Download ET 34 Answers - 15 June 2010 PDF, 126 KB

ET 34 - Electrician Theory Examination
Marking Schedule
Notes:1.
(1 mark) means that the preceding statement/answer earns 1 mark.
2.
This schedule sets out the accepted answers to the examination questions. A marker can
exercise their discretion and decide on the overall accuracy of any answer that is presented in
the candidate’s own words.
3.
Symbols and terms - alternatives
Power
W or P
Voltage
V or E or U
Phase
Active
Marks
Question 1
(a)
Any TWO of:
●
It ensures that the voltage between
P/E and P/N never rises above 230v
under fault conditions.
●
It ensures that between the general
mass of earth and any earthed
metal there will always be 0V
potential.
●
Provides a parallel path in the event
of the loss of the main neutral
●
Provides a low impedance path for
the fast operation of the protection
The power factor of the motor will be
poor
(2 marks)
(c)
Any ONE of:
(2 marks)
●
An open circuit in the L3 winding.
●
One of the links could be missing or
open-circuited.
(i)
Any ONE of:
(ii)
(½ mark)
●
Earth continuity (protective
earthing conductor) test
●
Visual test
●
Any ONE of:
●
Marking notes
(2 marks)
(b)
(d)
Reference
(½ mark)
-
In-line ammeter
-
Clamp-on ammeter
Turn on the appliance and
measure the current in the
protective earthing conductor.
(1 mark)
ET34 ANSWERS – 15 JUNE 2010
1
Marks
Question 1
(e)
(f)
Any TWO of:
(g)
(h)
Marking notes
(2 marks)
●
Lower the line current drawn from
the supply
●
●
Reduce system losses
Reduce voltage drop in lines and
cables
●
When current is flowing there is an
internal volt drop
Due to the impedance of the
windings.
(1 mark)
=
(75 ÷100)2 x 1200
(1 mark)
=
675W
(1 mark)
R
=
●
Reference
V2
(1 mark)
(½ mark)
P
=
230 x 230
(½ mark)
2000
=
P
=
26.45Ω
(½ mark)
2
V
R
=
240 x 240
26.45
=
2180W
(½ mark)
Alternative solution
( V2 / V1) 2 x W
=
( 240 / 230 ) 2 x 2000
=
( 240 / 230 ) 2 x 2000
=
(i)
IPSC
2177.6W
=
233
(1 mark)
0.175Ω
=
(j)
1331 A
(1 mark)
(i)
1MΩ
(1 mark)
(ii)
Any ONE of:
(1 mark)
●
1MΩ
●
0.01MΩ
ET34 ANSWERS – 15 JUNE 2010
2
Marks
Question 2
(a)
To prevent movement if a
prospective short-circuit current
occurs.
(b)
Any TWO of:
high
fault
(2 marks)
Load current
●
Resistivity of the conductors
●
Cross-sectional area
●
Ambient temperature
●
The insulation deteriorates at
temperatures over 75ºC.
Any ONE of:
●
The cable may become a fire
hazard.
●
The cable may become a
shock hazard.
(1 mark)
(d)
The XLPE cable insulation is able to
withstand a much higher continuous
temperature
(2 marks)
(e)
R2
●
=
Marking notes
(2 marks)
●
(c)
Reference
(1 mark)
R1 x L1
L2
=
600 x 140
(½ mark)
450
=
186.67MΩ
(½ mark)
=
186.67 – 140
(½ mark)
=
46.67 MΩ
(½ mark)
ET34 ANSWERS – 15 JUNE 2010
3
Marks
Question 3
(a)
AS/NZS 3000
Reference
Marking notes
(1 mark)
(b)
AS/NZS 3112
IEC 60309
Earth
Phase or
active
Neutral
Phase or
active
Neutral
Earth
(c)
●
Correct polarity on AS/NZS 3112
(1½ marks)
To gain marks
the
polarity
must
be
completely
correct
●
Correct polarity on IEC 60309
(1½ marks)
To gain marks
the
polarity
must
be
completely
correct
(i)
A-N
0V
(1½ marks)
To gain marks
the
polarity
must
be
completely
correct
A-E
0V
N-E
0V
A-N
230V
(1½ marks)
To gain marks
the
polarity
must
be
completely
correct
A-E
230V
N-E
0V
(ii)
ET34 ANSWERS – 15 JUNE 2010
4
Question 3
Marks
(d)
(2 marks)
(e)
Any TWO of:
●
The protective earthing conductor
is open-circuited.
●
There is no earth to the socket
●
The Earth terminal is broken.
Active and neutral are transposed (in the
final subcircuit or socket outlet).
Reference
Marking notes
(1 mark)
ET34 ANSWERS – 15 JUNE 2010
5
Marks
Question 4
(a)
(i)
IPH
=
Reference
Marking notes
(½ mark)
VPH
RPH
=
230
(½ mark)
30
IPH (IL)
=
7.66A
(ii)
=
VPH
IPH
(1 mark)
RPH
=
400
(½ mark)
30
IL
=
13.33A
(1 mark)
=
IPH x √3
(½ mark)
=
13.33 x √3
(½ mark)
=
23.1A
(1 mark)
(iii) Star is the most suitable connection
arrangement.
(b)
(½ mark)
In star
P
=
VL x IL x √3 x pf
(½ mark)
=
400 x 7.66 x √3 x 1
(½ mark)
=
5307W
(1 mark)
In delta
P
=
VL x IL x √3 x pf
=
400 x 23.1 x √3 x 1
=
16003W
Difference in power consumed
=
16003 – 5307
=
10696W
(1 mark)
(½ mark)
(½ mark)
ET34 ANSWERS – 15 JUNE 2010
6
Marks
Question 5
(a)
Reference
Marking notes
Total kW of the installation
=
10+ (12 x 0.25) + 20
(1 mark)
=
33 kW
(1 mark)
(b)
10,000
3,000
0.8 = 36.870
20,000
0.5 = 600
Q1
Q2
QT
S
=
ą x tan ǿ
(½ mark)
=
3000 x tan 36.87
(½ mark)
=
2250 VAr
=
ą x tan ǿ
=
20000 x tan 60
=
34641 VAr
=
34642 + 2250
(½ mark)
=
36891
(½ mark)
=
√ 330002 + 368912
(½ mark)
=
49496.9 VA
(1 mark)
(½ mark)
(1 mark)
(1 mark)
ET34 ANSWERS – 15 JUNE 2010
7
Marks
Question 5
(b)
Reference
Marking notes
Alternative solution
Phase angle of lighting load
=
cos-1 0.8 lag
=
-36.87 degrees
Phase angle of machines load
=
cos-1 0.5 lag
=
-60 degrees
Vertical component of lighting load
=
3kW tan -36.87 degrees
=
2.25KVAr
Vertical component of machines load
=
20kW tan -60 degrees
=
34.64 KVAr
Total load
=
(10kW + j0) + (3kW - j2.25)
+ (20kW - j34.64) KVA
(Rectangular Form.....R +jX )
Total load
=
(33kW- j36.89)
(Rectangular Form Total)
Total load = 49.5KVA
/_ 48.1 degrees
(Polar Form Total)
Alternative solution 2
Lighting
Cos-1 0.8 = 36.86 0
Tan 36.86 0 x 3 Kw
Q1 =
2,250 Kvar
Machines
Cos-1 0.5 = 60 0
Tan 60 0 x 20 Kw.
34.65 Kvar
Q2 =
Qt = Q1 + Q2
=
2,250 + 34.65
=
36 . 9 Kvar
Kw t from (a) = 33 Kw.
S
(c)
I
2
2
=
√P
=
√33
=
49 .5 Kvar
=
S
2
+ Q
+ 36 .9
2
(½ mark)
√3 x 400
=
49496.9
(½ mark)
√3 x 400
=
71.44A
(1 mark)
ET34 ANSWERS – 15 JUNE 2010
8
Question 6
Marks
(a)
(1 mark)
(b)
(i)
Protective earthing conductor test.
Reference
Marking notes
(ii) The multimeter set on the lowest
ohms scale
(½mark)
(iii) Test between the earth connection
of the isolating switch to the
vibrator frame.
(1 mark)
(iv) 1Ω or less
(1 mark)
(i)
Earth leakage test
(1 mark)
With this test,
it is presumed
that
the
isolation
is
removed
and
the
plant
relivened
(ii)
The clamp-type ammeter set on the
lowest scale
(½mark)
To gain this
mark the use of
the
ammeter
has
to
be
qualified.
(iii) ●
●
Start the motor running and
Measure the current in the
protective earthing conductor
(iv) Considerably more than 5 mA
(½mark)
(½mark)
(1 mark)
Alternative solution
(i) Continuity test from each phase to
frame of plant.
(ii)
To gain this
mark the use of
the multimeter
has
to
be
qualified.
Multimeter on low ohms.
(iii) Test resistance from each phase to
the proven earth conductor (or
plant) on both incoming and
outgoing phases of motor starter.
ET34 ANSWERS – 15 JUNE 2010
This
test
“mirrors” an IR
test. That is, it
is seeking to
locate a fault
that would be
more
easily
found by an IR
tester (which is
not available)
9
Marks
Question 6
Reference
Marking notes
(iv) A low enough resistance to indicate
a possible connection of a live
conductor to the frame.
(c)
(d)
●
To ensure the PEC is continuous
(½ mark)
●
And of low resistance
(½ mark)
Any TWO of:
(2 marks)
●
Intermittent open circuit in the
windings on one phase.
●
Intermittent open circuit in one
phase of the starter or isolating
switch contacts.
●
Faulty bearings
●
Intermittent
driven load
mechanical
fault
in
ET34 ANSWERS – 15 JUNE 2010
10
Question 7
(a)
(b)
Any TWO of:
●
Smoother running
●
More even torque
●
Less noise
●
To reduce vibration
●
Increased torque
●
Star/delta
●
Auto transformer
●
Primary resistance
Marks
Reference
Marking notes
(2 marks)
(3 marks)
ET34 ANSWERS – 15 JUNE 2010
11
Marks
Question 7
(c)
Reference
Marking notes
Pull-out torque
or
Break-down torque
Starting
torque
Running torque
Torque
Full-load torque
Speed
Rotor speed - NR
●
Correct axis
(1 mark)
●
Correctly shaped speed torque
characteristic
(1 mark)
●
Starting
identified
correctly
(½ mark)
●
Pull-out or break-down torque
correctly identified
(½ mark)
●
Full-load
identified
torque
correctly
(½ mark)
●
Running
identified
torque
correctly
(½ mark)
●
Rotor speed – NR correctly
identified
(½ mark)
●
Synchronous speed
correctly identified
(½ mark)
torque
–
NS
ET34 ANSWERS – 15 JUNE 2010
Synchronous
speed – NS
The
running
torque can be
shown
anywhere
between
the
break-down
torque
and
synchronous
speed
12
Marks
Question 8
Reference
Marking notes
(a)
N L L L
E
Tripping Device
Test
button
Sensing
coil/toroid
Three-phase socket
(b)
(1½ marks) ●
●
(½ mark)
●
●
Correctly connected phases, neutral
and earth.
●
Correctly connected test button and
resistance
●
(½ mark)
●
Correctly
connected
sensing
coil/toroid
Correctly connected tripping circuit
●
Correctly connected socket.
(½ mark)
●
Working circuit
●
There is an imbalance between the
neutral current and the sum of the
phase currents
(1 mark)
●
A magnetic field is induced into the
iron core.
(1 mark)
●
(½ mark)
(1½ marks) ●
●
ET34 ANSWERS – 15 JUNE 2010
A short circuit
No test resistance in the test
circuit.
The socket outlet is not
earthed
The active of the test button
circuit is connected to the
supply side of the main
contacts
The neutral is not switched
No neutral conductor shown
13
Marks
Question 8
(c)
●
The induced magnetic field induces
a current in the sensing coil
(1 mark)
●
The tripping coil is energised,
opening the RCD contacts
(1 mark)
Any TWO of:
●
Earth leakage
●
Overload
●
Short circuit
Reference
Marking notes
(1 mark)
ET34 ANSWERS – 15 JUNE 2010
14
Marks
Question 9
(a)
(i)
NS
=
Reference
Marking notes
(½ mark)
NP x VS
VP
=
2000 x 100
(½ mark)
230
(ii)
IP
=
870 turns
=
IS x VS
(1 mark)
(½ mark)
VP
=
15 x 100
(½ mark)
230
=
6.52A
(1 mark)
(iii) Natural air convection
(b)
(1 mark)
●
When the load is increased, the
secondary current (by Lenz’s law)
produces a greater demagnetising
force.
(1½ marks)
●
This reduces the primary flux,
reducing the back emf induced in
the primary
(1 mark)
●
When the back emf in the primary
reduces, the supply voltage is
increased
More current flows in the primary
(1½ marks)
●
(1 mark)
ET34 ANSWERS – 15 JUNE 2010
15