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) 
  
• Downcut milling  (2) 
  

5.3 Calculate: Gib head key:
5.3.1 

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

5.3.2 

• Thickness   =  D  
                          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 

∑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 

• 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 

•    N_INPUT       = T_B ×  T_D
     N_OUTPUT      T_A × T_C
  N_OUTPUT = T_A × T_C  ×  N_INPUT
                     T_B × T_D
  N_IOTPUT = 18 × 16  × 1660
                   36 ×  46
  = 288,70 r/min   (3) 

9.2.2 Velocity Ratio 

• V_R =     N_INPUT     
            N_OUTPUT
  =     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 

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

9.3.2 Effort on piston A  

• A_A = πD² 
  = π × 0.04²
         4
  = 1,256 × 10^-3 m²
• P_A = F_A
          A_A
  F_A = P_A × A_A
  =(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