Power System Fundamentals and Equipment
Power System Fundamentals and Equipment tests system topology, per-unit calculations, transformer and machine ratings, fault levels, and coordination. The exam expects a technician to move between actual ohms, per-unit values, and three-phase quantities without losing the base.
Core rules
- Per-unit: choose S_base and V_base; Z_base = V_base²/S_base. Convert actual ohms to pu by dividing by Z_base.
- Three-phase power: S = √3 × V_LL × I_L; P = S × power factor.
- A transformer changes voltage and current inversely; impedance in pu is approximately constant across the transformer when expressed on its own base.
- Fault level is limited by source and transformer impedance; lower impedance gives higher fault current.
- Coordination is time-current based: upstream devices should be slower than downstream devices for the same fault.
Worked example
A 100 MVA, 13.8 kV base is used. Z_base = 13.8²/100 = 1.9044 Ω. A feeder has an actual impedance of 0.50 Ω. In per unit: 0.50/1.9044 = 0.2625 pu. On this base, a three-phase fault at the feeder end with a 1.0 pu source would be approximately 1/0.2625 = 3.81 pu, or 381 MVA. The actual current at 13.8 kV is 381 MVA/(√3 × 13.8 kV) ≈ 15.9 kA. The calculation ignores resistance and source impedance, so it is a conservative estimate.
Traps
Using line-to-neutral voltage with √3; forgetting to change the base across a transformer; treating per-unit impedance as ohms; neglecting motor contribution to fault current; assuming coordination from pickup alone without time curves.