# EE3124 — Formula & Tips Sheet

Lectures 4–6 · AC Machines · Synchronous Generator · Synchronous Motor · Induction Machine

## LECTURE 4 — AC Machine Fundamentals

### 4.1 Synchronous Speed

nsync = 120 fe / P

nsync = synchronous speed (r/min), fe = electrical frequency (Hz), P = number of poles.

fe = nm · P / 120 P = 120 fe / nsync

### 4.2 Induced Voltage

Flux: φ = B · A = B · d · l (d = stator diameter, l = coil length)
Angular velocity: ω = 2π · n / 60 (r/min → rad/s)
Peak EMF: Ê = N · φ · ω
RMS EMF: Erms = Ê / √2
General per-phase RMS: EA = √2 · π · NC · φ · f

### 4.3 Three-Phase Connections

| Y-Connection | Δ-Connection
--- | --- | ---
Line Voltage | VL = √3 · Vφ | VL = Vφ
Line Current | IL = Iφ | IL = √3 · Iφ
Power | √3 · VL · IL · cos θ = 3 · Vφ · Iφ · cos θ

### 4.4 Key Concepts

- Swapping any two phase connections reverses the rotating field direction.
- Fractional-pitch windings reduce fundamental voltage slightly, attenuate harmonics drastically.
- Two-pole machine: mechanical speed (r/s) = electrical frequency (Hz).
- Four-pole: fe = 2 × fmech. General: fe = (P/2) × fmech.
- Armature windings → stator; field windings → rotor.
- Slot harmonics → parasitic torques → vibration and noise.

## LECTURE 5 — Synchronous Generator

### 5.1 Phasor Equation (Generator)

EA = Vφ + IA · RA + j · IA · XS = Vφ + IA · ZS

EA = internal generated voltage, Vφ = phase terminal voltage, IA = armature current, RA = armature resistance, XS = synchronous reactance, ZS = RA + j·XS.

### 5.2 Finding ZS from OC/SC Tests

- ① OC test at If,OC → VOC,line → Eo = VOC,line / √3 (Y-connected)
- ② Scale to SC field current: Eo(at If,SC) = Eo(at If,OC) × (If,SC / If,OC)
- ③ |ZS| = Eo / IA,rated
- ④ ZS = |ZS| ∠ cos⁻¹(RA / |ZS|), XS = √(|ZS|² − RA²)

### 5.3 Rated Current & Voltage Regulation

Irated = Srated / (√3 · VL,rated)
VR = (|EA| − |Vφ|) / |Vφ| × 100%
VR is positive for lagging loads, near zero for unity PF, and can be negative for leading loads.

### 5.4 Power & Torque

Pout = √3 · VL · IL · cos θ = 3 · Vφ · IA · cos θ
P = 3 · Vφ · EA · sin δ / XS (when RA ≈ 0)
τind = Pconv / ωsync, ωsync = 2π · nsync / 60

### 5.5 Power Flow (Generator)

Pin = Pout + 3·IA²·RA + Pcore + PF&W + Pstray
η = Pout / Pin × 100%

### 5.6 Phasor Diagram Steps

- ① Draw Vφ as reference (horizontal).
- ② Draw IA at angle −θ (lagging) or +θ (leading) from Vφ.
- ③ Add IA·RA drop (in phase with IA) to the tip of Vφ.
- ④ Add j·IA·XS drop (90° ahead of IA) → reach EA.
- ⑤ Angle from Vφ to EA = torque angle δ.

### 5.7 Connection Notes

Δ-connected: Vφ = VL, IA = IL/√3. Y-connected: Vφ = VL/√3, IA = IL.

### 5.8 Load Disconnection

IA → 0, so VT rises to |EA| (field current unchanged).

### 5.9 Paralleling — 4 Conditions

- ① Same RMS line voltages. ② Same phase sequence.
- ③ Same phase angles. ④ Oncoming generator frequency slightly higher.

## LECTURE 5 (cont.) — Synchronous Motor

### 5M.1 Phasor Equation (Motor)

EA = Vφ − IA · RA − j · IA · XS = Vφ − IA · ZS

Key difference from generator: In the motor, EA lags Vφ (δ is negative). The phasor equation subtracts the impedance drop.

### 5M.2 Power & Torque

Pin = √3 · VL · IL · cos θ = 3 · Vφ · IA · cos θ
Pconv = 3 · Vφ · EA · sin|δ| / XS
τind = Pconv / ωsync
τmax = 3 · Vφ · EA / (ωm · XS)

### 5M.3 Changing Excitation at Constant Load

- If P = const → EA · sin δ = const.
- Over-excitation (↑EA) → leading PF; under-excitation (↓EA) → lagging PF.

New torque angle: δ2 = sin⁻¹( (EA1 / EA2) · sin δ1 )

### 5M.4 Solving Steps

① nsync = 120f/P → ω. ② IL = P/(√3·VL·PF). For Δ: IA=IL/√3; Y: IA=IL. ③ Angle of IA: lag → −cos⁻¹(PF), lead → +cos⁻¹(PF). ④ EA = Vφ − j·XS·IA (if RA ≈ 0).

## LECTURE 6 — Induction Machine

### 6.1 Slip & Rotor Frequency

s = (nsync − nm) / nsync
nm = (1 − s) · nsync
nslip = s · nsync = nsync − nm
fr = s · fe (also: fr = nslip · P / 120)

### 6.2 Equivalent Circuit (per phase)

Stator: R1 + j·X1 → parallel j·XM → Rotor: R2/s + j·X2

R2/s = R2 + R2·(1−s)/s (copper loss + mechanical power conversion)

### 6.3 Power Flow

Pin = 3 · Vφ · IA · cos θ (= √3 · VL · IL · cos θ)
PSCL = 3 · IA² · R1 (stator copper loss)
PAG = Pin − PSCL − Pcore (air-gap power) [also = 3 · I2² · R2 / s]
PRCL = s · PAG (rotor copper loss)
Pconv = (1 − s) · PAG (converted mechanical power)
Pout = Pconv − PF&W − Pmisc

### 6.4 Torque

τind = PAG / ωsync ← uses synchronous speed, NOT rotor speed!
τload = Pout / ωm (uses actual rotor speed)
Common exam mistake: Using ωm instead of ωsync for τind.

### 6.5 Torque Expression & Maximum Torque

τind = [3 VTH² · (R2/s)] / { ωsync · [(RTH + R2/s)² + (XTH + X2)²] }

If IM = 0: VTH = Vφ, RTH = R1, XTH = X1.

Max torque when: R2 / smax = √(RTH² + (XTH + X2)²)
Max starting torque (s=1): R2 + Rext = √(R1² + (X1 + X2)²)

### 6.6 Equivalent Circuit — Solution Steps

① Vφ = VL/√3. ② Z2 = R2/s + j·X2. ③ ZF = (j·XM ∥ Z2). ④ Ztot = R1 + j·X1 + ZF. ⑤ IA = Vφ / Ztot. ⑥ PF = cos(∠IA). ⑦ Power flow chain: Pin → PAG → Pconv → Pout.

### 6.7 Key Properties

- Low-slip region: τ ∝ s. If torque doubles → slip doubles.
- Wound rotor + Rext: shifts τmax to higher slip; τmax value unchanged; improves starting torque & reduces starting current.
- η = Pout / Pin × 100%

## Quick Reference

Item | Formula
--- | ---
r/min → rad/s | ω = n × 2π / 60
hp → W | 1 hp = 746 W
Vφ (Y) | Vφ = VL / √3
Vφ (Δ) | Vφ = VL
IA (Y) | IA = IL
IA (Δ) | IA = IL / √3
3φ power | P = √3 · VL · IL · cos θ = 3 · Vφ · IA · cos θ
3φ apparent | S = √3 · VL · IL
Lagging PF | θ = −cos⁻¹(PF), I lags V
Leading PF | θ = +cos⁻¹(PF), I leads V

Exam Strategy: ① Identify machine type (generator/motor, sync/induction). ② Identify connection (Y/Δ). ③ Write the correct phasor equation. ④ Convert line ↔ phase quantities. ⑤ Compute step by step — never skip. ⑥ Always check units.