or in per-unit megawatts per hertz when a per-unit base is stated; the two differ only by the power base.

Substituting the load model into the swing equation and taking Laplace transforms with zero initial conditions gives the generator–load block, sometimes called the power-system block:

The generator–load transfer function
\[ \frac{2H}{f_0}\,s\,\Delta F(s) = \Delta P_m(s) - \Delta P_L(s) - D\,\Delta F(s) \]
\[ \Delta F(s) = \big[\Delta P_m(s) - \Delta P_L(s)\big]\cdot\underbrace{\frac{K_p}{1+sT_p}}_{G_p(s)}, \qquad K_p = \frac{1}{D}, \quad T_p = \frac{2H}{f_0 D} \]

The result is a simple first-order lag, and both of its parameters have a plain meaning. The gain \(K_p = 1/D\) says that a permanent \(1\) MW surplus, left entirely to the load's own frequency sensitivity, raises frequency until the load has grown by that same \(1\) MW. The time constant \(T_p = 2H/(f_0 D)\) is the ratio of stored kinetic energy to damping; with \(H \approx 5\) s and a damping of about \(1\%\) of rating per hertz at \(50\) Hz it comes out near \(20\) s. That is a slow block, and it is the reason a governor is needed at all: left to itself the system drifts a long way before the load's own damping arrests it.

Inertia is the first responder. In the first instant after a load step, before any valve has moved, the entire imbalance is supplied by the rotors. Setting \(\Delta P_m = 0\) and \(\Delta f = 0\) in the swing equation gives the initial rate of change of frequency, \(\mathrm{RoCoF} = -f_0\,\Delta P_L /(2H)\). A system with less spinning inertia — one carrying a large share of inverter-connected wind and solar, the subject of Chapter 39 — reaches any given frequency threshold sooner after the same disturbance, which is why RoCoF has become a planning quantity in its own right.
Section 33-3

Turbine and Governor Models

The generator–load block responds to \(\Delta P_m\), the mechanical power at the turbine shaft. That power does not change the instant a valve moves. Steam admitted to a turbine must fill the steam chest and the inlet piping before it does work on the blades, and this storage is well represented by a single lag.

Non-reheat turbine
\[ G_T(s) = \frac{\Delta P_m(s)}{\Delta P_v(s)} = \frac{1}{1+sT_T}, \qquad T_T \approx 0.2\text{–}0.5\ \text{s} \]

where \(\Delta P_v\) is the change in valve or gate position expressed in power units. A reheat unit has a second, much slower storage in the reheater, and a fraction \(K_r\) of the power appearing promptly in the high-pressure cylinder:

Reheat turbine
\[ G_T(s) = \frac{1 + s K_r T_r}{(1+sT_T)(1+sT_r)}, \qquad T_r \approx 5\text{–}10\ \text{s},\quad K_r \approx 0.3 \]

A hydro turbine is different in kind. Accelerating the water column in the penstock takes time, and closing the gate momentarily raises the pressure and hence the power before the flow falls — a non-minimum-phase behaviour represented by a right-half-plane zero, \(G_T(s) = (1-sT_w)/(1+0.5\,sT_w)\) with a water starting time \(T_w\) of about a second. Hydro governors must therefore be detuned with transient droop compensation, and hydro plant is slower on primary response than its reputation suggests.

Now the governor. Its job is to sense speed and move the valve. In the classical Watt arrangement, flyweights driven from the turbine shaft rise as speed rises and lower a sleeve; the sleeve works a pilot valve that admits oil to a hydraulic servomotor, and the servomotor moves the steam valve. Modern units replace the flyweights with an electronic speed transducer and the linkage with a controller, but the transfer function is the same. The oil servomotor integrates flow into position, and with its own position feedback it behaves as a lag:

Governor and hydraulic amplifier
\[ G_g(s) = \frac{\Delta P_v(s)}{\Delta P_g(s)} = \frac{1}{1+sT_g}, \qquad T_g \approx 0.08\text{–}0.3\ \text{s} \]

The signal \(\Delta P_g\) that drives it is the commanded power change, and constructing it is the interesting part.

Section 33-4

Droop and the Speed Regulation R

Imagine a governor that simply held speed constant: any deviation, however small, drives the valve until the deviation vanishes. This is called isochronous control, and on a single machine feeding an isolated load it works perfectly. Put two such machines in parallel and it fails immediately. Both are trying to enforce their own idea of exactly \(50.000\) Hz. Whichever setting is a hair higher drives its unit to full load while the other backs off to zero, and the load split is decided by measurement error rather than by intention. Two integrators cannot share one controlled variable.

The remedy is to give each governor a proportional characteristic instead: let the commanded power depend on the frequency error rather than its integral. The unit's steady-state characteristic then becomes a straight line of negative slope on a plot of frequency against output power,

Governor characteristic
\[ \Delta P_g = \Delta P_{\text{ref}} - \frac{1}{R}\,\Delta f \]

\(\Delta P_{\text{ref}}\) is the setting of the speed changer — the operator's or AGC's command, which slides the whole characteristic up and down without changing its slope. The constant \(R\) is the speed regulation or droop, with units of hertz per megawatt. Its reciprocal \(1/R\) is the megawatts of primary response the unit contributes per hertz of frequency depression.

P (MW) → f (Hz) f₀ original setting after ΔP_ref P₀ P₀+ΔP ΔP Δf slope = −R = Δf / ΔP f at no load f at full load
The droop characteristic — and the parallel shift produced by the speed changer

In practice \(R\) is quoted as a percentage, and the definition must be read carefully. A droop of \(R\%\) means that a change of \(R\%\) in frequency drives the unit from no load to full load. Converting to engineering units,

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Droop in percent and in MW/Hz
\[ R\% = \frac{\Delta f / f_0}{\Delta P / P_{\text{rated}}}\times 100, \qquad\qquad \frac{1}{R} = \frac{P_{\text{rated}}}{(R\%/100)\,f_0}\ \ \text{MW/Hz} \]

A \(500\) MW unit with \(4\%\) droop on a \(50\) Hz system responds with \(1/R = 500/(0.04\times 50) = 250\) MW/Hz. Typical settings are \(3\%\) to \(6\%\); a smaller droop means a stiffer, more responsive unit, and a droop of zero is the isochronous case that cannot share load.

Droop solves the load-sharing problem completely. Because every unit sees the same \(\Delta f\), and each contributes \(-\Delta f/R_i\), the increment is divided among the units in proportion to \(1/R_i\). If all units are set to the same percentage droop, that proportion is the ratio of their ratings — each machine picks up the same fraction of its own capacity, which is exactly the fair and safe outcome. Choosing unequal percentage droops deliberately biases the sharing, which is sometimes done to keep a particular unit on base load.

Droop is the price of cooperation. The proportional characteristic makes many governors coexist, but a proportional controller cannot eliminate its own error: if \(\Delta f\) returned to zero, the term \(-\Delta f/R\) would vanish and the extra generation would disappear with it. A permanent frequency offset is therefore not a defect of the primary loop but a structural necessity of it. Removing that offset is a separate job, done by a separate loop, on a separate time scale — Section 33-7.
Section 33-5

The Isolated Area: Block Diagram and Steady-State Frequency Drop

Assemble the three blocks. The speed changer setting and the droop feedback form the command \(\Delta P_g\); the governor lag turns that into valve position; the turbine lag turns valve position into mechanical power; the mechanical power meets the disturbance \(\Delta P_L\) at a summing point; and the generator–load block converts the net imbalance into a frequency deviation, which is fed back through \(1/R\).

ΔP_ref(s) Σ governor 1/(1+sT_g) turbine 1/(1+sT_T) Σ ΔP_L(s) generator–load K_p/(1+sT_p) Δf(s) 1 / R primary loop (droop) K_I / s supplementary loop (AGC)
The load frequency control loop of a single area — primary droop feedback and the supplementary integral loop

Reading the diagram gives the closed-loop response to a load disturbance. With the supplementary loop open and \(\Delta P_{\text{ref}} = 0\), the forward path from the summing junction to \(\Delta f\) is \(G_g G_T G_p\), and the feedback path is \(1/R\):

Closed-loop response of one area to a load change
\[ \Delta F(s) = \frac{-\,G_p(s)}{1 + \dfrac{1}{R}\,G_g(s)G_T(s)G_p(s)}\;\Delta P_L(s) \]
\[ = \frac{-\,\dfrac{K_p}{1+sT_p}}{1 + \dfrac{K_p}{R\,(1+sT_g)(1+sT_T)(1+sT_p)}}\;\Delta P_L(s) \]

For a step load increase \(\Delta P_L(s) = \Delta P_L/s\), the final-value theorem gives the steady-state frequency deviation. Every lag contributes unity at \(s=0\), so the whole dynamic structure collapses:

Final value
\[ \Delta f_{ss} = \lim_{s\to 0} s\,\Delta F(s) = \frac{-K_p\,\Delta P_L}{1 + K_p/R} = \frac{-\Delta P_L}{1/K_p + 1/R} = \frac{-\Delta P_L}{D + 1/R} \]

The same result follows in two lines from the steady-state power balance, and it is worth seeing it that way because the physics is then transparent. In the new steady state the frequency has settled, so the rotors are neither gaining nor losing kinetic energy and \(\Delta P_m = \Delta P_e\). The governors have supplied \(\Delta P_m = -\Delta f/R\); the load has fallen by its own frequency sensitivity, giving relief \(-D\,\Delta f\); together these must cover the disturbance \(\Delta P_L\).

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The central result of primary control
\[ \Delta f_{ss} = \frac{-\Delta P_L}{\beta}, \qquad \boxed{\ \beta = D + \frac{1}{R}\ } \]

\(\beta\) is the area frequency response characteristic (AFRC), also called the area stiffness or the composite regulation, measured in MW/Hz. It is the sum of the governor response of every unit on droop control and the natural frequency sensitivity of the load. In a real interconnection \(1/R\) dominates \(D\) by one to two orders of magnitude, but \(D\) is retained because it is what remains when every unit is at its limit and no governor can respond further.

When an area contains \(n\) units on governor control, the individual regulations combine in parallel because each unit sees the same \(\Delta f\):

Composite regulation of an area
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}, \qquad \beta = D + \sum_{i=1}^{n}\frac{1}{R_i} \]

A unit that is at its maximum output, or that has been taken off governor control to run at a fixed setpoint, contributes nothing to this sum. This is why system operators track responsive spinning reserve rather than reserve alone: capacity that is spinning but blocked from responding does not appear in \(\beta\), and the frequency dip after a loss of generation is set by \(\beta\), not by the reserve on paper.

Section 33-6

Dynamics of the Primary Loop

The steady-state answer is the one examinations ask for, but the transient shape is what an operator actually sees on the frequency trace, and it is instructive to extract it from the same block diagram. The governor and turbine time constants are of the order of a few tenths of a second, while \(T_p\) is tens of seconds. Setting \(T_g = T_T = 0\) as a first approximation leaves a single lag in the loop:

First-order approximation of the primary loop
\[ \Delta F(s) = \frac{-K_p\,\Delta P_L/s}{1+sT_p+K_p/R} = \frac{-\Delta P_L}{s}\cdot\frac{K_p'}{1+sT_p'} \]
\[ K_p' = \frac{K_p}{1+K_p/R} = \frac{1}{\beta}, \qquad T_p' = \frac{T_p}{1+K_p/R} \]

so that

Time response to a step load increase
\[ \Delta f(t) = -\frac{\Delta P_L}{\beta}\left(1 - e^{-t/T_p'}\right) \]

Two features deserve comment. First, the governor loop has shortened the effective time constant by the factor \(1+K_p/R\), which is typically \(30\) to \(60\): a system that would have drifted for half a minute on load damping alone settles in a second or so. Second, the response in this approximation is a pure exponential with no overshoot, which is not what real traces show. Restoring \(T_g\) and \(T_T\) makes the characteristic polynomial cubic, and the two fast lags introduce enough phase lag for the frequency to undershoot and recover — a dip of perhaps twice the final offset, over five to ten seconds. Reheat units, with their extra \(5\)–\(10\) s lag, are slower still, and their contribution to the initial dip is limited to the high-pressure fraction \(K_r\).

The frequency nadir, not the settling point, sets the protection. Under-frequency load-shedding relays are armed at thresholds such as \(48.8\) Hz. What must clear those thresholds is the transient minimum, which depends on inertia (how fast the frequency falls initially), on \(\beta\) (where it will settle) and on the governor and turbine lags (how quickly the response arrives). Two systems with identical \(\beta\) can have very different nadirs if one is rich in fast-responding hydro and the other in reheat steam.
Section 33-7

Supplementary Control: Restoring the Frequency

Primary control leaves the system running at \(f_0 - \Delta P_L/\beta\). Left alone, successive load changes through the day would accumulate frequency offsets in whichever direction the load happened to move, and synchronous clocks would drift without bound. The remedy is to move the speed-changer settings — to slide the droop characteristics of Figure 33-1 bodily upward until the intersection with the new load falls back on \(f_0\).

How much should they move? From the steady-state balance with a non-zero reference change,

Steady state with a speed-changer command
\[ \Delta f_{ss} = \frac{\Delta P_{\text{ref}} - \Delta P_L}{\beta} \quad\Longrightarrow\quad \Delta f_{ss}=0 \ \text{ requires } \ \Delta P_{\text{ref}} = \Delta P_L \]

The reference must be advanced by the whole of the load increase. That is unsurprising: at nominal frequency the droop term contributes nothing and the load relief is zero, so the entire extra demand must come from the reference setting. What is less obvious is how to generate \(\Delta P_{\text{ref}}\) automatically, since \(\Delta P_L\) is never measured directly.

The answer is to let the frequency error itself generate the command, through an integrator. Take

Integral supplementary control
\[ \Delta P_{\text{ref}}(t) = -K_I\int_0^{t}\Delta f\,d\tau \qquad\Longleftrightarrow\qquad \Delta P_{\text{ref}}(s) = -\frac{K_I}{s}\,\Delta F(s) \]

The argument for why this works is short and completely general. Suppose the system reaches a steady state. Then every signal is constant, and in particular \(\Delta P_{\text{ref}}\) is constant. But \(\Delta P_{\text{ref}}\) is the output of an integrator whose input is \(\Delta f\); an integrator with a constant non-zero input produces a ramp, not a constant. The only constant input consistent with a constant output is zero. Hence \(\Delta f = 0\) in any steady state that exists at all. The integrator does not need to know \(\Delta P_L\); it simply keeps pushing until the error that drives it disappears, and at that moment \(\Delta P_{\text{ref}}\) is sitting at exactly \(\Delta P_L\).

The negative sign is essential and easy to get wrong. A load increase makes \(\Delta f\) negative; the integral of a negative quantity is negative; multiplying by \(-K_I\) with \(K_I>0\) makes \(\Delta P_{\text{ref}}\) positive, raising generation. Get the sign backwards and the loop drives the frequency away from nominal without limit.

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Why the loop must be integral, not proportional
A proportional supplementary controller \(\Delta P_{\text{ref}} = -K_c\,\Delta f\) merely adds \(K_c\) to \(\beta\), giving \(\Delta f_{ss} = -\Delta P_L/(\beta + K_c)\) — smaller, but never zero. Only an integrator has infinite d.c. gain and therefore zero steady-state error.

In the classical literature the two loops together are described as proportional-plus-integral control of frequency: the droop provides the proportional term, fast and distributed; the supplementary loop provides the integral term, slow and centralised. The split of duties is deliberate — the fast term must be proportional so that many units can share it, and the slow term must be integral so that the error is finally removed.

Adding the integrator raises the order of the loop by one and therefore raises a stability question. With the fast lags again neglected, the characteristic equation of the single area with both loops closed is obtained by substituting the supplementary command back into the block diagram:

Characteristic equation with integral control
\[ \Delta F(s) = \frac{-K_p\,\Delta P_L(s)}{(1+sT_p) + K_p\left(\dfrac{1}{R}+\dfrac{K_I}{s}\right)} = \frac{-s\,K_p\,\Delta P_L(s)}{T_p s^2 + \left(1+\dfrac{K_p}{R}\right)s + K_pK_I} \]
\[ \omega_n = \sqrt{\frac{K_pK_I}{T_p}}, \qquad 2\zeta\omega_n = \frac{1+K_p/R}{T_p} \]

The factor \(s\) in the numerator is the analytical statement that a step disturbance produces zero steady-state frequency error. The two parameters show the trade-off: raising \(K_I\) raises \(\omega_n\), which speeds up the restoration, but \(\zeta = (1+K_p/R)/(2T_p\omega_n)\) falls as \(\omega_n\) rises, so too large a gain makes the frequency hunt. In practice \(K_I\) is set well below the critically damped value, because the supplementary loop must be slower than the primary loop it is superimposed on, and slower still than the turbine's permitted rate of loading. Ten to thirty seconds for the bulk of the restoration is a realistic target; AGC cycles typically issue new setpoints every two to four seconds.

Section 33-8

Two Areas and the Tie Line

Real systems are not isolated. Neighbouring utilities interconnect because it lets them share reserve, exchange energy economically and ride through each other's disturbances. Interconnection also creates a new controlled quantity: the power flowing on the tie lines, which is contracted in advance and must be held to schedule.

Model two areas, each represented as in Section 33-5, joined by a tie line of reactance \(X_{12}\). From Chapter 26, the power transferred from area 1 to area 2 is

Tie-line power
\[ P_{12} = \frac{|V_1||V_2|}{X_{12}}\sin(\delta_1-\delta_2) \]

Load frequency control deals in small deviations about an operating point, so linearise. Holding the voltage magnitudes constant — legitimate because voltage control is a separate and faster loop, Chapter 34 — and writing \(\delta_i = \delta_i^0 + \Delta\delta_i\),

Linearised tie-line power and the synchronizing coefficient
\[ \Delta P_{12} = T_{12}\,(\Delta\delta_1 - \Delta\delta_2), \qquad T_{12} = \frac{|V_1||V_2|}{X_{12}}\cos\!\big(\delta_1^0-\delta_2^0\big) \]

\(T_{12}\) is the synchronizing power coefficient of the tie, in MW per radian — the same quantity that governed small-signal stability in Chapter 28. A strong tie (small \(X_{12}\)) has a large \(T_{12}\).

The block diagram needs \(\Delta P_{12}\) in terms of the frequency deviations it already carries, and angle is the integral of frequency. Since \(\delta\) is in electrical radians and \(f\) in hertz, \(d\delta/dt = 2\pi\,\Delta f\), so

Tie-line block
\[ \Delta\delta_i = 2\pi\int \Delta f_i\,dt \quad\Longrightarrow\quad \Delta P_{12}(s) = \frac{2\pi T_{12}}{s}\Big[\Delta F_1(s)-\Delta F_2(s)\Big] \]

The tie line is an integrator driven by the difference of the two area frequencies. That single fact explains most of the qualitative behaviour of interconnected systems. As long as the two areas are at different frequencies, the tie-line flow keeps changing; flow settles only when \(\Delta f_1 = \Delta f_2\), which is another way of saying that the interconnection is synchronous and there is only one steady-state frequency.

AREA 1 β₁ = D₁ + 1/R₁ H₁ , ΔP_L1 AREA 2 β₂ = D₂ + 1/R₂ H₂ , ΔP_L2 tie line 2πT₁₂ / s Δf₁ Δf₂ ΔP₁₂ = (2πT₁₂/s)(Δf₁ − Δf₂) +ΔP₁₂ −ΔP₁₂ Σ Σ B₁ Δf₁ B₂ Δf₂ ACE₁ = ΔP₁₂ + B₁Δf₁ ACE₂ = ΔP₂₁ + B₂Δf₂ each ACE drives −K_I/s into that area's speed changers
Two-area load frequency control — the tie line couples the areas, the ACE decouples the controllers

Consider what happens with primary control only — no supplementary loops — when a load step \(\Delta P_{L1}\) occurs in area 1. In the steady state both areas share one frequency, \(\Delta f_1=\Delta f_2=\Delta f\). Write the power balance of each area, taking \(\Delta P_{12}\) positive for flow out of area 1:

Steady-state balance of the two areas
\[ \text{Area 1:}\quad -\frac{\Delta f}{R_1} - \Delta P_{L1} - \Delta P_{12} = D_1\Delta f \]
\[ \text{Area 2:}\quad -\frac{\Delta f}{R_2} + \Delta P_{12} = D_2\Delta f \]

The second equation gives \(\Delta P_{12} = \beta_2\,\Delta f\) directly. Substituting into the first,

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Two areas on primary control alone
\[ \Delta f = \frac{-\Delta P_{L1}}{\beta_1+\beta_2}, \qquad \Delta P_{12} = \frac{-\beta_2\,\Delta P_{L1}}{\beta_1+\beta_2} \]

The frequency drop is governed by the combined stiffness of both areas, so interconnection immediately reduces the excursion — the benefit that motivates it. But \(\Delta P_{12}\) is negative: power flows into area 1 from area 2. Area 2 is supporting a disturbance it did not cause, its own generators have moved off their economic setpoints, and the tie flow no longer matches the contracted schedule.

Governor support across the tie is exactly what one wants for the first few seconds and exactly what one does not want after a minute. The supplementary loop must therefore be given a target that reflects both objectives: return the frequency to nominal and return the tie flow to schedule, with the correction made by the area that caused the imbalance.

Before leaving the tie line, one dynamic consequence is worth recording. Neglect governors and damping for a moment and let the two areas swing against each other through the tie. Combining the two swing equations with \(\Delta\ddot\delta = 2\pi\,d\Delta f/dt\) and eliminating gives, for \(\theta = \Delta\delta_1-\Delta\delta_2\),

The inter-area oscillation mode
\[ \ddot\theta + 2\pi f_0\,T_{12}\left(\frac{1}{2H_1}+\frac{1}{2H_2}\right)\theta = 0 \;\Longrightarrow\; \omega_{\text{osc}} = \sqrt{2\pi f_0 T_{12}\left(\frac{1}{2H_1}+\frac{1}{2H_2}\right)} \]

These inter-area modes lie between about \(0.1\) and \(0.8\) Hz — much slower than the local machine modes of Chapter 28, because the effective inertias are those of whole areas. They are lightly damped, and a supplementary controller tuned too aggressively will excite them, which is the practical reason AGC is deliberately slow.

Section 33-9

Area Control Error and the Frequency-Bias Setting

Each area needs one scalar error signal to drive its integrator. Two candidates are available to it locally: its own frequency deviation \(\Delta f_i\), and the deviation of its net tie-line interchange from schedule, \(\Delta P_{\text{tie},i}\). Neither alone is adequate. Frequency alone would make both areas respond to any disturbance anywhere, so the innocent area would keep generating for its neighbour indefinitely. Tie flow alone would leave the frequency permanently offset if both areas happened to be in balance with each other but jointly short of generation.

The classical solution, due to Nathan Cohn, is to use a weighted sum called the area control error:

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Tie-line bias control
\[ \mathrm{ACE}_i = \Delta P_{\text{tie},i} + B_i\,\Delta f, \qquad \Delta P_{\text{ref},i} = -K_{I,i}\int \mathrm{ACE}_i\,dt \]

\(B_i\) is the frequency-bias setting of area \(i\), in MW/Hz, and is conventionally entered as a positive number with the sign convention that \(\Delta P_{\text{tie},i}\) is positive for export. The tie-line term restores the schedule; the bias term restores the frequency; the integrator guarantees that both are eventually satisfied exactly.

The equilibrium argument of Section 33-7 applies to each integrator separately. In any steady state, \(\mathrm{ACE}_1 = 0\) and \(\mathrm{ACE}_2 = 0\). Writing these out with \(\Delta P_{\text{tie},1} = \Delta P_{12}\) and \(\Delta P_{\text{tie},2} = -\Delta P_{12}\), and remembering that the interconnection has one frequency,

The only possible steady state
\[ \Delta P_{12} + B_1\Delta f = 0, \qquad -\Delta P_{12} + B_2\Delta f = 0 \]
\[ \text{adding:}\quad (B_1+B_2)\,\Delta f = 0 \;\Longrightarrow\; \Delta f = 0 \;\Longrightarrow\; \Delta P_{12}=0 \]

Both objectives are met for any positive bias settings. This is a strong and slightly surprising result: the final state does not depend on the value of \(B\) at all. What does depend on \(B\) is the path taken to get there, and in particular whether the innocent area's controller is disturbed on the way.

To see this, evaluate the two ACEs immediately after primary control has settled but before the supplementary loops have acted, using the results of Section 33-8 for a disturbance \(\Delta P_{L1}\) in area 1. Choose the biases equal to the area frequency response characteristics, \(B_i = \beta_i\):

ACE with ideal bias, disturbance in area 1
\[ \mathrm{ACE}_1 = \Delta P_{12} + \beta_1\Delta f = \frac{-\beta_2\Delta P_{L1}}{\beta_1+\beta_2} + \beta_1\!\left(\frac{-\Delta P_{L1}}{\beta_1+\beta_2}\right) = -\Delta P_{L1} \]
\[ \mathrm{ACE}_2 = -\Delta P_{12} + \beta_2\Delta f = \frac{+\beta_2\Delta P_{L1}}{\beta_1+\beta_2} - \frac{\beta_2\Delta P_{L1}}{\beta_1+\beta_2} = 0 \]
🔑
Non-interactive control
With \(B_i=\beta_i\), the ACE of the disturbed area equals the disturbance itself, and the ACE of every other area is exactly zero. Each area sees only its own imbalance.

This is the whole design intent of tie-line bias control. Area 2 lends governor support for the first seconds — that is the value of the interconnection — and then its controller, seeing zero error, withdraws that support as area 1's controller takes up the load. No communication between the areas is required: the bias setting encodes the neighbour's expected response.

If the bias is set too low, the disturbed area's ACE understates the imbalance and the neighbour's ACE becomes non-zero, so the neighbour begins correcting a disturbance it did not cause and the two controllers work against each other for a while. If the bias is set too high, the ACE overstates the imbalance and the area overshoots, drawing frequency past nominal. Neither error is unstable — the final state is still \(\Delta f = 0\), \(\Delta P_{12}=0\) — but both waste control effort and worsen the transient. Real biases are reviewed annually against measured system response, and grid codes commonly require \(B\) to be at least a stated fraction of the measured \(\beta\).

One further practical refinement matters. The ACE as written contains a time-error and inadvertent-interchange component: because \(\Delta f\) is only driven to zero, not its integral, synchronous clocks accumulate a small offset. Systems therefore add a slow correction term that runs the frequency a few millihertz off nominal for a few hours to pay back the accumulated time error, and a similar term to pay back accumulated inadvertent energy interchange on the tie.

Finally, the supplementary command must be split among the units of the area. This is where Chapter 31 returns. The economic dispatch calculation, run every few minutes, produces not only the optimal outputs but their sensitivity to a change in total demand — the participation factors \(\alpha_i\), with \(\sum_i \alpha_i = 1\), given by

Participation factors from the dispatch of Chapter 31
\[ \alpha_i = \frac{\partial P_i}{\partial P_D} = \frac{1/\big(d^2C_i/dP_i^2\big)}{\displaystyle\sum_j 1/\big(d^2C_j/dP_j^2\big)}, \qquad \Delta P_{\text{ref},i} = \alpha_i\,\Delta P_{\text{ref}} \]

Two allocations therefore operate in sequence after every disturbance. In the first seconds the increment is divided by droop, in proportion to \(1/R_i\) — a purely technical split that ignores cost entirely. Over the following minutes AGC re-divides it in proportion to \(\alpha_i\) — an economic split that ignores droop entirely. The final state is the economic one; the transient state is the physical one.

LayerTime scaleActs onAchievesAllocation rule
Inertial response0–2 sRotor kinetic energyLimits RoCoFIn proportion to \(H_i\)
Primary (governor droop)2–30 sValve positionArrests the excursion; \(\Delta f = -\Delta P_L/\beta\)In proportion to \(1/R_i\)
Secondary (AGC)30 s – 15 minSpeed-changer setting\(\Delta f \to 0\), tie flow \(\to\) scheduleParticipation factors \(\alpha_i\)
Tertiary (re-dispatch)15 min – hoursUnit commitment and dispatchRestores reserve and economyChapters 31 and 32
Section 33-10

Worked Examples

1 Load sharing between two units on droop

Problem. Two generators run in parallel on a small isolated system at \(50\) Hz. Unit A is rated \(100\) MW with \(4\%\) droop and is carrying \(50\) MW; unit B is rated \(200\) MW with \(5\%\) droop and is carrying \(100\) MW. The load increases by \(30\) MW. Neglect load damping. Find the new frequency and the new output of each unit.

Solution. Convert each droop to megawatts per hertz using the definition of Section 33-4.

Regulation of each unit
\[ \frac{1}{R_A}=\frac{100}{0.04\times 50}=\frac{100}{2}=50\ \text{MW/Hz}, \qquad \frac{1}{R_B}=\frac{200}{0.05\times 50}=\frac{200}{2.5}=80\ \text{MW/Hz} \]

Both units see the same frequency, so their regulations add:

Composite response and frequency
\[ \frac{1}{R_{\text{eq}}}=50+80=130\ \text{MW/Hz}, \qquad \Delta f = \frac{-30}{130} = -0.2308\ \text{Hz} \]
\[ f = 50 - 0.2308 = 49.769\ \text{Hz} \]

Each unit's increment follows from its own droop line:

New outputs
\[ \Delta P_A = -\frac{\Delta f}{R_A} = 50\times 0.2308 = 11.54\ \text{MW} \;\Rightarrow\; P_A = 61.54\ \text{MW} \]
\[ \Delta P_B = -\frac{\Delta f}{R_B} = 80\times 0.2308 = 18.46\ \text{MW} \;\Rightarrow\; P_B = 118.46\ \text{MW} \]

The increments sum to \(30.00\) MW, as they must. Note the split: A takes \(38.5\%\) and B \(61.5\%\), whereas their ratings are in the ratio \(33.3 : 66.7\). Because A has the stiffer droop it is loaded slightly harder than its share of capacity — the direct consequence of unequal percentage settings.

2 Steady-state frequency drop of an isolated area

Problem. A control area has a total installed capacity of \(2000\) MW, all of it on governor control with an average droop of \(4\%\). It is supplying \(1000\) MW at \(50\) Hz. The load is such that a \(1\%\) change in frequency produces a \(1\%\) change in load. A block of \(50\) MW of load is switched on. Find the new steady-state frequency, and the split of the \(50\) MW between governor response and load relief.

Solution. The governor response of the whole area, from the droop definition applied to the total rating:

Area regulation
\[ \frac{1}{R}=\frac{2000}{0.04\times 50}=\frac{2000}{2}=1000\ \text{MW/Hz} \]

The damping constant is read from the load's own sensitivity: \(1\%\) of \(1000\) MW is \(10\) MW, and \(1\%\) of \(50\) Hz is \(0.5\) Hz, so

Load damping and area stiffness
\[ D = \frac{10\ \text{MW}}{0.5\ \text{Hz}} = 20\ \text{MW/Hz}, \qquad \beta = D+\frac{1}{R} = 20+1000 = 1020\ \text{MW/Hz} \]
Frequency deviation
\[ \Delta f = \frac{-\Delta P_L}{\beta} = \frac{-50}{1020} = -0.04902\ \text{Hz}, \qquad f = 49.951\ \text{Hz} \]

The two contributions to covering the \(50\) MW:

Where the 50 MW comes from
\[ \Delta P_m = -\frac{\Delta f}{R} = 1000\times 0.04902 = 49.02\ \text{MW (extra generation)} \]
\[ -D\,\Delta f = 20\times 0.04902 = 0.98\ \text{MW (load relief)} \]
\[ 49.02 + 0.98 = 50.00\ \text{MW}\ \checkmark \]

Load damping contributes under \(2\%\) of the correction here, which is typical. Had the disturbance been much larger — a \(500\) MW generator tripping, say, with no reserve left on the remaining units — the governor term would saturate and \(D\) alone would have to arrest the fall, giving \(\Delta f = -500/20 = -25\) Hz. That is of course not a real number; long before it is reached, under-frequency load shedding removes enough demand to restore a balance the generators cannot.

3 Restoring nominal frequency, and who supplies the correction

Problem. An isolated area contains three units on governor control: G1, \(500\) MW at \(4\%\) droop; G2, \(300\) MW at \(5\%\); G3, \(200\) MW at \(6\%\). Load damping is \(15\) MW/Hz and the system is at \(50\) Hz. A \(60\) MW load increase occurs. (a) Find the frequency after primary control settles and the output change of each unit. (b) AGC then restores \(50\) Hz, distributing the correction with participation factors \(0.5, 0.3, 0.2\). Find the final change in each unit's output and the required speed-changer commands.

Solution (a). Regulations in MW/Hz:

Unit regulations
\[ \frac{1}{R_1}=\frac{500}{0.04\times50}=250, \quad \frac{1}{R_2}=\frac{300}{0.05\times50}=120, \quad \frac{1}{R_3}=\frac{200}{0.06\times50}=66.67\ \text{MW/Hz} \]
\[ \sum \frac{1}{R_i} = 436.67, \qquad \beta = 15 + 436.67 = 451.67\ \text{MW/Hz} \]
Frequency and primary allocation
\[ \Delta f = \frac{-60}{451.67} = -0.13285\ \text{Hz}, \qquad f = 49.867\ \text{Hz} \]
\[ \Delta P_1 = 250(0.13285)=33.21,\quad \Delta P_2 = 120(0.13285)=15.94,\quad \Delta P_3 = 66.67(0.13285)=8.86\ \text{MW} \]

These sum to \(58.01\) MW; the load relief \(15\times0.13285 = 1.99\) MW makes up the remaining amount, and \(58.01+1.99=60.00\) MW.

Solution (b). With \(\Delta f\) driven back to zero the droop terms all vanish and the load relief disappears, so the units must supply the entire \(60\) MW from their reference settings:

Final allocation by participation factor
\[ \Delta P_1 = 0.5\times 60 = 30\ \text{MW},\quad \Delta P_2 = 0.3\times 60 = 18\ \text{MW},\quad \Delta P_3 = 0.2\times 60 = 12\ \text{MW} \]

Because the final frequency deviation is zero, \(\Delta P_{\text{ref},i}=\Delta P_i\) for each unit: the speed changers must be advanced by exactly \(30\), \(18\) and \(12\) MW. Compare the two allocations — G1 was carrying \(33.21\) MW under droop and ends at \(30\) MW; G3 was carrying \(8.86\) MW and ends at \(12\) MW. The units are unloaded and reloaded against each other while the total stays at \(60\) MW, because primary control divides by droop and secondary control divides by cost.

4 Choosing the integral gain

Problem. For the area of Example 2, take \(H = 5\) s on the \(2000\) MVA base. Working in per unit on that base with frequency in hertz, find \(K_p\) and \(T_p\); then find the integral gain \(K_I\) that gives critical damping with the governor and turbine lags neglected, and the peak frequency dip for the \(50\) MW step.

Solution. Convert the area constants to per unit on \(2000\) MW. The damping \(D = 20\) MW/Hz becomes \(0.01\) pu MW/Hz, and \(1/R = 1000\) MW/Hz becomes \(0.5\) pu MW/Hz, so \(R = 2\) Hz per pu MW. Then

Generator–load block
\[ K_p = \frac{1}{D} = \frac{1}{0.01} = 100\ \text{Hz/pu MW}, \qquad T_p = \frac{2H}{f_0 D} = \frac{2\times 5}{50\times 0.01} = 20\ \text{s} \]

From Section 33-7, with \(K_p/R = 100\times 0.5 = 50\):

Second-order parameters
\[ 2\zeta\omega_n = \frac{1+K_p/R}{T_p} = \frac{51}{20} = 2.55\ \text{s}^{-1}, \qquad \omega_n^2 = \frac{K_pK_I}{T_p} = 5K_I \]
\[ \zeta = 1 \;\Rightarrow\; \omega_n = 1.275\ \text{rad/s} \;\Rightarrow\; K_I = \frac{\omega_n^2}{5} = \frac{1.6256}{5} = 0.325 \]

With \(\zeta=1\) the response to \(\Delta P_L = 50\ \text{MW} = 0.025\) pu is the critically damped form

Frequency transient
\[ \Delta f(t) = -\frac{K_p\,\Delta P_L}{T_p}\;t\,e^{-\omega_n t} = -0.125\,t\,e^{-1.275t}\ \text{Hz} \]
\[ \frac{d(\Delta f)}{dt}=0 \;\Rightarrow\; t_{\max}=\frac{1}{\omega_n}=0.784\ \text{s}, \qquad \Delta f_{\max} = -0.125(0.784)e^{-1} = -0.0361\ \text{Hz} \]

The frequency dips to \(49.964\) Hz and returns to \(50.000\) Hz, against the permanent \(49.951\) Hz that droop alone would have left (Example 2). A gain of \(K_I = 0.325\) is aggressive for real plant — it would demand the full \(50\) MW within a couple of seconds, faster than a steam turbine may be loaded and fast enough to excite the inter-area modes of Section 33-8 in an interconnected system. A practical setting is several times smaller, trading a slower return for a gentler demand on the machines.

5 A two-area interconnection

Problem. Two areas are interconnected. Area 1: \(2000\) MW capacity, \(5\%\) droop, \(D_1 = 20\) MW/Hz. Area 2: \(4000\) MW capacity, \(4\%\) droop, \(D_2 = 30\) MW/Hz. The system is at \(50\) Hz with the tie line carrying its scheduled flow. A \(100\) MW load increase occurs in area 1. With supplementary control disabled, find the steady-state frequency, the change in tie-line flow, and the contribution of each area. Then, with \(V_1=V_2=1.0\) pu, \(X_{12}=2.0\) pu on a \(2000\) MVA base, \(\delta_1^0-\delta_2^0 = 10^\circ\), \(H_1 = 5\) s and \(H_2 = 10\) s (both on that base), estimate the inter-area oscillation frequency.

Solution. Area frequency response characteristics:

The two stiffnesses
\[ \frac{1}{R_1}=\frac{2000}{0.05\times50}=800, \qquad \beta_1 = 800+20 = 820\ \text{MW/Hz} \]
\[ \frac{1}{R_2}=\frac{4000}{0.04\times50}=2000, \qquad \beta_2 = 2000+30 = 2030\ \text{MW/Hz} \]
Frequency and tie flow
\[ \Delta f = \frac{-100}{820+2030} = \frac{-100}{2850} = -0.03509\ \text{Hz}, \qquad f = 49.965\ \text{Hz} \]
\[ \Delta P_{12} = \beta_2\,\Delta f = 2030\times(-0.03509) = -71.23\ \text{MW} \]

The negative sign, with \(\Delta P_{12}\) defined as flow from area 1 to area 2, means \(71.23\) MW now flows the other way: area 2 is supporting area 1. Check the books of each area:

Verification
\[ \text{Area 1: } 800(0.03509)+20(0.03509) = 28.07+0.70 = 28.77\ \text{MW};\quad 28.77+71.23 = 100\ \checkmark \]
\[ \text{Area 2: } 2000(0.03509)+30(0.03509) = 70.18+1.05 = 71.23\ \text{MW exported}\ \checkmark \]

Notice that area 1, where the disturbance occurred, supplies only \(28\%\) of it. The larger and stiffer neighbour carries the rest, which is precisely why the frequency fell by only \(0.035\) Hz instead of the \(100/820 = 0.122\) Hz it would have fallen had area 1 stood alone.

For the oscillation, first the synchronizing coefficient:

Synchronizing coefficient and inter-area mode
\[ T_{12} = \frac{|V_1||V_2|}{X_{12}}\cos(\delta_1^0-\delta_2^0) = \frac{1.0\times1.0}{2.0}\cos 10^\circ = 0.4924\ \text{pu MW/rad} \]
\[ \omega_{\text{osc}} = \sqrt{2\pi(50)(0.4924)\left(\frac{1}{10}+\frac{1}{20}\right)} = \sqrt{314.16\times0.4924\times0.15} = 4.82\ \text{rad/s} \]
\[ f_{\text{osc}} = \frac{4.82}{2\pi} = 0.77\ \text{Hz} \]

A \(0.77\) Hz mode is a classic inter-area oscillation. Any supplementary controller must be slow compared with it, which puts a hard ceiling on \(K_I\).

6 Area control error, and what a wrong bias does

Problem. For the two-area system of Example 5, a \(200\) MW load increase now occurs in area 2. (a) Find \(\Delta f\) and \(\Delta P_{12}\) after primary control settles. (b) Compute \(\mathrm{ACE}_1\) and \(\mathrm{ACE}_2\) with ideal biases \(B_i=\beta_i\). (c) Repeat with \(B_1\) mistakenly set to \(410\) MW/Hz, and comment.

Solution (a). By symmetry with Section 33-8, a disturbance in area 2 gives

Frequency and tie flow
\[ \Delta f = \frac{-\Delta P_{L2}}{\beta_1+\beta_2}=\frac{-200}{2850}= -0.07018\ \text{Hz}, \qquad f = 49.930\ \text{Hz} \]
\[ \Delta P_{12} = -\beta_1\,\Delta f = -820\times(-0.07018)= +57.54\ \text{MW} \]

Now the flow is positive: area 1 exports \(57.54\) MW to help its neighbour. Area 2 itself supplies \(2030\times0.07018 = 142.46\) MW, and \(57.54+142.46 = 200\) MW.

Solution (b). With \(B_1=820\) and \(B_2=2030\) MW/Hz:

Ideal bias
\[ \mathrm{ACE}_1 = \Delta P_{12}+B_1\Delta f = 57.54 + 820(-0.07018) = 57.54-57.54 = 0 \]
\[ \mathrm{ACE}_2 = -\Delta P_{12}+B_2\Delta f = -57.54 + 2030(-0.07018) = -57.54-142.46 = -200\ \text{MW} \]

Area 2's controller sees exactly its own \(200\) MW deficit and nothing else; area 1's controller sees zero and does not move its speed changers at all, even though its generators are at that moment producing an extra \(57.54\) MW. That extra output is withdrawn automatically as area 2's integrator raises its own references and the frequency returns to \(50\) Hz.

Solution (c). With \(B_1 = 410\) MW/Hz, half its correct value:

Under-set bias
\[ \mathrm{ACE}_1 = 57.54 + 410(-0.07018) = 57.54 - 28.77 = +28.77\ \text{MW} \]

A positive ACE is read as over-generation, so area 1's controller now reduces its own references by up to \(28.77\) MW while area 2 is still trying to make up a \(200\) MW shortfall. The frequency dips further before recovering, and both sets of turbines are moved further than necessary. The end state is unaffected — Section 33-9 showed that \(\Delta f=0\) and \(\Delta P_{12}=0\) is the only equilibrium for any positive biases — but the interconnection has spent extra regulating energy to reach it. This is the practical reason grid codes specify a minimum bias and require it to be verified against measured frequency-response data.

Review

Chapter Summary

Imbalance moves frequency

\(\dfrac{2H}{f_0}\dfrac{d\Delta f}{dt}=\Delta P_m-\Delta P_L-D\Delta f\); the generator–load block is \(K_p/(1+sT_p)\) with \(K_p=1/D\), \(T_p=2H/(f_0D)\).

Droop

\(\Delta P_g=\Delta P_{\text{ref}}-\Delta f/R\). Proportional action lets many governors share one frequency; isochronous action cannot.

Area stiffness

\(\beta=D+1/R\) in MW/Hz, with \(1/R_{\text{eq}}=\sum 1/R_i\) over the responsive units.

Primary result

\(\Delta f_{ss}=-\Delta P_L/\beta\) — a permanent offset, structurally unavoidable with proportional control.

Integral action

\(\Delta P_{\text{ref}}=-K_I\!\int\!\Delta f\,dt\) forces \(\Delta f=0\), because a constant output demands zero input to an integrator.

Tie line

\(\Delta P_{12}(s)=\dfrac{2\pi T_{12}}{s}\big(\Delta F_1-\Delta F_2\big)\) with \(T_{12}=\dfrac{V_1V_2}{X_{12}}\cos(\delta_1^0-\delta_2^0)\).

Two areas

Primary control alone gives \(\Delta f=-\Delta P_{L1}/(\beta_1+\beta_2)\) and \(\Delta P_{12}=-\beta_2\Delta P_{L1}/(\beta_1+\beta_2)\).

ACE and bias

\(\mathrm{ACE}_i=\Delta P_{\text{tie},i}+B_i\Delta f\); with \(B_i=\beta_i\) the undisturbed area's ACE is exactly zero.

Practice

Problems

Take \(f_0 = 50\) Hz unless stated otherwise. Convert every droop to MW/Hz before doing anything else — most errors in this topic are unit errors, not conceptual ones.

  1. A \(250\) MW unit has \(5\%\) droop and is running at \(150\) MW at \(50\) Hz. The system frequency falls to \(49.7\) Hz. Find the unit's new output, assuming the speed-changer setting is unchanged.
  2. Three units of \(400\), \(300\) and \(300\) MW, all with \(4\%\) droop, share an isolated system. The load damping is \(12\) MW/Hz. Find the frequency after a \(45\) MW load increase and the output change of each unit.
  3. An area with \(\beta = 1500\) MW/Hz loses a \(250\) MW generator. Find the steady-state frequency. If under-frequency load shedding is armed at \(49.5\) Hz, does it operate? What if only \(60\%\) of the units were on governor control?
  4. A control area has \(H = 4\) s on a \(3000\) MVA base, \(D = 0.008\) pu MW/Hz and an equivalent droop of \(3\) Hz per pu MW. Find \(K_p\), \(T_p\) and the time constant of the primary loop with the governor and turbine lags neglected.
  5. For the area of Problem 4, find the integral gain that gives a damping ratio of \(0.7\), and the resulting undamped natural frequency. Would you expect this setting to be usable if the area is interconnected through a tie with a \(0.5\) Hz inter-area mode?
  6. Two areas have \(\beta_1 = 900\) MW/Hz and \(\beta_2 = 1600\) MW/Hz. A \(120\) MW load increase occurs in area 2 and supplementary control is out of service. Find \(\Delta f\), the tie-line flow change with sign, and the megawatts each area supplies.
  7. For the system of Problem 6, compute \(\mathrm{ACE}_1\) and \(\mathrm{ACE}_2\) for ideal bias settings, and again for \(B_1 = 1400\) MW/Hz. Describe what each controller does in the two cases.
  8. A tie line of \(X_{12}=0.4\) pu joins two areas operating with \(V_1=V_2=1.0\) pu and an angle difference of \(20^\circ\). Find \(T_{12}\). If the areas have \(H_1 = 6\) s and \(H_2 = 6\) s on the same base, find the inter-area oscillation frequency, and state what happens to it if a second identical tie line is built in parallel.
Tip: almost every question in this chapter is answered by one of three lines — \(\Delta f = -\Delta P_L/\beta\) for a lone area, \(\Delta f = -\Delta P_L/(\beta_1+\beta_2)\) with \(\Delta P_{12}=\beta_2\Delta f\) for two areas on droop, and \(\mathrm{ACE}_i = \Delta P_{\text{tie},i}+B_i\Delta f\) once supplementary control is in service. Decide first which of the three situations the question describes, then which quantities are given in MW/Hz and which in percent, and the arithmetic follows in two steps.