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Transcript of N3 Electrotechnology April 2016 Memorandum › download › userupload › Lecturer … · 3.1...
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NATIONAL CERTIFICATE
APRIL EXAMINATION
ELECTRO-TECHNOLOGY N3
5 APRIL 2016
This marking guideline consists of 11 pages.
MARKING GUIDELINE
MARKING GUIDELINE -2- T500(E)(A5)T ELECTRO-TECHNOLOGY N3
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QUESTION 1 1.1 1.1.1 Yoke P (2) 1.1.2 Pole shoes P (2) 1.2 1.2.1 • By moving brushes backwards in the motor. P
• By moving brushes forwards in the generator. P
(2)
1.2.2 • Interpoles are smaller poles placed between the main poles. P • Connected in series with the armature and must have the same
polarity as the main poles - passed in the motor – to ensure sparkless commutation. P
(2)
1.2.3 • By making use of series winding on the main field poles. P • Varying the main field to the load condition. P
(2)
[10] QUESTION 2 2.1 • The number of pairs of poles used. P
• The strength of the magnetic field or flux. P • The rate at which the magnetic flux is cut by the moving conductor. P • The number of active conductors connected in series. P
(4)
2.2 Separately excited generator. P (1)
P P
For correct labelled sketch
(2) Self-excited generator P (1)
P P
For correct labelled sketch
(2) [10]
MARKING GUIDELINE -3- T500(E)(A5)T ELECTRO-TECHNOLOGY N3
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QUESTION 3 3.1 • Effective field flux is reduced due to armature reaction as the load
increases. P • Voltage drop due to the armature circuit resistance. P
(2) 3.2 Used as the generator in the Ward-Leonard motor generator system. P (1) 3.3 3.3.1 Shunt generator – it is used where a constant voltage is required.
P
(1) 3.3.2 Series generator – as a booster on DC transmission line. P (1) 3.4 • Flux (Ф) P
• Armature current (Ia) P
(2) 3.5 • Driving crane P
• Train P • Hoists P • Lifts • Trolley buses • Electric vehicle (Any 3 x 1 )
(3) [10]
MARKING GUIDELINE -4- T500(E)(A5)T ELECTRO-TECHNOLOGY N3
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QUESTION 4 Given: R = 300 mm = 0,3 m; N = 420 r/min; effective load = 425 N.m V = 0,21 kV = 210 V; I = 33 000 m A = 33A
4.1 Input Power [P] = IV
= 33 A x 210 VP = 6 930 WP = 6,93 kWP Answer
(3) 4.2 Output Power [P] =
= PP
= 5608,47 WP = 5,609 kW PAnswer
(4) 4.3 Efficiency =
= P
= 0, 80938 x 100%P = 80,938 % P Answer
(3) [10]
602 NWrx P
603,0425420142,32 ´´´´
%100´InputOutput
%10093,6609,5
´kWkW
MARKING GUIDELINE -5- T500(E)(A5)T ELECTRO-TECHNOLOGY N3
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QUESTION 5 Given: :- e = 20 V; e = 42 V; e = 83 V; e = 120 V; e = 95 V; e = 35 V; e = 18 V. 5.1 5.1.1 Alternating voltage. P (1) 5.1.2
Actual [E ] =
= P
= P
= = 69.721 VP Answer
(3) 5.1.3 Average Value [E ] =
= P
= P = 59 VP Answer
(3) 5.1.4 Form factor= =
= 1,182 Answer P
(1) 5.1.5 • Sine wave P
• Peak wave • Sinusoidal wave (Any 1 x 1)
(1) 5.2 Maximum value – is the maximum or peak value of an alternating voltage or
current. P
(1) [10]
1 2 3 4 5 6 7
RMS neeeeeee 27
26
25
24
23
22
21 ++++++
7183595120834220 2222222 ++++++
702734
8614
AVE 77654321 eeeeeee ++++++
( )7
183595120834220 ++++++
7413
AVE
RMSE
EV
V59
721,69
MARKING GUIDELINE -6- T500(E)(A5)T ELECTRO-TECHNOLOGY N3
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QUESTION 6
6.1 X X
= 2 x 3,142 x 60 Hz x 400 x 10 =
= 150,816 ΩP Answer = 53,045 Ω P Answer
Impedance of the circuit [Z] =
= P = P = 98,281 ΩP Answer
(5) 6.2
Circuit current [I ] =
= P
= 2,442 A P Answer
(2) 6.3 Phase angle: Tan θ =
= P
θ = Tan 9,777P = 84,160º laggingP Answer
(3) [10]
fLL P= 2fCC P
=21
3-6105060142,32
1-´´´´
22 )( CL XXR -+22 )045,53816,150(10 -+
168.9559100 +
t ZVt
W281,98240V
RXX CL -
10045,53816,150 -
1-
MARKING GUIDELINE -7- T500(E)(A5)T ELECTRO-TECHNOLOGY N3
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QUESTION 7 7.1 • With a star connection two voltages are available, namely V and V .P
• By earthing the neutral, earth leakage protection is simplified. P
(2) 7.2 7.2.1 V = 380 V P. (Given) (1) 7.2.2 V =
V = P
= 219,393 VP Answer
(2) 7.2.3 Input power =
= P
= 29 411,765 W P Answer Pin =
I = P
= 55,858 A Answer P Therefore I (In star connection) I = 55.858 A P Answer
(5) [10]
L ph
L
L PHV3
ph 3380V
100´h
Output
1008500025
´
qCosIV LL ´´´3
L 8.03803765,41129´´W
phC I=
ph
MARKING GUIDELINE -8- T500(E)(A5)T ELECTRO-TECHNOLOGY N3
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QUESTION 8 8.1 Pink P (1) 8.2 • Winding P
• Core P
(2) 8.3 8.3.1 Primary number of turns[N ] = 42 P
= 126 turns P Answer
(2)
8.3.2
I P
= 0,654 AP Answer
(2)
8.3.3 Secondary volt-ampere = VI
= 70 Volt x 0,654 A = 45,78 VAP Answer
(1)
[8]
1 VV
70210
´
1
2
2
1
II
VV
=
VAV
70218,0210
2´
=
MARKING GUIDELINE -9- T500(E)(A5)T ELECTRO-TECHNOLOGY N3
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QUESTION 9 9.1 FOUR marks for ANY RELEVANT correct labelling
THREE marks for correct sketch.
(7) 9.2 • A deflecting device P
• A controlling device P • A damping device P
(3) [10]
MARKING GUIDELINE -10- T500(E)(A5)T ELECTRO-TECHNOLOGY N3
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QUESTION 10 10.1 10.1.1
PP for correct diagram and labelling
(2)
10.1.2
P P for correct diagram and labelling
(2)
10.2 10.2.1
remainder 0 0,5 x 2 = 1 remainder 1 remainder 0 P remainder 1
10,5 = 1010,1 P Answer
2 10 2 5 2 2 2 1
(2)
10.2.2
remainder 0 0,25 x 2 = 0 remainder 1 0,5 x 2 = 1 remainder 1 remainder 1 P
14,25 = 1110,01 P Answer
2 14 2 7 2 3 2 1
(2)
10.2.3 -1010,10
1110,01 0011,11 PP Answer
(2)
10 2
10 2
2
MARKING GUIDELINE -11- T500(E)(A5)T ELECTRO-TECHNOLOGY N3
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10.3
PP
1 for correct sketch and 1 for labelling
(2) [12]
TOTAL: 100
