📘 Per Unit System – Advanced Network Problems
Now we apply per unit system to complete power networks. This is real GATE-level solving.
🔹 Problem 1 – Generator & Transformer Network
Given: Generator: 50 MVA, 11 kV, X = 0.2 pu Transformer: 50 MVA, 11/132 kV, X = 0.1 pu Choose base: S_base = 50 MVA V_base (generator side) = 11 kV Transformer side base voltage: = 132 kV Since base MVA same as rating, per unit values remain same.Generator reactance = 0.2 pu Transformer reactance = 0.1 pu
Total reactance: = 0.2 + 0.1 = 0.3 pu🔹 Problem 2 – Change of Base
Generator rating: 100 MVA, 11 kV Reactance = 0.15 pu Convert to new base: S_new = 50 MVA Voltage same. Formula: Z_new = Z_old × (S_new / S_old) = 0.15 × (50/100)Z_new = 0.075 pu
🔹 Problem 3 – Multi-Level Network
Generator: 20 MVA, 13.8 kV, X = 0.25 pu Transformer: 20 MVA, 13.8/69 kV, X = 0.1 pu Line impedance = 20 Ω Find per unit impedance of line. Choose: S_base = 20 MVA V_base (line side) = 69 kV Base impedance: Z_base = V² / S = (69²) / 20 = 4761 / 20 = 238.05 Ω Line per unit impedance: = 20 / 238.05Z_line ≈ 0.084 pu
Total system reactance: = 0.25 + 0.1 + 0.084 = 0.434 pu🔹 Problem 4 – Convert PU Back to Actual
Given: Z_pu = 0.2 Base impedance = 50 Ω Actual impedance: = 0.2 × 50Z_actual = 10 Ω
🔹 Problem 5 – Concept Question
Why per unit simplifies transformer calculations? Because: Turns ratio automatically adjusted when same MVA base used. Very common conceptual GATE question.🎯 GATE Focus
- Always select common MVA base
- Adjust voltage base according to transformer ratio
- Add reactances directly in pu
- Convert line impedance carefully
Per Unit System = Simplifies Entire Power Network
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