JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year ELECTRICAL MACHINES – I Previous Question Paper (March - 2021 - R16)

B.Tech | II Year | I Semester | R16 | March - 2021

Answer PaperPaper Code: 133AMMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

Explain the different methods of excitation of DC generators with suitable diagrams.

8 marks

Answer

DC generators are classified based on the method used to excite their field windings to produce magnetic flux. The two primary methods are:

1. Separately Excited DC Generator: The field winding is energized by an external independent DC source (e.g., a battery or another DC generator). The field current is independent of the armature voltage, allowing for a wide range of output voltage control.

2. Self-Excited DC Generator: The field winding is energized by the generator's own armature current. These are further categorized based on the connection of the field winding:

  • Shunt Wound: The field winding is connected in parallel with the armature. It is designed with many turns of fine wire to carry a small fraction of the total current.
  • Series Wound: The field winding is connected in series with the armature. It consists of a few turns of thick wire to carry the full load current.
  • Compound Wound: This combines both shunt and series field windings. In a long shunt connection, the shunt field is connected across the series combination of armature and series field. In a short shunt connection, the shunt field is connected directly across the armature terminals.
b

A long shunt d.c. compound generator has an armature, series field and shunt field resistances of 0.04Ω, 0.03Ω and 200Ω respectively. It supplies a load current of 180A at 400V. Calculate the generated emf. Assume contact drop/brush =1.2V.

7 marks

Answer

Given data:

Load current (I_L) = 180 A

Terminal voltage (V) = 400 V

Armature resistance (R_a) = 0.04 Ω

Series field resistance (R_se) = 0.03 Ω

Shunt field resistance (R_sh) = 200 Ω

Brush contact drop = 1.2 V

For a long shunt compound generator, the shunt field current is:

I_sh = (V + I_L * R_se) / R_sh = (400 + 180 * 0.03) / 200 = 405.4 / 200 = 2.027 A

The armature current is:

I_a = I_L + I_sh = 180 + 2.027 = 182.027 A

The generated EMF (E_g) is calculated as:

E_g = V + I_a * R_a + I_a * R_se + Brush Drop
E_g = 400 + (182.027 * 0.04) + (182.027 * 0.03) + 1.2
E_g = 400 + 7.281 + 5.461 + 1.2 = 413.942 V

The generated EMF is 413.94 V.

215 marks
a

Explain the following methods of improving commutation:

i) Resistance commutation ii) E.m.f. commutation

8 marks

Answer

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b

A 6 pole lap connected 230V dc shunt motor has 410 armature conductors. It takes 42 A on full load. The flux per pole is 0.05 weber. The armature and field resistances are 0.1Ω and 230Ω respectively. Contact drop per brush = 1.2V. Determine the speed of motor at full load.

7 marks

Answer

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315 marks
a

Derive the torque developed in a DC Motor.

8 marks

Answer

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b

A 220 V DC Shunt motor draws a no – load armature current of 3 A when running at 1400 rpm. Determine its speed when taking an armature current of 65A, if armature reaction weakens the flux by 3.2%. Take armature resistance as 0.25Ω.

7 marks

Answer

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415 marks
a

Explain the working of a three point starter with a neat connection diagram.

8 marks

Answer

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b

Derive the condition for maximum efficiency for a DC generator or DC Motor.

7 marks

Answer

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515 marks
a

Explain how the efficiency can be predetermined by using Swinburne’s Test on a DC Shunt machine.

8 marks

Answer

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b

The voltage per turn of a single phase transformer is 1.1 V. When the primary winding is connected to a 230V, 50Hz AC supply, the secondary voltage is found to be 550V. Find: i) the primary and secondary turns, ii) Core area if the maximum flux density is 1.2 Tesla.

7 marks

Answer

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615 marks
a

Distinguish between a core type and Shell type transformer.

8 marks

Answer

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b

Derive the condition for maximum efficiency of a Single Phase transformer.

7 marks

Answer

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7

The number of turns on the primary and secondary windings of a transformer are 1200 and 300 respectively. When the load current on the secondary is 120A at 0.85 power factor lagging, the primary current is 32 A at 0.72 power factor lagging. Determine the no-load current of the transformer and its phase with respect to the voltage.

15 marks

Answer

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815 marks
a

Explain the concept of parallel operation of a Single – phase transformers.

7 marks

Answer

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b

The primary and secondary resistances of a 1100/220V transformer are 0.32 Ω and 0.0198 Ω respectively. If iron loss amounts to 300W, determine the secondary current at which maximum efficiency occurs and find the maximum efficiency at 0.82 power factor.

8 marks

Answer

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About this paper

This page contains the question paper of ELECTRICAL MACHINES – I from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech EEE II Year I Semester, March - 2021, R16.

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