Chapter 4 — System-level stability and a sizing case

Grid-Forming Inverters and Power-System Stability · Results R34 to R40 · 50 Hz unless a passage states otherwise

The problem this chapter answers

A system operator has eight synchronous condensers. A synchronous condenser is a synchronous machine with no turbine and no load on its shaft. It supplies reactive power, short-circuit current and inertia; on average it generates no active power. The operator must decide whether grid-forming converters can replace the inertia of those eight machines, and how much converter plant to buy. That decision needs a number, and the number is not "1.6 GW of grid-forming plant". It is three numbers, in three different units, and only one of them binds. Reactive power and short-circuit current are two further services the condensers supply. This chapter sizes the inertia replacement only; §4.5 Constraint 3 shows why the other two cannot be ignored at the same instant.

Chapter 1 gave one machine against a stiff grid, Chapter 2 one grid-following converter against a Thevenin source, and Chapter 3 one grid-forming converter with its pair (H_eq, K_D,eq). A purchase decision is about a fleet inside a network.

This chapter builds that step, in four moves.

It reduces a network with load buses to one set of coupled swing equations (Definition 4.1, Model 4.2).
It assembles machines and grid-forming converters into one small-signal state-space model and reads stability from the eigenvalues (Definition 4.3, Model 4.4).
It derives the initial rate of change of frequency (Theorem 4.5) and the frequency nadir (Theorem 4.6).
It sizes a grid-forming fleet against energy, power headroom and the converter current limit, separately, and identifies which one binds (Example 4.2).

The answer for the case defined in this chapter is stated here, so that the reader can check the argument against it. Power headroom binds. The current limit binds at the same instant. Stored energy is not close to binding. A 1.6 GVA condenser fleet releases 179.2 MJ over a 0.8 Hz excursion. A 1120 MVA converter fleet with a one-hour store holds 4 032 000 MJ, which is 22 500 times as much. The same fleet must hold 224 MW of power above its dispatch. A converter running at 1.0 pu with a current limit of 1.2 pu has exactly 0.20 pu of current left. Every one of those numbers is derived in Example 4.2.

How to read the results, the statuses and the citations

Each result carries a local number (Definition 4.1, Theorem 4.5) and the global identifier from the book plan (R34, R35). Prose uses the local number. Prerequisites from earlier chapters are cited by their global identifier and linked to the result in the chapter that proves them. Each result also carries its epistemic status in words: proved here, computed here, cited, or assumed for this case. A cited result is never written as though this book proved it.

Every published number carries a citation of the form [S14, National Grid ESO 2019, Technical Report on the events of 9 August 2019, event summary — verify]. The draft stage could not open these sources. The location part therefore names the topic inside the source, not a page number, and every citation carries — verify. Only a human who has read the page removes that mark and replaces the topic with a page. A number this book computes carries no citation; it carries its example number instead.

Chapter notation

Every symbol below is from the book-wide notation table, section 3.5. The middle column gives the plain-text name used in that table, because this chapter renders Greek letters and deltas as glyphs. Sections that need a symbol outside the table declare it in a Local notation panel at the top of the section, and nowhere else.

Symbol	Plan name	Meaning	Unit
Y	Y	bus admittance matrix	pu
Y_red	Y_red	Kron-reduced admittance matrix at the internal nodes	pu
`B`_ij	B_ij	susceptance of the reduced branch between nodes `i` and `j`	pu
δ_COI	delta_COI	centre-of-inertia angle	rad
ω_COI	omega_COI	centre-of-inertia frequency	rad/s
`H`_i, `S`_i	H_i, S_i	inertia constant and rating of unit `i`	s, VA
`E`_kin,sys	E_kin_sys	system stored kinetic energy, Σ `H`_i`S`_i	MJ; 1 GVA·s = 1000 MJ
`H`_sys	H_sys	system inertia constant on the system base	s
`S`_sys	S_sys	system base apparent power	VA (GVA)
Δ`P`	dP	step loss of infeed	MW
RoCoF	RoCoF	rate of change of frequency, d`f`/d`t`	Hz/s
RoCoF₀	RoCoF_0	initial rate of change of frequency	Hz/s
`f`_nadir	f_nadir	lowest frequency reached after the loss	Hz
Δ`f`_nadir	df_nadir	frequency deviation at the nadir	Hz
`T`	T	time over which primary response is fully delivered	s
`R`_p	R_p	volume of primary response	MW
`D`_load	D_load	load damping	MW/Hz
x, u, y	x, u, y	small-signal state, input and output vectors	mixed
A, B, C, D	A, B, C, D	state, input, output and feedthrough matrices	mixed
λ_i, σ_i, ω_d,i	lambda_i, sigma_i, omega_d_i	eigenvalue `i` and its real and imaginary parts	1/s, 1/s, rad/s
ζ_i	zeta_i	damping ratio of mode `i`	—
`p`_ki	p_ki	participation factor of state `k` in mode `i`	—
`f`₀, ω₀	f_0, omega_0	rated frequency and rated electrical angular frequency	Hz, rad/s
`H`, `E`_kin, `K`_D	H, E_kin, K_D	inertia constant, stored energy and damping coefficient of one unit	s, MJ, pu
δ, Δω	delta, dw	rotor or internal angle, and per-unit speed deviation (ω − ω₀)/ω₀. Chapter 1 writes the deviation Δω̄ with a bar (R02); this chapter drops the bar.	rad, —
`P`_m, `P`_e	P_m, P_e	mechanical power in, electrical power out	pu
`K`_s	K_s	synchronizing torque coefficient	pu/rad
`H`_v, `H`_eq, `K`_D,eq	H_v, H_eq, K_D_eq	virtual and equivalent inertia constants, equivalent damping	s, s, pu
`I`_max	I_max	converter current limit	pu
`E`_res	E_res	usable energy reserve of a converter	MJ
`P`, `Q`	P, Q	active and reactive power	pu or MW, Mvar

Terms used in this chapter

Terms of art. Each term below is used in this chapter and is defined nowhere else in the book. Each definition names where the chapter uses it.

Synchronous condenser — a synchronous machine with no turbine and no mechanical load on its shaft. It runs connected to the grid and supplies reactive power, short-circuit current and the inertia of its rotor. On average it generates no active power; it draws a small amount to cover its losses. A flywheel is a mass added to the shaft to raise H. Used in Example 4.2.
Infeed — active power injected into the system by a generating unit or an import link. A loss of infeed is the sudden disconnection of that injection; ΔP is its size in MW. Used in Theorems 4.5 and 4.6.
Governor — the controller on a machine that changes mechanical power in response to a measured frequency deviation. A deadband is the band of frequency deviation around f₀ inside which the governor does not act. Used in §4.3 and §4.4.1.
Primary response — the active-power increase that governors, and equivalent converter controls, deliver in response to a frequency fall within seconds and without an operator instruction. R_p is its volume in MW and T the time by which it is fully delivered. On a grid-forming converter the droop law of R25 produces it. Used in Theorem 4.6.
Headroom — the active power a unit or a fleet can add above its present dispatch, in MW or in per unit of rating. For a converter it is bounded by I_max (R33). Used in Example 4.2 as P_headroom.
Low-frequency demand disconnection — a relay scheme that disconnects blocks of demand automatically when the frequency falls to preset levels, to stop the fall. In Great Britain it acted at 48.8 Hz on 9 August 2019 [S14, National Grid ESO 2019, Technical Report on the events of 9 August 2019, event summary — verify]. Used in Example 4.1 and Citation 4.7.
Embedded generation — generation connected to the distribution network rather than to the transmission network. The system operator does not measure its output directly, and it carries its own protection. Used in Citation 4.7.
Loss-of-mains protection — protection on embedded generation that disconnects the unit when it detects that its part of the network has separated from the main system. That separated state is called islanding. One detection method measures RoCoF and trips above a threshold. A system-wide frequency fall can exceed the same threshold, so the relay cannot tell islanding from a system event. Used in §4.3 and Exercise 4.1.

4.1 From one machine to a network

Local notation for §4.1. These symbols are not in the book-wide table. They are defined here and used with this meaning for the rest of the chapter.

t — time, in seconds, measured from the disturbance.
n — number of source units in the system.
E_i — internal voltage magnitude of unit i, in per unit. For a machine this is the constant emf behind transient reactance of R05. For a grid-forming converter it is the commanded internal voltage of R24.
δ_i, Δω_i — the angle δ and the per-unit speed deviation Δω of unit i. The unsubscripted symbols keep their book meaning.
K_D,i — damping coefficient of unit i, in per unit on its own base.
I_i — current phasor injected by unit i at its internal node, in per unit on S_sys.
δ_i,0 — equilibrium value of δ_i.
X_i — total reactance between the internal node of unit i and the load bus, in per unit on S_sys.
V_L — voltage phasor at load bus L, in per unit.
Y_L — constant-impedance load admittance at bus L, in per unit.
P_i, Q_i — active and reactive power injected by unit i at its internal node, at the equilibrium. P_e,i and P_m,i — electrical output and mechanical or commanded input of unit i in (4.3), on the system base; at the equilibrium P_e,i = P_m,i = P_i.
G_ij — conductance of the reduced branch between nodes i and j. The book table lists only B_ij, because a lossless reduction has no conductance. Reduction of a load that consumes real power produces one, so the symbol is needed.
N — number of buses in the unreduced network.
y_ij, y_i0 — series admittance of the branch between buses i and j, and shunt admittance from bus i to the reference, both in per unit.
V, I_bus — vectors of bus voltage phasors and injected currents. Subscripts A and B select the two partitions of §4.1.2: group A is the internal source nodes, group B every other bus.

A power system is a set of source nodes, a set of load nodes, and a mesh of branches. The swing equation of R04 applies to every source separately; the network couples them. This section writes that coupling as a matrix, then removes every node with no dynamics of its own, leaving one equation per unit.

4.1.1 The bus admittance matrix

Definition 4.1 — Bus admittance matrix stated hereR34, part 1

Number the buses 1 to N. For a network of branches with series admittance y_ij between buses i and j, and shunt admittance y_i0 from bus i to the reference, the bus admittance matrix Y has entries

Y_ii = y_i0 + Σ_{j ≠ i} y_ij,
Y_ij = −y_ij for i ≠ j,

and the network equation is I_bus = Y V, where V is the vector of bus voltage phasors and I_bus the vector of injected currents. All quantities are in per unit on the system base S_sys, using the base definitions of R01.

Two facts about Y matter here. It is symmetric, because a passive branch carries the same admittance both ways. And a constant-impedance load is not an injection: it is a shunt admittance, so it enters Y_ii and contributes zero to I_bus. The second fact makes the next step possible.

4.1.2 Kron reduction

Split the buses into two groups. Group A holds the n internal source nodes, one behind each unit's internal reactance. Group B holds every other bus, including bus L. Every load is modelled as a constant impedance and folded into Y, so no bus in group B injects current. Write the partitioned network equation.

[ I_A ; 0 ] = [ Y_AA Y_AB ; Y_BA Y_BB ] [ V_A ; V_B ]

The lower block reads Y_BAV_A + Y_BBV_B = 0. If Y_BB is invertible, solve it for V_B and substitute into the upper block.

Y_red = Y_AA − Y_AB Y_BB⁻¹ Y_BA, I_A = Y_red V_A

(4.1)

Equation (4.1) is Kron reduction. It is exact, not an approximation, under one condition: the eliminated buses inject no current. That condition holds here because every load is a constant impedance. It fails for a constant-power load, and it fails for any bus with its own dynamics. A constant-power load can be brought inside the condition: replace it with the constant impedance that draws the same power at the operating-point voltage. The replacement is exact at that point and approximate away from it, which is the same trade Model 4.4 makes when it linearises. The step is standard [S05, Machowski, Bialek and Bumby 2008, Power System Dynamics: Stability and Control, 2nd ed., network reduction for multi-machine models — verify].

Kron reduction removes structure and keeps behaviour. The reduced network has no load bus and no line: every pair of source nodes is connected directly, so a radial system becomes a complete mesh. The branches are not pure susceptances, because the real power drawn by the load appears as the conductances G_ij.

Model 4.2 — Reduced multi-unit power injection proved hereR34, part 1

Write Y_red,ij = G_ij + jB_ij and hold each internal voltage magnitude E_i constant. The active power leaving internal node i is

P_e,i = Σ_j E_iE_j [ B_ij sin(δ_i − δ_j) + G_ij cos(δ_i − δ_j) ]

(4.2)

Derivation.

Row i of (4.1) gives the current leaving internal node i: I_i = Σ_j Y_red,ij E_j e^jδ_j. Here j outside a subscript is the imaginary unit and j inside a subscript is a node index. The complex power leaving the node is S_i = E_i e^jδ_i conj(I_i). Substitute Y_red,ij = G_ij + jB_ij:

S_i = Σ_j E_iE_j (G_ij − jB_ij) e^{j(δ_i − δ_j)}.

Write the exponential as cos(δ_i − δ_j) + j sin(δ_i − δ_j) and take the real part. The product (G − jB)(cos + j sin) has real part G cos + B sin, because (−j)(j) = +1. That is (4.2). The imaginary part, G sin − B cos, is the reactive power Q_i, which this chapter does not use. ∎

Sign of B_ij. B_ij is an element of Y_red, not the susceptance of a physical branch. For a lossless branch of reactance X between nodes i and j, the branch admittance is −j/X, and by Definition 4.1 the matrix element is Y_ij = +j/X. So B_ij = +1/X, and the values printed in §4.1.3 are positive for that reason.

Assumptions. Balanced three-phase operation. Fundamental-frequency phasors. Every load a constant impedance. Every internal voltage magnitude constant, which is the classical machine model of R05 for a machine and the constant-E approximation of R24 for a grid-forming converter. The network is algebraic: it has no state.

Omissions. Network electromagnetic transients, so nothing below the fundamental-frequency timescale. Flux decay and excitation dynamics, so no voltage-stability conclusion. Constant-power and motor loads. Tap changers. The first is repaid in §4.6 as an open problem (R40); the rest are in the book scope statement.

Setting every G_ij to zero recovers R05. For two nodes with one susceptance B₁₂ = 1/X between them, (4.2) gives P_e,1 = (E₁E₂/X) sin(δ₁ − δ₂). Model 4.2 is a strict generalisation of the model the reader already has.

4.1.3 Test system TS4

One test system serves every figure and network exercise in this chapter, and no other network is introduced. TS4 is a design case defined by this book. It is not measured plant, and none of its numbers is a published value.

Unit	Type	`S`_i (MVA)	`H`_i or `H`_eq (s)	`K`_D or `K`_D,eq (pu, own base)	internal `X` (pu, own base)	dispatch (pu)	`X` to bus L (pu, system base)
1	synchronous machine	400	4.0	2	0.30	0.75	0.10
2	synchronous machine	300	3.0	2	0.35	0.60	0.12
3	grid-forming converter	200	2.0	20	0.20	0.50	0.08
4	grid-forming converter	100	2.0	20	0.20	0.50	0.15

System base S_sys = 1000 MVA. Rated frequency f₀ = 50 Hz, so ω₀ = 2π × 50 = 314.1593 rad/s. The load at bus L is 630 MW and 120 Mvar, constant impedance, so Y_L = (630 − j120)/1000 = 0.6300 − j0.1200 pu. Generation sums to 400(0.75) + 300(0.60) + 200(0.50) + 100(0.50) = 300 + 180 + 100 + 50 = 630 MW, so the case balances exactly. The converters carry H_eq and K_D,eq, which by R32 is what any of the four grid-forming families reduces to near the operating point. By R26 those two values are the droop law with m_p = 0.05 pu. The power-filter cutoff is then ω_c = 1/(2m_pH_eq) = 1/(2 × 0.05 × 2.0) = 5.0 rad/s, which is 0.80 Hz.

Stored kinetic energy follows from R03 applied unit by unit:

E_kin,sys = 400(4.0) + 300(3.0) + 200(2.0) + 100(2.0) = 1600 + 900 + 400 + 200 = 3100 MVA·s = 3.100 GVA·s, so H_sys = 3100/1000 = 3.100 s on the 1000 MVA base.

Each unit's internal reactance is on its own base. Converting to the system base multiplies by S_sys/S_i, and the connecting reactance is already on the system base. The total reactance from internal node to bus L is therefore

unit 1: 0.30 × 1000/400 + 0.10 = 0.7500 + 0.10 = 0.8500 pu;
unit 2: 0.35 × 1000/300 + 0.12 = 1.1667 + 0.12 = 1.2867 pu;
unit 3: 0.20 × 1000/200 + 0.08 = 1.0000 + 0.08 = 1.0800 pu;
unit 4: 0.20 × 1000/100 + 0.15 = 2.0000 + 0.15 = 2.1500 pu.

The equilibrium needs three assumptions, all choices of this case. Assumed for this case:

the load-bus voltage is V_L = 1.0∠0° pu;
each unit's reactive injection is proportional to its rating, Q_i = 120 × S_i/1000 Mvar, so the four sum to 120 Mvar;
each unit's active injection is its dispatch.

With these, the current injected by unit i is I_i = conj((P_i + jQ_i)/V_L), and the internal voltage phasor is V_L + jX_i I_i. The result is:

Unit	`X`_i (pu)	`P`_i (pu)	`Q`_i (pu)	`E`_i (pu)	δ_i (deg)
1	0.8500	0.3000	0.0480	1.0716	13.767
2	1.2867	0.1800	0.0360	1.0716	12.481
3	1.0800	0.1000	0.0240	1.0316	6.009
4	2.1500	0.0500	0.0120	1.0314	5.983

Reducing the five-bus network (four internal nodes and bus L) by (4.1) gives the transfer terms of the reduced mesh. The six transfer susceptances are B₁₂ = 0.2555, B₁₃ = 0.3043, B₁₄ = 0.1529, B₂₃ = 0.2011, B₂₄ = 0.1010 and B₃₄ = 0.1203 pu. The matching conductances are G₁₂ = 0.0465, G₁₃ = 0.0553, G₁₄ = 0.0278, G₂₃ = 0.0366, G₂₄ = 0.0184 and G₃₄ = 0.0219 pu. The diagonal of Y_red also carries a conductance at each internal node: G₁₁ = 0.0703, G₂₂ = 0.0307, G₃₃ = 0.0436 and G₄₄ = 0.0110 pu. The diagonal susceptances B_ii do not enter (4.2), because sin(δ_i − δ_i) = 0.

The conductances are not numerical noise. They are the load, redistributed onto every branch and every node of the reduced mesh. In (4.2) the j = i term is E_i²G_ii. For unit 1 it is 1.0716² × 0.0703 = 0.0807 pu, which is 27 % of that unit's dispatch. Substituting all sixteen entries into (4.2) returns P_e,1 = 0.3000, P_e,2 = 0.1800, P_e,3 = 0.1000 and P_e,4 = 0.0500 pu, which are the four dispatches. With the diagonal terms left out, the same sum returns 0.2193, 0.1448, 0.0537 and 0.0383 pu. Run both sums; the difference is the load. The reduction is therefore exact to the digits printed here.

Fig. 4.1 — Test system TS4 before and after Kron reduction, with every reactance marked. Panel (a) is the radial system as built: four units, each behind its own internal reactance and a connecting reactance, feeding one constant-impedance load at bus L. Panel (b) is the reduced equivalent of (4.1): bus L is gone, and every pair of internal nodes is joined directly. Reduction removes the load bus and the radial structure; it keeps the power transfer exactly, and it turns the real power of the load into the six transfer conductances G_ij and the four self-conductances G_ii listed in §4.1.3. The figure illustrates Definition 4.1 and Model 4.2 (R34). Numbers computed for TS4 in §4.1.3.

4.2 The mixed grid in state-space form

Local notation for §4.2.

K_ij — element (i, j) of the synchronizing matrix, ∂P_e,i/∂δ_j evaluated at the equilibrium, in pu/rad. It is the multi-unit generalisation of K_s from R06.
c — converter share: the fraction of the total 1000 MVA of rating that sits on grid-forming converters, dimensionless.
K — the n × n matrix of the K_ij; K_D — the diagonal matrix of damping coefficients on the system base; M — the diagonal matrix of 2H_iS_i/S_sys; I_n — the n × n identity matrix.
v_ki, w_ik — entry k of the right eigenvector of mode i, and entry k of the left eigenvector of mode i.
A prefixed Δ on a state, as in Δδ_i, marks a small deviation of that state from its equilibrium value. The speed state is already a deviation, so its small deviation is written ΔΔω_i.
Δω_COI — centre-of-inertia speed deviation, (ω_COI − ω₀)/ω₀, dimensionless. By (4.5) it equals Σ_i H_iS_i Δω_i / E_kin,sys. Used in Theorem 4.5.

Every unit in TS4 obeys the same two equations. That is the point of R32: a machine obeys them because it has a rotor, and a grid-forming converter because its control law was built to. The two equations, for unit i on the system base, are R04 unit by unit.

dδ_i/dt = ω₀ Δω_i, 2 H_i (S_i/S_sys) dΔω_i/dt = P_m,i − P_e,i − K_D,i (S_i/S_sys) Δω_i

(4.3)

The factor S_i/S_sys converts H and K_D from the unit's own base to the system base. A 400 MVA machine with H = 4.0 s has 4.0 × 400/1000 = 1.600 s on the 1000 MVA base, and K_D = 2 pu becomes 0.800 pu. Omitting the conversion overstates that unit's contribution to E_kin,sys by the factor S_sys/S_i, which is 2.5 here.

4.2.1 System stored energy and the centre of inertia

Definition 4.3 — System stored energy, system inertia constant, centre of inertia stated hereR34, part 2

For n units with inertia constants H_i and ratings S_i:

E_kin,sys = Σ_i=1ⁿ H_i S_i, H_sys = E_kin,sys / S_sys

(4.4)

The centre-of-inertia angle and frequency are the inertia-weighted means of the unit angles and speed deviations:

δ_COI = (Σ_i H_i S_i δ_i) / E_kin,sys, ω_COI = ω₀(1 + (Σ_i H_iS_i Δω_i) / E_kin,sys)

(4.5)

Three points of care. E_kin,sys in (4.4) is a sum of energies and does not depend on the base; H_sys does. A unit contributes in proportion to H_iS_i, so a large machine with modest H can outweigh a small one with large H. A grid-forming converter enters through H_eq, and R33 bounds what that entry is worth by E_res and I_max; the sum does not know those bounds, and §4.5 makes that gap concrete.

Status: definition, stated here; the centre-of-inertia construction is standard [S05, Machowski, Bialek and Bumby 2008, Power System Dynamics: Stability and Control, 2nd ed., centre-of-inertia reference — verify].

4.2.2 Linearisation and the state matrix

Model 4.4 — Small-signal state space of the mixed grid proved hereR34, part 2

Take the state vector x = [Δδ₁ … Δδ_n, ΔΔω₁ … ΔΔω_n]^T, where the prefixed Δ marks a small deviation from the equilibrium of §4.1.3. Differentiate (4.2) at the equilibrium to get the synchronizing matrix

K_ij = ∂P_e,i/∂δ_j evaluated at (δ_1,0 … δ_n,0)

(4.6)

Linearising (4.3) then gives dx/dt = A x + B u with the block state matrix

A = [ 0 ω₀I_n ; −M⁻¹K −M⁻¹K_D ], M = diag(2H_i S_i/S_sys)

(4.7)

The input u = [ΔP_m,1 … ΔP_m,n]^T holds the mechanical or commanded power deviations on the system base, and B = [0 ; M⁻¹]. The output y = C x + D u selects whatever is measured; for frequency work C picks the speed states and D = 0. Eigenvalues λ_i = σ_i + jω_d,i of A give the modes. The damping ratio is ζ_i = −σ_i / √(σ_i² + ω_d,i²). Scale the right and left eigenvectors of A so that Σ_k v_kiw_ik = 1 for each mode. The participation factor p_ki = |v_kiw_ik| then sums to 1 over k, and it says how much state k takes part in mode i. That is what labels a mode "machine" or "converter".

Assumptions. Everything Model 4.2 assumes, plus: deviations small enough that (4.2) is linear in them; constant P_m; constant H_eq and K_D,eq, which by R32 holds near the operating point only; and no current limiting. The last fails first in a large disturbance, and §4.6 names it as an open problem (R40).

One eigenvalue of A is always zero, for a structural reason: (4.2) depends on angle differences only, so adding the same constant to every δ_i changes nothing. That zero mode is the free rotation of the whole system, and it is why frequency stability is read against a reference such as δ_COI.

4.2.3 TS4 at the base case

Applying (4.6) to TS4 at the equilibrium of §4.1.3 gives the synchronizing matrix, in pu/rad:

∂`P`_e,i/∂δ_j	`j` = 1	2	3	4
`i` = 1	+0.7804	−0.2921	−0.3251	−0.1633
2	−0.2945	+0.6194	−0.2163	−0.1086
3	−0.3416	−0.2254	+0.6950	−0.1280
4	−0.1716	−0.1132	−0.1280	+0.4128

Every diagonal entry is positive and every off-diagonal entry is negative: each unit pulls back towards its neighbours. That is the multi-unit form of R06, which gives K_s > 0 for |δ₀| < 90°. The matrix is not quite symmetric, because the conductances G_ij break the symmetry a lossless network would have.

4.2.4 Sweeping the converter share

The question that matters for the purchase decision is not what TS4 does at one operating point. It is what happens as machines are replaced by converters. The sweep is defined by the six rules below. The rules are a choice of this case, not a result. Assumed for this case:

Total rating stays at 1000 MVA. A converter share c puts 1000c MVA on units 3 and 4, split 2:1 as in the base case, and 1000(1 − c) MVA on units 1 and 2, split 4:3.
Total dispatch stays at 630 MW. Each unit's dispatch starts as its rating times its base-case per-unit dispatch. One common factor then scales all four so that they sum to 630 MW. A unit's per-unit dispatch therefore changes with c. At c = 0.20 the unscaled sum is 648.6 MW and the factor is 630/648.6 = 0.9713. Unit 1 then runs at 0.75 × 0.9713 = 0.7285 pu instead of 0.75 pu.
Each unit keeps its H, its K_D and its internal reactance in per unit on its own base.
Each unit's reactive injection stays proportional to its rating, as in §4.1.3, so the four sum to 120 Mvar.
The connecting reactances on the system base do not change.
A unit of zero rating is removed from the model.

The base case is c = 0.30, where the factor in rule 2 is 1.

System stored energy follows directly. Machines contribute 1000(1 − c)(4 × 4/7 + 3 × 3/7) = 1000(1 − c) × 25/7 = 3571.4(1 − c) MVA·s. Converters contribute 2.0 × 1000c = 2000c MVA·s. Adding them: E_kin,sys(c) = 3571.4 − 1571.4c MVA·s. At c = 0.30 this is 3571.4 − 471.4 = 3100.0 MVA·s, which matches §4.1.3.

converter share `c`	`E`_kin,sys (MVA·s)	`H`_sys (s)	least-damped oscillatory ζ	modes present
0 %	3571.4	3.571	0.0134	1 pair (2 units)
20 %	3257.1	3.257	0.0136	3 pairs
40 %	2942.9	2.943	0.0141	3 pairs
60 %	2628.6	2.629	0.0152	3 pairs
80 %	2314.3	2.314	0.0187	3 pairs
100 %	2000.0	2.000	0.1393	1 pair (2 units)

Two things move in opposite directions, and reading only one gives the wrong conclusion. System stored energy falls by 44 %, from 3571.4 to 2000.0 MVA·s, because the converters carry H_eq = 2.0 s against the machines' 4.0 s and 3.0 s. The least-damped oscillatory mode gets better, from ζ = 0.0134 to 0.0187 while machines are still present. The reason is K_D,eq = 20 pu on the converters against K_D = 2 pu on the machines, which is the droop equivalence of R26: a 5 % droop gain gives K_D,eq = 1/m_p = 20 pu. The converters damp the machine mode that they do not replace.

A third mode changes the reading. At c = 20 % a mode sits at λ = −2.2151 + j17.3463 s⁻¹ with ζ = 0.1267 and its largest participation on converter 3. By c = 80 % the same branch of the locus has moved to λ = −0.7671 + j12.9359 s⁻¹ with ζ = 0.0592, and its largest participation has moved to machine 2. So one mode's damping falls by a factor of 2.1 as the converter share rises, while the least-damped mode's improves. Beyond 80 % the same branch keeps falling: ζ = 0.0458 at 85 %, 0.0319 at 90 % and 0.0206 at 95 %. It crosses ζ = 0.05 between c = 80 % and 85 %. It is never the least-damped mode, because the machine mode sits below it at every step. "Converter penetration damps the system" and "converter penetration undamps the system" are both false as general statements: the modes disagree.

Fig. 4.2 — Eigenvalues of TS4 in the upper half plane as the converter share c rises from 0 % to 100 % in 20 % steps. Complex conjugates are not drawn. Teal circles are modes whose largest speed-state participation is a synchronous machine; orange squares are modes whose largest participation is a converter. Dashed rays are lines of constant damping ratio. The cluster near σ = −0.2 s⁻¹ is the machine mode: it barely moves, and its damping ratio rises from 0.0134 to 0.0187. The orange mode at σ = −2.5 s⁻¹ is the converter mode set by K_D,eq = 20 pu. The mode travelling from (−2.2151, 17.3463) at 20 % to (−0.7671, 12.9359) at 80 % is the one that loses damping, from ζ = 0.1267 to ζ = 0.0592. The strip below the axis prints E_kin,sys at each step; it falls by 44 % across the sweep while the least-damped mode improves. At c = 0 and c = 100 % only two units exist, so only one conjugate pair does. The figure illustrates Model 4.4 (R34). Numbers computed for TS4 in §4.2.4.

Fig. 4.2 also shows what Model 4.4 cannot tell the operator. Every mode is stable at every share, and nothing in the eigenvalues says the system is short of stored energy. The next two sections answer that with a different tool.

4.3 Rate of change of frequency

A generator trips. One instant earlier, generation matched demand; one instant later it is short by ΔP. No governor and no converter control has responded yet. Only stored energy can supply the missing power in that instant, so the frequency falls at a rate that measures how much stored energy there is.

Theorem 4.5 — Initial rate of change of frequency proved hereR35

Let n units obey (4.3) and stay connected through a disturbance. Let E_kin,sys be their stored energy from (4.4), in MJ, counting those n units only and not any unit that trips. At t = 0 an infeed of ΔP MW is lost as a step. Assume:

no load damping and no primary response inside the first instant, so every P_m,i is unchanged at t = 0⁺;
no angle δ_i changes at t = 0⁺;
the total electrical load is unchanged at t = 0⁺, so the n units together supply the whole pre-event load.

Let Δω_COI = (ω_COI − ω₀)/ω₀ be the centre-of-inertia speed deviation declared in the §4.2 local notation panel. By (4.5) it equals Σ_i H_iS_i Δω_i / E_kin,sys. Then the centre-of-inertia frequency falls at

RoCoF₀ = − f₀ ΔP2 E_kin,sys Hz/s

(4.8)

The chapter quotes the magnitude f₀ΔP/(2E_kin,sys) when the direction is clear from the sentence.

Proof.

Multiply each unit's speed equation in (4.3) by S_sys and sum over the n connected units:

Σ_i 2H_iS_i dΔω_i/dt = Σ_i (P_m,i − P_e,i)S_sys − Σ_i K_D,iS_i Δω_i.

At t = 0⁺ every Δω_i is still zero, so the damping sum is zero. By the definition of Δω_COI, the left side is 2E_kin,sys dΔω_COI/dt. On the right, assumption 1 holds every P_m,i at its pre-event value, so Σ_i P_m,iS_sys is the pre-event output of the n units. That output is the pre-event load minus ΔP, because the lost infeed supplied ΔP of that load. By assumption 3 and the algebraic network of Model 4.2, the n units' electrical outputs sum to the whole pre-event load at t = 0⁺. The sum of mechanical minus electrical power is therefore −ΔP in MW. Hence

2E_kin,sys dΔω_COI/dt = −ΔP.

Frequency in hertz is f = f₀(1 + Δω_COI), so df/dt = f₀ dΔω_COI/dt. Substituting gives (4.8). ∎

Where the hypotheses bind. Assumption 3 is not exact for the constant-impedance load of Model 4.2. The bus voltages move at the instant of the trip, and the load power moves with them to first order in the voltage change. This theorem neglects that change. Assumption 2 fails after the first swing, which is why (4.8) is an initial rate and not a trajectory; Theorem 4.6 supplies the trajectory. If the tripped unit's stored energy is counted in E_kin,sys by mistake, (4.8) understates the rate by the ratio of the two energies.

Three readings of (4.8) matter.

One. RoCoF₀ is proportional to 1/E_kin,sys with no control parameter in it: no governor, no droop gain, no filter cutoff. That is why it measures system strength; it reports stored energy and nothing else.

Two. RoCoF₀ is a centre-of-inertia quantity. Equation (4.8) is exact for ω_COI and not for any single bus. Immediately after the loss, the units electrically closest to the lost infeed supply most of the missing power and slow fastest. A local measurement can therefore read a rate higher than (4.8) until the angles have redistributed the deficit. A relay measures a bus, not the centre of inertia, so the planner's number and the relay's number differ.

Three. RoCoF is also a protection setting, and here two published figures circulate that must never be merged.

Two rate-of-change-of-frequency figures, kept separate.

The ENTSO-E withstand guidance is reported as 2 Hz/s measured over 500 ms [S15, ENTSO-E 2018, Rate of Change of Frequency (RoCoF) Withstand Capability — Guidance document for national implementation, withstand requirement — verify]. This is a capability a unit must survive.
The 1 Hz/s figure is the rate-of-change-of-frequency standard adopted in Ireland under the DS3 programme [S16, EirGrid and SONI, DS3 programme documentation, rate-of-change-of-frequency standard — verify, including which document and which year]. It is a system operating limit and a relay setting. This book's source list has no entry for the corresponding Great Britain loss-of-mains setting, so no Great Britain value is printed here.

These are different quantities with different purposes. This book never writes "ENTSO-E 1 Hz/s". Where §4.5 uses 1 Hz/s, it uses it as a design input of that case and attributes it to S16.

The two uses pull the same number in opposite directions. A lower relay threshold disconnects embedded generation sooner during a genuine system event, which makes the event worse. A higher threshold permits less stored energy, and widens the window in which an island goes undetected. Exercise 4.1 asks for that argument in full.

TS4 makes the sensitivity concrete. Take a 100 MW loss. At the base case, RoCoF₀ = 50 × 100 / (2 × 3100) = 5000/6200 = 0.8065 Hz/s. At c = 100 %, E_kin,sys = 2000 MVA·s and RoCoF₀ = 50 × 100 / (2 × 2000) = 5000/4000 = 1.2500 Hz/s. The same loss on the same rating produces a rate 1.55 times higher, purely because E_kin,sys fell from 3100 to 2000 MVA·s.

4.4 The frequency nadir

Local notation for §4.4.

Δf(t) — frequency deviation f(t) − f₀, in hertz, as a function of time. The book table defines only its value at the nadir, Δf_nadir.
τ — the time constant 2E_kin,sys/(f₀ D_load), in seconds. Used only in the load-damping extension and in Exercise 4.3.

RoCoF₀ is the slope at one instant. The depth of the fall is set by how fast the response arrives, and the simplest useful model of that is a linear ramp.

Theorem 4.6 — Frequency nadir under a linear primary-response ramp proved hereR36

Take the aggregated system of Theorem 4.5. At t = 0 an infeed of ΔP MW is lost as a step. Primary response R_p is delivered as a linear ramp that starts at t = 0 and reaches its full value at t = T. Assume R_p = ΔP and no load damping. Then the frequency deviation is

Δf(t) = − f₀ ΔP2 E_kin,sys ( t − t²/(2T) ) for 0 ≤ t ≤ T

(4.9)

and the nadir is reached at t = T with

Δf_nadir = − f₀ ΔP T4 E_kin,sys

(4.10)

Proof.

The aggregated speed equation from the proof of Theorem 4.5, written in hertz, is

(2E_kin,sys/f₀) dΔf/dt = −ΔP + R_p t/T, for 0 ≤ t ≤ T.

Put R_p = ΔP and integrate once from 0 to t with Δf(0) = 0. That gives (4.9). Differentiate (4.9): the slope is zero when 1 − t/T = 0, that is at t = T. The slope is negative before T and zero after it, so t = T is the minimum. Substituting t = T into (4.9) gives −(f₀ΔP/(2E_kin,sys))(T − T/2) = −f₀ΔPT/(4E_kin,sys), which is (4.10). The factor 4 is 2 × 2: one 2 from the swing equation, one 2 from the triangular area of the ramp. ∎

Status: proved here. The low-order frequency-response model this simplifies is standard [S13, Anderson and Mirheydar 1990, "A Low-Order System Frequency Response Model", IEEE Trans. Power Systems 5(3), 720–729, nadir model and its assumptions — verify].

Remark 4.1 — When the response does not match the loss

Keep the ramp but let R_p ≠ ΔP. The slope of Δf is zero when R_pt/T = ΔP. If R_p ≥ ΔP, the nadir is at t* = TΔP/R_p ≤ T, with Δf_nadir = −f₀ΔP²T/(4E_kin,sysR_p). More response arrives sooner and the nadir is shallower by the factor ΔP/R_p. If R_p < ΔP, the slope never reaches zero. After t = T the frequency keeps falling at the constant rate f₀(ΔP − R_p)/(2E_kin,sys). There is no nadir until something else acts. For Example 4.1 with R_p = 1500 MW, t* = 6.667 s and Δf_nadir = −0.4167 Hz. A Runge-Kutta run returns −0.416667 Hz at 6.6667 s.

4.4.1 What the assumptions do to the answer

A reader may meet the claim that (4.10) is conservative, that is, that it overstates the depth. The formula alone does not support that claim. Two of its assumptions push in opposite directions.

No load damping makes the computed nadir too deep. Real demand falls as frequency falls. That is a real power term, −D_loadΔf, on the right of the aggregated equation, and it opposes the deviation. Omitting it removes help that exists.
A ramp that starts at t = 0 makes the computed nadir too shallow. Real primary response has a deadband, a measurement delay and a governor time constant (Terms panel) [S13, Anderson and Mirheydar 1990, "A Low-Order System Frequency Response Model", IEEE Trans. Power Systems 5(3), 720–729, governor model — verify]. It does not start at the instant of the loss. Omitting that delay supplies help sooner than it arrives.
A ramp is not a governor. A real response is closer to a first-order lag than to a straight line.
No current limiting. If any converter hits I_max the model of R33 no longer describes it, and (4.10) does not know. This is the assumption that breaks in a large event.

The net direction is therefore not decidable from the formula alone. Equation (4.10) is a screening formula whose two largest errors have opposite signs, not a bound.

4.4.2 Load damping as a named extension

Adding load damping gives (2E_kin,sys/f₀) dΔf/dt = −ΔP + ΔPt/T − D_loadΔf. The equation is first order with time constant τ = 2E_kin,sys/(f₀D_load), and the group that decides whether damping matters is T/τ = D_loadTf₀/(2E_kin,sys), with D_load in MW/Hz, T in s, f₀ in Hz and E_kin,sys in MJ. Units: (MW/Hz)(s)(Hz)/MJ = MW·s/MJ = 1. For T/τ ≪ 1 the damping term is negligible; for T/τ ≫ 1 it, not the ramp, sets the depth. Exercise 4.3 works this out.

4.4.3 Worked case

Example 4.1 — The textbook nadir computed hereR37

Inputs. A 50 Hz system. E_kin,sys = 200 GVA·s = 200 000 MJ. This is the stored energy of the units that stay connected after the loss, as Theorem 4.5 requires. This figure is an input of this example. It is of the order reported for Great Britain in 2019 [S14, National Grid ESO 2019, Technical Report on the events of 9 August 2019, system inertia at the time of the event — verify]. This book has not read the value from that report. Treat it as assumed for this case, and verify the figure and its year before quoting it elsewhere. Loss of infeed ΔP = 1000 MW. Primary response R_p = 1000 MW delivered as a linear ramp over T = 10 s. No load damping.

Step 1 — initial rate, from (4.8).

RoCoF₀ = f₀ΔP/(2E_kin,sys) = 50 × 1000 / (2 × 200 000) = 50 000 / 400 000 = 0.125 Hz/s.

Step 2 — nadir, from (4.10).

Δf_nadir = −f₀ΔPT/(4E_kin,sys) = −50 × 1000 × 10 / (4 × 200 000) = −500 000 / 800 000 = −0.625 Hz, reached at t = T = 10 s. Hence f_nadir = 50 − 0.625 = 49.375 Hz.

Check. Integrating the aggregated equation numerically with a fourth-order Runge-Kutta step of 1 × 10⁻⁴ s returns a minimum deviation of −0.625000 Hz at t = 10.0000 s. The difference from the closed form is 1.8 × 10⁻¹³ Hz, which is round-off.

Step 3 — how much energy that is. The stored energy released down to the nadir is 2E_kin,sys|Δf|/f₀ = 2 × 200 000 × 0.625 / 50 = 5000 MJ, which is 2.5 % of the system's stored energy. The released energy is 2.5 % of the store. The depth of the nadir is set by the 10 s delivery time T in (4.10), not by the size of the store.

Step 4 — halve the inertia. Take E_kin,sys = 100 GVA·s with everything else unchanged.

RoCoF₀ = 50 × 1000 / (2 × 100 000) = 0.250 Hz/s.

Δf_nadir = −50 × 1000 × 10 / (4 × 100 000) = −1.250 Hz, so f_nadir = 48.750 Hz.

Reading. Both quantities scale as 1/E_kin,sys, so halving stored energy doubles the initial rate and doubles the depth. The halved-inertia nadir of 48.750 Hz sits 0.050 Hz below 48.8 Hz. That is the frequency at which Great Britain's low-frequency demand disconnection acted on 9 August 2019 [S14, National Grid ESO 2019, Technical Report on the events of 9 August 2019, event summary — verify]. §4.5 restates it with the event's two other figures as Citation 4.7. That is a scale check, not a claim about the event, which had a different loss, system and response.

Fig. 4.3 — Frequency against time for the Example 4.1 disturbance at four values of system stored energy: 300, 200, 100 and 50 GVA·s. Each curve is equation (4.9) evaluated every 0.5 s and held flat after t = T = 10 s. The filled circles mark the nadirs: 49.583, 49.375, 48.750 and 47.500 Hz. The dashed orange line is 48.8 Hz. That is the frequency at which Great Britain's low-frequency demand disconnection acted in 2019 [S14, National Grid ESO 2019, Technical Report on the events of 9 August 2019, event summary — verify]. §4.5 restates it as Citation 4.7. Halving the stored energy doubles the depth every time, so the nadir deviation scales as 1/E_kin,sys, which is what (4.10) says. The figure illustrates Theorem 4.6 and Example 4.1 (R36, R37).

4.5 A sizing case: replacing a 1.6 GVA synchronous condenser fleet

Local notation for §4.5.

E_kin,fleet — stored kinetic energy of the condenser fleet, in MJ. It is (4.4) applied to that fleet alone.
S_gfm — total rating of the grid-forming fleet that replaces the condensers, in MVA.
E_released — stored energy released over a frequency excursion, in MJ; P_headroom — instantaneous power delivered at a given rate of change of frequency, in MW; P_inertial — the same power for one converter, in per unit of its own rating.
Δf — here a fixed frequency excursion, not the time function Δf(t) of §4.4.

Citation 4.7 — Great Britain, 9 August 2019 cited, not proved hereR38

This book takes three figures from the event report and no others. The frequency nadir was 48.8 Hz. Low-frequency demand disconnection acting at 48.8 Hz shed about 931 MW of demand. The total infeed loss was about 1878 MW, including embedded generation that disconnected on its own protection [S14, National Grid ESO 2019, Technical Report on the events of 9 August 2019, 6 September 2019, event summary — verify].

No clock time and no individual plant output is quoted here, because the draft stage could not open the report; any such figure must be read from S14 first. The structural point taken from the event is that the outcome followed from a sequence: a loss, a second loss including embedded generation that tripped on its own protection, a nadir, then demand disconnection.

Example 4.2 — Sizing the grid-forming replacement computed hereR39

The fleet is a design case defined by this book. It is not a published fleet and it is not measured plant. The project brief labels it "1.6 GW"; a synchronous condenser is rated in MVA, and this book therefore writes 1.6 GVA throughout. Take 8 condensers of 200 MVA each, and take H = 3.5 s for each, including its flywheel. That inertia constant is assumed for this case — verify against a manufacturer figure or a standard before quoting it elsewhere. Everything below is arithmetic on those assumptions.

Stored energy of the fleet. By (4.4), E_kin,fleet = 8 × 200 × 3.5 = 5600 MVA·s = 5.6 GVA·s = 5600 MJ.

Now size a grid-forming fleet against three constraints, separately. They have different units. They cannot be added and their maximum is not meaningful, because a megajoule, a megawatt and a per-unit current are not comparable quantities.

Constraint 1 — stored energy

Over a frequency excursion of Δf = 0.8 Hz at f₀ = 50 Hz, the energy the condensers release is

E_released = 2 E_kin,fleet Δf/f₀ = 2 × 5600 × 0.8 / 50 = 8960/50 = 179.2 MJ

(4.11)

In other units, 179.2 / 3.6 = 49.78 kWh. Equation (4.11) is the fleet form of the energy bound (3.30) of R33 (Corollary 3.8), with E_kin,fleet in place of H_eqS_base. Equation (4.11) is the linearisation of the exact expression E_kin,fleet(1 − (1 − Δf/f₀)²) = 5600(1 − 0.984²) = 177.77 MJ. The dropped second-order term is 5600 × 0.016² = 1.43 MJ, which is 0.80 % of (4.11). The linearisation is stated, not assumed silently.

Compare that with the store on the replacement fleet. The fleet rating S_gfm = 1120 MVA is the figure that Constraint 2 of this example returns. Assumed for this case: a one-hour store, so E_res = 1120 MWh = 1120 × 3600 = 4 032 000 MJ = 1.12 GWh. The ratio is 4 032 000 / 179.2 = 22 500. Stored energy is not the binding constraint. It is short of binding by a factor of 22 500, that is 10^4.35.

Constraint 2 — power headroom

At a rate of change of frequency of 1 Hz/s, the instantaneous power the condensers deliver is

P_headroom = 2 E_kin,fleet RoCoF / f₀ = 2 × 5600 × 1.0 / 50 = 11200/50 = 224 MW

(4.12)

The 1 Hz/s here is a design input of this case. It is the figure adopted in Ireland under DS3 [S16, EirGrid and SONI, DS3 programme documentation, rate-of-change-of-frequency standard — verify]. It is not the ENTSO-E withstand figure, which is reported as 2 Hz/s over 500 ms [S15, ENTSO-E 2018, Rate of Change of Frequency (RoCoF) Withstand Capability — Guidance document for national implementation, withstand requirement — verify]. The two are kept apart here as §4.3 requires.

A converter fleet must hold 224 MW above its dispatch to match that. Convert it to a fleet rating. By R33 (Corollary 3.8, inequality (3.31)), a converter with equivalent inertia constant H_eq delivers, in per unit of its own rating, 2H_eq RoCoF/f₀. For a virtual synchronous machine, R32 gives H_eq = H_v, the chosen virtual inertia constant of R28. So

P_inertial = 2 H_v RoCoF / f₀ = 2 × 5.0 × 1.0 / 50 = 10/50 = 0.20 pu

(4.13)

at H_v = 5.0 s, which is assumed for this case. Hence

S_gfm = 224 / 0.20 = 1120 MVA.

A consistency check falls out of this. 1120 MVA at H_eq = H_v = 5.0 s carries H_eqS_gfm = 1120 × 5.0 = 5600 MVA·s of equivalent stored energy, which is exactly E_kin,fleet. Plan rule §3.6 item 4 keeps H, H_v and H_eq apart. The equality H_eq = H_v is a result of R32 for this family, not an identification. That is not a coincidence: (4.12) and (4.13) are the same relation applied twice, so matching the headroom at one rate of change of frequency is the same as matching H×S. The rate 1 Hz/s cancels out.

Constraint 3 — the current limit

224 MW on 1120 MVA is 224/1120 = 0.20 pu of extra power. Assumed for this case: the terminal voltage is 1.0 pu and the power factor is unity, so per-unit current equals per-unit power and 0.20 pu of power is 0.20 pu of current. A converter with I_max = 1.2 pu running at 1.0 pu of dispatch has 1.2 − 1.0 = 0.20 pu of current left. The margin after the inertial current is therefore 0.20 − 0.20 = 0.00 pu.

Nothing is left. There is no current for voltage support at the same instant and none for a fault contribution. A converter running at its limit is no longer described by the linear model of R32.

What binds

Power headroom binds, and the current limit binds at the same instant because it is the headroom divided by the fleet rating. Stored energy does not bind: it is short by a factor of 22 500. Writing the requirement as "1.6 GW of inertia" answers none of the three.

What this case does not answer. Whether those converters stay stable while saturated at I_max. Model 4.4 assumes no current limiting and R32 holds near the operating point; a large disturbance breaks both. That is the first open problem of §4.6.

Fig. 4.4 — The three constraints of Example 4.2, each drawn as the ratio of what the requirement needs to what the 1120 MVA fleet has, on a logarithmic axis. The three quantities are of different kinds — megajoules, megawatts and per-unit current — so the bars may be compared with the dashed line at ratio 1, and with nothing else. They cannot be added and their maximum has no meaning. Stored energy is short of binding by a factor of 22 500. Power headroom and the current limit both sit exactly on the line, because the third constraint is the second divided by the fleet rating. The figure illustrates Example 4.2 (R39). Numbers computed in Example 4.2.

4.5.1 The 2019 event as a scale check

This section makes two comparisons and refuses a third.

Scale of the fleet. The condenser fleet holds 5600 MJ against the 200 000 MJ of the Example 4.1 system, that is 2.80 % of it. This replacement decision covers 2.8 % of that system's inertia, not all of it.

Scale of the deviation. Example 4.1 at half inertia gave a nadir of 48.750 Hz from a 1000 MW loss with a 10 s ramp. Citation 4.7 reports 48.8 Hz from a total loss of about 1878 MW. The two are within 0.05 Hz, which says Theorem 4.6 is of the right order for a system of this size, and nothing more.

What is refused. This book does not use Theorem 4.6 to reproduce the 2019 event. It does not know that event's stored energy, response volume or delivery time, and Citation 4.7 lists the only three figures taken from the report. Treating the agreement as a validation would need those inputs.

Fig. 4.5 — The reported sequence of the Great Britain event of 9 August 2019: a first loss, a further loss compounded by embedded generation disconnecting on its own protection, a nadir at 48.8 Hz, and low-frequency demand disconnection shedding about 931 MW at that frequency. The total infeed loss was about 1878 MW. No clock time is drawn, because none was read from the source. The order of the losses and of the embedded-generation disconnection is drawn as the draft understood it. It was not read from S14, so it carries verify like the values. The figure shows that the outcome followed from the sequence and not from any one of the three numbers. It illustrates Citation 4.7 (R38). Every value carries [S14 — verify].

4.6 What remains open

Citation 4.8 — Open problems and the standards landscape cited, not proved hereR40

Each item below is a question with a source. None is a prediction, and none carries a date.

1. What does a grid-forming converter do while it is saturated at I_max? Every model in this book assumes it is not. Model 4.4 assumes no current limiting; R32 holds near the operating point; R33 names I_max as a bound but does not describe behaviour beyond it. Example 4.2 puts a fleet at exactly 0.00 pu of remaining current. The question is what the angle dynamics do when the current controller clips, and whether the converter resynchronises when the clipping ends. The specification side of this question is written down [S17, National Grid ESO, Grid Code modification GC0137, Minimum Specification Required for Provision of GB Grid Forming Capability, current-limit and performance requirements — verify].

2. What fault current does a converter-dominated system produce, and will the protection see it? A synchronous machine delivers 5 to 7 times rated current into a close fault, set by its subtransient reactance [S01, Kundur 1994, Power System Stability and Control, Ch. 3 — verify]. Chapter 2 prints the same figure beside equation (2.7). A converter supplies a current close to I_max, which is 1.2 pu in Example 4.2. The question is what that does to the reach and grading of existing protection [S18, IEEE Std 2800-2022, Standard for Interconnection and Interoperability of Inverter-Based Resources, fault-ride-through and current-injection requirements — verify].

3. Do the four grid-forming families interoperate? R32 shows they agree to first order near the operating point. It does not show they agree away from it, and Chapter 3 states where they separate. The question is what a fleet of mixed droop, virtual synchronous machine, matching and dispatchable virtual oscillator plant does together under a large disturbance, when the plant comes from different vendors [S12, Tayyebi, Groß, Anta, Kupzog and Dörfler 2020, "Frequency Stability of Synchronous Machines and Grid-Forming Power Converters", IEEE J. Emerging and Selected Topics in Power Electronics 8(2), 1004–1018, side-by-side comparison of the families — verify].

4. Can a converter-dominated system be started from black? A black start needs a source that energises a dead network, holds voltage and frequency while load is picked up, and survives the inrush. Nothing here addresses inrush. The question is which grid-forming families can do it, and under what conditions [S20, ENTSO-E 2017, High Penetration of Power Electronic Interfaced Power Sources (HPoPEIPS), IGD, system-level consequences of converter penetration — verify].

5. Do the manufacturer models match the measured plant? Every eigenvalue in Fig. 4.2 came from parameters taken as given. A real study uses vendor models, often black-box. The question is the gap between those models and the plant as measured [S18, IEEE Std 2800-2022, Standard for Interconnection and Interoperability of Inverter-Based Resources, model-validation requirements — verify].

6. Is the classification of stability itself still complete? The 2021 revision added converter-driven and resonance stability to the three classical classes, which is R11. The question is whether the frequency-stability arithmetic of Example 4.2 can stand without a converter-driven-stability calculation beside it [S03, Hatziargyriou et al. 2021, "Definition and Classification of Power System Stability — Revisited and Extended", IEEE Trans. Power Systems 36(4), 3271–3281, converter-driven and resonance stability classes — verify].

4.6.1 What this chapter does not cover

Four boundaries, stated so that the reader does not carry a result past them.

Nothing below the fundamental-frequency timescale. Model 4.2 makes the network algebraic, so electromagnetic transients, travelling waves and subsynchronous resonance are outside it. Fig. 4.2 could not show a converter-driven oscillation set by the current controller and the filter, because the model has no state that could carry one.
No voltage stability. Internal voltage magnitudes are constant in (4.2), and excitation control, flux decay and reactive-power limits are absent.
No large-signal converter behaviour. Theorem 4.6 is a linear ramp on an aggregated first-order equation, and Example 4.2 stops at the current limit.
No protection engineering and no market design. §4.6 states these as questions with sources and goes no further.

Exercises

Exercise 4.1 (conceptual)

Explain why the rate of change of frequency is both a measure of system strength and a protection setting, and why those two uses can conflict.

Answer target

By Theorem 4.5, RoCoF₀ = f₀ΔP/(2E_kin,sys) contains no control parameter, so it reports stored energy directly: that is the system-strength use. The same quantity is the input to loss-of-mains protection on embedded generation, which disconnects when it measures a rate above its threshold: that is the protection use. The conflict has two directions. Raising the threshold lets the system run on less stored energy without spurious disconnection, but widens the window in which a genuine islanding event goes undetected. Lowering the threshold detects islanding sooner, but makes embedded generation disconnect during a system-wide event and so deepens the very event it measured, which is the compounding mechanism in the sequence of Citation 4.7. Add the point from §4.3: (4.8) is exact for ω_COI, while a relay measures one bus, so the two uses do not even read the same number.

Exercise 4.2 (derivational)

Derive Δf_nadir = −f₀ΔPT/(4E_kin,sys) from the aggregated swing equation with a linear-ramp response, and state where the factor 4 comes from.

Answer target

Integrate (2E_kin,sys/f₀)dΔf/dt = −ΔP + ΔPt/T from 0 to T with Δf(0) = 0. This gives (4.9). The slope vanishes at t = T, and the slope is negative before and zero after, so the nadir is at t = T. Substituting gives (4.10). The factor 4 is 2 × 2. One 2 is the 2 in 2E_kin,sys from the swing equation of R04. The other 2 is the ½ of the triangular area under the ramp, which appears as T − T/2 = T/2 in the integration.

Exercise 4.3 (derivational)

Add load damping D_load in MW/Hz, show that the nadir is shallower, and state the limit in which the damping term dominates the ramp.

Answer target

The aggregated equation becomes (2E_kin,sys/f₀)dΔf/dt = −ΔP + ΔPt/T − D_loadΔf. Because Δf < 0 during the fall, the new term is positive and opposes the deviation, so the nadir is shallower than (4.10) for every D_load > 0. The equation is first order with time constant τ = 2E_kin,sys/(f₀D_load), and the dimensionless group is T/τ = D_loadTf₀/(2E_kin,sys). Units: (MW/Hz)(s)(Hz)/MJ = MW·s/MJ = 1, dimensionless, so the grouping is consistent. Damping dominates the ramp when T/τ ≫ 1. Numerical check on the Example 4.1 system, by fourth-order Runge-Kutta with step 1 × 10⁻⁴ s. The three D_load values test the grouping only and come from no source.

`D`_load (MW/Hz)	`T`/τ	nadir deviation (Hz)	time of nadir (s)
0	0	−0.6250	10.000
500	0.6250	−0.4464	7.768
1000	1.2500	−0.3513	6.487
2000	2.5000	−0.2494	5.011

Against the undamped row, the nadir is both shallower and earlier, and it no longer occurs at t = T.

Exercise 4.4 (computational)

Integrate the Example 4.1 system with T swept over 5, 10, 20 and 30 s, and confirm that the nadir depth depends linearly on T.

Answer target

Equation (4.10) predicts −0.3125, −0.625, −1.25 and −1.875 Hz. Integrating the aggregated equation with a fourth-order Runge-Kutta step of 1 × 10⁻⁴ s over 3T returns −0.312500, −0.625000, −1.250000 and −1.875000 Hz, each at t = T. The largest absolute difference from the closed form across the four runs is 2.8 × 10⁻¹² Hz, which is round-off at double precision and not a modelling error. The dependence is exactly linear, because T enters (4.10) as a first power. Report the integration step and the tolerance, as done here; a run that reports only the four numbers has not shown that they came from an integration.

Exercise 4.5 (computational)

For TS4, find the converter share c at which the least-damped mode falls below ζ = 0.05, sweeping in 5 % steps.

Answer target

State the parameter set first: TS4 of §4.1.3, the sweep rule of §4.2.4, the equilibrium assumptions of §4.1.3, and Model 4.4 with no current limiting. Then report what the sweep returns and nothing else. For this parameter set the sweep returns no crossing. The least-damped oscillatory mode is already below ζ = 0.05 at c = 0, where it is 0.0134, and it stays below 0.05 at every step up to c = 95 %, where it is 0.0185. Along the way it rises to 0.0138 at 30 %, 0.0145 at 50 %, 0.0187 at 80 % and 0.0212 at 90 %, then falls back to 0.0185 at 95 %. The mode that Fig. 4.2 shows losing damping does cross ζ = 0.05: it is 0.0592 at 80 %, 0.0458 at 85 %, 0.0319 at 90 % and 0.0206 at 95 %. It is never the least-damped mode, so it does not answer the question as asked. Report both. At c = 100 % both machines have been removed and the only remaining conjugate pair has ζ = 0.1393, so the reported minimum jumps. The correct answer is therefore that the question's premise does not hold for this parameter set: the machine mode is lightly damped at every share, and the converters improve it rather than degrade it, because K_D,eq = 20 pu against the machines' 2 pu. Note also that the state dimension changes at the two endpoints, where two units disappear, so the mode count is 1 pair at c = 0 and at c = 100 %, and 3 pairs in between. Do not adjust the parameters to manufacture a crossing; report the sweep.

Exercise 4.6 (conceptual)

Example 4.2 concludes that power headroom binds and stored energy does not. State one change to the case that would reverse that conclusion.

Answer target

Any one of the following. (a) A required response duration of minutes instead of seconds. A rate-of-change-of-frequency requirement implies a few seconds. The energy term grows with duration while the headroom term does not. At a 224 MW draw the 4 032 000 MJ store lasts 4 032 000/224 = 18 000 s = 5.0 h, so the reversal needs a duration of that order or a much smaller store. (b) A converter fleet with no long-duration store, such as photovoltaic without a battery. Then E_res is whatever the DC link holds. R30 computes a DC-link store of 14.4 kJ on a 1 MVA base, that is 0.0144 MJ per MVA. The one-hour store assumed here holds 3600 MJ per MVA. The ratio is 3600/0.0144 = 250 000. (c) A requirement referenced to a deep and sustained frequency excursion rather than to a rate of change of frequency, because (4.11) scales with Δf and with duration while (4.12) scales with RoCoF only.

Chapter summary

Definition 4.1 and Model 4.2 (R34, part 1) reduced the network to its source nodes, and Definition 4.3 and Model 4.4 (R34, part 2) assembled machines and converters into one state matrix. Fig. 4.2 showed that the eigenvalues of TS4 do not all move one way: stored energy fell 44 % while the least-damped mode improved from ζ = 0.0134 to 0.0187 and another degraded from ζ = 0.1267 at 20 % to 0.0592 at 80 % and 0.0206 at 95 %. Theorem 4.5 (R35) gave RoCoF₀ = f₀ΔP/(2E_kin,sys), exact for the centre of inertia and not for any single bus. Theorem 4.6 (R36) gave Δf_nadir = −f₀ΔPT/(4E_kin,sys). Example 4.1 (R37) gave 0.125 Hz/s and 49.375 Hz, doubling to 0.250 Hz/s and 48.750 Hz at half the stored energy. Citation 4.7 (R38) recorded the three figures taken from the Great Britain event of 9 August 2019. Example 4.2 (R39) sized the replacement of a 1.6 GVA condenser fleet: 179.2 MJ needed against 4 032 000 MJ held, 224 MW of headroom against 224 MW, and 0.20 pu of current against 0.20 pu. Citation 4.8 (R40) listed six open questions, the first of which starts where Example 4.2 stops: at the current limit, with nothing in reserve.

The reader can now do what plan Outcome 4 asks. Given a mixed fleet, compute E_kin,sys by (4.4), the initial rate by (4.8) and the nadir by (4.10). Given a replacement decision, write it as three sizing problems in three units by the method of Example 4.2, and name the one that binds. Given an eigenvalue plot, read damping from the constant-ζ rays and label each mode by its participation factors.

Every number here was recomputed before printing, by draft_ch4_check.py in the draft stage's scratchpad. Every citation carries — verify: no source was opened.