JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year ELECTRICAL TECHNOLOGY Previous Question Paper (October - 2020 - R16)

B.Tech | II Year | I Semester | R16 | October - 2020

Answer PaperPaper Code: 133ANMaximum Marks: 75Time: 2 hours

Answer any five questions

All questions carry equal marks

Main

1

Explain the principle of operation of dc generator and derive its emf equation.

15 marks

Answer

A DC generator operates on the principle of Faraday's Law of Electromagnetic Induction. When a conductor is moved within a magnetic field such that it cuts the magnetic flux lines, an electromotive force (EMF) is induced in the conductor. If the circuit is closed, this induced EMF causes a current to flow.

The construction involves a stationary field system (stator) that provides the magnetic flux and a rotating armature (rotor) carrying conductors. As the armature rotates, the conductors cut the magnetic flux, inducing an alternating EMF. A commutator and brush assembly are used to convert this alternating EMF into unidirectional (DC) output at the terminals.

To derive the EMF equation, let:

  • Φ = flux per pole in Webers (Wb)
  • P = number of poles
  • Z = total number of armature conductors
  • N = speed of the armature in revolutions per minute (rpm)
  • A = number of parallel paths

The flux cut by one conductor in one revolution is ΦP. The time taken for one revolution is 60/N seconds. According to Faraday's law, the EMF induced in one conductor is the rate of change of flux:

e = (ΦP) / (60/N) = (ΦPN) / 60 volts

For a wave-wound generator, A = 2, and for a lap-wound generator, A = P. The total induced EMF (E) is the EMF per conductor multiplied by the number of conductors in series per parallel path (Z/A):

E = (ΦPNZ) / (60A) volts

This equation represents the generated EMF in a DC machine, where the voltage is directly proportional to the magnetic flux, the speed of rotation, and the number of armature conductors.

2

Explain different speed control methods of dc shunt motor.

15 marks

Answer

The speed of a DC shunt motor is given by the relation N ∝ (V - IaRa) / Φ, where N is speed, V is terminal voltage, Ia is armature current, Ra is armature resistance, and Φ is flux per pole. Based on this equation, speed can be controlled by varying the flux, the armature resistance, or the applied voltage.

The primary methods are:

  1. Flux Control Method: This method involves inserting a variable resistor (rheostat) in series with the shunt field winding. Increasing the resistance reduces the field current and consequently the flux Φ. Since speed is inversely proportional to flux, reducing the flux increases the motor speed above its rated value. This is an efficient method but is limited by the magnetic saturation of the core and commutation stability.
  2. Armature Resistance Control Method: This method involves inserting a variable resistor in series with the armature circuit. Increasing the resistance increases the voltage drop (IaRa), which reduces the back EMF (Eb = V - IaRa). As speed is directly proportional to back EMF, the motor speed decreases. This method allows for speed control below the rated speed but is inefficient due to high power loss in the external resistor.
  3. Voltage Control Method (Ward-Leonard System): This method involves supplying the armature of the shunt motor from a separate variable DC voltage source, while the field is kept at a constant excitation. By varying the applied voltage V, the speed can be adjusted smoothly over a wide range from zero to above rated speed. This method provides excellent speed regulation and is used in applications requiring precise control, such as elevators and steel rolling mills, though it is expensive due to the requirement of an additional motor-generator set.
315 marks
a

What do you understand by efficiency of a transformer?

8 marks

Answer

Sign in to view the complete answer paper.

b

Explain the working principle of 1-ϕ Transformer.

7 marks

Answer

Sign in to view the complete answer paper.

4

A single phase, 10 kVA, 11000/220 V, 50 Hz transformer has core loss of 300W at rated voltage and copper loss of 400W at full load.

15 marks
a

Find the efficiency of transformer feeding to a load of 8 kVA at 0.8 p.f. lagging.

8 marks

Answer

Sign in to view the complete answer paper.

b

What will be the maximum efficiency of the transformer.

7 marks

Answer

Sign in to view the complete answer paper.

515 marks
a

Explain working principle of 3-phase induction motor.

9 marks

Answer

Sign in to view the complete answer paper.

b

The frequency of emf in the stator of a 6-pole induction motor is 50Hz, and that of rotor is 1.5Hz. Calculate slip and at what speed will the motor run.

6 marks

Answer

Sign in to view the complete answer paper.

615 marks
a

Derive the expression for Torque-Slip characteristics of a 3-phase Induction Motor.

8 marks

Answer

Sign in to view the complete answer paper.

b

Determine the efficiency of a three phase, 11 kV, 50 Hz, 100 HP, 1440 rpm, induction motor having rotational loss of 2500W and stator copper loss of 2000W.

7 marks

Answer

Sign in to view the complete answer paper.

715 marks
a

Give the constructional details of alternator.

7 marks

Answer

Sign in to view the complete answer paper.

b

Derive EMF Equation of alternator.

8 marks

Answer

Sign in to view the complete answer paper.

8

Explain construction and working of moving iron instruments and derive the torque equation.

15 marks

Answer

Sign in to view the complete answer paper.

About this paper

This page contains the solved question paper with answers of ELECTRICAL TECHNOLOGY from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech ETM II Year I Semester, October - 2020, R16.

For the question paper without answers, switch to the Question Paper tab.

Other papers in ELECTRICAL TECHNOLOGY

Loading Question Papers

Fetching data from database...