Jadavpur University 2007 B.E Mechanical Engineering ELECTRICAL TECHNOLOGY-II - Question Paper

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FIRST ENGG. (Mechanical) EXAMINATION, 2007
(2nd Semester, Old Syllabus)
ELECTRICAL TECHNOLOGY–II
Time : 3 hours Full Marks : 100
(50 marks for every part)
Use a separate Answer-Script for every part.
PART–I
ans ques. No. one and 2 others from the rest.
1. a) describe Breadth Factor of an armature winding of a
Synchronous Generator. Derive the identical and discuss
its effect on the voltage induced. 4
b) Show that if a 3 phase balanced current is passed
through a 3 phase balanced winding a rotating
magnetic field of constant magnitude is established. 6
c) In an alternator the regulation was obtained to be negative.
What is your inference on the Power Factor of the
alternator. discuss with a phasor diagram. 4
d) Even with reduction / increase in excitation current of a
synchronous machine the armature current is larger
than its minimum value — discuss. 4
2. a) elaborate the conditions to be met before a 3 phase
alternator is paralleled to a 3 phase system. 4
Ex/ML12E/143/07(OLD)
7. a) Show that the balanced 3-phase lower consumed by
balanced 3-phase load can be measured by two
wattmeters only. 10
b) Also show that the nature of the load can be determined
from the readings of the 2 wattmeters. 6
8. a) Deduce the relationship ranging from phase and line
volumes of star and delta connected balanced 3-phase
system. 10
b) A 3-phase delta connected a.c. motor, when connected
to a 3-phase 440 V, 50 Hz system is developing 30 KW
at an efficiency of 86 percent and a p.f. of 0.83.
compute (i) the line current and (ii) the phase current.
6
9. a) define briefly the different methods of starting 3-
phase induction motors. 10
b) An 8-pole alternator runs at 750 r.p.m. It supplies power
to a 6-pole, 3-phase induction motor, which has a full
load slip of three percent. obtain the full load speed of the
induction motor. 6
––––––––×––––––––
( four )
b) 2 identical 3 phase alternators are coupled in
parallel to a total load of 1500 KW at 11 KV and 0.8
P.F. lagging. The per phase reactance 60O and
resistance 2.8O of the Synchronous machines. The
power supplied by every machine is maintained the
same. They excitation of the 1st alternator is adjusted
so that the armature current is 45A lagging.
compute :
(i) The armature current of the 2nd Alternator
(ii) The P.F. at which every operate
(iii) The e.m.f. of 1st alternator. 12
3. a) discuss the starting difficulties in a Synchronous motor.
How it can be overcome. 4
What is the nature of speed vs load current of a
Synchronous motor — sketch it. 2
b) A 100 h.p. 400 V three phase star connected Synchronous
motor resistance and Synchronous reactance ranging from
the lines 0.08O and 0.8O respectively. compute for full
load at 0.8 P.F. leading the open circuit e.m.f. per phase
and mechanical power developed. Draw the phasor
diagram. 10
4. a) A D.C. feeder is uniformly and is fed from 1 end and
a different identical feeder is fed from both ends at identical
potential.
i = current tapped off and r = resistance both per unit
length. l = total length of the feeder. Show that in first
case voltage drop at the furthest pt is four times the
voltage drop at mid point in 2nd case. 10
b) What is the advantages of D.C. supply over A.C. one.
6
PART–II
ans any 3 ques..
Two marks allotted for neatness.
5. a) define the open circuit test and short circuit test of
a 1-phase transformer with the help of circuit diagrams,
showing the connection of different meters used for
these tests. Draw the equivalent circuits. 12
b) What info we can get from the above 2 tests ?
4
6. a) describe voltage regulation of a transformer. obtain the
condition under which maximum voltage regulation
occurs. 8
b) A 6600/440 V, 1-phase transformer has an equivalent
resistance of 0.02 p.u. and an equivalent reactance
of 0.05 p.u. obtain the full-load voltage regulation of
the transformer at 0.8 p.f. lag, if the primary voltage is
6600 V. obtain also the secondary terminal voltage at
full load. 8
( two ) ( three )
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Earning:   Approval pending.