(SEM V) THEORY EXAMINATION 2023-24 ELECTRICAL MACHINES-II

SECTION A – Short Answers (2 × 10 = 20 Marks)

(a) A 6-pole alternator rotates at 1000 rpm. What is the frequency of generated voltage?

f=P×N120=6×1000120=50 Hzf = \frac{P \times N}{120} = \frac{6 \times 1000}{120} = 50\ \text{Hz}f=120P×N​=1206×1000​=50 Hz

Frequency = 50 Hz

(b) Methods for predetermining voltage regulation of a 3-phase alternator

EMF (Synchronous impedance) method                MMF (Ampere-turn) method

ZPF (Zero Power Factor) or Potier method            ASA (American Standards Association) method

(c) What are the V-curves of a synchronous motor?

The V-curves are plots of armature current vs field current at constant mechanical load.

They show that armature current is minimum at unity power factor and increases for both leading & lagging conditions (curve resembles “V” shape).

(d) Conditions for parallel operation of two single-phase alternators

Equal voltage magnitude                                        Equal frequency

Equal phase sequence                                             Equal phase angle (in phase)

(e) Why are rotor slots of an induction motor skewed?

To reduce magnetic hum and noise.                      To minimize cogging (magnetic locking).

To smooth torque and improve performance.

(f) Condition for producing maximum torque in a 3-phase induction motor

R2=X2R_2 = X_2R2​=X2​

(where R2R_2R2​ = rotor resistance per phase, X2X_2X2​ = rotor reactance per phase).
At this condition, torque is maximum and independent of rotor resistance.

(g) Define cogging in an induction motor

Cogging (Magnetic Locking):
When rotor and stator teeth align and magnetic locking occurs, preventing the motor from starting.
Occurs when stator & rotor slots are equal in number.

(h) Why is power factor of induction motor low at starting?

At starting, rotor speed is zero ⇒ slip s=1s = 1s=1.
Rotor current frequency is high ⇒ rotor reactance high ⇒ lagging current ⇒ low power factor.

(i) Different types of single-phase induction motors

Split-phase induction motor                                Capacitor start motor

Capacitor start–capacitor run motor                    Permanent split capacitor (PSC) motor

Shaded pole motor

(j) Modifications in a DC series motor to operate on AC supply

Use laminated core to reduce eddy current losses.

Add compensating winding to neutralize armature reaction.

Use high reactance field winding for reduced current.

Add interpoles for sparkless commutation.

SECTION B – Descriptive Questions (Any 3 × 10 = 30 Marks)

(a) Rotating Field System in Alternators

Preferred over stationary field because:       Slip rings carry low DC current only.

Lighter, smaller rotor.                                       Easier insulation and cooling.

Types of Rotors:

Salient pole rotor (Hydro generators) – large diameter, short axial length.

Non-salient (Cylindrical) rotor (Turbo alternators) – small diameter, long axial length.

(b) Working of Synchronous Motor

When connected to 3-phase supply, the stator creates a rotating magnetic field (RMF).

The rotor field (DC excitation) locks with RMF → synchronous speed.

Starting: Motor is not self-starting → use damper winding (acts as squirrel cage motor initially).

(c) Numerical – 3-Phase Induction Motor

Given:
R2=0.5Ω,X2=5Ω,6−pole,50HzR_2 = 0.5Ω, X_2 = 5Ω, 6-pole, 50HzR2​=0.5Ω,X2​=5Ω,6−pole,50Hz
(i) Synchronous speed Ns=120f/P=1000 rpmN_s = 120f/P = 1000\ rpmNs​=120f/P=1000 rpm
(ii) For maximum torque: s=R2/X2=0.1⇒N=900 rpms = R_2/X_2 = 0.1 \Rightarrow N = 900\ rpms=R2​/X2​=0.1⇒N=900 rpm.
(iii) Ratio Tmax/Tst=(sst/smax)=(1/0.1)=10T_{max}/T_{st} = (s_{st}/s_{max}) = (1/0.1) = 10Tmax​/Tst​=(sst​/smax​)=(1/0.1)=10.
For half of max torque → use Rext=R2(1/√2−1)R_{ext} = R_2(1/√2 - 1)Rext​=R2​(1/√2−1).

(d) Star–Delta and Autotransformer Starters

Star–Delta Starter: Reduces voltage per phase by √3, limits starting current to 1/3 of DOL.

Autotransformer Starter: Reduces applied voltage by tapping ratio; better torque per ampere of line current.

(e) Shaded Pole & Permanent Split Capacitor Motor

Shaded Pole Motor: Small shaded copper band creates rotating field; simple but low torque.

PSC Motor: Capacitor permanently connected in series with auxiliary winding; quiet, reliable, used in fans.

SECTION C – Long Questions (Any 1 from each)

3(a) Alternator EMF Calculation

Given data → use:

E=4.44×f×N×Φ×Kd×KpE = 4.44 \times f \times N \times \Phi \times K_d \times K_pE=4.44×f×N×Φ×Kd​×Kp​

where Kd=sin⁡(mα/2)msin⁡(α/2)K_d = \frac{\sin(m\alpha/2)}{m\sin(\alpha/2)}Kd​=msin(α/2)sin(mα/2)​, Kp=cos⁡(β/2)K_p = \cos(\beta/2)Kp​=cos(β/2).
Substitute f=50, m=2, β=180−150=30°f=50,\ m=2,\ \beta=180-150=30°f=50, m=2, β=180−150=30° to find flux per pole.

4(a) Load Sharing Between Parallel Generators

Load shared ∝ drop characteristics:

P1P2=Δf2Δf1\frac{P_1}{P_2} = \frac{Δf_2}{Δf_1}P2​P1​​=Δf1​Δf2​​

Use given frequency regulation to calculate 3000 kW load distribution and maximum load limit without overloading.

5(a) Torque–Slip Characteristic

At start (s = 1): Torque ∝ R₂.

At max torque: s=R2/X2s = R_2/X_2s=R2​/X2​.

Operating region: 0 < s < s_max → stable.
Graph: Starting → Rising → Peak → Falling torque curve.

6(a) Speed Control Methods of Induction Motor

Consequent Pole Method: Change stator pole number → changes synchronous speed.

V/f Method: Maintain constant flux by keeping V/fV/fV/f constant → used with variable frequency drives (VFDs).

7(a) Tests on Single Phase Induction Motor

No-load test: Determines no-load current, losses, power factor.

Blocked-rotor test: Determines equivalent circuit parameters (R₁, R₂, X₁, X₂).

7(b) Double Cage Induction Motor

Has two rotor cages:

Outer cage: High resistance, low reactance → better starting torque.

Inner cage: Low resistance, high reactance → good efficiency at running speed.

Equivalent circuit = two parallel rotor circuits.