Load Frequency Control and AGC

Convert the droop. A droop of 5% means that 5% of nominal frequency — that is, 2.5 Hz — moves the unit across its whole range of 250 MW:

Read it as a sensitivity: this machine offers 100 MW for every hertz the system frequency is depressed, until it runs out of capacity.

The new output. With \(\Delta P_{\text{ref}} = 0\) the governor characteristic is simply \(\Delta P = -\Delta f/R\):

Note that the unit does not ask why the frequency fell, nor where. It responds to the local measurement of a global quantity, which is what allows primary control to work without any communication.

The ends of the characteristic. Full load requires a further 100 MW above the present 150 MW, and no load requires 150 MW less:

The two ends are 2.50 Hz apart, which is the 5% of 50 Hz that the definition promised. The 150 MW operating point sits 60% of the way along, so the characteristic is not symmetric about 50 Hz — where it sits is decided entirely by the speed-changer setting.

The picture. Sliding the speed changer moves this line bodily, parallel to itself, without changing its slope:

Regulations, one unit at a time. Each uses its own rating with the common 4% and 50 Hz:

They add, because every unit sees the same frequency. That is the whole justification — the three governors are not in series with anything, they are three parallel responses to one measurement:

The frequency. One line:

Under 0.09 Hz for a 45 MW step, on a system of 1000 MW of installed capacity. That is what a stiff area looks like.

The allocation. Each unit moves along its own droop line by \(-\Delta f/R_i\):

The split 17.58 : 13.18 : 13.18 is exactly 400 : 300 : 300. Equal percentage droops produce sharing in proportion to rating, so every machine picks up the same 4.4% of its own capacity — the fair and safe outcome, and the reason grid codes specify a common droop band.

Load relief closes the books. The load itself is smaller at the lower frequency:

Load damping supplies 2.3% of the correction here. It is small, but it is what remains when every governor has run out of reserve — Problem 5 shows what that looks like.

Regulations. Different ratings and different percentages, so both must be converted separately:

The small machine is nearly as stiff as the large one. Its 3% setting compensates almost entirely for its being less than half the size.

Frequency and shares, with \(D = 0\) so that \(\beta = 216.67\) MW/Hz:

The comparison, which is the point of the problem:

Unit A picks up 12.3% of its own rating while unit B picks up 6.2% of its. The stiffer machine is worked more than twice as hard relative to its size, and would reach its ceiling first on a large disturbance.

The inversion. Proportional sharing requires the regulations themselves to be in the ratio of the ratings:

Equal to A's setting, as it had to be. The general statement is worth remembering: sharing is proportional to rating if and only if the percentage droops are equal, whatever the ratings are. Unequal percentages are used deliberately — a base-load nuclear unit given 8% droop moves less, an aeroderivative gas turbine given 3% moves more.

The two characteristics drawn together make the mechanism visible: one common frequency drop, two different horizontal excursions set by two different slopes.

Invert the central result. A generator trip is a load increase as far as the surviving system is concerned, so \(\Delta P_L = +250\) MW:

This is the only way \(\beta\) is ever really known. It is not computed from nameplates; it is measured from events, and every large interconnection maintains a running record of them.

What the data predicts, if the belief about the plant were true:

The system is half as stiff as the paperwork says. A factor of two is far too large for the damping constant to explain, so the discrepancy must be in the governor term.

Extract the governor response actually delivered. Damping is not in doubt:

Say it in megawatts, which is what an operator can act on. The capacity that behaved as 4%-droop plant is

So roughly 3100 MW of spinning plant contributed nothing. The usual reasons are all mundane and all common: units already at maximum output, units placed on a fixed megawatt setpoint for commercial reasons, units whose governor valves were fully open at the moment of the trip, and units whose governor deadband exceeded 0.17 Hz.

Why the distinction matters. Reserve on paper is not \(\beta\). Capacity that is spinning but blocked from responding appears in the reserve report and not in the frequency response, so an area can be fully reserved and still fall twice as far as expected. The quantity that must be tracked is responsive reserve, and this event has just measured it.

Turn the percentages into a slope. One per cent of frequency is 0.5 Hz; 1.5% of the load is 60 MW:

Positive, and it must be: motors driving fans and pumps slow with the system and demand less. It is a helpful term, and it is the only term that survives when the governors cannot move.

Area stiffness.

The settled frequency after losing 400 MW:

Where the 400 MW comes from:

Load relief supplies 5.7% here — larger than in Problem 2, because this area's damping is a larger fraction of its stiffness. In a real interconnection the fraction is usually between 2% and 8%.

The failure case. If every governor valve is already wide open, \(1/R\) drops out of \(\beta\) entirely and only the load's own sensitivity is left:

Which is not a number any system reaches, because long before it does, under-frequency load shedding removes demand that the generators cannot supply. Working backwards, damping alone can arrest the fall at a 48.8 Hz relay setting only for a disturbance of \(120 \times 1.2 = 144\) MW; anything larger sheds load.

The design conclusion. \(\beta\) falls by a factor of nearly eighteen when the governor term vanishes, so the whole of an area's frequency performance rests on keeping some plant off its ceiling. That is why responsive spinning reserve is scheduled explicitly and why an area running every unit flat out at peak is running without primary control, however much capacity is synchronised.

The block parameters follow directly from their definitions:

Both parameters have plain meanings. \(K_p = 1/D\) says that a permanent 1 pu MW surplus, left entirely to the load's own sensitivity, would raise the frequency until the load had grown by that same amount. \(T_p\) is the ratio of stored kinetic energy to damping, and 20 s is a very slow block — which is precisely why a governor is needed.

The first instant after the trip. Set \(\Delta P_m = 0\) and \(\Delta f = 0\) in the swing equation: no valve has moved and no damping has developed, so the whole imbalance comes out of the rotors:

Note what does not appear: droop, damping, turbine time constants. The first derivative of frequency is decided by inertia and the size of the event, and by nothing else.

Time to a threshold on inertia alone, holding the initial slope:

A bound, not a prediction — the real trace bends upward as governors act, as Problem 7 shows. But it is the right bound for the protection engineer: it says how long the governors have.

Halve the inertia. Nothing else in the calculation changes:

Twice as fast, and half the time available. The settled frequency, however, is untouched: \(\Delta f_{ss} = -\Delta P_L/\beta\) contains no \(H\) at all. Inertia sets the speed of the excursion; \(\beta\) sets its destination.

Why RoCoF became a planning quantity. Rate-of-change-of-frequency relays fitted to embedded generation trip at settings of the order of 0.5 to 1.0 Hz/s. On the original system a 180 MW loss produces 0.375 Hz/s and nothing trips; on the low-inertia system the same event produces 0.75 Hz/s and a tranche of distributed generation may disconnect, enlarging the disturbance while it is still developing.

Stiffness. Work in per unit and convert at the end:

The closed-loop time constant. With the two fast lags removed, the loop reduces to a single lag whose time constant is divided by the loop gain:

The governor loop has cut the open-loop 20 s down to under half a second — a factor of 42.7. A system left to its own load damping would drift for the better part of a minute; with droop it is settled in about a second and a half.

Times off the exponential:

The consistency check, which is the useful part of this problem. Differentiate the response at \(t = 0\):

Exactly the RoCoF of Problem 6, computed there from \(H\) alone with no mention of \(R\) or \(D\).

Why it must agree. At \(t = 0^+\) the valve has not moved and \(\Delta f\) is still zero, so neither the droop term nor the damping term contributes anything, and the swing equation reduces to the inertial statement. Algebraically:

Both \(\beta\) and \(D\) cancel identically. Any closed-loop model that gets this wrong has an error in it, which makes the check worth two lines of anybody's time.

What the approximation costs. Restoring \(T_g\) and \(T_T\) makes the characteristic polynomial cubic, and the two fast lags supply enough phase lag for the frequency to undershoot its final value and recover — a dip of roughly twice the settled offset over five to ten seconds. The single-lag model gets the destination exactly right and the worst point badly wrong, which is the distinction Problem 15 turns on.

Close the loop. The supplementary command enters at the same summing point as the droop feedback, so the two feedback paths add:

The factor \(s\) in the numerator is the whole argument for integral action written as algebra: for a step \(\Delta P_L(s) = \Delta P_L/s\) that \(s\) cancels the pole at the origin, so the final-value theorem returns zero. The steady-state frequency error is gone for any \(K_I > 0\).

Standard second-order form. Dividing through by \(T_p\):

Only \(\omega_n\) contains \(K_I\); the damping term is fixed by the primary loop. So raising the integral gain speeds the restoration and lowers \(\zeta\) at the same time — one knob, two opposed effects.

The two gains. From \(\omega_n = 2.1333/(2\zeta)\) and \(K_I = \omega_n^2 T_p/K_p = \omega_n^2/6.25\):

Relaxing from critical damping to \(\zeta = 0.7\) doubles the gain and buys a faster return at the cost of a small overshoot past nominal.

The transient with \(\zeta = 0.7\). Inverting the transfer function for a step, with \(\Delta P_L = 0.06\) pu:

The leading coefficient 0.375 Hz/s is again the RoCoF — the initial slope cannot be changed by any controller, only by inertia.

The peak. Differentiating, the extremum is at \(\omega_d t = \cos^{-1}\zeta\):

The comparison. Droop alone (Problem 7) left the system permanently at 49.824 Hz. With the supplementary loop the worst point is 49.887 Hz and the final point is nominal:

Better on both counts, which looks like a free lunch and is not. The integrator demands the full 180 MW from the plant inside a couple of seconds — faster than a steam turbine may be loaded — and a loop this fast will excite the inter-area modes of Problem 9. Real settings are several times smaller, and the practical target is that the bulk of the restoration takes ten to thirty seconds.

The coefficient is the slope of the power-angle curve at the operating point — the same quantity that governed small-signal stability in Set 25:

The last form is the operationally useful one. At the stated 20° the tie is carrying \((1000/1.5)\sin20^\circ = 228\) MW, and one further degree adds \((1000/1.5)(\sin21^\circ-\sin20^\circ) = 10.9\) MW — the linearised coefficient, evaluated exactly.

Why the tie behaves as an integrator. Angle is the integral of frequency, and with \(\delta\) in electrical radians and \(f\) in hertz, \(d\delta/dt = 2\pi\,\Delta f\):

So the flow keeps changing for as long as the two areas are at different frequencies, and settles only when \(\Delta f_1 = \Delta f_2\). That single statement is what "synchronous interconnection" means, and it is why the two-area steady state of Problem 10 has one frequency and not two.

The inter-area mode. Two inertias swinging against one spring:

Within the 0.1–0.8 Hz band that characterises inter-area modes. They are far slower than the local machine modes of Set 25 because the effective inertias here are those of whole areas rather than of single rotors.

The second circuit. Two equal reactances in parallel halve \(X_{12}\), so:

A stiffer spring between unchanged masses: the mode rises by \(\sqrt2\). The interconnection is stronger in every steady-state sense — more transfer capability, a smaller angle for the same flow — and its oscillation is faster and, in general, better damped by the same power system stabilisers.

The consequence for control. Any supplementary controller must be slow compared with the inter-area mode, or it will excite it. At 0.79 Hz the mode period is 1.27 s, so an AGC loop with \(\omega_n = 1.52\) rad/s (0.24 Hz, from Problem 8) sits comfortably below it — but only by a factor of three, which is the practical reason AGC gains are set several times lower than the critically damped value.

The two stiffnesses.

Set up the steady state. The tie has stopped changing, so the two areas are at one frequency. Write each area's power balance, taking \(\Delta P_{12}\) positive for flow out of area 1:

The second equation alone gives \(\Delta P_{12} = \beta_2\,\Delta f\), which is the useful shortcut: area 2 has no disturbance of its own, so everything it does appears on the tie.

Solve.

Negative, with the stated sign convention, means the flow is into area 1: area 2 is sending 101.13 MW to help a neighbour whose load it does not serve.

Who supplies what.

Area 1, where the load actually appeared, supplies only 32.6% of it. The split is \(\beta_1 : \beta_2\) and has nothing to do with which area is at fault.

Standing alone, area 1 would have had only its own stiffness:

Three times the excursion. This is the benefit that motivates interconnection in the first place, and it is delivered automatically by the physics, with no controller and no communication of any kind.

And the problem it creates. Area 2's generators have moved off their economic setpoints, area 2 is exporting 101 MW it never contracted to sell, and the tie flow no longer matches schedule. Governor support across a tie is exactly what one wants for the first few seconds and exactly what one does not want after a minute — which is the entire brief for the supplementary loop.

Evaluate the two errors from the numbers already obtained, \(\Delta f = -0.03974\) Hz and \(\Delta P_{12} = -101.13\) MW, with export counted positive:

Read what those two numbers say. Area 1's controller sees exactly its own 150 MW deficit — not the 48.87 MW its governors happened to supply, and not the 101.13 MW that appeared on the tie, but the disturbance itself. Area 2's controller sees zero, and therefore does not move a single speed changer, even though its generators are at that moment producing an extra 101.13 MW.

That extra output is withdrawn automatically as area 1's integrator raises its own references and the frequency comes back to nominal. Area 2 lent support for the first seconds and is repaid without ever having been told what happened.

The general result behind it. With \(B_i = \beta_i\) and a disturbance \(\Delta P_{L1}\) in area 1:

Identically, for any \(\beta_1\) and \(\beta_2\). This is what "non-interactive control" means: the bias setting encodes the neighbour's expected response so exactly that the tie term and the frequency term cancel in every undisturbed area.

The equilibrium argument — and it needs no numbers at all. Each \(\Delta P_{\text{ref},i}\) is the output of an integrator driven by \(\mathrm{ACE}_i\). In a steady state every signal is constant; an integrator with a constant non-zero input produces a ramp, not a constant; so the only steady state that can exist has \(\mathrm{ACE}_1 = \mathrm{ACE}_2 = 0\):

Both objectives are met for any positive pair of biases, and the biases have vanished from the answer. The final state does not depend on them at all.

So what does the bias decide? The path, not the destination. With \(B_i = \beta_i\) the innocent area is never disturbed and the whole correction is made by the area that caused it. With any other setting the innocent area's controller is given a non-zero error and starts moving plant for no reason, as Problem 12 shows in numbers.

The primary steady state, by symmetry with Problem 10 — the disturbed area is now the second one:

Positive this time: area 1 exports 81.46 MW to help its neighbour. Area 2 supplies \(2545\times0.06623 = 168.54\) MW itself, and \(81.46 + 168.54 = 250\) MW. ✓

With ideal biases \(B_1 = 1230\), \(B_2 = 2545\) MW/Hz:

The mirror image of Problem 11, as it must be. The area that caused the disturbance owns the whole of it; the innocent area is left alone.

Bias set too low: \(B_1 = 600\) MW/Hz, less than half its correct value.

A positive ACE is read as over-generation. Area 1's controller therefore starts winding its own references down by up to 41.7 MW while area 2 is still trying to make up a 250 MW shortfall. The two controllers work against each other, the frequency dips further before recovering, and both sets of turbines are moved further than necessary.

Bias set too high: \(B_1 = 2000\) MW/Hz.

Now the error is negative, so area 1 raises its own generation by about 51 MW that nobody asked it for. Frequency is pushed back through nominal from below, area 2's integrator has to unwind some of its own correction, and the pair hunt briefly before settling.

Collect the three cases:

Neither error is unstable. Problem 11 showed that \(\Delta f = 0\), \(\Delta P_{12} = 0\) is the only equilibrium for any positive bias, so all three cases end in the same place. What differs is the regulating energy spent getting there and the depth of the transient.

The asymmetry that matters in practice. Of the two errors, the under-set bias is worse. It makes the innocent area actively oppose the correction, so the frequency recovery is slowed at the moment it matters most. That is why grid codes specify a minimum bias — commonly a stated fraction of the measured \(\beta\) — rather than a tolerance band, and why the measurement of Problem 4 is repeated annually.

Stage one — the droop allocation. Convert, add, divide:

The primary shares:

A split of 125 : 100 : 75, which is 250 : 200 : 150 — the ratings, because the percentage droops are equal. Cost has played no part whatever.

Stage two — the participation factors. They come from the second derivatives of the cost curves, exactly as Set 27 derived them:

The flattest incremental-cost curve takes the largest share, because a unit whose incremental cost rises slowly can absorb more megawatts before it becomes the expensive one.

The final allocation. AGC drives \(\Delta f\) to zero, so the droop terms all vanish and the load relief disappears with them. The units must supply the entire 90 MW from their reference settings:

Compare the two stages, which is the object of the problem:

Unit 3 is unloaded during the second stage even though the system's total generation is rising, because it is the expensive machine at the margin. The extra 2.90 MW in the totals is the load relief being repaid as the frequency comes back to nominal — the load grows again on the way up.

The general shape. Two allocations run in sequence after every disturbance, and they answer different questions:

The transient state is the physical one and the final state is the economic one. Units are unloaded and reloaded against each other while the total stays fixed, and that reshuffling is invisible on the frequency trace.

The isochronous case. An isochronous governor drives its valve until its own frequency error is zero, so its steady-state characteristic is a horizontal line at whatever frequency its instrument believes to be 50.00 Hz:

These two demands cannot both be met. The system runs at one frequency, so at least one governor sees a non-zero error and keeps integrating.

Follow it through. Suppose the system settles anywhere below 50.02 Hz. Unit B sees a negative error and keeps raising its valve; unit A, seeing the frequency now above its own 50.00 Hz target, keeps lowering its valve. Neither stops:

The load split is decided entirely by which instrument reads high, and by nothing else. A 0.02 Hz calibration difference — 0.04% — produces a 300 MW difference in output. Two integrators cannot share one controlled variable.

The droop case. A proportional characteristic converts the same reference error into a finite, small offset. With 4% on a 300 MW machine:

Three megawatts out of three hundred — one per cent, and stable. The units share the load in the intended proportion plus a small bias, and neither runs away.

The scaling. The sharing error is \(\Delta f_{\text{instr}}/R\), so it grows as the droop is tightened:

Zero droop is the isochronous case, and the table shows it as the limit of a continuous trend rather than as a separate phenomenon. A stiffer unit is more responsive and more sensitive to reference error at the same time.

The principle, and it is not confined to power systems: a shared variable can be regulated by at most one integrator. Droop is the price of cooperation — it guarantees the permanent frequency offset of Problems 2 to 7, and in exchange it lets an unlimited number of governors act on one measurement without any of them needing to know the others exist.

Where isochronous control is still used: a single machine feeding an isolated load — a ship, an emergency generator, an island with one unit — where the sharing problem does not arise. It is also used deliberately on exactly one unit of a small island system, with every other unit on droop, so that one machine takes all the residual error. That is the same architecture as AGC, with the integrator placed at a machine rather than at a control centre.

Set up the swing equation with the ramp. Writing the area's stored energy as \(2HS/f_0\) with \(S\) the MVA base:

Integrate. The deficit shrinks linearly and vanishes at \(T_r\), which is therefore where the frequency stops falling:

A compact and useful result: the nadir is the initial RoCoF multiplied by half the response time. Half, because the deficit is triangular — the area under a triangle, not a rectangle.

Evaluate for the stated area.

The settled frequency, which involves none of the same quantities:

The transient minimum is twice the permanent offset. Load shedding at 48.8 Hz does not operate — but the margin is 0.2 Hz, not the 0.7 Hz that the settling point alone would suggest.

Halve the inertia. Nothing in \(\beta\) changes, so the destination is untouched; only the path changes:

Load shedding now operates and sheds a block of demand the system did not actually need to lose — the generators were always going to cover the deficit by \(t = 5\) s. The event is made worse by the protection, and the settled-state calculation gave no warning at all.

What would have to change to survive it. Requiring the nadir to stay above 48.8 Hz means \(K'T_r/2 \le 1.2\):

The governors would have to deliver the whole 800 MW in three seconds instead of five. That is achievable for hydro and open-cycle gas, marginal for a non-reheat steam unit and out of the question for a reheat unit (Problem 16). The alternative levers are more inertia, a faster-than-governor resource such as a battery, or a lower shedding threshold.

The two traces make the point better than the algebra does. Same disturbance, same \(\beta\), same destination, different worst point:

Expand in partial fractions. For a unit step, \(\Delta P_m(s) = G_T(s)/s\) with poles at \(0\), \(-1/T_T\) and \(-1/T_r\):

Check the sum: \(1 - 0.27273 - 0.72727 = 0\), so \(\Delta P_m(0) = 0\) as a strictly proper transfer function requires. The two residues are the two halves of the response, and their relative size is the whole story.

Read the structure. The fast term dies in about a second; after that the response is

So the high-pressure cylinder delivers about 27% of the eventual response within a second, and the remaining 73% arrives with the reheater's 8 s time constant. The fraction \(K_r\) is exactly the share of the total power developed in the HP cylinder, which is what the parameter means physically.

Tabulate against the non-reheat unit:

The two curves on one axis:

The consequence for the nadir. Problem 15's ramp model assumed the full response arrived by \(T_r = 5\) s. A fleet of reheat units delivers only 61% by then, so the effective ramp is much longer and the nadir correspondingly deeper. Crudely, replacing \(T_r = 5\) s by an effective 12 s in \(-KT_r/2\) multiplies the dip by 2.4.

The settling point is untouched — the reheat unit does eventually deliver its whole \(1/R\) contribution, so it appears in \(\beta\) at full value. It is only the nadir that suffers, which is exactly the quantity \(\beta\) does not describe.

The system-design conclusion. Two areas with identical \(\beta\) and identical inertia can have nadirs a hertz apart if one is rich in hydro and open-cycle gas and the other in reheat steam. Frequency-response markets have accordingly moved from paying for \(\beta\) to paying for delivery within a stated window — typically the megawatts sustained between 1 and 10 s. A reheat unit sells very little of that product, and a battery sells almost nothing else.

The step response. With \(T_w = 1\):

Evaluate it, and the shape is startling:

Opening the gate to demand one more per-unit of power produces two per-unit less at the first instant. The power does not even change sign until \(t = \tfrac{T_w}{2}\ln 3 = 0.55\) s, and does not reach 90% of its target until 1.70 s.

The physics. The power developed is \(P \propto Hq\) — head times flow. Opening the gate raises the flow area instantly, but the water column in the penstock has mass and cannot accelerate instantly:

The same effect works in reverse on closing: shutting a gate momentarily raises the power, and it is the reason a penstock needs a surge tank. The right-half-plane zero at \(s = +1/T_w\) is the transfer-function statement of a mechanical fact about water.

Why the governor must be detuned. A right-half-plane zero contributes phase lag while contributing gain. Close a fast proportional loop around it and the loop responds to the initial wrong-way power by demanding more gate, which makes the power fall further — a positive feedback that shows up as sustained hunting. The remedy is transient droop: a governor whose regulation is large immediately after a movement and relaxes to the permanent value as the water column settles:

Typical values are a permanent droop of 4% and a transient droop of 20–40% with a reset time of several seconds. The governor is deliberately made five to ten times weaker on the time scale that matters for stability.

What that costs in frequency response. During the first second or two the machine behaves as though it had 25% droop rather than 4%, so its contribution to \(1/R\) is roughly a fifth of the nameplate value:

Which qualifies the folklore. Hydro plant is often called the fastest primary responder, and on a 10–30 s view it is: no thermal storage, no reheater, and full output available. On a 1–3 s view — the view the nadir takes — it is handicapped by its own water column and by the detuning that column forces on its governor.

The comparison worth carrying away. Over the first two seconds, a non-reheat steam unit delivers essentially all of its response, a hydro unit delivers a fraction of its response and starts by delivering the wrong sign, and a reheat unit delivers about 43% (Problem 16). Over sixty seconds all three deliver their full \(1/R\). Every one of them appears identically in \(\beta\).

A synchronous clock counts cycles, so its error is the integral of the fractional frequency deviation:

Every mains-timed clock in the interconnection is 8.64 seconds slow. The deviation was only 0.04% and would pass unremarked on any frequency trace; it is the integration over six hours that makes it visible.

Paying it back requires the opposite deviation for the same accumulated area:

Symmetric, because the correction offset is equal and opposite. Real time-error corrections use a smaller offset — a few millihertz — over a longer window, so that the correction is invisible to everything except the clocks.

The inadvertent interchange, which accumulates in exactly the same way but in energy:

Energy delivered but not contracted. It is repaid by running the tie 30 MW below schedule for six hours, or 10 MW below for eighteen — payback in kind rather than in cash, which is the standard arrangement between interconnected utilities.

Why zero ACE does not prevent either. The supplementary loop drives \(\mathrm{ACE} \to 0\), which forces \(\Delta f \to 0\) and \(\Delta P_{\text{tie}} \to 0\) — but says nothing about their integrals:

Whatever error accumulated while the loop was correcting stays on the books for ever, and there is nothing in \(\mathrm{ACE}_i = \Delta P_{\text{tie},i} + B_i\Delta f\) that can remove it.

What is actually done. A slow outer correction is added to the ACE:

Two further integrators, deliberately made very slow — hours, not minutes — so that they never interact with the frequency-control loop. They are switched in by agreement between the interconnected areas, usually when the accumulated error passes a stated threshold such as ten seconds.

The plant's ramp capability:

Nearly six minutes to answer a disturbance that primary control had arrested inside two seconds. That gap between the two time scales is not a defect of AGC; it is the reason the two loops exist separately.

The number of cycles:

The controller issues 86 successive setpoints, each of which the plant only partially achieves before the next arrives. The AGC cycle is far faster than the plant, which is deliberate — it means the loop is always working from a fresh measurement and never waiting.

The reconciliation. Problem 8's gain of \(K_I = 0.372\) gave \(\omega_n = 1.52\) rad/s, a settling time of a few seconds. That controller would demand the whole correction inside about two seconds:

A demand 170 times what the plant can deliver. The valves would saturate, the integrator would wind up against the limit, and the loop's actual behaviour would bear no relation to the linear design.

So what really limits \(K_I\)? Three ceilings, and the linear stability calculation is the loosest of them:

Setting the gain from criterion 3 alone gives a controller that is stable on paper and useless in service, because it spends every disturbance in saturation.

Design practice. The gain is set so that the loop's demand is comfortably inside the fleet's ramp capability — typically the bulk of the restoration in ten to thirty seconds for a small disturbance, with the loop simply rate-limited for a large one. Anti-windup on the integrator is mandatory, because without it the ACE integral accumulated during a five-minute ramp would drive a large overshoot the moment the plant catches up.

A design lever the problem hints at. If six minutes is too slow, the answer is not more gain but more regulating units. Adding a third 300 MW machine on regulation raises the fleet ramp to 30 MW/min and cuts the time to 4.0 min. Regulating capability is a scheduled quantity, procured like reserve, and it is what actually determines an area's AGC performance.

Today's system. Three quantities, three independent formulas:

The nadir clears 48.8 Hz by 75 mHz. That is a thin margin, and it is the margin the system is actually designed to, not the comfortable 0.7 Hz that the settling point suggests.

After the displacement. \(\beta\) is maintained by assumption, so the destination does not move; only the path does:

Load shedding operates, and at 47.75 Hz so do the under-frequency protections of the generators themselves on many machines. A system that maintained every steady-state indicator has lost its ability to survive its own largest infeed.

Size the battery. Fast frequency response delivered essentially instantaneously reduces the deficit that inertia and governors must cover, so the nadir formula applies to \(\Delta P_L - P_b\):

Check: with 500 MW injected, \(K = 50\times400/60\,000 = 0.333\) Hz/s and the nadir is \(0.333\times3 = 1.00\) Hz, giving 49.00 Hz. ✓

The energy, which is the part that surprises people. The battery holds 500 MW for roughly the six seconds until the governors have taken over, then ramps out:

A very high power rating on a very small store. Frequency-response batteries are sized by megawatts and by C-rate, not by megawatt-hours, which is why they are physically much smaller than an energy-arbitrage installation of the same power.

What the battery does not fix. Three things, and each needs a different remedy:

The first is the sharpest limitation. A battery responding within a cycle still responds after the frequency has begun to fall, so it cannot flatten the initial slope, and RoCoF relays on embedded generation see the full 0.75 Hz/s regardless. Only rotating mass — retained synchronous plant, synchronous condensers, or grid-forming converters emulating inertia — changes that number.

The three-lever summary for a low-inertia area, in the order the disturbance meets them: