Download EELEC Bachelor of Engineering in Electrical Engineering Exam - Summer 2008 and more Exams Electrical Engineering in PDF only on Docsity! Cork Institute of Technology Bachelor of Engineering in Electrical Engineering - Award (EELEC_7_Y3) Summer 2008 Electrical Engineering (Time: 3 Hours) Instructions Answer any Five questions. All questions carry equal marks. The use of non-programmable calculator is permitted. In marking scripts, the examiner will take account of clarity of expression and communication skills. Examiners: Mr. M. Hennessy Prof. E. McQuade Mr. Joe Buckley Q1. (a) “All electrical equipment using alternating current is designed to use a voltage with a clean and regular sine wave. However, in present day networks, this type of curve is extremely rare. Harmonic frequencies create distortions in the sine wave, causing interference to equipment connected to the network.”. Discuss the above in the context of the most common harmonics which stress networks i.e. the third harmonic, the fifth harmonic and the seventh harmonic. (10 marks) (b) An e.m.f. is represented by e = 100 sin θ + 30 sin (3θ + 20°) + 10 sin (5θ + 10°). The fundamental frequency is 50 Hz. (i) Determine the r.m.s. value of the current in a circuit in which R = 5Ω, L = 0.02H. (5 marks) (ii) Derive the equation for the current wave. (5 marks) Q2. (a) Compare and contrast squirrel cage motors and slipring motors currently being used in industry under the following headings: (i) starting torque (ii) stall torque (iii) speed control (iv) industrial applications (v) maintenance. (10 marks) (b) A 200kW, 400V, 4 pole, 3 phase 50Hz slipring induction motor has a slip of 4% and the starting torque is equal to the full load torque. Calculate: (i) the starting torque if the supply voltage falls to 380V (ii) the slip when developing full load torque at this voltage. (10 marks) Q3. A switchboard is fed via a 1500 kVA distribution transformer as shown in Fig. 1 attached. (a) Calculate the maximum short circuit current that can arise in the system. (6 marks) (b) Calculate the short circuit current that would arise in the event of a short circuit occurring at the entry to the cable box of the 250kW squirrel cage motor. You may assume: (i) the primary voltage is 12kV, 3-phase, 50Hz (ii) the secondary voltage is 500V, 3-phase, 50Hz (iii) the percentage impedance of the transformer is 5% (iv) the impedance of the 800 A switchfuse is (0.651 x 10-3 + j0) Ω (v) the impedance of the 350 A starter is (1.341 x 10-3 + j0) Ω (vi) the impedance of the 2000A ACB is (0.032 x 10-3 + j0) Ω (14 marks) Figued i
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Current ratings.
The PEX cable ca" continacnsty ba loaded to
a conductor teenparature of BIS 6
there Is a rssh af drying Out the parroune
may rest © tharnal instebeity. For gucunce curred
fatings for s conductor temperature of BS°C ew
an as weil Thee GurTent ratings ate cakculsted
according to IEC, Publ 287, ard the tollowing
coraitions
One multi-core codls crane thea phase group »
of engl care ostles.
Temperature of ground roe
Temperature of ambent or 25
1-24 KVcables orm
S245 KY cables om
waidin
O"C Mw
— Correct ratings for single-core PEX cables, A.
Rated voltage 12-24 KV. Aluminium conmuctors:
fang gran
Avhene ee
Benepe teres at
_megeaaat_|
lal estes
I
hee
Vihen the conditions differ fram these, the
actual current ratings are obtainad fy multiplying
from tables 7-13 by raling factors trom tables
see example an page 2.
Screens bonded ata srgle poirt
reens are connected ard earth
ord of the cable route. Scroors bondae
moans that the screens are
at both ends of the cabie saute.
nears that
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aaiGiavecae
a eit»
Rating factors for cables in ground
Currgnt ratings in tabies 7-13 are vabd at
conditions spactiad on p. 26 Atother condhines
detual current ratings are oblained by multiptying
Johoas from taties 7-13 by rating factors from
tables 14-20,
Example 00 using rating factors.
12 KY PEX cable with duminum conductor
HOv35,
iim? in graund ih trefeod formation with scroees bonded
at both ee
14 Rating factors for depth of layin.
(eae simran
o agra
BOC conductar Orrent
temperature rang6ssA (ible)
Laying dapin OF m Rating tector 10 (lable 14)
Ground temp, 29°C Hating factor 22 {table 15)
Trsetnat rebistivity
ground toms
Two paraileé groups.
0.07 mapacra Rating tactor 0.85 (tate 17)
Reaut: The current rating at SO"C conductor ternmera-
ture for each group in the cable route:
B55 Ax 1 OXOBI*084x0.85 "4358
Rating factor 0.84 [table 16)
biter
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os Rating factors for ground twmperatu
(Prem hs Wah oes -
wespraese
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16 Rating tactors for thermal resistivity of around. .
a aE Le haa Se, eC
veo [oss | ome am 08
7 snap rating tactors tor muti-core PEX cables in fat formation in round. _
yo pees Ta sicathse pew sea s
ap tases oe
are aa ost | (am of Sa
ar | ea fae cw are 253
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qire
nirgin-core cables in the same trecch
18 Rating factors for cables In pipes.
The current ratings in tabies 7-13 tor coblest laid
in ground shal de reduced by the rating tactor 08D
Hf the coble # lac in an untilsd pipe of PYE or PE in
ground. Pips diameter is asaumed 2-5 tenes that ot
the cable. Its howsnnr poasinks to eimmate tis
nstierg (6OUc70n by filing Ihe ppo otter the cable has
4 used, factors accarcing to table 17 are not applicable
I factors in thia table
Alors @ivan in the above table can alto be anplod to weewea! cours of threw
bean pulled in, with « material which os thermaby oqusl
to the ambiont ground,
‘ot paralel piper exch containing a three-cara
ois of # eu OT three ziyga-core cables and with
the cables equaly o2ded, the curmat ratings are de
crpased by using the following rating ta:
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of ow | Om oss |
aro | o@ | ce | ob