# EE3124 — Sample Examination Paper

Lectures 4–6 | Total: 100 marks | Time: 2 hours
Answer ALL questions. Show all working clearly.

## Section A — AC Machine Fundamentals [20 marks]

Question 1 [10 marks]
A three-phase Y-connected two-pole synchronous generator has a stator with 4000 turns per phase. The rotor flux is φ = 0.008 Wb and the rotor spins at 3600 r/min.
(a) What is the electrical frequency fe? [2 marks]
(b) Calculate the RMS phase voltage EA. [3 marks]
(c) What is the RMS terminal (line-to-line) voltage VT? [2 marks]
(d) If re-connected in Δ, what would VT be? [3 marks]
Solution 1
(a)
fe = nsync · P / 120 = 3600 × 2 / 120 = 60 Hz
(b) Using EA = √2 · π · NC · φ · f
EA = √2 × π × 4000 × 0.008 × 60
= 1.4142 × 3.1416 × 4000 × 0.008 × 60
= 1.4142 × 6031.9
= 8529 V ≈ 8.53 kV
(c) Y-connected: VT = √3 · EA
VT = √3 × 8529 = 14,772 V ≈ 14.77 kV
(d) In Δ-connection, phase voltage = line voltage, and EA is the phase voltage.
∴ VT = VL = Vφ = EA = 8529 V ≈ 8.53 kV
The Δ terminal voltage is lower than Y by a factor of √3.

Question 2 [10 marks]
A three-phase Δ-connected synchronous generator rotates at 1200 r/min. The rotor peak flux density B = 0.5 T, stator diameter d = 0.4 m, coil length l = 0.3 m, N = 25 turns per coil.
(a) How many poles does this generator have if fe = 60 Hz? [2 marks]
(b) Calculate the flux per pole φ. [2 marks]
(c) Find the peak induced voltage per coil Ê. [3 marks]
(d) Calculate the RMS terminal voltage VT. [3 marks]
Solution 2
(a)
P = 120 fe / nsync = 120 × 60 / 1200 = 6 poles
(b)
φ = B · d · l = 0.5 × 0.4 × 0.3 = 0.06 Wb
(c)
ω = 2π × 1200/60 = 125.66 rad/s
Ê = N · φ · ω = 25 × 0.06 × 125.66 = 188.5 V
(d)
Erms = Ê / √2 = 188.5 / 1.4142 = 133.3 V
Δ-connected → VT = Vline = Vφ = 133.3 V

## Section B — Synchronous Generator [30 marks]

Question 3 [15 marks]
A 13.8-kV, 50-MVA, 0.9-PF-lagging, 60-Hz, four-pole, Y-connected synchronous generator has XS = 2.5 Ω and RA = 0.2 Ω. Friction & windage losses = 1 MW, core losses = 1.5 MW.
(a) What is the synchronous speed? [2 marks]
(b) Calculate the rated armature current IA. [2 marks]
(c) Calculate EA (magnitude and angle) at rated conditions. [5 marks]
(d) What is the voltage regulation? [2 marks]
(e) Calculate input power from the prime mover and overall efficiency. [4 marks]
Solution 3
(a)
nsync = 120 × 60 / 4 = 1800 r/min
(b)
IA = S / (√3 · VL) = 50 × 10⁶ / (√3 × 13800) = 2092 A
(c) Phase voltage and current phasor:
Vφ = 13800 / √3 = 7967 V
PF = 0.9 lag → θ = −cos⁻¹(0.9) = −25.84°
IA = 2092 ∠ −25.84° A
Generator phasor equation:
EA = Vφ + IA · (RA + j·XS)
= 7967∠0° + (2092∠−25.84°)(0.2 + j·2.5)
|ZS| = √(0.2² + 2.5²) = 2.508 Ω, ∠ZS = tan⁻¹(2.5/0.2) = 85.43°
= 7967 + (2092 × 2.508) ∠(−25.84° + 85.43°)
= 7967 + 5247 ∠ 59.59°
= 7967 + 2663 + j·4526
= 10630 + j·4526
|EA| = √(10630² + 4526²) = 11,554 V
δ = tan⁻¹(4526 / 10630) = 23.1°
(d)
VR = (|EA| − Vφ) / Vφ × 100% = (11554 − 7967) / 7967 × 100% = 45.0%
(e)
Pout = S × PF = 50 × 0.9 = 45 MW
Pelec loss = 3 · IA² · RA = 3 × 2092² × 0.2 = 2.63 MW
Pin = 45 + 2.63 + 1.5 + 1.0 = 50.13 MW
η = 45 / 50.13 × 100% = 89.7%

Question 4 [15 marks]
A 480-V, 60-Hz, Δ-connected, four-pole synchronous generator has XS = 0.12 Ω, RA = 0.018 Ω. At full load it supplies 1000 A (line) at 0.8 PF lagging. F&W losses = 30 kW, core losses = 25 kW.
(a) Speed of rotation? [2 marks]
(b) Find Vφ and IA. [2 marks]
(c) Calculate EA (magnitude and angle). [4 marks]
(d) Output power, input power, and efficiency. [5 marks]
(e) If load removed (IF unchanged), what is VT? [2 marks]
Solution 4
(a)
nsync = 120 × 60 / 4 = 1800 r/min
(b) Δ-connected:
Vφ = VT = 480 V
IA = IL / √3 = 1000 / √3 = 577.4 A
(c) θ = cos⁻¹(0.8) = 36.87° (lagging → negative angle for current)
IA = 577.4 ∠ −36.87° A
EA = Vφ + IA(RA + jXS)
|ZS| = √(0.018² + 0.12²) = 0.1213 Ω, ∠ZS = tan⁻¹(0.12/0.018) = 81.47°
= 480∠0° + (577.4∠−36.87°)(0.1213∠81.47°)
= 480 + 70.04 ∠ 44.60°
= 480 + 49.91 + j·49.14
= 529.9 + j·49.14
|EA| = √(529.9² + 49.14²) = 532.2 V
δ = tan⁻¹(49.14 / 529.9) = 5.30°
(d)
Pout = √3 · VL · IL · cos θ = √3 × 480 × 1000 × 0.8 = 665 kW
Pelec = 3 × 577.4² × 0.018 = 18.0 kW
Pin = 665 + 18.0 + 25 + 30 = 738.0 kW
η = 665 / 738 × 100% = 90.1%
(e) Load removed → IA = 0 → VT = EA
VT rises from 480 V to 532.2 V

## Section C — Synchronous Motor [20 marks]

Question 5 [20 marks]
A 480-V, 60-Hz, 300-hp, 0.8-PF-lagging, eight-pole, Δ-connected synchronous motor has XS = 0.5 Ω and RA ≈ 0. Ignore all losses. Assume EA ∝ IF and EA = 480 V when IF = 4 A.
(a) Synchronous speed in r/min and rad/s. [2 marks]
(b) At rated load (300 hp, 0.8 PF lagging): find IL, IA, and EA (magnitude & angle). [8 marks]
(c) Calculate τind and τmax at this excitation. [4 marks]
(d) If EA increased by 30% (load constant): new δ, new IA, and new PF. [6 marks]
Solution 5
(a)
nsync = 120 × 60 / 8 = 900 r/min
ωsync = 2π × 900 / 60 = 94.25 rad/s
(b) No losses → Pin = Pout
Pin = 300 × 746 = 223,800 W = 223.8 kW
IL = P / (√3 · VL · PF) = 223800 / (√3 × 480 × 0.8) = 336.5 A
Δ: IA = IL / √3 = 336.5 / √3 = 194.3 A
PF = 0.8 lagging → IA = 194.3 ∠ −36.87° A
Motor phasor equation (Δ: Vφ = 480 V):
EA = Vφ − j·XS·IA
= 480∠0° − (0.5∠90°)(194.3∠−36.87°)
= 480 − 97.15 ∠ 53.13°
= 480 − 58.29 − j·77.72
= 421.7 − j·77.72
|EA| = √(421.7² + 77.72²) = 428.8 V
δ = tan⁻¹(−77.72 / 421.7) = −10.45° (EA lags Vφ, as expected for motor)
(c)
τind = Pconv / ωsync = 223800 / 94.25 = 2374 N·m
τmax = 3 · Vφ · EA / (ωsync · XS)
= 3 × 480 × 428.8 / (94.25 × 0.5)
= 617,472 / 47.125 = 13,104 N·m
Ratio: τind / τmax = 2374 / 13104 ≈ 18.1% of maximum torque.
(d) EA2 = 1.30 × 428.8 = 557.4 V
Since P = const → EA · sin δ = const:
δ2 = sin⁻¹( (EA1 / EA2) · sin δ1 )
= sin⁻¹( (428.8 / 557.4) × sin(−10.45°) )
= sin⁻¹( 0.7694 × (−0.1814) )
= sin⁻¹(−0.1395) = −8.02°
New armature current:
IA2 = (Vφ − EA2) / (j·XS)
= (480∠0° − 557.4∠−8.02°) / (0.5∠90°)
Numerator = 480 − (551.9 − j·77.72) = −71.9 + j·77.72
|num| = √(71.9² + 77.72²) = 105.9
∠num = 180° − tan⁻¹(77.72/71.9) = 180° − 47.2° = 132.8°
IA2 = (105.9 / 0.5) ∠(132.8° − 90°) = 211.8 ∠ 42.8° A
|IA2| = 211.8 A (phase), IL2 = √3 × 211.8 = 366.8 A (line)
PF = cos(42.8°) = 0.734 leading (positive angle → I leads V)
Increasing excitation changed PF from 0.8 lagging to 0.734 leading (over-excited).

## Section D — Induction Machine [30 marks]

Question 6 [10 marks]
A 400-V, three-phase, four-pole, 50-Hz, Y-connected induction motor runs at 1440 r/min. It delivers 18 kW of mechanical output. Mechanical losses = 600 W.
(a) Calculate nsync, s, nslip, and fr. [4 marks]
(b) Calculate Pconv and PAG. [3 marks]
(c) Calculate τind and τload. [3 marks]
Solution 6
(a)
nsync = 120 × 50 / 4 = 1500 r/min
s = (1500 − 1440) / 1500 = 60 / 1500 = 0.04 (4%)
nslip = s · nsync = 0.04 × 1500 = 60 r/min
fr = s · fe = 0.04 × 50 = 2 Hz
(b)
Pconv = Pout + Pmech loss = 18 + 0.6 = 18.6 kW
PAG = Pconv / (1 − s) = 18.6 / 0.96 = 19.375 kW
(c)
ωsync = 2π × 1500 / 60 = 157.08 rad/s
τind = PAG / ωsync = 19375 / 157.08 = 123.3 N·m
ωm = 2π × 1440 / 60 = 150.80 rad/s
τload = Pout / ωm = 18000 / 150.80 = 119.4 N·m

Question 7 [12 marks]
A 460-V, 60-Hz, four-pole, Y-connected induction motor has: R1 = 0.20 Ω, R2 = 0.15 Ω, X1 = 0.80 Ω, X2 = 1.00 Ω, XM = 22 Ω.
PF&W = 300 W, Pcore = 350 W, Pmisc = 100 W. Slip s = 0.025.
(a) Line current IL. [3 marks]
(b) Stator power factor. [1 mark]
(c) Air-gap power PAG. [2 marks]
(d) Pconv and Pout. [2 marks]
(e) τind and τload. [2 marks]
(f) Overall efficiency. [2 marks]
Solution 7
Preliminary:
Vφ = 460 / √3 = 265.6 V
nsync = 120 × 60 / 4 = 1800 r/min, ωsync = 188.5 rad/s
nm = (1 − 0.025) × 1800 = 1755 r/min, ωm = 183.8 rad/s
(a) Build the equivalent circuit:
Z2 = R2/s + j·X2 = 0.15/0.025 + j·1.00 = 6.0 + j·1.0 Ω
|Z2| = √(36 + 1) = 6.083 Ω, ∠Z2 = tan⁻¹(1/6) = 9.46°
Parallel combination ZF = j·XM ∥ Z2:
ZF = (j·22 × (6.0 + j·1.0)) / (j·22 + 6.0 + j·1.0)
Numerator = (22∠90°)(6.083∠9.46°) = 133.8 ∠ 99.46°
Denominator = 6.0 + j·23.0 → |denom| = 23.77, ∠denom = 75.38°
ZF = 133.8 / 23.77 ∠(99.46° − 75.38°) = 5.63 ∠ 24.08° = 5.14 + j·2.30 Ω
Total impedance:
Ztot = R1 + j·X1 + ZF = 0.20 + j·0.80 + 5.14 + j·2.30 = 5.34 + j·3.10 Ω
|Ztot| = √(5.34² + 3.10²) = 6.18 Ω, ∠Ztot = tan⁻¹(3.10/5.34) = 30.1°
IA = Vφ / Ztot = 265.6 / 6.18 = 43.0 A ∠ −30.1°
(b)
PF = cos(30.1°) = 0.865 lagging
(c)
Pin = 3 × 265.6 × 43.0 × 0.865 = 29.65 kW
PSCL = 3 × 43.0² × 0.20 = 1.109 kW
PAG = 29.65 − 1.11 − 0.35 = 28.19 kW
(d)
Pconv = (1 − s) · PAG = 0.975 × 28.19 = 27.49 kW
Pout = 27.49 − 0.30 − 0.10 = 27.09 kW
(e)
τind = PAG / ωsync = 28190 / 188.5 = 149.5 N·m
τload = Pout / ωm = 27090 / 183.8 = 147.4 N·m
(f)
η = Pout / Pin × 100% = 27.09 / 29.65 × 100% = 91.4%

Question 8 [8 marks]
A three-phase, 380-V, 50-Hz, six-pole, Y-connected wound-rotor induction motor has stator impedance (0.12 + j·0.50) Ω/phase and rotor impedance (0.18 + j·0.50) Ω/phase (referred to stator). Neglect IM.
(a) Find Rext (referred to stator) to achieve maximum starting torque. [4 marks]
(b) Calculate this maximum starting torque. [4 marks]
Solution 8
Preliminary:
nsync = 120 × 50 / 6 = 1000 r/min
ωsync = 2π × 1000 / 60 = 104.72 rad/s
Vφ = 380 / √3 = 219.4 V
(a) For max starting torque at s = 1:
R2' = √(R1² + (X1 + X2)²)
R2 + Rext = √(0.12² + (0.50 + 0.50)²)
= √(0.0144 + 1.0) = √1.0144 = 1.0072 Ω
Rext = 1.0072 − 0.18 = 0.8272 Ω
(b) With R2' = 1.0072 Ω, s = 1:
τind = 3 · Vφ² · (R2'/s) / { ωsync · [(R1 + R2'/s)² + (X1 + X2)²] }
Numerator = 3 × 219.4² × 1.0072 = 3 × 48136.4 × 1.0072 = 145,456
(R1 + R2')² = (0.12 + 1.0072)² = 1.1272² = 1.2706
(X1 + X2)² = 1.0² = 1.0
Denominator = 104.72 × (1.2706 + 1.0) = 104.72 × 2.2706 = 237.8
τmax,start = 145456 / 237.8 = 611.7 N·m