Power Systems and Electrical Machines
Power Systems and Electrical Machines tests per-unit analysis, three-phase relationships, transformer and machine models, and the fault and protection calculations that follow from them.
Three-phase and per-unit
- S = √3 V_L I_L for a balanced load; P = √3 V_L I_L cosφ and Q = √3 V_L I_L sinφ. In a Y connection the line voltage leads the phase voltage by 30°.
- Per-unit: choose S_base and V_base, then Z_base = V_base²/S_base and I_base = S_base/(√3 V_base); convert across transformer zones using the turns ratio.
- Delta–wye conversion: Z_Y = Z_Δ/3.
Transformers, machines and lines
- Ideal transformer: V₁/V₂ = N₁/N₂ = I₂/I₁; referred impedance Z₂′ = a²Z₂. The short-circuit test gives series impedance; the open-circuit test gives core loss.
- Induction motor: n_s = 120f/P, slip s = (n_s − n)/n_s, rotor frequency s·f. Torque varies with V², so a 10 % dip cuts torque by about 19 %.
- Synchronous machine: P = (EV/X_s)sinδ, with δ = 90° the steady-state stability limit.
- Line models: short lines use series impedance only, medium lines add shunt capacitance in a nominal π, long lines need hyperbolic (exact) parameters.
- Symmetrical faults use the positive-sequence network alone; asymmetrical faults need sequence networks combined at the fault point.
Worked example
A 4-pole, 50 Hz induction motor runs at 1440 rpm. Synchronous speed is n_s = 120 × 50/4 = 1500 rpm, so s = (1500 − 1440)/1500 = 0.04 and rotor frequency is 0.04 × 50 = 2 Hz. Separately, a 10 MVA load at 11 kV and 0.8 pf lagging draws I = 10 × 10⁶/(√3 × 11 × 10³) ≈ 525 A with P = 8 MW.
Traps
Applying √3 twice in a per-phase calculation; forgetting that per-unit impedance changes when crossing a transformer; mixing line and phase quantities in delta; assuming slip stays constant as load torque changes; treating a three-phase fault current as three times the per-phase value.