JNTU Previous Papers-Electrical & Electronic Engineering-Electrical Machines-I-Nov-2008
JNTU II B.Tech I Semester Regular Examinations, November 2008
(Electrical & Electronic Engineering)
1. (a) Describe the principle of energy conversion. From a consideration of the various energies involved, develop the model of an electro mechanical conversion device.
(b) Show that the torque developed in a doubly excited magnetic system is equal to the rate of increase of field energy with respect to the displacement at constant currents .
2. (a) Distinguish between lap and wave windings.
(b) The armature of a 2 pole, 200V generator has 400 conductors and runs at 300
rpm. Calculate the useful flux or pole. If the number of turns in each field coil is 1200, what is the average value of the emf induced in each coil on breaking the field, if the flux dies away completely in 0.1s.
3. (a) How are demagnetizing and cross magnetizing ampere-turns/pole in a D.C Machines calculated?
(b) Determine AT/pole for each interpole of a 4 pole generator with 88 slots each containing 900 amp – conductors. The interpole air gap is 0.01 m and flux density in the interpole air gap is 0.3 T . The effects of iron parts of iron parts of the circuits and leakage may neglected.
4. (a) How are the series and shunt windings arranged on the pole of a dc compound
(b) How will you distinguish between series and shunt windings of a dc compound machine?
5. Two short-shunt compound generators A and B running in parallel supply a load current of 140A at a terminal voltage 100 V. An equalizig bar connects the two machines. The data regarding the machines are:
Generator A: Ra=0.02 ohm; Rsh=80ohm; Rse=0.02 ohm.
Generator B: Ra=0.05 ohm; Rsh=100 ohm; Rse=0.05 ohm; e.m.f generator B, 105
(a) current in series windings
(b) armature currents
(c) current in equalizer
(d) e.m.f generated by generator A.
6. (a) Explain the basic performance equations for a dc motor.
(b) A 4-pole, 250 V series motor has a wave-connected armature with 1254 conductors. The flux per pole is 22 mWb when the motor is taking 50 A. Iron and friction losses amount to 1.0 Kw. Armature resistance is 0.2 ohm andseries field resistance is 0.2 ohm. Calculate:
i. the speed
ii. the BHP
iii. the shaft torque and
iv. the efficiency at this load.
7. (a) Draw and explain the speed-torque characteristics and torque characteristics of dc shunt motor.
(b) A 240V, 50A,800 rpm dc shunt motor has armature circuit resistance of 0.2. If load torque is reduced to 60% of its full-load value and a resistance of 2 is inserted in series with armature circuit, find the motor speed. Armature reaction weakens the field flux by 4% at full load and by 2% at 60% of full load.
8. A 10kW 900 rpm, 400V dc shunt motor has armature circuit resistance (including
brushes) of 1 and shunt field resistance of 400. If efficiency at rated load is 85%, then calculate:
(a) The no-laoad armature current,
(b) The speed when motor draws 20A from the mains and
(c) The armature, eurrent, when the total(or internal) torque developed is 98.5
Assume the flux remain constant.