Contents
Introduction
Objectives
- Establish the equivalent circuit of a line, a cable and a transformer from manufacturer data.
- Convert physical quantities to per-unit quantities and back.
- Calculate a voltage drop and losses; check a voltage profile.
- Write the load-flow equations, do one Gauss-Seidel iteration by hand and interpret the convergence of the three methods.
- Calculate three-phase, phase-to-phase and phase-to-earth short-circuit currents according to IEC 60909 and deduce the breaking capacity.
Prerequisites: level 1, elementary matrix calculus. Duration: 20 to 25 h.
2.1 Lines and cables
A line or cable is characterised by four per-length quantities (per km): resistance r, inductance l (reactance x = lω), capacitance c and conductance g (neglected). Orders of magnitude:
| Asset | r (Ω/km) | x (Ω/km) | c (nF/km) | Remark |
|---|---|---|---|---|
| MV overhead line, AAAC 148 mm² | 0.22 to 0.27 | 0.35 to 0.40 | 9 to 10 | R/X ≈ 0.7 |
| MV overhead line, AAAC 54.6 mm² | 0.60 | 0.40 | 9 | R/X ≈ 1.5 |
| MV underground cable, Al 150 mm² | 0.21 | 0.11 | 250 to 400 | large capacitance |
| LV copper cable 4 × 95 mm² | 0.24 | 0.08 | — | R/X ≈ 3 |
Models according to the electrical length: short line (< 80 km overhead): R + jX in series; medium line (π model): R + jX in series and C/2 to earth at each end; long line: distributed-parameter model. NRJXpert offers the R, L, RL and RLC π models (Network palette) with R = r·ℓ, L = l·ℓ, C = c·ℓ.
Zero-sequence quantities. For earth faults, r₀, x₀ (return through the earth, about 3 times x for overhead lines) and c₀ are also needed: « Zero sequence » section of the lines.
↑ Back to top2.2 Transformers
The Kapp equivalent circuit reduces the transformer to a series impedance Zsc = Rsc + jXsc and a magnetising branch (Rfe ∥ jXm), from the short-circuit test (usc, Psc) and the no-load test (i₀, P₀):
The vector group (Dyn, YNd, Yyn…) plays no role in the positive sequence but determines the zero-sequence path: a delta winding blocks the zero-sequence current, an earthed star neutral lets it through. Dyn11, very common in distribution, isolates the MV network from LV earth faults and provides the LV network with an earthed neutral.
Example 2.1 — Distribution transformer
250 kVA, 30 kV / 400 V, usc = 4 %, Psc = 3.25 kW. Calculate Zsc, Rsc, Xsc referred to the secondary, the rated current and the voltage drop at full load with cos φ = 0.8.
Show the solution
Zsc = 0.04 × 400²/250 000 = 25.6 mΩ; Rsc = 3 250 × 400²/250 000² = 8.32 mΩ; Xsc = 24.21 mΩ. In = 250 000/(√3 × 400) = 360.8 A.
ΔU ≈ √3·I·(R cos φ + X sin φ) = √3 × 360.8 × (0.00832 × 0.8 + 0.02421 × 0.6) = 13.2 V, i.e. 3.3 %.
These are exactly the values computed by the parameter dialog of the NRJXpert three-phase transformer (Network palette).
2.3 Per-unit system
A base power Sb (common to the whole network, e.g. 100 MVA or 10 MVA) and a base voltage Ub per voltage level (transformer ratios) are chosen. Every quantity is divided by its base:
Advantages: transformers disappear (ratio 1 in pu), voltages are close to 1 pu, orders of magnitude can be compared between networks (the usc of a transformer = its impedance in pu on its own base). The NRJXpert server works in pu for the load flow (Sb = 100 MVA, Ub = network voltage entered in the calculation settings).
↑ Back to top2.4 Voltage drop and losses
For a line of impedance R + jX carrying P + jQ at line voltage U:
Practical consequences: the voltage drop depends mainly on P in resistive networks (LV, MV with small cross-sections) and on Q in inductive networks (HV); hence voltage control through reactive power in transmission, and the role of controlled inverters in distribution (level 5). Losses vary as the square of the power and the inverse square of the voltage: doubling the voltage divides losses by 4.
Example 2.2 — MV feeder
A 10 km, 20 kV line (r = 0.3 Ω/km, x = 0.35 Ω/km) supplies 2 MW at cos φ = 0.9 at its end. Voltage drop and losses?
Show the solution
R = 3 Ω, X = 3.5 Ω, Q = 2 × tan(25.84°) = 0.969 Mvar. ΔU/U = (3 × 2 + 3.5 × 0.969)/20² = 2.35 % (470 V); losses = 3 × (2² + 0.969²)/20² = 0.037 MW = 37 kW (1.9 %).
2.5 Load flow: formulation
The load flow (power flow) calculates the steady state of a network from the generated and consumed powers. Since powers rather than impedances are imposed, the equations are non-linear.
Bus types
| Type | Data | Unknowns | In NRJXpert |
|---|---|---|---|
| Slack (swing) | |V|, θ = 0 | P, Q | « Slack » generator: the source substation, which covers the losses |
| Generation (PV) | P, |V| | Q, θ | Voltage-controlled generator |
| Load (PQ) | P, Q | |V|, θ | Load bus |
Admittance matrix and equations
Ybus is built element by element: for a branch of admittance yik between i and k, yik is added to Yii and Ykk and subtracted from Yik and Yki; admittances to earth (C/2 of the lines) are added to the diagonal. At node i:
Pi = Σk |Vi||Vk||Yik| cos(θik − δi + δk) Qi = −Σk |Vi||Vk||Yik| sin(θik − δi + δk)
2.6 Solution methods
Gauss-Seidel
For each node the most recent voltages are used. PV node: Qi is first calculated with the current voltages, then |Vi| is reset to its set-point after the update. Simple, but slow convergence (tens of iterations); an acceleration factor (1.4 to 1.6) helps.
Example 2.3 — One iteration by hand
Two-node network in pu: node 1 slack V₁ = 1∠0°; line z = 0.02 + j0.06; node 2 load 0.8 + j0.4 pu. Start from V₂ = 1∠0° and iterate down to 10⁻⁴.
Show the solution
y = 1/z = 5 − j15; Y₂₂ = 5 − j15, Y₂₁ = −5 + j15. S₂ = −0.8 − j0.4 (load = negative injection).
V₂(1) = [(−0.8 + j0.4)/1 + (5 − j15)·1]/(5 − j15) = 1 + (−0.8 + j0.4)/(5 − j15) = 1 − 0.04 − j0.04 = 0.96 − j0.04 → 0.9608∠−2.39°.
Then: 0.9575∠−2.39°; 0.95736∠−2.395°; 0.95735∠−2.395° (difference < 10⁻⁴ at the 4th iteration). The final voltage is 0.9573 pu ∠−2.39°.
Newton-Raphson
The power equations are linearised around the current state and solved for the corrections:
Quadratic convergence: 3 to 6 iterations whatever the size of the network, but each iteration costs a factorisation of the Jacobian. It is the reference method.
Fast decoupled (Stott)
Assumptions: R ≪ X, small angles, |V| ≈ 1. Then ∂P/∂|V| ≈ 0 and ∂Q/∂δ ≈ 0, and the sub-matrices become constant (B' and B''): they are factorised only once. Very fast on transmission networks, but slow or divergent when R/X is large (distribution networks): see the lab below and the distribution load flow of level 5.
Lab 3 — Load flow of a three-node 30 kV network
Data (Un = 30 kV, AAAC lines, r = 0.27 Ω/km, x = 0.38 Ω/km):
| Element | Data |
|---|---|
| P1 — source substation (slack) | V = 31.5 kV, phase 0 |
| G2 — power plant (PV bus) | Pgen = 3 MW, V = 31.4 kV |
| Ch3 — load (PQ bus) | Pload = 5 MW, Qload = 2.5 Mvar |
| Lines | L12 = 15 km (4.05 + j5.70 Ω), L13 = 20 km (5.40 + j7.60 Ω), L23 = 10 km (2.70 + j3.80 Ω) |
- Calculation settings: network voltage 30 kV, frequency 50 Hz.
- Network (load flow) palette, single-line variant: place two generators and a load bus; set P1 as slack, G2 as PV bus, Ch3 as PQ load with the values above (voltages in kV, powers in MW and Mvar).
- Connect the three nodes with three impedances (Passive elements section) of the R + jX values above, or with three RL lines from the palette (length and per-length characteristics; l = 0.38/314.16 = 1.21 mH/km).
- Calculation Load flow, method Newton-Raphson, F5.
- Compare with the results below (
exemples/loadflow_30kV_3noeuds.xmlcan also be opened with Help › Open a tutorial example: the « Calculate a network file (XML) » window sends this network to the server):
| Node | V (kV) | Angle | Pgen (MW) | Qgen (Mvar) |
|---|---|---|---|---|
| P1 | 31.500 | 0° | 2.060 | 1.429 |
| G2 | 31.400 | +0.006° | 3.000 | 1.156 |
| Ch3 | 30.937 | −0.476° | — | — |
Currents: L12 = 8.3 A, L13 = 38.4 A, L23 = 66.0 A; load current 104.3 A. Losses = 2.060 + 3.000 − 5.000 = 60 kW (check: Σ 3·R·I² = 0.8 + 23.9 + 35.3 kW).
To do: run again with Gauss-Seidel (20 iterations) then Fast decoupled (35 iterations) against 3 for Newton-Raphson. Why is fast decoupled the slowest here? (R/X = 0.71: the assumption R ≪ X does not hold.) Set G2 to 30.5 kV: G2 then absorbs 6 Mvar. Conclusion: in a meshed network, small voltage set-point differences cause large reactive power flows.
2.7 Short circuit according to IEC 60909-0
Principle: equivalent voltage source
All sources are replaced by a single source c·Un/√3 at the fault point; loads are neglected (except motors) and the network is reduced to the impedance seen from the fault. The voltage factor c covers voltage variations and approximations:
| Nominal voltage | cmax (breaking capacity, withstand) | cmin (protection settings) |
|---|---|---|
| Un ≤ 1 kV, tolerance +6 % | 1.05 | 0.95 |
| Un ≤ 1 kV, tolerance +10 % | 1.10 | 0.95 |
| Un > 1 kV | 1.10 | 1.00 |
Symmetrical components
An unbalanced three-phase system is decomposed (Fortescue) into three balanced systems: positive (1), negative (2) and zero (0) sequence, with a = ej2π/3:
Each element has three impedances Z₁, Z₂, Z₀ (static elements: Z₂ = Z₁); the impedance seen from the fault is calculated in each of the three networks, which are then combined according to the type of fault:
| Fault | Initial symmetrical current I"k | Combination of the networks |
|---|---|---|
| Three-phase | c·Un / (√3·|Z₁|) | positive sequence only |
| Phase-to-phase | c·Un / |Z₁ + Z₂| | positive and negative in opposition |
| Two-phase-to-earth | earth current I"kE2E = √3·c·Un·|Z₂| / |Z₁Z₂ + Z₁Z₀ + Z₂Z₀| | negative ∥ zero |
| Phase-to-earth | √3·c·Un / |Z₁ + Z₂ + Z₀| | all three in series |
Peak current: ip = κ·√2·I"k with κ = 1.02 + 0.98·e−3R/X. Fault far from generators: breaking current Ib and steady-state current Ik equal to I"k. Short-circuit power: S"k = √3·Un·I"k.
Example 2.4 — Faults at the end of a 20 kV feeder
The upstream network is represented at the substation by Z₁ = Z₂ = 0.1 + j2 Ω and Z₀ = 0.05 + j1 Ω (earthed neutral). A 10 km line has r = 0.3, x = 0.35, r₀ = 0.45, x₀ = 1.2 Ω/km. Calculate I"k at the end of the line for the four types of fault (maximum Isc).
Show the solution
Z₁ = Z₂ = 0.1 + 3 + j(2 + 3.5) = 3.1 + j5.5 Ω (|Z₁| = 6.314 Ω); Z₀ = 0.05 + 4.5 + j(1 + 12) = 4.55 + j13 Ω; c = 1.1.
| Fault | Calculation | I"k |
|---|---|---|
| Three-phase | 1.1 × 20 000/(√3 × 6.314) | 2 012 A |
| Phase-to-phase | 22 000/|6.2 + j11| | 1 742 A |
| Two-phase-to-earth | largest phase current | 1 907 A |
| Phase-to-earth | √3 × 22 000/|10.75 + j24| | 1 449 A |
κ = 1.02 + 0.98·e−3 × 0.5636 = 1.201; ip = 1.201 × √2 × 2 012 = 3.42 kA; S"k = 69.7 MVA. Note: the phase-to-earth fault is the weakest here because Z₀ > Z₁; near a Dyn transformer with an earthed neutral, it is often the opposite.
Lab 4 — Short circuit in NRJXpert
- Calculation settings: network voltage 20 kV.
- Place a generator (single-line variant); Short-circuit column: Rd = 0.1, Xd = 2 Ω, Ri = Xi = 0 (Z₂ = Z₁), earthed neutral, R0 = 0.05, X0 = 1 Ω.
- Place an RL line (Network palette): 10 km, r = 300 mΩ/km, l = 1.114 mH/km (0.35 Ω/km at 50 Hz); Zero sequence section: r0 = 0.45, x0 = 1.2 Ω/km. Place a load bus at the end (it is neglected in the calculation) and connect.
- Double-click on the terminal at the end of the line: three-phase Fault, maximum current, Un = 20 kV. Choose the Short circuit calculation and run.
- Read the summary (I"k, ip, κ, S"k, Z1, Z2, Z0, I"k of the four types) and compare with example 2.4. Repeat with a phase-to-earth fault: earth current 1 449 A (file
exemples/courtcircuit_20kV_monophase.xml). - Right-click on the line › Short circuit at this point…: plot I"k against the distance (0, 2, 4, …, 10 km). Why is the curve not in 1/d?
2.8 Protection: what short-circuit currents are used for
- Breaking capacity of the device ≥ maximum I"k (and ip ≤ making capacity) at the point of installation.
- Sensitivity: the minimum short-circuit current (cmin, furthest fault, often phase-to-earth or phase-to-phase) must trip the protection within the required time: this is the « minimum Isc » calculation of NRJXpert.
- Thermal withstand of cables: I²·t ≤ k²·S² (k ≈ 143 for XLPE-insulated copper).
- Selectivity: only the protection immediately upstream of the fault must trip (current, time or logic selectivity).
- Inverters: their contribution is limited to 1.1 to 1.5 times their rated current; a network supplied mainly by inverters has low fault currents, which makes detection harder (level 5).
Exercises
Exercise 1 — Short circuit at the secondary of a transformer
The transformer of example 2.1 is supplied by a network of infinite short-circuit power. Calculate the three-phase I"k at the secondary (c = 1.05), κ and ip.
Show the solution
I"k = 1.05 × 400/(√3 × 0.0256) = 9.47 kA; R/X = 0.344; κ = 1.02 + 0.98·e−1.031 = 1.370; ip = 1.370 × √2 × 9.47 = 18.3 kA. The LV main breaker must have a breaking capacity of at least 10 kA (15 or 25 kA is chosen to account for the actual short-circuit power of the upstream network).
Exercise 2 — Change of base
A 2 MVA, 30/0.4 kV transformer has usc = 6 %. Express its impedance in pu on a 100 MVA base, then in ohms on the 30 kV side.
Show the solution
z = 0.06 × 100/2 = 3 pu; Zb = 30²/100 = 9 Ω; Z = 27 Ω (= 0.06 × 30²/2).
Exercise 3 — Voltage profile
In lab 3, the load becomes 8 MW, 4 Mvar. Without detailed calculation, what will happen to V3, to Qgen of G2 and to the losses? Check with NRJXpert.
Show the solution
Forecast: V3 decreases (≈ (R·P + X·Q)/U² increases); G2, which holds its voltage, supplies more reactive power; losses grow roughly as the square of the flows. Calculation (Newton-Raphson, 3 iterations): V3 = 30.629 kV (−1.23°) instead of 30.937 kV; Qgen of G2 = 3.71 Mvar instead of 1.16; the substation supplies 5.17 MW and 0.53 Mvar; losses = 5.172 + 3 − 8 = 172 kW, i.e. 2.9 times more for a load 1.6 times larger.
Exercise 4 — Iterative methods
Explain why Newton-Raphson converges in 3 iterations from any reasonable initialisation (« flat » voltages 1∠0°), while Gauss-Seidel needs 20. What happens if the load becomes too large (beyond the maximum transmissible power)?
Show the solution
NR uses the derivatives (tangent): near the solution the error is squared at each iteration. GS corrects each node separately, with linear convergence. Beyond the voltage collapse point (nose of the P-V curve) there is no solution any more: no method converges and the server returns a non-convergence message.
Level 2 quiz
1. In the π model of a line, the total capacitance C is distributed:
- entirely at the sending end
- C/2 at each end
- in series with R
- C at each end
2. A transformer has usc = 4 %. Its impedance on its own base is:
- 4 Ω
- 0.4 pu
- 0.04 pu
- 25 pu
3. On a highly resistive low-voltage network, the voltage drop depends mainly on:
- the active power
- the reactive power
- the frequency
- the cable capacitance
4. In a load flow, the unknowns of a PV node are:
- P and Q
- |V| and θ
- P and θ
- Q and θ
5. The slack node is used to:
- set the frequency
- provide the angle reference and cover the unknown losses
- represent the loads
- control the reactive power of the PV nodes
6. The fast decoupled method is poorly suited to:
- large transmission networks
- meshed networks
- distribution networks with high R/X
- networks without a PV node
7. To size the breaking capacity of a circuit breaker, one uses:
- the maximum I"k (cmax)
- the minimum I"k (cmin)
- the rated current
- the overload current
8. A phase-to-earth fault connects the sequence networks:
- positive only
- positive and negative in opposition
- negative and zero in parallel
- positive, negative and zero in series
9. A delta winding:
- lets zero-sequence current through to the upstream network
- blocks the zero-sequence current
- doubles the positive-sequence impedance
- suppresses the three-phase fault current
10. For R/X = 0.1, the factor κ is about:
- 1.02
- 1.30
- 1.75
- 2.00
Author
November 2026