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19.3 Load Flow, Economic Dispatch & Power System Stability
PRC Registered Electrical Engineer REE
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19.3 Load Flow, Economic Dispatch & Power System Stability
Key Takeaways Explain This Topic AI
Load flow classifies every bus as slack (V and δ specified), PV or generator (P and |V| specified), or PQ or load (P and Q specified).
Gauss-Seidel load flow converges linearly with iteration count rising with system size, while Newton-Raphson converges quadratically in three to five iterations almost independently of size.
Real power transferred over a line is approximately P = (V1·V2/X)·sin δ, so maximum steady-state transfer occurs at a power angle of 90°.
Economic dispatch neglecting losses equalises the incremental cost of every generator, dC/dP = λ, subject to each unit's output limits.
The swing equation is (2H/ωs)·d²δ/dt² = Pm − Pe, and the equal-area criterion judges first-swing transient stability without integrating it.
Last updated: August 2026
19.3 Load Flow, Economic Dispatch & Power System Stability
The third pillar of Enhanced TOS topic K, Power System Analysis (PRBEE Resolution No. 40, s. 2024) is the steady-state and dynamic behaviour of an interconnected network: where power flows, what it costs, and whether the machines stay in step after a disturbance.
1. The Load-Flow Problem
Each bus has four quantities — ∣ V ∣ |V| ∣ V ∣ , δ \delta δ , P P P and Q Q Q — of which two are specified and two are computed.
Bus type Specified Computed Count
Slack / swing ∣ V ∣ \lvert V \rvert ∣ V ∣ , δ \delta δ (reference, δ = 0 \delta = 0 δ = 0 ) P P P , Q Q Q Exactly one per island
- PV / generator / voltage-controlled P P P , ∣ V ∣ \lvert V \rvert ∣ V ∣ Q Q Q , δ \delta δ Each generator bus
- PQ / load P P P , Q Q Q ∣ V ∣ \lvert V \rvert ∣ V ∣ , δ \delta δ All remaining buses
The slack bus exists because total system losses are unknown until the solution converges; the slack generator absorbs the mismatch. There is exactly one per electrical island, and it also sets the angle reference.
A PV bus reverts to a PQ bus whenever its reactive output hits the machine's Q Q Q limit — the voltage schedule can no longer be held and ∣ V ∣ |V| ∣ V ∣ becomes a computed quantity. This limit switching is why real load-flow programs iterate on bus type as well as on voltage.
Bus Admittance Matrix Y i i = ∑ admittances connected to bus i , Y i j = − y i j ( i ≠ j ) Y_{ii} = \sum \text{admittances connected to bus } i, \qquad Y_{ij} = -\,y_{ij} \ (i \ne j) Y ii = ∑ admittances connected to bus i , Y ij = − y ij ( i = j )
Y b u s Y_{bus} Y b u s is symmetric for a network without phase shifters, highly sparse (a bus connects to only a few others), and diagonally dominant — the property that guarantees Gauss-Seidel convergence, as established in the numerical methods chapter.
Power-Flow Equations P i = ∑ k = 1 n ∣ V i ∣ ∣ V k ∣ ∣ Y i k ∣ cos ( θ i k + δ k − δ i ) P_i = \sum_{k=1}^{n} |V_i||V_k||Y_{ik}|\cos(\theta_{ik} + \delta_k - \delta_i) P i = k = 1 ∑ n ∣ V i ∣∣ V k ∣∣ Y ik ∣ cos ( θ ik + δ k − δ i ) Q i = − ∑ k = 1 n ∣ V i ∣ ∣ V k ∣ ∣ Y i k ∣ sin ( θ i k + δ k − δ i ) Q_i = -\sum_{k=1}^{n} |V_i||V_k||Y_{ik}|\sin(\theta_{ik} + \delta_k - \delta_i) Q i = − k = 1 ∑ n ∣ V i ∣∣ V k ∣∣ Y ik ∣ sin ( θ ik + δ k − δ i )
Solution Methods
- Method Convergence Iterations Memory Notes
- Gauss-Seidel Linear 50–150, grows with system size Low Simple, robust from a poor start, slow
- Newton-Raphson Quadratic 3–5, nearly independent of size High (Jacobian) Industry standard
Fast decoupled Linear but fast 5–15 Low Exploits P P P – δ \delta δ and Q Q Q – ∣ V ∣ \vert V\vert ∣ V ∣ decoupling in high- X / R X/R X / R networks
Acceleration factors of roughly 1.4 to 1.6 are applied to Gauss-Seidel to reduce the iteration count.
2. Power Transfer and Voltage Stability
For a short line of reactance X X X between buses of voltage V 1 V_1 V 1 and V 2 V_2 V 2 separated by power angle δ \delta δ , with resistance neglected: P = V 1 V 2 X sin δ Q = V 1 ( V 1 − V 2 cos δ ) X P = \frac{V_1 V_2}{X}\sin\delta \qquad\qquad Q = \frac{V_1\left(V_1 - V_2\cos\delta\right)}{X} P = X V 1 V 2 sin δ Q = X V 1 ( V 1 − V 2 cos δ )
Two engineering rules fall straight out and are examined constantly:
Real power flows from the leading to the lagging bus and is controlled by the angle ; maximum steady-state transfer occurs at δ = 90 ° \delta = 90° δ = 90° ;
Reactive power flows from the higher-magnitude to the lower-magnitude bus and is controlled by the voltage magnitude .
This decoupling is the basis of the fast-decoupled load flow and of every voltage-control scheme: to raise a bus voltage, inject VARs locally rather than trying to import them across a reactive network.
Voltage collapse appears on the P–V (nose) curve : as load increases, voltage falls slowly, then plunges at the nose point where the Jacobian becomes singular and no solution exists. Operating margins are stated as a distance from that nose.
3. Economic Dispatch
Each thermal unit has a cost curve, conventionally quadratic: C i = a i + b i P i + c i P i 2 ( PHP/h ) ⟹ d C i d P i = b i + 2 c i P i ( PHP/MWh ) C_i = a_i + b_iP_i + c_iP_i^{2} \quad (\text{PHP/h}) \qquad \Longrightarrow \qquad \frac{dC_i}{dP_i} = b_i + 2c_iP_i \quad (\text{PHP/MWh}) C i = a i + b i P i + c i P i 2 ( PHP/h ) ⟹ d P i d C i = b i + 2 c i P i ( PHP/MWh )
Neglecting losses, minimum total cost occurs when every unit runs at the same incremental cost : d C 1 d P 1 = d C 2 d P 2 = ⋯ = λ , subject to ∑ P i = P D , P i , min ≤ P i ≤ P i , max \frac{dC_1}{dP_1} = \frac{dC_2}{dP_2} = \cdots = \lambda, \qquad \text{subject to} \quad \sum P_i = P_D, \quad P_{i,\min} \le P_i \le P_{i,\max} d P 1 d C 1 = d P 2 d C 2 = ⋯ = λ , subject to ∑ P i = P D , P i , m i n ≤ P i ≤ P i , m a x
A unit driven outside its limits is fixed at the violated limit and removed from the equal- λ \lambda λ set, and λ \lambda λ is re-solved for the remaining units.
Including transmission losses, the coordination equations introduce penalty factors : d C i d P i × L i = λ , L i = 1 1 − ∂ P L ∂ P i \frac{dC_i}{dP_i} \times L_i = \lambda, \qquad L_i = \frac{1}{1 - \dfrac{\partial P_L}{\partial P_i}} d P i d C i × L i = λ , L i = 1 − ∂ P i ∂ P L 1
A generator electrically remote from load has a penalty factor above unity, so it must be cheaper at the busbar to be worth dispatching — the reason a distant coal plant may be backed down in favour of nearer gas.
Note the distinction from unit commitment , which decides which units run over a horizon accounting for start-up cost and minimum up/down times. Economic dispatch allocates load among units already running.
4. Power System Stability
Type Disturbance Question answered
Steady-state stability Small, gradual load change Can the machine follow slow changes?
Transient stability Large — fault, line trip, sudden load rejection Does the machine survive the first swing ?
Dynamic stability Small, with controls modelled Do governor and exciter dynamics damp or amplify oscillation?
The Swing Equation 2 H ω s d 2 δ d t 2 = P m − P e (per unit) \frac{2H}{\omega_s}\frac{d^{2}\delta}{dt^{2}} = P_m - P_e \quad \text{(per unit)} ω s 2 H d t 2 d 2 δ = P m − P e (per unit)
where H H H is the inertia constant in MJ/MVA (typically 3–10 s for thermal units), P m P_m P m the mechanical input and P e P_e P e the electrical output. A large H H H means large stored kinetic energy and a slow, stable response — which is precisely why grids with high inverter-based renewable penetration are concerned about declining system inertia.
Equal-Area Criterion
For a single machine against an infinite bus, the system remains stable through the first swing if the decelerating area available after fault clearing is at least equal to the accelerating area accumulated during the fault: A accelerating ≤ A decelerating A_{\text{accelerating}} \le A_{\text{decelerating}} A accelerating ≤ A decelerating
The critical clearing angle δ c r \delta_{cr} δ cr is where the two areas are exactly equal, and the critical clearing time is how long the protection has to clear the fault before that angle is exceeded. This is the direct engineering link between stability studies and relay/breaker speed specifications: faster clearing buys stability margin, which is why high-speed relaying and single-pole tripping exist on critical transmission lines. Steady-state stability limit: P max = ∣ E ∣ ∣ V ∣ X s at δ = 90 ° \text{Steady-state stability limit: } P_{\max} = \frac{|E||V|}{X_s} \text{ at } \delta = 90° Steady-state stability limit: P m a x = X s ∣ E ∣∣ V ∣ at δ = 90°
- Solved Board Exam Examples
- Example 1: Power Angle and Transfer Limit
A generator with E = 1.20 E = 1.20 E = 1.20 pu behind a synchronous reactance of j 0.80 j0.80 j 0.80 pu feeds an infinite bus at 1.00 1.00 1.00 pu. Find the maximum steady-state power transfer and the power angle at 1.0 pu output.
Solution. P max = ( 1.20 ) ( 1.00 ) 0.80 = 1.50 pu at δ = 90 ° P_{\max} = \frac{(1.20)(1.00)}{0.80} = \boxed{1.50 \text{ pu}} \quad \text{at } \delta = 90° P m a x = 0.80 ( 1.20 ) ( 1.00 ) = 1.50 pu at δ = 90°
For P = 1.0 P = 1.0 P = 1.0 pu: 1.0 = 1.5 sin δ ⟹ sin δ = 0.6667 ⟹ δ = 41.8 ° 1.0 = 1.5\sin\delta \implies \sin\delta = 0.6667 \implies \delta = \boxed{41.8°} 1.0 = 1.5 sin δ ⟹ sin δ = 0.6667 ⟹ δ = 41.8°
The machine operates well inside the 90° limit, leaving a healthy stability margin.
Example 2: Economic Dispatch of Two Units
Two units have incremental costs d C 1 / d P 1 = 8.0 + 0.024 P 1 dC_1/dP_1 = 8.0 + 0.024P_1 d C 1 / d P 1 = 8.0 + 0.024 P 1 and d C 2 / d P 2 = 6.4 + 0.040 P 2 dC_2/dP_2 = 6.4 + 0.040P_2 d C 2 / d P 2 = 6.4 + 0.040 P 2 (₱/MWh, P P P in MW). Dispatch a total load of 300 MW economically, ignoring losses and limits.
Solution. Set the incremental costs equal: 8.0 + 0.024 P 1 = 6.4 + 0.040 P 2 , P 1 + P 2 = 300 8.0 + 0.024P_1 = 6.4 + 0.040P_2, \qquad P_1 + P_2 = 300 8.0 + 0.024 P 1 = 6.4 + 0.040 P 2 , P 1 + P 2 = 300
Substituting P 2 = 300 − P 1 P_2 = 300 - P_1 P 2 = 300 − P 1 : 8.0 + 0.024 P 1 = 6.4 + 0.040 ( 300 − P 1 ) = 6.4 + 12 − 0.040 P 1 8.0 + 0.024P_1 = 6.4 + 0.040(300 - P_1) = 6.4 + 12 - 0.040P_1 8.0 + 0.024 P 1 = 6.4 + 0.040 ( 300 − P 1 ) = 6.4 + 12 − 0.040 P 1 0.064 P 1 = 10.4 ⟹ P 1 = 162.5 MW , P 2 = 137.5 MW 0.064P_1 = 10.4 \implies P_1 = \boxed{162.5 \text{ MW}}, \qquad P_2 = \boxed{137.5 \text{ MW}} 0.064 P 1 = 10.4 ⟹ P 1 = 162.5 MW , P 2 = 137.5 MW
Check: λ 1 = 8.0 + 0.024 ( 162.5 ) = 11.90 \lambda_1 = 8.0 + 0.024(162.5) = 11.90 λ 1 = 8.0 + 0.024 ( 162.5 ) = 11.90 ₱/MWh and λ 2 = 6.4 + 0.040 ( 137.5 ) = 11.90 \lambda_2 = 6.4 + 0.040(137.5) = 11.90 λ 2 = 6.4 + 0.040 ( 137.5 ) = 11.90 ₱/MWh — equal, as required. Note that the unit with the lower intercept does not simply take all the load; the slopes decide the split.
Example 3: Rotor Acceleration from the Swing Equation
A 60 Hz turbine-generator of H = 5.0 H = 5.0 H = 5.0 MJ/MVA suffers a fault that drops its electrical output from 1.0 pu to 0.30 pu while mechanical input stays at 1.0 pu. Find the initial rotor acceleration in electrical degrees per second squared.
Solution. The accelerating power is P a = 1.0 − 0.30 = 0.70 P_a = 1.0 - 0.30 = 0.70 P a = 1.0 − 0.30 = 0.70 pu, and ω s = 360 f = 21 600 \omega_s = 360f = 21\,600 ω s = 360 f = 21 600 electrical degrees per second. d 2 δ d t 2 = ω s 2 H P a = 21 600 2 ( 5.0 ) ( 0.70 ) = 2160 ( 0.70 ) = 1512 elec. deg/s 2 \frac{d^{2}\delta}{dt^{2}} = \frac{\omega_s}{2H}P_a = \frac{21\,600}{2(5.0)}(0.70) = 2160(0.70) = \boxed{1512 \text{ elec. deg/s}^{2}} d t 2 d 2 δ = 2 H ω s P a = 2 ( 5.0 ) 21 600 ( 0.70 ) = 2160 ( 0.70 ) = 1512 elec. deg/s 2
At this rate the rotor advances roughly 7.6 electrical degrees in the first 0.1 s, so a five-cycle (83 ms) breaker keeps the swing small — the quantitative justification for high-speed clearing.