Electrical Machines

Syllabus

Single-phase transformer (equivalent circuit, phasor diagram, open and short circuit tests, regulation, efficiency); three-phase transformers (connections, vector groups, parallel operation); auto-transformer; electromechanical energy conversion; DC machines (types, characteristics, speed control); three-phase induction machines (principle, types, torque-speed characteristics, tests, equivalent circuit, starting and speed control); single-phase induction motors; synchronous machines (cylindrical and salient pole, regulation, parallel operation, starting); losses and efficiency.

Key concepts and formulas

Transformer.

• EMF equation E = 4.44fNΦ_m; turns ratio a = N₁/N₂ = V₁/V₂ = I₂/I₁.
• The open circuit test gives the core (iron) loss and the shunt branch; the short circuit test gives the full-load copper loss and the series impedance.
• Copper loss scales with the square of the load fraction x: P_cu = x²P_cu,FL. Maximum efficiency occurs when copper loss = core loss, at x = √(P_core/P_cu,FL).
• Efficiency η = output/(output + core loss + copper loss).
• Approximate regulation ≈ (R cos φ ± X sin φ) in per unit (+ for lagging).
• Auto-transformer: with K = V_low/V_high, it needs (1 − K) times the copper of a two-winding transformer of the same rating, so the fraction of copper saved is K.

DC machines. Generated EMF E = PΦZN/(60A) (A = 2 for wave, A = P for lap winding). Motor: V = E_b + I_aR_a; torque T = PΦZI_a/(2πA); speed N ∝ E_b/Φ. A series motor must never be started without load.

Three-phase induction motor.

• Synchronous speed N_s = 120f/P; slip s = (N_s − N)/N_s; rotor frequency = sf.
• Power flow: air-gap power P_ag; rotor copper loss = sP_ag; mechanical power = (1 − s)P_ag. So P_ag : P_cu,r : P_m = 1 : s : (1 − s).
• Maximum torque occurs at slip s_m = R₂/X₂ and its value is independent of rotor resistance. Starting torque is maximum when R₂ = X₂.
• Torque ∝ V² (at a given slip). Star-delta starting reduces the starting current and torque to 1/3 of direct-on-line values.

Synchronous machines. Speed N_s = 120f/P. Power of a cylindrical rotor machine P = (EV/X_s) sin δ, maximum at δ = 90°. An over-excited synchronous motor draws leading current (used as a synchronous condenser). Regulation by EMF, MMF and ZPF methods. Conditions for paralleling alternators: same voltage, frequency, phase sequence and phase.

Worked example

A 4-pole, 50 Hz induction motor runs at 1440 rpm with an air-gap power of 20 kW. Find the slip, the rotor copper loss and the mechanical power. A 100 kVA transformer has core loss 1 kW and full-load copper loss 1.5 kW: find the load at maximum efficiency and the maximum efficiency at unity power factor.

import math

P, f, N, Pag = 4, 50, 1440, 20.0
Ns = 120 * f / P
s = (Ns - N) / Ns
print(f"Ns = {Ns:.0f} rpm, slip = {s:.3f}, rotor Cu loss = {s * Pag:.2f} kW, Pm = {(1 - s) * Pag:.2f} kW")
S, core, cu = 100, 1.0, 1.5
x = math.sqrt(core / cu)
out = x * S
eta = out / (out + core + x * x * cu)
print(f"max efficiency at {x * 100:.2f}% load ({out:.2f} kVA), eta = {eta * 100:.3f}%")
print(f"EMF: 4.44 * 50 * 200 * 0.01 = {4.44 * 50 * 200 * 0.01:.1f} V")

Ns = 1500 rpm, slip = 0.040, rotor Cu loss = 0.80 kW, Pm = 19.20 kW
max efficiency at 81.65% load (81.65 kVA), eta = 97.609%
EMF: 4.44 * 50 * 200 * 0.01 = 444.0 V

Practice questions