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JNTU B.Tech III ELECTRICAL MACHINES-III Previous Question Paper (March-2021 - R13/R15)

B.Tech | III | I | R13/R15 | March-2021

Answer PaperPaper Code: 125AE/115AEMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

By means of a neat diagram, describe the main parts of an alternator with their functions.

8 marks

Answer

An alternator consists of two primary components: the stator and the rotor. The stator is the stationary part that houses the armature winding in slots on its inner periphery. It is constructed from laminated steel stampings to minimize eddy current losses. The rotor is the rotating part that provides the necessary magnetic field. It is typically one of two types: the salient pole rotor, used for low-speed machines, featuring projecting poles; or the cylindrical rotor, used for high-speed turbo-alternators, featuring a smooth, solid steel cylinder with slots for the field winding. Other essential parts include the exciter, which provides DC current to the rotor field windings, and the stator frame, which provides mechanical support and protection.

b

Calculate the r.m.s value of induced emf per phase of 8-pole, 3-phase, 50Hz alternator with 3 slots per pole per phase and 6 conductors per slot in two layers. The coil span is 150°. The flux per pole has a fundamental component of 0.2Wb and 10% third harmonic component.

7 marks

Answer

Given: P=8, f=50Hz, slots per pole per phase (m)=3, conductors per slot=6, coil span=150°, Φ₁=0.2Wb, Φ₃=0.1Φ₁=0.02Wb. Total slots = 8 poles × 3 slots/pole/phase × 3 phases = 72 slots. Conductors per phase (Zph) = (72 slots × 6 conductors/slot) / 3 phases = 144. Turns per phase (Tph) = 144 / 2 = 72.

Distribution factor (Kd): For m=3, Kd = sin(mβ/2) / (m sin(β/2)). Slot angle β = 180° / (slots/pole) = 180° / (72/8) = 20°. Kd = sin(3×20°/2) / (3×sin(20°/2)) = sin(30°)/(3×sin(10°)) = 0.5 / 0.5209 = 0.9598.

Pitch factor (Kp): Kp = cos(α/2). Coil span is 150°, so pitch deficiency α = 180° - 150° = 30°. Kp = cos(15°) = 0.9659. For 3rd harmonic, Kp3 = cos(3×15°) = cos(45°) = 0.7071. Kd3 = sin(3×3×20°/2) / (3×sin(3×20°/2)) = sin(90°)/(3×sin(30°)) = 1 / 1.5 = 0.6667.

Fundamental EMF (E1) = 4.44 × f × Φ1 × Tph × Kp × Kd = 4.44 × 50 × 0.2 × 72 × 0.9659 × 0.9598 = 2943.4V. 3rd Harmonic EMF (E3) = 4.44 × (3f) × Φ3 × Tph × Kp3 × Kd3 = 4.44 × 150 × 0.02 × 72 × 0.7071 × 0.6667 = 448.6V.

E_rms = sqrt(E_1^2 + E_3^2) = sqrt(2943.4^2 + 448.6^2) = 2977.4V
215 marks
a

Derive emf equation for an alternator from fundamentals.

8 marks

Answer

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b

Draw the load characteristics of 3-φ alternator.

7 marks

Answer

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3

A 30KVA, 440V, 50Hz, 3-phase, Star-connected alternator gave the following test data:

Field current (A)24678101214
Terminal voltage (V)155287395440475530570592
S.C Current (A)1122344046576980

Resistance between any two terminals is 0.3 ohms. Find the regulation at full load, 0.8 P.f lagging, by MMF method.

15 marks

Answer

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415 marks
a

Explain the two reaction theory of salient pole alternator. Give the procedure for drawing phasor diagram for salient pole machine.

8 marks

Answer

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b

Deduce the relation for voltage regulation of salient pole alternator.

7 marks

Answer

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5

Analyse the instantaneous short circuit currents of an alternator and explain how sub-transient, transient and steady state reactance are determined.

15 marks

Answer

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615 marks
a

Explain the necessary condition for synchronization of an alternator.

7 marks

Answer

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b

Two 50MVA, 3-phase alternators operate in parallel. The settings of the governors are such that the rise in speed from full-load to no-load in 2% in one machine and 3% in the other, the characteristics being straight lines in both cases. If each machine is fully loaded when the total load is 100MW, what will be the load on each machine when the total load reduced to 60MW?

8 marks

Answer

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715 marks
a

Explain the procedure for starting of synchronous motor.

8 marks

Answer

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b

A 220V, 3-phase, star connected synchronous motor has a resistance of 0.22 per phase and a synchronous reactance of 2.4Ω per phase. The motor is operating at 0.6 power factor leading with a line current of 180A. Determine the internal induced emf.

7 marks

Answer

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815 marks
a

Explain the construction details of single-phase induction motors.

8 marks

Answer

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b

Explain the concept of double revolving field theory with the help of neat diagrams.

7 marks

Answer

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About this paper

This page contains the question paper of ELECTRICAL MACHINES-III from JNTU for B.Tech EEE III I, March-2021, R13/R15.

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