MECHANICAL TECHNOLOGY
GRADE 12
MEMORANDUM
NOVEMBER 2017
NATIONAL SENIOR CERTIFICATE
QUESTION 1: MULTIPLE-CHOICE QUESTIONS
1.1 B ✓ (1)
1.2 C ✓ (1)
1.3 D ✓ (1)
1.4 B ✓ (1)
1.5 C ✓ (1)
1.6 D ✓ (1)
1.7 D ✓ (1)
1.8 C ✓ (1)
1.9 B ✓ (1)
1.10 B ✓ (1)
1.11 A ✓ (1)
1.12 C ✓ (1)
1.13 B ✓ (1)
1.14 B ✓ (1)
1.15 B ✓ (1)
1.16 D ✓ (1)
1.17 B✓ (1)
1.18 B ✓ (1)
1.19 B ✓ (1)
1.20 A ✓ (1)
[20]
QUESTION 2: SAFETY
2.1 Surface grinder:
2.2 Hydraulic press:
2.3 MIG/MAGS welding:
2.4 Spring compressor:
[10]
QUESTION 3: TOOLS AND EQUIPMENT
3.1 Volt and ammeter:
3.2 Uses of the multimeter:
3.3 Compression Test:
3.4 Tests:
3.4.1 A beam bending test is to investigate the deflection / bend ✓✓ of beams. (2)
3.4.2 A cylinder leakage tester is to check whether gases or air leaks ✓✓ from the cylinders / valve leak. (2)
[12]
QUESTION 4: MATERIALS
4.1 Properties of structures:
4.1.1 Cementite: hard ✓ and brittle ✓ (2)
4.1.2 Ferrite: soft ✓ and ductile ✓ (2)
4.2 Heating process of carbon steel:
4.2.1 Iron-Carbon ✓ Equilibrium ✓ Diagram (2)
4.2.2
4.2.3 700 – 800 °C ✓✓ (2)
[13]
QUESTION 5: TERMINOLOGY
5.1 Screw thread terms:
5.1.3: NOTE: Any other corresponding point on the screw thread (5)
5.2 Milling processes:
5.2.1 Up-cut milling ✓ (1)
5.2.2 Down-cut milling ✓ (1)
5.3 Indexing:
Indexing = 40
A
= 40
22
= 1 18 x 3
22 3
= 1 54
66
1 full turn and 54 holes on the 66-hole circle (6)✓✓✓✓✓✓
5.4 Dividing head:
5.4.1 The sector arm save time and removes the possibility of error in counting the number of holes for each move of the index pin. ✓ (2)
5.4.2 The index plate is equipped with accurate spaced holes on different-diameter circles. Each circle has a different number of holes. These circles allow the crank handle to be given an accurate part of a turn to obtain the desired spacing. ✓✓ (2)
5.4.3 The index pin can be set in the crank handle so that it can be dropped into calculated hole and lock the crank the hole circles. ✓✓ (2)
5.4.4 Ratio between worm and worm gear: 40:1 ✓✓ (2)
5.5 Gear terminology:
5.5.1 The pitch-circle diameter 'PCD'
module(m) = PCD
T
PCD = m x T
= 3 x 94
PCD = 282mm (3)✓✓✓
5.5.2 The outside diameter:
outside diameter = PCD + 2m
OD = 282+2(3)
OD = 288mm (2)✓✓
5.5.3 The dedendum:
Dedendum b = 1.157m
b = 1.157 x 3
b = 3.47mm
or
b = 1.25m
b = 1.25 x 3
b = 3.75mm (2)✓✓
5.5.4 The cutting depth:
Cutting dept = 2.157 x m
= 2.157 x 3
= 6.47 mm
or
Cutting depth = 2.25 x m
= 2.25 x 3
= 6.75mm (2)✓✓
[30]
QUESTION 6: JOINING METHODS
6.1 Causes of undercutting:
6.2 Prevention of slag inclusion:
6.3 Liquid dye penetrant test:
6.4 Advantages of using a MIGS/MAGS welding:
6.5 Gas flow meter:
Control the flow of rate of shielding gas ✓ and measure the flow rate. ✓ (2)
6.6 MIGS/MAGS welding process:
6.7 Shielding gas in MIGS/MAGS:
6.8 Earth cable:
6.9 THREE types of gasses used for MIGS/MAGS welding:
[25]
QUESTION 7: FORCES
7.1
(13)
7.2 Stress and Strain:
7.2.1 Stress in the bar:
A = πD2
4
= π x 0.0562
4
= 2.46 x 10-3m-2
σ = F
A
= 40 x 10 3
2.46 x 10-3
= 16260162.6Pa
= 16.26 x 106Pa
= 16.26MPa (5)✓✓✓✓✓
7.2.2 Strain:
ε = σ
E
ε = 16.26 x 106
90 x 109
= 0.18 x 10-3(3)✓✓✓
7.2.3 Change in length:
ε = Δl
ol
Δl = ε x ol
= (0.18 x 10-3) x 0.85
= 0.15 x 10-3m
OR
= 0.15 mm(3)✓✓✓
7.3
Calculate A. Moments about B:
∑RHM = ∑LHM
(A x 12) = (960 x 6) + (750 x 8)
12A = 5760 + 6000
12 12
A = 980N
Calculate B. Moments about A:
∑RHM = ∑LHM
(B x 12) = (750 x 4) + (960 x 6) + (300 x 12)
12B = 3000 + 5760 + 3600
12B = 12360
12 12
B = 1030N (6)✓✓✓✓✓✓
[30]
QUESTION 8: MAINTENANCE
8.1 Pour point:
The lowest temperature ✓ at which a liquid will flow. ✓ (2)
8.2 Advantages of cutting fluids:
8.3 'ATF':
Automatic transmission fluid ✓✓ (2)
8.4 Main parts of a clutch:
Pressure plate ✓ clutch plate ✓ release bearing (Thrust bearing) ✓ (3)
8.5 Results of a stretched chain:
8.6 Causes of belt slip:
[15]
QUESTION 9: SYSTEM AND CONTROLS
9.1 Gear drives:
9.1.1 Rotation frequency of the output shaft:
NF = TA x TC x TE
NA TBx TD x TF
NF = TA x TC x TE x NA
TB x TD x TF
NF = 30 x 20 x 50 x 2300
40 x 60 x 70
=410.71r/min (3)✓✓✓
9.1.2 Velocity Ratio:
VR = NINPUT
NOUTPUT
= 2300
410.71
= 5.6:1 (2)✓✓
or
VR = NOUTPUT
NINPUT
= 410.71
2300
= 1:0.178
9.2 Belt Drives:
9.2.1 Rotation frequency of the driven pulley:
V = πDn
n = V
πD
= 32
πx(0.26)
nr/min = 39.18 x 60
nr/min = 2350.6r/min (3)✓✓✓
9.2.2 Tensile force in the tight side:
T1 = 2.5
T2
T1 = 2.5 x T2
=2.5 x 140
=350N (2)
9.2.3 Power transmitted:
P = (T1 - T2)v
P = (350 - 140)x 32
=6720 watts (3)✓✓✓
9.3 Hydraulics:
9.3.1 Fluid pressure:
AA = πD2
4
= π0.022
4
= 0.31 x 10-3m2
PA = F
AA
= 300 Pa
0.31 x 10-3
= 967741.94Pa
= 0.97 x 106Pa
= 0.967 MPa (4)✓✓✓✓
9.3.2 Stroke at piston B:
AB = πD2
4
= π0.0752
4
= 4.42 x 10-3m2
VB = VA
\AB x LB = AB x LA
LB = AA x LA
AB
= (0.31 x 10-3) x 185
4.42 x 10-3
= 12.98mm (4)✓✓✓✓
9.4 Traction control:
It prevents the wheels from spinning ✓✓(2)
9.5 Safety belt:
Safety belts need to be activated (buckle up) by the driver/passenger ✓✓ (2)
[25]
QUESTION 10: TURBINES
10.1 Water turbine:
10.2 Runaway speed of a water turbine:
Runaway speed of a water turbine is its speed at full flow ✓ and with no shaft load ✓ (2)
10.3 Water turbine:
10.3.1 Type of turbine:
10.3.2
10.3.3 Advantages of water turbine:
10.4 Function of turbo and superchargers:
To increase ✓ volumetric efficiency ✓ of an internal combustion engine. (2)
10.5 Compressor used in a turbocharger:
Centrifugal ✓ (1)
10.6 Turbocharger:
Exhaust gasses ✓ (1)
10.7 Advantage of a turbocharger:
10.8 Advantage of a steam turbine:
[20]
GRAND TOTAL: 200