JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year ELECTRICAL MACHINES – I Previous Question Paper (March - 2021 - R16)
Answer any five questions. All questions carry equal marks.
Main
Explain the different methods of excitation of DC generators with suitable diagrams.
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.
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.
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:
The armature current is:
The generated EMF (E_g) is calculated as:
The generated EMF is 413.94 V.
Explain the following methods of improving commutation:
i) Resistance commutation ii) E.m.f. commutation
Answer
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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.
Answer
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Derive the torque developed in a DC Motor.
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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.
Answer
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Explain the working of a three point starter with a neat connection diagram.
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Derive the condition for maximum efficiency for a DC generator or DC Motor.
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Explain how the efficiency can be predetermined by using Swinburne’s Test on a DC Shunt machine.
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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.
Answer
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Distinguish between a core type and Shell type transformer.
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Derive the condition for maximum efficiency of a Single Phase transformer.
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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.
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Explain the concept of parallel operation of a Single – phase transformers.
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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.
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