Per-Unit System and Load Flow
Per-unit quantities and change of base, single-line diagrams, bus admittance matrix formation, bus types, Gauss-Seidel and Newton-Raphson load flow, and voltage control.
Large networks are analysed in per unit on common bases, and their steady-state operation is found by load flow (power flow) studies.
Per-unit system
- Choose S_base (three-phase MVA) and V_base (line kV) in one zone; V_base in other zones follows the transformer ratios.
- Z_base = V_base²/S_base; I_base = S_base/(√3 V_base).
- Change of base: Z_new = Z_old (S_new/S_old)(V_old/V_new)².
- Advantages: transformer impedances are the same on both sides, values fall in a narrow range, and √3 factors disappear.
Bus admittance matrix (Y_bus)
For a network with line admittances y_ij:
- Diagonal Y_ii = sum of all admittances connected to bus i (including shunt charging).
- Off-diagonal Y_ij = −y_ij (zero if not connected).
- Y_bus is sparse and symmetric (without phase shifters). Z_bus = Y_bus⁻¹ is full.
Bus types
| Bus | Specified | Unknown |
|---|---|---|
| Slack (swing) | V, δ (= 0) | P, Q |
| PV (generator) | P, V | Q, δ |
| PQ (load) | P, Q | V, δ |
Power flow equations
P_i − jQ_i = V_i* Σ Y_ij V_j
These are non-linear, so they are solved iteratively:
- Gauss–Seidel: simple, little memory per iteration, but many iterations (convergence slows with system size); acceleration factor about 1.6.
- Newton–Raphson: uses the Jacobian; quadratic convergence, few iterations (3–5) independent of size; standard for large systems.
- Fast decoupled: uses P–δ and Q–V decoupling with constant matrices; very fast.
- DC load flow: lossless, flat voltages, P = B′δ; linear and approximate.
Voltage and frequency control
- Real power ↔ frequency (governors, load frequency control).
- Reactive power ↔ voltage (generator excitation, tap-changing transformers, shunt capacitors and reactors, SVCs).
Worked example
A 50 MVA, 11 kV generator has X = 0.2 pu. On a 100 MVA, 11 kV base:
- X_new = 0.2 × (100/50) × (11/11)² = 0.4 pu.
Practice
What is the base impedance (Ω) for 100 MVA and 132 kV? (Give 2 decimal places.)
Show answer and explanation
Answer: 174.24Z_base = kV²/MVA = 17,424/100 = 174.24 Ω.
A bus connects to two lines with admittances −j5 and −j10 pu. What is the diagonal element Y_ii (imaginary part, in pu)?
Show answer and explanation
Answer: -15Y_ii = −j5 − j10 = −j15 pu.
A 20 MVA, 33 kV transformer has X = 0.1 pu. What is its reactance on a 100 MVA, 33 kV base (pu)?
Show answer and explanation
Answer: 0.5X_new = 0.1 × 100/20 = 0.5 pu.
At a PV bus, the quantities specified are:
Show answer and explanation
Answer: BA generator bus holds its real power output and voltage magnitude; Q and δ are found by the load flow.