Load Flow Analysis (Gauss-Seidel) — help

Overview

Performs multi-bus power system load flow analysis using the Gauss-Seidel iterative method. Supports PQ (load), PV (generator), and Slack (reference) bus types. Assembles the bus admittance matrix (Y-bus) from line impedances, solves for bus voltages, and computes line power flows and system losses. Optional Successive Over-Relaxation (SOR) acceleration and generator reactive power limit enforcement are supported.

Applicable to radial and mesh networks with up to 50 buses. All quantities are internally converted to the per-unit system using a user-specified base MVA and base kV.

Input Parameters

System Parameters

Parameter	Type	Default	Range	Description
`num_buses`	Integer	3	2 – 50	Total number of buses in the system
`base_mva`	Number	100.0	1.0 – 10000.0	System base power in MVA for per-unit calculations
`base_kv`	Number	11.0	0.1 – 1000.0	System base voltage in kV
`frequency`	Enum: 50 / 60	50	—	System frequency (Hz)

Bus Data (array)

Each element in the bus_data array defines one bus. At least one bus must be type Slack.

Parameter	Type	Default	Range	Description
`bus_number`	Integer	—	0 – 49	Bus index (zero-based)
`bus_type`	Enum: Slack / PV / PQ	PQ	—	Bus type — Slack (reference), PV (generator), or PQ (load)
`v_magnitude`	Number	1.0	—	Specified voltage magnitude in per-unit (Slack and PV buses)
`v_angle`	Number	0.0	—	Specified voltage angle in degrees (Slack bus only)
`p_gen`	Number	0.0	—	Generator real power output (MW)
`q_gen`	Number	0.0	—	Generator reactive power output (MVAr)
`p_load`	Number	0.0	—	Load real power consumption (MW)
`q_load`	Number	0.0	—	Load reactive power consumption (MVAr)
`q_min`	Number	-999.0	—	Minimum generator reactive capability in MVAr (PV buses)
`q_max`	Number	999.0	—	Maximum generator reactive capability in MVAr (PV buses)

Line Data (array)

Each element in the line_data array defines a transmission line or cable between two buses. Impedances are specified in per-unit.

Parameter	Type	Default	Range	Description
`from_bus`	Integer	—	0 – 49	Sending-end bus index
`to_bus`	Integer	—	0 – 49	Receiving-end bus index
`z_real`	Number	—	—	Line resistance in per-unit
`z_imag`	Number	—	—	Line reactance in per-unit
`b_shunt`	Number	0.0	—	Half-line charging susceptance in per-unit

Solver Settings

Parameter	Type	Default	Range	Description
`tolerance`	Number	1×10^-6	1×10^-10 – 1×10^-3	Maximum voltage change for convergence
`max_iterations`	Integer	100	10 – 1000	Maximum number of Gauss-Seidel iterations
`use_sor`	Boolean	false	—	Enable Successive Over-Relaxation acceleration
`sor_factor`	Number	1.6	1.0 – 2.0	SOR acceleration factor (ω); 1.0 = no acceleration
`enforce_q_limits`	Boolean	true	—	Convert PV to PQ bus when reactive limits are violated
`calculate_line_flows`	Boolean	true	—	Compute power flow through each line after convergence
`calculate_losses`	Boolean	true	—	Compute total real and reactive system losses

Output Results

Bus Voltages (per bus)

Output	Unit	Description
`magnitude`	pu	Bus voltage magnitude
`angle_deg`	°	Bus voltage angle in degrees
`angle_rad`	rad	Bus voltage angle in radians
`real`	pu	Real component of bus voltage
`imag`	pu	Imaginary component of bus voltage

Power Injections (per bus)

Output	Unit	Description
`p_specified_mw`	MW	Specified net real power injection (generation − load)
`q_specified_mvar`	MVAr	Specified net reactive power injection
`p_calculated_mw`	MW	Calculated real power injection from converged voltages
`q_calculated_mvar`	MVAr	Calculated reactive power injection from converged voltages
`p_mismatch_mw`	MW	Absolute difference between specified and calculated real power
`q_mismatch_mvar`	MVAr	Absolute difference for reactive power (PQ buses only)

Line Flows (per line)

Output	Unit	Description
`p_from_to_mw`	MW	Real power flow from sending to receiving end
`q_from_to_mvar`	MVAr	Reactive power flow from sending to receiving end
`p_to_from_mw`	MW	Real power flow from receiving to sending end
`q_to_from_mvar`	MVAr	Reactive power flow from receiving to sending end
`p_loss_mw`	MW	Real power loss in the line
`q_loss_mvar`	MVAr	Reactive power loss in the line
`i_from_pu`	pu	Current magnitude at sending end
`i_to_pu`	pu	Current magnitude at receiving end

System Summary

Output	Unit	Description
`converged`	Boolean	Whether the solution converged within the tolerance
`iterations`	—	Number of Gauss-Seidel iterations performed
`p_loss_mw`	MW	Total system real power losses (sum of all line losses)
`q_loss_mvar`	MVAr	Total system reactive power losses
`y_bus_real`	pu	Real part of Y-bus admittance matrix
`y_bus_imag`	pu	Imaginary part of Y-bus admittance matrix

Formulas

Y-bus Admittance Matrix

Series admittance of each line:

\[ y_{ij} = \frac{1}{z_{ij}} = \frac{1}{R_{ij} + jX_{ij}} \]

Diagonal elements (self-admittance):

\[ Y_{ii} = \sum_{k \neq i} y_{ik} + \sum_{k} y_{\text{shunt},ik} \]

Off-diagonal elements (mutual admittance):

\[ Y_{ij} = -y_{ij} \quad (i \neq j) \]

Where \(z_{ij} = R_{ij} + jX_{ij}\) is the series impedance and \(y_{\text{shunt}}\) is the half-line charging susceptance (pi-model).

Gauss-Seidel Voltage Update (PQ Bus)

\[ V_i^{(k+1)} = \frac{1}{Y_{ii}} \left( \frac{S_i^*}{\left(V_i^{(k)}\right)^*} - \sum_{j \neq i} Y_{ij} V_j \right) \]

Where:

\(S_i = P_i + jQ_i\) is the net scheduled complex power injection (generation − load) at bus \(i\)
\(V_i^{(k)}\) is the voltage at bus \(i\) at iteration \(k\)
For \(j < i\), the already-updated \(V_j^{(k+1)}\) is used; for \(j > i\), \(V_j^{(k)}\) is used (Gauss-Seidel property)

PV Bus Voltage Enforcement

After computing a temporary voltage update, the magnitude is enforced:

\[ V_i^{(k+1)} = |V_i|_{\text{spec}} \cdot e^{j\angle V_i^{\text{temp}}} \]

The reactive power is then recalculated from the updated voltages:

\[ Q_i^{(k+1)} = \text{Im}\left\{ V_i^{(k+1)} \left( \sum_j Y_{ij} V_j \right)^* \right\} \]

If \(Q_i\) exceeds generator limits (\(Q_{\min}\) or \(Q_{\max}\)), the bus is converted from PV to PQ and the reactive power is clamped at the violated limit.

Successive Over-Relaxation (SOR)

\[ V_i^{(k+1)} \leftarrow (1 - \omega)\, V_i^{(k)} + \omega\, V_i^{(k+1)} \]

Where \(\omega\) is the acceleration factor. Typical values are 1.0 – 1.2 for well-conditioned systems (up to 2.0 maximum). Setting \(\omega = 1.0\) reduces to standard Gauss-Seidel.

Power Flow Equations

Net complex power injection at bus \(i\):

\[ S_i = P_i + jQ_i = V_i \cdot I_i^* = V_i \left( \sum_{j=0}^{n-1} Y_{ij}\, V_j \right)^* \]

Expanding into real and reactive components:

\[ P_i = \text{Re}\left\{ V_i \left( \sum_{j} Y_{ij}\, V_j \right)^* \right\} \] \[ Q_i = \text{Im}\left\{ V_i \left( \sum_{j} Y_{ij}\, V_j \right)^* \right\} \]

Line Power Flow and Losses

Current from bus \(i\) to bus \(j\):

\[ I_{ij} = (V_i - V_j)\, y_{ij} + V_i\, y_{\text{shunt},ij} \]

Power flow from bus \(i\) to bus \(j\):

\[ S_{ij} = V_i \cdot I_{ij}^* \]

Line losses are the algebraic sum of flows in both directions:

\[ S_{\text{loss},ij} = S_{ij} + S_{ji} \]

Total system losses:

\[ P_{\text{loss,total}} = \sum_{\text{all lines}} P_{\text{loss},ij} \]

Convergence Criterion

\[ \max_i \left| V_i^{(k+1)} - V_i^{(k)} \right| < \varepsilon \]

Where \(\varepsilon\) is the convergence tolerance (default: \(1 \times 10^{-6}\) pu). If the maximum voltage change across all buses drops below this threshold, the solution is considered converged.

Per-Unit Conversion

Power quantities are converted between MW/MVAr and per-unit using the system base:

\[ P_{\text{pu}} = \frac{P_{\text{MW}}}{S_{\text{base}}} \qquad Q_{\text{pu}} = \frac{Q_{\text{MVAr}}}{S_{\text{base}}} \]

Where \(S_{\text{base}}\) is the base MVA (default 100 MVA).

Worked Examples

These examples are run by the fe_calc_load_flow MCP tool, which takes an inline network in SI units (ohms, watts, kV, with transformers, taps, shunts and branch ratings) and solves it with the PSAS engine's Newton-Raphson or Gauss-Seidel solver or with pandapower. The per-unit conversion, Y-bus, power-flow equations and branch-flow formulas are the same as above; two-bus cases are also solved in closed form, and meshed cases are checked against published textbook solutions.

In the tool, every solver enforces PV reactive limits on the generator's output \(Q_G = Q_{\text{net}} + Q_{\text{load}}\): a generator that cannot hold its set-point within \([Q_{\min}, Q_{\max}]\) is held at the limit and its bus solved as PQ, and it returns to PV control only once its voltage crosses back over the set-point (Glover, Sarma & Overbye §6.5). The Gauss-Seidel acceleration factor moves PQ voltages and PV angles, never a PV magnitude; it stops on the largest voltage change per sweep and also reports the power mismatch of the result. Branch losses are the series \(|I|^2(R + jX)\) only; line charging and shunt capacitors/reactors are reported separately as shunt reactive power:

\[ S_{ij} + S_{ji} = |I_{\text{series}}|^2 (R + jX) \; - \; j\tfrac{B}{2}\left(|V_i|^2 + |V_j|^2\right) \]

The 28 examples below are the validation cases for this calculation. Each one was produced by the calculation engine from the inputs shown, every figure was checked against the independent reference quoted at the end of the example, and the same cases are run against every engine build (see validation scorecard, tool fe_calc_load_flow; worked examples for every calculation).

#	Case	Buses	Solver	V_min (pu)	V_max (pu)	Max loading (%)	Losses (kW)	Result
1	Two-bus lossless line, lagging load (closed form)	2	Newton-Raphson	0.9782 (L)	1.0000 (S)	–	0.0	PASS
2	Two-bus lossy line on the pandapower solver	2	pandapower (NR)	0.9518 (L)	1.0000 (S)	70.7 (LN1)	160.0	PASS
3	Two-bus Gauss-Seidel, 50 MW + 20 Mvar (under-voltage)	2	Gauss-Seidel	0.9390 (B1)	1.0000 (B0)	–	1,644.4	FAIL
4	Unity power-factor load on a 33 kV line	2	Newton-Raphson	0.9832 (LOAD)	1.0000 (GXP)	–	132.4	PASS
5	Leading load raises the receiving voltage (over-voltage)	2	Newton-Raphson	1.0000 (S)	1.0517 (L)	–	90.4	FAIL
6	Ferranti rise on a 300 km open-ended 132 kV line	2	Newton-Raphson	1.0000 (A)	1.0530 (B)	–	71.6	FAIL
7	33 kV cable loaded past its rating	2	Newton-Raphson	0.9960 (PLANT)	1.0000 (SUB)	112.3 (FDR1)	18.1	FAIL
8	Project limits 0.90–1.10 pu and 80 % loading	2	Newton-Raphson	0.9468 (R)	1.0000 (S)	75.4 (SWER)	139.4	PASS
9	415 V cable from per-km data on a 1 MVA base	2	Newton-Raphson	0.9711 (DB1)	1.0000 (MSB)	69.1 (SUBMAIN)	3.9	PASS
10	1.5 MVA 11/0.433 kV transformer from nameplate %Z	2	Newton-Raphson	0.9618 (LV)	1.0000 (HV)	98.1 (TX1)	10.7	PASS
11	Same transformer on the −5 % HV tap	2	Newton-Raphson	1.0000 (HV)	1.0166 (LV)	97.7 (TX1)	9.6	PASS
12	Off-nominal tap solved by pandapower	2	pandapower (NR)	1.0000 (HV)	1.0166 (LV)	97.7 (TX1)	9.6	PASS
13	3 Mvar shunt capacitor lifts a sagging bus	2	Newton-Raphson	0.9600 (L)	1.0000 (S)	–	143.9	PASS
14	PV generator bus on a reactive tie (power-angle equation)	2	Newton-Raphson	1.0000 (GRID)	1.0200 (GEN)	–	0.0	PASS
15	PV bus with local load on a lossy line	2	Newton-Raphson	1.0000 (GRID)	1.0300 (MILL)	–	54.7	PASS
16	PV generator inside its Q limits (Gauss-Seidel)	2	Gauss-Seidel	1.0000 (GRID)	1.0500 (GEN)	–	0.0	PASS
17	PV generator held at Q_max (Gauss-Seidel)	2	Gauss-Seidel	1.0000 (GRID)	1.0275 (GEN)	–	0.0	PASS
18	Generator Q limit with a local reactive load (Newton-Raphson)	2	Newton-Raphson	1.0000 (GRID)	1.0087 (GEN)	–	0.0	PASS
19	Generator held at Q_min (pandapower)	2	pandapower (NR)	0.9880 (GEN)	1.0000 (GRID)	–	0.0	PASS
20	Bus load with a motor and an embedded generator	2	Newton-Raphson	0.9638 (PLANT)	1.0000 (S)	–	81.5	PASS
21	Slack at 1.02 pu ∠30°, charged cable, 60 Hz (pandapower)	2	pandapower (NR)	1.0086 (L)	1.0200 (S)	–	37.5	PASS
22	Saadat three-bus system, Gauss-Seidel	3	Gauss-Seidel	0.9818 (2)	1.0500 (1)	–	14,300.0	PASS
23	Saadat three-bus system with a PV bus, Newton-Raphson	3	Newton-Raphson	0.9717 (2)	1.0500 (1)	–	18,422.8	PASS
24	Three-bus slack + PV generator + PQ load	3	Newton-Raphson	0.9501 (2)	1.0500 (0)	–	5,190.6	PASS
25	Grainger & Stevenson four-bus 230 kV system	4	Newton-Raphson	0.9690 (3)	1.0200 (4)	–	4,809.1	PASS
26	Glover, Sarma & Overbye five-bus system with transformers	5	Newton-Raphson	0.8338 (2)	1.0500 (3)	66.8 (T1)	34,838.8	FAIL
27	Five-bus system on the pandapower solver	5	pandapower (NR)	0.8338 (2)	1.0500 (3)	66.8 (T1)	34,838.8	FAIL
28	Gauss-Seidel stopped by a 5-iteration cap	3	Gauss-Seidel	0.9823 (2)	1.0500 (1)	–	–	FAIL

Example 1 — Two-bus lossless line, lagging load (closed form)

A purely reactive line feeding a lagging PQ load: the receiving-end quartic gives the voltage, and all the reactive loss is I²X.

Given: S_base = 10 MVA, 50 Hz. Buses: S (11 kV, SLACK at 11 kV (1 pu) ∠0°); L (11 kV, PQ, load 5 MW + j2 Mvar). Branches: LN1 LINE S–L: 0 + j1.21 Ω. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	11 kV: Z_base = 11² / 10 = 12.1 Ω; powers ÷ 10 MVA
Branch impedances	LN1: z = (0.0000 + j1.2100) / 12.1 = 0 + j0.1 pu; y = 1/z = 0.0000 − j10.0000 pu
Scheduled injections	Net injection = generation − load (machines included): L: S = (−5.000 − j2.000)/10 = −0.50000 − j0.20000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_S,S = 0.0000 − j10.0000; Y_L,L = 0.0000 − j10.0000; Y_S,L = 0.0000 + j10.0000
Closed-form voltage	V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 with P = 0.5, Q = 0.2 pu, R + jX = 0.000000 + j0.100000 pu, E = 1.00000: V⁴ − 0.960000V² + 0.0029 = 0 → V² = (0.96 + √0.910000)/2 → V = 0.97825 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (0.05)/(0.97697) → θ = 2.9298° behind E; δ₂ = 0.0000 − 2.9298 = −2.9298°
Load-flow result	Newton-Raphson: V_L = 0.97825∠−2.9298° pu = 10.7607 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: LN1: S_S→ = 5.0000 + j2.3030 MVA, S_L→ = −5.0000 − j2.0000 MVA; loss = sum = \|I\|²(R + jX) = 0.00 kW + j303.04 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 0.00 kW + j303.04 kvar; slack supplies P = 5.0000 MW, Q = 2.3030 Mvar. P balance: P_slack + ΣP_gen = 5.0000 + 0.0000 = 5.0000 MW = ΣP_load + P_loss = 5.0000 + 0.0000 = 5.0000 MW. Q balance: ΣQ_inj = 303.04 kvar = Q_loss + Q_shunt = 303.04 + (0.00) = 303.04 kvar
Convergence	converged after 4 iterations; largest power mismatch 1.69e-10 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus L at 0.97825 pu → 0.97825 vs limit 0.95000 pu PASS
Maximum voltage	highest bus S at 1.00000 pu → 1.00000 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 0.9782–1.0000 pu, losses 0.0 kW; governing: minimum voltage. No rating_mva on LN1: loading not checked.
Reference	Hand calculation (closed-form two-bus): Z_base = 11²/10 = 12.1 Ω; z = 0 + j0.1 pu; load P = 0.5, Q = 0.2 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 0.960000V² + 0.0029 = 0 → V = 0.97825 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle −2.9298°; S_from = 5.0000 + j2.3030 MVA; series loss \|I\|²z = 0.00 + j303.04 (kW, kvar); slack supplies 5.00000 + j2.30304 (MW, Mvar) (validation case `psas-2bus-lossless-nr`)

Example 2 — Two-bus lossy line on the pandapower solver

The same feeder with R/X = 0.5 solved by pandapower: the closed-form voltage, I²R loss and the slack's supply are reproduced, and the 8 MVA rating is checked.

Given: S_base = 10 MVA, 50 Hz. Buses: S (11 kV, SLACK at 11 kV (1 pu) ∠0°); L (11 kV, PQ, load 5 MW + j2 Mvar). Branches: LN1 LINE S–L: 0.605 + j1.21 Ω; rated 8 MVA. Solver pandapower (NR); limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	11 kV: Z_base = 11² / 10 = 12.1 Ω; powers ÷ 10 MVA
Branch impedances	LN1: z = (0.6050 + j1.2100) / 12.1 = 0.05 + j0.1 pu; y = 1/z = 4.0000 − j8.0000 pu
Scheduled injections	Net injection = generation − load (machines included): L: S = (−5.000 − j2.000)/10 = −0.50000 − j0.20000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_S,S = 4.0000 − j8.0000; Y_L,L = 4.0000 − j8.0000; Y_S,L = −4.0000 + j8.0000
Closed-form voltage	V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 with P = 0.5, Q = 0.2 pu, R + jX = 0.050000 + j0.100000 pu, E = 1.00000: V⁴ − 0.910000V² + 0.003625 = 0 → V² = (0.91 + √0.813600)/2 → V = 0.95184 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (0.04)/(0.950999) → θ = 2.4085° behind E; δ₂ = 0.0000 − 2.4085 = −2.4085°
Load-flow result	pandapower (NR): V_L = 0.95184∠−2.4085° pu = 10.4702 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: LN1: S_S→ = 5.1600 + j2.3201 MVA, S_L→ = −5.0000 − j2.0000 MVA; loss = sum = \|I\|²(R + jX) = 160.04 kW + j320.09 kvar; loading \|S_from\|/rating = 5.6576/8 = 70.72 %
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 160.04 kW + j320.09 kvar; slack supplies P = 5.1600 MW, Q = 2.3201 Mvar. P balance: P_slack + ΣP_gen = 5.1600 + 0.0000 = 5.1600 MW = ΣP_load + P_loss = 5.0000 + 0.1600 = 5.1600 MW. Q balance: ΣQ_inj = 320.09 kvar = Q_loss + Q_shunt = 320.09 + (0.00) = 320.09 kvar
Convergence	pandapower Newton-Raphson converged in 3 iterations on its own mismatch tolerance PASS
Minimum voltage	lowest bus L at 0.95184 pu → 0.95184 vs limit 0.95000 pu PASS
Maximum voltage	highest bus S at 1.00000 pu → 1.00000 vs limit 1.05000 pu PASS
Branch loading	most loaded branch LN1 → 70.72 vs limit 100.00 % PASS
Result	4/4 checks pass → PASS; V 0.9518–1.0000 pu, losses 160.0 kW; governing: branch loading
Reference	Hand calculation (closed-form two-bus): Z_base = 11²/10 = 12.1 Ω; z = 0.05 + j0.1 pu; load P = 0.5, Q = 0.2 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 0.910000V² + 0.003625 = 0 → V = 0.95184 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle −2.4085°; S_from = 5.1600 + j2.3201 MVA; series loss \|I\|²z = 160.04 + j320.09 (kW, kvar); loading 5.6576/8 = 70.72 %; slack supplies 5.16004 + j2.32009 (MW, Mvar) (validation case `psas-2bus-lossy-pandapower`)

Example 3 — Two-bus Gauss-Seidel, 50 MW + 20 Mvar (under-voltage)

The page's former two-bus Gauss-Seidel example: the first iteration by hand, the converged 0.939 pu, line flow and losses — and the 0.95 pu under-voltage check it fails.

Given: S_base = 100 MVA, 50 Hz. Buses: B0 (100 kV, SLACK at 100 kV (1 pu) ∠0°); B1 (100 kV, PQ, load 50 MW + j20 Mvar). Branches: L01 LINE B0–B1: 5 + j15 Ω. Solver Gauss-Seidel, tolerance 1e-08 pu, ≤ 200 iterations, acceleration α = 1.6; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	100 kV: Z_base = 100² / 100 = 100 Ω; powers ÷ 100 MVA
Branch impedances	L01: z = (5.0000 + j15.0000) / 100 = 0.05 + j0.15 pu; y = 1/z = 2.0000 − j6.0000 pu
Scheduled injections	Net injection = generation − load (machines included): B1: S = (−50.000 − j20.000)/100 = −0.50000 − j0.20000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_B0,B0 = 2.0000 − j6.0000; Y_B1,B1 = 2.0000 − j6.0000; Y_B0,B1 = −2.0000 + j6.0000
Gauss-Seidel sweep 1	V_B1⁽¹⁾ = [S/V⁽⁰⁾ − ΣY_ikV_k] / Y_ii = [(−0.5000 + j0.2000)/1 − (−2.0000 + j6.0000)(1.0000 + j0.0000)] / (2.0000 − j6.0000) = 0.94500 − j0.06500 pu; accelerated V ← 1 + 1.6 × (−0.05500 − j0.06500) = 0.91200 − j0.10400 pu; sweeps repeat until the largest \|ΔV\| < 1e-08 pu
Closed-form voltage	V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 with P = 0.5, Q = 0.2 pu, R + jX = 0.050000 + j0.150000 pu, E = 1.00000: V⁴ − 0.890000V² + 0.00725 = 0 → V² = (0.89 + √0.763100)/2 → V = 0.93903 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (0.065)/(0.936778) → θ = 3.9692° behind E; δ₂ = 0.0000 − 3.9692 = −3.9692°
Load-flow result	Gauss-Seidel: V_B1 = 0.93903∠−3.9692° pu = 93.9030 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: L01: S_B0→ = 51.6444 + j24.9332 MVA, S_B1→ = −50.0000 − j20.0000 MVA; loss = sum = \|I\|²(R + jX) = 1,644.40 kW + j4,933.21 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 1,644.40 kW + j4,933.22 kvar; slack supplies P = 51.6444 MW, Q = 24.9332 Mvar. P balance: P_slack + ΣP_gen = 51.6444 + 0.0000 = 51.6444 MW = ΣP_load + P_loss = 50.0000 + 1.6444 = 51.6444 MW. Q balance: ΣQ_inj = 4,933.21 kvar = Q_loss + Q_shunt = 4,933.22 + (0.00) = 4,933.22 kvar
Convergence	converged after 57 iterations; largest \|ΔV\| in the last sweep 9.15e-09 vs tolerance 1e-08 pu; power mismatch of the returned voltages 1.86e-08 pu PASS
Minimum voltage	lowest bus B1 at 0.93903 pu → 0.93903 vs limit 0.95000 pu FAIL
Maximum voltage	highest bus B0 at 1.00000 pu → 1.00000 vs limit 1.05000 pu PASS
Result	2/3 checks pass → FAIL; V 0.9390–1.0000 pu, losses 1,644.4 kW; governing: minimum voltage. No rating_mva on L01: loading not checked.
Reference	Hand calculation (closed-form two-bus): Z_base = 100²/100 = 100 Ω; z = 0.05 + j0.15 pu; load P = 0.5, Q = 0.2 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 0.890000V² + 0.00725 = 0 → V = 0.93903 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle −3.9692°; S_from = 51.6444 + j24.9332 MVA; series loss \|I\|²z = 1644.40 + j4933.21 (kW, kvar); slack supplies 51.64440 + j24.93321 (MW, Mvar) (validation case `psas-2bus-gauss-seidel-undervoltage`)

Example 4 — Unity power-factor load on a 33 kV line

With Q = 0 the voltage drop is set by the resistive term PR and the angle by PX.

Given: S_base = 10 MVA, 50 Hz. Buses: GXP (33 kV, SLACK at 33 kV (1 pu) ∠0°); LOAD (33 kV, PQ, load 8 MW + j0 Mvar). Branches: OHL LINE GXP–LOAD: 2.178 + j4.356 Ω. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	33 kV: Z_base = 33² / 10 = 108.9 Ω; powers ÷ 10 MVA
Branch impedances	OHL: z = (2.1780 + j4.3560) / 108.9 = 0.02 + j0.04 pu; y = 1/z = 10.0000 − j20.0000 pu
Scheduled injections	Net injection = generation − load (machines included): LOAD: S = (−8.000 + j0.000)/10 = −0.80000 + j0.00000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_GXP,GXP = 10.0000 − j20.0000; Y_LOAD,LOAD = 10.0000 − j20.0000; Y_GXP,LOAD = −10.0000 + j20.0000
Closed-form voltage	V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 with P = 0.8, Q = 0 pu, R + jX = 0.020000 + j0.040000 pu, E = 1.00000: V⁴ − 0.968000V² + 0.00128 = 0 → V² = (0.968 + √0.931904)/2 → V = 0.98320 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (0.032)/(0.982676) → θ = 1.8651° behind E; δ₂ = 0.0000 − 1.8651 = −1.8651°
Load-flow result	Newton-Raphson: V_LOAD = 0.98320∠−1.8651° pu = 32.4455 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: OHL: S_GXP→ = 8.1324 + j0.2648 MVA, S_LOAD→ = −8.0000 + j0.0000 MVA; loss = sum = \|I\|²(R + jX) = 132.41 kW + j264.83 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 132.41 kW + j264.82 kvar; slack supplies P = 8.1324 MW, Q = 0.2648 Mvar. P balance: P_slack + ΣP_gen = 8.1324 + 0.0000 = 8.1324 MW = ΣP_load + P_loss = 8.0000 + 0.1324 = 8.1324 MW. Q balance: ΣQ_inj = 264.82 kvar = Q_loss + Q_shunt = 264.82 + (0.00) = 264.82 kvar
Convergence	converged after 4 iterations; largest power mismatch 1.58e-11 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus LOAD at 0.98320 pu → 0.98320 vs limit 0.95000 pu PASS
Maximum voltage	highest bus GXP at 1.00000 pu → 1.00000 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 0.9832–1.0000 pu, losses 132.4 kW; governing: minimum voltage. No rating_mva on OHL: loading not checked.
Reference	Hand calculation (closed-form two-bus): Z_base = 33²/10 = 108.9 Ω; z = 0.02 + j0.04 pu; load P = 0.8, Q = 0 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 0.968000V² + 0.00128 = 0 → V = 0.98320 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle −1.8651°; S_from = 8.1324 + j0.2648 MVA; series loss \|I\|²z = 132.41 + j264.83 (kW, kvar); slack supplies 8.13241 + j0.26483 (MW, Mvar) (validation case `psas-2bus-unity-pf`)

Example 5 — Leading load raises the receiving voltage (over-voltage)

A leading (negative-Q) load makes PR + QX negative, so the receiving end rises above the source to 1.052 pu and fails the 1.05 pu limit.

Given: S_base = 10 MVA, 50 Hz. Buses: S (11 kV, SLACK at 11 kV (1 pu) ∠0°); L (11 kV, PQ, load 2 MW − j6 Mvar). Branches: C1 CABLE S–L: 0.3025 + j1.21 Ω. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	11 kV: Z_base = 11² / 10 = 12.1 Ω; powers ÷ 10 MVA
Branch impedances	C1: z = (0.3025 + j1.2100) / 12.1 = 0.025 + j0.1 pu; y = 1/z = 2.3529 − j9.4118 pu
Scheduled injections	Net injection = generation − load (machines included): L: S = (−2.000 + j6.000)/10 = −0.20000 + j0.60000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_S,S = 2.3529 − j9.4118; Y_L,L = 2.3529 − j9.4118; Y_S,L = −2.3529 + j9.4118
Closed-form voltage	V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 with P = 0.2, Q = −0.6 pu, R + jX = 0.025000 + j0.100000 pu, E = 1.00000: V⁴ − 1.110000V² + 0.00425 = 0 → V² = (1.11 + √1.215100)/2 → V = 1.05174 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (0.035)/(1.05116) → θ = 1.9071° behind E; δ₂ = 0.0000 − 1.9071 = −1.9071°
Load-flow result	Newton-Raphson: V_L = 1.05174∠−1.9071° pu = 11.5691 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: C1: S_S→ = 2.0904 − j5.6384 MVA, S_L→ = −2.0000 + j6.0000 MVA; loss = sum = \|I\|²(R + jX) = 90.40 kW + j361.61 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 90.40 kW + j361.61 kvar; slack supplies P = 2.0904 MW, Q = −5.6384 Mvar. P balance: P_slack + ΣP_gen = 2.0904 + 0.0000 = 2.0904 MW = ΣP_load + P_loss = 2.0000 + 0.0904 = 2.0904 MW. Q balance: ΣQ_inj = 361.61 kvar = Q_loss + Q_shunt = 361.61 + (0.00) = 361.61 kvar
Convergence	converged after 4 iterations; largest power mismatch 1.39e-09 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus S at 1.00000 pu → 1.00000 vs limit 0.95000 pu PASS
Maximum voltage	highest bus L at 1.05174 pu → 1.05174 vs limit 1.05000 pu FAIL
Result	2/3 checks pass → FAIL; V 1.0000–1.0517 pu, losses 90.4 kW; governing: maximum voltage. No rating_mva on C1: loading not checked.
Reference	Hand calculation (closed-form two-bus): Z_base = 11²/10 = 12.1 Ω; z = 0.025 + j0.1 pu; load P = 0.2, Q = −0.6 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 1.110000V² + 0.00425 = 0 → V = 1.05174 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle −1.9071°; S_from = 2.0904 − j5.6384 MVA; series loss \|I\|²z = 90.40 + j361.61 (kW, kvar); slack supplies 2.09040 − j5.63839 (MW, Mvar) (validation case `psas-2bus-leading-overvoltage`)

Example 6 — Ferranti rise on a 300 km open-ended 132 kV line

An unloaded long line modelled as a nominal π: the far-end charging current through the series impedance lifts the open end to 1.053 pu.

Given: S_base = 100 MVA, 50 Hz. Buses: A (132 kV, SLACK at 132 kV (1 pu) ∠0°); B (132 kV, PQ). Branches: OHL1 LINE A–B: 300 km × (0.07 + j0.4) Ω/km; B_total 0.00084 S. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	132 kV: Z_base = 132² / 100 = 174.24 Ω; powers ÷ 100 MVA
Branch impedances	OHL1: (0.07 + j0.4) × 300 = 21.0000 + j120.0000 Ω; z = (21.0000 + j120.0000) / 174.24 = 0.120523 + j0.688705 pu; y = 1/z = 0.2465 − j1.4089 pu; B_total = 0.00084 × 174.24 = 0.146362 pu (B/2 = 0.0731808 at each end)
Scheduled injections	Net injection = generation − load (machines included): B: S = (0.000 + j0.000)/100 = 0.00000 + j0.00000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_A,A = 0.2465 − j1.3357; Y_B,B = 0.2465 − j1.3357; Y_A,B = −0.2465 + j1.4089
Thevenin source	E_th = −Y₂₁V₁/Y₂₂ = −(−0.2465 + j1.4089) × 1∠0° / (0.2465 − j1.3357) = 1.05303∠−0.5322° pu; Z_th = 1/Y₂₂ = 0.133645 + j0.724017 pu
Closed-form voltage	V⁴ + [2(PR + QX) − E_th²]V² + (R² + X²)(P² + Q²) = 0 with P = 0, Q = 0 pu, R + jX = 0.133645 + j0.724017 pu, E_th = 1.05303: V⁴ − 1.108871V² + 0 = 0 → V² = (1.10887 + √1.229595)/2 → V = 1.05303 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (0)/(1.10887) → θ = 0.0000° behind E_th; δ₂ = −0.5322 − 0.0000 = −0.5322°
Load-flow result	Newton-Raphson: V_B = 1.05303∠−0.5322° pu = 138.9999 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: OHL1: S_A→ = 0.0716 − j15.0239 MVA, S_B→ = 0.0000 + j0.0000 MVA; sum = series loss \|I\|²(R + jX) 71.57 kW + j408.99 kvar + charging j−15,432.89 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 71.57 kW + j408.99 kvar; shunt Q (SHUNT elements, line charging) −15,432.89 kvar, reported separately; slack supplies P = 0.0716 MW, Q = −15.0239 Mvar. P balance: P_slack + ΣP_gen = 0.0716 + 0.0000 = 0.0716 MW = ΣP_load + P_loss = 0.0000 + 0.0716 = 0.0716 MW. Q balance: ΣQ_inj = −15,023.90 kvar = Q_loss + Q_shunt = 408.99 + (−15,432.89) = −15,023.90 kvar
Convergence	converged after 4 iterations; largest power mismatch 1.07e-10 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus A at 1.00000 pu → 1.00000 vs limit 0.95000 pu PASS
Maximum voltage	highest bus B at 1.05303 pu → 1.05303 vs limit 1.05000 pu FAIL
Result	2/3 checks pass → FAIL; V 1.0000–1.0530 pu, losses 71.6 kW; governing: maximum voltage. No rating_mva on OHL1: loading not checked.
Reference	Hand calculation (closed-form two-bus): Z_base = 132²/100 = 174.24 Ω; z = 0.120523 + j0.688705 pu; Thevenin at the load bus E_th = 1.05303∠−0.5322° pu, Z_th = 0.133645 + j0.724017 pu; load P = 0, Q = 0 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 1.108871V² + 0 = 0 → V = 1.05303 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle −0.5322°; S_from = 0.0716 − j15.0239 MVA; series loss \|I\|²z = 71.57 + j408.99 (kW, kvar); shunt Q (charging/capacitor) −15432.89 kvar, not a loss; slack supplies 0.07157 − j15.02390 (MW, Mvar) (validation case `psas-ferranti-open-line`)

Example 7 — 33 kV cable loaded past its rating

Voltage is fine but the 5 MVA cable carries 5.6 MVA, so branch loading fails and governs.

Given: S_base = 10 MVA, 50 Hz. Buses: SUB (33 kV, SLACK at 33 kV (1 pu) ∠0°); PLANT (33 kV, PQ, load 5 MW + j2.5 Mvar). Branches: FDR1 CABLE SUB–PLANT: 0.625 + j0.5 Ω; rated 5 MVA. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	33 kV: Z_base = 33² / 10 = 108.9 Ω; powers ÷ 10 MVA
Branch impedances	FDR1: z = (0.6250 + j0.5000) / 108.9 = 0.00573921 + j0.00459137 pu; y = 1/z = 106.2439 − j84.9951 pu
Scheduled injections	Net injection = generation − load (machines included): PLANT: S = (−5.000 − j2.500)/10 = −0.50000 − j0.25000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_SUB,SUB = 106.2439 − j84.9951; Y_PLANT,PLANT = 106.2439 − j84.9951; Y_SUB,PLANT = −106.2439 + j84.9951
Closed-form voltage	V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 with P = 0.5, Q = 0.25 pu, R + jX = 0.005739 + j0.004591 pu, E = 1.00000: V⁴ − 0.991965V² + 1.6881e-05 = 0 → V² = (0.991965 + √0.983927)/2 → V = 0.99597 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (0.000860882)/(0.995966) → θ = 0.0495° behind E; δ₂ = 0.0000 − 0.0495 = −0.0495°
Load-flow result	Newton-Raphson: V_PLANT = 0.99597∠−0.0495° pu = 32.8669 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: FDR1: S_SUB→ = 5.0181 + j2.5145 MVA, S_PLANT→ = −5.0000 − j2.5000 MVA; loss = sum = \|I\|²(R + jX) = 18.08 kW + j14.46 kvar; loading \|S_from\|/rating = 5.6128/5 = 112.26 %
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 18.08 kW + j14.46 kvar; slack supplies P = 5.0181 MW, Q = 2.5145 Mvar. P balance: P_slack + ΣP_gen = 5.0181 + 0.0000 = 5.0181 MW = ΣP_load + P_loss = 5.0000 + 0.0181 = 5.0181 MW. Q balance: ΣQ_inj = 14.46 kvar = Q_loss + Q_shunt = 14.46 + (0.00) = 14.46 kvar
Convergence	converged after 3 iterations; largest power mismatch 3.50e-08 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus PLANT at 0.99597 pu → 0.99597 vs limit 0.95000 pu PASS
Maximum voltage	highest bus SUB at 1.00000 pu → 1.00000 vs limit 1.05000 pu PASS
Branch loading	most loaded branch FDR1 → 112.26 vs limit 100.00 % FAIL
Result	3/4 checks pass → FAIL; V 0.9960–1.0000 pu, losses 18.1 kW; governing: branch loading
Reference	Hand calculation (closed-form two-bus): Z_base = 33²/10 = 108.9 Ω; z = 0.00573921 + j0.00459137 pu; load P = 0.5, Q = 0.25 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 0.991965V² + 1.6881e−05 = 0 → V = 0.99597 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle −0.0495°; S_from = 5.0181 + j2.5145 MVA; series loss \|I\|²z = 18.08 + j14.46 (kW, kvar); loading 5.6128/5 = 112.26 %; slack supplies 5.01808 + j2.51446 (MW, Mvar) (validation case `psas-2bus-branch-overload`)

Example 8 — Project limits 0.90–1.10 pu and 80 % loading

A 0.947 pu voltage that would fail the default 0.95 pu limit passes the project's 0.90 pu limit, and 75 % loading passes the 80 % cap.

Given: S_base = 10 MVA, 50 Hz. Buses: S (22 kV, SLACK at 22 kV (1 pu) ∠0°); R (22 kV, PQ, load 4 MW + j3 Mvar). Branches: SWER LINE S–R: 2.42 + j4.84 Ω; rated 7 MVA. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.9–1.1 pu, loading ≤ 80 %.

Per-unit bases	22 kV: Z_base = 22² / 10 = 48.4 Ω; powers ÷ 10 MVA
Branch impedances	SWER: z = (2.4200 + j4.8400) / 48.4 = 0.05 + j0.1 pu; y = 1/z = 4.0000 − j8.0000 pu
Scheduled injections	Net injection = generation − load (machines included): R: S = (−4.000 − j3.000)/10 = −0.40000 − j0.30000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_S,S = 4.0000 − j8.0000; Y_R,R = 4.0000 − j8.0000; Y_S,R = −4.0000 + j8.0000
Closed-form voltage	V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 with P = 0.4, Q = 0.3 pu, R + jX = 0.050000 + j0.100000 pu, E = 1.00000: V⁴ − 0.900000V² + 0.003125 = 0 → V² = (0.9 + √0.797500)/2 → V = 0.94684 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (0.025)/(0.946514) → θ = 1.5130° behind E; δ₂ = 0.0000 − 1.5130 = −1.5130°
Load-flow result	Newton-Raphson: V_R = 0.94684∠−1.5130° pu = 20.8306 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: SWER: S_S→ = 4.1394 + j3.2789 MVA, S_R→ = −4.0000 − j3.0000 MVA; loss = sum = \|I\|²(R + jX) = 139.43 kW + j278.86 kvar; loading \|S_from\|/rating = 5.2807/7 = 75.44 %
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 139.43 kW + j278.86 kvar; slack supplies P = 4.1394 MW, Q = 3.2789 Mvar. P balance: P_slack + ΣP_gen = 4.1394 + 0.0000 = 4.1394 MW = ΣP_load + P_loss = 4.0000 + 0.1394 = 4.1394 MW. Q balance: ΣQ_inj = 278.86 kvar = Q_loss + Q_shunt = 278.86 + (0.00) = 278.86 kvar
Convergence	converged after 4 iterations; largest power mismatch 1.12e-09 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus R at 0.94684 pu → 0.94684 vs limit 0.90000 pu PASS
Maximum voltage	highest bus S at 1.00000 pu → 1.00000 vs limit 1.10000 pu PASS
Branch loading	most loaded branch SWER → 75.44 vs limit 80.00 % PASS
Result	4/4 checks pass → PASS; V 0.9468–1.0000 pu, losses 139.4 kW; governing: branch loading
Reference	Hand calculation (closed-form two-bus): Z_base = 22²/10 = 48.4 Ω; z = 0.05 + j0.1 pu; load P = 0.4, Q = 0.3 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 0.900000V² + 0.003125 = 0 → V = 0.94684 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle −1.5130°; S_from = 4.1394 + j3.2789 MVA; series loss \|I\|²z = 139.43 + j278.86 (kW, kvar); loading 5.2807/7 = 75.44 %; slack supplies 4.13943 + j3.27886 (MW, Mvar) (validation case `psas-custom-limits`)

Example 9 — 415 V cable from per-km data on a 1 MVA base

Per-km impedance × length, a 1 MVA base so the tolerance suits a 134 kVA load, and the 2.9 % LV drop the load flow returns.

Given: S_base = 1 MVA, 50 Hz. Buses: MSB (0.415 kV, SLACK at 0.415 kV (1 pu) ∠0°); DB1 (0.415 kV, PQ, load 0.12 MW + j0.06 Mvar). Branches: SUBMAIN CABLE MSB–DB1: 0.15 km × (0.232 + j0.0721) Ω/km; rated 0.2 MVA. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	0.415 kV: Z_base = 0.415² / 1 = 0.172225 Ω; powers ÷ 1 MVA
Branch impedances	SUBMAIN: (0.232 + j0.0721) × 0.15 = 0.0348 + j0.0108 Ω; z = (0.0348 + j0.0108) / 0.172225 = 0.202061 + j0.0627958 pu; y = 1/z = 4.5131 − j1.4026 pu
Scheduled injections	Net injection = generation − load (machines included): DB1: S = (−0.120 − j0.060)/1 = −0.12000 − j0.06000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_MSB,MSB = 4.5131 − j1.4026; Y_DB1,DB1 = 4.5131 − j1.4026; Y_MSB,DB1 = −4.5131 + j1.4026
Closed-form voltage	V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 with P = 0.12, Q = 0.06 pu, R + jX = 0.202061 + j0.062796 pu, E = 1.00000: V⁴ − 0.943970V² + 0.000805897 = 0 → V² = (0.94397 + √0.887855)/2 → V = 0.97114 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (−0.00458818)/(0.97113) → θ = −0.2707° behind E; δ₂ = 0.0000 − (−0.2707) = 0.2707°
Load-flow result	Newton-Raphson: V_DB1 = 0.97114∠0.2707° pu = 0.4030 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: SUBMAIN: S_MSB→ = 0.1239 + j0.0612 MVA, S_DB1→ = −0.1200 − j0.0600 MVA; loss = sum = \|I\|²(R + jX) = 3.86 kW + j1.20 kvar; loading \|S_from\|/rating = 0.1382/0.2 = 69.08 %
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 3.86 kW + j1.20 kvar; slack supplies P = 0.1239 MW, Q = 0.0612 Mvar. P balance: P_slack + ΣP_gen = 0.1239 + 0.0000 = 0.1239 MW = ΣP_load + P_loss = 0.1200 + 0.0039 = 0.1239 MW. Q balance: ΣQ_inj = 1.20 kvar = Q_loss + Q_shunt = 1.20 + (0.00) = 1.20 kvar
Convergence	converged after 4 iterations; largest power mismatch 2.53e-12 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus DB1 at 0.97114 pu → 0.97114 vs limit 0.95000 pu PASS
Maximum voltage	highest bus MSB at 1.00000 pu → 1.00000 vs limit 1.05000 pu PASS
Branch loading	most loaded branch SUBMAIN → 69.08 vs limit 100.00 % PASS
Result	4/4 checks pass → PASS; V 0.9711–1.0000 pu, losses 3.9 kW; governing: branch loading
Reference	Hand calculation (closed-form two-bus): Z_base = 0.415²/1 = 0.172225 Ω; z = 0.202061 + j0.0627958 pu; load P = 0.12, Q = 0.06 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 0.943970V² + 0.000805897 = 0 → V = 0.97114 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle 0.2707°; S_from = 0.1239 + j0.0612 MVA; series loss \|I\|²z = 3.86 + j1.20 (kW, kvar); loading 0.1382/0.2 = 69.08 %; slack supplies 0.12386 + j0.06120 (MW, Mvar) (validation case `psas-lv-cable-per-km`)

Example 10 — 1.5 MVA 11/0.433 kV transformer from nameplate %Z

Nameplate 6 % with X/R 8 on a 1.5 MVA base is the per-unit impedance directly; the LV voltage and the 98 % loading follow.

Given: S_base = 1.5 MVA, 50 Hz. Buses: HV (11 kV, SLACK at 11 kV (1 pu) ∠0°); LV (0.433 kV, PQ, load 1.2 MW + j0.75 Mvar). Branches: TX1 XFMR HV–LV: 6 % on 1.5 MVA, X/R 8; rated 1.5 MVA. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	11 kV: Z_base = 11² / 1.5 = 80.6667 Ω; 0.433 kV: Z_base = 0.433² / 1.5 = 0.124993 Ω; powers ÷ 1.5 MVA
Branch impedances	TX1: z = 6/100 × 1.5/1.5 = 0.06 pu; R = z/√(1 + 8²) = 0.00744208, X = 8R = 0.0595367 pu; y = 1/z = 2.0672 − j16.5380 pu
Scheduled injections	Net injection = generation − load (machines included): LV: S = (−1.200 − j0.750)/1.5 = −0.80000 − j0.50000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_HV,HV = 2.0672 − j16.5380; Y_LV,LV = 2.0672 − j16.5380; Y_HV,LV = −2.0672 + j16.5380
Closed-form voltage	V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 with P = 0.8, Q = 0.5 pu, R + jX = 0.007442 + j0.059537 pu, E = 1.00000: V⁴ − 0.928556V² + 0.003204 = 0 → V² = (0.928556 + √0.849400)/2 → V = 0.96182 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (0.0439083)/(0.960815) → θ = 2.6165° behind E; δ₂ = 0.0000 − 2.6165 = −2.6165°
Load-flow result	Newton-Raphson: V_LV = 0.96182∠−2.6165° pu = 0.4165 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: TX1: S_HV→ = 1.2107 + j0.8359 MVA, S_LV→ = −1.2000 − j0.7500 MVA; loss = sum = \|I\|²(R + jX) = 10.74 kW + j85.92 kvar; loading \|S_from\|/rating = 1.4713/1.5 = 98.08 %
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 10.74 kW + j85.92 kvar; slack supplies P = 1.2107 MW, Q = 0.8359 Mvar. P balance: P_slack + ΣP_gen = 1.2107 + 0.0000 = 1.2107 MW = ΣP_load + P_loss = 1.2000 + 0.0107 = 1.2107 MW. Q balance: ΣQ_inj = 85.92 kvar = Q_loss + Q_shunt = 85.92 + (0.00) = 85.92 kvar
Convergence	converged after 4 iterations; largest power mismatch 1.17e-09 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus LV at 0.96182 pu → 0.96182 vs limit 0.95000 pu PASS
Maximum voltage	highest bus HV at 1.00000 pu → 1.00000 vs limit 1.05000 pu PASS
Branch loading	most loaded branch TX1 → 98.08 vs limit 100.00 % PASS
Result	4/4 checks pass → PASS; V 0.9618–1.0000 pu, losses 10.7 kW; governing: branch loading
Reference	Hand calculation (closed-form two-bus): Z_base = 11²/1.5 = 80.6667 Ω; z = 0.00744208 + j0.0595367 pu; load P = 0.8, Q = 0.5 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 0.928556V² + 0.003204 = 0 → V = 0.96182 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle −2.6165°; S_from = 1.2107 + j0.8359 MVA; series loss \|I\|²z = 10.74 + j85.92 (kW, kvar); loading 1.4713/1.5 = 98.08 %; slack supplies 1.21074 + j0.83592 (MW, Mvar) (validation case `psas-xfmr-impedance-percent`)

Example 11 — Same transformer on the −5 % HV tap

An off-nominal ratio a = 0.95 on the HV side makes the LV bus see 1/a = 1.053 pu behind the transformer impedance, lifting it about 5 %.

Given: S_base = 1.5 MVA, 50 Hz. Buses: HV (11 kV, SLACK at 11 kV (1 pu) ∠0°); LV (0.433 kV, PQ, load 1.2 MW + j0.75 Mvar). Branches: TX1 XFMR HV–LV: 6 % on 1.5 MVA, X/R 8; tap a = 0.95 on the HV side; rated 1.5 MVA. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	11 kV: Z_base = 11² / 1.5 = 80.6667 Ω; 0.433 kV: Z_base = 0.433² / 1.5 = 0.124993 Ω; powers ÷ 1.5 MVA
Branch impedances	TX1: z = 6/100 × 1.5/1.5 = 0.06 pu; R = z/√(1 + 8²) = 0.00744208, X = 8R = 0.0595367 pu; y = 1/z = 2.0672 − j16.5380 pu; tap a = 0.95: Y_ff = y/a² = 2.2906 − j18.3246, Y_ft = Y_tf = −y/a = −2.1760 + j17.4084, Y_tt = y
Scheduled injections	Net injection = generation − load (machines included): LV: S = (−1.200 − j0.750)/1.5 = −0.80000 − j0.50000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_HV,HV = 2.2906 − j18.3246; Y_LV,LV = 2.0672 − j16.5380; Y_HV,LV = −2.1760 + j17.4084
Thevenin source	E_th = −Y₂₁V₁/Y₂₂ = −(−2.1760 + j17.4084) × 1∠0° / (2.0672 − j16.5380) = 1.05263∠0.0000° pu; Z_th = 1/Y₂₂ = 0.007442 + j0.059537 pu
Closed-form voltage	V⁴ + [2(PR + QX) − E_th²]V² + (R² + X²)(P² + Q²) = 0 with P = 0.8, Q = 0.5 pu, R + jX = 0.007442 + j0.059537 pu, E_th = 1.05263: V⁴ − 1.036589V² + 0.003204 = 0 → V² = (1.03659 + √1.061701)/2 → V = 1.01661 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (0.0439083)/(1.06921) → θ = 2.3516° behind E_th; δ₂ = 0.0000 − 2.3516 = −2.3516°
Load-flow result	Newton-Raphson: V_LV = 1.01661∠−2.3516° pu = 0.4402 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: TX1: S_HV→ = 1.2096 + j0.8269 MVA, S_LV→ = −1.2000 − j0.7500 MVA; loss = sum = \|I\|²(R + jX) = 9.61 kW + j76.91 kvar; loading \|S_from\|/rating = 1.4652/1.5 = 97.68 %
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 9.61 kW + j76.91 kvar; slack supplies P = 1.2096 MW, Q = 0.8269 Mvar. P balance: P_slack + ΣP_gen = 1.2096 + 0.0000 = 1.2096 MW = ΣP_load + P_loss = 1.2000 + 0.0096 = 1.2096 MW. Q balance: ΣQ_inj = 76.91 kvar = Q_loss + Q_shunt = 76.91 + (0.00) = 76.91 kvar
Convergence	converged after 4 iterations; largest power mismatch 6.31e-11 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus HV at 1.00000 pu → 1.00000 vs limit 0.95000 pu PASS
Maximum voltage	highest bus LV at 1.01661 pu → 1.01661 vs limit 1.05000 pu PASS
Branch loading	most loaded branch TX1 → 97.68 vs limit 100.00 % PASS
Result	4/4 checks pass → PASS; V 1.0000–1.0166 pu, losses 9.6 kW; governing: branch loading
Reference	Hand calculation (closed-form two-bus): Z_base = 11²/1.5 = 80.6667 Ω; z = 0.00744208 + j0.0595367 pu; Thevenin at the load bus E_th = 1.05263∠−0.0000° pu, Z_th = 0.00744208 + j0.0595367 pu; load P = 0.8, Q = 0.5 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 1.036589V² + 0.003204 = 0 → V = 1.01661 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle −2.3516°; S_from = 1.2096 + j0.8269 MVA; series loss \|I\|²z = 9.61 + j76.91 (kW, kvar); loading 1.4652/1.5 = 97.68 %; slack supplies 1.20961 + j0.82691 (MW, Mvar) (validation case `psas-xfmr-off-nominal-tap`)

Example 12 — Off-nominal tap solved by pandapower

The tapped transformer through the pandapower backend (an HV tap changer at +(a − 1) × 100 %) lands on the same closed-form LV voltage.

Given: S_base = 1.5 MVA, 50 Hz. Buses: HV (11 kV, SLACK at 11 kV (1 pu) ∠0°); LV (0.433 kV, PQ, load 1.2 MW + j0.75 Mvar). Branches: TX1 XFMR HV–LV: 6 % on 1.5 MVA, X/R 8; tap a = 0.95 on the HV side; rated 1.5 MVA. Solver pandapower (NR); limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	11 kV: Z_base = 11² / 1.5 = 80.6667 Ω; 0.433 kV: Z_base = 0.433² / 1.5 = 0.124993 Ω; powers ÷ 1.5 MVA
Branch impedances	TX1: z = 6/100 × 1.5/1.5 = 0.06 pu; R = z/√(1 + 8²) = 0.00744208, X = 8R = 0.0595367 pu; y = 1/z = 2.0672 − j16.5380 pu; tap a = 0.95: Y_ff = y/a² = 2.2906 − j18.3246, Y_ft = Y_tf = −y/a = −2.1760 + j17.4084, Y_tt = y
Scheduled injections	Net injection = generation − load (machines included): LV: S = (−1.200 − j0.750)/1.5 = −0.80000 − j0.50000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_HV,HV = 2.2906 − j18.3246; Y_LV,LV = 2.0672 − j16.5380; Y_HV,LV = −2.1760 + j17.4084
Thevenin source	E_th = −Y₂₁V₁/Y₂₂ = −(−2.1760 + j17.4084) × 1∠0° / (2.0672 − j16.5380) = 1.05263∠0.0000° pu; Z_th = 1/Y₂₂ = 0.007442 + j0.059537 pu
Closed-form voltage	V⁴ + [2(PR + QX) − E_th²]V² + (R² + X²)(P² + Q²) = 0 with P = 0.8, Q = 0.5 pu, R + jX = 0.007442 + j0.059537 pu, E_th = 1.05263: V⁴ − 1.036589V² + 0.003204 = 0 → V² = (1.03659 + √1.061701)/2 → V = 1.01661 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (0.0439083)/(1.06921) → θ = 2.3516° behind E_th; δ₂ = 0.0000 − 2.3516 = −2.3516°
Load-flow result	pandapower (NR): V_LV = 1.01661∠−2.3516° pu = 0.4402 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: TX1: S_HV→ = 1.2096 + j0.8269 MVA, S_LV→ = −1.2000 − j0.7500 MVA; loss = sum = \|I\|²(R + jX) = 9.61 kW + j76.91 kvar; loading \|S_from\|/rating = 1.4652/1.5 = 97.68 %
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 9.61 kW + j76.91 kvar; slack supplies P = 1.2096 MW, Q = 0.8269 Mvar. P balance: P_slack + ΣP_gen = 1.2096 + 0.0000 = 1.2096 MW = ΣP_load + P_loss = 1.2000 + 0.0096 = 1.2096 MW. Q balance: ΣQ_inj = 76.91 kvar = Q_loss + Q_shunt = 76.91 + (0.00) = 76.91 kvar
Convergence	pandapower Newton-Raphson converged in 3 iterations on its own mismatch tolerance PASS
Minimum voltage	lowest bus HV at 1.00000 pu → 1.00000 vs limit 0.95000 pu PASS
Maximum voltage	highest bus LV at 1.01661 pu → 1.01661 vs limit 1.05000 pu PASS
Branch loading	most loaded branch TX1 → 97.68 vs limit 100.00 % PASS
Result	4/4 checks pass → PASS; V 1.0000–1.0166 pu, losses 9.6 kW; governing: branch loading
Reference	Hand calculation (closed-form two-bus): Z_base = 11²/1.5 = 80.6667 Ω; z = 0.00744208 + j0.0595367 pu; Thevenin at the load bus E_th = 1.05263∠−0.0000° pu, Z_th = 0.00744208 + j0.0595367 pu; load P = 0.8, Q = 0.5 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 1.036589V² + 0.003204 = 0 → V = 1.01661 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle −2.3516°; S_from = 1.2096 + j0.8269 MVA; series loss \|I\|²z = 9.61 + j76.91 (kW, kvar); loading 1.4652/1.5 = 97.68 %; slack supplies 1.20961 + j0.82691 (MW, Mvar) (validation case `psas-xfmr-tap-pandapower`)

Example 13 — 3 Mvar shunt capacitor lifts a sagging bus

Without the capacitor the load bus sits at 0.930 pu (fails); a 3 Mvar constant-admittance capacitor brings it back inside limits. Its output is shunt Q, reported apart from the line's I²X loss.

Given: S_base = 10 MVA, 50 Hz. Buses: S (11 kV, SLACK at 11 kV (1 pu) ∠0°); L (11 kV, PQ, load 5 MW + j4 Mvar). Branches: LN1 LINE S–L: 0.605 + j1.21 Ω; CAP1 SHUNT at L: 0 − j40.3333 Ω to earth. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	11 kV: Z_base = 11² / 10 = 12.1 Ω; powers ÷ 10 MVA
Branch impedances	LN1: z = (0.6050 + j1.2100) / 12.1 = 0.05 + j0.1 pu; y = 1/z = 4.0000 − j8.0000 pu CAP1: z_sh = (0.0000 − j40.3333) / 12.1 = 0.00000 − j3.33333 pu → y_sh = 0.00000 + j0.30000 pu (3.000 Mvar of capacitive output at 1 pu)
Scheduled injections	Net injection = generation − load (machines included): L: S = (−5.000 − j4.000)/10 = −0.50000 − j0.40000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_S,S = 4.0000 − j8.0000; Y_L,L = 4.0000 − j7.7000; Y_S,L = −4.0000 + j8.0000
Thevenin source	E_th = −Y₂₁V₁/Y₂₂ = −(−4.0000 + j8.0000) × 1∠0° / (4.0000 − j7.7000) = 1.03080∠−0.8859° pu; Z_th = 1/Y₂₂ = 0.053128 + j0.102271 pu
Closed-form voltage	V⁴ + [2(PR + QX) − E_th²]V² + (R² + X²)(P² + Q²) = 0 with P = 0.5, Q = 0.4 pu, R + jX = 0.053128 + j0.102271 pu, E_th = 1.03080: V⁴ − 0.927613V² + 0.00544561 = 0 → V² = (0.927613 + √0.838684)/2 → V = 0.96005 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (0.0298844)/(0.989177) → θ = 1.7305° behind E_th; δ₂ = −0.8859 − 1.7305 = −2.6164°
Load-flow result	Newton-Raphson: V_L = 0.96005∠−2.6164° pu = 10.5606 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: LN1: S_S→ = 5.1439 + j1.5227 MVA, S_L→ = −5.0000 − j1.2349 MVA; loss = sum = \|I\|²(R + jX) = 143.89 kW + j287.78 kvar CAP1 (shunt at L): S = 0.0000 − j2.7651 MVA (generates 2.7651 Mvar at the solved voltage)
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 143.89 kW + j287.78 kvar; shunt Q (SHUNT elements, line charging) −2,765.11 kvar, reported separately; slack supplies P = 5.1439 MW, Q = 1.5227 Mvar. P balance: P_slack + ΣP_gen = 5.1439 + 0.0000 = 5.1439 MW = ΣP_load + P_loss = 5.0000 + 0.1439 = 5.1439 MW. Q balance: ΣQ_inj = −2,477.33 kvar = Q_loss + Q_shunt = 287.78 + (−2,765.11) = −2,477.33 kvar
Convergence	converged after 4 iterations; largest power mismatch 8.76e-10 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus L at 0.96005 pu → 0.96005 vs limit 0.95000 pu PASS
Maximum voltage	highest bus S at 1.00000 pu → 1.00000 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 0.9600–1.0000 pu, losses 143.9 kW; governing: minimum voltage. No rating_mva on LN1: loading not checked.
Reference	Hand calculation (closed-form two-bus): Z_base = 11²/10 = 12.1 Ω; z = 0.05 + j0.1 pu; Thevenin at the load bus E_th = 1.03080∠−0.8859° pu, Z_th = 0.0531279 + j0.102271 pu; load P = 0.5, Q = 0.4 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 0.927613V² + 0.00544561 = 0 → V = 0.96005 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle −2.6164°; S_from = 5.1439 + j1.5227 MVA; series loss \|I\|²z = 143.89 + j287.78 (kW, kvar); shunt Q (charging/capacitor) −2765.12 kvar, not a loss; without the capacitor V = 0.92955 pu (quartic with E = 1, R + jX = z); slack supplies 5.14389 + j1.52267 (MW, Mvar) (validation case `psas-shunt-capacitor`)

Example 14 — PV generator bus on a reactive tie (power-angle equation)

A 6 MW generator holding 1.02 pu: δ from P = V₁V₂ sin δ / X, then the reactive power each end supplies.

Given: S_base = 10 MVA, 50 Hz. Buses: GRID (33 kV, SLACK at 33 kV (1 pu) ∠0°); GEN (33 kV, PV holding 33.66 kV (1.02 pu), P_gen 6 MW). Branches: TIE LINE GRID–GEN: 0 + j10.89 Ω. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	33 kV: Z_base = 33² / 10 = 108.9 Ω; powers ÷ 10 MVA
Branch impedances	TIE: z = (0.0000 + j10.8900) / 108.9 = 0 + j0.1 pu; y = 1/z = 0.0000 − j10.0000 pu
Scheduled injections	Net injection = generation − load (machines included): GEN: P = 6.000/10 = 0.6 pu at \|V\| = 1.02 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_GRID,GRID = 0.0000 − j10.0000; Y_GEN,GEN = 0.0000 − j10.0000; Y_GRID,GEN = 0.0000 + j10.0000
Power-angle equation	P₂ = V₂²G₂₂ + V₁V₂(G₂₁ cos δ + B₂₁ sin δ): 0.6 = 1.02² × 0 + 1 × 1.02 × (0 cos δ + 10 sin δ) → 0 cos δ + 10.2 sin δ = 0.6 → δ = atan2(10.2, 0) − arccos(0.6/10.2) = 3.3723° (relative to the slack)
Reactive power	Q₂ = −V₂²B₂₂ + V₁V₂(G₂₁ sin δ − B₂₁ cos δ) = −1.02² × (−10) + 1.02 × (0 × 0.0588235 − 10 × 0.998268) = 0.221662 pu = 2.2166 Mvar; engine Q_inj = 2.2166 Mvar
Load-flow result	Newton-Raphson: V_GEN = 1.02000∠3.3723° pu = 33.6600 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: TIE: S_GRID→ = −6.0000 − j1.8234 MVA, S_GEN→ = 6.0000 + j2.2166 MVA; loss = sum = \|I\|²(R + jX) = 0.00 kW + j393.25 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 0.00 kW + j393.25 kvar; slack supplies P = −6.0000 MW, Q = −1.8234 Mvar. P balance: P_slack + ΣP_gen = −6.0000 + 6.0000 = 0.0000 MW = ΣP_load + P_loss = 0.0000 + 0.0000 = 0.0000 MW. Q balance: ΣQ_inj = 393.25 kvar = Q_loss + Q_shunt = 393.25 + (0.00) = 393.25 kvar
Convergence	converged after 3 iterations; largest power mismatch 3.46e-10 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus GRID at 1.00000 pu → 1.00000 vs limit 0.95000 pu PASS
Maximum voltage	highest bus GEN at 1.02000 pu → 1.02000 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 1.0000–1.0200 pu, losses 0.0 kW; governing: minimum voltage. No rating_mva on TIE: loading not checked.
Reference	Hand calculation (closed-form two-bus): Z_base = 33²/10 = 108.9 Ω; z = 0 + j0.1 pu; PV bus P = 0.6 pu at \|V\| = 1.02: P = V²G₂₂ + V₁V₂(G₂₁cos δ + B₂₁sin δ) gives δ = 3.3723°; Q = −V²B₂₂ + V₁V₂(G₂₁sin δ − B₂₁cos δ) = 0.221662 pu = 2.2166 Mvar; S_from = −6.0000 − j1.8234 MVA; series loss \|I\|²z = 0.00 + j393.25 (kW, kvar); slack supplies −6.00000 − j1.82338 (MW, Mvar) (validation case `psas-pv-bus-lossless`)

Example 15 — PV bus with local load on a lossy line

Net 3 MW export from a bus held at 1.03 pu through R + jX: the PV angle from G and B terms, and the generator's reactive output.

Given: S_base = 10 MVA, 50 Hz. Buses: GRID (11 kV, SLACK at 11 kV (1 pu) ∠0°); MILL (11 kV, PV holding 11.33 kV (1.03 pu), P_gen 4 MW, load 1 MW + j0.5 Mvar). Branches: LN1 LINE GRID–MILL: 0.605 + j1.21 Ω. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	11 kV: Z_base = 11² / 10 = 12.1 Ω; powers ÷ 10 MVA
Branch impedances	LN1: z = (0.6050 + j1.2100) / 12.1 = 0.05 + j0.1 pu; y = 1/z = 4.0000 − j8.0000 pu
Scheduled injections	Net injection = generation − load (machines included): MILL: P = 3.000/10 = 0.3 pu at \|V\| = 1.03 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_GRID,GRID = 4.0000 − j8.0000; Y_MILL,MILL = 4.0000 − j8.0000; Y_GRID,MILL = −4.0000 + j8.0000
Power-angle equation	P₂ = V₂²G₂₂ + V₁V₂(G₂₁ cos δ + B₂₁ sin δ): 0.3 = 1.03² × 4 + 1 × 1.03 × (−4 cos δ + 8 sin δ) → −4.12 cos δ + 8.24 sin δ = −3.9436 → δ = atan2(8.24, −4.12) − arccos(−3.9436/9.2126) = 1.2202° (relative to the slack)
Reactive power	Q₂ = −V₂²B₂₂ + V₁V₂(G₂₁ sin δ − B₂₁ cos δ) = −1.03² × (−8) + 1.03 × (−4 × 0.0212944 − 8 × 0.999773) = 0.161336 pu = 1.6134 Mvar; engine Q_inj = 1.6134 Mvar
Load-flow result	Newton-Raphson: V_MILL = 1.03000∠1.2202° pu = 11.3300 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: LN1: S_GRID→ = −2.9453 − j1.5040 MVA, S_MILL→ = 3.0000 + j1.6134 MVA; loss = sum = \|I\|²(R + jX) = 54.68 kW + j109.37 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 54.68 kW + j109.37 kvar; slack supplies P = −2.9453 MW, Q = −1.5040 Mvar. P balance: P_slack + ΣP_gen = −2.9453 + 4.0000 = 1.0547 MW = ΣP_load + P_loss = 1.0000 + 0.0547 = 1.0547 MW. Q balance: ΣQ_inj = 109.37 kvar = Q_loss + Q_shunt = 109.37 + (0.00) = 109.37 kvar
Convergence	converged after 3 iterations; largest power mismatch 2.46e-08 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus GRID at 1.00000 pu → 1.00000 vs limit 0.95000 pu PASS
Maximum voltage	highest bus MILL at 1.03000 pu → 1.03000 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 1.0000–1.0300 pu, losses 54.7 kW; governing: minimum voltage. No rating_mva on LN1: loading not checked.
Reference	Hand calculation (closed-form two-bus): Z_base = 11²/10 = 12.1 Ω; z = 0.05 + j0.1 pu; PV bus P = 0.3 pu at \|V\| = 1.03: P = V²G₂₂ + V₁V₂(G₂₁cos δ + B₂₁sin δ) gives δ = 1.2202°; Q = −V²B₂₂ + V₁V₂(G₂₁sin δ − B₂₁cos δ) = 0.161336 pu = 1.6134 Mvar; S_from = −2.9453 − j1.5040 MVA; series loss \|I\|²z = 54.68 + j109.37 (kW, kvar); slack supplies −2.94532 − j1.50399 (MW, Mvar) (validation case `psas-pv-bus-lossy-with-load`)

Example 16 — PV generator inside its Q limits (Gauss-Seidel)

Gauss-Seidel checks the generator's reactive limits each sweep: holding 1.05 pu needs 5.42 Mvar, inside ±6 Mvar, so the bus stays PV and sits exactly on the 1.05 pu limit (a pass).

Given: S_base = 10 MVA, 50 Hz. Buses: GRID (33 kV, SLACK at 33 kV (1 pu) ∠0°); GEN (33 kV, PV holding 34.65 kV (1.05 pu), P_gen 6 MW, Q limits −6 … 6 Mvar). Branches: TIE LINE GRID–GEN: 0 + j10.89 Ω. Solver Gauss-Seidel, tolerance 1e-08 pu, ≤ 500 iterations, acceleration α = 1.6; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	33 kV: Z_base = 33² / 10 = 108.9 Ω; powers ÷ 10 MVA
Branch impedances	TIE: z = (0.0000 + j10.8900) / 108.9 = 0 + j0.1 pu; y = 1/z = 0.0000 − j10.0000 pu
Scheduled injections	Net injection = generation − load (machines included): GEN: P = 6.000/10 = 0.6 pu at \|V\| = 1.05 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_GRID,GRID = 0.0000 − j10.0000; Y_GEN,GEN = 0.0000 − j10.0000; Y_GRID,GEN = 0.0000 + j10.0000
Gauss-Seidel start	PV bus GEN from a flat start: Q = −Im{V*_iΣY_ikV_k} = 0.52500 pu; each sweep updates V like a PQ bus with that Q, accelerates it, V ← V + 1.6(V_new − V), and only then resets \|V\| to 1.05 pu (α moves the angle, never the magnitude). If the generator's Q_G = Q + Q_load leaves its limits the bus is held PQ at the limit until its voltage crosses back over the set-point
Power-angle equation	P₂ = V₂²G₂₂ + V₁V₂(G₂₁ cos δ + B₂₁ sin δ): 0.6 = 1.05² × 0 + 1 × 1.05 × (0 cos δ + 10 sin δ) → 0 cos δ + 10.5 sin δ = 0.6 → δ = atan2(10.5, 0) − arccos(0.6/10.5) = 3.2758° (relative to the slack)
Reactive power	Q₂ = −V₂²B₂₂ + V₁V₂(G₂₁ sin δ − B₂₁ cos δ) = −1.05² × (−10) + 1.05 × (0 × 0.0571429 − 10 × 0.998366) = 0.542157 pu = 5.4216 Mvar; engine Q_inj = 5.4216 Mvar
Q limits	Q_G = 5.422 Mvar vs −6 … 6 Mvar: inside the limits, so the bus stays PV at 1.05 pu
Load-flow result	Gauss-Seidel: V_GEN = 1.05000∠3.2758° pu = 34.6500 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: TIE: S_GRID→ = −6.0000 − j4.8284 MVA, S_GEN→ = 6.0000 + j5.4216 MVA; loss = sum = \|I\|²(R + jX) = 0.00 kW + j593.14 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 0.00 kW + j593.14 kvar; slack supplies P = −6.0000 MW, Q = −4.8284 Mvar. P balance: P_slack + ΣP_gen = −6.0000 + 6.0000 = 0.0000 MW = ΣP_load + P_loss = 0.0000 + 0.0000 = 0.0000 MW. Q balance: ΣQ_inj = 593.14 kvar = Q_loss + Q_shunt = 593.14 + (0.00) = 593.14 kvar
Convergence	converged after 26 iterations; largest \|ΔV\| in the last sweep 7.66e-09 vs tolerance 1e-08 pu; power mismatch of the returned voltages 2.62e-08 pu PASS
Minimum voltage	lowest bus GRID at 1.00000 pu → 1.00000 vs limit 0.95000 pu PASS
Maximum voltage	highest bus GEN at 1.05000 pu → 1.05000 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 1.0000–1.0500 pu, losses 0.0 kW; governing: minimum voltage. No rating_mva on TIE: loading not checked.
Reference	Hand calculation (closed-form two-bus): Z_base = 33²/10 = 108.9 Ω; z = 0 + j0.1 pu; PV bus P = 0.6 pu at \|V\| = 1.05: P = V²G₂₂ + V₁V₂(G₂₁cos δ + B₂₁sin δ) gives δ = 3.2758°; Q = −V²B₂₂ + V₁V₂(G₂₁sin δ − B₂₁cos δ) = 0.542157 pu = 5.4216 Mvar; S_from = −6.0000 − j4.8284 MVA; series loss \|I\|²z = 0.00 + j593.14 (kW, kvar); Q = 5.4216 Mvar lies inside Q_min −6 / Q_max +6 Mvar, so the bus stays PV at 1.05 pu; slack supplies −6.00000 − j4.82843 (MW, Mvar) (validation case `psas-pv-qlimits-gauss-seidel`)

Example 17 — PV generator held at Q_max (Gauss-Seidel)

The same generator with Q_max = 3 Mvar: holding 1.05 pu needs 5.42 Mvar, so it is held at 3 Mvar and the bus is solved as PQ — it settles at 1.027 pu, below its set-point, which keeps it at the limit.

Given: S_base = 10 MVA, 50 Hz. Buses: GRID (33 kV, SLACK at 33 kV (1 pu) ∠0°); GEN (33 kV, PV holding 34.65 kV (1.05 pu), P_gen 6 MW, Q limits −6 … 3 Mvar). Branches: TIE LINE GRID–GEN: 0 + j10.89 Ω. Solver Gauss-Seidel, tolerance 1e-08 pu, ≤ 500 iterations, acceleration α = 1.6; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	33 kV: Z_base = 33² / 10 = 108.9 Ω; powers ÷ 10 MVA
Branch impedances	TIE: z = (0.0000 + j10.8900) / 108.9 = 0 + j0.1 pu; y = 1/z = 0.0000 − j10.0000 pu
Scheduled injections	Net injection = generation − load (machines included): GEN: P = 6.000/10 = 0.6 pu at \|V\| = 1.05 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_GRID,GRID = 0.0000 − j10.0000; Y_GEN,GEN = 0.0000 − j10.0000; Y_GRID,GEN = 0.0000 + j10.0000
Gauss-Seidel start	PV bus GEN from a flat start: Q = −Im{V*_iΣY_ikV_k} = 0.52500 pu; each sweep updates V like a PQ bus with that Q, accelerates it, V ← V + 1.6(V_new − V), and only then resets \|V\| to 1.05 pu (α moves the angle, never the magnitude). If the generator's Q_G = Q + Q_load leaves its limits the bus is held PQ at the limit until its voltage crosses back over the set-point
Power-angle equation	P₂ = V₂²G₂₂ + V₁V₂(G₂₁ cos δ + B₂₁ sin δ): 0.6 = 1.05² × 0 + 1 × 1.05 × (0 cos δ + 10 sin δ) → 0 cos δ + 10.5 sin δ = 0.6 → δ = atan2(10.5, 0) − arccos(0.6/10.5) = 3.2758° (relative to the slack)
Reactive power	Q₂ = −V₂²B₂₂ + V₁V₂(G₂₁ sin δ − B₂₁ cos δ) = −1.05² × (−10) + 1.05 × (0 × 0.0571429 − 10 × 0.998366) = 0.542157 pu = 5.4216 Mvar
Q limits	Q_G = 5.422 Mvar vs −6 … 3 Mvar: above Q_max, so the generator is held at 3.000 Mvar and the bus is solved as PQ with Q_net = 3.000 − 0.000 = 3.000 Mvar = 0.3 pu
Closed-form voltage	V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 with P = −0.6, Q = −0.3 pu, R + jX = 0.000000 + j0.100000 pu, E = 1.00000: V⁴ − 1.060000V² + 0.0045 = 0 → V² = (1.06 + √1.105600)/2 → V = 1.02749 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (−0.06)/(1.02574) → θ = −3.3477° behind E; δ₂ = 0.0000 − (−3.3477) = 3.3477°
Hold	at Q_max the bus sits below its set-point (1.02749 vs 1.05 pu), so it stays PQ: holding 1.05 pu would need a Q_G outside the limit (Glover, Sarma & Overbye §6.5)
Load-flow result	Gauss-Seidel: V_GEN = 1.02749∠3.3477° pu = 33.9072 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: TIE: S_GRID→ = −6.0000 − j2.5738 MVA, S_GEN→ = 6.0000 + j3.0000 MVA; loss = sum = \|I\|²(R + jX) = 0.00 kW + j426.24 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 0.00 kW + j426.24 kvar; slack supplies P = −6.0000 MW, Q = −2.5738 Mvar. P balance: P_slack + ΣP_gen = −6.0000 + 6.0000 = 0.0000 MW = ΣP_load + P_loss = 0.0000 + 0.0000 = 0.0000 MW. Q balance: ΣQ_inj = 426.24 kvar = Q_loss + Q_shunt = 426.24 + (0.00) = 426.24 kvar
Convergence	converged after 43 iterations; largest \|ΔV\| in the last sweep 8.33e-09 vs tolerance 1e-08 pu; power mismatch of the returned voltages 3.19e-08 pu PASS
Minimum voltage	lowest bus GRID at 1.00000 pu → 1.00000 vs limit 0.95000 pu PASS
Maximum voltage	highest bus GEN at 1.02749 pu → 1.02749 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 1.0000–1.0275 pu, losses 0.0 kW; governing: minimum voltage. PV bus(es) held at a reactive limit, so not at their voltage set-point: GEN (Q_max). No rating_mva on TIE: loading not checked.
Reference	Hand calculation (closed-form two-bus): Z_base = 33²/10 = 108.9 Ω; z = 0 + j0.1 pu; at its set-point \|V\| = 1.05 pu the generator would need Q_G = Q_net + Q_load = 5.4216 + 0 = 5.4216 Mvar, above Q_max 3 Mvar, so it is held at the limit and the bus is PQ with Q_net = 3 − 0 = 3 Mvar; load P = −0.6, Q = −0.3 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 1.060000V² + 0.0045 = 0 → V = 1.02749 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle 3.3477°; S_from = −6.0000 − j2.5738 MVA; series loss \|I\|²z = 0.00 + j426.24 (kW, kvar); V = 1.02749 pu is below the set-point, so the bus stays held (PV→PQ switching with hold, Glover, Sarma & Overbye §6.5); slack supplies −6.00000 − j2.57376 (MW, Mvar). (Textbook edition and example/figure numbers pending verification.) (validation case `psas-pv-qmax-binding-gauss-seidel`)

Example 18 — Generator Q limit with a local reactive load (Newton-Raphson)

The limit applies to the generator, not the bus: holding 1.02 pu needs 2.16 Mvar net, inside 3 Mvar, but 4.16 Mvar from the machine once the 2 Mvar local load is added — so the generator is held at 3 Mvar and the bus sags to 1.009 pu.

Given: S_base = 10 MVA, 50 Hz. Buses: GRID (33 kV, SLACK at 33 kV (1 pu) ∠0°); GEN (33 kV, PV holding 33.66 kV (1.02 pu), P_gen 6 MW, Q limits −3 … 3 Mvar, load 1 MW + j2 Mvar). Branches: TIE LINE GRID–GEN: 0 + j10.89 Ω. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	33 kV: Z_base = 33² / 10 = 108.9 Ω; powers ÷ 10 MVA
Branch impedances	TIE: z = (0.0000 + j10.8900) / 108.9 = 0 + j0.1 pu; y = 1/z = 0.0000 − j10.0000 pu
Scheduled injections	Net injection = generation − load (machines included): GEN: P = 5.000/10 = 0.5 pu at \|V\| = 1.02 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_GRID,GRID = 0.0000 − j10.0000; Y_GEN,GEN = 0.0000 − j10.0000; Y_GRID,GEN = 0.0000 + j10.0000
Power-angle equation	P₂ = V₂²G₂₂ + V₁V₂(G₂₁ cos δ + B₂₁ sin δ): 0.5 = 1.02² × 0 + 1 × 1.02 × (0 cos δ + 10 sin δ) → 0 cos δ + 10.2 sin δ = 0.5 → δ = atan2(10.2, 0) − arccos(0.5/10.2) = 2.8097° (relative to the slack)
Reactive power	Q₂ = −V₂²B₂₂ + V₁V₂(G₂₁ sin δ − B₂₁ cos δ) = −1.02² × (−10) + 1.02 × (0 × 0.0490196 − 10 × 0.998798) = 0.216262 pu = 2.1626 Mvar
Q limits	Q_G = Q₂ + Q_load = 2.163 + 2.000 = 4.163 Mvar vs −3 … 3 Mvar: above Q_max, so the generator is held at 3.000 Mvar and the bus is solved as PQ with Q_net = 3.000 − 2.000 = 1.000 Mvar = 0.1 pu
Closed-form voltage	V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 with P = −0.5, Q = −0.1 pu, R + jX = 0.000000 + j0.100000 pu, E = 1.00000: V⁴ − 1.020000V² + 0.0026 = 0 → V² = (1.02 + √1.030000)/2 → V = 1.00868 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (−0.05)/(1.00744) → θ = −2.8413° behind E; δ₂ = 0.0000 − (−2.8413) = 2.8413°
Hold	at Q_max the bus sits below its set-point (1.00868 vs 1.02 pu), so it stays PQ: holding 1.02 pu would need a Q_G outside the limit (Glover, Sarma & Overbye §6.5)
Load-flow result	Newton-Raphson: V_GEN = 1.00868∠2.8413° pu = 33.2866 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: TIE: S_GRID→ = −5.0000 − j0.7445 MVA, S_GEN→ = 5.0000 + j1.0000 MVA; loss = sum = \|I\|²(R + jX) = 0.00 kW + j255.54 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 0.00 kW + j255.54 kvar; slack supplies P = −5.0000 MW, Q = −0.7445 Mvar. P balance: P_slack + ΣP_gen = −5.0000 + 6.0000 = 1.0000 MW = ΣP_load + P_loss = 1.0000 + 0.0000 = 1.0000 MW. Q balance: ΣQ_inj = 255.54 kvar = Q_loss + Q_shunt = 255.54 + (0.00) = 255.54 kvar
Convergence	converged after 6 iterations; largest power mismatch 1.52e-07 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus GRID at 1.00000 pu → 1.00000 vs limit 0.95000 pu PASS
Maximum voltage	highest bus GEN at 1.00868 pu → 1.00868 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 1.0000–1.0087 pu, losses 0.0 kW; governing: minimum voltage. PV bus(es) held at a reactive limit, so not at their voltage set-point: GEN (Q_max). No rating_mva on TIE: loading not checked.
Reference	Hand calculation (closed-form two-bus): Z_base = 33²/10 = 108.9 Ω; z = 0 + j0.1 pu; at its set-point \|V\| = 1.02 pu the generator would need Q_G = Q_net + Q_load = 2.1626 + 2 = 4.1626 Mvar, above Q_max 3 Mvar, so it is held at the limit and the bus is PQ with Q_net = 3 − 2 = 1 Mvar; load P = −0.5, Q = −0.1 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 1.020000V² + 0.0026 = 0 → V = 1.00868 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle 2.8413°; S_from = −5.0000 − j0.7445 MVA; series loss \|I\|²z = 0.00 + j255.54 (kW, kvar); V = 1.00868 pu is below the set-point, so the bus stays held (PV→PQ switching with hold, Glover, Sarma & Overbye §6.5); slack supplies −5.00000 − j0.74446 (MW, Mvar). (Textbook edition and example/figure numbers pending verification.) (validation case `psas-pv-qmax-local-load-nr`)

Example 19 — Generator held at Q_min (pandapower)

A 0.95 pu set-point would make the generator absorb 4.56 Mvar, beyond its −1 Mvar limit: held at Q_min the bus floats up to 0.988 pu instead.

Given: S_base = 10 MVA, 50 Hz. Buses: GRID (33 kV, SLACK at 33 kV (1 pu) ∠0°); GEN (33 kV, PV holding 31.35 kV (0.95 pu), P_gen 6 MW, Q limits −1 … 6 Mvar). Branches: TIE LINE GRID–GEN: 0 + j10.89 Ω. Solver pandapower (NR); limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	33 kV: Z_base = 33² / 10 = 108.9 Ω; powers ÷ 10 MVA
Branch impedances	TIE: z = (0.0000 + j10.8900) / 108.9 = 0 + j0.1 pu; y = 1/z = 0.0000 − j10.0000 pu
Scheduled injections	Net injection = generation − load (machines included): GEN: P = 6.000/10 = 0.6 pu at \|V\| = 0.95 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_GRID,GRID = 0.0000 − j10.0000; Y_GEN,GEN = 0.0000 − j10.0000; Y_GRID,GEN = 0.0000 + j10.0000
Power-angle equation	P₂ = V₂²G₂₂ + V₁V₂(G₂₁ cos δ + B₂₁ sin δ): 0.6 = 0.95² × 0 + 1 × 0.95 × (0 cos δ + 10 sin δ) → 0 cos δ + 9.5 sin δ = 0.6 → δ = atan2(9.5, 0) − arccos(0.6/9.5) = 3.6211° (relative to the slack)
Reactive power	Q₂ = −V₂²B₂₂ + V₁V₂(G₂₁ sin δ − B₂₁ cos δ) = −0.95² × (−10) + 0.95 × (0 × 0.0631579 − 10 × 0.998004) = -0.456034 pu = −4.5603 Mvar
Q limits	Q_G = −4.560 Mvar vs −1 … 6 Mvar: below Q_min, so the generator is held at −1.000 Mvar and the bus is solved as PQ with Q_net = −1.000 − 0.000 = −1.000 Mvar = −0.1 pu
Closed-form voltage	V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 with P = −0.6, Q = 0.1 pu, R + jX = 0.000000 + j0.100000 pu, E = 1.00000: V⁴ − 0.980000V² + 0.0037 = 0 → V² = (0.98 + √0.945600)/2 → V = 0.98803 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (−0.06)/(0.98621) → θ = −3.4815° behind E; δ₂ = 0.0000 − (−3.4815) = 3.4815°
Hold	at Q_min the bus sits above its set-point (0.98803 vs 0.95 pu), so it stays PQ: holding 0.95 pu would need a Q_G outside the limit (Glover, Sarma & Overbye §6.5)
Load-flow result	pandapower (NR): V_GEN = 0.98803∠3.4815° pu = 32.6051 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: TIE: S_GRID→ = −6.0000 + j1.3790 MVA, S_GEN→ = 6.0000 − j1.0000 MVA; loss = sum = \|I\|²(R + jX) = 0.00 kW + j379.02 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 0.00 kW + j379.02 kvar; slack supplies P = −6.0000 MW, Q = 1.3790 Mvar. P balance: P_slack + ΣP_gen = −6.0000 + 6.0000 = 0.0000 MW = ΣP_load + P_loss = 0.0000 + 0.0000 = 0.0000 MW. Q balance: ΣQ_inj = 379.02 kvar = Q_loss + Q_shunt = 379.02 + (0.00) = 379.02 kvar
Convergence	pandapower Newton-Raphson converged in 3 iterations on its own mismatch tolerance PASS
Minimum voltage	lowest bus GEN at 0.98803 pu → 0.98803 vs limit 0.95000 pu PASS
Maximum voltage	highest bus GRID at 1.00000 pu → 1.00000 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 0.9880–1.0000 pu, losses 0.0 kW; governing: minimum voltage. PV bus(es) held at a reactive limit, so not at their voltage set-point: GEN (Q_min). No rating_mva on TIE: loading not checked.
Reference	Hand calculation (closed-form two-bus): Z_base = 33²/10 = 108.9 Ω; z = 0 + j0.1 pu; at its set-point \|V\| = 0.95 pu the generator would need Q_G = Q_net + Q_load = −4.5603 + 0 = −4.5603 Mvar, below Q_min −1 Mvar, so it is held at the limit and the bus is PQ with Q_net = −1 − 0 = −1 Mvar; load P = −0.6, Q = 0.1 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 0.980000V² + 0.0037 = 0 → V = 0.98803 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle 3.4815°; S_from = −6.0000 + j1.3790 MVA; series loss \|I\|²z = 0.00 + j379.02 (kW, kvar); V = 0.98803 pu is above the set-point, so the bus stays held (PV→PQ switching with hold, Glover, Sarma & Overbye §6.5); slack supplies −6.00000 + j1.37902 (MW, Mvar). (Textbook edition and example/figure numbers pending verification.) (validation case `psas-pv-qmin-binding-pandapower`)

Example 20 — Bus load with a motor and an embedded generator

Machines aggregate into the bus injection — the motor adds to load, the generator offsets it — before the two-bus solution.

Given: S_base = 10 MVA, 50 Hz. Buses: S (11 kV, SLACK at 11 kV (1 pu) ∠0°); PLANT (11 kV, PQ, load 3 MW + j1 Mvar, induction motor 2 MW + j1.2 Mvar, generator 1.5 MW + j0.5 Mvar). Branches: LN1 LINE S–PLANT: 0.605 + j1.21 Ω. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	11 kV: Z_base = 11² / 10 = 12.1 Ω; powers ÷ 10 MVA
Branch impedances	LN1: z = (0.6050 + j1.2100) / 12.1 = 0.05 + j0.1 pu; y = 1/z = 4.0000 − j8.0000 pu
Scheduled injections	Net injection = generation − load (machines included): PLANT: S = (−3.500 − j1.700)/10 = −0.35000 − j0.17000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_S,S = 4.0000 − j8.0000; Y_PLANT,PLANT = 4.0000 − j8.0000; Y_S,PLANT = −4.0000 + j8.0000
Closed-form voltage	V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 with P = 0.35, Q = 0.17 pu, R + jX = 0.050000 + j0.100000 pu, E = 1.00000: V⁴ − 0.931000V² + 0.0018925 = 0 → V² = (0.931 + √0.859191)/2 → V = 0.96383 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (0.0265)/(0.963463) → θ = 1.5755° behind E; δ₂ = 0.0000 − 1.5755 = −1.5755°
Load-flow result	Newton-Raphson: V_PLANT = 0.96383∠−1.5755° pu = 10.6021 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: LN1: S_S→ = 3.5815 + j1.8630 MVA, S_PLANT→ = −3.5000 − j1.7000 MVA; loss = sum = \|I\|²(R + jX) = 81.49 kW + j162.98 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 81.49 kW + j162.98 kvar; slack supplies P = 3.5815 MW, Q = 1.8630 Mvar. P balance: P_slack + ΣP_gen = 3.5815 + 1.5000 = 5.0815 MW = ΣP_load + P_loss = 5.0000 + 0.0815 = 5.0815 MW. Q balance: ΣQ_inj = 162.98 kvar = Q_loss + Q_shunt = 162.98 + (0.00) = 162.98 kvar
Convergence	converged after 4 iterations; largest power mismatch 1.08e-10 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus PLANT at 0.96383 pu → 0.96383 vs limit 0.95000 pu PASS
Maximum voltage	highest bus S at 1.00000 pu → 1.00000 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 0.9638–1.0000 pu, losses 81.5 kW; governing: minimum voltage. No rating_mva on LN1: loading not checked.
Reference	Hand calculation (closed-form two-bus): Z_base = 11²/10 = 12.1 Ω; z = 0.05 + j0.1 pu; load P = 0.35, Q = 0.17 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 0.931000V² + 0.0018925 = 0 → V = 0.96383 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle −1.5755°; S_from = 3.5815 + j1.8630 MVA; series loss \|I\|²z = 81.49 + j162.98 (kW, kvar); slack supplies 3.58149 + j1.86298 (MW, Mvar) (validation case `psas-machines-gen-motor`)

Example 21 — Slack at 1.02 pu ∠30°, charged cable, 60 Hz (pandapower)

A slack set-point above nominal and a 30° reference angle carry straight through; the cable's charging susceptance enters the Thevenin source.

Given: S_base = 10 MVA, 60 Hz. Buses: S (13.8 kV, SLACK at 14.076 kV (1.02 pu) ∠30°); L (13.8 kV, PQ, load 4 MW + j2 Mvar). Branches: CBL CABLE S–L: 0.38088 + j0.3809 Ω; B_total 0.0025 S. Solver pandapower (NR); limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	13.8 kV: Z_base = 13.8² / 10 = 19.044 Ω; powers ÷ 10 MVA
Branch impedances	CBL: z = (0.3809 + j0.3809) / 19.044 = 0.02 + j0.0200011 pu; y = 1/z = 24.9987 − j25.0000 pu; B_total = 0.0025 × 19.044 = 0.04761 pu (B/2 = 0.023805 at each end)
Scheduled injections	Net injection = generation − load (machines included): L: S = (−4.000 − j2.000)/10 = −0.40000 − j0.20000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_S,S = 24.9987 − j24.9762; Y_L,L = 24.9987 − j24.9762; Y_S,L = −24.9987 + j25.0000
Thevenin source	E_th = −Y₂₁V₁/Y₂₂ = −(−24.9987 + j25.0000) × 1.02∠30° / (24.9987 − j24.9762) = 1.02049∠29.9727° pu; Z_th = 1/Y₂₂ = 0.020019 + j0.020001 pu
Closed-form voltage	V⁴ + [2(PR + QX) − E_th²]V² + (R² + X²)(P² + Q²) = 0 with P = 0.4, Q = 0.2 pu, R + jX = 0.020019 + j0.020001 pu, E_th = 1.02049: V⁴ − 1.017376V² + 0.000160161 = 0 → V² = (1.01738 + √1.034412)/2 → V = 1.00857 pu
Angle	tan θ = (PX − QR)/(V² + PR + QX) = (0.00399661)/(1.02923) → θ = 0.2225° behind E_th; δ₂ = 29.9727 − 0.2225 = 29.7502°
Load-flow result	pandapower (NR): V_L = 1.00857∠29.7502° pu = 13.9183 kV, matching the closed form
Branch flows	S_ij = V_iI*_ij at each end: CBL: S_S→ = 4.0375 + j1.5477 MVA, S_L→ = −4.0000 − j2.0000 MVA; sum = series loss \|I\|²(R + jX) 37.53 kW + j37.54 kvar + charging j−489.82 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 37.53 kW + j37.54 kvar; shunt Q (SHUNT elements, line charging) −489.82 kvar, reported separately; slack supplies P = 4.0375 MW, Q = 1.5477 Mvar. P balance: P_slack + ΣP_gen = 4.0375 + 0.0000 = 4.0375 MW = ΣP_load + P_loss = 4.0000 + 0.0375 = 4.0375 MW. Q balance: ΣQ_inj = −452.28 kvar = Q_loss + Q_shunt = 37.54 + (−489.82) = −452.28 kvar
Convergence	pandapower Newton-Raphson converged in 5 iterations on its own mismatch tolerance PASS
Minimum voltage	lowest bus L at 1.00857 pu → 1.00857 vs limit 0.95000 pu PASS
Maximum voltage	highest bus S at 1.02000 pu → 1.02000 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 1.0086–1.0200 pu, losses 37.5 kW; governing: minimum voltage. No rating_mva on CBL: loading not checked.
Reference	Hand calculation (closed-form two-bus): Z_base = 13.8²/10 = 19.044 Ω; z = 0.02 + j0.0200011 pu; Thevenin at the load bus E_th = 1.02049∠29.9727° pu, Z_th = 0.0200191 + j0.020001 pu; load P = 0.4, Q = 0.2 pu: V⁴ + [2(PR + QX) − E²]V² + (R² + X²)(P² + Q²) = 0 → V⁴ − 1.017376V² + 0.000160161 = 0 → V = 1.00857 pu; tan θ = (PX − QR)/(V² + PR + QX) → angle 29.7502°; S_from = 4.0375 + j1.5477 MVA; series loss \|I\|²z = 37.53 + j37.54 (kW, kvar); shunt Q (charging/capacitor) −489.82 kvar, not a loss; slack supplies 4.03753 + j1.54772 (MW, Mvar) (validation case `psas-slack-setpoint-60hz-pandapower`)

Example 22 — Saadat three-bus system, Gauss-Seidel

A published meshed example: the solution lands on V₂ = 0.98 − j0.06 and V₃ = 1.00 − j0.05 pu, with the slack supplying 409.5 MW + j189 Mvar.

Given: S_base = 100 MVA, 50 Hz. Buses: 1 (100 kV, SLACK at 105 kV (1.05 pu) ∠0°); 2 (100 kV, PQ, load 256.6 MW + j110.2 Mvar); 3 (100 kV, PQ, load 138.6 MW + j45.2 Mvar). Branches: L12 LINE 1–2: 2 + j4 Ω; L13 LINE 1–3: 1 + j3 Ω; L23 LINE 2–3: 1.25 + j2.5 Ω. Solver Gauss-Seidel, tolerance 1e-08 pu, ≤ 500 iterations, acceleration α = 1.6; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	100 kV: Z_base = 100² / 100 = 100 Ω; powers ÷ 100 MVA
Branch impedances	L12: z = (2.0000 + j4.0000) / 100 = 0.02 + j0.04 pu; y = 1/z = 10.0000 − j20.0000 pu L13: z = (1.0000 + j3.0000) / 100 = 0.01 + j0.03 pu; y = 1/z = 10.0000 − j30.0000 pu L23: z = (1.2500 + j2.5000) / 100 = 0.0125 + j0.025 pu; y = 1/z = 16.0000 − j32.0000 pu
Scheduled injections	Net injection = generation − load (machines included): 2: S = (−256.600 − j110.200)/100 = −2.56600 − j1.10200 pu; 3: S = (−138.600 − j45.200)/100 = −1.38600 − j0.45200 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_1,1 = 20.0000 − j50.0000; Y_2,2 = 26.0000 − j52.0000; Y_3,3 = 26.0000 − j62.0000; Y_1,2 = −10.0000 + j20.0000; Y_1,3 = −10.0000 + j30.0000; Y_2,3 = −16.0000 + j32.0000
Gauss-Seidel sweep 1	V₂⁽¹⁾ = [S/V⁽⁰⁾ − ΣY_ikV_k] / Y_ii = [(−2.5660 + j1.1020)/1 − [(−10.0000 + j20.0000)(1.0500 + j0.0000) + (−16.0000 + j32.0000)(1.0000 + j0.0000)]] / (26.0000 − j52.0000) = 0.98254 − j0.03100 pu; accelerated V ← 1 + 1.6 × (−0.01746 − j0.03100) = 0.97206 − j0.04960 pu; sweeps repeat until the largest \|ΔV\| < 1e-08 pu
Mismatch check	Solved voltages 1 1.05000∠0.0000°, 2 0.98184∠−3.5035°, 3 1.00125∠−2.8624° pu; S_i = V_i(ΣY_ikV_k)* from them: 1 (slack): S = 409.493 + j188.987 MVA → generation 409.500 MW + j189.000 Mvar; 2: S = −2.5659 − j1.1018 vs −2.5660 − j1.1020 pu scheduled; 3: S = −1.3861 − j0.4521 vs −1.3860 − j0.4520 pu scheduled (differences in the 4th decimal are the rounding of the printed voltages)
Branch flows	S_ij = V_iI*_ij at each end: L12: S_1→ = 199.5000 + j84.0000 MVA, S_2→ = −191.0000 − j67.0000 MVA; loss = sum = \|I\|²(R + jX) = 8,500.00 kW + j17,000.00 kvar L13: S_1→ = 210.0000 + j105.0000 MVA, S_3→ = −205.0000 − j90.0000 MVA; loss = sum = \|I\|²(R + jX) = 5,000.00 kW + j15,000.00 kvar L23: S_2→ = −65.6000 − j43.2000 MVA, S_3→ = 66.4000 + j44.8000 MVA; loss = sum = \|I\|²(R + jX) = 800.00 kW + j1,600.00 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 14,300.00 kW + j33,600.00 kvar; slack supplies P = 409.5000 MW, Q = 189.0000 Mvar. P balance: P_slack + ΣP_gen = 409.5000 + 0.0000 = 409.5000 MW = ΣP_load + P_loss = 395.2000 + 14.3000 = 409.5000 MW. Q balance: ΣQ_inj = 33,600.00 kvar = Q_loss + Q_shunt = 33,600.00 + (0.00) = 33,600.00 kvar
Convergence	converged after 40 iterations; largest \|ΔV\| in the last sweep 6.96e-09 vs tolerance 1e-08 pu; power mismatch of the returned voltages 1.44e-07 pu PASS
Minimum voltage	lowest bus 2 at 0.98184 pu → 0.98184 vs limit 0.95000 pu PASS
Maximum voltage	highest bus 1 at 1.05000 pu → 1.05000 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 0.9818–1.0500 pu, losses 14,300.0 kW; governing: minimum voltage. No rating_mva on L12, L13, L23: loading not checked.
Reference	Saadat, Power System Analysis (2nd ed., McGraw-Hill), Ch. 6 three-bus Gauss-Seidel example (100 MVA base, z12 = 0.02 + j0.04, z13 = 0.01 + j0.03, z23 = 0.0125 + j0.025 pu, here on a 100 kV base so Z_base = 100 Ω): published V2 = 0.98 − j0.06 pu, V3 = 1.00 − j0.05 pu, S1 = 409.5 MW + j189.0 Mvar. Reproduced by an independent rectangular Newton: V2 = 0.98184∠−3.5035°, V3 = 1.00125∠−2.8624° pu; total losses 14.3000 MW. (Textbook edition and example/figure numbers pending verification.) (validation case `psas-saadat-3bus-gauss-seidel`)

Example 23 — Saadat three-bus system with a PV bus, Newton-Raphson

The same lines with bus 3 voltage-controlled at 1.04 pu and 200 MW: the published V₂ = 0.9717∠−2.696°, δ₃ = −0.499° and Q₃ = 146.2 Mvar.

Given: S_base = 100 MVA, 50 Hz. Buses: 1 (100 kV, SLACK at 105 kV (1.05 pu) ∠0°); 2 (100 kV, PQ, load 400 MW + j250 Mvar); 3 (100 kV, PV holding 104 kV (1.04 pu), P_gen 200 MW). Branches: L12 LINE 1–2: 2 + j4 Ω; L13 LINE 1–3: 1 + j3 Ω; L23 LINE 2–3: 1.25 + j2.5 Ω. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	100 kV: Z_base = 100² / 100 = 100 Ω; powers ÷ 100 MVA
Branch impedances	L12: z = (2.0000 + j4.0000) / 100 = 0.02 + j0.04 pu; y = 1/z = 10.0000 − j20.0000 pu L13: z = (1.0000 + j3.0000) / 100 = 0.01 + j0.03 pu; y = 1/z = 10.0000 − j30.0000 pu L23: z = (1.2500 + j2.5000) / 100 = 0.0125 + j0.025 pu; y = 1/z = 16.0000 − j32.0000 pu
Scheduled injections	Net injection = generation − load (machines included): 2: S = (−400.000 − j250.000)/100 = −4.00000 − j2.50000 pu; 3: P = 200.000/100 = 2 pu at \|V\| = 1.04 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_1,1 = 20.0000 − j50.0000; Y_2,2 = 26.0000 − j52.0000; Y_3,3 = 26.0000 − j62.0000; Y_1,2 = −10.0000 + j20.0000; Y_1,3 = −10.0000 + j30.0000; Y_2,3 = −16.0000 + j32.0000
Mismatch check	Solved voltages 1 1.05000∠0.0000°, 2 0.97168∠−2.6965°, 3 1.04000∠−0.4988° pu; S_i = V_i(ΣY_ikV_k)* from them: 1 (slack): S = 218.424 + j140.850 MVA → generation 218.423 MW + j140.852 Mvar; 2: S = −4.0000 − j2.5000 vs −4.0000 − j2.5000 pu scheduled; 3: P = 2.0000 vs 2.0000 pu scheduled, Q = 1.4617 pu (146.175 Mvar) (differences in the 4th decimal are the rounding of the printed voltages)
Branch flows	S_ij = V_iI*_ij at each end: L12: S_1→ = 179.3618 + j118.7339 MVA, S_2→ = −170.9684 − j101.9472 MVA; loss = sum = \|I\|²(R + jX) = 8,393.36 kW + j16,786.72 kvar L13: S_1→ = 39.0610 + j22.1176 MVA, S_3→ = −38.8783 − j21.5693 MVA; loss = sum = \|I\|²(R + jX) = 182.76 kW + j548.29 kvar L23: S_2→ = −229.0316 − j148.0528 MVA, S_3→ = 238.8783 + j167.7462 MVA; loss = sum = \|I\|²(R + jX) = 9,846.71 kW + j19,693.42 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 18,422.83 kW + j37,028.43 kvar; slack supplies P = 218.4228 MW, Q = 140.8515 Mvar. P balance: P_slack + ΣP_gen = 218.4228 + 200.0000 = 418.4228 MW = ΣP_load + P_loss = 400.0000 + 18.4228 = 418.4228 MW. Q balance: ΣQ_inj = 37,028.43 kvar = Q_loss + Q_shunt = 37,028.43 + (0.00) = 37,028.43 kvar
Convergence	converged after 4 iterations; largest power mismatch 1.17e-09 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus 2 at 0.97168 pu → 0.97168 vs limit 0.95000 pu PASS
Maximum voltage	highest bus 1 at 1.05000 pu → 1.05000 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 0.9717–1.0500 pu, losses 18,422.8 kW; governing: minimum voltage. No rating_mva on L12, L13, L23: loading not checked.
Reference	Saadat, Power System Analysis (2nd ed.), Ch. 6 three−bus example with bus 3 voltage−controlled (1.04 pu, 200 MW; bus 2 load 400 MW + j250 Mvar), solved there by Gauss−Seidel and Newton−Raphson: published V2 = 0.97168∠−2.696°, V3 = 1.04∠−0.4988°, S1 = 218.423 MW + j140.852 Mvar, Q3 = 146.177 Mvar. Independent Newton: V2 = 0.97168∠−2.6965°, δ3 = −0.4988°, Q3 = 146.177 Mvar. (Textbook edition and example/figure numbers pending verification.) (validation case `psas-saadat-3bus-pv-nr`)

Example 24 — Three-bus slack + PV generator + PQ load

The page's former three-bus example: a 100 MW generator at 1.04 pu pushes power back toward the slack on line 0–1 while the load bus sags to 0.950 pu.

Given: S_base = 100 MVA, 50 Hz. Buses: 0 (100 kV, SLACK at 105 kV (1.05 pu) ∠0°); 1 (100 kV, PV holding 104 kV (1.04 pu), P_gen 100 MW, Q limits −50 … 150 Mvar); 2 (100 kV, PQ, load 100 MW + j50 Mvar). Branches: L01 LINE 0–1: 2 + j6 Ω; L02 LINE 0–2: 8 + j24 Ω; L12 LINE 1–2: 6 + j18 Ω. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	100 kV: Z_base = 100² / 100 = 100 Ω; powers ÷ 100 MVA
Branch impedances	L01: z = (2.0000 + j6.0000) / 100 = 0.02 + j0.06 pu; y = 1/z = 5.0000 − j15.0000 pu L02: z = (8.0000 + j24.0000) / 100 = 0.08 + j0.24 pu; y = 1/z = 1.2500 − j3.7500 pu L12: z = (6.0000 + j18.0000) / 100 = 0.06 + j0.18 pu; y = 1/z = 1.6667 − j5.0000 pu
Scheduled injections	Net injection = generation − load (machines included): 1: P = 100.000/100 = 1 pu at \|V\| = 1.04 pu; 2: S = (−100.000 − j50.000)/100 = −1.00000 − j0.50000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_0,0 = 6.2500 − j18.7500; Y_1,1 = 6.6667 − j20.0000; Y_2,2 = 2.9167 − j8.7500; Y_0,1 = −5.0000 + j15.0000; Y_0,2 = −1.2500 + j3.7500; Y_1,2 = −1.6667 + j5.0000
Mismatch check	Solved voltages 0 1.05000∠0.0000°, 1 1.04000∠1.4243°, 2 0.95008∠−4.1459° pu; S_i = V_i(ΣY_ikV_k)* from them: 0 (slack): S = 5.191 + j61.135 MVA → generation 5.191 MW + j61.133 Mvar; 1: P = 1.0000 vs 1.0000 pu scheduled, Q = 0.0444 pu (4.441 Mvar); 2: S = −1.0000 − j0.5000 vs −1.0000 − j0.5000 pu scheduled (differences in the 4th decimal are the rounding of the printed voltages)
Branch flows	S_ij = V_iI*_ij at each end: L01: S_0→ = −35.2954 + j29.8274 MVA, S_1→ = 35.6827 − j28.6653 MVA; loss = sum = \|I\|²(R + jX) = 387.38 kW + j1,162.14 kvar L02: S_0→ = 40.4859 + j31.3054 MVA, S_2→ = −38.5854 − j25.6039 MVA; loss = sum = \|I\|²(R + jX) = 1,900.51 kW + j5,701.54 kvar L12: S_1→ = 64.3173 + j33.1042 MVA, S_2→ = −61.4146 − j24.3961 MVA; loss = sum = \|I\|²(R + jX) = 2,902.70 kW + j8,708.09 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 5,190.59 kW + j15,571.77 kvar; slack supplies P = 5.1906 MW, Q = 61.1329 Mvar. P balance: P_slack + ΣP_gen = 5.1906 + 100.0000 = 105.1906 MW = ΣP_load + P_loss = 100.0000 + 5.1906 = 105.1906 MW. Q balance: ΣQ_inj = 15,571.77 kvar = Q_loss + Q_shunt = 15,571.77 + (0.00) = 15,571.77 kvar
Convergence	converged after 4 iterations; largest power mismatch 3.31e-08 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus 2 at 0.95008 pu → 0.95008 vs limit 0.95000 pu PASS
Maximum voltage	highest bus 0 at 1.05000 pu → 1.05000 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 0.9501–1.0500 pu, losses 5,190.6 kW; governing: minimum voltage. No rating_mva on L01, L02, L12: loading not checked.
Reference	Hand calculation (help-page three-bus example; y01 = 5 − j15, y02 = 1.25 − j3.75, y12 = 1.6667 − j5 pu): independent Newton gives V1 = 1.04000∠1.4243°, V2 = 0.95008∠−4.1459° pu, Q1 = 4.439 Mvar; L01 -35.30 MW / loss 0.387 MW, L02 40.49 MW / loss 1.901 MW, L12 64.32 MW / loss 2.903 MW; total loss 5.191 MW (the former page figures 0.9501∠−4.15°, −35.30/40.49/64.32 MW, 5.19 MW agree) (validation case `psas-3bus-slack-pv-pq`)

Example 25 — Grainger & Stevenson four-bus 230 kV system

A published meshed 230 kV system with line charging and a PV generator: V₂ = 0.982∠−0.976°, V₃ = 0.969∠−1.872°, δ₄ = 1.523° and the slack's 186.8 MW + j114.5 Mvar.

Given: S_base = 100 MVA, 50 Hz. Buses: 1 (230 kV, SLACK at 230 kV (1 pu) ∠0°, load 50 MW + j30.99 Mvar); 2 (230 kV, PQ, load 170 MW + j105.35 Mvar); 3 (230 kV, PQ, load 200 MW + j123.94 Mvar); 4 (230 kV, PV holding 234.6 kV (1.02 pu), P_gen 318 MW, load 80 MW + j49.58 Mvar). Branches: 1-2 LINE 1–2: 5.33232 + j26.6616 Ω; B_total 0.000193762 S; 1-3 LINE 1–3: 3.93576 + j19.6788 Ω; B_total 0.000146503 S; 2-4 LINE 2–4: 3.93576 + j19.6788 Ω; B_total 0.000146503 S; 3-4 LINE 3–4: 6.72888 + j33.6444 Ω; B_total 0.000241021 S. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	230 kV: Z_base = 230² / 100 = 529 Ω; powers ÷ 100 MVA
Branch impedances	1-2: z = (5.3323 + j26.6616) / 529 = 0.01008 + j0.0504 pu; y = 1/z = 3.8156 − j19.0781 pu; B_total = 0.000193762 × 529 = 0.1025 pu (B/2 = 0.05125 at each end) 1-3: z = (3.9358 + j19.6788) / 529 = 0.00744 + j0.0372 pu; y = 1/z = 5.1696 − j25.8478 pu; B_total = 0.000146503 × 529 = 0.0775001 pu (B/2 = 0.03875 at each end) 2-4: z = (3.9358 + j19.6788) / 529 = 0.00744 + j0.0372 pu; y = 1/z = 5.1696 − j25.8478 pu; B_total = 0.000146503 × 529 = 0.0775001 pu (B/2 = 0.03875 at each end) 3-4: z = (6.7289 + j33.6444) / 529 = 0.01272 + j0.0636 pu; y = 1/z = 3.0237 − j15.1185 pu; B_total = 0.000241021 × 529 = 0.1275 pu (B/2 = 0.0637501 at each end)
Scheduled injections	Net injection = generation − load (machines included): 2: S = (−170.000 − j105.350)/100 = −1.70000 − j1.05350 pu; 3: S = (−200.000 − j123.940)/100 = −2.00000 − j1.23940 pu; 4: P = 238.000/100 = 2.38 pu at \|V\| = 1.02 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_1,1 = 8.9852 − j44.8360; Y_2,2 = 8.9852 − j44.8360; Y_3,3 = 8.1933 − j40.8638; Y_4,4 = 8.1933 − j40.8638; Y_1,2 = −3.8156 + j19.0781; Y_1,3 = −5.1696 + j25.8478; Y_2,4 = −5.1696 + j25.8478; Y_3,4 = −3.0237 + j15.1185
Mismatch check	Solved voltages 1 1.00000∠0.0000°, 2 0.98242∠−0.9761°, 3 0.96900∠−1.8722°, 4 1.02000∠1.5231° pu; S_i = V_i(ΣY_ikV_k)* from them: 1 (slack): S = 136.812 + j83.525 MVA → generation 186.809 MW + j114.501 Mvar; 2: S = −1.7000 − j1.0535 vs −1.7000 − j1.0535 pu scheduled; 3: S = −2.0001 − j1.2396 vs −2.0000 − j1.2394 pu scheduled; 4: P = 2.3800 vs 2.3800 pu scheduled, Q = 1.3186 pu (131.859 Mvar) (differences in the 4th decimal are the rounding of the printed voltages)
Branch flows	S_ij = V_iI*_ij at each end: 1-2: S_1→ = 38.6915 + j22.2984 MVA, S_2→ = −38.4648 − j31.2363 MVA; sum = series loss \|I\|²(R + jX) 226.71 kW + j1,133.54 kvar + charging j−10,071.41 kvar 1-3: S_1→ = 98.1175 + j61.2124 MVA, S_3→ = −97.0861 − j63.5687 MVA; sum = series loss \|I\|²(R + jX) 1,031.44 kW + j5,157.19 kvar + charging j−7,513.52 kvar 2-4: S_2→ = −131.5352 − j74.1137 MVA, S_4→ = 133.2507 + j74.9195 MVA; sum = series loss \|I\|²(R + jX) 1,715.48 kW + j8,577.39 kvar + charging j−7,771.52 kvar 3-4: S_3→ = −102.9139 − j60.3713 MVA, S_4→ = 104.7493 + j56.9301 MVA; sum = series loss \|I\|²(R + jX) 1,835.45 kW + j9,177.27 kvar + charging j−12,618.50 kvar
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 4,809.08 kW + j24,045.39 kvar; shunt Q (SHUNT elements, line charging) −37,974.94 kvar, reported separately; slack supplies P = 186.8091 MW, Q = 114.5008 Mvar. P balance: P_slack + ΣP_gen = 186.8091 + 318.0000 = 504.8091 MW = ΣP_load + P_loss = 500.0000 + 4.8091 = 504.8091 MW. Q balance: ΣQ_inj = −13,929.55 kvar = Q_loss + Q_shunt = 24,045.39 + (−37,974.94) = −13,929.56 kvar
Convergence	converged after 4 iterations; largest power mismatch 1.07e-09 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus 3 at 0.96900 pu → 0.96900 vs limit 0.95000 pu PASS
Maximum voltage	highest bus 4 at 1.02000 pu → 1.02000 vs limit 1.05000 pu PASS
Result	3/3 checks pass → PASS; V 0.9690–1.0200 pu, losses 4,809.1 kW; governing: minimum voltage. No rating_mva on 1-2, 1-3, 2-4, 3-4: loading not checked.
Reference	Grainger & Stevenson, Power System Analysis (McGraw-Hill 1994), Ch. 9 four-bus example (Tables 9.2/9.3: 230 kV, 100 MVA base, line charging Y/2 = 0.05125/0.03875/0.03875/0.06375 pu): published solution V2 = 0.982∠−0.976°, V3 = 0.969∠−1.872°, V4 = 1.02∠1.523°, slack generation 186.81 MW + j114.50 Mvar. Independent Newton: V2 = 0.98242∠−0.9761°, V3 = 0.96900∠−1.8722°, δ4 = 1.5231°, slack 186.809 MW + j114.501 Mvar. (Textbook edition and example/figure numbers pending verification.) (validation case `psas-stevenson-4bus-nr`)

Example 26 — Glover, Sarma & Overbye five-bus system with transformers

A published 15/345 kV system with two generator transformers and charged lines: bus 2 falls to 0.834 pu, so the under-voltage check fails exactly as the textbook's heavily loaded bus does.

Given: S_base = 100 MVA, 50 Hz. Buses: 1 (15 kV, SLACK at 15 kV (1 pu) ∠0°); 2 (345 kV, PQ, load 800 MW + j280 Mvar); 3 (15 kV, PV holding 15.75 kV (1.05 pu), P_gen 520 MW, load 80 MW + j40 Mvar); 4 (345 kV, PQ); 5 (345 kV, PQ). Branches: L24 LINE 2–4: 10.7122 + j119.025 Ω; B_total 0.00144507 S; rated 1,200 MVA; L25 LINE 2–5: 5.35612 + j59.5125 Ω; B_total 0.00073934 S; rated 1,200 MVA; L45 LINE 4–5: 2.67806 + j29.7563 Ω; B_total 0.00036967 S; rated 1,200 MVA; T1 XFMR 5–1: 1.78537 + j23.805 Ω; rated 600 MVA; T2 XFMR 4–3: 0.892687 + j11.9025 Ω; rated 1,000 MVA. Solver Newton-Raphson, tolerance 1e-06 pu, ≤ 50 iterations; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	345 kV: Z_base = 345² / 100 = 1190.25 Ω; 15 kV: Z_base = 15² / 100 = 2.25 Ω; powers ÷ 100 MVA
Branch impedances	L24: z = (10.7122 + j119.0250) / 1190.25 = 0.009 + j0.1 pu; y = 1/z = 0.8928 − j9.9197 pu; B_total = 0.00144507 × 1190.25 = 1.72 pu (B/2 = 0.86 at each end) L25: z = (5.3561 + j59.5125) / 1190.25 = 0.0045 + j0.05 pu; y = 1/z = 1.7855 − j19.8393 pu; B_total = 0.00073934 × 1190.25 = 0.88 pu (B/2 = 0.44 at each end) L45: z = (2.6781 + j29.7563) / 1190.25 = 0.00225 + j0.025 pu; y = 1/z = 3.5711 − j39.6786 pu; B_total = 0.00036967 × 1190.25 = 0.44 pu (B/2 = 0.22 at each end) T1: z = (1.7854 + j23.8050) / 1190.25 = 0.0015 + j0.02 pu; y = 1/z = 3.7290 − j49.7203 pu T2: z = (0.8927 + j11.9025) / 1190.25 = 0.00075 + j0.01 pu; y = 1/z = 7.4580 − j99.4406 pu
Scheduled injections	Net injection = generation − load (machines included): 2: S = (−800.000 − j280.000)/100 = −8.00000 − j2.80000 pu; 3: P = 440.000/100 = 4.4 pu at \|V\| = 1.05 pu; 4: S = (0.000 + j0.000)/100 = 0.00000 + j0.00000 pu; 5: S = (0.000 + j0.000)/100 = 0.00000 + j0.00000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_1,1 = 3.7290 − j49.7203; Y_2,2 = 2.6783 − j28.4590; Y_3,3 = 7.4580 − j99.4406; Y_4,4 = 11.9219 − j147.9589; Y_5,5 = 9.0856 − j108.5782; Y_1,5 = −3.7290 + j49.7203; Y_2,4 = −0.8928 + j9.9197; Y_2,5 = −1.7855 + j19.8393; Y_3,4 = −7.4580 + j99.4406; Y_4,5 = −3.5711 + j39.6786
Mismatch check	Solved voltages 1 1.00000∠0.0000°, 2 0.83377∠−22.4064°, 3 1.05000∠−0.5973°, 4 1.01930∠−2.8340°, 5 0.97429∠−4.5479° pu; S_i = V_i(ΣY_ikV_k)* from them: 1 (slack): S = 394.840 + j114.275 MVA → generation 394.839 MW + j114.283 Mvar; 2: S = −8.0000 − j2.8000 vs −8.0000 − j2.8000 pu scheduled; 3: P = 4.4001 vs 4.4000 pu scheduled, Q = 2.9750 pu (297.503 Mvar); 4: S = −0.0002 − j0.0004 vs 0.0000 + j0.0000 pu scheduled; 5: S = 0.0000 + j0.0002 vs 0.0000 + j0.0000 pu scheduled (differences in the 4th decimal are the rounding of the printed voltages)
Branch flows	S_ij = V_iI*_ij at each end: L24: S_2→ = −291.8411 − j139.1054 MVA, S_4→ = 303.6824 + j121.5384 MVA; sum = series loss \|I\|²(R + jX) 11,841.26 kW + j131,569.56 kvar + charging j−149,136.55 kvar; loading \|S_from\|/rating = 323.2979/1,200 = 26.94 % L25: S_2→ = −508.1589 − j140.8946 MVA, S_5→ = 525.6621 + j263.0209 MVA; sum = series loss \|I\|²(R + jX) 17,503.22 kW + j194,480.19 kvar + charging j−72,353.90 kvar; loading \|S_from\|/rating = 527.3298/1,200 = 43.94 % L45: S_4→ = 134.3986 + j150.3545 MVA, S_5→ = −133.3577 − j182.5297 MVA; sum = series loss \|I\|²(R + jX) 1,040.90 kW + j11,565.55 kvar + charging j−43,740.73 kvar; loading \|S_from\|/rating = 201.6667/1,200 = 16.81 % T1: S_5→ = −392.3044 − j80.4913 MVA, S_1→ = 394.8388 + j114.2829 MVA; loss = sum = \|I\|²(R + jX) = 2,534.37 kW + j33,791.65 kvar; loading \|S_from\|/rating = 400.4767/600 = 66.75 % T2: S_4→ = −438.0810 − j271.8929 MVA, S_3→ = 440.0000 + j297.4796 MVA; loss = sum = \|I\|²(R + jX) = 1,919.01 kW + j25,586.77 kvar; loading \|S_from\|/rating = 515.5974/1,000 = 51.56 %
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 34,838.76 kW + j396,993.71 kvar; shunt Q (SHUNT elements, line charging) −265,231.18 kvar, reported separately; slack supplies P = 394.8388 MW, Q = 114.2829 Mvar. P balance: P_slack + ΣP_gen = 394.8388 + 520.0000 = 914.8388 MW = ΣP_load + P_loss = 880.0000 + 34.8388 = 914.8388 MW. Q balance: ΣQ_inj = 131,762.52 kvar = Q_loss + Q_shunt = 396,993.71 + (−265,231.18) = 131,762.53 kvar
Convergence	converged after 6 iterations; largest power mismatch 1.47e-10 vs tolerance 1e-06 pu PASS
Minimum voltage	lowest bus 2 at 0.83377 pu → 0.83377 vs limit 0.95000 pu FAIL
Maximum voltage	highest bus 3 at 1.05000 pu → 1.05000 vs limit 1.05000 pu PASS
Branch loading	most loaded branch T1 → 66.75 vs limit 100.00 % PASS
Result	3/4 checks pass → FAIL; V 0.8338–1.0500 pu, losses 34,838.8 kW; governing: minimum voltage
Reference	Glover, Sarma & Overbye, Power System Analysis and Design, Ch. 6 five-bus example (Fig. 6.2, Tables 6.1–6.3; 100 MVA base, 15/345 kV): published V2 = 0.834∠−22.407°, V3 = 1.050∠−0.597°, V4 = 1.019∠−2.834°, V5 = 0.974∠−4.548°. Independent Newton: V2 = 0.83377∠−22.4064°, V3 = 1.05000∠−0.5973°, V4 = 1.01930∠−2.8340°, V5 = 0.97429∠−4.5479°; Q_G3 = 337.48 Mvar; total loss 34.839 MW. Branch ratings (1200 MVA lines, 600/1000 MVA transformers) are this case's own. (Textbook edition and example/figure numbers pending verification.) (validation case `psas-glover-5bus-nr`)

Example 27 — Five-bus system on the pandapower solver

The same transformer-coupled system through pandapower: identical voltages, flows and transformer loadings to the native Newton-Raphson.

Given: S_base = 100 MVA, 50 Hz. Buses: 1 (15 kV, SLACK at 15 kV (1 pu) ∠0°); 2 (345 kV, PQ, load 800 MW + j280 Mvar); 3 (15 kV, PV holding 15.75 kV (1.05 pu), P_gen 520 MW, load 80 MW + j40 Mvar); 4 (345 kV, PQ); 5 (345 kV, PQ). Branches: L24 LINE 2–4: 10.7122 + j119.025 Ω; B_total 0.00144507 S; rated 1,200 MVA; L25 LINE 2–5: 5.35612 + j59.5125 Ω; B_total 0.00073934 S; rated 1,200 MVA; L45 LINE 4–5: 2.67806 + j29.7563 Ω; B_total 0.00036967 S; rated 1,200 MVA; T1 XFMR 5–1: 1.78537 + j23.805 Ω; rated 600 MVA; T2 XFMR 4–3: 0.892687 + j11.9025 Ω; rated 1,000 MVA. Solver pandapower (NR); limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	345 kV: Z_base = 345² / 100 = 1190.25 Ω; 15 kV: Z_base = 15² / 100 = 2.25 Ω; powers ÷ 100 MVA
Branch impedances	L24: z = (10.7122 + j119.0250) / 1190.25 = 0.009 + j0.1 pu; y = 1/z = 0.8928 − j9.9197 pu; B_total = 0.00144507 × 1190.25 = 1.72 pu (B/2 = 0.86 at each end) L25: z = (5.3561 + j59.5125) / 1190.25 = 0.0045 + j0.05 pu; y = 1/z = 1.7855 − j19.8393 pu; B_total = 0.00073934 × 1190.25 = 0.88 pu (B/2 = 0.44 at each end) L45: z = (2.6781 + j29.7563) / 1190.25 = 0.00225 + j0.025 pu; y = 1/z = 3.5711 − j39.6786 pu; B_total = 0.00036967 × 1190.25 = 0.44 pu (B/2 = 0.22 at each end) T1: z = (1.7854 + j23.8050) / 1190.25 = 0.0015 + j0.02 pu; y = 1/z = 3.7290 − j49.7203 pu T2: z = (0.8927 + j11.9025) / 1190.25 = 0.00075 + j0.01 pu; y = 1/z = 7.4580 − j99.4406 pu
Scheduled injections	Net injection = generation − load (machines included): 2: S = (−800.000 − j280.000)/100 = −8.00000 − j2.80000 pu; 3: P = 440.000/100 = 4.4 pu at \|V\| = 1.05 pu; 4: S = (0.000 + j0.000)/100 = 0.00000 + j0.00000 pu; 5: S = (0.000 + j0.000)/100 = 0.00000 + j0.00000 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_1,1 = 3.7290 − j49.7203; Y_2,2 = 2.6783 − j28.4590; Y_3,3 = 7.4580 − j99.4406; Y_4,4 = 11.9219 − j147.9589; Y_5,5 = 9.0856 − j108.5782; Y_1,5 = −3.7290 + j49.7203; Y_2,4 = −0.8928 + j9.9197; Y_2,5 = −1.7855 + j19.8393; Y_3,4 = −7.4580 + j99.4406; Y_4,5 = −3.5711 + j39.6786
Mismatch check	Solved voltages 1 1.00000∠0.0000°, 2 0.83377∠−22.4064°, 3 1.05000∠−0.5973°, 4 1.01930∠−2.8340°, 5 0.97429∠−4.5479° pu; S_i = V_i(ΣY_ikV_k)* from them: 1 (slack): S = 394.840 + j114.275 MVA → generation 394.839 MW + j114.283 Mvar; 2: S = −8.0000 − j2.8000 vs −8.0000 − j2.8000 pu scheduled; 3: P = 4.4001 vs 4.4000 pu scheduled, Q = 2.9750 pu (297.503 Mvar); 4: S = −0.0002 − j0.0004 vs 0.0000 + j0.0000 pu scheduled; 5: S = 0.0000 + j0.0002 vs 0.0000 + j0.0000 pu scheduled (differences in the 4th decimal are the rounding of the printed voltages)
Branch flows	S_ij = V_iI*_ij at each end: L24: S_2→ = −291.8411 − j139.1054 MVA, S_4→ = 303.6824 + j121.5384 MVA; sum = series loss \|I\|²(R + jX) 11,841.26 kW + j131,569.56 kvar + charging j−149,136.55 kvar; loading \|S_from\|/rating = 323.2979/1,200 = 26.94 % L25: S_2→ = −508.1589 − j140.8946 MVA, S_5→ = 525.6621 + j263.0209 MVA; sum = series loss \|I\|²(R + jX) 17,503.22 kW + j194,480.19 kvar + charging j−72,353.90 kvar; loading \|S_from\|/rating = 527.3298/1,200 = 43.94 % L45: S_4→ = 134.3986 + j150.3545 MVA, S_5→ = −133.3577 − j182.5297 MVA; sum = series loss \|I\|²(R + jX) 1,040.90 kW + j11,565.55 kvar + charging j−43,740.73 kvar; loading \|S_from\|/rating = 201.6667/1,200 = 16.81 % T1: S_5→ = −392.3044 − j80.4913 MVA, S_1→ = 394.8388 + j114.2829 MVA; loss = sum = \|I\|²(R + jX) = 2,534.37 kW + j33,791.65 kvar; loading \|S_from\|/rating = 400.4767/600 = 66.75 % T2: S_4→ = −438.0810 − j271.8929 MVA, S_3→ = 440.0000 + j297.4796 MVA; loss = sum = \|I\|²(R + jX) = 1,919.01 kW + j25,586.77 kvar; loading \|S_from\|/rating = 515.5974/1,000 = 51.56 %
Losses and balance	Total series losses Σ\|I\|²(R + jX) = 34,838.76 kW + j396,993.71 kvar; shunt Q (SHUNT elements, line charging) −265,231.18 kvar, reported separately; slack supplies P = 394.8388 MW, Q = 114.2829 Mvar. P balance: P_slack + ΣP_gen = 394.8388 + 520.0000 = 914.8388 MW = ΣP_load + P_loss = 880.0000 + 34.8388 = 914.8388 MW. Q balance: ΣQ_inj = 131,762.52 kvar = Q_loss + Q_shunt = 396,993.71 + (−265,231.18) = 131,762.53 kvar
Convergence	pandapower Newton-Raphson converged in 5 iterations on its own mismatch tolerance PASS
Minimum voltage	lowest bus 2 at 0.83377 pu → 0.83377 vs limit 0.95000 pu FAIL
Maximum voltage	highest bus 3 at 1.05000 pu → 1.05000 vs limit 1.05000 pu PASS
Branch loading	most loaded branch T1 → 66.75 vs limit 100.00 % PASS
Result	3/4 checks pass → FAIL; V 0.8338–1.0500 pu, losses 34,838.8 kW; governing: minimum voltage
Reference	Glover, Sarma & Overbye, Power System Analysis and Design, Ch. 6 five-bus example (Fig. 6.2, Tables 6.1–6.3; 100 MVA base, 15/345 kV): published V2 = 0.834∠−22.407°, V3 = 1.050∠−0.597°, V4 = 1.019∠−2.834°, V5 = 0.974∠−4.548°. Independent Newton: V2 = 0.83377∠−22.4064°, V3 = 1.05000∠−0.5973°, V4 = 1.01930∠−2.8340°, V5 = 0.97429∠−4.5479°; Q_G3 = 337.48 Mvar; total loss 34.839 MW. Branch ratings (1200 MVA lines, 600/1000 MVA transformers) are this case's own. (Textbook edition and example/figure numbers pending verification.) (validation case `psas-glover-5bus-pandapower`)

Example 28 — Gauss-Seidel stopped by a 5-iteration cap

Gauss-Seidel on the Saadat system cannot settle in 5 sweeps: the tool reports non-convergence, fails the convergence check and warns that the voltages are not an operating point.

Given: S_base = 100 MVA, 50 Hz. Buses: 1 (100 kV, SLACK at 105 kV (1.05 pu) ∠0°); 2 (100 kV, PQ, load 256.6 MW + j110.2 Mvar); 3 (100 kV, PQ, load 138.6 MW + j45.2 Mvar). Branches: L12 LINE 1–2: 2 + j4 Ω; L13 LINE 1–3: 1 + j3 Ω; L23 LINE 2–3: 1.25 + j2.5 Ω. Solver Gauss-Seidel, tolerance 1e-06 pu, ≤ 5 iterations, acceleration α = 1.6; limits 0.95–1.05 pu, loading ≤ 100 %.

Per-unit bases	100 kV: Z_base = 100² / 100 = 100 Ω; powers ÷ 100 MVA
Branch impedances	L12: z = (2.0000 + j4.0000) / 100 = 0.02 + j0.04 pu; y = 1/z = 10.0000 − j20.0000 pu L13: z = (1.0000 + j3.0000) / 100 = 0.01 + j0.03 pu; y = 1/z = 10.0000 − j30.0000 pu L23: z = (1.2500 + j2.5000) / 100 = 0.0125 + j0.025 pu; y = 1/z = 16.0000 − j32.0000 pu
Scheduled injections	Net injection = generation − load (machines included): 2: S = (−256.600 − j110.200)/100 = −2.56600 − j1.10200 pu; 3: S = (−138.600 − j45.200)/100 = −1.38600 − j0.45200 pu
Y-bus	Y_ii = Σ y_ik + shunts, Y_ik = −y_ik (pu): Y_1,1 = 20.0000 − j50.0000; Y_2,2 = 26.0000 − j52.0000; Y_3,3 = 26.0000 − j62.0000; Y_1,2 = −10.0000 + j20.0000; Y_1,3 = −10.0000 + j30.0000; Y_2,3 = −16.0000 + j32.0000
Gauss-Seidel sweep 1	V₂⁽¹⁾ = [S/V⁽⁰⁾ − ΣY_ikV_k] / Y_ii = [(−2.5660 + j1.1020)/1 − [(−10.0000 + j20.0000)(1.0500 + j0.0000) + (−16.0000 + j32.0000)(1.0000 + j0.0000)]] / (26.0000 − j52.0000) = 0.98254 − j0.03100 pu; accelerated V ← 1 + 1.6 × (−0.01746 − j0.03100) = 0.97206 − j0.04960 pu; sweeps repeat until the largest \|ΔV\| < 1e-06 pu
Iterations	stopped at the 5-iteration cap with voltages 1 1.05000∠0.0000°, 2 0.98231∠−3.9529°, 3 1.00261∠−3.1699° pu — not an operating point, so flows and losses are not reported here
Convergence	NOT converged after 5 iterations; largest \|ΔV\| in the last sweep 1.66e-02 vs tolerance 1e-06 pu; power mismatch of the returned voltages 2.45e-01 pu FAIL
Minimum voltage	lowest bus 2 at 0.98231 pu → 0.98231 vs limit 0.95000 pu PASS
Maximum voltage	highest bus 1 at 1.05000 pu → 1.05000 vs limit 1.05000 pu PASS
Result	2/3 checks pass → FAIL; not converged in 5 iterations; governing: convergence. Not converged after 5 iterations — results are not a valid operating point. No rating_mva on L12, L13, L23: loading not checked.
Reference	Hand calculation: accelerated Gauss-Seidel (α = 1.6) from a flat start on the Saadat three-bus system; the largest voltage change per sweep is 7.31e-02, 4.07e-02, 3.88e-02, 1.66e-02, 1.66e-02 pu — still far above 1e-6 pu after 5 sweeps, so the run cannot converge. (Textbook edition and example/figure numbers pending verification.) (validation case `psas-gs-iteration-cap`)

Standards & References

IEEE Std 399 (Brown Book) — IEEE Recommended Practice for Industrial and Commercial Power Systems Analysis. Provides guidance on load flow study methods and assumptions.
IEC 60909 — Short-circuit currents in three-phase AC systems. Y-bus construction and per-unit conventions are consistent with this standard.
AS/NZS 3000:2018 — Electrical installations (Wiring Rules). General voltage regulation requirements for electrical networks.
Glover, Sarma & Overbye — Power Systems Analysis and Design, 6th Edition. Gauss-Seidel and SOR methods for load flow.
Stevenson & Grainger — Power Systems Analysis. Foundational reference for bus admittance matrix construction and iterative methods.
Saadat — Power System Analysis, 2nd Edition. Three-bus Gauss-Seidel and Newton-Raphson examples used in the worked examples.

Provisional citations: the expected values of the textbook-system worked examples (Saadat; Glover, Sarma & Overbye; Grainger & Stevenson) come from an independent hand solution. The edition and example/figure numbers cited for them are pending verification against the books.

Tips & Limitations

Tips

Start with a flat-start initial guess (1.0 pu at all PQ buses). The calculator does this automatically.
For systems that converge slowly, enable SOR with \(\omega = 1.2\) – \(1.6\). Values above 1.6 may cause oscillation.
Use the default tolerance of \(10^{-6}\) for engineering accuracy. Looser tolerances (e.g. \(10^{-4}\)) converge faster but sacrifice precision.
For PV buses, set realistic \(Q_{\min}\) and \(Q_{\max}\) limits based on generator capability curves. The defaults of ±999 MVAr are effectively unconstrained.
Line impedances must be entered in per-unit on the system base. Convert from Ω using: \(Z_{\text{pu}} = Z_{\Omega} / Z_{\text{base}}\) where \(Z_{\text{base}} = V_{\text{base}}^2 / S_{\text{base}}\).
The Slack bus absorbs all system losses and any power imbalance. Place it at the strongest source in the network (e.g. utility interconnection point).
Bus voltage magnitudes between 0.95 and 1.05 pu are typical. Warnings are generated for voltages outside 0.90 – 1.10 pu.

Limitations

Gauss-Seidel only — this calculator does not implement Newton-Raphson or Fast Decoupled methods. Gauss-Seidel converges more slowly and may fail on ill-conditioned systems (high R/X ratios, large networks).
Balanced three-phase — only positive-sequence balanced loads are modelled. Unbalanced or single-phase loads are not supported.
No transformer tap modelling — all lines are modelled as simple pi-sections. Off-nominal transformer taps, phase-shifting transformers, and FACTS devices are not included.
Maximum 50 buses — for larger systems, Newton-Raphson methods with sparse matrix techniques are recommended.
Steady-state only — transient stability, motor starting, and dynamic phenomena are not considered.
No automatic base conversion — all impedances must be entered in per-unit on the specified system base. The calculator does not convert between different voltage bases or equipment ratings.

Developer Guide

Architecture

File layout:

components/c0025_load_flow/component.py — Component registration, parameter schema (LOAD_FLOW_SCHEMA), SVG diagram generator
components/c0025_load_flow/calculations.py — LoadFlowCalculator class with all numerical logic (no Django dependencies)
components/c0025_load_flow/equations.py — build_load_flow_equations() — generates formatted HTML with LaTeX equations from results
components/c0025_load_flow/excel_generator.py — create_c0025_excel_report() — Excel export
components/c0025_load_flow/templates/calcs/components/c0025_load_flow/sidebar_form.html — sidebar input form template

Calculator class: LoadFlowCalculator

Key data structures:

Y_bus — NumPy complex matrix (np.ndarray, shape n×n)
voltages — NumPy complex array of bus voltages (np.ndarray, shape n)
bus_types — list of strings ("Slack", "PV", "PQ")
p_specified, q_specified — NumPy arrays of scheduled power injections in per-unit

Code Paths

Key methods in LoadFlowCalculator (calculations.py):

Function	Purpose
`calculate()`	Main entry point — orchestrates the full load flow solution
`_parse_bus_data()`	Parse bus array, identify Slack/PV/PQ types, convert MW/MVAr to per-unit
`_build_ybus()`	Assemble Y-bus admittance matrix from line impedances
`_initialize_voltages()`	Set initial voltage guesses (flat start for PQ, specified for Slack/PV)
`_gauss_seidel_iterations()`	Iterative solver with PV enforcement, SOR, and Q-limit checking
`_calculate_power_injections()`	Compute actual \(P\) and \(Q\) at each bus from converged voltages
`_calculate_line_flows()`	Compute power flow through each line using pi-model currents
`_calculate_losses(line_flows)`	Sum line losses to get total system losses
`_format_bus_voltages()`	Convert complex voltages to magnitude/angle/real/imag dict format
`_generate_warnings()`	Check for non-convergence, voltage violations (< 0.90 or > 1.10 pu)
`_generate_recommendations()`	Suggest SOR, higher iteration counts, or capacitor banks

Key functions in component.py:

Function	Purpose
`generate_load_flow_diagram_svg(parameters, results)`	Generate SVG network topology diagram with bus positions, line impedances, and voltage results
`register_component(registry)`	Register calc type, schema, calculator class, and SVG generators with the component registry

Key function in equations.py:

Function	Purpose
`build_load_flow_equations(results)`	Generate HTML with LaTeX equations and result tables for the calculation report

Testing

Running tests:

pytest calcs/test_c0025_load_flow.py -v

Test coverage:

Component registration verification
3-bus load flow (Slack + PV + PQ) — convergence, voltage magnitudes, power mismatches, losses
4-bus load flow (Slack + PV + 2×PQ) — larger network validation
SVG network diagram generation and structure validation
LaTeX equation builder content validation

How to extend:

Add new bus types (e.g. PV with droop) by extending the _gauss_seidel_iterations() sweep logic and adding the type to the bus_type enum in the schema.
Add transformer modelling by extending _build_ybus() to include tap-ratio modified admittances.
Add shunt compensation elements (capacitor banks, reactors) by extending the Y-bus diagonal elements with fixed shunt admittances at specific buses.
For Newton-Raphson support, implement a separate solver method alongside the existing Gauss-Seidel and expose it via a solver_method schema parameter.
Write tests covering edge cases: isolated buses, zero-impedance lines, all PQ buses (no PV), and systems that fail to converge.