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JNTU B.Tech III ELECTRICAL MACHINES-III Previous Question Paper (August - 2022 - R13)

B.Tech | III | III-I | R13 | August - 2022

Answer PaperPaper Code: 115AEMaximum Marks: 75Time: 3 hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a

Explain in detail about the constructional features of round rotor synchronous machines.

7 marks

Answer

A round rotor (or non-salient pole) synchronous machine is designed for high-speed operation, typically driven by steam turbines. Its construction features include:

  • Stator: Similar to an induction motor, the stator consists of a cylindrical frame made of cast iron or steel, housing a core built from silicon steel laminations to reduce eddy current losses. The inner periphery contains slots to accommodate the three-phase armature winding.
  • Rotor: The rotor is a solid steel cylinder with slots cut along its outer periphery, extending over about two-thirds of the pole pitch. These slots house the field windings, which are secured by heavy steel wedges to withstand high centrifugal forces.
  • Air Gap: The air gap is uniform throughout the circumference, which minimizes flux harmonics and ensures a sinusoidal induced EMF.
  • Cooling: Due to high rotational speeds and large physical size, these machines require sophisticated forced-air or hydrogen cooling systems to dissipate heat generated in the windings and core.
b

A three phase star connected alternator has an open circuit voltage of 7000V. The armature resistance and synchronous resistance are 0.5 Ω and 5 Ω per phase respectively. Find the terminal voltage and the phase difference between terminal voltage and open circuit EMF at a power factor of 0.85 lagging. Given load current is 170A.

8 marks

Answer

Given: Phase voltage E = 7000 / √3 = 4041.45 V, R = 0.5 Ω, Xs = 5 Ω, I = 170 A, cos φ = 0.85 (lagging), sin φ = 0.5268.

V cos δ = E - I(R cos φ + Xs sin φ)
V cos δ = 4041.45 - 170(0.5 × 0.85 + 5 × 0.5268) = 4041.45 - 495.23 = 3546.22 V
V sin δ = I(Xs cos φ - R sin φ)
V sin δ = 170(5 × 0.85 - 0.5 × 0.5268) = 170(4.25 - 0.2634) = 677.72 V

Terminal phase voltage V = √(3546.22² + 677.72²) = 3610.4 V. The line-to-line terminal voltage is 3610.4 × √3 = 6254.5 V. The phase difference δ is:

δ = tan⁻¹(677.72 / 3546.22) = 10.81°
215 marks
a

Explain in detail about pitch and winding factors in synchronous machines.

8 marks

Answer

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b

A 100 kVA, 400V, 3 phase star connected alternator has the following test results: OCC: Line to line voltage is 400V, Field current is 3.5 A; Air-gap line: Line to line voltage is 300V, Field current is 2.5A; SCC: Field current is 2.5A, Armature current is 144.34 A. The armature resistance is negligible. Calculate the rated phase voltage, rated phase current and base impedance.

7 marks

Answer

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315 marks
a

Explain in detail about the experimental determination of Xd and Xq.

7 marks

Answer

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b

The OCC of a 6-pole, 440V, 50Hz, three phase star connected alternator is given. A field current of 7.5 A is needed to circulate the full load rated current of 40A under short circuit conditions. The field current for rated terminal voltage under full load zero power conditions is 15A. The armature resistance is 0.25 Ω per phase. Find the regulation at full load current of 40A at 0.85 lagging power factor using synchronous impedance method.

8 marks

Answer

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415 marks
a

Explain in detail about finding the regulation of synchronous generators using ZPF method.

9 marks

Answer

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b

A three phase 11 kV star connected alternator has armature resistance of 1 Ω and synchronous reactance of 20 Ω per phase. Find regulation for a load of 1000 kW at power factor of 0.9 leading.

6 marks

Answer

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515 marks
a

Explain the effect of change of excitation on load sharing in synchronous generators.

8 marks

Answer

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b

A 1 MVA, three phase star connected, 8 pole, 1500 rpm alternator is operating on 6 kV bus bars. Synchronous reactance is 5 Ω per phase. Find synchronizing power and torque per mechanical degree of displacement for full load 0.9 power factor lagging.

7 marks

Answer

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615 marks
a

Give the detailed analysis of short circuit current wave form in synchronous generators.

8 marks

Answer

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b

Two three phase star connected alternators supply a load of 2000 kW at 0.9 power factor lagging and share the load equally. The excitation of the second machine is so adjusted that it supplies 100A at a lagging power factor. The synchronous impedances of alternators are (0.3+j9) Ω and (0.4+j7) Ω respectively. Find current and power factor of each machine. The terminal voltage is 5 kV.

7 marks

Answer

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715 marks
a

Discuss hunting and its suppression in case of synchronous motor.

7 marks

Answer

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b

A 5 MVA, three phase star connected, 6 kV, 16-pole, 50 Hz, salient pole synchronous motor has xd = 5 Ω and xq = 3 Ω per phase. The motor is working at full load unity power factor and rated voltage. Find (i) Current (ii) Back EMF.

8 marks

Answer

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815 marks
a

Draw the schematic diagram of a shaded pole motor and explain the operation.

7 marks

Answer

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b

Explain in detail about the constructional features of single phase induction motors.

8 marks

Answer

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This page contains the question paper of ELECTRICAL MACHINES-III from JNTU for B.Tech EEE III III-I, August - 2022, R13.

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