# Cochin University Previous Year Question Papers

# BE EE Electrical Machines Nov 2008

# 5th Semester

**EE 504 Electrical Machines II**

**(2002 Scheme)**

(All questions carry EQUAL marks)

I. (a) Derive the emf equation of an alternator and explain coil span factor and distribution factor.

(b) What are the advantages of stationary armature and rotating field in an alternator?

**OR**

II. (a) Distinguish between cylindrical rotor and salient pole synchronous machines.

(b) A3 phase star connected alternator has the following data: voltage required to be generated on open circuit = 4000 V; speed = 500 rpm; slots/pole/phase = 3; conductors/slot = 12. Calculate (i) Number of poles and (ii) useful flux/pole.

III. (a) Explain synchronous impedance method of determining regulation of alternator.

(b) A 3300 V, 3 phase star connected alternator has a full load current of 100 A. On short circuit a field current of 5 amperes was necessary to produce full load current. The emf on open circuit for the same excitation was 900 volts. The armature resistance is

0.8 Q /phase. Determine the full load voltage regulation for :

(i) 0.8 pf lagging (ii) 0.8 pf leading.

**OR**

IV. (a) Describe the slip test method for the measurement of Xd and Xq of synchronous machines.

(b) A 3 phase star connected salient pole alternator is driven at a speed near synchronous with field circuit open, and the stator is supplied from a balanced 3 — phase supply. Voltmeter connected across the line gave minimum and maximum readings of 2800 and 2820 volts. The line current fluctuated between 360 and 275 amperes. Find the direct and quadrature axis synchronous reactances per phase. Neglect armature resistance.

V. (a) What are the advantages of connecting alternators in parallel? What are the conditions to be fulfilled for the parallel operation of alternators,?

(b) Two 3 phase alternators operate in parallel on the same load. Determine the KW output and power factor of each machine under the following conditions. Synchronous impedance of each generator0.2 + y‘2Q/ phase, impedance of load : 3 + _/4Q./ phase. Induced emf/ phase : 2000 + j o volts for machine 1 and 2200 + j\ 00 for machine 2.

**OR**

VI. (a) Why synchronous motor is not self starting? Explain the starting methods.

(b) A 6600 V, 3 phase star connected synchronous motor draws a full load current of 80 A at 0.8 pf leading. The armature resistance is 2.2Q and reactance 22Q per phase. If the stray losses of the machine are 3200 W, find : (i) emf induced, (ii) output power and (iii) efficiency of the machine.

VII. (a) Explain the effect of varying the excitation of a synchronous generator connected to infinite bus on the power factor, armature current and load angle.

(b) A 3 phase 11 KV, 10 MV, star connected synchronous generator has a synchronous impedance of (0.8 + j 8.0) Ohms/phase. If the excitation is such that the open circuit voltage is 14 KV, determine (i) the maximum output of the generator (ii) the current and power factor at the maximum output.

VIII. (a) What are V – curves? How are they determined experimentally?

(b)An alternator supplying 600 KW at 0.8 p.f lagging has its power factor raised to unity by means of an over excited synchronous motor. With the armature current unchanged, how much power can the alternator thus supply to the synchronous motor and what power can the latter develop when operating at an efficiency of 86%.

IX. (a) Find the power factor at which synchronous motor runs.

(b) Discuss the phenomenon of sudden 3 – phase short circuit at the armature terminals of an alternator. Draw a typical wave shape of current and mark the different regions.

X. (a) What is meant by hunting in synchronous machines? How this can be avoided?

(b) Derive an expression for the natural frequency of undamped oscillation.

A 10000 KVA, 4 pole, 6600 V, 50 Hz, 3 phase star connected alternator has a synchronous reactance of 25% and operates on constant voltage, constant frequency bus bars. If the natural time period of oscillation while operating at full load and unity power factor is to be limited to 1.5 seconds, calculate the moment of inertia of the rotating system.