JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year ELECTRICAL MACHINES-1 Previous Question Paper (March - 2021 - R18)

B.Tech | II Year | II-I | R18 | March - 2021

Answer PaperPaper Code: 153AQMaximum Marks: 75Time: 3 hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a

Explain the working principle of DC generator in detail.

7 marks

Answer

A DC generator operates on the principle of Faraday's Law of Electromagnetic Induction, which states that whenever a conductor cuts magnetic flux, an electromotive force (EMF) is induced in it. The magnitude of this induced EMF is given by e = Blv sin(θ).

The construction consists of a stationary magnetic field (stator) and a rotating armature (rotor). When the armature is rotated by a prime mover within the magnetic field, the conductors on the armature cut the magnetic lines of force. This induces an alternating EMF in the armature conductors. To convert this alternating EMF into a unidirectional (DC) voltage, a commutator and brushes are used. The commutator acts as a mechanical rectifier, reversing the connections of the armature coils at the exact moment the current reverses, ensuring that the current collected by the brushes always flows in the same direction.

b

The OCC of a separately excited DC generator driven at 600 rpm is as follows:

Field current (A)0.20.40.60.81.01.21.41.6
EMF (V)16.031.040.048.051.055.060.062.0

If the machine is connected as shunt generator and driven at 600 rpm and has a field resistance of 100Ω, find the open circuit voltage and exciting current.

8 marks

Answer

To find the open circuit voltage and exciting current for a shunt generator, we must identify the intersection of the Open Circuit Characteristic (OCC) curve and the Field Resistance Line (FRL). The FRL is defined by the equation V = I_f × R_f, where R_f = 100 Ω.

We calculate the voltage values for the field resistance line:

  • For I_f = 0.4 A, V = 0.4 × 100 = 40 V. (OCC value is 31 V)
  • For I_f = 0.6 A, V = 0.6 × 100 = 60 V. (OCC value is 40 V)
  • For I_f = 0.5 A, V = 0.5 × 100 = 50 V.

By plotting the OCC and the field resistance line V = 100 × I_f, the intersection point represents the steady-state operating point. Based on the provided data, the intersection occurs where the field resistance line crosses the OCC curve. By linear interpolation between I_f = 0.4 A and 0.6 A, the intersection point is approximately at I_f ≈ 0.52 A and V ≈ 52 V. Thus, the open circuit voltage is approximately 52 V and the exciting current is 0.52 A.

215 marks
a

Explain the differences between self and separately excited DC generators.

7 marks

Answer

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b

Draw a developed diagram of 2 layer wave winding for a 8 pole DC generator with 60 armature conductors.

8 marks

Answer

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315 marks
a

Derive the condition for maximum efficiency in a DC shunt motor.

7 marks

Answer

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b

A 150V DC shunt motor with armature resistance of 0.3Ω runs at 500 rpm at full load and takes an armature current of 12A. If the resistance of 2.5Ω is placed in the armature circuit, find the speeds at full load torque and half full load torques.

8 marks

Answer

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415 marks
a

Discuss in detail about the applications of DC compound motors.

7 marks

Answer

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b

A 100V DC shunt motor runs at 1000 rpm at no load and takes 10A. The total armature and shunt field resistances are given by 0.25Ω and 250Ω respectively. Calculate the speed when loaded and taking 40A. Assume the flux to be constant.

8 marks

Answer

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515 marks
a

How to separate stray losses in DC motors? Explain.

7 marks

Answer

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b

Discuss in detail about Field’s test with necessary equations.

8 marks

Answer

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615 marks
a

Draw the phasor diagram of single phase transformer under leading power factor conditions and explain.

7 marks

Answer

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b

A 2000/100 V transformer draws a no-load primary current of 0.5A and absorbs 100 watts. Find the magnetizing and iron loss currents.

8 marks

Answer

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715 marks
a

Explain in detail about Sumpner’s test on transformers.

7 marks

Answer

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b

A 10kVA, 2000/200V, single-phase transformer gave the following test results:

Open-circuit test: 200V, 0.9A, 50W

Short-circuit test: 70V, 5A, 45W

Calculate the efficiency at half-load and full-load at 0.8 p.f.

8 marks

Answer

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8

Explain the procedure of conducting Swinburne’s test as d.c. machine.

15 marks

Answer

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About this paper

This page contains the solved question paper with answers of ELECTRICAL MACHINES-1 from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech EEE II Year II-I, March - 2021, R18.

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