MECHANICAL TECHNOLOGY
GRADE 12
MEMORANDUM
NATIONAL SENIOR CERTIFICATE
NOVEMBER 2016
QUESTION 1: MULTIPLE-CHOICE QUESTIONS
1.1 A✓ (1)
1.2 A✓ (1)
1.3 C✓ (1)
1.4 C✓ (1)
1.5 B✓ (1)
1.6 C✓ (1)
1.7 D✓ (1)
1.8 C✓ (1)
1.9 A✓ (1)
1.10 C✓ (1)
1.11 C✓ (1)
1.12 B✓ (1)
1.13 C✓ (1)
1.14 D✓ (1)
1.15 C✓ (1)
1.16 C✓ (1)
1.17 C✓ (1)
1.18 D✓ (1)
1.19 A✓ (1)
1.20 C✓ (1)
[20]
QUESTION 2: SAFETY
2.1 Surface Grinder
2.2 Arc welding
2.3 Bearing puller
2.4 3 mm✓ (1)
2.5 Rockwell tester
[10]
QUESTION 3: TOOLS AND EQUIPMENT
3.1 Beam Bending Test
3.2 CO reading (Gas Analyser)
3.3 Advantages of MIG/MAGS welding
3.4 Multimeter
To test:
3.5 Compression tester:
3.5.1 High tension lead
3.5.2 Fuel injectors disconnected
3.5.3 Throttle valve fully open
3.5.4 Recording the readings
[12]
QUESTION 4: MATERIALS
4.1 Characteristics of cementite
4.2 Case hardening on camshaft
4.3 Tempering hardened steel
4.4 Iron carbon equilibrium diagram
(7)
[13]
QUESTION 5: TERMINOLOGY
5.1 Gear calculation
5.1.1
Module = PCD
T
=90 ✓
30
= 3 ✓
(2)
5.1.2
OD =m(T+2)
= 3(90+2) ✓
= 276 mm ✓
(2)
5.1.3
PCD = m x T
= 3 x 90
= 270 mm
(2)
5.1.4
Dedenum = 1,25 x m or 1,157 x m
= 1,25 x 3 or 1,157 x 3 ✓
= 3,75 mm or 2,471mm ✓
(2)
5.1.5
Y = PCD A + PCDB ✓
2 2
= 270 + 90 ✓
2 2
= 135 + 45
= 180mm ✓
(3)
5.1.6
Indexing = 40
n
=40 ✓
33
=1 7 x 2
33 2
= 1 44 ✓
66
Indexing: = one full turn of the crank and 14 holes in a 66 hole plate✓ (3)
5.2 Key calculations
5.2.1
Width of key = Diameter
4
= 92 ✓
4
=23 mm ✓
(2)
5.2.2
Length of key = 1,5 x Diameter
= 1,5 x 92 ✓
= 15,33 mm ✓
(2)
5.2.3
Thickness of key = Diameter
6
= 92 ✓
6
= 15,33 mm ✓ (2)
5.2.4
T2 = T1 - 138 ✓✓
100
= 15,33 - 1,38 ✓
= 13,95 mm ✓
(4)
5.3 Milling Machine components
5.3.1 Functions of the index plate
5.3.2 Function of the sector arms
5.4 External V-screw thread
[30]
QUESTION 6: JOINING METHODS
6.1 Causes of welding defects
6.1.1 Undercutting
6.1.2 Lack of fusion
6.2 Ultrasonic test
6.3 Ultrasonic vs X-ray testing
6.4 X-ray test on welded joint
6.5 Dye penetrant test
6.6 Destructive tests
6.7 MIG/MAGS shielding gases
6.8 MIG/MAGS welding
[25]
QUESTION 7: FORCES
7.1 Resultant
∑HC = 300cos45° - 200cos25°+185✓✓✓
= 212,13 - 181,26 + 185
= 215,87N✓
∑VC = 200sin25° - 100 + 300sin45° ✓✓✓
= 84,52 - 100 + 212,13
= 196,65N ✓
Horizontal component | Magnitude | Vertical component | Magnitude |
-200Cos25° ✓ | - 181,26 N | 200Sin25°✓ | 84,52 N |
300Cos45°✓ | 212,13 N | 300Sin45°✓ | 212,13 N |
185✓ | 185 N | 0 | 0 N |
0 | 0 N | -100 | - 100 N |
TOTAL | 215,87 N ✓ | TOTAL | 196,65 N✓ |
R2 = HC2 +VC2
R = √215,872 + 196,652 ✓
R = 292,01N ✓
Tanθ = VC ✓
HC
= 196,65 ✓
215,87
θ = 42,33°
R = 292,01 N at 42,33° (42°20') north of east ✓
OR bearing 47,67° (47°40') East
7.2 Stress and Strain
7.2.1
(5)
7.2.2 Strain
ε = σ ✓
E
ε = 20 x 106 ✓
90 x 109
= 0,22 x 10-3 ✓ (3)
7.2.3 Change in length
ε = Δl ✓
ol
Δl = ε x ol
= (0,22 x 10-3) x 0,3✓
= 0,07 x 10-3m
= 0,07 mm✓
(3)
7.3 Moments
Calculate A: Moments about B
ΣRHM = ΣLHM
(A x 8) + (300 x 2) = (550 x 6) + (800 x 3) ✓✓
8A = 3300 + 2400 - 600
8A = 8100
8 8
A = 637,5 N ✓
Calculate B: Moments about A
ΣLHM = ΣRHM
(B x 8) = (550 x 2) + (800 x 5) + (300 x 10) ✓✓
8B = 1100 + 4000 + 3000
8B = 8100
8 8
B = 1012,5 N ✓
ΣUpward = ΣDownward
637,5 + B = 300 + 550 + 800
B = 300 + 550 + 800 - 637,5
B = 1012,5N
(6)
[30]
QUESTION 8: MAINTENANCE
8.1 Effects – routine maintenance
8.2 Preventive maintenance
8.3 Properties of oil
8.3.1 Pour point
Lowest temperature ✓at which a liquid remains pourable. ✓ (2)
8.3.2 Flashpoint
Lowest temperature✓ at which the oil gives off vapour which can ignite.✓(2)
8.4 Belt drive maintenance
8.5 Cutting fluid
8.6 Chain-drive maintenance
By doing the following regularly:
[15]
QUESTION 9: SYSTEMS AND CONTROL
9.1 Belt Drives
9.1.1 Rotation frequency of the driver pulley
V = πDn
n = V
πD
= 36
π(0,23)
= 49,82 r.s-1 x 60
= 2989,35r/min✓✓✓✓
(4)
9.1.2 Power transmitted
T1= 2,5
T2
T1 = 2,5 x T2
= 2,5 x 140
= 350 N
P = (T1 - T2)V
P= (350 - 140)x36
= 7560 Watt or 7,56 kW✓✓✓✓
(4)
9.2 Gear drives
9.2.1 Rotation frequency of the output shaft✓✓✓
(3)
9.2.2 Velocity Ratio
VR = NINPUT
NOUTPUT
= 1380
160
= 8,625 : 1
= 8,63:1✓✓
(2)
9.3 Hydraulics
9.3.1 Fluid pressure
AA = πDA2
4
= π0.042
4
= 1,26 x 10-3m2
PA = FA
AA
= 275 Pa
1,2566 x 10-3
= 218844 Pa
= 218,84 kPa✓✓✓✓
(4)
9.3.2 Load on piston B in kg
AB = πD2
4
= π0.0752
4
= 4,42 x 10-3
PB = FB
AB
FB = PB x AB
= (218,85 x 103) x (4,42 x 10-3)
= 967,32N
Mass = 967,32N
10
= 96,73kg✓✓✓✓
(4)
9.4 Traction control
Traction control system prevents ✓ the wheels from spinning. ✓ (2)
9.5 Air bags
[25]
QUESTION 10: TURBINES
10.1 Supercharger
10.2 Supercharger function
10.3 Advantages of supercharger
10.4 Turbocharger
By the exhaust gases. ✓ (1)
10.5 Runaway speed
The runaway speed of a water turbine is its speed at full flow ✓ and with no shaft load. ✓ (2)
10.6 Water turbines
10.6.1 Reaction principle ✓✓
OR
Francis ✓✓
OR
Description of operation:
Reaction turbines are acted on by water which changes pressure as it moves through the turbine and gives up its energy. The turbine must be encased to contain the water pressure or suction or the must be fully submerged in the water flow. ✓✓
Any 1 x 2 (2)
10.6.2 Impulse principle ✓✓
OR
Pelton ✓✓
OR
Description of operation:
The impulse turbine changes the velocity of the water jet. The jet pushes on the turbine’s curved blades which changes the direction of the flow. The resulting change in momentum (impulse) causes a force on the turbine blades. ✓✓
Any 1 x 2 (2)
10.6.3 Reaction principle ✓✓
OR
Kaplan ✓✓
OR
Description of operation:
Reaction turbines are acted on by water which changes pressure as it moves through the turbine and gives up its energy. The turbine must be encased to contain the water pressure or suction or the must be fully submerged in the water flow. ✓✓
Any 1 x 2 (2)
10.7 Gas turbine
[20]
TOTAL: 200