Load Flow Analysis

Both Gauss–Seidel (G-S) and Newton–Raphson (N-R) solve the nonlinear power-flow equations iteratively. G-S updates one bus voltage at a time using $Y_{bus}$; N-R linearises all power equations together using the Jacobian matrix.

Comparison

Advantages of N-R over G-S

1. Quadratic convergence: a highly accurate solution in 3–5 iterations.
2. Number of iterations does not grow with system size, so it is faster overall for large systems.
3. More reliable; converges where G-S diverges (heavy loading, series capacitors, high R/X).
4. Not sensitive to the choice of slack bus; no acceleration factor needed.
5. The Jacobian can be reused for sensitivity analysis, contingency studies and optimal power flow.

Disadvantages of N-R compared with G-S

1. Programming logic is more complex.
2. Each iteration takes longer, since the Jacobian is recalculated and a linear system solved.
3. Needs more memory (reduced by sparsity techniques and decoupled versions).
4. Needs a reasonably good initial guess (flat start usually works); a poor start may diverge.

The decoupled load-flow method is a simplification of Newton–Raphson in which the weak coupling between P and $|V|$ and between Q and $\delta$ is neglected, so the large N-R system splits into two smaller ones.

Assumptions

1. Transmission lines have high X/R ratio ($X \gg R$), so $G_{ik} \ll B_{ik}$.
2. Angle differences between adjacent buses are small, so $\sin(\delta_i - \delta_k) \approx 0$ (small) and $\cos(\delta_i - \delta_k) \approx 1$.
3. Hence real power depends mainly on angles and reactive power mainly on voltage magnitudes: $\partial P/\partial |V| \approx 0$ (submatrix $J_2 = 0$) and $\partial Q/\partial \delta \approx 0$ (submatrix $J_3 = 0$).

Derivation

Full N-R equations:

$$\begin{bmatrix} \Delta P \\ \Delta Q \end{bmatrix} = \begin{bmatrix} J_1 & J_2 \\ J_3 & J_4 \end{bmatrix} \begin{bmatrix} \Delta\delta \\ \Delta|V| \end{bmatrix}$$

Setting $J_2 = J_3 = 0$:

$$\Delta P = J_1\,\Delta\delta = \left[\frac{\partial P}{\partial \delta}\right]\Delta\delta,\qquad \Delta Q = J_4\,\Delta|V| = \left[\frac{\partial Q}{\partial |V|}\right]\Delta|V|$$

where

$$\begin{aligned} \frac{\partial P_i}{\partial \delta_i} &= \sum_{k \ne i} |V_i||V_k||Y_{ik}|\sin(\theta_{ik} - \delta_i + \delta_k) \\ \frac{\partial P_i}{\partial \delta_k} &= -|V_i||V_k||Y_{ik}|\sin(\theta_{ik} - \delta_i + \delta_k) \\ \frac{\partial Q_i}{\partial |V_i|} &= -2|V_i||Y_{ii}|\sin\theta_{ii} - \sum_{k \ne i} |V_k||Y_{ik}|\sin(\theta_{ik} - \delta_i + \delta_k) \\ \frac{\partial Q_i}{\partial |V_k|} &= -|V_i||Y_{ik}|\sin(\theta_{ik} - \delta_i + \delta_k) \end{aligned}$$

$J_1$ is of order $(n-1)$ (all buses except slack) and $J_4$ is of order $(n-1-m)$ (PQ buses only, $m$ = number of PV buses).

Algorithm

1. Form $Y_{bus}$. Set flat start: $|V_i| = 1.0$ at PQ buses, specified $|V|$ at PV buses, $\delta_i = 0$.
2. Compute $P_i^{cal}$ (all non-slack buses) and $Q_i^{cal}$ (PQ buses).
3. Find mismatches $\Delta P_i = P_i^{sp} - P_i^{cal}$, $\Delta Q_i = Q_i^{sp} - Q_i^{cal}$.
4. If all $|\Delta P|, |\Delta Q| < \varepsilon$, go to step 8.
5. Form $J_1$ and solve $\Delta\delta = J_1^{-1}\Delta P$; update $\delta \leftarrow \delta + \Delta\delta$.
6. Form $J_4$ and solve $\Delta|V| = J_4^{-1}\Delta Q$; update $|V| \leftarrow |V| + \Delta|V|$. Check PV-bus Q limits.
7. Go to step 2.
8. Compute slack bus power, line flows and losses.

The Gauss–Seidel (G-S) method solves the load-flow equations by rewriting each bus equation to give $V_i$ in terms of the other voltages and updating the buses one by one, using the newest values immediately.

Example network

Consider a 3-bus system: bus 1 slack ($V_1 = |V_1|\angle 0$ given), bus 2 PV (generator, $P_2$, $|V_2|$ given, $Q_{2,min} \le Q_2 \le Q_{2,max}$), bus 3 PQ (load, $P_3$, $Q_3$ given).

   G1                      G2
   |  slack       PV bus   |
  (1)----------------------(2)
     \                    /
      \                  /
       ---------(3)------
                 |
               Load (PQ bus)

Basic G-S equations

From $I_i = \sum_k Y_{ik}V_k$ and $P_i - jQ_i = V_i^*I_i$:

$$V_i^{(r+1)} = \frac{1}{Y_{ii}}\left[\frac{P_i - jQ_i}{V_i^{*(r)}} - \sum_{k=1}^{i-1}Y_{ik}V_k^{(r+1)} - \sum_{k=i+1}^{n}Y_{ik}V_k^{(r)}\right]$$ $$Q_i = -\operatorname{Im}\left\{V_i^*\sum_{k=1}^{n}Y_{ik}V_k\right\}$$

Algorithm

1. Read data and form $Y_{bus}$. Net injections $P_i = P_{Gi} - P_{Di}$, $Q_i = Q_{Gi} - Q_{Di}$.
2. Initial values (flat start): slack $V_1$ as given; PV buses $|V_i| = |V_i|^{sp}$, $\delta_i = 0$; PQ buses $V_i = 1.0\angle 0^\circ$. Choose tolerance $\varepsilon$ and acceleration factor $\alpha$ (about 1.4–1.6). Set iteration count $r = 0$.
3. For each bus $i = 2, \ldots, n$ (slack bus skipped):
   • If PQ bus: compute $V_i^{(r+1)}$ from the G-S equation with specified $P_i$, $Q_i$.
   • If PV bus:
     1. Compute $Q_i^{(r+1)} = -\operatorname{Im}\{V_i^{*(r)}\sum_k Y_{ik}V_k\}$ using the latest voltages.
     2. Check Q limits:
        • If $Q_{i,min} \le Q_i^{(r+1)} \le Q_{i,max}$: compute $V_i$ from the G-S equation with this $Q_i$, keep its angle $\delta_i$, and reset its magnitude to $|V_i|^{sp}$: $V_i = |V_i|^{sp}\angle\delta_i$.
        • If $Q_i > Q_{i,max}$: set $Q_i = Q_{i,max}$; if $Q_i < Q_{i,min}$: set $Q_i = Q_{i,min}$. Treat the bus as a PQ bus for this iteration: compute $V_i$ from the G-S equation and do not reset its magnitude.
   • Acceleration (PQ buses): $V_{i,acc}^{(r+1)} = V_i^{(r)} + \alpha\left(V_i^{(r+1)} - V_i^{(r)}\right)$.
   • Use the new $V_i$ immediately for the following buses.
4. Convergence check: $\Delta V_{max} = \max_i |V_i^{(r+1)} - V_i^{(r)}|$. If $\Delta V_{max} > \varepsilon$, set $r = r + 1$ and repeat step 3. (A bus that was switched to PQ may return to PV if its voltage later allows Q to come back within limits.)
5. After convergence:
   • Slack bus power: $P_1 - jQ_1 = V_1^*\sum_k Y_{1k}V_k$.
   • Q at PV buses.
   • Line flows: $S_{ik} = V_i\left[(V_i - V_k)y_{ik} + V_i\frac{y'_{ik}}{2}\right]^*$.
   • Losses: $S_{L,ik} = S_{ik} + S_{ki}$; total loss $= \sum S_{L}$.
6. Print results.

Flow chart (summary)

 Form Ybus -> flat start -> r = 0
        |
        v
  for i = 2..n:
    PQ? -> new Vi (G-S), accelerate
    PV? -> find Qi -> within limits?
           yes: new Vi, reset |Vi|
           no : Qi = limit, treat as PQ
        |
        v
  max|dV| < eps ? --no--> r = r+1, repeat
        | yes
        v
  slack power, line flows, losses

The Newton–Raphson (N-R) method linearises the nonlinear power equations about the current estimate by a first-order Taylor series and solves for the corrections in bus voltage angles and magnitudes, repeating until the power mismatches vanish.

Power equations

With $Y_{kn} = |Y_{kn}|\angle\theta_{kn}$ and $V_k = |V_k|\angle\delta_k$:

$$\begin{aligned} P_k &= \sum_{n=1}^{N}|V_k||V_n||Y_{kn}|\cos(\theta_{kn} + \delta_n - \delta_k) \\ Q_k &= -\sum_{n=1}^{N}|V_k||V_n||Y_{kn}|\sin(\theta_{kn} + \delta_n - \delta_k) \end{aligned}$$

So $P_k$ and $Q_k$ are functions of all $\delta$ and $|V|$.

Unknowns and equations

Let bus 1 be the slack bus, buses $2 \ldots m+1$ PV buses ($m$ of them), and the rest PQ buses.

• Unknowns: $\delta_k$ at all $(N-1)$ non-slack buses; $|V_k|$ at the $(N-1-m)$ PQ buses.
• Equations: $P_k$ at the $(N-1)$ non-slack buses; $Q_k$ at the $(N-1-m)$ PQ buses.
• Total: $2N - 2 - m$ equations and unknowns.

Linearisation (Taylor series)

Let $\delta_k^{(0)}$, $|V_k|^{(0)}$ be initial estimates, and $\Delta\delta_k$, $\Delta|V_k|$ the corrections. Expanding $P_k$ about the estimate and neglecting higher-order terms:

$$P_k^{sp} = P_k^{(0)} + \sum_{n=2}^{N}\frac{\partial P_k}{\partial \delta_n}\Delta\delta_n + \sum_{n \in PQ}\frac{\partial P_k}{\partial |V_n|}\Delta|V_n|$$

and similarly for $Q_k$. Writing the mismatches $\Delta P_k = P_k^{sp} - P_k^{(0)}$ and $\Delta Q_k = Q_k^{sp} - Q_k^{(0)}$:

$$\begin{bmatrix} \Delta P \\ \Delta Q \end{bmatrix} = \begin{bmatrix} J_1 & J_2 \\ J_3 & J_4 \end{bmatrix} \begin{bmatrix} \Delta\delta \\ \Delta|V| \end{bmatrix}, \quad J_1 = \frac{\partial P}{\partial \delta},\ J_2 = \frac{\partial P}{\partial |V|},\ J_3 = \frac{\partial Q}{\partial \delta},\ J_4 = \frac{\partial Q}{\partial |V|}$$

General expression for corrections

$$\begin{bmatrix} \Delta\delta \\ \Delta|V| \end{bmatrix}^{(r)} = [J^{(r)}]^{-1} \begin{bmatrix} \Delta P \\ \Delta Q \end{bmatrix}^{(r)}, \qquad \begin{aligned} \delta_k^{(r+1)} &= \delta_k^{(r)} + \Delta\delta_k^{(r)} \\ |V_k|^{(r+1)} &= |V_k|^{(r)} + \Delta|V_k|^{(r)} \end{aligned}$$

In practice the linear system is solved by triangular factorisation (Gaussian elimination), not by explicit inversion.

Elements of the Jacobian

Differentiating the given expressions (write $\phi_{kn} = \theta_{kn} + \delta_n - \delta_k$):

$J_1$ ($\partial P/\partial\delta$), order $(N-1)\times(N-1)$:

$$\frac{\partial P_k}{\partial \delta_k} = \sum_{n \ne k}|V_k||V_n||Y_{kn}|\sin\phi_{kn},\qquad \frac{\partial P_k}{\partial \delta_n} = -|V_k||V_n||Y_{kn}|\sin\phi_{kn}\ (n \ne k)$$

$J_2$ ($\partial P/\partial|V|$), order $(N-1)\times(N-1-m)$:

$$\frac{\partial P_k}{\partial |V_k|} = 2|V_k||Y_{kk}|\cos\theta_{kk} + \sum_{n \ne k}|V_n||Y_{kn}|\cos\phi_{kn},\qquad \frac{\partial P_k}{\partial |V_n|} = |V_k||Y_{kn}|\cos\phi_{kn}$$

$J_3$ ($\partial Q/\partial\delta$), order $(N-1-m)\times(N-1)$:

$$\frac{\partial Q_k}{\partial \delta_k} = \sum_{n \ne k}|V_k||V_n||Y_{kn}|\cos\phi_{kn},\qquad \frac{\partial Q_k}{\partial \delta_n} = -|V_k||V_n||Y_{kn}|\cos\phi_{kn}$$

$J_4$ ($\partial Q/\partial|V|$), order $(N-1-m)\times(N-1-m)$:

$$\frac{\partial Q_k}{\partial |V_k|} = -2|V_k||Y_{kk}|\sin\theta_{kk} - \sum_{n \ne k}|V_n||Y_{kn}|\sin\phi_{kn},\qquad \frac{\partial Q_k}{\partial |V_n|} = -|V_k||Y_{kn}|\sin\phi_{kn}$$

Off-diagonal elements are zero where $Y_{kn} = 0$, so the Jacobian is as sparse as $Y_{bus}$. It is not symmetric and must be recomputed every iteration.

N-R algorithm

1. Read bus and line data; form $Y_{bus}$ and express its elements in polar form.
2. Flat start: $\delta_k^{(0)} = 0$ at all non-slack buses; $|V_k|^{(0)} = 1.0$ at PQ buses; $|V_k| = |V_k|^{sp}$ at PV buses. Set $r = 0$, tolerance $\varepsilon$.
3. Compute $P_k^{(r)}$ (non-slack buses) and $Q_k^{(r)}$ (PQ buses) from the power equations.
4. For PV buses, compute $Q_k^{(r)}$ and check limits. If violated, fix $Q_k$ at the limit and treat the bus as a PQ bus (its $|V_k|$ becomes an unknown, adding one row/column).
5. Compute mismatches $\Delta P_k^{(r)} = P_k^{sp} - P_k^{(r)}$, $\Delta Q_k^{(r)} = Q_k^{sp} - Q_k^{(r)}$.
6. If $\max(|\Delta P|, |\Delta Q|) < \varepsilon$, go to step 10.
7. Evaluate the Jacobian elements $J_1$–$J_4$ with the present $\delta$ and $|V|$.
8. Solve $[J][\Delta\delta;\ \Delta|V|] = [\Delta P;\ \Delta Q]$ for the corrections.
9. Update $\delta_k^{(r+1)} = \delta_k^{(r)} + \Delta\delta_k$, $|V_k|^{(r+1)} = |V_k|^{(r)} + \Delta|V_k|$; set $r = r + 1$ and go to step 3.
10. Compute slack bus power, PV-bus reactive powers, line flows $S_{ik}$ and losses; print results.

 Ybus -> flat start -> P,Q calc -> dP, dQ
                         ^             |
                         |      converged? --yes--> flows
                         |             | no
   update d,|V| <- solve J*dx <- form Jacobian

N-R converges quadratically, usually in 3–5 iterations regardless of system size.

In load-flow analysis each bus $i$ has four quantities: real power $P_i$, reactive power $Q_i$, voltage magnitude $|V_i|$ and voltage angle $\delta_i$. Two are specified and the other two are found by the load flow. On this basis buses are of three types.

1. Slack (swing / reference) bus

• Usually the largest generating station; only one in the system.
• Specified: $|V|$ and $\delta$ (normally $\delta = 0^\circ$, the reference angle).
• Found: $P$ and $Q$.
• Its generation supplies the difference between total load plus the (initially unknown) losses and the scheduled generation of other units.

2. PV (generator / voltage-controlled) bus

• Buses with generators, synchronous condensers or SVCs that can hold voltage.
• Specified: $P$ (set by turbine) and $|V|$ (set by AVR); limits $Q_{min}$, $Q_{max}$.
• Found: $Q$ and $\delta$.
• If computed Q goes outside its limits, Q is fixed at the limit and the bus becomes a PQ bus.

3. PQ (load) bus

• Buses with only loads (or fixed generation); about 80–90% of all buses.
• Specified: $P$ and $Q$ (net injection $P_G - P_D$, $Q_G - Q_D$).
• Found: $|V|$ and $\delta$.

Summary

In N-R analysis this gives $(N-1)$ P-equations (all except slack) and $(N-1-m)$ Q-equations (PQ buses only).

Data and Ybus

Each line: $z = j0.12$, $y = 1/j0.12 = -j8.3333$ pu. Every bus has two lines:

$$Y_{bus} = \begin{bmatrix} -j16.6667 & j8.3333 & j8.3333 \\ j8.3333 & -j16.6667 & j8.3333 \\ j8.3333 & j8.3333 & -j16.6667 \end{bmatrix}$$

Bus 1: slack, $V_1 = 1.0\angle 0^\circ$. Buses 2 and 3 are PQ buses with net injections $S_2 = -1.5 - j0.4$ and $S_3 = -1.2 - j0.5$ pu. Flat start: $V_2^{(0)} = V_3^{(0)} = 1.0\angle 0^\circ$.

G-S equation (latest values used immediately):

$$V_i^{(r+1)} = \frac{1}{Y_{ii}}\left[\frac{P_i - jQ_i}{V_i^{*(r)}} - \sum_{k \ne i}Y_{ik}V_k\right]$$

(i) Iteration 1

$$\begin{aligned} V_2^{(1)} &= \frac{1}{-j16.6667}\left[\frac{-1.5 + j0.4}{1.0} - j8.3333(1.0) - j8.3333(1.0)\right] \\ &= \frac{-1.5 - j16.2667}{-j16.6667} = 0.9760 - j0.0900 = 0.9801\angle -5.2685^\circ \\ V_3^{(1)} &= \frac{1}{-j16.6667}\left[\frac{-1.2 + j0.5}{1.0} - j8.3333(1.0) - j8.3333(0.9760 - j0.0900)\right] \\ &= \frac{-1.95 - j15.9667}{-j16.6667} = 0.9580 - j0.1170 = 0.9651\angle -6.9630^\circ \end{aligned}$$

Iteration 2

$$\begin{aligned} \frac{-1.5 + j0.4}{V_2^{*(1)}} &= -1.4865 + j0.5469 \\ V_2^{(2)} &= \frac{(-1.4865 + j0.5469) - (0.9750 + j16.3167)}{-j16.6667} \\ &= 0.9462 - j0.1477 = 0.9576\angle -8.8715^\circ \\ \frac{-1.2 + j0.5}{V_3^{*(1)}} &= -1.1714 + j0.6650 \\ V_3^{(2)} &= \frac{(-1.1714 + j0.6650) - (1.2307 + j16.2182)}{-j16.6667} \\ &= 0.9332 - j0.1441 = 0.9443\angle -8.7797^\circ \end{aligned}$$

Here $\sum_{k \ne 2}Y_{2k}V_k = j8.3333(V_1 + V_3^{(1)}) = 0.9750 + j16.3167$ and $\sum_{k \ne 3}Y_{3k}V_k = j8.3333(V_1 + V_2^{(2)}) = 1.2307 + j16.2182$.

(ii) Slack bus power (with iteration-2 voltages)

$$\begin{aligned} I_1 &= Y_{11}V_1 + Y_{12}V_2 + Y_{13}V_3 = 2.4318 - j1.0052 \\ S_1 &= V_1I_1^* = 1.0(2.4318 + j1.0052) \end{aligned}$$

So $P_1 = 2.4318$ pu and $Q_1 = 1.0052$ pu.

(iii) Network losses

Line flows $S_{ik} = V_i[y_{ik}(V_i - V_k)]^*$:

Answer:

• $V_2 = 0.9576\angle -8.87^\circ$ pu, $V_3 = 0.9443\angle -8.78^\circ$ pu after 2 iterations.
• Slack bus: $P_1 = 2.432$ pu, $Q_1 = 1.005$ pu.
• Losses: real power loss $= 0$ (lines have no resistance); reactive power loss $= 0.418$ pu.

Note: after only two iterations the solution has not fully converged, so $S_1$ minus the loads ($2.432 - 2.7 = -0.27$) does not yet equal the losses; with more iterations $P_1$ tends to 2.7 pu.

Specified and unspecified quantities

The slack bus supplies the losses and balances generation and load; a PV bus whose Q hits a limit is converted to a PQ bus.

Assumptions in decoupled load flow

1. $X \gg R$ for transmission lines, so $G_{ik} \ll B_{ik}$.
2. Angle differences across lines are small: $\sin(\delta_i - \delta_k) \approx 0$, $\cos(\delta_i - \delta_k) \approx 1$.
3. So P depends mainly on $\delta$ and Q mainly on $|V|$: $J_2 = \partial P/\partial|V| \approx 0$ and $J_3 = \partial Q/\partial\delta \approx 0$.

The N-R equations then split into:

$$\Delta P = J_1\,\Delta\delta,\qquad \Delta Q = J_4\,\Delta|V|$$

Steps

1. Form $Y_{bus}$; flat start ($\delta = 0$, $|V| = 1.0$ at PQ buses, $|V|^{sp}$ at PV buses).
2. Calculate $P_i$ for all non-slack buses and $Q_i$ for PQ buses.
3. Find mismatches $\Delta P$ and $\Delta Q$; stop if all are below tolerance.
4. Form $J_1$ and solve $\Delta\delta = J_1^{-1}\Delta P$; update $\delta$.
5. Form $J_4$ and solve $\Delta|V| = J_4^{-1}\Delta Q$; update $|V|$.
6. Check Q limits of PV buses; repeat from step 2.
7. After convergence, compute slack power, line flows and losses.

The Fast Decoupled Load Flow (FDLF) method (Stott and Alsac) simplifies the decoupled N-R method further so that the two Jacobian submatrices become constant matrices $B'$ and $B''$, formed once.

Approximations

1. Decoupling: $X \gg R$, so $\partial P/\partial|V| \approx 0$ and $\partial Q/\partial\delta \approx 0$ ($J_2 = J_3 = 0$).
2. Small angle differences: $\cos(\delta_i - \delta_k) \approx 1$, $\sin(\delta_i - \delta_k) \approx 0$; with $G_{ik}\sin(\delta_i - \delta_k) \ll B_{ik}$.
3. $Q_i \ll B_{ii}|V_i|^2$, so the $Q_i$ term in the diagonal elements is neglected.
4. Voltages near 1.0 pu, so $|V_k| \approx 1$ in the Jacobian terms.

With these, $J_1 \approx -|V|B'|V|$ and $J_4 \approx -|V|B''$, giving:

$$\frac{\Delta P}{|V|} = -B'\,\Delta\delta,\qquad \frac{\Delta Q}{|V|} = -B''\,\Delta|V|$$

• $B'$: imaginary part of $Y_{bus}$ for non-slack buses (order $N-1$). Usually shunts, line charging, off-nominal taps and series resistance are omitted from it.
• $B''$: imaginary part of $Y_{bus}$ for PQ buses only (order $N-1-m$).

Why it is more efficient than N-R

1. $B'$ and $B''$ are constant: formed and factorised (LU) only once; full N-R recomputes and refactorises the Jacobian each iteration.
2. The two matrices are real, symmetric and sparse and about half the size of the full Jacobian, so less memory and fewer operations.
3. Each iteration is only forward/back substitution, so it is several times faster per iteration.
4. More iterations are needed (linear-type convergence), but total time is much less; very suitable for on-line and contingency analysis (many repeated load flows).
5. Accuracy of the final answer is not lost, since the mismatches $\Delta P$, $\Delta Q$ are still computed exactly.

Algorithm

1. Read data; form $Y_{bus}$, then form $B'$ and $B''$ and factorise them.
2. Flat start: $\delta = 0$; $|V| = 1.0$ at PQ buses and $|V|^{sp}$ at PV buses.
3. Compute $P_i$ at non-slack buses; $\Delta P_i = P_i^{sp} - P_i$; form $\Delta P/|V|$.
4. Solve $-B'\Delta\delta = \Delta P/|V|$; update $\delta \leftarrow \delta + \Delta\delta$.
5. Compute $Q_i$ at PQ buses with the new angles; $\Delta Q_i = Q_i^{sp} - Q_i$; form $\Delta Q/|V|$. (Check PV-bus Q limits.)
6. Solve $-B''\Delta|V| = \Delta Q/|V|$; update $|V| \leftarrow |V| + \Delta|V|$.
7. If $\max|\Delta P|$ and $\max|\Delta Q|$ are less than tolerance, go to step 8; else go to step 3.
8. Compute slack bus power, line flows and losses.

Decoupled load-flow equations

In a transmission system, real power depends mainly on voltage angles and reactive power mainly on voltage magnitudes. Decoupled load-flow equations exploit this by dropping the weak couplings in the N-R equations.

Full N-R:

$$\begin{bmatrix} \Delta P \\ \Delta Q \end{bmatrix} = \begin{bmatrix} J_1 & J_2 \\ J_3 & J_4 \end{bmatrix} \begin{bmatrix} \Delta\delta \\ \Delta|V| \end{bmatrix}$$

Since $J_2 = \partial P/\partial|V|$ and $J_3 = \partial Q/\partial\delta$ are small, they are set to zero, giving two separate (decoupled) sets:

$$\Delta P = \left[\frac{\partial P}{\partial \delta}\right]\Delta\delta,\qquad \Delta Q = \left[\frac{\partial Q}{\partial |V|}\right]\Delta|V|$$

These are solved alternately: the P–δ equations give angle corrections, the Q–V equations give magnitude corrections. Each system is about a quarter the size of the full Jacobian.

Assumptions in the Fast Decoupled load flow method

1. Lines have high X/R ratio, so $G_{ik} \ll B_{ik}$ and $J_2$, $J_3$ are neglected.
2. Angle differences between connected buses are small: $\cos(\delta_i - \delta_k) \approx 1$, $\sin(\delta_i - \delta_k) \approx 0$.
3. $Q_i \ll B_{ii}|V_i|^2$, so $Q_i$ is neglected in the diagonal Jacobian terms.
4. $|V_i| \approx 1.0$ pu in the Jacobian terms.
5. In $B'$, elements that mainly affect Q (shunt reactances, line charging, off-nominal taps) and series resistances are omitted.

The result is

$$\frac{\Delta P}{|V|} = -B'\Delta\delta,\qquad \frac{\Delta Q}{|V|} = -B''\Delta|V|$$

with constant matrices $B'$ and $B''$ that are formed and factorised only once.

Basis: In transmission networks $X \gg R$ and the angle difference across lines is small. So real power flow depends mainly on voltage angles and reactive power mainly on voltage magnitudes ($P = \frac{V_1V_2}{X}\sin\delta$, $Q \approx \frac{V_1(V_1 - V_2)}{X}$). Thus $\partial P/\partial|V|$ and $\partial Q/\partial\delta$ are small and are set to zero in the N-R Jacobian, giving $\Delta P = J_1\Delta\delta$ and $\Delta Q = J_4\Delta|V|$.

Advantages gained:

• Two small systems instead of one large one, so less memory and computation per iteration.
• In the fast decoupled form, $B'$ and $B''$ are constant: factorised only once, making it very fast.
• Simple programming; suitable for on-line control and contingency studies with repeated load flows.
• Final accuracy is unchanged because exact mismatches are used.

Difference between G-S and N-R methods

Assumptions for decoupled load flow

1. Transmission lines have high X/R ratio, so $G_{ik} \ll B_{ik}$.
2. Voltage angle differences between connected buses are small: $\sin(\delta_i - \delta_k) \approx 0$, $\cos(\delta_i - \delta_k) \approx 1$.
3. Hence $\partial P/\partial|V| \approx 0$ and $\partial Q/\partial\delta \approx 0$, so the P–δ and Q–V problems are solved separately: $\Delta P = J_1\Delta\delta$, $\Delta Q = J_4\Delta|V|$.

A load flow (power flow) study finds the steady-state voltage magnitude and angle at every bus of a network for a given load and generation, and from these the real and reactive power flows and losses in every line.

When and why it is needed

• Planning and expansion: to check that a proposed line, transformer or generator keeps all bus voltages within limits (about ±5%) and no line is overloaded.
• Daily operation: to schedule generation, set transformer taps and switch capacitors/reactors for the forecast load.
• Contingency studies: to see what happens if a line or generator trips (N−1 security).
• Input to other studies: economic dispatch, loss calculation, and the pre-fault voltages needed for fault and stability studies.

Load flow equations for the N-R method

With $V_i = |V_i|\angle\delta_i$ and $Y_{ij} = |Y_{ij}|\angle\theta_{ij}$:

$$\begin{aligned} P_i &= \sum_{j=1}^{n}|V_i||V_j||Y_{ij}|\cos(\theta_{ij} - \delta_i + \delta_j) \\ Q_i &= -\sum_{j=1}^{n}|V_i||V_j||Y_{ij}|\sin(\theta_{ij} - \delta_i + \delta_j) \end{aligned}$$

Linearising about the present estimate gives

$$\begin{bmatrix} \Delta P \\ \Delta Q \end{bmatrix} = \begin{bmatrix} J_1 & J_2 \\ J_3 & J_4 \end{bmatrix} \begin{bmatrix} \Delta\delta \\ \Delta|V| \end{bmatrix}$$

where $J_1 = \partial P/\partial\delta$, $J_2 = \partial P/\partial|V|$, $J_3 = \partial Q/\partial\delta$, $J_4 = \partial Q/\partial|V|$. The full Jacobian is rebuilt and solved every iteration.

Assumptions for decoupled load flow

1. Line resistance is much smaller than reactance ($R \ll X$), so $\theta_{ij} \approx 90^\circ$.
2. Angle differences across lines are small, so $\cos(\delta_i - \delta_j) \approx 1$ and $\sin(\delta_i - \delta_j) \approx 0$.
3. Hence $P$ depends mainly on $\delta$ and $Q$ mainly on $|V|$, so $J_2 \approx 0$ and $J_3 \approx 0$:

$$\begin{aligned} \Delta P &= J_1\,\Delta\delta \\ \Delta Q &= J_4\,\Delta|V| \end{aligned}$$

4. For the fast decoupled method, also take $Q_i \ll B_{ii}|V_i|^2$ and $|V_i| \approx 1$, which gives $\dfrac{\Delta P}{|V|} = -B'\,\Delta\delta$ and $\dfrac{\Delta Q}{|V|} = -B''\,\Delta|V|$, where $B'$ and $B''$ are constant matrices built from $B_{bus}$.

Computational advantages over N-R

• Two small matrices replace one large Jacobian, so memory is roughly halved.
• In the fast decoupled form, $B'$ and $B''$ are constant and are factorised only once.
• Each iteration is much faster. It needs a few more iterations, but total time is lower, which suits on-line and contingency studies.

Load flow equations relate the power injected at each bus to the bus voltages through the bus admittance matrix. They follow from the nodal equation $I_{bus} = Y_{bus}V_{bus}$.

Step 1: Current injection at bus i

For an $n$-bus network,

$$\begin{bmatrix} I_1 \\ I_2 \\ \vdots \\ I_n \end{bmatrix} = \begin{bmatrix} Y_{11} & Y_{12} & \cdots & Y_{1n} \\ Y_{21} & Y_{22} & \cdots & Y_{2n} \\ \vdots & & & \vdots \\ Y_{n1} & Y_{n2} & \cdots & Y_{nn} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ \vdots \\ V_n \end{bmatrix}$$

so the current injected at bus $i$ is

$$I_i = \sum_{k=1}^{n} Y_{ik}V_k$$

Step 2: Complex power injection

The net complex power injected at bus $i$ is $S_i = P_i + jQ_i = V_iI_i^{*}$. Taking the conjugate,

$$\begin{aligned} P_i - jQ_i &= V_i^{*}I_i \\ &= V_i^{*}\sum_{k=1}^{n} Y_{ik}V_k \end{aligned}$$

Here $P_i = P_{Gi} - P_{Li}$ and $Q_i = Q_{Gi} - Q_{Li}$.

Step 3: Polar form (used in N-R)

Put $V_i = |V_i|\angle\delta_i$ and $Y_{ik} = |Y_{ik}|\angle\theta_{ik}$:

$$P_i - jQ_i = \sum_{k=1}^{n}|V_i||V_k||Y_{ik}|\angle(\theta_{ik} - \delta_i + \delta_k)$$

Equating real and imaginary parts:

$$\begin{aligned} P_i &= \sum_{k=1}^{n}|V_i||V_k||Y_{ik}|\cos(\theta_{ik} - \delta_i + \delta_k) \\ Q_i &= -\sum_{k=1}^{n}|V_i||V_k||Y_{ik}|\sin(\theta_{ik} - \delta_i + \delta_k) \end{aligned}$$

These are the static load flow equations: $2n$ real, non-linear equations in $|V|$ and $\delta$, solved by iteration.

Step 4: Form used in Gauss-Seidel

Separating the $k = i$ term in Step 2:

$$V_i = \frac{1}{Y_{ii}}\left[\frac{P_i - jQ_i}{V_i^{*}} - \sum_{k\ne i} Y_{ik}V_k\right]$$

At each bus two of the four quantities ($P$, $Q$, $|V|$, $\delta$) are given and the other two are found by solving these equations.