Mechanical Technology Grade 12 NSC November 2012 QP (Eng)

Downloaded from hlayiso.com GRAAD 12 NATIONAL SENIOR CERTIFICATE GRADE 12 MECHANICAL TECHNOLOGY NOVEMBER 2012 MARKS: 200 TIME: 3 hours This question paper consists of 17 pages and a 5-page formula sheet. Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 2 DBE/November 2012 NSC INSTRUCTIONS AND INFORMATION 1. Write your centre number and examination number in the spaces provided on the ANSWER BOOK. 2. Read ALL the questions carefully. 3. Answer ALL the questions. 4. Answer QUESTION 1 on the inside of the front cover page of the ANSWER BOOK. Choose an answer and make a cross (X) in the block (A–D) next to the question number (1.1–1.20) in the ANSWER BOOK. 5. Number the answers correctly according to the numbering system used in this question paper. 6. Start EACH question on a NEW page. 7. Show ALL calculations and units. Round off final answers to TWO decimal places. 8. Candidates may use non-programmable scientific calculators and drawing instruments. 9. The value of gravitational force should be taken as 10 m/s2. 10. All dimensions are in millimetres, unless stated otherwise in the question. 11. Write neatly and legibly. 12. Use the criteria below to assist you in managing your time. QUESTION CONTENT COVERED MARKS TIME 1 Multiple-choice questions 20 18 minutes 2 Tools and Equipment 20 18 minutes 3 Materials 20 18 minutes 4 Safety, Terminology and Joining Methods 50 45 minutes 5 Maintenance and Turbines 40 36 minutes 6 Forces, Systems and Control 50 45 minutes TOTAL 200 180 minutes Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 3 DBE/November 2012 NSC QUESTION 1: MULTIPLE-CHOICE QUESTIONS Various options are provided as possible answers to the following questions. Choose the answer and make a cross (X) in the block (A–D) next to the question number (1.1–1.20) on the inside of the front cover page of the ANSWER BOOK. Example: QUESTION OPTIONS 1.21 A B C D 1.1 What is the advantage of down-cut milling? A The strain on the arbor and cutter is less. B Finish obtained is finer. C More vibration is experienced. D Built-up pressure is developed. (1) 1.2 Which ONE of the following safety procedures does NOT relate to the cylinder leakage tester? A Use water to remove dust around the spark-plug area. B Clean and remove the oil filler cap. C Do not exceed the prescribed pressure in the test cylinder. D The tester must fit properly and be well tightened in the spark-plug hole or injector hole. (1) 1.3 What is the function of a gas analyser? A Analyse inlet gases B Analyse oil gases C Analyse fuel gases D Analyse exhaust gases (1) 1.4 The main reason for using a torsion tester is to measure the ability of metal to … without breaking. A bend B twist C elongate D shorten (1) 1.5 Alloys can … A reduce electrical conductivity. B increase resistance to corrosion. C reduce strength and hardness. D alter insulation properties. (1) Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 4 DBE/November 2012 NSC 1.6 Thermosetting plastic materials are materials that … A can be stretched and return rapidly to their original shape. B cannot be softened and reshaped by reheating. C soften under heat and become hard when cooled. D stay soft after reheating. (1) 1.7 Identify the type of milling cutter shown in FIGURE 1.1 below. FIGURE 1.1 A Dovetail slot cutter B Convex cutter C Cylindrical/Helical cutter D Single corner rounding cutter (1) 1.8 Which lathe operation is shown in FIGURE 1.2 below? FIGURE 1.2 A Diameter turning B Thread cutting C Facing off D Boring (1) 1.9 Upcut milling is also known as … A gang milling. B slot milling. C climb milling. D conventional milling. (1) 1.10 What is the reason for using a free-bend test? A To measure the ductility of the weld metal B To test the skill of the welder C To train welders D To approve welds according to certain standards (1) Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 5 DBE/November 2012 NSC 1.11 What is the advantage of a helical cutter? A Vibration experienced by the machine is more. B Narrow cuts may be made. C No side thrusts due to sliding action of the cutter. D Wider cutters may be used. (1) 1.12 Tensile stress can be defined or described as an internal force in a material resisting … A a shearing load. B a pulling load. C a pushing load. D any load. (1) 1.13 What is understood by the term Young's modulus? A The force value required to produce a unit area in a tensile test specimen B The ratio between stress and strain in a metal, provided that the limit of elasticity is not exceeded C A measurement of the extension or contraction of material due to the load experienced D A ratio of the deformation by application of an external force (1) 1.14 Which step in adjusting the tension in the chain of a chain drive does FIGURE 1.3 below show? Centre distance S S 'S–S' min = 3% 'S–S' max = 5% FIGURE 1.3 A Check the distance between the centres. B Slacken the clamp bolts and screw the adjusting bolts in or out. C Check the amount of slack in the chain. If it is between 3% and 5% it is acceptable. D Tighten the clamping bolts. (1) Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 6 DBE/November 2012 NSC 1.15 Which procedure for removing a bearing is shown in FIGURE 1.4 below? Ram Inner ring Bearing Outer ring Support strip Shaft FIGURE 1.4 A Applying force to remove the shaft B Applying force to the bearing C Removing any rust from the shaft with emery paper D Measuring the position of the bearing (1) 1.16 Which ONE of the following is an advantage of V-belts? A Can be used on loose pulleys B May be used over long distances C Able to absorb shock loads D Operate at a high bearing pressure (1) 1.17 What is the purpose of the linkage shown in FIGURE 1.5 below? FIGURE 1.5 A Converts rotary motion into oscillating motion B Converts oscillating motion into rotary motion C Converts reciprocating motion into rotary motion D Converts rotary motion into reciprocating motion (1) Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 7 DBE/November 2012 NSC 1.18 Determine the distance 'D' moved by the follower for one full revolution of the pear-shaped cam shown in FIGURE 1.6 below, given that d 1 = 5 mm and d 2 = 12 mm. Cam D Follower d2 d1 FIGURE 1.6 A 7 mm B 17 mm C 5 mm D 12 mm (1) 1.19 What is understood by the term pressure ratio in turbochargers? A The pressure that the turbocharger generates in the intake manifold B The pressure that the turbocharger generates in the exhaust manifold C The inlet pressure compared to the outlet pressure D Decreased pressure in the exhaust manifold (1) 1.20 The volumetric efficiency of a blower is the … A ability to heat intake air. B ability to cool exhaust air. C ability to drive the blower. D ability to fill the cylinder with air above atmospheric pressure. (1) [20] Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 8 DBE/November 2012 NSC QUESTION 2: TOOLS AND EQUIPMENT 2.1 When a dry compression test was conducted on cylinder number one of a four-stroke, four-cylinder, spark-ignition engine the result was very low compared to the engine specifications. State THREE possible causes of the low compression in cylinder number one. (3) 2.2 Define torsion. (2) 2.3 FIGURE 2.1 below indicates a multimeter that is used to test different electrical components and circuits. Label parts A–E on the multimeter. A B C D E FIGURE 2.1 (5) 2.4 The function of the cylinder leakage tester is to check whether the engine leaks gases from the cylinder during compression strokes. State THREE possible leakages during the cylinder leakage test and state the cause of each. (6) 2.5 What does the abbreviation MAGS/MIGS stand for in welding equipment? (2) 2.6 Name TWO gases that can be used in MAGS/MIGS welding. (2) [20] Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 9 DBE/November 2012 NSC QUESTION 3: MATERIALS 3.1 Name ONE difference between non-ferrous metals and composites. (2) 3.2 FIGURE 3.1 below shows the results of a compressive strength test on two materials, A and B. Materials before test A B Force = Force Materials after test A B FIGURE 3.1 3.2.1 Which of the materials, A or B, has the highest compressive strength? (1) 3.2.2 Give a reason for the answer to QUESTION 3.2.1. (2) 3.3 Name THREE main classes/groups of carbon steel. (3) 3.4 Name FOUR properties of carbon steel that will change when the carbon content is increased. (4) 3.5 Choosing the most suitable material for a particular product requires you to have good knowledge of materials. State TWO uses and TWO properties of the following engineering materials: 3.5.1 Duralumin (2 + 2) (4) 3.5.2 Polyvinyl chloride (PVC) (2 + 2) (4) [20] Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 10 DBE/November 2012 NSC QUESTION 4: SAFETY, TERMINOLOGY AND JOINING METHODS 4.1 State THREE safety precautions to be observed when operating a beam bending tester. (3) 4.2 State THREE safety precautions to be observed when storing gas cylinders. (3) 4.3 Give TWO advantages of each of the following milling operations: 4.3.1 Upcut milling (2) 4.3.2 Down-cut milling (2) 4.4 A gear with 63 teeth has to be machined on a milling machine. Use A = 60 or N = 60. 4.4.1 Calculate the indexing required. (4) 4.4.2 Calculate the change gears required. (5) 4.5 Calculate the feed in millimetres per minute of an 80 mm diameter cutter with 14 teeth, operating at a cutter speed of 20 metres per minute and a feed of 0,08 mm per tooth. (7) 4.6 Mary is a quality controller who inspects welded joints on workpieces. She uses an ultrasonic test to determine whether the welds have any welding defects. Explain the principle of operation of the ultrasonic test on welded joints. (5) 4.7 State TWO possible causes and ONE preventive action for each of the following weld defects: 4.7.1 Slag inclusion (2 + 1) (3) 4.7.2 Undercutting (2 + 1) (3) 4.8 Which milling cutter can be used to cut the following profiles? 4.8.1 Spur gear (1) 4.8.2 Groove (1) 4.8.3 Rack (1) 4.8.4 Blind hole (1) Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 11 DBE/November 2012 NSC 4.9 FIGURE 4.1 below shows a dividing head of a milling machine. Label parts A–E. E A D FIGURE 4.1 (5) 4.10 State the TWO classes of milling cutters and give ONE example of each. (4) [50] Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 12 DBE/November 2012 NSC QUESTION 5: MAINTENANCE AND TURBINES 5.1 FIGURE 5.1 below shows a clutch and pressure-plate assembly. B A C D E F FIGURE 5.1 5.1.1 Label parts A–F in the clutch and pressure-plate assembly. (6) 5.1.2 State TWO functions of the clutch plate in this clutch and pressure- plate assembly. (4) 5.1.3 Name THREE causes of slip in this clutch and pressure-plate assembly. (3) 5.1.4 Name TWO types of clutches used in the engineering environment. (2) 5.2 State THREE functions of lubricating oil. (3) 5.3 State THREE guidelines for the maintenance of cutting fluid used in a lathe. (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 13 DBE/November 2012 NSC 5.4 The performance of smaller engines can be improved by using superchargers (FIGURE 5.2). C C D D FIGURE 5.2 5.4.1 Identify the type of supercharger shown in FIGURE 5.2. (1) 5.4.2 Label parts A–D. (4) 5.4.3 Explain the operation of the supercharger shown in FIGURE 5.2. (5) 5.4.4 Name THREE advantages of superchargers. (3) 5.5 Name THREE classifications of steam turbines. (3) 5.6 State THREE advantages of gas turbines. (3) [40] Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 14 DBE/November 2012 NSC QUESTION 6: FORCES, SYSTEMS AND CONTROL 6.1 A hydraulic system is being used to lift a machine part into position. The specifications of the system are diagrammatically presented in FIGURE 6.1 below. Ø 38 mm Ø 175 mm 300 N Load = ? Piston A Piston B FIGURE 6.1 Determine the following by means of calculations: 6.1.1 The fluid pressure in the hydraulic system when in equilibrium (5) 6.1.2 The load that can be lifted by piston B when a force of 300 N is applied to piston A (6) 6.2 A steel spoke of a bicycle wheel has a cross-sectional area of 2,2 mm2 and a length of 300 mm. The nut of the spoke is tightened until the length increases by 0,2 mm. Determine, by means of calculations, the tensile force in the spoke material. Take Young's elasticity modulus for the spoke material as 245 GPa. (9) Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 15 DBE/November 2012 NSC 6.3 FIGURE 6.2 shows a belt-drive system. A pulley with a diameter of 0,25 m drives a driven pulley with a diameter of 0,35 m. The driver pulley rotates at 1 000 r/min. T 1 = 200 N and T 2 =90 N. NDN = ? r/min DDN = 0,35 m NDR = 1 000 r/min DDR = 0,25 m FIGURE 6.2 Determine by means of calculations: 6.3.1 The rotation frequency of the driven pulley in r/min (revolutions per minute) (3) 6.3.2 The power transmitted (3) 6.3.3 The belt speed of the system in metres per second (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 16 DBE/November 2012 NSC 6.4 FIGURE 6.3 shows a gear drive system. A driver gear on the shaft of an electric motor has 24 teeth and meshes with a gear on a counter shaft with 40 teeth. On this counter shaft is another driver gear with 20 teeth that meshes with a gear with 48 teeth on a second counter shaft. The second counter shaft has a driver gear with 42 teeth which drives a gear with 90 teeth on the output shaft. Counter Electrical shaft motor 48 42 24 Counter shaft Output shaft 40 20 90 FIGURE 6.3 Calculate: 6.4.1 The rotation frequency of the output shaft if the electric motor rotates at 1 440 r/min (4) 6.4.2 The velocity ratio between the input and output shaft (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mechanical Technology 17 DBE/November 2012 NSC 6.5 A differential wheel-and-axle lifting machine is shown in FIGURE 6.4 below. A load of 2,4 kN is lifted when an effort of 400 N is applied. The diameters of the pulleys are 210 mm, 160 mm and 140 mm respectively. d1 = 140 d2 = 160 D = 210 400 N 2,4 kN FIGURE 6.4 Determine by means of calculations: 6.5.1 The mechanical advantage (2) 6.5.2 The velocity ratio (3) 6.5.3 The mechanical efficiency of the machine (2) 6.6 A single-plate friction clutch transmits a maximum torque of 336 Nm. The clutch plate has friction material on both sides with a friction co-efficient of 0,4. The total applied force on the friction surfaces is 3,5 kN. Determine by means of calculations: 6.6.1 The effective diameter of the clutch plate in millimetres (5) 6.6.2 The power transmitted at 3 200 r/min (3) [50] TOTAL: 200 Copyright reserved

Downloaded from hlayiso.com Mechanical Technology 1 DBE/November 2012 NSC FORMULA SHEET FOR MECHANICAL TECHNOLOGY – GRADE 12 1. BELT DRIVES πDN 1.1 Belt speed = 60 π ( D + t )× N 1.2 Belt speed = (t = belt thickness) 60 1.3 Belt mass = Area × length × density (A = thickness × width) Diameter of driven pulley 1.4 Speed ratio = Diameter of driver pulley 1.5 N 1 D1 = N 2 D2 π( D + d ) ( D - d )2 1.6 Open-belt length = + + 2c 2 4c π( D + d ) ( D + d )2 1.7 Crossed -belt length = + + 2c 2 4c 2π NT 1.8 Power ( P ) = 60 T1 1.9 Ratio of tight side to slack side = T2 (T1 − T2 ) π D N 1.10 Power ( P ) = where T1 = force in the tight side 60 T1 1.11 Width = permissible tensile force Copyright reserved

Downloaded from hlayiso.com Mechanical Technology 2 DBE/November 2012 NSC 2. FRICTION CLUTCHES 2.1 Torque ( T ) = µWnR µ = coefficient of friction W = total force n = number of friction surfaces R = effective radius 2.2 2 π NT Power ( P ) = 60 3. STRESS AND STRAIN Force F 3.1 Stress = or ( σ = ) Area A change in length ( ΔL ) 3.2 Strain ( ε ) = original length ( L ) stress σ 3.3 Young' s mod ulus ( E ) = or ( ) strain ε πd 2 3.4 Ashaft = 4 π(D 2 − d 2 ) 3.5 Apipe = 4 4. HYDRAULICS Force ( F ) 4.1 Pressure ( P ) = Area ( A ) F1 F2 4.2 = A1 A2 4.3 Work done = force × distance 4.4 Volume = Cross-sectional area × stroke length ( l or s ) Copyright reserved

Downloaded from hlayiso.com Mechanical Technology 3 DBE/November 2012 NSC 5. WHEEL AND AXLE effort distance 2D 5.1 Velocity ratio ( VR ) = = load distance d 2 − d 1 Load ( W ) 5.2 Mechanical advantage ( MA ) = Effort ( F ) MA 5.3 Mechanical efficiency ( ηmech ) = × 100% VR 6. LEVERS Load ( W ) 6.1 Mechanical advantage ( MA ) = Effort ( F ) 6.2 Input movement ( IM ) = Effort × distance moved by effort 6.3 Output movement ( OM ) = Load × distance moved by load Input movement 6.4 Velocity ratio ( VR ) = Output movement 7. SCREW THREADS 7.1 Pitch diameter = Outside diameter − ½ pitch 7.2 Pitch circumference = π × pitch diameter 7.3 Lead = pitch × number of starts Lead 7.4 Helix angle : tan θ = Pitch circumference 7.5 Leading tool angle = 90° – (helix angle + clearance angle) 7.6 Following/Trailing angle = 90° + (helix angle – clearance angle) height 7.7 Number of turns = lead Copyright reserved

Downloaded from hlayiso.com Mechanical Technology 4 DBE/November 2012 NSC 8. GEAR DRIVES 2 π NT 8.1 Power ( P ) = 60 Product of the number of teeth on driven gear 8.2 Gear ratio = Product of the number of teeth on driver gear N Input Product of the number of teeth on driven gear 8.3 = N Output Product of the number of teeth on driver gear 8.4 Torque = force × radius 8.5 Torque transmitted = gear ratio × input torque Pitch-circle diameter ( PCD ) 8.6 Module ( m ) = Number of teeth ( T ) 8.7 N 1T1 = N 2T2 and N1 D1 = N 2 D2 circular pitch ( CP ) × number of teeth ( T ) 8.8 Pitch-circle diameter ( PCD ) = π 8.9 Outside diameter ( OD ) = PCD + 2 module 8.10 Addendum ( a ) = module ( m ) 8.11 Dedendum ( b ) = 1,157 m or Dedendum ( b ) = 1,25 m 8.12 Cutting depth ( h ) = 2,157 m or Cutting depth ( h ) = 2,25 m 8.13 Clearance ( c ) = 0,157 m or Clearance ( c ) = 0,25 8.14 Circular pitch ( CP ) = m × π N input 8.15 VR (Velocity ratio) = N output Copyright reserved

Downloaded from hlayiso.com Mechanical Technology 5 DBE/November 2012 NSC 9. CINCINNATI DIVIDING HEAD TABLE FOR THE MILLING MACHINE Hole circles Side 1 24 25 28 30 34 37 38 39 41 42 43 Side 2 46 47 49 51 53 54 57 58 59 62 66 Standard change gears 24 x 2 28 32 40 44 48 56 64 72 86 100 40 9.1 Simple indexing = ( where n = number of divisions ) n 9.2 Change gears: Dr 40 Dr (A − n) 40 Dr 40 = ( A − n )× or = × or = (N − n) × Dv A Dv A 1 Dv N 10. CALCULATIONS OF FEED 10.1 Feed (f) = f 1 × T × N Where: f = feed in millimetres per minute f 1 = feed per tooth in millimetres T = number of teeth on cutter N = number of revolutions of cutter per minute 10.2 Cutting speed (V) = π × D × N Where: D = diameter of the cutter in metres Copyright reserved