MECHANICAL TECHNOLOGY
GRADE 12 
NSC PAST PAPERS AND MEMOS
FEBRUARY/MARCH 2017

MEMORANDUM

QUESTION 1: MULTIPLE-CHOICE QUESTIONS 
1.1 B ✔ (1)
1.2 D ✔ (1)
1.3 C ✔ (1)
1.4 C ✔ (1)
1.5 B ✔ (1)
1.6 D ✔ (1)
1.7 A ✔ (1)
1.8 B ✔ (1)
1.9 A ✔ (1)
1.10 B ✔ (1)
1.11 B ✔ (1)
1.12 B ✔ (1)
1.13 D ✔ (1)
1.14 D ✔ (1)
1.15 A ✔ (1)
1.16 C ✔ (1)
1.17 A ✔ (1)
1.18 B ✔ (1)
1.19 B ✔ (1) 
1.20 A ✔ (1)

[20]

QUESTION 2: SAFETY 
2.1 Safety – Coil spring compressor: 

  • Make certain that the diameter of the compressor bolts can take the  pressure of the coil spring. ✔
  • Do not exceed the maximum pressure. ✔
  • Make sure the compressors are clean and free from oil. ✔
  • Ensure that the compressors are in a good working condition. ✔
    (Any 2 x 1) (2) 

2.2 Safety – Hydraulic Press: 

  • Take notice of the predetermined pressure of the hydraulic press. ✔
  • Ensure the pressure gauge is in a good working order. ✔
  • Platform on which the work piece rests must be rigid and square with the  cylinder of the press. ✔ 
  • The prescribed equipment must be used. ✔ 
  • Check for oil leaks. ✔
    (Any 3 x 1) (3) 

2.3 Safety – beam bender: 

  • Ensure the beam is clamped parallel to the backboard. ✔
  • Do not leave plastic beams loaded for any length of time, they tend to  creep. ✔
  • All the weight must be gently dropped onto the hanger as to reduce  inaccuracies due to friction. ✔ 
  • Do not exceed the tester's maximum load. ✔ 
  • Make sure the tester is stable. ✔
    (Any 2 x 1) (2) 

2.4 Testers: 
2.4.1 Brinell Tester: 

  • The tester must be mounted rigidly on a worktable. ✔ (1) 

2.4.2 Bearing and gear Puller: 

  • Make sure that the puller is at 90° to the work piece before you  start to pull. ✔
  • ∙ Ensure that the clamps are tight and will not slip from the work  piece. ✔
    (Any 1 x 1) (1) 

2.4.3 Torsion tester: 

  • Get specification (torsion) of the different materials and the  size of rods you would like to test. ✔ (1) 

[10]

QUESTION 3: TOOLS AND EQUIPMENT 
3.1 Fuel pressure: 

  • Faulty diaphragm ✔ 
  • Clogged fuel filter ✔
  • Faulty non return valves ✔ 
  • Worn gasket ✔ 
    (Any 2 x 1) (2) 

3.2 Precision measuring instruments: 
3.2.1 Depth micro-meter ✔ 

  • Vernier calliper ✔
    (Any 1 x 1) (1) 

3.2.2 Screw-thread micro-meter ✔ (1)  
3.3 Depth micro-meter reading: 

  • Reading = 50 + 1,5+ 0,49 ✔ 
    = 51,99 mm. ✔ (2) 

3.4 Multimeter measurements: 

  • DC current measurement ✔
  • DC voltage measurement ✔
  • AC measurement ✔ 
  • Resistance measurement ✔ 
  • Diode measurement ✔
  • Continuity measurement ✔
    (Any 2 x 1) (2) 

3.5 Trace the cylinder leakage in an engine: 

  • Listen to at the carburettor for a hissing noise. ✔ 
  • Listen at the exhaust pipe for a hissing noise. ✔ 
  • Listen for hissing noise in the dipstick hole. ✔ 
  • Listen to hissing noise by removing the filler cap on the tappet cover. ✔
  • By checking whether there are bubbles in the radiator water for blown  cylinder head gasket or cracked cylinder block. ✔
    (Any 2 x 1) (2) 

3.6 Uses of cooling pressure tester: 

  • To test if the pressure cap on the cooling system operates according to  the prescribed pressure of the system. ✔ 
  • To pump compressed air into the cooling system to determine whether  they are any water leakage in the system. ✔ (2)

[12]

QUESTION 4: MATERIALS 
4.1 Properties/characteristics: 
4.1.1 Cementite: 

  • Hard and brittle ✔✔ (2) 

4.1.2 Pearlite: 

  • Good ductility ✔ 
  • Very hard ✔ 
  • Strong and tough ✔ 
  • Resistance to deformation ✔
    (Any 2 x 1) (2) 

4.2 Iron –carbon equilibrium diagram 
4.2.1 Iron –carbon equilibrium diagram ✔ (1) 
4.2.2 

  1. – Ferrite + Pearlite ✔ 
  2. – Austenite + Ferrite ✔ 
  3. – Austenite ✔ 
  4. – Austenite + Cementite ✔ 
  5. – Ferrite + Cementite ✔ (5) 

4.2.3 Austenite: 

  • Soft, ✔ grain structure fine ✔ (2) 

4.3 720 °C ✔ (1)

[13]

QUESTION 5: TERMINOLOGY 
5.1 Indexing: 

  • Indexing = 40      ✔ 
                      n
    = 40 ÷ 2     ✔ 
      118   2
    = 20         ✔   (3) 
       59
    No fullturnsand 20 holes in a 59 -hole plate

5.2 Milling processes: 

  • Up-cut milling (2) 
    up cut milling
  • Downcut milling  (2) 
    down cut milling

5.3 Calculate: Gib head key:
5.3.1 

  • Width  =  
                   4
    = 102   ✔ 
         4
    = 25,5 mm   (2)   ✔ 

5.3.2 

  • Thickness   =  
                            6
    = 102   ✔ 
         6
    = 17 mm   (2)   ✔ 

5.3.3 

  • Length   = D × 1.5
    = 102  × 1.5 ✔ 
    = 153 mm   (2)   ✔ 

5.3.4 

  •  Thickness at smallend(t ) T = −    L    
                                                        100
    = 17 -  153    ✔ 
                100
    t = 17 - 1,53
    = 15,47 mm   (4)   ✔ 

5.4 Calculate – Spur gear: 
5.4.1

  • Addendum = m 
    = 3 mm ✔ (1) 

5.4.2

  • Dedendum = 1,157m      or        = 1,25m 
                        = 1,157 x 3 ✔         = 1,25 x 3 ✔
                         = 3,47 mm ✔         = 3,75 mm ✔ (2) 

5.4.3

  • Clearance = 0,157m            or             = 0,25m
                       = 0,157 x 3 ✔                    = 0,25 x 3 ✔
                       = 0,47 mm ✔                     = 0,75 mm ✔ (2) 

5.4.4 

  •  Module  = PCD
    T
    PCD = m × T ✔  
    = 3  ×  60
    = 180mm ✔(2) 

5.4.5 

  • OD = PCD + 2m 
    = 180 + 2(3) 
    = 180 + 6 ✔ 
    = 186 mm ✔ (2) 

5.4.6 

  • Cutting depth = 2,157 m             or                     = 2,25 m 
                          = 2,157 x 3 ✔                               = 2,25 x 3 ✔ 
                          = 6,47 mm ✔                                = 6,75 mm ✔ (2) 

5.4.7

  • Circular pitch = m x  π
    = 3 x  π ✔
    = 9,43 mm ✔ (2) 

[30]

QUESTION 6: JOINING METHODS 
6.1 Slag inclusion ✔ (1) 

6.2 Visual inspection defects 

  • Shape of profile ✔
  • Uniformity of surface ✔ 
  • Overlap ✔ 
  • Undercutting ✔ 
  • Penetration bead ✔ 
  • Root groove ✔ 
  • Crack free ✔
    (Any 4 x 1) (4) 

6.3 Causes of incomplete penetration: 

  • Weld speed too fast ✔ 
  • Joint design faulty ✔ 
  • Electrode too large ✔
  • Current too low ✔
    (Any 2 x 1) (2) 

6.4 Prevention of lack of fusion 

  • Adjust electrode size ✔ 
  • Correct preparation of joint ✔ 
  • Correct weld current ✔
  • Correct arc length ✔ 
  • Correct weld speed ✔
    (Any 2 x 1) (2) 

6.5 Destructive test 

Related Items

6.5.1 Machinability test ✔ (1)
6.5.2 Nick-break test ✔ (1)
6.5.3 Bend test ✔ (1) 

6.6 Dye penetration test 

  • Clean the weld that needs to be tested. ✔
  • Spray dye onto the surface and leave to penetrate. ✔✔
  • Excess dye is cleaned away with a cleaning agent. ✔ 
  • Allow surface to dry. ✔ 
  • Spray a developer onto the surface to bring out the dye trapped in the  crack. ✔ 
  • The dye will show all the surface defects. ✔ (7)

6.7 Functions of MIG/MAGS components 
6.7.1 Wire feed controller 

  • Feeds the consumable electrode wire to the welding gun at a  constant predetermined speed. ✔✔ (2) 

6.7.2 Welding gun 

  • Activates the supply of gas, power and wire feed✔✔ (2) 

6.8 Purpose of inert gas 

  • The inert gas shields the molten pool from the atmospheric gases. ✔✔ (2)

[25]

QUESTION 7: FORCES 
7.1 Forces 
forces
∑HC = 3,4 (√) + 1,8cos70º  (√) - 1,2cos40º  (√) + 2cos50º (√)
= 3,4 + 0,62 - 0,92 + 1,29
= 4,39 kN  (√)

∑VC = 1,2 sin40 (√) + 2sin50º  (√) - 1,8sin70º  (√) 
= 0,77 + 1,53 - 1,69
= 0,61 kN  (√)

OR 

Horizontal  component

Magnitudes 

Vertical  component

Magnitudes

-1,2cos40°✔ 

-0,92 kN

1,2sin40✔ 

0,77

3,4 ✔ 

3,4kN 

0

2cos50°✔ 

1,29kN

2sin50°✔ 

1,53

1,8cos70°✔ 

0,62kN 

-1,8sin70°✔ 

1,69

TOTAL 

4,39kN ✔ 

TOTAL 

0,61kN ✔

  • R2 = HC2 + VC2
    R = √4,392 + 0.612
    R = 4,43 kN 
  • Tanθ = VC
                HC
    = 0,61
      4,39
    θ = 7,91º
    R = 4,43 N  at  7,91º northe of east   (13) 

7.2 Stress and Strain 
7.2.1 Stress: 

  • A = π(D2 - d2)
                  4
    A = π(0,0982 - 0,0672)
                         4
    = 4,02 × 10-3 m2
    σ = F
          A
    σ =      40000      
            4,02 × 10-3
    σ = 9950248,76Pa
    σ = 9,95 MPa        (5) 

7.2.2 Strain: 

  • ε =    σ    
             E
    ε  = 9,95 × 106
           90 × 109
    = 0,11 × 10-3     or 1,11 × 10-4      (3) 

7.2.3 Change in length 

  • ε =   ΔI  
           oI
    ΔI =  ε  ×  ol
    = (0,11  × 10-3) × 0,08
    =8,8 ×  10-6m
    = 8,8 × 10-3 mm   (3) 

7.3 Moments 
Calculate A. Moments about B 

  • ∑RHM =∑ LHM
    (A × 11,6) = (200 × 5,8) + (928 × 5,8) + (600 × 2,8)
    11,6A = 1160 + 5382,4 + 1680
    11,6A = 8222,4
     11,6       11,6
    A = 708,83 N

Calculate B. Moments about A 

  • ∑LHM =∑ RHM
    (B × 11,6) = (600 × 8,8) + (928 × 5,8) + (200 × 5,8)
    11,6B = 5280 + 5382,4 + 1160
    11,6B = 11822,40
     11,6        11,6
    B = 1019,17 N    (6) 

[30] 

QUESTION 8: MAINTENANCE  
8.1 Preventative maintenance 

  • Can be described as maintenance of equipment or system before a fault  occurs.✔ ✔  (2) 

8.2 Lock out 

  • Locking out means that the machine's start switch cannot be activated without  the knowledge of a servicing technician otherwise an accident would  occur.✔ ✔  (2) 

8.3 Clutch free-play 

  • The distance the pedal moves before the slack is taken from the linkage and  release bearing. ✔ ✔  (2) 

8.4 Viscosity index 

  • Viscosity index is a measure of how much the oil's viscosity changes as  temperature changes. ✔  (1) 

8.5 Replace clutch plate: 

  • Worn friction linings. ✔ 
  • Weak or broken springs. ✔ 
  • Glazed friction linings due to overheating. ✔ 
  • Oil on friction linings. ✔
    (Any 2 x 1) (2) 

8.6 Grease – high viscosity 

  • To ensure that the grease coats and sticks ✔  to the bearing surfaces it is  lubricating. ✔  (2) 

8.7 Cutting fluid 

  • Mixture of soluble oil ✔  and water. ✔  (2) 

8.8 Viscosity of cutting fluid 

  • Has a low viscosity to allow easy flow ✔  and effective dissipation of excess  heat. ✔  (2)

[15]

QUESTION 9: SYSTEMS AND CONTROL 
9.1 Gear drives  
9.1.1 Rotation frequency of the output shaft 

  •    NINPUT       = TB ×  TD
       NOUTPUT      TA × TC
    NOUTPUT = TA × TC  ×  NINPUT
                       TB × TD
    NIOTPUT = 18 × 16  × 1660
                     36 ×  46
    = 288,70 r/min   (3) 

9.2.2 Velocity Ratio 

  • VR   NINPUT     
              NOUTPUT
       1660     
          288,70
    =   5,75: 1    (2) 

9.2 Belt Drives 
9.2.1 Rotation frequency of the driver pulley π(D t) N 

  • V = π( D + t) × N
                  60
    N =     V × 60    
            π(D + t)
    N =      36   ×    60     
         π(230 + 12) × 10-3
    2841,11r/min (4) 

9.2.2 Power transmitted 

  • T1 = 2,5
    T2
    T1 = 2,5 × T2 
    =2,5  × 110
    = 275 N
  • P = (T1 - T2)V
    P = (275 - 110) × 36
    = 5940W
    5,94kW (4) 

9.3 Hydraulics 
9.3.1 Fluid pressure 

  • AB = πD2
              4
    = π × 0.0752
             4
    = 4.42 × 10-3 m2
  • PB = F
    AB
       700 × 10   Pa
       4.42 × 10-3 
    = 1583710, 41Pa
    = 1583, 71 kPa ✔ (4) 

9.3.2 Effort on piston A  

  • AA = πD
    π × 0.042
           4
    = 1,256 × 10-3 m2
  • PA = FA
            AA
    FA = PA × AA
    =(1583,71 × 103) × (1,256 × 10-3)
    = 1990,10 N
    = 1,99 kN  (4) 

9.4 ABS 

  • Prevents wheel from locking during heavy breaking. ✔✔ (2) 

9.5 Seat belt 

  • A seat belt has to be activated for its safety to be functional. ✔✔ (2)

[25] 

QUESTION 10: TURBINES 
10.1 Impulse Turbine 

  • Waterwheel ✔ 
  • Pelton ✔ 
  • Turgo ✔ 
  • Michell – Banki/Crossflow/Ossberger✔ 
  • Jonval turbine ✔ 
  • Reverse overshot waterwheel ✔ 
  • Archimedes' screw turbine ✔ 
    (Any 2 x 1) (2) 

10.2 Water turbine 
10.2.1

  • Water turbine ✔
  • Kaplan-turbine ✔
  • Reaction turbine ✔
    (Any 1 x 1) (1) 

10.2.2 Parts 

  1. – Wicked gate ✔ 
  2. – Rotor ✔ 
  3. – Stator ✔ 
  4. – Shaft ✔ 
  5. – Water-flow ✔ 
  6. – Blades ✔  (6) 

10.2.3 Advantages of water turbine 

  • Low maintenance ✔ 
  • No need for lubrication ✔ 
  • Fewer moving parts ✔ 
  • Environmental friendly ✔ 
  • Cost effective ✔ 
    (Any 2 x 1) (2) 

10.3 Turbines 
10.3.1 Advantage of supercharger:  

  • Increases the output power of the engine. ✔ 
  • A smaller engine fitted with a centrifugal blower delivers the  same power as a larger engine. ✔   
  • It eliminates lack of oxygen above sea level. ✔ 
  • Increases the volumetric efficiency of the engine. ✔ 
  • With the aid of the intercooler both the power and the torque  output of the engine are increased. ✔ 
    (Any 2 x 1) (2)

10.3.2 Advantages of steam turbines:  

  • It is compact. ✔ 
  • No lubrication is required. ✔ 
  • Steam turbine speeds can be more accurately regulated. ✔ 
  • A variety of fuels can be used to obtain steam. ✔ 
  • Steam turbines are more economical. ✔ 
  • Higher speeds can be obtained as compared to internal  combustion engine. ✔ 
  • Convert heat energy into mechanical energy. ✔ 
    (Any 2 x 1) (2) 

10.3.3 Advantages of gas turbines:  

  • Very high power to weight ratio ✔ 
  • Smaller than most reciprocating engines of the same power  rate ✔ 
  • Moves in one direction only, with far less vibration ✔ 
  • Low operating pressures ✔ 
  • High operating speeds ✔ 
  • Low lubricating oil cost and consumption ✔   
    (Any 2 x 1) (2) 

10.4 Turbo lag 

  • It is a delay ✔  between pushing on the accelerator ✔  and feeling turbo  kick in. ✔  (3)

[20] 
TOTAL: 200

Last modified on Wednesday, 07 July 2021 08:32